In Determinant Tic-Tac-Toe, Player 1 enters a 1 in an empty 3 × 3 matrix. Player 0 counters with a 0
in a vacant position, and play continues in turn until the 3 × 3 matrix is completed with five 1’s and
four 0’s. Player 0 wins if the determinant is 0 and player 1 wins otherwise.
(a) If Player 1 goes first and enters a 1 in the middle square, is there a strategy that can give player 2 a guaranteed win?
Note: I have seen a similar question at https://mathoverflow.net/questions/312034/matrix-tic-tac-toe, however this is based on the first number entered in the top left.
Note: I have seen a similar question and solution at http://math.ucr.edu/~muralee/p4sols.pdf but I'm not quite sure how the proof provided extends to Player 1 starting in the middle.
Perhaps the above could be used as a starting point?
A website I have been using to visualise this is http://textbooks.math.gatech.edu/ila/demos/tictactoe/tictactoe.html (default set to 2 x 2 and Player 0 first but this can be changed).
Thanks!
Best Answer
In the first place, it doesn't matter where ONE starts; the nine squares are all equivalent. (Permuting rows and columns can only change the sign of the determinant, not whether its zero or nonzero.) Personally, for neatness, I'd start with a $1$ in a corner, but it's your question so I'll do it your way.
Anyway, the game is a win for player ZERO; he can force not only the determinant but even the permanent of the $3\times3$ matrix to be zero.
Notation. Let me write $a_{i,j}=1$ to mean that player ONE write a $1$ in the $(i,j)$-square of the matix, $a_{i,j}=0$ to mean that player ZERO writes a $0$ in that square.
The game starts with the move $a_{2,2}=1$. I claim that ZERO can win by replying with $a_{1,1}=0$. Now, because of symmetry, there are only four choices for ONE's next move: $a_{2,1}=1,a_{3,1}=1,\ a_{3,2}=1,\ a_{3,3}=1$.
First variation. $a_{2,2}=1,\ a_{1,1}=0,\ a_{2,1}=1,\ a_{1,3}=0,\ a_{1,2}=1,\ a_{3,3}=0$
with a double threat of $a_{2,3}=0$ and $a_{3,1}=0$.
Second variation. $a_{2,2}=1,\ a_{1,1}=0,\ a_{3,1}=1,\ a_{1,3}=0,\ a_{1,2}=1,\ a_{2,3}=0$
threatening $a_{2,1}=0$ and $a_{3,3}=0$.
Third variation. $a_{2,2}=1,\ a_{1,1}=0,\ a_{3,2}=1,\ a_{1,3}=0,\ a_{1,2}=1,\ a_{2,1}=0$
threatening $a_{2,3}=0$ and $a_{3,1}=0$.
Fourth variation. $a_{2,2}=1,\ a_{1,1}=0,\ a_{3,3}=1,\ a_{1,2}=0,\ a_{1,3}=1,\ a_{3,1}=0$
threatening $a_{2,1}=0$ and $a_{3,2}=0$.