HexWiki - User contributions [en]

Puzzles

2021-02-20T21:24:31Z

Tompo1: /* Puzzle 6 */ Fixed position orientation (incorrect edges) and added a link and a comment

Solving puzzles is a very good way of becoming a stronger player. Solve as many as possible! And feel free to post your own puzzles here.

== [[Piet Hein]]'s puzzles ==

See article [[Piet Hein's puzzles]]

== [[Claude Berge]]'s puzzles ==

See article [[Claude Berge's puzzles]]

== Bert Enderton ==

=== Puzzle 1 ===

Red to play and win.

<hex>R7 C7 Q1
Hc3 Hf3
Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

Red to play and win.

<hex>R7 C7 Q1
Hc3
Hf5
Vb6</hex>

=== Puzzle 3 ===

Red to play and win.

This is a very difficult puzzle whose complete solution is extremely complex.

<hexboard size="6x6"
contents="B c2 R f4"
/>

=== Puzzle 4 ===

Red to play and win.

<hexboard size="7x7"
xcontents="B b4 c4 d4 R b3 b6"
contents="B f4 e4 d4 R f5 f2"
/>

== Other authors ==

=== Puzzle 1 ===
By [[John Tromp]]

Blue to play and win.

<hex> R10 C10 Q1
Ri1
Bg4 Rh4
Bf5 Rg5
Be6 Rf6
Bd7
Rc8
Rd9 </hex>

=== Puzzle 2 ===
By ''lazyplayer''. Blue to play and win.

<hex> R7 C7 Q1
Vb2
He3
Vb5
</hex>

=== Puzzle 3 ===
By [[David J Bush]]. Taken from a game on [[Playsite]] in 2003. Red to play and win.

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9"
/>

Source: this [[Little Golem]] forum [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=14 thread].

=== Puzzle 4 ===
Designed by ''Door1'', helped by ''David J Bush''. Inspired by a game on [[Kurnik]] in May 2005. Blue to play and win.

<hexboard size="10x10"
coords="show"
contents="
R g3 b4 e4 c6 d7 d8
B b2 g2 f3 f4 e5"
/>

Source: this [[Little Golem]] forum [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=212 thread].

=== Puzzle 5 ===
Designed by ''Arek Kulczycki''. Game where both players play very close to perfect. Move 7.b9 was the first mistake, however it was very hard to refute it. 
Blue to move and win (there is just 1 winning sequence).

<hexboard size="11x11"
coords="show"
contents="B 4:h2 R 1:a4 R 5:b5 R 3:c6 B 6:e7 B 2:g8 R 7:b9"
/>

=== Puzzle 6 ===
From Ryan B. Hayward, "A puzzling Hex primer" (https://webdocs.cs.ualberta.ca/~hayward/papers/puzzlingHexPrimer.pdf). Red to play and win.

<hexboard size="6x6"
coords="show"
contents="R a4 a6 B a5 b5"
/>

Note that this position is [[Equivalent patterns|equivalent]] to the position with the pieces at a5 and a6 removed. This could arise in response to the [[Small_boards#Winner_depending_on_the_first_move|winning opening move a4]].

== See also ==

* [[Solutions to puzzles]]
* [[Ladder puzzles]]
* [[Puzzle server]]

[[Cameron Browne]] offers a lot of original puzzles in his book [[Hex Strategy Making the Right Connections]]

[[Matthew Seymour]] has created a website with 500 interactive Hex puzzles at http://www.mseymour.ca/hex_puzzle/hexpuzzle.html

[[Ryan Hayward]] and [[Bjarne Toft]] include several sets of puzzles in their book [[Hex: The Full Story]], including 49 puzzles originally published in Politiken, 28 unpublished puzzles by Jens Lindhard, 99 puzzles by Henderson, and the 4 puzzles by Bert Enderton.

[[category:Puzzle]]

Multiple threat

2021-02-19T17:42:57Z

Tompo1: /* Overlapping threats */ typo

A '''multiple threat''' is when a player threatens to connect in two or more different ways. Ideally, the opponent cannot defend against all of them simultaneously; or at least the presence of multiple threats severely constrains the opponent's options.

Whenever possible, a player should make each move achieve at least two different goals. Moves that contain only a single threat are generally not hard to meet. Moves that contain multiple threats are more difficult, and sometimes impossible, to stop.

It is sometimes called fork in Hex and other games, and called [[miai]] in go.

== Connection by double threat ==

The most common example of a multiple treat is when two of a player's groups are connected to each other by a double threat. In the following position, Red's two groups are connected by double threat at the two cells marked "*".
<hexboard size="4x5"
coords="none"
edges="none"
contents="R a3 b2 c4 d3 e2 B c3 E *:b4 *:d1"
/>
If Blue moves at one of the cells marked "*" or its neighbors, Red can respond at the other one, thus guaranteeing a connection between the two groups.

== Overlapping threats ==

In the following situation, Red can connect the two groups by moving at any one of the cells marked "*".
<hexboard size="4x5"
coords="none"
edges="none"
contents="R d1 b4 c4 B b2 e2 E *:c2 *:d2 *:c3"
/>
However, these threats are '''overlapping'''; each of the three threatened connections passes through the cell c3. Therefore, by moving at c3, Blue can neutralize all three threats at the same time, denying Red the connection.
<hexboard size="4x5"
coords="none"
edges="none"
contents="R d1 b4 c4 B b2 e2 B c3"
/>

== Ladder escape forks ==

A [[ladder escape fork]] is a typical example of a multiple threat.

[[category:Basic Strategy]]
[[category:Definition]]
{{stub}}

Solutions to puzzles

2021-02-19T11:35:21Z

Tompo1: /* Puzzle 4 */ Additional explanation - perhaps this could be generalised?

== Piet Hein ==
See [[Solutions to Piet Hein's puzzles]]
== Claude Berge ==
See [[Solutions to Claude Berge's puzzles]]
== Bert Enderton ==
=== Puzzle 1 ===

The unique winning first move is Red b4.
(Blue can easily defend the bridges at e3/e4 and g2/g3. Red d3 is defeated by blue b5, red b5 is defeated by blue d3, and all other red moves are defeated by blue b4).

<hex>R7 C7 Q1
Sd1
Sb3 Hc3 Sd3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

Notice that the red stone at a6 is already connected south via the ladder escape at e5. Also, red threatens to connect north via a double threat at b3 and d3; each of these would connect to the northern edge via [[Template IIIa]]. The two templates overlap at d1, but blue d1 is defeated via the following sequence of forcing moves:

<hex>R7 C7 Q1
H2d1
H4b2 V7c2 V5e2
V3b3 Hc3 H6d3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

A winning move is Red e3. This connects to the top edge via [[Template IIIa]], and to the bottom edge via [[Edge template J5|Template J5]]. The two templates overlap at one point d3, but if Blue plays there, Red can reply f3 to seal the connection up and down.

<hexboard size="7x7"
coords="show"
contents="B c3 E *:d3 R 1:e3 B f5 R b6"
/>

=== Puzzle 3 ===

The unique winning first move is Red c3!
(e2 is defeated by e3, d3 is defeated by e1, e3 is defeated by e2, and b4 is defeated by d3).

<hex>R6 C6 Q1 Vf4 Hc2 V1c3</hex>

The following seem like horizontal's (Blue's) best tries from the above position.

{| cellspacing="0" style="width: 500px"
| 1. d2 || e2 || 2. d5 || c5 || 3. d4 || b3 || 4. c4 || a5
|-
| 1. d3 || b3 || 2. b5 || e3
|-
| 1. e1 || d2 || 2. d1 || d1 || 3. c4 || b3 || 4. e3 || a5 || or || 4. b5 || e3
|-
| ........ || .... || 2. c4 || b3 || 3. c5 || e3 || 4. e4 || f3 || 5. e6 || d5
|-
| 1. d5 || b3 || 2. d2 || b2 || 3. c4 || a5 || 4. a6 || c5 || 5. b5 || e3 || 6. d5 || f5
|-
| ........ || .... || ........ || .... || 3. b5 || c5 || 4. c4 || e3
|-
| ........ || .... || ........ || .... || 3. b6 || a6 || 4. b4 || c5 || 5. c4 || e3 || 6. e6 || d4
|-
| 1. c4 || e3 || 2. e2 || b3 || 3. d3 || a5
|-
| 1. b5 || d4 || 2. d3 || f2 || 3. f1 || d2 || 4. c4 || e2 || 5. e4 || e3 || 6. c5 || e5
|-
| ........ || .... || 2. e1 || d2 || 3. d1 || f1 || 4. e2 || f2...
|-
| 1. b4 || d2 || 2. d5 || c5 || 3. c4 || e3 || 4. d4 || e5
|}

=== Puzzle 4 ===

<hexboard size="7x7"
contents="B f4 e4 d4 R f5 f2 1:c4"
/>

Red 1. c4 is the only way to prevent Blue from connecting immediately by b5, c3 or c4. Red threatens to connect c4 to the top and bottom edges with assistance from the [[ladder escape]]s at f2 and f5.

Blue 2. c5 is met by Red 3. a6, which connects to the bottom edge with assistance from the ladder escape at f5 and threatens to connect to c4 or [[climbing|climb]] to a4 and then connect to the top edge with assistance from the ladder escape at f2.

Blue may attempt a [[ladder escape fork]] by playing in the [[edge template]] between c4 and the top edge, but 2. c3 is met by 3. b4 and 2. c2 is met by 3. b3, which maintain the connection to the top edge and block Blue's ladder escape.

== Other authors ==

=== Puzzle 1 ===
Not posted yet...

=== Puzzle 2 ===
Note that Red is connected upwards by [[Edge_template_J5|Template J5]]. The only way to prevent Red from connecting downwards is to play in the cell marked *.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 E *:b6"
/>
Therefore, the only possible move for Blue is 1.b6. Perhaps surprisingly, it is also a winning move.
If Red responds as expected at 2.c5, Blue's response is to intrude Red's upward connection at 3.b4. This leaves Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:c5 B 3:b4"
/>
If Red instead plays 2.g4, Blue can respond with 3.d5, again leaving Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:g4 B 3:d5"
/>,
Other possible moves by Red also have a strong response, for example 2.c6 3.c4, or 2.d4 3.d5, or 2.d5 3.c4, or 2.e4 3.e5.

=== Puzzle 3 ===

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9
R 1:e9 3:f8 5:g7 7:i5 9:j6
B 2:e8 4:f7 6:g6 8:h6"
/>

=== Puzzle 4 ===

<hexboard size="10x10"
coords="show"
contents="
R g3 b4 e4 c6 d7 d8
B b2 g2 f3 f4 e5
B 1:c4
E *:line(g4,h4,j2,j3)
"
/>

Blue 1. c4 threatens to connect to the left edge by b5 or c3 and to the group of 4 Blue pieces by d5, d4 or e3. If Red replies in the region a2-c2-c4-a6, other than at c3 to threaten c4's connection to the left edge, then Blue 3. c3 connects to the left edge, and to the group of 4 pieces via e2 or d5. The only reply outside the region a2-c2-c4-a6 that prevents the group of 4 Blue pieces connecting immediately to the left edge by d5, d4 or e3 is Red 2. d4. However, Red 2. d4 and 2. c3 can be answered by 3. e2.

Blue also threatens to connect the group of 4 pieces to the right edge by 3. g5, connecting via [[Edge template IV1d]] and by the [[ladder escape fork]] 3. i1 4. j1 5. i3 6. i2 7. g4. Red can only meet both these threats by playing in overlap of these templates, in the marked cells. However these replies can be met by:

2. i3 3. i2 4. h2 5. g4 6. h3 7. h5

2. j3 3. i2 4. h2 5. g4 6. h3 7. h4 8. i3 9. i5

2. g4 3. i1 4. j1 5. i2 6. j2 7. i3 8. j3 9. i5 10. i4 11. h5 12. h4 13. f6

2. h4 then either 3. g4 threatening 5. i2 and 5. g6 - or simply 3. i2 4. h2 5. g4 6. h3 7. g6 - forming a [[crescent]] and connecting with the assistance of the second row [[ladder escape]] at i2

2. j2 3. h2 4. i2 ([[Dominated_cell#Capture-domination|capture-dominates]] the alternatives i1 & j1) 5. h5 - a double [[Ziggurat]] connection

[Note that although we used [[Edge template IV1d]] from g5 in the above explanation to minimise the overlap and the resulting case analysis, we don't need the whole template or the method of connection used for that. If however we use the smaller and simpler [[Edge template IV1a]] from g5, then the overlap with the ladder escape fork template is larger and we need to consider the additional replies h3, i2 and j1 (although j1 is [[Dominated_cell#Capture-domination|capture-dominated]] by i2, so any winning variations against i2 also work against j1). Although 3. g5, connecting via Edge template IV1d, works against all these replies, there are simpler and quicker responses, e.g.: 2. h3 3. i4; 2. i2 3. h3 4. i3 5. h5 or 4.j3 5. i5; 2. j1 3. i3. So the analysis above only demonstrates that Blue is connected, rather than showing the smallest connection template or the quickest or simplest method of connection.]

=== Puzzle 5 ===
part 1:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 B 3:d7 B e7 R 4:b8 B g8 R b9 R 2:c9 R 8:e9 R 10:f9 B 1:b10 R 6:c10 B 9:e10 B 7:b11 B 5:c11"
/>
part 2:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 R 10:j6 B d7 B e7 B 9:i7 R 6:j7 R b8 B g8 B 7:i8 R 8:j8 R b9 R c9 R e9 R f9 R 2:g9 B 5:h9 R 4:i9 B b10 R c10 B e10 B 1:f10 B 3:g10 B b11 B c11"
/>
part 3:
<hexboard size="11x11"
coords="show"
contents="B h2 R 2:j2 R a4 R b5 B 1:g5 R c6 R j6 B d7 B e7 B 3:f7 B i7 R j7 R b8 B g8 B i8 R j8 R b9 R c9 R e9 R f9 R g9 B h9 R i9 B b10 R c10 B e10 B f10 B g10 B b11 B c11"
/>

== See Also ==

[[Puzzles|Back to puzzles page]]

[[category:Puzzle]]
{{stub}}

Solutions to puzzles

2021-02-15T08:56:12Z

Tompo1: /* Puzzle 4 */ another simpler instructive variation

== Piet Hein ==
See [[Solutions to Piet Hein's puzzles]]
== Claude Berge ==
See [[Solutions to Claude Berge's puzzles]]
== Bert Enderton ==
=== Puzzle 1 ===

The unique winning first move is Red b4.
(Blue can easily defend the bridges at e3/e4 and g2/g3. Red d3 is defeated by blue b5, red b5 is defeated by blue d3, and all other red moves are defeated by blue b4).

<hex>R7 C7 Q1
Sd1
Sb3 Hc3 Sd3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

Notice that the red stone at a6 is already connected south via the ladder escape at e5. Also, red threatens to connect north via a double threat at b3 and d3; each of these would connect to the northern edge via [[Template IIIa]]. The two templates overlap at d1, but blue d1 is defeated via the following sequence of forcing moves:

<hex>R7 C7 Q1
H2d1
H4b2 V7c2 V5e2
V3b3 Hc3 H6d3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

A winning move is Red e3. This connects to the top edge via [[Template IIIa]], and to the bottom edge via [[Edge template J5|Template J5]]. The two templates overlap at one point d3, but if Blue plays there, Red can reply f3 to seal the connection up and down.

<hexboard size="7x7"
coords="show"
contents="B c3 E *:d3 R 1:e3 B f5 R b6"
/>

=== Puzzle 3 ===

The unique winning first move is Red c3!
(e2 is defeated by e3, d3 is defeated by e1, e3 is defeated by e2, and b4 is defeated by d3).

<hex>R6 C6 Q1 Vf4 Hc2 V1c3</hex>

The following seem like horizontal's (Blue's) best tries from the above position.

{| cellspacing="0" style="width: 500px"
| 1. d2 || e2 || 2. d5 || c5 || 3. d4 || b3 || 4. c4 || a5
|-
| 1. d3 || b3 || 2. b5 || e3
|-
| 1. e1 || d2 || 2. d1 || d1 || 3. c4 || b3 || 4. e3 || a5 || or || 4. b5 || e3
|-
| ........ || .... || 2. c4 || b3 || 3. c5 || e3 || 4. e4 || f3 || 5. e6 || d5
|-
| 1. d5 || b3 || 2. d2 || b2 || 3. c4 || a5 || 4. a6 || c5 || 5. b5 || e3 || 6. d5 || f5
|-
| ........ || .... || ........ || .... || 3. b5 || c5 || 4. c4 || e3
|-
| ........ || .... || ........ || .... || 3. b6 || a6 || 4. b4 || c5 || 5. c4 || e3 || 6. e6 || d4
|-
| 1. c4 || e3 || 2. e2 || b3 || 3. d3 || a5
|-
| 1. b5 || d4 || 2. d3 || f2 || 3. f1 || d2 || 4. c4 || e2 || 5. e4 || e3 || 6. c5 || e5
|-
| ........ || .... || 2. e1 || d2 || 3. d1 || f1 || 4. e2 || f2...
|-
| 1. b4 || d2 || 2. d5 || c5 || 3. c4 || e3 || 4. d4 || e5
|}

=== Puzzle 4 ===

<hexboard size="7x7"
contents="B f4 e4 d4 R f5 f2 1:c4"
/>

Red 1. c4 is the only way to prevent Blue from connecting immediately by b5, c3 or c4. Red threatens to connect c4 to the top and bottom edges with assistance from the [[ladder escape]]s at f2 and f5.

Blue 2. c5 is met by Red 3. a6, which connects to the bottom edge with assistance from the ladder escape at f5 and threatens to connect to c4 or [[climbing|climb]] to a4 and then connect to the top edge with assistance from the ladder escape at f2.

Blue may attempt a [[ladder escape fork]] by playing in the [[edge template]] between c4 and the top edge, but 2. c3 is met by 3. b4 and 2. c2 is met by 3. b3, which maintain the connection to the top edge and block Blue's ladder escape.

== Other authors ==

=== Puzzle 1 ===
Not posted yet...

=== Puzzle 2 ===
Note that Red is connected upwards by [[Edge_template_J5|Template J5]]. The only way to prevent Red from connecting downwards is to play in the cell marked *.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 E *:b6"
/>
Therefore, the only possible move for Blue is 1.b6. Perhaps surprisingly, it is also a winning move.
If Red responds as expected at 2.c5, Blue's response is to intrude Red's upward connection at 3.b4. This leaves Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:c5 B 3:b4"
/>
If Red instead plays 2.g4, Blue can respond with 3.d5, again leaving Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:g4 B 3:d5"
/>,
Other possible moves by Red also have a strong response, for example 2.c6 3.c4, or 2.d4 3.d5, or 2.d5 3.c4, or 2.e4 3.e5.

=== Puzzle 3 ===

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9
R 1:e9 3:f8 5:g7 7:i5 9:j6
B 2:e8 4:f7 6:g6 8:h6"
/>

=== Puzzle 4 ===

<hexboard size="10x10"
coords="show"
contents="
R g3 b4 e4 c6 d7 d8
B b2 g2 f3 f4 e5
B 1:c4
E *:line(g4,h4,j2,j3)
"
/>

Blue 1. c4 threatens to connect to the left edge by b5 or c3 and to the group of 4 Blue pieces by d5, d4 or e3. If Red replies in the region a2-c2-c4-a6, other than at c3 to threaten c4's connection to the left edge, then Blue 3. c3 connects to the left edge, and to the group of 4 pieces via e2 or d5. The only reply outside the region a2-c2-c4-a6 that prevents the group of 4 Blue pieces connecting immediately to the left edge by d5, d4 or e3 is Red 2. d4. However, Red 2. d4 and 2. c3 can be answered by 3. e2.

Blue also threatens to connect the group of 4 pieces to the right edge by 3. g5, connecting via [[Edge template IV1d]] and by the [[ladder escape fork]] 3. i1 4. j1 5. i3 6. i2 7. g4. Red can only meet both these threats by playing in overlap of these templates, in the marked cells. However these replies can be met by:

2. i3 3. i2 4. h2 5. g4 6. h3 7. h5

2. j3 3. i2 4. h2 5. g4 6. h3 7. h4 8. i3 9. i5

2. g4 3. i1 4. j1 5. i2 6. j2 7. i3 8. j3 9. i5 10. i4 11. h5 12. h4 13. f6

2. h4 then either 3. g4 threatening 5. i2 and 5. g6 - or simply 3. i2 4. h2 5. g4 6. h3 7. g6 - forming a [[crescent]] and connecting with the assistance of the second row [[ladder escape]] at i2

2. j2 3. h2 4. i2 ([[Dominated_cell#Capture-domination|capture dominates]] the alternatives i1 & j1) 5. h5 - a double [[Ziggurat]] connection

=== Puzzle 5 ===
part 1:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 B 3:d7 B e7 R 4:b8 B g8 R b9 R 2:c9 R 8:e9 R 10:f9 B 1:b10 R 6:c10 B 9:e10 B 7:b11 B 5:c11"
/>
part 2:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 R 10:j6 B d7 B e7 B 9:i7 R 6:j7 R b8 B g8 B 7:i8 R 8:j8 R b9 R c9 R e9 R f9 R 2:g9 B 5:h9 R 4:i9 B b10 R c10 B e10 B 1:f10 B 3:g10 B b11 B c11"
/>
part 3:
<hexboard size="11x11"
coords="show"
contents="B h2 R 2:j2 R a4 R b5 B 1:g5 R c6 R j6 B d7 B e7 B 3:f7 B i7 R j7 R b8 B g8 B i8 R j8 R b9 R c9 R e9 R f9 R g9 B h9 R i9 B b10 R c10 B e10 B f10 B g10 B b11 B c11"
/>

== See Also ==

[[Puzzles|Back to puzzles page]]

[[category:Puzzle]]
{{stub}}

Solutions to Claude Berge's puzzles

2021-02-14T19:46:43Z

Tompo1: Harmonised the section headings

==Puzzle Solutions==

=== Puzzle 1 ===

<hexboard size="5x5"
contents="B a1 E *:e1 R b2 E +:c2 B d2 R e2 B 1:b3 R c3 B d3 R e3 E +:b4 B c4 R a5 B b5 E *:d5"
/>

Blue plays at b3. This piece is connected to the left border by [[miai|a3/a4]] and to the long blue chain by c2/b4. The long blue chain is connected to the right border by e1/d5.

=== Puzzle 2 ===
This is identical to [[Piet Hein's puzzles#Puzzle 7|Peit Hein's Puzzle 7]]. You may wish to try the other puzzles on that page before consulting the solutions.

=== Puzzle 3 ===
Red threatens to play at 2, 4 and 6.
<hexboard size="14x14"
coords="show"
contents="B d2 B f2 R b3 B d3 R g3 B i3 R a4 B b4 B e4 B f4 R g4 B j4 R a5 B c5 R g5 B l5 B a6 B b6 R c6 B d6 B j6 R c7 B e7 B h7 R c8 B f8 R h8 R j8 R c9 R c10 R c11 R c12 R c13
R 1:d5 3:d4 5:e3 7:f3 B 2:e6 4:c4 6:e2"
/>

=== Puzzle 4 ===
The simplest solution is perhaps the sequence of [[forcing move]]s by Red below, although 1 or 3 at 5 may be quicker. The solutions to puzzles 3 and 4 are examples of [[climbing]].

<hexboard size="14x14"
coords="show"
contents="B k1 R g2 B h2 R l2 R g3 B i3 B k3 R l3 R g4 B i4 R l4 R g5 B j5 R l5 R g6 B h6 B k6 R l6 R g7 B i7 R l7 R g8 B j8 R l8 R g9 B h9 B k9 R l9 R g10 B i10 R l10 R g11 B j11 R m11 B g12 B h12 R i12 B k12 R n12 B e13 B i13 B l13 B a14 B b14 B c14 B d14 B g14 B i14 B l14 B m14 B n14
R 1:f13 3:k13 5:j12 7:i11 9:j10 11:j9 13:i9 15:i8 17:j7 19:j6 21:i6 23:i5 25:j4 27:j3 29:j2 31:i2 33:h3 35:h13 37:g13
B 2:f12 4:l12 6:k11 8:h11 10:k10 12:k8 14:h10 16:h8 18:k7 20:k5 22:h7 24:h5 26:k4 28:k2 30:j1 32:i1 34:j13 36:h14
"
/>

=== Puzzle 5 ===

This puzzle is usually presented as "Blue to play and win", but this is too easy, since Blue can win by playing at k3, leaving the bottom part of the puzzle irrelevant. But this is likely not what Berge had in mind, since in his manuscript "L'Art subtil du Hex", he instead outlines a solution to the bottom part of the puzzle. Berge's intention is a bit difficult to discern, since different drafts of his manuscript contain slightly different versions of this puzzle, and a final definitive version was never published. He also refers to this as a "study" rather than a puzzle. Regardless of what Berge's original intention was, since it is in fact a second move win for Blue, "Red to move and Blue to win" seems like a good interpretation of it.

The puzzle decomposes into two disjoint parts. In order to win, Red would need to connect k4 to the top edge while also connecting l7 to the large red group that is already connected to the bottom edge. However, Red cannot achieve both of these goals simultaneously. If Red plays in the bottom part of the board, Blue can play k3. This leads to a 4th row ladder that Red cannot win. For example:

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R 1:k9 B 2:k3 R 3:j4 B 4:j3"
/>

On the other hand, if Red plays in the top part of the board, Blue can play k8.

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R 1:k3 B 2:k8"
/>

With this move, Blue threatens to immediately connect right via [[Edge template IIIa]]. A typical sequence of follow-up moves is:

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R k3 B k8 R 3:l8 B 4:k9 R 5:l9 B 6:k11 R 7:k10 B 8:j11 E *:i12 *:j10"
/>

Now the blue stone at k11 is safely connected to the right via [[Edge template IV2a]], and is also connected left via the double threat at the hexes marked *.

== See also ==
* Back to [[Claude Berge's puzzles]]
* Main page: [[Puzzles]]

[[category:puzzle]]
{{stub}}

Solutions to Claude Berge's puzzles

2021-02-14T19:39:13Z

Tompo1: /* Puzzle 4 */ Additional explanation with a link

== Puzzle 1 ==

<hexboard size="5x5"
contents="B a1 E *:e1 R b2 E +:c2 B d2 R e2 B 1:b3 R c3 B d3 R e3 E +:b4 B c4 R a5 B b5 E *:d5"
/>

Blue plays at b3. This piece is connected to the left border by [[miai|a3/a4]] and to the long blue chain by c2/b4. The long blue chain is connected to the right border by e1/d5.

=== Puzzle 2 ===
This is identical to [[Piet Hein's puzzles#Puzzle 7|Peit Hein's Puzzle 7]]. You may wish to try the other puzzles on that page before consulting the solutions.

=== Puzzle 3 ===
Red threatens to play at 2, 4 and 6.
<hexboard size="14x14"
coords="show"
contents="B d2 B f2 R b3 B d3 R g3 B i3 R a4 B b4 B e4 B f4 R g4 B j4 R a5 B c5 R g5 B l5 B a6 B b6 R c6 B d6 B j6 R c7 B e7 B h7 R c8 B f8 R h8 R j8 R c9 R c10 R c11 R c12 R c13
R 1:d5 3:d4 5:e3 7:f3 B 2:e6 4:c4 6:e2"
/>

=== Puzzle 4 ===
The simplest solution is perhaps the sequence of [[forcing move]]s by Red below, although 1 or 3 at 5 may be quicker. The solutions to puzzles 3 and 4 are examples of [[climbing]].

<hexboard size="14x14"
coords="show"
contents="B k1 R g2 B h2 R l2 R g3 B i3 B k3 R l3 R g4 B i4 R l4 R g5 B j5 R l5 R g6 B h6 B k6 R l6 R g7 B i7 R l7 R g8 B j8 R l8 R g9 B h9 B k9 R l9 R g10 B i10 R l10 R g11 B j11 R m11 B g12 B h12 R i12 B k12 R n12 B e13 B i13 B l13 B a14 B b14 B c14 B d14 B g14 B i14 B l14 B m14 B n14
R 1:f13 3:k13 5:j12 7:i11 9:j10 11:j9 13:i9 15:i8 17:j7 19:j6 21:i6 23:i5 25:j4 27:j3 29:j2 31:i2 33:h3 35:h13 37:g13
B 2:f12 4:l12 6:k11 8:h11 10:k10 12:k8 14:h10 16:h8 18:k7 20:k5 22:h7 24:h5 26:k4 28:k2 30:j1 32:i1 34:j13 36:h14
"
/>

== Puzzle 5 ==

This puzzle is usually presented as "Blue to play and win", but this is too easy, since Blue can win by playing at k3, leaving the bottom part of the puzzle irrelevant. But this is likely not what Berge had in mind, since in his manuscript "L'Art subtil du Hex", he instead outlines a solution to the bottom part of the puzzle. Berge's intention is a bit difficult to discern, since different drafts of his manuscript contain slightly different versions of this puzzle, and a final definitive version was never published. He also refers to this as a "study" rather than a puzzle. Regardless of what Berge's original intention was, since it is in fact a second move win for Blue, "Red to move and Blue to win" seems like a good interpretation of it.

The puzzle decomposes into two disjoint parts. In order to win, Red would need to connect k4 to the top edge while also connecting l7 to the large red group that is already connected to the bottom edge. However, Red cannot achieve both of these goals simultaneously. If Red plays in the bottom part of the board, Blue can play k3. This leads to a 4th row ladder that Red cannot win. For example:

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R 1:k9 B 2:k3 R 3:j4 B 4:j3"
/>

On the other hand, if Red plays in the top part of the board, Blue can play k8.

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R 1:k3 B 2:k8"
/>

With this move, Blue threatens to immediately connect right via [[Edge template IIIa]]. A typical sequence of follow-up moves is:

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R k3 B k8 R 3:l8 B 4:k9 R 5:l9 B 6:k11 R 7:k10 B 8:j11 E *:i12 *:j10"
/>

Now the blue stone at k11 is safely connected to the right via [[Edge template IV2a]], and is also connected left via the double threat at the hexes marked *.

== See also ==
* Back to [[Claude Berge's puzzles]]
* Main page: [[Puzzles]]

[[category:puzzle]]
{{stub}}

Solutions to puzzles

2021-02-14T15:50:54Z

Tompo1: /* Puzzle 4 */ possibly simpler more instructive variation

== Piet Hein ==
See [[Solutions to Piet Hein's puzzles]]
== Claude Berge ==
See [[Solutions to Claude Berge's puzzles]]
== Bert Enderton ==
=== Puzzle 1 ===

The unique winning first move is Red b4.
(Blue can easily defend the bridges at e3/e4 and g2/g3. Red d3 is defeated by blue b5, red b5 is defeated by blue d3, and all other red moves are defeated by blue b4).

<hex>R7 C7 Q1
Sd1
Sb3 Hc3 Sd3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

Notice that the red stone at a6 is already connected south via the ladder escape at e5. Also, red threatens to connect north via a double threat at b3 and d3; each of these would connect to the northern edge via [[Template IIIa]]. The two templates overlap at d1, but blue d1 is defeated via the following sequence of forcing moves:

<hex>R7 C7 Q1
H2d1
H4b2 V7c2 V5e2
V3b3 Hc3 H6d3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

A winning move is Red e3. This connects to the top edge via [[Template IIIa]], and to the bottom edge via [[Edge template J5|Template J5]]. The two templates overlap at one point d3, but if Blue plays there, Red can reply f3 to seal the connection up and down.

<hexboard size="7x7"
coords="show"
contents="B c3 E *:d3 R 1:e3 B f5 R b6"
/>

=== Puzzle 3 ===

The unique winning first move is Red c3!
(e2 is defeated by e3, d3 is defeated by e1, e3 is defeated by e2, and b4 is defeated by d3).

<hex>R6 C6 Q1 Vf4 Hc2 V1c3</hex>

The following seem like horizontal's (Blue's) best tries from the above position.

{| cellspacing="0" style="width: 500px"
| 1. d2 || e2 || 2. d5 || c5 || 3. d4 || b3 || 4. c4 || a5
|-
| 1. d3 || b3 || 2. b5 || e3
|-
| 1. e1 || d2 || 2. d1 || d1 || 3. c4 || b3 || 4. e3 || a5 || or || 4. b5 || e3
|-
| ........ || .... || 2. c4 || b3 || 3. c5 || e3 || 4. e4 || f3 || 5. e6 || d5
|-
| 1. d5 || b3 || 2. d2 || b2 || 3. c4 || a5 || 4. a6 || c5 || 5. b5 || e3 || 6. d5 || f5
|-
| ........ || .... || ........ || .... || 3. b5 || c5 || 4. c4 || e3
|-
| ........ || .... || ........ || .... || 3. b6 || a6 || 4. b4 || c5 || 5. c4 || e3 || 6. e6 || d4
|-
| 1. c4 || e3 || 2. e2 || b3 || 3. d3 || a5
|-
| 1. b5 || d4 || 2. d3 || f2 || 3. f1 || d2 || 4. c4 || e2 || 5. e4 || e3 || 6. c5 || e5
|-
| ........ || .... || 2. e1 || d2 || 3. d1 || f1 || 4. e2 || f2...
|-
| 1. b4 || d2 || 2. d5 || c5 || 3. c4 || e3 || 4. d4 || e5
|}

=== Puzzle 4 ===

<hexboard size="7x7"
contents="B f4 e4 d4 R f5 f2 1:c4"
/>

Red 1. c4 is the only way to prevent Blue from connecting immediately by b5, c3 or c4. Red threatens to connect c4 to the top and bottom edges with assistance from the [[ladder escape]]s at f2 and f5.

Blue 2. c5 is met by Red 3. a6, which connects to the bottom edge with assistance from the ladder escape at f5 and threatens to connect to c4 or [[climbing|climb]] to a4 and then connect to the top edge with assistance from the ladder escape at f2.

Blue may attempt a [[ladder escape fork]] by playing in the [[edge template]] between c4 and the top edge, but 2. c3 is met by 3. b4 and 2. c2 is met by 3. b3, which maintain the connection to the top edge and block Blue's ladder escape.

== Other authors ==

=== Puzzle 1 ===
Not posted yet...

=== Puzzle 2 ===
Note that Red is connected upwards by [[Edge_template_J5|Template J5]]. The only way to prevent Red from connecting downwards is to play in the cell marked *.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 E *:b6"
/>
Therefore, the only possible move for Blue is 1.b6. Perhaps surprisingly, it is also a winning move.
If Red responds as expected at 2.c5, Blue's response is to intrude Red's upward connection at 3.b4. This leaves Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:c5 B 3:b4"
/>
If Red instead plays 2.g4, Blue can respond with 3.d5, again leaving Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:g4 B 3:d5"
/>,
Other possible moves by Red also have a strong response, for example 2.c6 3.c4, or 2.d4 3.d5, or 2.d5 3.c4, or 2.e4 3.e5.

=== Puzzle 3 ===

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9
R 1:e9 3:f8 5:g7 7:i5 9:j6
B 2:e8 4:f7 6:g6 8:h6"
/>

=== Puzzle 4 ===

<hexboard size="10x10"
coords="show"
contents="
R g3 b4 e4 c6 d7 d8
B b2 g2 f3 f4 e5
B 1:c4
E *:line(g4,h4,j2,j3)
"
/>

Blue 1. c4 threatens to connect to the left edge by b5 or c3 and to the group of 4 Blue pieces by d5, d4 or e3. If Red replies in the region a2-c2-c4-a6, other than at c3 to threaten c4's connection to the left edge, then Blue 3. c3 connects to the left edge, and to the group of 4 pieces via e2 or d5. The only reply outside the region a2-c2-c4-a6 that prevents the group of 4 Blue pieces connecting immediately to the left edge by d5, d4 or e3 is Red 2. d4. However, Red 2. d4 and 2. c3 can be answered by 3. e2.

Blue also threatens to connect the group of 4 pieces to the right edge by 3. g5, connecting via [[Edge template IV1d]] and by the [[ladder escape fork]] 3. i1 4. j1 5. i3 6. i2 7. g4. Red can only meet both these threats by playing in overlap of these templates, in the marked cells. However these replies can be met by:

2. i3 3. i2 4. h2 5. g4 6. h3 7. h5

2. j3 3. i2 4. h2 5. g4 6. h3 7. h4 8. i3 9. i5

2. g4 3. i1 4. j1 5. i2 6. j2 7. i3 8. j3 9. i5 10. i4 11. h5 12. h4 13. f6

2. h4 3. g4 threatening 5. i2 and 5. g6

2. j2 3. h2 4. i2 5. h5 - a double [[Ziggurat]] connection

=== Puzzle 5 ===
part 1:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 B 3:d7 B e7 R 4:b8 B g8 R b9 R 2:c9 R 8:e9 R 10:f9 B 1:b10 R 6:c10 B 9:e10 B 7:b11 B 5:c11"
/>
part 2:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 R 10:j6 B d7 B e7 B 9:i7 R 6:j7 R b8 B g8 B 7:i8 R 8:j8 R b9 R c9 R e9 R f9 R 2:g9 B 5:h9 R 4:i9 B b10 R c10 B e10 B 1:f10 B 3:g10 B b11 B c11"
/>
part 3:
<hexboard size="11x11"
coords="show"
contents="B h2 R 2:j2 R a4 R b5 B 1:g5 R c6 R j6 B d7 B e7 B 3:f7 B i7 R j7 R b8 B g8 B i8 R j8 R b9 R c9 R e9 R f9 R g9 B h9 R i9 B b10 R c10 B e10 B f10 B g10 B b11 B c11"
/>

== See Also ==

[[Puzzles|Back to puzzles page]]

[[category:Puzzle]]
{{stub}}

Solutions to puzzles

2021-02-14T15:28:13Z

Tompo1: /* Other authors */ Added solution to Puzzle 4

== Piet Hein ==
See [[Solutions to Piet Hein's puzzles]]
== Claude Berge ==
See [[Solutions to Claude Berge's puzzles]]
== Bert Enderton ==
=== Puzzle 1 ===

The unique winning first move is Red b4.
(Blue can easily defend the bridges at e3/e4 and g2/g3. Red d3 is defeated by blue b5, red b5 is defeated by blue d3, and all other red moves are defeated by blue b4).

<hex>R7 C7 Q1
Sd1
Sb3 Hc3 Sd3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

Notice that the red stone at a6 is already connected south via the ladder escape at e5. Also, red threatens to connect north via a double threat at b3 and d3; each of these would connect to the northern edge via [[Template IIIa]]. The two templates overlap at d1, but blue d1 is defeated via the following sequence of forcing moves:

<hex>R7 C7 Q1
H2d1
H4b2 V7c2 V5e2
V3b3 Hc3 H6d3 Hf3
V1b4 Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

A winning move is Red e3. This connects to the top edge via [[Template IIIa]], and to the bottom edge via [[Edge template J5|Template J5]]. The two templates overlap at one point d3, but if Blue plays there, Red can reply f3 to seal the connection up and down.

<hexboard size="7x7"
coords="show"
contents="B c3 E *:d3 R 1:e3 B f5 R b6"
/>

=== Puzzle 3 ===

The unique winning first move is Red c3!
(e2 is defeated by e3, d3 is defeated by e1, e3 is defeated by e2, and b4 is defeated by d3).

<hex>R6 C6 Q1 Vf4 Hc2 V1c3</hex>

The following seem like horizontal's (Blue's) best tries from the above position.

{| cellspacing="0" style="width: 500px"
| 1. d2 || e2 || 2. d5 || c5 || 3. d4 || b3 || 4. c4 || a5
|-
| 1. d3 || b3 || 2. b5 || e3
|-
| 1. e1 || d2 || 2. d1 || d1 || 3. c4 || b3 || 4. e3 || a5 || or || 4. b5 || e3
|-
| ........ || .... || 2. c4 || b3 || 3. c5 || e3 || 4. e4 || f3 || 5. e6 || d5
|-
| 1. d5 || b3 || 2. d2 || b2 || 3. c4 || a5 || 4. a6 || c5 || 5. b5 || e3 || 6. d5 || f5
|-
| ........ || .... || ........ || .... || 3. b5 || c5 || 4. c4 || e3
|-
| ........ || .... || ........ || .... || 3. b6 || a6 || 4. b4 || c5 || 5. c4 || e3 || 6. e6 || d4
|-
| 1. c4 || e3 || 2. e2 || b3 || 3. d3 || a5
|-
| 1. b5 || d4 || 2. d3 || f2 || 3. f1 || d2 || 4. c4 || e2 || 5. e4 || e3 || 6. c5 || e5
|-
| ........ || .... || 2. e1 || d2 || 3. d1 || f1 || 4. e2 || f2...
|-
| 1. b4 || d2 || 2. d5 || c5 || 3. c4 || e3 || 4. d4 || e5
|}

=== Puzzle 4 ===

<hexboard size="7x7"
contents="B f4 e4 d4 R f5 f2 1:c4"
/>

Red 1. c4 is the only way to prevent Blue from connecting immediately by b5, c3 or c4. Red threatens to connect c4 to the top and bottom edges with assistance from the [[ladder escape]]s at f2 and f5.

Blue 2. c5 is met by Red 3. a6, which connects to the bottom edge with assistance from the ladder escape at f5 and threatens to connect to c4 or [[climbing|climb]] to a4 and then connect to the top edge with assistance from the ladder escape at f2.

Blue may attempt a [[ladder escape fork]] by playing in the [[edge template]] between c4 and the top edge, but 2. c3 is met by 3. b4 and 2. c2 is met by 3. b3, which maintain the connection to the top edge and block Blue's ladder escape.

== Other authors ==

=== Puzzle 1 ===
Not posted yet...

=== Puzzle 2 ===
Note that Red is connected upwards by [[Edge_template_J5|Template J5]]. The only way to prevent Red from connecting downwards is to play in the cell marked *.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 E *:b6"
/>
Therefore, the only possible move for Blue is 1.b6. Perhaps surprisingly, it is also a winning move.
If Red responds as expected at 2.c5, Blue's response is to intrude Red's upward connection at 3.b4. This leaves Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:c5 B 3:b4"
/>
If Red instead plays 2.g4, Blue can respond with 3.d5, again leaving Blue in a strong position.
<hexboard size="7x7"
coords="show"
contents="R b2 B e3 R b5 B 1:b6 R 2:g4 B 3:d5"
/>,
Other possible moves by Red also have a strong response, for example 2.c6 3.c4, or 2.d4 3.d5, or 2.d5 3.c4, or 2.e4 3.e5.

=== Puzzle 3 ===

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9
R 1:e9 3:f8 5:g7 7:i5 9:j6
B 2:e8 4:f7 6:g6 8:h6"
/>

=== Puzzle 4 ===

<hexboard size="10x10"
coords="show"
contents="
R g3 b4 e4 c6 d7 d8
B b2 g2 f3 f4 e5
B 1:c4
E *:line(g4,h4,j2,j3)
"
/>

Blue 1. c4 threatens to connect to the left edge by b5 or c3 and to the group of 4 Blue pieces by d5, d4 or e3. If Red replies in the region a2-c2-c4-a6, other than at c3 to threaten c4's connection to the left edge, then Blue 3. c3 connects to the left edge, and to the group of 4 pieces via e2 or d5. The only reply outside the region a2-c2-c4-a6 that prevents the group of 4 Blue pieces connecting immediately to the left edge by d5, d4 or e3 is Red 2. d4. However, Red 2. d4 and 2. c3 can be answered by 3. e2.

Blue also threatens to connect the group of 4 pieces to the right edge by 3. g5, connecting via [[Edge template IV1d]] and by the [[ladder escape fork]] 3. i1 4. j1 5. i3 6. i2 7. g4. Red can only meet both these threats by playing in overlap of these templates, in the marked cells. However these replies can be met by:

2. i3 3. i2 4. h2 5. g4 6. h3 7. h5

2. j3 3. i2 4. h2 5. g4 6. h3 7. h4 8. i3 9. i5

2. g4 3. i1 4. j1 5. i2 6. j2 7. i3 8. j3 9. i5 10. i4 11. h5 12. h4 13. f6

2. h4 3. g4 threatening 5. i2 and 5. g6

2. j2 3. h2 4. i2 5. h4. 6. i4 7. h6

=== Puzzle 5 ===
part 1:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 B 3:d7 B e7 R 4:b8 B g8 R b9 R 2:c9 R 8:e9 R 10:f9 B 1:b10 R 6:c10 B 9:e10 B 7:b11 B 5:c11"
/>
part 2:
<hexboard size="11x11"
coords="show"
contents="B h2 R a4 R b5 R c6 R 10:j6 B d7 B e7 B 9:i7 R 6:j7 R b8 B g8 B 7:i8 R 8:j8 R b9 R c9 R e9 R f9 R 2:g9 B 5:h9 R 4:i9 B b10 R c10 B e10 B 1:f10 B 3:g10 B b11 B c11"
/>
part 3:
<hexboard size="11x11"
coords="show"
contents="B h2 R 2:j2 R a4 R b5 B 1:g5 R c6 R j6 B d7 B e7 B 3:f7 B i7 R j7 R b8 B g8 B i8 R j8 R b9 R c9 R e9 R f9 R g9 B h9 R i9 B b10 R c10 B e10 B f10 B g10 B b11 B c11"
/>

== See Also ==

[[Puzzles|Back to puzzles page]]

[[category:Puzzle]]
{{stub}}

Puzzles

2021-02-14T12:43:12Z

Tompo1: /* Puzzle 4 */ Updated to new diagram format

Solving puzzles is a very good way of becoming a stronger player. Solve as many as possible! And feel free to post your own puzzles here.

== [[Piet Hein]]'s puzzles ==

See article [[Piet Hein's puzzles]]

== [[Claude Berge]]'s puzzles ==

See article [[Claude Berge's puzzles]]

== Bert Enderton ==

=== Puzzle 1 ===

Red to play and win.

<hex>R7 C7 Q1
Hc3 Hf3
Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

Red to play and win.

<hex>R7 C7 Q1
Hc3
Hf5
Vb6</hex>

=== Puzzle 3 ===

Red to play and win.

This is a very difficult puzzle whose complete solution is extremely complex.

<hexboard size="6x6"
contents="B c2 R f4"
/>

=== Puzzle 4 ===

Red to play and win.

<hexboard size="7x7"
xcontents="B b4 c4 d4 R b3 b6"
contents="B f4 e4 d4 R f5 f2"
/>

== Other authors ==

=== Puzzle 1 ===
By [[John Tromp]]

Blue to play and win.

<hex> R10 C10 Q1
Ri1
Bg4 Rh4
Bf5 Rg5
Be6 Rf6
Bd7
Rc8
Rd9 </hex>

=== Puzzle 2 ===
By ''lazyplayer''. Blue to play and win.

<hex> R7 C7 Q1
Vb2
He3
Vb5
</hex>

=== Puzzle 3 ===
By [[David J Bush]]. Taken from a game on [[Playsite]] in 2003. Red to play and win.

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9"
/>

Source: this [[Little Golem]] forum [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=14 thread].

=== Puzzle 4 ===
Designed by ''Door1'', helped by ''David J Bush''. Inspired by a game on [[Kurnik]] in May 2005. Blue to play and win.

<hexboard size="10x10"
coords="show"
contents="
R g3 b4 e4 c6 d7 d8
B b2 g2 f3 f4 e5"
/>

Source: this [[Little Golem]] forum [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=212 thread].

=== Puzzle 5 ===
Designed by ''Arek Kulczycki''. Game where both players play very close to perfect. Move 7.b9 was the first mistake, however it was very hard to refute it. 
Blue to move and win (there is just 1 winning sequence).

<hexboard size="11x11"
coords="show"
contents="B 4:h2 R 1:a4 R 5:b5 R 3:c6 B 6:e7 B 2:g8 R 7:b9"
/>

=== Puzzle 6 ===
From Ryan B. Hayward, "A puzzling Hex primer". Red to play and win.

<hexboard size="6x6"
coords="show"
contents="R d1 B e1 R f1 B e2"
/>

== See also ==

* [[Solutions to puzzles]]
* [[Ladder puzzles]]
* [[Puzzle server]]

[[Cameron Browne]] offers a lot of original puzzles in his book [[Hex Strategy Making the Right Connections]]

[[Matthew Seymour]] has created a website with 500 interactive Hex puzzles at http://www.mseymour.ca/hex_puzzle/hexpuzzle.html

[[Ryan Hayward]] and [[Bjarne Toft]] include several sets of puzzles in their book [[Hex: The Full Story]], including 49 puzzles originally published in Politiken, 28 unpublished puzzles by Jens Lindhard, 99 puzzles by Henderson, and the 4 puzzles by Bert Enderton.

[[category:Puzzle]]

Solutions to puzzles

2021-02-13T14:21:25Z

Tompo1: /* Bert Enderton */ Solution to puzzle 4 added

Puzzles

2021-02-03T20:25:18Z

Tompo1: /* Puzzle 3 */ Updated to new diagram format

Solving puzzles is a very good way of becoming a stronger player. Solve as many as possible! And feel free to post your own puzzles here.

== [[Piet Hein]]'s puzzles ==

See article [[Piet Hein's puzzles]]

== [[Claude Berge]]'s puzzles ==

See article [[Claude Berge's puzzles]]

== Bert Enderton ==

=== Puzzle 1 ===

Red to play and win.

<hex>R7 C7 Q1
Hc3 Hf3
Vc4 Hd4
Hc5 Ve5
Va6</hex>

=== Puzzle 2 ===

Red to play and win.

<hex>R7 C7 Q1
Hc3
Hf5
Vb6</hex>

=== Puzzle 3 ===

Red to play and win.

This is a very difficult puzzle whose complete solution is extremely complex.

<hexboard size="6x6"
contents="B c2 R f4"
/>

=== Puzzle 4 ===

Red to play and win.

<hexboard size="7x7"
xcontents="B b4 c4 d4 R b3 b6"
contents="B f4 e4 d4 R f5 f2"
/>

== Other authors ==

=== Puzzle 1 ===
By [[John Tromp]]

Blue to play and win.

<hex> R10 C10 Q1
Ri1
Bg4 Rh4
Bf5 Rg5
Be6 Rf6
Bd7
Rc8
Rd9 </hex>

=== Puzzle 2 ===
By ''lazyplayer''. Blue to play and win.

<hex> R7 C7 Q1
Vb2
He3
Vb5
</hex>

=== Puzzle 3 ===
By [[David J Bush]]. Taken from a game on [[Playsite]] in 2003. Red to play and win.

<hexboard size="10x10"
coords="show"
contents="R a3 e3 f3 g3 a4 b4 c4 e4 g4 h4 c5 d5 f5 e6 e7
B e2 f2 g2 h2 i2 b3 d3 d4 f4 b5 e5 d6 g8 c9 d9"
/>

Source: this [[Little Golem]] forum [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=14 thread].

=== Puzzle 4 ===
Designed by ''Door1'', helped by ''David J Bush''. Inspired by a game on [[Kurnik]] in May 2005. Blue to play and win.

<hex> R10 C10 Q1
Hb2 Hg2
Hf3 Vg3
Vb4 Ve4 Hf4
He5
Vc6
Vd7
Vd8
</hex>

Source: this [[Little Golem]] forum [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=212 thread].

=== Puzzle 5 ===
Designed by ''Arek Kulczycki''. Game where both players play very close to perfect. Move 7.b9 was the first mistake, however it was very hard to refute it. 
Blue to move and win (there is just 1 winning sequence).

<hexboard size="11x11"
coords="show"
contents="B 4:h2 R 1:a4 R 5:b5 R 3:c6 B 6:e7 B 2:g8 R 7:b9"
/>

=== Puzzle 6 ===
From Ryan B. Hayward, "A puzzling Hex primer". Red to play and win.

<hexboard size="6x6"
coords="show"
contents="R d1 B e1 R f1 B e2"
/>

== See also ==

* [[Solutions to puzzles]]
* [[Ladder puzzles]]
* [[Puzzle server]]

[[Cameron Browne]] offers a lot of original puzzles in his book [[Hex Strategy Making the Right Connections]]

[[Matthew Seymour]] has created a website with 500 interactive Hex puzzles at http://www.mseymour.ca/hex_puzzle/hexpuzzle.html

[[Ryan Hayward]] and [[Bjarne Toft]] include several sets of puzzles in their book [[Hex: The Full Story]], including 49 puzzles originally published in Politiken, 28 unpublished puzzles by Jens Lindhard, 99 puzzles by Henderson, and the 4 puzzles by Bert Enderton.

[[category:Puzzle]]

Solutions to puzzles

2021-02-03T19:13:15Z

Tompo1: /* Other authors */ Solution to puzzle 3 added

Solutions to Claude Berge's puzzles

2021-02-01T11:14:56Z

Tompo1: /* Puzzles 2-4 */ Added the solutions (a link for puzzle 2 - should this really be identical to Hein's #7?)

== Puzzle 1 ==

<hexboard size="5x5"
contents="B a1 E *:e1 R b2 E +:c2 B d2 R e2 B 1:b3 R c3 B d3 R e3 E +:b4 B c4 R a5 B b5 E *:d5"
/>

Blue plays at b3. This piece is connected to the left border by [[miai|a3/a4]] and to the long blue chain by c2/b4. The long blue chain is connected to the right border by e1/d5.

=== Puzzle 2 ===
This is identical to [[Piet Hein's puzzles#Puzzle 7|Peit Hein's Puzzle 7]]. You may wish to try the other puzzles on that page before consulting the solutions.

=== Puzzle 3 ===
Red threatens to play at 2, 4 and 6.
<hexboard size="14x14"
coords="show"
contents="B d2 B f2 R b3 B d3 R g3 B i3 R a4 B b4 B e4 B f4 R g4 B j4 R a5 B c5 R g5 B l5 B a6 B b6 R c6 B d6 B j6 R c7 B e7 B h7 R c8 B f8 R h8 R j8 R c9 R c10 R c11 R c12 R c13
R 1:d5 3:d4 5:e3 7:f3 B 2:e6 4:c4 6:e2"
/>

=== Puzzle 4 ===
The simplest solution is perhaps the sequence of [[forcing move]]s by Red below, although 1 or 3 at 5 may be quicker.

<hexboard size="14x14"
coords="show"
contents="B k1 R g2 B h2 R l2 R g3 B i3 B k3 R l3 R g4 B i4 R l4 R g5 B j5 R l5 R g6 B h6 B k6 R l6 R g7 B i7 R l7 R g8 B j8 R l8 R g9 B h9 B k9 R l9 R g10 B i10 R l10 R g11 B j11 R m11 B g12 B h12 R i12 B k12 R n12 B e13 B i13 B l13 B a14 B b14 B c14 B d14 B g14 B i14 B l14 B m14 B n14
R 1:f13 3:k13 5:j12 7:i11 9:j10 11:j9 13:i9 15:i8 17:j7 19:j6 21:i6 23:i5 25:j4 27:j3 29:j2 31:i2 33:h3 35:h13 37:g13
B 2:f12 4:l12 6:k11 8:h11 10:k10 12:k8 14:h10 16:h8 18:k7 20:k5 22:h7 24:h5 26:k4 28:k2 30:j1 32:i1 34:j13 36:h14
"
/>

== Puzzle 5 ==

This puzzle is usually presented as "Blue to play and win", but this is too easy, since Blue can win by playing at k3, leaving the bottom part of the puzzle irrelevant. But this is likely not what Berge had in mind, since in his manuscript "L'Art subtil du Hex", he instead outlines a solution to the bottom part of the puzzle. Berge's intention is a bit difficult to discern, since different drafts of his manuscript contain slightly different versions of this puzzle, and a final definitive version was never published. He also refers to this as a "study" rather than a puzzle. Regardless of what Berge's original intention was, since it is in fact a second move win for Blue, "Red to move and Blue to win" seems like a good interpretation of it.

The puzzle decomposes into two disjoint parts. In order to win, Red would need to connect k4 to the top edge while also connecting l7 to the large red group that is already connected to the bottom edge. However, Red cannot achieve both of these goals simultaneously. If Red plays in the bottom part of the board, Blue can play k3. This leads to a 4th row ladder that Red cannot win. For example:

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R 1:k9 B 2:k3 R 3:j4 B 4:j3"
/>

On the other hand, if Red plays in the top part of the board, Blue can play k8.

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R 1:k3 B 2:k8"
/>

With this move, Blue threatens to immediately connect right via [[Edge template IIIa]]. A typical sequence of follow-up moves is:

<hexboard size="14x14"
coords="show"
contents="B n1 B m2 R n2 B l3 R m3 R n3 R k4 R l4 B m4 R n4 B e5 B f5 B g5 B h5 B i5 B j5 B k5 B l5 R m5 B c6 R e6 R f6 R g6 R h6 R i6 B j6 R k6 R l6 R c7 R d7 B e7 B f7 B g7 B h7 B i7 B k7 R l7 R b8 B c8 B d8 R e8 R f8 R g8 R h8 B i8 R b9 R d9 B e9 R f9 B g9 R h9 R c10 B d10 R e10 B f10 R g10 B h10 R c11 B d11 R e11 B f11 R g11 R h11 R i11 B b12 B c12 B d12 B e12 R f12 B g12 R j12 B k12 B g13 B h13 R j13 R k13 B e14 B i14 R k3 B k8 R 3:l8 B 4:k9 R 5:l9 B 6:k11 R 7:k10 B 8:j11 E *:i12 *:j10"
/>

Now the blue stone at k11 is safely connected to the right via [[Edge template IV2a]], and is also connected left via the double threat at the hexes marked *.

== See also ==
* Back to [[Claude Berge's puzzles]]
* Main page: [[Puzzles]]

[[category:puzzle]]
{{stub}}

Hex theory

2021-01-23T12:09:14Z

Tompo1: /* Winning strategy for non-square boards */

Unlike many other games, it is possible to say certain things about '''[[Hex]]''', with absolute certainty. Whether this makes Hex a [[why did you start playing Hex|better game]] is of course debatable, but many find this attribute charming.

The most important properties of Hex are the following:

== Winning Strategy ==

* When the [[Swap rule|swap option]] is not used, the [[Red (player)|first player]] has a [[winning strategy]].
* When playing with the swap option, the second player has a [[winning strategy]].
* On non-square boards, i.e., boards of size ''n''×''m'', where ''n''≠''m'', the player with the shorter distance to cover has a [[winning strategy]] regardless of who starts.

These first and third statements are proved below. The second statement is a simply consequence of the swap rule: since Hex has no draws, each move is either winning or losing. If the opening move would be winning without the swap rule, the second player swaps and inherits the win. If the opening move would be losing, the second player declines to swap and goes on to win. Thus, the second player can always win.

=== No winning strategy for Blue ===

In [http://en.wikipedia.org/wiki/Chess chess], while nobody seriously believes that black has a [[winning strategy]], nobody has been able to disprove it. On the other hand, in Hex, a simple strategy-stealing argument shows that the [[Blue (player)|second player]] cannot have a [[winning strategy]], and therefore the [[Red (player)|first player]] must have one.

In fact, we can prove a more general statement: for boards of size ''n''×''n'', any position that is symmetric (i.e., invariant by reflection about the short or long diagonal and inverting the color of the pieces) is a winning position for the next player to move under optimal play. This follows from the fact that Hex is a monotone game: a position with additional pieces of a player's color is always at least as good for that player as the position without the additional pieces. If "passing" were allowed, it would therefore never be to a player's advantage to pass. If the second player to move had a winning strategy for a symmetric position, then the first player to move could simply steal that strategy by passing and therefore themselves becoming the second player to move. Since passing does not help the player, they also have a winning strategy without passing, contradicting the assumption that the other player was winning.

=== Winning strategy for non-square boards ===

Consider a board of size ''(n+1)'' × ''n''.
In this case, Blue has a shorter distance to cover than Red. Blue has the following second-player [[winning strategy]]. Divide the board into two triangles as shown.

<hexboard size="6x5"
contents="S red:all blue:area(a6,e6,e2)
E 1:a1 2:b1 3:c1 4:d1 5:e1 6:a2 7:b2 8:c2 9:d2 10:a3 11:b3 12:c3 13:a4 14:b4 15:a5
E 1:e6 2:e5 3:e4 4:e3 5:e2 6:d6 7:d5 8:d4 9:d3 10:c6 11:c5 12:c4 13:b6 14:b5 15:a6"
/>

Now whenever Red plays in the red triangle, Blue responds by playing in the cell of the same number in the blue triangle, and vice versa. After filling all of the cells in this way, Red cannot have a winning path as outlined below.

Assume for a contradiction that Red does have a winning path, then let R be a minimal subset of the Red piece positions that connects the Red edges (i.e. no subset of R connects the Red edges) and set all other cells to Blue. It can be shown that R is a chain of connected cells ''{r(1),r(2),...,r(k)}'' with ''r(1)'' on the top row, ''r(k)'' on the bottom row and ''r(i)'' adjacent to ''r(i+1)'' for ''0<i<k''.

Let ''r(j)'' be the first cell of this chain in the blue triangle above, then ''r(j-1)'' is in the red triangle and must be in the 9 o'clock direction from ''r(j)'' because the cell in the 11 o'clock direction has the same number label as ''r(j)'', and so must be Blue.

<hexboard size="3x2"
coords="none"
edges="none"
visible="area(b1,a2,a3,b2)"
contents="R arrow(12):a2 j:b2 B j:b1 arrow(2):a3 S red:(b1 a2) blue:(b2 a3) "/>

Now ''{r(1),r(2),...,r(j-1)}'' lies entirely within the red triangle, and so the cells in the blue triangle with corresponding labels ''{b(1),b(2),...,b(j-1)}'' must all be Blue and form a connected chain from the right edge to ''b(j-1)'' since the adjacency relationships are the same within the 2 triangles, and so ''{r(1),r(2),...,r(j-1),b(j-1),b(j-2),...,b(2),b(1)}'' forms a connected chain from the top edge to the right edge. Notice that ''{r(j),r(j+1),...,r(k)}'' cannot contain any of the cells ''{r(1),r(2),...,r(j-1)}'' by minimality of R (so intuitively ''r(j)'' is cut off from bottom edge giving a contradiction as we shall show). Also ''r(1)'' is the only Red cell on the top row, by minimality of R, so if we set the cells ''{r(1),r(2),...,r(j-1)}'' to Blue, this produces a chain of Blue pieces entirely within the red triangle connecting ''r(j-1)'' to the left edge via ''{r(1),r(2),...,r(j-1)}'' and the top row, and does not alter the Red chain ''{r(j),r(j+1),...,r(k)}'', which is assumed to connect ''r(j)'' to the bottom edge.

<hexboard size="6x5"
coords="none"
contents="S red:all blue:area(a6,e6,e2)
B arrow(8):c3 arrow(2):c4
R arrow(6):d3 "
/>

Now, if we add a short diagonal between the two triangles, making an ''(n+1)x(n+1)'' board, and set one cell on this short diagonal to Blue, linking ''r(j-1)'' to ''b(j-1)'' (the 2 Blue arrowed cells) and the rest to Red, then any connected Red group from the smaller board, together with adjacent Red pieces on the short diagonal is still connected on the larger board, so ''r(j)'' is still connected to the bottom edge. Also, since ''r(j-1)'' and ''b(j-1)'' are connected to the left and right edges respectively by Blue chains entirely within the red and blue triangles respectively, they are still connected on the larger board, and so there is a Blue chain linking the left and right edges via ''r(j-1)'', the Blue piece on the short diagonal and ''b(j-1)''.

<hexboard size="6x6"
coords="none"
contents="S red:area(a1,a5,e1) blue:area(b6,f6,f2)
B arrow(8):c3 arrow(2):d4 d3
R arrow(6):e3 f1 e2 c4 b5 a6 "
/>

But ''r(j)'' is now connected to the top edge via the short diagonal, so there is also a Red chain linking the top and bottom edges on the larger board. This gives the required contradiction.

Alternatively, (on the smaller board without changing ''{r(1),r(2),...,r(j-1)}'' from Red to Blue) we can consider the directed boundary between cells containing Red and Blue pieces (in a similar manner to the Gayle proof of no draws) beginning at the boundary between ''r(j-1)'' and ''b(j-1)'' (the arrowed pieces in the 4-cell diagram above) in the direction away from ''r(j)''. The cells to the left bank of this directed boundary are all connected to ''b(j-1)'' and hence to the right edge. The cells to the right bank of this directed boundary are ''{r(j-1),r(j-2),...}'', which are all within the red triangle. If this directed boundary meets the left edge, this gives a connection from ''b(j-1)'' to the left edge, otherwise the directed boundary meets the top edge, and this gives a connection from ''b(j-1)'' to the cell on the top row to the immediate left of ''r(1)''. However ''r(1)'' is the only Red cell on the top row, by minimality of R, so the Blue cell to its left, and hence also ''b(j-1)'', is connected to the left edge, giving the required contradiction.

This shows that Red cannot have a winning path, so Blue must have a winning path. ∎

An analogous strategy works for all boards of size ''n''×''m'' with ''n'' > ''m''. If the difference between ''n'' and ''m'' is greater than 1, Blue can simply ignore the additional rows, say at the bottom of the board, i.e., pretend that they have already been filled with Red pieces. If Red moves in the ignored area, Blue can [[passing|pass]] (or in case passing is not permitted, Blue can move anywhere).

Since we showed that Blue has a second-player winning strategy, it follows that Blue also has a first-player winning strategy, since the additional move cannot hurt Blue.

For symmetric reasons, Red has a winning strategy when ''n'' < ''m''.

See [[parallelogram boards]] for an analysis of how much headstart the player with the larger distance needs to win, for different non-square boards.

== No draw ==

If a Hex board is full then there is one and only one player connecting their edges. See also [[draw]]. The proof idea is quite simple. On a full Hex board, consider the set ''A'' of all red cells that are connected to Red's top edge. If this set contains a cell on Red's bottom edge, then Red is the winner. Otherwise, Blue has a winning path by going along the boundary of ''A''.

Links to more detailed proofs are on [[Jack van Rijswijck|Javhar]]'s page [http://javhar1.googlepages.com/hexcannotendinadraw "Hex cannot end in a draw"].

== Complexity ==

* The problem of determining the winner of a given Hex position (on a board of size ''n''×''n'') is '''PSPACE-complete'''. The fact that it is a member of PSPACE is not surprising, because one can decide the winner by simply exploring the entire game tree, i.e., by playing every possible sequence of moves, which takes only a polynomial amount of memory. The fact that this decision problem is PSPACE-hard was first proved by [http://academic.timwylie.com/17CSCI4341/hex_acta.pdf Stefan Reisch] in 1979.

Several other related decision problems are also PSPACE-complete. For example:

* The detection of [[virtual connection]]s is PSPACE-complete. Clearly, this problem is in PSPACE, since one can decide the validity of a virtual connection by exploring every possible sequence of moves within the given carrier set. The fact that it is PSPACE-hard follows from the PSPACE-hardness of detecting a winning position, since a board position is winning (for the second player) if and only if it gives a virtual connection between the player's two edges.

* The detection of [[dominated cell]]s is PSPACE-complete. More specifically, given a board size, a position on that board, a player to move, and two empty cells X and Y, the problem of deciding whether X dominates Y is PSPACE-complete. To see why it is in PSPACE, note that it is sufficient to check for each of X and Y whether it is a winning move for the player in question. X dominates Y if and only if X is a winning move or Y is not a winning move. For PSPACE-hardness, consider the following position on a board of size (''n''+2)×(''n''+2), where the cells marked "*" denote some arbitrary position of an ''n''×''n'' board:<hexboard size="6x6" contents="R c2 d2 e2 f2 a2 a3 a4 a5 a6 B a1 d1 e1 f1 b2 b3 b4 b5 b6 E *:c3 *:c4 *:c5 *:c6 *:d3 *:d4 *:d5 *:d6 *:e3 *:e4 *:e5 *:e6 *:f3 *:f4 *:f5 *:f6"/>For Red, the move at b1 is clearly winning, whereas the move at c1 is winning if and only if Red has a first-player win in the game marked "*". Therefore, b1 dominates c1. Moreover, c1 dominates b1 if and only if Red has a first-player winning strategy for "*". Moreover, b1 ''strictly'' dominates c1 if and only if Red has no such strategy. Since answering the latter question on the ''n''×''n'' board is PSPACE-hard, it follows that both domination and strict domination are PSPACE-hard problems.

There are some computational problems in Hex that are easier than PSPACE.

* The problem of deciding whether a given cell is [[dead cell|dead]] is in co-NP. Equivalently, the problem of deciding whether a given cell is [[dead cell|alive]] is in NP. This is because an empty cell is alive if and only if it belongs to some minimal winning path, relative to the given board position. Given such a path, it is checkable in polynomial time whether it is winning and minimal. [https://webdocs.cs.ualberta.ca/~hayward/papers/bergeParis.pdf Björnsson et al.] showed that recognizing alive nodes is NP-complete in the [[Shannon game]], a generalization of Hex. It is not known whether recognizing alive cells in Hex is also NP-complete or whether it is easier.

== Solving Hex ==

* Hex has been solved on [[small boards]].

== See also ==

[[Open problems]]

== External links ==

* Stefan Reisch. [http://academic.timwylie.com/17CSCI4341/hex_acta.pdf Hex is PSPACE-complete], 1979.
* Stefan Kiefer. [https://web.archive.org/web/20070625134953/http://www.fmi.uni-stuttgart.de/szs/publications/info/kiefersn.Kie03.shtml Die Menge der virtuellen Verbindungen im Spiel Hex ist PSPACE-vollständig]. Studienarbeit Nr. 1887, Universität Stuttgart, Juli 2003. In German.
* Thomas Maarup. [http://maarup.net/thomas/hex/ Hex]. Master's thesis, 2005.
* Yngvi Björnsson, Ryan Hayward, Michael Johanson, Jack Van Rijswijck. [https://webdocs.cs.ualberta.ca/~hayward/papers/bergeParis.pdf Dead cell analysis in Hex and the Shannon game]. Trends in Mathematics, pp.45–59, 2006.

[[category:Theory]]

Hex theory

2021-01-22T19:45:50Z

Tompo1: /* Winning strategy for non-square boards */ Fixed proof error - connection symmetry is only for groups within one of the triangles

Unlike many other games, it is possible to say certain things about '''[[Hex]]''', with absolute certainty. Whether this makes Hex a [[why did you start playing Hex|better game]] is of course debatable, but many find this attribute charming.

The most important properties of Hex are the following:

== Winning Strategy ==

* When the [[Swap rule|swap option]] is not used, the [[Red (player)|first player]] has a [[winning strategy]].
* When playing with the swap option, the second player has a [[winning strategy]].
* On non-square boards, i.e., boards of size ''n''×''m'', where ''n''≠''m'', the player with the shorter distance to cover has a [[winning strategy]] regardless of who starts.

These first and third statements are proved below. The second statement is a simply consequence of the swap rule: since Hex has no draws, each move is either winning or losing. If the opening move would be winning without the swap rule, the second player swaps and inherits the win. If the opening move would be losing, the second player declines to swap and goes on to win. Thus, the second player can always win.

=== No winning strategy for Blue ===

In [http://en.wikipedia.org/wiki/Chess chess], while nobody seriously believes that black has a [[winning strategy]], nobody has been able to disprove it. On the other hand, in Hex, a simple strategy-stealing argument shows that the [[Blue (player)|second player]] cannot have a [[winning strategy]], and therefore the [[Red (player)|first player]] must have one.

In fact, we can prove a more general statement: for boards of size ''n''×''n'', any position that is symmetric (i.e., invariant by reflection about the short or long diagonal and inverting the color of the pieces) is a winning position for the next player to move under optimal play. This follows from the fact that Hex is a monotone game: a position with additional pieces of a player's color is always at least as good for that player as the position without the additional pieces. If "passing" were allowed, it would therefore never be to a player's advantage to pass. If the second player to move had a winning strategy for a symmetric position, then the first player to move could simply steal that strategy by passing and therefore themselves becoming the second player to move. Since passing does not help the player, they also have a winning strategy without passing, contradicting the assumption that the other player was winning.

=== Winning strategy for non-square boards ===

Consider a board of size ''(n+1)'' × ''n''.
In this case, Blue has a shorter distance to cover than Red. Blue has the following second-player [[winning strategy]]. Divide the board into two triangles as shown.

<hexboard size="6x5"
contents="S red:all blue:area(a6,e6,e2)
E 1:a1 2:b1 3:c1 4:d1 5:e1 6:a2 7:b2 8:c2 9:d2 10:a3 11:b3 12:c3 13:a4 14:b4 15:a5
E 1:e6 2:e5 3:e4 4:e3 5:e2 6:d6 7:d5 8:d4 9:d3 10:c6 11:c5 12:c4 13:b6 14:b5 15:a6"
/>

Now whenever Red plays in the red triangle, Blue responds by playing in the cell of the same number in the blue triangle, and vice versa. After filling all of the cells in this way, Red cannot have a winning path as outlined below.

Assume for a contradiction that Red does have a winning path, then let R be a minimal subset of the Red piece positions that connects the Red edges (i.e. no subset of R connects the Red edges) and set all other cells to Blue. It can be shown that R is a chain of connected cells ''{r(1),r(2),...,r(k)}'' with ''r(1)'' on the top row, ''r(k)'' on the bottom row and ''r(i)'' adjacent to ''r(i+1)'' for ''0<i<k''.

Let ''r(j)'' be the first cell of this chain in the blue triangle above, then ''r(j-1)'' is in the red triangle and must be in the 9 o'clock direction from ''r(j)'' because the cell in the 11 o'clock direction has the same number label as ''r(j)'', and so must be Blue.

<hexboard size="3x2"
coords="none"
edges="none"
visible="area(b1,a2,a3,b2)"
contents="R arrow(12):a2 j:b2 B j:b1 arrow(2):a3 S red:(b1 a2) blue:(b2 a3) "/>

Now ''{r(1),r(2),...,r(j-1)}'' lies entirely within the red triangle, and so the cells in the blue triangle with corresponding labels ''{b(1),b(2),...,b(j-1)}'' must all be Blue and form a connected chain from the right edge to ''b(j-1)'' since the adjacency relationships are the same within the 2 triangles, and so ''{r(1),r(2),...,r(j-1),b(j-1),b(j-2),...,b(2),b(1)}'' forms a connected chain from the top edge to the right edge. Notice that ''{r(j),r(j+1),...,r(k)}'' cannot contain any of the cells ''{b(1),b(2),...,b(j-1)}'' because they are Blue, or the cells ''{r(1),r(2),...,r(j-1)}'' by minimality of R, and so cannot connect ''r(j)'' to the bottom edge (although this may be difficult to prove).

Alternatively, (in a similar manner to the Gayle proof of no draws) we can consider the directed boundary between cells containing Red and Blue pieces beginning at the boundary between ''r(j-1)'' and ''b(j-1)'' (the arrowed pieces in the diagram) in the direction away from ''r(j)''. The cells to the left bank of this directed boundary are all connected to ''b(j-1)'' and hence to the right edge. The cells to the right bank of this directed boundary are ''{r(j-1),r(j-2),...}'', which are all within the red triangle. If this directed boundary meets the left edge, this gives a connection from ''b(j-1)'' to the left edge, otherwise the directed boundary meets the top edge, and this gives a connection from ''b(j-1)'' to the cell on the top row to the immediate left of ''r(1)''. However ''r(1)'' is the only Red cell on the top row, by minimality of R, so the Blue cell to its left, and hence also ''b(j-1)'', is connected to the left edge, giving the required contradiction.

This shows that Red cannot have a winning path, so Blue must have a winning path. ∎

An analogous strategy works for all boards of size ''n''×''m'' with ''n'' > ''m''. If the difference between ''n'' and ''m'' is greater than 1, Blue can simply ignore the additional rows, say at the bottom of the board, i.e., pretend that they have already been filled with Red pieces. If Red moves in the ignored area, Blue can [[passing|pass]] (or in case passing is not permitted, Blue can move anywhere).

Since we showed that Blue has a second-player winning strategy, it follows that Blue also has a first-player winning strategy, since the additional move cannot hurt Blue.

For symmetric reasons, Red has a winning strategy when ''n'' < ''m''.

See [[parallelogram boards]] for an analysis of how much headstart the player with the larger distance needs to win, for different non-square boards.

== No draw ==

If a Hex board is full then there is one and only one player connecting their edges. See also [[draw]]. The proof idea is quite simple. On a full Hex board, consider the set ''A'' of all red cells that are connected to Red's top edge. If this set contains a cell on Red's bottom edge, then Red is the winner. Otherwise, Blue has a winning path by going along the boundary of ''A''.

Links to more detailed proofs are on [[Jack van Rijswijck|Javhar]]'s page [http://javhar1.googlepages.com/hexcannotendinadraw "Hex cannot end in a draw"].

== Complexity ==

* The problem of determining the winner of a given Hex position (on a board of size ''n''×''n'') is '''PSPACE-complete'''. The fact that it is a member of PSPACE is not surprising, because one can decide the winner by simply exploring the entire game tree, i.e., by playing every possible sequence of moves, which takes only a polynomial amount of memory. The fact that this decision problem is PSPACE-hard was first proved by [http://academic.timwylie.com/17CSCI4341/hex_acta.pdf Stefan Reisch] in 1979.

Several other related decision problems are also PSPACE-complete. For example:

* The detection of [[virtual connection]]s is PSPACE-complete. Clearly, this problem is in PSPACE, since one can decide the validity of a virtual connection by exploring every possible sequence of moves within the given carrier set. The fact that it is PSPACE-hard follows from the PSPACE-hardness of detecting a winning position, since a board position is winning (for the second player) if and only if it gives a virtual connection between the player's two edges.

* The detection of [[dominated cell]]s is PSPACE-complete. More specifically, given a board size, a position on that board, a player to move, and two empty cells X and Y, the problem of deciding whether X dominates Y is PSPACE-complete. To see why it is in PSPACE, note that it is sufficient to check for each of X and Y whether it is a winning move for the player in question. X dominates Y if and only if X is a winning move or Y is not a winning move. For PSPACE-hardness, consider the following position on a board of size (''n''+2)×(''n''+2), where the cells marked "*" denote some arbitrary position of an ''n''×''n'' board:<hexboard size="6x6" contents="R c2 d2 e2 f2 a2 a3 a4 a5 a6 B a1 d1 e1 f1 b2 b3 b4 b5 b6 E *:c3 *:c4 *:c5 *:c6 *:d3 *:d4 *:d5 *:d6 *:e3 *:e4 *:e5 *:e6 *:f3 *:f4 *:f5 *:f6"/>For Red, the move at b1 is clearly winning, whereas the move at c1 is winning if and only if Red has a first-player win in the game marked "*". Therefore, b1 dominates c1. Moreover, c1 dominates b1 if and only if Red has a first-player winning strategy for "*". Moreover, b1 ''strictly'' dominates c1 if and only if Red has no such strategy. Since answering the latter question on the ''n''×''n'' board is PSPACE-hard, it follows that both domination and strict domination are PSPACE-hard problems.

There are some computational problems in Hex that are easier than PSPACE.

* The problem of deciding whether a given cell is [[dead cell|dead]] is in co-NP. Equivalently, the problem of deciding whether a given cell is [[dead cell|alive]] is in NP. This is because an empty cell is alive if and only if it belongs to some minimal winning path, relative to the given board position. Given such a path, it is checkable in polynomial time whether it is winning and minimal. [https://webdocs.cs.ualberta.ca/~hayward/papers/bergeParis.pdf Björnsson et al.] showed that recognizing alive nodes is NP-complete in the [[Shannon game]], a generalization of Hex. It is not known whether recognizing alive cells in Hex is also NP-complete or whether it is easier.

== Solving Hex ==

* Hex has been solved on [[small boards]].

== See also ==

[[Open problems]]

== External links ==

* Stefan Reisch. [http://academic.timwylie.com/17CSCI4341/hex_acta.pdf Hex is PSPACE-complete], 1979.
* Stefan Kiefer. [https://web.archive.org/web/20070625134953/http://www.fmi.uni-stuttgart.de/szs/publications/info/kiefersn.Kie03.shtml Die Menge der virtuellen Verbindungen im Spiel Hex ist PSPACE-vollständig]. Studienarbeit Nr. 1887, Universität Stuttgart, Juli 2003. In German.
* Thomas Maarup. [http://maarup.net/thomas/hex/ Hex]. Master's thesis, 2005.
* Yngvi Björnsson, Ryan Hayward, Michael Johanson, Jack Van Rijswijck. [https://webdocs.cs.ualberta.ca/~hayward/papers/bergeParis.pdf Dead cell analysis in Hex and the Shannon game]. Trends in Mathematics, pp.45–59, 2006.

[[category:Theory]]

Captured cell

2021-01-17T17:42:04Z

Tompo1: /* Examples */ changed wording

An area of the board (empty or not) has been ''captured'' by a player if all of the opponent's pieces in that area are [[dead cell|dead]], and for any possible move by the opponent in the area, the player has a counter-strategy that kills all of the opponent's pieces in that area.

It is never advantageous for a player to move in an area that has been captured by the opponent. A captured area may as well be assumed to have been filled with the capturing player's pieces, as this does not change the strategic value of the position.

== Examples ==

The most common example of capture is the second row edge template

<hexboard size="5x5"
coords="none"
edges="bottom"
visible="area(c3,a5,d5,d3)"
contents="R c4 S red:(b5 c5) E a:b5 b:c5"/>

The two shaded cells are captured by Red. If Blue plays at a, Red can play at b, [[dead cell|killing]] a.
Conversely, if Blue plays at b, Red can play at a, killing b. Therefore, both cells are captured and the above position is strategically equivalent to the following.

<hexboard size="5x5"
coords="none"
edges="bottom"
visible="area(c3,a5,d5,d3)"
contents="R c4 R b5 c5"/>

These sequences of moves demonstrate that a and b are captured in the original position, however if Blue actually plays at a or b in the 5-cell edge template below, then Red can play at c capturing all 5 cells. This response [[dominated cell|capture-dominates]] the solid connection at a or b (although it may be inferior with regard to the secondary objective of minimizing the number of moves needed to win).

<hexboard size="3x3"
coords="none"
edges="bottom"
visible="area(b2,a3,c3,c2)"
contents="R b2 S red:(a3 b3) E a:a3 b:b3 c:c2"/>

Here are some other examples of cells captured by Red:

<hexboard size="5x6"
coords="hide"
visible="b4 c4 d4 d2 c3 d3"
contents="R b4 c4 d4 d2 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 e2 d2 c3 d3"
contents="R b4 c4 e2 d2 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 d2 e3 c3 d3"
contents="R b4 c4 d2 B e3 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 d4 b3 e3 c3 d3"
contents="R b4 c4 d4 b3 e3 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 d4 e3 c2 c3 d3"
contents="R b4 c4 d4 e3 B c2 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b3 c4 d4 e3 c2 c3 d3"
contents="R c4 d4 e3 B b3 c2 S red:(c3 d3)"
/>

== Usage ==

Consider the following position, with Blue to move:

<hexboard size="7x7"
contents="R f2 g2 g3 B d3 d4 S red:(e3 f3)"
/>

The two shaded cells are captured by Red, because if Blue plays at either one of them, Red can play the other, killing Blue's piece. Since each Red-captured cell can be treated as a red piece, it follows that Red is connected to the bottom edge by [[Edge_templates_with_two_adjacent_stones#edge_template_V2a|edge template V2a]], even though Red does not have an actual piece at f3.

== Captured cells and dead cells ==

Any cell in which a player actually has a piece is trivially captured by that player. Moreover, since [[dead cell]]s can be treated as cells of either color, an empty dead cell is captured by both players. (Dead cells containing an opponent's piece may also sometimes be captured, but when considering such cells as part of a captured area, beware of the [[Dead_cell#Interaction_between_multiple_dead_cells|interaction between multiple dead cells]]).

The analysis of dead cells and captured cells may sometimes go through multiple iterations: as some cells are discovered to be captured, they create other dead cells, which in turns may create additional captured cells, and so on.

For example, consider the effect of a red piece at b2:

<hexboard size="5x5"
edges="top left"
visible="area(a1,a5,c5,e1)"
contents="R b2 E +:a1 +:a2 *:b1 *:c1 S red:(a1,a2,b1,c1)"
/>

First, b2 captures the two cells marked "*". Then, because the cells marked "*" can be treated as if they were red pieces, the cells marked "+" become dead, and therefore also captured. Thus, a single red piece at b2 has captured four other cells.

Moreover, if there is an additional blue piece at a4, Red b2 actually captures five cells:

<hexboard size="5x5"
edges="top left"
visible="area(a1,a5,c5,e1)"
contents="R b2 B a4 E +:a1 +:a2 *:b1 *:c1 +:a3 S red:(a1,a2,b1,c1,a3)"
/>

First, b1 and c1 are Red-captured and a1 and a2 are dead as in the previous example. Finally, since a2 can be treated as a red piece, it also kills a3.

== Captured is not the same as connected ==

Based on the example of the 2nd row edge template above, one may wonder whether cells that are part of a template are automatically captured. This is not the case. To see why not, consider the following position containing an interior bridge template, with Blue to move:

<hexboard coords="show" size="5x4" contents="R c2 R b4 E *:b3 E *:c3 B a5 B b5 R d4"/>

The two cells marked "*" form part of a bridge, but they are not captured. Indeed, if Blue intrudes into the bridge at c3, Red will lose because she cannot simultaneously defend the bridge and prevent Blue from connecting at c4. On the other hand, had even one of the cells marked "*" been occupied by Red (it does not matter which one), the position would have been winning for Red. This shows that neither of the cells marked "*" is captured by Red.

Analogous things can be said about other interior templates as well. Each of the following positions contains an interior template whose [[carrier]] is marked "*". With Blue to move, each position is winning for Blue. But if any one of the cells marked "*" is replaced by a red piece, the position becomes winning for Red.

The [[mouth]]:
<hexboard size="5x5" contents="R b4 R b2 R d2 R c4 E *:c2 *:b3 *:c3 *:d3 B a5 B b5 B c5 B a1 B b1 B c1 R e4" />

The [[box]]:
<hexboard size="5x6" contents="R c2 R b4 R e2 R d4 E *:d2 E *:b3 *:c3 *:d3 *:e3 *:c4 R f4 B c1 B d1 B a5 B b5 B c5 B d5"/>

The [[open box]]:
<hexboard size="5x6" contents="R b4 R b3 R e2 R d4 E *:d2 E *:c2 *:c3 *:d3 *:e3 *:c4 B a5 B b5 B c5 B d5 B b1 R f4" />

== Generous capture ==

In some situations, it can happen that an area is not technically captured by a player, but the area ''would'' be captured if the ''opponent'' had additional pieces on the board. For example, consider the following situation:
<hexboard size="7x7" contents="R f2 g2 g3 E *:e3 *:f3 E +:d3 E +:d4"/>
The two cells marked "*" are not captured by Red. However, as we already saw above, they would become captured (by Red) if Blue occupied the cells marked "+". Red can sometimes take advantage of such a situation by mentally "giving" the additional cells to Blue, i.e., playing as if Blue already had pieces there. In other words, if Red promises never to move in the cells marked "+", then she can treat the cells marked "*" as captured. We refer to this as "generous" capture, because to capture the cells, the player must generously (albeit only mentally) give additional cells to the opponent.

In the above example, the red group is connected to the bottom edge by generous capture and [[Edge_templates_with_two_adjacent_stones#edge_template_V2a|edge template V2a]]. Note that it is important that this template does not overlap the "generous" cells d3 and d4, i.e., it would still be valid if Blue actually occupied these cells.

It may seem paradoxical that Red can gain an advantage by mentally surrendering some cells to Blue. Normally, additional Blue pieces can only be bad for Red. So what is the catch? Wny can't we just consider the cells marked "*" above as captured without the mental contortion of giving additional pieces to Blue? The answer is that in some situations, generous capture may help Red connect in one direction, while interfering with Red's connection in the other. As an example of this, consider the following position, with Blue to move:
<hexboard size="7x7" contents="R f2 g2 g3 E *:e3 *:f3 E +:d3 R c1 R c2 R c3 B a6 B b5 B c4 B d2 B e1 B d4 B g1"/>
Red is connected to the top edge via [[double threat]] at f1 and d3. Red is also connected to the bottom edge by generous capture: a generous blue piece at d3 captures f3 for Red, and therefore Red is connected down by [[Edge_templates_with_two_adjacent_stones#edge_template_V2a|edge template V2a]]. However, the catch is that Red cannot achieve both of these things simultaneously: the generous blue piece at d3 invalidates Red's connection to the top — even though this piece only exists in Red's imagination! And indeed, this position is winning for Blue: a possible winning move for Blue is e3, which requires Red to defend the upward and downward connections at the same time.

== More examples of generous capture ==

We've seen [[#Captured is not the same as connected|above]] that many templates, such as the bridge, the mouth, the box, and the open box, do not capture any of the cells in their carrier. However, all of these templates ''generously'' capture their carrier if certain blue pieces are added. The following diagrams show which blue pieces can be added to each template to capture the cells marked "*":

The [[bridge]]:
<hexboard size="5x4"
coords="none"
edges="none"
visible="c2 b4 b3 c3 a3 d3"
contents="R c2 R b4 E *:b3 E *:c3 B a3 B d3"
/>

The [[mouth]]:
<hexboard size="5x5"
coords="none"
edges="none"
visible="b4 b2 d2 c4 c2 b3 c3 d3 e3"
contents="R b4 R b2 R d2 R c4 E *:c2 *:b3 *:c3 *:d3 B e3"
/>

The [[box]]:
<hexboard size="5x6"
coords="none"
edges="none"
visible="c2 b4 e2 d4 d2 b3 c3 d3 e3 c4 a3 f3"
contents="R c2 R b4 R e2 R d4 E *:d2 E *:b3 *:c3 *:d3 *:e3 *:c4 B a3 B f3"
/>

The [[open box]]:
<hexboard size="5x6"
coords="none"
edges="none"
visible="c2 b3 e2 d4 d2 b4 c3 d3 e3 c4 a5 b5 f3"
contents="R c2 R b3 R e2 R d4 E *:d2 E *:b4 *:c3 *:d3 *:e3 *:c4 B a5 B b5 B f3"
/>

== See also ==

*[[Equivalent patterns]]

*[[Computer Hex]]

*[[Dominated cell]]s

*[[Dead cell]]s

== External links ==

Henderson and Hayward, [https://webdocs.cs.ualberta.ca/~hayward/papers/revDom.pdf "Captured-reversible moves and star decomposition domination in Hex"].

[[Ryan Hayward]]'s [http://www.cs.ualberta.ca/~hayward/publications.html publication page] contains research articles on dead, vulnerable, captured, and dominated cells.

[[category:Theory]]
[[category:Computer Hex]]
[[category:Intermediate Strategy]]

Captured cell

2021-01-17T17:19:42Z

Tompo1: /* Examples */ Extended discussion of the second row template with another example

An area of the board (empty or not) has been ''captured'' by a player if all of the opponent's pieces in that area are [[dead cell|dead]], and for any possible move by the opponent in the area, the player has a counter-strategy that kills all of the opponent's pieces in that area.

It is never advantageous for a player to move in an area that has been captured by the opponent. A captured area may as well be assumed to have been filled with the capturing player's pieces, as this does not change the strategic value of the position.

== Examples ==

The most common example of capture is the second row edge template

<hexboard size="5x5"
coords="none"
edges="bottom"
visible="area(c3,a5,d5,d3)"
contents="R c4 S red:(b5 c5) E a:b5 b:c5"/>

The two shaded cells are captured by Red. If Blue plays at a, Red can play at b, [[dead cell|killing]] a.
Conversely, if Blue plays at b, Red can play at a, killing b. Therefore, both cells are captured and the above position is strategically equivalent to the following.

<hexboard size="5x5"
coords="none"
edges="bottom"
visible="area(c3,a5,d5,d3)"
contents="R c4 R b5 c5"/>

These sequences of moves demonstrate that a and b are captured in the original position, however if Blue actually plays at a or b in the 5-cell edge template below, then Red can play at c capturing all 5 cells. This response [[dominated cell|capture-dominates]] the solid connection at a or b (although it may not meet the secondary objective of minimizing the number of moves needed to win).

<hexboard size="3x3"
coords="none"
edges="bottom"
visible="area(b2,a3,c3,c2)"
contents="R b2 S red:(a3 b3) E a:a3 b:b3 c:c2"/>

Here are some other examples of cells captured by Red:

<hexboard size="5x6"
coords="hide"
visible="b4 c4 d4 d2 c3 d3"
contents="R b4 c4 d4 d2 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 e2 d2 c3 d3"
contents="R b4 c4 e2 d2 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 d2 e3 c3 d3"
contents="R b4 c4 d2 B e3 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 d4 b3 e3 c3 d3"
contents="R b4 c4 d4 b3 e3 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b4 c4 d4 e3 c2 c3 d3"
contents="R b4 c4 d4 e3 B c2 S red:(c3 d3)"
/>
<hexboard size="5x6"
coords="hide"
visible="b3 c4 d4 e3 c2 c3 d3"
contents="R c4 d4 e3 B b3 c2 S red:(c3 d3)"
/>

== Usage ==

Consider the following position, with Blue to move:

<hexboard size="7x7"
contents="R f2 g2 g3 B d3 d4 S red:(e3 f3)"
/>

The two shaded cells are captured by Red, because if Blue plays at either one of them, Red can play the other, killing Blue's piece. Since each Red-captured cell can be treated as a red piece, it follows that Red is connected to the bottom edge by [[Edge_templates_with_two_adjacent_stones#edge_template_V2a|edge template V2a]], even though Red does not have an actual piece at f3.

== Captured cells and dead cells ==

Any cell in which a player actually has a piece is trivially captured by that player. Moreover, since [[dead cell]]s can be treated as cells of either color, an empty dead cell is captured by both players. (Dead cells containing an opponent's piece may also sometimes be captured, but when considering such cells as part of a captured area, beware of the [[Dead_cell#Interaction_between_multiple_dead_cells|interaction between multiple dead cells]]).

The analysis of dead cells and captured cells may sometimes go through multiple iterations: as some cells are discovered to be captured, they create other dead cells, which in turns may create additional captured cells, and so on.

For example, consider the effect of a red piece at b2:

<hexboard size="5x5"
edges="top left"
visible="area(a1,a5,c5,e1)"
contents="R b2 E +:a1 +:a2 *:b1 *:c1 S red:(a1,a2,b1,c1)"
/>

First, b2 captures the two cells marked "*". Then, because the cells marked "*" can be treated as if they were red pieces, the cells marked "+" become dead, and therefore also captured. Thus, a single red piece at b2 has captured four other cells.

Moreover, if there is an additional blue piece at a4, Red b2 actually captures five cells:

<hexboard size="5x5"
edges="top left"
visible="area(a1,a5,c5,e1)"
contents="R b2 B a4 E +:a1 +:a2 *:b1 *:c1 +:a3 S red:(a1,a2,b1,c1,a3)"
/>

First, b1 and c1 are Red-captured and a1 and a2 are dead as in the previous example. Finally, since a2 can be treated as a red piece, it also kills a3.

== Captured is not the same as connected ==

Based on the example of the 2nd row edge template above, one may wonder whether cells that are part of a template are automatically captured. This is not the case. To see why not, consider the following position containing an interior bridge template, with Blue to move:

<hexboard coords="show" size="5x4" contents="R c2 R b4 E *:b3 E *:c3 B a5 B b5 R d4"/>

The two cells marked "*" form part of a bridge, but they are not captured. Indeed, if Blue intrudes into the bridge at c3, Red will lose because she cannot simultaneously defend the bridge and prevent Blue from connecting at c4. On the other hand, had even one of the cells marked "*" been occupied by Red (it does not matter which one), the position would have been winning for Red. This shows that neither of the cells marked "*" is captured by Red.

Analogous things can be said about other interior templates as well. Each of the following positions contains an interior template whose [[carrier]] is marked "*". With Blue to move, each position is winning for Blue. But if any one of the cells marked "*" is replaced by a red piece, the position becomes winning for Red.

The [[mouth]]:
<hexboard size="5x5" contents="R b4 R b2 R d2 R c4 E *:c2 *:b3 *:c3 *:d3 B a5 B b5 B c5 B a1 B b1 B c1 R e4" />

The [[box]]:
<hexboard size="5x6" contents="R c2 R b4 R e2 R d4 E *:d2 E *:b3 *:c3 *:d3 *:e3 *:c4 R f4 B c1 B d1 B a5 B b5 B c5 B d5"/>

The [[open box]]:
<hexboard size="5x6" contents="R b4 R b3 R e2 R d4 E *:d2 E *:c2 *:c3 *:d3 *:e3 *:c4 B a5 B b5 B c5 B d5 B b1 R f4" />

== Generous capture ==

In some situations, it can happen that an area is not technically captured by a player, but the area ''would'' be captured if the ''opponent'' had additional pieces on the board. For example, consider the following situation:
<hexboard size="7x7" contents="R f2 g2 g3 E *:e3 *:f3 E +:d3 E +:d4"/>
The two cells marked "*" are not captured by Red. However, as we already saw above, they would become captured (by Red) if Blue occupied the cells marked "+". Red can sometimes take advantage of such a situation by mentally "giving" the additional cells to Blue, i.e., playing as if Blue already had pieces there. In other words, if Red promises never to move in the cells marked "+", then she can treat the cells marked "*" as captured. We refer to this as "generous" capture, because to capture the cells, the player must generously (albeit only mentally) give additional cells to the opponent.

In the above example, the red group is connected to the bottom edge by generous capture and [[Edge_templates_with_two_adjacent_stones#edge_template_V2a|edge template V2a]]. Note that it is important that this template does not overlap the "generous" cells d3 and d4, i.e., it would still be valid if Blue actually occupied these cells.

It may seem paradoxical that Red can gain an advantage by mentally surrendering some cells to Blue. Normally, additional Blue pieces can only be bad for Red. So what is the catch? Wny can't we just consider the cells marked "*" above as captured without the mental contortion of giving additional pieces to Blue? The answer is that in some situations, generous capture may help Red connect in one direction, while interfering with Red's connection in the other. As an example of this, consider the following position, with Blue to move:
<hexboard size="7x7" contents="R f2 g2 g3 E *:e3 *:f3 E +:d3 R c1 R c2 R c3 B a6 B b5 B c4 B d2 B e1 B d4 B g1"/>
Red is connected to the top edge via [[double threat]] at f1 and d3. Red is also connected to the bottom edge by generous capture: a generous blue piece at d3 captures f3 for Red, and therefore Red is connected down by [[Edge_templates_with_two_adjacent_stones#edge_template_V2a|edge template V2a]]. However, the catch is that Red cannot achieve both of these things simultaneously: the generous blue piece at d3 invalidates Red's connection to the top — even though this piece only exists in Red's imagination! And indeed, this position is winning for Blue: a possible winning move for Blue is e3, which requires Red to defend the upward and downward connections at the same time.

== More examples of generous capture ==

We've seen [[#Captured is not the same as connected|above]] that many templates, such as the bridge, the mouth, the box, and the open box, do not capture any of the cells in their carrier. However, all of these templates ''generously'' capture their carrier if certain blue pieces are added. The following diagrams show which blue pieces can be added to each template to capture the cells marked "*":

The [[bridge]]:
<hexboard size="5x4"
coords="none"
edges="none"
visible="c2 b4 b3 c3 a3 d3"
contents="R c2 R b4 E *:b3 E *:c3 B a3 B d3"
/>

The [[mouth]]:
<hexboard size="5x5"
coords="none"
edges="none"
visible="b4 b2 d2 c4 c2 b3 c3 d3 e3"
contents="R b4 R b2 R d2 R c4 E *:c2 *:b3 *:c3 *:d3 B e3"
/>

The [[box]]:
<hexboard size="5x6"
coords="none"
edges="none"
visible="c2 b4 e2 d4 d2 b3 c3 d3 e3 c4 a3 f3"
contents="R c2 R b4 R e2 R d4 E *:d2 E *:b3 *:c3 *:d3 *:e3 *:c4 B a3 B f3"
/>

The [[open box]]:
<hexboard size="5x6"
coords="none"
edges="none"
visible="c2 b3 e2 d4 d2 b4 c3 d3 e3 c4 a5 b5 f3"
contents="R c2 R b3 R e2 R d4 E *:d2 E *:b4 *:c3 *:d3 *:e3 *:c4 B a5 B b5 B f3"
/>

== See also ==

*[[Equivalent patterns]]

*[[Computer Hex]]

*[[Dominated cell]]s

*[[Dead cell]]s

== External links ==

Henderson and Hayward, [https://webdocs.cs.ualberta.ca/~hayward/papers/revDom.pdf "Captured-reversible moves and star decomposition domination in Hex"].

[[Ryan Hayward]]'s [http://www.cs.ualberta.ca/~hayward/publications.html publication page] contains research articles on dead, vulnerable, captured, and dominated cells.

[[category:Theory]]
[[category:Computer Hex]]
[[category:Intermediate Strategy]]

Equivalent patterns

2021-01-17T15:19:10Z

Tompo1: /* Example 5a (Opening a1+b1 loses) */ Typo

We say that two Hex [[pattern]]s (subsets of a board) are '''equivalent patterns''' if, when one of them occurs embedded in ''any'' Hex board, it could be replaced by the other and the side who has winning strategy does not change.

== Edge equivalence==

Edges are equivalent to edge-connected rows of pieces on a larger board.

=== Example 0 ===

Any position on an nxm board is equivalent to the same position on the interior (i.e. the cells non-adjacent to the edges) of an (n+2)x(m+2) board with the first rows next to each edge filled with friendly pieces and [[dead cell]]s at the vertex cells (or equivalently either red or blue pieces at any of the vertex cells)

<hexboard size="4x4"
float="inline"
coords="none"
contents="S area(a1,d1,d4,a4)"
/> and <hexboard size="6x6"
float="inline"
coords="none"
contents="S area(b2,e2,e5,b5) R line(b1,e1) line(b6,e6) B line(a2,a5) line(f2,f5) E *:a1 *:f1 *:f6 *:a6"
/>

This allows the examples below to be applied at the edges: either formally, by converting to the larger board, applying the equivalence in the example, then converting back to the smaller board; or informally, by considering an edge to be a row of friendly pieces.

== Equivalence by capture ==

Cells that are [[Captured cell|captured]] by one player can be filled in with stones of that player's color.

=== Example 1 ===
For example, the two patterns below are equivalent, and an example of what is known as the [[useless triangle]].

<hexboard size="2x3"
float="inline"
coords="none"
edges="none"
visible="area(a2,c2,c1,b1)"
contents="R a2 b1 c1 c2 B b2"
/> and <hexboard size="2x3"
float="inline"
coords="none"
edges="none"
visible="area(a2,c2,c1,b1)"
contents="R a2 b1 c1 c2 R b2"
/>

The blue stone in the left pattern is [[captured cell|captured]] by Red. Therefore, the position is equivalent to one where this stone is actually red.

The knowledge of equivalent patterns turns out to be very useful for playing Hex, because it allows the player to disregard some pieces in the board, or prune the analysis tree. Some positions are much clearer than other equivalent positions. In the examples below, the simpler pattern is usually written on the right. One can make use of these equivalent patterns by mentally always replacing the left pattern with the right one.

Here are several examples of pairs of equivalent positions, and a short explanation on the way to prove them so.

=== Example 2 ===
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 b3 d3"
contents="R a2 b1 c1 d1 d2 c3 B b2 c2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 b3 d3"
contents="R a2 b1 c1 d1 d2 c3 R b2 c2"
/>

The equivalence is obtained by applying example 1 twice.

Both above examples are instances of the following rule to produce equivalence pairs. Given a [[chain]] G, let the '''neighborhood''' of G, neigh(G), be the set of cells that are adjacent to G but do not belong to it. In a given pattern P1, suppose that G is a red chain, and that C is a cell in neigh(G), which may be empty or occupied by Blue. Let P2 be the result of placing a red piece in C (removing a Blue stone, if necessary). Therefore P2 contains a chain G' containing G and C. If neigh(G')=neigh(G) then P1 and P2 are equivalent. Of course the dual argument also holds for blue chains.

This rule justifies the following equivalent pairs:

=== Example 3 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 c2 c3"
contents="R a2 b1 c1 B b2 b3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 c2 c3"
contents="R a2 b1 c1 R b2 B b3"
/>

=== Example 4 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 c2 c3"
contents="B a3 R b1 c1 B b2 b3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 c2 c3"
contents="B a3 R b1 c1 R b2 B b3"
/>

Another rule producing equivalent patterns: If there are two empty cells C1 and C2 in a pattern, such that if Blue occupies one of them, Red can occupy the other capturing the former, then an equivalent position is obtained if both C1 and C2 are occupied by Red.

Equivalent pairs obtained with this rule:

=== Example 5 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="area(a1,a3,c1)"
contents="R a1--c1 a3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="area(a1,a3,c1)"
contents="R all"
/>

This is a very instructive pattern because it shows that playing 2 rows out from a [[friendly]] edge with 2 free cells on the first row below the play is equivalent to playing at all three cells simultaneously, and hence at least as good as playing at either cell on the first row. So as a rule, one should never play on the first row in such a situation. We can see this by viewing the 3 upper red stones as part of a friendly edge.

This pattern, along with Example 4, can be used to show that the 2-move opening a1 + b1 (or either single move) for Red loses on a board 3x3 or bigger:

=== Example 5a (Opening a1+b1 loses) ===

<hexboard size="4x3"
float="inline"
coords="top left"
edges="top left"
contents="R a1 b1 B b2"
/> and <hexboard size="4x3"
float="inline"
coords="top left"
edges="top left"
contents="B a1 b1 B b2 a2 a3"
/>

In response to a1+b1 by Red, Blue can play at b2. This captures a2 and a3 by Example 5, and then a1 and b1 by applying Example 4 twice (use of these examples at the edges is justified by Example 0 above). Thus, the left position is equivalent to the right one. This includes the cells a1 and a2, so the original position is losing for Red by a strategy stealing argument.

In fact, making the opening play b2 is equivalent to playing 5 stones at a1, a2, b1, b2, and c1. However this is known to produce a losing position on boards of size 4, 7 & 8 (see [[small boards]]).

This example also shows that the opening moves a1 and b1 should never be swapped. Playing at b2 is always at least as good for Blue as swapping (but there are even better moves for Blue available elsewhere on the board).

=== Example 6 ===
<hexboard size="3x2"
float="inline"
coords="none"
edges="none"
contents="R a1 a3 b1 b3"
/> and <hexboard size="3x2"
float="inline"
coords="none"
edges="none"
contents="R a1 a2 a3 b1 b2 b3"
/>

=== Example 7 ===
<hexboard size="2x4"
float="inline"
coords="none"
edges="none"
visible="-a1"
contents="R a2--b1--d1--d2"
/> and <hexboard size="2x4"
float="inline"
coords="none"
edges="none"
visible="-a1"
contents="R all"
/>
Using both rules together:

=== Example 8 ===
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="area(a2,b3,d2,c1)"
contents="R c1 b3 B a2 d2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="area(a2,b3,d2,c1)"
contents="R c1 b3 B a2 d2 R b2 c2"
/>

=== Example 9 ===
''(generalization of the Example 5, for any horizontal length)''

<hexboard size="3x5"
float="inline"
coords="none"
edges="none"
visible="area(a3,e3,e1,c1)"
contents="R a3--e3 c1--e1 B c2 d2"
/> and <hexboard size="3x5"
float="inline"
coords="none"
edges="none"
visible="area(a3,e3,e1,c1)"
contents="R all"
/>

Here is a third (rather obvious) rule for equivalent patterns: Any area surrounded by a single chain of the opponent may be randomly filled. This happens because the outcome of the game does not depend on it at all.

== Equivalence for reasons other than capture ==

Capture is not the only way in which equivalent patterns arise. The following is an example of this.

=== Example 10 ===

The following two patterns are equivalent:
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 b3--d3"
contents="R b1--d1 d2 B a3 c2 E a:b2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 b3--d3"
contents="R b1--d1 d2 B a3 b2 E a:c2"
/>
This is easily seen by noticing that if the cells marked "a" are both blue, the patterns become equal, whereas if the cells marked "a" are both red, then the blue piece next to each of them may as well be red by Examples 1 and 3, respectively, so the patterns are equivalent. So whenever one player's strategy calls for playing in the cell marked "a" in one of these two patterns, the same player can play in the cell marked "a" in the other pattern.

== Practical examples ==

=== First example game ===

Let us see a practical example. In game [http://www.littlegolem.net/jsp/game/game.jsp?gid=206040&nmove=55 #206040] at [[Little Golem]], the situation after 55. m2 is shown in the board below.

<hex>R13 C13 Q1 Hc1 Hd6 Hd8 Hd9 Hd10 Hd11 Hd12 He6 He12 Hf5 Hf10 Hf12 Hg5 Hg6 Hg12 Hh6 Hh8 Hh9 Hh12 Hi5 Hi11 Hj4 Hj6 Hj11 Hk3 Hk11 Hl11 Vc8 Vc9 Vc10 Vc11 Vc12 Vd7 Ve7 Ve8 Ve10 Ve11 Vf4 Vf6 Vf8 Vf11 Vg11 Vh5 Vh7 Vh11 Vi6 Vi8 Vi10 Vj5 Vj10 Vk4 Vk10 Vl10 Vm2</hex>

Clearly, m2 is strongly connected to the top, because the stone in f4 is a [[ladder escape]]. On the other hand, it is strongly connected to the bottom exactly if blue cannot connect k3 with the right, using j6 and maybe the [[group]] in h8-h9-f10 as a ladder escape. In fact he cannot do it, and it is much clearer if some patterns are locally replaced by other equivalent ones, rendering:

<hex>R13 C13 Q1 Hc1 Hd6 Vd8 Vd9 Vd10 Vd11 Vd12 He6 Ve12 Hf5 Vf10 Vf12 Hg5 Hg6 Vg12 Hh6 Vh8 Vh9 Vh12 Hi5 Vi11 Hj4 Hj6 Vj11 Hk3 Vk11 Vl11 Vc8 Vc9 Vc10 Vc11 Vc12 Vd7 Ve7 Ve8 Ve10 Ve11 Vf4 Hf6 Vf8 Vf11 Vg11 Hh5 Vh7 Vh11 Vi6 Vi8 Vi10 Vj5 Vj10 Vk4 Vk10 Vl10 Vm2 Ve9 Vf7 Vf9 Vg7 Vg8 Vg9 Vg10 Vh10 Vi7 Vi9 Vb13 Vc13 Vm10 Vm11 Vi12 Vj12 Vk12 Vl12 Vm12 Vd13 Ve13 Vf13 Vg13 Vh13 Vi13 Vj13 Vk13 Vl13 Vm13</hex>

The changes have been:

* f6 and h5 swap color, as in Example 1.

* The blue group from d8 to l11 is completely wrapped by a strongly connected group belonging to the opponent, except for a narrow section (typically, 2 cells). This kind of groups typically are captured, exactly in the same reason as in Example 1.

* If blue moves to i7, red moves to i9 and conversely, in both cases enclosing the blue group h8-h9-f10 as before. So, we can use the second rule for detecting equivalent patterns, capturing it.

* The equivalence in Example 5 can be used for the stone in c12.

* The equivalence in Example 4 can be used, adding a stone for Red at m11.

* The area in the bottom of the board is now surrounded by a red chain and a blue chain. Therefore, it may be filled as we please.

The blue stone in j6 remained, completely alone and too near to the red group to be of any use in such a small region, so it is obvious that Red has won.

=== Second example game ===

In the following game, Red's c9 is connected to the top edge via [[Fifth_row_edge_templates#V-2-a|edge template V-2a]]. A 3rd row ladder is about to form along the bottom edge. Red wonders whether his other stones are enough to escape the ladder.
<hexboard size="11x11"
contents="R a11 B d6 B f6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10"
/>

The analysis of this position seems complicated at first. It can be simplified by using the equivalence in [[#Example 10|Example 10]] to shift the blue stone from f6 to e6.
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10"
/>
The key observation is that the single stone i6 on the 6th row is enough for a 3rd-to-6th row switchback (see [[Switchback#B6_switchback|B6 switchback]]) or even for a 3rd-to-5th row switchback (see [[Switchback#C6_switchback|C6 switchback]]). The resulting ladder is then escaped by g6 via interior ziggurat. Specifically, if Red pushes the ladder all the way to g9, the red stone at g6 will be connected to this ladder by an interior ziggurat:
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R 1:d9 B 2:d10 R 3:e9 B 4:e10 R 5:f9 B 6:f10 R 7:g9 B 8:g10"
/>
Now Red can play the 3rd-to-6th row switchback by breaking the ladder and playing one more piece. Note how i6 becomes the ladder stone for a 3rd row ladder in the opposite direction, which the ziggurat escapes. Therefore, Red is connected.
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R d9 B d10 R e9 B e10 R f9 B f10 R g9 B g10
R 9:i9 B 10:h9 R 11:j7"
/>
One may wonder whether it would have helped Blue to [[Ladder_handling#Defending|yield]] to a 2nd row ladder. This would not have helped here. For example:
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R 1:d9 B 2:d10 R 3:e9 B 4:e10 R 5:f9 B 6:f11 R 7:f10 B 8:e11 R 9:h10 B 10:g10
R 11:h9 B 12:g9 R 13:h7"
/>
This time, Black's 13 acts as a ladder stone for a 2nd row ladder stone in the opposite direction, which the ziggurat again escapes.
[[Category: Strategy]]
[[Category: Computer Hex]]
[[Category: Theory]]

Equivalent patterns

2021-01-17T13:42:33Z

Tompo1: /* Example 5b (Opening A1+B1 loses) */ application of Example 4 seems more obvious than Example 1

We say that two Hex [[pattern]]s (subsets of a board) are '''equivalent patterns''' if, when one of them occurs embedded in ''any'' Hex board, it could be replaced by the other and the side who has winning strategy does not change.

== Edge equivalence==

Edges are equivalent to edge-connected rows of pieces on a larger board.

=== Example 0 ===

Any position on an nxm board is equivalent to the same position on the interior (i.e. the cells non-adjacent to the edges) of an (n+2)x(m+2) board with the first rows next to each edge filled with friendly pieces and [[dead cell]]s at the vertex cells (or equivalently either red or blue pieces at any of the vertex cells)

<hexboard size="4x4"
float="inline"
coords="none"
contents="S area(a1,d1,d4,a4)"
/> and <hexboard size="6x6"
float="inline"
coords="none"
contents="S area(b2,e2,e5,b5) R line(b1,e1) line(b6,e6) B line(a2,a5) line(f2,f5) E *:a1 *:f1 *:f6 *:a6"
/>

This allows the examples below to be applied at the edges: either formally, by converting to the larger board, applying the equivalence in the example, then converting back to the smaller board; or informally, by considering an edge to be a row of friendly pieces.

== Equivalence by capture ==

Cells that are [[Captured cell|captured]] by one player can be filled in with stones of that player's color.

=== Example 1 ===
For example, the two patterns below are equivalent, and an example of what is known as the [[useless triangle]].

<hexboard size="2x3"
float="inline"
coords="none"
edges="none"
visible="area(a2,c2,c1,b1)"
contents="R a2 b1 c1 c2 B b2"
/> and <hexboard size="2x3"
float="inline"
coords="none"
edges="none"
visible="area(a2,c2,c1,b1)"
contents="R a2 b1 c1 c2 R b2"
/>

The blue stone in the left pattern is [[captured cell|captured]] by Red. Therefore, the position is equivalent to one where this stone is actually red.

The knowledge of equivalent patterns turns out to be very useful for playing Hex, because it allows the player to disregard some pieces in the board, or prune the analysis tree. Some positions are much clearer than other equivalent positions. In the examples below, the simpler pattern is usually written on the right. One can make use of these equivalent patterns by mentally always replacing the left pattern with the right one.

Here are several examples of pairs of equivalent positions, and a short explanation on the way to prove them so.

=== Example 2 ===
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 b3 d3"
contents="R a2 b1 c1 d1 d2 c3 B b2 c2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 b3 d3"
contents="R a2 b1 c1 d1 d2 c3 R b2 c2"
/>

The equivalence is obtained by applying example 1 twice.

Both above examples are instances of the following rule to produce equivalence pairs. Given a [[chain]] G, let the '''neighborhood''' of G, neigh(G), be the set of cells that are adjacent to G but do not belong to it. In a given pattern P1, suppose that G is a red chain, and that C is a cell in neigh(G), which may be empty or occupied by Blue. Let P2 be the result of placing a red piece in C (removing a Blue stone, if necessary). Therefore P2 contains a chain G' containing G and C. If neigh(G')=neigh(G) then P1 and P2 are equivalent. Of course the dual argument also holds for blue chains.

This rule justifies the following equivalent pairs:

=== Example 3 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 c2 c3"
contents="R a2 b1 c1 B b2 b3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 c2 c3"
contents="R a2 b1 c1 R b2 B b3"
/>

=== Example 4 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 c2 c3"
contents="B a3 R b1 c1 B b2 b3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 c2 c3"
contents="B a3 R b1 c1 R b2 B b3"
/>

Another rule producing equivalent patterns: If there are two empty cells C1 and C2 in a pattern, such that if Blue occupies one of them, Red can occupy the other capturing the former, then an equivalent position is obtained if both C1 and C2 are occupied by Red.

Equivalent pairs obtained with this rule:

=== Example 5 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="area(a1,a3,c1)"
contents="R a1--c1 a3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="area(a1,a3,c1)"
contents="R all"
/>

This is a very instructive pattern because it shows that playing 2 rows out from a [[friendly]] edge with 2 free cells on the first row below the play is equivalent to playing at all three cells simultaneously, and hence at least as good as playing at either cell on the first row. So as a rule, one should never play on the first row in such a situation. We can see this by viewing the 3 upper red stones as part of a friendly edge.

This pattern, along with Example 4, can be used to show that the 2-move opening a1 + b1 (or either single move) for Red loses on a board 3x3 or bigger:

=== Example 5a (Opening a1+b1 loses) ===

<hexboard size="4x3"
float="inline"
coords="top left"
edges="top left"
contents="R a1 b1 B b2"
/> and <hexboard size="4x3"
float="inline"
coords="top left"
edges="top left"
contents="B a1 b1 B b2 a2 a3"
/>

In response to a1+b1 by Red, Blue can play at b2. This captures a2 and a3 by Example 5, and then a1 and b1 by applying Example 4 twice (use of these examples at the edges is justified by Example 0 above). Thus, the left position is equivalent to the right one. This includes the cells a1 and a2, so the original position is losing for Red by a strategy stealing argument.

In fact, making the opening play b2 is equivalent to playing 5 stones at a1, a2, b1, b1, and c1. However this is known to produce a losing position on boards of size 4, 7 & 8 (see [[small boards]]).

This example also shows that the opening moves a1 and b1 should never be swapped. Playing at b2 is always at least as good for Blue as swapping (but there are even better moves for Blue available elsewhere on the board).

=== Example 6 ===
<hexboard size="3x2"
float="inline"
coords="none"
edges="none"
contents="R a1 a3 b1 b3"
/> and <hexboard size="3x2"
float="inline"
coords="none"
edges="none"
contents="R a1 a2 a3 b1 b2 b3"
/>

=== Example 7 ===
<hexboard size="2x4"
float="inline"
coords="none"
edges="none"
visible="-a1"
contents="R a2--b1--d1--d2"
/> and <hexboard size="2x4"
float="inline"
coords="none"
edges="none"
visible="-a1"
contents="R all"
/>
Using both rules together:

=== Example 8 ===
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="area(a2,b3,d2,c1)"
contents="R c1 b3 B a2 d2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="area(a2,b3,d2,c1)"
contents="R c1 b3 B a2 d2 R b2 c2"
/>

=== Example 9 ===
''(generalization of the Example 5, for any horizontal length)''

<hexboard size="3x5"
float="inline"
coords="none"
edges="none"
visible="area(a3,e3,e1,c1)"
contents="R a3--e3 c1--e1 B c2 d2"
/> and <hexboard size="3x5"
float="inline"
coords="none"
edges="none"
visible="area(a3,e3,e1,c1)"
contents="R all"
/>

Here is a third (rather obvious) rule for equivalent patterns: Any area surrounded by a single chain of the opponent may be randomly filled. This happens because the outcome of the game does not depend on it at all.

== Equivalence for reasons other than capture ==

Capture is not the only way in which equivalent patterns arise. The following is an example of this.

=== Example 10 ===

The following two patterns are equivalent:
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 b3--d3"
contents="R b1--d1 d2 B a3 c2 E a:b2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 b3--d3"
contents="R b1--d1 d2 B a3 b2 E a:c2"
/>
This is easily seen by noticing that if the cells marked "a" are both blue, the patterns become equal, whereas if the cells marked "a" are both red, then the blue piece next to each of them may as well be red by Examples 1 and 3, respectively, so the patterns are equivalent. So whenever one player's strategy calls for playing in the cell marked "a" in one of these two patterns, the same player can play in the cell marked "a" in the other pattern.

== Practical examples ==

=== First example game ===

Let us see a practical example. In game [http://www.littlegolem.net/jsp/game/game.jsp?gid=206040&nmove=55 #206040] at [[Little Golem]], the situation after 55. m2 is shown in the board below.

<hex>R13 C13 Q1 Hc1 Hd6 Hd8 Hd9 Hd10 Hd11 Hd12 He6 He12 Hf5 Hf10 Hf12 Hg5 Hg6 Hg12 Hh6 Hh8 Hh9 Hh12 Hi5 Hi11 Hj4 Hj6 Hj11 Hk3 Hk11 Hl11 Vc8 Vc9 Vc10 Vc11 Vc12 Vd7 Ve7 Ve8 Ve10 Ve11 Vf4 Vf6 Vf8 Vf11 Vg11 Vh5 Vh7 Vh11 Vi6 Vi8 Vi10 Vj5 Vj10 Vk4 Vk10 Vl10 Vm2</hex>

Clearly, m2 is strongly connected to the top, because the stone in f4 is a [[ladder escape]]. On the other hand, it is strongly connected to the bottom exactly if blue cannot connect k3 with the right, using j6 and maybe the [[group]] in h8-h9-f10 as a ladder escape. In fact he cannot do it, and it is much clearer if some patterns are locally replaced by other equivalent ones, rendering:

<hex>R13 C13 Q1 Hc1 Hd6 Vd8 Vd9 Vd10 Vd11 Vd12 He6 Ve12 Hf5 Vf10 Vf12 Hg5 Hg6 Vg12 Hh6 Vh8 Vh9 Vh12 Hi5 Vi11 Hj4 Hj6 Vj11 Hk3 Vk11 Vl11 Vc8 Vc9 Vc10 Vc11 Vc12 Vd7 Ve7 Ve8 Ve10 Ve11 Vf4 Hf6 Vf8 Vf11 Vg11 Hh5 Vh7 Vh11 Vi6 Vi8 Vi10 Vj5 Vj10 Vk4 Vk10 Vl10 Vm2 Ve9 Vf7 Vf9 Vg7 Vg8 Vg9 Vg10 Vh10 Vi7 Vi9 Vb13 Vc13 Vm10 Vm11 Vi12 Vj12 Vk12 Vl12 Vm12 Vd13 Ve13 Vf13 Vg13 Vh13 Vi13 Vj13 Vk13 Vl13 Vm13</hex>

The changes have been:

* f6 and h5 swap color, as in Example 1.

* The blue group from d8 to l11 is completely wrapped by a strongly connected group belonging to the opponent, except for a narrow section (typically, 2 cells). This kind of groups typically are captured, exactly in the same reason as in Example 1.

* If blue moves to i7, red moves to i9 and conversely, in both cases enclosing the blue group h8-h9-f10 as before. So, we can use the second rule for detecting equivalent patterns, capturing it.

* The equivalence in Example 5 can be used for the stone in c12.

* The equivalence in Example 4 can be used, adding a stone for Red at m11.

* The area in the bottom of the board is now surrounded by a red chain and a blue chain. Therefore, it may be filled as we please.

The blue stone in j6 remained, completely alone and too near to the red group to be of any use in such a small region, so it is obvious that Red has won.

=== Second example game ===

In the following game, Red's c9 is connected to the top edge via [[Fifth_row_edge_templates#V-2-a|edge template V-2a]]. A 3rd row ladder is about to form along the bottom edge. Red wonders whether his other stones are enough to escape the ladder.
<hexboard size="11x11"
contents="R a11 B d6 B f6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10"
/>

The analysis of this position seems complicated at first. It can be simplified by using the equivalence in [[#Example 10|Example 10]] to shift the blue stone from f6 to e6.
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10"
/>
The key observation is that the single stone i6 on the 6th row is enough for a 3rd-to-6th row switchback (see [[Switchback#B6_switchback|B6 switchback]]) or even for a 3rd-to-5th row switchback (see [[Switchback#C6_switchback|C6 switchback]]). The resulting ladder is then escaped by g6 via interior ziggurat. Specifically, if Red pushes the ladder all the way to g9, the red stone at g6 will be connected to this ladder by an interior ziggurat:
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R 1:d9 B 2:d10 R 3:e9 B 4:e10 R 5:f9 B 6:f10 R 7:g9 B 8:g10"
/>
Now Red can play the 3rd-to-6th row switchback by breaking the ladder and playing one more piece. Note how i6 becomes the ladder stone for a 3rd row ladder in the opposite direction, which the ziggurat escapes. Therefore, Red is connected.
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R d9 B d10 R e9 B e10 R f9 B f10 R g9 B g10
R 9:i9 B 10:h9 R 11:j7"
/>
One may wonder whether it would have helped Blue to [[Ladder_handling#Defending|yield]] to a 2nd row ladder. This would not have helped here. For example:
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R 1:d9 B 2:d10 R 3:e9 B 4:e10 R 5:f9 B 6:f11 R 7:f10 B 8:e11 R 9:h10 B 10:g10
R 11:h9 B 12:g9 R 13:h7"
/>
This time, Black's 13 acts as a ladder stone for a 2nd row ladder stone in the opposite direction, which the ziggurat again escapes.
[[Category: Strategy]]
[[Category: Computer Hex]]
[[Category: Theory]]

Equivalent patterns

2021-01-17T13:24:30Z

Tompo1: Added Edge equivalence section

We say that two Hex [[pattern]]s (subsets of a board) are '''equivalent patterns''' if, when one of them occurs embedded in ''any'' Hex board, it could be replaced by the other and the side who has winning strategy does not change.

== Edge equivalence==

Edges are equivalent to edge-connected rows of pieces on a larger board.

=== Example 0 ===

Any position on an nxm board is equivalent to the same position on the interior (i.e. the cells non-adjacent to the edges) of an (n+2)x(m+2) board with the first rows next to each edge filled with friendly pieces and [[dead cell]]s at the vertex cells (or equivalently either red or blue pieces at any of the vertex cells)

<hexboard size="4x4"
float="inline"
coords="none"
contents="S area(a1,d1,d4,a4)"
/> and <hexboard size="6x6"
float="inline"
coords="none"
contents="S area(b2,e2,e5,b5) R line(b1,e1) line(b6,e6) B line(a2,a5) line(f2,f5) E *:a1 *:f1 *:f6 *:a6"
/>

This allows the examples below to be applied at the edges: either formally, by converting to the larger board, applying the equivalence in the example, then converting back to the smaller board; or informally, by considering an edge to be a row of friendly pieces.

== Equivalence by capture ==

Cells that are [[Captured cell|captured]] by one player can be filled in with stones of that player's color.

=== Example 1 ===
For example, the two patterns below are equivalent, and an example of what is known as the [[useless triangle]].

<hexboard size="2x3"
float="inline"
coords="none"
edges="none"
visible="area(a2,c2,c1,b1)"
contents="R a2 b1 c1 c2 B b2"
/> and <hexboard size="2x3"
float="inline"
coords="none"
edges="none"
visible="area(a2,c2,c1,b1)"
contents="R a2 b1 c1 c2 R b2"
/>

The blue stone in the left pattern is [[captured cell|captured]] by Red. Therefore, the position is equivalent to one where this stone is actually red.

The knowledge of equivalent patterns turns out to be very useful for playing Hex, because it allows the player to disregard some pieces in the board, or prune the analysis tree. Some positions are much clearer than other equivalent positions. In the examples below, the simpler pattern is usually written on the right. One can make use of these equivalent patterns by mentally always replacing the left pattern with the right one.

Here are several examples of pairs of equivalent positions, and a short explanation on the way to prove them so.

=== Example 2 ===
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 b3 d3"
contents="R a2 b1 c1 d1 d2 c3 B b2 c2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 b3 d3"
contents="R a2 b1 c1 d1 d2 c3 R b2 c2"
/>

The equivalence is obtained by applying example 1 twice.

Both above examples are instances of the following rule to produce equivalence pairs. Given a [[chain]] G, let the '''neighborhood''' of G, neigh(G), be the set of cells that are adjacent to G but do not belong to it. In a given pattern P1, suppose that G is a red chain, and that C is a cell in neigh(G), which may be empty or occupied by Blue. Let P2 be the result of placing a red piece in C (removing a Blue stone, if necessary). Therefore P2 contains a chain G' containing G and C. If neigh(G')=neigh(G) then P1 and P2 are equivalent. Of course the dual argument also holds for blue chains.

This rule justifies the following equivalent pairs:

=== Example 3 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 c2 c3"
contents="R a2 b1 c1 B b2 b3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a3 c2 c3"
contents="R a2 b1 c1 R b2 B b3"
/>

=== Example 4 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 c2 c3"
contents="B a3 R b1 c1 B b2 b3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 c2 c3"
contents="B a3 R b1 c1 R b2 B b3"
/>

Another rule producing equivalent patterns: If there are two empty cells C1 and C2 in a pattern, such that if Blue occupies one of them, Red can occupy the other capturing the former, then an equivalent position is obtained if both C1 and C2 are occupied by Red.

Equivalent pairs obtained with this rule:

=== Example 5 ===
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="area(a1,a3,c1)"
contents="R a1--c1 a3"
/> and <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
visible="area(a1,a3,c1)"
contents="R all"
/>

This is a very instructive pattern because it shows that playing 2 rows out from a [[friendly]] edge with 2 free cells on the first row below the play is equivalent to playing at all three cells simultaneously, and hence at least as good as playing at either cell on the first row. So as a rule, one should never play on the first row in such a situation. We can see this by viewing the 3 upper red stones as part of a friendly edge.

This pattern, along with Example 4, can be used to show that the 2-move opening a1 + b1 (or either single move) for Red loses on a board 3x3 or bigger:

=== Example 5b (Opening A1+B1 loses) ===

<hexboard size="4x3"
float="inline"
coords="top left"
edges="top left"
contents="R a1 b1 B b2"
/> and <hexboard size="4x3"
float="inline"
coords="top left"
edges="top left"
contents="B a1 b1 B b2 a2 a3"
/>

In response to a1+b1 by Red, Blue can play at b2. This captures a2 and a3 by Example 5, and then a1 and b1 by applying Example 1 twice. Thus, the left position is equivalent to the right one. This includes the cells a1 and a2, so the original position is losing for Red by a strategy stealing argument.

In fact, making the opening play b2 is equivalent to playing 5 stones at a1, a2, b1, b1, and c1. However this is known to produce a losing position on boards of size 4, 7 & 8 (see [[small boards]]).

This example also shows that the opening moves a1 and b1 should never be swapped. Playing at b2 is always at least as good for Blue as swapping (but there are even better moves for Blue available elsewhere on the board).

=== Example 6 ===
<hexboard size="3x2"
float="inline"
coords="none"
edges="none"
contents="R a1 a3 b1 b3"
/> and <hexboard size="3x2"
float="inline"
coords="none"
edges="none"
contents="R a1 a2 a3 b1 b2 b3"
/>

=== Example 7 ===
<hexboard size="2x4"
float="inline"
coords="none"
edges="none"
visible="-a1"
contents="R a2--b1--d1--d2"
/> and <hexboard size="2x4"
float="inline"
coords="none"
edges="none"
visible="-a1"
contents="R all"
/>
Using both rules together:

=== Example 8 ===
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="area(a2,b3,d2,c1)"
contents="R c1 b3 B a2 d2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="area(a2,b3,d2,c1)"
contents="R c1 b3 B a2 d2 R b2 c2"
/>

=== Example 9 ===
''(generalization of the Example 5, for any horizontal length)''

<hexboard size="3x5"
float="inline"
coords="none"
edges="none"
visible="area(a3,e3,e1,c1)"
contents="R a3--e3 c1--e1 B c2 d2"
/> and <hexboard size="3x5"
float="inline"
coords="none"
edges="none"
visible="area(a3,e3,e1,c1)"
contents="R all"
/>

Here is a third (rather obvious) rule for equivalent patterns: Any area surrounded by a single chain of the opponent may be randomly filled. This happens because the outcome of the game does not depend on it at all.

== Equivalence for reasons other than capture ==

Capture is not the only way in which equivalent patterns arise. The following is an example of this.

=== Example 10 ===

The following two patterns are equivalent:
<hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 b3--d3"
contents="R b1--d1 d2 B a3 c2 E a:b2"
/> and <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 a2 b3--d3"
contents="R b1--d1 d2 B a3 b2 E a:c2"
/>
This is easily seen by noticing that if the cells marked "a" are both blue, the patterns become equal, whereas if the cells marked "a" are both red, then the blue piece next to each of them may as well be red by Examples 1 and 3, respectively, so the patterns are equivalent. So whenever one player's strategy calls for playing in the cell marked "a" in one of these two patterns, the same player can play in the cell marked "a" in the other pattern.

== Practical examples ==

=== First example game ===

Let us see a practical example. In game [http://www.littlegolem.net/jsp/game/game.jsp?gid=206040&nmove=55 #206040] at [[Little Golem]], the situation after 55. m2 is shown in the board below.

<hex>R13 C13 Q1 Hc1 Hd6 Hd8 Hd9 Hd10 Hd11 Hd12 He6 He12 Hf5 Hf10 Hf12 Hg5 Hg6 Hg12 Hh6 Hh8 Hh9 Hh12 Hi5 Hi11 Hj4 Hj6 Hj11 Hk3 Hk11 Hl11 Vc8 Vc9 Vc10 Vc11 Vc12 Vd7 Ve7 Ve8 Ve10 Ve11 Vf4 Vf6 Vf8 Vf11 Vg11 Vh5 Vh7 Vh11 Vi6 Vi8 Vi10 Vj5 Vj10 Vk4 Vk10 Vl10 Vm2</hex>

Clearly, m2 is strongly connected to the top, because the stone in f4 is a [[ladder escape]]. On the other hand, it is strongly connected to the bottom exactly if blue cannot connect k3 with the right, using j6 and maybe the [[group]] in h8-h9-f10 as a ladder escape. In fact he cannot do it, and it is much clearer if some patterns are locally replaced by other equivalent ones, rendering:

<hex>R13 C13 Q1 Hc1 Hd6 Vd8 Vd9 Vd10 Vd11 Vd12 He6 Ve12 Hf5 Vf10 Vf12 Hg5 Hg6 Vg12 Hh6 Vh8 Vh9 Vh12 Hi5 Vi11 Hj4 Hj6 Vj11 Hk3 Vk11 Vl11 Vc8 Vc9 Vc10 Vc11 Vc12 Vd7 Ve7 Ve8 Ve10 Ve11 Vf4 Hf6 Vf8 Vf11 Vg11 Hh5 Vh7 Vh11 Vi6 Vi8 Vi10 Vj5 Vj10 Vk4 Vk10 Vl10 Vm2 Ve9 Vf7 Vf9 Vg7 Vg8 Vg9 Vg10 Vh10 Vi7 Vi9 Vb13 Vc13 Vm10 Vm11 Vi12 Vj12 Vk12 Vl12 Vm12 Vd13 Ve13 Vf13 Vg13 Vh13 Vi13 Vj13 Vk13 Vl13 Vm13</hex>

The changes have been:

* f6 and h5 swap color, as in Example 1.

* The blue group from d8 to l11 is completely wrapped by a strongly connected group belonging to the opponent, except for a narrow section (typically, 2 cells). This kind of groups typically are captured, exactly in the same reason as in Example 1.

* If blue moves to i7, red moves to i9 and conversely, in both cases enclosing the blue group h8-h9-f10 as before. So, we can use the second rule for detecting equivalent patterns, capturing it.

* The equivalence in Example 5 can be used for the stone in c12.

* The equivalence in Example 4 can be used, adding a stone for Red at m11.

* The area in the bottom of the board is now surrounded by a red chain and a blue chain. Therefore, it may be filled as we please.

The blue stone in j6 remained, completely alone and too near to the red group to be of any use in such a small region, so it is obvious that Red has won.

=== Second example game ===

In the following game, Red's c9 is connected to the top edge via [[Fifth_row_edge_templates#V-2-a|edge template V-2a]]. A 3rd row ladder is about to form along the bottom edge. Red wonders whether his other stones are enough to escape the ladder.
<hexboard size="11x11"
contents="R a11 B d6 B f6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10"
/>

The analysis of this position seems complicated at first. It can be simplified by using the equivalence in [[#Example 10|Example 10]] to shift the blue stone from f6 to e6.
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10"
/>
The key observation is that the single stone i6 on the 6th row is enough for a 3rd-to-6th row switchback (see [[Switchback#B6_switchback|B6 switchback]]) or even for a 3rd-to-5th row switchback (see [[Switchback#C6_switchback|C6 switchback]]). The resulting ladder is then escaped by g6 via interior ziggurat. Specifically, if Red pushes the ladder all the way to g9, the red stone at g6 will be connected to this ladder by an interior ziggurat:
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R 1:d9 B 2:d10 R 3:e9 B 4:e10 R 5:f9 B 6:f10 R 7:g9 B 8:g10"
/>
Now Red can play the 3rd-to-6th row switchback by breaking the ladder and playing one more piece. Note how i6 becomes the ladder stone for a 3rd row ladder in the opposite direction, which the ziggurat escapes. Therefore, Red is connected.
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R d9 B d10 R e9 B e10 R f9 B f10 R g9 B g10
R 9:i9 B 10:h9 R 11:j7"
/>
One may wonder whether it would have helped Blue to [[Ladder_handling#Defending|yield]] to a 2nd row ladder. This would not have helped here. For example:
<hexboard size="11x11"
contents="R a11 B d6 B e6 R c4 B d7 R i6 B i5 R g6 B h4 R g5 B g4 R f5 B f4 R e5
B e3 R b5 B d4 R c6 B b6 R c5 B d5 R b8 B b9 R c8 B c10 R c9 B b10
R 1:d9 B 2:d10 R 3:e9 B 4:e10 R 5:f9 B 6:f11 R 7:f10 B 8:e11 R 9:h10 B 10:g10
R 11:h9 B 12:g9 R 13:h7"
/>
This time, Black's 13 acts as a ladder stone for a 2nd row ladder stone in the opposite direction, which the ziggurat again escapes.
[[Category: Strategy]]
[[Category: Computer Hex]]
[[Category: Theory]]