Hex is usually played on a parallelogram shaped board. What if you play it on a Torus?

One thing I notice is that the idea of connecting opposite sides doesn't make much sense anymore, since a torus has no sides.

What you can do is assign the players target "loops", where two loops are considered equivalent if they can continuously deformed into each other. If two loops are different, they will intersect, unless one of the loops is contractible. That means that if the two players set of "victory" loops is disjoint, one player winning stops the other player from winning.

Has there been any research done on Torus-shaped-board Hex games?

  • $\begingroup$ I removed the "general topology" and "homotopy theory" tags as this question doesn't really fall under those fields. $\endgroup$ – j.c. Sep 26 '17 at 23:11
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    $\begingroup$ @j.c. not necessarily. You have to prove there are no ties first. Also, I included those tags since loop equivalence is usually studied in those fields, but if you don't think they fit, that's okay. $\endgroup$ – PyRulez Sep 26 '17 at 23:13
  • $\begingroup$ I agree that I was wrong about the strategy stealing argument, I edited my comment just as you were replying. $\endgroup$ – j.c. Sep 26 '17 at 23:15
  • $\begingroup$ There are infinitely many different homotopy classes of loops on the torus. I presume you're thinking about what could be called (1,0)- and (0,1)-loops, but what if a player achieves e.g. a (1,1)-loop? $\endgroup$ – Steven Stadnicki Sep 27 '17 at 0:03
  • $\begingroup$ @StevenStadnicki that's why I was talking about sets of loops. You basically need to partition the set of loops without the null loop. If you don't, and a player achieves a loop not in the sets, it will definitely be a tie. $\endgroup$ – PyRulez Sep 27 '17 at 0:05

If every homology class is a winning loop for one of the two players, and each player has at least one winning loop, then the game cannot end in a draw.

Proof: If one player has a particular loop, then the other player cannot have any of the other loops, so they either have no loops or the same loop. If both players have the same loops, one player wins. So either one player wins or one player has no loops. It suffices to show that when one player has no loops, the other player wins.

If that player has any closed loops that are homologous to zero, then they bound a disc, and we can remove all of the other player's hexes in that disc. So we can assume that player's hexes are actually a union of discs. Then the other player's hexes are a punctured torus, which maps surjectively on first homology to a torus, so the other player has every loop and wins.

Using strategy stealing & symmetry we can prove the existence of winning strategies in some cases:

If there is an automorphism of the board that sends all the second player's loops to some of the first player's loops, then the first player has a strategy to guarantee at least a draw. (Proof: If not, then the second player has a winning strategy, which the first player can steal.)

If there is an involution of the board with no fixed points that sends the first player's loops to some of the second-player's loops, the second player can guarantee at least a draw. (Proof: Each turn, apply that involution to the first player's move and then play it.)

If a draw can be ruled out, then these scenarios imply the existence of winning strategies.


Far as I can tell... Hex is a "Princeton" phenomenon. It originated in their Math department as invented by John Nash / Piet Hein. The game is not widely available (but these days with Amazon, relatively easy to find). And... there's not much activity now (in favor of of games like Scrabble or Checkers or Mancala).

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Hex strategy is discussed pretty thoroughly in the books by Cameron Browne:

He stopped revising is page in 2007.

As for research level math there was a series called Games of No Chance as published by MSRI

Also the Game Theory and the AI research and look very different. One strategy is to evaluate game positions in terms of surreal numbers (as Conway might have) another could be to read through the entire decision tree and estimate a probablity of winning. My suspicion is the second option is far less nuanced and yet we have done effectivly so with Chess and Go.

As for the Torus case, we could defining winning position for A if they complete the [A] homology cycle. We could define winning position for B if they win the [B] homology cycle. This almost reduces to classical Hex.

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Who wins if an [A+B] cycle is completed? There are also game theory textbooks one should consider:

Additionally, there was a random process called the Harmonic Explorer which looks oddly like a game of Hex and the winning paths converge to $SLE_4$

enter image description here


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