On optimal Play in the Game of Hex

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On optimal Play in the Game of Hex

• By Chris Senkler and Frank Bergmann

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Topics

• Short Reminder
• What is Hex?
• History of Hex
• Optimal Play in the Game of Hex
• An Automatic Theorem Proving Approach to Game
  Programming

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What is Hex?

• Rules
• The game starts with an empty board of hexagonal
  tiles. Each player owns 2 opposite sides of
  the board. Players alternate turns placing
  pieces on the tiles, one player taking white
  pieces and the other taking black.
• The first player to create a string to touching
  tiles of their color connecting their two sides
  wins.
• It is also known that the Game of Hex cannot end
  in a draw.
• Nash proved in 1949 that player I always has a
  winning strategy for symmetric boards.

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History Of Hex

• The game was first presented in 1959 in
  Scientific American.
• Invented in 1942 by Piet Hein.
• Independently rediscovered by Nash in 1948.
• Important to Mathematics mostly because of its
  relation to the Brower Fixed Point Theorem.

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Optimal Play in the Game of Hex
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Introduction

• This part of the presentation will concentrate on
  two points
• What is the shortest path with which player one
  can guarantee a win?
• What is the minimal number of moves player one
  must make to guarantee a win?

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The Shortest Path

• Let Hex(n,l) be a Hex game of size n with the
  added constrain that player one wins only by
  constructing a path of length less then or equal
  to l.
• In such a game a draw is possible.
• Define ?(n) min l player I has a winning
  strategy for Hex(n,l)
• Instead of determining ?(n) exactly, determine a
  lower bound for ?(n).

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Lower bound for ?(n)

• If ?(n) gt n, then ?(n1) gt n1
• Inductive proof
• All Hex(n1,n1) winning paths for player one
  must either
• Contain a Hex(n,n) winning path for player one
  through the standard n x n sub-board or
• Contain both 1,n and 1,(n1)
• Assume that ?(n) gt n, this means that player two
  can prevent a winning Hex(n,n) path for player
  one on the n x n sub-board.
• Whenever I plays on the n x n sub-board II
  follows his strategy, otherwise he can play on
  1,n or 1,(n1).
• ? Player I cant have a winning path in
  Hex(n1,n1)
• For the initial case consider ?(4) gt 4

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Proving ?(4) gt 4Categories

• There are up to symmetry four categories of
  distinct first moves.
• In order to show ?(4) gt 4 each opening move will
  be considered along with continuing paths.

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Proving ?(4) gt 4Category I
Player I Blue Player II Red

• Blue cannot prevent a connection between the top
  red edge and the bottom red edge
• ? blue cannot win Hex(4,4) with an opening move
  from category I

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Proving ?(4) gt 4Category II
Player I Blue Player II Red

• If blue begins with an opening move in category
  II, red immediately occupies the corner point and
  thus eliminated the possibility of any winning
  path of length 4.
• ? blue cannot win Hex(4,4) with an opening move
  from category II

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Proving ?(4) gt 4Category III
Player I Blue Player II Red

• Blue cannot prevent a connection between the top
  red edge
• ? blue cannot win Hex(4,4) with an opening move
  from category III

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Proving ?(4) gt 4Category III
Player I Blue Player II Red

• Blue cannot prevent a connection between the top
  red edge and the bottom red edge
• ? blue cannot win Hex(4,4) with an opening move
  from category III

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Proving ?(4) gt 4Category IV
Player I Blue Player II Red

• Blue cannot reach the opposite border with a path
  of length 4
• ? blue cannot win Hex(4,4) with an opening move
  from category IV
• ? There exists no opening move that allows player
  one to construct a winning path with length 4 or
  less.

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Minimal number of Counters

• Let Hexd(n) be the game of Hex(n) with the added
  constraint that player one must win by playing at
  most d counters.
• Define d(n) min d player one can guarantee
  a win at Hexd(n)
• Again only a lower bound will be established. (A
  trivial lower bound would be d(n) ?(n))

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Lower bound for d(n)

• d(n) n ?n/4?
• Proof
• Divide the n x n board into nx4 sub-boards Hk and
  one n x (n-4?n/4? ) sub-board H.
• In order to win Hex(n) player one must have a
  winning path across H and each Hk.
• Player one can complete a path across H with
  (n-4?n/4? ) counters but for each k, player one
  needs at least 5 counters.

• ? Player one needs at least
• 5?n/4? (n-4?n/4? ) n ?n/4?
• counters to complete a winning path in Hex(n)

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Summary

• A lower bound for the shortest path length and
  the minimal number of counters could be
  established for Hex games of size n
• Shortest path length ?(n) gt n for all n 4
• Minimal number of counters d(n) n ?n/4?

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The Game of Hex

• An Automatic Theorem Proving Approach to Game
  Programming

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Connections
Two pieces are connected if they are of the same
color and share an edge. These 2 white
pieces are connected. The black pieces are not.
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Virtual Connections
These two white pieces are virtually connected
because even if it is blacks turn White can
connect. If black A, white B, and visa versa
A
B
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The Problem

• Computers evaluate a game tree, which is hard for
  a game like chess, where there are an average of
  about 40 legal moves from a given game position.
  Hex has an average of over 100 on most sizes of
  board, which makes a complete analysis
  impossible.
• Without a concrete way of evaluating a position,
  a computer has to play out every branch to the
  end before knowing if it was a good move or not.

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Common Virtual Connections
Two Bridge connects 2 stones
Edge Connection from fourth row
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Ladder connection
Connects either stone to the bottom edge.
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Why Virtual connections are Good
The computer can reduce the game tree because it
no longer has to consider moves at the X points,
because the responses to these are wrote.
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Circuits

• In order to solve the other problem mentioned
  earlier we need a concrete way to evaluate game
  positions in which one player has not yet won.
  To do this we put the game in terms of something
  that the computer can understand. We equate it
  to electrical resistance in a circuit.

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We apply opposite charges to the sides of the
board. We Evaluate the position for Black first,
and then for white. For black We assign the
value rb(c)1 if c is empty, rb(c)0 if c has a
black piece, and rb(c)infinity if c is occupied
by a white piece. For each pair of neighboring
cells we associate the electrical link with the
resistance rb(c1,c2)rb(c1)rb(c2). Virtual
Connections are treated as having resistance 0 if
they are yours, and infinity if they belong to
the other player.
X
Y
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We repeat this operation for white, using similar
algorithms. The overall position is now given by
the formula ERb/Rw. If E ever reaches 0, black
has a clear path to victory, and if E ever goes
to infinity white has a clear path to victory.
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Using this algorithm, the computer can

• Evaluate the current game position
• Evaluate the relative value of different moves
• Stop an evaluation of a branch of the game tree
  once the equation E reaches 0 or infinity.

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Summary

• In order to make it possible for a computer to
  play intelligently in hex it is necessary to
  reduce the branches of the game tree that are
  analyzed and to evaluate the position of a game
  even if one player has not yet won. This can be
  done by the use of virtual connections and by
  treating the board as an electrical circuit whose
  positional value is measured as the electrical
  resistance.

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References

• 1 Garikai Campbell, on optimal play in the game
  of Hex, INTEGERS Electronic Journal of
  Combinatorial Number Theory, vol. 4, G2, 2004
• 2 Vadim V. Anshelevich, The Game of Hex An
  Automatic Theorem Proving Approach to Game
  Programming, Proceedings of the Seventeenth
  National Conference on Artificial Intelligence
  (AAAI-2000), pp.189-194, 2000