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I was teaching myself how to play a hex board game by reading some books. I learned how to do $2$ x $2$ and $3$ x $3$ hex games by starting at the principal diagonal. I wanted to know what the winning strategy would be for player one (white) in a $4$ x $4$ Hex game starting from the principal diagonal.

Consider a $4$ x $4$ Hex.

Show that White has a winning strategy, starting anywhere on the principal diagonal that is in any of the hexagons $1,6, 11,$ or $16$.

Here is what the setup would be:

Let the Hexagons be represented by numbers such as:

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

Let White have the first move. Let black have the second move.

White opens up at 6 (principal diagonal).

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what game are you talking about? And what do the dollar digns indicate in your question? – warren Jun 11 '14 at 19:37
@warren I believe that the OP is talking about this game of Hex. – ghoppe Jun 11 '14 at 22:32
@ghoppe - hope so ... I thought it was about general game-theory on a hex board (as tagged and worded) :) – warren Jun 11 '14 at 22:40
@warren Yes, I was similarly confused at first. The question is poorly worded. :) – ghoppe Jun 11 '14 at 22:47
The dollar signs are used elsewhere to make mathematical expressions display properly. I'm guessing they're not activated on this SE since they'd be rarely necessary. – ConMan Jun 12 '14 at 5:36

The winning strategy for such a small Hex board is shown in this basic strategy guide.

Like tic-tac-toe, on a 4x4 board white will always win by opening on the main diagonal, because for every counter that black can make, there is another way for white to force the win. Once white can form a "two-bridge" by placing the second piece in a non-adjacent space but with two empty neighbouring hexes in common, there is no way to block white from completing the chain, since there are two empty adjacent hexes to complete the link.

Here's an example position after white's second turn:

4x4 Hex Grid

As you could see by playing out all the scenarios for white along the main diagonal, it's impossible to stop white from forming a "two-bridge"

  • If white plays a1, there are two-bridges at b3 and c2. Black can't stop both.
  • If white plays b2, there are two-bridge plays at a3, c4, d3, and c1.
  • If white plays c3, it's a mirror position of b2.
  • If white plays d4, it's a mirror position of a1.

Black can't win because there is always a way for white to orphan a black piece on this size board, making it impossible for black to form a "two-bridge".

I'm not that familiar with the game but my understanding is that the strategy only really becomes deep and interesting once the board gets to a 10x10 size.

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