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To the right is the problem illustration from last month.
It shows a classical tic-tac-toe game after the first five moves.
Player X has a fork, and O cannot win.
The question is, can X set up a classical fork in Quantum Tic-Tac-Toe,
and does it work?
The first answer is that if the first two moves needed for the
fork are already collapsed to classical marks, then it is not
possible to produce the standard fork in just one more move without
a fortuitous external situation. For instance, if O had
one spooky mark in the fork square (square 5 in this example)
and then X played over her pair, one of the two collapses
would give the square to X and produce the fork, but the other would give it to O, blocking the fork before it happened;
so this case would require the cooperation of the opponent. However,
if the spooky mark was one of X's, then he could self-collapse,
but this requires an extra move and means that O can never
capitalize on this strategy; the game is over too soon for her
(the setup requires 4 moves all by itself).
The second answer is that it might be that X's first three moves are entangled with each other
and self-collapse directly to the fork situation. If O
tries to block, by playing in both squares of the fork (6 and 9),
then X wins on the next move by playing
there also; it doesn't matter which way O collapses the cyclic
entanglement (see the PDF file) . Her stronger move is to play somewhere else, so
that X requires a second move to complete the fork. X could get both 3-rows in this situation; but if O were able to engineer
her own 3-row she might still prevail. Note that this last
strategy requires 5 moves from the protagonist, so this strategy
is also unavailable to O when she is on the offensive.
X can usually thwart this last defensive strategy of O
however, by anticipating where she would have to set up her 3-row
and making sure that his 3-row is earlier in time (see
the PDF file). At least she eked out a half point.
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