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What I've found is:

First Player If N Is Odd (non-misere): The player takes the center piece and symmetrically imitates every one of the opponent's moves.

Second Player If N Is Even (non-misere): Player copies opponent's moves symmetrically. You will eventually take the last piece and win.

But what should I do when N is even and I go first, or when I go second and N is odd?

My best guess is I should count amount of rectangles and alone "chips" and, based on that, make moves, but the problem is that I'm not sure that this will work and maybe you have a better strategy for this game.

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The game is solved. The "tactic" for playing the losing side is to hope the other player makes a mistake. With perfect play, there are no moves that are better than others for the losing side : they all lose.

If the winning-side player fails to play optimally, then your move should attempt to restore the position to one it could have reached if you had been the winning-side all along.

For instance, on a 3x3, if your opponent goes first and takes a side square instead of the center, then you should take the 2x1 region of the center and the opposite side square.

Note though that the winning-side player has other equally-winning moves beyond what the winning algorithm prescribes, it's just that the algorithm reduces it to a sufficient move. On a 3x3, for instance, the first player could take the whole center row and still be winning.

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