How to find the winning strategy in this variation of chomp?
We have an $m \times n$ board where $m \ge 3$ and $n \ge 3$, this time the top left corner is the poisonous piece and the player that picks a square eats all of the chocolate pieces to the left and down of the chosen piece.
I know what that the winning strategy in chomp uses symmetry but I can't figure out how to find the winning strategy using symmetry in this variation. I think I am misinterpreting the question because if player one always chooses the square that is one below the top right corner on his first move then player one will always win. Unless in this version the square chocolates diagonally from the chosen square are not eaten?
Best Answer
Just pick the rightmost piece in the second row. Then all that's left is the top row, so you win.
(If you just want to know which player wins, the same strategy-stealing argument as in ordinary Chomp tells you the first player wins. If picking the bottom left square is a winning move, the first player does that. Otherwise, the second player has a winning move after the first player picks the bottom left square, and the first player can instead use whatever that winning move is.)