HexWiki - User contributions [en]

User:StillYetAnother11

2026-03-20T18:58:57Z

Selinger: Minor typos

== log2(1.5) ratio ==

I was thinking about why katahex evaluates a7 as the most balanced opening on 12x12 which is very anomalous because openings in the middle of the 1st column tend to be quite strong, although in Hayward's 2003 paper, a4 on 7x7 has the highest number of nodes of all solved openings. The smallest board size after 12x12 where such an opening seems quite playable is 17x17 since katahex evaluates a10 as more balanced than its neighbors and is still one of the most balanced openings on the first column. A pattern starts to emerge where if you encode the acute corner as 0 and the obtuse corner as 1 you'll find that these openings match ratios that well approximate a particular ratio namely log2(1.5). The first semiconvergents of log2(1.5) are 1/1, 1/2, 2/3, 3/5, 4/7, 7/12, 10/17, 17/29, 24/41, 31/53 and 55/94. The next fraction after 10/17 is 17/29 so is a17 on 29x29 more balanced than its neighbors? I've run 200 engine games for each opening: a16, a17 and a18 and the results were 110 wins for a16, 102 wins for a17 and 112 wins for a18. I also ran 100 games on 41x41 for each of the following openings. The results were 57 wins for a23, 49 wins for a24 and 56 wins for a25 so the pattern seems to be consistent. This has a benefit when it comes to opening variety because most balanced openings in hex are predominantly on your own edge and not on your opponent's edge. I believe that having the denominators of semiconvergents of log2(1.5) as board sizes is the most logical and consistent way of deriving board sizes in hex.

== The best way for red to fight in c2 opening on 11x11 ==

<hexboard size="11x11"
coords="hide"
contents="R 1:c2 E *:c4 E *:d5 R 3:b8 B 2:d8 R 5:c9 B 4:f9"
/>

I prefer to play 5.c9 here even though it's not the best move according to katahex. Moves marked with * are the only moves winning for blue.

<hexboard size="11x11"
coords="hide"
contents="R 1:c2 R 7:j4 B 6:d5 E *:f5 E *:f7 R 3:b8 B 2:d8 R 5:c9 B 4:f9"
/>

7.j4 is again not the top engine move but here is why I prefer it. Moves marked with * are the only winning moves for blue but there are still ways for blue to lose if they don't play this very concrete position correctly.

<hexboard size="11x11"
coords="hide"
contents="R 1:c2 R 7:j4 B 6:d5 B 8:f5 E +:b7 R 3:b8 B 2:d8 R 5:c9 B 4:f9"
/>

At this point red can try a bunch of different moves. Blue needs to know that the cell marked with + is the only winning move to a lot of red's responses. You should continue to analyze this position yourself with katahex.

== g3 opening on 13x13 ==

Additional tests with katahex showed that after d10 i7 is the only winning move for red.

<hexboard size="13x13"
coords="hide"
contents="R 1:g3 R 3:i7 B 2:d10"
/>

While j9 is losing for red. If red plays * then + is the only winning response for blue.

<hexboard size="13x13"
coords="hide"
contents="R 1:g3 E +:g4 B 6:i5 E *:d8 R 3:j9 B 2:d10 R 5:i10 B 4:j10"
/>

Computer Hex

2026-02-18T05:16:00Z

Selinger: /* Front End */ Added Hexata.

This page lists some software programs and programming topics that may be of interest to [[Hex]] players. The programs include AI opponents and tools for analysing completed games.

More complete or up-to-date information is welcome.

== AI techniques used in Hex ==

*[[Minimax (computer)|Minimax]] and alpha-beta search were used by [[Queenbee]].
*[[UCT]] is used in MoHex.

== Programs with AI ==

There are several computer programs which play Hex.

=== Available programs ===

{| class="wikitable"
|-
! Program !! Platforms !! Remarks
|-
| [[KataHex]] || Linux, Windows || By HZY. As of 2024, the strongest available Hex program. Very strong AI, open source.
|-
| [[MoHex]] || Linux || Was the strongest available Hex program in 2010. It uses the UCT-Monte Carlo approach and is developed at the University of Alberta by Philip Henderson, Broderick Arneson and Ryan Hayward. Also has an efficient solver.
|-
| [[Hexy]] || Windows || The second strongest program available. It was the first program to use virtual connections and was champion of the 5th Computer Olympiad in 2000.
|-
| [[Wolve]] || Linux || Gold medallist of 2008 Computer Olympiads.
|-
| [[Six]] || Linux, Unix, Windows || by Gábor Melis.
|-
| [http://www.mattesmedjan.se/hexilla/ Hexilla] || Java || By Jonatan Rydh, released in October 2009.
|-
| [https://play.google.com/store/apps/details?id=com.game.hex Hex] || Android || By Five Factorial, released in January 2017. It uses MoHex engine on Expert level.
|}

=== Mac platform ===

No known programs for the Mac are available. There is a work around by using an emulator such as BlueStacks that allows Android programs to be run.

=== Unavailable programs ===

{| class="wikitable"
|-
! Program !! Platforms !! Remarks
|-
| [[Mongoose]] || || By [[Yngvi Björnsson]], [[Ryan Hayward]], Mike Johanson, Morgan Kan, and Nathan Po.
|-
| [[Queenbee]] || || By [[Jack van Rijswijck]]. Won silver at the London 2000 CGO.
|-
| [[Hexy (iPhone)|Hexy]] || iPhone || Despite using the same name, this program has no relation to [[Hexy]]. It was released in November 2008, offers an AI opponent; the AI appears to be a custom design and hasn't been rated. As of 26 December 2019 it is no longer available.
|-
| [https://itunes.apple.com/app/id423845369 Hexatious] || iPad, iPhone || Released in August 2009, appears to offer a stronger AI than the iPhone Hexy app (in particular, Hexatious easily beats the other iPhone app in head-to-head competition). As of 26 December 2019 it is no longer available.
|-
| [https://itunes.apple.com/app/id397349481 Hex Nash] || iPad, iPhone || Released February 2011, no AI but supports online asynchronous play and local play. As of 26 December 2019 it is no longer available.
|}

== Non playing programs ==

=== Front End ===
* [[HexGui]] is a graphical user interface designed by [[Broderick Arneson]] ("ab"). It can be used as an interactive game board to try out plays and variations, and it can also be used as a front end for any computer Hex program that can communicate via [[GTP]]. It works well as a front-end to [[MoHex]]. HexGui can read and write the [[Smart Game Format]]. An up-to-date version of HexGui is available from [https://github.com/selinger/hexgui GitHub].
* Hexata [https://github.com/hexanna1/hexata] is a lightweight, keyboard-first GUI designed by [[User:Hexanna|Hexanna]]. It is written in Python and can interact with [[KataHex]].

=== Reviewing and Editing Programs ===

* [https://minortriad.com/ahex.html AHex] by [[User:Tom239|Tom Ace]] lets you analyze Hex and Havannah games and can import games from littlegolem.net.
* [http://canyon23.net/jgame/README_hex.html JHex] by Kevin lets you analyse a game, and databases of games.
* [http://www.drking.org.uk/hexagons/hex KHex] by David King is a tool for reviewing games. Very well suited for sharing commented games (it exports games in [[Smart Game Format]]).
* [http://www.drking.org.uk/hexagons/hex KHex18] by David King is an online app for reviewing games, which can read LittleGolem game text.

== Protocols ==

* [[GTP]] is a text-based protocol for interacting with Hex software. It is based on the Go Text Protocol, and allows Hex software to interact with Hex strategy engines.

== File formats ==

* The [[Smart Game Format]] (SGF) is a file format for storing annotated game trees. The format nor only stores a sequence of moves comprising a game, but can also contain variations (several different games played out from the same position), as well as comments on every move in the game.

== External link ==

=== Articles ===

*Anshelevich, Vadim V. [http://home.earthlink.net/~vanshel/VAnshelevich-ARTINT.pdf A hierarchical approach to computer Hex].
*van Rijswijck, Jack. [http://home.fuse.net/swmeyers/y-hex.pdf Search and evaluation in Hex].
*Rasmussen, Rune K. and Maire, Frederic D. and Hayward, Ross F. (2006) [http://eprints.qut.edu.au/5121/1/5121_1.pdf A Move Generating Algorithm for Hex Solvers].
*Rasmussen, Rune K. (2008) [http://eprints.qut.edu.au/18616/1/01Thesis.pdf Algorithmic approaches for playing and solving Shannon games] (PhD Thesis).

== See also ==
[[History of computer Hex]]

The [[ICGA|International Computer Games Association]] also has some [http://www.cs.unimaas.nl/icga/games/hex/ information on Hex]. They organize an annual [[Computer Olympiad]], which also covers Hex.

[[category:Computer Hex]]

Building KataHex

2026-01-25T16:42:08Z

Selinger: Some copy-editing. Added an introductory paragraph.

This page contains detailed instructions for building the [[KataHex]] software from source code. KataHex is a free and open-source computer Hex program. For general information about KataHex, see [[KataHex|the main KataHex page]].

There is more than one version of KataHex out there, so the instructions on this page have variants.

----
== Ubuntu Linux ==
You can get Selinger's version of KataHex with the following command, which connects you to the Git history, and enables you to contribute edits:
<pre>
git clone https://github.com/selinger/katahex.git
</pre>

This creates a katahex directory wherever you execute that command. If you want to build multiple variants, you will have to do a separate load for each one
(trying to reuse a build directory creates chaos). Just rename the one you just created to reflect the variant name, and do another git clone.

You may have to install additional software to work with KataHex, using <code>apt-get</code> or <code>synaptic</code> to install packages.
For example, the TCMALLOC option protects you from problems related to memory management when you run more than one instance at at time. You may have to install one or more packages.
* <code>libtcmalloc-minimal4t64</code>
* <code>librust-tcmalloc-dev</code>
* <code>librust-tcmalloc-sys-dev</code>

The code works with at least 4 backends (software where the real work is done), EIGEN, OPENCL, CUDA, AND TENSORRT. Which of these backends you should choose depends on what hardware you own. If you do not have a GPU, use the EIGEN backend.

==== EIGEN (CPU only) ====

The following BASH commands then may do the trick, but will produce code that may only work on the exact CPU model you use for the build. See comments in the KataHex source code, as well as the KataGo source code (on with KataHex is based, and which contains more details).

<pre>
cmake . -DUSE_BACKEND=EIGEN -DUSE_AVX2=1 -DUSE_TCMALLOC=1 -DCMAKE_CXX_FLAGS='-march=native' -DMAX_BOARD_LEN=19
make -j 4
</pre>

==== OPENCL ====
You may be able to use almost the same commands as for EIGEN, just by changing to -DUSE_BACKEND=OPENCL.

==== CUDA ====
I have not yet been able to build with the CUDA backend.

==== TENSORRT ====
I have not yet been able to build with the TENSORRT backend.

Building KataHex

2026-01-25T16:34:30Z

Selinger: Selinger moved page Buillding KataHex to Building KataHex: Fixing a typo in the page name

There is more than one version of KataHex out there, so these instructions have variants

----
== Ubuntu Linux ==
You can get Selinger's version of KataHex with the following command, which connects you to the Git history, and enables you to contribute edits:
<pre>
git clone https://github.com/selinger/katahex.git
</pre>

This creates a katahex directory wherever you execute that command. If you want to build multiple variants, you will have to do a separate load for each one
(trying to reuse a build directory creates chaos). Just rename the one you just created to reflect the variant name, and do another git clone.

You may have to install additional software to work with KataHex, using <code>apt-get</code> or <code>synaptic</code> to install packages.
For example, the TCMALLOC option protects from problems related to memory management when you run more than one instance at at time. You may have to install one or more packages
* <code>libtcmalloc-minimal4t64</code>
* <code>librust-tcmalloc-dev</code>
* <code>librust-tcmalloc-sys-dev</code>

The code works with at least 4 backends (software where the real work is done), EIGEN, OPENCL, CUDA, AND TENSORRT

==== EIGEN (CPU only) ====

The following BASH commands then may do the trick, but will produce code that may only work on the exact CPU model you use for the build. See comments in the KataHex source code, as well as the KataGo source code (on with KataHex is based, and which contains more details).

<pre>
cmake . -DUSE_BACKEND=EIGEN -DUSE_AVX2=1 -DUSE_TCMALLOC=1 -DCMAKE_CXX_FLAGS='-march=native' -DMAX_BOARD_LEN=19
make -j 4
</pre>

==== OPENCL ====
You may be able to use almost the same commands as for EIGEN, just by changing to -DUSE_BACKEND=OPENCL

==== CUDA ====
I have not yet been able to build with the CUDA backend

==== TENSORRT ====
I have not yet been able to build with the TENSORRT backend

Buillding KataHex

2026-01-25T16:34:30Z

Selinger: Selinger moved page Buillding KataHex to Building KataHex: Fixing a typo in the page name

#REDIRECT [[Building KataHex]]

Combinatorial game theory

2026-01-12T23:16:13Z

Selinger: /* Combinatorial game notation */ typo

''This page is about the combinatorial game theory of Hex. For combinatorial game theory in general, see [https://en.wikipedia.org/wiki/Combinatorial_game_theory Wikipedia].''

Combinatorial game theory is a formalism that can be used to analyze sequential two-player perfect information games. It was developed by Berlekamp, Conway, and Guy starting in the 1960s, and has since been applied to a variety of games, including Hex. One of the central ideas of combinatorial game theory is that games can be split into multiple components (for example, separate regions of the Hex board), and each component analyzed separately. Combinatorial game theory can be used to reason about such concepts as [[domination]], reversibility, [[inferiority|inferior moves]], and [[equivalent patterns|equivalence of positions]].

== Concepts ==

In this section, we introduce some of the main concepts of combinatorial game theory, as it relates to Hex. We do this mostly by example. A more mathematical treatment follows later.

=== How the game ends ===

Normally, a game of Hex ends either when a player has connected their edges, or when a player resigns. However, for the game theoretic analysis, it is easier if we imagine that the game continues until the board is completely filled with stones. Notice that after one player has a winning path, filling the rest of board with stones can no longer change who the winner is. So we are free to assume that the game continues until there are no more empty cells on the board. We will make this assumption throughout this article.

=== Regions ===

By a ''region'' of the Hex board, we just mean a set of one or more cells on the board. The region may include special features such as board edges, and it can optionally be pre-populated with some stones.

'''Examples.'''

* The whole board is a region.
*: <hexboard size="5x5"
coords="none"
/>

* A single empty cell is a region.
*: <hexboard size="1x1"
coords="none"
edges="none"
/>

* The 3-cell corner is a region.
*: <hexboard size="2x2"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
/>

* The following region is completely surrounded by red and blue stones, with three groups or red stones (and therefore three groups of blue stones) in the boundary. This kind of region is called a ''3-terminal region''.
*: <hexboard size="5x5"
coords="none"
edges="none"
visible="-a1 area(d5,e5,e3)"
contents="R b1 a2 a3 B a4 a5 R b5 c5 d4 B d3 e2 R e1 d1 B c1"
/>

=== Outcomes ===

By an ''outcome'' for a region, we mean one of the possible ways of completely filling it with stones. Since we assume that the game continues until the board is completely filled, the outcome describes what the region may look like at the end of the game. We say that two outcomes for the same region are ''equivalent'' if for every way of filling the rest of the board (i.e., the part outside of the region) with stones, the first outcome produces the same winner as the second one. Since equivalent outcomes produce the same winner in all situations, we often consider them to be equal, i.e., when we say "the same outcome", we mean "equivalent outcomes".

'''Examples.'''

* When the region is the whole Hex board, there are only two possible outcomes, namely, "Red wins" and "Blue wins".

* When the region is a single empty cell, there are two possible outcomes, namely, "Red gets the cell" and "Blue gets the cell".

* For the 3-cell corner, there are 8 possible ways of filling it with red and blue stones. However, most of these are equivalent, and we end up with only 3 non-equivalent outcomes:

*: '''Outcome 1: Red gets everything.''' This outcome can be achieved in 3 possible ways; notice that they are equivalent because the blue stone is [[dead cell|dead]] in all cases.
*: <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="R a2 b1 b2"
/> <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="B a2 R b1 b2"
/> <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="R a2 b1 B b2"
/>
*: '''Outcome 2: Blue gets everything.''' This outcome can also be achieved in 3 possible ways.
*: <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="B a2 b1 b2"
/> <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="B a2 R b1 B b2"
/> <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="B a2 b1 R b2"
/>
*: '''Outcome 3: Red and Blue each get half of the corner.''' This outcome can be achieved in 2 possible ways, because the corner cell is [[dead cell|dead]].
*: <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="R a2 B b1 R b2"
/> <hexboard size="2x2"
float="inline"
coords="none"
edges="bottom right"
visible="area(b1,a2,b2)"
contents="R a2 B b1 B b2"
/>

* For a 3-terminal region, there are 5 possible outcomes. Although there are many ways of filling the region with red and blue stones, at the end of the game, all that matters is which terminals are connected to each other. Suppose Red's terminals have been numbered 1, 2, and 3. Then from Red's point of view, the 5 outcomes are:
** Red connects all 3 terminals.
**: <hexboard size="5x5"
coords="none"
edges="none"
visible="-a1 area(d5,e5,e3)"
contents="R b1 1:a2 a3 B a4 a5 R b5 3:c5 d4 B d3 e2 R e1 2:d1 B c1
B b2 R b3 b4 c2 B c3 c4 d2"
/>
** Red connects only terminals 1 and 2.
**: <hexboard size="5x5"
coords="none"
edges="none"
visible="-a1 area(d5,e5,e3)"
contents="R b1 1:a2 a3 B a4 a5 R b5 3:c5 d4 B d3 e2 R e1 2:d1 B c1
B b2 R b3 B b4 R c2 B c3 c4 d2"
/>
** Red connects only terminals 1 and 3.
**: <hexboard size="5x5"
coords="none"
edges="none"
visible="-a1 area(d5,e5,e3)"
contents="R b1 1:a2 a3 B a4 a5 R b5 3:c5 d4 B d3 e2 R e1 2:d1 B c1
B b2 R b3 b4 B c2 c3 c4 d2"
/>
** Red connects only terminals 2 and 3.
**: <hexboard size="5x5"
coords="none"
edges="none"
visible="-a1 area(d5,e5,e3)"
contents="R b1 1:a2 a3 B a4 a5 R b5 3:c5 d4 B d3 e2 R e1 2:d1 B c1
B b2 b3 b4 c2 R c3 c4 d2"
/>
** Red connects none of her terminals.
**: <hexboard size="5x5"
coords="none"
edges="none"
visible="-a1 area(d5,e5,e3)"
contents="R b1 1:a2 a3 B a4 a5 R b5 3:c5 d4 B d3 e2 R e1 2:d1 B c1
B b2 b3 b4 c2 c3 c4 d2"
/>

=== Positions and options ===

Consider a region of the Hex board. By a ''position'' in the region, we mean the state of the region after zero or more moves have been played, or in other words, a version of the region in which zero or more of its emtpy hexes have been filled with stones (of any color).

Players change positions by making moves. In a given position, we say that ''Red's options'' are all positions that Red can reach by making a single move in the region, and ''Blue's options'' are all the positions that Blue can reach by making a single move in the region.

'''Example.''' Consider the position
G = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R a2 c1 b3 B b1 a3 d1 d2 c3"
/>
In this position, Red has two options:
G₁ = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R a2 c1 b3 B b1 a3 d1 d2 c3 R b2"
/> and G₂ = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R a2 c1 b3 B b1 a3 d1 d2 c3 R c2"
/>
Also, Blue has two options:
H₁ = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R a2 c1 b3 B b1 a3 d1 d2 c3 B b2"
/> and H₂ = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R a2 c1 b3 B b1 a3 d1 d2 c3 B c2"
/>

=== Combinatorial game notation ===

In combinatorial game theory, the players are called Left and Right. In Hex, Red is Left and Blue is Right. We therefore call Red's options ''left options'' and Blue's options ''right options''.

In combinatorial game theory, the word ''game'' is used to mean ''position''. Thus, an unfinished game has left and right options that are themselves games.

The notation for a combinatorial game is G = {G₁, …, Gₙ | H₁, ..., Hₘ}, where G₁, …, Gₙ are the left options and H₁, ..., Hₘ are the right options. Note that the game is enclosed in curly braces, and a vertical bar is used to separate the left options from the right options.

'''Example.''' Consider the position from the previous example:
G = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R a2 c1 b3 B b1 a3 d1 d2 c3"
/>
As we saw, G has two left options G₁ and G₂ and two right options H₁ and H₂. We can write

G = {G₁, G₂ | H₁, H₂} = [[File:CGT-example1.png|x65px]]

Recursively, each of the positions G₁, G₂, H₁, H₂ is also a combinatorial game with a left and right option. For example,

G₁ = [[File:CGT-example2.png|x65px]] = [[File:CGT-example3.png|x65px]]

By recursively decomposing all options, we arrive at this representation of the game G:

G = [[File:CGT-example4.png|x65px]]

When a position has no empty cells, it is called an ''atomic'' position. Atomic positions have no left or right options and cannot be further decomposed. Instead, we can replace each atomic position by the associated outcome. For example, the position G in this example is a 3-terminal position. Let us write "⊤" (pronounced "top") for the outcome "Red connects all three terminals", "⊥" (pronounced "bottom") for the outcome "Red connects no terminals", and "a" for the outcome "Red connects the right two terminals, but not the left one". With these notations, the combinatorial game G can be more succinctly described as:

G = {{⊤|⊤}, {⊤|a} | {a|⊥}, {⊤|⊥}}.

'''Remark.''' In a combinatorial game, the players do not always alternate. For example, in the game G, Red can move to the left option G₂ = {⊤|a}. Then Red can move again to the left option ⊤. The reason that players are not necessarily alternating is that the combinatorial game is meant to represent play ''in a region''. For example, Red might move in the region, Blue might move somewhere outside the region, and then Red might move in the region again. Thus, although the players were alternating in the game at large, within the region, Red has made two consecutive moves.

=== Simplification of combinatorial games ===

The combinatorial game notation we introduced in the previous section is not very compact. In fact, it is basically a representation of the entire game tree, i.e., a description of all moves that both players could make in every possible position of the game. For a position with more than a few empty cells, the game tree is much too large to write down in its entirety.

Fortunately, combinatorial games can be simplified. Much like the rules by which school children learn to simplify (5 + 8) * 2 to 13 * 2 and then to 26, there is a specific set of rules for simplifying combinatorial games and calculating their ''value''. These rules are a bit complicated and we will describe them in more detail in the mathematical section below. However, the main point is that the rules are mechanical and universal: every finite combinatorial game can be simplified to a unique value, and this can be done either by hand or by using a "CGT calculator". Moreover, when a Hex game has been decomposed into several disjoint regions, we can in principle calculate and simplify the value of each region independently, and then combine them to determine the value, and therefore the winner, of the entire game. In practice, this calculation is too hard when there is lots of empty space on the board, but can be feasible for certain end game positions.

'''Remark.''' Unlike in commerce, where the value of everything can be expressed in terms of money, the value of a Hex position is not simply a number. Rather, it is a specially simplified combinatorial game that is a concise description of what each player can get from the position. When two positions have the same value, then they are equivalent, even if they look very different on the Hex board. The following example shows the value of a Hex position, and also explains what that value means in English.

'''Example.''' Once again, we consider the position from the previous example:
G = <hexboard size="3x4"
float="inline"
coords="none"
edges="none"
visible="-a1 d3"
contents="R 1:a2 2:c1 3:b3 B b1 a3 d1 d2 c3 E x:b2 y:c2"
/>
As we saw, the combinatorial notation for this position is

: G = {{⊤|⊤}, {⊤|a} | {a|⊥}, {⊤|⊥}}.

Using the simplification rules we will discuss in more detail later, this can be simplified (or "evaluated") to

: G = {⊤ | {a | ⊥}}.

The English explanation of this value is something every Hex player will agree with: If Red moves first in the region, Red moves at x and connects all three terminals. After this, the region is [[settled region|settled]], i.e., neither Blue nor Red wants to make another move there. If Blue moves first in the region, Blue moves at x and ''threatens'' to connect all of Blue's terminals, but Red can reply at y to achieve outcome a. In this latter case, Red connects terminals 2 and 3, but not terminal 1.

== Comparison games ==

=== Comparing positions ===

When we consider two different Hex positions in the same region, we often ask questions like: are the positions equivalent? Is one of them better for Red than the other? We can formalize this notion by defining an ''order'' on positions.

Given two positions P and Q in the same Hex region, we say that P ≤ Q if Q is at least as good for Red as P. The symbol "≤" is pronounced "less than or equal", and is called an ''order relation''. Note that the order is always defined from Red's point of view, i.e., by "better", we mean "better for Red". But of course Blue has the opposite point of view: from Blue's point of view, P ≤ Q means that P is at least as good for Blue as Q.

'''Remark.''' The ordering of positions is not total: one can have positions P and Q such that neither P ≤ Q nor Q ≤ P holds. For example, P could be better for Red in some situations, and better for Blue in others. Such positions are called ''incomparable''.

When two positions are such that both P ≤ Q and Q ≤ P hold, then P is at least as good for Red as Q and vice versa. In that case, we say that the positions are ''equivalent''.

So far, we have not been very precise about what we mean by "at least as good". One way to define this is by considering the region to be part of a larger game, and then check to see who is winning that game. Specifically, P ≤ Q means that for every way of embedding the region inside a larger game, and no matter whose turn it is, if P is winning for Red, then Q is also winning for Red.

=== The comparison game ===

In practice, we do not have time to think about all of the infinitely many ways that the rest of the game could be affecting our region. It turns out that there is a much simpler way to prove that P ≤ Q, which only requires us to think about P and Q, and not about the rest of the board. This is done by playing the so-called comparison game, which we now describe.

Let P and Q be two positions in the same Hex region. The ''comparison game'' is a game played by two players, who are called ''Truthifier'' and ''Falsifier''. Truthifier's goal is to prove that P ≤ Q and Falsifier's goal is to prove that P ≰ Q. Therefore, to prove P ≤ Q, we must describe a winning strategy for Truthifier.

The game is played as follows. The positions P and Q are set up next to each other. The players alternate, with Falsifier going first.
A move by Falsifier consists of either placing a red stone in P or placing a blue stone in Q. A move by Truthifier consists of either placing a blue stone in P or placing a red stone in Q. Thus, whenever Falsifier plays in one component (P or Q), Truthifier has the option to either respond in the same component (using the opposite color) or in the opposite component (using the same color). The game continues until both P and Q are completely filled with stones, i.e., there are no more empty cells. At this point, it is easy to check whether P ≤ Q, because it amounts to checking that for every red path through P, there is a corresponding red path through Q. If this is the case at the end of the game, Truthifier wins; otherwise, Falsifier wins.

'''Remark.''' The above description of the comparison game works for Hex. For some other classes of combinatorial games, the description might be slightly more complicated, depending on such details as if and when passing is allowed. We will get back to this point in the mathematical section below.

'''Example.''' The page [[theorems about templates]] claims that B is at least as good for Red as A.
A: <hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="B b1 b2 E x:a2"
/> B: <hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="B a2 R b2 E y:b1"
/>
We now show how to prove A ≤ B using the comparison game. We must describe a winning strategy for Truthifier. Falsifier makes the first move. If Falsifier plays a red stone at x, Truthifier responds with a red stone at y. Since this kills the blue stone, the outcome of B is better for Red than that of A, so Truthifier wins. On the other hand, if Falsifier plays a blue stone at y, Truthifier responds with a blue stone at x. Since B has a red stone and A doesn't, B again has a better outcome than A, so Truthifier wins. Since Truthifier wins the comparison game no matter how Falsifier plays, we have proved A ≤ B.

Note that in this example, Truthifier always responds to Falsifier's move by playing in the opposite component. There are no other choices, since each component only has a single empty cell. Let us now consider a more complicated example.

'''Example.''' Here is another claim from [[theorems about templates]]. We claim that A ≤ C, where
A: <hexboard size="3x4"
float="inline"
edges="none"
coords="none"
visible="area(a2,b3,d1,c1)"
contents="R c1 b2 b3 B a2 d1 E w:c2"
/> C: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a2,b3,d1,c1)"
contents="R c1 b3 B a2 E x:b2 y:c2 z:d1"
/>
To prove it, we consider every possible move by Falsifier.
* If Falsifier plays Red w, Truthifier responds with Red y, killing x, and leaving A ≤ C (since C has an empty cell where A has a blue stone).
* If Falsifier plays Blue x, Truthifier responds with Red y, killing x and leaving A ≤ C.
* If Falsifier plays Blue z, Truthifier responds with Red x, leaving the two regions equal.
* If Falsifier plays Blue y, Truthifier responds with Red x. Now if Falsifier plays Blue z, Truthifier responds with Blue w, leaving the two regions equal. If Falsifier instead plays Red w, Truthifier responds with Red z, capturing y and leaving C better than A.

'''Remark.''' In the last example, we did not always play the game to the end. Indeed, as soon as Truthifier reaches a position in which it is clear that A ≤ C, we can stop the comparison game, as we already know that Truthifier has a winning strategy from that point onwards.

=== Proof that the comparison game works ===

Let P and Q be two positions in the same Hex region, and suppose that Truthifier has a winning strategy for the comparison game. We claim that P ≤ Q in the sense that for any position of the rest of the board, if Red has a winning strategy for P in that position, then Red has a winning strategy for Q in that position. Indeed, Red can win in Q by following the following procedure.
Red sets up two Hex boards. The left board contains the game with position P in the region, and the right board contains the same game, but with position Q in the region. It is the same color's turn on both boards. Red's goal is to win against Blue on the right board. The left board is only for Red's private use.

Case 1: It is Blue's turn. Then Blue makes a move on the right board. Case 1.1: Blue's move is outside the region. Then Red copies the same move, using a blue stone, on the left board. Now it is Red's turn on both boards. Case 1.2: Blue's move is within the region Q. Then Blue just played one of Falsifier's moves in the comparison game. By assumption, Truthifier has a winning strategy for the comparison game. If Truthifier's strategy is to make a red move in Q, then Red makes the corresponding move in Q. It is now Blue's turn again on both boards. If Truthifier's strategy is to make a blue move in P, then Red makes the corresponding move, using a blue stone, in P. If is now Red's turn on both boards. Case 2: It is Red's turn. By assumption, Red has a winning strategy on the left board. So Red plays a winning move on the left board. Case 2.1: Red's move is outside the region. Then Red makes the same move on the right board. It is now Blue's turn on both boards. Case 2.2: Red's move is in the region P. Then Red just played one of Falsifier's move in the comparison game. Truthifier has a winning strategy in the comparison game. If Truthifier's winning strategy is to make a blue move in P, then Red makes that move (using a blue stone) on the left board. It is now Red's turn again on both boards. If Truthifier's winning strategy is to make a red move in Q, then Red makes that move on the right board. It is now Blue's turn on both boards.

Red can continue playing on both boards in the above manner until both boards are filled. Since Red followed a winning strategy on the left board, Red has a winning path. Since both boards are identical outside the region, and Q (now being completely filled) is at least as good as P, Red also has a winning path on the right board. Therefore Red wins the game against Blue.

'''Remark.''' We showed that if Truthifier has a winning strategy for the comparison game, then P ≤ Q. The converse is not always true. In other words, it is possible for Truthifier to lose the comparison game, even though Q is at least as good as P in all contexts of larger Hex games. However, such situations are relatively rare.

== Mathematical details ==

To be written.

== References ==

* J. H. Conway, "On Numbers and Games". 1st edition 1976. 2nd edition, A. K. Peters, 2001.
* P. Selinger, "On the combinatorial value of Hex positions", [http://math.colgate.edu/~integers/vol22.html#g3 Integers 22:G3], 2022. Also available from [https://arxiv.org/abs/2101.06694 arxiv:2101.06694]

[[Category: Theory]]

Board Game Arena

2026-01-10T23:45:39Z

Selinger: Updates

Board Game Arena ([https://www.boardgamearena.com www.boardgamearena.com]), often abbreviated BGA, is a website offering online play for a large number of board games, including many commercially published games. It was created in 2010 by Grégory Isabelli and Emmanuel Colin, and has more than 10 million users in 2026.

Until approximately 2024, Board Game Arena was the most popular [[online playing|site]] for playing real-time Hex, but it has now been overtaken by [[PlayHex]]. (The other most popular Hex site, [[Little Golem]], only offers turn-based play).

== Hex on Board Game Arena ==

=== Board sizes ===

Board Game Arena offers six different board sizes for Hex: 6x6 (beginner), 11x11 (classic), 12x12 (twelve), 13x13 (expert), 14x14 (Nash), 15x15 (big), and 19x19 (huge). The 6x6 board size is only available in training mode, i.e., players cannot earn or lose [[Elo rating|Elo points]] for playing 6x6.

=== Time controls ===

There are several time control options:

* Real-time play, at three different speeds: fast, normal, and slow. The actual time per move changes dynamically based on past games, but a typical value for normal speed is 3 minutes initially, with 40 additional seconds per move (up to a maximum of 3 minutes).

* Turn-based play, ranging from one move per 2 days (slowest option) to 24 moves per day (fastest option).

* Training mode (no time control). Training mode games do not affect [[Elo rating]].

When a player goes over time, the game may continue, but at some point the opponent gets the opportunity to kick the player out.

=== Passing ===

Hex on Board Game Arena includes a passing move. Three consecutive passes are not allowed, so if a player passes, the opponent can effectively reject the pass by passing themselves, forcing the player to make a move.

=== Statistics ===

In October 2021, 1643 games of Hex were played on Board Game Arena (53 games per day). Of these, 1352 (82%) were real-time games. The most popular board size was 11x11 (93%). During the site's lifetime to October 2021, 26815 players have played at least one game of Hex, and 424 players have played at least 50 games.

In January 2026, 426 games of Hex were played (13 games per day). Of these, 251 (59%) were real-time games. The most popular board size was 11x11 (77%). During the site's lifetime to January 2026, 38679 players have played at least one game of Hex, and 648 players have played at least 50 games.

== Rating system ==

Board Game Arena uses an [[Elo rating]] system. The K-factor is k = 60 for the first 10 games, k = 40 for the next 10 games, and k = 20 thereafter. There are special rules for ratings under 100: Players rated below 100 can never lose Elo points, and win at least 1 Elo point for their first game. Once a player has reached Elo 100, they cannot go below 100 again, so 100 points is basically the baseline rating. The strongest Hex players are rated in the 900s.

== Tournaments ==

Any player with a premium membership can create a tournament. Recent Hex competitions of the [https://mindsportsolympiad.com/ Mind Sports Olympiad] have been hosted at BGA.

=== Difficulties ===

During tournaments and in arena mode, games are sometimes ended by force when the next round is about to start or when the tournament or arena season ends. In these cases, the player with the least time remaining on their clock is declared the loser (rather the player who first ran out of time).

This has led to unexpected results in some tournaments, notably in the [https://boardgamearena.com/table?table=197416475 final game of the 2021 Mind Sports Olympiad for 14x14 Hex]. In this game, Stanley Kozera was in the lead when his opponent, Paweł Guz, ran out of time. The players decided to continue the game, but then the system unexpectely ended the game and Paweł Guz was declared the winner. This resulted in Stanley Kozera, who had been undefeated at the tournament, being awarded 3rd place by the Board Game Arena tournament system, finishing behind Paweł Guz and Maciej Brzeski, each of whom had lost one game. Mind Sports Olympiad later corrected the [https://mindsportsolympiad.com/2021-medals/ tournament results], awarding gold medals to both Stanley Kozera and Maciej Brzeski.

== External links ==

* The address is [https://www.boardgamearena.com www.boardgamearena.com].

[[category: hex community]]
[[category: other games]]
[[category: online play]]

Question

2026-01-08T01:54:18Z

Selinger: /* Example: U-turn */ Typo and editing

A '''question''' is a move that forces a player to choose between two or more available options in an area of the board. The player can either '''answer''' the question, by playing one of the available responses, or '''not answer''' and [[tenuki|play elsewhere]]. However, if the original question is sufficiently [[forcing move|forcing]], it must be answered right away.

Sometimes a player has a choice of accomplishing one of several different things in a given region, such as: choosing to connect one stone vs. another, connecting more strongly vs. denying the opponent a ladder escape, etc. In such situations, it is often in the player's interest to postpone the choice as long as possible, to keep their options open until they know more about what is going on on the rest of the board. By playing a question, the opponent can sometimes force them to make the choice earlier than they would have liked.

Of course, in some very general sense, ''every'' move is a question, namely the question: "How will you respond to this move?". But usually the term is applied more narrowly in some region of the board that is not quite [[settled region|settled]], where one player plays a move that forces the other player to settle it, or at least significantly simplify it, in one of several ways.

A [[joseki]] can be viewed as a sequence of questions and answers.

== Example: U-turn ==

Perhaps the simplest example of a question is playing in the center of a [[wheel#U-turn|U-turn]]. The U-turn is the following position, consisting of two [[Multiple_threat#Overlapping_threats|overlapping]] bridges:
<hexboard size="3x3"
edges="none"
coords="none"
visible="-a1 c3 c2"
contents="R A:a2 B:c1 C:b3 E *:b2"
/>
With Blue to move, Red has a choice between connecting A to B or connecting A to C, but Red cannot achieve both of these things with a single move. By playing in the cell marked "*", Blue asks the question "which of B or C do you want to connect to?"

The following is a position where asking this question is the only winning move for Blue:
<hexboard size="4x3"
edges="all"
coords="top left"
contents="R a2 B:a3 C:c2 A:b4 B a1 b2 c4 E *:b3"
/>
With Blue to move, in the upper part, ''Blue'' gets to choose whether Red connects B or C to the top edge. In the lower part, Blue can force ''Red'' to choose whether to connect B or C to the bottom edge. By playing the question at "*", Blue forces Red to make this choice. If Red chooses B, Blue blocks B on top and vice versa.

== Example: Edge template with two stones ==

Consider the following edge template:
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="area(e1,a5,h5,h1)"
contents="R A:e1 B:g1 E *:f1 *:f3"
/>
This template has the curious property that, with Blue to move, Red can choose to connect either A or B to the edge, but cannot guarantee to connect them both. (See edge templates [[Fifth_row_edge_templates#V-2-g|V-2g]] and [[Fifth_row_edge_templates#V-2-g|V-2h]]).

Blue can ask the question "Which stone to you want to connect?" by playing at either of the cells marked "*". Red does not need to decide right away. For example, in the below diagram, if Blue intrudes at 1 and Red responds at 2, Red still retains the potential to connect either stone. But if Blue then plays at 3, then Blue threatens to cut off A at x or B at y. Red cannot defend both.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="area(e1,a5,h5,h1)"
contents="R A:e1 B:g1 B 1:f1 R 2:d3 B 3:e2 E x:d2 y:f3"
/>
Similar, if the game proceeds like in the next diagram, then after move 7, Blue threatens to cut off A at x or B at y, and Red cannot save both connections.
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="area(e1,a5,h5,h1)"
contents="R A:e1 B:g1 B 1:f1 R 2:e2 B 3:f2 R 4:g2 B 5:e4 R 6:e3 B 7:f3 E x:d4 y:g3"
/>
Of course, there are many other lines of play to consider. If Blue is not careful at each move, Red might be able to connect both A and B. But if Blue plays correctly, Red is eventually forced to answer the question.

== Example: Obtuse corner ==

It is relatively common for a player to occupy the fourth stone on the short diagonal:
<hexboard size="11x5"
edges="left bottom"
coords="left bottom"
visible="area(a7,a11,e11,e7)"
contents="R d8"
/>
One way to ask this stone a question is to play e9:
<hexboard size="11x5"
edges="left bottom"
coords="left bottom"
visible="area(a7,a11,e11,e7)"
contents="R d8 B 1:e9"
/>
Assuming Red wants to reconnect her stone to the edge, she has two main choices. She can either play at c8, gaining more strength on top, or at c10, gaining more strength on the bottom. (There are also several other moves that would reconnect Red, but they are generally no better than c8 or c10).

If Red plays the [[minimax|minimaxing move]] at c8, she connects to the edge by [[edge template IV2a]] and gaining significant [[territory|strength]] toward the top:
<hexboard size="11x5"
edges="left bottom"
coords="left bottom"
visible="area(a7,a11,e11,e7)"
contents="R d8 B 1:e9 R 2:c8 S area(c8,b9,a11,d11,d8)"
/>
In this situation, Red has a 2nd row [[ladder escape]] in the corner, and Blue has a potential 2nd row [[ladder escape fork]]. However, Blue can decide to take away Red's 2nd row ladder escape at the expense of also giving up Blue's ladder escape, by playing like this:
<hexboard size="11x5"
edges="left bottom"
coords="left bottom"
visible="area(a7,a11,e11,e7)"
contents="R d8 B 1:e9 R 2:c8 B 3:c9 R 4:d9 B 5:d10 R 6:b9"
/>

If Red plays at c10, Red gets less strength towards the top, but gains a 2nd row ladder escape that Blue cannot take away. Blue still gets a 2nd row ladder escape by playing at c9.
<hexboard size="11x5"
edges="left bottom"
coords="left bottom"
visible="area(a7,a11,e11,e7)"
contents="R d8 B 1:e9 R 2:c10 S area(d8,c9,b11,c11,d9)"
/>

== See also ==

* [[Peep]]

[[category: Advanced Strategy]]
[[category: Definition]]

Bottleneck

2025-11-30T16:26:00Z

Selinger: Added politigaarden

The '''bottleneck''' is a strategic formation in the game of Hex where one player blocks the opponent's progress by occupying key positions on the board, creating a narrow passage that is difficult for the opponent to break through. The board fragment shown below illustrates this concept, with Blue occupying the key positions in a bottleneck formation, hindering Red's ability to connect to the bottom of the board.
<hexboard size="4x7"
coords="none"
edges="bottom"
visible="area(d1,a4,g4,g1)"
contents="R e1 e2 B d2 f2 d4"
/>
This formation often leads to a [[ladder]], with Red getting ladders in both directions, as shown in the second board fragment.
<hexboard size="4x7"
coords="none"
edges="bottom"
visible="area(d1,a4,g4,g1)"
contents="R e1 e2 B d2 f2 d4 R 1:d3 B 2:c4 R 3:c3 B 4:b4 R 5:e3 B 6:e4 R 7:f3 B 8:f4"
/>

The bottleneck is often a result of a defensive maneuver, for example when a player plays a [[Blocking#The near block|near block]] followed by an [[Blocking#The adjacent block|adjacent block]]. It is also a common strategy for leaving a [[Bridge ladder]].

It's important to note that, although the bottleneck can be an effective defense strategy, it also has its own weaknesses and can be exploited by a skilled opponent. Knowing how to both create and counter bottleneck formations is an important aspect of mastering the game of Hex.

== History ==

According to Hayward and Toft, the bottleneck was known to [[Piet Hein]] and Jens Lindhard in 1943, and they called it the Politigården, after the police headquarters in Copenhagen. The building of the police headquarters is such that upon entering it, one has to choose to go left or right.

[[File:Politigaarden.png|center|alt=xyz|350px|The police headquarters in Copenhagen]]

== References ==

* [[Ryan Hayward]] and [[Bjarne Toft]]. [[Hex: The Full Story]]. CRC Press, 2019. ISBN 978-0367144227.

== See also ==

* [[Blocking]]
* [[Bridge ladder]]

[[category:strategy]]
[[category:definition]]

File:Politigaarden.png

2025-11-30T16:17:44Z

Selinger: The police headquarters in Copenhagen, from Google maps.

The police headquarters in Copenhagen, from Google maps.

Bridge ladder

2025-11-30T00:37:21Z

Selinger: /* Example */ typo

A ''bridge ladder'' is a sequence of moves such as the following:
<hexboard size="7x8"
edges="bottom"
coords="none"
contents="B a3 R c2 B 1:b4 R 2:d3 B 3:c5 R 4:e4 B 5:d6 R 6:f5 B 7:e7 R 8:g6"
/>
Here, Red is the ''attacker'', Blue is the ''defender'', and both players play a sequence of [[bridge]]s that approach the attacker's edge at a 30 degree angle, with the defender being closer to the edge than the attacker. Bridge ladders sometimes happen when the defender repeatedly tries to [[blocking|block]] the attacker with a [[Blocking#The_near_block|near block]], and the attacker repeatedly [[bridge]]s to one side.

In the above example, Red ''wins'' the bridge ladder (i.e., Red connects to the edge). However, if the ladder starts closer to a blue edge, the outcome can be different:
<hexboard size="7x7"
edges="bottom right"
coords="none"
contents="B a2 R c1 B 1:b3 R 2:d2 B 3:c4 R 4:e3 B 5:d5 R 6:f4 B 7:e6 R 8:g5 B 9:f7"
/>
This time Blue wins the ladder. Generally, when a bridge ladder moves towards an empty acute corner, whichever player is closer to the [[board#Diagonals|long diagonal]] wins the ladder. Also note that from the point of view of the red edge, Red is the attacker and Blue is the defender, but from the point of view of the blue edge, Blue is the attacker and Red is the defender. This is typical for bridge ladders approaching an acute corner.

== Bottlenecking from a bridge ladder ==

Let us call the player who would lose a bridge ladder if it continued until the end the ''underdog''. So Blue is the underdog in the first example above, and Red is the underdog in the second example.

Since the underdog stands to lose the bridge ladder, the onus is usually on them to do something about it, typically by creating a [[bottleneck]].

=== Example ===

Consider a bridge ladder starting on the 6th row. Blue is the underdog.
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B 1:c4 R 2:e3 B 3:d5 R 4:f4 B 5:e6 R 6:g5 B 7:f7 R 8:h6"
/>
Instead of continuing the ladder to the end, Blue has the choice to create a [[bottleneck]] on the 5th row, 4th row, or 3rd row:

'''5th row bottleneck:'''
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B 1:d3 R 2:c3 B 3:b5"
/>
Red gets a pair of 4th row ladders.

'''4th row bottleneck:'''
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B 1:c4 R 2:e3 B 3:e4 R 4:d4 B 5:c6"
/>
Red gets a pair of 3rd row ladders.

'''3rd row bottleneck:'''
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B 1:c4 R 2:e3 B 3:d5 R 4:f4 B 5:f5 R 6:e5 B 7:d7"
/>
Red gets a pair of 2nd row ladders.

Blue must choose carefully when to bottleneck. One might think that it is good for Blue to bottleneck as soon as possible, because this results in a ladder further from the red edge. But on the other hand, especially when the bridge ladder is approaching an acute corner, bottlenecking sooner also keeps Blue further from the blue edge. For example, in each of the above scenarios, Red may try to pivot as follows:

'''5th row bottleneck:'''
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B d3 R c3 B b5 R 1:c4 B 2:c5 R 3:e4 B 4:d4 R 5:f2 B 6:e2"
/>
Red pivots at 3. Assuming 3 connects to the bottom edge, Red gets a 4th row ladder along the bottom edge, and Blue gets a 4th row ladder along the right edge.

'''4th row bottleneck:'''
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B c4 R e3 B e4 R d4 B c6 R 1:d5 B 2:d6 R 3:f5 B 4:e5 R 5:g3 B 6:f3"
/>
Red pivots at 3. Red gets a 3rd row ladder along the bottom edge, and Blue gets a 3rd row ladder along the right edge.

'''3rd row bottleneck:'''
<hexboard size="7x8"
edges="bottom right"
coords="none"
contents="B b3 R d2 B c4 R e3 B d5 R f4 B f5 R e5 B d7 R 1:e6 B 2:e7 R 3:g6 B 4:f6 R 5:h4 B 6:g4"
/>
Red pivots at 3. Red gets a 2nd row ladder along the bottom edge, and Blue gets a 2nd row ladder along the right edge.

== Bridge ladder approaching an obtuse corner ==

When a bridge ladder approaches an obtuse corner, the situation is in principle similar, but there are some differences depending on who is the underdog.

For example, consider the following:
<hexboard size="5x9"
coords="none"
edges="bottom left"
contents="R h1 B g3 R 1:f2 B 2:e4 R 3:d3 B 4:c5 R 5:b4"
/>
Here, Red wins the ladder, and Blue's last opportunity to [[bottleneck]] was move 2, which would have given Red a 2nd row ladder. On the other hand, when the bridge ladder starts further to the left, the situation is different:
<hexboard size="5x9"
coords="none"
edges="bottom left"
contents="R g1 B f3 R 1:e2 B 2:d4 R 3:c3 B 4:b5 R 5:a4 B 6:a5"
/>
If the bridge ladder continues to the end, Blue connects. Red can't create a bottleneck, but Red can turn the ladder around, for example like this, resulting in a 2nd row ladder for Blue:
<hexboard size="5x9"
coords="none"
edges="bottom left"
contents="R g1 B f3 R 1:e2 B 2:d4 R 3:c3 B 4:b5 R 5:a5 B 6:b4"
/>
or like this, resulting in a 4th row ladder for Blue:
<hexboard size="5x9"
coords="none"
edges="bottom left"
contents="R g1 B f3 R 1:e2 B 2:d4 R 3:c4 B 4:d3"
/>
or even like this, resulting in no ladder for Blue:
<hexboard size="5x9"
coords="none"
edges="bottom left"
contents="R g1 B f3 R 1:d2 B 2:e2 R 3:c2"
/>

== Application: last opportunity to pivot from a ladder ==

Consider an (ordinary) [[ladder]] moving parallel to an edge. In the absence of a ladder escape, the attacker must at some point do something, typically [[Ladder_handling#Attacking|pivot]] or play a [[cornering]] move. One may ask when is the last possible opportunity to pivot. A useful heuristic is to consider the bridge ladder that would result if the defender yielded after the pivot. For example, consider a 4th row ladder approaching from the left. If Red pivots at 5, then Blue is the underdog in the resulting bridge ladder, so Blue has to do something else (like [[bottleneck|bottlenecking]]).
<hexboard size="5x8"
edges="bottom right"
coords="none"
contents="R a2 B a3 R b2 B b3 R 1:c2 B 2:c3 R 3:d2 B 4:d3 R 5:f2 B 6:e4 R 7:g3 B 8:f5 R 9:h4 B 9:g6"
/>
On the other hand, if Red waits until 7 to pivot, ''Red'' ends up being the underdog, and cannot connect.
<hexboard size="5x8"
edges="bottom right"
coords="none"
contents="R a2 B a3 R b2 B b3 R 1:c2 B 2:c3 R 3:d2 B 4:d3 R 5:e2 B 6:e3 R 7:g2 B 8:f4 R 9:h3 B 10:g5"
/>
Therefore, generally speaking, the last opportunity to pivot from a ladder approaching an [[Board#Corners|acute corner]] is before the ladder has reached the long diagonal. A similar analysis applies to ladders approaching an [[Board#Corners|obtuse corner]].

[[category:Intermediate Strategy]]
[[category:Definition]]

Theory of ladder escapes

2025-11-27T04:01:13Z

Selinger: /* Sixth row ladders and up */ Updated the section in light of the existence of template IV1b.

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this, we first formalise what it means to be a ladder.

Informally, a ladder escape (say, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, no matter how far away from the ladder escape the ladder starts. So strictly speaking, to check that a pattern is a 4th row ladder escape, we must check that the attacker can connect to the edge from an ''infinite set'' of positions. This raises the issue of how one can check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page for 2nd, 3rd, 4th, and 5th row ladders. It might be possible to resolve it for 6th row ladders but this has not yet been done, partly because such ladders are of little practical use. For 7th row ladders we run into a new difficulty – Blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which is currently unresolved. It may in theory be that there are no 7th row ladders at all.

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

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

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

== Algebraic notation ==

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

=== Open patterns ===

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

=== Addition ===

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

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

=== The shift operation ===

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

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

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

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

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

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

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

=== The reduce operation ===

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

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

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

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

== Second row ladders ==

=== Definition of ladder ===

There is no issue at all with defining a 2nd row ladder. Informally, a 2nd row ladder looks like this:
<hexboard size="2x8"
coords="hide"
edges="bottom"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>
Note that at each point in the ladder, Blue's move is forced. Red can choose to continue pushing the ladder as long as she wants to. We formally define a second row ladder as follows:

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

=== Definition of ladder escape ===

Before we give the formal definition of a second row ladder escape, let us consider an example. The following pattern is an example of a second row ladder escape.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The cells marked "+" indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is,
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:e1 E *:f1 R g1 R h1 E *:a2 E *:b2 E *:c2 E *:d2 E *:e2 R 1:a3 E *:h3 E *:h4"
/>
Red can guarantee a connection from the ladder stone (marked 1) to the bottom edge.

Let us clarify what the hexes marked "+" in the ladder escape pattern mean. They indicate the last point where the 2nd row ladder is allowed to start. So for example, saying that the pattern above is a second row ladder escape means (among other things) that Red must win the following position:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here, Red's ladder stone is marked "1", and the claim (easily verified) is that even with Blue to play, Red can connect the ladder stone to the bottom:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>
The reason that the pattern is a second row ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 B h3 B h4"
/>
Even here, Red can force a connection to the edge, even if it is Blue's move, because Blue must keep defending on the first row and Red keeps attacking on the second row,
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 R 3:b3 R 5:c3 R 7:d3 R 9:e3 B h3 B 2:a4 B 4:b4 B 6:c4 B 6:d4 B h4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that Red can break through to the edge. We can now give a more formal definition of a second row ladder escape.

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

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

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

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

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

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

=== Characterization of ladder escapes ===

The definition of a 2nd row ladder escape allows the ladder to be an ''arbitrary'' distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However, this means that we cannot directly use the definition to check that something is a 2nd row ladder escape, because this would require checking that infinitely many patterns are virtual connections. Can we find some finite criterion for checking 2nd row ladder escapes? Fortunately, as every Hex player knows, the answer is yes. We have the following theorem:

'''Theorem 1.''' Consider a candidate P for a 2nd row ladder escape. Schematically:
<hexboard size="4x4"
coords="hide"
edges="bottom"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(Here the asterisks indicate the [[carrier]] of P, which can contain any stones at all, and can be of any shape or size, as long as it includes no cells to the left of the cells marked "+"). Then P is a valid 2nd row ladder escape if and only if L2+P is a virtual connection for Red.
<hexboard size="4x4"
coords="hide"
edges="bottom"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>

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

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

=== Examples ===

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

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

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

== Third row ladders ==

=== Definition of ladder ===

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

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

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

=== Definition of ladder escape ===

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because Blue can defend against a third row ladder in more than one way: Blue can at some stage decide to [[ladder handling|yield]] to the second row. The following definition is unsurprising.

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

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

In pictures, for the pattern
<hexboard size="3x3"
coords="hide"
edges="bottom"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(where the carrier is schematically indicated by stars) to be a 3rd row ladder escape, it must give rise to a virtual connection when we attach a 3rd row ladder at distance 0,
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1,
<hexboard size="3x5"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6,
<hexboard size="3x10"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any other distance.

=== Characterization of ladder escapes ===

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

=== Examples ===

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

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

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

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

The following third row ladder escapes also escape 2nd row ladders at distance 2 or greater (but not at distance 0 or 1):
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="R b1 E *:d1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R c1 E *:d1"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R b1 R c1 E *:d1 S b2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-area(b3,c2,d2,d4,b4)"
contents="R ↓:d1 E +:a2 +:a3 +:a4"
/>
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 b2 d2 e2 a3 b3 d3 e3 a4 b4 d4 e4"
contents="R ↓:e1 E *:a2 E *:b2 E +:c2 E *:d2 E *:e2 E *:a3 E *:b3 E +:c3 E *:d3 E *:e3 E *:a4 E *:b4 E +:c4 E *:d4 E *:e4"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 e2 e3 e4 e5"
contents="E +:a3 +:a4 +:a5 E *:a1 *:b1 *:c1 *:e2 *:e3 *:e4 *:e5 R ↓:e1 S d2"
/>
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 g1 h1 h2"
contents="E *:a1 E +:a3 E +:a4 E +:a5 E *:b1 E *:c1 R e1 E *:g1 E *:h1 E *:h2"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1"
contents="E +:(a3 a4 a5) R c1 d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 d1 d2"
contents="E +:(a3 a4 a5) R b1 c1"
/><hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1"
contents="E *:a1 E *:a2 E +:a4 E +:a5 E +:a6 E *:b1 R c1 R d2"
/>
<hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2"
contents="E *:a1 E *:b1 R c1 E *:a2 R b2 E +:a4 E +:a5 E +:a6"
/>
The version of this last pattern on David King's website has the cells marked "+" (he uses arrows) sloping in the other direction; the location that is shown here makes the template minimal.

== Fourth row ladders ==

=== Definition of ladder ===

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

The key part of the definition is that the 6 hexes forming a triangle below the ladder stone are all vacant. Note that even filling in one of these can invalidate the ladder: even if we fill in the bottom left corner of the triangle,
<hexboard size="5x5"
coords="hide"
edges="bottom"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then Blue has this move,
<hexboard size="5x5"
coords="hide"
edges="bottom"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To establish the ladder, Red needs at a minimum those 6 vacant hexes under her ladder stone.

=== Definition of ladder escape ===

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

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

The five moves marked 2 below all lose instantly to Red 3:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:d1 B 2:e1 E *:a2 E *:b2 B 2:e2 E *:a3 B 2:e3 B 2:e4 R 3:c2"
/>
The two moves marked 2 below also lose instantly:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:b3 B 2:a4 E *:a2 E *:b2 E *:a3 R 3:d2"
/>
The move marked 2 below can be answered by Red 3, moving us to position L4+''n''+P, which is a virtual connection by the induction hypothesis.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3"
/>
Move 2 below leads us to a 2nd row ladder, which ↑+↑+P escapes, so 5 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 B 4:b4"
/>
Both moves marked 2 below lead us to a 3rd row ladder, which ↑+P escapes, so 3 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 E *:a3 B 2:c3 B 2:b4"
/>
Move 2 below also leads to a 3rd row ladder (note Blue 4 must be in the triangle left and below from Red 3; Blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to a 2nd row ladder:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 B 4:b4 B 2:c4"
/>
The final choice for move 2 below also gives a 2nd row ladder.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the induction, so P is a 4th row ladder escape. □

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

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

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

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

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

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

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

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

=== Examples ===

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

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

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

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

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

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

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

== Fifth row ladders ==

=== Definition of ladder ===

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

Unlike in the case of 2nd, 3rd, and 4th row ladders, this time it is not sufficient for a triangle of cells below and to the right of the ladder stone to be empty. We also need three additional empty cells to the left of this triangle. This is a minimal requirement; if even one of these cells is occupied by Blue, for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 R o4 B a5 R o5 B a6 R o6"
/>
then Blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 R o4 B a5 R o5 B a6 R o6 B 1:e5"
/>
The main line is complex; see for example [http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 this Little Golem discussion thread]. Many of the main lines of defense involve Blue playing an upside-down version of [[Tom's move]], for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 R o4 B a5 R o5 B a6 R o6 B 1:e5 R 2:e4 B 3:e3 R 4:f2 B 5:f3 R 6:g2 B 7:h4 E *:d5 *:f4"
/>
Note that Blue's 1 is connected to 5 by double threat at "*", and 7 is Tom's move upside-down, i.e., with the top line of blue stones serving as the "edge". Therefore, to establish the ladder, Red needs at minimum the specified 13 vacant hexes under the ladder stone.

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

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

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

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

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

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

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

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

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

=== Examples ===

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

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

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

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

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

== Sixth row ladders and up ==

Because of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick, 6th and higher row ladders do not exist in the usual sense. More specifically, even if we allow an arbitrary amount of empty space under the ladder stone, it is not possible for the attacker to keep pushing the ladder. Consider the following situation:
<hexboard size="7x13"
coords="show"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2"
/>
Let's assume there is an arbitrary amount of empty space in the bottom 4 rows to the left of this diagram. The stone marked "1" is connected to the top, and looks like it could be the ladder stone for a potential 6th row ladder. If such a ladder were possible, the red stones on the M-file should certainly escape it.

From Blue's point of view, Blue is the attacker in an upside-down 2nd row ladder. Blue can therefore use an upside-down version of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick. To do so, Blue plays at 2:
<hexboard size="7x13"
coords="show"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5"
/>
If both Red and Blue keep playing [[optimal play|optimally]], the best that Red can get is a pair of parallel 2nd and 4th row ladders in the opposite direction:
<hexboard size="7x13"
coords="show"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5 R 3:d3 B 4:c5 R 5:c4 B 6:b5 R 7:e4 B 8:e3 R 9:d4 B 10:e6 R 11:d5 B 12:c7 R 13:c6 B 14:b7"
/>
Here, stone 10 is [[virtual connection|connected]] to the line of blue stones along the top, so Red has no way of connecting right. Red can now push a 4th row ladder from 5, and/or a 2nd row ladder from 13. There is not enough space for Red to immediately perform [[Tom's move]]. So unless Red has a ladder escape somewhere to the left of this diagram, or unless there's enough space on the 5th row somewhere to the left of this diagram to perform Tom's move, Red fails to connect to the edge.

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

The situation for 7th row ladders is similar. The existence of [[edge template VI1b]] guarantees a red connection if Blue just ignores the ladder and plays elsewhere. But this requires a large amount of space. Just like for 6th row ladders, Blue has many options besides pushing the ladder, and it is not known whether a viable notion of 7th row ladder exists.

For 8th row ladders the situation is even worse. There is currently no known 8th row template that can guarantee a red connection even if Blue ignores the ladder. Thus, it is possible that 8th row ladders do not even exist in theory. Of course they do not occur in practice either.

== Second-to-fourth row switchbacks ==

=== Definition of switchback ===

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

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

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

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

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

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

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

=== Characterization of switchbacks ===

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

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

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

=== Examples ===

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

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

== Third-to-fifth row switchbacks ==

=== Definition of switchback ===

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

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

=== Characterization of switchbacks ===

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

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

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

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

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

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

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

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

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

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

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

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

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

=== Examples ===

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

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

== Second and fourth row parallel ladders ==

=== Definition of ladder ===

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

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

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

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

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

=== Examples ===

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

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

== Third and fifth row parallel ladders ==

=== Definition of ladder ===

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

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

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

=== Examples ===

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

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

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

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

== Second and fourth row terraced ladders ==

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

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

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

=== Definition of ladder ===

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

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

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

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

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Next, consider this pattern:

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

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

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

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

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

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

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

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

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

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

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

=== Non-examples ===

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

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

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

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

Tom's move

2025-11-23T18:53:34Z

Selinger: /* Bridge-first variant */ Copy-edit

== Introduction ==

'''Tom's move''' is a trick that enables a player to make a connection from a 2nd-and-4th row [[parallel ladder]]. It can also be used to connect a 2nd row [[ladder]] using a single stone on the 4th row, or to connect a single stone on the 4th row to the edge. Its name originates from Tom Ace (player [[User:Tom239|Tom239]]), who devised it during a game against dj11, on 15 December 2002 on [[Playsite]]. This was not its first use ever, but it is how it came to be known among Hex players on Playsite.

== Description ==

Suppose Red has a 2nd-and-4th row [[parallel ladder]] and the amount of space shown here:
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 B b3 a5 E *:d3"
/>
Then Red can connect by playing at "*", the so-called "Tom's move".

== Usage examples ==

=== Connecting a 2nd row ladder using an isolated stone on the 4th row ===

Red to move and win:

<hexboard size="6x9"
edges="all"
coords="none"
contents="R a5 a4 d3 b2 B a6 b4 c3 d2 e1 f1 g1 h1 i1 h2 i2 i3"
/>

The solution is to [[ladder handling|push]] the [[ladder]] to 3 and then play Tom's move:

<hexboard size="6x9"
edges="all"
coords="none"
contents="R a5 a4 d3 b2 B a6 b4 c3 d2 e1 f1 g1 h1 i1 h2 i2 i3 R 1:b5 B 2:b6 R 3:c5 B 4:c6 R 5:f4"
/>

=== A single stone on the 4th row is connected ===

Consider a single stone on the 4th row, with the amount of space shown:
<hexboard size="5x9"
edges="bottom"
coords="none"
visible="-area(a1,a4,d1) h1 i1 i2"
contents="R arrow(12):d2"
/>

Then Red can connect as follows:
<hexboard size="5x9"
edges="bottom"
coords="none"
visible="-area(a1,a4,d1) h1 i1 i2"
contents="R arrow(12):d2 B 1:d3 R 2:c3 B 3:b5 R 4:c4 B 5:c5 R 6:f3"
/>
Red squeezes through the [[bottleneck]] at 2, starts a 2nd row ladder at 4, then plays Tom's move at 6. Note that all of Blue's moves are forced; if Blue plays differently, Red connects outright. This is [[edge template IV1d]].

=== In a game ===
Red to move:
<hexboard size="11x11"
edges="all"
coords="show"
contents="R b7 c6 d4 d5 d6 d8 f5 g3 g5 h5 i4 i6 j5 k2 k3 B c5 c7 d7 e4 e5 e6 f6 g6 h4 h6 h7 i3 i7 j3 j4 j6"
/>
Red's d4 [[group]] is already connected to the top edge by [[edge template IV1a|edge template IV1-a]]. To connect to the bottom, Red plays as follows:
<hexboard size="11x11"
edges="all"
coords="show"
contents="R b7 c6 d4 d5 d6 d8 f5 g3 g5 h5 i4 i6 j5 k2 k3 B c5 c7 d7 e4 e5 e6 f6 g6 h4 h6 h7 i3 i7 j3 j4 j6
R 1:b8 B 2:c9 R 3:a10 B 4:a11 R 5:b10 B 6:b11 R 7:c10 B 8:c11 R 9:f9"
/>
Now Red is connected by Tom's move. Note that d8 is already connected to Red's group by [[double threat]] at c8 and d9.

== Why Tom's move is connected ==

To compute Blue's [[mustplay region]], we consider two red [[threat]]s:

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:b4 4:c4 6:e4 B b3 a5 3:b5 5:c5 S b4 area(b5,d3,e3,e5)"
/>

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:c2 B b3 a5 S c2 c3 d2 area(b5,d3,e3,e5)"
/>

These leaves only blue moves in the [[ziggurat]].

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 B b3 a5 S area(b5,d3,e3,e5) E a:c4"
/>

If Blue plays there other than at a, then Red plays at a.
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:c4 B b3 a5 E *:c2 *:b4 z:b5 y:c5 x:(e5 d5 e4 d4 e3) S area(b5,d3,e3,e5)"
/>
In that case, Red's 2 connects back via either of the cells marked "*", and since the piece Blue just played is in only one of the x,y,z regions, Red's 2 also connects down via either of the remaining two of those three regions.

Thus Blue's only remaining hope is to play at a.

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 B b3 a5 1:c4"
/>

Red responds like this:

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:b4 4:e2 B b3 a5 1:c4 3:b5 E *:d1 *:c3"
/>

Red's 4 is now connected to the bottom via [[Edge_template_IV2b|edge template IV2b]], and to
Red's main group by double threat at the cells marked "*". Note that 2 and 3 do not actually need to be played; these moves have been included for clarity.

== Pushing the 4th row ladder first ==

Sometimes, there is not enough room to play Tom's move right away, but enough room can be created by first pushing the 4th row ladder. For a typical example, consider the following situation. It is Blue's turn, and Red wants to connect her stone to the bottom edge.
<hexboard size="6x11"
edges="bottom right"
coords="none"
visible="-area(a1,a5,e1)"
contents="R d3 B e2 f2 j2 k3 g1--j1"
/>
If Red tries to play Tom's move immediately, it doesn't work, because 8 does not connect back to Red's main group.
<hexboard size="6x11"
edges="bottom right"
coords="none"
visible="-area(a1,a5,e1)"
contents="R d3 B e2 f2 j2 k3 g1--j1 B 1:d4 R 2:c4 B 3:b6 R 4:c5 B 5:c6 R 6:f4 B 7:e5 R 8:g3 B 9:e4"
/>
What Red can do instead is start by pushing the 4th row ladder twice.
<hexboard size="6x11"
edges="bottom right"
coords="none"
visible="-area(a1,a5,e1)"
contents="R d3 B e2 f2 j2 k3 g1--j1 B 1:d4 R 2:e3 B 3:e4 R 4:f3 B 5:f4 R 6:c4 B 7:b6 R 8:c5 B 9:c6 R 10:d5 B 11:d6 R 12:e5 B 13:e6 R 14:h4 B 15:g5 R 16:i3 E *:g4 *:h2"
/>
Of course this only works if after pushing the ladder, there is still enough room for Tom's move.

It is not actually necessary to push the 2nd row ladder (moves 6–13 can be omitted), but they have been included for clarity.

Note that when Red pushes the 4th row ladder, Blue cannot [[ladder handling|yield]], as this would give Red a [[ladder escape fork]] for the below 2nd row ladder. Also, as explained in more detail in the article on [[parallel ladder]]s, the 4th row ladder must be pushed ''before'' the 2nd row ladder has caught up to it. If Red starts by first pushing the 2nd row ladder, then it is too late to push the 4th row ladder.

== Variants ==

=== Alternative connection up ===

Tom's move also works when the hex marked "a" is not empty, provided that "b" connects to Red's main group.
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 B b3 a5 E a:c1 b:d1"
/>

For example, Tom's move works in this situation:
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-g2 h2 h3 f1--h1 -area(a1,a5,c1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R e4 E b:e2 R d1 B a:d2 R c2"
/>

This is a special case of a [[Theorems_about_templates#Alternative_connection_up|general theorem]].

=== Tall variant ===

If "d" is empty, there is a variant of Tom's move that does not require a connection via "b", or even for "b" to be empty; it merely requires "c" and "e" to threaten to connect to Red's main group. An example is this situation:
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-h2 h3 g1--h1 -area(a1,a5,c2,d1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R e4 E c:f2 d:g2 e:e3 R e1 B b:e2 R d2"
/>
In this version of Tom's move, Red's 4 is different than usual (the usual move 4 does not work here). As before, moves 2 and 3 don't need to be played, but make it easier to see how the connection works.
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-h2 h3 g1--h1 -area(a1,a5,c2,d1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R x:e4 R e1 B e2 R d2
B 1:d5 R 2:c5 B 3:c6 R 4:f2 E y:f3"
/>
Note that "x" is connected to Red's main group without requiring "y", and "4" is also connected to Red's main group without requiring "y". (However, Red cannot guarantee to connect both "x" and "4" to her main group without requiring "y"). If Blue tries to cut Red off from the edge, Red responds as follows:
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-h2 h3 g1--h1 -area(a1,a5,c2,d1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R x:e4 R e1 B e2 R d2
B 1:d5 R 2:c5 B 3:c6 R 4:f2 E y:f3 B 5:e5 R 6:g4"
/>
Now no matter how Blue plays, Red can connect to the edge by a sequence of forcing moves. The situation is analogous to the [[Interior_template#The_hammock|hammock template]].

=== Bridge-first variant ===
If the end of Red's second row ladder is not yet directly beneath the end of their 4th row ladder, Red can opt to play the bridge to "*" first, instead of playing at "*".
<hexboard size="5x8"
edges="bottom"
coords="none"
visible="-(a1--a3 b2 b1 c1 g1 h1 h2)"
contents="R arrow(12):c2 b3 e3 1:e1 B c3 a5 d4 E *:f2"
/>

Unless Blue themselves plays at "*", Red can respond to any intrusion Blue makes by playing at "*". This will result either in Red being connected outright via the bridges and [[Edge_template_IV2b|edge template IV2b]], connecting with a double threat similar to the normal Tom's move, or connecting through a variation of the following double threat:

<hexboard size="5x8"
edges="bottom"
coords="none"
visible="-(a1--a3 b2 b1 c1 g1 h1 h2)"
contents="R arrow(12):c2 b3 e3 e1 2:f2 4:c4 B c3 a5 d4 1:e4 3:f4 S b4 d3"
/>

If Blue plays at "*", Red responds as follows, and has a double threat similar the normal Tom's move.
<hexboard size="5x8"
edges="bottom"
coords="none"
visible="-(a1--a3 b2 b1 c1 g1 h1 h2)"
contents="R arrow(12):c2 b3 e3 e1 2:e4 B c3 a5 d4 1:f2"
/>

This variation should be used if you don't want the opponent to get the territory from intruding into the bridge of a standard Tom's move. But what if the opponent plays that move before you do to ensure they get it? In this case it is often good to respond as follows, and use the tall variant to connect "2".

<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-area(a1,a4,d1) h2 h3 g1--h1"
contents="R arrow(12):c3 b4 e4 2:e1 B c4 a6 1:e2"
/>

Here is a specific example position where the bridge-first variant is necessary. Red's only winning move is Tom's move (a), and if Blue plays at (b), Red's only winning response is (c).
<hexboard size="7x7"
contents="R d1 a2--b2--b4--a5 B a1--c1 e1--g1 b5 E a:d5 b:c6 c:d3"
/>

== Tom's move for 3rd-and-5th row parallel ladders ==

Main article: [[Tom's move for 3rd and 5th row parallel ladders]].

There is a version of Tom's move that works for parallel ladders on the 3rd and 5th rows. It requires a large amount of space:
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="area(d1,b3,b6,l6,l4,j2,f1)"
contents="R arrow(12):c2 b3 b4 B b5 c3 R 1:e3"
/>
By playing at "1", Red can connect to the edge. Verifying this requires a lot of steps, but here is the basic idea:
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="area(d1,b3,b6,l6,l4,j2,f1)"
contents="R arrow(12):c2 b3 b4 B b5 c3 E x:i4
R 1:e3 B 2:d4 R 3:c4 B 4:c5 R 5:f2 E *:d3 *:e1"
/>
Notice that Red's 3 is connected left by double threat at the two cells marked "*", and connected right by [[Fifth_row_edge_templates#V-2-m|edge template V2m]]. The latter template is itself based on Tom's move at "x". It works, for example, like this:
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="area(d1,b3,b6,l6,l4,j2,f1)"
contents="R arrow(12):c2 b3 b4 B b5 c3
R 1:e3 B 2:d4 R 3:c4 B 4:c5 R 5:f2 E *:d3 *:e1
B 6:f4 R 7:g3 B 8:g4 R 9:e4 B 10:d6 R 11:e5 B 12:e6 R 13:f5 B 14:f6 R 15:i4"
/>
Now Red is connected by the (ordinary) Tom's move.

See [[Tom's move for 3rd and 5th row parallel ladders]] for more details.

=== Tall variant ===

Tom's move for 3rd-and-5th row parallel ladders also has a tall version. It works in the same way as the tall version of Tom's move for 2nd-and-4th row parallel ladders above, given the appropriate amount of space.

<hexboard size="7x12"
coords="hide"
edges="bottom"
visible="area(d2,b4,b7,l7,l5,j3,g2,f1,e1)"
contents="R arrow(12):e1 d2 c3 b4 b5 B b6 c4 e2 R 1:e4 B 2:d5 R 3:f2"
/>

== See also ==

* [[Parallel ladder]]
* [[Edge template IV1d]]
* [[Fifth_row_edge_templates#V-2-m|Edge template V2m]]

[[category:ladder]]
[[category:Advanced Strategy]]

Tom's move

2025-11-23T18:46:00Z

Selinger: /* Bridge-first variant */ Added an example where bridge-first variant is the only winning play.

== Introduction ==

'''Tom's move''' is a trick that enables a player to make a connection from a 2nd-and-4th row [[parallel ladder]]. It can also be used to connect a 2nd row [[ladder]] using a single stone on the 4th row, or to connect a single stone on the 4th row to the edge. Its name originates from Tom Ace (player [[User:Tom239|Tom239]]), who devised it during a game against dj11, on 15 December 2002 on [[Playsite]]. This was not its first use ever, but it is how it came to be known among Hex players on Playsite.

== Description ==

Suppose Red has a 2nd-and-4th row [[parallel ladder]] and the amount of space shown here:
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 B b3 a5 E *:d3"
/>
Then Red can connect by playing at "*", the so-called "Tom's move".

== Usage examples ==

=== Connecting a 2nd row ladder using an isolated stone on the 4th row ===

Red to move and win:

<hexboard size="6x9"
edges="all"
coords="none"
contents="R a5 a4 d3 b2 B a6 b4 c3 d2 e1 f1 g1 h1 i1 h2 i2 i3"
/>

The solution is to [[ladder handling|push]] the [[ladder]] to 3 and then play Tom's move:

<hexboard size="6x9"
edges="all"
coords="none"
contents="R a5 a4 d3 b2 B a6 b4 c3 d2 e1 f1 g1 h1 i1 h2 i2 i3 R 1:b5 B 2:b6 R 3:c5 B 4:c6 R 5:f4"
/>

=== A single stone on the 4th row is connected ===

Consider a single stone on the 4th row, with the amount of space shown:
<hexboard size="5x9"
edges="bottom"
coords="none"
visible="-area(a1,a4,d1) h1 i1 i2"
contents="R arrow(12):d2"
/>

Then Red can connect as follows:
<hexboard size="5x9"
edges="bottom"
coords="none"
visible="-area(a1,a4,d1) h1 i1 i2"
contents="R arrow(12):d2 B 1:d3 R 2:c3 B 3:b5 R 4:c4 B 5:c5 R 6:f3"
/>
Red squeezes through the [[bottleneck]] at 2, starts a 2nd row ladder at 4, then plays Tom's move at 6. Note that all of Blue's moves are forced; if Blue plays differently, Red connects outright. This is [[edge template IV1d]].

=== In a game ===
Red to move:
<hexboard size="11x11"
edges="all"
coords="show"
contents="R b7 c6 d4 d5 d6 d8 f5 g3 g5 h5 i4 i6 j5 k2 k3 B c5 c7 d7 e4 e5 e6 f6 g6 h4 h6 h7 i3 i7 j3 j4 j6"
/>
Red's d4 [[group]] is already connected to the top edge by [[edge template IV1a|edge template IV1-a]]. To connect to the bottom, Red plays as follows:
<hexboard size="11x11"
edges="all"
coords="show"
contents="R b7 c6 d4 d5 d6 d8 f5 g3 g5 h5 i4 i6 j5 k2 k3 B c5 c7 d7 e4 e5 e6 f6 g6 h4 h6 h7 i3 i7 j3 j4 j6
R 1:b8 B 2:c9 R 3:a10 B 4:a11 R 5:b10 B 6:b11 R 7:c10 B 8:c11 R 9:f9"
/>
Now Red is connected by Tom's move. Note that d8 is already connected to Red's group by [[double threat]] at c8 and d9.

== Why Tom's move is connected ==

To compute Blue's [[mustplay region]], we consider two red [[threat]]s:

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:b4 4:c4 6:e4 B b3 a5 3:b5 5:c5 S b4 area(b5,d3,e3,e5)"
/>

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:c2 B b3 a5 S c2 c3 d2 area(b5,d3,e3,e5)"
/>

These leaves only blue moves in the [[ziggurat]].

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 B b3 a5 S area(b5,d3,e3,e5) E a:c4"
/>

If Blue plays there other than at a, then Red plays at a.
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:c4 B b3 a5 E *:c2 *:b4 z:b5 y:c5 x:(e5 d5 e4 d4 e3) S area(b5,d3,e3,e5)"
/>
In that case, Red's 2 connects back via either of the cells marked "*", and since the piece Blue just played is in only one of the x,y,z regions, Red's 2 also connects down via either of the remaining two of those three regions.

Thus Blue's only remaining hope is to play at a.

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 B b3 a5 1:c4"
/>

Red responds like this:

<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 2:b4 4:e2 B b3 a5 1:c4 3:b5 E *:d1 *:c3"
/>

Red's 4 is now connected to the bottom via [[Edge_template_IV2b|edge template IV2b]], and to
Red's main group by double threat at the cells marked "*". Note that 2 and 3 do not actually need to be played; these moves have been included for clarity.

== Pushing the 4th row ladder first ==

Sometimes, there is not enough room to play Tom's move right away, but enough room can be created by first pushing the 4th row ladder. For a typical example, consider the following situation. It is Blue's turn, and Red wants to connect her stone to the bottom edge.
<hexboard size="6x11"
edges="bottom right"
coords="none"
visible="-area(a1,a5,e1)"
contents="R d3 B e2 f2 j2 k3 g1--j1"
/>
If Red tries to play Tom's move immediately, it doesn't work, because 8 does not connect back to Red's main group.
<hexboard size="6x11"
edges="bottom right"
coords="none"
visible="-area(a1,a5,e1)"
contents="R d3 B e2 f2 j2 k3 g1--j1 B 1:d4 R 2:c4 B 3:b6 R 4:c5 B 5:c6 R 6:f4 B 7:e5 R 8:g3 B 9:e4"
/>
What Red can do instead is start by pushing the 4th row ladder twice.
<hexboard size="6x11"
edges="bottom right"
coords="none"
visible="-area(a1,a5,e1)"
contents="R d3 B e2 f2 j2 k3 g1--j1 B 1:d4 R 2:e3 B 3:e4 R 4:f3 B 5:f4 R 6:c4 B 7:b6 R 8:c5 B 9:c6 R 10:d5 B 11:d6 R 12:e5 B 13:e6 R 14:h4 B 15:g5 R 16:i3 E *:g4 *:h2"
/>
Of course this only works if after pushing the ladder, there is still enough room for Tom's move.

It is not actually necessary to push the 2nd row ladder (moves 6–13 can be omitted), but they have been included for clarity.

Note that when Red pushes the 4th row ladder, Blue cannot [[ladder handling|yield]], as this would give Red a [[ladder escape fork]] for the below 2nd row ladder. Also, as explained in more detail in the article on [[parallel ladder]]s, the 4th row ladder must be pushed ''before'' the 2nd row ladder has caught up to it. If Red starts by first pushing the 2nd row ladder, then it is too late to push the 4th row ladder.

== Variants ==

=== Alternative connection up ===

Tom's move also works when the hex marked "a" is not empty, provided that "b" connects to Red's main group.
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 d3 B b3 a5 E a:c1 b:d1"
/>

For example, Tom's move works in this situation:
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-g2 h2 h3 f1--h1 -area(a1,a5,c1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R e4 E b:e2 R d1 B a:d2 R c2"
/>

This is a special case of a [[Theorems_about_templates#Alternative_connection_up|general theorem]].

=== Tall variant ===

If "d" is empty, there is a variant of Tom's move that does not require a connection via "b", or even for "b" to be empty; it merely requires "c" and "e" to threaten to connect to Red's main group. An example is this situation:
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-h2 h3 g1--h1 -area(a1,a5,c2,d1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R e4 E c:f2 d:g2 e:e3 R e1 B b:e2 R d2"
/>
In this version of Tom's move, Red's 4 is different than usual (the usual move 4 does not work here). As before, moves 2 and 3 don't need to be played, but make it easier to see how the connection works.
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-h2 h3 g1--h1 -area(a1,a5,c2,d1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R x:e4 R e1 B e2 R d2
B 1:d5 R 2:c5 B 3:c6 R 4:f2 E y:f3"
/>
Note that "x" is connected to Red's main group without requiring "y", and "4" is also connected to Red's main group without requiring "y". (However, Red cannot guarantee to connect both "x" and "4" to her main group without requiring "y"). If Blue tries to cut Red off from the edge, Red responds as follows:
<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-h2 h3 g1--h1 -area(a1,a5,c2,d1)"
contents="R arrow(12):b4 b5 c3 B a6 b6 c4 R x:e4 R e1 B e2 R d2
B 1:d5 R 2:c5 B 3:c6 R 4:f2 E y:f3 B 5:e5 R 6:g4"
/>
Now no matter how Blue plays, Red can connect to the edge by a sequence of forcing moves. The situation is analogous to the [[Interior_template#The_hammock|hammock template]].

=== Bridge-first variant ===
If the end of Red's second row ladder is not yet directly beneath the end of their 4th row ladder, Red can opt to play the bridge to "*" first, instead of playing at "*".
<hexboard size="5x8"
edges="bottom"
coords="none"
visible="-(a1--a3 b2 b1 c1 g1 h1 h2)"
contents="R arrow(12):c2 b3 e3 1:e1 B c3 a5 d4 E *:f2"
/>

Unless Blue themselves plays at "*", Red can respond to any intrusion Blue makes by playing at "*". This will result either in Red being connected outright via the bridges and [[Edge_template_IV2b|edge template IV2b]], connecting with a double threat similar to the normal Tom's move, or connecting through a variation of the following double threat:

<hexboard size="5x8"
edges="bottom"
coords="none"
visible="-(a1--a3 b2 b1 c1 g1 h1 h2)"
contents="R arrow(12):c2 b3 e3 e1 2:f2 4:c4 B c3 a5 d4 1:e4 3:f4 S b4 d3"
/>

If Blue plays at "*", Red responds as follows, and has a double threat similar the normal Tom's move.
<hexboard size="5x8"
edges="bottom"
coords="none"
visible="-(a1--a3 b2 b1 c1 g1 h1 h2)"
contents="R arrow(12):c2 b3 e3 e1 2:e4 B c3 a5 d4 1:f2"
/>

This variation should be used if you don't want the opponent to get the territory from intruding into the bridge of a standard Tom's move. But what if the opponent plays that move before you do to ensure they get it? In this case it is often good to respond as follows, and use the tall variant to connect "2".

<hexboard size="6x8"
edges="bottom"
coords="none"
visible="-area(a1,a4,d1) h2 h3 g1--h1"
contents="R arrow(12):c3 b4 e4 2:e1 B c4 a6 1:e2"
/>

Here is a specific example position where the only winning move for Red is Tom's move (a), and if Blue plays at (b), Red's only winning response is the bridge-first move (c).
<hexboard size="7x7"
contents="R d1 a2--b2--b4--a5 B a1--c1 e1--g1 b5 E a:d5 b:c6 c:d3"
/>

== Tom's move for 3rd-and-5th row parallel ladders ==

Main article: [[Tom's move for 3rd and 5th row parallel ladders]].

There is a version of Tom's move that works for parallel ladders on the 3rd and 5th rows. It requires a large amount of space:
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="area(d1,b3,b6,l6,l4,j2,f1)"
contents="R arrow(12):c2 b3 b4 B b5 c3 R 1:e3"
/>
By playing at "1", Red can connect to the edge. Verifying this requires a lot of steps, but here is the basic idea:
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="area(d1,b3,b6,l6,l4,j2,f1)"
contents="R arrow(12):c2 b3 b4 B b5 c3 E x:i4
R 1:e3 B 2:d4 R 3:c4 B 4:c5 R 5:f2 E *:d3 *:e1"
/>
Notice that Red's 3 is connected left by double threat at the two cells marked "*", and connected right by [[Fifth_row_edge_templates#V-2-m|edge template V2m]]. The latter template is itself based on Tom's move at "x". It works, for example, like this:
<hexboard size="6x12"
coords="hide"
edges="bottom"
visible="area(d1,b3,b6,l6,l4,j2,f1)"
contents="R arrow(12):c2 b3 b4 B b5 c3
R 1:e3 B 2:d4 R 3:c4 B 4:c5 R 5:f2 E *:d3 *:e1
B 6:f4 R 7:g3 B 8:g4 R 9:e4 B 10:d6 R 11:e5 B 12:e6 R 13:f5 B 14:f6 R 15:i4"
/>
Now Red is connected by the (ordinary) Tom's move.

See [[Tom's move for 3rd and 5th row parallel ladders]] for more details.

=== Tall variant ===

Tom's move for 3rd-and-5th row parallel ladders also has a tall version. It works in the same way as the tall version of Tom's move for 2nd-and-4th row parallel ladders above, given the appropriate amount of space.

<hexboard size="7x12"
coords="hide"
edges="bottom"
visible="area(d2,b4,b7,l7,l5,j3,g2,f1,e1)"
contents="R arrow(12):e1 d2 c3 b4 b5 B b6 c4 e2 R 1:e4 B 2:d5 R 3:f2"
/>

== See also ==

* [[Parallel ladder]]
* [[Edge template IV1d]]
* [[Fifth_row_edge_templates#V-2-m|Edge template V2m]]

[[category:ladder]]
[[category:Advanced Strategy]]

Online playing

2025-11-05T01:39:54Z

Selinger: /* Services for game analysis */ Fixed HexWorld info.

Hex can be played online either using a web-based format or via e-mail.

=Web-based games=

Web-based games can be played either in ''real-time'', where moves are made within minutes (or even seconds), or ''turn-based'', where a player has days for one move.

==Realtime playing sites==

The most popular sites (approximatively ordered by the number of hex games played on them daily) are:
* [[Board Game Arena]] supports real-time as well as turn-based play
* [[Kurnik|PlayOK (Kurnik)]] for real-time play with rankings, game records, tournaments
* [https://playhex.org/ PlayHex] by Alcalyn. Real-time and turn-based play.
* [[igGameCenter]] for real-time play, with time settings, and ranking (it has many connection games).
* [[boardspace]] for real-time play
* http://www.ludoteka.com/ for real-time play
* [https://hex.plus Hex+] for high speed real-time play

==Turn-based playing sites==

* [[Little Golem]]
* [[Board Game Arena]]
* [https://playhex.org/ PlayHex]
* See also [http://www.gamerz.net/pbmserv Richard's server]. One can play completely by e-mail, but it also has a [http://www.gamerz.net/pbmserv/List.php?Hex graphical interface] now. Furthermore any sized board is supported.
* [https://play.abstractplay.com/ Abstract Play]
* Hex may also be played over e-mail, in a turn-based fashion, with [[Unicode and ASCII boards]].

==Services for game analysis==

There are some services in the net, which help one play out different variations, analyze the games and share game records.

* [https://hexworld.org/board HexWorld] has an interactive board that can be shared as a link. It supports board sizes up to 53. It can import games from [[PlayHex]], [[Little Golem]], [[Board Game Arena]], and [[PlayOK]].

* [https://www.trmph.com/hex/board TRMPH] has 3 board sizes: 11, 13 and 19. It can import games from [[Little Golem]].

* [https://minortriad.com/ahex.html AHex] is a service for analyzing games, including variations. See the extensive help page for more info. [[Little Golem]] provides a link to analyze games using this service.

== See also ==

* [[Hex clubs]]

[[category: Hex community]]
[[category: online play]]

Online playing

2025-11-05T01:11:10Z

Selinger: Deleting hexmaster link (it does not exist).

Hex can be played online either using a web-based format or via e-mail.

=Web-based games=

Web-based games can be played either in ''real-time'', where moves are made within minutes (or even seconds), or ''turn-based'', where a player has days for one move.

==Realtime playing sites==

The most popular sites (approximatively ordered by the number of hex games played on them daily) are:
* [[Board Game Arena]] supports real-time as well as turn-based play
* [[Kurnik|PlayOK (Kurnik)]] for real-time play with rankings, game records, tournaments
* [https://playhex.org/ PlayHex] by Alcalyn. Real-time and turn-based play.
* [[igGameCenter]] for real-time play, with time settings, and ranking (it has many connection games).
* [[boardspace]] for real-time play
* http://www.ludoteka.com/ for real-time play
* [https://hex.plus Hex+] for high speed real-time play

==Turn-based playing sites==

* [[Little Golem]]
* [[Board Game Arena]]
* [https://playhex.org/ PlayHex]
* See also [http://www.gamerz.net/pbmserv Richard's server]. One can play completely by e-mail, but it also has a [http://www.gamerz.net/pbmserv/List.php?Hex graphical interface] now. Furthermore any sized board is supported.
* [https://play.abstractplay.com/ Abstract Play]
* Hex may also be played over e-mail, in a turn-based fashion, with [[Unicode and ASCII boards]].

==Services for game analysis==

There are some services in the net, which help one play out different variations, analyze the games and share game records.

* [https://hexworld.org/board HexWorld] has an interactive board that can be shared as a link. It has all board sizes up to 31.

* [https://www.trmph.com/hex/board TRMPH] has 3 board sizes: 11, 13 and 19. It can import games from [[Little Golem]].

* [https://minortriad.com/ahex.html AHex] is a service for analyzing games, including variations. See the extensive help page for more info. [[Little Golem]] provides a link to analyze games using this service.

== See also ==

* [[Hex clubs]]

[[category: Hex community]]
[[category: online play]]

Liberty

2025-10-25T14:56:09Z

Selinger: Updated the article on liberties, which was a stub.

A '''liberty''' of an [[isolated piece|isolated stone]] or a [[group]] is an unoccupied [[Hex (board element)|cell]] [[chain|adjacent]] to it. A stone or group with few liberties is generally weaker than one with many liberties.

== Advice for players ==

=== Old advice ===

An older version of this page, dated 2005, contained the following advice:

: ''Do not play an isolated piece with three or fewer liberties. Such a move is always a bad move — it is always possible to find a better one.''

The advice is not accurate. We now know that there are some situations where playing a stone with 3 liberties is the unique winning move ([[Wheel#The intrusion at B|see here for an example]]). Nevertheless, it is still a useful guideline in most situations that arise in real games.

=== Updated advice ===

Below, we analyze groups with up to 3 liberties. Based on the analysis, we can formulate the following principle:

: ''Do not play a stone that creates an interior group with 2 or fewer liberties, or with 3 liberties when two of them are adjacent or bolstered.''

== Analysis of groups with up to 3 liberties ==

=== Definitions ===

By an '''interior group''' we mean a group of [[friendly]] stones that is not adjacent to an edge.

A liberty L of a group X is said to be '''bolstered''' if it matches one of the following patterns (up to rotation and symmetry):
<hexboard size="3x3"
float="inline"
edges="none"
coords="none"
visible="-a1 c3 -a3,b3"
contents="R X:b1,c1 B a2 c2 E L:b2"
/><hexboard size="3x3"
float="inline"
edges="none"
coords="none"
visible="-a1 c3 -a3,b3"
contents="R X:b1 B a2 c1 c2 E L:b2"
/><hexboard size="3x3"
float="inline"
edges="none"
coords="none"
visible="-a1 c3 -a3,c2"
contents="R X:b1 b3 B a2 c1 E L:b2"
/>

=== Groups with one liberty ===

If an interior group has only a single liberty, the group is [[dead cell|dead]]. It can never become part of a shortest winning path for the group's owner.

Examples:
<hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B c2--e2--e3--d4--b4 R c3 d3"
visible="area(b3,c2,e2,e3,d4,b4)"
/> <hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B b4--b3--c2--e2--e3--d4 R c3 d3"
visible="area(b3,c2,e2,e3,d4,b4)"
/>
In both cases, the red group has a single liberty and is dead.

=== Groups with two adjacent liberties ===

If an interior group has only two liberties, and these liberties are adjacent to each other, the group is [[dead cell|dead]].

Examples:
<hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B b3--c2--e2--e3--d4 R c3 d3"
visible="area(b3,c2,e2,e3,d4,b4)"
/> <hexboard size="5x7"
float="inline"
edges="none"
coords="none"
contents="B b4--b3--c2--f2--f3--e4 R c3 d3 e3"
visible="area(b3,c2,f2,f3,e4,b4)"
/>
In both cases, the red group has two adjacent liberties and is dead.

=== Groups with two bolstered liberties ===

If an interior group has two non-adjacent liberties, and both of these liberties are bolstered, the group is [[captured cell|captured]]. If the group's owner plays in one liberty, the opponent can play in the other, [[dead cell|killing]] the group.

Examples:
<hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B b4--b3--c2 e2--e3--d4 R c3 d3"
visible="area(b3,c2,e2,e3,d4,b4)"
/><hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B b3--c2--e2 d4--c4 R c3 d3 a5 f3"
visible="area(b3,c2,e2,e3,d4,b4) a5 f3"
/>

=== Groups with two general liberties ===

If an interior group has two non-adjacent liberties that are not both bolstered, then the group is not in general captured. If one of the liberties is bolstered and the other is not, playing in the non-bolstered liberty [[domination|dominates]] playing in the bolstered one for both players.

Examples: Both players prefer "a" to "b".
<hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B c4--b4--b3--c2 e2--e3 R c3 d3 E a:d4 b:d2"
visible="area(b3,c2,e2,e3,d4,b4)"
/><hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B b3--c2--e2 d4--c4 R c3 d3 f3 E a:b4 b:e3"
visible="area(b3,c2,e2,e3,d4,b4) f3"
/>

For all groups with two general liberties, if it is the opponent's turn, they can [[dead cell|kill]] the group by taking away one of the liberties. Therefore, no player should play a move that creates an interior group with only 2 liberties.

=== Groups with three liberties ===

If an interior group has 3 liberties, it is not in general captured.

Examples: Three non-adjacent liberties, zero or one of which are bolstered.
<hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B all R b2 c2 E b1 a3 c3"
visible="area(a2,b1,d1,d2,c3,a3)"
/><hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B all R b2 c2 E c1 a3 c3"
visible="area(a2,b1,d1,d2,c3,a3)"
/>

However, if two of the liberties are adjacent or bolstered, then if it is the opponent's turn, the opponent can [[dead cell|kill]] the group by playing in the remaining liberty. Therefore, no player should play a move that creates an interior group with only 3 liberties of which two are adjacent or bolstered.

Examples: Two of three liberties are adjacent or bolstered. Blue can kill the red group by playing at "a".
<hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B c4--b4--b3--c2 e2--e3 R c3 d3 E a:d4 c2"
visible="area(b3,c2,e2,e3,d4,b4)"
/><hexboard size="5x6"
float="inline"
edges="none"
coords="none"
contents="B b3--c2--e2 d4--c4 R c3 d3 f3 E a:b4 d2"
visible="area(b3,c2,e2,e3,d4,b4) f3"
/>

== See also ==

* [[Bolstered template]]
* [[Dead cell]]
* [[Captured cell]]
* [[Domination]]

[[Category:Definition]]
[[Category:Theory]]

Wheel

2025-10-25T12:58:43Z

Selinger: Updated witness for intrusion B.

The '''wheel''' is a 3-stone [[interior template]]. It consists of three same-coloured pieces around a central point, along with the four empty cells marked "A" and "B".
<hexboard size="3x3"
coords="none"
edges="none"
visible="-a1 c3"
contents="R a3 b1 c2 E B:b2 A:(a2 b3 c1)"
/>

The red pieces are strongly [[Connection|connected]] to each other. [[Blue (player)|Blue]] cannot break the connection between them, and his available forcing moves are not very good. The wheel also has a large circumference, so that it is easy to connect to it from other parts of the board. The drawback is that it requires three pieces and is rather compact, so that it is often not [[efficiency|efficient]] enough to cover a large area.

Blue has three possible [[forcing move]]s, labelled A. If Blue plays in one of them, [[Red (player)|Red]] can restore the connection by playing in any of the other cells labelled A. (She can also play at B, but it's usually not as good.) By doing this, Red strengthens her connection, and the influence her pieces have on the rest of the board. It is therefore usually better for Blue not to [[Intrusion|intrude]].

Blue should not usually attempt to intrude by playing at B. Red can then restore the connection between her pieces by playing at any A. Red's position improves, while Blue's piece is not very useful. Although Blue still has to opportunity to intrude in one of the remaining cells marked A, it would have been better for Blue to do this in the first place.

==U-turn==

The wheel should never be confused with the ''U-turn'', shown here:

<hexboard size="3x3"
coords="none"
edges="none"
visible="-a1 c3 b3"
contents="R a3 b1 c2 E B:b2 A:(a2 c1)"
/>
If only two of the cells marked "A" are empty, and not all three of them, it is not a wheel and it is not a template. The U-turn is a very weak position. The U-turn sometimes arises when a player tries to connect via [[bridge]]s and takes a sharp turn. The problem is that Blue can play at B, leaving Red unable to defend both bridges.

==The intrusion at B==

While in most cases, it is not a good idea to [[intrusion|intrude]] into the wheel at "B", there are some positions where a move at "B" is winning and all other moves are losing. Such positions are quite rare. Here is an example, with Blue to move.

<hexboard size="7x7"
contents="B a1--e1 g1 a2 b2 a3 f3 e4 b5 c5 b6 c6 d6 g6 b7 c7 g7--e7 f5 R f1 d2 g2--g5 e3 a4--a7 d7 c4 d5
E A:(c3 e2 d4) B:d3 S area(c3,c4,d4,e3,e2,d2)"
/>
Note that "B" is the only winning move, both inside and outside the wheel. This position comes from the paper "On 3-terminal positions in Hex".

== References ==

* Eric Demer and Peter Selinger (2025). On 3-terminal positions in Hex. Available from <https://arxiv.org/abs/2507.08247>.

[[category:interior templates]]

Y

2025-08-27T23:13:51Z

Selinger: /* References */ Added Browne reference

The game of Y is a [[connection game]] first invented in the early 1950s by John Milnor, and independently discovered in 1953 by Craige Schensted (later known as Ea Ea) and Charles Titus. In its original form, it is played on a triangular grid of hexagons, but Schensted and Titus preferred grids the include some pentagons, see [[#Variations|Variations]] below. There are two [[player]]s, who have one colour each, and a move consists of placing a stone of your colour in one of the hexagons on the board. The winner is the first player to complete a [[chain]] connecting all three sides of the board. Y is a kind of generalisation of [[Hex]], perhaps the one closest to it, but there are some strategic peculiarities, such as [[corner template]]s.

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

== No draws ==

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

=== At most one winner ===

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

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

''First step'': if there is a group of stones that is completely surrounded by the opponent, let's consider the board with the surrounded group replaced by opponent's stones. The new board has the same status as the old one, as the surrounded group was not winning and the new big group is winning if and only if it was winning on the old board. Also note that there is still at least one group left.

Repeat this step until there are no more completely surrounded groups of either color. The resulting board has the same status as the original.

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

Repeat this step.

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

Repeat this step.

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

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

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

For another proof, see [[#Y-Reduction|Y-Reduction]] below.

== The first player wins ==

In Y the [[strategy-stealing argument]] can be applied. It proves that the second player has no winning strategy. The argument is that if the second player had a winning strategy, then the first player could chose a random first move and then pretend that she is the second player and apply the strategy. An important point is that an extra stone is not a disadvantage in Y. Since Y is a perfect information game without draws, there is a winning strategy for one player. The second player has no winning strategy so the first player has one.

== Relationship to [[Hex]] ==

=== Embedding Hex in Y ===

Hex game can be seen as a special case of a Y game. For instance consider the following Y board of size 7.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>
We can simulate a Hex game of [[size]] 4 on it.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
contents="R area(g1,e3,g3) B area(c5,a7,c7)"
/>
Now the only way to win for Blue is to cross the board horizontally, whereas the only way for Red to do so is to cross the board vertically.

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

The above proof of the Y theorem (that there is exactly one winner in Y) therefore implies the Hex theorem (that there is exactly one winner in Hex).

=== Embedding Y in Hex ===

The game of Y cannot be embedded in Hex in such a way that each single cell in Y corresponds to a single Hex cell. However, perhaps surprisingly, every Y position can be converted to a Hex position such that some Y cells correspond to two Hex cells. The technical term for this is that we express the Y board as a ''quotient'' of the Hex board.

Consider a Y position with numbered cells:
<hexboard size="4x4"
coords="none"
edges="none"
visible="area(d1,a4,d4)"
contents="B 1:d1 2:c2 R 3:d2 4:b3 5:c3 B 6:d3 7:a4 8:b4 R 9:c4 B 10:d4"
/>
We convert this to a Hex position by mirroring cells across the short diagonal:
<hexboard size="4x4"
coords="none"
edges="all"
contents="B 1:d1 2:c2 R 3:d2,c1 4:b3 5:c3,b2 B 6:d3,b1 7:a4 8:b4,a3 R 9:c4,a2 B 10:d4,a1"
/>
The winner of the Y position is also the winner of the Hex position, and vice versa. Proof: to win the Y position, a player must have a [[group]] that is connected to the short diagonal and also to the right and bottom edges. By mirror symmetry, the same stone of the short diagonal is also connected to the left and top edges, so it is a winning path in Hex (either top to bottom for Red, or left to right for Blue). Conversely, consider the winner of the Hex position, say Red. Then Red has a group of stones that connects the top edge to the bottom edge. It must pass through the short diagonal. By symmetry, that same stone on the short diagonal is also connected to the right and left edges, and therefore it is a winning group in Y.

This way of seeing a Y board as a quotient of a Hex board appears in Cameron Browne's book "Connection Games", Figure 7.14.

== Y-Reduction ==

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

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

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

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

== Swap ==

The [[swap rule]] can be used in Y. Opening moves in the center are good, and opening moves in the corners are bad, so there may well exist average opening moves. Further information, see [[Where to swap (y)]].

== Variations ==

As an alternative to the [[swap rule]]. Alternatively, one can play Double-Move Y, also known as Master Y: The first player places one stone on the board, and each subsequent move consists of placing two stones on the board. This is a pretty challenging variant, even on small boards.

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

[[Image:Y93_bent.gif]]

== See also ==

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

Help for [[programming the bent Y board]]

Try this [[Y puzzle]].

== On the web ==

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

== References ==

* John F. Nash. Some games and machines for playing them. RAND Corporation Report D-1164, February 2, 1952. https://www.rand.org/pubs/documents/D1164.html Credits John Milnor as the inventor of the game.

* Cameron Browne, [http://www.amazon.com/Connection-Games-Variations-Cameron-Browne/dp/1568812248/ref=pd_bbs_sr_1/104-1532904-9846317?ie=UTF8&s=books&qid=1177663469&sr=8-1 "Connection Games: Variations on a Theme"]. A K Peters/CRC Press, 2005.

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

Y

2025-08-27T23:11:56Z

Selinger: /* Embedding Y in Hex */ Added a reference

The game of Y is a [[connection game]] first invented in the early 1950s by John Milnor, and independently discovered in 1953 by Craige Schensted (later known as Ea Ea) and Charles Titus. In its original form, it is played on a triangular grid of hexagons, but Schensted and Titus preferred grids the include some pentagons, see [[#Variations|Variations]] below. There are two [[player]]s, who have one colour each, and a move consists of placing a stone of your colour in one of the hexagons on the board. The winner is the first player to complete a [[chain]] connecting all three sides of the board. Y is a kind of generalisation of [[Hex]], perhaps the one closest to it, but there are some strategic peculiarities, such as [[corner template]]s.

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

== No draws ==

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

=== At most one winner ===

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

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

''First step'': if there is a group of stones that is completely surrounded by the opponent, let's consider the board with the surrounded group replaced by opponent's stones. The new board has the same status as the old one, as the surrounded group was not winning and the new big group is winning if and only if it was winning on the old board. Also note that there is still at least one group left.

Repeat this step until there are no more completely surrounded groups of either color. The resulting board has the same status as the original.

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

Repeat this step.

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

Repeat this step.

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

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

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

For another proof, see [[#Y-Reduction|Y-Reduction]] below.

== The first player wins ==

In Y the [[strategy-stealing argument]] can be applied. It proves that the second player has no winning strategy. The argument is that if the second player had a winning strategy, then the first player could chose a random first move and then pretend that she is the second player and apply the strategy. An important point is that an extra stone is not a disadvantage in Y. Since Y is a perfect information game without draws, there is a winning strategy for one player. The second player has no winning strategy so the first player has one.

== Relationship to [[Hex]] ==

=== Embedding Hex in Y ===

Hex game can be seen as a special case of a Y game. For instance consider the following Y board of size 7.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>
We can simulate a Hex game of [[size]] 4 on it.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
contents="R area(g1,e3,g3) B area(c5,a7,c7)"
/>
Now the only way to win for Blue is to cross the board horizontally, whereas the only way for Red to do so is to cross the board vertically.

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

The above proof of the Y theorem (that there is exactly one winner in Y) therefore implies the Hex theorem (that there is exactly one winner in Hex).

=== Embedding Y in Hex ===

The game of Y cannot be embedded in Hex in such a way that each single cell in Y corresponds to a single Hex cell. However, perhaps surprisingly, every Y position can be converted to a Hex position such that some Y cells correspond to two Hex cells. The technical term for this is that we express the Y board as a ''quotient'' of the Hex board.

Consider a Y position with numbered cells:
<hexboard size="4x4"
coords="none"
edges="none"
visible="area(d1,a4,d4)"
contents="B 1:d1 2:c2 R 3:d2 4:b3 5:c3 B 6:d3 7:a4 8:b4 R 9:c4 B 10:d4"
/>
We convert this to a Hex position by mirroring cells across the short diagonal:
<hexboard size="4x4"
coords="none"
edges="all"
contents="B 1:d1 2:c2 R 3:d2,c1 4:b3 5:c3,b2 B 6:d3,b1 7:a4 8:b4,a3 R 9:c4,a2 B 10:d4,a1"
/>
The winner of the Y position is also the winner of the Hex position, and vice versa. Proof: to win the Y position, a player must have a [[group]] that is connected to the short diagonal and also to the right and bottom edges. By mirror symmetry, the same stone of the short diagonal is also connected to the left and top edges, so it is a winning path in Hex (either top to bottom for Red, or left to right for Blue). Conversely, consider the winner of the Hex position, say Red. Then Red has a group of stones that connects the top edge to the bottom edge. It must pass through the short diagonal. By symmetry, that same stone on the short diagonal is also connected to the right and left edges, and therefore it is a winning group in Y.

This way of seeing a Y board as a quotient of a Hex board appears in Cameron Browne's book "Connection Games", Figure 7.14.

== Y-Reduction ==

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

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

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

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

== Swap ==

The [[swap rule]] can be used in Y. Opening moves in the center are good, and opening moves in the corners are bad, so there may well exist average opening moves. Further information, see [[Where to swap (y)]].

== Variations ==

As an alternative to the [[swap rule]]. Alternatively, one can play Double-Move Y, also known as Master Y: The first player places one stone on the board, and each subsequent move consists of placing two stones on the board. This is a pretty challenging variant, even on small boards.

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

[[Image:Y93_bent.gif]]

== See also ==

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

Help for [[programming the bent Y board]]

Try this [[Y puzzle]].

== On the web ==

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

== References ==

* John F. Nash. Some games and machines for playing them. RAND Corporation Report D-1164, February 2, 1952. https://www.rand.org/pubs/documents/D1164.html Credits John Milnor as the inventor of the game.

* Felix Weilacher, "I guess you can get Y as a quotient of normal Hex by mirroring moves across the short diagonal". In a post on the [[Hex forums|Hex Discord]], August 27, 2025.

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

Y

2025-08-27T21:08:55Z

Selinger: Added Y as a quotient of Hex

The game of Y is a [[connection game]] first invented in the early 1950s by John Milnor, and independently discovered in 1953 by Craige Schensted (later known as Ea Ea) and Charles Titus. In its original form, it is played on a triangular grid of hexagons, but Schensted and Titus preferred grids the include some pentagons, see [[#Variations|Variations]] below. There are two [[player]]s, who have one colour each, and a move consists of placing a stone of your colour in one of the hexagons on the board. The winner is the first player to complete a [[chain]] connecting all three sides of the board. Y is a kind of generalisation of [[Hex]], perhaps the one closest to it, but there are some strategic peculiarities, such as [[corner template]]s.

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

== No draws ==

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

=== At most one winner ===

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

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

''First step'': if there is a group of stones that is completely surrounded by the opponent, let's consider the board with the surrounded group replaced by opponent's stones. The new board has the same status as the old one, as the surrounded group was not winning and the new big group is winning if and only if it was winning on the old board. Also note that there is still at least one group left.

Repeat this step until there are no more completely surrounded groups of either color. The resulting board has the same status as the original.

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

Repeat this step.

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

Repeat this step.

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

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

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

For another proof, see [[#Y-Reduction|Y-Reduction]] below.

== The first player wins ==

In Y the [[strategy-stealing argument]] can be applied. It proves that the second player has no winning strategy. The argument is that if the second player had a winning strategy, then the first player could chose a random first move and then pretend that she is the second player and apply the strategy. An important point is that an extra stone is not a disadvantage in Y. Since Y is a perfect information game without draws, there is a winning strategy for one player. The second player has no winning strategy so the first player has one.

== Relationship to [[Hex]] ==

=== Embedding Hex in Y ===

Hex game can be seen as a special case of a Y game. For instance consider the following Y board of size 7.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
/>
We can simulate a Hex game of [[size]] 4 on it.
<hexboard size="7x7"
coords="none"
edges="none"
visible="area(g1,a7,g7)"
contents="R area(g1,e3,g3) B area(c5,a7,c7)"
/>
Now the only way to win for Blue is to cross the board horizontally, whereas the only way for Red to do so is to cross the board vertically.

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

The above proof of the Y theorem (that there is exactly one winner in Y) therefore implies the Hex theorem (that there is exactly one winner in Hex).

=== Embedding Y in Hex ===

The game of Y cannot be embedded in Hex in such a way that each single cell in Y corresponds to a single Hex cell. However, perhaps surprisingly, every Y position can be converted to a Hex position such that some Y cells correspond to two Hex cells. The technical term for this is that we express the Y board as a ''quotient'' of the Hex board.

Consider a Y position with numbered cells:
<hexboard size="4x4"
coords="none"
edges="none"
visible="area(d1,a4,d4)"
contents="B 1:d1 2:c2 R 3:d2 4:b3 5:c3 B 6:d3 7:a4 8:b4 R 9:c4 B 10:d4"
/>
We convert this to a Hex position by mirroring cells across the short diagonal:
<hexboard size="4x4"
coords="none"
edges="all"
contents="B 1:d1 2:c2 R 3:d2,c1 4:b3 5:c3,b2 B 6:d3,b1 7:a4 8:b4,a3 R 9:c4,a2 B 10:d4,a1"
/>
The winner of the Y position is also the winner of the Hex position, and vice versa. Proof: to win the Y position, a player must have a [[group]] that is connected to the short diagonal and also to the right and bottom edges. By mirror symmetry, the same stone of the short diagonal is also connected to the left and top edges, so it is a winning path in Hex (either top to bottom for Red, or left to right for Blue). Conversely, consider the winner of the Hex position, say Red. Then Red has a group of stones that connects the top edge to the bottom edge. It must pass through the short diagonal. By symmetry, that same stone on the short diagonal is also connected to the right and left edges, and therefore it is a winning group in Y.

The fact that a Y board can be seen as a quotient of a Hex board was first observed by Felix Weilacher on the [[Hex_forums|Hex Discord]], following a similar observation by the user Bobson that Y is a quotient of Projective Hex.

== Y-Reduction ==

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

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

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

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

== Swap ==

The [[swap rule]] can be used in Y. Opening moves in the center are good, and opening moves in the corners are bad, so there may well exist average opening moves. Further information, see [[Where to swap (y)]].

== Variations ==

As an alternative to the [[swap rule]]. Alternatively, one can play Double-Move Y, also known as Master Y: The first player places one stone on the board, and each subsequent move consists of placing two stones on the board. This is a pretty challenging variant, even on small boards.

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

[[Image:Y93_bent.gif]]

== See also ==

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

Help for [[programming the bent Y board]]

Try this [[Y puzzle]].

== On the web ==

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

== References ==

* John F. Nash. Some games and machines for playing them. RAND Corporation Report D-1164, February 2, 1952. https://www.rand.org/pubs/documents/D1164.html Credits John Milnor as the inventor of the game.

* Felix Weilacher, "I guess you can get Y as a quotient of normal Hex by mirroring moves across the short diagonal". In a post on the [[Hex forums|Hex Discord]], August 27, 2025.

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

Hex clubs

2025-08-12T03:41:37Z

Selinger: Updates

This page lists Hex clubs, by which we mean communities of people who get together to play Hex (in-person or online) regularly or at specific times. Maybe there is a Hex club in your area, or you can join an online one. Maybe you would like to start a Hex club in your city! If you do, please add it here.

=== North America ===

* '''Halifax Hex Club.''' In Halifax, Canada. Meets in-person every other Friday. Active since February 2020. Open to the public. Website: https://www.mathstat.dal.ca/~selinger/hex-club/.

* '''Toronto Hex Club.''' In Toronto, Canada. <del>Currently dormant. Was active from January to June 2023. Meets Mondays, 8-11pm. Open to the public. Location: 281 Jedburgh Rd.</del> No longer dormant, but currently figuring out a consistent time and place. Contact: The #toronto channel on the [https://discord.gg/QXu9UqDdAC Hex Discord server].

=== Europe ===

* '''Go Club Darmstadt.''' In Darmstadt, Germany. Meets Tuesdays 7pm. This is an open Go club where some people play Hex sometimes. Location: Hobbit, Lauteschlägerstr. 3. Open to the public. Website: https://www.dgob.de/spielen/spielabende/ and search for Darmstadt. Contact: Tim Unverzagt, [mailto:go.darmstadt@gmail.com go.darmstadt@gmail.com]. May be dormant; received no reply from the above address.

=== Online ===

* '''Friday Hex Training Club.''' Meets first Friday of each month on [[Board Game Arena]] at 21:00 Netherlands time. May be dormant. Was active from April 2022 until September 2024. Open to the public. Website: https://boardgamearena.com/group?id=11592014.

* '''PlayOK Hex Meetings.''' Every Friday and Sunday at 6pm (London time) on [[PlayOK]]. Active since early 2024.

== See also ==

* [[Online playing]]
* [[Hex forums]]

[[category: Hex community]]
[[category: Online play]]

Smart Game Format

2025-08-12T03:30:22Z

Selinger: Another small typo

The Smart Game Format (SGF) is a file format for game records of 2-player board games. It's a text-only, tree-based format that was originally designed for the game of [[Go]], but has been adapted for a number of other games including Hex. Games stored in the SGF format can easily be emailed, posted or processed with text-based tools. SGF files usually use the filename extension <code>.sgf</code>.

The main purpose of SGF is to store records of completed games, and to provide features for storing games that have been analyzed and annotated (e.g., board markup, comments, and variations).

== Description of the file format ==

SGF is a text-based file format (not a binary format). The format can be described in two steps:

1. Syntax rules. This governs how SGF files are parsed. At this level, the SGF format provides a very generic representation of abstract trees of key-value dictionaries. It can be used for many kinds of tree-like data and is not necessarily limited to game trees.

2. Semantic rules. This governs how specific keys and values should be interpreted, often in game-specific ways.

The [http://www.red-bean.com/sgf/ official SGF specification] mixes syntactic and semantics concepts; for example, it specifies how property values must be parsed depending on what kind of property they belong to. By contrast, here we give an (equivalent) description that strictly separates syntax from semantics. This allows the file format to be parsed without any semantic knowledge, and it allows semantic properties to be checked without any knowledge of parsing.

=== Lexical structure ===

When reading SGF, the text is first converted to a sequence of lexical tokens. There are 8 different kinds of token:

* left parenthesis '('
* right parenthesis ')'
* semicolon ';'
* left square bracket '['
* right square bracket ']'
* colon ':'
* a property name, which is a sequence of one or more upper-case ASCII letters
* a literal string, which is any sequence of:
** any characters except ':', ']', and '\'
** two-character escape sequences, which consist of '\' followed by any character

A literal string starts immediately after a '[' or ':' token, and extends until the next unescaped ']' or ':'. No characters except ':', ']', and '\' have special syntactic meanings in literal strings, and in particular, if '[', '(', ')', or ';' appear in a literal string, they are not interpreted as separate tokens. The escape character is '\', and any character following '\' is added to the literal string unchanged, even if that character is ':', ']', or '\'. The only exception is that if '\' is immediately followed by a newline, both are removed. Literal strings may contain whitespace characters, including newlines, and these are preserved. Whitespace is also preserved at the beginning or end of literal strings.

All of the tokens are expressed in the ASCII character set, except for literal string data, which can use any character set. White space before or after tokens is ignored except when it is part of a literal string. Newlines can be encoded as NL, CR, CRNL, or NLCR. The interpretation of literal string data, including what kinds of strings can be used in specified contexts, is further defined by semantic rules, but plays no role in parsing.

In current applications, property names consist of one or two upper-case ASCII letters, and some legacy implementations may not recognize property values that are longer than 2 letters.

=== Syntactic structure ===

An SGF file describes one or more finitely branching ordered trees. Moreover, each node of the tree holds a dictionary, which is a mapping from property names to certain kinds of structured values. We begin by describing the encoding of dictionaries.

==== Dictionaries ====

A ''tuple'' consists of the token '[', zero or more literal strings that are separated by ':', and the token ']'. Examples of tuples are:

[]
[value]
[value1:value2]
[value1:value2:value3]
[Values may be arbitrary strings of characters,
including newlines and other whitespace.
Be aware\: the characters '\:', '\]', and '\\' must be escaped.
Other characters \m\a\y be escaped but this is optional. Sequences
such as \n have no special meaning; this is just another way to
write the letter n.
]

A ''binding'' consists of a property name followed by one or more tuples. Examples are:

FF[4]
AP[HexGui:0.9]
AB[a1][a2][a3]
C[This is a comment!]

Where a property name is followed by more than one tuple, the data is intended to be unordered. In other words, the following are two ways of expressing exactly the same data:

AB[a1][a2]
AB[a2][a1]

The data within each tuple is ordered. For example, the following are distinct:

AP[name:version]
AP[version:name]

The semantic rules place further restrictions on how many tuples are allowed after certain property names, and how many components are allowed in certain tuples.

A ''dictionary'' consists of zero more more bindings. The property names in any one dictionary must be distinct, and their ordering is not significant (they may and often will be reordered by an application). Example:

AP[HexGui:0.9]FF[4]GM[11]SZ[11]

==== Tree structure ====

A ''node'' in the tree consists of the token ';' followed by a dictionary (remember that dictionaries can be empty).
Here are some examples of nodes:

;AP[HexGui:0.9]FF[4]GM[11]SZ[11]
;B[i3]
;
;AB[f4][g2]PL[B]

A ''tree'' is given by the following grammar:

tree ::= '(' node+ tree* ')'

Here, <code>node+</code> means a sequence of one or more nodes, and <code>tree*</code> means a sequence of zero or more trees. Trees are interpreted as follows: the tree

( node₁ node₂ node₃ ... nodeₙ tree₁ ... treeₖ )

has a single root <code>node₁</code> with a single child <code>node₂</code>, which has a single child <code>node₃</code> and so on until <code>nodeₙ</code>, which has ''k'' children <code>tree₁ ... treeₖ</code>:

[[Image: Tree1.png|x101px]]

Note that it is possible that ''k'' = 0, in which case <code>nodeₙ</code> is a leaf; it is also possible that ''n'' = 1, in which case the entire tree is a leaf. Here are some examples:

{| class="wikitable"
|-
!Code
!Tree
|-
| <code>(a)</code>
| [[Image: Tree2.png|x23px]]
|-
| <code>(a b c)</code>
| [[Image: Tree3.png|x23px]]
|-
| <code>(a (b (c)))</code>
| [[Image: Tree3.png|x23px]]
|-
| <code>(a (b) (c))</code>
| [[Image: Tree4.png|x44px]]
|-
| <code>(a b (c) (d e (f (g) (h)) (i)))</code>
| [[Image: Tree5.png|x53px]]
|}

Finally, an SGF file holds a sequence of one or more trees. (The idea of this is that a single file may hold more than one game record, each with its own root. However, in practice, most SGF files contain exactly one tree, and most software that reads SGF files will ignore all but the first tree in it).

To conclude this section, here is an example of a syntactically (but not semantically) well-formed SGF file representing the tree

[[Image: Tree5.png|x53px]].

Each node has a dictionary with a single property <code>NN</code> holding the node's label.

(;NN[a];NN[b](;NN[c])(;NN[d];NN[e](;NN[f](;NN[g])(;NN[h]))(;NN[i])))

=== Semantic rules ===

Each property accepts specific types of values that are described below. Some properties may only appear in root nodes, and other properties may appear in any node. Some properties are mutually exclusive, i.e., cannot be used together in the same node.

A tuple with 1 component, such as <code>[11]</code>, is referred to as a ''simple value'', and a tuple with more than 1 component, such as <code>[name:version]</code>, is referred to as a ''composite value''. The SGF format does not permit composite values with more than 2 components.

As a special case, if a property requires a simple value, but a composite value is specified, it is converted to a simple value by concatenating all of its literal strings into a single string separated by ':'. This is because the SGF specification stipulates that ':' may appear unescaped in literal strings for properties whose semantics expects a simple value. Since semantic information is not available at parsing time, we re-construct such values during semantic interpretation.

Some common value types are:

* Number. Example: <code>[11]</code>.
* Point. This is the name of a cell on the Hex board. Following standard Hex [[conventions]], a cell is named by a column label (one or more letters) followed by a row label (one or more digits). If there are more than 26 columns, they are labeled by base-26 alphabet numbers, i.e., the next columns after 'z' are 'aa', 'ab', 'ac', etc. Cell names are case insensitive. Examples: <code>[a1]</code>, <code>[f6]</code>, <code>[ab28]</code>.
* Move. This is either a point, or one of the special moves 'swap-sides', 'swap-pieces', 'pass', 'resign', 'forfeit'.
* Text. This is arbitrary text, except that all whitespace characters other than newlines (example: tab, page break) are converted to spaces.
* Simpletext. This is arbitrary text, except that all whitespace characters including newlines are converted to spaces.

The SGF format defines a large number of property names, but many are rarely used, not relevant to Hex, or not supported by current software. We only list the most common property values. Others can be found in the [http://www.red-bean.com/sgf/ official SGF specification].

Users and applications are permitted to define their own private property names, as long as they do not clash with existing ones. A useful convention is for private property names to start with 'X'. Applications that read SGF files should ignore property names that they do not know about, and if possible, should preserve them (i.e., when writing the same file again).

In SGF, the players are always called B and W (black and white), regardless of which actual player colors were used in the original game. See [[conventions]] for more information on player colors and cell numbering.

==== Root properties ====

The following properties may appear at root nodes. They describe global attributes of the game, such as its board size.

* '''AP'''. Value: composite name : version. This identifies the name and version of the software application that generated the SGF file. Example: <code>AP[HexGUI:0.10]</code>.
* '''FF'''. Value: integer. This identifies the version of the SGF file format, currently 4. Example: <code>FF[4]</code>.
* '''GM'''. Value: integer. This identifies the game. The value for Hex is 11. Example: <code>GM[11]</code>.
* '''SZ'''. Value: integer, or composite integer : integer. This identifies the board size. For non-square boards, the number of columns is given before the number of rows, e.g. <code>SZ[6:7]</code> for a board with 6 columns and 7 rows (i.e., the distance between the white edges is smaller than the distance between the black edges). If the number of rows and columns is equal, it must be given as a single integer, e.g. <code>SZ[11]</code>.

The following properties are not currently supported by [[HexGui]]:

* '''PB''', '''PW'''. Value: simpletext. The name of the black player and white player, respectively. Example: <code>PB[Bill LeBoeuf]</code>.
* '''RE'''. Value: simpletext. The result of the game. If given, it must be one of the following: 'B+' or 'W+' for a black or white win, respectively, 'Void' for no result (such as suspended play), '?' for an unknown result. Optionally, the method of winning can be specified after '+', as follows: 'B+R', 'B+Resign', 'W+R', or 'W+Resign' for win by [[resigning]], 'B+T', 'B+Time', 'W+T', 'W+Time' for a win on time, 'B+F', 'B+Forfeit', 'W+F', or 'W+Forfeit' for a win by forfeit.
* '''DT'''. Value: date. The date on which the game was played, in the format 'YYYY-MM-DD'. There is support for partial dates and date ranges; see the [http://www.red-bean.com/sgf/ official specification site] for details.
* '''EV'''. Value: simpletext. The name of the event, e.g., tournament. Example: <code>EV[2022 Mind Sports Olympiad]</code>
* '''GC'''. Value: text. Background information on the game, or a summary of the game itself. This free-form text not usually interpreted by software.
* '''SO'''. Value: simpletext. The source of the game (e.g., book). This can be used to identify the website and table number for games played online. Example: <code>SO[BGA 123456789]</code>

==== Node properties ====

The following properties may appear at any node in a game tree. They describe attributes of the particular move or node. There are two kinds of nodes: move nodes and setup nodes. A ''move node'' holds a single move by one player, including special moves such as 'swap-sides' or 'resign'. A ''setup node'' exists to set up a board position, for example, a special starting position for a game or puzzle, or a position that is used to explain some point in a game comment. A node is a setup node if it does not contain the property B or W. The root node is always a setup node.

* '''B''', '''W'''. Value: move. A move by the black, respectively white, player. There can be at most one B or W property at a given node. Examples: <code>B[a3]</code>, <code>W[swap-pieces]</code>, <code>B[resign]</code>. A node that has no B or W property is a setup node.
* '''AB''', '''AW''', '''AE'''. Value: list of cells. These properties cannot be combined with the B or W properties. In other words, they are only permitted in setup nodes (including the root node of the game). The values of AB, AW, and AE are lists of cells to be occupied by black, white, or empty, respectively. The cell contents overwrite whatever was there before. In particular, AE can be used to empty a previously occupied cell. A setup node usually also has a PL property to define whose turn it is. Example: <code>AB[e7][e8][e9]AW[a6][b6]AE[f8][g6][g7]PL[B]</code>.
* '''PL'''. Value: 'B' or 'W'. Sets the player whose turn it is after the current move or setup. This is especially useful in conjunction with setup nodes, but can also be used for move nodes, say in certain [[handicap]] situations where a player gets two moves in a row. The SGF format does not enforce that moves are alternating, nor that the player who makes the next move is actually the player whose turn it is. The PL property is mostly used as a display hint, for example, to set the color of the cursor, or to tell the user of a Hex puzzle whose turn it is.
* '''C'''. Value: text. A human-readable comment for the given node. Comments are free-form, but it is good style to avoid referring to physical board directions since it is not known how the board is oriented for the viewer. So instead of "left edge" or "bottom right corner", it might be better to refer to the "A-edge" or the "k11 corner". It probably makes sense to refer to the players as Black and White, regardless of what the original game colors were, since most SGF viewers tend to use black and white. Example: <code>C[White is already connected to the K-edge.]</code>

The following property is only partially supported by [[HexGui]]:

* '''LB'''. Value: list of composite cell : simpletext. This assigns (preferably short) labels to cells. Example: <code>LB[a1:x][a2:y][a3:z]</code>

=== Example ===

Here is a small but complete example of a game with two branches, some comments, and a setup node:

(;AP[HexGui:0.10.GIT]FF[4]GM[11]SZ[7]C[Example game]
;B[c5]C[This opening is too strong. White will definitely swap it.]
;W[swap-pieces];B[c4];W[c5];B[a6];W[c6]C[Good.]
;B[a7];W[b5];B[a5];W[b3]
(;B[d2]C[See the next variation for what happens if Black plays b4.]
;W[b4];B[d4];W[e5];B[resign];)
(;B[b4];W[d2]
;AB[a2][b2][c1][d1][d4][d5][e1][e5][f1][f5][g5]
C[Note that White is connected by templates, requiring only the area shown.])
)

== See Also ==

* [[Coordinates]]
* [[Conventions]]

== External links ==

You can find more info in the [http://www.red-bean.com/sgf/ Official Specification Site].

[[category:computer Hex]]
[[category:game record]]

Smart Game Format

2025-08-12T03:25:18Z

Selinger: /* Semantic rules */ Minor typos

The Smart Game Format (SGF) is a file format for game records of 2-player board games. It's a text-only, tree-based format that was originally designed for the game of [[Go]], but has been adapted for a number of other games including Hex. Games stored in the SGF format can easily be emailed, posted or processed with text-based tools. SGF files usually use the filename extension <code>.sgf</code>.

The main purpose of SGF is to store records of completed games, and to provide features for storing games that have been analyzed and annotated (e.g., board markup, comments, and variations).

== Description of the file format ==

SGF is a text-based file format (not a binary format). The format can be described in two steps:

1. Syntax rules. This governs how SGF files are parsed. At this level, the SGF format provides a very generic representation of abstract trees of key-value dictionaries. It can be used for many kinds of tree-like data and is not necessarily limited to game trees.

2. Semantic rules. This governs how specific keys and values should be interpreted, often in game-specific ways.

The [http://www.red-bean.com/sgf/ official SGF specification] mixes syntactic and semantics concepts; for example, it specifies how property values must be parsed depending on what kind of property they belong to. By contrast, here we give an (equivalent) description that strictly separates syntax from semantics. This allows the file format to be parsed without any semantic knowledge, and it allows semantic properties to be checked without any knowledge of parsing.

=== Lexical structure ===

When reading SGF, the text is first converted to a sequence of lexical tokens. There are 8 different kinds of token:

* left parenthesis '('
* right parenthesis ')'
* semicolon ';'
* left square bracket '['
* right square bracket ']'
* colon ':'
* a property name, which is a sequence of one or more upper-case ASCII letters
* a literal string, which is any sequence of:
** any characters except ':', ']', and '\'
** two-character escape sequences, which consist of '\' followed by any character

A literal string starts immediately after a '[' or ':' token, and extends until the next unescaped ']' or ':'. No characters except ':', ']', and '\' have special syntactic meanings in literal strings, and in particular, if '[', '(', ')', or ';' appear in a literal string, they are not interpreted as separate tokens. The escape character is '\', and any character following '\' is added to the literal string unchanged, even if that character is ':', ']', or '\'. The only exception is that if '\' is immediately followed by a newline, both are removed. Literal strings may contain whitespace characters, including newlines, and these are preserved. Whitespace is also preserved at the beginning or end of literal strings.

All of the tokens are expressed in the ASCII character set, except for literal string data, which can use any character set. White space before or after tokens is ignored except when it is part of a literal string. Newlines can be encoded as NL, CR, CRNL, or NLCR. The interpretation of literal string data, including what kinds of strings can be used in specified contexts, is further defined by semantic rules, but plays no role in parsing.

In current applications, property names consist of one or two upper-case ASCII letters, and some legacy implementations may not recognize property values that are longer than 2 letters.

=== Syntactic structure ===

An SGF file describes one or more finitely branching ordered trees. Moreover, each node of the tree holds a dictionary, which is a mapping from property names to certain kinds of structured values. We begin by describing the encoding of dictionaries.

==== Dictionaries ====

A ''tuple'' consists of the token '[', zero or more literal strings that are separated by ':', and the token ']'. Examples of tuples are:

[]
[value]
[value1:value2]
[value1:value2:value3]
[Values may be arbitrary strings of characters,
including newlines and other whitespace.
Be aware\: the characters '\:', '\]', and '\\' must be escaped.
Other characters \m\a\y be escaped but this is optional. Sequences
such as \n have no special meaning; this is just another way to
write the letter n.
]

A ''binding'' consists of a property name followed by one or more tuples. Examples are:

FF[4]
AP[HexGui:0.9]
AB[a1][a2][a3]
C[This is a comment!]

Where a property name is followed by more than one tuple, the data is intended to be unordered. In other words, the following are two ways of expressing exactly the same data:

AB[a1][a2]
AB[a2][a1]

The data within each tuple is ordered. For example, the following are distinct:

AP[name:version]
AP[version:name]

The semantic rules place further restrictions on how many tuples are allowed after certain property names, and how many components are allowed in certain tuples.

A ''dictionary'' consists of zero more more bindings. The property names in any one dictionary must be distinct, and their ordering is not significant (they may and often will be reordered by an application). Example:

AP[HexGui:0.9]FF[4]GM[11]SZ[11]

==== Tree structure ====

A ''node'' in the tree consists of the token ';' followed by a dictionary (remember that dictionaries can be empty).
Here are some examples of nodes:

;AP[HexGui:0.9]FF[4]GM[11]SZ[11]
;B[i3]
;
;AB[f4][g2]PL[B]

A ''tree'' is given by the following grammar:

tree ::= '(' node+ tree* ')'

Here, <code>node+</code> means a sequence of one or more nodes, and <code>tree*</code> means a sequence of zero or more trees. Trees are interpreted as follows: the tree

( node₁ node₂ node₃ ... nodeₙ tree₁ ... treeₖ )

has a single root <code>node₁</code> with a single child <code>node₂</code>, which has a single child <code>node₃</code> and so on until <code>nodeₙ</code>, which has ''k'' children <code>tree₁ ... treeₖ</code>:

[[Image: Tree1.png|x101px]]

Note that it is possible that ''k'' = 0, in which case <code>nodeₙ</code> is a leaf; it is also possible that ''n'' = 1, in which case the entire tree is a leaf. Here are some examples:

{| class="wikitable"
|-
!Code
!Tree
|-
| <code>(a)</code>
| [[Image: Tree2.png|x23px]]
|-
| <code>(a b c)</code>
| [[Image: Tree3.png|x23px]]
|-
| <code>(a (b (c)))</code>
| [[Image: Tree3.png|x23px]]
|-
| <code>(a (b) (c))</code>
| [[Image: Tree4.png|x44px]]
|-
| <code>(a b (c) (d e (f (g) (h)) (i)))</code>
| [[Image: Tree5.png|x53px]]
|}

Finally, an SGF file holds a sequence of one or more trees. (The idea of this is that a single file may hold more than one game record, each with its own root. However, in practice, most SGF files contain exactly one tree, and most software that reads SGF files will ignore all but the first tree in it).

To conclude this section, here is an example of a syntactically (but not semantically) well-formed SGF file representing the tree

[[Image: Tree5.png|x53px]].

Each node has a dictionary with a single property <code>NN</code> holding the node's label.

(;NN[a];NN[b](;NN[c])(;NN[d];NN[e](;NN[f](;NN[g])(;NN[h]))(;NN[i])))

=== Semantic rules ===

Each property accepts specific types of values that are described below. Some properties may only appear in root nodes, and other properties may appear in any node. Some properties are mutually exclusive, i.e., cannot be used together in the same node.

A tuple with 1 component, such as <code>[11]</code>, is referred to as a ''simple value'', and a tuple with more than 1 component, such as <code>[name:version]</code>, is referred to as a ''composite value''. The SGF format does not permit composite values with more than 2 components.

As a special case, if a property requires a simple value, but a composite value is specified, it is converted to a simple value by concatenating all of its literal strings into a single string separated by ':'. This is because the SGF specification stipulates that ':' may appear unescaped in literal strings for properties whose semantics expects a simple value. Since semantic information is not available at parsing time, we re-construct such values during semantic interpretation.

Some common value types are:

* Number. Example: <code>[11]</code>.
* Point. This is the name of a cell on the Hex board. Following standard Hex [[conventions]], a cell is named by a column label (one or more letters) followed by a row label (one or more digits). If there are more than 26 columns, they are labeled by base-26 alphabet numbers, i.e., the next columns after 'z' are 'aa', 'ab', 'ac', etc. Cell names are case insensitive. Examples: <code>[a1]</code>, <code>[f6]</code>, <code>[ab28]</code>.
* Move. This is either a point, or one of the special moves 'swap-sides', 'swap-pieces', 'pass', 'resign', 'forfeit'.
* Text. This is arbitrary text, except that all whitespace characters other than newlines (example: tab, page break) are converted to spaces.
* Simpletext. This is arbitrary text, except that all whitespace characters including newlines are converted to spaces.

The SGF format defines a large number of property names, but many are rarely used, not relevant to Hex, or not supported by current software. We only list the most common property values. Others can be found in the [http://www.red-bean.com/sgf/ official SGF specification].

Users and applications are permitted to define their own private property names, as long as they do not clash with existing ones. A useful convention is for private property names to start with 'X'. Applications that read SGF files should ignore property names that they do not know about, and if possible, should preserve them (i.e., when writing the same file again).

In SGF, the players are always called B and W (black and white), regardless of which actual player colors were used in the original game. See [[conventions]] for more information on player colors and cell numbering.

==== Root properties ====

The following properties may appear at root nodes. They describe global attributes of the game, such as its board size.

* '''AP'''. Value: composite name : version. This identifies the name and version of the software application that generated the SGF file. Example: <code>AP[HexGUI:0.10]</code>.
* '''FF'''. Value: integer. This identifies the version of the SGF file format, currently 4. Example: <code>FF[4]</code>.
* '''GM'''. Value: integer. This identifies the game. The value for Hex is 11. Example: <code>GM[11]</code>.
* '''SZ'''. Value: integer, or composite integer : integer. This identifies the board size. For non-square boards, the number of columns is given before the number of rows, e.g. <code>SZ[6:7]</code> for a board with 6 columns and 7 rows (i.e., the distance between the white edges is smaller than the distance between the black edges). If the number of rows and columns is equal, it must be given as a single integer, e.g. <code>SZ[11]</code>.

The following properties are not currently supported by [[HexGui]]:

* '''PB''', '''PW'''. Value: simpletext. The name of the black player and white player, respectively. Example: <code>PB[Bill LeBoeuf]</code>.
* '''RE'''. Value: simpletext. The result of the game. If given, it must be one of the following: 'B+' or 'W+' for a black or white win, respectively, 'Void' for no result (such as suspended play), '?' for an unknown result. Optionally, the method of winning can be specified after '+', as follows: 'B+R', 'B+Resign', 'W+R', or 'W+Resign' for win by [[resigning]], 'B+T', 'B+Time', 'W+T', 'W+Time' for a win on time, 'B+F', 'B+Forfeit', 'W+F', or 'W+Forfeit' for a win by forfeit.
* '''DT'''. Value: date. The date on which the game was played, in the format 'YYYY-MM-DD'. There is support for partial dates and date ranges; see the [http://www.red-bean.com/sgf/ official specification site] for details.
* '''EV'''. Value: simpletext. The name of the event, e.g., tournament. Example: <code>EV[2022 Mind Sports Olympiad]</code>
* '''GC'''. Value: text. Background information on the game, or a summary of the game itself. This free-form text not usually interpreted by software.
* '''SO'''. Value: simpletext. The source of the game (e.g., book). This can be used to identify the website and table number for games played online. Example: <code>SO[BGA 123456789]</code>

==== Node properties ====

The following properties may appear at any node in a game tree. They describe attributes of the particular move or node. There are two kinds of nodes: move nodes and setup nodes. A ''move node'' holds a single move by one player, including special moves such as 'swap-sides' or 'resign'. A ''setup node'' exists to set up a board position, for example, a special starting position for a game or puzzle, or a position that is used to explain some point in a game comment. A node is a setup node if it does not contain the property B or W. The root node is always a setup node.

* '''B''', '''W'''. Value: move. A move by the black, respectively white, player. There can be at most one B or W property at a given node. Examples: <code>B[a3]</code>, <code>W[swap-pieces]</code>, <code>B[resign]</code>. A node that has no B or W property is a setup node.
* '''AB''', '''AW''', '''AE'''. Value: list of cells. These properties cannot be combined with the B or W properties. In other words, they are only permitted in setup nodes (including the root node of the game). The values of AB, AW, and AE are lists of cells to be occupied by black, white, or empty, respectively. The cell contents overwrite whatever was there before. In particular, AE can be used to empty a previously occupied cell. A setup node usually also has a PL property to define whose turn it is. Example: <code>AB[e7][e8][e9]AW[a6][b6]AE[f8][g6][g7]PL[B]</code>.
* '''PL'''. Value: 'B' or 'W'. Sets the player whose turn it is after the current move or setup. This is especially useful in conjunction with setup nodes, but can also be used for move nodes, say in certain [[handicap]] situations where a player gets two moves in a row. The SGF format does not enforce that moves are alternating, nor that the player who makes the next move is actually the player whose turn it is. The PL property is mostly used as a display hint, for example, to set the color of the cursor, or to tell the user of a Hex puzzle whose turn it is.
* '''C'''. Value: text. A human-readable comment for the given node. Comments are free-form, but it is good style to avoid referring to physical board directions since it is not known how the board is oriented for the viewer. So instead of "left edge" or "bottom right corner", it might be better to refer to the "A-edge" or the "k11 corner". It probably makes sense to refer to the players as Black and White, regardless of what the original game colors were, since most SGF viewers tend to use black and white. Example: <code>C[White is already connected to the K-edge.]</code>

The following property is only partially supported by [[HexGui]]:

* '''LB'''. Value: list of composite cell : simpletext. This assigns (preferably short) labels to cells. Example: <code>LB[a1:x][a2:y][a3:z]</code>

=== Example ===

Here is a small but complete example of a game with two branches, some comments, and a setup node:

(;AP[HexGui:0.10.GIT]FF[4]GM[11]SZ[7]C[Example game]
;B[c5]C[This opening is too strong. White will definitely swap it.]
;W[swap-pieces];B[c4];W[c5];B[a6];W[c6]C[Good.]
;B[a7];W[b5];B[a5];W[b3]
(;B[d2]C[See the next variation for what happens if Black plays b4.]
;W[b4];B[d4];W[e5];B[resign];)
(;B[b4];W[d2]
;AB[a2][b2][c1][d1][d4][d5][e1][e5][f1][f5][g5]
C[Note that White is connected by templates, requiring only the area shown.])
)

== See Also ==

* [[Coordinates]]
* [[Conventions]]

== External links ==

You can find more info in the [http://www.red-bean.com/sgf/ Official Specifications Site].

[[category:computer Hex]]
[[category:game record]]

Hex Strategy Making the Right Connections

2025-06-30T00:39:31Z

Selinger: Updated time interval

"Hex Strategy: Making the Right Connections" is a valuable resource for understanding the game of [[Hex]], written by Cameron Browne and published in 2000. The book covers a wide range of topics, including [[strategy]], [[Hex theory|theory]], and [[computer Hex]].

In writing the book, Browne compiled the advice of dozens of expert players at the time, using Richard Rognlie's server and forum. However, given that the book was written over 25 years ago, some of the information may no longer be current. The opening chapter, which focuses mainly on non-[[swap rule|swap]] Hex, is now outdated. While the remaining strategy advice may still be relevant, some of it has been superseded. This is due to the significant advancements in the game that have taken place in recent years, as well as the emergence of highly skilled computer Hex players, who have added to our understanding of Hex strategy since the book's publication. Despite this, the book still offers valuable insights and is a worthwhile read for those interested in the game of Hex.

== External links ==

Cameron Browne's [http://www.cameronius.com/ website]

{{stub}}
[[category:strategy]]
[[category:Hex community]]

Domination

2025-06-29T16:52:50Z

Selinger: Added two links

In game theory, a move ''dominates'' another move if it is at least as good. In Hex, we say that a cell X dominates another cell Y in a given position (and from a particular player's point of view) if playing at X is at least as good as playing at Y for that player. If there is a set of cells in which one cell dominates all of the others, the player can eliminate the dominated cells from consideration, because moving in the dominating cell will be at least as good. This can often simplify the analysis of Hex positions.

Technically, every winning move dominates every other winning move (as well as every losing move), since any winning move is as good as any other. However, it is often possible to figure out domination ''locally'', i.e., by looking at a few nearby cells, rather than having to consider the whole board. In particular, it is often possible to figure out whether one move dominates another without actually knowing whether either of these moves is winning or losing.

If a player already knows that they are winning, domination is no longer important. In that case, the player might prefer to play locally dominated moves that win [[efficiency|efficiently]], i.e., that minimize the number of remaining moves until the game is won.

== Capture-domination ==

In general, it can be [[Hex theory#Complexity|difficult]] to determine whether one move dominates another. But there are many situations where the concept of [[captured cell|capturing]] can be used to reason about domination.

If moving at X would capture Y, then X always dominates Y. In this case, we say that X ''capture-dominates'' Y. It is often possible to figure this out locally, i.e., by looking at a few nearby cells.

== Examples of dominated patterns ==

In all of the following examples, we assume that Red is the player to move. In each case, the move marked "*" capture-dominates the moves marked "+", because moving at "*" would [[captured cell|capture]] the cells marked "+". The cells that are left white can be any color.

<hexboard size="4x5"
coords="none"
edges="bottom"
visible="area(c2,a4,d4,d2)"
contents="E +:b4 *:c3 +:c4"
/>
<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R b4 c4 d4 E +:c3 +:d3 *:d2"
/>
<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R b4 c4 e2 E +:c3 +:d3 *:d2"
/>
<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R c4 d2 B e3 E +:c3 +:d3 *:b4"
/>
<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R b4 c4 d4 B c2 E +:c3 +:d3 *:e3"
/>
<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R c4 d4 B c2 b3 E +:c3 +:d3 *:e3"
/>
<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R c4 d4 B b3 E +:c3 +:d3 *:d2"/>

== Mutually dominating moves ==

It is possible for two or more cells to dominate each other. In this case, they are all equally good moves. However, the player can still eliminate all but one of these moves from consideration.

For example, in the following position, each of the cells marked "*" dominates the other two. So any of these moves are equally good for Red (but of course there may be other moves on the board that are even better). When Red considers possible moves, she can concentrate on whichever of the three marked cells is the most convenient, and ignore the two others.

<hexboard size="5x6"
coords="none"
edges="none"
visible="area(c1,a3,a5,d5,f3,f1)"
contents="R b3 b4 d4 e3 E *:c3 *:d3 *:c4"
/>

== Usage examples ==

=== Example 1 ===

Consider the following position, with Red to move. What is the best way for Red to connect her two groups?

<hexboard size="7x7"
coords="none"
edges="none"
visible="area(e2,c4,b4,b5,c6,e5,f4,f3)"
contents="R b4 b5 e5 f4 e3 B e2 d3 c6 E a:f3 b:c4 c:d4 d:e4 f:c5 g:d5"
/>

The answer is c. A move at f would also connect Red's groups, but c dominates f. In fact, Red c [[captured cell|captures]] b, d, f, and g, and therefore dominates all of these moves.

<hexboard size="7x7"
coords="none"
edges="none"
visible="area(e2,c4,b4,b5,c6,e5,f4,f3)"
contents="R b4 b5 e5 f4 e3 B e2 d3 c6 E +:c4 +:c5 *:d4 +:d5 +:e4"
/>

To see why c is strictly better than f, note that if Red connects using f, then Blue still has a [[forcing move]] at a; but if Red connects using c, Blue has no such forcing move.

=== Example 2 ===

Consider the following position, where Red has [[captured cell|captured]] cells a and b. If Blue actually plays in one of the cells a or b, what should Red do? Red could certainly play in the other of the two cells, enforcing the capture. However, it is better for Red to play at c, [[captured cell|capturing]] all 5 cells. This response capture-dominates the solid connection at a or b.
<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"/>
In fact, if there is additional space, Red can even play at d, which captures all 7 cells and therefore dominates c (as well as a and b):
<hexboard size="3x4"
coords="none"
edges="bottom"
visible="area(b2,a3,d3,d2)"
contents="R b2 S red:(a3 b3) E a:a3 b:b3 c:c2 d:d2"/>

== Types of domination ==

There are various different ways in which we can figure out that one move dominates another. Capture-domination, already discussed above, is by far the most common. However, there are many other kinds of domination. We list several of them here. Of these, fillin-domination and switch-domination are the most general; all of the other kinds of domination listed here are special cases of these.

Unless noted otherwise, we discuss all of the following types of domination from Red's point of view. Of course, the analogous notions also exist for Blue.

=== Fillin-domination ===

In a given position, we say that an empty cell Y can be ''filled-in'' by Red if having a red stone at Y does not help Red; in other words, if the position with Y being empty is equivalent to the position with Y being red. If Y can be filled-in by Red, we also say that Y is Red ''fillin''.

Fillin is more general than capture: every captured cell is fillin, but not every fillin is captured. For example, it can be shown that in the following situation, Y is red fillin (see [[#Red.27s_move_at_h|the 10-cell acute corner]] below). But Y is not red-captured.

<hexboard size="5x5"
float="inline"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)-e2"
contents="b5 c5 d5 e5 c4 d4 e4 d3 e3 e2 R d3 E Y:e3"
/>

We say that X ''fillin-dominates'' Y (for Red) if a red stone at X makes Y into fillin. In other words, if having a red stone at X ''and'' Y is no better for Red than just having red stone at X.

Proof of domination: suppose that X and Y are empty, and moving at Y is winning for Red. Then certainly occupying both X and Y in a single move would also be winning for Red. On the other hand, due to fillin, moving at X is just as good for Red as moving at both X and Y. Therefore, moving at X is also winning.

A note of caution about fillin: unlike capture, fillin is not closed under unions. In other words, if some set S of cells is fillin, and some other set T of cells is fillin, it does not follow that S ∪ T is fillin.

=== Capture-domination ===

Capture-domination was already described above. Since every captured cell is also fillin, capture-domination is a special case of fillin-domination.

=== Star decomposition domination ===

Suppose that some region of the Hex board is completely surrounded by red and blue stones (or board edges), in such a way that the boundary consists of exactly two groups of red stones and two groups of blue stones. Suppose, moreover, that the region is such that whoever moves next can connect their two groups. [https://webdocs.cs.ualberta.ca/~hayward/papers/revDom.pdf Henderson and Hayward] call such a region a "star". If coloring some cell in the region does not alter the star-ness of the region, then that cell is fillin.

Star decomposition domination is the special case of fillin-domination where the fillin is due to a star region.
As an example of this, consider the following position, with Blue to move. Which of the marked moves dominates the others?
<hexboard size="5x6"
coords="none"
edges="bottom"
visible="area(c2,a4,a5,e5,f4,f2)"
contents="B b4 R d3 E d:b5 a:c4 b:d4 e:d5 c:e4"
/>

The answer is c. We reason as follows. If Blue plays at c, then the shaded region is a star:
<hexboard size="5x6"
coords="none"
edges="bottom"
visible="area(c2,a4,a5,e5,f4,f2)"
contents="B b4 R d3 S b5 c4 d4 d5 c5 B c:e4 E d:b5 a:c4 b:d4 e:d5 E O:c5"
/>
Moreover, each of a, b, d, e can be individually colored blue without changing that the region is a star. Therefore, each one of a, b, d, and e is individually fillin (note that not all of these cells can be filled-in simultaneously, or else the region would no longer be a star). It follows that c dominates each of a, b, d, and e (but not O).

=== Switch-domination ===

Consider a position with two empty cells X and Y. We say that X ''switch-dominates'' Y if the following holds: For every way of filling the remaining empty cells of the board with red and blue stones, if X=blue and Y=red is winning, then X=red and Y=blue is also winning.

Proof of domination: Suppose that in some position, X switch-dominates Y, and Red has a winning strategy S that starts by Red moving at Y. We claim that Red also has a winning strategy that starts by moving at X. In fact, let S' be the strategy that is exactly like S, but with the cells X and Y switched. In other words, every time Blue moves in one of X or Y, Red pretends that Blue has moved in the other of these cells. And every time the strategy S tells Red to move in X or Y, Red moves in the other of these cells. The strategy S' starts with Red moving at X. Assume that Red follows the strategy S' and that the game continues until the entire board is filled with stones (if the game finishes earlier, just let the players continue playing). At the end of the game, we reach a position where X=red. Case 1: Y=red. In that case, X and Y are the same color, so the position is exactly the same as would have been reached if Red had followed strategy S (and therefore a win for Red). Case 2: Y=blue. If Red had followed the strategy S, the game would have ended in exactly the same position, but with X=blue and Y=red. This would have been winning for Red. By the switch-domination assumption, X=red and Y=blue is then also a win for Red.

Here is an example:

<hexboard size="3x3"
edges="none"
coords="none"
visible="-a1,c3"
contents="R b1 a2 a3 E X:c2 Y:b2"
/>

We claim that X switch-dominates Y. To see why, fill the board with stones such that X=blue and Y=red and assume Red is the winner. Then the blue stone at X [[dead cell|kills]] the red stone at Y, and therefore this position is equivalent to X=blue and Y=blue. Since it is winning for Red, it follows that X=red and Y=blue is also winning for Red, showing that X switch-dominates Y.

We note that switch-domination does not imply fillin-domination (and therefore does not imply capture-domination either). In fact, in the example we just gave, X does not fillin-dominate Y. To see why, consider X=red. Then with Blue to move, the left position is winning for Blue, and the right position is winning for Red. This shows that they are not equivalent, i.e., Y is not fillin.

<hexboard size="5x4"
float="inline"
edges="all"
coords="none"
contents="B a1 a2 b1 c1 c4 c5 d3 d4 d5 R a3 a4 b2 b5 X:c3 d1 E Y:b3 S area(b2,a3,a4,b4,c3,c2)"
/>
<hexboard size="5x4"
float="inline"
edges="all"
coords="none"
contents="B a1 a2 b1 c1 c4 c5 d3 d4 d5 R a3 a4 b2 b5 X:c3 d1 R Y:b3 S area(b2,a3,a4,b4,c3,c2)"
/>

Conversely, fillin-domination does not imply switch-domination. For example, consider the following 2x2 position with Red to move:
<hexboard size="2x2"
edges="all"
coords="none"
contents="R a1 E X:b1 Y:a2"
/>
Here, X captures the entire bottom row, so X capture-dominates Y. But X does not switch-dominate Y; for example, if we fill the rest of the board like this,
<hexboard size="2x2"
edges="all"
coords="none"
contents="R a1 B b2 E X:b1 Y:a2"
/>
then (X,Y)=(blue,red) is winning and (X,Y)=(red,blue) is losing for Red.

We also note that switch-domination is the same for both players: If X switch-dominates Y from Red's point of view, then X also switch-dominates Y from Blue's point of view. Fillin-domination does not have this property.

=== Kill-domination ===

From Red's point of view, we say that X ''kill-dominates'' Y if a blue stone at X would kill Y. In fact, the example of switch-domination we gave above was also an example of kill-domination.

Kill-domination is a special case of switch-domination. Proof: Suppose X kill-dominates Y. Consider a completely-filled board position with X=blue and Y=red that is winning for Red. Since Y is dead, X=blue and Y=blue is also winning for Red, therefore X=red and Y=blue is also winning for Red. Therefore X switch-dominates Y.

However, not every example of switch-domination arises from kill-domination. Switch-domination is the same from both players' point of view, but kill-domination is not.

An example of kill-domination is a [[Peep#Bridge_peep|bridge peep]]:
<hexboard size="3x3"
edges="none"
coords="none"
visible="-a1 a3 b1 c3"
contents="B c1 b3 R a2 E X:c2 Y:b2"
/>
Here, Blue X would kill Y, and therefore from both Red and Blue's point of view, X dominates Y.

=== Path-domination ===

From Red's point of view, we say that X ''path-dominates'' Y if every minimal red winning path that passes though Y also passes through X. By a "minimal winning path", we mean a set of stones, added to the given position, that yields a [[chain]] between the player's edges, and such that no proper subset of the stones has this property.

Path-domination is a special case of switch-domination. Proof: Suppose X path-dominates Y. Consider a completely-filled board position with X=blue and Y=red that is winning for Red. Change the red stone at Y to a blue stone. If the resulting position is winning for Red, then so is X=red and Y=blue, because the extra red stone at X can only help Red. If the resulting position is losing, then Y must have been on some minimal red winning path. But we assumed that every such path contains X, a contradiction.

Perhaps surprisingly, path-domination and kill-domination are actually the same thing. To see why, it is helpful to know that an empty cell Y is dead if and only if there is no minimal winning path containing Y. Now X path-dominates Y from Red's point of view if and only if every minimal red winning path through Y also contains X. This is the case if and only if there is no minimal red winning path through Y when X is blue. But that is equivalent to saying that Y is dead when X is blue, i.e., X kill-dominates Y from Red's point of view.

=== Neighborhood-domination ===

From Red's point of view, define a ''neighbor'' of a cell X to be a cell that is either X itself, adjacent to X, or connected to X by an unbroken [[chain]] of red stones. We say that X ''neighborhood-dominates'' Y if both X and Y are empty, and every empty neighbor of Y is also an empty neighbor of X.

Neighborhood-domination is a special case of switch-domination. Proof: Suppose X neighborhood-dominates Y. Consider a completely-filled board position with X=blue and Y=red that is winning for Red. Consider a minimal winning path for Red. If the path does not pass through Y, then Y=blue would still be winning and there is nothing to show. If the path passes through Y, then the originally-empty cells on the path immediately before and after Y are neighbors of Y. By assumption, they are therefore also neighbors of X. Hence, the position with X=red and Y=blue is still winning for Red, with the winning path going through X instead of Y.

In fact, neighborhood-domination is also a special case of capture-domination. Proof: Suppose x neighborhood-dominates y from Red's point of view. Now put a red stone on x. Then all of y's neighbors are already connected to each other via the red stone on x, so that y can no longer be part of any minimal winning path. Therefore y is dead. Since dead cells are captured, y is captured by x, so that x captured-dominates y.

== Example: the 10-cell acute corner ==

Consider the following cells in the acute corner:
<hexboard size="5x5"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)"
contents="a:b5 b:c5 c:d5 d:e5 e:c4 f:d4 g:e4 h:d3 i:e3 j:e2"
/>

Assume that the 10 indicated corner cells are still empty, and Red wants to make a move in the corner. Also assume that it is not the first move of the game (so Red does not need to worry about her move being swapped). Where should Red consider playing? It turns out that there are only 3 moves that dominate all other possibilities. Red should move at e, h, or j.

Caveat: Please note that this does not mean that Red should never make a move other than e, h, or j in the acute corner. It only means that if the corner is empty, Red's ''first'' move in the corner should be e, h, or j. In the presence of other (red or blue) stones, Red might of course need to move elsewhere in the corner. By symmetry, Blue's first move in the empty corner should be at a, e, or h.

=== Red's move at e ===

A red stone at e captures a, b, c, d, f, and g.
<hexboard size="5x5"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)"
contents="a:b5 b:c5 c:d5 d:e5 e:c4 f:d4 g:e4 h:d3 i:e3 j:e2 R e:c4 S red:area(b5,e5,e4,c4)"
/>
Therefore, e capture-dominates a, b, c, d, f, and g.

=== Red's move at h ===

Red's move at h fillin-dominates i. To show this, we must show that the following two corner positions are equivalent:

<hexboard size="5x5"
float="inline"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)-e2"
contents="a:b5 b:c5 c:d5 d:e5 e:c4 f:d4 g:e4 h:d3 i:e3 j:e2 R h:d3"
/> is equivalent to <hexboard size="5x5"
float="inline"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)-e2"
contents="a:b5 b:c5 c:d5 d:e5 e:c4 f:d4 g:e4 h:d3 i:e3 j:e2 R h:d3 i:e3"
/>
Clearly the right position is at least as good for Red as the left one, since Red has an additional stone. To see why the left position is at least as good as the right one, we consider two cases. Case 1: Red is the next player to make a move in the region. In this case, Red will play at e, capturing the entire region. Since the whole region (including i) is red-captured, the red stone at i no longer matters. Case 2: Blue is the next player to make a move in the region. If Blue moves anywhere other than e, Red responds at e, capturing the entire region (including i) and [[dead cell|killing]] Blue's stone. If Blue moves at e, then the two positions are:
<hexboard size="5x5"
float="inline"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)-e2"
contents="a:b5 b:c5 c:d5 d:e5 e:c4 f:d4 g:e4 h:d3 i:e3 j:e2 R h:d3 B e:c4"
/> and <hexboard size="5x5"
float="inline"
edges="bottom right"
coords="none"
visible="area(b5,e5,e2)-e2"
contents="a:b5 b:c5 c:d5 d:e5 e:c4 f:d4 g:e4 h:d3 i:e3 j:e2 R h:d3 i:e3 B e:c4"
/>

In this case, the only thing left to determine the outcome of the region {b,c,d,e,f,g,h,i} is whether Red connects h to the red edge or Blue connects e to the blue edge. In both the left and the right region, whichever player moves next achieves this outcome, so the two regions are equivalent.

Since having h is as good for Red as having both h and i, the move at h fillin-dominates i.

=== Red's move at j ===

The remaining possibility for Red in the empty 10-cell corner is to move at j.

=== e, h, and j are not dominated ===

We have seen above that all red moves other than e, h, and j in the corner are dominated. Could it be that one of these three remaining moves is also dominated? This is not the case. To see that e, h, and j are not dominated, it suffices to give examples of positions in which e (respectively h and j) is the only winning move.

Here are such positions. In each position, it is Red's turn, and the only winning move is indicated by the letter e, h, or j:

<hexboard size="5x5"
float="inline"
coords="none"
edges="all"
contents="B area(a1,a5,e1)-b4,c3 R line(b1,b3) S area(b5,e5,e2) E e:c4"
/>
<hexboard size="5x5"
float="inline"
coords="none"
edges="all"
contents="B area(a1,a5,e1) R line(e1,d2) S area(b5,e5,e2) E h:d3"
/>
<hexboard size="5x5"
float="inline"
coords="none"
edges="all"
contents="B area(a1,a5,e1) R line(e1,a5)-d2 S area(b5,e5,e2) E j:e2"
/>

== See also ==

*[[Equivalent patterns]]

*[[Computer Hex]]

*[[Dead cell]]

*[[Captured cell]]

== External links ==

Philip T. Henderson. [https://era.library.ualberta.ca/items/dd8ce116-183f-4ad0-b7e6-618d38f132ff "Playing and solving the game of Hex"]. Ph.D. thesis, University of Alberta, 2010.

Philip T. Henderson and Ryan B. Hayward, [https://webdocs.cs.ualberta.ca/~hayward/papers/revDom.pdf "Captured-reversible moves and star decomposition domination in Hex"]. Integers Journal, 2013.

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

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

Ziggurat

2025-06-29T16:02:07Z

Selinger: Supplied a missing link

The '''ziggurat''', also known as '''edge template III1-a''' or '''template 4-3-2''', is a 3rd row [[edge template]] with one stone.

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

It is one of the most basic non-trivial edge templates. The [[carrier]] is very small so this template occurs in real games quite often. The small size of the carrier makes it efficient as a threat of [[Mustplay_region#Verification_of_templates|template reduction]] when building other templates. It also appears sometimes in the middle of the board as an [[interior template]].

This ziggurat can be used to prove very easily that a move in the center of a size 5 board is winning.
<hexboard size="5x5"
coords="hide"
contents="R c3 S blue:area(c3,e1,b1,b3) red:area(c3,a5,d5,d3)"
/>

== Defending the ziggurat ==

Red has two main threats by playing at "A" or "B":
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="area(a3,d3,d1,c1)"
contents="R c1 R A:b2 S a3 b3 c1 b2"
/>
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="area(a3,d3,d1,c1)"
contents="R c1 R B:d2 S c1 d1 c2 d2 c3 d3"
/>
These moves lead to easy direct connection. Because there is no common empty [[hex (board element)|hex]] used in both threats, Blue cannot prevent Red from connecting to the bottom.

== Origin of the name ==

[[File:Chogha_Zanbil,_Ziggurat_(model).jpg|right|200px]]
A ziggurat is a type of flat-topped pyramid built in ancient Mesopotamia. The use of this name for the Hex template was apparently coined by Kevin O'Gorman in a post to this [https://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=48 Little Golem thread] on October 2, 2003. Incidentally it is the very same thread in which the creation of this HexWiki was proposed. O'Gorman wrote:

''"So here are two names for consideration. We already have the “2-bridge”, immortalized by Cameron. I think the next most common and useful one is the “temple” (I actually call it the “ziggurat” but that may be asking too much) [...]. The shape reminds me of a mesoamerican temple."''

== See also ==
* Some [[puzzles]] directly involve the ziggurat!
* [[Edge templates everybody should know]]
* [[Edge templates with one stone]]
* [[Multiple threats]]

[[category:edge templates]]
[[category:Basic Strategy]]

About HexWiki

2025-06-13T21:30:20Z

Selinger: Fixed date order

== News and history ==
* November 18, 2024: HexWiki has moved to a new server! The old server had deteriorated over time, becoming slow and sometimes unresponsive. Please let us know if you encounter any problems with the new server.
* December 6, 2020: Many new features were added to the [[New board diagrams|board diagrams]], including the ability to show partial boards.
* September 28, 2017: The site has been moved to a new hosting service. Please let us know if something on the site is not working.
* March 2017: Boards are now rendered as SVG images! Thanks to [[User:Tom239|Tom239]] for writing the code for this feature.
* August 2015: HexWiki is back! Please see the article [[the new HexWiki]] for information.

== Things to be done ==
* We need hex boards to “float” around text to make pages shorter and more readable
* Make it possible to visualize "live" boards for showing off variations
* Boring but easy: make [[Printable Y boards]] more readable, like [[Printable boards]].
* Add forum
* Update [[Tournaments]] with information on recent or current tournaments

[[category: HexWiki]]

About HexWiki

2025-06-13T21:29:40Z

Selinger: /* News and history */ Added November 2024 update

== News and history ==
* December 6, 2020: Many new features were added to the [[New board diagrams|board diagrams]], including the ability to show partial boards.
* September 28, 2017: The site has been moved to a new hosting service. Please let us know if something on the site is not working.
* March 2017: Boards are now rendered as SVG images! Thanks to [[User:Tom239|Tom239]] for writing the code for this feature.
* August 2015: HexWiki is back! Please see the article [[the new HexWiki]] for information.
* November 18, 2024: HexWiki has moved to a new server! The old server had deteriorated over time, becoming slow and sometimes unresponsive. Please let us know if you encounter any problems with the new server.

== Things to be done ==
* We need hex boards to “float” around text to make pages shorter and more readable
* Make it possible to visualize "live" boards for showing off variations
* Boring but easy: make [[Printable Y boards]] more readable, like [[Printable boards]].
* Add forum
* Update [[Tournaments]] with information on recent or current tournaments

[[category: HexWiki]]

Main Page

2025-06-13T21:28:55Z

Selinger: Removed "new server" warning

Welcome to HexWiki, the comprehensive resource dedicated to the board game of [[Hex]].

Hex is a [[connection game]] invented in the 1940s that has gained popularity in recent years on online gaming platforms. With its simple [[rules]] and deep [[strategy|strategic elements]], it has captivated a growing audience of players who find the game both challenging and engaging.

[[Image:Br2.png|thumb|]]
[[Image:Br6.png|thumb|]]

Depending on your interest, some good pages to start exploring this site are:

* An overview of the basic [[rules]] of Hex.
* A look at the [[History of Hex|historical background]] of the game.
* The [[strategy roadmap]], which lists skills in the approximate order in which players should learn them.
* The [[strategy]] page, which includes further links to specific topics on Hex strategy.
* Practice your skills by solving some challenging [[Puzzles|puzzles]].
* Study [[Commented games|commented games]] to learn from experienced players.
* Information on [[computer Hex|computer-based versions]] of the game.
* Information on where to play Hex, including [[online playing|online options]] and [[Hex clubs|local clubs]].
* Information on [[tournaments]] and competitions for Hex players.
* Information on [[physical hex sets|making your own physical Hex set]].
* Information on [[variants using the same equipment|variations of the game]].
* Tips for [[typesetting Hex|typesetting Hex diagrams and boards]].
* Learn more about HexWiki itself by visiting [[About HexWiki|this page]].
* Join the Hex community by visiting the [[Hex forums|Hex discussion forums]].
* A [[Hex Bibliography|bibliography]] of books and articles about Hex.

In addition to the above pages, you can also browse our [[Special:Categories|categories]] and [[Special:Allpages|pages]] for more information on Hex.

Conventions

2025-06-06T17:38:05Z

Selinger: Updated introductory paragraph, to indicate that the conventions are now more stable than when the article was first written

This page lists possible conventions for Hex games. These conventions concern the color of the players, who goes first, the orientation of the board, and the numbering of the cells on the board. Unlike much older games such as Chess and Go, Hex has no governing body that could set standardized rules, and for most of its history, different people, game sites, books, and computer programs sometimes use different conventions. However, in the last few years, most sites have converged on a compatible set of conventions, making game records more easily interchangeable.

We distinguish ''logical'' from ''physical'' conventions. Logical conventions are concerned with the abstract rules of Hex, irrespectively of how the players visualize the game. Physical conventions relate to the specific appearance of the game board. When the game is played remotely, for example on an internet game server or over email, it is in principle possible for the two players to follow different physical conventions; for example, each player may choose their own preferred orientation of the board, or their own preferred colors for the pieces. However, both players should follow the same logical conventions.

== Universal vs. local conventions ==

There are several aspects of a game of Hex that are arbitrary and do not affect game play in an essential way. This includes the orientation (rotation and reflection) of the board, the colors of the players, who goes first, which edge belongs to which player, and whether and how the cells on the board are numbered.

The same also applies to other games; for example, the game of Chess would not change in an essential way if the white square were in the bottom left corner instead of the bottom right one, if black went first instead of white, if the white queen started on a black square instead of a white one, if the ranks were lettered and the files numbered instead of the other way around, and so on.

In the case of established games such as Chess and Go, there are universal conventions that all players have agreed on. This has certain advantages. For example, every chess player understands what it means to move a white pawn from d2 to d4, without requiring further explanation.

In Hex, there were no universal conventions prior to ca. 2020, and different players, books, game sites, tournaments, or Hex programs often used different conventions. Fortunately, some universal conventions seem to have crystallized in recent years, especially for the logical aspects of the game (such as the coordinate system and who goes first). Some physical conventions also seem to have become standard; for example, the colors are now almost universally black/white or red/blue, with black or red going first. Some conventions are unsettled; for example, both swap methods (swap-pieces and swap-sides) seem to be in common use. This is not a problem as long as game records specify which method is used. Other conventions, such as the board orientation, seem to depend on user preference and probably don't need to be standardized, although there is a strong preference for positive orientations.

== Logical conventions ==

=== Cell naming ===

Cells are arranged in a grid, with each cell named by a letter and a number. Letters can be uppercase or lowercase. The following convention seems to be nearly universally used:

[[Image:Logical-Board.png|right|]]

'''Convention: coordinate origin'''

Possible values:

* '''Acute:''' The cell A1 is an an acute corner of the board.

In particular, this means that the three cells A1, B1, and A2 are all adjacent to each other.

For the purpose of the following exposition, it is useful to refer to the cells that share a common number as a "rank", and to the cells that share a common letter as a "file", as in Chess. For example, the 1-rank consists of the cells A1, B1, C1, etc., and the A-file consists of the cells A1, A2, A3, etc.

Note that in some games, such as Go, it is customary to omit certain letters from the alphabet, especially the letter I. This is done "to avoid confusion between I and J", and presumably dates from a time when typesetting was uncommon and people had terrible handwriting. However, in Hex, the standard 26-letter alphabet is used. If more than 26 letters are needed, alphabet numbers can be used: the next file after Z is AA, then AB, and so on.

=== Edge coloring ===

To avoid referencing specific physical attributes of the game, we will refer to the color of the first piece played in the game as "color 1", and to the other color as "color 2". One pair of opposite edges "belongs to" color 1; these are the edges that the player who is playing color 1 is trying to connect. The other pair of opposite edges belongs to color 2.

'''Convention: edge coloring'''

Possible values:

* '''Normal:''' The edge that is adjacent to the 1-rank belongs to color 1, and the edge that is adjacent to the A-file belongs to color 2.

In other words, the normal edge coloring convention states that the color 1 edges are parallel to ranks, and the color 2 edges are parallel to files. In a typical rendering of the board, the letters (naming files) are written along the color 1 edge and the numbers (naming ranks) are written along the color 2 edge.

I am not aware of anybody who has used the opposite convention. However, there are some authors who use no edge coloring convention at all.

=== Swapping ===

There are two different ways of implementing the [[swap rule]]. Which convention is chosen will affect the notation for games. It is also possible to permit both methods of swapping; then it is up to the player to decide which method to use.

'''Convention: swap method'''

Possible values:

* '''Swap sides:''' Upon playing a swap move, the board position stays the same and the players change colors.

* '''Swap pieces:''' Upon playing a swap move, the players keep their colors and the board position is mirrored (ranks and files interchanged) and the color of the pieces is inverted. For example, a black piece at A2 would be replaced by a white piece at B1.

Here is an example using the swap sides convention: Player 1 plays a black piece at g4. Player 2 swaps sides. The board state remains unchanged, and immediately after the swap, it is player 1's turn to play a white piece.

Here is an example using the swap pieces convention: Player 1 plays a black piece at g4. Player 2 swaps pieces, and replaces the black piece at g4 with a white piece at d7. Immediately after the swap, it is player 1's turn to play a black piece.

When using algebraic notation for a sequence of moves, it is important to know which swap method was used. The swap method should either be defined for the context in which it is used, or else the notation should state explicitly which method was used. For example, with the swap sides convention, the notation "g4 swap f7 e7" means player 1 plays color 1 at g4, player 2 swaps, player 1 plays color 2 at f7, and player 2 plays color 1 at e7. With the swap pieces convention, the same game would be described as "g4 swap g6 g5". Here, player 1 plays color 1 at g4, player 2 replaces this by a piece by color 2 at d7, then player 1 plays color 1 at g6 and player 2 plays color 2 at g5. It is best not to use ambiguous notation at all; the two games can then be denoted "g4 swap-sides f7 e7" and "g4 swap-pieces g6 g5".

Note that the above logical conventions are purely symbolic. They do not depend on any particular board layout.

=== Passing ===

Although passing (skipping a move) is not always considered part of the classic Hex rules, allowing it does not change the nature of the game and has certain advantages. See the page on [[passing]] for more details.

'''Convention: passing'''

Possible values:

* '''Explicit''': Players can pass and there is an explicit passing move in the game record, e.g.: "Black passed".

* '''Implicit''': There is no explicit passing move, but a player can move twice in a row (presumably when the other player allows it).

* '''None''': Moves must be strictly alternating and passing is not allowed.

== Physical conventions ==

=== Colors ===

The most common color schemes are:

* Black and white.
* Red and blue.
* V and H.

With each color scheme, there are two possible conventions to which color is color 1. The more common conventions seem to be "black goes first", "red goes first", and "V goes first", but the opposite conventions also exist.

=== Board orientation ===

The board can be oriented in a number of different ways. It can be oriented in the ''positive'' or ''negative'' senses, and rotated to various angles.

'''Convention: board sense'''

Possible values:

* '''Positive:''' The cells A1, B1, A2 form a clockwise triangle in that order.

* '''Negative:''' The cells A1, B1, A2 form a counterclockwise triangle in that order.

The positive sense can also be described by saying that if the board is rotated so that A1 points west, then the ranks run from southwest to northeast, and the files run from northwest to southeast. This convention can also be described as "letters above numbers", because if A1 points left, the letters A,B,C marking the files appear above the numbers 1,2,3 marking the ranks.

'''Convention: board rotation'''

There is probably no need for a preferred board rotation; indeed, players using a physical game board may freely move around the board and look at it from any angle they want.

In computer hex, the most common board rotations are:

* '''Diamond:''' the A1 corner points west, i.e., in the direction of 9 o'clock on an analog clock.

* '''Flat:''' the A1 corner points northwest, i.e., in the direction of 10 o'clock.

* '''Flat II:''' the A1 corner points southwest, i.e., in the direction of 8 o'clock.

The following table illustrates some common board layouts. Here, color 1 is shown as black and color 2 is shown as white.

{| class="wikitable" style="text-align: center"
|-
| [[Image:Positive-Diamond.png|250px]]

Positive diamond
| [[Image:Positive-Flat.png|250px]]

Positive flat
| [[Image:Positive-Flat2.png|250px]]

Positive flat II
|-
| [[Image:Negative-Diamond.png|250px]]

Negative diamond
| [[Image:Negative-Flat.png|250px]]

Negative flat
| [[Image:Negative-Flat2.png|250px]]

Negative flat II
|}

== List of conventions used ==

{| class="wikitable"
|-
!
! Origin
! Edge coloring
! Swap
! Passing
! Colors
! First player
! Orientation
|-
| Browne
| acute
| normal
| sides
| N/A
| black/white
| black
| positive diamond
|-
| Seymour
| acute
| normal
| pieces
| explicit in puzzles, none in book
| black/white
| black
| positive diamond
|-
| Berge
| acute
| normal
| ?
| N/A
| black/white
| white
| negative flat II
|-
| Hayward and Toft
| acute
| ?
| sides
| N/A
| black/white
| any
| diamond
|-
| Hexy
| acute
| normal
| pieces
| none
| red/blue
| red
| positive diamond / positive flat / negative flat II
|-
| Mohex
| acute
| normal
| N/A
| implicit
| black/white
| black
| positive flat
|-
| HexGui
| acute
| normal
| both
| implicit
| black/white
| black
| any
|-
| Board Game Arena
| acute
| normal
| pieces
| explicit
| red/blue
| red
| positive flat
|-
| Little Golem
| acute
| normal
| pieces
| none
| black/white
| black
| positive flat
|-
| igGameCenter
| acute
| normal
| sides
| none
| red/blue or black/white
| red or black
| positive flat
|-
| PlayOK
| acute
| normal
| pieces
| none
| black/white
| black
| positive flat
|-
| PlayHex
| acute
| normal
| pieces
| explicit
| red/blue
| red
| any positive
|-
| AbstractPlay
| acute
| normal
| pieces
| none
| red/blue
| red
| positive flat
|-
| HexWorld
| acute
| normal
| both
| explicit
| black/white or red/blue
| black or red
| any positive
|-
| TRMPH
| acute
| normal
| N/A
| none
| black/white
| black
| positive flat
|-j
| HexWiki
| acute
| normal
| both
| N/A
| red/blue
| red
| positive flat
|}

Notes:

* Browne states that there is no universal convention for which color goes first, but black goes first in all examples in the book.

* Berge states that any player can go first, but white goes first in his example games.

* TRMPH implements swapping incorrectly. It changes the color of the piece, but not its location.

* MoHex does not implement a swap-pieces move. It implements swap-sides, which it incorrectly (and confusingly) calls swap-pieces.

* BoardGameArena has an explicit passing move, but three consecutive passes are not allowed. Therefore, if a player passes, the opponent can effectively reject the pass by passing too.

* AbstractPlay uses red and blue as the default colors, but users can set custom colors for themselves (these are not seen by the opponent). It uses the positive flat orientation and the board can be rotated in increments of 90 degrees.

[[category: Rules and Conventions]]

Ladder puzzle 1/Solution

2025-05-26T02:29:35Z

Selinger: Terminology: prefer "pivot" to "break"

== Solution 1: e5 then c7 ==

Red's main threat is the ladder starting at c7. If Red plays out this ladder, Blue can block it. So Red needs a helping stone somewhere on the right and on the second line from the bottom. Red 1.e5 threatens the follow-up moves at *, which Blue must defend.

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Ve5 *d7 *f6 +e6 +d8</hex>

The potential connection at d7 passes through e6, c7, c8, and d8. The potential connection at f6 passes through e6, f5, and [[Template IIIa]]. These two potential connections only overlap at the two hexes marked +, so Blue must play there. However, if Blue plays 2.d8, Red wins with the following sequence of forcing moves:

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Ve5 Hd8 Vd7 Hc8 Vf7 He6 Vf5 Hf6 Vh5</hex>

Therefore, Blue must play 2.e6. Red can now play the ladder from c7, pivot at g7, and win:

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Ve5 He6 Vc7 Hc8 Vd7 Hd8 Ve7 He8 Vg7 Hf7 Vg6 Hf6 Vg4</hex>

== Solution 2: e5 then f5 ==

Alternatively, after 1.e5 2.e6, Red could have continued the 4th row ladder with 3.f5, to which Blue may reply 4.f6. This allows Red to play a double threat 5.g7. This stone is the ladder helper, and it also threatens to connect along the top. Either way, Red wins.

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Ve5 He6 Vf5 Hf6 Vg7</hex>

However, this play is more complicated to analyze, because Blue may also respond in a number of other places instead of f6. If Blue plays 4.e8, then 5.g7 still works, albeit for slightly different reasons:

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Ve5 He6 Vf5 He8 Vg7 Hg6 Vc7 Hc8 Ve7 *d7 *f6 +d8 +f7</hex>

Note that in the final position, the red stone at e7 is connected up via * and down via +, so it is a winning position.

If Blue plays 4.f7 or anything to the right of it, Red gets their ladder escape at 5.e7:

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Ve5 He6 Vf5 Hf7 Ve7 Hf6 Vc7</hex>

Again, e7 is connected upwards via *.

== Solution 3: f6 ==

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 N:on Vf6 He7</hex>

The stone at f6 (essentially [[Tom's move]], but requiring a bit less space than usual due to h4) is a ladder escape for the ladder starting at c7, and it also threatens to connect via e5. Blue has no choice but to defend at e7. Red continues like this:

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 Vf6 He7 V3g6 H4f7 V5g7 H6e6</hex>

Now Red plays f4 completing the win. f4 is connected to the bottom edge via a [[trapezoid]], and to Red's central group via double threat at f3 and e5. Therefore, all of Red's pieces form a single group which is connected to both the top and bottom.

== Solution 4: switchback ==

Red can also use h4 to do a [[switchback]]. To set this up, Red starts playing the 2nd row ladder and pivots at f7:

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 V1c7 H2c8 V3d7 H4d8 V5f7 H6e7</hex>

Now the killer move is f5, which connects to Red's main group by double threat, and is also connected to h4. Blue is forced to play at f6, but Red reconnects with h5 via [[Fourth_row_edge_templates#IV-2-d|edge template IV-2-d]].

<hex>R8 C8 Q1 Vc6 Vd5 Vd4 Ve3 Vh4 Hd6 Hb8 Hc1 He4 Hg3 Vc7 Hc8 Vd7 Hd8 Vf7 He7 V7f5 H8f6 V9h5</hex>

This method works no matter the size of the board or how far away the 5th row stone is. If h5 is further to the right, the 2nd row ladder will just turn into a 4th row ladder going the opposite way, eventually connecting with d5.

[[category:ladder puzzle]]

Climbing

2025-05-26T02:24:52Z

Selinger: Terminology: prefer "pivot" to "break"

'''Climbing''' means playing a series of [[forcing move]]s by which a player gains significant distance across the board and potentially connects to the opposite edge, by repeatedly threatening to connect to the player's nearby pieces.

Here is an idealized example. Red to move.
<hexboard size="9x9"
contents="R e3 e4 d5 d6 c7 c8 B a9 b9 c9 d7 e5 f3 f8 g7 g6 h5 h4 i3"
/>
Red wins by climbing from e8.
<hexboard size="9x9"
contents="R e3 e4 d5 d6 c7 c8 B a9 b9 c9 d7 e5 f3 f8 g7 g6 h5 h4 i3
R 1:e8 B 2:d8 R 3:f6 B 4:e6 R 5:g4 B 6:f4 R 7:h2"
/>
Note that every single one of Blue's moves is forced. Although Blue could intrude into some of Red's bridges or other templates, this does not help.

Note that climbing was possible even though Blue seemed to have more strength on the right side of the board than Red. What makes climbing work is the exposed [[flank]] of unprotected Red pieces that Red can repeatedly threaten to connect to. The potential for climbing is often difficult for beginners to spot, and can lead to swift and unexpected defeat. It is therefore a good idea to try to deny the opponent opportunities to climb.

== Example ==

Climbing does not always have to proceed by bridges. A combination of bridges and adjacent moves is common. Here is an example from an actual game. Red to move.
<hexboard size="11x11"
contents="R i2 g7 g4 f8 f6 d7 c7 B h5 g8 e10 e8 e4 c10 c9"
/>
Red starts a 3rd row ladder, then immediately pivots and climbs.
<hexboard size="11x11"
contents="R i2 g7 g4 f8 f6 d7 c7 B h5 g8 e10 e8 e4 c10 c9
R 1:f9 B 2:f10 R 3:h9 B 4:g9 R 5:i7 B 6:h7 R 7:i6 B 8:h6 R 9:j4 S area(h4,j4,k3,k1)"
/>
At this point, Red is connected by an [[interior_template#Interior_templates_from_edge_templates|interior ziggurat]], shaded in the above diagram, to [[edge template IV2d]]. Blue [[resigning|resigns]].

== Zipper ==

[[Image: Zipper.png|right|70px]]
'''Zippering''' is a special case of climbing where the player's threatened connections are all on one side, and the attacker mostly proceeds by bridges. This is called a "zipper" because it vaguely looks like an actual zipper (see the illustration on the right). For example, consider the following position, with Red to move:
<hexboard size="11x11"
contents="R h2 g3 g4 g5 f6 f7 f8 e9 B f3 h3 f5 h7 f9 d9 d11"
/>
Red pushes the ladder, pivots, and zippers all the way to the opposite edge.
<hexboard size="11x11"
contents="R h2 g3 g4 g5 f6 f7 f8 e9 B f3 h3 f5 h7 f9 d9 d11
R 1:d10 B 2:c11 R 3:b10 B 4:c10 R 5:c8 B 6:e7 R 7:d6 B 8:e6 R 9:e4 B 10:f4 R 11:e3"
/>
Move 1 was actually unnecessary; we have shown it to make it more obvious why 3 was forcing. Red could have immediately started with 3.

== Climbing from a ladder ==

Climbing often starts from a ladder. The attacker pushes the ladder to a certain point, then ''pivots'', often by playing one hex ahead of the ladder. The defender must close the gap between the ladder and the pivot piece, which gives the attacker an opportunity to climb. To find good climbing opportunities, it is useful to consider how far the attacker can climb "unassisted", starting from various ladders. After that, the attacker can potentially climb even further if there are additional forcing moves available.

=== 2nd row ladder ===

'''Scenario 1:''' In this scenario, Red's space is limited. Red can climb to the 4th row, potentially [[bridge|bridging]] to a stone on the 6th row. The shaded cells are not needed for this and can be occupied by Blue.
<hexboard size="6x8"
coords="hide"
edges="bottom"
visible="-area(a1,a2,h2,h1)"
contents="R a5 b5 c5 B a6 b6 c6 d4 f5 R 1:e5 B 2:d5 R 3:f3
S area(a1,a4,d4,g1,g4,f5,f6,h6,h1)"
/>

'''Scenario 2:''' If Red has slightly more space, Red can climb to the 5th row, potentially bridging to a stone on the 7th row.
<hexboard size="6x8"
coords="hide"
edges="bottom"
visible="-a1--h1"
contents="R a5 b5 c5 B a6 b6 c6 f5 R 1:e5 B 2:d5 R 3:e4 B 4:d4 R 5:f2
S area(a1,a4,c4,f1,g1,g3,f4,f6,h6,h1)"
/>

'''Scenario 3:''' In this scenario, Red's 2nd row ladder comes with a [[Switchback#Switchback threat|switchback threat]], i.e., a 2nd-to-4th row switchback would allow Red to connect. In this case, Red can climb to the 6th row, potentially bridging to a stone on the 8th row. This kind of play is also called a ''switchback fork''.
<hexboard size="6x8"
coords="hide"
edges="bottom"
contents="R a5 b5 c5 a4 B a6 b6 c6 f5 R 1:e5 B 2:d5 R 3:f3 B 4:e3 R 5:g1
S area(a1,a2,e2,f1) area(h1,f5,f6,h6)"
/>

'''Scenario 4:''' Finally, if Red has a [[Switchback#Switchback threat|switchback threat]] and slightly more space on the right, Red can climb all the way to the 7th row, potentially bridging to a stone on the 9th row. This is highly threatening; note that on an 11×11 board, the 9th row is almost on the opposite side of he board.
<hexboard size="7x8"
coords="hide"
edges="bottom"
contents="R a5 a6 b6 c6 B a7 b7 c7 h5 h6 R 1:e6 B 2:d6 R 3:e4 B 4:e5 R 5:g4 B 6:f4 R 7:g3 B 8:f3 R 9:h1
S area(a1,a3,e3,g1) area(h3,h7)"
/>

Of course, there are many variations of these basic scenarios, depending on what other pieces Red and Blue have on the board. But the four scenarios shown above are common, and are good starting points for planning more complex ladder escape forks.

'''Scenario 5:''' If Red has enough space, Red can also yield to the 3rd row and then climb to the 6th row, even without a switchback threat. This is basically the same as Scenario 2 for 3rd row ladders below.

<hexboard size="8x8"
edges="bottom"
coords="hide"
visible="-area(a1,a2,h2,h1)"
contents="R h1 R 1:a7 B 2:a8 R 3:b7 B 4:b8 R 5:d6 B 6:d7 R 7:f6 B 8:e6 R 9:f5 B 10:e5 R 11:g3
S area(a1,a6,b6,c5,d5,f3,f2,g1) area(h3,h8) g5"
/>

'''Scenario 6:''' Given a different amount of space and no switchback threat, Red can still climb to the 6th row as follows:

<hexboard size="6x8"
edges="bottom"
coords="hide"
contents="R 1:a5 B 2:a6 R 3:b5 B 4:b6 R 5:f4 B 6:e5 R 7:d5 B 8:c5 R 9:d4 B 10:c4 R 11:f1
E *:e4 *:f5
S area(a1,a4,b4,e1) area(h1,h6)"
/>
Note that 5 is connected to the edge by [[double threat]]s marked "*", and 11 is connected to 9 and 5 by an [[interior_template#Interior_templates_from_edge_templates|interior ziggurat]].

'''Scenario 7:''' Given an enormous amount of space, Red can climb to the 7th row without a switchback threat:

<hexboard size="7x11"
edges="bottom"
coords="hide"
contents="R 1:a6 B 2:a7 R 3:b6 B 4:b7 R 5:c6 B 6:c7 R 7:d6 B 8:d7 R 9:g5 B 10:e6 R 11:f4 B 12:e5 R 13:f3 B 14:d4
R 15:h1 B 16:g2 R 17:h2 B 18:g4 R 19:i3
S area(a1,a5,b5,e2,f2,g1) area(i1,k4,k1)"
/>
Note that 19 is connected to the edge by [[Fifth_row_edge_templates#V-2-f|edge template V2-f]]. Also, move 9 is similar to a [[cornering|cornering move]], but requires more space in the situation shown here.

=== 3rd row ladder ===

The situation for 3rd row ladders is largely similar to that of 2nd row ladders. Scenarios 1—3 work without much modification.

'''Scenario 1:''' In the most constrained scenario, Red can climb to the 5th row, potentially bridging to a stone on the 7th row.
<hexboard size="7x9"
coords="hide"
edges="bottom"
visible="-a1--i1 a2--i2"
contents="R a5 b5 c5 B a6 b6 c6 d4 g5 R 1:e5 B 2:d5 R 3:f3
S area(a1,a4,d4,g1,g7,i7,i1)"
/>

'''Scenario 2:''' If Red has slightly more space, Red can climb to the 6th row, potentially bridging to a stone on the 8th row.
<hexboard size="7x9"
coords="hide"
edges="bottom"
visible="-a1--i1"
contents="R a5 b5 c5 B a6 b6 c6 g5 R 1:e5 B 2:d5 R 3:e4 B 4:d4 R 5:f2
S area(a1,a4,c4,f1,g1,g7,i7,i1) f4"
/>

However, if the ladder starts further away and Blue yields at just the right moment, Red needs more space to climb to the 6th row.
<hexboard size="7x9"
coords="hide"
edges="bottom"
visible="-a1--i1"
contents="R a5 B a6 R b5 B b7 R 1:b6 B 2:a7 R 3:f5 B 4:e6 R 5:d6 B 6:c6 R 7:d5 B 8:c5 R 9:f2
S area(a1,a4,c4,f1,h1,h7,i7,i1)"
/>
After the optional 1-2 exchange, 3 and 9 can be played in either order; they are the only two winning moves for Red within this space.

'''Scenario 3:''' If Red's 3rd row ladder comes with a [[Switchback#Switchback threat|switchback threat]], Red can play a switchback fork and climb to the 7th row, potentially bridging to a stone on the 9th row.
<hexboard size="7x9"
coords="hide"
edges="bottom"
contents="R a5 b5 c5 a4 B a6 b6 c6 g5 R 1:e5 B 2:d5 R 3:f3 B 4:e3 R 5:g1
S area(a1,a2,e2,f1) area(h1,g3,g7,i7,i1)"
/>

'''Scenario 4:''' If Red has a [[Switchback#Switchback threat|switchback threat]] and significantly more space on the right, Red can climb all the way to the 8th row, potentially bridging to a stone on the 10th row. The shaded cells are not required to be empty.
<hexboard size="8x9"
coords="hide"
edges="bottom"
contents="R a5 a6 b6 c6 B a7 b7 c7 i5 R 1:e6 B 2:d6 R 3:e4 B 4:e5 R 5:g4 B 6:f4 R 7:g3 B 8:f3 R 9:h1
S area(a1,a4,b3,e3,g1) i1--i4 h3"
/>

'''Yielding:''' [[Yielding]] to the 2nd row does not help Blue in any of these scenarios. If Blue yields at the last possible moment in scenarios 1–4, Red can use a few extra moves to achieve the same outcome as without yielding, and actually requires slightly less space. For example, this is how scenario 1 plays out if Blue yields:
<hexboard size="7x9"
coords="hide"
edges="bottom"
visible="-a1--i1 a2--i2"
contents="R a5 b5 c5 B a6 b6 c7 d4 g5 R 1:c6 B 2:b7 R 3:e6 B 4:d6 R 5:e5 B 6:d5 R 7:f3
S area(a1,a4,d4,g1,g7,i7,i1)"
/>
If Blue tries to yield earlier in scenarios 1–3, Red can play, respectively, scenarios 2–4 for 2nd row ladders to achieve the same outcome, and does not even require the 3rd-to-5th row [[Switchback#Switchback threat|switchback threat]]. For example, this is how scenario 3 plays out if Blue yields early:
<hexboard size="7x9"
coords="hide"
edges="bottom"
contents="R a5 b5 B a6 b7 g5 R 1:b6 B 2:a7 R 3:d6 B 4:c6 R 5:d4 B 6:d5 R 7:f4 B 8:e4 R 9:f3 B 10:e3 R 11:g1
S area(a1,a2,e2,f1) area(h1,g3,g7,i7,i1)"
/>
In scenario 4, if Blue yields any earlier than the second-to-last opportunity, Red can simply [[ladder handling|jump]] back to the 3rd row. The final and most interesting case is when Blue yields exactly at the second-to-last opportunity. In that case, after optionally invading Blue's bridge, the unique winning move is 3:
<hexboard size="8x9"
coords="hide"
edges="bottom"
contents="R a5 a6 b6 B a7 b8 i5 R 1:b7 B 2:a8 R 3:f6
S area(a1,a4,b3,e3,g1) i1--i4 h3"
/>
After this, there are several possibilities, depending on how Blue responds. The main line is as follows:
<hexboard size="8x9"
coords="hide"
edges="bottom"
contents="R a5 a6 b6 B a7 b8 i5 R 1:b7 B 2:a8 R 3:f6
B 4:c7 R 5:d6 B 6:c6 R 7:d4 B 8:d5 R 9:e4 B 10:e5 R 11:g4 B 12:f4 R 13:g3 B 14:f3 R 15:h1
S area(a1,a4,b3,e3,g1) i1--i4 h3"
/>

'''Climbing from a 3rd row ladder in an obtuse corner:''' Another special case to consider is when the 3rd row ladder is approaching an obtuse corner and there is very little space. Consider the following example, with Red's ladder approaching from the right:
<hexboard size="5x8"
coords="hide"
edges="bottom left"
contents="B e2 f2 g2 g4 R h2 R g3 B f4 R f3 B e4"/>
There's not enough room for Red to [[ladder handling#Attacking|push]] one more time, as this will give Blue a 2nd row ladder:
<hexboard size="5x8"
coords="hide"
edges="bottom left"
contents="B e2 f2 g2 g4 R h2 R g3 B f4 R f3 B e4 R 1:e3 B 2:d4 R 3:c3 B 4:b5 R 5:a5 B 6:b4 R 7:a4 B 8:b3"
/>
The obvious solution is for Red to pivot immediately and hold Blue to a 5th row ladder:
<hexboard size="5x8"
coords="hide"
edges="bottom left"
contents="B e2 f2 g2 g4 R h2 R g3 B f4 R f3 B e4 R 1:d3 B 2:e3 R 3:d2 B 4:e1 E x:b4"
/>
However, this is not optimal, because Blue has a forcing move at x, giving Blue 2nd and 3rd row ladder escapes:
A slightly better solution is the following:
<hexboard size="5x8"
coords="hide"
edges="bottom left"
contents="B e2 f2 g2 g4 R h2 R g3 B f4 R f3 B e4 R 1:d4 B 2:e3 R 3:d2 B 4:e1 E x:b4 y:c3 S area(d2,a5,d5)"
/>
Note that Red has formed [[edge template IV2d]], still holding Blue to a 5th row ladder. Blue still has a forcing move at x, but because Red can respond at y, this only gives Blue a 2nd row ladder escape, rather than a 3rd row escape.

However, even this solution is not optimal for Red. It turns out that playing a different move 3 is even better for Red:
<hexboard size="5x8"
coords="hide"
edges="bottom left"
contents="B e2 f2 g2 g4 R h2 R g3 B f4 R f3 B e4 R 1:d4 B 2:e3 R 3:b2
E a:c2 b:d2 c:b3 d:c3 e:a4 f:b4"
/>
Move 3 is named ''Eric's move'' after Eric Demer, who discovered it. If Red plays there, Blue in theory no longer gets a 5th row ladder, nor any kind of ladder. If Blue plays at a or c, Red responds at b; if Blue plays at b, Red responds at a; if Blue plays at d, Red responds at e; if Blue plays at f, Red responds at d. This works in essentially the same way as blocking a 5th row ladder when there's not enough space under the ladder stone; see [[Theory_of_ladder_escapes#Definition_of_ladder_4|theory of ladder escapes]].

=== 4th row and higher ladders ===

The situation for 4th row and higher ladders is essentially similar, provided that the attacker can guarantee that the pivot piece connects to the edge. For example, in the following situation, the pivot piece "1" is not connected to the edge, and Blue could [[foiling|foil]] by playing at "a":
<hexboard size="6x9"
coords="hide"
edges="bottom"
contents="R d3 e3 B c4 d4 e4 b2 i2 R 1:g3 E a:g4 E *:c3 *:i5"
/>
However, if Red had, for example, one more piece at either of the locations marked "*" (or pretty much anywhere else near the bottom edge), then the pivot piece would be sufficiently connected for the pivot to work in the same way as for 2nd or 3rd row ladders.

[[Category:Definition]]
[[Category:Intermediate Strategy]]

Open problems

2025-03-15T16:11:58Z

Selinger: Better sectioning; added some context

== Currently open problems ==

* Are there cells other than a1 and b1 which are theoretically losing first moves?

* Is it true that for every cell (defined in terms of direction and distance from an [[Board#Corners|acute corner]]) there is an ''n'' such that for any [[Board]] of size at least ''n'' that cell is a losing [[opening|opening move]]?

* Conversely, is it true that, for example, c3 is a winning first move on every Hex board of size at least 5?

* Is the [[center opening|center hex]] on every Hex board of [[Board_size|odd size]] a winning opening move?

* On boards of all [[board size|sizes]], is every opening move on the [[Board#Diagonals|short diagonal]] winning?

* Is the following true? Assume one player is in a winning position (will win with [[optimal play]]) and the opponent plays in a hex X. Let the set A consist of all empty hexes that are members of any path between opponents edges that uses the stone at X. If A is non-empty, A contains a winning move. Otherwise any move is winning, even [[passing]] the turn. (This problem was posed by [[Jory]] in the [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=167 Little Golem forum].)

:*The generalization of the above statement to monotone set-coloring games is false, as shown in [[#Failure_for_maker-breaker_games|section 3]] below.

== Formerly open problems ==

=== [[Sixth row template problem]] ===

Does there exist an [[edge template]] which guarantees a secure [[connection]] for a [[piece]] on the sixth row?

'''Answer:''' Yes, [[edge template VI1a]] is such a template.

=== Triangle template problem ===

Are the templates below valid in their generalization to larger sizes? (This problem was posed by [[Jory]] in the [http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=167 Little Golem forum].)

<hexboard size="1x1"
float="inline"
edges="bottom"
coords="none"
contents="R a1"/><hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="area(b1,a2,b2)"
contents="R b1"/><hexboard size="3x3"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,c3)"
contents="R c1 a3"/><hexboard size="4x4"
float="inline"
edges="bottom"
coords="none"
visible="area(d1,a4,d4)"
contents="R d1 b3"/><hexboard size="5x5"
float="inline"
edges="bottom"
coords="none"
visible="area(e1,a5,e5)"
contents="R e1 c3 a5"/><hexboard size="6x6"
float="inline"
edges="bottom"
coords="none"
visible="area(f1,a6,f6)"
contents="R f1 d3 b5"/>

'''Answer:''' No. The first one in the sequence that is not connected is the one of height 8.

In fact, using a variant of [[Tom's move]], it is easy to see that even the following triangle, which has more red stones, is not an edge template:
<hexboard size="8x8"
float="inline"
edges="bottom"
coords="none"
visible="area(h1,a8,h8)"
contents="R h1 f3 d5 b7 c7,a8--e8"/>

To see why, imagine that the right edge is a blue edge and that all cells outside the carrier are occupied by Blue. Note that Blue gets a 2nd-and-4th row [[parallel ladder]]. Blue wins by playing the [[Tom's_move#Tall_variant|tall variant of Tom's move]]:
<hexboard size="8x8"
float="inline"
edges="bottom right"
coords="none"
visible="area(h1,a8,h8) g1--a7"
contents="R h1 f3 d5 b7 c7,a8--e8 B g1--a7 B 1:g2 R 2:h2 B 3:f5 R 4:g4 B 5:g3 R 6:h3 B 7:d6 R 8:g5 B 9:f7"/>

There is in fact a template of height 8 continuing the above sequence, but it requires slightly more space:

<hexboard size="8x9"
float="inline"
edges="bottom"
coords="none"
visible="area(h1,a8,i8,i6)"
contents="R h1 f3 d5 b7"/>
The corresponding template of height 9 requires this much space:

<hexboard size="9x11"
float="inline"
edges="bottom"
coords="none"
visible="area(i1,a9,k9,k7,i5)"
contents="R i1 g3 e5 c7 a9"/>

=== Seventh row template problem ===

Does there exist an [[edge template]] which guarantees a secure [[connection]] for a [[piece]] on the seventh row?

'''Answer:''' Yes. See [[Seventh row edge templates]].

== Failure for maker-breaker games ==

This section contains two counterexamples to a ''generalization'' of the following problem.

* Is the following true? Assume one player is in a winning position (will win with [[optimal play]]) and the opponent plays in a hex X. Let the set A consist of all empty hexes that are members of any path between opponents edges that uses the stone at X. If A is non-empty, A contains a winning move. Otherwise any move is winning, even [[passing]] the turn.

The generalization is to [https://en.wikipedia.org/wiki/Maker-Breaker_game maker-breaker games]. In these games, the players alternatively color a cell. Some subsets of the cells are designated as ''winning sets''. One player is called the Maker and the other the Breaker. The Maker's goal is to color all of the cells in a winning set, and the Breaker's goal is to prevent this. Hex is a maker-breaker game, where the winning sets are exactly the paths connecting Maker's edges.

I give two examples. The first is simple and highly-symmetric. The second goes 3-of-3: There are exactly 3 legal moves outside the set, and all 3 of them win.

simple and highly-symmetric:

Take a hexagonal [https://en.wikipedia.org/wiki/Bipyramid bipyramid], and give each equatorial edge an arrow pointing to a pole, such that the directions of the arrows alternate. Consider the maker-breaker game on the vertices of the resulting object, where the winning sets are the vertices of faces such that the equatorial edge's arrow points to the face's polar vertex.

Breaker is in a winning position: No matter what Maker's first move is, Breaker's first move is a pole. (If Maker's was not a pole, then there is symmetry between the poles until Breaker chooses one.) If neither of Maker's first two moves was a pole, then Breaker's second move is the other pole, winning for Breaker. Otherwise, Breaker pretends Maker's pole move was Maker's first move, and wins by [[Pairing_strategy|pairing]] using the equatorial edges whose arrows point to the pole Maker played.

Now, assume Maker plays a pole X. The analogue to the set A, is the set of vertices that are members of any minimal winning set which uses the vertex X. These are exactly the equatorial vertices, so in particular this set is non-empty.

However, if Breaker responds on the equator, then Breaker loses: Let Y be where Breaker just played. Maker responds on the equator, either opposite of Y, or the cell 120 degrees from Y whose immediate threat is adjacent to - rather than opposite from - Y. Breaker must defend against that threat, after which Maker plays the other equatorial vertex adjacent to where Maker just played. Lastly, Maker wins by playing either the pole Maker didn't already play, or continuing in the same direction on the equator.

going three-of-three:

This one is a maker-breaker game whose underlying set is {0,1L,1R,2L,2R,3L,3R,4}. There are exactly 6 minimal winning sets, and they are {0,1L,2L},{2L,3L,3R},{2L,3L,4} and the three formed by interchanging L with R.

Breaker is in a winning position: By symmetry, assume Maker does not play a R element. 0 and 2L each [[Dominated_cell#Switch-domination|switch-dominate]] 1L, so this leaves 0,2L,3L,4 as candidates for Maker's first move. If Maker's first move is 3L or 4, then Breaker can play 2L and win with the [[Pairing_strategy|pairing]] {0,1R},{2R,3R}. If Maker's first move is 2L or 0, then Breaker wins by playing any of 3L,3R,4 and using the [[Pairing_strategy|pairs]] {0,1L} and {1R,2R} and whichever of {3L,3R,4} Breaker hasn't yet played.

Now, assume Maker plays 0. The analogue to the set A, is the set of vertices that are members of any minimal winning set which uses the element 0. This is exactly {1L,1R,2L,2R}.

If Breaker responds in {1L,1R,2L,2R}, then Breaker loses: By symmetry, assume Breaker plays 1L or 2L. Maker responds at 2R, threatening to win immediately with 1R. Breaker must defend against that threat, after which Maker plays 3R, and wins with 3L or 4.

(As noted in the "Breaker is in a winning position" part here, all three of Breaker's moves outside of the analogue-of-A win for Breaker.)

[[category: Open problems]]
[[category: Forums]]

KataHex

2025-01-31T17:41:28Z

Selinger: /* Links */ Added newest pre-trained networks

'''KataHex''' is a free and open-source computer Hex program, capable of defeating top-level human players. It implements Monte Carlo tree search with a convolutional neural network providing position evaluation and policy guidance.

== History and versions ==

KataHex is based on [https://en.wikipedia.org/wiki/KataGo KataGo], a computer Go program developed by David Wu that was first released on 27 February 2019. It was adapted for Hex by "HZY" between February 2020 and May 2022. While initially unnamed, the Hex-adaptation of KataGo quickly became known as "KataHex" among Hex players.

The HZY implementation of KataHex speaks a non-standard dialect of [[GTP]] and can only interact with a specially modified version of the Go GUI known as LizzieYzy. A further adaptation of KataHex that is capable of interfacing with [[Hexgui]] was made by [[User:Selinger|Selinger]].

== Pre-trained networks ==

HZY initially trained the neural network model on two NVIDIA GeForce RTX 2080 Ti GPUs for about 20 days on 13x13. They then up-trained the network on 19x19 for one day, and on 27x27 for an additional day. [https://zhuanlan-zhihu-com.translate.goog/p/476464087?_x_tr_sl=auto&_x_tr_tl=en According to HZY], the up-trained 19x19 network is relatively reliable, but the 27x27-network is not.

The newest model, 20240812, is trained on two RTX 4090 GPUs for 2 months on 15x15, 15 days on 19x19, and 3 days on 27x27.

Running KataHex requires both a neural net model (the "weights"), and an engine to load the weights. Each neural net is able to play Hex not only at the size it was trained on, but also smaller and larger sizes, as long as the engine is compiled to support the size. (However, the net will not play particularly well on sizes larger than it was trained on.)

"KataHex" refers generally to the KataHex program, but often more specifically to the pre-trained 19x19 neural network made by HZY.

== Limitations ==

Since KataHex was only trained on self-play, it does not always do well when asked to play from an arbitrary starting position. Older nets are particularly thrown off if the starting position does not have the same number of black and white stones, but the issue seems to have been mitigated with the 20240812 net.

== Links ==

'''Program:'''

* David Wu's original KataGo: <https://github.com/lightvector/KataGo>
* HZY's KataHex: <https://github.com/hzyhhzy/KataGo/tree/Hex2024>
* Selinger's modifications: <https://github.com/selinger/katahex/tree/Hex2024>

'''Pre-trained network models:'''

* 20250131 (current strongest net): [https://github.com/hzyhhzy/KataGomo/releases/download/Hex_20250131/hex3_27x_b28.bin.gz hex3_27x_b28.bin.gz]
* 20250131 (supports move limit mode, but slightly weaker): [https://github.com/hzyhhzy/KataGomo/releases/download/Hex_20250131/hex3_mm19x_b28.bin.gz hex3_mm19x_b28.bin.gz]
* 20240812: [https://drive.usercontent.google.com/download?id=1YeqRvAYs7YjtPh0xBbDnHxeo2xrLEOdX hex27x3.bin.gz]
* 20220618: [https://drive.google.com/file/d/1xMvP_75xgo0271nQbmlAJ40rvpKiFTgP/view katahex_model_20220618.bin.gz]

'''GUIs:'''

* The 20240812 net is bundled with a modified LizzieYzy GUI: <https://github.com/hzyhhzy/KataGo/releases/tag/Hex_20240812>
* An older net with the modified LizzieYzy GUI: [https://drive.google.com/file/d/1qbTTmPFiUkM_346DeKS1E9gJR-roNH63/view KataHex_LizzieYZY_20220313.zip].
* Selinger's KataHex works with HexGUI: <https://github.com/selinger/hexgui>

== See also ==

* [[Strategic advice from KataHex]]

[[Category:Computer Hex]]
[[category:Hex playing program]]

Commented games

2025-01-26T00:33:28Z

Selinger: Added a game with video comments by Florian Jamain

The best ways for getting better at Hex are to learn strategies, solve problems and to replay games of stronger players. The following collection of games are intended to help you get stronger.

{| class="wikitable" border="1" cellpadding="2" cellspacing="0"
|-
! Game !! Date !! Commented by !! Intended audience
|-
| [[V vs. H game 1|Vertical vs. Horizontal]] || ca. 1994|| David Boll || ---
|-
| [[Glenn C. Rhoads vs. unknown]] || ca. 2001 || [[Glenn C. Rhoads]] || advanced
|-
| [[Bill LeBoeuf vs. Universidad de Oviedo]] || November 2003 || Bill LeBoeuf || ---
|-
| [[Niall vs. Halladba, November 2010 | Halladba vs. Niall]] || November 2010 || [[user:Niall | Niall]] || ---
|-
| [[Nietsabes vs. Niall, November 2010 | Nietsabes vs. Niall]] || November 2010 || Niall || ---
|-
| [[James A. Cook vs. Niall, December 2010 | James A. Cook vs. Niall]] || December 2010 || Niall || ---
|-
| [[Niall vs Daniel Sepczuk - Dec 2011 | Niall vs. Daniel Sepczuk]] || December 2011 || Niall || ---
|-
| [[FIFI25_vs._murasawa,_October_2021 | FIFI25 vs. murasawa]] || October 2021 || Demer || ---
|-
| [[Image:Icon-video.png|16px|baseline|link=https://youtu.be/RljkP0FejQA|]] [https://youtu.be/RljkP0FejQA Daniel Sepczuk vs. Arek Kulczycki] || September 2024 || Florian Jamain || ---
|-
| [[Image:Icon-video.png|16px|baseline|link=https://youtu.be/nIMm2qjXIto|]] [https://youtu.be/nIMm2qjXIto emerytszachowy vs. realefab] || September 2024 || Florian Jamain || ---
|}

Remarks:
* The '''Game''' is a link to the commented game.
* The '''Commented by''' field contains the name of the person who did the commentary.
* The '''Intended audience''' can be ''beginners'', ''advanced'', ''experts'', whatever.
* The icon [[Image:Icon-video.png|16px|baseline|]] indicates '''video comments'''.

Some more commented games are available in the forum of littlegolem.net.

[[category:Game record]]

Commented games

2025-01-13T06:12:19Z

Selinger: Added a link to Florian Jamain's video commentary.

The best ways for getting better at Hex are to learn strategies, solve problems and to replay games of stronger players. The following collection of games are intended to help you get stronger.

{| class="wikitable" border="1" cellpadding="2" cellspacing="0"
|-
! Game !! Date !! Commented by !! Intended audience
|-
| [[V vs. H game 1|Vertical vs. Horizontal]] || ca. 1994|| David Boll || ---
|-
| [[Glenn C. Rhoads vs. unknown]] || ca. 2001 || [[Glenn C. Rhoads]] || advanced
|-
| [[Bill LeBoeuf vs. Universidad de Oviedo]] || November 2003 || Bill LeBoeuf || ---
|-
| [[Niall vs. Halladba, November 2010 | Halladba vs. Niall]] || November 2010 || [[user:Niall | Niall]] || ---
|-
| [[Nietsabes vs. Niall, November 2010 | Nietsabes vs. Niall]] || November 2010 || Niall || ---
|-
| [[James A. Cook vs. Niall, December 2010 | James A. Cook vs. Niall]] || December 2010 || Niall || ---
|-
| [[Niall vs Daniel Sepczuk - Dec 2011 | Niall vs. Daniel Sepczuk]] || December 2011 || Niall || ---
|-
| [[FIFI25_vs._murasawa,_October_2021 | FIFI25 vs. murasawa]] || October 2021 || Demer || ---
|-
| [[Image:Icon-video.png|16px|baseline|link=https://youtu.be/RljkP0FejQA|]] [https://youtu.be/RljkP0FejQA Daniel Sepczuk vs. Arek Kulczycki] || September 2024 || Florian Jamain || ---
|}

Remarks:
* The '''Game''' is a link to the commented game.
* The '''Commented by''' field contains the name of the person who did the commentary.
* The '''Intended audience''' can be ''beginners'', ''advanced'', ''experts'', whatever.
* The icon [[Image:Icon-video.png|16px|baseline|]] indicates '''video comments'''.

Some more commented games are available in the forum of littlegolem.net.

[[category:Game record]]

File:Icon-video.png

2025-01-13T05:56:00Z

Selinger: An icon to use when linking to a video.

An icon to use when linking to a video.

Dead cell

2025-01-11T00:53:27Z

Selinger: Added a dead cell pattern.

A dead cell is a cell whose colour does not affect the outcome of the game. More formally, in a given Hex position, a cell is ''dead'' if for every way of filling all empty cells of the board with red and blue pieces, the winner remains the same when the colour of the dead cell is changed from red to blue or vice versa.

A cell that is not dead is ''alive''.

== Examples ==

[[Useless triangle]]s are examples of dead cells.

In the following [[pattern]]s, the dead cell is labelled with a star. Some patterns include empty cells; however, these cells have only been included for context; they do not need to be empty. They can be occupied by stones of either color or even be outside the board.

=== Dead cells in the interior ===

<hexboard size="2x3"
coords="none"
edges="none"
visible="-a1"
contents="R a2 b1 c1 c2 E *:b2"
/>

<hexboard size="3x3"
coords="none"
edges="none"
visible="-a1 a2 b3 c3"
contents="B a3 R b1 c1 c2 E *:b2"
/>

<hexboard size="3x3"
coords="none"
edges="none"
visible="-a1 b1 b3 c3"
contents="B a2 a3 R c1 c2 E *:b2"
/>

<hexboard size="3x4"
coords="none"
edges="none"
visible="-a1 d3 c1 b3"
contents="R b1 a2 a3 B d1 d2 c3 E *:b2 *:c2"
/>

=== Dead cells near edges and corners ===

<hexboard size="3x4"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="R c2 b3 E *:c3"
/>

<hexboard size="2x5"
coords="none"
edges="bottom"
visible="-a1"
contents="R b2 d2 E *:c2"
/>

<hexboard size="3x5"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="B c2 R d3 E *:c3"
/>

<hexboard size="3x5"
coords="none"
edges="bottom"
visible="-a1 a2 b1"
contents="B c2 d2 E *:c3"
/>

<hexboard size="3x4"
coords="none"
edges="bottom right"
visible="-a1 a2 b1"
contents="R c3 E *:d3"
/>

== Usage ==

Because the colour does not affect the result of the game, dead cells can be treated as if they contained a blue piece or a red piece, without changing the strategic value of the position. This often simplifies the analysis of Hex positions. In particular, dead cells are [[captured cell|captured]] by both players.

For example, in the following hypothetical situation (with Red to move), Red might wonder whether the piece at e2 could somehow be useful as a [[ladder escape]].
<hexboard size="5x5"
coords="show"
contents="R a2 R a3 R a4 B a5 B c3 B d2 B e1 R e2"
/>
However, the cell e2 is dead, and therefore the position is strategically equivalent to the one where e2 is blue. It follows that e2 cannot possibly be useful to Red as a ladder escape, or for any other purpose.

It is never advantageous to move in a dead empty cell. It is also never advantageous to move in a cell that the opponent can kill (i.e., turn into a dead cell) with the next move. Such a cell is called ''vulnerable'' for the player who should not move there.

This concept can be useful in determining which side of a [[bridge]] it is better to intrude upon. In the following example, b3 is vulnerable for Blue. If Blue plays b3, Red can respond at c3, killing b3. Since b3 is now dead, it can be treated as a red piece. Effectively, Blue has gained nothing, and Red has gained two new pieces at b3 and c3. It follows that if Blue wants to intrude into the bridge, she should do so at c3. See also the article on [[bolstered template]]s.

<hex>R5 C5 Q1 Ha3 Vb4 Vc2</hex>

Dead cell analysis often plays a role in determining cells that are [[captured cell|captured]] or [[dominated cell|dominated]].

Since dead cells, vulnerable cells, captured cells, and [[dominated cell]]s are never good candiates for a player's next move, dead cell analysis can significantly speed up [[computer Hex]] algorithms, since it can reduce the number of possibilities that must be explored.

Dead cell analysis is also used in the proof that [[A1 opening|a1 is a losing opening]].

== Interaction between multiple dead cells ==

While the colour of a dead cell does not affect whether a position is winning or losing, it can affect whether other cells are dead. For example, in the following position, both c3 and d3 are dead.

<hexboard size="4x6" coords="show" contents="B a4 b4 d4 e4 f4 R c2 c3 d2 d3 e2"/>

Therefore we can colour c3 blue without changing the strategic value of the position. Alternatively, we can colour d3 blue without changing its value. However, we cannot colour ''both'' c3 and d3 blue, as this would actually change the outcome of the game. In other words, if we change the colour of c3, then d3 is no longer dead, and vice versa.

An extreme example of this is the position where the entire Hex board is filled with red pieces. This is obviously a win for Red. Also, assuming the board size is at least 2x2, every single cell on the board is dead. However, if we were to change all of them to blue, it would clearly change the winner of the game.

However, if a dead cell is currently empty, then it will remain dead no matter what colour it or other empty dead cells are filled with. In particular, a set of several empty dead cells is captured, as a whole, by both players.

== See Also ==
*[[Equivalent patterns]]

*[[Computer Hex]]

*[[Dominated cell]]s

*[[Captured cell]]s

== External link ==

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

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

Theorems about templates

2024-12-17T18:21:38Z

Selinger: Fixed numbering of theorems

There are a number of theorems about templates, some of which can be useful in play. Some of these theorems concern how to construct new templates from existing ones. Others concern how to play when templates overlap. Others explain why templates tend to have particular shapes.

== New templates from old ==

When we have theorems that allow us to construct new templates from known ones, there are fewer templates to memorize.

=== Corner clipping ===

We begin by observing that the following two positions are strategically equivalent:
A: <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(d1,a4,c4,d3)"
contents="B d1--d3 E x:c2 y:b3 z:c3"
/> B: <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(d1,a4,c4,d3)"
contents="B d1--d3 R c2 B c3 E w:b3"
/>
Indeed, if Red plays first in the region, then ''x'' [[captured cell|captures]] the entire triangle (''x'',''y'',''z'') in A, and ''w'' captures the corresponding triangle in B. Therefore, under [[optimal play]], Red achieves exactly the same thing in A as in B. Similarly, if Blue plays first in the region, ''y'' captures the whole triangle in A and ''w'' [[dead cell|kills]] the red stone and therefore captures the whole triangle in B. Therefore, under optimal play, Blue achieves the same thing in A as in B. It follows that A and B are equivalent.

Then we have the following theorem about edge templates:

'''Theorem 1 (corner clipping).''' Suppose an edge template has a corner of the form
A: <hexboard size="3x3"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1--c3"
/>
Here, the blue-shaded cells must not be part of the template, i.e., they must be outside of its [[carrier]]. Then the pattern where this corner has been replaced by
B: <hexboard size="3x3"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1--c3--b3 R b2"
/>
is also an edge template. The converse is also true, i.e., if some template has a corner of shape B, the corresponding pattern with shape A is also a template.

'''Proof.''' Since the shaded hexes are outside the carrier, they may as well be blue stones. Then by the above observation, A and B are completely equivalent, so if some pattern containing A is connected, then so is the corresponding pattern containing B. Moreover, it is easy to see that if removing one cell from the carrier of A would yield a connected shape, then the same could be achieved by removing one cell or the red stone from B, and vice versa. Therefore, the template containing A is minimal if and only if the template containing B is minimal. □

'''Examples'''

Corner clipping shows that the ziggurat is equivalent to edge templates [[Edge template III2b|III2b]] and [[Edge template III2g|III2g]], as well as a very compact 3-stone template:

<hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="R c1"
/> ⇔ <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 d3"
contents="R c1 d2"
/> ⇔ <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 a3"
contents="R c1 b2"
/> ⇔ <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 a3 d3"
contents="R c1 b2 d2"
/>

[[Edge template IV2a]]:

<hexboard size="4x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1--a3 b1"
contents="R c1 d1"
/> ⇔ <hexboard size="4x4"
float="inline"
coords="none"
edges="bottom"
visible="-a1--a3 b1 d4"
contents="R c1 d1 d3"
/>

'''Remark.''' For the clipped corner theorem to hold, the corner must be of this shape
<hexboard size="3x3"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1--c3"
/>
and not merely that one:
<hexboard size="3x3"
edges="bottom"
coords="none"
visible="-a1 a2 b1 c1"
contents="S blue:c1--c3"
/>
In other words, the cell on the 3rd row should not be part of the carrier. However, if the corner is merely of the latter form, the clipped template is still valid. It may not be minimal. For example, consider [[edge template IV2b]]:
<hexboard size="4x5"
coords="none"
edges="bottom"
visible="-a1--a3 b1 e1"
contents="R c1 b2"
/>
Clipping the right corner is no problem. If we clip the left corner, the resulting pattern is connected, but not minimal:
<hexboard size="4x5"
coords="none"
edges="bottom"
visible="-a1--a3 b1 e1 a4"
contents="R c1 b2 b3 E *:(e2--e4)"
/>
The three hexes marked "*" could be removed from the carrier while still remaining connected.

=== Large corner clipping ===

Observe that the following positions are equivalent:
A: <hexboard size="4x2"
float="inline"
edges="none"
coords="none"
visible="-b4"
contents="B a1--b1--b3--a4 R a3"
/> B: <hexboard size="4x2"
float="inline"
edges="none"
coords="none"
visible="-b4"
contents="B a1--b1--b3--a4 R a2"
/>
Indeed, if Red moves in the region, A and B become identical. If Blue moves in the region, Blue [[dead cell|kills]] Red's stone, so again A and B also become identical.

Then we have the following theorem about edge templates:

'''Theorem 2 (large corner clipping).''' Suppose an edge template has a corner of the form
A: <hexboard size="4x4"
edges="bottom"
coords="none"
visible="area(b4,d4,d1,c1)"
contents="S blue:c1--d1--d4"
/>
Here, the blue-shaded cells must not be part of the template, i.e., they must be outside of its [[carrier]]. Then the pattern where this corner has been replaced by
B: <hexboard size="4x4"
edges="bottom"
coords="none"
visible="area(b4,d4,d1,c1)"
contents="S blue:c1--d1--d4--c4 R c2"
/>
is also an edge template. The converse is also true, i.e., if some template has a corner of shape B, the corresponding pattern with shape A is also a template.

'''Proof.''' By ordinary corner clipping, A is equivalent to
<hexboard size="4x4"
edges="bottom"
coords="none"
visible="area(b4,d4,d1,c1)"
contents="S blue:c1--d1--d4--c4 R c3"
/>
which is in turns equivalent to B by the above observation. □

'''Examples'''

Large corner clipping shows that the ziggurat is equivalent to [[edge template III2a]]:

<hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="R c1"
/> ⇔ <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 d3"
contents="R c1 d1"
/>

Similarly, we can construct several new templates from [[edge template IV2a]]:
<hexboard size="4x7"
float="inline"
coords="none"
edges="bottom"
visible="area(e1,c2,a4,g4,g2,f1)"
contents="R e1"
/> ⇔ <hexboard size="4x7"
float="inline"
coords="none"
edges="bottom"
visible="area(e1,c2,a4,g4,g2,f1)-a4"
contents="R arrow(12):e1 c2"
/> ⇔ <hexboard size="4x7"
float="inline"
coords="none"
edges="bottom"
visible="area(e1,c2,a4,g4,g2,f1)-g4"
contents="R arrow(12):e1 g2"
/>
There are of course additional possibilities, such as clipping both corners, combining large and ordinary corner clipping, etc.

=== Corner bending ===

The idea of corner bending is similar to that of corner clipping. We again start with an observation about two positions. This time, we claim that B is at least as good for Red as A.
A: <hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="B b1 b2 E x:a2"
/> B: <hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="B a2 R b2 E y:b1"
/>
Indeed, if Red plays first in B, then ''y'' [[dead cell|kills]] the blue stone and therefore [[captured cell|captures]] the whole region, which is certainly at least as good as anything that Red could achieve in A. On the other hand, if Blue plays first in A, then Blue occupies the whole region, which is certainly at least as bad for Red as anything Blue could achieve in B. Therefore, if any position containing A is winning for Red, then so is the corresponding position containing B.

We obtain the following theorem:

'''Theorem 3 (corner bending).''' Suppose an edge template has a corner of the form
A: <hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="S blue:(b1 b2) E x:a2"
/>
Here, the blue-shaded cell must not be part of the template, i.e., it must be outside of its [[carrier]]. Then the pattern where this corner has been replaced by
B: <hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="S blue:a2 R b2 E y:b1"
/>
is still connected. (It may fail to be an edge template only because it may fail to be minimal).

'''Proof.''' Since the shaded hexes are outside the carrier, they may as well be blue stones. Then by the above observation, B is at least as good for Red as A, so if some region containing A is connected for Red, then so is the same region containing B. □

'''Examples'''

The following is a [[ziggurat]], followed by ziggurats with one or two bent corners.

<hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="R c1"
/> <hexboard size="3x5"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 d3 e1 d4"
contents="R c1 e3"
/> <hexboard size="3x5"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 c1 b3"
contents="R d1 a3"
/> <hexboard size="3x6"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 c1 b3 f1 e3"
contents="R d1 a3 f3"
/>

=== Crescenting ===

We begin by observing that the following three positions are strategically equivalent:

A: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(b1,a3,c4,d3,d1)"
contents="B all R a3 z:b3 B x:b2 y:c2 w:c3"
/> B: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(b1,a3,c4,d3,d1)"
contents="B all R a3 z:b3 x:b2 y:c2 w:c3"
/> C: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(b1,a3,c4,d3,d1)"
contents="B all R a3 x:b2 y:c2 E z:b3 w:c3"
/>

Indeed, A and B are equivalent because x, y, and w are [[dead cell|dead]], and B and C are equivalent because z and w are [[captured cell|captured]]. We then have the following theorem about templates:

'''Theorem 4 (crescenting).''' Suppose some (edge or interior) template has a piece of boundary of the form
<hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(b1,a3,c4,d3,d1)"
contents="S blue:all none:a3,b3 R a3 b3"
/>
Here, as usual, the blue-shaded cells are not part of the template. Then the pattern where this area has been replaced by
<hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(b1,a3,c4,d3,d1)"
contents="S blue:all none:a3,b3,b2,c2,c3 R a3 b2 c2"
/>
is also a template. Moreover, the converse also holds. (Caveat: the construction preserves minimality with respect to empty cells in the carrier. In some boundary cases, it is possible that some of the red stones are not actually necessary in one or the other template).

'''Proof.''' Since the shaded hexes are outside the carrier, they may as well be blue stones. Then the equivalence follows by the above observation. □

'''Examples'''

Many templates with two adjacent stones have crescented versions. Indeed, the crescent itself is such a template:
<hexboard size="3x2"
float="inline"
edges="none"
coords="none"
visible="-b3"
contents="R a1 a3 b1"
/> ⇔ <hexboard size="4x3"
float="inline"
edges="none"
coords="none"
visible="area(a2,a4,c2,c1,b1)"
contents="R a2 a4 b1 c1"
/>
Here is a crescented version of [[edge template IV2a]]. Note that crescenting can also be applied recursively, resulting in templates that resemble a [[Interior_template#The_long_crescent|long crescent]].
<hexboard size="4x4"
float="inline"
coords="none"
edges="bottom"
visible="area(c1,b2,a4,d4,d1)"
contents="R c1 d1"
/> ⇔ <hexboard size="5x5"
float="inline"
coords="none"
edges="bottom"
visible="area(d1,b3,a5,d5,d3,e2,e1)"
contents="R c2 d1 e1"
/> ⇔ <hexboard size="6x6"
float="inline"
coords="none"
edges="bottom"
visible="area(e1,b4,a6,d6,d4,f2,f1)"
contents="R c3 d2 e1 f1"
/>

Often, the crescent-like shape can be more generally replaced with any capped [[flank]]. See also [[Flank#Edge_templates_from_capped_flanks|edge templates from capped flanks]].

=== Alternative connection up ===

Some templates, such as [[Tom's move]], have an "alternative connection up". There is a general theorem about this. We begin by observing that from Red's point of view, C is at least as good as B, and B is at least as good as A:

A: <hexboard size="3x4"
float="inline"
edges="none"
coords="none"
visible="area(a2,b3,d1,c1)"
contents="R c1 b2 b3 B a2 d1 E c2"
/> B: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a2,b3,d1,c1)"
contents="R c1 b3 B a2 R b2 B w:c2 E z:d1"
/> C: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a2,b3,d1,c1)"
contents="R c1 b3 B a2 E x:b2 y:c2 z:d1"
/>

To see that B is at least as good for Red as A, note that if Red plays first in the region, then Red plays z in B, which [[dead cell|kills]] w and is at least as good as anything Red could do in A. If Blue plays first in the region, A and B become identical. To see that C is at least as good for Red as B, note that Red can get at least one of x and y by defending the [[bridge]]. No matter which way the bridge goes, the result is identical or better for Red than B.

'''Theorem 5 (alternative connection up).''' Suppose some (edge or interior) template has a piece of boundary of the form

A: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a3,b4,d2,d1,c1)"
contents="S blue:all none:b4,c3 R arrow(12):b4"
/>

Here, as usual, the blue-shaded cells are not part of the template, but the stone marked "↑" must of course be connected up. Then the patterns where this area has been replaced by

B: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a3,b4,d2,d1,c1)"
contents="S blue:all none:b4,c3,b3,d2,c2 R b4 arrow(12):c2 R b3 B c3"
/> or C: <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a3,b4,d2,d1,c1)"
contents="S blue:all none:b4,c3,b3,d2,c2 R b4 arrow(12):c2"
/>

are also connected. (They may fail to be templates only because they may fail to be minimal).

'''Proof.''' Consider the template containing A. Since the shaded hexes are outside the carrier, they may as well be blue stones, except that we need to connect the stone marked "↑" to something, which we can without loss of generality do like this:
<hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a3,b4,d2,d1,c1)"
contents="B a3 b2 d1 d2 R a:b4 b:b3 c:c2 arrow(12):c1"
/>
The fact that the stones "b" and "c" are adjacent to cells in the template does not matter, because "a" is adjacent to the same cells anyway. By the above observation, each of the following is at least as good for Red, and therefore also connected:
<hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a3,b4,d2,d1,c1)"
contents="B a3 b2 d1 R b4 c2 arrow(12):c1 R b3 B c3"
/> and <hexboard size="4x4"
float="inline"
edges="none"
coords="none"
visible="area(a3,b4,d2,d1,c1)"
contents="B a3 b2 d1 R b4 c2 arrow(12):c1"
/>
Therefore, the patterns using B and C are connected, proving the theorem. □

'''Example'''

[[Tom's move]] ensures that the following is connected:
<hexboard size="5x7"
edges="bottom"
coords="none"
visible="-(a2 a1 b1 f1 g1 g2)"
contents="R arrow(12):b2 a3 a4 B b3"
/>
By Theorem 5, the following are therefore also connected:
<hexboard size="6x8"
float="inline"
edges="bottom"
coords="none"
visible="-g2 h2 h3 f1--h1 -area(a1,a6,c1)"
contents="R b4 b5 c3 B a6 c4 R arrow(12):d1 R c2 B d2"
/> and <hexboard size="6x8"
float="inline"
edges="bottom"
coords="none"
visible="-g2 h2 h3 f1--h1 -area(a1,a6,c1)"
contents="R b4 b5 c3 B a6 c4 R arrow(12):d1"
/>
This is exactly the "[[Tom's_move#Alternative_connection_up|alternative connection up]]" of Tom's move.

== Ladder creation templates from templates ==

By an argument similar to the first observation above, we observe that the following three positions are strategically equivalent:
A: <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,d3,d1)"
contents="B c1--c3,d1--d3 E x:b2 y:a3 z:b3"
/> B: <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,d3,d1)"
contents="B c1--c3,d1--d3 R b2 B a3 E w:b3"
/> C: <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,d3,d1)"
contents="B c1,b3,d1--d3 R c2 E x:b2 y:a3 z:c3"
/>
Indeed, whoever plays first in each region captures the entire region: Red by playing at ''x'' or ''w'', and Blue by playing at ''y'' or ''w''.

An interesting application of this is getting a 2nd row [[ladder creation template]] from an ordinary template.

'''Theorem 6 (ladder creation template from template).''' Suppose an edge template has a corner of the form
A: <hexboard size="3x4"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1--c3,d1--d3"
/>
where as usual, the blue-shaded cells are not part of the template. Then the pattern where this corner has been replaced by either one of
B: <hexboard size="3x4"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1--c3,d1--d3 B a3 E arrow(3):b2,b3"
/>
C: <hexboard size="3x4"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1,d1--d3 B b3 E arrow(3):c2,c3"
/>
is a 2nd row ladder creation template. The converse is also true, i.e., if some pattern with a corner of shape B or C is a 2nd row ladder creation template, then the corresponding pattern with shape A is an edge template.

'''Proof.''' For the equivalence of A and B, note that by the above observation, A is connected if and only if B' is connected:
B': <hexboard size="3x4"
edges="bottom"
coords="none"
visible="-a1 a2 b1"
contents="S blue:c1--c3,d1--d3 B a3 R b2 E b3"
/>
By essentially Theorem 1 of the article on the [[theory of ladder escapes]], B' is connected if and only if B creates a 2nd row ladder. Finally, since it is an "if and only if", it follows that if A is minimal, so is B, and vice versa. The argument for the equivalence of A and C is analogous. □

'''Example'''

From the [[ziggurat]], we get the following ladder creation templates:
<hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,d3,d1)"
contents="R c1 B c3 E arrow(3):d2,d3"
/> <hexboard size="3x5"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,e3,e2,d1)"
contents="R c1 B d3 E arrow(3):e2,e3"
/> <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="area(c1,a3,d3,d1)"
contents="R c1 B b3 E arrow(9):b2,a3"
/> <hexboard size="3x5"
float="inline"
edges="bottom"
coords="none"
visible="area(d1,b2,a3,e3,e1)"
contents="R d1 B b3 E arrow(9):b2,a3"
/>

What we learn from this is that a Blue [[intrusion]] into the very corner of Red's template, or the cell right next to the corner, is not usually a good idea. Red can reconnect by creating a 2nd row ladder escape, potentially far away. This will often allow Red to play a [[minimaxing]] response. In particular, if Red already has a 2nd row ladder escape, the intrusion is not even valid (it does not even threaten to disconnect Red).

It is worth remarking that if the edge template's corner is merely of this shape,
<hexboard size="3x4"
edges="bottom"
coords="none"
visible="-a1 a2 b1 c1 d1"
contents="S blue:c1--c3,d1--d3"
/>
then the left-to-right direction of Theorem 6 is still valid: Given an edge template, we still obtain a ladder creation template in two different ways. However, in this case, the latter may not be minimal.

== Overlapping templates ==

When templates [[Template#Overlapping_templates|overlap]], they are usually not both valid. However, there are some exceptions where templates can overlap and still be valid. It is useful to know them.

=== Edge template II in the overlap ===

'''Theorem 7.''' If the region in which two edge templates overlap is [[edge template II]], then both templates remain valid.
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="b3 c2 c3"
contents="R c2"
/>

'''Proof.''' The two empty hexes in edge template II are [[captured cell|captured]], and therefore they can be replaced by red stones without changing the strategic value.
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="b3 c2 c3"
contents="R c2 b3 c3"
/>
Overlapping templates are only invalid if there are empty cells in the overlap. □

'''Example'''

<hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="area(d1,a4,d4)"
contents="R d1 d3"
/> + <hexboard size="4x4"
float="inline"
coords="hide"
edges="bottom"
visible="area(d1,a4,d4)"
contents="R d1 b3"
/> = <hexboard size="4x6"
float="inline"
coords="hide"
edges="bottom"
visible="area(d1,a4,f4,f1) - e1"
contents="R d1 d3 f1"
/>
Both templates remain valid, i.e., all three red stones can be connected to the edge simultaneously.

=== The ziggurat theorem ===

The following theorem is due to Eric Demer.

'''Theorem 8 (ziggurat theorem).''' Consider a [[ziggurat]].
<hexboard size="3x4"
visible="area(a3,d3,d1,c1)"
edges="bottom"
coords="none"
contents="R 1:c1 E x:a3 y:d3 z:c2"
/>
If the ziggurat overlaps another edge template in the cell ''x'', and/or overlaps another edge template in the cell ''y'', all templates (i.e., the ziggurat itself and its neighboring templates) remain valid. If Blue plays in the overlap at ''x'' or ''y'', Red can restore all templates by playing at ''z''. Moreover, this even remains true if the ziggurat has not been completed yet (i.e., if the template stone 1 has not yet been played).

'''Proof.''' If Blue plays at ''x'' or ''y'' (or both), clearly Red playing at ''z'' defends the ziggurat. What we must show is that it defends the neighboring templates as well. But if Red plays at ''z'', then the two hexes just below ''z'' are [[captured cell|captured]].
<hexboard size="3x4"
visible="area(a3,d3,d1,c1)"
edges="bottom"
coords="none"
contents="R 1:c1 B x:a3 y:d3 R z:c2 R b3 c3"
/>
Then the neighboring templates are valid by Theorem 3 above (the corner bending theorem).

The other thing we must show is that if Blue starts by playing in the ziggurat anywhere other than at ''x'' or ''y'', then Red can always reconnect in a way that either captures or no longer needs ''x'' and ''y''. Indeed, if Blue plays anywhere on the right, Red can play like this:
<hexboard size="3x4"
visible="area(a3,d3,d1,c1)"
edges="bottom"
coords="none"
contents="R 1:c1 E x:a3 y:d3 B d1 c2 d2 c3 R b2"
/>
which captures ''x'' and no longer needs ''y''. And if Blue plays on the left, Red responds like this, which captures ''y'' and no longer needs ''x'':
<hexboard size="3x4"
visible="area(a3,d3,d1,c1)"
edges="bottom"
coords="none"
contents="R 1:c1 E x:a3 y:d3 B b2 b3 R d2"
/>
□

'''Example'''

The classic application of the ziggurat theorem is [[edge template IV2c]]:
<hexboard size="4x7"
coords="hide"
edges="bottom"
visible="-area(a1,a3,d1) -g1"
contents="R e1 f1 B e2 E a:d2 b:f2 c:d4"
/>
Red is threatening to play at ''a'' or ''b'', getting a ziggurat each way. Blue's only hope is to play in the overlap. Alas, by the ziggurat theorem, this does not work. Red knows that she should play at 2 (the symmetric move would of course also have worked):
<hexboard size="4x7"
coords="hide"
edges="bottom"
visible="-area(a1,a3,d1) -g1"
contents="R e1 f1 B e2 B 1:d4 R 2:c3"
/>
Now Blue is sure to lose:
<hexboard size="4x7"
coords="hide"
edges="bottom"
visible="-area(a1,a3,d1) -g1"
contents="R e1 f1 B e2 B d4 R c3 B 3:d2 R 4:f2 B 5:f3 R 6:e3 B 7:e4 R 8:d3"
/>
There are other ways for Red to connect here; for one, Red's moves 2, 4, 6 could have been played in a different order. But by using the ziggurat theorem, Red can easily know what to do without having to think hard, and can concentrate on other trickier areas of the board.

'''Generalizations.''' There are many other edge templates (besides the ziggurat) for which a version of the ziggurat theorem holds, but it is not known whether it holds for all templates of the appropriate shape. Perhaps a list of such templates could be added here at some point.

== The shape of templates ==

You may have noticed that many edge templates resemble each other: their boundaries tend to follow a relatively small number of possible shapes. This is not a coincidence. Some of it is due to the following theorems.

Observation: the two empty cells in the following pattern are [[captured cell|captured]] by Blue.
<hexboard size="4x3"
edges="none"
coords="none"
visible="-a1--a3 b4 c4"
contents="R a4 B b1 c1--c3"
/>
Indeed, if Red plays in one of these cells, Blue can play in the other, [[dead cell|killing]] Red's stone. From this, we immediately get the following theorem:

'''Theorem 9.''' There is no edge template with an empty corner of height 2, i.e., with a corner of this shape:
<hexboard size="3x2"
edges="bottom"
coords="none"
contents="S blue:(a1 b1--b3)"
/>
Here, as in previous theorems, the blue-shaded cells indicate the outside of the template, i.e., they are not part of the carrier.

'''Proof.''' Since the shaded hexes are outside the carrier, they may as well be blue stones. But then the two empty cells are captured by the previous observation, which means they can also be filled in with blue stones, contradicting the minimality of the template. □

We note that the theorem only says that the corner of an edge template cannot be of height 2 if that corner is empty. When there are stones in the corner, height 2 is possible, as in the following examples:
<hexboard size="2x2"
float="inline"
edges="bottom"
coords="none"
visible="-a1"
contents="R b1"
/> <hexboard size="3x4"
float="inline"
edges="bottom"
coords="none"
visible="-a1 a2 b1 d1"
contents="R arrow(12):c1 d3"
/>

Observation: The following two regions are equivalent:
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
contents="B a1--c1--c3 E x:a3"
/> <hexboard size="3x3"
float="inline"
coords="none"
edges="none"
contents="B a1--c1--c3 y:b2 E x:a3 "
/>
To see why, first assume that Blue plays first in the region. Then Blue x captures the entire region, so the stone at y no longer matters. Next, assume that Red plays first in the region. If Red plays anywhere other than x, then Blue x [[dead cell|kills]] the red stone. If Red plays at x, then y is [[dead cell|dead]], so the stone at y no longer matters. As a consequence, we get the following theorem:

'''Theorem 10.''' There is no edge template with a piece of boundary of this shape:
<hexboard size="3x3"
float="inline"
coords="none"
edges="none"
contents="S blue:a1--c1--c3 E y:b2"
/>
Here, as in previous theorems, the blue-shaded cells indicate the outside of the template, i.e., they are not part of the carrier.

'''Proof.''' Since the shaded hexes are outside the carrier, they may as well be blue stones. Then by the previous observation, y might as well be blue, contradicting the minimality of the template. □

Once again, we remark that Theorem 10 presupposes that there are no stones in the relevant portion of the template. For example, the [[ziggurat]] is an edge template that ends in two columns of height 3, but this does not contradict Theorem 10 due to the presence of a red stone in one of these columns.

'''Theorem 11.''' There is no edge template with a corner of this shape:
<hexboard size="4x3"
float="inline"
coords="none"
edges="bottom"
visible="-a1 a2 b1 c4"
contents="S blue:(c1--c3 b4) E y:b3"
/>
As always, the blue-shaded cells indicate the outside of the template, i.e., they are not part of the carrier.

'''Proof.''' This is very similar to Theorem 10. By the previous observation, y might as well be blue, contradicting the minimality of the template. □

Theorems 9–11 explain why the rightmost few columns of edge templates frequently have one of these shapes
<hexboard size="5x3"
float="inline"
coords="none"
edges="bottom"
visible="area(a1,a5,c5,c3)"
/> <hexboard size="5x2"
float="inline"
coords="none"
edges="bottom"
visible="area(a1,a5,b5,b3)"
/>
and never one of these, unless the template contains stones in those regions:
<hexboard size="5x3"
float="inline"
coords="none"
edges="bottom"
visible="area(a1,a5,c5,c3)-b2"
/> <hexboard size="5x3"
float="inline"
coords="none"
edges="bottom"
visible="area(a1,a5,c5,c3)+b1"
/> <hexboard size="5x3"
float="inline"
coords="none"
edges="bottom"
visible="area(a2,a5,c5,c4)"
/> <hexboard size="4x2"
float="inline"
coords="none"
edges="bottom"
visible="-b4"
/>

[[Category: Theory]]

Unicode and ASCII boards

2024-11-07T22:03:26Z

Selinger: Added convention for X and O.

The board can be represented in ASCII or Unicode (with fixed-width fonts) using either the full or compact formats below.

For the stones, the usual ASCII convention is to use "X" for Black and "O" for White. Another convention is to use "V" for the vertical player (Black) and "H" for the horizontal player (White). With Unicode, the symbols "●", "○", and "⋅" can also be used for a black stone, white stone, and empty cell, respectively.

== Full layouts ==

<pre>
A B C D E F G H I J K
__ __ __ __ __ __ __ __ __ __ __
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 1
1 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 2
2 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 3
3 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 4
4 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 5
5 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 6
6 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 7
7 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 8
8 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 9
9 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 10
10 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/_
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \ 11
11 \__/\__/\__/\__/\__/\__/\__/\__/\__/\__/\__/

A B C D E F G H I J K
</pre>

<pre>
__
__/ \__
__/ \__/ \__
__/ \__/ \__/ \__
__/ \__/ \__/ \__/ \__
X __/ \__/ \__/ \__/ \__/ \__ 0
__/ \__/ \__/ \__/ \__/ \__/ \__
__/ \__/ \__/ \__/ \__/ \__/ \__/ \__
__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__
__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__
__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__
/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \
\__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/
\__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/
\__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/
\__/ \__/ \__/ \__/ \__/ \__/ \__/ \__/
\__/ \__/ \__/ \__/ \__/ \__/ \__/
\__/ \__/ \__/ \__/ \__/ \__/
0 \__/ \__/ \__/ \__/ \__/ X
\__/ \__/ \__/ \__/
\__/ \__/ \__/
\__/ \__/
\__/
</pre>

== Compact layouts ==

<pre>
A B C D E F G H I J K
1 . . . . . . . . . . .
2 . . . . . . . V . . .
3 . . V . . . . V . . .
4 . . . V . . . . . . .
5 . . . . . . V . . . .
6 . . V H H H V . . . .
7 . . . . . . H . . . .
8 . H . V . H . . . . .
9 . H . . . . . . . . .
10 . . . . . . . . . . .
11 . . . . . . . . . . .

</pre>

<pre>
A B C D E F G H I J K
■■■■■■■■■■■■■■■■■■■■■■■■
1 \ · · · · · · · · · · · \
2 \ · · · · · · · · · · · \
3 \ ● · · · · · · · · · · \
4 \ · · · · · · · · · · · \
5 \ · · · · · · ○ ○ · · · \
6 \ · · · · ● ○ ● · · · · \
7 \ · · · · ○ · · · · · · \
8 \ · · ● · · · · · ● ● · \
9 \ · · · · · · · ● ○ ○ · \
10 \ · · · · · ○ · · · · · \
11 \ · · · · · · ● · · · · \
■■■■■■■■■■■■■■■■■■■■■■■■
</pre>

<pre>
/.\
/. .\
/. . .\
/. . . .\
O /. . . . .\ X
/. . . O . .\
/. . O . . . .\
/. . . . . . . .\
/. . X O X . . . .\
/. . . . . . . . . .\
. . . . . . . . . . .
\. . . . X . . . . ./
\. . . . . . . . ./
\. . . . . . . ./
\. . . . . . ./
\. . . . . ./
X \. . . . ./ O
\. . . ./
\. . ./
\. ./
\./
</pre>

[[Category: Resources]]

MoHex

2024-11-02T12:19:25Z

Selinger: Added "links" header

MoHex is a Hex program based on Monte Carlo tree search developed at the University of Alberta by [[Philip Henderson]], [[Broderick Arneson]] and [[Ryan Hayward]].

MoHex is Free Software and released under the terms of the Lesser General Public License.

MoHex won the gold medal at the Computer Olympiad 2009.

== Links ==

* [https://github.com/cgao3/benzene-vanilla-cmake Current version of Benzene, containing MoHex, updated 2020]
* [http://sourceforge.net/projects/benzene/ Old project page at SourceForge, last updated in 2010]
* [http://webdocs.cs.ualberta.ca/~hayward/hex/#MoHex Information at Ryan Hayward's page]
* [http://www.grappa.univ-lille3.fr/icga/program.php?id=555 MoHex results at ICGA tournaments]
* [http://webdocs.cs.ualberta.ca/~hayward/papers/rptPamplona.pdf MoHex Wins Hex Tournament (14th Computer Olympiad 2009 Pamplona) (PDF)]

[[Category:Computer Hex]]
[[category:Hex playing program]]

KataHex

2024-11-01T00:38:16Z

Selinger: /* Links */ Updated github links to Hex2024 branch.

'''KataHex''' is a free and open-source computer Hex program, capable of defeating top-level human players. It implements Monte Carlo tree search with a convolutional neural network providing position evaluation and policy guidance.

== History and versions ==

KataHex is based on [https://en.wikipedia.org/wiki/KataGo KataGo], a computer Go program developed by David Wu that was first released on 27 February 2019. It was adapted for Hex by "HZY" between February 2020 and May 2022. While initially unnamed, the Hex-adaptation of KataGo quickly became known as "KataHex" among Hex players.

The HZY implementation of KataHex speaks a non-standard dialect of [[GTP]] and can only interact with a specially modified version of the Go GUI known as LizzieYzy. A further adaptation of KataHex that is capable of interfacing with [[Hexgui]] was made by [[User:Selinger|Selinger]].

== Pre-trained networks ==

HZY initially trained the neural network model on two NVIDIA GeForce RTX 2080 Ti GPUs for about 20 days on 13x13. They then up-trained the network on 19x19 for one day, and on 27x27 for an additional day. [https://zhuanlan-zhihu-com.translate.goog/p/476464087?_x_tr_sl=auto&_x_tr_tl=en According to HZY], the up-trained 19x19 network is relatively reliable, but the 27x27-network is not.

The newest model, 20240812, is trained on two RTX 4090 GPUs for 2 months on 15x15, 15 days on 19x19, and 3 days on 27x27.

Running KataHex requires both a neural net model (the "weights"), and an engine to load the weights. Each neural net is able to play Hex not only at the size it was trained on, but also smaller and larger sizes, as long as the engine is compiled to support the size. (However, the net will not play particularly well on sizes larger than it was trained on.)

"KataHex" refers generally to the KataHex program, but often more specifically to the pre-trained 19x19 neural network made by HZY.

== Limitations ==

Since KataHex was only trained on self-play, it does not always do well when asked to play from an arbitrary starting position. Older nets are particularly thrown off if the starting position does not have the same number of black and white stones, but the issue seems to have been mitigated with the 20240812 net.

== Links ==

'''Program:'''

* David Wu's original KataGo: <https://github.com/lightvector/KataGo>
* HZY's KataHex: <https://github.com/hzyhhzy/KataGo/tree/Hex2024>
* Selinger's modifications: <https://github.com/selinger/katahex/tree/Hex2024>

'''Pre-trained network models:'''

* 20240812 (current strongest net): [https://drive.usercontent.google.com/download?id=1YeqRvAYs7YjtPh0xBbDnHxeo2xrLEOdX hex27x3.bin.gz]
* 20220618: [https://drive.google.com/file/d/1xMvP_75xgo0271nQbmlAJ40rvpKiFTgP/view katahex_model_20220618.bin.gz]

'''GUIs:'''

* The 20240812 net is bundled with a modified LizzieYzy GUI: <https://github.com/hzyhhzy/KataGo/releases/tag/Hex_20240812>
* An older net with the modified LizzieYzy GUI: [https://drive.google.com/file/d/1qbTTmPFiUkM_346DeKS1E9gJR-roNH63/view KataHex_LizzieYZY_20220313.zip].
* Selinger's KataHex works with HexGUI: <https://github.com/selinger/hexgui>

== See also ==

* [[Strategic advice from KataHex]]

[[Category:Computer Hex]]
[[category:Hex playing program]]

KataHex

2024-10-31T00:17:39Z

Selinger: /* Links */ Removed link to Chinese website that has disappeared

'''KataHex''' is a free and open-source computer Hex program, capable of defeating top-level human players. It implements Monte Carlo tree search with a convolutional neural network providing position evaluation and policy guidance.

== History and versions ==

KataHex is based on [https://en.wikipedia.org/wiki/KataGo KataGo], a computer Go program developed by David Wu that was first released on 27 February 2019. It was adapted for Hex by "HZY" between February 2020 and May 2022. While initially unnamed, the Hex-adaptation of KataGo quickly became known as "KataHex" among Hex players.

The HZY implementation of KataHex speaks a non-standard dialect of [[GTP]] and can only interact with a specially modified version of the Go GUI known as LizzieYzy. A further adaptation of KataHex that is capable of interfacing with [[Hexgui]] was made by [[User:Selinger|Selinger]].

== Pre-trained networks ==

HZY initially trained the neural network model on two NVIDIA GeForce RTX 2080 Ti GPUs for about 20 days on 13x13. They then up-trained the network on 19x19 for one day, and on 27x27 for an additional day. [https://zhuanlan-zhihu-com.translate.goog/p/476464087?_x_tr_sl=auto&_x_tr_tl=en According to HZY], the up-trained 19x19 network is relatively reliable, but the 27x27-network is not.

The newest model, 20240812, is trained on two RTX 4090 GPUs for 2 months on 15x15, 15 days on 19x19, and 3 days on 27x27.

Running KataHex requires both a neural net model (the "weights"), and an engine to load the weights. Each neural net is able to play Hex not only at the size it was trained on, but also smaller and larger sizes, as long as the engine is compiled to support the size. (However, the net will not play particularly well on sizes larger than it was trained on.)

"KataHex" refers generally to the KataHex program, but often more specifically to the pre-trained 19x19 neural network made by HZY.

== Limitations ==

Since KataHex was only trained on self-play, it does not always do well when asked to play from an arbitrary starting position. Older nets are particularly thrown off if the starting position does not have the same number of black and white stones, but the issue seems to have been mitigated with the 20240812 net.

== Links ==

'''Program:'''

* David Wu's original KataGo: <https://github.com/lightvector/KataGo>
* HZY's KataHex: <https://github.com/hzyhhzy/KataGo/tree/Hex2022>
* Selinger's modifications: <https://github.com/selinger/katahex>

'''Pre-trained network models:'''

* 20240812 (current strongest net): [https://drive.usercontent.google.com/download?id=1YeqRvAYs7YjtPh0xBbDnHxeo2xrLEOdX hex27x3.bin.gz]
* 20220618: [https://drive.google.com/file/d/1xMvP_75xgo0271nQbmlAJ40rvpKiFTgP/view katahex_model_20220618.bin.gz]

'''GUIs:'''

* The 20240812 net is bundled with a modified LizzieYzy GUI: <https://github.com/hzyhhzy/KataGo/releases/tag/Hex_20240812>
* An older net with the modified LizzieYzy GUI: [https://drive.google.com/file/d/1qbTTmPFiUkM_346DeKS1E9gJR-roNH63/view KataHex_LizzieYZY_20220313.zip].
* Selinger's KataHex works with HexGUI: <https://github.com/selinger/hexgui>

== See also ==

* [[Strategic advice from KataHex]]

[[Category:Computer Hex]]
[[category:Hex playing program]]

KataHex

2024-10-31T00:15:55Z

Selinger: /* Links */ Provided a direct download link for the 20240812 model

'''KataHex''' is a free and open-source computer Hex program, capable of defeating top-level human players. It implements Monte Carlo tree search with a convolutional neural network providing position evaluation and policy guidance.

== History and versions ==

KataHex is based on [https://en.wikipedia.org/wiki/KataGo KataGo], a computer Go program developed by David Wu that was first released on 27 February 2019. It was adapted for Hex by "HZY" between February 2020 and May 2022. While initially unnamed, the Hex-adaptation of KataGo quickly became known as "KataHex" among Hex players.

The HZY implementation of KataHex speaks a non-standard dialect of [[GTP]] and can only interact with a specially modified version of the Go GUI known as LizzieYzy. A further adaptation of KataHex that is capable of interfacing with [[Hexgui]] was made by [[User:Selinger|Selinger]].

== Pre-trained networks ==

HZY initially trained the neural network model on two NVIDIA GeForce RTX 2080 Ti GPUs for about 20 days on 13x13. They then up-trained the network on 19x19 for one day, and on 27x27 for an additional day. [https://zhuanlan-zhihu-com.translate.goog/p/476464087?_x_tr_sl=auto&_x_tr_tl=en According to HZY], the up-trained 19x19 network is relatively reliable, but the 27x27-network is not.

The newest model, 20240812, is trained on two RTX 4090 GPUs for 2 months on 15x15, 15 days on 19x19, and 3 days on 27x27.

Running KataHex requires both a neural net model (the "weights"), and an engine to load the weights. Each neural net is able to play Hex not only at the size it was trained on, but also smaller and larger sizes, as long as the engine is compiled to support the size. (However, the net will not play particularly well on sizes larger than it was trained on.)

"KataHex" refers generally to the KataHex program, but often more specifically to the pre-trained 19x19 neural network made by HZY.

== Limitations ==

Since KataHex was only trained on self-play, it does not always do well when asked to play from an arbitrary starting position. Older nets are particularly thrown off if the starting position does not have the same number of black and white stones, but the issue seems to have been mitigated with the 20240812 net.

== Links ==

'''Program:'''

* David Wu's original KataGo: <https://github.com/lightvector/KataGo>
* HZY's KataHex: <https://github.com/hzyhhzy/KataGo/tree/Hex2022>
* Selinger's modifications: <https://github.com/selinger/katahex>

'''Pre-trained network models:'''

* 20240812 (current strongest net): [https://drive.usercontent.google.com/download?id=1YeqRvAYs7YjtPh0xBbDnHxeo2xrLEOdX hex27x3.bin.gz]
* 20220618: [https://drive.google.com/file/d/1xMvP_75xgo0271nQbmlAJ40rvpKiFTgP/view katahex_model_20220618.bin.gz]

'''GUIs:'''

* The 20240812 net is bundled with a modified LizzieYzy GUI: <https://github.com/hzyhhzy/KataGo/releases/tag/Hex_20240812>
* An older net with the modified LizzieYzy GUI: [https://drive.google.com/file/d/1qbTTmPFiUkM_346DeKS1E9gJR-roNH63/view KataHex_LizzieYZY_20220313.zip].
* Selinger's KataHex works with HexGUI: <https://github.com/selinger/hexgui>

'''Analysis:'''

* HZY's analysis of win-rates for opening moves: [https://zhuanlan.zhihu.com/p/476464087 Chinese] [https://zhuanlan-zhihu-com.translate.goog/p/476464087?_x_tr_sl=auto&_x_tr_tl=en English]
* [[Strategic advice from KataHex]]

[[Category:Computer Hex]]
[[category:Hex playing program]]

GTP

2024-10-24T16:14:19Z

Selinger: /* Backends */ Added KataHex

Hex GTP (or simply GTP) is a text-based interface for interacting with Hex software. A typical use is to separate front-end software, which provides graphical display of game boards, from back-end software, which implements a Hex strategy engine. The use of a standardized protocol in principle permits any strategy engine, such as [[MoHex]] or [[KataHex]], to be plugged into any graphical front-end, such as [[HexGui]].

Moreover, GTP can also be used by humans to interact with a strategy engine directly, i.e., without the use of a graphical front-end.

== History ==

Hex GTP is based on the Go Text Protocol (GTP), which was originally developed for Go as part of the [https://www.gnu.org/software/gnugo/gnugo_19.html GNU Go] software.

== Software supporting GTP ==

=== Frontends ===

* [[HexGui]]
* Todo: add more

=== Backends ===

* [[Six]]
* [[MoHex]]
* [[Wolve]]
* [[KataHex]]

== Protocol specification ==

The protocol follows a client-server model. It uses textual commands and responses expressed in the ASCII character set. Commands are sent by the client to the server, and responses are sent by the server to the client. When the server is a local process, it reads commands from its standard input stream, writes machine readable responses to its standard output stream, and may also write optional diagonstic and progress information to its standard error stream.

=== Command format ===

Commands consist of a single line of text, and are terminated by a newline character. Each command consists of an optional identity number, a keyword, and zero or more arguments, all separated by whitespace.

=== Response format ===

Responses consist of one or more lines of text, and terminated by two consecutive newline characters. A successful response starts with the character '=', followed immediately by the optional identity number of the command the response refers to, whitespace, and the rest of the response (whose format depends on the command in question).
A failure response starts with the character '?', followed immediately by the optional identity number of the command the response refers to, whitespace, and an error message.

=== Command arguments ===

Several of the commands take one or more arguments. Some common argument types are:

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Argument
! Possible values
|-
| ''n''
| An integer.
|-
| ''player''
| 'black' or 'white'.
|-
| ''move''
| A cell such as 'a7' or 'c4', or one of the special moves 'resign' or 'swap-pieces'.
|}

=== Command reference ===

There are a large number of commands, and the command 'list_commands' lists all of them. Some commands are server specific. The following lists some of the common commands:

==== Board handling ====

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Command
! Meaning
|-
| boardsize ''n''
| Set the board size to ''n'' × ''n'' and clear the board.
|-
| boardsize ''n'' ''m''
| Set the board size to ''n'' × ''m'' and clear the board.
|-
| clear_board
| Clear the current board.
|-
| showboard
| Display the current board in a format that is both human readable and machine readable.
|-
| final_score
| Determine the winner of a completed game. One player must have a solid chain connecting their two edges. The response is 'B+' for a black win, 'W+' for a white win, or 'cannot score' if there is no solid chain. Since most Hex games result in one player resigning before a solid chain is formed, the 'final_score' command is of limited use in practice.
|}

==== Making moves ====

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Command
! Meaning
|-
| play ''player'' ''move''
| Play the given ''move'' for the given ''player''.
|-
| genmove ''player''
| Automatically generate a good move for the given ''player'' and play it.
|-
| reg_genmove ''player''
| Automatically generate a good move for the given ''player'' and show it, but do not play it.
|-
| undo
| Undo the most recent move.
|-
| all_legal_moves
| Show the list of all legal moves for the current position.
|}

==== Interacting with the server ====

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Command
! Meaning
|-
| name
| Show the name of the server program.
|-
| version
| Show the version of the server program.
|-
| protocol_version
| Show the version of the GTP protocol that the server program understands.
|-
| list_commands
| Show a list of all commands known to the server program.
|-
| known_command ''cmd''
| Return 'true' if ''cmd'' is a known command, and 'false' otherwise.
|-
| quit
| Close the connection to the server.
|}

== Example session ==

boardsize 7 7
=

showboard
=

dcf37c659c5632f9
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . . . . . .\3
4\. . . . . . .\4
5\. . . . . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

play black c3
=

showboard
=

63d78dc21cfe93dc
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . . . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

play white d5
=

showboard
=

7edf0d0dfa16e804
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . W . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

genmove black
= b6

showboard
=

b3d2f7b4917a6c30
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . W . . .\5
6\. B . . . . .\6
7\. . . . . . .\7
a b c d e f g

undo
=

showboard
=

7edf0d0dfa16e804
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . W . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

quit
=

[[Category:Computer Hex]]

GTP

2024-10-24T16:13:55Z

Selinger: Replaced Six and Wolve by KataHex

Hex GTP (or simply GTP) is a text-based interface for interacting with Hex software. A typical use is to separate front-end software, which provides graphical display of game boards, from back-end software, which implements a Hex strategy engine. The use of a standardized protocol in principle permits any strategy engine, such as [[MoHex]] or [[KataHex]], to be plugged into any graphical front-end, such as [[HexGui]].

Moreover, GTP can also be used by humans to interact with a strategy engine directly, i.e., without the use of a graphical front-end.

== History ==

Hex GTP is based on the Go Text Protocol (GTP), which was originally developed for Go as part of the [https://www.gnu.org/software/gnugo/gnugo_19.html GNU Go] software.

== Software supporting GTP ==

=== Frontends ===

* [[HexGui]]
* Todo: add more

=== Backends ===

* [[Six]]
* [[MoHex]]
* [[Wolve]]
* Todo: add more

== Protocol specification ==

The protocol follows a client-server model. It uses textual commands and responses expressed in the ASCII character set. Commands are sent by the client to the server, and responses are sent by the server to the client. When the server is a local process, it reads commands from its standard input stream, writes machine readable responses to its standard output stream, and may also write optional diagonstic and progress information to its standard error stream.

=== Command format ===

Commands consist of a single line of text, and are terminated by a newline character. Each command consists of an optional identity number, a keyword, and zero or more arguments, all separated by whitespace.

=== Response format ===

Responses consist of one or more lines of text, and terminated by two consecutive newline characters. A successful response starts with the character '=', followed immediately by the optional identity number of the command the response refers to, whitespace, and the rest of the response (whose format depends on the command in question).
A failure response starts with the character '?', followed immediately by the optional identity number of the command the response refers to, whitespace, and an error message.

=== Command arguments ===

Several of the commands take one or more arguments. Some common argument types are:

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Argument
! Possible values
|-
| ''n''
| An integer.
|-
| ''player''
| 'black' or 'white'.
|-
| ''move''
| A cell such as 'a7' or 'c4', or one of the special moves 'resign' or 'swap-pieces'.
|}

=== Command reference ===

There are a large number of commands, and the command 'list_commands' lists all of them. Some commands are server specific. The following lists some of the common commands:

==== Board handling ====

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Command
! Meaning
|-
| boardsize ''n''
| Set the board size to ''n'' × ''n'' and clear the board.
|-
| boardsize ''n'' ''m''
| Set the board size to ''n'' × ''m'' and clear the board.
|-
| clear_board
| Clear the current board.
|-
| showboard
| Display the current board in a format that is both human readable and machine readable.
|-
| final_score
| Determine the winner of a completed game. One player must have a solid chain connecting their two edges. The response is 'B+' for a black win, 'W+' for a white win, or 'cannot score' if there is no solid chain. Since most Hex games result in one player resigning before a solid chain is formed, the 'final_score' command is of limited use in practice.
|}

==== Making moves ====

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Command
! Meaning
|-
| play ''player'' ''move''
| Play the given ''move'' for the given ''player''.
|-
| genmove ''player''
| Automatically generate a good move for the given ''player'' and play it.
|-
| reg_genmove ''player''
| Automatically generate a good move for the given ''player'' and show it, but do not play it.
|-
| undo
| Undo the most recent move.
|-
| all_legal_moves
| Show the list of all legal moves for the current position.
|}

==== Interacting with the server ====

{| class="wikitable" style="width: 100%"
! style="width: 150pt; vertical-align: top" | Command
! Meaning
|-
| name
| Show the name of the server program.
|-
| version
| Show the version of the server program.
|-
| protocol_version
| Show the version of the GTP protocol that the server program understands.
|-
| list_commands
| Show a list of all commands known to the server program.
|-
| known_command ''cmd''
| Return 'true' if ''cmd'' is a known command, and 'false' otherwise.
|-
| quit
| Close the connection to the server.
|}

== Example session ==

boardsize 7 7
=

showboard
=

dcf37c659c5632f9
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . . . . . .\3
4\. . . . . . .\4
5\. . . . . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

play black c3
=

showboard
=

63d78dc21cfe93dc
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . . . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

play white d5
=

showboard
=

7edf0d0dfa16e804
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . W . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

genmove black
= b6

showboard
=

b3d2f7b4917a6c30
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . W . . .\5
6\. B . . . . .\6
7\. . . . . . .\7
a b c d e f g

undo
=

showboard
=

7edf0d0dfa16e804
a b c d e f g
1\. . . . . . .\1
2\. . . . . . .\2
3\. . B . . . .\3
4\. . . . . . .\4
5\. . . W . . .\5
6\. . . . . . .\6
7\. . . . . . .\7
a b c d e f g

quit
=

[[Category:Computer Hex]]

HexGui

2024-10-20T14:16:41Z

Selinger: Added an explicit Links section

HexGui is an application by [[Broderick Arneson]] to display a Hex game board, keep track of a game tree (a sequence of moves with variations), load and save games, attach a Hex strategy engine, and debug it. HexGui is written in Java and works on any operating system. HexGui can connect to any Hex strategy engine that supports the [[GTP]] protocol.

== History ==

HexGui is a modified version of the open source application [http://gogui.sourceforge.net/ GoGui] written in Java by Markus Enzenberger.

== Availability ==

HexGui is no longer available from its original [http://webdocs.cs.ualberta.ca/~broderic/hex/ download site].
A more recent version is available from [https://github.com/selinger/hexgui GitHub].

== Other software with the same name ==

There was also another program named HexGui and derived from GoGui, which used to be available from [http://mgame99.mg.funpic.de/hex.php HexGui homepage]. However, that link is now broken.

== Links ==

* [https://github.com/selinger/hexgui HexGUI]

[[Category:Computer Hex]]
[[Category:Hex playing program]]

Theory of ladder escapes

2024-10-19T16:00:51Z

Selinger: /* Examples */ Removed the "strange" example, which wasn't minimal and illustrated no point.

The object of this page is to formalise precisely what it means for a pattern to be a ladder escape. To do this, we first formalise what it means to be a ladder.

Informally, a ladder escape (say, a 4th row ladder escape) is supposed to give the attacker a guarantee that their 4th row ladder will be able to connect to the edge, no matter how far away from the ladder escape the ladder starts. So strictly speaking, to check that a pattern is a 4th row ladder escape, we must check that the attacker can connect to the edge from an ''infinite set'' of positions. This raises the issue of how one can check in a finite time whether a given pattern is a 4th row ladder escape.

This issue is resolved on this page for 2nd, 3rd, 4th, and 5th row ladders. It might be possible to resolve it for 6th row ladders but this has not yet been done, partly because such ladders are of little practical use. For 7th row ladders we run into a new difficulty – Blue can simply ignore the ladder and play near the escape, because no appropriate 6th row edge template seems to be known which will connect an ignored 7th row ladder to the edge. This presents a theoretical obstruction which is currently unresolved. It may in theory be that there are no 7th row ladders at all.

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

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

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

== Algebraic notation ==

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

=== Open patterns ===

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

=== Addition ===

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

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

=== The shift operation ===

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

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

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

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

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

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

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

=== The reduce operation ===

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

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

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

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

== Second row ladders ==

=== Definition of ladder ===

There is no issue at all with defining a 2nd row ladder. Informally, a 2nd row ladder looks like this:
<hexboard size="2x8"
coords="hide"
edges="bottom"
contents="R a1 R b1 R c1 R d1 R 2:e1 R 4:f1 B a2 B b2 B c2 B 1:d2 B 3:e2"
/>
Note that at each point in the ladder, Blue's move is forced. Red can choose to continue pushing the ladder as long as she wants to. We formally define a second row ladder as follows:

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

=== Definition of ladder escape ===

Before we give the formal definition of a second row ladder escape, let us consider an example. The following pattern is an example of a second row ladder escape.
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="E *:a1 E *:b1 R c1 R d1 E *:a2 E +:a3 E *:d3 E +:a4 E *:d4"
/>
Of course directly underneath the pattern is the bottom (red) edge. The cells marked "+" indicate where the ladder connects. The reason this is a second row ladder escape is that however far away the ladder is,
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:e1 E *:f1 R g1 R h1 E *:a2 E *:b2 E *:c2 E *:d2 E *:e2 R 1:a3 E *:h3 E *:h4"
/>
Red can guarantee a connection from the ladder stone (marked 1) to the bottom edge.

Let us clarify what the hexes marked "+" in the ladder escape pattern mean. They indicate the last point where the 2nd row ladder is allowed to start. So for example, saying that the pattern above is a second row ladder escape means (among other things) that Red must win the following position:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="B a1 B b1 R c1 R d1 B a2 R 1:a3 B d3 B d4"
/>
Here, Red's ladder stone is marked "1", and the claim (easily verified) is that even with Blue to play, Red can connect the ladder stone to the bottom:
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 d3 d4"
contents="B a1 B b1 R c1 R d1 B a2 R 5:b2 R 1:a3 B 4:b3 R 3:c3 B d3 B 2:a4 B d4"
/>
The reason that the pattern is a second row ladder escape is that this escape sequence works even if the ladder is a long way away:
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 B h3 B h4"
/>
Even here, Red can force a connection to the edge, even if it is Blue's move, because Blue must keep defending on the first row and Red keeps attacking on the second row,
<hexboard size="4x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 d1 e1 f1 a2 b2 c2 d2 e2 h3 h4"
contents="B a1 B b1 B c1 B d1 B e1 B f1 R g1 R h1 B a2 B b2 B c2 B d2 B e2 R 1:a3 R 3:b3 R 5:c3 R 7:d3 R 9:e3 B h3 B 2:a4 B 4:b4 B 6:c4 B 6:d4 B h4"
/>
and now we are back at the previous position with the ladder right next to the escape, where we have already seen that Red can break through to the edge. We can now give a more formal definition of a second row ladder escape.

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

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

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

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

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

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

=== Characterization of ladder escapes ===

The definition of a 2nd row ladder escape allows the ladder to be an ''arbitrary'' distance away from the escape, which is of course what we want in practice; there is no reason that the escape should be right next to the ladder. However, this means that we cannot directly use the definition to check that something is a 2nd row ladder escape, because this would require checking that infinitely many patterns are virtual connections. Can we find some finite criterion for checking 2nd row ladder escapes? Fortunately, as every Hex player knows, the answer is yes. We have the following theorem:

'''Theorem 1.''' Consider a candidate P for a 2nd row ladder escape. Schematically:
<hexboard size="4x4"
coords="hide"
edges="bottom"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 E +:a3 E *:b3 E *:c3 E *:d3 E +:a4 E *:b4 E *:c4 E *:d4"
/>
(Here the asterisks indicate the [[carrier]] of P, which can contain any stones at all, and can be of any shape or size, as long as it includes no cells to the left of the cells marked "+"). Then P is a valid 2nd row ladder escape if and only if L2+P is a virtual connection for Red.
<hexboard size="4x4"
coords="hide"
edges="bottom"
contents="E *:a1 E *:b1 E *:c1 E *:d1 E *:a2 E *:b2 E *:c2 E *:d2 R a3 E *:b3 E *:c3 E *:d3 E *:b4 E *:c4 E *:d4"
/>

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

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

=== Examples ===

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

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

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

== Third row ladders ==

=== Definition of ladder ===

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

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

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

=== Definition of ladder escape ===

We have seen a lot of the formalism of ladder escapes in the above section on second row escapes. However there is a new twist with third row ladder escapes, because Blue can defend against a third row ladder in more than one way: Blue can at some stage decide to [[ladder handling|yield]] to the second row. The following definition is unsurprising.

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

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

In pictures, for the pattern
<hexboard size="3x3"
coords="hide"
edges="bottom"
contents="E +:a1 E *:b1 E *:c1 E +:a2 E *:b2 E *:c2 E +:a3 E *:b3 E *:c3"
/>
(where the carrier is schematically indicated by stars) to be a 3rd row ladder escape, it must give rise to a virtual connection when we attach a 3rd row ladder at distance 0,
<hexboard size="3x4"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="B a1 R 1:b1 E *:c1 E *:d1 B a2 E *:c2 E *:d2 E *:c3 E *:d3"
/>
or at distance 1,
<hexboard size="3x5"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="B a1 R b1 E *:d1 E *:e1 B a2 E *:d2 E *:e2 E *:d3 E *:e3"
/>
or at distance 6,
<hexboard size="3x10"
coords="hide"
edges="bottom"
visible="-a1 a2"
contents="B a1 R b1 E *:i1 E *:j1 B a2 E *:i2 E *:j2 E *:i3 E *:j3"
/>
or at any other distance.

=== Characterization of ladder escapes ===

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

=== Examples ===

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

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

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

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

The following third row ladder escapes also escape 2nd row ladders at distance 2 or greater (but not at distance 0 or 1):
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="R b1 E *:d1 E +:a2 E +:a3 E +:a4"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R c1 E *:d1"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-d1"
contents="E +:a2 E +:a3 E +:a4 R b1 R c1 E *:d1 S b2"
/>
<hexboard size="4x4"
coords="hide"
edges="bottom"
visible="-a1 area(b3,c2,d2,d4,b4)"
contents="R ↓:d1 E +:a2 +:a3 +:a4"
/>
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1 b1 a2 b2 d2 e2 a3 b3 d3 e3 a4 b4 d4 e4"
contents="R ↓:e1 E *:a2 E *:b2 E +:c2 E *:d2 E *:e2 E *:a3 E *:b3 E +:c3 E *:d3 E *:e3 E *:a4 E *:b4 E +:c4 E *:d4 E *:e4"
/>
<hexboard size="5x5"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 e2 e3 e4 e5"
contents="E +:a3 +:a4 +:a5 E *:a1 *:b1 *:c1 *:e2 *:e3 *:e4 *:e5 R ↓:e1 S d2"
/>
<hexboard size="5x8"
coords="hide"
edges="bottom"
visible="-a1 b1 c1 g1 h1 h2"
contents="E *:a1 E +:a3 E +:a4 E +:a5 E *:b1 E *:c1 R e1 E *:g1 E *:h1 E *:h2"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1"
contents="E +:(a3 a4 a5) R c1 d1"
/>
<hexboard size="5x4"
coords="hide"
edges="bottom"
visible="-a1 d1 d2"
contents="E +:(a3 a4 a5) R b1 c1"
/><hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 a2 b1"
contents="E *:a1 E *:a2 E +:a4 E +:a5 E +:a6 E *:b1 R c1 R d2"
/>
<hexboard size="6x4"
coords="hide"
edges="bottom"
visible="-a1 b1 a2"
contents="E *:a1 E *:b1 R c1 E *:a2 R b2 E +:a4 E +:a5 E +:a6"
/>
The version of this last pattern on David King's website has the cells marked "+" (he uses arrows) sloping in the other direction; the location that is shown here makes the template minimal.

== Fourth row ladders ==

=== Definition of ladder ===

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

The key part of the definition is that the 6 hexes forming a triangle below the ladder stone are all vacant. Note that even filling in one of these can invalidate the ladder: even if we fill in the bottom left corner of the triangle,
<hexboard size="5x5"
coords="hide"
edges="bottom"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B a4 B a5"
/>
then Blue has this move,
<hexboard size="5x5"
coords="hide"
edges="bottom"
contents="B a1 B b1 R c1 B d1 B e1 B a2 B b2 R c2 B a3 B b3 B 1:d3 B a4 B a5"
/>
which is easily seen to stop the ladder. To establish the ladder, Red needs at a minimum those 6 vacant hexes under her ladder stone.

=== Definition of ladder escape ===

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

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

The five moves marked 2 below all lose instantly to Red 3:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:d1 B 2:e1 E *:a2 E *:b2 B 2:e2 E *:a3 B 2:e3 B 2:e4 R 3:c2"
/>
The two moves marked 2 below also lose instantly:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 B 2:b3 B 2:a4 E *:a2 E *:b2 E *:a3 R 3:d2"
/>
The move marked 2 below can be answered by Red 3, moving us to position L4+''n''+P, which is a virtual connection by the induction hypothesis.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 R 3:d1 E *:a2 E *:b2 B 2:c2 E *:a3"
/>
Move 2 below leads us to a 2nd row ladder, which ↑+↑+P escapes, so 5 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 B 2:d2 E *:a3 R 5:c3 B 4:b4"
/>
Both moves marked 2 below lead us to a 3rd row ladder, which ↑+P escapes, so 3 (and therefore 1) is connected to the edge:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 E *:a3 B 2:c3 B 2:b4"
/>
Move 2 below also leads to a 3rd row ladder (note Blue 4 must be in the triangle left and below from Red 3; Blue can also play out the bridge between 1 and 3 but this doesn't help):
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:d2 R 5:e2 E *:a3 B 4:c3 B 2:d3"
/>
Move 2 below leads us to a 2nd row ladder:
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 B 4:b4 B 2:c4"
/>
The final choice for move 2 below also gives a 2nd row ladder.
<hexboard size="4x5"
coords="hide"
edges="bottom"
visible="-a1--a3 b1 b2"
contents="E *:a1 E *:b1 R 1:c1 E *:a2 E *:b2 R 3:c2 E *:a3 R 5:d3 R 7:e3 B 4:b4 B 6:c4 B 2:d4"
/>

This completes the induction, so P is a 4th row ladder escape. □

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

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

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

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

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

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

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

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

=== Examples ===

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

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

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

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

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

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

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

== Fifth row ladders ==

=== Definition of ladder ===

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

Unlike in the case of 2nd, 3rd, and 4th row ladders, this time it is not sufficient for a triangle of cells below and to the right of the ladder stone to be empty. We also need three additional empty cells to the left of this triangle. This is a minimal requirement; if even one of these cells is occupied by Blue, for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 R o4 B a5 R o5 B a6 R o6"
/>
then Blue can block the ladder with this move:
<hexboard size="6x15"
coords="hide"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 R o4 B a5 R o5 B a6 R o6 B 1:e5"
/>
The main line is complex; see for example [http://littlegolem.net/jsp/forum/topic2.jsp?forum=50&topic=669 this Little Golem discussion thread]. Many of the main lines of defense involve Blue playing an upside-down version of [[Tom's move]], for example like this:
<hexboard size="6x15"
coords="hide"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 B n1 B o1 R a2 R b2 R c2 R d2 R e2 R o2 B a3 B b3 B c3 B d3 R o3 B a4 B b4 R o4 B a5 R o5 B a6 R o6 B 1:e5 R 2:e4 B 3:e3 R 4:f2 B 5:f3 R 6:g2 B 7:h4 E *:d5 *:f4"
/>
Note that Blue's 1 is connected to 5 by double threat at "*", and 7 is Tom's move upside-down, i.e., with the top line of blue stones serving as the "edge". Therefore, to establish the ladder, Red needs at minimum the specified 13 vacant hexes under the ladder stone.

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

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

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

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

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

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

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

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

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

=== Examples ===

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

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

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

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

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

== Sixth row ladders and up ==

Because of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick, 6th and higher row ladders do not exist in the usual sense. More specifically, even if we allow an arbitrary amount of empty space under the ladder stone, it is not possible for the attacker to keep pushing the ladder. Consider the following situation:
<hexboard size="7x13"
coords="show"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2"
/>
Let's assume there is an arbitrary amount of empty space in the bottom 4 rows to the left of this diagram. The stone marked "1" is connected to the top, and looks like it could be the ladder stone for a potential 6th row ladder. If such a ladder were possible, the red stones on the M-file should certainly escape it.

From Blue's point of view, Blue is the attacker in an upside-down 2nd row ladder. Blue can therefore use an upside-down version of the [[Switchback#2nd-to-6th_row_switchback|2nd-to-6th row switchback]] trick. To do so, Blue plays at 2:
<hexboard size="7x13"
coords="show"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5"
/>
If both Red and Blue keep playing [[optimal play|optimally]], the best that Red can get is a pair of parallel 2nd and 4th row ladders in the opposite direction:
<hexboard size="7x13"
coords="show"
edges="bottom"
contents="R a1 B b1 B c1 B d1 B e1 B f1 B g1 B h1 B i1 B j1 B k1 B l1 B m1 R m2 R m3 R m4 R m5 R m6 R m7 R a2 B a3 R b2 B b3 R 1:c2 B 2:e5 R 3:d3 B 4:c5 R 5:c4 B 6:b5 R 7:e4 B 8:e3 R 9:d4 B 10:e6 R 11:d5 B 12:c7 R 13:c6 B 14:b7"
/>
Here, stone 10 is [[virtual connection|connected]] to the line of blue stones along the top, so Red has no way of connecting right. Red can now push a 4th row ladder from 5, and/or a 2nd row ladder from 13. There is not enough space for Red to immediately perform [[Tom's move]]. So unless Red has a ladder escape somewhere to the left of this diagram, or unless there's enough space on the 5th row somewhere to the left of this diagram to perform Tom's move, Red fails to connect to the edge.

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

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

== Second-to-fourth row switchbacks ==

=== Definition of switchback ===

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

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

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

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

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

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

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

=== Characterization of switchbacks ===

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

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

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

=== Examples ===

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

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

== Third-to-fifth row switchbacks ==

=== Definition of switchback ===

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

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

=== Characterization of switchbacks ===

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

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

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

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

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

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

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

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

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

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

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

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

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

=== Examples ===

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

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

== Second and fourth row parallel ladders ==

=== Definition of ladder ===

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

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

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

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

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

=== Examples ===

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

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

== Third and fifth row parallel ladders ==

=== Definition of ladder ===

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

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

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

=== Examples ===

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

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

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

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

== Second and fourth row terraced ladders ==

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

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

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

=== Definition of ladder ===

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

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

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

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

=== Definition of ladder escape ===

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

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

=== Characterization of ladder escapes ===

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Next, consider this pattern:

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

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

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

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

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

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

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

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

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

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

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

=== Non-examples ===

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

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

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

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

Handicap

2024-10-07T00:55:29Z

Selinger: Added progressive board size

Playing with '''handicap''' means to give one of the [[player]]s (preferably the weaker one) an advantage at the start of the game. The point of this is to make the game more even, so that it will be challenging for both players.

There are several ways a handicap could be implemented in Hex. One of the main issues is that the advantage given to one player should be predictable and somewhat quantifiable. A system that is reasonably well-motivated and seems to be gaining popularity is the Demer handicap system, described below. Some other proposals for handicapping systems are also discussed below.

== Demer handicap system ==

The Demer handicap system is based on the idea of giving the weaker player a certain number of free moves at the beginning of the game. By selectively using the swap rule, the handicap can be given in increments of 0.5 moves. This system was proposed by Eric Demer.

=== Measuring advantage by number of moves ===

The idea behind measuring the handicap in terms of fractions of moves is the following. Consider a game in which Red goes first and the swap rule is not used. Compare this to a game in which Blue goes first and the swap rule is not used. How much better is the first game for Red? The only difference between the two games is that Red plays one extra move at the beginning of the first game. We therefore say that Red has a 1 move advantage in the first game, compared to the second game. It is also clear that the second game is equally good for Blue as the first is for Red. Compared to a theoretically fair game, it therefore makes sense to say that a game without the swap rule gives exactly a 0.5 move advantage to the player who moves first.

If the swap rule is used, the game is very close to fair. (In theory, it gives a slight advantage to the second player, but this advantage is small and we will ignore it.) In summary, we now have a game that gives a 0.5 move advantage to Red (Red goes first without swap), and a game that gives a 0 move advantage (the swap rule is used). We can increase any player's advantage by 1 move by giving that player an extra move at the beginning. In games where the swap rule is used, the extra move should be given just after the swap decision has been made (since swapping when there are already two pieces on the board would give a large advantage to the second player).

We therefore arrive at the following handicap system.

=== Description of the Demer handicap system ===

For the purpose of the following description, we assume that the the [[Conventions#Swapping|swap-pieces convention]] is used, i.e., when a player swaps, the player keeps the same color, but the board position is mirrored. If the [[Conventions#Swapping|swap-sides convention]] is used instead, the method remains the same but the description must be adjusted accordingly.

* 0 move advantage for Red (no handicap): Red starts and the swap rule is used. Symbolically: (Red, Blue swaps, Red, Blue, ...) or (Red, Blue, Red, Blue, ...)

* 0.5 move advantage for Red: Red starts and the swap rule is not used. Symbolically: (Red, Blue, Red, Blue, ...)

* 1 move advantage for Red: Red gets one additional move before Blue's first non-swap move. Explicitly: Red plays the first piece, then Blue decides whether to swap or not. If Blue swaps, Red gets two consecutive moves. If Blue does not swap, Red gets one additional move. Symbolically: (Red, Blue swaps, Red, Red, Blue, ...) or (Red, Red, Blue, Red, Blue, ...)

* 1.5 move advantage for Red: Red plays the first two pieces and the swap rule is not used. Symbolically: (Red, Red, Blue, Red, ...)

* 2 move advantage for Red: Red gets two additional moves before Blue's first non-swap move. Symbolically: (Red, Blue swaps, Red, Red, Red, Blue, ...) or (Red, Red, Red, Blue, Red, Blue, ...)

For the integral handicaps, i.e., those where the swap rule is used, it is also possible to give the advantage to Blue. This can be done as follows:

* 0 move advantage for Blue (no handicap): Red starts and the swap rule is used. Symbolically: (Red, Blue swaps, Red, Blue, ...) or (Red, Blue, Red, Blue, ...)

* 1 move advantage for Blue: Blue gets one additional move before Red's second move. Explicitly: Red plays the first piece, then Blue decides whether to swap or not. If Blue swaps, Blue gets an additional move. If Blue does not swap, Blue gets two consecutive moves. Symbolically: (Red, Blue swaps, Blue, Red, ...) or (Red, Blue, Blue, Red, ...)

* 2 move advantage for Blue: Blue gets two additional moves before Red's second move. Symbolically: (Red, Blue swaps, Blue, Blue, Red, ...) or (Red, Blue, Blue, Blue, Red, ...)

The system can theoretically also be used for larger handicaps (2.5 moves, 3 moves, etc.), but such large handicaps probably do not make much sense on small board sizes. For example, on an 11 × 11 board, Red only needs 3 pieces to connect her edges by [[bridge]]s and [[edge template]]s.

=== Sign convention ===

By convention, a handicap that benefits Red is specified as a positive number, and a handicap that benefits Blue is specified as a negative number.

=== Relation of handicap to player strengths ===

One may ask what the appropriate handicap amount is, given two players' [[Elo rating]]s. There are currently no reliable statistics on this, as handicap games are rare (or even non-existent) on game servers where Elo-rated players play. A very ballpark estimate, based on limited anecdotal evidence, is that a 0.5 move handicap corresponds to a difference of about 250 Elo points on 11 × 11 boards. This means that a 0.5 move advantage increases the odds of winning by a factor of approximately 4. On larger boards, the effect of handicap moves is probably somewhat smaller.

=== Drawbacks ===

A drawback of the Demer handicap system is that even a 0.5 move handicap gives the player a relatively large advantage, especially on [[Small boards|smaller boards]].

=== Handicap 0.5 vs. no swap ===

Sometimes players choose to play without the swap rule, but without the intention to give an advantage to the weaker player. This typically happens because the players are novices and either don't know about the swap rule, or do not perceive the swap rule as making much difference at their level of play. Some game servers, such as [[PlayOK]], actually offer "no swap" as a game option, but then alternate player colors in subsequent games.

Although playing without the swap rule is equivalent to playing with handicap 0.5 in the Demer handicap system, such games should not be marked as handicap 0.5 in game records. Instead, the notation "no swap" or "N/S" can be used.

== Other suggestions for handicap systems ==

Various other methods for handicapping games have been suggested. The potential advantage of these methods is that they might be able to produce more fine-grained handicaps than the Demer system (i.e., handicaps in increments smaller than 0.5 moves). The disadvantage is that without large-scale testing, it would be difficult to quantify these handicaps, i.e., to figure out exactly how many fractional "moves" each handicap corresponds to.

=== Non-rhombic boards ===

A seemingly natural way to give an advantage to a player is to decrease the distance between the player's edges, i.e., to play on an ''m × n'' board where ''m'' is distinct from ''n''. Unfortunately, this doesn't work very well, since there exists an easy, explicit winning strategy for the player with the shorter distance. See [[Hex_theory#Winning_strategy_for_non-square_boards|winning strategy for non-square boards]].

However, the idea of a non-rhombic board can perhaps be combined with Demer handicaps to arrive at more fine-grained handicaps (i.e., handicaps of less than 0.5 moves). For example, it may make sense to give Red a 1.5 move advantage in exchange for slightly decreasing the distance between Blue's edges. But doing so would require careful calibration, and there is no obvious way to quantify the resulting advantage or disadvantage.

See [[parallelogram boards]] for an analysis of how much headstart the player with the longer distance needs, for various small non-rhombic board sizes.

=== Fixed openings ===

Another possible way to give more fine-grained handicaps is to play without the swap rule, but to place Red's first piece in a pre-defined position (rather than allowing Red to place the piece freely). A piece placed in the center of the board would give Red an advantage of 0.5 moves, whereas a fairly-placed piece (i.e., a piece that Blue would be equally likely to swap or not to swap) would give 0 moves of advantage. If the initial piece is placed somewhere between these two extremes, handicaps between 0 and 0.5 moves can be achieved, although it is difficult to quantify the exact advantage conferred by any particular opening move.

This method could also be adapted to give handicaps greater than 0.5. For very large handicaps, one could experiment by having a central red piece plus the first blue piece in a bad position as part of the setup (with Red to move), placing two red pieces near the edges and letting Blue go first, etc. By setting up a position beforehand and deciding who is to move, one can in principle create positions that are arbitrarily balanced towards one or the other player.

The price is that one gets a slightly different game, so that it's possible that a player might become especially good at certain common handicap positions. But this should be of no more concern than it is in Go (and much less than in Shogi or Chess). It might be worthwhile to work out a standard ladder of handicap positions, sorted according to their bias to Red.

Some evidence (from the win rate evaluations of AI bots) suggests that c3 and the 2-2 obtuse corner (shown below) give Red approximately a 0.25 move advantage, at least for board sizes 13×13 to 19×19.

<hexboard size="13x13"
coords="show"
contents="E *:(c3 b12 k11 l2)"
/>

=== First to win N games ===

When playing matches, rather than individual games, a possible handicap method is to play "First to win N games" to win the match, with different values of N for each player. The weaker player would be expected to win fewer games than the stronger player, to win the match. This method may make sense if the players are relatively similar, but not equal, in strength. For example, for players whose [[Elo rating]] differs by 100, the odds of winning are approximately 7 : 4, so it may make sense to play "you win the match if you win 7 games, but I win the match if I win 4 games".

(However, the specific numbers one should use for such a handicap are more complicated than that. For example, with 7 : 4 game-odds, "you win the match if you win 5 games, but I win the match if I win 3 games" both is faster and makes the match-odds closer to 1 : 1.)

== Progressive board size ==

For beginners, it is often difficult to win against experienced players even with a large handicap. [[User:Mason|Mason Mackaman]] suggested the following procedure to keep the game enjoyable: Play with handicap 0.5 (i.e., without swap), starting from the smallest board size (even 1x1, if it is available). Each time the beginner wins, increase the board size by 1. The players will typically move through the smallest board sizes very quickly, and then slow down somewhere near 6x6 or 7x7. This ensures that the beginner is always on the verge of winning, and can clearly measure their improvement. This method might be more fun than starting with a large handicap on 11x11.

[[category:Rules and Conventions]]