If played randomly, is there an advantage to either player? Is there an obvious strategy to play? Is there an advantage to picking a parity or going first? Can you find final configurations for which the "Odd" player has 8,7,6,5, and 4 sums?
Source: Original. Sorry it doesn't have an obvious "puzzle answer," but I thought I'd share it anyway.
As player 1, if odd, place a 0 in the center; if even, place a 1 in the
center. If player 2 places "k", player 1 places "1-k" in the opposite
location, making the 4 sums through the center equal the desired parity, a
guaranteed tie. Player 2 can only tie if all four side sums are the
same parity. But two of the sums are "a+b+c" and "3-(a+b+c)". Since those
are opposite parity (and similarly for the other two sides), player 2 will
lose 6 to 2.
Joe DeVincentis sent these configurations with various numbers of odd rows:
8 rows 111 111 111 |
7 rows 100 001 010 |
6 rows 111 110 111 |
5 rows 111 111 110 |
4 rows 111 101 111 |