HexWiki - User contributions [en]

User:Bobson8

2025-05-09T01:54:37Z

Bobson8:

<hexboard size="11x11"
coords="hide"
contents="R e4 R h4 R c7 R g7 R c8 R d8 R e8 R f8"
/>

=== 7th row single-stone pre-template ===

The following pattern guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. Whether this connection is minimal is not known. This partially answers the single-stone 7th row edge template problem (The template exists but the shape is unknown).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1"
/>

To start with, one notice that Red can connect to the edge with a variant of [[edge template V1b]] from both sides.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

=== A conjecture on Trapezoid boards ===

Conjecture. For an isosceles trapezoid board of edge lengths a (Red), b (Blue), a+b-1 (Red), b (Blue),

(1) "Red always loses regardless of who goes first" if and only if b >= 2a+1
(In other words, Red's average edge length <= Blue's-1).

(2) "Red always wins regardless of who goes first" if and only if b <= 2a-3
(In other words, Red's average edge length >= Blue's+1).

Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true.

To prove this conjecture one need to verify

(1) When b=2a-3, Red wins even if Blue goes first.

(2) When b=2a-2, Red loses if Blue goes first.

(3) When b=2a, Red wins if Red goes first.

(4) When b=2a+1, Red loses even if Red goes first.

We demo some small cases (a=1,2,3).

(a,b)=(1,2): Red wins with a bridge.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R 1:b1 a3"
/>

(a,b)=(1,3): Blue edges form a span template.

<hexboard size="3x4"
coords="hide"
edges="none"
visible="area(a2,a3,c3,d2,d1,b1)"
contents="B b1 d1 a2 d2"
/>

(a,b)=(2,1): Red edges form a bridge template.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R b1 a3"
/>

(a,b)=(2,2): Blue wins with a chain of two bridges.

<hexboard size="2x4"
coords="hide"
edges="none"
visible="area(a1,a2,c2,d1)"
contents="B a1 d1 1:b2"
/>

(a,b)=(2,4): Red wins with a chain of a bridge and a ziggurat.

<hexboard size="6x5"
coords="hide"
edges="none"
visible="area(a5,a6,d6,e5,e1)"
contents="R e1 a6 b6 c6 d6 1:d3 S gray:area(e1,a5,a6,c6,d5,e2)"
/>

(a,b)=(2,5): Blue edges form a template. One possible defence for Red:

<hexboard size="5x7"
coords="hide"
edges="none"
visible="area(a4,a5,f5,g4,g1,d1)"
contents="B a4 b3 c2 d1 g1 g2 g3 g4 2:d4 R 1:e2"
/>

(a,b)=(3,3): Red edges are connected by the edge template IV2a.

<hexboard size="5x5"
coords="hide"
edges="none"
visible="area(a4,a5,d5,e4,e1,d1)"
contents="R a5 b5 c5 d5 e1 d1 S gray:area(a4,a5,c5,d4,e2,e1,d1)"
/>

(a,b)=(3,4): Blue wins with a chain of two ziggurats.

<hexboard size="4x7"
coords="hide"
edges="none"
visible="area(a3,a4,f4,g3,g1,c1)"
contents="B a3 b2 c1 g1 g2 g3 1:d3 S gray:area(a3,a4,f4,g3,g1,f1,e2,d2,d1,c1)"
/>

(a,b)=(3,6): Red can win with this template:

<hexboard size="8x8"
coords="hide"
edges="none"
visible="area(a7,a8,g8,h7,h1,g1)"
contents="R h1 g1 a8 b8 c8 d8 e8 f8 g8 1:f4 S gray:area(a7,a8,f8,g7, g4,h3,h1,g1)"
/>

(a,b)=(3,7): Blue edges form a pre-template. There are two template within it, removing cells either with * or with + (due to Quasar). When Red plays 1, Blue can play at a for the * case, or b for the + case:

<hexboard size="7x10"
coords="hide"
edges="none"
visible="area(a6,a7,i7,j6,j1,f1)"
contents="+:(d7 e7 f7) *:h1 a:f5 b:f4 B a6 b5 c4 d3 e2 f1 j6 j5 j4 j3 j2 j1 R 1:g3"
/>

== Some interior templates ==

These templates assume that one has to connect all the red pieces on the boundaries. (The middle pieces are also gauranteed to be connected, though insignificant in practice.)

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,e3,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1 e3"
/>

This template can be verified by the disjoint spanning tree theorem (the criterion being used in the Shannon switching game).

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,d4,e2,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1"
/>

Removing a Red terminal on a corner from the previous template still gives a template.

<hexboard size="7×7"
coords="hide"
edges="none"
visible="area(a5,e5,g3,g2,f1,d1,b2,a3)"
contents="R a3 a5 c3 c5 e3 e5 g3"
/>

A strategy for this template could be intepreted as the Shannon switching game on this graph of 7 vertices and 12 edges. There is an exceptional rule for this game: Red can save the dotted edge only after the yellow and green edges have been played by any player.

[[Image: TripleV_SSG.png|center|500px]]

When the dotted edge have not been played, Red can save the yellow edge by playing a red piece at the yellow spot and connect the two parts with a bridge. If Blue breaks the yellow edge by playing a blue piece on it, then Red can also automatically peep at the yellow spot. The same applies to green.

In any other cases, If Blue plays inside the tunnel, then they are considered breaking the dotted edge. Red can save the dotted edge by playing a red piece on the cyan spot and connect to the yellow and green spots with bridges.

This switching game itself is not obviously winning for Red. One has to check case by case.

User:Bobson8

2025-05-05T12:09:19Z

Bobson8:

<hexboard size="11x11"
coords="hide"
contents="R e4 R h4 R c7 R g7 R c8 R d8 R e8 R f8"
/>

=== 7th row single-stone pre-template ===

The following pattern guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. Whether this connection is minimal is not known. This partially answers the single-stone 7th row edge template problem (The template exists but the shape is unknown).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1"
/>

To start with, one notice that Red can connect to the edge with a variant of [[edge template V1b]] from both sides.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

=== A conjecture on Trapezoid boards ===

Conjecture. For an isosceles trapezoid board of edge lengths a (Red), b (Blue), a+b-1 (Red), b (Blue),

(1) "Red always loses regardless of who goes first" if and only if b >= 2a+1
(In other words, Red's average edge length <= Blue's-1).

(2) "Red always wins regardless of who goes first" if and only if b <= 2a-3
(In other words, Red's average edge length >= Blue's+1).

Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true.

To prove this conjecture one need to verify

(1) When b=2a-3, Red wins even if Blue goes first.

(2) When b=2a-2, Red loses if Blue goes first.

(3) When b=2a, Red wins if Red goes first.

(4) When b=2a+1, Red loses even if Red goes first.

We demo some small cases (a=1,2,3).

(a,b)=(1,2): Red wins with a bridge.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R 1:b1 a3"
/>

(a,b)=(1,3): Blue edges form a span template.

<hexboard size="3x4"
coords="hide"
edges="none"
visible="area(a2,a3,c3,d2,d1,b1)"
contents="B b1 d1 a2 d2"
/>

(a,b)=(2,1): Red edges form a bridge template.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R b1 a3"
/>

(a,b)=(2,2): Blue wins with a chain of two bridges.

<hexboard size="2x4"
coords="hide"
edges="none"
visible="area(a1,a2,c2,d1)"
contents="B a1 d1 1:b2"
/>

(a,b)=(2,4): Red wins with a chain of a bridge and a ziggurat.

<hexboard size="6x5"
coords="hide"
edges="none"
visible="area(a5,a6,d6,e5,e1)"
contents="R e1 a6 b6 c6 d6 1:d3 S gray:area(e1,a5,a6,c6,d5,e2)"
/>

(a,b)=(2,5): Blue edges form a template. One possible defence for Red:

<hexboard size="5x7"
coords="hide"
edges="none"
visible="area(a4,a5,f5,g4,g1,d1)"
contents="B a4 b3 c2 d1 g1 g2 g3 g4 2:d4 R 1:e2"
/>

(a,b)=(3,3): Red edges are connected by the edge template IV2a.

<hexboard size="5x5"
coords="hide"
edges="none"
visible="area(a4,a5,d5,e4,e1,d1)"
contents="R a5 b5 c5 d5 e1 d1 S gray:area(a4,a5,c5,d4,e2,e1,d1)"
/>

(a,b)=(3,4): Blue wins with a chain of two ziggurats.

<hexboard size="4x7"
coords="hide"
edges="none"
visible="area(a3,a4,f4,g3,g1,c1)"
contents="B a3 b2 c1 g1 g2 g3 1:d3 S gray:area(a3,a4,f4,g3,g1,f1,e2,d2,d1,c1)"
/>

(a,b)=(3,6): Red can win with this template:

<hexboard size="8x8"
coords="hide"
edges="none"
visible="area(a7,a8,g8,h7,h1,g1)"
contents="R h1 g1 a8 b8 c8 d8 e8 f8 g8 1:f4 S gray:area(a7,a8,f8,g7, g4,h3,h1,g1)"
/>

(a,b)=(3,7): Blue edges form a pre-template. There are two template within it, removing cells either with * or with + (due to Quasar). When Red plays 1, Blue can play at a for the * case, or b for the + case:

<hexboard size="7x10"
coords="hide"
edges="none"
visible="area(a6,a7,i7,j6,j1,f1)"
contents="+:(d7 e7 f7) *:h1 a:f5 b:f4 B a6 b5 c4 d3 e2 f1 j6 j5 j4 j3 j2 j1 R 1:g3"
/>

== Some interior templates ==

These templates assume that one has to connect all the red pieces on the boundaries. (The middle pieces are also gauranteed to be connected, though insignificant in practice.)

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,e3,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1 e3"
/>

This template can be verified by the disjoint spanning tree theorem (the criterion being used in the Shannon switching game).

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,d4,e2,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1"
/>

Removing a Red terminal on a corner from the previous template still gives a template.

<hexboard size="7×7"
coords="hide"
edges="none"
visible="area(a5,e5,g3,g2,f1,d1,b2,a3)"
contents="R a3 a5 c3 c5 e3 e5 g3"
/>

A strategy for this template could be intepreted as the Shannon switching game on this graph of 7 vertices and 12 edges. There is an exceptional rule for this game: Red can save the dotted edge only after the yellow and green edges have been played by any player.

[[Image: TripleV_SSG.png|center|500px]]

When the dotted edge is not broken, Red saves the yellow edge by playing a red piece at the yellow spot and connect the two parts with a bridge. If Blue breaks the yellow edge by playing a blue piece on it, then Red can also automatically peep at the yellow spot. The same applies to green.

In any other cases, If Blue plays inside the tunnel, then they are considered breaking the dotted edge. Red saves the dotted edge by playing a red piece on the cyan spot and connect to the yellow and green spots with bridges.

This switching game itself is not obviously winning for Red. One has to check case by case.

User:Bobson8

2025-05-05T12:01:34Z

Bobson8: Add image

<hexboard size="11x11"
coords="hide"
contents="R e4 R h4 R c7 R g7 R c8 R d8 R e8 R f8"
/>

=== 7th row single-stone pre-template ===

The following pattern guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. Whether this connection is minimal is not known. This partially answers the single-stone 7th row edge template problem (The template exists but the shape is unknown).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1"
/>

To start with, one notice that Red can connect to the edge with a variant of [[edge template V1b]] from both sides.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

=== A conjecture on Trapezoid boards ===

Conjecture. For an isosceles trapezoid board of edge lengths a (Red), b (Blue), a+b-1 (Red), b (Blue),

(1) "Red always loses regardless of who goes first" if and only if b >= 2a+1
(In other words, Red's average edge length <= Blue's-1).

(2) "Red always wins regardless of who goes first" if and only if b <= 2a-3
(In other words, Red's average edge length >= Blue's+1).

Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true.

To prove this conjecture one need to verify

(1) When b=2a-3, Red wins even if Blue goes first.

(2) When b=2a-2, Red loses if Blue goes first.

(3) When b=2a, Red wins if Red goes first.

(4) When b=2a+1, Red loses even if Red goes first.

We demo some small cases (a=1,2,3).

(a,b)=(1,2): Red wins with a bridge.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R 1:b1 a3"
/>

(a,b)=(1,3): Blue edges form a span template.

<hexboard size="3x4"
coords="hide"
edges="none"
visible="area(a2,a3,c3,d2,d1,b1)"
contents="B b1 d1 a2 d2"
/>

(a,b)=(2,1): Red edges form a bridge template.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R b1 a3"
/>

(a,b)=(2,2): Blue wins with a chain of two bridges.

<hexboard size="2x4"
coords="hide"
edges="none"
visible="area(a1,a2,c2,d1)"
contents="B a1 d1 1:b2"
/>

(a,b)=(2,4): Red wins with a chain of a bridge and a ziggurat.

<hexboard size="6x5"
coords="hide"
edges="none"
visible="area(a5,a6,d6,e5,e1)"
contents="R e1 a6 b6 c6 d6 1:d3 S gray:area(e1,a5,a6,c6,d5,e2)"
/>

(a,b)=(2,5): Blue edges form a template. One possible defence for Red:

<hexboard size="5x7"
coords="hide"
edges="none"
visible="area(a4,a5,f5,g4,g1,d1)"
contents="B a4 b3 c2 d1 g1 g2 g3 g4 2:d4 R 1:e2"
/>

(a,b)=(3,3): Red edges are connected by the edge template IV2a.

<hexboard size="5x5"
coords="hide"
edges="none"
visible="area(a4,a5,d5,e4,e1,d1)"
contents="R a5 b5 c5 d5 e1 d1 S gray:area(a4,a5,c5,d4,e2,e1,d1)"
/>

(a,b)=(3,4): Blue wins with a chain of two ziggurats.

<hexboard size="4x7"
coords="hide"
edges="none"
visible="area(a3,a4,f4,g3,g1,c1)"
contents="B a3 b2 c1 g1 g2 g3 1:d3 S gray:area(a3,a4,f4,g3,g1,f1,e2,d2,d1,c1)"
/>

(a,b)=(3,6): Red can win with this template:

<hexboard size="8x8"
coords="hide"
edges="none"
visible="area(a7,a8,g8,h7,h1,g1)"
contents="R h1 g1 a8 b8 c8 d8 e8 f8 g8 1:f4 S gray:area(a7,a8,f8,g7, g4,h3,h1,g1)"
/>

(a,b)=(3,7): Blue edges form a pre-template. There are two template within it, removing cells either with * or with + (due to Quasar). When Red plays 1, Blue can play at a for the * case, or b for the + case:

<hexboard size="7x10"
coords="hide"
edges="none"
visible="area(a6,a7,i7,j6,j1,f1)"
contents="+:(d7 e7 f7) *:h1 a:f5 b:f4 B a6 b5 c4 d3 e2 f1 j6 j5 j4 j3 j2 j1 R 1:g3"
/>

== Some interior templates ==

These templates assume that one has to connect all the red pieces on the boundaries. (The middle pieces are also gauranteed to be connected, though insignificant in practice.)

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,e3,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1 e3"
/>

This template can be verified by the disjoint spanning tree theorem (the criterion being used in the Shannon switching game).

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,d4,e2,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1"
/>

Removing a Red terminal on a corner from the previous template still gives a template.

<hexboard size="7×7"
coords="hide"
edges="none"
visible="area(a5,e5,g3,g2,f1,d1,b2,a3)"
contents="R a3 a5 c3 c5 e3 e5 g3"
/>

A strategy for this template could be intepreted as the Shannon switching game on this graph of 7 vertices and 12 edges. There is an exceptional rule for this game: Red can save the dotted edge only after the yellow and green edges have been played by any player.

[[Image: TripleV_SSG.png|center|500px]]

Red saves the yellow edge by playing a red piece at the yellow spot and connect the two parts with a bridge. If Blue breaks the yellow edge by playing a blue piece on it, then Red can also automatically peep at the yellow spot. The same applies to green.

In any other cases, If Blue plays inside the tunnel, then they are considered breaking the dotted edge. Red saves the dotted edge by playing a red piece on the cyan spot and connect to the yellow and green spots with bridges.

This switching game itself is not obviously winning for Red. One has to check case by case.

File:TripleV SSG.png

2025-05-05T11:57:41Z

Bobson8: An interior template and its Shannon switching game intepretation.

An interior template and its Shannon switching game intepretation.

User:Bobson8

2025-05-05T11:54:41Z

Bobson8: Some interior templates and descriptions

<hexboard size="11x11"
coords="hide"
contents="R e4 R h4 R c7 R g7 R c8 R d8 R e8 R f8"
/>

=== 7th row single-stone pre-template ===

The following pattern guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. Whether this connection is minimal is not known. This partially answers the single-stone 7th row edge template problem (The template exists but the shape is unknown).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1"
/>

To start with, one notice that Red can connect to the edge with a variant of [[edge template V1b]] from both sides.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

=== A conjecture on Trapezoid boards ===

Conjecture. For an isosceles trapezoid board of edge lengths a (Red), b (Blue), a+b-1 (Red), b (Blue),

(1) "Red always loses regardless of who goes first" if and only if b >= 2a+1
(In other words, Red's average edge length <= Blue's-1).

(2) "Red always wins regardless of who goes first" if and only if b <= 2a-3
(In other words, Red's average edge length >= Blue's+1).

Indeed, this conjecture is an analog of the results on parallelogram boards. I have verified this conjecture for a=1,2,3, and KataHex suggested that a=4,5,6 are possibly true.

To prove this conjecture one need to verify

(1) When b=2a-3, Red wins even if Blue goes first.

(2) When b=2a-2, Red loses if Blue goes first.

(3) When b=2a, Red wins if Red goes first.

(4) When b=2a+1, Red loses even if Red goes first.

We demo some small cases (a=1,2,3).

(a,b)=(1,2): Red wins with a bridge.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R 1:b1 a3"
/>

(a,b)=(1,3): Blue edges form a span template.

<hexboard size="3x4"
coords="hide"
edges="none"
visible="area(a2,a3,c3,d2,d1,b1)"
contents="B b1 d1 a2 d2"
/>

(a,b)=(2,1): Red edges form a bridge template.

<hexboard size="3x2"
coords="hide"
edges="none"
visible="area(a3,b2,b1,a2)"
contents="R b1 a3"
/>

(a,b)=(2,2): Blue wins with a chain of two bridges.

<hexboard size="2x4"
coords="hide"
edges="none"
visible="area(a1,a2,c2,d1)"
contents="B a1 d1 1:b2"
/>

(a,b)=(2,4): Red wins with a chain of a bridge and a ziggurat.

<hexboard size="6x5"
coords="hide"
edges="none"
visible="area(a5,a6,d6,e5,e1)"
contents="R e1 a6 b6 c6 d6 1:d3 S gray:area(e1,a5,a6,c6,d5,e2)"
/>

(a,b)=(2,5): Blue edges form a template. One possible defence for Red:

<hexboard size="5x7"
coords="hide"
edges="none"
visible="area(a4,a5,f5,g4,g1,d1)"
contents="B a4 b3 c2 d1 g1 g2 g3 g4 2:d4 R 1:e2"
/>

(a,b)=(3,3): Red edges are connected by the edge template IV2a.

<hexboard size="5x5"
coords="hide"
edges="none"
visible="area(a4,a5,d5,e4,e1,d1)"
contents="R a5 b5 c5 d5 e1 d1 S gray:area(a4,a5,c5,d4,e2,e1,d1)"
/>

(a,b)=(3,4): Blue wins with a chain of two ziggurats.

<hexboard size="4x7"
coords="hide"
edges="none"
visible="area(a3,a4,f4,g3,g1,c1)"
contents="B a3 b2 c1 g1 g2 g3 1:d3 S gray:area(a3,a4,f4,g3,g1,f1,e2,d2,d1,c1)"
/>

(a,b)=(3,6): Red can win with this template:

<hexboard size="8x8"
coords="hide"
edges="none"
visible="area(a7,a8,g8,h7,h1,g1)"
contents="R h1 g1 a8 b8 c8 d8 e8 f8 g8 1:f4 S gray:area(a7,a8,f8,g7, g4,h3,h1,g1)"
/>

(a,b)=(3,7): Blue edges form a pre-template. There are two template within it, removing cells either with * or with + (due to Quasar). When Red plays 1, Blue can play at a for the * case, or b for the + case:

<hexboard size="7x10"
coords="hide"
edges="none"
visible="area(a6,a7,i7,j6,j1,f1)"
contents="+:(d7 e7 f7) *:h1 a:f5 b:f4 B a6 b5 c4 d3 e2 f1 j6 j5 j4 j3 j2 j1 R 1:g3"
/>

== Some interior templates ==

These templates assume that one has to connect all the red pieces on the boundaries. (The middle pieces are also gauranteed to be connected, though insignificant in practice.)

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,e3,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1 e3"
/>

This template can be verified by the disjoint spanning tree theorem (the criterion being used in the Shannon switching game).

<hexboard size="5×5"
coords="hide"
edges="none"
visible="area(a5,c5,d4,e2,e1,c1,a3)"
contents="R a3 a5 c1 c3 c5 e1"
/>

Removing a Red terminal on a corner from the previous template still gives a template.

<hexboard size="7×7"
coords="hide"
edges="none"
visible="area(a5,e5,g3,g2,f1,d1,b2,a3)"
contents="R a3 a5 c3 c5 e3 e5 g3"
/>

A strategy for this template could be intepreted as the Shannon switching game on this graph of 7 vertices and 12 edges. There is an exceptional rule for this game: Red can save the dotted edge only after the yellow and green edges have been played by any player.

(Image Placeholder)

Red saves the yellow edge by playing a red piece at the yellow spot and connect the two parts with a bridge. If Blue breaks the yellow edge by playing a blue piece on it, then Red can also automatically peep at the yellow spot. The same applies to green.

In any other cases, If Blue plays inside the tunnel, then they are considered breaking the dotted edge. Red saves the dotted edge by playing a red piece on the cyan spot and connect to the yellow and green spots with bridges.

This switching game itself is not obviously winning for Red. One has to check case by case.

Theory of ladder escapes

2024-10-19T18:14:15Z

Bobson8: Fix a 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 formalise what it means to be a ladder.

Informally, a ladder escape (say, 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, no matter how far away from the ladder escape the ladder starts. So strictly speaking, to check that a pattern is a 4th row ladder escape, we must check that the attacker can connect to the edge from an ''infinite set'' of positions. This raises the issue of how one can 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, 4th, and 5th row ladders. It might be possible to resolve it for 6th row ladders but this has not yet been done, partly because such ladders are of little practical use. 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 is currently unresolved. It may in theory be that there are no 7th row ladders at all.

For the purpose of our analysis, we assume that all ladders move from left to right along the red bottom edge, with Red being the attacker. Of course, the analogous analysis also applies to ladders moving in the opposite direction or along different edges.

The analysis of 2nd–4th row ladders on this page was originally contributed by the user [[User:Wccanard|Wccanard]] in 2016.

'''A note on terminology.''' The usual definition of a ''template'' is a pattern that has a stated property (for example, being [[virtual connection|connected]]) and is also minimal with that property. In other words, a template is usually defined by two properties: validity and minimality. For the purpose of ''this'' page, we are mostly concerned with validity. Since it would be awkward to write "template but not necessarily minimal" throughout all of the definitions and proofs on this page, we adopt the convention, on this page only, that "template" means a pattern that is valid but not necessarily minimal. We will then speak of a "minimal template" when necessary.

== Algebraic notation ==

Before we start, let us introduce some notation that will be useful.

=== Open patterns ===

A ''pattern'' is a set of cells, each of which may be empty or occupied by a stone of either color. In this article, we will only be concerned with patterns that include a red board edge. A pattern is ''open on the left'' if it comes with some cells marked "+" on its left side. No cells to the left of those "+"s may be part of the pattern. A pattern is ''open on the right'' if it comes with some cells marked "−" on its right side. No cells to the right of those "−"s may be part of the pattern. A pattern is ''open on both sides'' if it is open on the left and on the right. A pattern is ''closed'' if it is not open on either side. For example, the following four patterns are open on the right, open on both sides, open on the left, and closed, respectively:
<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1 c2 c3)"
/> <hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="all"
contents="R b1 B c1 E +:(a1 a2 a3) -:(c2 c3)"
/> <hexboard size="3x2"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1"
contents="R b3 E +:(a2 a3)"
/> <hexboard size="3x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 c1 d3 B d1"
/>
The cells labelled "+" (if any) are called the ''left boundary'' of the pattern, the cells labelled "−" are called its ''right boundary'', and the ''carrier'' of a pattern consists of all cells that are part of the pattern (empty or not), except the boundaries.

=== Addition ===

Suppose P is a pattern that is open on the right, Q is a pattern that is open on the left, and the right boundary of P has the same number of cells and shape as the left boundary of Q. Then we write P+Q for the pattern obtained by joining P and Q along their common boundary. More specifically, P+Q is obtained as follows: delete the right boundary from P and the left boundary from Q. Then glue the patterns P and Q together along the line where the boundaries were deleted from each. For example:
<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1 c2 c3)"
/> + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x5"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 a3 b1 e3 e4"
contents="R b2 d1 e1"
/>
It is important to note that the carrier of P+Q consists of just the carriers of P and Q, ''without'' the boundary cells that have been deleted. The purpose of the boundary cells "+" and "−" is just to indicate where the patterns will be attached. It is possible to add more than two patterns, for example:
<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1 c2 c3)"
/> + <hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="all"
contents="R b1 B c1 E +:(a1 a2 a3) -:(c2 c3)"
/> + <hexboard size="3x2"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1"
contents="R b3 E +:(a2 a3)"
/> = <hexboard size="3x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 c1 d3 B d1"
/>

Note that addition is only well-defined if P and Q can be glued together without their carriers overlapping. We will be careful to ensure that this is always the case. However, the addition is associative, i.e., (P + Q) + R is well-defined if and only if P + (Q + R) is well-defined, and in that case, they are equal.

=== The shift operation ===

If P is a pattern that is open on the left, we write 1 + P for the pattern obtained from P by shifting its left boundary one column to the left, and adding a column of empty cells where the boundary used to be. More generally, for any integer ''n'' ≥ 0, we write ''n'' + P for iterating this operation ''n'' times, i.e., for shifting the left boundary of P to the left by ''n'' columns and filling the additional space with empty cells. For example:

1 + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x5"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1 e3 e4"
contents="E +:(a2 a3 a4) R d1 e1"
/>

4 + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x8"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1 c1 d1 e1 h3 h4"
contents="E +:(a2 a3 a4) R g1 h1"
/>

Note that empty cells are only added to the height of the boundary. For a pattern that is open on the right, we can do exactly the same thing on the other side, i.e., P + ''n'' is obtained by shifting the right boundary to the right by ''n'' columns, filling the additional space with empty cells. For example:

<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="R c1 E -:(c2 c3)"
/> + 1     = <hexboard size="3x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1 d1"
contents="R c1 E -:(d2 d3)"
/>

<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="R c1 E -:(c2 c3)"
/> + 4     = <hexboard size="3x7"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1 d1 e1 f1 g1"
contents="R c1 E -:(g2 g3)"
/>

We call this the ''shift operation''. Note that it is associative: If P and Q have matching boundaries, then (P + ''n'') + Q = P + (''n'' + Q).

=== The reduce operation ===

If P is a pattern that is open on the left, we write ↑ + P for the pattern obtained from P by erasing the topmost "+" cell from its left boundary. The cell that formerly contained the "+" is no longer part of the pattern (i.e., it is not replaced by an empty cell). For example:

↑ + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 a2 d3 d4"
contents="E +:(a3,a4) R c1 d1"
/>

We call this the ''reduce operation''. The shift and reduce operations can be combined with each other and with addition of patterns. For instance, ''n'' + ↑ + ''m'' + P is the pattern obtained from P by first shifting its left boundary by ''m'' cells to the left, then reducing the size of that boundary by one cell, and then shifting it by another ''n'' cells to the left. For example:

2 + ↑ + 3 + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x9"
float="inline"
coords="none"
edges="bottom"
visible="-a1--f1 a2--c2 i3 i4"
contents="E +:(a3 a4) R h1 i1"
/>

== Second row ladders ==

=== Definition of ladder ===

There is no issue at all with defining a 2nd row ladder. Informally, a 2nd row ladder looks like this:
<hexboard size="2x8"
coords="hide"
edges="bottom"
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"
/>
Note that at each point in the ladder, Blue's move is forced. Red can choose to continue pushing the ladder as long as she wants to. We formally define a second row ladder as follows:

'''Definition.''' A ''second row ladder'' is a pattern like this:
L2: <hexboard size="2x2"
coords="hide"
edges="bottom"
contents="R a1 E -:(b1 b2)"
/>
Here, the red stone is on the second row, and we call it the ''ladder stone''. Red's goal is to connect the ladder stone to the bottom edge. The cell immediately below and to the right of the ladder stone is empty. We denote this pattern by L2.

=== Definition of ladder escape ===

Before we give the formal definition of a second row ladder escape, let us consider an example. The following pattern is an example of a second row ladder escape.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
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 cells marked "+" indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is,
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:e1 E *:f1 R g1 R h1 E *:a2 E *:b2 E *:c2 E *:d2 E *:e2 R 1:a3 E *:h3 E *:h4"
/>
Red can guarantee a connection from the ladder stone (marked 1) to the bottom edge.

Let us clarify what the hexes marked "+" in the ladder escape pattern 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 (among other things) that Red must win the following position:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here, Red's ladder stone is marked "1", and the claim (easily verified) is that even with Blue to play, Red can connect the ladder stone to the bottom:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
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 row ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 B h3 B h4"
/>
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="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 R 3:b3 R 5:c3 R 7:d3 R 9:e3 B h3 B 2:a4 B 4:b4 B 6:c4 B 6:d4 B h4"
/>
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. We can now give a more formal definition of a second row ladder escape.

'''Definition.''' A ''second row ladder escape template'' (or simply ''second row ladder escape'') is given by the following data. It is a pattern P that is open on the left (see [[#Algebraic notation|algebraic notation]] above), with a boundary of this shape:
<hexboard size="2x1"
coords="hide"
edges="bottom"
contents="E +:(a1 a2)"
/>
subject to the following axiom: for all ''n'' ≥ 0, the position L2 + ''n'' + P is a [[strong connection|virtual connection]] from the ladder stone (marked 1) to the edge.

Concretely, this means that any position consisting of a second row ladder L2,
<hexboard size="2x1"
coords="hide"
edges="bottom"
contents="R 1:a1"
/>
followed directly to the right by an arbitrary number (zero or more) of pairs of vacant hexes on the first and second rows,
<hexboard size="2x1"
coords="hide"
edges="bottom"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder or rightmost column of vacant hexes onto the hexes marked "+"), allows Red to connect the ladder stone to the edge.

Terminology and notation: If we have a left-open pattern whose boundary is of the correct shape, but we are not sure whether it satisfies the axiom of a second row ladder escape, then we refer to it as a ''candidate for a second row ladder escape'' (or simply ''candidate'' if the rest is clear from the context). A candidate is ''valid'' if it is actually a ladder escape.

We also define what it means for a ladder escape template to be minimal.

'''Definition.''' A 2nd row ladder escape template is ''minimal'' if the following two things are true. First, 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 second, if the two hexes directly to the right of the two cells marked "+" are both vacant hexes in the pattern, then moving the cells marked "+" one hex to the right results in a new pattern which is not a 2nd row ladder escape.

Below, we will use analogous terminology and notations for ladders and ladder escapes on the 3rd and higher rows.

=== Characterization of ladder escapes ===

The definition of a 2nd row ladder escape 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 we cannot directly use the definition to check that something is a 2nd row ladder escape, because this would require checking that infinitely many patterns are virtual connections. Can we find some finite criterion for checking 2nd row ladder escapes? Fortunately, as every Hex player knows, the answer is yes. We have the following theorem:

'''Theorem 1.''' Consider a candidate P for a 2nd row ladder escape. Schematically:
<hexboard size="4x4"
coords="hide"
edges="bottom"
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"
/>
(Here the asterisks indicate the [[carrier]] of P, which can contain any stones at all, and can be of any shape or size, as long as it includes no cells to the left of the cells marked "+"). Then P is a valid 2nd row ladder escape if and only if L2+P is a virtual connection for Red.
<hexboard size="4x4"
coords="hide"
edges="bottom"
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"
/>

'''Proof.''' If the ladder escape is valid, then by definition, L2+''n''+P is a virtual connection for all ''n'' ≥ 0, and in particular for ''n'' = 0. This proves the left-to-right implication.

To go the other way we actually have to play some Hex, but it's pretty trivial. We must show that L2+''n''+P is a virtual connection for all ''n''. This is an easy induction on ''n''. For ''n'' = 0, the claim is true by assumption. If ''n'' > 0, then Blue must play directly below Red's ladder stone (or else Red will connect to the edge immediately), and now Red can play a ladder stone at distance ''n''−1 on the second row, which is a virtual connection by induction hypothesis. □

=== Examples ===

We can use Theorem 1 to prove that all of the following patterns are minimal second row ladder escapes. Most of these templates are taken from [http://www.drking.org.uk/hexagons/hex/templates.html David King's website], and there are several more there. For several of the templates, the corresponding pattern on David King's site is not minimal by our definition; for these templates, we have moved the cells marked "+" to the right to make the template minimal.

<hexboard size="2x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
edges="bottom"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d3 d4"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E *:b1 R d1 R d2 E *:d3 E *:d4"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d3 d4"
contents="E +:a3 E +:a4 R c1 R d1"
/>
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 f1 a2"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 f1 a2"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d4 d5"
contents="E *:a1 *:a2 *:b1 *:d4 *:d5 +:a4 +:a5 R c1 R d2"
/>
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 b1 b2 c1 c2 d1 e1 i1 j1 j2"
contents="E *:a1 E *:a2 E *:a3 E +:a4 E +:a5 E *:b1 E *:b2 E *:c1 E *:c2 E *:d1 E *:e1 R g1 E *:i1 E *:j1 E *:j2"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="E +:(a4 a5) R c1 d1"
/>
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 b1 b2 c1 c2 d1 e1 i1 j1 j2"
contents="E *:a1 E *:a2 E *:a3 E +:a4 E +:a5 E *:b1 E *:b2 E *:c1 E *:c2 E *:d1 E *:e1 R g1 E *:i1 E *:j1 E *:j2 R h1 S g2"
/>
The shaded cell is not part of the template, and can be occupied by Blue.

In the below templates, the stone marked "↓" indicates a stone connected to the bottom edge, but the connection is not shown. The connection from 10 to the edge must not use any of the empty hexes in the pattern.
<hexboard size="2x3"
coords="hide"
edges="bottom"
visible="-b2 c2"
contents="E +:a1 R ↓:c1 E +:a2 E *:b2 E *:c2"
/>
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 b3 c3 d3"
contents="E *:a1 E +:a2 R ↓:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 a2 d2 c3 d3 b4 c4 d4"
contents="E *:a1 E *:b1 E *:c1 R ↓:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>

== Third row ladders ==

=== Definition of ladder ===

There is a minor issue with defining ladders on the 3rd and higher rows. We want a definition that 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"
edges="bottom"
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 few blue stones on the first row. It is intuitively clear (and also provably true) that these blue stones 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 to have no stones on the first three rows to the right of the ladder (until we reach the escape), we do not want to also guarantee that there are no stones on the first row to the left of the ladder. We formally define third row ladders as follows.

'''Definition.''' A ''third row ladder'' is a pattern like this:
L3: <hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1--c3)"
/>
The red stone is again called the ''ladder stone'', and Red's goal is to connect the ladder stone to the bottom edge. We denote this pattern by L3.

The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red ladder stone. This is a minimal requirement, because for example if one of these cells were filled,
<hexboard size="3x2"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this,
<hexboard size="4x6"
coords="hide"
edges="all"
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"
edges="all"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>

=== Definition of ladder escape ===

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 [[ladder handling|yield]] to the second row. The following definition is unsurprising.

'''Definition.''' A ''third row ladder escape'' is given by the following data. It is a pattern P that is open on the left, with a boundary of the shape
<hexboard size="3x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a3)"
/>
To be a third row ladder escape, the pattern must satisfy the property that for all ''n'' ≥ 0, L3 + ''n'' + P is a virtual connection from the Red ladder stone to the bottom edge, with Blue to move.

Like for second row escapes, a pattern that has the required shape for a ladder escape, but it is not (yet) known to be a valid ladder escape, is called a ''candidate''.

In pictures, for the pattern
<hexboard size="3x3"
coords="hide"
edges="bottom"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(where the carrier is schematically indicated by stars) to be a 3rd row ladder escape, it must give rise to a virtual connection when we attach a 3rd row ladder at distance 0,
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
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"
edges="bottom"
visible="-a1 a2"
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"
edges="bottom"
visible="-a1 a2"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any other distance.

=== Characterization of ladder escapes ===

Just like for second row ladder escapes, we again find ourselves in the situation that trying to use the definition to check that something is a 3rd row ladder escape involves checking that infinitely many positions are virtual connections. Once again, we have a theorem that allows us to replace this by a finite condition.

'''Theorem 2.''' Consider a candidate P for a 3rd row ladder escape. Assume that (a) L2+↑+P is a virtual connection and (b) L3+P is a virtual connection, each from the ladder stone to the bottom edge. Then P is a valid third row ladder escape.

'''Proof.''' First note that by Theorem 1, because L2+↑+P is a virtual connection, P escapes all 2nd row ladders. Now under the assumptions of the theorem, we must show that L3+''n''+P is a virtual connection for all ''n'' ≥ 0. We prove this by induction on ''n''. For ''n'' = 0, the claim is true by assumption (b). Now suppose the claim is true for some ''n''. To show the claim for ''n''+1, consider the position L3+''n''+1+P. The first three columns of this position look like this:
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R 1:b1 E *:a2"
/>
(This is followed by ''n'' more columns of three empty hexes and by the pattern P). Blue has three possible moves in a triangle under stone 1, and Blue needs to play one of these or he will lose instantly. We analyze all three moves in turn.

For the first, Red pushes the ladder and will connect to the edge because by induction hypothesis, L3+''n''+P connects to the edge, so stone 3 connects to the edge, and so stone 1 does too.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
For the second, Red just wins outright, i.e., we do not need to use the induction hypothesis.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
And for the third, Red responds like this. Since stone 3 is a 2nd row ladder stone, it is connected to the edge because, as we noted above, ↑+P is a 2nd row ladder escape. Therefore stone 1 is also connected.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:b1 B 2:b3 R 3:c2"
/>

The induction is now complete, showing that P is a 3rd row ladder escape. □

Remark: in more concrete terms, Theorem 2 states that a pattern P is a 3rd row ladder escape if the pattern becomes a virtual connection (from the ladder stone to the edge) when we attach each of the following two patterns to its left boundary:

A:
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:b1 E -:(c1--c3)"
/>
B:
<hexboard size="3x2"
coords="hide"
edges="bottom"
visible="-a1"
contents="R 1:a2 E -:(b1--b3)"
/>

In contrast to the situation with 2nd row ladders, while Theorem 2 is ''sufficient'' to show that a position is a 3rd row ladder escape, it is not ''necessary''. For example, consider the following third row ladder escape template P:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="R b1 E +:a2 E +:a3 E +:a4"
/>
One can check directly that L3+2+P and L2+↑+2+P are both virtual connections, so that 2+P is a 3rd row ladder escape by Theorem 2. In particular, L3+''n''+P is a virtual connection for all ''n'' ≥ 2. Moreover, one can check that L3+P and L3+1+P are also virtual connections, so that P is a valid 3rd row ladder escape.

It is, however, not a valid 2nd row ladder escape for ladders at distance 0, because in the position L2+↑+P,
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1 a2"
contents="R b1 R 1:a3"
/>
the ladder stone marked "1" cannot connect to the edge.

Theorem 2 is therefore not sufficient to check that a given pattern is a 3rd row ladder escape. We need to work a little harder to get a necessary and sufficient condition for 3rd row ladder escapes.

'''Lemma 3 (2nd to 3rd row jump).''' Any 3rd row ladder escape also escapes 2nd row ladders that start at distance 2 or greater. More specifically, if L3+P is a virtual connection, then so is L2+↑+2+P.

The lemma is perhaps easier understood in pictures: given any 3rd row ladder escape, replacing the three cells marked "+"
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="c1 c2 c3"
contents="E *:a1 E *:a2 E *:a3 E *:b1 E *:b2 E *:b3 E +:c1 E +:c2 E +:c3"/>
by the pattern
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 E +:a2 E +:a3"/>
yields a 2nd row ladder escape.

'''Proof.''' Assume L3+P is a virtual connection. We must show that L2+↑+2+P is a virtual connection. It looks as follows, with P added to it:
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R 1:a2"/>
But Blue must play 2, and Red can jump to 3. Then 3 is a 3rd row ladder stone, and is connected to the edge because L3+P is a virtual connection by assumption. Therefore, 1 is also connected.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R 1:a2 B 2:a3 R 3:c1"/>
□

We now finally get a necessary and sufficient condition for 3rd row ladder escapes in the following theorem.

'''Theorem 4.''' Consider a candidate P for a 3rd row ladder escape. Then P is a valid 3rd row ladder escape if and only if L3+P, L3+1+P, and L3+2+P are virtual connections.

'''Proof.''' The left-to-right implication is trivial, since by definition, if P is valid then L3+''n''+P is a virtual connection for all ''n'', including ''n'' = 0, 1, 2. For the opposite implication, assume that L3+P, L3+1+P, and L3+2+P are virtual connections. By Lemma 3, L2+↑+2+P is a virtual connection. By Theorem 2 and the assumption about L3+2+P, 2+P is a 3rd row ladder escape. It follows that L3+''n''+P is a virtual connection for all ''n'' ≥ 2. Since we additionally assumed this to be the case for ''n'' = 0 and ''n'' = 1, P is a valid third row ladder escape. □

As a matter of fact, Theorem 4 is not tight. We can get the following better result. However, the proof of Theorem 4 generalizes more easily to 4th row and higher ladders, which is why it is of interest.

'''Theorem 5.''' Consider a candidate P for a 3rd row ladder escape. Then P is a valid 3rd row ladder escape if and only if L3+P and L3+1+P are virtual connections.

'''Proof.''' The left-to-right implication is trivial. For the right-to-left implication, by Theorem 4, it suffices to show that L3+2+P is a virtual connection. Indeed, consider Blue's options in the position L3+2+P:
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:b1"/>
As usual, there are only two possible moves for Blue to avoid losing immediately. If Blue moves at 2, then Red can respond at 3, which connects to the edge because L3+1+P is a virtual connection by assumption.
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 E *:a2 R 1:b1 B 2:b2 R 3:c1"/>
If Blue instead moves at 2, then Red responds as follows, which connects to the edge because L3+P is a virtual connection by assumption.
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 E *:a2 R 1:b1 B 2:b3 R 3:b2 B 4:a3 R 5:d1"/>
□

=== Examples ===

Here are some examples of third row ladder escapes. Again most of these are taken from [http://www.drking.org.uk/hexagons/hex/templates.html David King's website]. For several of the ladder escape templates, the version shown on David King's website is not minimal by our definition; in these cases, we have moved the cells marked "+" to the right to make the template minimal. All of the templates in this section have been proven to be third row ladder escapes using Theorem 5. All of them are minimal. As before, a stone marked "↓" is assumed to be connected to the bottom row, but the connection is not shown. Any shaded cells are not part of the pattern and can be occupied by Blue.

The following third row ladder escapes also escape 2nd row ladders at distance 0 or greater:
<hexboard size="3x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-b2 c2 b3 c3"
contents="E +:a1 R ↓:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 d2 b3 c3 d3 b4 c4 d4"
contents="E *:a1 E *:b1 R ↓:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>

<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d3"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E +:a5 E *:b1 R d1 R d2 E *:d3 R d4"
/>
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 g1 g2"
contents="E *:a1 E +:a3 E +:a4 E +:a5 E *:b1 E *:c1 R e1 R f3 E *:g1 E *:g2 S e3"
/>
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1 f3 f4 f5"
contents="E +:a3 +:a4 +:a5 R e3 R f2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/>

The following third row ladder escapes also escape 2nd row ladders at distance 1 or greater (but not at distance 0):
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 f1 g1 g2"
contents="R b2 E +:a3 +:a4 +:a5"
/>

The following third row ladder escapes also escape 2nd row ladders at distance 2 or greater (but not at distance 0 or 1):
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="R b1 E *:d1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R c1 E *:d1"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R b1 R c1 E *:d1 S b2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-area(b3,c2,d2,d4,b4)"
contents="R ↓:d1 E +:a2 +:a3 +:a4"
/>
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 b2 d2 e2 a3 b3 d3 e3 a4 b4 d4 e4"
contents="R ↓: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"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 e2 e3 e4 e5"
contents="E +:a3 +:a4 +:a5 E *:a1 *:b1 *:c1 *:e2 *:e3 *:e4 *:e5 R ↓:e1 S d2"
/>
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 g1 h1 h2"
contents="E *:a1 E +:a3 E +:a4 E +:a5 E *:b1 E *:c1 R e1 E *:g1 E *:h1 E *:h2"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1"
contents="E +:(a3 a4 a5) R c1 d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 d1 d2"
contents="E +:(a3 a4 a5) R b1 c1"
/><hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1"
contents="E *:a1 E *:a2 E +:a4 E +:a5 E +:a6 E *:b1 R c1 R d2"
/>
<hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2"
contents="E *:a1 E *:b1 R c1 E *:a2 R b2 E +:a4 E +:a5 E +:a6"
/>
The version of this last pattern on David King's website has the cells marked "+" (he uses arrows) sloping in the other direction; the location that is shown here makes the template minimal.

== Fourth row ladders ==

=== Definition of ladder ===

'''Definition.''' A ''fourth row ladder'' is a pattern like this:
L4: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R c1 E -:(d1--d4)"
/>
Again, the red stone is called the ''ladder stone'' and Red wants to connect the ladder stone to the bottom edge. We denote this pattern by L4.

The key part of the definition is that the 6 hexes forming a triangle below the ladder stone are all vacant. Note that even filling in one of these can invalidate the ladder: even if we fill in the bottom left corner of the triangle,
<hexboard size="5x5"
coords="hide"
edges="bottom"
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"
edges="bottom"
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 establish the ladder, Red needs at a minimum those 6 vacant hexes under her ladder stone.

=== Definition of ladder escape ===

The definition of a 4th row ladder escape is entirely analogous to that of 2nd and 3rd row ladder escapes.

'''Definition.''' A ''fourth row ladder escape'' is given by a pattern P that is open on the left with a boundary of this shape.
<hexboard size="4x1"
coords="none"
edges="bottom"
contents="E +:(a1--a4)"
/>
Moreover, it must satisfy that for all ''n'' ≥ 0, L4 + ''n'' + P is a virtual connection from the Red ladder stone to the bottom edge.

As before, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

=== Characterization of ladder escapes ===

We have already encountered all of the relevant ideas. 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!

'''Theorem 6.''' Consider a candidate P for a 4th row ladder escape. If L2+↑+↑+P, L3+↑+P, L4+P, and L4+1+P are virtual connections, then P is a 4th row ladder escape.

'''Proof.''' The idea of the proof is the same as for 3rd row ladders. First observe that by Theorems 1 and 2, since L2+↑+↑+P and L3+↑+P are virtual connections, ↑+P escapes all 3rd row ladders and ↑+↑+P escapes all 2nd row ladders. We must prove that L4+''n''+P is a virtual connection for all ''n'' ≥ 0. We proceed by induction on ''n''. The base cases ''n'' = 0 and ''n'' = 1 hold by assumption. For the induction step, assume the claim is true for ''n'' ≥ 1. We need to prove the claim for ''n''+1.

Consider the position L4+''n''+1+P. It looks like this, with ''n''−1 additional columns of four vacant hexes and the pattern P attached on the right:

<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R 1:c1"
/>
We need to prove that the ladder stone 1 is connected to the edge.

The five moves marked 2 below all lose instantly to Red 3:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:d1 B 2:e1 E *:a2 E *:b2 B 2:e2 E *:a3 B 2:e3 B 2:e4 R 3:c2"
/>
The two moves marked 2 below also lose instantly:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:b3 B 2:a4 E *:a2 E *:b2 E *:a3 R 3:d2"
/>
The move marked 2 below can be answered by Red 3, moving us to position L4+''n''+P, which is a virtual connection by the induction hypothesis.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3"
/>
Move 2 below leads us to a 2nd row ladder, which ↑+↑+P escapes, so 5 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 B 4:b4"
/>
Both moves marked 2 below lead us to a 3rd row ladder, which ↑+P escapes, so 3 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 E *:a3 B 2:c3 B 2:b4"
/>
Move 2 below also leads to a 3rd row ladder (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"
edges="bottom"
visible="-a1--a3 b1 b2"
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 a 2nd row ladder:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 B 4:b4 B 2:c4"
/>
The final choice for move 2 below also gives a 2nd row ladder.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
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 induction, so P is a 4th row ladder escape. □

Remark: In more concrete terms, Theorem 6 states that P is a 4th row ladder escape if adding each of the following patterns to P gives a virtual connection.
A: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R 1:c1 E -:(d1--d4)"
/>
B:<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R 1:c1 E -:(e1--e4)"
/>
C:<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="R 1:b2 E -:(c1--c4)"
/>
D:<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1--a2"
contents="R 1:a3 E -:(b1--b4)"
/>

Remark: Theorem 6 is analogous to Theorem 2. It gives a sufficient, but not a necessary condition for a candidate to be a 4th row ladder escape. Once again, the criterion in Theorem 6 can be checked in a finite amount of time. To get a theorem with a necessary and sufficient condition, we need another "jump lemma":

'''Lemma 7 (3rd to 4th row jump).''' Any 4th row ladder escape also escapes 3rd row ladders that start at distance 3 or greater.
More specifically, if L4+P and L4+1+P are virtual connections, then so is L3+↑+3+P.

'''Proof.''' Consider the position L3+↑+3+P, which looks as follows, with P added to it:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="R 1:b2"
/>
There are only two possible moves for Blue that don't lose immediately. If Blue moves at 2, then Red can respond at 3, which is a 4th row ladder stone and connects to the edge because L4+1+P is a virtual connection by assumption.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="E *:a1 E *:a2 E *:a3 E *:b1 R 1:b2 B 2:b3 R 3:d1"/>
In Blue moves instead at 2 in the following diagram, then Red can respond as shown.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="R 1:b2 B 2:b4 R 3:b3 B 4:a4 R 5:d3 B 6:c3 R 7:d1 B 8:d2 R 9:e1"/>
Now Red's stone 9 is a 4th row ladder stone. Although the additional red stone 5 does not belong in the L4 template, this stone can only help Red. By assumption, L4+P is a virtual connection, and so stone 9, and therefore stone 1, is connected to the edge. □

We then arrive at a necessary and sufficient condition for fourth row ladder escapes.

'''Theorem 8.''' Given a candiate P for a 4th row ladder escape. Then P is a valid 4rd row ladder escape if and only if L4+P, L4+1+P, L4+2+P, ..., L4+6+P are virtual connections.

'''Proof.''' The proof is similar to that of Theorem 4. Again, the left-to-right implication is trivial. For the right-to-left implication, assume that L4+P, L4+1+P, L4+2+P, ..., L4+6+P are virtual connections. By Lemma 7 applied to P and 2+P, we know that L3+↑+3+P and L3+↑+5+P are virtual connections. By Lemma 3 applied to ↑+3+P, we know that L2+↑+2+↑+3+P is a virtual connection, and therefore also L2+↑+↑+5+P, which differs from L2+↑+2+↑+3+P only in that it contains two additional empty hexes. Since L2+↑+↑+5+P, L3+↑+5+P, L4+(5+P), and L4+(6+P) are virtual connections, we know by Theorem 6 that 5+P is a valid 4th row ladder escape. Therefore, L4+''n''+P is a virtual connection for all ''n'' ≥ 5. Since we assumed this to be also true for ''n'' = 0, 1, 2, 3, 4, it follows that P is a valid 4th row ladder escape. □

Remark: Like Theorem 4, it is likely that Theorem 8 is not tight, in the sense that there probably exists an even simpler condition that is necessary and sufficient for 4th row ladder escapes (perhaps analogous to Theorem 5). Also, in practice, Theorem 6 is often easier to check since it involves fewer conditions.

=== Examples ===

Here are some examples of fourth row ladder escapes. Most are taken from [http://www.drking.org.uk/hexagons/hex/templates.html David King's website]. In each case we have moved the column of "+"s as far as possible to the right to yield a minimal template. The validity of all of these escapes has been proved using Theorem 8.

The following fourth row ladder escape templates also escape 2nd and 3rd row ladders at distance 0 or greater:
<hexboard size="4x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b1 R b3"
/>
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c1"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b2 E *:c1"
/>
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b3 R c1 E *:c3 E *:c4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 b1"
contents="E +:a2 E +:a3 E +:a4 E +:a5 R b4 R c1"
/>
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R d2 E *:e1 E *:f1 E *:f2"
/>

The following fourth row ladder escape templates also escape 2nd row ladders and 3rd row ladders at distance 1 and greater.
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b1 R c1"
/>

<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b1 R c2"
/>

The following fourth row ladder escape templates also escape 2nd row ladders at distance 2 and greater, as well as 3rd row ladders at distance 1 or greater. The stone marked "↓" is assumed to be connected to the bottom edge, although the connection is now shown.
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b1 R c2"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c1 R c2"
/>
<hexboard size="5x1"
coords="hide"
edges="bottom"
contents="R ↓:a1 E +:a2 +:a3 +:a4 +:a5"
/>

The following fourth row ladder escape templates also escape 2nd row ladders at distance 1 and greater, as well as 3rd row ladders at distance 0 or greater.
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 b1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 E *:b1 R b2 R d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 b1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 E *:b1 R b2 R c1"
/>
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1"
contents="E +:a3 E +:a4 E +:a5 E +:a6 R c2 R e1"
/>
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1"
contents="E +:a3 E +:a4 E +:a5 E +:a6 R c2 R d1"
/>

The following fourth row ladder escape template also escapes 2nd row ladders at distance 2 and greater, as well as 3rd row ladders at distance 0 or greater.
<hexboard size="6x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1 f3"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E +:a5 E +:a6 E *:b1 E *:c1 R e3 R f2 E *:f3"
/>

== Fifth row ladders ==

=== Definition of ladder ===

'''Definition.''' A ''fifth row ladder'' is a pattern like this:
L5: <hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R e1 E -:(f1--f5)"
/>
As usual, the red stone is called the ''ladder stone'' and Red's goal is to connect it to the bottom edge. We denote this pattern by L5.

Unlike in the case of 2nd, 3rd, and 4th row ladders, this time it is not sufficient for a triangle of cells below and to the right of the ladder stone to be empty. We also need three additional empty cells to the left of this triangle. This is a minimal requirement; if even one of these cells is occupied by Blue, for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
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 R o4 B a5 R o5 B a6 R o6"
/>
then Blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
edges="bottom"
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 R o4 B a5 R o5 B a6 R o6 B 1:e5"
/>
The main line is complex; see for example [http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 this Little Golem discussion thread]. Many of the main lines of defense involve Blue playing an upside-down version of [[Tom's move]], for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
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 R o4 B a5 R o5 B a6 R o6 B 1:e5 R 2:e4 B 3:e3 R 4:f2 B 5:f3 R 6:g2 B 7:h4 E *:d5 *:f4"
/>
Note that Blue's 1 is connected to 5 by double threat at "*", and 7 is Tom's move upside-down, i.e., with the top line of blue stones serving as the "edge". Therefore, to establish the ladder, Red needs at minimum the specified 13 vacant hexes under the ladder stone.

=== Definition of ladder escape ===

The definition of a 5th row ladder escape is as expected.

'''Definition.''' A ''fifth row ladder escape'' is a pattern P that is open on the left with boundary of this shape:
<hexboard size="5x1"
coords="none"
edges="bottom"
contents="E +:(a1--a5)"
/>
It must satisfy the following axiom: for all ''n'' ≥ 0, L5 + ''n'' + P connects the red ladder stone to the bottom edge, with Blue to move. As usual, a ''candiate'' is such a pattern that satisfies everything except perhaps the axiom.

=== Characterization of ladder escapes ===

'''Theorem 9.'''
Consider a candiate P for a fifth row ladder escape. Assume L5+P, L5+1+P, L5+2+P, L4+↑+P, L4+↑+1+P, L3+↑+↑+P, and L2+↑+↑+↑+P are all virtual connections. Then P is a 5th row ladder escape.

'''Proof.''' The proof idea is the same as for 3rd and 4th row ladders, but there are a lot more cases to consider. First, note that by previous theorems, ↑+P escapes all 4th row ladders, ↑+↑+P escapes all 3rd row ladders, and ↑+↑+↑+P escapes all 2nd row ladders. We prove by induction on ''n'' that L5+''n''+P is a virtual connection for all ''n'' ≥ 0. The base cases ''n'' = 0, 1, 2 are true by assumption. For the induction step, assume the claim is true for ''n'' ≥ 2. We need to prove the claim for ''n''+1.

Consider the position L5+''n''+1+P, which looks like this (followed by an additional ''n''−2 columns of five empty hexes and the pattern P):
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4"
/>
The eight moves marked 2 below all lose instantly to Red 3 by [[Fourth_row_edge_templates#IV-1-a|edge template IV-1-a]]:
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f1 B 2:g1 B 2:g2 B 2:h1 B 2:h2 B 2:h3 B 2:h4 B 2:h5 R 3:e2"
/>
The three moves marked 2 below also lose instantly by [[Fourth_row_edge_templates#IV-1-a|edge template IV-1-a]]:
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:a5 B 2:b4 B 2:c3 R 3:f2"
/>
The six moves marked 2 below give a 4th row ladder, which ↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:b5 B 2:c4 B 2:c5 B 2:d3 B 2:d4 B 2:e3 R 3:f2"
/>
This leaves us with 11 more moves to consider.
If Blue pushes the ladder by making the move marked 2 below, Red can answer 3, moving us to position L5+''n''+P, which is a virtual connection by the induction hypothesis.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e2 R 3:f1"
/>
Move 2 below gives a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f2 R 3:e2 B 4:d4 R 5:e3"
/>
If Blue plays move 2 below, Red can answer 3, forcing Blue to respond in the [[ziggurat]] below and to the left of stone 3. If Blue plays in any of the cells marked 4, Red plays 5 and gets a 4th row ladder, which ↑+P escapes. Blue could have also first intruded upon the bridge between 1 and 3, but this does not help. From now on, we tacitly ignore bridge intrusions that are not helpful to Blue.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f3 R 3:f2 B 4:b5 B 4:c4 B 4:c5 B 4:d3 B 4:d4 B 4:d5 B 4:e3 B 4:e4 R 5:g2"
/>
If, on the other hand, Blue plays 4 below, then Red gets a 2nd row ladder, which ↑+↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f3 R 3:f2 B 4:e5 R 5:e3 B 6:d4 R 7:e4 B 8:d5 R 9:f4"
/>
If Blue plays move 2 below, Red gets a 2nd row ladder, which ↑+↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g3 R 3:e2 B 4:d4 R 5:f3 B 6:e5 R 7:f4"
/>
If Blue plays move 2 below, Red can answer 3, forcing Blue to respond in one of the hexes marked 4. Then Red can play 5 and gets a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e4 R 3:e2 B 4:c5 B 4:d3 B 4:d3 B 4:d4 B 4:d5 R 5:f3"
/>
Similarly, if Blue plays move 2 below, Red can answer 3, forcing Blue to respond in one of the hexes marked 4. Then Red can play 5 and gets a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f4 R 3:f2 B 4:d5 B 4:e3 B 4:e3 B 4:e4 B 4:e5 R 5:g3"
/>
If Blue plays move 2 below, Red can answer 3. There are only four hexes where Blue can respond without losing outright. If Blue moves in one of the three hexes marked 4, then Red gets a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g4 R 3:f2 B 4:d5 B 4:e3 B 4:e4 R 5:g3 B 6:f4 R 7:h3"
/>
If Blue instead moves in the hex marked 4 below, then the sequence plays out slightly differently, but Red still gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g4 R 3:f2 B 4:e5 R 5:d4 B 6:e3 R 7:g3 B 8:f4 R 9:h3"
/>
If Blue plays move 2 below, Red can answer 3. Then Blue must respond in any of the hexes marked "+", or else Blue will immediately lose to [[Edge_templates_with_one_stone#Edge_template_III1b|edge template III1b]] or a [[ziggurat]].
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 E +:e5 E +:f3 E +:f4 E +:f5"
/>
If Blue responds in one of the two hexes marked 4, Red gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 B 4:e5 B 4:f4 R 5:g3"
/>
If Blue instead responds with move 4 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 B 4:f3 R 5:e3 B 6:d4 R 7:e4"
/>
If Blue instead responds with move 4 below, Red still gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 B 4:f5 R 5:e3 B 6:d4 R 7:f4 B 8:e5 R 9:g4"
/>

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in one of the hexes marked "+" to avoid immediately losing to [[Edge_templates_with_one_stone#Edge_template_III1b|edge template III1b]].
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 E +:c5 E +:d4 E +:d5 E +:e3 E +:e4 E +:f3 E +:f4 E +:f5 E +:g3 E +:g4 E +:g5"
/>
If Blue responds in one of the hexes marked 4 below, Red gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:c5 B 4:d4 B 4:d5 B 4:e3 B 4:e4 B 4:f4 R 5:g3"
/>
If Blue instead responds in one of the hexes marked 4 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:f3 B 4:g3 R 5:e3 B 6:d4 R 7:e4 B 8:d5 R 9:f4"
/>
If Blue instead responds at 4, Red gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:g4 R 5:d4 B 6:e3 R 7:g3 B 8:f4 R 9:h3"
/>
If Blue instead responds at 4, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:f5 R 5:f3 B 6:e4 R 7:g4"
/>
If Blue instead responds at 4, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:g5 R 5:f3 B 6:e4 R 7:g4 B 8:f5 R 9:h4"
/>

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in one of the hexes marked "+" to avoid immediately losing to [[Edge_templates_with_one_stone#Edge_template_III1b|edge template III1b]] or a [[bridge]].
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f5 R 3:e2 E +:c5 E +:d3 E +:d4 E +:d5"
/>
If Blue responds in one of the hexes marked 4 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f5 R 3:e2 B 4:c5 B 4:d3 B 4:d4 R 5:f3 B 6:e4 R 7:g4"
/>
If Blue instead responds at 4 below, Red also gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f5 R 3:e2 B 4:d5 R 5:e3 B 6:d4 R 7:f4 B 8:e5 R 9:g4"
/>

Finally, if Blue plays move 2 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g5 R 3:e2 B 4:d4 R 5:f3 B 6:e5 R 7:g4 B 8:f5 R 9:h4"
/>
This completes the induction, so P is a 5th row ladder escape. □

Remark: In more concrete terms, Theorem 9 states that P is a 5th row ladder escape if adding each of the following patterns to P gives a virtual connection.
A: <hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R 1:e1 E -:(f1--f5)"
/>
B: <hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R 1:e1 E -:(g1--g5)"
/>
C: <hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R 1:e1 E -:(h1--h5)"
/>
D: <hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1--b3 c1"
contents="R 1:c2 E -:(d1--d5)"
/>
E: <hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a4 b1--b3 c1"
contents="R 1:c2 E -:(e1--e5)"
/>
F: <hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-a1--a4 b1--b2"
contents="R 1:b3 E -:(c1--c5)"
/>
G: <hexboard size="5x2"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 E -:(b1--b5)"
/>

Like Theorems 2 and 5, Theorem 9 gives a sufficient, but not necessary condition for 5th row ladder escapes. We do not currently have a necessary and sufficient condition. One problem is that we have no appropriate "jump lemma" from 4th to 5th row ladders. In fact, we can prove that no such jump lemma is possible.

'''Lemma 10 (No jumping from 4th to 5th row).''' Suppose Red is the attacker in a 4th row ladder. Given enough Blue pieces on the 6th row, and enough space on the right, jumping is not an option for Red. If Red tries to jump, Blue can block the ladder, and Red will get at most a 2nd row ladder in the opposite direction.

'''Proof.''' If Red tries to jump, Blue can play as follows.
<hexboard size="6x13"
coords="hide"
edges="bottom"
contents="R a1 a2 a3 B a4 R 1:b3 B 2:b4 R 3:c3 B 4:c4 R 5:e2 B 6:e5
B b1 c1 d1 e1 f1 g1 h1 i1 j1 k1 l1 m1 E *:m3 *:m4 *:m5 *:m6 *:l4 *:l5 *:l6 *:k5 *:k6 *:j5 *:j6 *:i6 *:h6"
/>
This is exactly an upside-down version of the situation in Theorem 16 below. No matter where Red plays next, Blue can prevent Red from connecting. The hexes marked "*" are not required by Blue (i.e., they could be occupied by Red). Under [[optimal play]], Red gets at most a 2nd row ladder in the opposite direction as follows:
<hexboard size="6x13"
coords="hide"
edges="bottom"
contents="R a1 a2 a3 B a4 R 1:b3 B 2:b4 R 3:c3 B 4:c4 R 5:e2 B 6:e5
B b1 c1 d1 e1 f1 g1 h1 i1 j1 k1 l1 m1 E *:m3 *:m4 *:m5 *:m6 *:l4 *:l5 *:l6 *:k5 *:k6 *:j5 *:j6 *:i6 *:h6
R 7:e3 B 8:e4 R 9:d4 B 10:c6 R 11:c5"
/>
See the proof of Theorem 16 for a detailed discussion of all the possible moves. □

Lemma 10 is a significant obstacle to establishing a necessary and sufficient criterion for 5th row ladder escapes. We do have the following generalization of Theorem 9, which gives a weaker sufficient condition (it is perhaps also necessary, but this has not been shown):

'''Theorem 11.''' Given a candiate P for a 5th row ladder escape. If there is some ''n'' ≥ 0 such that L5+P, L5+1+P, ..., L5+''n''+P, L5+''n''+1+P, L5+''n''+2+P, as well as L4+↑+''n''+P, L4+↑+''n''+1+P, L3+↑+↑+''n''+P and L2+↑+↑+↑+''n''+P, are virtual connections, then P is a valid 5th row ladder escape.

'''Proof.''' This follows directly from Theorem 9 applied to ''n''+P. □

=== Examples ===

Here are some examples of fifth row ladder escapes. The validity of these escapes has been proved using Theorem 11. These escapes are minimal.

The following fifth row ladder escapes also escape 2nd to 4th row ladders at distance 0 or greater:
<hexboard size="5x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b2 R b4"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c1 R b3"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-c4 c5"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c2 R b4 E *:c4 *:c5"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-c4 c5"
contents="E +:(a1--a5) R b1 R b4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c1 R b4"
/>

The following fifth row ladder escape also escapes 2nd row ladders at distance 1 or greater, and 3rd and 4th row ladders at distance 0 or greater:
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-c1 c5"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b2 R c3 E *:c1 E *:d1 E *:d2"
/>

The following fifth row ladder escape also escapes 2nd row ladders at distance 2 or greater, 3rd row ladders at distance 0 or greater, and 4th row ladders at distance 1 or greater:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-c1 d1 d2"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b1 R b2 E *:c1 E *:d1 E *:d2"
/>

The following fifth row ladder escape also escapes 2nd row ladders at distance 2 or greater, 3rd row ladders at distance 0 or greater, and 4th row ladders at distance 2 or greater:
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a2--a6 e1"
contents="E *:a2 E *:a3 E *:a4 E *:a5 E *:a6 E +:b2 E +:b3 E +:b4 E +:b5 E +:b6 R c3 E *:e1"
/>

== Sixth row ladders and up ==

Because of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick, 6th and higher row ladders do not exist in the usual sense. More specifically, even if we allow an arbitrary amount of empty space under the ladder stone, it is not possible for the attacker to keep pushing the ladder. Consider the following situation:
<hexboard size="7x13"
coords="show"
edges="bottom"
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 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2"
/>
Let's assume there is an arbitrary amount of empty space in the bottom 4 rows to the left of this diagram. The stone marked "1" is connected to the top, and looks like it could be the ladder stone for a potential 6th row ladder. If such a ladder were possible, the red stones on the M-file should certainly escape it.

From Blue's point of view, Blue is the attacker in an upside-down 2nd row ladder. Blue can therefore use an upside-down version of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick. To do so, Blue plays at 2:
<hexboard size="7x13"
coords="show"
edges="bottom"
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 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5"
/>
If both Red and Blue keep playing [[optimal play|optimally]], the best that Red can get is a pair of parallel 2nd and 4th row ladders in the opposite direction:
<hexboard size="7x13"
coords="show"
edges="bottom"
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 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5 R 3:d3 B 4:c5 R 5:c4 B 6:b5 R 7:e4 B 8:e3 R 9:d4 B 10:e6 R 11:d5 B 12:c7 R 13:c6 B 14:b7"
/>
Here, stone 10 is [[virtual connection|connected]] to the line of blue stones along the top, so Red has no way of connecting right. Red can now push a 4th row ladder from 5, and/or a 2nd row ladder from 13. There is not enough space for Red to immediately perform [[Tom's move]]. So unless Red has a ladder escape somewhere to the left of this diagram, or unless there's enough space on the 5th row somewhere to the left of this diagram to perform Tom's move, Red fails to connect to the edge.

Note that this argument does not show that 6th row ladders are categorically impossible. It only shows that the "usual" notion of ladder does not work. It is conceivable that 6th row ladders are possible under additional assumptions. For example, there might be a notion of 6th row ladder that requires additional space on the 7th row to its right, or on the 5th row to its left. It is currently unknown whether any viable notion of 6th row ladder exists.

For 7th row ladders the situation is even worse. As explained in [[open problems about edge templates]], no amount of space under the ladder (even if we demand that the entire 5th row is clear) is known to guarantee a red connection if Blue just ignores the ladder and plays elsewhere. Thus, it is possible that 7th row ladders do not even exist in theory. Of course they do not occur in practice either.

== Second-to-fourth row switchbacks ==

=== Definition of switchback ===

Informally, a 2nd-to-4th row [[switchback]] is a pattern that allows the attacker to turn around a 2nd row ladder into a ladder on the 4th row in the opposite direction. For example, in the following situation, suppose ladder stone marked "1" is connected to the top, with Blue to move.
<hexboard size="4x7"
coords="show"
edges="bottom"
contents="R b1 a2 a3 1:b3 B b2 a4 R g1"
/>
Red pushes the 2nd row ladder to d3, the breaks at f3:
<hexboard size="4x7"
coords="show"
edges="bottom"
contents="R b1 a2 a3 1:b3 B b2 a4 R g1 B 2:b4 R 3:c3 B 4:c4 R 5:d3 B 6:d4 R 7:f3"
/>
At this point, Blue is forced to play 8, and then a new ladder starts in the opposite direction:
<hexboard size="4x7"
coords="show"
edges="bottom"
contents="R b1 a2 a3 b3 B b2 a4 R g1 B b4 R c3 B c4 R d3 B d4 R f3 B 8:e3 R 9:f1 B 10:e2 R 11:e1 B 12:d2 R 13:d1"
/>
In this example, the ladder reconnects to Red's original group, although in general this does not need to be the case (even if the switchback doesn't connect, Red has just created a parallel edge 4 cells from the original edge - a large advantage for Red in any case).

To formalize the concept of a 2nd-to-4th row switchback, consider a 2nd row ladder.

L2: <hexboard size="2x2"
coords="hide"
edges="bottom"
contents="R a1 E -:(b1--b2)"
/>

This is the same ladder as defined in the section of second-row ladders above; only this time, Red's goal will be slightly different. To explain Red's goal, we show a slightly larger area around L2:
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2 c2 c3 c4"
contents="R a3 E *:a1 *:a2 *:b1 *:b3 *:b4 *:c2 *:c3 *:c4 a:c1 b:b2 b:b1 E -:(b3--b4)"
/>
This time, Red's goal will be to do at least one of the following two things: either connect the red ladder stone to the edge, or else, occupy the cell marked "a" with a red stone that is connected to the edge, without using the cells marked "b" or any cells to their left. We refer to this as the ''switchback condition''. We also call "a" the ''switchback cell'' and "b" the ''gap cells''. With this in mind, we now give the definition of a 2nd-to-4th row switchback template.

'''Definition.''' A ''2nd-to-4th row switchback template'' (or simply 2-to-4 switchback) is given by the following data. It is a pattern P with a left boundary of shape
<hexboard size="4x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a4)"
/>
and satisfying the following axiom: L2+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0.

As usual, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid switchback template.

As in previous sections, we write L2+↑+↑+''n''+P for the pattern obtained from P by moving the four hexes marked "+" to the left by ''n'' columns (leaving 4 rows of empty space), then removing the top two cells marked "+" (they are not part of the pattern) and replacing the remaining cells marked "+" by L2. Note that the switchback cell "a" and gap cells "b" are not part of L2. They are simply three cells on the board whose position is defined relative to L2. Depending on the value of ''n'', they may or may not end up being inside the pattern P.

=== Characterization of switchbacks ===

The following theorem gives a finite necessary and sufficient condition for a pattern to be a switchback.

'''Theorem 12.''' Given a candidate P for a 2-to-4 switchback. Then P is a valid 2-to-4 switchback if and only if L2+↑+↑+P satisfies the switchback condition.

'''Proof.''' The left-to-right implication is trivial. For the opposite implication, we will show, by induction on ''n'', that L2+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0. The base case ''n'' = 0 holds by assumption. Now suppose that the claim is true for some ''n''. To show the claim for ''n''+1, consider the position L2+↑+↑+''n''+1+P. It looks as follows, with ''n'' more empty columns and P appended on the right:
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:a3"
/>
Here, the ladder stone is marked "1". Blue has no choice but to push the ladder, and Red also pushes:
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:a3 B 2:a4 R 3:b3"
/>
At this point, the induction hypothesis guarantees that Red can either connect 3 to the edge, or else that Red can occupy and connect the switchback cell "a" while keeping "b" empty:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4 d2 d3 d4"
contents="R 1:a3 B 2:a4 R 3:b3 E a:d1 b:c2 b:c1 b:b2 b:b1"
/>
If 3 is connected to the edge, then so is 1, and we are done. Otherwise, "a" is connected to the edge and "b" is empty. Thus, the board looks like this, with "a" now acting as a ladder stone:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4 d2 d3 d4"
contents="E *:a1 *:a2 R a3 B a4 R b3 R 1:d1"
/>
Since Red's stones on the 2nd row are already connected to the top, and 1 is connected to the bottom, Blue has no choice but to respond at 2. Then Red can play 3.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4 d2 d3 d4"
contents="E *:a1 *:a2 R a3 B a4 R b3 R 1:d1 B 2:c2 R 3:c1"
/>
Now the switchback condition for L2+↑+↑+''n''+1+P is satisfied, proving the theorem. □

=== Examples ===

Any 2nd row ladder escape template trivially also works as a switchback template (with the location of the cells marked "+" adjusted as necessary; they may need to be moved to the left if there isn't space for the two additional "+"s in the pattern). Since such a template escapes 2nd row ladders outright, there is no need for the second part of the switchback condition.

The following are examples of 2nd-to-4th row switchback templates that are not second row ladder escapes. They are minimal.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d3 d4"
contents="E +:a1 +:a2 +:a3 +:a4 R d2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d3 d4"
contents="E +:a1 +:a2 +:a3 +:a4 R d1"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--c1"
contents="E +:a2 +:a3 +:a4 +:a5 R e1"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--c1"
contents="E +:a2 +:a3 +:a4 +:a5 R d1"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--c1"
contents="E +:a2 +:a3 +:a4 +:a5 R e1 R d1 S d2"
/>
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="area(a2,a5,g5,g3,f1,d1)"
contents="E +:a2--a5 R f2 S d5"
/>
In the last two templates, the shaded hex is not part of the template, and can be occupied by Blue.
The following template is useful for obtuse corners:
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1--c1 area(d5,f5,f3)"
contents="E +:a2 +:a3 +:a4 +:a5 R e1 f1"
/>
In the following template, the stone marked "↓" is assumed to be connected to the bottom edge, but the connection is not shown. The blue stone is not technically part of the pattern; however, if this cell were empty, the pattern would already work as a 2nd row ladder escape.
<hexboard size="4x7"
coords="hide"
edges="bottom"
visible="-g3 g4 f4"
contents="E +:a1 +:a2 +:a3 +:a4 B b4 R ↓:g2"
/>

== Third-to-fifth row switchbacks ==

=== Definition of switchback ===

The definition of 3rd-to-5th row switchbacks is similar to that of 2nd-to-4th row switchbacks.
Consider a 3rd row ladder.
L3: <hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1--c3)"
/>
We define the locations of the switchback cell "a" and gap cells "b" relative to L3:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 d2--d5"
contents="R b3 E b:c1 b:c2 a:d1 -:(c3--c5)"
/>
Again, the ''switchback condition'' states that with Blue to move, Red can either connect the ladder stone to the edge, or else Red can occupy the switchback cell and connect it to the edge, without using the gap cells or anything to their left.

'''Definition.''' A ''3nd-to-5th row switchback template'' (or simply 3-to-5 switchback) is given by the following data. It is a pattern P, open on the left with boundary
<hexboard size="5x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a5)"
/>
and subject to the requirement that L3+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0.

=== Characterization of switchbacks ===

The following theorem gives a finite necessary and sufficient condition for a pattern to be a switchback. It is analogous to the corresponding theorem for 3rd row ladders.

'''Theorem 13.''' Given a candidate P for a 3-to-5 switchback. Then P is a valid 3-to-5 switchback if and only if L3+↑+↑+P and L3+↑+↑+1+P satisfy the switchback condition.

'''Proof.''' The left-to-right implication is trivial. For the opposite implication, we will show, by induction on ''n'', that L3+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0. The base cases ''n'' = 0 and ''n'' = 1 hold by assumption. Now suppose that the claim is true for ''n'' and ''n''+1. To show the claim for ''n''+2, consider the position L3+↑+↑+''n''+2+P. It looks as follows, with ''n'' more empty columns and P appended on the right:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R 1:b3"
/>

The ladder stone is marked "1". As usual for 3rd row ladders, Blue must either push or yield, or else Red will connect to the edge outright. If Blue pushes, then so does Red:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R 1:b3 B 2:b4 R 3:c3"
/>
By induction hypothesis, L3+↑+↑+''n''+1+P satisfies the switchback condition, so either Red can connect to the edge, or else Red can get a new ladder stone 1 in the switchback cell for L3+↑+↑+''n''+1+P, which allows Red to satisfy the switchback condition for L3+↑+↑+''n''+2+P with stone 3.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R b3 B b4 R c3 R 1:e1 B 2:d2 R 3:d1 E *:(c4--c5 d3--d5 e2--e5)"
/>
The other option is for Blue to yield. (We will see later that when ''n'' is large enough, yielding in this situation is actually a terrible idea for Blue, since it will allow Red to use P to connect to the edge. But this is not relevant for the current proof).
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="E *:a3 R 1:b3 E *:a4 *:a1 *:a2 *:b1 *:b2 B 2:b5 R 3:b4 B 4:a5 R 5:d3"
/>
Now by the induction hypothesis, L3+↑+↑+''n''+P satisfies the switchback condition, so either Red can connect to the edge, or else Red can get a new ladder stone 1 in the switchback cell for L3+↑+↑+''n''+P. This allows Red to satisfy the switchback condition for L3+↑+↑+''n''+2+P with stone 5.
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R b3 B b5 R b4 B a5 R d3 R 1:f1 B 2:e2 R 3:e1 B 4:d2 R 5:d1 E *:(d4--d5 e3--e5 f2--f5)"
/>
This finishes the proof of the theorem. □

One may ask whether every 3-to-5 switchback template also works as a 2-to-4 switchback template. This is indeed the case at sufficient distance, due to the following jumping lemma.

'''Lemma 14 (2nd to 3rd row switchback jump).''' Any 3-to-5 switchback template is also a 2-to-4 switchback template at distance 4 or greater. More specifically, if P is a 3-to-5 switchback template, then ↑+4+P is a 2-to-4 switchback template.

'''Proof.''' Suppose P is a 3-to-5 switchback template, and consider Q = ↑+4+P. By Theorem 12, we must show that L2+↑+↑+Q satisfies the switchback condition. The position L2+↑+↑+Q looks like this, with P attached on the right:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4"
/>
After Blue pushes the ladder at 2, Red plays 3, which is essentially [[Tom's move]]. While this move is not sufficient to connect Red to the edge, it creates enough trouble to allow Red to get the desired switchback in the presence of P.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3"
/>
Let us consider Blue's options. If Blue moves outside the area marked "x", Red simply pushes the ladder and connects, using 3 as a ladder escape.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 E x:b4 x:b5 x:c4 x:c5 x:d4 x:d5 x:e3 x:e4 x:e5"
/>
If Blue moves in any of the cells marked 4, Red gets the switchback without using P:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:d4 B 4:d5 B 4:e3 B 4:e4 B 4:e5 R 5:c4 B 6:b4 R 7:c2"
/>
If Blue moves at 4, Red also gets the switchback without using P:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:b4 R 5:c2"
/>
If Blue moves at 4, Red also gets the switchback without using P. Note that 3 is connected by [[edge template III2b]].
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:b5 R 5:c4 B 6:b4 R 7:c2"
/>
If Blue moves at 4, Red also gets the switchback without using P. Note that 3 is connected to the edge by double threat.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:c5 R 5:b5 B 6:b4 R 7:c2"
/>
This leaves only one option for Blue. If Blue moves at 4, then Red responds as follows:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="E *:a1 *:a2 *:a3 R 1:a4 B 2:a5 R 3:d3 B 4:c4 R 5:b4 B 6:b5 R 7:d2"
/>
By hypothesis, since P is a 3-to-5 switchback template, Red can either connect 3 to the edge, or else get a connected red stone at "a", with "b" empty:
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="E *:a1 *:a2 *:a3 R 1:a4 B 2:a5 R 3:d3 B 4:c4 R 5:b4 B 6:b5 R 7:d2 E a:f1 b:e1 b:e2"
/>
In either case, 7 is connected to the edge, so Red has the desired switchback. □

'''Corollary 15.''' In a 3rd row ladder at distance 5 or greater to a 3-to-5 switchback, Blue cannot yield.

'''Proof.''' If Blue yields, then Red can switch back the resulting 2nd row ladder to the 4th row by the previous lemma. This will reconnect to Red's original 3rd row ladder, and therefore connect Red to the edge. In a diagram:
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R 1:b3 B 2:b5 R 3:b4 B 4:a5 R 5:e3 B 6:d4 R 7:c4 B 8:c5 R 9:e2 E a:g1 b:f1 b:f2"
/>
Note that Blue must play 6 for the same reason as in the lemma. Since Red will either connect 5 or "a" to the edge, 7 is also connected. Rather than just giving Red a switchback, 7 is actually connected to 1 by a [[Interior template#The crescent|crescent]]. □

Here is another interesting fact about 3-to-5 switchbacks. Given enough space, the defender of a 3rd row ladder cannot yield without giving the attacker a switchback.

'''Theorem 16.''' Given enough space to the right of a 3rd row ladder and two empty rows above it, if the defender tries to yield, the attacker can achieve a 3-to-5 switchback without requiring any addtional stones.

'''Proof.''' Let 1 be the ladder stone of a 3rd row ladder, and assume there is at least as much space as indicated in the following diagram.
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R 1:b3"
/>
If Blue yields at 2, then Red can play as follows:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R 1:b3 B 2:b5 R 3:b4 B 4:a5 R 5:e2
E x:c2 x:c3 x:c4 x:c5 x:d2 x:d3 x:d4 x:d5 x:e3"
/>
If Blue's next move is outside of the area marked "x", then Red connects to the edge outright, as follows:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
R 7:d4 B 8:c4 R 9:d2"
/>
Therefore, Blue must move in one of the hexes marked "x" above. This leaves nine possible moves for Blue.

If Blue moves at one of the hexes marked 6 below, then Red connects by [[edge template IV2b]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c2 6:d2 R 7:d3"
/>
If Blue moves at 6, then Red pushes the second row ladder twice and connects by [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c3 R 7:c4 B 8:c5 R 9:d4 B 10:d5 R 11:g3"
/>
If Blue moves at 6, then Red gets 2nd and 4th row parallel ladders, which connect by [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:d3 R 7:d2 B 8:e3 R 9:c4 B 10:c5 R 11:d4 B 12:d5 R 13:g3"
/>
If Blue moves at 6, then Red connects by a [[Interior template#The crescent|crescent]] and [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:e3 R 7:c4 B 8:c5 R 9:d4 B 10:d5 R 11:g3"
/>
If Blue moves at 6, then Red connects by [[Interior template#The crescent|crescent]] and [[ziggurat]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:d4 R 7:c4 B 8:c5 R 9:f3"
/>
If Blue moves at 6, then Red connects by [[Interior template#The shopping cart|shopping cart]] and [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c5 R 7:d4 B 8:d5 R 9:g3"
/>
If Blue moves at 6, the situation is almost identical:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:d5 R 7:d4 B 8:c5 R 9:g3"
/>
Note that in all cases so far, Red connected outright, i.e., didn't need a switchback. The final remaining possibility is for Blue to move at 6 in the following diagram. Then Red gets the switchback. Note that 7 is connected to the edge by [[edge template IV2b]].
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c4 R 7:d3 B 8:c3 R 9:d1"
/>
This finishes the proof of the theorem. □

=== Examples ===

Every 3rd row ladder escape template is also a 3-to-5 switchback template (possibly with the location of the column of "+"s adjusted), but it need not be minimal. Here are some examples of 3-to-5 switchback templates that are not 3rd row ladder escapes.

<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-e1"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 *:e1 R e3"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 R d2"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 R d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 R d1 R c1 S c2"
/>
The shaded cell is not part of the template and can be occupied by Blue.

== Second and fourth row parallel ladders ==

=== Definition of ladder ===

[[Parallel ladder]]s, especially on the 2nd and 4th rows, are quite common in Hex. For example, consider this situation, with Blue to move and the Red stone connected to the top:
<hexboard size="4x7"
coords="hide"
edges="bottom"
contents="B b2 R d1"
/>
Play may proceed as follows:
<hexboard size="4x7"
coords="hide"
edges="bottom"
contents="B b2 R d1 B 1:d2 R 2:e1 B 3:e2 R 4:c2 B 5:b4 R 6:c3 B 7:c4"
/>
Now Red has a choice: she can either continue pushing the 4th row ladder from 2, or the 2nd row ladder from 6. However, having parallel ladders puts Red in a stronger position than having a 2nd row ladder or a 4th row ladder alone. As we will see, there exist ladder escape templates than can escape a parallel ladder, but can neither escape a 2nd row ladder nor a 4th row ladder on its own.

'''Note.''' Unlike with single-row ladders, in the case of a parallel ladder, Red actually has a choice whether to push the 2nd row ladder or the 4th row ladder. For this reason, our formal definition of a parallel ladder follows a slightly different approach than that we took for single-row ladders above. Whereas above, we always assumed that ''Blue'' was next to move (and the ladder stone was already in a pushing position), here, we will assume that ''Red'' is next to move. This affects the definition of the ladder pattern, in that the ladder stones do not yet have empty space below them.

'''Definition.''' A ''parallel ladder on the 2nd and 4th rows'', or ''2-4 parallel ladder'' for short, is a pattern like this:
L24: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 E -:(d1--d4)"
/>
The red stones on the second and fourth rows are called the ladder stones, and Red's goal is to connect at least one of these stones to the bottom edge, with ''Red'' to move first. (We can assume that both ladder stones are already connected to the top). We denote this pattern by L24. There is also a variant of L24 that looks like this:
L24a: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 a3"
contents="R b2 R c1 B a4 c2 E -:(d1--d4)"
/>
L24 and L24a are equivalent, and for simplicity we will only use L24.

=== Definition of ladder escape ===

'''Definition.''' An ''escape template for parallel ladders on the 2nd and 4th rows'', or ''2-4 parallel ladder escape'' for short, is a pattern P with a left boundary of shape
<hexboard size="4x1"
coords="hide"
edges="bottom"
contents="+:(a1--a4)"
/>
satisfying the following axiom: adding a 2-4 parallel ladder at distance ''n'', for any ''n'' ≥ 0, allows Red to connect, in the sense that it guarantees the connection of at least one of the red ladder stones to the bottom edge, with Red to move.

As before, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

=== Characterization of ladder escapes ===

Fortunately, 2-4 parallel ladders are easy to analyze; they are almost as simple as 2nd row ladders. The reason is that, just as for 2nd row ladders, the defender has no choice; he must always push, because as we will see, yielding is not an option. We get a simple and clean theorem with a necessary and sufficient condition for 2-4 parallel ladder escapes.

'''Theorem 17.''' Consider a candidate P for a 2-4 parallel ladder escape. Then P is a valid 2-4 parallel ladder escape if and only if L24+P, L24+1+P, and L24+2+P allow Red to connect (with Red to move).

'''Proof.''' If the ladder escape is valid, then by definition, L24+''n''+P allows Red to connect for all ''n'', including ''n'' = 0, 1, 2. So the left-to-right implication is trivial. To prove the right-to-left implication, assume L24+P, L24+1+P, and L24+2+P allow Red to connect. We prove by induction that L24+''n''+P allows Red to connect for all ''n'' ≥ 0. The cases ''n'' = 0, 1, 2 are true by assumption. Now suppose the claim is true for some ''n'' ≥ 2. We must show the claim for ''n''+1. To do so, consider the position L24+''n''+1+P. The first six columns of this position look like this:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2"
/>
This is followed by ''n''−2 more columns of four empty hexes and by the pattern P. Red starts by pushing the 4th row ladder.
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1"
/>
Now consider Blue's possible moves. If Blue moves in any of the hexes marked 2 below (or elsewhere on the board), Red wins outright (i.e., without using the induction hypothesis).
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:e1 2:e2 2:e3 2:e4 2:f1 2:f2 2:f3 2:f4 2:d3 2:d4 R 3:c3"
/>
If Blue moves in any of the hexes marked 2 below, Red also wins outright:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:b3 2:b4 2:c3 R 3:e2"
/>
If Blue moves at 2 below, Red also wins outright:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:c4 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:e3"
/>
This means that the only possible move that is not immediately losing for Blue is to push the 4th row ladder. In this case, Red can respond as follows:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:d2 R 3:b3 B 4:b4"
/>
This position allows Red to connect by induction hypothesis, finishing the proof. □

It is clear that every 2nd row ladder escape and every 4th row ladder escape is also an escape for 2nd-and-4th row parallel ladders, since Red can decide to push only the 2nd row ladder, or only the 4th row ladder. In addition, 2nd-to-4th row switchback templates also work as 2-4 parallel ladder escapes. This is intuitively clear, as Red can simply push the 2nd row ladder and switch it back to the 4th row, where it will connect with the 4th row of the parallel ladder. The following theorem proves this more formally, using the definitions.

'''Theorem 18.''' Every 2nd-to-4th row switchback template is also a 2-4 parallel ladder escape.

'''Proof.''' Assume P is a 2nd-to-4th row switchback template. To show that P is a 2-4 parallel ladder escape, we must show that L24+''n''+P allows Red to connect with Red to move, for all ''n'' ≥ 0. Consider the position L24+''n''+P, which looks as follows, with an additional ''n'' blank columns and P on the right:
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2"
/>
Red plays as follows. At this point, since ''n''+P is a 2nd-to-4th row switchback template, Red can either connect 3 to the edge, or get a connected stone at "a" with "b" empty.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:b3 B 2:b4 R 3:c3 E a:e1 E b:d1 b:d2"
/>
This allows Red to connect at least one of the ladder stones, as required. □

=== Examples ===

As mentioned above, every 2nd row ladder escape, every 4th row ladder escape, and every 2nd-to-4th row switchback template works as a 2-4 parallel ladder escape. But there are some examples of 2-4 parallel ladder escapes that are none of the above. The most famous of these is [[Tom's move]], which states that a sufficient amount of empty space is enough for a 2-4 parallel ladder to connect to the edge. Specifically, the following is a 2-4 parallel ladder escape template:
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E +:a2 +:a3 +:a4 +:a5"
/>
Other examples:
<hexboard size="5x5"
visible="-a1 a2 b1 e4 e5 a3--a5 e3"
edges="bottom"
coords="none"
contents="E +:b2--b5 R d1 e1 B d2"
/>
Here are some other examples of 2-4 parallel ladder escapes that are neither 2nd nor 4th row ladder escapes nor 2nd-to-4th row switchbacks. They can be shown to be valid by Theorem 17, and are minimal. Unlike Tom's move, these ladder escapes don't require space on the 5th row.

While the following two patterns aren't switchbacks at distance 0 or 1, they do work as 2nd-to-4th row switchbacks at distance 2 or greater.
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-b3 b4"
contents="E +:a1 +:a2 +:a3 +:a4 R b2"
/>
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-b3 b4"
contents="E +:a1 +:a2 +:a3 +:a4 R b1"
/>

== Third and fifth row parallel ladders ==

=== Definition of ladder ===

Parallel ladders on the 3rd and 5th rows are less common than those on the 2nd and 4th rows, but they can occur. Pushing such ladders is less straightforward, as the defender has more options. Basically, as we will show, if the defender refuses to push, then the attacker can at least get a 2nd row ladder. Moreover, a 2nd-to-4th row switchback template is in that case sufficient for the attacker to connect.

'''Definition.''' A ''parallel ladder on the 3nd and 5th rows'', or ''3-5 parallel ladder'' for short, is a pattern like this:
L35: <hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 E -:(d1--d5)"
/>
The red stones on the third and fifth rows are again called the ladder stones, and Red's goal is to connect at least one of these stones to the bottom edge, with ''Red'' to move first. We denote this pattern by L35. Just like for 2-4 parallel ladders, there is an equivalent pattern for L35 that looks like this:
L35a: <hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 a3"
contents="R b2 R c1 B a4 c2 E -:(d1--d5)"
/>

=== Definition of ladder escape ===

'''Definition.''' An ''escape template for parallel ladders on the 3rd and 5th rows'', or ''3-5 parallel ladder escape'' for short, is a pattern P with a left boundary of shape
<hexboard size="5x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a5)"
/>
satisfying the following axiom: adding a 3-5 parallel ladder at distance ''n'', for any ''n'' ≥ 0, allows Red to connect, in the sense that it guarantees the connection of at least one of the red ladder stones to the bottom edge, with Red to move.

As always, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

=== Characterization of ladder escapes ===

3-5 parallel ladder escapes are not quite as easy to characterize as those for 2-4 parallel ladders, because the defender has more options. We get the following theorem, which only contains a sufficient condition for a pattern to be a 3-5 parallel ladder escape.

'''Theorem 19.''' Consider a candidate P for a 3-5 parallel ladder escape. If L35+P, L35+1+P, ..., L35+3+P allow Red to connect (with Red to move), and if ↑+P is a 2nd-to-4th row switchback template, then P is a valid 3-5 parallel ladder escape.

'''Proof.''' We prove by induction that L35+''n''+P allows Red to connect for all ''n'' ≥ 0. The cases ''n'' = 0, ..., 3 are true by assumption. Now suppose the claim is true for 0, ..., ''n'', where ''n'' ≥ 3. We must show the claim for ''n''+1. To do so, consider the position L35+(''n''+1)+P. The position looks like this, followed by ''n''−3 additional empty columns and P:
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2"
/>
Red starts by pushing the 5th row ladder.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 E x:a5 x:b4 x:b5 x:c3 x:c4 x:c5 x:d2 x:d3 x:d4 x:d5 x:e1 x:e2 x:e3 x:e4 x:e5 x:f2 x:f3 x:f4 x:f5 x:g3 x:g4 x:g5"
/>
Now consider Blue's possible moves. If Blue moves anywhere except the hexes marked "x", then Red wins outright by bridging from 1 to [[edge template IV1a|edge template IV-1a]].

If Blue moves in any of the hexes marked 2 below, Red moves at 3 and connects by [[ziggurat]] and [[double threat]].
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:e1 2:e2 2:e3 2:e4 2:e5 2:f2 2:f3 2:f4 2:f5 2:g3 2:g4 2:g5 R 3:c3"
/>
This leaves 10 more moves to consider.

If Blue moves at 2, Red pushes the 3rd row ladder.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d2 R 3:b3 E x:b4 y:b5"
/>
Now Blue must either push at "x" or yield at "y" (or else Red will connect immediately). If Blue pushes at "x", then Red has a 3-5 parallel ladder at distance ''n'', which connects by induction hypothesis.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d2 R 3:b3 B 4:b4"
/>
If Blue instead yields at "y", then Red can push the 2nd row ladder and use the switchback to either connect 7 to the edge or get a connected stone at "a". Note that "a" is connected to either 1 or 7 by double threat, so Red connects. (As a matter of fact, Red can do better in this case and get a 2-4 parallel ladder, but it is not needed for the current proof).
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d2 R 3:b3 B 4:b5 R 5:c4 B 6:c5 R 7:d4 E a:f2"
/>
If Blue moves at 2, then Red plays as follows and connects by [[edge template III2e]]:
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:c3 R 3:b4 B 4:b3 R 5:e2 B 6:e3 R 7:d3"
/>
If Blue moves at 2, then Red gets a 2nd row ladder that connects via the 2-to-4 switchback at "a". The moves 4 and 5 can also be played in the opposite order without changing the result.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d3 R 3:c3 B 4:d2 R 5:b3 B 6:b5 R 7:c4 B 8:c5 R 9:d4 E a:f2"
/>
If Blue moves at 2, then Red gets a 2nd row ladder that connects via the 2-to-4 switchback at "a". (In fact, Red can get a 2-4 parallel ladder, but it is not needed in this proof).
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b4 R 3:e2 B 4:e3 R 5:d3 B 6:c5 R 7:d4 B 8:d5 R 9:e4 E a:g2"
/>
If Blue moves at 2, Red connects.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:c4 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:e3"
/>
If Blue moves at 2, Red connects
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d4 R 3:b4 B 4:b3 R 5:e2 B 6:d3 R 7:f3"
/>
If Blue moves at 2, Red connects by [[edge template IV2d]].
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:a5 R 3:c4 B 4:c3 R 5:e2"
/>
If Blue moves at 2, Red can respond at 3.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b5 R 3:e2 E x:e3 y:d5"
/>
Now Blue must respond at "x" or "y", or else Red will connect immediately. If Blue plays at "x", Red gets a 2nd row ladder which connects via switchback at "a". Note that 5 is connected to at least one ladder stone by double threat.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b5 R 3:e2 B 4:e3 R 5:c4 B 6:c5 R 7:d4 B 8:d5 R 9:e4 E a:g2"
/>
If Blue instead plays at "y", Red also gets a 2nd row ladder which connects via switchback at "a".
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b5 R 3:e2 B 4:d5 R 5:d4 B 6:c5 R 7:e4 E a:g2"
/>
If Blue moves at 2, Red connects.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:c5 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:d3 B 8:c4 R 9:e4"
/>
Finally, if Blue moves at 2, Red also connects.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d5 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:e3 B 8:f4 R 9:d4"
/>
This finishes the proof. □

=== Examples ===

Like 2-4 parallel ladders, 3-5 parallel ladders have the property that they can connect to the edge outright if given enough space. There is an analog of [[Tom's move]] for 3-5 parallel ladders. The following diagram shows the amount of space required. If Red moves in the cell marked "x", Red can guarantee to connect at least one of the ladder stones marked "1" to the edge. The cell marked "x" is essentially the unique winning move (the only other winning option for Red is to push the 3rd row ladder one more hex before playing "x").
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 b1 b2 b3 c1 g1 h1 i1 j1 k1 k2 l1 l2 l3"
contents="R 1:a4 1:c2 E x:e3 B a5 c3"
/>
We note that this particular pattern is not technically a 3-5 parallel ladder escape. Without additional empty space on the 6th row, it only escapes 3-5 parallel ladders at distance 0 (as shown) and at distance 1. If the ladder starts further away, Blue has the option of yielding to a 2nd row ladder for which Red would need a 2-to-4 switchback template to connect.

Every 3rd row ladder escape, every 5th row ladder escape, and every 3-to-5 switchback template is also a 3-5 parallel ladder escape. Examples of 3-5 parallel ladder escapes that aren't one of the above are relatively rare, but they do exist. The following are some examples. They have been proved correct using Theorem 19, and they are minimal.

The following template escapes 3-5 parallel ladders, but it is not a 3rd-to-5th row switchback, nor does it escape 3rd, 4th, or 5th row ladders.
<hexboard size="6x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E +:(a2--a6) R f6"
/>

The following template escapes 3-5 parallel ladders, but it is not a 3rd-to-5th row switchback, nor does it escape 2nd, 3rd, 4th, or 5th row ladders.
<hexboard size="6x8"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2 g1 g2 h1 h2 h5 h6"
contents="E +:(a2--a6) R h4"
/>

== Second and fourth row terraced ladders ==

Sometimes it can happen that a ladder forms on top of another ladder, with the two rows of attacking stones not yet connected to the edge nor to each other. We call this a ''terraced ladder''. The following is an example of a terraced ladder on the 2nd and 4th rows, with Blue to move:
<hexboard size="5x10"
coords="hide"
edges="bottom"
contents="R a2 B a3 R b2 B b3 R c2 B c3 R d2 B a4 B a5 R b4 B b5 R c4 B c5 R d4 B d5 R e4 B e5 R f4"
/>
Although terraced ladders look superficially similar to parallel ladders, they should not be confused. There are two important differences: (1) in a parallel ladder, the two rows of attacking stones are connected to each other, whereas in a terraced ladder, they are not, and (2) in a parallel ladder, the upper ladder is "ahead" of the lower one, whereas in a terraced ladders, the upper ladder is at the same level or behind the lower ladder.

In fact, as we noted above, from the attacker's point of view, having 2nd and 4th row parallel ladders is ''stronger'' than having only a 2nd row ladder or only a 4th row ladder. For terraced ladders, the opposite is true: a 2nd and 4th row terraced ladder is ''weaker'' than having only a 2nd row ladder or only a 4th row ladder. Nevertheless, despite being relatively weak, terraced ladders can be pushed, and there is a notion of terraced ladder escape at arbitrary distance.

Before we develop the theory of terraced ladders, it is worth noting that terraced ladders from Red's point of view are parallel ladders from Blue's point of view, and vice versa. This can be seen by putting a row of blue stones on top, giving Blue an "edge":
<hexboard size="5x10"
coords="hide"
edges="bottom"
contents="R a2 B a3 R b2 B b3 R c2 B c3 R d2 B a4 B a5 R b4 B b5 R c4 B c5 R d4 B d5 R e4 B e5 R f4 B a1 b1 c1 d1 e1 f1 g1 h1 i1 j1"
/>
Indeed, from Red's point of view, Red has terraced ladders trying to connect to the bottom edge, whereas from Blue's point of view, Blue has parallel ladders trying to connect to the top edge. The fact that parallel ladders are better for Blue than individual ladders explains why terraced ladders are worse for Red than individual ladders.

=== Definition of ladder ===

In a terraced ladder, it is the defender, not the attacker, who decides whether to push the 2nd or 4th row ladder. Since the 4th row ladder can lag behind the 2nd row ladder by an arbitrary distance, there isn't just a single ladder template, but a family of them.

'''Definition.''' A 2nd and 4th row terraced ladder is any one of the following patterns:

T(0):
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4"
contents="R a3 b1 E -:(c1 c2 b3 b4)"
/>
T(1):
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="R a1 a3 E -:(c1 c2 b3 b4)"
/>
T(2):
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a4 d3 d4"
contents="R a1 a3 R b3 E -:(d1 d2 c3 c4)"
/>
T(3):
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a4 b4 e3 e4"
contents="R a1 a3 R b3 R c3 E -:(e1 e2 d3 d4)"
/>
and so on. In general, for ''k'' ≥ 1, the pattern T(''k'') looks like
<hexboard size="4x1"
float="inline"
coords="hide"
edges="bottom"
visible="-a4"
contents="R a1 a3"
/> followed by ''k''−1 columns of <hexboard size="4x1"
float="inline"
coords="hide"
edges="bottom"
visible="-a4"
contents="R a3"
/> followed by <hexboard size="4x3"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 c3 c4"
contents="R a3 E -:(c1 c2 b3 b4)"
/>

In these patterns, we refer to Red's stone on the 4th row as the ''top ladder stone'', and to Red's rightmost stone on the 2nd row as the ''bottom ladder stone''. Red's goal is to connect the top ladder stone to the bottom edge, assuming it is Blue's turn first. We can assume that the top ladder stone is already connected to the top edge, but we do not assume that the top and bottom ladder stones are connected to each other.

=== Definition of ladder escape ===

'''Definition.''' A ladder escape template for 2nd and 4th row terraced ladders, or 2-4 terraced ladder escape for short, is a pattern P with left boundary shaped like this,
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1 a2 b3 b4"
contents="E +:(a3 a4 b1 b2)"
/>
This pattern P must satisfy the following axiom: for all ''k'' ≥ 0 and all ''n'' ≥ 0, T(''k'')+''n''+P guarantees a connection of the top ladder stone to the bottom edge, with Blue to move.

As always, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid escape. If P is such a candidate, schematically of the form
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E +:(a3 a4 b1 b2) *:b3 *:b4 *:c1 *:c2 *:c3 *:c4"
/>
then we write ∗+P to denote the pattern obtained from P by replacing the top two cells marked "+" by empty cells, and adding two new cells marked "+" just to their left. The resulting template is then of the shape required for 4th row ladder escapes.
<hexboard size="4x3"
coords="hide"
edges="bottom"
contents="E +:(a1--a4) *:b3 *:b4 *:c1 *:c2 *:c3 *:c4"
/>

=== Characterization of ladder escapes ===

We will reduce the problem of establishing a terraced ladder escape to finitely many cases. This is done by two lemmas. Lemma 20 states that we only need to consider finitely many values of ''n'' (the distance from the bottom ladder stone to the escape). Lemma 21 states that we only need to consider finitely many values of ''k'' (the distance from the top ladder stone to the bottom ladder stone). .

'''Lemma 20.''' Suppose P is a candidate for a 2-4 terraced ladder escape. Assume T(''k'')+P, T(''k'')+1+P, and T(''k'')+2+P are virtual connections for all ''k'' ≥ 0 (i.e., they guarantee a connection of the top ladder stone to the edge, with Blue to move). Moreover, assume that ∗+P escapes 2nd, 3rd, and 4th row ladders (or more precisely, ∗+P escapes 4th row ladders, ↑+∗+P (which has 3 cells marked "+") escapes 3rd row ladders, and ↑+↑+∗+P escapes 2nd row ladders). Then P is a valid 2-4 terraced ladder escape.

'''Proof.''' We need to show that T(''k'')+''n''+P is a virtual connection for the top ladder stone, for all ''k'',''n'' ≥ 0. We prove this by nested induction, with the outer induction being on ''n'', and the inner induction on ''k''. The base cases ''n'' = 0, 1, 2 are true by assumption. Now consider some ''n'' ≥ 3, and suppose the claim is true up to ''n''−1. We need to show the claim for ''n''. Consider the position T(''k'')+''n''+P, which looks like this:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 E x:b1 x:c1 x:d1 x:e1 x:a2 x:b2 x:c2 x:d2 x:e2 x:f2 x:e3 x:f3 x:d4 x:e4 x:f4"
/>
followed by ''n''−3 additional empty columns and P. Here, our diagram illustrates the case ''k'' = 4, but the following arguments are valid for all ''k'' ≥ 0. The first observation is that if Blue's next move is outside of the area marked "x", Red connects to the edge immediately by a [[Interior_template#The_long_crescent|long crescent]] and [[edge template III2b]]:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 R e2 "
/>
Note that this works for all ''k'' ≥ 0, although for ''k'' = 0 and ''k'' = 1, the connection is simpler and does not require a long crescent. Therefore, Blue must move in the area marked "x".

We consider each of Blue's options in turn. If Blue moves just below the top ladder stone, then Red responds by pushing the 4th row ladder. In case ''k'' > 0, this leads to the position T(''k''−1)+''n''+P, and he claim holds by the inner induction hypothesis:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:a2 R 2:b1"
/>
In case ''k'' = 0, the situation is worse for Blue: in this case, Red gets a bona fide 4th row ladder, which ∗+P escapes by assumption:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 a2 e3 e4"
contents="R a3 b1 B 1:b2 R 2:c1"
/>
If Blue moves anywhere else on the 3rd or 4th row, then Red connects the two ladders and gets a second row ladder, which ↑+↑+∗+P escapes by assumption. This works for all ''k'' ≥ 0, although for illustration, we show only the case ''k'' = 4:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:b1 1:b2 1:c1 1:c2 1:d1 1:d2 1:e1 1:e2 1:f2 R 2:a2"
/>
This leaves only 5 possible Blue moves to consider.

If Blue moves at 1, Red gets a 3rd row ladder, which ↑+∗+P escapes. Note that 2 is connected to the top ladder stone by a [[Interior_template#The_long_crescent|long crescent]] (for ''k'' ≥ 2) or directly (for k = 0, 1).
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:e3 R 2:e2 B 3:d4 R 4:f2"
/>

If Blue moves at 1, Red gets a 3rd row ladder, which ↑+∗+P escapes by assumption:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:f3 R 2:f2 B 3:e2 R 4:a2 B 5:d4 R 6:e3 B 7:e4 R 8:g2"
/>
Note that there are several alternatives to Blue's move 3, but they all result in a 3rd row ladder for Red.

If Blue moves at 1, Red simply pushes the 2nd row ladder, and we are now in position T(''k''+1)+''n''−1+P, which is a virtual connection by the outer induction hypothesis.
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:d4 R 2:e3"
/>

If Blue moves at 1, Red gets a 2nd row ladder, which ↑+↑+∗+P escapes by assumption. Note again that 2 is connected to the top ladder stone.
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:e4 R 2:e2 B 3:d4 R 4:f3"
/>

If Blue moves at 1, Red gets a 2nd row ladder.
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:f4 R 2:e2 B 3:d4 R 4:f3 B 5:e4 R 6:g3"
/>
This finishes the proof of the lemma. □

Having reduced the distance ''n'' to finitely many cases, we would now like to reduce the parameter ''k'' to finitely many cases as well. The following lemma does this.

'''Lemma 21.''' Suppose P is a candidate for a 2-4 terraced ladder escape. Assume T(0)+P, T(1)+P, and T(2)+P are virtual connections (i.e., they guarantee a connection of the top ladder stone to the edge, with Blue to move). Then T(''k'')+P is a virtual connection for all ''k'' ≥ 0.

'''Proof.''' The first step in the proof is to show that the following two interior patterns are equivalent. By "interior pattern", we mean that the bottom row of red stones does not have to be a board edge.

B(1): <hexboard size="4x3"
coords="none"
edges="none"
visible="-a1 b1 d4"
contents="B c1 R a2 a4--c4 E x:c2 y:c3 z:a3 -:(d1--d3)"
/>

B(2): <hexboard size="4x4"
coords="none"
edges="none"
visible="-a1 b1 e4"
contents="B c1--d1 R a2 a4--d4 E x:d2 y:d3 z:a3 -:(e1--e3)"
/>

If Red plays first in the region, a red move at x [[captured cell|captures]] the entire region, so x is the only move that Red needs to consider, and its outcome is the same in B(1) and B(2).

If Blue moves first in the region, all of the interior moves (i.e., in unmarked cells) are [[Dominated_cell#Star_decomposition_domination|star-decomposition dominated]] by x. Therefore, Blue only needs to consider the moves x, y, and z. One can show that each of these three moves (x, y, and z) in region B(2) is equivalent to the corresponding move in region B(1). For example, after Blue moves at x, z dominates all of the interior moves and whoever plays there [[captured cell|captures]] the interior, regardless of whether the region is B(1) or B(2).

A consequence of the fact that regions B(1) and B(2) are equivalent is that all "longer" versions of these regions are also equivalent to B(1), B(2), and each other, i.e.,

B(3): <hexboard size="4x5"
coords="none"
edges="none"
visible="-a1 b1 f4"
contents="B c1--e1 R a2 a4--e4 E -:(f1--f3)"
/>

B(4): <hexboard size="4x6"
coords="none"
edges="none"
visible="-a1 b1 g4"
contents="B c1--f1 R a2 a4--f4 E -:(g1--g3)"
/>
and so on. This is easily proved by induction, because each longer region is obtained from the previous one by replacing a subregion of the form B(1) by B(2), which we already showed to be equivalent.

Next, consider this pattern:

B(0): <hexboard size="4x2"
coords="none"
edges="none"
visible="-a1 b1 c4"
contents="R a2 a4--b4 E x:b2 y:b3 z:a3 -:(c1--c3)"
/>

We claim that B(1) is at least as good as B(0) for Red, in the sense that anything Red can achieve with B(0), Red can also achieve with B(1). (In fact, B(1) is strictly better for Red than B(0), but that fact is not required for this proof). If Red moves first in the region B(0), the move at x again captures the whole region, and therefore achieves everything Red might hope to achieve in the region. In this case, B(0) and B(1) are equivalent. If Blue moves first, the situation is slightly more complicated. We must show that B(0) is at least as good for Blue as B(1). If Blue plays at x in B(1), then Blue has the corresponding option to move at x in B(0), which works for the same reason as in the proof of the equivalence of B(1) and B(2) above. If Blue plays at z in B(1), Red can respond by pushing the ladder, which creates a position that is literally B(0). If Blue plays at y in B(1), Red can respond at x, and a case distinction shows that no matter how the remaining 3 cells are filled, filling them in the same way in B(0) gives an equivalent position.

Finally, let C(0), C(1), C(2), ... be the same patterns as B(0), B(1), B(2), ..., except with the blue stones removed from the carrier. I.e.:

C(0): <hexboard size="4x2"
coords="none"
edges="none"
visible="-a1 b1 c4"
contents="R a2 a4--b4 E -:(c1--c3)"
/>

C(1): <hexboard size="4x3"
coords="none"
edges="none"
visible="-a1 b1 d4 c1"
contents="R a2 a4--c4 E -:(d1--d3)"
/>

C(2): <hexboard size="4x4"
coords="none"
edges="none"
visible="-a1 b1 e4 c1--d1"
contents="R a2 a4--d4 E -:(e1--e3)"
/>
etc. Note that for all ''k'' ≥ 0, C(''k'') is at least as good for Red as B(''k''). Because if the neighboring cells we removed from the templates are in fact occupied by Blue, then C(''k'') is the same as B(''k''); otherwise, if they are empty or Red, it can only help Red.

In particular, since each C(''k'') is at least as good for Red as B(''k''), and each B(''k'') is at least as good as B(0) = C(0), it follows that if Red wins any position containing C(0), then Red also wins the corresponding position containing C(''k'').

The final step in the proof is now easy. Simply observe that each T(''k''+2) is obtained from T(2) by replacing a subpattern of the form C(0) by C(''k''). Therefore, in any context P where T(2)+P is winning for Red, T(''k''+2)+P is also winning for Red. Combining this with the remaining two base cases T(0)+P and T(1)+P, we get the lemma. □

By combining the previous two lemmas, we obtain a sufficient condition for the validity of terraced ladder escapes.

'''Theorem 22.''' Consider a candidate P for a 2-4 terraced ladder escape. Assume T(''k'')+''n''+P is a virtual connection for ''n''=1,2,3 and ''k''=1,2,3 (nine possibilities). Moreover, assume that ∗+P escapes 2nd, 3rd, and 4th row ladders (or more precisely, ∗+P escapes 4th row ladders, ↑+∗+P (which has 3 cells marked "+") escapes 3rd row ladders, and ↑+↑+∗+P escapes 2nd row ladders). Then P is a valid 2-4 terraced ladder escape.

'''Proof.''' By Lemma 21, T(''k'')+''n''+P is a virtual connection for ''n''=1,2,3 and all ''k'' ≥ 0. Therefore, the hypothesis of Lemma 20 is satisfied, and thus P is valid. □

=== Non-examples ===

Since terraced ladders are weaker than 4th row ladders, any terraced ladder escape is also a 4th row ladder escape. The question then becomes: which 4th row ladder escapes are ''not'' terraced ladder escapes? Most, but not all, of the examples of 4th row ladder escapes given above also escape terraced ladders.

The following patterns escape 4th row ladders but do not escape terraced ladders:

<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R d2 E *:e1 E *:f1 E *:f2"
/>
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E +:a5 E +:a6 E *:b1 E *:c1 R c2 R d1"
/>

[[category: Theory]]
[[category: Ladder]]

Theory of ladder escapes

2024-10-18T05:21:41Z

Bobson8: Fix the property of a 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 formalise what it means to be a ladder.

Informally, a ladder escape (say, 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, no matter how far away from the ladder escape the ladder starts. So strictly speaking, to check that a pattern is a 4th row ladder escape, we must check that the attacker can connect to the edge from an ''infinite set'' of positions. This raises the issue of how one can 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, 4th, and 5th row ladders. It might be possible to resolve it for 6th row ladders but this has not yet been done, partly because such ladders are of little practical use. 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 is currently unresolved. It may in theory be that there are no 7th row ladders at all.

For the purpose of our analysis, we assume that all ladders move from left to right along the red bottom edge, with Red being the attacker. Of course, the analogous analysis also applies to ladders moving in the opposite direction or along different edges.

The analysis of 2nd–4th row ladders on this page was originally contributed by the user [[User:Wccanard|Wccanard]] in 2016.

'''A note on terminology.''' The usual definition of a ''template'' is a pattern that has a stated property (for example, being [[virtual connection|connected]]) and is also minimal with that property. In other words, a template is usually defined by two properties: validity and minimality. For the purpose of ''this'' page, we are mostly concerned with validity. Since it would be awkward to write "template but not necessarily minimal" throughout all of the definitions and proofs on this page, we adopt the convention, on this page only, that "template" means a pattern that is valid but not necessarily minimal. We will then speak of a "minimal template" when necessary.

== Algebraic notation ==

Before we start, let us introduce some notation that will be useful.

=== Open patterns ===

A ''pattern'' is a set of cells, each of which may be empty or occupied by a stone of either color. In this article, we will only be concerned with patterns that include a red board edge. A pattern is ''open on the left'' if it comes with some cells marked "+" on its left side. No cells to the left of those "+"s may be part of the pattern. A pattern is ''open on the right'' if it comes with some cells marked "−" on its right side. No cells to the right of those "−"s may be part of the pattern. A pattern is ''open on both sides'' if it is open on the left and on the right. A pattern is ''closed'' if it is not open on either side. For example, the following four patterns are open on the right, open on both sides, open on the left, and closed, respectively:
<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1 c2 c3)"
/> <hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="all"
contents="R b1 B c1 E +:(a1 a2 a3) -:(c2 c3)"
/> <hexboard size="3x2"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1"
contents="R b3 E +:(a2 a3)"
/> <hexboard size="3x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 c1 d3 B d1"
/>
The cells labelled "+" (if any) are called the ''left boundary'' of the pattern, the cells labelled "−" are called its ''right boundary'', and the ''carrier'' of a pattern consists of all cells that are part of the pattern (empty or not), except the boundaries.

=== Addition ===

Suppose P is a pattern that is open on the right, Q is a pattern that is open on the left, and the right boundary of P has the same number of cells and shape as the left boundary of Q. Then we write P+Q for the pattern obtained by joining P and Q along their common boundary. More specifically, P+Q is obtained as follows: delete the right boundary from P and the left boundary from Q. Then glue the patterns P and Q together along the line where the boundaries were deleted from each. For example:
<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1 c2 c3)"
/> + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x5"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 a3 b1 e3 e4"
contents="R b2 d1 e1"
/>
It is important to note that the carrier of P+Q consists of just the carriers of P and Q, ''without'' the boundary cells that have been deleted. The purpose of the boundary cells "+" and "−" is just to indicate where the patterns will be attached. It is possible to add more than two patterns, for example:
<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1 c2 c3)"
/> + <hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="all"
contents="R b1 B c1 E +:(a1 a2 a3) -:(c2 c3)"
/> + <hexboard size="3x2"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1"
contents="R b3 E +:(a2 a3)"
/> = <hexboard size="3x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2"
contents="R b1 c1 d3 B d1"
/>

Note that addition is only well-defined if P and Q can be glued together without their carriers overlapping. We will be careful to ensure that this is always the case. However, the addition is associative, i.e., (P + Q) + R is well-defined if and only if P + (Q + R) is well-defined, and in that case, they are equal.

=== The shift operation ===

If P is a pattern that is open on the left, we write 1 + P for the pattern obtained from P by shifting its left boundary one column to the left, and adding a column of empty cells where the boundary used to be. More generally, for any integer ''n'' ≥ 0, we write ''n'' + P for iterating this operation ''n'' times, i.e., for shifting the left boundary of P to the left by ''n'' columns and filling the additional space with empty cells. For example:

1 + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x5"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1 e3 e4"
contents="E +:(a2 a3 a4) R d1 e1"
/>

4 + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x8"
float="inline"
coords="none"
edges="bottom"
visible="-a1 b1 c1 d1 e1 h3 h4"
contents="E +:(a2 a3 a4) R g1 h1"
/>

Note that empty cells are only added to the height of the boundary. For a pattern that is open on the right, we can do exactly the same thing on the other side, i.e., P + ''n'' is obtained by shifting the right boundary to the right by ''n'' columns, filling the additional space with empty cells. For example:

<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="R c1 E -:(c2 c3)"
/> + 1     = <hexboard size="3x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1 d1"
contents="R c1 E -:(d2 d3)"
/>

<hexboard size="3x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="R c1 E -:(c2 c3)"
/> + 4     = <hexboard size="3x7"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1 d1 e1 f1 g1"
contents="R c1 E -:(g2 g3)"
/>

We call this the ''shift operation''. Note that it is associative: If P and Q have matching boundaries, then (P + ''n'') + Q = P + (''n'' + Q).

=== The reduce operation ===

If P is a pattern that is open on the left, we write ↑ + P for the pattern obtained from P by erasing the topmost "+" cell from its left boundary. The cell that formerly contained the "+" is no longer part of the pattern (i.e., it is not replaced by an empty cell). For example:

↑ + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 a2 d3 d4"
contents="E +:(a3,a4) R c1 d1"
/>

We call this the ''reduce operation''. The shift and reduce operations can be combined with each other and with addition of patterns. For instance, ''n'' + ↑ + ''m'' + P is the pattern obtained from P by first shifting its left boundary by ''m'' cells to the left, then reducing the size of that boundary by one cell, and then shifting it by another ''n'' cells to the left. For example:

2 + ↑ + 3 + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/> = <hexboard size="4x9"
float="inline"
coords="none"
edges="bottom"
visible="-a1--f1 a2--c2 i3 i4"
contents="E +:(a3 a4) R h1 i1"
/>

== Second row ladders ==

=== Definition of ladder ===

There is no issue at all with defining a 2nd row ladder. Informally, a 2nd row ladder looks like this:
<hexboard size="2x8"
coords="hide"
edges="bottom"
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"
/>
Note that at each point in the ladder, Blue's move is forced. Red can choose to continue pushing the ladder as long as she wants to. We formally define a second row ladder as follows:

'''Definition.''' A ''second row ladder'' is a pattern like this:
L2: <hexboard size="2x2"
coords="hide"
edges="bottom"
contents="R a1 E -:(b1 b2)"
/>
Here, the red stone is on the second row, and we call it the ''ladder stone''. Red's goal is to connect the ladder stone to the bottom edge. The cell immediately below and to the right of the ladder stone is empty. We denote this pattern by L2.

=== Definition of ladder escape ===

Before we give the formal definition of a second row ladder escape, let us consider an example. The following pattern is an example of a second row ladder escape.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
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 cells marked "+" indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is,
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:e1 E *:f1 R g1 R h1 E *:a2 E *:b2 E *:c2 E *:d2 E *:e2 R 1:a3 E *:h3 E *:h4"
/>
Red can guarantee a connection from the ladder stone (marked 1) to the bottom edge.

Let us clarify what the hexes marked "+" in the ladder escape pattern 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 (among other things) that Red must win the following position:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here, Red's ladder stone is marked "1", and the claim (easily verified) is that even with Blue to play, Red can connect the ladder stone to the bottom:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
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 row ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 B h3 B h4"
/>
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="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 R 3:b3 R 5:c3 R 7:d3 R 9:e3 B h3 B 2:a4 B 4:b4 B 6:c4 B 6:d4 B h4"
/>
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. We can now give a more formal definition of a second row ladder escape.

'''Definition.''' A ''second row ladder escape template'' (or simply ''second row ladder escape'') is given by the following data. It is a pattern P that is open on the left (see [[#Algebraic notation|algebraic notation]] above), with a boundary of this shape:
<hexboard size="2x1"
coords="hide"
edges="bottom"
contents="E +:(a1 a2)"
/>
subject to the following axiom: for all ''n'' ≥ 0, the position L2 + ''n'' + P is a [[strong connection|virtual connection]] from the ladder stone (marked 1) to the edge.

Concretely, this means that any position consisting of a second row ladder L2,
<hexboard size="2x1"
coords="hide"
edges="bottom"
contents="R 1:a1"
/>
followed directly to the right by an arbitrary number (zero or more) of pairs of vacant hexes on the first and second rows,
<hexboard size="2x1"
coords="hide"
edges="bottom"
contents=""
/>
followed by the second row ladder escape pattern (where the ladder slots into the escape by putting the ladder or rightmost column of vacant hexes onto the hexes marked "+"), allows Red to connect the ladder stone to the edge.

Terminology and notation: If we have a left-open pattern whose boundary is of the correct shape, but we are not sure whether it satisfies the axiom of a second row ladder escape, then we refer to it as a ''candidate for a second row ladder escape'' (or simply ''candidate'' if the rest is clear from the context). A candidate is ''valid'' if it is actually a ladder escape.

We also define what it means for a ladder escape template to be minimal.

'''Definition.''' A 2nd row ladder escape template is ''minimal'' if the following two things are true. First, 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 second, if the two hexes directly to the right of the two cells marked "+" are both vacant hexes in the pattern, then moving the cells marked "+" one hex to the right results in a new pattern which is not a 2nd row ladder escape.

Below, we will use analogous terminology and notations for ladders and ladder escapes on the 3rd and higher rows.

=== Characterization of ladder escapes ===

The definition of a 2nd row ladder escape 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 we cannot directly use the definition to check that something is a 2nd row ladder escape, because this would require checking that infinitely many patterns are virtual connections. Can we find some finite criterion for checking 2nd row ladder escapes? Fortunately, as every Hex player knows, the answer is yes. We have the following theorem:

'''Theorem 1.''' Consider a candidate P for a 2nd row ladder escape. Schematically:
<hexboard size="4x4"
coords="hide"
edges="bottom"
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"
/>
(Here the asterisks indicate the [[carrier]] of P, which can contain any stones at all, and can be of any shape or size, as long as it includes no cells to the left of the cells marked "+"). Then P is a valid 2nd row ladder escape if and only if L2+P is a virtual connection for Red.
<hexboard size="4x4"
coords="hide"
edges="bottom"
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"
/>

'''Proof.''' If the ladder escape is valid, then by definition, L2+''n''+P is a virtual connection for all ''n'' ≥ 0, and in particular for ''n'' = 0. This proves the left-to-right implication.

To go the other way we actually have to play some Hex, but it's pretty trivial. We must show that L2+''n''+P is a virtual connection for all ''n''. This is an easy induction on ''n''. For ''n'' = 0, the claim is true by assumption. If ''n'' > 0, then Blue must play directly below Red's ladder stone (or else Red will connect to the edge immediately), and now Red can play a ladder stone at distance ''n''−1 on the second row, which is a virtual connection by induction hypothesis. □

=== Examples ===

We can use Theorem 1 to prove that all of the following patterns are minimal second row ladder escapes. Most of these templates are taken from [http://www.drking.org.uk/hexagons/hex/templates.html David King's website], and there are several more there. For several of the templates, the corresponding pattern on David King's site is not minimal by our definition; for these templates, we have moved the cells marked "+" to the right to make the template minimal.

<hexboard size="2x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 R b2"
/>
<hexboard size="2x2"
coords="hide"
edges="bottom"
contents="E +:a1 R b1 E +:a2"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R b1 E +:a2 E +:a3"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R c1 E +:a2 E +:a3"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d3 d4"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E *:b1 R d1 R d2 E *:d3 E *:d4"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d3 d4"
contents="E +:a3 E +:a4 R c1 R d1"
/>
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 f1 a2"
contents="E *:a1 E *:b1 E *:c1 R d1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 f1 a2"
contents="E *:a1 E *:b1 E *:c1 R e1 E *:f1 E *:a2 E +:a3 E +:a4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d4 d5"
contents="E *:a1 *:a2 *:b1 *:d4 *:d5 +:a4 +:a5 R c1 R d2"
/>
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 b1 b2 c1 c2 d1 e1 i1 j1 j2"
contents="E *:a1 E *:a2 E *:a3 E +:a4 E +:a5 E *:b1 E *:b2 E *:c1 E *:c2 E *:d1 E *:e1 R g1 E *:i1 E *:j1 E *:j2"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="E +:(a4 a5) R c1 d1"
/>
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 b1 b2 c1 c2 d1 e1 i1 j1 j2"
contents="E *:a1 E *:a2 E *:a3 E +:a4 E +:a5 E *:b1 E *:b2 E *:c1 E *:c2 E *:d1 E *:e1 R g1 E *:i1 E *:j1 E *:j2 R h1 S g2"
/>
The shaded cell is not part of the template, and can be occupied by Blue.

In the below templates, the stone marked "↓" indicates a stone connected to the bottom edge, but the connection is not shown. The connection from 10 to the edge must not use any of the empty hexes in the pattern.
<hexboard size="2x3"
coords="hide"
edges="bottom"
visible="-b2 c2"
contents="E +:a1 R ↓:c1 E +:a2 E *:b2 E *:c2"
/>
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 b3 c3 d3"
contents="E *:a1 E +:a2 R ↓:d2 E +:a3 E *:b3 E *:c3 E *:d3"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 a2 d2 c3 d3 b4 c4 d4"
contents="E *:a1 E *:b1 E *:c1 R ↓:d1 E *:a2 E *:d2 E +:a3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>

== Third row ladders ==

=== Definition of ladder ===

There is a minor issue with defining ladders on the 3rd and higher rows. We want a definition that 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"
edges="bottom"
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 few blue stones on the first row. It is intuitively clear (and also provably true) that these blue stones 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 to have no stones on the first three rows to the right of the ladder (until we reach the escape), we do not want to also guarantee that there are no stones on the first row to the left of the ladder. We formally define third row ladders as follows.

'''Definition.''' A ''third row ladder'' is a pattern like this:
L3: <hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1--c3)"
/>
The red stone is again called the ''ladder stone'', and Red's goal is to connect the ladder stone to the bottom edge. We denote this pattern by L3.

The key part of the definition is that we are guaranteeing the triangle of three empty hexes under the red ladder stone. This is a minimal requirement, because for example if one of these cells were filled,
<hexboard size="3x2"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R b1 B a2 B a3"
/>
then in reality the game could look like this,
<hexboard size="4x6"
coords="hide"
edges="all"
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"
edges="all"
contents="R a1 B b1 B c1 B d1 B e1 B f1 R a2 R b2 B a3 B 1:c3 B a4"
/>

=== Definition of ladder escape ===

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 [[ladder handling|yield]] to the second row. The following definition is unsurprising.

'''Definition.''' A ''third row ladder escape'' is given by the following data. It is a pattern P that is open on the left, with a boundary of the shape
<hexboard size="3x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a3)"
/>
To be a third row ladder escape, the pattern must satisfy the property that for all ''n'' ≥ 0, L3 + ''n'' + P is a virtual connection from the Red ladder stone to the bottom edge, with Blue to move.

Like for second row escapes, a pattern that has the required shape for a ladder escape, but it is not (yet) known to be a valid ladder escape, is called a ''candidate''.

In pictures, for the pattern
<hexboard size="3x3"
coords="hide"
edges="bottom"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(where the carrier is schematically indicated by stars) to be a 3rd row ladder escape, it must give rise to a virtual connection when we attach a 3rd row ladder at distance 0,
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
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"
edges="bottom"
visible="-a1 a2"
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"
edges="bottom"
visible="-a1 a2"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any other distance.

=== Characterization of ladder escapes ===

Just like for second row ladder escapes, we again find ourselves in the situation that trying to use the definition to check that something is a 3rd row ladder escape involves checking that infinitely many positions are virtual connections. Once again, we have a theorem that allows us to replace this by a finite condition.

'''Theorem 2.''' Consider a candidate P for a 3rd row ladder escape. Assume that (a) L2+↑+P is a virtual connection and (b) L3+P is a virtual connection, each from the ladder stone to the bottom edge. Then P is a valid third row ladder escape.

'''Proof.''' First note that by Theorem 1, because L2+↑+P is a virtual connection, P escapes all 2nd row ladders. Now under the assumptions of the theorem, we must show that L3+''n''+P is a virtual connection for all ''n'' ≥ 0. We prove this by induction on ''n''. For ''n'' = 0, the claim is true by assumption (b). Now suppose the claim is true for some ''n''. To show the claim for ''n''+1, consider the position L3+''n''+1+P. The first three columns of this position look like this:
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R 1:b1 E *:a2"
/>
(This is followed by ''n'' more columns of three empty hexes and by the pattern P). Blue has three possible moves in a triangle under stone 1, and Blue needs to play one of these or he will lose instantly. We analyze all three moves in turn.

For the first, Red pushes the ladder and will connect to the edge because by induction hypothesis, L3+''n''+P connects to the edge, so stone 3 connects to the edge, and so stone 1 does too.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R 1:b1 R 3:c1 E *:a2 B 2:b2"
/>
For the second, Red just wins outright, i.e., we do not need to use the induction hypothesis.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 R 1:b1 E *:a2 R 3:c2 B 2:a3"
/>
And for the third, Red responds like this. Since stone 3 is a 2nd row ladder stone, it is connected to the edge because, as we noted above, ↑+P is a 2nd row ladder escape. Therefore stone 1 is also connected.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:b1 B 2:b3 R 3:c2"
/>

The induction is now complete, showing that P is a 3rd row ladder escape. □

Remark: in more concrete terms, Theorem 2 states that a pattern P is a 3rd row ladder escape if the pattern becomes a virtual connection (from the ladder stone to the edge) when we attach each of the following two patterns to its left boundary:

A:
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:b1 E -:(c1--c3)"
/>
B:
<hexboard size="3x2"
coords="hide"
edges="bottom"
visible="-a1"
contents="R 1:a2 E -:(b1--b3)"
/>

In contrast to the situation with 2nd row ladders, while Theorem 2 is ''sufficient'' to show that a position is a 3rd row ladder escape, it is not ''necessary''. For example, consider the following third row ladder escape template P:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="R b1 E +:a2 E +:a3 E +:a4"
/>
One can check directly that L3+2+P and L2+↑+2+P are both virtual connections, so that 2+P is a 3rd row ladder escape by Theorem 2. In particular, L3+''n''+P is a virtual connection for all ''n'' ≥ 2. Moreover, one can check that L3+P and L3+1+P are also virtual connections, so that P is a valid 3rd row ladder escape.

It is, however, not a valid 2nd row ladder escape for ladders at distance 0, because in the position L2+↑+P,
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1 a2"
contents="R b1 R 1:a3"
/>
the ladder stone marked "1" cannot connect to the edge.

Theorem 2 is therefore not sufficient to check that a given pattern is a 3rd row ladder escape. We need to work a little harder to get a necessary and sufficient condition for 3rd row ladder escapes.

'''Lemma 3 (2nd to 3rd row jump).''' Any 3rd row ladder escape also escapes 2nd row ladders that start at distance 2 or greater. More specifically, if L3+P is a virtual connection, then so is L2+↑+2+P.

The lemma is perhaps easier understood in pictures: given any 3rd row ladder escape, replacing the three cells marked "+"
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="c1 c2 c3"
contents="E *:a1 E *:a2 E *:a3 E *:b1 E *:b2 E *:b3 E +:c1 E +:c2 E +:c3"/>
by the pattern
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 E +:a2 E +:a3"/>
yields a 2nd row ladder escape.

'''Proof.''' Assume L3+P is a virtual connection. We must show that L2+↑+2+P is a virtual connection. It looks as follows, with P added to it:
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R 1:a2"/>
But Blue must play 2, and Red can jump to 3. Then 3 is a 3rd row ladder stone, and is connected to the edge because L3+P is a virtual connection by assumption. Therefore, 1 is also connected.
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R 1:a2 B 2:a3 R 3:c1"/>
□

We now finally get a necessary and sufficient condition for 3rd row ladder escapes in the following theorem.

'''Theorem 4.''' Consider a candidate P for a 3rd row ladder escape. Then P is a valid 3rd row ladder escape if and only if L3+P, L3+1+P, and L3+2+P are virtual connections.

'''Proof.''' The left-to-right implication is trivial, since by definition, if P is valid then L3+''n''+P is a virtual connection for all ''n'', including ''n'' = 0, 1, 2. For the opposite implication, assume that L3+P, L3+1+P, and L3+2+P are virtual connections. By Lemma 3, L2+↑+2+P is a virtual connection. By Theorem 2 and the assumption about L3+2+P, 2+P is a 3rd row ladder escape. It follows that L3+''n''+P is a virtual connection for all ''n'' ≥ 2. Since we additionally assumed this to be the case for ''n'' = 0 and ''n'' = 1, P is a valid third row ladder escape. □

As a matter of fact, Theorem 4 is not tight. We can get the following better result. However, the proof of Theorem 4 generalizes more easily to 4th row and higher ladders, which is why it is of interest.

'''Theorem 5.''' Consider a candidate P for a 3rd row ladder escape. Then P is a valid 3rd row ladder escape if and only if L3+P and L3+1+P are virtual connections.

'''Proof.''' The left-to-right implication is trivial. For the right-to-left implication, by Theorem 4, it suffices to show that L3+2+P is a virtual connection. Indeed, consider Blue's options in the position L3+2+P:
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:b1"/>
As usual, there are only two possible moves for Blue to avoid losing immediately. If Blue moves at 2, then Red can respond at 3, which connects to the edge because L3+1+P is a virtual connection by assumption.
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 E *:a2 R 1:b1 B 2:b2 R 3:c1"/>
If Blue instead moves at 2, then Red responds as follows, which connects to the edge because L3+P is a virtual connection by assumption.
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E *:a1 E *:a2 R 1:b1 B 2:b3 R 3:b2 B 4:a3 R 5:d1"/>
□

=== Examples ===

Here are some examples of third row ladder escapes. Again most of these are taken from [http://www.drking.org.uk/hexagons/hex/templates.html David King's website]. For several of the ladder escape templates, the version shown on David King's website is not minimal by our definition; in these cases, we have moved the cells marked "+" to the right to make the template minimal. All of the templates in this section have been proven to be third row ladder escapes using Theorem 5. All of them are minimal. As before, a stone marked "↓" is assumed to be connected to the bottom row, but the connection is not shown. Any shaded cells are not part of the pattern and can be occupied by Blue.

The following third row ladder escapes also escape 2nd row ladders at distance 0 or greater:
<hexboard size="3x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 R b2 E +:a3"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
contents="E +:a1 R b1 E +:a2 E +:a3"
/>
<hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-b2 c2 b3 c3"
contents="E +:a1 R ↓:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 d2 b3 c3 d3 b4 c4 d4"
contents="E *:a1 E *:b1 R ↓:d1 E +:a2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 d3"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E +:a5 E *:b1 R d1 R d2 E *:d3 R d4"
/>
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 g1 g2"
contents="E *:a1 E +:a3 E +:a4 E +:a5 E *:b1 E *:c1 R e1 R f3 E *:g1 E *:g2 S e3"
/>
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1 f3 f4 f5"
contents="E +:a3 +:a4 +:a5 R e3 R f2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 d3 d4"
contents="E +:(a2,a3,a4) R c1 d1"
/>

The following third row ladder escapes also escape 2nd row ladders at distance 1 or greater (but not at distance 0):
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 R b1 R c1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 f1 g1 g2"
contents="R b2 E +:a3 +:a4 +:a5"
/>

The following third row ladder escapes also escape 2nd row ladders at distance 2 or greater (but not at distance 0 or 1):
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="R b1 E *:d1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R c1 E *:d1"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R b1 R c1 E *:d1 S b2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 area(b3,c2,d2,d4,b4)"
contents="R ↓:d1 E +:a2 +:a3 +:a4"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 e2 e3 e4 e5"
contents="E +:a3 +:a4 +:a5 E *:a1 *:b1 *:c1 *:e2 *:e3 *:e4 *:e5 R ↓:e1 S d2"
/>
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 g1 h1 h2"
contents="E *:a1 E +:a3 E +:a4 E +:a5 E *:b1 E *:c1 R e1 E *:g1 E *:h1 E *:h2"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1"
contents="E +:(a3 a4 a5) R c1 d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 d1 d2"
contents="E +:(a3 a4 a5) R b1 c1"
/><hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1"
contents="E *:a1 E *:a2 E +:a4 E +:a5 E +:a6 E *:b1 R c1 R d2"
/>
<hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2"
contents="E *:a1 E *:b1 R c1 E *:a2 R b2 E +:a4 E +:a5 E +:a6"
/>
The version of this last pattern on David King's website has the cells marked "+" (he uses arrows) sloping in the other direction; the location that is shown here makes the template minimal.

The following is a rather strange example of a third row ladder escape:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a2 b2 d2 e2 a3 b3 d3 e3 a4 b4 d4 e4"
contents="R ↓:a1 R ↓: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"
/>
One of the reasons this example is strange is that if the ladder starts at distance 1 or greater, the pattern is no longer minimal. Since the ladder must come from the left, this template therefore does not come up in practice. Nevertheless, it satisfies our formal definition of a 3rd row ladder escape. This template also escapes 2nd row ladders at distance 2 or greater (but not at distance 0 or 1).

== Fourth row ladders ==

=== Definition of ladder ===

'''Definition.''' A ''fourth row ladder'' is a pattern like this:
L4: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R c1 E -:(d1--d4)"
/>
Again, the red stone is called the ''ladder stone'' and Red wants to connect the ladder stone to the bottom edge. We denote this pattern by L4.

The key part of the definition is that the 6 hexes forming a triangle below the ladder stone are all vacant. Note that even filling in one of these can invalidate the ladder: even if we fill in the bottom left corner of the triangle,
<hexboard size="5x5"
coords="hide"
edges="bottom"
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"
edges="bottom"
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 establish the ladder, Red needs at a minimum those 6 vacant hexes under her ladder stone.

=== Definition of ladder escape ===

The definition of a 4th row ladder escape is entirely analogous to that of 2nd and 3rd row ladder escapes.

'''Definition.''' A ''fourth row ladder escape'' is given by a pattern P that is open on the left with a boundary of this shape.
<hexboard size="4x1"
coords="none"
edges="bottom"
contents="E +:(a1--a4)"
/>
Moreover, it must satisfy that for all ''n'' ≥ 0, L4 + ''n'' + P is a virtual connection from the Red ladder stone to the bottom edge.

As before, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

=== Characterization of ladder escapes ===

We have already encountered all of the relevant ideas. 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!

'''Theorem 6.''' Consider a candidate P for a 4th row ladder escape. If L2+↑+↑+P, L3+↑+P, L4+P, and L4+1+P are virtual connections, then P is a 4th row ladder escape.

'''Proof.''' The idea of the proof is the same as for 3rd row ladders. First observe that by Theorems 1 and 2, since L2+↑+↑+P and L3+↑+P are virtual connections, ↑+P escapes all 3rd row ladders and ↑+↑+P escapes all 2nd row ladders. We must prove that L4+''n''+P is a virtual connection for all ''n'' ≥ 0. We proceed by induction on ''n''. The base cases ''n'' = 0 and ''n'' = 1 hold by assumption. For the induction step, assume the claim is true for ''n'' ≥ 1. We need to prove the claim for ''n''+1.

Consider the position L4+''n''+1+P. It looks like this, with ''n''−1 additional columns of four vacant hexes and the pattern P attached on the right:

<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R 1:c1"
/>
We need to prove that the ladder stone 1 is connected to the edge.

The five moves marked 2 below all lose instantly to Red 3:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:d1 B 2:e1 E *:a2 E *:b2 B 2:e2 E *:a3 B 2:e3 B 2:e4 R 3:c2"
/>
The two moves marked 2 below also lose instantly:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:b3 B 2:a4 E *:a2 E *:b2 E *:a3 R 3:d2"
/>
The move marked 2 below can be answered by Red 3, moving us to position L4+''n''+P, which is a virtual connection by the induction hypothesis.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3"
/>
Move 2 below leads us to a 2nd row ladder, which ↑+↑+P escapes, so 5 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 B 4:b4"
/>
Both moves marked 2 below lead us to a 3rd row ladder, which ↑+P escapes, so 3 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 E *:a3 B 2:c3 B 2:b4"
/>
Move 2 below also leads to a 3rd row ladder (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"
edges="bottom"
visible="-a1--a3 b1 b2"
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 a 2nd row ladder:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 B 4:b4 B 2:c4"
/>
The final choice for move 2 below also gives a 2nd row ladder.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
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 induction, so P is a 4th row ladder escape. □

Remark: In more concrete terms, Theorem 6 states that P is a 4th row ladder escape if adding each of the following patterns to P gives a virtual connection.
A: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R 1:c1 E -:(d1--d4)"
/>
B:<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="R 1:c1 E -:(e1--e4)"
/>
C:<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="R 1:b2 E -:(c1--c4)"
/>
D:<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1--a2"
contents="R 1:a3 E -:(b1--b4)"
/>

Remark: Theorem 6 is analogous to Theorem 2. It gives a sufficient, but not a necessary condition for a candidate to be a 4th row ladder escape. Once again, the criterion in Theorem 6 can be checked in a finite amount of time. To get a theorem with a necessary and sufficient condition, we need another "jump lemma":

'''Lemma 7 (3rd to 4th row jump).''' Any 4th row ladder escape also escapes 3rd row ladders that start at distance 3 or greater.
More specifically, if L4+P and L4+1+P are virtual connections, then so is L3+↑+3+P.

'''Proof.''' Consider the position L3+↑+3+P, which looks as follows, with P added to it:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="R 1:b2"
/>
There are only two possible moves for Blue that don't lose immediately. If Blue moves at 2, then Red can respond at 3, which is a 4th row ladder stone and connects to the edge because L4+1+P is a virtual connection by assumption.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="E *:a1 E *:a2 E *:a3 E *:b1 R 1:b2 B 2:b3 R 3:d1"/>
In Blue moves instead at 2 in the following diagram, then Red can respond as shown.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1"
contents="R 1:b2 B 2:b4 R 3:b3 B 4:a4 R 5:d3 B 6:c3 R 7:d1 B 8:d2 R 9:e1"/>
Now Red's stone 9 is a 4th row ladder stone. Although the additional red stone 5 does not belong in the L4 template, this stone can only help Red. By assumption, L4+P is a virtual connection, and so stone 9, and therefore stone 1, is connected to the edge. □

We then arrive at a necessary and sufficient condition for fourth row ladder escapes.

'''Theorem 8.''' Given a candiate P for a 4th row ladder escape. Then P is a valid 4rd row ladder escape if and only if L4+P, L4+1+P, L4+2+P, ..., L4+6+P are virtual connections.

'''Proof.''' The proof is similar to that of Theorem 4. Again, the left-to-right implication is trivial. For the right-to-left implication, assume that L4+P, L4+1+P, L4+2+P, ..., L4+6+P are virtual connections. By Lemma 7 applied to P and 2+P, we know that L3+↑+3+P and L3+↑+5+P are virtual connections. By Lemma 3 applied to ↑+3+P, we know that L2+↑+2+↑+3+P is a virtual connection, and therefore also L2+↑+↑+5+P, which differs from L2+↑+2+↑+3+P only in that it contains two additional empty hexes. Since L2+↑+↑+5+P, L3+↑+5+P, L4+(5+P), and L4+(6+P) are virtual connections, we know by Theorem 6 that 5+P is a valid 4th row ladder escape. Therefore, L4+''n''+P is a virtual connection for all ''n'' ≥ 5. Since we assumed this to be also true for ''n'' = 0, 1, 2, 3, 4, it follows that P is a valid 4th row ladder escape. □

Remark: Like Theorem 4, it is likely that Theorem 8 is not tight, in the sense that there probably exists an even simpler condition that is necessary and sufficient for 4th row ladder escapes (perhaps analogous to Theorem 5). Also, in practice, Theorem 6 is often easier to check since it involves fewer conditions.

=== Examples ===

Here are some examples of fourth row ladder escapes. Most are taken from [http://www.drking.org.uk/hexagons/hex/templates.html David King's website]. In each case we have moved the column of "+"s as far as possible to the right to yield a minimal template. The validity of all of these escapes has been proved using Theorem 8.

The following fourth row ladder escape templates also escape 2nd and 3rd row ladders at distance 0 or greater:
<hexboard size="4x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b1 R b3"
/>
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c1"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b2 E *:c1"
/>
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b3 R c1 E *:c3 E *:c4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 b1"
contents="E +:a2 E +:a3 E +:a4 E +:a5 R b4 R c1"
/>
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R d2 E *:e1 E *:f1 E *:f2"
/>

The following fourth row ladder escape templates also escape 2nd row ladders and 3rd row ladders at distance 1 and greater.
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b1 R c1"
/>

<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="E +:a1 E +:a2 E +:a3 E +:a4 R b1 R c2"
/>

The following fourth row ladder escape templates also escape 2nd row ladders at distance 2 and greater, as well as 3rd row ladders at distance 1 or greater. The stone marked "↓" is assumed to be connected to the bottom edge, although the connection is now shown.
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b1 R c2"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-a1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c1 R c2"
/>
<hexboard size="5x1"
coords="hide"
edges="bottom"
contents="R ↓:a1 E +:a2 +:a3 +:a4 +:a5"
/>

The following fourth row ladder escape templates also escape 2nd row ladders at distance 1 and greater, as well as 3rd row ladders at distance 0 or greater.
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 b1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 E *:b1 R b2 R d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 b1"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 E *:b1 R b2 R c1"
/>
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1"
contents="E +:a3 E +:a4 E +:a5 E +:a6 R c2 R e1"
/>
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1"
contents="E +:a3 E +:a4 E +:a5 E +:a6 R c2 R d1"
/>

The following fourth row ladder escape template also escapes 2nd row ladders at distance 2 and greater, as well as 3rd row ladders at distance 0 or greater.
<hexboard size="6x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1 f3"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E +:a5 E +:a6 E *:b1 E *:c1 R e3 R f2 E *:f3"
/>

== Fifth row ladders ==

=== Definition of ladder ===

'''Definition.''' A ''fifth row ladder'' is a pattern like this:
L5: <hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R e1 E -:(f1--f5)"
/>
As usual, the red stone is called the ''ladder stone'' and Red's goal is to connect it to the bottom edge. We denote this pattern by L5.

Unlike in the case of 2nd, 3rd, and 4th row ladders, this time it is not sufficient for a triangle of cells below and to the right of the ladder stone to be empty. We also need three additional empty cells to the left of this triangle. This is a minimal requirement; if even one of these cells is occupied by Blue, for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
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 R o4 B a5 R o5 B a6 R o6"
/>
then Blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
edges="bottom"
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 R o4 B a5 R o5 B a6 R o6 B 1:e5"
/>
The main line is complex; see for example [http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 this Little Golem discussion thread]. Many of the main lines of defense involve Blue playing an upside-down version of [[Tom's move]], for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
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 R o4 B a5 R o5 B a6 R o6 B 1:e5 R 2:e4 B 3:e3 R 4:f2 B 5:f3 R 6:g2 B 7:h4 E *:d5 *:f4"
/>
Note that Blue's 1 is connected to 5 by double threat at "*", and 7 is Tom's move upside-down, i.e., with the top line of blue stones serving as the "edge". Therefore, to establish the ladder, Red needs at minimum the specified 13 vacant hexes under the ladder stone.

=== Definition of ladder escape ===

The definition of a 5th row ladder escape is as expected.

'''Definition.''' A ''fifth row ladder escape'' is a pattern P that is open on the left with boundary of this shape:
<hexboard size="5x1"
coords="none"
edges="bottom"
contents="E +:(a1--a5)"
/>
It must satisfy the following axiom: for all ''n'' ≥ 0, L5 + ''n'' + P connects the red ladder stone to the bottom edge, with Blue to move. As usual, a ''candiate'' is such a pattern that satisfies everything except perhaps the axiom.

=== Characterization of ladder escapes ===

'''Theorem 9.'''
Consider a candiate P for a fifth row ladder escape. Assume L5+P, L5+1+P, L5+2+P, L4+↑+P, L4+↑+1+P, L3+↑+↑+P, and L2+↑+↑+↑+P are all virtual connections. Then P is a 5th row ladder escape.

'''Proof.''' The proof idea is the same as for 3rd and 4th row ladders, but there are a lot more cases to consider. First, note that by previous theorems, ↑+P escapes all 4th row ladders, ↑+↑+P escapes all 3rd row ladders, and ↑+↑+↑+P escapes all 2nd row ladders. We prove by induction on ''n'' that L5+''n''+P is a virtual connection for all ''n'' ≥ 0. The base cases ''n'' = 0, 1, 2 are true by assumption. For the induction step, assume the claim is true for ''n'' ≥ 2. We need to prove the claim for ''n''+1.

Consider the position L5+''n''+1+P, which looks like this (followed by an additional ''n''−2 columns of five empty hexes and the pattern P):
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4"
/>
The eight moves marked 2 below all lose instantly to Red 3 by [[Fourth_row_edge_templates#IV-1-a|edge template IV-1-a]]:
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f1 B 2:g1 B 2:g2 B 2:h1 B 2:h2 B 2:h3 B 2:h4 B 2:h5 R 3:e2"
/>
The three moves marked 2 below also lose instantly by [[Fourth_row_edge_templates#IV-1-a|edge template IV-1-a]]:
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:a5 B 2:b4 B 2:c3 R 3:f2"
/>
The six moves marked 2 below give a 4th row ladder, which ↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:b5 B 2:c4 B 2:c5 B 2:d3 B 2:d4 B 2:e3 R 3:f2"
/>
This leaves us with 11 more moves to consider.
If Blue pushes the ladder by making the move marked 2 below, Red can answer 3, moving us to position L5+''n''+P, which is a virtual connection by the induction hypothesis.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e2 R 3:f1"
/>
Move 2 below gives a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f2 R 3:e2 B 4:d4 R 5:e3"
/>
If Blue plays move 2 below, Red can answer 3, forcing Blue to respond in the [[ziggurat]] below and to the left of stone 3. If Blue plays in any of the cells marked 4, Red plays 5 and gets a 4th row ladder, which ↑+P escapes. Blue could have also first intruded upon the bridge between 1 and 3, but this does not help. From now on, we tacitly ignore bridge intrusions that are not helpful to Blue.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f3 R 3:f2 B 4:b5 B 4:c4 B 4:c5 B 4:d3 B 4:d4 B 4:d5 B 4:e3 B 4:e4 R 5:g2"
/>
If, on the other hand, Blue plays 4 below, then Red gets a 2nd row ladder, which ↑+↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f3 R 3:f2 B 4:e5 R 5:e3 B 6:d4 R 7:e4 B 8:d5 R 9:f4"
/>
If Blue plays move 2 below, Red gets a 2nd row ladder, which ↑+↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g3 R 3:e2 B 4:d4 R 5:f3 B 6:e5 R 7:f4"
/>
If Blue plays move 2 below, Red can answer 3, forcing Blue to respond in one of the hexes marked 4. Then Red can play 5 and gets a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e4 R 3:e2 B 4:c5 B 4:d3 B 4:d3 B 4:d4 B 4:d5 R 5:f3"
/>
Similarly, if Blue plays move 2 below, Red can answer 3, forcing Blue to respond in one of the hexes marked 4. Then Red can play 5 and gets a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f4 R 3:f2 B 4:d5 B 4:e3 B 4:e3 B 4:e4 B 4:e5 R 5:g3"
/>
If Blue plays move 2 below, Red can answer 3. There are only four hexes where Blue can respond without losing outright. If Blue moves in one of the three hexes marked 4, then Red gets a 3rd row ladder, which ↑+↑+P escapes.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g4 R 3:f2 B 4:d5 B 4:e3 B 4:e4 R 5:g3 B 6:f4 R 7:h3"
/>
If Blue instead moves in the hex marked 4 below, then the sequence plays out slightly differently, but Red still gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g4 R 3:f2 B 4:e5 R 5:d4 B 6:e3 R 7:g3 B 8:f4 R 9:h3"
/>
If Blue plays move 2 below, Red can answer 3. Then Blue must respond in any of the hexes marked "+", or else Blue will immediately lose to [[Edge_templates_with_one_stone#Edge_template_III1b|edge template III1b]] or a [[ziggurat]].
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 E +:e5 E +:f3 E +:f4 E +:f5"
/>
If Blue responds in one of the two hexes marked 4, Red gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 B 4:e5 B 4:f4 R 5:g3"
/>
If Blue instead responds with move 4 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 B 4:f3 R 5:e3 B 6:d4 R 7:e4"
/>
If Blue instead responds with move 4 below, Red still gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:d5 R 3:f2 B 4:f5 R 5:e3 B 6:d4 R 7:f4 B 8:e5 R 9:g4"
/>

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in one of the hexes marked "+" to avoid immediately losing to [[Edge_templates_with_one_stone#Edge_template_III1b|edge template III1b]].
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 E +:c5 E +:d4 E +:d5 E +:e3 E +:e4 E +:f3 E +:f4 E +:f5 E +:g3 E +:g4 E +:g5"
/>
If Blue responds in one of the hexes marked 4 below, Red gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:c5 B 4:d4 B 4:d5 B 4:e3 B 4:e4 B 4:f4 R 5:g3"
/>
If Blue instead responds in one of the hexes marked 4 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:f3 B 4:g3 R 5:e3 B 6:d4 R 7:e4 B 8:d5 R 9:f4"
/>
If Blue instead responds at 4, Red gets a 3rd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:g4 R 5:d4 B 6:e3 R 7:g3 B 8:f4 R 9:h3"
/>
If Blue instead responds at 4, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:f5 R 5:f3 B 6:e4 R 7:g4"
/>
If Blue instead responds at 4, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:e5 R 3:f2 B 4:g5 R 5:f3 B 6:e4 R 7:g4 B 8:f5 R 9:h4"
/>

If Blue plays move 2 below, Red can answer 3. Then Blue must respond in one of the hexes marked "+" to avoid immediately losing to [[Edge_templates_with_one_stone#Edge_template_III1b|edge template III1b]] or a [[bridge]].
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f5 R 3:e2 E +:c5 E +:d3 E +:d4 E +:d5"
/>
If Blue responds in one of the hexes marked 4 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f5 R 3:e2 B 4:c5 B 4:d3 B 4:d4 R 5:f3 B 6:e4 R 7:g4"
/>
If Blue instead responds at 4 below, Red also gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:f5 R 3:e2 B 4:d5 R 5:e3 B 6:d4 R 7:f4 B 8:e5 R 9:g4"
/>

Finally, if Blue plays move 2 below, Red gets a 2nd row ladder.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="E *:a1 E *:b1 E *:c1 E *:d1 R 1:e1 E *:a2 E *:b2 E *:c2 E *:d2 E *:a3 E *:b3 E *:a4 B 2:g5 R 3:e2 B 4:d4 R 5:f3 B 6:e5 R 7:g4 B 8:f5 R 9:h4"
/>
This completes the induction, so P is a 5th row ladder escape. □

Remark: In more concrete terms, Theorem 9 states that P is a 5th row ladder escape if adding each of the following patterns to P gives a virtual connection.
A: <hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R 1:e1 E -:(f1--f5)"
/>
B: <hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R 1:e1 E -:(g1--g5)"
/>
C: <hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-area(d1,a1,a4)-d2"
contents="R 1:e1 E -:(h1--h5)"
/>
D: <hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1--b3 c1"
contents="R 1:c2 E -:(d1--d5)"
/>
E: <hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a4 b1--b3 c1"
contents="R 1:c2 E -:(e1--e5)"
/>
F: <hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-a1--a4 b1--b2"
contents="R 1:b3 E -:(c1--c5)"
/>
G: <hexboard size="5x2"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 E -:(b1--b5)"
/>

Like Theorems 2 and 5, Theorem 9 gives a sufficient, but not necessary condition for 5th row ladder escapes. We do not currently have a necessary and sufficient condition. One problem is that we have no appropriate "jump lemma" from 4th to 5th row ladders. In fact, we can prove that no such jump lemma is possible.

'''Lemma 10 (No jumping from 4th to 5th row).''' Suppose Red is the attacker in a 4th row ladder. Given enough Blue pieces on the 6th row, and enough space on the right, jumping is not an option for Red. If Red tries to jump, Blue can block the ladder, and Red will get at most a 2nd row ladder in the opposite direction.

'''Proof.''' If Red tries to jump, Blue can play as follows.
<hexboard size="6x13"
coords="hide"
edges="bottom"
contents="R a1 a2 a3 B a4 R 1:b3 B 2:b4 R 3:c3 B 4:c4 R 5:e2 B 6:e5
B b1 c1 d1 e1 f1 g1 h1 i1 j1 k1 l1 m1 E *:m3 *:m4 *:m5 *:m6 *:l4 *:l5 *:l6 *:k5 *:k6 *:j5 *:j6 *:i6 *:h6"
/>
This is exactly an upside-down version of the situation in Theorem 16 below. No matter where Red plays next, Blue can prevent Red from connecting. The hexes marked "*" are not required by Blue (i.e., they could be occupied by Red). Under [[optimal play]], Red gets at most a 2nd row ladder in the opposite direction as follows:
<hexboard size="6x13"
coords="hide"
edges="bottom"
contents="R a1 a2 a3 B a4 R 1:b3 B 2:b4 R 3:c3 B 4:c4 R 5:e2 B 6:e5
B b1 c1 d1 e1 f1 g1 h1 i1 j1 k1 l1 m1 E *:m3 *:m4 *:m5 *:m6 *:l4 *:l5 *:l6 *:k5 *:k6 *:j5 *:j6 *:i6 *:h6
R 7:e3 B 8:e4 R 9:d4 B 10:c6 R 11:c5"
/>
See the proof of Theorem 16 for a detailed discussion of all the possible moves. □

Lemma 10 is a significant obstacle to establishing a necessary and sufficient criterion for 5th row ladder escapes. We do have the following generalization of Theorem 9, which gives a weaker sufficient condition (it is perhaps also necessary, but this has not been shown):

'''Theorem 11.''' Given a candiate P for a 5th row ladder escape. If there is some ''n'' ≥ 0 such that L5+P, L5+1+P, ..., L5+''n''+P, L5+''n''+1+P, L5+''n''+2+P, as well as L4+↑+''n''+P, L4+↑+''n''+1+P, L3+↑+↑+''n''+P and L2+↑+↑+↑+''n''+P, are virtual connections, then P is a valid 5th row ladder escape.

'''Proof.''' This follows directly from Theorem 9 applied to ''n''+P. □

=== Examples ===

Here are some examples of fifth row ladder escapes. The validity of these escapes has been proved using Theorem 11. These escapes are minimal.

The following fifth row ladder escapes also escape 2nd to 4th row ladders at distance 0 or greater:
<hexboard size="5x2"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b2 R b4"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c1 R b3"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-c4 c5"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c2 R b4 E *:c4 *:c5"
/>
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-c4 c5"
contents="E +:(a1--a5) R b1 R b4"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R c1 R b4"
/>

The following fifth row ladder escape also escapes 2nd row ladders at distance 1 or greater, and 3rd and 4th row ladders at distance 0 or greater:
<hexboard size="5x3"
coords="hide"
edges="bottom"
visible="-c1 c5"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b2 R c3 E *:c1 E *:d1 E *:d2"
/>

The following fifth row ladder escape also escapes 2nd row ladders at distance 2 or greater, 3rd row ladders at distance 0 or greater, and 4th row ladders at distance 1 or greater:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-c1 d1 d2"
contents="E +:a1 E +:a2 E +:a3 E +:a4 E +:a5 R b1 R b2 E *:c1 E *:d1 E *:d2"
/>

The following fifth row ladder escape also escapes 2nd row ladders at distance 2 or greater, 3rd row ladders at distance 0 or greater, and 4th row ladders at distance 2 or greater:
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a2--a6 e1"
contents="E *:a2 E *:a3 E *:a4 E *:a5 E *:a6 E +:b2 E +:b3 E +:b4 E +:b5 E +:b6 R c3 E *:e1"
/>

== Sixth row ladders and up ==

Because of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick, 6th and higher row ladders do not exist in the usual sense. More specifically, even if we allow an arbitrary amount of empty space under the ladder stone, it is not possible for the attacker to keep pushing the ladder. Consider the following situation:
<hexboard size="7x13"
coords="show"
edges="bottom"
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 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2"
/>
Let's assume there is an arbitrary amount of empty space in the bottom 4 rows to the left of this diagram. The stone marked "1" is connected to the top, and looks like it could be the ladder stone for a potential 6th row ladder. If such a ladder were possible, the red stones on the M-file should certainly escape it.

From Blue's point of view, Blue is the attacker in an upside-down 2nd row ladder. Blue can therefore use an upside-down version of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick. To do so, Blue plays at 2:
<hexboard size="7x13"
coords="show"
edges="bottom"
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 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5"
/>
If both Red and Blue keep playing [[optimal play|optimally]], the best that Red can get is a pair of parallel 2nd and 4th row ladders in the opposite direction:
<hexboard size="7x13"
coords="show"
edges="bottom"
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 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5 R 3:d3 B 4:c5 R 5:c4 B 6:b5 R 7:e4 B 8:e3 R 9:d4 B 10:e6 R 11:d5 B 12:c7 R 13:c6 B 14:b7"
/>
Here, stone 10 is [[virtual connection|connected]] to the line of blue stones along the top, so Red has no way of connecting right. Red can now push a 4th row ladder from 5, and/or a 2nd row ladder from 13. There is not enough space for Red to immediately perform [[Tom's move]]. So unless Red has a ladder escape somewhere to the left of this diagram, or unless there's enough space on the 5th row somewhere to the left of this diagram to perform Tom's move, Red fails to connect to the edge.

Note that this argument does not show that 6th row ladders are categorically impossible. It only shows that the "usual" notion of ladder does not work. It is conceivable that 6th row ladders are possible under additional assumptions. For example, there might be a notion of 6th row ladder that requires additional space on the 7th row to its right, or on the 5th row to its left. It is currently unknown whether any viable notion of 6th row ladder exists.

For 7th row ladders the situation is even worse. As explained in [[open problems about edge templates]], no amount of space under the ladder (even if we demand that the entire 5th row is clear) is known to guarantee a red connection if Blue just ignores the ladder and plays elsewhere. Thus, it is possible that 7th row ladders do not even exist in theory. Of course they do not occur in practice either.

== Second-to-fourth row switchbacks ==

=== Definition of switchback ===

Informally, a 2nd-to-4th row [[switchback]] is a pattern that allows the attacker to turn around a 2nd row ladder into a ladder on the 4th row in the opposite direction. For example, in the following situation, suppose ladder stone marked "1" is connected to the top, with Blue to move.
<hexboard size="4x7"
coords="show"
edges="bottom"
contents="R b1 a2 a3 1:b3 B b2 a4 R g1"
/>
Red pushes the 2nd row ladder to d3, the breaks at f3:
<hexboard size="4x7"
coords="show"
edges="bottom"
contents="R b1 a2 a3 1:b3 B b2 a4 R g1 B 2:b4 R 3:c3 B 4:c4 R 5:d3 B 6:d4 R 7:f3"
/>
At this point, Blue is forced to play 8, and then a new ladder starts in the opposite direction:
<hexboard size="4x7"
coords="show"
edges="bottom"
contents="R b1 a2 a3 b3 B b2 a4 R g1 B b4 R c3 B c4 R d3 B d4 R f3 B 8:e3 R 9:f1 B 10:e2 R 11:e1 B 12:d2 R 13:d1"
/>
In this example, the ladder reconnects to Red's original group, although in general this does not need to be the case (even if the switchback doesn't connect, Red has just created a parallel edge 4 cells from the original edge - a large advantage for Red in any case).

To formalize the concept of a 2nd-to-4th row switchback, consider a 2nd row ladder.

L2: <hexboard size="2x2"
coords="hide"
edges="bottom"
contents="R a1 E -:(b1--b2)"
/>

This is the same ladder as defined in the section of second-row ladders above; only this time, Red's goal will be slightly different. To explain Red's goal, we show a slightly larger area around L2:
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2 c2 c3 c4"
contents="R a3 E *:a1 *:a2 *:b1 *:b3 *:b4 *:c2 *:c3 *:c4 a:c1 b:b2 b:b1 E -:(b3--b4)"
/>
This time, Red's goal will be to do at least one of the following two things: either connect the red ladder stone to the edge, or else, occupy the cell marked "a" with a red stone that is connected to the edge, without using the cells marked "b" or any cells to their left. We refer to this as the ''switchback condition''. We also call "a" the ''switchback cell'' and "b" the ''gap cells''. With this in mind, we now give the definition of a 2nd-to-4th row switchback template.

'''Definition.''' A ''2nd-to-4th row switchback template'' (or simply 2-to-4 switchback) is given by the following data. It is a pattern P with a left boundary of shape
<hexboard size="4x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a4)"
/>
and satisfying the following axiom: L2+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0.

As usual, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid switchback template.

As in previous sections, we write L2+↑+↑+''n''+P for the pattern obtained from P by moving the four hexes marked "+" to the left by ''n'' columns (leaving 4 rows of empty space), then removing the top two cells marked "+" (they are not part of the pattern) and replacing the remaining cells marked "+" by L2. Note that the switchback cell "a" and gap cells "b" are not part of L2. They are simply three cells on the board whose position is defined relative to L2. Depending on the value of ''n'', they may or may not end up being inside the pattern P.

=== Characterization of switchbacks ===

The following theorem gives a finite necessary and sufficient condition for a pattern to be a switchback.

'''Theorem 12.''' Given a candidate P for a 2-to-4 switchback. Then P is a valid 2-to-4 switchback if and only if L2+↑+↑+P satisfies the switchback condition.

'''Proof.''' The left-to-right implication is trivial. For the opposite implication, we will show, by induction on ''n'', that L2+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0. The base case ''n'' = 0 holds by assumption. Now suppose that the claim is true for some ''n''. To show the claim for ''n''+1, consider the position L2+↑+↑+''n''+1+P. It looks as follows, with ''n'' more empty columns and P appended on the right:
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:a3"
/>
Here, the ladder stone is marked "1". Blue has no choice but to push the ladder, and Red also pushes:
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R 1:a3 B 2:a4 R 3:b3"
/>
At this point, the induction hypothesis guarantees that Red can either connect 3 to the edge, or else that Red can occupy and connect the switchback cell "a" while keeping "b" empty:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4 d2 d3 d4"
contents="R 1:a3 B 2:a4 R 3:b3 E a:d1 b:c2 b:c1 b:b2 b:b1"
/>
If 3 is connected to the edge, then so is 1, and we are done. Otherwise, "a" is connected to the edge and "b" is empty. Thus, the board looks like this, with "a" now acting as a ladder stone:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4 d2 d3 d4"
contents="E *:a1 *:a2 R a3 B a4 R b3 R 1:d1"
/>
Since Red's stones on the 2nd row are already connected to the top, and 1 is connected to the bottom, Blue has no choice but to respond at 2. Then Red can play 3.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4 d2 d3 d4"
contents="E *:a1 *:a2 R a3 B a4 R b3 R 1:d1 B 2:c2 R 3:c1"
/>
Now the switchback condition for L2+↑+↑+''n''+1+P is satisfied, proving the theorem. □

=== Examples ===

Any 2nd row ladder escape template trivially also works as a switchback template (with the location of the cells marked "+" adjusted as necessary; they may need to be moved to the left if there isn't space for the two additional "+"s in the pattern). Since such a template escapes 2nd row ladders outright, there is no need for the second part of the switchback condition.

The following are examples of 2nd-to-4th row switchback templates that are not second row ladder escapes. They are minimal.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d3 d4"
contents="E +:a1 +:a2 +:a3 +:a4 R d2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d3 d4"
contents="E +:a1 +:a2 +:a3 +:a4 R d1"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--c1"
contents="E +:a2 +:a3 +:a4 +:a5 R e1"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--c1"
contents="E +:a2 +:a3 +:a4 +:a5 R d1"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--c1"
contents="E +:a2 +:a3 +:a4 +:a5 R e1 R d1 S d2"
/>
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="area(a2,a5,g5,g3,f1,d1)"
contents="E +:a2--a5 R f2 S d5"
/>
In the last two templates, the shaded hex is not part of the template, and can be occupied by Blue.
The following template is useful for obtuse corners:
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1--c1 area(d5,f5,f3)"
contents="E +:a2 +:a3 +:a4 +:a5 R e1 f1"
/>
In the following template, the stone marked "↓" is assumed to be connected to the bottom edge, but the connection is not shown. The blue stone is not technically part of the pattern; however, if this cell were empty, the pattern would already work as a 2nd row ladder escape.
<hexboard size="4x7"
coords="hide"
edges="bottom"
visible="-g3 g4 f4"
contents="E +:a1 +:a2 +:a3 +:a4 B b4 R ↓:g2"
/>

== Third-to-fifth row switchbacks ==

=== Definition of switchback ===

The definition of 3rd-to-5th row switchbacks is similar to that of 2nd-to-4th row switchbacks.
Consider a 3rd row ladder.
L3: <hexboard size="3x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="R b1 E -:(c1--c3)"
/>
We define the locations of the switchback cell "a" and gap cells "b" relative to L3:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 d2--d5"
contents="R b3 E b:c1 b:c2 a:d1 -:(c3--c5)"
/>
Again, the ''switchback condition'' states that with Blue to move, Red can either connect the ladder stone to the edge, or else Red can occupy the switchback cell and connect it to the edge, without using the gap cells or anything to their left.

'''Definition.''' A ''3nd-to-5th row switchback template'' (or simply 3-to-5 switchback) is given by the following data. It is a pattern P, open on the left with boundary
<hexboard size="5x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a5)"
/>
and subject to the requirement that L3+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0.

=== Characterization of switchbacks ===

The following theorem gives a finite necessary and sufficient condition for a pattern to be a switchback. It is analogous to the corresponding theorem for 3rd row ladders.

'''Theorem 13.''' Given a candidate P for a 3-to-5 switchback. Then P is a valid 3-to-5 switchback if and only if L3+↑+↑+P and L3+↑+↑+1+P satisfy the switchback condition.

'''Proof.''' The left-to-right implication is trivial. For the opposite implication, we will show, by induction on ''n'', that L3+↑+↑+''n''+P satisfies the switchback condition for all ''n'' ≥ 0. The base cases ''n'' = 0 and ''n'' = 1 hold by assumption. Now suppose that the claim is true for ''n'' and ''n''+1. To show the claim for ''n''+2, consider the position L3+↑+↑+''n''+2+P. It looks as follows, with ''n'' more empty columns and P appended on the right:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R 1:b3"
/>

The ladder stone is marked "1". As usual for 3rd row ladders, Blue must either push or yield, or else Red will connect to the edge outright. If Blue pushes, then so does Red:
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R 1:b3 B 2:b4 R 3:c3"
/>
By induction hypothesis, L3+↑+↑+''n''+1+P satisfies the switchback condition, so either Red can connect to the edge, or else Red can get a new ladder stone 1 in the switchback cell for L3+↑+↑+''n''+1+P, which allows Red to satisfy the switchback condition for L3+↑+↑+''n''+2+P with stone 3.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R b3 B b4 R c3 R 1:e1 B 2:d2 R 3:d1 E *:(c4--c5 d3--d5 e2--e5)"
/>
The other option is for Blue to yield. (We will see later that when ''n'' is large enough, yielding in this situation is actually a terrible idea for Blue, since it will allow Red to use P to connect to the edge. But this is not relevant for the current proof).
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="E *:a3 R 1:b3 E *:a4 *:a1 *:a2 *:b1 *:b2 B 2:b5 R 3:b4 B 4:a5 R 5:d3"
/>
Now by the induction hypothesis, L3+↑+↑+''n''+P satisfies the switchback condition, so either Red can connect to the edge, or else Red can get a new ladder stone 1 in the switchback cell for L3+↑+↑+''n''+P. This allows Red to satisfy the switchback condition for L3+↑+↑+''n''+2+P with stone 5.
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R b3 B b5 R b4 B a5 R d3 R 1:f1 B 2:e2 R 3:e1 B 4:d2 R 5:d1 E *:(d4--d5 e3--e5 f2--f5)"
/>
This finishes the proof of the theorem. □

One may ask whether every 3-to-5 switchback template also works as a 2-to-4 switchback template. This is indeed the case at sufficient distance, due to the following jumping lemma.

'''Lemma 14 (2nd to 3rd row switchback jump).''' Any 3-to-5 switchback template is also a 2-to-4 switchback template at distance 4 or greater. More specifically, if P is a 3-to-5 switchback template, then ↑+4+P is a 2-to-4 switchback template.

'''Proof.''' Suppose P is a 3-to-5 switchback template, and consider Q = ↑+4+P. By Theorem 12, we must show that L2+↑+↑+Q satisfies the switchback condition. The position L2+↑+↑+Q looks like this, with P attached on the right:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4"
/>
After Blue pushes the ladder at 2, Red plays 3, which is essentially [[Tom's move]]. While this move is not sufficient to connect Red to the edge, it creates enough trouble to allow Red to get the desired switchback in the presence of P.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3"
/>
Let us consider Blue's options. If Blue moves outside the area marked "x", Red simply pushes the ladder and connects, using 3 as a ladder escape.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 E x:b4 x:b5 x:c4 x:c5 x:d4 x:d5 x:e3 x:e4 x:e5"
/>
If Blue moves in any of the cells marked 4, Red gets the switchback without using P:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:d4 B 4:d5 B 4:e3 B 4:e4 B 4:e5 R 5:c4 B 6:b4 R 7:c2"
/>
If Blue moves at 4, Red also gets the switchback without using P:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:b4 R 5:c2"
/>
If Blue moves at 4, Red also gets the switchback without using P. Note that 3 is connected by [[edge template III2b]].
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:b5 R 5:c4 B 6:b4 R 7:c2"
/>
If Blue moves at 4, Red also gets the switchback without using P. Note that 3 is connected to the edge by double threat.
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="R 1:a4 B 2:a5 R 3:d3 B 4:c5 R 5:b5 B 6:b4 R 7:c2"
/>
This leaves only one option for Blue. If Blue moves at 4, then Red responds as follows:
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="E *:a1 *:a2 *:a3 R 1:a4 B 2:a5 R 3:d3 B 4:c4 R 5:b4 B 6:b5 R 7:d2"
/>
By hypothesis, since P is a 3-to-5 switchback template, Red can either connect 3 to the edge, or else get a connected red stone at "a", with "b" empty:
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1--a3"
contents="E *:a1 *:a2 *:a3 R 1:a4 B 2:a5 R 3:d3 B 4:c4 R 5:b4 B 6:b5 R 7:d2 E a:f1 b:e1 b:e2"
/>
In either case, 7 is connected to the edge, so Red has the desired switchback. □

'''Corollary 15.''' In a 3rd row ladder at distance 5 or greater to a 3-to-5 switchback, Blue cannot yield.

'''Proof.''' If Blue yields, then Red can switch back the resulting 2nd row ladder to the 4th row by the previous lemma. This will reconnect to Red's original 3rd row ladder, and therefore connect Red to the edge. In a diagram:
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2"
contents="R 1:b3 B 2:b5 R 3:b4 B 4:a5 R 5:e3 B 6:d4 R 7:c4 B 8:c5 R 9:e2 E a:g1 b:f1 b:f2"
/>
Note that Blue must play 6 for the same reason as in the lemma. Since Red will either connect 5 or "a" to the edge, 7 is also connected. Rather than just giving Red a switchback, 7 is actually connected to 1 by a [[Interior template#The crescent|crescent]]. □

Here is another interesting fact about 3-to-5 switchbacks. Given enough space, the defender of a 3rd row ladder cannot yield without giving the attacker a switchback.

'''Theorem 16.''' Given enough space to the right of a 3rd row ladder and two empty rows above it, if the defender tries to yield, the attacker can achieve a 3-to-5 switchback without requiring any addtional stones.

'''Proof.''' Let 1 be the ladder stone of a 3rd row ladder, and assume there is at least as much space as indicated in the following diagram.
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R 1:b3"
/>
If Blue yields at 2, then Red can play as follows:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R 1:b3 B 2:b5 R 3:b4 B 4:a5 R 5:e2
E x:c2 x:c3 x:c4 x:c5 x:d2 x:d3 x:d4 x:d5 x:e3"
/>
If Blue's next move is outside of the area marked "x", then Red connects to the edge outright, as follows:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
R 7:d4 B 8:c4 R 9:d2"
/>
Therefore, Blue must move in one of the hexes marked "x" above. This leaves nine possible moves for Blue.

If Blue moves at one of the hexes marked 6 below, then Red connects by [[edge template IV2b]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c2 6:d2 R 7:d3"
/>
If Blue moves at 6, then Red pushes the second row ladder twice and connects by [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c3 R 7:c4 B 8:c5 R 9:d4 B 10:d5 R 11:g3"
/>
If Blue moves at 6, then Red gets 2nd and 4th row parallel ladders, which connect by [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:d3 R 7:d2 B 8:e3 R 9:c4 B 10:c5 R 11:d4 B 12:d5 R 13:g3"
/>
If Blue moves at 6, then Red connects by a [[Interior template#The crescent|crescent]] and [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:e3 R 7:c4 B 8:c5 R 9:d4 B 10:d5 R 11:g3"
/>
If Blue moves at 6, then Red connects by [[Interior template#The crescent|crescent]] and [[ziggurat]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:d4 R 7:c4 B 8:c5 R 9:f3"
/>
If Blue moves at 6, then Red connects by [[Interior template#The shopping cart|shopping cart]] and [[Tom's move]]:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c5 R 7:d4 B 8:d5 R 9:g3"
/>
If Blue moves at 6, the situation is almost identical:
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:d5 R 7:d4 B 8:c5 R 9:g3"
/>
Note that in all cases so far, Red connected outright, i.e., didn't need a switchback. The final remaining possibility is for Blue to move at 6 in the following diagram. Then Red gets the switchback. Note that 7 is connected to the edge by [[edge template IV2b]].
<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="-a1--a4 b1 b2 i1 j1 j2"
contents="E *:a1 *:a2 *:a3 *:a4 *:b1 *:b2 *:i1 *:j1 *:j2 R b3 B b5 R b4 B a5 R e2
B 6:c4 R 7:d3 B 8:c3 R 9:d1"
/>
This finishes the proof of the theorem. □

=== Examples ===

Every 3rd row ladder escape template is also a 3-to-5 switchback template (possibly with the location of the column of "+"s adjusted), but it need not be minimal. Here are some examples of 3-to-5 switchback templates that are not 3rd row ladder escapes.

<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-e1"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 *:e1 R e3"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 R d2"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 R d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
contents="E +:a1 +:a2 +:a3 +:a4 +:a5 R d1 R c1 S c2"
/>
The shaded cell is not part of the template and can be occupied by Blue.

== Second and fourth row parallel ladders ==

=== Definition of ladder ===

[[Parallel ladder]]s, especially on the 2nd and 4th rows, are quite common in Hex. For example, consider this situation, with Blue to move and the Red stone connected to the top:
<hexboard size="4x7"
coords="hide"
edges="bottom"
contents="B b2 R d1"
/>
Play may proceed as follows:
<hexboard size="4x7"
coords="hide"
edges="bottom"
contents="B b2 R d1 B 1:d2 R 2:e1 B 3:e2 R 4:c2 B 5:b4 R 6:c3 B 7:c4"
/>
Now Red has a choice: she can either continue pushing the 4th row ladder from 2, or the 2nd row ladder from 6. However, having parallel ladders puts Red in a stronger position than having a 2nd row ladder or a 4th row ladder alone. As we will see, there exist ladder escape templates than can escape a parallel ladder, but can neither escape a 2nd row ladder nor a 4th row ladder on its own.

'''Note.''' Unlike with single-row ladders, in the case of a parallel ladder, Red actually has a choice whether to push the 2nd row ladder or the 4th row ladder. For this reason, our formal definition of a parallel ladder follows a slightly different approach than that we took for single-row ladders above. Whereas above, we always assumed that ''Blue'' was next to move (and the ladder stone was already in a pushing position), here, we will assume that ''Red'' is next to move. This affects the definition of the ladder pattern, in that the ladder stones do not yet have empty space below them.

'''Definition.''' A ''parallel ladder on the 2nd and 4th rows'', or ''2-4 parallel ladder'' for short, is a pattern like this:
L24: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 E -:(d1--d4)"
/>
The red stones on the second and fourth rows are called the ladder stones, and Red's goal is to connect at least one of these stones to the bottom edge, with ''Red'' to move first. (We can assume that both ladder stones are already connected to the top). We denote this pattern by L24. There is also a variant of L24 that looks like this:
L24a: <hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 a3"
contents="R b2 R c1 B a4 c2 E -:(d1--d4)"
/>
L24 and L24a are equivalent, and for simplicity we will only use L24.

=== Definition of ladder escape ===

'''Definition.''' An ''escape template for parallel ladders on the 2nd and 4th rows'', or ''2-4 parallel ladder escape'' for short, is a pattern P with a left boundary of shape
<hexboard size="4x1"
coords="hide"
edges="bottom"
contents="+:(a1--a4)"
/>
satisfying the following axiom: adding a 2-4 parallel ladder at distance ''n'', for any ''n'' ≥ 0, allows Red to connect, in the sense that it guarantees the connection of at least one of the red ladder stones to the bottom edge, with Red to move.

As before, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

=== Characterization of ladder escapes ===

Fortunately, 2-4 parallel ladders are easy to analyze; they are almost as simple as 2nd row ladders. The reason is that, just as for 2nd row ladders, the defender has no choice; he must always push, because as we will see, yielding is not an option. We get a simple and clean theorem with a necessary and sufficient condition for 2-4 parallel ladder escapes.

'''Theorem 17.''' Consider a candidate P for a 2-4 parallel ladder escape. Then P is a valid 2-4 parallel ladder escape if and only if L24+P, L24+1+P, and L24+2+P allow Red to connect (with Red to move).

'''Proof.''' If the ladder escape is valid, then by definition, L24+''n''+P allows Red to connect for all ''n'', including ''n'' = 0, 1, 2. So the left-to-right implication is trivial. To prove the right-to-left implication, assume L24+P, L24+1+P, and L24+2+P allow Red to connect. We prove by induction that L24+''n''+P allows Red to connect for all ''n'' ≥ 0. The cases ''n'' = 0, 1, 2 are true by assumption. Now suppose the claim is true for some ''n'' ≥ 2. We must show the claim for ''n''+1. To do so, consider the position L24+''n''+1+P. The first six columns of this position look like this:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2"
/>
This is followed by ''n''−2 more columns of four empty hexes and by the pattern P. Red starts by pushing the 4th row ladder.
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1"
/>
Now consider Blue's possible moves. If Blue moves in any of the hexes marked 2 below (or elsewhere on the board), Red wins outright (i.e., without using the induction hypothesis).
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:e1 2:e2 2:e3 2:e4 2:f1 2:f2 2:f3 2:f4 2:d3 2:d4 R 3:c3"
/>
If Blue moves in any of the hexes marked 2 below, Red also wins outright:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:b3 2:b4 2:c3 R 3:e2"
/>
If Blue moves at 2 below, Red also wins outright:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:c4 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:e3"
/>
This means that the only possible move that is not immediately losing for Blue is to push the 4th row ladder. In this case, Red can respond as follows:
<hexboard size="4x6"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 R 1:d1 B a4 c2 B 2:d2 R 3:b3 B 4:b4"
/>
This position allows Red to connect by induction hypothesis, finishing the proof. □

It is clear that every 2nd row ladder escape and every 4th row ladder escape is also an escape for 2nd-and-4th row parallel ladders, since Red can decide to push only the 2nd row ladder, or only the 4th row ladder. In addition, 2nd-to-4th row switchback templates also work as 2-4 parallel ladder escapes. This is intuitively clear, as Red can simply push the 2nd row ladder and switch it back to the 4th row, where it will connect with the 4th row of the parallel ladder. The following theorem proves this more formally, using the definitions.

'''Theorem 18.''' Every 2nd-to-4th row switchback template is also a 2-4 parallel ladder escape.

'''Proof.''' Assume P is a 2nd-to-4th row switchback template. To show that P is a 2-4 parallel ladder escape, we must show that L24+''n''+P allows Red to connect with Red to move, for all ''n'' ≥ 0. Consider the position L24+''n''+P, which looks as follows, with an additional ''n'' blank columns and P on the right:
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2"
/>
Red plays as follows. At this point, since ''n''+P is a 2nd-to-4th row switchback template, Red can either connect 3 to the edge, or get a connected stone at "a" with "b" empty.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:b3 B 2:b4 R 3:c3 E a:e1 E b:d1 b:d2"
/>
This allows Red to connect at least one of the ladder stones, as required. □

=== Examples ===

As mentioned above, every 2nd row ladder escape, every 4th row ladder escape, and every 2nd-to-4th row switchback template works as a 2-4 parallel ladder escape. But there are some examples of 2-4 parallel ladder escapes that are none of the above. The most famous of these is [[Tom's move]], which states that a sufficient amount of empty space is enough for a 2-4 parallel ladder to connect to the edge. Specifically, the following is a 2-4 parallel ladder escape template:
<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E +:a2 +:a3 +:a4 +:a5"
/>
Other examples:
<hexboard size="5x5"
visible="-a1 a2 b1 e4 e5 a3--a5 e3"
edges="bottom"
coords="none"
contents="E +:b2--b5 R d1 e1 B d2"
/>
Here are some other examples of 2-4 parallel ladder escapes that are neither 2nd nor 4th row ladder escapes nor 2nd-to-4th row switchbacks. They can be shown to be valid by Theorem 17, and are minimal. Unlike Tom's move, these ladder escapes don't require space on the 5th row.

While the following two patterns aren't switchbacks at distance 0 or 1, they do work as 2nd-to-4th row switchbacks at distance 2 or greater.
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-b3 b4"
contents="E +:a1 +:a2 +:a3 +:a4 R b2"
/>
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-b3 b4"
contents="E +:a1 +:a2 +:a3 +:a4 R b1"
/>

== Third and fifth row parallel ladders ==

=== Definition of ladder ===

Parallel ladders on the 3rd and 5th rows are less common than those on the 2nd and 4th rows, but they can occur. Pushing such ladders is less straightforward, as the defender has more options. Basically, as we will show, if the defender refuses to push, then the attacker can at least get a 2nd row ladder. Moreover, a 2nd-to-4th row switchback template is in that case sufficient for the attacker to connect.

'''Definition.''' A ''parallel ladder on the 3nd and 5th rows'', or ''3-5 parallel ladder'' for short, is a pattern like this:
L35: <hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 E -:(d1--d5)"
/>
The red stones on the third and fifth rows are again called the ladder stones, and Red's goal is to connect at least one of these stones to the bottom edge, with ''Red'' to move first. We denote this pattern by L35. Just like for 2-4 parallel ladders, there is an equivalent pattern for L35 that looks like this:
L35a: <hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 a3"
contents="R b2 R c1 B a4 c2 E -:(d1--d5)"
/>

=== Definition of ladder escape ===

'''Definition.''' An ''escape template for parallel ladders on the 3rd and 5th rows'', or ''3-5 parallel ladder escape'' for short, is a pattern P with a left boundary of shape
<hexboard size="5x1"
coords="hide"
edges="bottom"
contents="E +:(a1--a5)"
/>
satisfying the following axiom: adding a 3-5 parallel ladder at distance ''n'', for any ''n'' ≥ 0, allows Red to connect, in the sense that it guarantees the connection of at least one of the red ladder stones to the bottom edge, with Red to move.

As always, a ''candidate'' is a pattern that has the correct shape, but is not (yet) known to be a valid escape.

=== Characterization of ladder escapes ===

3-5 parallel ladder escapes are not quite as easy to characterize as those for 2-4 parallel ladders, because the defender has more options. We get the following theorem, which only contains a sufficient condition for a pattern to be a 3-5 parallel ladder escape.

'''Theorem 19.''' Consider a candidate P for a 3-5 parallel ladder escape. If L35+P, L35+1+P, ..., L35+3+P allow Red to connect (with Red to move), and if ↑+P is a 2nd-to-4th row switchback template, then P is a valid 3-5 parallel ladder escape.

'''Proof.''' We prove by induction that L35+''n''+P allows Red to connect for all ''n'' ≥ 0. The cases ''n'' = 0, ..., 3 are true by assumption. Now suppose the claim is true for 0, ..., ''n'', where ''n'' ≥ 3. We must show the claim for ''n''+1. To do so, consider the position L35+(''n''+1)+P. The position looks like this, followed by ''n''−3 additional empty columns and P:
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2"
/>
Red starts by pushing the 5th row ladder.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 E x:a5 x:b4 x:b5 x:c3 x:c4 x:c5 x:d2 x:d3 x:d4 x:d5 x:e1 x:e2 x:e3 x:e4 x:e5 x:f2 x:f3 x:f4 x:f5 x:g3 x:g4 x:g5"
/>
Now consider Blue's possible moves. If Blue moves anywhere except the hexes marked "x", then Red wins outright by bridging from 1 to [[edge template IV1a|edge template IV-1a]].

If Blue moves in any of the hexes marked 2 below, Red moves at 3 and connects by [[ziggurat]] and [[double threat]].
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:e1 2:e2 2:e3 2:e4 2:e5 2:f2 2:f3 2:f4 2:f5 2:g3 2:g4 2:g5 R 3:c3"
/>
This leaves 10 more moves to consider.

If Blue moves at 2, Red pushes the 3rd row ladder.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d2 R 3:b3 E x:b4 y:b5"
/>
Now Blue must either push at "x" or yield at "y" (or else Red will connect immediately). If Blue pushes at "x", then Red has a 3-5 parallel ladder at distance ''n'', which connects by induction hypothesis.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d2 R 3:b3 B 4:b4"
/>
If Blue instead yields at "y", then Red can push the 2nd row ladder and use the switchback to either connect 7 to the edge or get a connected stone at "a". Note that "a" is connected to either 1 or 7 by double threat, so Red connects. (As a matter of fact, Red can do better in this case and get a 2-4 parallel ladder, but it is not needed for the current proof).
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d2 R 3:b3 B 4:b5 R 5:c4 B 6:c5 R 7:d4 E a:f2"
/>
If Blue moves at 2, then Red plays as follows and connects by [[edge template III2e]]:
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:c3 R 3:b4 B 4:b3 R 5:e2 B 6:e3 R 7:d3"
/>
If Blue moves at 2, then Red gets a 2nd row ladder that connects via the 2-to-4 switchback at "a". The moves 4 and 5 can also be played in the opposite order without changing the result.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d3 R 3:c3 B 4:d2 R 5:b3 B 6:b5 R 7:c4 B 8:c5 R 9:d4 E a:f2"
/>
If Blue moves at 2, then Red gets a 2nd row ladder that connects via the 2-to-4 switchback at "a". (In fact, Red can get a 2-4 parallel ladder, but it is not needed in this proof).
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b4 R 3:e2 B 4:e3 R 5:d3 B 6:c5 R 7:d4 B 8:d5 R 9:e4 E a:g2"
/>
If Blue moves at 2, Red connects.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:c4 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:e3"
/>
If Blue moves at 2, Red connects
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d4 R 3:b4 B 4:b3 R 5:e2 B 6:d3 R 7:f3"
/>
If Blue moves at 2, Red connects by [[edge template IV2d]].
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:a5 R 3:c4 B 4:c3 R 5:e2"
/>
If Blue moves at 2, Red can respond at 3.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b5 R 3:e2 E x:e3 y:d5"
/>
Now Blue must respond at "x" or "y", or else Red will connect immediately. If Blue plays at "x", Red gets a 2nd row ladder which connects via switchback at "a". Note that 5 is connected to at least one ladder stone by double threat.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b5 R 3:e2 B 4:e3 R 5:c4 B 6:c5 R 7:d4 B 8:d5 R 9:e4 E a:g2"
/>
If Blue instead plays at "y", Red also gets a 2nd row ladder which connects via switchback at "a".
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:b5 R 3:e2 B 4:d5 R 5:d4 B 6:c5 R 7:e4 E a:g2"
/>
If Blue moves at 2, Red connects.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:c5 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:d3 B 8:c4 R 9:e4"
/>
Finally, if Blue moves at 2, Red also connects.
<hexboard size="5x7"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 b2"
contents="R a3 R c1 B a4 c2 R 1:d1 B 2:d5 R 3:b4 B 4:b3 R 5:d2 B 6:c3 R 7:e3 B 8:f4 R 9:d4"
/>
This finishes the proof. □

=== Examples ===

Like 2-4 parallel ladders, 3-5 parallel ladders have the property that they can connect to the edge outright if given enough space. There is an analog of [[Tom's move]] for 3-5 parallel ladders. The following diagram shows the amount of space required. If Red moves in the cell marked "x", Red can guarantee to connect at least one of the ladder stones marked "1" to the edge. The cell marked "x" is essentially the unique winning move (the only other winning option for Red is to push the 3rd row ladder one more hex before playing "x").
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 b1 b2 b3 c1 g1 h1 i1 j1 k1 k2 l1 l2 l3"
contents="R 1:a4 1:c2 E x:e3 B a5 c3"
/>
We note that this particular pattern is not technically a 3-5 parallel ladder escape. Without additional empty space on the 6th row, it only escapes 3-5 parallel ladders at distance 0 (as shown) and at distance 1. If the ladder starts further away, Blue has the option of yielding to a 2nd row ladder for which Red would need a 2-to-4 switchback template to connect.

Every 3rd row ladder escape, every 5th row ladder escape, and every 3-to-5 switchback template is also a 3-5 parallel ladder escape. Examples of 3-5 parallel ladder escapes that aren't one of the above are relatively rare, but they do exist. The following are some examples. They have been proved correct using Theorem 19, and they are minimal.

The following template escapes 3-5 parallel ladders, but it is not a 3rd-to-5th row switchback, nor does it escape 3rd, 4th, or 5th row ladders.
<hexboard size="6x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E +:(a2--a6) R f6"
/>

The following template escapes 3-5 parallel ladders, but it is not a 3rd-to-5th row switchback, nor does it escape 2nd, 3rd, 4th, or 5th row ladders.
<hexboard size="6x8"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2 g1 g2 h1 h2 h5 h6"
contents="E +:(a2--a6) R h4"
/>

== Second and fourth row terraced ladders ==

Sometimes it can happen that a ladder forms on top of another ladder, with the two rows of attacking stones not yet connected to the edge nor to each other. We call this a ''terraced ladder''. The following is an example of a terraced ladder on the 2nd and 4th rows, with Blue to move:
<hexboard size="5x10"
coords="hide"
edges="bottom"
contents="R a2 B a3 R b2 B b3 R c2 B c3 R d2 B a4 B a5 R b4 B b5 R c4 B c5 R d4 B d5 R e4 B e5 R f4"
/>
Although terraced ladders look superficially similar to parallel ladders, they should not be confused. There are two important differences: (1) in a parallel ladder, the two rows of attacking stones are connected to each other, whereas in a terraced ladder, they are not, and (2) in a parallel ladder, the upper ladder is "ahead" of the lower one, whereas in a terraced ladders, the upper ladder is at the same level or behind the lower ladder.

In fact, as we noted above, from the attacker's point of view, having 2nd and 4th row parallel ladders is ''stronger'' than having only a 2nd row ladder or only a 4th row ladder. For terraced ladders, the opposite is true: a 2nd and 4th row terraced ladder is ''weaker'' than having only a 2nd row ladder or only a 4th row ladder. Nevertheless, despite being relatively weak, terraced ladders can be pushed, and there is a notion of terraced ladder escape at arbitrary distance.

Before we develop the theory of terraced ladders, it is worth noting that terraced ladders from Red's point of view are parallel ladders from Blue's point of view, and vice versa. This can be seen by putting a row of blue stones on top, giving Blue an "edge":
<hexboard size="5x10"
coords="hide"
edges="bottom"
contents="R a2 B a3 R b2 B b3 R c2 B c3 R d2 B a4 B a5 R b4 B b5 R c4 B c5 R d4 B d5 R e4 B e5 R f4 B a1 b1 c1 d1 e1 f1 g1 h1 i1 j1"
/>
Indeed, from Red's point of view, Red has terraced ladders trying to connect to the bottom edge, whereas from Blue's point of view, Blue has parallel ladders trying to connect to the top edge. The fact that parallel ladders are better for Blue than individual ladders explains why terraced ladders are worse for Red than individual ladders.

=== Definition of ladder ===

In a terraced ladder, it is the defender, not the attacker, who decides whether to push the 2nd or 4th row ladder. Since the 4th row ladder can lag behind the 2nd row ladder by an arbitrary distance, there isn't just a single ladder template, but a family of them.

'''Definition.''' A 2nd and 4th row terraced ladder is any one of the following patterns:

T(0):
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2 c3 c4"
contents="R a3 b1 E -:(c1 c2 b3 b4)"
/>
T(1):
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-c3 c4"
contents="R a1 a3 E -:(c1 c2 b3 b4)"
/>
T(2):
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a4 d3 d4"
contents="R a1 a3 R b3 E -:(d1 d2 c3 c4)"
/>
T(3):
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a4 b4 e3 e4"
contents="R a1 a3 R b3 R c3 E -:(e1 e2 d3 d4)"
/>
and so on. In general, for ''k'' ≥ 1, the pattern T(''k'') looks like
<hexboard size="4x1"
float="inline"
coords="hide"
edges="bottom"
visible="-a4"
contents="R a1 a3"
/> followed by ''k''−1 columns of <hexboard size="4x1"
float="inline"
coords="hide"
edges="bottom"
visible="-a4"
contents="R a3"
/> followed by <hexboard size="4x3"
float="inline"
coords="hide"
edges="bottom"
visible="-a1 a2 a3 c3 c4"
contents="R a3 E -:(c1 c2 b3 b4)"
/>

In these patterns, we refer to Red's stone on the 4th row as the ''top ladder stone'', and to Red's rightmost stone on the 2nd row as the ''bottom ladder stone''. Red's goal is to connect the top ladder stone to the bottom edge, assuming it is Blue's turn first. We can assume that the top ladder stone is already connected to the top edge, but we do not assume that the top and bottom ladder stones are connected to each other.

=== Definition of ladder escape ===

'''Definition.''' A ladder escape template for 2nd and 4th row terraced ladders, or 2-4 terraced ladder escape for short, is a pattern P with left boundary shaped like this,
<hexboard size="4x2"
coords="hide"
edges="bottom"
visible="-a1 a2 b3 b4"
contents="E +:(a3 a4 b1 b2)"
/>
This pattern P must satisfy the following axiom: for all ''k'' ≥ 0 and all ''n'' ≥ 0, T(''k'')+''n''+P guarantees a connection of the top ladder stone to the bottom edge, with Blue to move.

As always, a candidate is a pattern that has the correct shape, but is not (yet) known to be a valid escape. If P is such a candidate, schematically of the form
<hexboard size="4x3"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="E +:(a3 a4 b1 b2) *:b3 *:b4 *:c1 *:c2 *:c3 *:c4"
/>
then we write ∗+P to denote the pattern obtained from P by replacing the top two cells marked "+" by empty cells, and adding two new cells marked "+" just to their left. The resulting template is then of the shape required for 4th row ladder escapes.
<hexboard size="4x3"
coords="hide"
edges="bottom"
contents="E +:(a1--a4) *:b3 *:b4 *:c1 *:c2 *:c3 *:c4"
/>

=== Characterization of ladder escapes ===

We will reduce the problem of establishing a terraced ladder escape to finitely many cases. This is done by two lemmas. Lemma 20 states that we only need to consider finitely many values of ''n'' (the distance from the bottom ladder stone to the escape). Lemma 21 states that we only need to consider finitely many values of ''k'' (the distance from the top ladder stone to the bottom ladder stone). .

'''Lemma 20.''' Suppose P is a candidate for a 2-4 terraced ladder escape. Assume T(''k'')+P, T(''k'')+1+P, and T(''k'')+2+P are virtual connections for all ''k'' ≥ 0 (i.e., they guarantee a connection of the top ladder stone to the edge, with Blue to move). Moreover, assume that ∗+P escapes 2nd, 3rd, and 4th row ladders (or more precisely, ∗+P escapes 4th row ladders, ↑+∗+P (which has 3 cells marked "+") escapes 3rd row ladders, and ↑+↑+∗+P escapes 2nd row ladders). Then P is a valid 2-4 terraced ladder escape.

'''Proof.''' We need to show that T(''k'')+''n''+P is a virtual connection for the top ladder stone, for all ''k'',''n'' ≥ 0. We prove this by nested induction, with the outer induction being on ''n'', and the inner induction on ''k''. The base cases ''n'' = 0, 1, 2 are true by assumption. Now consider some ''n'' ≥ 3, and suppose the claim is true up to ''n''−1. We need to show the claim for ''n''. Consider the position T(''k'')+''n''+P, which looks like this:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 E x:b1 x:c1 x:d1 x:e1 x:a2 x:b2 x:c2 x:d2 x:e2 x:f2 x:e3 x:f3 x:d4 x:e4 x:f4"
/>
followed by ''n''−3 additional empty columns and P. Here, our diagram illustrates the case ''k'' = 4, but the following arguments are valid for all ''k'' ≥ 0. The first observation is that if Blue's next move is outside of the area marked "x", Red connects to the edge immediately by a [[Interior_template#The_long_crescent|long crescent]] and [[edge template III2b]]:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 R e2 "
/>
Note that this works for all ''k'' ≥ 0, although for ''k'' = 0 and ''k'' = 1, the connection is simpler and does not require a long crescent. Therefore, Blue must move in the area marked "x".

We consider each of Blue's options in turn. If Blue moves just below the top ladder stone, then Red responds by pushing the 4th row ladder. In case ''k'' > 0, this leads to the position T(''k''−1)+''n''+P, and he claim holds by the inner induction hypothesis:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:a2 R 2:b1"
/>
In case ''k'' = 0, the situation is worse for Blue: in this case, Red gets a bona fide 4th row ladder, which ∗+P escapes by assumption:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 a2 e3 e4"
contents="R a3 b1 B 1:b2 R 2:c1"
/>
If Blue moves anywhere else on the 3rd or 4th row, then Red connects the two ladders and gets a second row ladder, which ↑+↑+∗+P escapes by assumption. This works for all ''k'' ≥ 0, although for illustration, we show only the case ''k'' = 4:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:b1 1:b2 1:c1 1:c2 1:d1 1:d2 1:e1 1:e2 1:f2 R 2:a2"
/>
This leaves only 5 possible Blue moves to consider.

If Blue moves at 1, Red gets a 3rd row ladder, which ↑+∗+P escapes. Note that 2 is connected to the top ladder stone by a [[Interior_template#The_long_crescent|long crescent]] (for ''k'' ≥ 2) or directly (for k = 0, 1).
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:e3 R 2:e2 B 3:d4 R 4:f2"
/>

If Blue moves at 1, Red gets a 3rd row ladder, which ↑+∗+P escapes by assumption:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:f3 R 2:f2 B 3:e2 R 4:a2 B 5:d4 R 6:e3 B 7:e4 R 8:g2"
/>
Note that there are several alternatives to Blue's move 3, but they all result in a 3rd row ladder for Red.

If Blue moves at 1, Red simply pushes the 2nd row ladder, and we are now in position T(''k''+1)+''n''−1+P, which is a virtual connection by the outer induction hypothesis.
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:d4 R 2:e3"
/>

If Blue moves at 1, Red gets a 2nd row ladder, which ↑+↑+∗+P escapes by assumption. Note again that 2 is connected to the top ladder stone.
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:e4 R 2:e2 B 3:d4 R 4:f3"
/>

If Blue moves at 1, Red gets a 2nd row ladder.
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a4 b4 c4 h3 h4"
contents="R a1 a3 b3 c3 d3 B 1:f4 R 2:e2 B 3:d4 R 4:f3 B 5:e4 R 6:g3"
/>
This finishes the proof of the lemma. □

Having reduced the distance ''n'' to finitely many cases, we would now like to reduce the parameter ''k'' to finitely many cases as well. The following lemma does this.

'''Lemma 21.''' Suppose P is a candidate for a 2-4 terraced ladder escape. Assume T(0)+P, T(1)+P, and T(2)+P are virtual connections (i.e., they guarantee a connection of the top ladder stone to the edge, with Blue to move). Then T(''k'')+P is a virtual connection for all ''k'' ≥ 0.

'''Proof.''' The first step in the proof is to show that the following two interior patterns are equivalent. By "interior pattern", we mean that the bottom row of red stones does not have to be a board edge.

B(1): <hexboard size="4x3"
coords="none"
edges="none"
visible="-a1 b1 d4"
contents="B c1 R a2 a4--c4 E x:c2 y:c3 z:a3 -:(d1--d3)"
/>

B(2): <hexboard size="4x4"
coords="none"
edges="none"
visible="-a1 b1 e4"
contents="B c1--d1 R a2 a4--d4 E x:d2 y:d3 z:a3 -:(e1--e3)"
/>

If Red plays first in the region, a red move at x [[captured cell|captures]] the entire region, so x is the only move that Red needs to consider, and its outcome is the same in B(1) and B(2).

If Blue moves first in the region, all of the interior moves (i.e., in unmarked cells) are [[Dominated_cell#Star_decomposition_domination|star-decomposition dominated]] by x. Therefore, Blue only needs to consider the moves x, y, and z. One can show that each of these three moves (x, y, and z) in region B(2) is equivalent to the corresponding move in region B(1). For example, after Blue moves at x, z dominates all of the interior moves and whoever plays there [[captured cell|captures]] the interior, regardless of whether the region is B(1) or B(2).

A consequence of the fact that regions B(1) and B(2) are equivalent is that all "longer" versions of these regions are also equivalent to B(1), B(2), and each other, i.e.,

B(3): <hexboard size="4x5"
coords="none"
edges="none"
visible="-a1 b1 f4"
contents="B c1--e1 R a2 a4--e4 E -:(f1--f3)"
/>

B(4): <hexboard size="4x6"
coords="none"
edges="none"
visible="-a1 b1 g4"
contents="B c1--f1 R a2 a4--f4 E -:(g1--g3)"
/>
and so on. This is easily proved by induction, because each longer region is obtained from the previous one by replacing a subregion of the form B(1) by B(2), which we already showed to be equivalent.

Next, consider this pattern:

B(0): <hexboard size="4x2"
coords="none"
edges="none"
visible="-a1 b1 c4"
contents="R a2 a4--b4 E x:b2 y:b3 z:a3 -:(c1--c3)"
/>

We claim that B(1) is at least as good as B(0) for Red, in the sense that anything Red can achieve with B(0), Red can also achieve with B(1). (In fact, B(1) is strictly better for Red than B(0), but that fact is not required for this proof). If Red moves first in the region B(0), the move at x again captures the whole region, and therefore achieves everything Red might hope to achieve in the region. In this case, B(0) and B(1) are equivalent. If Blue moves first, the situation is slightly more complicated. We must show that B(0) is at least as good for Blue as B(1). If Blue plays at x in B(1), then Blue has the corresponding option to move at x in B(0), which works for the same reason as in the proof of the equivalence of B(1) and B(2) above. If Blue plays at z in B(1), Red can respond by pushing the ladder, which creates a position that is literally B(0). If Blue plays at y in B(1), Red can respond at x, and a case distinction shows that no matter how the remaining 3 cells are filled, filling them in the same way in B(0) gives an equivalent position.

Finally, let C(0), C(1), C(2), ... be the same patterns as B(0), B(1), B(2), ..., except with the blue stones removed from the carrier. I.e.:

C(0): <hexboard size="4x2"
coords="none"
edges="none"
visible="-a1 b1 c4"
contents="R a2 a4--b4 E -:(c1--c3)"
/>

C(1): <hexboard size="4x3"
coords="none"
edges="none"
visible="-a1 b1 d4 c1"
contents="R a2 a4--c4 E -:(d1--d3)"
/>

C(2): <hexboard size="4x4"
coords="none"
edges="none"
visible="-a1 b1 e4 c1--d1"
contents="R a2 a4--d4 E -:(e1--e3)"
/>
etc. Note that for all ''k'' ≥ 0, C(''k'') is at least as good for Red as B(''k''). Because if the neighboring cells we removed from the templates are in fact occupied by Blue, then C(''k'') is the same as B(''k''); otherwise, if they are empty or Red, it can only help Red.

In particular, since each C(''k'') is at least as good for Red as B(''k''), and each B(''k'') is at least as good as B(0) = C(0), it follows that if Red wins any position containing C(0), then Red also wins the corresponding position containing C(''k'').

The final step in the proof is now easy. Simply observe that each T(''k''+2) is obtained from T(2) by replacing a subpattern of the form C(0) by C(''k''). Therefore, in any context P where T(2)+P is winning for Red, T(''k''+2)+P is also winning for Red. Combining this with the remaining two base cases T(0)+P and T(1)+P, we get the lemma. □

By combining the previous two lemmas, we obtain a sufficient condition for the validity of terraced ladder escapes.

'''Theorem 22.''' Consider a candidate P for a 2-4 terraced ladder escape. Assume T(''k'')+''n''+P is a virtual connection for ''n''=1,2,3 and ''k''=1,2,3 (nine possibilities). Moreover, assume that ∗+P escapes 2nd, 3rd, and 4th row ladders (or more precisely, ∗+P escapes 4th row ladders, ↑+∗+P (which has 3 cells marked "+") escapes 3rd row ladders, and ↑+↑+∗+P escapes 2nd row ladders). Then P is a valid 2-4 terraced ladder escape.

'''Proof.''' By Lemma 21, T(''k'')+''n''+P is a virtual connection for ''n''=1,2,3 and all ''k'' ≥ 0. Therefore, the hypothesis of Lemma 20 is satisfied, and thus P is valid. □

=== Non-examples ===

Since terraced ladders are weaker than 4th row ladders, any terraced ladder escape is also a 4th row ladder escape. The question then becomes: which 4th row ladder escapes are ''not'' terraced ladder escapes? Most, but not all, of the examples of 4th row ladder escapes given above also escape terraced ladders.

The following patterns escape 4th row ladders but do not escape terraced ladders:

<hexboard size="5x6"
coords="hide"
edges="bottom"
visible="-a1 e1 f1 f2"
contents="E *:a1 E +:a2 E +:a3 E +:a4 E +:a5 R d2 E *:e1 E *:f1 E *:f2"
/>
<hexboard size="6x5"
coords="hide"
edges="bottom"
visible="-a1 a2 b1 c1"
contents="E *:a1 E *:a2 E +:a3 E +:a4 E +:a5 E +:a6 E *:b1 E *:c1 R c2 R d1"
/>

[[category: Theory]]
[[category: Ladder]]

User:Bobson8

2024-08-26T04:42:42Z

Bobson8: Fix a wrong strategy

Connection game

2024-06-21T03:45:09Z

Bobson8:

A '''connection game''' is a [[game]] where the goal is to use the [[Piece (general)|piece]]s to create a connection between some parts of the [[Board (general)|board]]. The first connection game was [[Hex]], which was invented in 1942. Several others have been created since then.

== Timeline ==

;[[Hex]] ([[Piet Hein]] 1942 and [[John Nash]] 1948)
:The original connection game. Played on a [[Printable boards|rhombic hex grid]].
;[[Y]] ([[John Milnor]] 1950s, [[Craige Schensted]] and [[Charles Titus]] 1953)
:Played on a [[Printable Y boards|triangluar grid of hexagons]]
;[[Twixt]] ([[Alex Randolph]], 1960s)
:Played on a [[square grid]] of holes into which the players place [[peg]]s. The pegs can be connected via [[Bridge (general)|bridges]]; a bridge connects two holes that are separated by a knight's move.
;[[Havannah]] ([[Christian Freeling]], 1980)
:Multiple goals, two connection-oriented and one shape-oriented.
;[http://www.di.fc.ul.pt/~jpn/gv/quax.htm Quax] (Bill Taylor?, 2000?)
:Played on a square grid with the possibility of diagonal connections.
;[[Onyx]] ([[Larry Back]], 2000)
:Played on an original grid consisting of both triangles and squares. It is the first connection game with a [[capturing rule]].
;[[Gonnect]] ([[João Pedro Neto]], 2000)
:This game is simply [[Go]], but with a different goal, namely to create a connection between any two opposite sides.
;[[Unlur]] ([[Jorge Gómez Arrausi]], 2001)
:Played on a [[hexagonal hex grid]]. Unique in the way that the players have [[different objectives]].
;[[Bridg-It]] ([[David Gale]], ca. 1958)
:Played on an interlaced square board.
;[[MindNinja]] (Nicholas Bentley, 2006)
:Generalized game of shape-building and connection, where board and win conditions are decided with help of a [[pie rule]].
;[[Atoll]] (Mark Steere, 2008)
:A generalization of Hex to boards with four or more perimeter segments. With four segments, it is identical to Hex.
:See [http://www.marksteeregames.com/Atoll_rules.pdf the rule sheet] for more information.

== References ==

;[[Cameron Browne]], [http://www.amazon.com/Connection-Games-Variations-Cameron-Browne/dp/1568812248/ref=pd_bbs_sr_1/104-1532904-9846317?ie=UTF8&s=books&qid=1177663469&sr=8-1 "Connection Games: Variations on a Theme"]

[[Category: Other games]]

Connection game

2024-06-21T03:40:10Z

Bobson8: Crediting Milnor

A '''connection game''' is a [[game]] where the goal is to use the [[Piece (general)|piece]]s to create a connection between some parts of the [[Board (general)|board]]. The first connection game was [[Hex]], which was invented in 1942. Several others have been created since then.

== Timeline ==

;[[Hex]] ([[Piet Hein]] 1942 and [[John Nash]] 1948)
:The original connection game. Played on a [[Printable boards|rhombic hex grid]].
;[[Y]] ([[John Milnor]], [[Craige Schenstead]] and [[Charles Titus]], 1950s)
:Played on a [[Printable Y boards|triangluar grid of hexagons]]
;[[Twixt]] ([[Alex Randolph]], 1960s)
:Played on a [[square grid]] of holes into which the players place [[peg]]s. The pegs can be connected via [[Bridge (general)|bridges]]; a bridge connects two holes that are separated by a knight's move.
;[[Havannah]] ([[Christian Freeling]], 1980)
:Multiple goals, two connection-oriented and one shape-oriented.
;[http://www.di.fc.ul.pt/~jpn/gv/quax.htm Quax] (Bill Taylor?, 2000?)
:Played on a square grid with the possibility of diagonal connections.
;[[Onyx]] ([[Larry Back]], 2000)
:Played on an original grid consisting of both triangles and squares. It is the first connection game with a [[capturing rule]].
;[[Gonnect]] ([[João Pedro Neto]], 2000)
:This game is simply [[Go]], but with a different goal, namely to create a connection between any two opposite sides.
;[[Unlur]] ([[Jorge Gómez Arrausi]], 2001)
:Played on a [[hexagonal hex grid]]. Unique in the way that the players have [[different objectives]].
;[[Bridg-It]] ([[David Gale]], ca. 1958)
:Played on an interlaced square board.
;[[MindNinja]] (Nicholas Bentley, 2006)
:Generalized game of shape-building and connection, where board and win conditions are decided with help of a [[pie rule]].
;[[Atoll]] (Mark Steere, 2008)
:A generalization of Hex to boards with four or more perimeter segments. With four segments, it is identical to Hex.
:See [http://www.marksteeregames.com/Atoll_rules.pdf the rule sheet] for more information.

== References ==

;[[Cameron Browne]], [http://www.amazon.com/Connection-Games-Variations-Cameron-Browne/dp/1568812248/ref=pd_bbs_sr_1/104-1532904-9846317?ie=UTF8&s=books&qid=1177663469&sr=8-1 "Connection Games: Variations on a Theme"]

[[Category: Other games]]

Y

2024-06-20T04:26:15Z

Bobson8:

The game of Y is a [[connection game]] invented by [[Craige Schensted]] and [[Charles Titus]]. In its original form, it is played on a [[triangular grid of hexagons]]. There are two [[Player (general)|players]], who have one colour each, and a move consists of placing a piece of your colour in one of the hexagons on the board. The winner is the first player to complete a [[chain]] connecting all three sides of the board. Y is a kind of generalisation of [[Hex]], perhaps the one the nearest from it, but there are some strategic peculiarities, such as [[corner template]]s

<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>

== No draws ==

Y cannot end in a draw. That is, once the board is complete there must be one and only one winner.

=== Less than two winners ===
There cannot be two winners at the same time. If there were, each player would have a region of the board touching all three sides of the triangle as well as the opponent's region. Considering the three sides as regions themselves, this gives a map of five regions, each of which is adjacent to the other four. However, this is impossible, as the graph K5 is non-planar.

=== At least one winner ===
It can be proved by an algorithm that once a board is complete there is at least one player linking the 3 sides. Let the "state" of a board refer to the answer to the question "Is there at least one winner?" We want to prove that the state of every board is "Yes".

''First step'': if there is a pawn group (red for instance) completely surrounded by the opponent (blue for instance) let's consider the board with this pawn group replaced by opponent's pawns (blue ones). The new board has the same status as the older one as the remote group was not winning and the new big (blue) one is winning iff it was in the former board. Also note that there is still at least one group left.

Repeat this step until there is no completely surrounded group more (of either color). The board obtained has the same state as the original.

''Second step'': if there is a pawn group surrounded by the opponent and a side, removing it does not change the state of the board (for similar reasons as in step 1) and there is still at least one group left.

Repeat this step.

''Third step'': if there is a pawn group surrounded by the opponent and two sides removing it does not change the state of the board (for similar reasons as in step 1) and there is still at least one group left.

Repeat this step.

It is quite clear that after applying this algorithm there is no group connected to more than 1 opponent's groups. No group is connected to zero sides and one opponent's group, no group is connected to one side and one opponent's group, no group is connected to two sides and one opponent's group. No group can be connected to 0 1 or 2 sides without connecting an opponent group. Moreover there is at least one group left. Hence this group left is connected to 3 sides.

So the state of the board is "yes"; as it is the same as the state of the beginning board, there was a winner to begin with.

Note that this algorithm ends because the number of different groups is finite.

=== Extension to [[Hex]] ===

The proof above extends to Hex because a Hex game can be seen as a subset of a Y game.

For instance consider the following Y board of size 7.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>

We can simulate a Hex game of [[size]] 4 on it.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
contents="R area(g1,e3,g3) B area(c5,a7,c7)"
/>

Now the only way to win for Blue is to cross the board horizontally, whereas the only way for Red to do so is to cross the board vertically.

Each game of Hex on a board of size ''n'' can be played on a Y board of size 2''n''−1 with the rules of Y. The players just need to place some stones to "construct" the Hex board.

== Y-Reduction ==

Given a Y board of size n filled with red or blue stones, there is an operation of replacing each of the upper triangle of size 2 with a hex at its center, and with a stone of color representing the majority of the three hexes being replaced. The result would be a Y board of size n-1. This operation is called the Y-reduction, introduced by Craige Schensted.

<hexboard size="5x24"
coords="none"
edges="none"
visible="area(e1,a5,e5) area(j2,g5,j5) area(n3,l5,n5) q4 p5 q5 s5"
contents="R b4 c3 c4 d4 d5 e2 e5 h4 i4 j5 i5 m4 m5 n5 p5 q5 s5 B a5 b5 c5 d2 d3 e1 e3 e4 j2 i3 j3 g5 h5 j4 n3 n4 l5 q4"
/>

For example, the above Y board of size 5 can be reduced to size 4, and so on.

An important property of this operation is that, one color has a winning chain if and only if the color has a winning chain for its Y-reduction. As a consequence, one can repeatedly reduce the board until size 1 to determine the winner. This can also be seen as a proof of exactly one winner for Y.

== The first player wins ==
In Y the [[strategy-stealing argument]] can be applied. It proves that the second player has no winning strategy. The argument is that if the second player had a winning strategy, then the first player could chose a random first move and then pretend that she is the second player and apply the strategy. An important point is that an extra pawn is not a disadvantage in Y. Y is a complete and perfect information game in which no draw can be conceived, so there is a winning strategy for one player. The second player has no winning strategy so the first player has one.

== Swap ==

The [[Swap rule]] can be used in Y too, the corner are bad moves to be played so there may well exist average moves to begin with. Further information on [[where to swap (y)|where to swap]].

== Variations ==

The game is usually played with the [[swap rule]]. Alternatively, one can play Double-move Y, also known as [[Master Y]]: The first player places one piece on the board, and each subsequent move consists of placing two pieces on the board. This is a pretty challenging variant, even on small boards.

The inventors tried out a number of alternative playing grids, and eventually concluded that the most suitable one is the following "bent" version. The pieces are placed on the intersections (like in [[Go]]).

[[Image:Y93_bent.gif]]

== See also ==

You can find more boards here: [[Printable Y boards]]

Help for [[programming the bent Y board]]

Try this [[Y puzzle]].

== On the web ==

* http://www.gamepuzzles.com/gameofy.htm
* http://www.gamepuzzles.com/revugy.htm (Games magazine reviews)
* http://home.flash.net/~markthom/html/the_game_of_y.html (Dead link?)
* http://www.neutreeko.net/y.htm
* http://www.iggamecenter.com/ ('''igGameCenter''' - play "Y" online with other opponents from your iGoogle homepage)

[[Category: Y]]
[[Category: Theory]]

Y

2024-06-20T04:10:04Z

Bobson8: Adding Y-reduction

The game of Y is a [[connection game]] invented by [[Craige Schensted]] and [[Charles Titus]]. In its original form, it is played on a [[triangular grid of hexagons]]. There are two [[Player (general)|players]], who have one colour each, and a move consists of placing a piece of your colour in one of the hexagons on the board. The winner is the first player to complete a [[chain]] connecting all three sides of the board. Y is a kind of generalisation of [[Hex]], perhaps the one the nearest from it, but there are some strategic peculiarities, such as [[corner template]]s

<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>

== No draws ==

Y cannot end in a draw. That is, once the board is complete there must be one and only one winner.

=== Less than two winners ===
There cannot be two winners at the same time. If there were, each player would have a region of the board touching all three sides of the triangle as well as the opponent's region. Considering the three sides as regions themselves, this gives a map of five regions, each of which is adjacent to the other four. However, this is impossible, as the graph K5 is non-planar.

=== At least one winner ===
It can be proved by an algorithm that once a board is complete there is at least one player linking the 3 sides. Let the "state" of a board refer to the answer to the question "Is there at least one winner?" We want to prove that the state of every board is "Yes".

''First step'': if there is a pawn group (red for instance) completely surrounded by the opponent (blue for instance) let's consider the board with this pawn group replaced by opponent's pawns (blue ones). The new board has the same status as the older one as the remote group was not winning and the new big (blue) one is winning iff it was in the former board. Also note that there is still at least one group left.

Repeat this step until there is no completely surrounded group more (of either color). The board obtained has the same state as the original.

''Second step'': if there is a pawn group surrounded by the opponent and a side, removing it does not change the state of the board (for similar reasons as in step 1) and there is still at least one group left.

Repeat this step.

''Third step'': if there is a pawn group surrounded by the opponent and two sides removing it does not change the state of the board (for similar reasons as in step 1) and there is still at least one group left.

Repeat this step.

It is quite clear that after applying this algorithm there is no group connected to more than 1 opponent's groups. No group is connected to zero sides and one opponent's group, no group is connected to one side and one opponent's group, no group is connected to two sides and one opponent's group. No group can be connected to 0 1 or 2 sides without connecting an opponent group. Moreover there is at least one group left. Hence this group left is connected to 3 sides.

So the state of the board is "yes"; as it is the same as the state of the beginning board, there was a winner to begin with.

Note that this algorithm ends because the number of different groups is finite.

=== Extension to [[Hex]] ===

The proof above extends to Hex because a Hex game can be seen as a subset of a Y game.

For instance consider the following Y board of size 7.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>

We can simulate a Hex game of [[size]] 4 on it.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
contents="R area(g1,e3,g3) B area(c5,a7,c7)"
/>

Now the only way to win for Blue is to cross the board horizontally, whereas the only way for Red to do so is to cross the board vertically.

Each game of Hex on a board of size ''n'' can be played on a Y board of size 2''n''−1 with the rules of Y. The players just need to place some stones to "construct" the Hex board.

== Y-Reduction ==

Given a triangular board of size n filled with red or blue stones, there is an operation of replacing each of the upper triangle of size 2 with a hex at its center, and with a stone of color representing the majority of the three hexes being replaced. The result would be a triangular board of size n-1. This operation is called the Y-reduction, introduced by Craige Schensted.

<hexboard size="5x24"
coords="none"
edges="none"
visible="area(e1,a5,e5) area(j2,g5,j5) area(n3,l5,n5) q4 p5 q5 s5"
contents="R b4 c3 c4 d4 d5 e1 e5 h4 i4 j5 i5 m4 m5 n5 p5 q5 s5 B a5 b5 c5 d2 d3 e2 e3 e4 j2 i3 j3 g5 h5 j4 n3 n4 l5 q4"
/>

For example, the above triangular board of size 5 can be reduced to size 4, and so on.

An important property of this operation is that, one color has a winning chain of Y if and only if the color has a winning chain for its Y-reduction. As a consequence, one can repeatedly reduce the board until size 1 to determine the winner. This can also be seen as a proof of exactly one winner for Y.

== The first player wins ==
In Y the [[strategy-stealing argument]] can be applied. It proves that the second player has no winning strategy. The argument is that if the second player had a winning strategy, then the first player could chose a random first move and then pretend that she is the second player and apply the strategy. An important point is that an extra pawn is not a disadvantage in Y. Y is a complete and perfect information game in which no draw can be conceived, so there is a winning strategy for one player. The second player has no winning strategy so the first player has one.

== Swap ==

The [[Swap rule]] can be used in Y too, the corner are bad moves to be played so there may well exist average moves to begin with. Further information on [[where to swap (y)|where to swap]].

== Variations ==

The game is usually played with the [[swap rule]]. Alternatively, one can play Double-move Y, also known as [[Master Y]]: The first player places one piece on the board, and each subsequent move consists of placing two pieces on the board. This is a pretty challenging variant, even on small boards.

The inventors tried out a number of alternative playing grids, and eventually concluded that the most suitable one is the following "bent" version. The pieces are placed on the intersections (like in [[Go]]).

[[Image:Y93_bent.gif]]

== See also ==

You can find more boards here: [[Printable Y boards]]

Help for [[programming the bent Y board]]

Try this [[Y puzzle]].

== On the web ==

* http://www.gamepuzzles.com/gameofy.htm
* http://www.gamepuzzles.com/revugy.htm (Games magazine reviews)
* http://home.flash.net/~markthom/html/the_game_of_y.html (Dead link?)
* http://www.neutreeko.net/y.htm
* http://www.iggamecenter.com/ ('''igGameCenter''' - play "Y" online with other opponents from your iGoogle homepage)

[[Category: Y]]
[[Category: Theory]]

Template VI1/Intrusion on the 3rd row

2024-05-17T05:02:34Z

Bobson8: wording

This article deals with a special case in the defense of [[edge template VI1a]], namely the intrusion on the 3rd that is not eliminated by [[sub-templates threat]]s.

== Basic situation ==

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 E a:h2 b:k2 c:l3"
/>

In this situation, there are only 3 possible winning moves for Red, and they are "a", "b", and "c". Of these, "a" is the easiest to verify, so we will assume Red plays there.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2"
/>

Before continuing the analysis, we first note that Red can escape all 2nd row ladders coming from the left, as follows:
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
R 1:g5 B 2:g6 R 3:h5 B 4:h6 R 5:j5 B 6:i5 R 7:j4 B 8:i4 R 9:k2"
/>
Apart from attacking the bridge, which Red defends, Blue's next move must be in the shaded blue area, or else Red plays at d and connects.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
S blue:area(g3,d6,g6,g4,h3)
E a:g3 b:h3 c:f4 d:g4 e:e5 f:f5 g:g5 h:d6 i:e6 j:f6 k:g6 x:f3"
/>
If Blue plays at c, e, h, j or k, Red responds at d and gets a 2nd row ladder, which connects.
If Blue plays at b, Red plays at x and connects by [[edge template IV1a]].
If Blue plays at d, Red plays at x and gets a 2nd row ladder, which connects.
This leaves a, f, g, i.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 R 1:h3
E a:g3 d:g4 f:f5 g:g5 k:g6 y:i4"
/>
If Blue plays at a, f or g, then Red plays at b.
Note that Red connect down from the left by playing d, and connect down from the right by playing y.
The only way for Blue to block both is to play k, but Red with the same strategy on the left would at worst produce a 2nd row ladder toward the right.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 B e6 R 1:g4
E a:g3 b:h3 y:i4 g:g5 j:f6 z:d5"
/>

Finally if Blue plays at i, then Red plays at d. Apart from the bridges, Blue is forced to play g, and then Red plays b, forcing Blue to block on the right part (against Red y), and then Red wins with z.

If Blue intrudes at a, then Red responds at b, forcing Blue to block on the right part, and then Red wins with j.

If Blue intrudes at b, then Red responds at a. Blue is still forced to play g, and then Red wins with z.

[[category:edge templates]]

Edge template VI1a

2024-05-10T05:15:21Z

Bobson8: wording

Template VI1-a is a 6th row [[edge template]] with one stone.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2"
/>

This template is the first one stone 6th row [[edge template|template]] for which a proof of validity has been written out. The template has been verified by computer, and also verified to be minimal.

== Elimination of irrelevant Blue moves ==

Red has a number of direct threats to connect, using smaller templates. Blue must play in the carrier of these threats in order to counter them. To prevent Red from connecting Blue must play in the intersection of Red's threats carriers.

=== [[Edge template IV1a]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h6 h7 i3 i5 i6 i7 j3 j5 j6 j7"
/>

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 e7 f6 f7 g5 g6 g7 h5 h6 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== [[Edge template IV1b]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== Using [[Tom's move]] ===

6 intrusions can furthermore be discarded thanks to [[Tom's move]], also known as the [[parallel ladder]] trick. Of course, symmetry will cut our work in half!

If Blue moves in any of the cells marked "1" on the left (and, using mirror symmetry, in the corresponding 3 cells on the right), Red can respond as follows:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 4:h5 6:h6 2:i4 j2 B 5:g7 6:h7 3:i5 B 1:(e7 f6 g5)"
/>

At this point, Red can use [[Tom's move]] to connect:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h5 h6 i4 4:i6 j2 8:k3 2:k5 6:l4 B g7 h7 i5 5:i7 7:j5 3:j6 B 1:(e7 f6 g5)"
/>

=== Remaining intrusions ===

The only possible remaining intrusions for Blue are the following:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2
S f7 g6 g7 h5 h7 i3 i4 i5 i6 i7 j3
E a:f7 b:g7 c:g6 d:h5 e:i4 f:i3"
/>
By symmetry, if is sufficient to consider the six possible intrusions at a – f.

== Specific defense ==

For the moves that intersect all the carriers, Red has to find specific answers. Let's deal with the remaining intrusions!

=== Intrusion at a ===

If Blue intrudes at a, Red has several winning responses. For example, Red can play at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 E x:i5 y:h7"
/>
Apart from intrusion into the bridge, which Red defends, Blue's only possible moves are at x and y. If Blue plays at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]].
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:i5 R 4:g5 B 5:e6 R 6:g6 B 7:g7 R 8:h6 B 9:h7 R 10:k5"
/>
If Blue plays at y, Red has the following simple win, using the [[trapezoid]] template:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:h7 R 4:i5 B 5:j6 R 6:g5 B 7:e6 R 8:g7"
/>

=== Intrusion at b ===

If Blue intrudes at b, Red can respond at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 E x:i5 y:i6 z:i7 w:h7"
/>
Apart from intrusions into the bridge, which Red defends, Blue has only four possible moves x,y,z,w, because if Blue moves anywhere else, Red connects with either the [[ziggurat]] or [[edge template III1b]].

If Blue intrudes at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]]:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:i5 R 4:h5 B 5:f6 R 6:h6 B 7:h7 R 8:k5"
/>
If Blue intrudes at y or z (both shown simultaneously in the following diagram), Red can set up Tom's move on the opposite side:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:(i6 i7) R 4:h6 B 5:h7 R 8:f5"
/>
Finally, if Blue intrudes at w, Red can connect by the following variant of Tom's move:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:h7 R 4:h5 B 5:f6 R 6:k5"
/>

=== Intrusion at c ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g6"
/>

Red may play here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

Note that if Red plays at c, then in the blue area there is a strategy such that both Red 2 and c connect down without choice, unless Blue first plays at d. Also, with this strategy the paths for 2 connecting down would not pass c or d in any case.
Therefore, Blue must spend one move at either a,b,c or d in order to block Red on the right side, while Red 2 is always guaranteed to connect down. Thus we have this forcing sequence:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6 R 4:i3 B 5:i4 R 6:h4 B 7:h5 R 8:f5
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

(Assume that Blue 3 is played at either a,b,c or d, and there were no extra turns in the blue area.)

=== Intrusion at d ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:h5"
/>

Red may go here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:h3 B 1:h5"
/>

Details to follow. See more details [[Template_VI1/Intrusion_on_the_3rd_row|here]].

=== Intrusion at e ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i4"
/>

Red should move here (or the equivalent mirror-image move at "+"):

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 2:h3 j2 B 1:i4 E +:k3"
/>

Now the shaded area is a [[ladder creation template]], giving Red at least a 3rd row ladder as indicated.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) E arrow(3):(h5 h6 h7)"
/>

Red can escape both 2nd and 3rd row ladders using a [[ladder escape fork]] via "+". Specifically, Red escapes a third row ladder like this, and is connected by a [[ziggurat]] and double threat at "+":

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h6 R 3:j5 E +:i5"
/>

If Blue [[ladder handling|yields]], or Red starts out with a 2nd row ladder, the escape fork works anyway:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h7 R 3:h6 B 4:g7 R 5:j6 B 6:i6 R 7:j5 E +:i5"
/>

=== Intrusion at f ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i3"
/>

First establish a [[parallel ladder]] on the right.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 8:i4 j2 2:j3 6:j5 4:k4 B 9:h5 1:i3 3:i5 7:i7 5:k5"
/>

Then use [[Tom's move]]:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 12:f4 10:f5 14:h3 i4 j2 j3 j5 k4 B 11:f6 13:g5 h5 i3 i5 i7 k5"
/>

There are two marginal cases. If Blue 3 blocks at a or b, then Red 4 can reduce the situation to "Intrusion at a" or "Intrusion at b" as before, since intruding the vertical bridge is irrelevant in these two cases.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:j3 4:i4 B 1:i3 E a:i7 b:g7"
/>

[[category:edge templates]]
[[category:theory]]

Edge template V1d

2024-05-04T02:19:21Z

Bobson8: Adding category

Template V1-d is a 5th row [[edge template]] with 1 stone.

<hexboard size="5x13"
coords="hide"
edges="bottom"
visible="area(a5,m5,m3,k1,g1,c3)"
contents="R g1"
/>

It was discovered by [[User:Bobson8]] and can be regarded as a 5th row version of [[edge template VI1b]].

[[category:edge templates]]

Edge template VI1c

2024-05-04T02:18:42Z

Bobson8: Adding category

Template V1-c is a 6th row [[edge template]] with 1 stone.

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1"
/>

It was discovered by [[User:Bobson8]].

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1 S gray:area(a6,o6,o4,m2,k1,i1,e3,c4) E *:h3 A:j2"
/>

As mentioned in [[edge template VI1b]], Red's stone can connect to the edge within the shaded area unless Blue block at *. With the extra cell on the top left, Red can play at A and connect down with a ladder escape fork on the left against *.

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1 S gray:area(a6,o6,o4,m2,j1,h1,e3,c4) E +:(g4 h4) B:j3"
/>
On the other hand, Blue intruding + cells are the only two cases requiring the top right cell, and B is a correct response for Red in both cases.

[[category:edge templates]]

Category:Edge templates

2024-05-02T03:03:54Z

Bobson8: Adding a link, and also replacing a link

An '''edge template''' is a [[pattern]] which guarantees a [[connection]] to the [[edge]].

Here is an example of a [[Third row edge templates|third-row edge template]] ([[template IIIa]], also known as the ''Ziggurat''):

<hexboard size="3x4"
visible="area(a3,d3,d1,c1)"
edges="bottom"
coords="none"
contents="R c1"
/>

The red stone has a certain connection to the bottom, using only the shaded hexagons, even if Blue moves first.

Countless edge templates can be constructed, all of them of varying degrees of usefulness and frequency of occurence.

You could start by learning about [[Edge templates everybody should know]]. There are some templates that every good player will know.

Template lists by row number:

* [[Third row edge templates]]
* [[Fourth row edge templates]]
* [[Fifth row edge templates]]
* [[Sixth row edge templates]]
* [[Seventh row edge templates]]

[[category:templates]]

Edge templates with one stone

2024-05-02T02:57:48Z

Bobson8: Adding recent templates

This page lists some [[Edge template|edge templates]] with one stone to be connected to the bottom row. Not all of these templates are useful in practice. The fifth row to seventh row templates rarely occur in real play.

There is some overlap with the article [[Edge templates everybody should know]].

== First row edge template ==
<hexboard size="1x1"
coords="none"
edges="bottom"
contents="R a1"
/>

== Second row edge template ==

=== [[Edge template II]] ===

<hexboard size="2x2"
coords="none"
edges="bottom"
visible="-a1"
contents="R b1"
/>

== Third row edge templates ==

=== [[Ziggurat|Edge template III1a]] ===

<hexboard size="3x4"
coords="none"
edges="bottom"
visible="area(c1,a3,d3,d1)"
contents="R c1"
/>
Also known as the '''ziggurat'''.

=== [[Edge template III1b]] ===
The shaded cell is not part of the template and can be occupied by Blue.
<hexboard size="3x5"
coords="none"
edges="bottom"
visible="area(c1,a3,e3,e1)"
contents="R d1 S c3"
/>

== Fourth row edge templates ==
=== [[Edge template IV1a]] ===
<hexboard size="4x7"
coords="none"
edges="bottom"
visible="area(e1,c2,a4,g4,g2,f1)"
contents="R e1"
/>
=== [[Edge template IV1b]] ===
The shaded cell is not part of the template and can be occupied by Blue.
<hexboard size="4x8"
coords="none"
edges="bottom"
visible="area(e1,c2,a4,h4,h2,g1)"
contents="R f1 S e3"
/>
=== [[Edge template IV1c]] ===
The shaded cell is not part of the template and can be occupied by Blue.
<hexboard size="5x11"
coords="none"
edges="bottom"
visible="area(g1,c3,a5,k5,k3,i1)"
contents="R f2 S c5"
/>

=== [[Edge template IV1d]] ===
<hexboard size="5x11"
coords="none"
edges="bottom"
visible="area(g1,c5,k5,k3,i1)"
contents="R f2"
/>

== Fifth row edge templates ==

=== [[Template_Va|Edge template V1a]] ===

<hexboard size="5x10"
coords="hide"
edges="bottom"
visible="area(f1,c3,a5,j5,j3,h1)"
contents="R g1"
/>

=== [[Edge_template_V1b|Edge template V1b]] ===

<hexboard size="6x14"
coords="hide"
edges="bottom"
visible="area(f1,a6,n6,n4,l2,h1)"
contents="R e2"
/>

=== [[Edge_template_V1c|Edge template V1c]] ===

<hexboard size="6x16"
coords="hide"
edges="bottom"
visible="area(f2,c4,a6,p6,p4,n2,j1,h1)"
contents="R f2"
/>

=== [[Edge_template_V1d|Edge template V1d]] ===

<hexboard size="5x13"
coords="hide"
edges="bottom"
visible="area(a5,m5,m3,k1,g1,c3)"
contents="R g1"
/>

== Sixth row edge templates ==

=== [[Edge_template_VI1a|Edge template VI1a]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2"
/>

=== [[Edge_template_VI1b|Edge template VI1b]] ===

<hexboard size="6x19"
coords="none"
edges="bottom"
visible="area(a6,s6,s4,q2,m1,i1,c4)"
contents="R i1"
/>

=== [[Edge_template_VI1c|Edge template VI1c]] ===

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1"
/>

== Seventh row edge templates ==

=== [[Edge_template_VII1a|Edge template VII1a]] ===

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1"
/>

=== [[Edge_template_VII1b|Edge template VII1b]] ===

<hexboard size="7x19"
coords="none"
edges="bottom"
visible="area(a7,s7,s5,p2,n1,l1,i2,c5)"
contents="R m1"
/>

=== [[Edge_template_VII1c|Edge template VII1c]] ===

<hexboard size="7x21"
coords="none"
edges="bottom"
visible="area(a7,u7,u5,s3,q3,o1,m1,i3,g3,c5)"
contents="R n1"
/>

[[Category: Edge templates]]

User:Bobson8

2024-04-21T18:54:40Z

Bobson8: winning move with a smaller template

User:Bobson8

2024-04-21T18:41:24Z

Bobson8:

User:Bobson8

2024-04-21T18:40:47Z

Bobson8: typo

User:Bobson8

2024-04-21T10:59:48Z

Bobson8: typo

User:Bobson8

2024-04-21T07:00:06Z

Bobson8: A conjecture on trapezoid boards

Seventh row template problem

2023-12-01T14:25:51Z

Bobson8: updating verification status

Until October 2023, it was an open problem whether there exists a 7th row edge template with a single stone. During October and November 2023, the users Bobson, Comonoid, Mason, and Quasar of the [[Hex forums|Hex Discord]] collaborated on finding such a template, and ended up finding three different ones. Curiously, all of these templates are symmetric, although it is possible and quite likely that asymmetric templates also exist.

== Discovery of the templates ==

In chronological order:

=== Edge template VII-1a ===

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1"
/>

This template has width 17, and its [[carrier]] has size 84, including the red stone.

While checking the minimality of [[edge template VII1c]], Comonoid first noticed on November 4 that [[KataHex]] seemed to indicate that the red stone was connected within this carrier. The validity and minimality of the template was subsequently verified using [[MoHex]], and the minimality proof was finished on November 6.

=== Edge template VII-1c ===

<hexboard size="7x21"
coords="none"
edges="bottom"
visible="area(a7,u7,u5,s3,q3,o1,m1,i3,g3,c5)"
contents="R n1"
/>

This template has width 21, and its [[carrier]] has size 98, including the red stone.

After Eric Demer's discovery of [[edge template VI1b]] on October 6, Bobson noticed that this could likely be turned into a single-stone 7th row template. He found the first such pre-template of arbitrary width using HexProver on October 13. The pre-template was soon reduced to width 23 on October 14, as he realized that using edge template VI1b was not necessary.

To explain how the width 23 pre-template works, notice that Red can try to connect the 7th row stone to the edge with a variant of [[edge template V1b]] from both sides:

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

Bobson continued to reduce the size of this pre-template. He first considered the following width 19 carrier (the gray area):

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E *:m4 *:n4 +:(a7 b6 c5 e4 g3 w7 w6 w5 v4 u3) S gray:area(c7,u7,u5,s3,r3,p1,n1,j3,i3,e5)"
/>

MoHex shows that every intrusions other than * are solvable within this carrier. On the other hand, Bobson found that intrusions at * can be solved within width 21 (removing + cells), so the pre-template was then reduced to width 21 on October 27. The minimality was checked by a collaborative effort using [[MoHex]] (patched version made by Comonoid and Quasar so that it can be run on a 23×23 board), and was finally verified on November 8.

=== Edge template VII-1b ===

<hexboard size="7x19"
coords="none"
edges="bottom"
visible="area(a7,s7,s5,p2,n1,l1,i2,c5)"
contents="R m1"
/>

This template has width 19, and its [[carrier]] has size 90, including the red stone.

While checking the minimality of [[edge template VII1c]], Comonoid first proposed that there might be a width-19 template if the carrier has more vertical space. Comonoid, Quasar, and Mason came up with different carriers, and Mason described what turned out to be the final shape of the template on November 4 and proved its validity. Minimality was subsequently verified by a collaboration among several users, and the final case was finished by Quasar on November 17.

== Methodology ==

=== Validity ===

The validity of all templates was checked by Mason and Quasar using [[MoHex]], in some cases manually supplying the winning move to Mohex. Bobson also checked the validity of the three templates using the interactive [[HexProver]] software.

=== Minimality ===

Checking minimality is far more work than checking validity. In principle, one must sequentially remove each cell from the carrier (by placing a blue stone in it) and check that the resulting pattern is not connected, i.e., i.e., is a first-player win for Blue. However, due to symmetries and [[domination]], not all cells need to be checked. Specifically, if x dominates y from Blue's point of view, then if removing x from the carrier kept the template connected for Red, then removing y from the carrier would also keep the template connected. Therefore, if minimality has been confirmed with respect to y, then minimality does not need to be checked with respect to x.

For example, to verify the minimality of [[edge template VII1a]], only the shaded cells needed to be checked:

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1 S blue:area(l2,k3,i7,o7,n6,o5),n1,p3,q7
E a:m1 b:m2 c:n2 d:o2 e:n3 f:o3 g:o4 h:p4 i:p5 j:q5 k:o6 l:p6 m:q6 n:p7 x:n1 y:p3 z:q7"
/>
Specifically, a capture-dominates x, b kill-dominates x, c capture-dominates x, d fillin-dominates y, e kill-dominates d, f capture-dominates y, g kill-dominates y, h capture-dominates y, and each of i–n capture-dominates z. Therefore, if x, y, and z are necessary in the carrier, then so are a–n. The cells on the left do not need to be checked due to symmetry.

[[Category: Edge templates]]

Edge template VI1c

2023-11-21T06:09:56Z

Bobson8:

Edge template VI1c

2023-11-21T06:06:35Z

Bobson8: Added descriptions

Template V1-c is a 6th row [[edge template]] with 1 stone.

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1"
/>

It was discovered by [[User:Bobson8]].

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1 S gray:area(a6,o6,o4,m2,k1,i1,e3,c4) E *:h3 A:j2"
/>

As mentioned in [[edge template VI1b]], Red's stone can connect to the edge within the shaded area unless Blue block at *. With the extra cell on the top left, Red can play at A and connect down with a ladder escape fork on the left against *.

<hexboard size="6x15"
coords="none"
edges="bottom"
visible="area(a6,o6,o4,m2,k1,h1,e3,c4)"
contents="R i1 S gray:area(a6,o6,o4,m2,j1,h1,e3,c4) E +:(g4 h4) b:j3"
/>
On the other hand, Blue intruding + cells are the only two cases requiring the top right cell, and b is a correct response for Red in both cases.

Edge template VI1a

2023-11-19T01:22:56Z

Bobson8:

Template VI1-a is a 6th row [[edge template]] with one stone.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2"
/>

This template is the first one stone 6th row [[edge template|template]] for which a proof of validity has been written out. The template has been verified by computer, and also verified to be minimal.

== Elimination of irrelevant Blue moves ==

Red has a number of direct threats to connect, using smaller templates. Blue must play in the carrier of these threats in order to counter them. To prevent Red from connecting Blue must play in the intersection of Red's threats carriers.

=== [[Edge template IV1a]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h6 h7 i3 i5 i6 i7 j3 j5 j6 j7"
/>

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 e7 f6 f7 g5 g6 g7 h5 h6 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== [[Edge template IV1b]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== Using [[Tom's move]] ===

6 intrusions can furthermore be discarded thanks to [[Tom's move]], also known as the [[parallel ladder]] trick. Of course, symmetry will cut our work in half!

If Blue moves in any of the cells marked "1" on the left (and, using mirror symmetry, in the corresponding 3 cells on the right), Red can respond as follows:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 4:h5 6:h6 2:i4 j2 B 5:g7 6:h7 3:i5 B 1:(e7 f6 g5)"
/>

At this point, Red can use [[Tom's move]] to connect:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h5 h6 i4 4:i6 j2 8:k3 2:k5 6:l4 B g7 h7 i5 5:i7 7:j5 3:j6 B 1:(e7 f6 g5)"
/>

=== Remaining intrusions ===

The only possible remaining intrusions for Blue are the following:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2
S f7 g6 g7 h5 h7 i3 i4 i5 i6 i7 j3
E a:f7 b:g7 c:g6 d:h5 e:i4 f:i3"
/>
By symmetry, if is sufficient to consider the six possible intrusions at a – f.

== Specific defense ==

For the moves that intersect all the carriers, Red has to find specific answers. Let's deal with the remaining intrusions!

=== Intrusion at a ===

If Blue intrudes at a, Red has several winning responses. For example, Red can play at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 E x:i5 y:h7"
/>
Apart from intrusion into the bridge, which Red defends, Blue's only possible moves are at x and y. If Blue plays at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]].
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:i5 R 4:g5 B 5:e6 R 6:g6 B 7:g7 R 8:h6 B 9:h7 R 10:k5"
/>
If Blue plays at y, Red has the following simple win, using the [[trapezoid]] template:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:h7 R 4:i5 B 5:j6 R 6:g5 B 7:e6 R 8:g7"
/>

=== Intrusion at b ===

If Blue intrudes at b, Red can respond at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 E x:i5 y:i6 z:i7 w:h7"
/>
Apart from intrusions into the bridge, which Red defends, Blue has only four possible moves x,y,z,w, because if Blue moves anywhere else, Red connects with either the [[ziggurat]] or [[edge template III1b]].

If Blue intrudes at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]]:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:i5 R 4:h5 B 5:f6 R 6:h6 B 7:h7 R 8:k5"
/>
If Blue intrudes at y or z (both shown simultaneously in the following diagram), Red can set up Tom's move on the opposite side:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:(i6 i7) R 4:h6 B 5:h7 R 8:f5"
/>
Finally, if Blue intrudes at w, Red can connect by the following variant of Tom's move:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:h7 R 4:h5 B 5:f6 R 6:k5"
/>

=== Intrusion at c ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g6"
/>

Red may play here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

Note that if Red plays at c, then in the blue area both Red 2 and c connect down without choice, unless Blue first plays at d. Also, the paths for 2 connecting down would not pass c or d.
Therefore, Blue must spend one move at either a,b,c or d in order to block Red on the right side, while Red 2 is always guaranteed to connect down. Thus we have this forcing sequence:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6 R 4:i3 B 5:i4 R 6:h4 B 7:h5 R 8:f5
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

(Assume that Blue 3 is played at either a,b,c or d, and there were no extra turns in the blue area.)

=== Intrusion at d ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:h5"
/>

Red may go here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:h3 B 1:h5"
/>

Details to follow. See more details [[Template_VI1/Intrusion_on_the_3rd_row|here]].

=== Intrusion at e ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i4"
/>

Red should move here (or the equivalent mirror-image move at "+"):

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 2:h3 j2 B 1:i4 E +:k3"
/>

Now the shaded area is a [[ladder creation template]], giving Red at least a 3rd row ladder as indicated.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) E arrow(3):(h5 h6 h7)"
/>

Red can escape both 2nd and 3rd row ladders using a [[ladder escape fork]] via "+". Specifically, Red escapes a third row ladder like this, and is connected by a [[ziggurat]] and double threat at "+":

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h6 R 3:j5 E +:i5"
/>

If Blue [[ladder handling|yields]], or Red starts out with a 2nd row ladder, the escape fork works anyway:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h7 R 3:h6 B 4:g7 R 5:j6 B 6:i6 R 7:j5 E +:i5"
/>

=== Intrusion at f ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i3"
/>

First establish a [[parallel ladder]] on the right.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 8:i4 j2 2:j3 6:j5 4:k4 B 9:h5 1:i3 3:i5 7:i7 5:k5"
/>

Then use [[Tom's move]]:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 12:f4 10:f5 14:h3 i4 j2 j3 j5 k4 B 11:f6 13:g5 h5 i3 i5 i7 k5"
/>

There are two marginal cases. If Blue 3 blocks at a or b, then Red 4 can reduce the situation to "Intrusion at a or b", as intruding the vertical bridge is irrelevant in this two cases.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:j3 4:i4 B 1:i3 E a:i7 b:g7"
/>

[[category:edge templates]]
[[category:theory]]

Edge template VI1a

2023-11-19T01:18:30Z

Bobson8: completing intrusion at f

Template VI1-a is a 6th row [[edge template]] with one stone.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2"
/>

This template is the first one stone 6th row [[edge template|template]] for which a proof of validity has been written out. The template has been verified by computer, and also verified to be minimal.

== Elimination of irrelevant Blue moves ==

Red has a number of direct threats to connect, using smaller templates. Blue must play in the carrier of these threats in order to counter them. To prevent Red from connecting Blue must play in the intersection of Red's threats carriers.

=== [[Edge template IV1a]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h6 h7 i3 i5 i6 i7 j3 j5 j6 j7"
/>

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 e7 f6 f7 g5 g6 g7 h5 h6 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== [[Edge template IV1b]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== Using [[Tom's move]] ===

6 intrusions can furthermore be discarded thanks to [[Tom's move]], also known as the [[parallel ladder]] trick. Of course, symmetry will cut our work in half!

If Blue moves in any of the cells marked "1" on the left (and, using mirror symmetry, in the corresponding 3 cells on the right), Red can respond as follows:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 4:h5 6:h6 2:i4 j2 B 5:g7 6:h7 3:i5 B 1:(e7 f6 g5)"
/>

At this point, Red can use [[Tom's move]] to connect:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h5 h6 i4 4:i6 j2 8:k3 2:k5 6:l4 B g7 h7 i5 5:i7 7:j5 3:j6 B 1:(e7 f6 g5)"
/>

=== Remaining intrusions ===

The only possible remaining intrusions for Blue are the following:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2
S f7 g6 g7 h5 h7 i3 i4 i5 i6 i7 j3
E a:f7 b:g7 c:g6 d:h5 e:i4 f:i3"
/>
By symmetry, if is sufficient to consider the six possible intrusions at a – f.

== Specific defense ==

For the moves that intersect all the carriers, Red has to find specific answers. Let's deal with the remaining intrusions!

=== Intrusion at a ===

If Blue intrudes at a, Red has several winning responses. For example, Red can play at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 E x:i5 y:h7"
/>
Apart from intrusion into the bridge, which Red defends, Blue's only possible moves are at x and y. If Blue plays at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]].
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:i5 R 4:g5 B 5:e6 R 6:g6 B 7:g7 R 8:h6 B 9:h7 R 10:k5"
/>
If Blue plays at y, Red has the following simple win, using the [[trapezoid]] template:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:h7 R 4:i5 B 5:j6 R 6:g5 B 7:e6 R 8:g7"
/>

=== Intrusion at b (stub) ===

If Blue intrudes at b, Red can respond at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 E x:i5 y:i6 z:i7 w:h7"
/>
Apart from intrusions into the bridge, which Red defends, Blue has only four possible moves x,y,z,w, because if Blue moves anywhere else, Red connects with either the [[ziggurat]] or [[edge template III1b]].

If Blue intrudes at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]]:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:i5 R 4:h5 B 5:f6 R 6:h6 B 7:h7 R 8:k5"
/>
If Blue intrudes at y or z (both shown simultaneously in the following diagram), Red can set up Tom's move on the opposite side:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:(i6 i7) R 4:h6 B 5:h7 R 8:f5"
/>
Finally, if Blue intrudes at w, Red can connect by the following variant of Tom's move:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:h7 R 4:h5 B 5:f6 R 6:k5"
/>

=== Intrusion at c ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g6"
/>

Red may play here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

Note that if Red plays at c, then in the blue area both Red 2 and c connect down without choice, unless Blue first plays at d. Also, the paths for 2 connecting down would not pass c or d.
Therefore, Blue must spend one move at either a,b,c or d in order to block Red on the right side, while Red 2 is always guaranteed to connect down. Thus we have this forcing sequence:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6 R 4:i3 B 5:i4 R 6:h4 B 7:h5 R 8:f5
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

(Assume that Blue 3 is played at either a,b,c or d, and there were no extra turns in the blue area.)

=== Intrusion at d ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:h5"
/>

Red may go here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:h3 B 1:h5"
/>

Details to follow. See more details [[Template_VI1/Intrusion_on_the_3rd_row|here]].

=== Intrusion at e ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i4"
/>

Red should move here (or the equivalent mirror-image move at "+"):

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 2:h3 j2 B 1:i4 E +:k3"
/>

Now the shaded area is a [[ladder creation template]], giving Red at least a 3rd row ladder as indicated.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) E arrow(3):(h5 h6 h7)"
/>

Red can escape both 2nd and 3rd row ladders using a [[ladder escape fork]] via "+". Specifically, Red escapes a third row ladder like this, and is connected by a [[ziggurat]] and double threat at "+":

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h6 R 3:j5 E +:i5"
/>

If Blue [[ladder handling|yields]], or Red starts out with a 2nd row ladder, the escape fork works anyway:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h7 R 3:h6 B 4:g7 R 5:j6 B 6:i6 R 7:j5 E +:i5"
/>

=== Intrusion at f ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i3"
/>

First establish a [[parallel ladder]] on the right.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 8:i4 j2 2:j3 6:j5 4:k4 B 9:h5 1:i3 3:i5 7:i7 5:k5"
/>

Then use [[Tom's move]]:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 12:f4 10:f5 14:h3 i4 j2 j3 j5 k4 B 11:f6 13:g5 h5 i3 i5 i7 k5"
/>

There are two marginal cases. If Blue 3 blocks at a or b, then Red 4 can reduce the situation to "intrusion at a or b", as intruding the vertical bridge is irrelevant in this two cases.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:j3 4:i4 B 1:i3 E a:i7 b:g7"
/>

[[category:edge templates]]
[[category:theory]]

Seventh row template problem

2023-11-18T10:04:42Z

Bobson8:

Until October 2023, it was an open problem whether there exists a 7th row edge template with a single stone. During October and November 2023, the users Bobson, Comonoid, Mason, and Quasar of the [[Hex forums|Hex Discord]] collaborated on finding such a template, and ended up finding three different ones. Curiously, all of these templates are symmetric, although it is possible and quite likely that asymmetric templates also exist.

== Discovery of the templates ==

In chronological order:

=== Edge template VII-1c ===

<hexboard size="7x21"
coords="none"
edges="bottom"
visible="area(a7,u7,u5,s3,q3,o1,m1,i3,g3,c5)"
contents="R n1"
/>

This template has width 21.

After Eric Demer's discovery of [[edge template VI1b]] on October 6, Bobson noticed that this could likely be turned into a single-stone 7th row template. He found the first such pre-template of arbitrary width using HexProver on October 13. The pre-template was soon reduced to width 23 on October 14, as he realized that using edge template VI1b was not necessary.

To explain how the width 23 pre-template works, notice that Red can try to connect the 7th row stone to the edge with a variant of [[edge template V1b]] from both sides:

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

Bobson continued to reduce the size of this pre-template. He first considered the following width 19 carrier (contained in the width 23 pre-template):

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E *:m4 *:n4 S gray:area(c7,u7,u5,s3,r3,p1,n1,j3,i3,e5)"
/>

MoHex shows that every intrusions other than * are solvable within this carrier. On the other hand, Bobson found that intrusions at * can be solved within width 21, so the pre-template was then reduced to width 21 on October 27. The minimality was checked by a collaborative effort using [[MoHex]] (patched version made by Comonoid and Quasar so that it can be run on a 23×23 board), and was finally verified on November 8.

=== Edge template VII-1a ===

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1"
/>

This template has width 17.

While checking the minimality of [[edge template VII-1c]], Comonoid first noticed on November 4 that [[Strategic advice from KataHex|KataHex]] seemed to indicate that the red stone was connected within this carrier. The validity and minimality of the template was subsequently verified using [[MoHex]], and the minimality proof was finished on November 6.

=== Edge template VII-1b ===

<hexboard size="7x19"
coords="none"
edges="bottom"
visible="area(a7,s7,s5,p2,n1,l1,i2,c5)"
contents="R m1"
/>

This template has width 19.

While checking the minimality of [[edge template VII-1c]], Comonoid first proposed that there might be a width-19 template if the carrier is wide enough. Comonoid, Quasar, and Mason came up with different carriers, and Mason described what turned out to be the final shape of the template on November 4 and proved its validity. Minimality was subsequently verified by a collaboration among several users, and the final case was finished by Quasar on November 17.

== Methodology ==

=== Validity ===

The validity of VII1b and VII1a was checked by Mason using [[MoHex]], in some cases manually supplying the winning move to Mohex. Bobson also checked the validity of VII1c and VII1b using the interactive [[HexProver]] software.

=== Minimality ===

Checking minimality is far more work than checking validity. In principle, one must sequentially remove each cell from the carrier (by placing a blue stone in it) and check that the resulting pattern is not connected, i.e., i.e., is a first-player win for Blue. However, due to symmetries and [[domination]], not all cells need to be checked. Specifically, if x dominates y from Blue's point of view, then if removing x from the carrier kept the template connected for Red, then removing y from the carrier would also keep the template connected. Therefore, if minimality has been confirmed with respect to y, then minimality does not need to be checked with respect to x.

For example, to verify the minimality of [[edge template VII-1a]], only the shaded cells needed to be checked:

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1 S blue:area(l2,k3,i7,o7,n6,o5),n1,p3,q7
E a:m1 b:m2 c:n2 d:o2 e:n3 f:o3 g:o4 h:p4 i:p5 j:q5 k:o6 l:p6 m:q6 n:p7 x:n1 y:p3 z:q7"
/>
Specifically, a capture-dominates x, b kill-dominates x, c capture-dominates x, d fillin-dominates y, e kill-dominates d, f capture-dominates y, g kill-dominates y, h capture-dominates y, and each of i–n capture-dominates z. Therefore, if x, y, and z are necessary in the carrier, then so are a–n. The cells on the left do not need to be checked due to symmetry.

[[Category: Edge templates]]

Seventh row template problem

2023-11-18T09:42:11Z

Bobson8:

Until October 2023, it was an open problem whether there exists a 7th row edge template with a single stone. During October and November 2023, the users Bobson, Comonoid, Mason, and Quasar of the [[Hex forums|Hex Discord]] collaborated on finding such a template, and ended up finding three different ones. Curiously, all of these templates are symmetric, although it is possible and quite likely that asymmetric templates also exist.

== Discovery of the templates ==

In chronological order:

=== Edge template VII-1c ===

<hexboard size="7x21"
coords="none"
edges="bottom"
visible="area(a7,u7,u5,s3,q3,o1,m1,i3,g3,c5)"
contents="R n1"
/>

After Eric Demer's discovery of [[edge template VI1b]] on October 6, Bobson noticed that this could likely be turned into a single-stone 7th row template. He found the first such pre-template of arbitrary width using HexProver on October 13. The pre-template was soon reduced to width 23 on October 14, as he realized that using edge template VI1b was not necessary.

To explain how the width 23 pre-template works, notice that Red can try to connect the 7th row stone to the edge with a variant of [[edge template V1b]] from both sides:

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

Bobson continued to reduce the size of this pre-template. He first considered the following width 19 carrier (contained in the width 23 pre-template):

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E *:m4 *:n4 S gray:area(c7,u7,u5,s3,r3,p1,n1,j3,i3,e5)"
/>

MoHex shows that every intrusions other than * are solvable within this carrier. On the other hand, Bobson found that intrusions at * can be solved within width 21, so the pre-template was then reduced to width 21 on October 27. The minimality was checked by a collaborative effort using [[MoHex]] (patched version made by Comonoid and Quasar so that it can be run on a 23×23 board), and was finally verified on November 8.

=== Edge template VII-1a ===

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1"
/>

While checking the minimality of [[edge template VII-1c]], Comonoid first noticed on November 4 that [[Strategic advice from KataHex|KataHex]] seemed to indicate that the red stone was connected within this carrier. The validity and minimality of the template was subsequently verified using [[MoHex]], and the minimality proof was finished on November 6.

=== Edge template VII-1b ===

<hexboard size="7x19"
coords="none"
edges="bottom"
visible="area(a7,s7,s5,p2,n1,l1,i2,c5)"
contents="R m1"
/>

While checking the minimality of [[edge template VII-1c]], Comonoid first proposed that there might be a width-19 template if the carrier is wide enough. Comonoid, Quasar, and Mason came up with different carriers, and Mason described what turned out to be the final shape of the template on November 4 and proved its validity. Minimality was subsequently verified by a collaboration among several users, and the final case was finished by Quasar on November 17.

== Methodology ==

=== Validity ===

The validity of VII1b and VII1a was checked by Mason using [[MoHex]], in some cases manually supplying the winning move to Mohex. Bobson also checked the validity of VII1c and VII1b using the interactive [[HexProver]] software.

=== Minimality ===

Checking minimality is far more work than checking validity. In principle, one must sequentially remove each cell from the carrier (by placing a blue stone in it) and check that the resulting pattern is not connected, i.e., i.e., is a first-player win for Blue. However, due to symmetries and [[domination]], not all cells need to be checked. Specifically, if x dominates y from Blue's point of view, then if removing x from the carrier kept the template connected for Red, then removing y from the carrier would also keep the template connected. Therefore, if minimality has been confirmed with respect to y, then minimality does not need to be checked with respect to x.

For example, to verify the minimality of [[edge template VII-1a]], only the shaded cells needed to be checked:

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1 S blue:area(l2,k3,i7,o7,n6,o5),n1,p3,q7
E a:m1 b:m2 c:n2 d:o2 e:n3 f:o3 g:o4 h:p4 i:p5 j:q5 k:o6 l:p6 m:q6 n:p7 x:n1 y:p3 z:q7"
/>
Specifically, a capture-dominates x, b kill-dominates x, c capture-dominates x, d fillin-dominates y, e kill-dominates d, f capture-dominates y, g kill-dominates y, h capture-dominates y, and each of i–n capture-dominates z. Therefore, if x, y, and z are necessary in the carrier, then so are a–n. The cells on the left do not need to be checked due to symmetry.

[[Category: Edge templates]]

Seventh row template problem

2023-11-18T08:34:10Z

Bobson8: Added descriptions to VII1c

Until October 2023, it was an open problem whether there exists a 7th row edge template with a single stone. During October and November 2023, the users Bobson, Comonoid, Mason, and Quasar of the [[Hex forums|Hex Discord]] collaborated on finding such a template, and ended up finding three different ones. Curiously, all of these templates are symmetric, although it is possible and quite likely that asymmetric templates also exist.

== Discovery of the templates ==

In chronological order:

=== Edge template VII-1c ===

<hexboard size="7x21"
coords="none"
edges="bottom"
visible="area(a7,u7,u5,s3,q3,o1,m1,i3,g3,c5)"
contents="R n1"
/>

After Eric Demer's discovery of [[edge template VI1b]] on October 6, Bobson noticed that this could likely be turned into a single-stone 7th row template. He found the first such pre-template of arbitrary width using HexProver on October 13. The pre-template was soon reduced to width 23 on October 14, as he noticed that using edge template VI1b was not necessary. To explain how the width 23 pre-template works, notice that Red can try to connect the 7th row stone to the edge with a variant of [[edge template V1b]] from both sides:

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

Bobson continued to reduce the size of this pre-template. He first considered the following width 19 carrier (contained in the width 23 pre-template):

<hexboard size="7x19"
coords="none"
edges="bottom"
visible="area(a7,s7,s5,q3,p3,n1,l1,h3,g3,c5)"
contents="R m1 E *:k4 *:l4"
/>

With hints from MoHex, he solved every intrusions other than * within this carrier using HexProver. On the other hand, he found that intrusions at * can be solved within width 21, so the pre-template was then reduced to width 21 on October 27, and later it turned out to be minimal. The minimality was checked by a collaborative effort using [[MoHex]] (patched version made by Comonoid and Quasar so that it can run 23×23), and was finally verified on November 8.

=== Edge template VII-1a ===

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1"
/>

While checking the minimality of [[edge template VII-1c]], Comonoid first noticed on November 4 that [[Strategic advice from KataHex|KataHex]] seemed to indicate that the red stone was connected within this carrier. The validity and minimality of the template was subsequently verified using [[MoHex]], and the minimality proof was finished on November 6.

=== Edge template VII-1b ===

<hexboard size="7x19"
coords="none"
edges="bottom"
visible="area(a7,s7,s5,p2,n1,l1,i2,c5)"
contents="R m1"
/>

While checking the minimality of [[edge template VII-1c]], Comonoid first proposed that there might be a width-19 template if the carrier is wide enough. Comonoid, Quasar, and Mason came up with different carriers, and Mason described what turned out to be the final shape of the template on November 4 and proved its validity. Minimality was subsequently verified by a collaboration among several users, and the final case was finished by Quasar on November 17.

== Methodology ==

=== Validity ===

The validity of VII1b and VII1a was checked by Mason using [[MoHex]], in some cases manually supplying the winning move to Mohex. Bobson also checked the validity of VII1c and VII1b using the interactive [[HexProver]] software.

=== Minimality ===

Checking minimality is far more work than checking validity. In principle, one must sequentially remove each cell from the carrier (by placing a blue stone in it) and check that the resulting pattern is not connected, i.e., i.e., is a first-player win for Blue. However, due to symmetries and [[domination]], not all cells need to be checked. Specifically, if x dominates y from Blue's point of view, then if removing x from the carrier kept the template connected for Red, then removing y from the carrier would also keep the template connected. Therefore, if minimality has been confirmed with respect to y, then minimality does not need to be checked with respect to x.

For example, to verify the minimality of [[edge template VII-1a]], only the shaded cells needed to be checked:

<hexboard size="7x17"
coords="none"
edges="bottom"
visible="area(a7,q7,q5,p3,n1,j1,f3,c5)"
contents="R l1 S blue:area(l2,k3,i7,o7,n6,o5),n1,p3,q7
E a:m1 b:m2 c:n2 d:o2 e:n3 f:o3 g:o4 h:p4 i:p5 j:q5 k:o6 l:p6 m:q6 n:p7 x:n1 y:p3 z:q7"
/>
Specifically, a capture-dominates x, b kill-dominates x, c capture-dominates x, d fillin-dominates y, e kill-dominates d, f capture-dominates y, g kill-dominates y, h capture-dominates y, and each of i–n capture-dominates z. Therefore, if x, y, and z are necessary in the carrier, then so are a–n. The cells on the left do not need to be checked due to symmetry.

[[Category: Edge templates]]

User:Bobson8

2023-10-26T02:37:46Z

Bobson8: Replacing terms to avoid confusion

<hexboard size="11x11"
coords="hide"
contents="R e4 R h4 R c7 R g7 R c8 R d8 R e8 R f8"
/>

The following pattern guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. Whether this connection is minimal is not known. This partially answers the single-stone 7th row edge template problem (The template exists but the shape is unknown).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1"
/>

To start with, one notice that Red can connect to the edge with a variant of [[edge template V1b]] from both sides.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

User:Bobson8

2023-10-14T10:22:23Z

Bobson8: 7th row single stone template

<hexboard size="11x11"
coords="hide"
contents="R e4 R h4 R c7 R g7 R c8 R d8 R e8 R f8"
/>

The following template guarantees a stone on 7th row to connect to the edge. It was verified by me using HexProver developped by mirefek. This answers the single-stone 7th row edge template problem. Whether this template is minimal is not known.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1"
/>

To start with, one notice that Red can connect to the edge with a variant of [[edge template V1b]] from both sides.

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 1:n3 S blue:area(j7,w7,w5,u3,r3,p1,o1,n2,m4)"
/>

Thus Blue has these possible blocks (a~j, mirrored positions are removed).

<hexboard size="7x23"
coords="none"
edges="bottom"
visible="area(a7,w7,w5,u3,r3,p1,n1,j3,g3,c5)"
contents="R o1 E a:n2 b:n3 c:m4 d:l5 e:m5 f:k6 g:l6 h:j7 i:k7 j:l7 X:o2 Y:m2 Z:n4"
/>

For Blue playing a,b,c,d, Red can respond with X,Y,Y,Z respectively. For the rest Red can respond with b.

Template VI1/Intrusion on the 3rd row

2023-09-20T04:40:41Z

Bobson8:

This article deals with a special case in the defense of [[edge template VI1a]], namely the intrusion on the 3rd that is not eliminated by [[sub-templates threat]]s.

== Basic situation ==

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 E a:h2 b:k2 c:l3"
/>

In this situation, there are only 3 possible winning moves for Red, and they are "a", "b", and "c". Of these, "a" is the easiest to verify, so we will assume Red plays there.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2"
/>

Before continuing the analysis, we first note that Red can escape all 2nd row ladders coming from the left, as follows:
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
R 1:g5 B 2:g6 R 3:h5 B 4:h6 R 5:j5 B 6:i5 R 7:j4 B 8:i4 R 9:k2"
/>
Apart from attacking the bridge, which Red defends, Blue's next move must be in the shaded blue area, or else Red plays at d and connects.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
S blue:area(g3,d6,g6,g4,h3)
E a:g3 b:h3 c:f4 d:g4 e:e5 f:f5 g:g5 h:d6 i:e6 j:f6 k:g6 x:f3"
/>
If Blue plays at c, e, h, j or k, Red responds at d and gets a 2nd row ladder, which connects.
If Blue plays at b, Red plays at x and connects by [[edge template IV1a]].
If Blue plays at d, Red plays at x and gets a 2nd row ladder, which connects.
This leaves a, f, g, i.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 R 1:h3
E a:g3 d:g4 f:f5 g:g5 k:g6 y:i4"
/>
If Blue plays at a, f or g, then Red plays at b.
Note that Red connect down from the left by playing d, and connect down from the right by playing y.
The only way for Blue to block both is to play k, but Red with the same strategy on the left would produce a 2nd row ladder toward the right.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 B e6 R 1:g4
E a:g3 b:h3 y:i4 g:g5 j:f6 z:d5"
/>

Finally if Blue plays at i, then Red plays at d. Apart from the bridges, Blue is forced to play g, and then Red plays b, forcing Blue to block on the right part (against Red y), and then Red wins with z.

If Blue intrudes at a, then Red responds at b, forcing Blue to block on the right part, and then Red wins with j.

If Blue intrudes at b, then Red responds at a. Blue is still forced to play g, and then Red wins with z.

{{stub}}
[[category:edge templates]]

Template VI1/Intrusion on the 3rd row

2023-09-20T04:06:28Z

Bobson8: The last case simplified

This article deals with a special case in the defense of [[edge template VI1a]], namely the intrusion on the 3rd that is not eliminated by [[sub-templates threat]]s.

== Basic situation ==

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 E a:h2 b:k2 c:l3"
/>

In this situation, there are only 3 possible winning moves for Red, and they are "a", "b", and "c". Of these, "a" is the easiest to verify, so we will assume Red plays there.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2"
/>

Before continuing the analysis, we first note that Red can escape all 2nd row ladders coming from the left, as follows:
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
R 1:g5 B 2:g6 R 3:h5 B 4:h6 R 5:j5 B 6:i5 R 7:j4 B 8:i4 R 9:k2"
/>
Apart from attacking the bridge, which Red defends, Blue's next move must be in the shaded blue area, or else Red plays at d and connects.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
S blue:area(g3,d6,g6,g4,h3)
E a:g3 b:h3 c:f4 d:g4 e:e5 f:f5 g:g5 h:d6 i:e6 j:f6 k:g6 x:f3"
/>
If Blue plays at c, e, h, j or k, Red responds at d and gets a 2nd row ladder, which connects.
If Blue plays at b, Red plays at x and connects by [[edge template IV1a]].
If Blue plays at d, Red plays at x and gets a 2nd row ladder, which connects.
This leaves a, f, g, i.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 R 1:h3
E a:g3 d:g4 f:f5 g:g5 k:g6 y:i4"
/>
If Blue plays at a, f or g, then Red plays at b.
Note that Red connect down from the left by playing d, and connect down from the right by playing y.
The only way for Blue to block both is to play k, but Red with the same strategy on the left would produce a 2nd row ladder toward the right.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 B e6 R 1:g4
E a:g3 b:h3 y:i4 f:f5 g:g5 j:f6"
/>

Finally if Blue plays at i, then Red plays at d. Apart from the bridges, Blue is forced to play g, and then Red plays b, forcing Blue to block on the right part (against Red y). Red then wins with f, making a capped flank on the left.

If Blue intrudes at a, then Red responds at b, forcing Blue to block on the right part, and then Red wins with j.

If Blue intrudes at b, then Red responds at a. Blue is still forced to play g, and then Red wins with f.

{{stub}}
[[category:edge templates]]

Template VI1/Intrusion on the 3rd row

2023-09-19T15:49:18Z

Bobson8: Trying to complete the proof. The last case should be simplified.

This article deals with a special case in the defense of [[edge template VI1a]], namely the intrusion on the 3rd that is not eliminated by [[sub-templates threat]]s.

== Basic situation ==

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 E a:h2 b:k2 c:l3"
/>

In this situation, there are only 3 possible winning moves for Red, and they are "a", "b", and "c". Of these, "a" is the easiest to verify, so we will assume Red plays there.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2"
/>

Before continuing the analysis, we first note that Red can escape all 2nd row ladders coming from the left, as follows:
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
R 1:g5 B 2:g6 R 3:h5 B 4:h6 R 5:j5 B 6:i5 R 7:j4 B 8:i4 R 9:k2"
/>
Apart from attacking the bridge, which Red defends, Blue's next move must be in the shaded blue area, or else Red plays at d and connects.
<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2
S blue:area(g3,d6,g6,g4,h3)
E a:g3 b:h3 c:f4 d:g4 e:e5 f:f5 g:g5 h:d6 i:e6 j:f6 k:g6 x:f3"
/>
If Blue plays at c, e, h, j or k, Red responds at d and gets a 2nd row ladder, which connects.
If Blue plays at b, Red plays at x and connects by [[edge template IV1a]].
If Blue plays at d, Red plays at x and gets a 2nd row ladder, which connects.
This leaves a, f, g, i.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 R 1:h3
E a:g3 d:g4 f:f5 g:g5 k:g6 y:i4"
/>
If Blue plays at a,f or g, then Red plays at b.
Note that Red connect down from the left by playing d, and connect down from the right by playing y.
The only way for Blue to block both is to play k, but Red with the same strategy on the left would produce a 2nd row ladder toward the right.

<hexboard size="6x14"
coords="none"
edges="bottom"
visible="area(i1,c4,a6,o6,o4,k1)"
contents="R j1 B h4 R h2 B e6 R 1:g4 B 2:g5 R 3:f5 B 4:f6 R 5:d5 B 6:e5 R 7:e4 B 8:f4 R 9:f3
E a:g3 b:h3 y:i4"
/>

Finally if Blue plays at i, then Red plays at d. Suppose Blue don't intrude the bridge at a or b, then Red has the forcing sequence shown in the picture.

Now if the bridge was intruded by blue at a at some moment before 9, then Red should respond at b, forcing Blue to block on the right part (against Red y) and Red can win with one more move on the left.
If the bridge was intruded at b, then Red should respond at a and can win with a capped flank on the left.
These claims have to be checked on each turn.

{{stub}}
[[category:edge templates]]

Edge template VI1a

2023-09-19T15:21:33Z

Bobson8: Capitalization and fixing a typo

Template VI1-a is a 6th row [[edge template]] with one stone.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2"
/>

This template is the first one stone 6th row [[edge template|template]] for which a proof of validity has been written out. The template has been verified by computer, and also verified to be minimal.

== Elimination of irrelevant Blue moves ==

Red has a number of direct threats to connect, using smaller templates. Blue must play in the carrier of these threats in order to counter them. To prevent Red from connecting Blue must play in the intersection of Red's threats carriers.

=== [[Edge template IV1a]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h6 h7 i3 i5 i6 i7 j3 j5 j6 j7"
/>

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 e7 f6 f7 g5 g6 g7 h5 h6 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== [[Edge template IV1b]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== Using [[Tom's move]] ===

6 intrusions can furthermore be discarded thanks to [[Tom's move]], also known as the [[parallel ladder]] trick. Of course, symmetry will cut our work in half!

If Blue moves in any of the cells marked "1" on the left (and, using mirror symmetry, in the corresponding 3 cells on the right), Red can respond as follows:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 4:h5 6:h6 2:i4 j2 B 5:g7 6:h7 3:i5 B 1:(e7 f6 g5)"
/>

At this point, Red can use [[Tom's move]] to connect:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h5 h6 i4 4:i6 j2 8:k3 2:k5 6:l4 B g7 h7 i5 5:i7 7:j5 3:j6 B 1:(e7 f6 g5)"
/>

=== Remaining intrusions ===

The only possible remaining intrusions for Blue are the following:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2
S f7 g6 g7 h5 h7 i3 i4 i5 i6 i7 j3
E a:f7 b:g7 c:g6 d:h5 e:i4 f:i3"
/>
By symmetry, if is sufficient to consider the six possible intrusions at a – f.

== Specific defense ==

For the moves that intersect all the carriers, Red has to find specific answers. Let's deal with the remaining intrusions!

=== Intrusion at a ===

If Blue intrudes at a, Red has several winning responses. For example, Red can play at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 E x:i5 y:h7"
/>
Apart from intrusion into the bridge, which Red defends, Blue's only possible moves are at x and y. If Blue plays at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]].
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:i5 R 4:g5 B 5:e6 R 6:g6 B 7:g7 R 8:h6 B 9:h7 R 10:k5"
/>
If Blue plays at y, Red has the following simple win, using the [[trapezoid]] template:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:h7 R 4:i5 B 5:j6 R 6:g5 B 7:e6 R 8:g7"
/>

=== Intrusion at b (stub) ===

If Blue intrudes at b, Red can respond at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 E x:i5 y:i6 z:i7 w:h7"
/>
Apart from intrusions into the bridge, which Red defends, Blue has only four possible moves x,y,z,w, because if Blue moves anywhere else, Red connects with either the [[ziggurat]] or [[edge template III1b]].

If Blue intrudes at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]]:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:i5 R 4:h5 B 5:f6 R 6:h6 B 7:h7 R 8:k5"
/>
If Blue intrudes at y or z (both shown simultaneously in the following diagram), Red can set up Tom's move on the opposite side:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:(i6 i7) R 4:h6 B 5:h7 R 8:f5"
/>
Finally, if Blue intrudes at w, Red can connect by the following variant of Tom's move:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:h7 R 4:h5 B 5:f6 R 6:k5"
/>

=== Intrusion at c (stub) ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g6"
/>

Red may play here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

Note that if Red plays at c, then in the blue area both Red 2 and c connect down without choice, unless Blue first plays at d. Also, the paths for 2 connecting down would not pass c or d.
Therefore, Blue must spend one move at either a,b,c or d in order to block Red on the right side, while Red 2 is always guaranteed to connect down. Thus we have this forcing sequence:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6 R 4:i3 B 5:i4 R 6:h4 B 7:h5 R 8:f5
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

(Assume that Blue 3 is played at either a,b,c or d, and there were no extra moves in the blue area.)

=== Intrusion at d (stub) ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:h5"
/>

Red may go here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:h3 B 1:h5"
/>

Details to follow. See more details [[Template_VI1/Intrusion_on_the_3rd_row|here]].

=== Intrusion at e ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i4"
/>

Red should move here (or the equivalent mirror-image move at "+"):

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 2:h3 j2 B 1:i4 E +:k3"
/>

Now the shaded area is a [[ladder creation template]], giving Red at least a 3rd row ladder as indicated.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) E arrow(3):(h5 h6 h7)"
/>

Red can escape both 2nd and 3rd row ladders using a [[ladder escape fork]] via "+". Specifically, Red escapes a third row ladder like this, and is connected by a [[ziggurat]] and double threat at "+":

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h6 R 3:j5 E +:i5"
/>

If Blue [[ladder handling|yields]], or Red starts out with a 2nd row ladder, the escape fork works anyway:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h7 R 3:h6 B 4:g7 R 5:j6 B 6:i6 R 7:j5 E +:i5"
/>

=== Intrusion at f ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i3"
/>

First establish a [[parallel ladder]] on the right.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 8:i4 j2 2:j3 6:j5 4:k4 B 9:h5 1:i3 3:i5 7:i7 5:k5"
/>

Then use [[Tom's move]]:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 12:f4 10:f5 14:h3 i4 j2 j3 j5 k4 B 11:f6 13:g5 h5 i3 i5 i7 k5"
/>

[[category:edge templates]]
[[category:theory]]

Edge template VI1a

2023-09-19T14:09:45Z

Bobson8: A route for intrusion at c, also minor edit on intrusion at d

Template VI1-a is a 6th row [[edge template]] with one stone.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2"
/>

This template is the first one stone 6th row [[edge template|template]] for which a proof of validity has been written out. The template has been verified by computer, and also verified to be minimal.

== Elimination of irrelevant Blue moves ==

Red has a number of direct threats to connect, using smaller templates. Blue must play in the carrier of these threats in order to counter them. To prevent Red from connecting Blue must play in the intersection of Red's threats carriers.

=== [[Edge template IV1a]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h6 h7 i3 i5 i6 i7 j3 j5 j6 j7"
/>

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 e7 f6 f7 g5 g6 g7 h5 h6 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== [[Edge template IV1b]] ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R i4 j2 S i4 d7 e6 e7 f5 f6 f7 g5 g6 g7 h4 h5 h7 i3 i5 i6 i7 j3 j4 j5 j6 j7 k5 k6 k7"
/>

=== Using [[Tom's move]] ===

6 intrusions can furthermore be discarded thanks to [[Tom's move]], also known as the [[parallel ladder]] trick. Of course, symmetry will cut our work in half!

If Blue moves in any of the cells marked "1" on the left (and, using mirror symmetry, in the corresponding 3 cells on the right), Red can respond as follows:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 4:h5 6:h6 2:i4 j2 B 5:g7 6:h7 3:i5 B 1:(e7 f6 g5)"
/>

At this point, Red can use [[Tom's move]] to connect:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h5 h6 i4 4:i6 j2 8:k3 2:k5 6:l4 B g7 h7 i5 5:i7 7:j5 3:j6 B 1:(e7 f6 g5)"
/>

=== Remaining intrusions ===

The only possible remaining intrusions for Blue are the following:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2
S f7 g6 g7 h5 h7 i3 i4 i5 i6 i7 j3
E a:f7 b:g7 c:g6 d:h5 e:i4 f:i3"
/>
By symmetry, if is sufficient to consider the six possible intrusions at a – f.

== Specific defense ==

For the moves that intersect all the carriers, Red has to find specific answers. Let's deal with the remaining intrusions!

=== Intrusion at a ===

If Blue intrudes at a, Red has several winning responses. For example, White can play at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 E x:i5 y:h7"
/>
Apart from intrusion into the bridge, which Red defends, Blue's only possible moves are at x and y. If Blue plays at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]].
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:i5 R 4:g5 B 5:e6 R 6:g6 B 7:g7 R 8:h6 B 9:h7 R 10:k5"
/>
If Blue plays at y, Red has the following simple win, using the [[trapezoid]] template:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:f7 R 2:i4 B 3:h7 R 4:i5 B 5:j6 R 6:g5 B 7:e6 R 8:g7"
/>

=== Intrusion at b (stub) ===

If Blue intrudes at b, Red can respond at 2:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 E x:i5 y:i6 z:i7 w:h7"
/>
Apart from intrusions into the bridge, which Red defends, Blue has only four possible moves x,y,z,w, because if Blue moves anywhere else, Red connects with either the [[ziggurat]] or [[edge template III1b]].

If Blue intrudes at x, Red can set up a [[parallel ladder]] and connect using [[Tom's move]]:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:i5 R 4:h5 B 5:f6 R 6:h6 B 7:h7 R 8:k5"
/>
If Blue intrudes at y or z (both shown simultaneously in the following diagram), Red can set up Tom's move on the opposite side:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:(i6 i7) R 4:h6 B 5:h7 R 8:f5"
/>
Finally, if Blue intrudes at w, Red can connect by the following variant of Tom's move:
<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g7 R 2:i4 B 3:h7 R 4:h5 B 5:f6 R 6:k5"
/>

=== Intrusion at c (stub) ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:g6"
/>

Red may play here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

Note that if red plays at c, then in the blue area both red 2 and c connect down without choice, unless blue first plays at d. Also, the paths for red 2 connecting down would not pass c or d.
Therefore, blue must spend one move at either a,b,c or d in order to block red on the right side, while red 2 is always guaranteed to connect down. Thus we have this forcing sequence:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:i5 B 1:g6 R 4:i3 B 5:i4 R 6:h4 B 7:h5 R 8:f5
S blue:area(g7 m7 m5 l5 l3 k3)
E a:k2 b:j3 c:k3 d:j4"
/>

(Assume that blue 3 was played at either a,b,c or d, and there were no extra moves in the blue area.)

=== Intrusion at d (stub) ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:h5"
/>

Red may go here:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 2:h3 B 1:h5"
/>

Details to follow. See more details [[Template_VI1/Intrusion_on_the_3rd_row|here]].

=== Intrusion at e ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i4"
/>

Red should move here (or the equivalent mirror-image move at "+"):

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 2:h3 j2 B 1:i4 E +:k3"
/>

Now the shaded area is a [[ladder creation template]], giving Red at least a 3rd row ladder as indicated.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) E arrow(3):(h5 h6 h7)"
/>

Red can escape both 2nd and 3rd row ladders using a [[ladder escape fork]] via "+". Specifically, Red escapes a third row ladder like this, and is connected by a [[ziggurat]] and double threat at "+":

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h6 R 3:j5 E +:i5"
/>

If Blue [[ladder handling|yields]], or Red starts out with a 2nd row ladder, the escape fork works anyway:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R h3 j2 B i4 E +:k3 S area(h3,g3,e4,c5,a7,h7) R 1:h5 B 2:h7 R 3:h6 B 4:g7 R 5:j6 B 6:i6 R 7:j5 E +:i5"
/>

=== Intrusion at f ===

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R j2 B 1:i3"
/>

First establish a [[parallel ladder]] on the right.

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 8:i4 j2 2:j3 6:j5 4:k4 B 9:h5 1:i3 3:i5 7:i7 5:k5"
/>

Then use [[Tom's move]]:

<hexboard size="7x14"
coords="none"
edges="bottom"
visible="area(a7,n7,n5,k2,i2,c5)"
contents="R 12:f4 10:f5 14:h3 i4 j2 j3 j5 k4 B 11:f6 13:g5 h5 i3 i5 i7 k5"
/>

[[category:edge templates]]
[[category:theory]]

User:Bobson8

2023-01-21T01:01:12Z

Bobson8: Test