The initial question is:
Alan and Barbara play a game in which they take turns filling entries of an initially empty $2008\times 2008$ array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?
Answer: Barbara has a winning strategy by a basic parity argument. Label the rows $r_1,\ldots,r_{2008}$ and partition them into 'slices' of two, say $\{ \{r_1,r_2\},\{r_3,r_4\},\ldots,\{r_{2007},r_{2008}\}\}$. If Alan places an entry in some spot, Barbara places the same entry in the same column in that slice; for instance, if Alan plays $5$ in $a_{332,98}$, Barbara plays $5$ in $a_{331,98}$. This ensures the rows in each slice are equivalent, so the determinant of the matrix is zero.
- My question: How can this be modified if instead of a $2008\times 2008$ array, they play on a square array with odd sides ($3\times 3,5\times 5,$ etc., with the $1\times 1$ case discarded)?
I don't know of a winning strategy even for the $3\times 3$ case. I think Barbara's strategy breaks down because Alan isn't forced to complete a row in the same way as he was in the even case, so it isn't clear to me that Barbara can force one row to be a linear combination of another. I also think the strategy of, 'Barbara tries to get a column/row of zeroes,' is easily disrupted, so I don't see that as viable either.
