Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.