# Reverse Hex board game winning strategy

I just wanted to know the winning strategy to this question:

In a reverse Hex board game I know it means where the player who first forms a path between his/her edges loses. Find a winning strategy for Black in a $3$ x $3$ reverse Hex.

Here White (player one) moves up and down and Black (player two) moves left to right.

The Hex would look like:

    1 2 3

4 5 6

7 8 9


where 3 is the uppermost corner and 7 is the lower most corner. It sliding down to the left (northeast to southwest).

I just started to learn how to play this game. I wanted to find a convincing strategy for Black. I was telling my friend that Black has a winning strategy because if he does not play in the middle he has a chance of winning. If White does play in the middle then Black can play opposite it. It seemed like a good strategy at the time. Can someone please help me to provide a convincing argument to see the winning strategy for Black?

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Assuming you are black (the second player) and you play from left to right, so you must avoid creating a left-right connection. I'll borrow your cell numbering.

In your first move you take either 3 or 7. This will always succeed and by symmetry we may assume you have taken 7.

In your second move you take either 3, 4 or 6. Your opponent only has taken two squares, so again, this will always succeed. We handle the cases separately.

Case 3: now your third move is irrelevant, because whatever you take there will be at most one way to finish a left-right connection at your fourth move. Since you have two choices at your fourth move, you can avoid to make that connection. (e.g. if your third move takes 8, then 9 is the only possible way to finish the connection).

Case 4: your third move takes one of $\{1,2,3,6\}$. This will always succeed because your opponent has taken at most 3 cells. Again, whichever one of these four you take, there is at most one possibility to finish a connection at your fourth move and you can avoid it.

Case 6: your third move takes one of $\{2,3,4,9\}$. Same argument.

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The same argument used to show that white has a winning strategy in regular hex can be used to show that black has a winning strategy in reverse hex. In regular hex, the absence of a possible draw together with a strategy-stealing argument that appeals to the symmetric nature of the game shows white can always force a win. Applying the same logic to reverse hex shows that black can force a win.

The key reason is in the argument by contradiction: in regular hex, you assume black has winning strategy (for the purpose of contradiction), and then observe the white could play the analogous symmetric strategy 1 move ahead (and since more hex's on the board is good here) that means white would have a winning strategy (contradiction: we assumed black had winning strategy) hence we conclude black does not have a winning strategy, and since there are no ties, white must have such a strategy.

In reverse hex, we assume white has a winning strategy (for contradiction), and then observe the black could steal that strategy with 1 fewer pieces on the board, and since in reverse hex, fewer pieces is better this would give black a winning strategy. Hence by analogous logic as in regular hex, we conclude black must have a winning strategy in reverse hex.

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