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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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2 Answers 2

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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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  • $\begingroup$ This argument is incorrect. If it were valid, it should work for all board sizes, but in 2x2 reverse Hex (or more generally for any even-sized board), it is the first player who wins. The problem is that "more hexes on the board is good" is not necessarily true. Specifically, when the only remaining move is a losing one, having one less hex on the board would actually be a good thing. $\endgroup$
    – Peter Selinger
    Jan 21, 2022 at 4:11
  • $\begingroup$ My 3rd paragraph addresses this, in reverse hex, having 1 fewer pieces on the board is always a good thing (if having $x$ pieces on the board means you haven't lost, then you necessarily haven't lost w/ $x-1$ pieces on the board). Since white moves first, black could steal a supposed winning strategy for white (playing its symmetric analog) with 1 fewer pieces on the board. $\endgroup$
    – Justin Benfield
    Jan 21, 2022 at 9:04
  • $\begingroup$ Sorry, you are quite right that I mangled things when trying to pinpoint the flaw in your argument. The wrong statement is in the third paragraph: "in reverse hex, fewer pieces is better". This is not true. For example, consider 2x2 Hex, with the accute corners occupied by Black and White, respectively. White has a winning move. However, with one less white piece on the board, White does not have a winning move.Therefore, fewer pieces is not always better in reverse Hex. The proof is in the pudding: you "proved" that 2x2 reverse Hex is a 2nd player win, whereas it is actually a 1st player win. $\endgroup$
    – Peter Selinger
    Jan 22, 2022 at 18:29
  • $\begingroup$ Ok, so there is a problem with my logic. However, your claim that white (1st player to move) has a winning strategy on 2x2 reverse hex is wrong. Black can always 'play parallel' to whites first move, blocking the only 2nd move for white that they could play that wouldn't lose them the game. The situation you describe above is never reached in optimal play, but it does undermine the logic of my argument for the general case. The problem being that having 1 fewer piece on the board may allow your opponent to make a move that doesn't lose them the game (as shown in your example game state). $\endgroup$
    – Justin Benfield
    Jan 23, 2022 at 17:09
  • $\begingroup$ Excellent, this is a question that can actually be resolved by duelling. I challenge you to a game of 2x2 Reverse Hex. I will go first. If I win, I was right. OK? I play at a1 (acute corner). What is your move? $\endgroup$
    – Peter Selinger
    Jan 24, 2022 at 18:23

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