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One quantum fork is shown in the pdf file
where X has set up
a fork down the main diagonal (squares 1,5,9) and down the first column (squares 1,4,7).
O has three possible responses: she can try to defend by playing both of her spooky marks
within those five squares, she can ignore X's threat by playing both spooky marks in
the other four squares (2,3,6,8), or she can place one spooky mark in each set.
If she chooses to defend, she has ten possible moves: four where one spooky mark is in the
common square (square 1) and six where she does not play in square 1 at all. For every one
of them, X also has ten possible responses within the same five squares, so there are
100 permutations. Let's explore them a bit.
There are six moves for O where neither spooky mark is in the common square; in all six, X wins on the next move
(move 7) as long as he stays within those five squares (1, 4, 5, 7 and 9). In two of the six possibilities
O makes a cyclic entanglement that X gets to collapse. But X makes another cycle on move
7, and even though O gets to collapse that one, it does her no good.
For some permutations, regardless of O's choice,
X actually gets his 3-row by move 5 of the classical game.
One can think of this from the point of view of the classical game
as if O played very badly and in trying to avoid the fork,
missed a force that occurred prior to it.
There are four variations where O plays one spooky mark in the common square;
in two of them she makes a cyclic entanglement, which X gets to collapse.
He must force her out of the common square or he may lose. Having done that, X
can make another cycle on move 7 and O gets to collapse it.
If there was not a collapse on move 6, then X must play very carefully so that
O's spooky mark is not on the cyclic part of the cyclic entanglement.
He can always do this. However, since she placed one spooky mark in the common square,
he cannot achieve a win by move 5 of the game unless O collapses the move 7 cyclic
entanglement foolishly.
If O ignores his threat, X cannot get a collapse by move 7
since his moves form two entanglements, and it takes one move to merge them
and another to make a cycle. However, O cannot get a cycle by move 8 without
entangling with him, so she cannot prevent move 9 from happening by getting her own 3-row.
If he now makes the move shown in
this pdf file,
she cannot stop him from getting a 3-row.
As before she has 10 possible moves, 4 with a spooky mark in the common square,
and 6 with none, but now her move creates a cyclic entanglement so X gets to
choose the collapse, so no way will she be able to block both 3-rows. She can, however,
get a three row by move 8, earning half a point if she plays in squares 1 and 9.
However, if neither spooky mark is in the common square, then X always has
a collapse that lets him win on move 5.
Finally, if she plays one spooky mark in one of his 5 squares, and the other in one of the remaining 4 squares, there are 20 permutations and X has 10 responses for each. We believe that this situation works out essentially the same as for the case above, but given the 200 possibilities, it has not been exhaustively analyzed. Any takers?
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