HexWiki - User contributions [en]

Talk:Edge template V1b

2016-05-28T07:22:44Z

Wccanard:

Re: "(Note: As I am writing this I have only seen the claim of this being a valid template on the little golem thread. I have not checked it yet and also not if it is minimal. However, as this came from a very competent player I have no reson to doubt it.)"

This comment seems to cast some suspicion on whether the template is a template. It only takes benzene (computer program) around 10 seconds to check that this is definitely a template so I just wanted to stress that in my mind there is no doubt at all.

--wccanard

Re: e3 defence (currently not written) -- benzene (who might be playing a bit randomly as white) says that one main line is

e11 f10 f11 g10 g11 d11 c13 h10 h11 i10 i11 d12 d13 e12 e13 f12 f13 g12 g13 h12 h13 m12 j10

[Take away 8 from all the numbers (and leave the letters alone) to see the moves using the convention you're using (I'm on a standard 14x14 board so bottom row is 14 not 6)]

-- wccanard

Edge template V1b

2016-05-28T07:21:54Z

Wccanard: /* Defense against 1. e3 */

== The template ==

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 E *:a5"
/>

(From the [https://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 little golem forum])

Even though the diagram shows two pieces and one of them on the sixth row, it makes sense to say this is a fifth row template with one stone, as the stone on the sixth row plays no role whatsoever in connecting to the bottom. It could be removed and the stone on the fifth row will still be connected to the bottom. It is only shown so there is a possibility to connect to the top too, as without that it would not be of much use in a practical sense.

(Note: As I am writing this I have only seen the claim of this being a valid template on the little golem thread. I have not checked it yet and also not if it is minimal. However, as this came from a very competent player I have no reson to doubt it.)

== Defense against intrusions ==

Red has 3 main threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 E *:n3 E *:a4 E *:b4 E +:c4 R 1:d4 E *:a5 E +:b5 E +:c5 E +:d5 E +:a6 E +:b6 E +:c6 E +:d6"
/>

using the [[ziggurat]],

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 E *:n3 E *:a4 E *:b4 E +:c4 R 1:d4 E +:e4 E *:a5 E +:b5 E +:c5 E +:d5 E +:e5 E +:a6 E +:b6 E +:d6 E +:e6"
/>

using [[Defending_against_intrusions_in_template_1-IIIb|III-1-b]] and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E +:h2 E +:i2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:e3 R 1:f3 E +:g3 E +:h3 E +:i3 E +:j3 E *:n3 E *:a4 E *:b4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E *:a5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6"
/>

using [[Edge_template_IV1d|IV-1-d]].

For a blocking attempt, Blue has to play on the overlap:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:e3 E *:n3 E *:a4 E *:b4 E *:a5 E +:d5 E +:d6"
/>

=== Defense against 1. e3 ===

Details yet to come...

=== Defense against 1. d5 ===
Red can start like this:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 4:g3 E *:n3 E *:a4 E *:b4 R 2:d4 E *:a5 B 3:c5 B 1:d5"
/>

Continuation:

<div class="toccolours mw-collapsible mw-collapsed">
Red has this threat:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 R 3:e5 E +:f5 R 5:g5 B 4:d6 B 2:e6 E +:f6 E +:g6"
/>

So all these fields need to be considered:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 E +:e5 E +:f5 E +:g5 E +:d6 E +:e6 E +:f6 E +:g6"
/>

==== ... 5. f3 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 E *:a5 B c5 B d5"
/>

Red has 2 threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 E *:a5 B c5 B d5 R 1:e5 E +:d6 E +:e6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E +:h3 E *:n3 E *:a4 E *:b4 R d4 R e4 R 1:f4 E +:g4 E +:h4 E +:i4 E *:a5 B c5 B d5 E +:f5 E +:g5 E +:h5 E +:i5 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6"
/>

using [[Edge_template_IV2b|IV-2-b]].

Blue has to play on the only overlapping field:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 R 6:i4 E *:a5 B c5 B d5 R 2:e5 R 4:f5 B 3:d6 B 1:e6 B 5:f6"
/>

Red 6 is connected to the bottom and to the left with [[Tom's move]].

==== ... 5. e4 ====

Yet to come ...

==== ... 5. f4 ====

Yet to come ...

==== ... 5. g4 ====

Yet to come ...

==== ... 5. e5 ====

Yet to come ...

==== ... 5. f5 ====

Yet to come ...

==== ... 5. g5 ====

Yet to come ...

==== ... 5. d6 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 R 2:f4 E *:a5 B c5 B d5 B 1:d6"
/>

The group with 2 is connected to the top in two non-overlapping ways (see area marked with +) and to the bottom with [[Edge_template_IV2b|IV-2-b]].

==== ... 5. e6 ====

Red can respond here:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R 2:f5 B 1:e6"
/>

Continuation:
<div class="toccolours mw-collapsible mw-collapsed">

Now Red has two threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 R 1:e5 R f5 E +:d6 B e6 E +:f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 R 1:g4 E +:h4 E *:a5 B c5 B d5 E +:e5 R f5 E +:g5 E +:h5 B e6 E +:f6 E +:g6 E +:h6"
/>

Blue has to play on the overlap:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 E +:e5 R f5 B e6 E +:f6"
/>

''7. f2''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 1:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 R 8:h4 E *:a5 B c5 B d5 R 4:e5 R f5 B 7:g5 R 6:h5 B 5:d6 B e6 B 3:f6"
/>

''7. g2''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 1:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 6:h4 E *:a5 B c5 B d5 R f5 B 5:g5 R 4:h5 B e6 B 3:f6"
/>

''7. f3''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 R 6:i4 E *:a5 B c5 B d5 R 4:e5 R f5 B 5:d6 B e6 B 2:f6"
/>
6 is now connected to the left and to the bottom by [[Tom's move]].

''7. e4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 B 1:e4 R 4:i4 E *:a5 B c5 B d5 R f5 B e6 B 3:f6"
/>
4 is again connected to the left and to the bottom by [[Tom's move]].

''7. f4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 1:f4 E *:a5 B c5 B d5 E +:e5 R f5 E +:d6 B e6 E +:f6"
/>

Blue has to go on one of the 3 marked fields. However, d6 can't be any better than 35, so it's enough to look at e5 and g6. Let's have a look at g6 first:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 B f4 R 4:i4 E *:a5 B c5 B d5 R 2:e5 R f5 B 3:d6 B e6 B 1:f6"
/>

4 is now connected to the bottom and to the left in a similar way as in [[Tom's move]]. (Just the piece on g3 is connected to the left in a slightly different way.)

The other possible move was e5:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 2:h2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 B f4 E *:a5 B c5 B d5 B 1:e5 R f5 B e6"
/>

Red 2 is connected to the left by two non-overlapping ways and to the bottom by a 5th row template that has yet to be added to this wiki.

''7. g4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 1:g4 R 6:i4 E *:a5 B c5 B d5 R 4:e5 R f5 B 5:d6 B e6 B 3:f6"
/>

Once again 6 is now connected to the bottom and to the left in a similar way as in [[Tom's move]].

''7. e5''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 4:i4 E *:a5 B c5 B d5 B 1:e5 R f5 B e6 B 3:f6"
/>

And [[Tom's move]] at the end again.

''7. f6''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:i4 E *:a5 B c5 B d5 R f5 B e6 B 1:f6"
/>

Red 2 is connected to the bottom and to at least one of the red pieces in the middle by [[Tom's move]]. Red now has three threats to connect both these pieces to the top:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 1:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 R 1:f4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E +:f4 E +:g4 R i4 E *:a5 B c5 B d5 E +:e5 R f5 B e6 B f6"
/>

Blue has to play on the overlap:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

First move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 3:f4 R i4 E *:a5 B c5 B d5 R 4:e5 R f5 R 6:g5 B 5:d6 B e6 B f6"
/>

Second move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 R 4:h3 E *:n3 E *:a4 E *:b4 R d4 B 1:f4 B 3:g4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>
</div>
==== ... 5. f6 ====

Red can start like this:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R 2:e5 B 1:f6"
/>

Red now has these threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E *:a5 B c5 B d5 R e5 E +:d6 E +:e6 B f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 1:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 E +:h4 E *:a5 B c5 B d5 R e5 E +:f5 E +:g5 E +:h5 E +:d6 E +:e6 B f6 E +:g6 E +:h6"
/>

using [[Edge_template_IV2e|IV-2-e]].

Blue has to play on the overlap:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R e5 E +:d6 E +:e6 B f6"
/>

First move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 4:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 B 3:e4 B 5:f4 R 6:g4 E *:a5 B c5 B d5 R e5 B 7:f5 B 8:h5 B 1:d6 R 2:e6 B f6"
/>

You get the the defense against the other move by just swapping 1 and 2 in the diagram above.

==== ... 5. g6 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:f4 R 6:h4 E *:a5 B c5 B d5 B 5:f5 B 7:g5 R 8:i5 B 3:e6 R 4:f6 B 1:g6"
/>

Note that 2 is safely connected to the top, so 3 is forced.

</div>

=== Defense against 1. d6 ===

Red has this line:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R 2:d4 R 8:h4 E *:a5 B 3:c5 R 4:d5 R 6:e5 B 5:c6 B 1:d6 B 7:e6"
/>

Blue 3, 5 and 7 are forced.

Continuation:
<div class="toccolours mw-collapsible mw-collapsed">

Red has these threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 R h4 E +:i4 E *:a5 B c5 R d5 R e5 R 1:f5 R 3:g5 E +:h5 R 5:i5 B c6 B d6 B e6 B 2:f6 B 4:g6 E +:h6 E +:i6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 E +:g4 R h4 E *:a5 B c5 R d5 R e5 R 1:f5 E +:g5 R 3:h5 B c6 B d6 B e6 B 2:f6 E +:g6 E +:h6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R 1:g3 E +:h3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 R h4 E +:i4 E *:a5 B c5 R d5 R e5 E +:g5 E +:h5 E +:i5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6 E +:i6"
/>

The overlap in which Blue has to play is:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 R h4 E *:a5 B c5 R d5 R e5 E +:g5 E +:h5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6"
/>

Four of these five possible moves can be analysed in one line. In the following diagram assume Blue has played 1 on any of the fields marked with +:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 6:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 4:g3 E *:n3 E *:a4 E *:b4 R d4 B 5:f4 R h4 E *:a5 B c5 R d5 R e5 B 3:f5 R 2:g5 E +:h5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6"
/>

After Red 2 that group is safely connected to them bottom, now matter which of the pluses Blue chose before.

That leaves only one Blue move to deal with:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 7:g2 R 6:h2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 10:f3 R 8:g3 R 4:i3 E *:n3 E *:a4 E *:b4 R d4 B 9:f4 B 5:g4 R h4 E *:a5 B c5 R d5 R e5 R 2:f5 B 1:g5 B c6 B d6 B e6 B 3:f6"
/>

Note that Red 4 connects to the bottom with [[Edge_template_IV2b|IV-2-b]].
</div>
[[category:edge templates]]

Edge template V1b

2016-05-28T07:20:41Z

Wccanard: /* Defense against 1. e3 */

== The template ==

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 E *:a5"
/>

(From the [https://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 little golem forum])

Even though the diagram shows two pieces and one of them on the sixth row, it makes sense to say this is a fifth row template with one stone, as the stone on the sixth row plays no role whatsoever in connecting to the bottom. It could be removed and the stone on the fifth row will still be connected to the bottom. It is only shown so there is a possibility to connect to the top too, as without that it would not be of much use in a practical sense.

(Note: As I am writing this I have only seen the claim of this being a valid template on the little golem thread. I have not checked it yet and also not if it is minimal. However, as this came from a very competent player I have no reson to doubt it.)

== Defense against intrusions ==

Red has 3 main threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 E *:n3 E *:a4 E *:b4 E +:c4 R 1:d4 E *:a5 E +:b5 E +:c5 E +:d5 E +:a6 E +:b6 E +:c6 E +:d6"
/>

using the [[ziggurat]],

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 E *:n3 E *:a4 E *:b4 E +:c4 R 1:d4 E +:e4 E *:a5 E +:b5 E +:c5 E +:d5 E +:e5 E +:a6 E +:b6 E +:d6 E +:e6"
/>

using [[Defending_against_intrusions_in_template_1-IIIb|III-1-b]] and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E +:h2 E +:i2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:e3 R 1:f3 E +:g3 E +:h3 E +:i3 E +:j3 E *:n3 E *:a4 E *:b4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E *:a5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6"
/>

using [[Edge_template_IV1d|IV-1-d]].

For a blocking attempt, Blue has to play on the overlap:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:e3 E *:n3 E *:a4 E *:b4 E *:a5 E +:d5 E +:d6"
/>

=== Defense against 1. e3 ===

Details yet to come -- benzene (who might be playing a bit randomly as white) says that one main line is

e11 f10 f11 g10 g11 d11 c13 h10 h11 i10 i11 d12 d13 e12 e13 f12 f13 g12 g13 h12 h13 m12 j10

[Take away 8 from all the numbers (and leave the letters alone) to see the moves using the convention you're using (I'm on a standard 14x14 board so bottom row is 14 not 6)]

=== Defense against 1. d5 ===
Red can start like this:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 4:g3 E *:n3 E *:a4 E *:b4 R 2:d4 E *:a5 B 3:c5 B 1:d5"
/>

Continuation:

<div class="toccolours mw-collapsible mw-collapsed">
Red has this threat:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 R 3:e5 E +:f5 R 5:g5 B 4:d6 B 2:e6 E +:f6 E +:g6"
/>

So all these fields need to be considered:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 E +:e5 E +:f5 E +:g5 E +:d6 E +:e6 E +:f6 E +:g6"
/>

==== ... 5. f3 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 E *:a5 B c5 B d5"
/>

Red has 2 threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 E *:a5 B c5 B d5 R 1:e5 E +:d6 E +:e6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E +:h3 E *:n3 E *:a4 E *:b4 R d4 R e4 R 1:f4 E +:g4 E +:h4 E +:i4 E *:a5 B c5 B d5 E +:f5 E +:g5 E +:h5 E +:i5 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6"
/>

using [[Edge_template_IV2b|IV-2-b]].

Blue has to play on the only overlapping field:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 R 6:i4 E *:a5 B c5 B d5 R 2:e5 R 4:f5 B 3:d6 B 1:e6 B 5:f6"
/>

Red 6 is connected to the bottom and to the left with [[Tom's move]].

==== ... 5. e4 ====

Yet to come ...

==== ... 5. f4 ====

Yet to come ...

==== ... 5. g4 ====

Yet to come ...

==== ... 5. e5 ====

Yet to come ...

==== ... 5. f5 ====

Yet to come ...

==== ... 5. g5 ====

Yet to come ...

==== ... 5. d6 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 R 2:f4 E *:a5 B c5 B d5 B 1:d6"
/>

The group with 2 is connected to the top in two non-overlapping ways (see area marked with +) and to the bottom with [[Edge_template_IV2b|IV-2-b]].

==== ... 5. e6 ====

Red can respond here:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R 2:f5 B 1:e6"
/>

Continuation:
<div class="toccolours mw-collapsible mw-collapsed">

Now Red has two threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 R 1:e5 R f5 E +:d6 B e6 E +:f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 R 1:g4 E +:h4 E *:a5 B c5 B d5 E +:e5 R f5 E +:g5 E +:h5 B e6 E +:f6 E +:g6 E +:h6"
/>

Blue has to play on the overlap:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 E +:e5 R f5 B e6 E +:f6"
/>

''7. f2''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 1:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 R 8:h4 E *:a5 B c5 B d5 R 4:e5 R f5 B 7:g5 R 6:h5 B 5:d6 B e6 B 3:f6"
/>

''7. g2''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 1:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 6:h4 E *:a5 B c5 B d5 R f5 B 5:g5 R 4:h5 B e6 B 3:f6"
/>

''7. f3''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 R 6:i4 E *:a5 B c5 B d5 R 4:e5 R f5 B 5:d6 B e6 B 2:f6"
/>
6 is now connected to the left and to the bottom by [[Tom's move]].

''7. e4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 B 1:e4 R 4:i4 E *:a5 B c5 B d5 R f5 B e6 B 3:f6"
/>
4 is again connected to the left and to the bottom by [[Tom's move]].

''7. f4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 1:f4 E *:a5 B c5 B d5 E +:e5 R f5 E +:d6 B e6 E +:f6"
/>

Blue has to go on one of the 3 marked fields. However, d6 can't be any better than 35, so it's enough to look at e5 and g6. Let's have a look at g6 first:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 B f4 R 4:i4 E *:a5 B c5 B d5 R 2:e5 R f5 B 3:d6 B e6 B 1:f6"
/>

4 is now connected to the bottom and to the left in a similar way as in [[Tom's move]]. (Just the piece on g3 is connected to the left in a slightly different way.)

The other possible move was e5:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 2:h2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 B f4 E *:a5 B c5 B d5 B 1:e5 R f5 B e6"
/>

Red 2 is connected to the left by two non-overlapping ways and to the bottom by a 5th row template that has yet to be added to this wiki.

''7. g4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 1:g4 R 6:i4 E *:a5 B c5 B d5 R 4:e5 R f5 B 5:d6 B e6 B 3:f6"
/>

Once again 6 is now connected to the bottom and to the left in a similar way as in [[Tom's move]].

''7. e5''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 4:i4 E *:a5 B c5 B d5 B 1:e5 R f5 B e6 B 3:f6"
/>

And [[Tom's move]] at the end again.

''7. f6''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:i4 E *:a5 B c5 B d5 R f5 B e6 B 1:f6"
/>

Red 2 is connected to the bottom and to at least one of the red pieces in the middle by [[Tom's move]]. Red now has three threats to connect both these pieces to the top:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 1:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 R 1:f4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E +:f4 E +:g4 R i4 E *:a5 B c5 B d5 E +:e5 R f5 B e6 B f6"
/>

Blue has to play on the overlap:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

First move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 3:f4 R i4 E *:a5 B c5 B d5 R 4:e5 R f5 R 6:g5 B 5:d6 B e6 B f6"
/>

Second move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 R 4:h3 E *:n3 E *:a4 E *:b4 R d4 B 1:f4 B 3:g4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>
</div>
==== ... 5. f6 ====

Red can start like this:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R 2:e5 B 1:f6"
/>

Red now has these threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E *:a5 B c5 B d5 R e5 E +:d6 E +:e6 B f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 1:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 E +:h4 E *:a5 B c5 B d5 R e5 E +:f5 E +:g5 E +:h5 E +:d6 E +:e6 B f6 E +:g6 E +:h6"
/>

using [[Edge_template_IV2e|IV-2-e]].

Blue has to play on the overlap:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R e5 E +:d6 E +:e6 B f6"
/>

First move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 4:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 B 3:e4 B 5:f4 R 6:g4 E *:a5 B c5 B d5 R e5 B 7:f5 B 8:h5 B 1:d6 R 2:e6 B f6"
/>

You get the the defense against the other move by just swapping 1 and 2 in the diagram above.

==== ... 5. g6 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:f4 R 6:h4 E *:a5 B c5 B d5 B 5:f5 B 7:g5 R 8:i5 B 3:e6 R 4:f6 B 1:g6"
/>

Note that 2 is safely connected to the top, so 3 is forced.

</div>

=== Defense against 1. d6 ===

Red has this line:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R 2:d4 R 8:h4 E *:a5 B 3:c5 R 4:d5 R 6:e5 B 5:c6 B 1:d6 B 7:e6"
/>

Blue 3, 5 and 7 are forced.

Continuation:
<div class="toccolours mw-collapsible mw-collapsed">

Red has these threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 R h4 E +:i4 E *:a5 B c5 R d5 R e5 R 1:f5 R 3:g5 E +:h5 R 5:i5 B c6 B d6 B e6 B 2:f6 B 4:g6 E +:h6 E +:i6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 E +:g4 R h4 E *:a5 B c5 R d5 R e5 R 1:f5 E +:g5 R 3:h5 B c6 B d6 B e6 B 2:f6 E +:g6 E +:h6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R 1:g3 E +:h3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 R h4 E +:i4 E *:a5 B c5 R d5 R e5 E +:g5 E +:h5 E +:i5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6 E +:i6"
/>

The overlap in which Blue has to play is:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 R h4 E *:a5 B c5 R d5 R e5 E +:g5 E +:h5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6"
/>

Four of these five possible moves can be analysed in one line. In the following diagram assume Blue has played 1 on any of the fields marked with +:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 6:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 4:g3 E *:n3 E *:a4 E *:b4 R d4 B 5:f4 R h4 E *:a5 B c5 R d5 R e5 B 3:f5 R 2:g5 E +:h5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6"
/>

After Red 2 that group is safely connected to them bottom, now matter which of the pluses Blue chose before.

That leaves only one Blue move to deal with:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 7:g2 R 6:h2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 10:f3 R 8:g3 R 4:i3 E *:n3 E *:a4 E *:b4 R d4 B 9:f4 B 5:g4 R h4 E *:a5 B c5 R d5 R e5 R 2:f5 B 1:g5 B c6 B d6 B e6 B 3:f6"
/>

Note that Red 4 connects to the bottom with [[Edge_template_IV2b|IV-2-b]].
</div>
[[category:edge templates]]

Edge template V1b

2016-05-28T07:08:35Z

Wccanard: /* Defense against intrusions */

== The template ==

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 E *:a5"
/>

(From the [https://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 little golem forum])

Even though the diagram shows two pieces and one of them on the sixth row, it makes sense to say this is a fifth row template with one stone, as the stone on the sixth row plays no role whatsoever in connecting to the bottom. It could be removed and the stone on the fifth row will still be connected to the bottom. It is only shown so there is a possibility to connect to the top too, as without that it would not be of much use in a practical sense.

(Note: As I am writing this I have only seen the claim of this being a valid template on the little golem thread. I have not checked it yet and also not if it is minimal. However, as this came from a very competent player I have no reson to doubt it.)

== Defense against intrusions ==

Red has 3 main threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 E *:n3 E *:a4 E *:b4 E +:c4 R 1:d4 E *:a5 E +:b5 E +:c5 E +:d5 E +:a6 E +:b6 E +:c6 E +:d6"
/>

using the [[ziggurat]],

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 E *:n3 E *:a4 E *:b4 E +:c4 R 1:d4 E +:e4 E *:a5 E +:b5 E +:c5 E +:d5 E +:e5 E +:a6 E +:b6 E +:d6 E +:e6"
/>

using [[Defending_against_intrusions_in_template_1-IIIb|III-1-b]] and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E +:h2 E +:i2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:e3 R 1:f3 E +:g3 E +:h3 E +:i3 E +:j3 E *:n3 E *:a4 E *:b4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E *:a5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6"
/>

using [[Edge_template_IV1d|IV-1-d]].

For a blocking attempt, Blue has to play on the overlap:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:e3 E *:n3 E *:a4 E *:b4 E *:a5 E +:d5 E +:d6"
/>

=== Defense against 1. e3 ===

Yet to come ...

=== Defense against 1. d5 ===
Red can start like this:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 4:g3 E *:n3 E *:a4 E *:b4 R 2:d4 E *:a5 B 3:c5 B 1:d5"
/>

Continuation:

<div class="toccolours mw-collapsible mw-collapsed">
Red has this threat:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 R 3:e5 E +:f5 R 5:g5 B 4:d6 B 2:e6 E +:f6 E +:g6"
/>

So all these fields need to be considered:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 E +:e5 E +:f5 E +:g5 E +:d6 E +:e6 E +:f6 E +:g6"
/>

==== ... 5. f3 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 E *:a5 B c5 B d5"
/>

Red has 2 threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 E *:a5 B c5 B d5 R 1:e5 E +:d6 E +:e6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E +:h3 E *:n3 E *:a4 E *:b4 R d4 R e4 R 1:f4 E +:g4 E +:h4 E +:i4 E *:a5 B c5 B d5 E +:f5 E +:g5 E +:h5 E +:i5 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6"
/>

using [[Edge_template_IV2b|IV-2-b]].

Blue has to play on the only overlapping field:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 R 6:i4 E *:a5 B c5 B d5 R 2:e5 R 4:f5 B 3:d6 B 1:e6 B 5:f6"
/>

Red 6 is connected to the bottom and to the left with [[Tom's move]].

==== ... 5. e4 ====

Yet to come ...

==== ... 5. f4 ====

Yet to come ...

==== ... 5. g4 ====

Yet to come ...

==== ... 5. e5 ====

Yet to come ...

==== ... 5. f5 ====

Yet to come ...

==== ... 5. g5 ====

Yet to come ...

==== ... 5. d6 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 R 2:f4 E *:a5 B c5 B d5 B 1:d6"
/>

The group with 2 is connected to the top in two non-overlapping ways (see area marked with +) and to the bottom with [[Edge_template_IV2b|IV-2-b]].

==== ... 5. e6 ====

Red can respond here:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R 2:f5 B 1:e6"
/>

Continuation:
<div class="toccolours mw-collapsible mw-collapsed">

Now Red has two threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 R 1:e5 R f5 E +:d6 B e6 E +:f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 R 1:g4 E +:h4 E *:a5 B c5 B d5 E +:e5 R f5 E +:g5 E +:h5 B e6 E +:f6 E +:g6 E +:h6"
/>

Blue has to play on the overlap:

<hexboard size="6x14"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 E +:f4 E +:g4 E *:a5 B c5 B d5 E +:e5 R f5 B e6 E +:f6"
/>

''7. f2''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 1:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 R 8:h4 E *:a5 B c5 B d5 R 4:e5 R f5 B 7:g5 R 6:h5 B 5:d6 B e6 B 3:f6"
/>

''7. g2''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 1:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 6:h4 E *:a5 B c5 B d5 R f5 B 5:g5 R 4:h5 B e6 B 3:f6"
/>

''7. f3''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 R 6:i4 E *:a5 B c5 B d5 R 4:e5 R f5 B 5:d6 B e6 B 2:f6"
/>
6 is now connected to the left and to the bottom by [[Tom's move]].

''7. e4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 B 1:e4 R 4:i4 E *:a5 B c5 B d5 R f5 B e6 B 3:f6"
/>
4 is again connected to the left and to the bottom by [[Tom's move]].

''7. f4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 1:f4 E *:a5 B c5 B d5 E +:e5 R f5 E +:d6 B e6 E +:f6"
/>

Blue has to go on one of the 3 marked fields. However, d6 can't be any better than 35, so it's enough to look at e5 and g6. Let's have a look at g6 first:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 B f4 R 4:i4 E *:a5 B c5 B d5 R 2:e5 R f5 B 3:d6 B e6 B 1:f6"
/>

4 is now connected to the bottom and to the left in a similar way as in [[Tom's move]]. (Just the piece on g3 is connected to the left in a slightly different way.)

The other possible move was e5:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 2:h2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R e4 B f4 E *:a5 B c5 B d5 B 1:e5 R f5 B e6"
/>

Red 2 is connected to the left by two non-overlapping ways and to the bottom by a 5th row template that has yet to be added to this wiki.

''7. g4''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 1:g4 R 6:i4 E *:a5 B c5 B d5 R 4:e5 R f5 B 5:d6 B e6 B 3:f6"
/>

Once again 6 is now connected to the bottom and to the left in a similar way as in [[Tom's move]].

''7. e5''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 4:i4 E *:a5 B c5 B d5 B 1:e5 R f5 B e6 B 3:f6"
/>

And [[Tom's move]] at the end again.

''7. f6''

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:i4 E *:a5 B c5 B d5 R f5 B e6 B 1:f6"
/>

Red 2 is connected to the bottom and to at least one of the red pieces in the middle by [[Tom's move]]. Red now has three threats to connect both these pieces to the top:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 1:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:e4 R 1:f4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E +:f4 E +:g4 R i4 E *:a5 B c5 B d5 E +:e5 R f5 B e6 B f6"
/>

Blue has to play on the overlap:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>

First move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 B 1:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:e4 B 3:f4 R i4 E *:a5 B c5 B d5 R 4:e5 R f5 R 6:g5 B 5:d6 B e6 B f6"
/>

Second move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 2:f3 R g3 R 4:h3 E *:n3 E *:a4 E *:b4 R d4 B 1:f4 B 3:g4 R i4 E *:a5 B c5 B d5 R f5 B e6 B f6"
/>
</div>
==== ... 5. f6 ====

Red can start like this:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R 2:e5 B 1:f6"
/>

Red now has these threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:d3 E +:e3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 1:e4 E *:a5 B c5 B d5 R e5 E +:d6 E +:e6 B f6"
/>

and

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 1:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R g3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 E +:h4 E *:a5 B c5 B d5 R e5 E +:f5 E +:g5 E +:h5 E +:d6 E +:e6 B f6 E +:g6 E +:h6"
/>

using [[Edge_template_IV2e|IV-2-e]].

Blue has to play on the overlap:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 E *:a5 B c5 B d5 R e5 E +:d6 E +:e6 B f6"
/>

First move:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 4:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 B 3:e4 B 5:f4 R 6:g4 E *:a5 B c5 B d5 R e5 B 7:f5 B 8:h5 B 1:d6 R 2:e6 B f6"
/>

You get the the defense against the other move by just swapping 1 and 2 in the diagram above.

==== ... 5. g6 ====

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R g3 E *:n3 E *:a4 E *:b4 R d4 R 2:f4 R 6:h4 E *:a5 B c5 B d5 B 5:f5 B 7:g5 R 8:i5 B 3:e6 R 4:f6 B 1:g6"
/>

Note that 2 is safely connected to the top, so 3 is forced.

</div>

=== Defense against 1. d6 ===

Red has this line:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R 2:d4 R 8:h4 E *:a5 B 3:c5 R 4:d5 R 6:e5 B 5:c6 B 1:d6 B 7:e6"
/>

Blue 3, 5 and 7 are forced.

Continuation:
<div class="toccolours mw-collapsible mw-collapsed">

Red has these threats:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 R h4 E +:i4 E *:a5 B c5 R d5 R e5 R 1:f5 R 3:g5 E +:h5 R 5:i5 B c6 B d6 B e6 B 2:f6 B 4:g6 E +:h6 E +:i6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 E +:g4 R h4 E *:a5 B c5 R d5 R e5 R 1:f5 E +:g5 R 3:h5 B c6 B d6 B e6 B 2:f6 E +:g6 E +:h6"
/>

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E +:f2 E +:g2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E +:f3 R 1:g3 E +:h3 E *:n3 E *:a4 E *:b4 R d4 E +:f4 E +:g4 R h4 E +:i4 E *:a5 B c5 R d5 R e5 E +:g5 E +:h5 E +:i5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6 E +:i6"
/>

The overlap in which Blue has to play is:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 E *:n3 E *:a4 E *:b4 R d4 R h4 E *:a5 B c5 R d5 R e5 E +:g5 E +:h5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6"
/>

Four of these five possible moves can be analysed in one line. In the following diagram assume Blue has played 1 on any of the fields marked with +:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 R 6:f2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 4:g3 E *:n3 E *:a4 E *:b4 R d4 B 5:f4 R h4 E *:a5 B c5 R d5 R e5 B 3:f5 R 2:g5 E +:h5 B c6 B d6 B e6 E +:f6 E +:g6 E +:h6"
/>

After Red 2 that group is safely connected to them bottom, now matter which of the pluses Blue chose before.

That leaves only one Blue move to deal with:

<hexboard size="6x14"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:i1 E *:j1 E *:k1 E *:l1 E *:m1 E *:n1 E *:a2 E *:b2 E *:c2 E *:d2 R e2 B 7:g2 R 6:h2 E *:m2 E *:n2 E *:a3 E *:b3 E *:c3 R 10:f3 R 8:g3 R 4:i3 E *:n3 E *:a4 E *:b4 R d4 B 9:f4 B 5:g4 R h4 E *:a5 B c5 R d5 R e5 R 2:f5 B 1:g5 B c6 B d6 B e6 B 3:f6"
/>

Note that Red 4 connects to the bottom with [[Edge_template_IV2b|IV-2-b]].
</div>
[[category:edge templates]]

Talk:Edge template V1b

2016-05-28T07:08:00Z

Wccanard: comments about computer check

Theory of ladder escapes

2016-05-22T00:10:50Z

Wccanard:

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern, or removing a red stone from the pattern (and replacing it with an empty hex) gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 in the above template indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter pattern above -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of the above pattern on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of the above pattern on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder '''but it is not necessary'''. For example, here is a third row ladder escape:

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 E *:a2 E *:b2 E +:c2 E *:d2 E *:e2 E *:a3 E *:b3 E +:c3 E *:d3 E *:e3 E *:a4 E *:b4 E +:c4 E *:d4 E *:e4"
/>
It's a bit of a weird one, but one can check directly that (using notation as in the theorem) if P is this pattern then A+1+P and B+1+P are both edge templates, so the induction kicks in from there. Furthermore A+P is also an edge template. However B+P is the position

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 B a2 B b2 B d2 B e2 R a3 R b3 R 1:c3 B d3 B e3 B a4 B b4 B d4 B e4"
/>
and the 1 piece (connected to the top) cannot connect to the bottom or the 10 pieces. Our proposition is hence not sufficient to check that a given pattern is a 3rd row ladder escape. A weaker version of the proposition would however suffice: if P is a pattern and there exists some N>=0 such that B+N+P and all of A+N+P, A+(N-1)+P, A+(N-2)+P, ..., A+1+P, A+P are edge templates, then P is a 3rd row ladder escape pattern. This modified proposition is necessary and sufficient (because the attacker can choose to move up a row if they are laddering on the 2nd row).

My second remark is that despite all this, a 3rd row escape is '''not''' in general a 2nd row escape. More formally, if P (with its three plusses) is a third row ladder escape template, it may not be the case that removing the top plus gives us a second row ladder template. The simplest counterexample is the following:

<hexboard size="4x2"
coords="hide"
contents="E *:a1 R 10:b1 E +:a2 E *:b2 E +:a3 E *:b3 E +:a4 E *:b4"
/>

In words, this is a stone on the 4th row connected to the bottom edge. A perhaps clearer way of thinking about it is this:
<hexboard size="4x3"
coords="hide"
contents="B a1 R b1 R c1 E +:a2 B b2 R c2 E +:a3 B b3 R c3 E +:a4 B b4 R c4"
/>
A stone on the 4th row, connected to the bottom edge, is a 3rd row ladder escape, because red just ladders over to it on either the 2nd or 3rd row and then leaps up to connect to it. However removing the top plus gives us a pattern that is not a second row ladder escape, because for a second row ladder blue is allowed to completely occupy the third row (the third row is not part of the escape):
<hexboard size="4x7"
coords="hide"
contents="R a1 B e1 R f1 R g1 R a2 B b2 B c2 B d2 B e2 B f2 R g2 R a3 R 1:b3 R 3:c3 B f3 R g3 B a4 B 2:b4 B f4 R g4"
/>
and red cannot connect.

An even weaker question one could ask: what if we have a third row ladder escape, and we replace the top plus sign with an empty hex considered as part of the pattern? Is this always a 2nd row template? Again the answer is no and the weird example above (with A+n+P an edge template and B+n+P for all n>=1 an edge template, but not B+P) is a counterexample.

Conclusion: a third row ladder escape template is not always a second row ladder escape template!

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T23:34:18Z

Wccanard:

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern, or removing a red stone from the pattern (and replacing it with an empty hex) gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 in the above template indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter pattern above -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of the above pattern on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of the above pattern on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder *but it is not necessary*. For example, here is a third row ladder escape:

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 E *:a2 E *:b2 E +:c2 E *:d2 E *:e2 E *:a3 E *:b3 E +:c3 E *:d3 E *:e3 E *:a4 E *:b4 E +:c4 E *:d4 E *:e4"
/>
It's a bit of a weird one, but one can check directly that (using notation as in the theorem) if P is this pattern then A+2+P and B+2+P are both edge templates, so the induction kicks in from there. Furthermore A+1+P and A+P are also both edge templates. However B+P is the position

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 B a2 B b2 B d2 B e2 R a3 R b3 R 1:c3 B d3 B e3 B a4 B b4 B d4 B e4"
/>
and the 1 piece (connected to the top) cannot connect to the bottom or the 10 pieces. Our proposition is hence not sufficient to check that a given pattern is a 3rd row ladder escape. A weaker version of the proposition would however suffice: if P is a pattern and there exists some N>=0 such that B+N+P and all of A+N+P, A+(N-1)+P, A+(N-2)+P, ..., A+1+P, A+P are edge templates, then P is a 3rd row ladder escape pattern. This modified proposition is necessary and sufficient (because the attacker can choose to move up a row if they are laddering on the 2nd row).

My second remark is that despite all this, a 3rd row escape is '''not''' in general a 2nd row escape. More formally, if P (with its three plusses) is a third row ladder escape template, it may not be the case that removing the top plus gives us a second row ladder template. The simplest counterexample is the following:

<hexboard size="4x2"
coords="hide"
contents="E *:a1 R 10:b1 E +:a2 E *:b2 E +:a3 E *:b3 E +:a4 E *:b4"
/>

In words, this is a stone on the 4th row connected to the bottom edge. A perhaps clearer way of thinking about it is this:
<hexboard size="4x3"
coords="hide"
contents="B a1 R b1 R c1 E +:a2 B b2 R c2 E +:a3 B b3 R c3 E +:a4 B b4 R c4"
/>
A stone on the 4th row, connected to the bottom edge, is a 3rd row ladder escape, because red just ladders over to it on either the 2nd or 3rd row and then leaps up to connect to it. However removing the top plus gives us a pattern that is not a second row ladder escape, because for a second row ladder blue is allowed to completely occupy the third row (the third row is not part of the escape):
<hexboard size="4x7"
coords="hide"
contents="R a1 B e1 R f1 R g1 R a2 B b2 B c2 B d2 B e2 B f2 R g2 R a3 R 1:b3 R 3:c3 B f3 R g3 B a4 B 2:b4 B f4 R g4"
/>
and red cannot connect.

A question that has just occurred to me: what if we have a third row ladder escape, and we replace the top plus sign with an empty hex considered as part of the pattern? Is this always a 2nd row template? I will have to think about this.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T23:29:23Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 in the above template indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter pattern above -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of the above pattern on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of the above pattern on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder *but it is not necessary*. For example, here is a third row ladder escape:

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 E *:a2 E *:b2 E +:c2 E *:d2 E *:e2 E *:a3 E *:b3 E +:c3 E *:d3 E *:e3 E *:a4 E *:b4 E +:c4 E *:d4 E *:e4"
/>
It's a bit of a weird one, but one can check directly that (using notation as in the theorem) if P is this pattern then A+2+P and B+2+P are both edge templates, so the induction kicks in from there. Furthermore A+1+P and A+P are also both edge templates. However B+P is the position

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 B a2 B b2 B d2 B e2 R a3 R b3 R 1:c3 B d3 B e3 B a4 B b4 B d4 B e4"
/>
and the 1 piece (connected to the top) cannot connect to the bottom or the 10 pieces. Our proposition is hence not sufficient to check that a given pattern is a 3rd row ladder escape. A weaker version of the proposition would however suffice: if P is a pattern and there exists some N>=0 such that B+N+P and all of A+N+P, A+(N-1)+P, A+(N-2)+P, ..., A+1+P, A+P are edge templates, then P is a 3rd row ladder escape pattern. This modified proposition is necessary and sufficient (because the attacker can choose to move up a row if they are laddering on the 2nd row).

My second remark is that despite all this, a 3rd row escape is '''not''' in general a 2nd row escape. More formally, if P (with its three plusses) is a third row ladder escape template, it may not be the case that removing the top plus gives us a second row ladder template. The simplest counterexample is the following:

<hexboard size="4x2"
coords="hide"
contents="E *:a1 R 10:b1 E +:a2 E *:b2 E +:a3 E *:b3 E +:a4 E *:b4"
/>

In words, this is a stone on the 4th row connected to the bottom edge. A perhaps clearer way of thinking about it is this:
<hexboard size="4x3"
coords="hide"
contents="B a1 R b1 R c1 E +:a2 B b2 R c2 E +:a3 B b3 R c3 E +:a4 B b4 R c4"
/>
A stone on the 4th row, connected to the bottom edge, is a 3rd row ladder escape, because red just ladders over to it on either the 2nd or 3rd row and then leaps up to connect to it. However removing the top plus gives us a pattern that is not a second row ladder escape, because for a second row ladder blue is allowed to completely occupy the third row (the third row is not part of the escape):
<hexboard size="4x7"
coords="hide"
contents="R a1 B e1 R f1 R g1 R a2 B b2 B c2 B d2 B e2 B f2 R g2 R a3 R 1:b3 R 3:c3 B f3 R g3 B a4 B 2:b4 B f4 R g4"
/>
and red cannot connect.

A question that has just occurred to me: what if we have a third row ladder escape, and we replace the top plus sign with an empty hex considered as part of the pattern? Is this always a 2nd row template? I will have to think about this.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T23:27:24Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 in the above template indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter pattern above -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of the above pattern on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of the above pattern on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder *but it is not necessary*. For example, here is a third row ladder escape:

<hexboard size="4x5"
coords="hide"
contents="R 10:a1 R 10:e1 E *:a2 E *:b2 E +:c2 E *:d2 E *:e2 E *:a3 E *:b3 E +:c3 E *:d3 E *:e3 E *:a4 E *:b4 E +:c4 E *:d4 E *:e4"
/>
It's a bit of a weird one, but one can check directly that (using notation as in the theorem) if P is this pattern then A+2+P and B+2+P are both edge templates, so the induction kicks in from there. Furthermore A+1+P and A+P are also both edge templates. However B+P is the position

<hexboard size="4x4"
coords="hide"
contents="R 10:a1 R 10:d1 B a2 B c2 B d2 R a3 R 1:b3 B c3 B d3 B a4 B c4 B d4"
/>
and the 1 piece (connected to the top) cannot connect to the bottom or the 10 pieces. Our proposition is hence not sufficient to check that a given pattern is a 3rd row ladder escape. A weaker version of the proposition would however suffice: if P is a pattern and there exists some N>=0 such that B+N+P and all of A+N+P, A+(N-1)+P, A+(N-2)+P, ..., A+1+P, A+P are edge templates, then P is a 3rd row ladder escape pattern. This modified proposition is necessary and sufficient (because the attacker can choose to move up a row if they are laddering on the 2nd row).

My second remark is that despite all this, a 3rd row escape is '''not''' in general a 2nd row escape. More formally, if P (with its three plusses) is a third row ladder escape template, it may not be the case that removing the top plus gives us a second row ladder template. The simplest counterexample is the following:

<hexboard size="4x2"
coords="hide"
contents="E *:a1 R 10:b1 E +:a2 E *:b2 E +:a3 E *:b3 E +:a4 E *:b4"
/>

In words, this is a stone on the 4th row connected to the bottom edge. A perhaps clearer way of thinking about it is this:
<hexboard size="4x3"
coords="hide"
contents="B a1 R b1 R c1 E +:a2 B b2 R c2 E +:a3 B b3 R c3 E +:a4 B b4 R c4"
/>
A stone on the 4th row, connected to the bottom edge, is a 3rd row ladder escape, because red just ladders over to it on either the 2nd or 3rd row and then leaps up to connect to it. However removing the top plus gives us a pattern that is not a second row ladder escape, because for a second row ladder blue is allowed to completely occupy the third row (the third row is not part of the escape):
<hexboard size="4x7"
coords="hide"
contents="R a1 B e1 R f1 R g1 R a2 B b2 B c2 B d2 B e2 B f2 R g2 R a3 R 1:b3 R 3:c3 B f3 R g3 B a4 B 2:b4 B f4 R g4"
/>
and red cannot connect.

A question that has just occurred to me: what if we have a third row ladder escape, and we replace the top plus sign with an empty hex considered as part of the pattern? Is this always a 2nd row template? I will have to think about this.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T23:25:20Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 in the above template indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter pattern above -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of the above pattern on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of the above pattern on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder *but it is not necessary*. For example, here is a third row ladder escape:

<hexboard size="4x4"
coords="hide"
contents="R 10:a1 R 10:d1 E *:a2 E +:b2 E *:c2 E *:d2 E *:a3 E +:b3 E *:c3 E *:d3 E *:a4 E +:b4 E *:c4 E *:d4"
/>

It's a bit of a weird one, but one can check directly that (using notation as in the theorem) if P is this pattern then A+2+P and B+2+P are both edge templates, so the induction kicks in from there. Furthermore A+1+P and A+P are also both edge templates. However B+P is the position

<hexboard size="4x4"
coords="hide"
contents="R 10:a1 R 10:d1 B a2 B c2 B d2 R a3 R 1:b3 B c3 B d3 B a4 B c4 B d4"
/>
and the 1 piece (connected to the top) cannot connect to the bottom or the 10 pieces. Our proposition is hence not sufficient to check that a given pattern is a 3rd row ladder escape. A weaker version of the proposition would however suffice: if P is a pattern and there exists some N>=0 such that B+N+P and all of A+N+P, A+(N-1)+P, A+(N-2)+P, ..., A+1+P, A+P are edge templates, then P is a 3rd row ladder escape pattern. This modified proposition is necessary and sufficient (because the attacker can choose to move up a row if they are laddering on the 2nd row).

My second remark is that despite all this, a 3rd row escape is '''not''' in general a 2nd row escape. More formally, if P (with its three plusses) is a third row ladder escape template, it may not be the case that removing the top plus gives us a second row ladder template. The simplest counterexample is the following:

<hexboard size="4x2"
coords="hide"
contents="E *:a1 R 10:b1 E +:a2 E *:b2 E +:a3 E *:b3 E +:a4 E *:b4"
/>

In words, this is a stone on the 4th row connected to the bottom edge. A perhaps clearer way of thinking about it is this:
<hexboard size="4x3"
coords="hide"
contents="B a1 R b1 R c1 E +:a2 B b2 R c2 E +:a3 B b3 R c3 E +:a4 B b4 R c4"
/>
A stone on the 4th row, connected to the bottom edge, is a 3rd row ladder escape, because red just ladders over to it on either the 2nd or 3rd row and then leaps up to connect to it. However removing the top plus gives us a pattern that is not a second row ladder escape, because for a second row ladder blue is allowed to completely occupy the third row (the third row is not part of the escape):
<hexboard size="4x7"
coords="hide"
contents="R a1 B e1 R f1 R g1 R a2 B b2 B c2 B d2 B e2 B f2 R g2 R a3 R 1:b3 R 3:c3 B f3 R g3 B a4 B 2:b4 B f4 R g4"
/>
and red cannot connect.

A question that has just occurred to me: what if we have a third row ladder escape, and we replace the top plus sign with an empty hex considered as part of the pattern? Is this always a 2nd row template? I will have to think about this.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T23:00:02Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 in the above template indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter pattern above -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of the above pattern on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of the above pattern on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder *but it is not necessary*. For example, here is a third row ladder escape:

<hexboard size="4x4"
coords="hide"
contents="R 10:a1 R 10:d1 E *:a2 E +:b2 E *:c2 E *:d2 E *:a3 E +:b3 E *:c3 E *:d3 E *:a4 E +:b4 E *:c4 E *:d4"
/>

not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T22:58:11Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

I want to make some remarks about this argument, but first here are some examples of third row ladder escapes; in each case one can prove that the pattern is an escape using the proposition above. Again these are all taken from Dr King's website http://www.drking.org.uk/hexagons/hex/templates.html

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>

<hexboard size="3x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
[the 10 indicates a piece connected to the bottom row; note that I have moved the plusses one hex to the right from the version on Dr King's website, which is not minimal].
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 10:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
[again the 10 indicates a piece connected to the bottom row]
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 R d1 E *:f1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="3x2"
coords="hide"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
[this latter template -- a single stone on the second row -- is my favourite, because the proposition in this section can be used to show that it is a 3rd row ladder escape even though the template is so small.
<hexboard size="4x3"
coords="hide"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
[note that the version of this on Dr King's website is not minimal; I have moved the plusses one hex to the right]
<hexboard size="6x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 R d2 E *:a3 E *:b3 E +:a4 E +:a5 E +:a6"
/>
[note that the version of this on Dr King's website has the plusses (he uses arrows) sloping in the other direction; the location that I have put them makes the template minimal].

All of the patterns above can be proved to be third row ladder escapes using the proposition above, and I think they're all minimal.

However there are also some cautionary remarks to be made about 3rd row ladder escapes.

Firstly, in contrast to the situation with second row ladders, the proposition in this section is sufficient to show that a position is a third row ladder *but it is not necessary*. For example, here is a third row ladder escape:

<hexboard size="4x4"
coords="hide"
contents="R 10:a1 R 10:d1 E *:a2 E +:b2 E *:c2 E *:d2 E *:a3 E +:b3 E *:c3 E *:d3 E *:a4 E +:b4 E *:c4 E *:d4"
/>

not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T22:40:09Z

Wccanard:

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This may not be as easy as it looks. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More issues of this nature surely await anyone who tries to extend these results to 5th row ladders or higher. Because I am not sure that 5th row ladder escapes really ever come up in practice, I am not particularly motivated to pursue this.

Theory of ladder escapes

2016-05-21T22:38:12Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:37:13Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:36:36Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 3 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:35:19Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we have already dealt with.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 R 5:d2 E *:a3 B 2:c3 B 4:b4"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:33:04Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:32:26Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position B+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:31:30Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
Remember what we are doing -- this is the first two columns of position D at distance (m+1) from the pattern. From D+(m+1) blue must defend under the red ladder, and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:b3 R 5:c3 B 4:a4 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:29:13Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
From D+(m+1) blue must defend and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that there are only two moves which do not lose instantly for blue. We now deal with these two moves in turn. This one
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and this
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In either case we are done by the inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B and hence for all four templates.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:27:21Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
From D+(m+1) blue must defend and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:26:40Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
From D+(m+1) blue must defend and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the position C+(m+1). First note that all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:25:24Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
From D+(m+1) blue must defend and then red can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.

We next deal with the positionC+(m+1). First note that all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:24:03Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="R 1:a3 R 3:b3 B 2:a4"
/>
From D+(m+1) black must defend and then white can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.
The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:23:24Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: Note that position B is A+1, and that it's crucial that we use A+1 rather than A in this argument; this is a new phenomenon which appears for 4th row ladders.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The second easiest case is case D:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
From D+(m+1) black must defend and then white can play to the above position; we know by the inductive hypothesis that D+m is an edge template, and hence D+(m+1) is too.
The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:21:02Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:18:51Z

Wccanard: /* Fourth row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is the following data. It is a pattern, plus four plussed hexes (not part of the pattern) going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:17:17Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1)

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge, so piece three connects to the edge, so piece 1 does too. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by the inductive hypothesis B+m connects to the edge, which means that piece 5 connects to the edge, so piece 1 does too.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:15:28Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this -- the first three columns of position A+(m+1):

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:14:34Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move then red can reply thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
Now by our inductive hypothesis, position B+m is an edge template. This implies that piece 3 is connected to the edge, and hence piece 1 is too. So B+(m+1) is also an edge template.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:13:10Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider the left most two columns in position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:11:39Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above (still marked with stars), if it is a 3rd row ladder escape, must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:10:40Z

Wccanard: /* Third row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture, which is sitting on the bottom row of a hex board:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:08:57Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red can continue playing the ladder on the second row, and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:07:17Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with plusses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:06:00Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as moving right and approaching the escape from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:05:09Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as coming in from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:04:19Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means (amongst other things) that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B 10:g4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as coming in from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T22:03:45Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

The reason that the pattern is a second stone ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
Even here red can force a connection to the edge, even if it is blue's move, because blue must keep defending on the first row and red keeps attacking on the second row:
<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 R 3:d3 R 5:e3 R 7:f3 R 9:g3 B j3 B a4 B b4 B 2:c4 B 4:d4 B 6:e4 B 6:f4 B 10:g4 B j4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that red can break through to the edge.

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as coming in from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T21:58:35Z

Wccanard: /* 2nd row ladder escapes */

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is the following pattern.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as coming in from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Open problems about edge templates

2016-05-21T21:54:54Z

Wccanard: /* Is there a single-stone 7th row edge template? */

== Is there a single-stone 7th row edge template? ==

If one looks at the [[edge templates with one stone|single-stone edge templates]] on this site, you see templates where the single stone is getting further and further away from the edge. One might then jump to the conclusion that (as long as you have enough columns) you can create templates with one stone as high up as you like. This may be true, but at the time of writing (May 2016) this seems to be an '''open problem''' -- nobody knows how to do construct these templates. In fact when Cameron Browne wrote his book "Connection games" in 2005 he gave what at the time was apparently a complete list of known minimal single-stone edge templates (see the [[edge template|edge templates]] page for an explanation of the terminology used here), and there were only 12 such templates (counting reflections as different templates). There were some 5th row single-stone templates, but no 6th row template at all seemed to be known at the time.

In 2008 Art Duval announced the existence of a single-stone 6th row template on the Littlegolem forum here: https://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=339

Unfortunately, many of the links and the pictures in that thread no longer work; however the 6th row template lives on:
<hex>
R6 C14
Sa1 Sb1 Sc1 Sd1 Se1 Sf1 Sg1 Sh1 Rj1 Sl1 Sm1 Sn1
Sa2 Sb2 Sc2 Sd2 Se2 Sf2 Sm2 Sn2
Sa3 Sb3 Sc3 Sd3 Sn3
Sa4 Sb4
Sa5
</hex>
The template is also recorded on this site's page about [[edge templates with one stone]].

For simple 3rd row templates like the Ziggurat, the stone to be connected is on the 3rd row and there are no hexes on the 4th or higher rows in the template at all. However for a general single-stone n'th row template one could allow for the possibility that there are vacant hexes in the template on rows higher than n. For example this 4th row single-stone edge template
<hex> R5 C9
Sa1 Sb1 Sc1 Sd1 Sh1 Si1 Si2
Sa2 Sb2 Sc2 Vd2
Sa3 Sb3
Sa4
</hex>
has vacant hexes on the 5th row in the template.

But even allowing for the possibility of vacant template hexes in higher rows than the stone, it still seems at the time of writing (May 2016) that '''no 7th row single stone edge template is known'''. Is there such a template? If so, is there a single-stone 8th row edge template? Is there a single-stone 1000th row edge template? Experts seem to be divided on this, if the Littlegolem forum thread is anything to go by.

== Single-stone templates to connect a 7th row ladder ==

Say that red is laddering along the 4th row and blue suddenly decides to play elsewhere (move 1 in the position below).

<hexboard size="6x10"
coords="hide"
contents="R a1 B 1:j1 R a2 R a3 R b3 R c3 R d3 B a4 B b4 B c4"
/>

Red can now instantly connect to the bottom by playing on the third row, because of the Ziggurat 3rd row single-stone edge template (marked with plus signs).
<hexboard size="6x10"
coords="hide"
contents="R a1 B 1:j1 R a2 R a3 R b3 R c3 R d3 B a4 B b4 B c4 R 2:d4 E +:e4 E +:c5 E +:d5 E +:e5 E +:b6 E +:c6 E +:d6 E +:e6"
/>

But because no single-stone 7th row template is known, if red is laddering along the 8th row and blue decides to ignore the ladder and play elsewhere, red might not be able to immediately connect! Red can play on the 7th row but perhaps blue can block the connection to the edge, even on a very large board.

In fact, this issue is even a problem for 7th row ladders. Because at the time of writing (May 2016) the only known minimal 6th row single-stone template is (modulo reflection) the one shown in the previous section above, and this template has vacant hexes to both the left and the right of the stone, the template is not suitable for connecting a 6th row stone placed adjacent to a 7th row ladder. Is red's stone 1 connected to the bottom edge here?
<hexboard size="7x15"
coords="hide"
contents="R a1 R b1 R c1 R d1 R e1 R f1 R g1 B a2 B b2 B c2 B d2 B e2 B f2 R 1:g2"
/>
Our 6th row single stone edge template cannot be used. This sort of problem seems very difficult at the time of writing, even using the publically available computer programs for hex analysis, although one might imagine that custom computer code could be used to analyse this sort of position, and of course computers get faster over time so perhaps this problem will one day soon be accessible.

Because this problem seems to be unsolved, this provides a theoretical barrier for analysing 7th row ladder escape templates, as the defender can play in the escape before the ladder has reached the escape. For example it may be theoretically possible that no 7th row ladder escape template exists, although such a possibility does not seem likely.

We finish this section with the remark that this is not an issue for 6th row ladders, because an appropriate 5th row template is known: the hexes marked with a plus and the piece marked 1 form a single-stone 5th row template.

<hexboard size="7x16"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 B p1 R a2 R b2 R c2 R d2 R e2 E +:f2 E +:g2 E +:h2 B a3 B b3 B c3 B d3 R 1:e3 E +:f3 E +:g3 E +:h3 E +:i3 E +:j3 E +:k3 E +:l3 E +:d4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E +:l4 E +:m4 E +:c5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:l5 E +:m5 E +:n5 E +:b6 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6 E +:l6 E +:m6 E +:n6 E +:a7 E +:b7 E +:c7 E +:d7 E +:e7 E +:f7 E +:g7 E +:h7 E +:i7 E +:j7 E +:k7 E +:l7 E +:m7 E +:n7"
/>
In particular black has to keep defending against a 6th row ladder until the template no longer fits (e.g. because we are approaching a ladder escape).

=== Single-stone templates with a 120 degree corner ===

One reason that the standard 4th row edge template
<hex> R4 C7
Sa1 Sb1 Sc1 Sd1 Ve1 Sg1
Sa2 Sb2
Sa3</hex>
is more useful than the standard 5th row edge template
<hex>R5 C10 Vg1 Sa1 Sa2 Sa3 Sa4 Sb1 Sb2 Sb3 Sc1 Sc2 Sd1 Sd2 Se1 Si1 Sj1 Sj2</hex>
is because the stone in the 4th row template has three "liberties", i.e. adjacent hexes on the board but not in the template, whereas the stone in the 5th row template only has two. However there is a single-stone 5th row template where the stone has three liberties:
<hexboard size="6x16"
coords="hide"
contents="E +:h1 E +:i1 E +:j1 R f2 E +:g2 E +:h2 E +:i2 E +:j2 E +:k2 E +:l2 E +:m2 E +:n2 E +:e3 E +:f3 E +:g3 E +:h3 E +:i3 E +:j3 E +:k3 E +:l3 E +:m3 E +:n3 E +:o3 E +:c4 E +:d4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E +:l4 E +:m4 E +:n4 E +:o4 E +:p4 E +:b5 E +:c5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:l5 E +:m5 E +:n5 E +:o5 E +:p5 E +:a6 E +:b6 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6 E +:l6 E +:m6 E +:n6 E +:o6 E +:p6"
/>
(I learnt this template from shalev at http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 ).

Is there a single-stone 6th row template where the stone has three liberties outside the template?

Open problems about edge templates

2016-05-21T20:43:58Z

Wccanard: fix typo

== Is there a single-stone 7th row edge template? ==

If one looks at the [[edge templates with one stone|single-stone edge templates]] on this site, one might jump to the conclusion that if you allow yourself more and more space, you can create templates with one stone as high up as you like. This might be true, but this is an open problem -- nobody knows how to do it. In fact when Cameron Browne wrote his book "Connection games" in 2005 he gave what at the time was apparently a complete list of known minimal single-stone edge templates (see the [[edge template|edge templates]] page for an explanation of the terminology used here). There were only 12 such templates, and this included reflections. There were some 5th row single-stone templates, but no 6th row template seemed to be known at the time.

In 2008 Art Duval announced the existence of a single-stone 6th row template on the Littlegolem forum here: https://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=339

Unfortunately, many of the links and the pictures in that thread no longer work; however the 6th row template lives on:
<hex>
R6 C14
Sa1 Sb1 Sc1 Sd1 Se1 Sf1 Sg1 Sh1 Rj1 Sl1 Sm1 Sn1
Sa2 Sb2 Sc2 Sd2 Se2 Sf2 Sm2 Sn2
Sa3 Sb3 Sc3 Sd3 Sn3
Sa4 Sb4
Sa5
</hex>
The template is also recorded on this site's page about [[edge templates with one stone]].

For simple 3rd row templates like the Ziggurat, the stone to be connected is on the 3rd row and there are no hexes on the 4th or higher rows in the template at all. However for a general single-stone n'th row template one could allow for the possibility that there are vacant hexes in the template on rows higher than n. For example this 4th row single-stone edge template
<hex> R5 C9
Sa1 Sb1 Sc1 Sd1 Sh1 Si1 Si2
Sa2 Sb2 Sc2 Vd2
Sa3 Sb3
Sa4
</hex>
has vacant hexes on the 5th row in the template.

But even allowing for the possibility of vacant template hexes in higher rows than the stone, it still seems at the time of writing (May 2016) that '''no 7th row single stone edge template is known'''. Is there such a template? If so, is there a single-stone 8th row edge template? Is there a single-stone 1000th row edge template? Experts seem to be divided on this, if the Littlegolem forum thread is anything to go by.

== Single-stone templates to connect a 7th row ladder ==

Say that red is laddering along the 4th row and blue suddenly decides to play elsewhere (move 1 in the position below).

<hexboard size="6x10"
coords="hide"
contents="R a1 B 1:j1 R a2 R a3 R b3 R c3 R d3 B a4 B b4 B c4"
/>

Red can now instantly connect to the bottom by playing on the third row, because of the Ziggurat 3rd row single-stone edge template (marked with plus signs).
<hexboard size="6x10"
coords="hide"
contents="R a1 B 1:j1 R a2 R a3 R b3 R c3 R d3 B a4 B b4 B c4 R 2:d4 E +:e4 E +:c5 E +:d5 E +:e5 E +:b6 E +:c6 E +:d6 E +:e6"
/>

But because no single-stone 7th row template is known, if red is laddering along the 8th row and blue decides to ignore the ladder and play elsewhere, red might not be able to immediately connect! Red can play on the 7th row but perhaps blue can block the connection to the edge, even on a very large board.

In fact, this issue is even a problem for 7th row ladders. Because at the time of writing (May 2016) the only known minimal 6th row single-stone template is (modulo reflection) the one shown in the previous section above, and this template has vacant hexes to both the left and the right of the stone, the template is not suitable for connecting a 6th row stone placed adjacent to a 7th row ladder. Is red's stone 1 connected to the bottom edge here?
<hexboard size="7x15"
coords="hide"
contents="R a1 R b1 R c1 R d1 R e1 R f1 R g1 B a2 B b2 B c2 B d2 B e2 B f2 R 1:g2"
/>
Our 6th row single stone edge template cannot be used. This sort of problem seems very difficult at the time of writing, even using the publically available computer programs for hex analysis, although one might imagine that custom computer code could be used to analyse this sort of position, and of course computers get faster over time so perhaps this problem will one day soon be accessible.

Because this problem seems to be unsolved, this provides a theoretical barrier for analysing 7th row ladder escape templates, as the defender can play in the escape before the ladder has reached the escape. For example it may be theoretically possible that no 7th row ladder escape template exists, although such a possibility does not seem likely.

We finish this section with the remark that this is not an issue for 6th row ladders, because an appropriate 5th row template is known: the hexes marked with a plus and the piece marked 1 form a single-stone 5th row template.

<hexboard size="7x16"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 B p1 R a2 R b2 R c2 R d2 R e2 E +:f2 E +:g2 E +:h2 B a3 B b3 B c3 B d3 R 1:e3 E +:f3 E +:g3 E +:h3 E +:i3 E +:j3 E +:k3 E +:l3 E +:d4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E +:l4 E +:m4 E +:c5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:l5 E +:m5 E +:n5 E +:b6 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6 E +:l6 E +:m6 E +:n6 E +:a7 E +:b7 E +:c7 E +:d7 E +:e7 E +:f7 E +:g7 E +:h7 E +:i7 E +:j7 E +:k7 E +:l7 E +:m7 E +:n7"
/>
In particular black has to keep defending against a 6th row ladder until the template no longer fits (e.g. because we are approaching a ladder escape).

=== Single-stone templates with a 120 degree corner ===

One reason that the standard 4th row edge template
<hex> R4 C7
Sa1 Sb1 Sc1 Sd1 Ve1 Sg1
Sa2 Sb2
Sa3</hex>
is more useful than the standard 5th row edge template
<hex>R5 C10 Vg1 Sa1 Sa2 Sa3 Sa4 Sb1 Sb2 Sb3 Sc1 Sc2 Sd1 Sd2 Se1 Si1 Sj1 Sj2</hex>
is because the stone in the 4th row template has three "liberties", i.e. adjacent hexes on the board but not in the template, whereas the stone in the 5th row template only has two. However there is a single-stone 5th row template where the stone has three liberties:
<hexboard size="6x16"
coords="hide"
contents="E +:h1 E +:i1 E +:j1 R f2 E +:g2 E +:h2 E +:i2 E +:j2 E +:k2 E +:l2 E +:m2 E +:n2 E +:e3 E +:f3 E +:g3 E +:h3 E +:i3 E +:j3 E +:k3 E +:l3 E +:m3 E +:n3 E +:o3 E +:c4 E +:d4 E +:e4 E +:f4 E +:g4 E +:h4 E +:i4 E +:j4 E +:k4 E +:l4 E +:m4 E +:n4 E +:o4 E +:p4 E +:b5 E +:c5 E +:d5 E +:e5 E +:f5 E +:g5 E +:h5 E +:i5 E +:j5 E +:k5 E +:l5 E +:m5 E +:n5 E +:o5 E +:p5 E +:a6 E +:b6 E +:c6 E +:d6 E +:e6 E +:f6 E +:g6 E +:h6 E +:i6 E +:j6 E +:k6 E +:l6 E +:m6 E +:n6 E +:o6 E +:p6"
/>
(I learnt this template from shalev at http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 ).

Is there a single-stone 6th row template where the stone has three liberties outside the template?

Theory of ladder escapes

2016-05-21T00:30:54Z

Wccanard:

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is a pattern that looks something like this.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as coming in from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

== 5th and 6th row ladder escapes ==

To push this theory to the 5th and 6th row, one first needs to come up with a rigorous definition of a ladder (the issue being how much space one needs under the ladder stone to stop blue from cutting off the ladder immediately). Once this is done one needs to come up with a finite list of templates analogous to, say, A to D in the 4th row argument above. This is not immediate. For example B above is A plus one extra row, and when I tried to prove the result initially I only had A, C and D, and I could not get the argument to work; the extra space was essential. More of this awaits for anyone who tries to extend these results to 5th row ladders or higher.

Theory of ladder escapes

2016-05-21T00:27:17Z

Wccanard: This gives an algorithm capable of checking in a finite time whether a position is a 2nd, 3rd or 4th row ladder escape.

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After we have done this, we can formalise what it means to be a ladder escape, but now there is a catch. A ladder escape (say, to make things precise, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, however far away it is. So strictly speaking to check that a pattern is a 4th row ladder escape we need to check that red can connect to the edge from an *infinite set* of positions (the ladder can be as far away as you like). This raises the issue of how one can possibly check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page, for 2nd, 3rd and 4th row ladders. It might be possible to resolve it for 5th and 6th row ladders but these have no practical use and it will take a lot of work, so I have not done this. For 7th row ladders we run into a new difficulty -- blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which I do not know how to resolve; I think that in theory there could be no 7th row ladder escapes at all!

== Ladders ==

There is no issue at all with defining a 2nd row ladder: it looks like this.
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>

There is a minor issue with defining 3rd and higher row ladders though. We want a definition which is useful in practice and not too restrictive. For example we surely want this to be a third row ladder:
<hexboard size="4x10"
coords="hide"
contents="R c1 B b2 R c2 R d2 R e2 R f2 R 2:g2 R 4:h2 B a3 B b3 B c3 B d3 B e3 B 1:f3 B 3:g3 B a4 B d4"
/>
even though there are a couple of blue pieces on the first row. It is intuitively clear (and also provably true) that these blue pieces cannot be of any help to blue (they can never play a crucial role in any blue connection). So although we want a 3rd row ladder (moving from left to right, say) to have no stones on the first three rows to the right of the ladder (until we reach the escape of course), we do not want to also guarantee that there are no stones on the first row to the left of the ladder.

Here is the definition I came up with. A '''third row ladder''' is a pattern like this:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2"
/>
The asterisk is not part of the pattern. The red piece can be thought of as being connected to the top. The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red piece. This is a minimal requirement, because for example if one of these pieces were filled:
<hexboard size="3x2"
coords="hide"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this:
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B a4"
/>
and blue can block the ladder with this move.
<hexboard size="4x6"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>
So that is my working definition of a third row ladder, and it seems to work in practice.

The problems compound themselves as we move up the board. Here is my working definition of a 4th row ladder:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R c1 E *:a2 E *:b2 E *:a3"
/>
You see that I have replaced the blue piece with an asterisk (indicating "not part of the pattern"). This is OK because hexes which are not part of the pattern may as well be blue pieces. What I am suggesting here is that for a 4th row ladder to become established we should demand that the triangle with 6 hexes below the laddering piece are all vacant. Note that even filling in one of these can break the ladder: even if we fill in the bottom left corner of the triangle:
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then blue has this move
<hexboard size="5x5"
coords="hide"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To to make the ladder established red needs at a minimum those 6 vacant hexes under their red stone.

I have said above that I will be restricting my attention to 2nd, 3rd and 4th row ladders, and one of the reasons for this is that I do not know a satisfactory definition of a 5th row ladder. What I do know is that this
<hexboard size="5x5"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:c3 E *:a4 E *:b4 E *:a5"
/>
will '''not''' do, because it turns out that if it is blue to play in the diagram below:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 R o5 B a6 R o6"
/>
then blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 B c4 R o4 B a5 B b5 B 1:e5 R o5 B a6 R o6"
/>
The main line is complex; see http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 for example. The conclusion from this is that red needs more space in order to ensure that the ladder is established, and because I am only an amateur hex player I do not really know the minimal amount required or even if one can expect a "natural" answer to this -- perhaps there is more than one kind of 5th row ladder.

A similar issue arises with 6th row ladders -- I do not know how much space to guarantee red under the ladder stone. And for 7th row ladders I have already explained in [[open problems about edge templates]] that the situation is even worse -- no amount of space under the ladder (even if we demand that the entire 5th row is clear) seems to guarantee a red connection if blue just ignores the ladder and plays elsewhere, from which I deduce that I am struggling to make sense of the notion of a 7th row ladder from a theoretical point of view whilst simultaneously being aware that there is probably essentially no use for it from a practical point of view either.

== 2nd row ladder escapes ==

I have only formulated the definition of an n'th row ladder for 2<=n<=4 above. Let me start this discussion of ladder escapes with a discussion of 2nd row ladder escapes, because here we can see the main points without any of the technicalities about how much space one is supposed to allow under a ladder -- there is no space under the ladder.

Let me start with an example. An example of a '''second row ladder escape''' is a pattern that looks something like this.
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The pluses indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is:

<hexboard size="4x10"
coords="hide"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 R i1 R j1 R a2 B b2 B c2 B d2 B e2 B f2 B g2 R a3 R b3 R 1:c3 B j3 B a4 B b4 B j4"
/>
red can guarantee a connection from the ladder stone (marked 1) to the bottom edge. Note that I have replaced asterisks with blue stones -- any hex not in the pattern or the ladder can be thought of as a blue stone, as red is not allowed to move there.

Let me clarify what the plussed hexes mean: they indicate the last point where the 2nd row ladder is allowed to start. So for example saying that the pattern above is a second row ladder escape means that red must win the following position:

<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here we regard stone 1 as connected to the top, and the claim (easily verified) is that even with blue to play, red can connect to the bottom:
<hexboard size="4x4"
coords="hide"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>

Now we can guess the general definition. Before I start I should say that for simplicity I won't allow mirror images and all 2nd row ladders should be thought of as coming in from the left. A '''second row ladder escape''' is the following data. It is a pattern, plus two hexes with pluses in them, such that one of the plussed hexes is on the first row, the other is on the second row up and directly to the left of the first hex, and such that neither of the plussed hexes nor any hex directly to the left of either of the plussed hexes are in the pattern. This data is subject to the following axiom: any position comprising of a second row ladder
<hexboard size="2x1"
coords="hide"
contents="R 1:a1"
/>
followed directly to the right by as many pairs of vacant hexes as you like on the first and second rows:
<hexboard size="2x1"
coords="hide"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder onto the plussed hexes) is an edge template, in the sense that even if it is blue to move, red can guarantee a connection from the ladder stone marked 1 to the edge.

Notation: let's say that our 2nd row ladder pattern followed by n (an integer >= 0) pairs of vacant hexes is called a "second row ladder at distance n". What we are demanding of our second row ladder escape template is that it becomes an edge template when you slot in a second row ladder at distance n, for all values of n>=0.

This is an interesting definition because it allows the ladder to be an *arbitrary* distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However this means that to check that something is a 2nd row ladder escape we need to check that *infinitely many* patterns are edge templates. How can we do this? Well of course as every hex player knows,

Lemma: a pattern
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(where here the asterisks can indicate any pieces at all)
is a second row ladder escape if, and only if, replacing the plussed hexes with a second row ladder (at distance zero)
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>
is an edge template for red.

Proof: If the pattern is a second row ladder escape then *by definition* replacing the plusses with a second row distance zero ladder right next to the escape gives an edge template (indeed a second row ladder escape means that this position and infinitely many other positions are an edge template). So this finishes the implication in one direction.

To go the other way we actually have to play some hex, but it's pretty trivial. Say the pattern becomes an edge template if we insert a second row ladder at distance zero. We now have to prove that the pattern becomes an edge template if we insert a second row ladder at distance n for all n, and this is an easy induction on n, because blue must play directly below red's ladder piece and now red plays along the ladder and by induction this is an edge template.

QED

Definition: A 2nd row ladder escape template is '''minimal''' if the following two things are true. Firstly, removing any hex from the pattern gives a new pattern which is not a 2nd row ladder escape template any more. And secondly, if the two hexes directly to the right of the two plussed stones are both vacant hexes in the pattern, then moving the plussed hexes one hex to the right results in a new pattern which again is not a 2nd row ladder escape.

Corollary [cf http://www.drking.org.uk/hexagons/hex/templates.html]: the following positions are minimal second row ladder escapes:
<hexboard size="2x2"
coords="hide"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
(note that the corresponding pattern on Dr King's site is not minimal; I have moved the plusses)
<hexboard size="2x3"
coords="hide"
contents="E +:a1 R 10:c1 E +:a2 E *:b2 E *:c2"
/>
(here and below the 10 indicates a stone connected to the bottom edge, but the connection is not shown)
<hexboard size="3x4"
coords="hide"
contents="E *:a1 E +:a2 R 10:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R 10:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
and so on and so on (these templates are all taken from Dr King's site, and there are several more there).

== Third row ladder escapes ==

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because blue can defend against a third row ladder in more than one way: blue can at some stage decide to drop to the second row.

The following definition is unsurprising: A '''third row ladder escape''' is the following data: it is a pattern, and three plussed hexes (none of these plussed hexes are part of the pattern) going from the first to the third row, up and left as in the below picture:
<hexboard size="3x3"
coords="hide"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(here the starred hexes can be anything, they are just part of a general pattern, which could certainly go above the third row if necessary). The data must have the property that no hex directly to the left of any of the plussed hexes can be part of the pattern. And the data is subject to the axiom that for any n>=0, if you insert a third row ladder at distance n onto the pattern at the three plussed hexes, the resulting position is an edge template connecting the ladder stone to the bottom edge.

In pictures, our pattern above must give rise to an edge template when we attach a 3rd row template at distance 0
<hexboard size="3x4"
coords="hide"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1
<hexboard size="3x5"
coords="hide"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6
<hexboard size="3x10"
coords="hide"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any distance you like.

Now here is the interesting question. Again we find ourselves in the situation that trying to check that something is a 3rd row ladder escape template involves checking that infinitely many positions are edge templates. How do we get round to this?

Proposition: If we have a pattern plus three plussed hexes with nothing to the left of the plusses, and if the pattern becomes an edge template (connecting piece numbered 1 to the edge) when we attach each of the following two patterns to it at the plusses:

<hexboard size="3x2"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
and
<hexboard size="3x1"
coords="hide"
contents="R 1:a2"
/>
then the pattern is a third row ladder escape.

Proof: Call the positions above A and B (so A is a 3rd row ladder and B is a second row ladder with an empty space on top). If n>=0 is an integer then by A+n I mean the position A at a distance n, that is, A followed by n columns of three empty hexes. Similarly by B+n I mean position B followed by n columns of three empty hexes. I claim if inserting A and B both result in edge templates, then inserting A+n and B+n result in edge templates for all n. The proof is by induction on n. By assumption it is true for n=0. Say we have established it for n=m>=0 (that is, we know that A+m+pattern and B+m+pattern are both edge templates) and we want to establish these facts for A+(m+1) and B+(m+1). Let us first consider position B+(m+1):

<hexboard size="3x2"
coords="hide"
contents="R 1:a2"
/>
(and imagine that this goes on to connect to the pattern). It is blue's move and blue only has one move which does not lose instantly. If blue plays this move and red replies thus:
<hexboard size="3x2"
coords="hide"
contents="R 1:a2 R 3:b2 B 2:a3"
/>
then we see that by the inductive hypothesis, piece 3 is connected to the edge, and hence piece 1 is too.

For position A+(m+1) we need to work a little more. The position we need to consider is this:

<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2"
/>
(and imagine that it goes on to connect to the pattern). This time blue has three moves in a triangle under piece 1 and blue needs to play one of them or blue will lose instantly. We analyse all three in turn:

For the first
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
red will connect to the edge because by induction A+m connects to the edge. For the second
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
red just wins outright. And for the third this sequence is forced for blue:
<hexboard size="3x3"
coords="hide"
contents="E *:a1 R 1:b1 E *:a2 R 3:b2 R 5:c2 B 4:a3 B 2:b3"
/>
and red is connected because by induction B+m connects to the edge.

The induction is complete, and in particular we can conclude that A+n+pattern connects to the edge for all n, so in particular the pattern is a 3rd row ladder escape.

QED

Here are some remarks about this argument.

(I will elaborate on these later)
1) necessary but not sufficient (weird example of 3rd row escape for which hypothesis of lemma doesn't hold)
2) 3rd row escape is not a 2nd row escape
3) Is a 3rd row escape a 2nd row escape if we demand an extra row somehow? Not sure this is interesting.

Examples: crib from King website. Fix the one where the plusses slope the wrong way.

== Fourth row ladder escapes ==

We have seen all the ideas now. If you have worked through the ideas in the second and third row escapes then this will be relatively easy, other than the actual hex, which this time is quite fun!

A fourth row ladder escape is a pattern plus four plussed hexes going up and left from the first to the 4th row, such that there is no hex in the pattern to the left of any plussed hex. This data has to satisfy the following axiom: adding the pattern to a 4th row ladder (which I carefully defined above) at distance n, for any n>=0, gives an edge template.

Proposition: If adding each of the following patterns to a pattern P gives an edge template:
A: <hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
B:<hexboard size="4x4"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>
C:<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3"
/>
D:<hexboard size="4x1"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3"
/>
then P is a 4th row ladder escape.

Remark: again we have given a criterion which can be checked in a finite time.

Remark: note that the 2nd and 3rd row ladders do *not* have vacant hexes above them, in contrast to the 3rd row ladder escape result where the 2nd row ladder had a vacant hex above it.

Proof: again the same idea. We prove simultaneously that all of A+n+P and B+n+P and C+n+P and D+n+P are edge templates, by induction on n. The case n=0 is our assumption. We need to play a little hex to get our induction going.

The A case is easiest: A+(m+1)+P equals B+m+P so we're done by our inductive hypothesis.

The B case is the most fun. We consider the following position:

<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 E *:a3"
/>

Imagine that this, possibly after some columns of four vacant hexes, attaches to our pattern. We need to prove that stone 1 is connected to the edge.

The five moves marked with a blue stone all lose instantly:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B d1 B e1 E *:a2 E *:b2 B e2 E *:a3 B e3 B e4"
/>
Any of the moves marked 2 below can be answered by red 3, moving us to position A+m+P which is an edge template by the inductive hypothesis.
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3 B 2:b3 B 2:a4"
/>
Move 2 below leads us to D+m:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 R 7:d3 R 9:e3 B 4:b4 B 6:c4 B 8:d4 B 10:e4"
/>
Move 2 below leads us to C+(m+1) which we will deal with later:
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 B 4:d1 E *:a2 E *:b2 R 5:c2 R 3:d2 E *:a3 B 2:c3"
/>
Move 2 below leads us to C+m (note blue 4 must be in the triangle left and below from red 3; blue can also play out the bridge between 1 and 2 but this doesn't help)
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to C+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:d3 B 2:b4"
/>
Move 2 below leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 6:d2 E *:a3 R 7:c3 R 5:d3 R 9:e3 B 4:b4 B 2:c4 B 8:d4"
/>
The final choice for move 2 below also leads us to D+m
<hexboard size="4x5"
coords="hide"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the analysis for B, modulo the fact that we need to deal with C+(m+1). Let's do this now.

From C+(m+1) all moves lose instantly other than the two lines below:
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 R 3:c2 E *:a3 B 2:b3"
/>
leads to C+m, and
<hexboard size="4x3"
coords="hide"
contents="E *:a1 E *:b1 E *:a2 R 1:b2 E *:a3 R 3:c3 B 2:b4"
/>
leads to D+m. In particular this finishes the proof for B and C.
Finally D, the easiest:
<hexboard size="4x2"
coords="hide"
contents="E *:a1 E *:a2 R 1:a3 R 3:b3 B 2:a4"
/>
D+(m+1) leads straight to D+m.

So the induction is finished and in particular in all cases A+m+P is an edge template, meaning that P is a 4th row ladder escape.

QED.

Theory of ladder escapes

2016-05-21T00:12:04Z

Wccanard:

Theory of ladder escapes

2016-05-21T00:01:36Z

Wccanard:

Theory of ladder escapes

2016-05-20T23:36:49Z

Wccanard: Created page with "The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this we first need to formalise what it means to be a ladder! After..."

Ladder escape template

2016-05-20T22:11:29Z

Wccanard: Adding link to technical page

A '''ladder escape template''' is a [[pattern]] which can be used as a [[ladder escape]], that is a way of connecting a [[ladder]] to an [[edge]]. For example a stone on a second row will escape a second row ladder. Let's see this in action.

<hexboard size="2x8"
coords="hide"
contents="R a1 R 2:b1 R 4:c1 R 6:d1 R 8:e1 R h1 B 1:a2 B 3:b2 B 5:c2 B 7:d2 B 9:e2"
/>

When red's ladder reaches the second row stone on the right, red will be able to connect to the bottom edge, as follows:
<hexboard size="2x8"
coords="hide"
contents="R a1 R b1 R c1 R d1 R e1 R 1:f1 R 3:g1 R h1 B a2 B b2 B c2 B d2 B e2 B 2:f2"
/>

Edge templates can often be used as ladder escape templates and most of the examples below will look familiar to people who know their edge templates.

There is a technical page on [[proving that a pattern is a ladder escape]]; this is not necessary for beginners, who just need to recognise basic ladder escapes and how to use them.

Below, '''*''' stands for template carrier, '''+''' stands for projected ladder path.

== [[Second row|Row-2]] ladders ==

All of the common edge templates are valid. For instance:

===[[Template II]]===
<hex> R3 C8
Vb2 Vc2 Vd2 Pe2 Pf2 Pg2 Vh2
Ha3 Hb3 Hc3 Hd3 Pe3 Pf3 Pg3 Sh3</hex>

===[[Template IVa]]===
<hex> R4 C11
Vi1 Sj1
Sg2 Sh2 Si2 Sj2 Sk2
Vb3 Vc3 Pd3 Pe3 Pf3 Sg3 Sh3 Si3 Sj3 Sk3
Ha4 Hb4 Hc4 Pd4 Pe4 Pf4 Sg4 Sh4 Si4 Sj4 Sk4</hex>

== [[Third row|Row-3]] ladders==

Templates [[Template II|II]], [[Template IIIa|IIIa]], and [[Template IVa|IVa]] are valid.

== [[Fourth row|Row-4]] ladders==

===[[Template IIIa]]===
Template IIIa is valid.
<hex> R5 C7
Vb2 Vc2 Vd2
Ha3 Hb3 Hc3 Hd3 Vf3 Sg3
Se4 Sf4 Sg4
Sd5 Se5 Sf5 Sg5</hex>

===[[Template IVa]]===
Also template IVa is valid if you can [[double bridge]] to the [[escape piece]] as follows.

<hex>R6 C9 Vb3 Vc3 Vd3 Sf2 Vg3 Ha4 Hb4 Hc4 Hd4</hex>

[[Red]] can [[jumping|jump]] ahead to the escape template by playing at the marked [[cell]].

== External links ==
* [http://www.drking.org.uk/hexagons/hex/templates.html David King's Hex template page]

[[category:ladder]]
[[category:templates]]
{{stub}}

Ladder escape

2016-05-20T22:01:12Z

Wccanard:

In the [[ladder]] situation below, the piece h8 is a '''ladder escape''', or an '''escape piece'''.

<hex>R9 C9 Q1 Ve3 Vd4 Vd5 Vc6 Vc7 Hb8 Hb9 Hd7 He6 Hf6 Vh8</hex>

It allows [[Red (player)|Red]] (the [[attacking player]]) to [[jumping|jump]] a move ahead of the ladder and [[win]] the game. Red can play along the ladder, forcing [[Blue (player)|Blue]]'s response at each step. After move 9 at g8, Red is [[connection|connected]] to the [[bottom edge|bottom]].

<hex>R9 C9 Q1 Ve3 Vd4 Vd5 Vc6 Vc7 Hb8 Hb9 Hd7 He6 Hf6 Vh8 Mc8 Mc9 Md8 Md9 Me8 Me9 Mf8 Mf9 Mg8</hex>

A ladder escape can be thought of as a little boost for the attacker, giving them just enough extra space to get past the defender's chain. The h8 piece is called a '''second row ladder escape piece''' because it lets the ladder escape to the bottom edge past blue's defences.

Here is another example.

The two red stones on the right in the following position can be used to escape red's third row ladder and enable red to win the game.

<hexboard size="9x9"
coords="hide"
contents="R c2 B d2 B e2 R c3 B i3 R d4 R e4 R c5 B d5 B h5 B e6 R h6 R i6 R b7 R 2:c7 B a8 B 1:b8 B 3:c8"
/>

However blue plays the ladder, red will somehow be able to use the stones to connect to the bottom edge and win the game. For example

<hexboard size="9x9"
coords="hide"
contents="R c2 B d2 B e2 R c3 B i3 R d4 R e4 R c5 B d5 B h5 B e6 R h6 R i6 R b7 R c7 R 1:d7 R 3:e7 R 5:f7 B a8 B b8 B c8 B 2:d8 B 4:e8 B 6:f8 R 7:h8"
/>

or

<hexboard size="9x9"
coords="hide"
contents="R c2 B d2 B e2 R c3 B i3 R d4 R e4 R c5 B d5 B h5 B e6 R h6 R i6 R b7 R c7 R 1:d7 R 3:e7 E *:g7 B a8 B b8 B c8 B 2:d8 R 5:e8 R 7:f8 E *:g8 R 9:h8 B 6:d9 B 4:e9 B 8:f9"
/>

(here red can connect piece 7 to the group on the right in two ways via the starred hexes, and blue cannot block both). The two red stones on the right form a '''third row ladder escape template'''. All players should know the basic [[ladder escape template|ladder escape templates]] for second and third row ladders. An external resource listing many ladder escape templates is http://www.drking.org.uk/hexagons/hex/templates.html .

=== Creating a ladder escape ===

In this example, Red has no third row ladder escape piece. Red 1 provides an escape by threatening to connect to the main group.
<hexboard size="13x13"
contents="R i2 B j2 R m2 B g3 B i3 R k3 R g4 B h4 R i4 R h5 B j5 B f7 R g7 B e8 R f8 R e9 B f9 B c10 R d10 B e10 R b11 B c11 R d11 R e11 R 1:g11 R b12 B c12 B d12 B e12"
/>
After the sequence to Red 3, Blue cannot stop both of Red's non-overlapping connections.
<hexboard size="13x13"
contents="R i2 B j2 R m2 B g3 B i3 R k3 R g4 B h4 R i4 R h5 B j5 B f7 R g7 B e8 R f8 E *:h8 R e9 B f9 B c10 R d10 B e10 E *:f10 R 3:g10 R b11 B c11 R d11 R e11 B 2:f11 R 1:g11 R b12 B c12 B d12 B e12"
/>

=== Another Example ===

In the example below, Red 1 threatens to connect to the main group either via Red's second row ladder or higher up.
<hexboard size="13x13"
contents="R i2 B j2 R m2 B g3 B i3 R k3 R g4 B h4 R i4 R h5 B j5 B f7 R g7 B e8 R f8 R e9 B f9 R d10 B e10 R c11 B d11 R 1:h11 R c12 R d12 R e12 B b13 B c13 B d13 B e13"
/>
After the sequence to Red 7, Red can now connect 5 back to the main group through either of the spaces marked with *.
<hexboard size="13x13"
contents="R i2 B j2 R m2 B g3 B i3 R k3 R g4 B h4 R i4 R h5 B j5 B f7 R g7 B e8 R f8 E *:h8 R e9 B f9 R 7:h9 R d10 B e10 E *:f10 R 5:i10 R c11 B d11 B 6:g11 R 1:h11 R c12 R d12 R e12 R 3:f12 B 2:g12 B b13 B c13 B d13 B e13 B 4:f13"
/>

== Related articles ==
* [[ladder escape template]]
* [[ladder escape fork]]

[[category:ladder]]