Quantum Tic-Tac-Toe

Can X win, if O echoes his every move?

If an entanglement involve n moves, there are almost n² ways to turn it into a cycle [it is n(n-1), to be exact]. When you increase the size of an entanglement, you give your opponent that many more ways to turn it into a cycle. If there is a 7 move entanglement on the 7th move of the game, O has 21 choices of how to make it cyclic, 14 choices of how to extend it, and only one choice of how to make a new independent entanglement. If O makes a cyclic entanglement, X has only two ways to collapse it, and no choice at all on his next quantum move. Furthermore, until there is a collapse somewhere on the board, every player has 36 possible quantum moves (9*8/2). This keeps the branching ratio of the game high, increasing the computational resources each player has to supply in order to reason their way to a good move.

In short, it is a reasonable strategy to consider collapsing early and often. The logical extreme of this strategy is for O to make a cycle on each of X's moves: if X plays in squares 1 and 2, O plays in 2 and 1; if he plays in 5 and 9, she plays in 9 and 5, and so on. If she does this, can X win?

Be sure to fill in your name and email address! We want to give credit to the person that sends the best solution.

Our congratulations to Peter Schueller, who sent us O's next move, and a brief analysis of X's possible responses, on 29 March.

O might be able to defeat X, by defending in the past. Where move 1 is, has not been resolved yet in the classical game. If she can force move 1 to be in square 1 instead of square 9, she has a chance to get a three-row by move 6 in the classical game; X cannot get a three-row until move 7 at the earliest. Before reading further, see if you can now deduce O's move given the above as a hint. The solution is shown in this pdf file.

She must place her next move in squares 6 and 9. Now X is faced with a choice: he can insist on getting a three-row, and lose, or he can give it up and not lose. If X plays in squares 6 and 9 himself, trying for that three-row, he must let O choose how the resulting cycle collapses, and she will choose to win by a half point. If he plays in squares 1 and 9, again O gets to choose the collapse, and wins. Only if X moves in squares 1 and 6, does he have a chance. That move forces O to choose whether they both get a three-row (but his is earlier), or neither do. The final result is a cat's game.