Hex theory

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Unlike many other games, it is possible to say certain things about Hex, with absolute certainty. Whether this makes Hex a better game is of course debatable, but many find this attribute charming.

The most important properties of Hex are the following:

Winning Strategy

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

No winning strategy for Blue

In chess, while nobody seriously believes that black has a winning strategy, nobody has been able to disprove it. On the other hand, in Hex, a simple strategy-stealing argument shows that the second player cannot have a winning strategy, and therefore the first player must have one.

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

Winning strategy for non-square boards

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


Now whenever Red plays in the pink triangle, Blue responds by playing in the cell of the same number in the blue triangle, and vice versa. After filling all of the cells in this way, Red cannot have a winning path. To see why, assume, for the sake of contradiction, that Red has a winning path. Let the "top component" be the set of all cells in the pink triangle that are occupied by a red stone and connected, by a path entirely within the pink triangle, to the top edge. Let the "pink diagonal" be the set of pink cells that are adjacent to the blue triangle. It is easy to see that there exists some cell on the pink diagonal that belongs to the top component (for example, consider the first time Red's winning path, starting from the top edge, leaves the pink triangle). Let X be the bottom-most cell on the pink diagonal that belongs to the top component. Let Y be the cell immediately below and to the right of X. Note that the cells X and Y are labelled with the same number, so Y is occupied by a blue stone. Since X is connected to the top edge by a path of red stones entirely within the pink triangle, by symmetry, Y is connected to the right edge by a path of blue stones entirely within the blue triangle. We consider two cases. If X is the bottom-most cell of the pink triangle, then Y touches the left edge. This gives a winning path for Blue, contradicting the existence of a winning path for Red. Otherwise, let Z be the cell immediately below and to the left of X. Since X was the bottom-most cell of the top component on the pink diagonal, Z must be occupied by a blue stone, and must be connected, by a path entirely within the pink triangle, to the left edge (for example, such a path can be constructed by following along the perimeter of the top component). Since Y is connected to the right edge, Z is connected to the left edge, and Y and Z are adjacent to each other, this gives a winning path for Blue, contradicting the existence of a winning path for Red and finishing the proof.

To illustrate the proof, the cells X, Y, and Z are marked in the following diagram:


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

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

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

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

No draw

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

Links to more detailed proofs are on Javhar's page "Hex cannot end in a draw".


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

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

  • The detection of virtual connections is PSPACE-complete. Clearly, this problem is in PSPACE, since one can decide the validity of a virtual connection by exploring every possible sequence of moves within the given carrier set. The fact that it is PSPACE-hard follows from the PSPACE-hardness of detecting a winning position, since a board position is winning (for the second player) if and only if it gives a virtual connection between the player's two edges.
  • The detection of dominated cells is PSPACE-complete. More specifically, given a board size, a position on that board, a player to move, and two empty cells X and Y, the problem of deciding whether X dominates Y is PSPACE-complete. To see why it is in PSPACE, note that it is sufficient to check for each of X and Y whether it is a winning move for the player in question. X dominates Y if and only if X is a winning move or Y is not a winning move. For PSPACE-hardness, consider the following position on a board of size (n+2)×(n+2), where the cells marked "*" denote some arbitrary position of an n×n board:


    For Red, the move at b1 is clearly winning, whereas the move at c1 is winning if and only if Red has a first-player win in the game marked "*". Therefore, b1 dominates c1. Moreover, c1 dominates b1 if and only if Red has a first-player winning strategy for "*". Moreover, b1 strictly dominates c1 if and only if Red has no such strategy. Since answering the latter question on the n×n board is PSPACE-hard, it follows that both domination and strict domination are PSPACE-hard problems.

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

  • The problem of deciding whether a given cell is dead is in co-NP. Equivalently, the problem of deciding whether a given cell is alive is in NP. This is because an empty cell is alive if and only if it belongs to some minimal winning path, relative to the given board position. Given such a path, it is checkable in polynomial time whether it is winning and minimal. Björnsson et al. showed that recognizing alive nodes is NP-complete in the Shannon game, a generalization of Hex. It is not known whether recognizing alive cells in Hex is also NP-complete or whether it is easier.

Solving Hex

See also

Open problems

External links