Quantum Tic-Tac-Toe

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This Month’s Challenge: November 2005

End Game Puzzle

Consider a Quantum Tic-Tac-Toe game where, at the end of move 4, a collapse has placed real marks in all four corner squares. Both X and O have a single open 2-row, but there are two possible cases: either they have 2-rows on opposite sides of the board, or they have 2-rows that share the center square.

In a classical game, from this position X has an easy win on move 5 in both cases. In a quantum game, can X win from either of these positions? Can he win by move 5? Can he prevent O from getting any points at all?

As a bonus problem, find the cyclic entanglement on move 4 that sets up these two situations (four real marks in the corners, both arrangements).


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Last Month’s Answer: October 2005
A Mid-Game Puzzle

Last month, you were given a game in progress where O has an opportunity to win, if she can find it. To the right is the quantum game and its classical ensemble, made from the move list given last month. Brute-force examination of the possibilities would be daunting; there are seven open squares, so O has 21 choices for move 6. Fortunately, she can use the ensemble to guide her choice.

X has open 2-rows in four of the six games in the ensemble, while O has open 2-rows in only two. It seems reasonable to eliminate as many of X's 2-rows as possible. O can eliminate four games where X has a 2-row, by placing move 6 in squares 5 and 8. X has no real choice about the resulting collapse, because one choice is an immediate win for O.

 

The result looks good for O: two games left in the ensemble, O has open 2-rows in both games, X has nothing in either game.

However, X can prevent O from winning in several ways. For instance, X can play move 7 in squares 3 and 6; O cannot move into 3 and 6 herself, because X will win. She wants to choose the next collapse, so her only choice is to move in squares 3 and 2. X's last move can be in squares 4 and 6, or 3 and 6; either way, it's a cat's game; X doesn't win, but neither does O.

 

O's strongest choice for move 6 does not look like a winning move, at first glance. However, if she places move 6 in squares 5 and 6, look at the classical ensemble: there are five games, and in all five O has an open 2-row, while X has a 2-row in only one game. Moreover, if O makes the correct choice at move 8, X can no longer win.

On move 7, X will either make a cycle, which will leave real marks in all squares but 2 and 3, or he will not make a cycle, and all squares but 7 and 9 are still available. If X makes a cycle, O must move in squares 2 and 3, and so must X, but O chooses the collapse and wins. If X does not make a cycle, O's correct move is in squares 3 and 8; this may or may not make a cycle, depending on where X had moved. If X gets to collapse a cycle after move 8, he does not have a choice that gives him a win. If there is no cycle to collapse after move 8, there will be one after move 9 – but O gets to choose the collapse and she wins.


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