I think everyone knows the combinatorial games that are of the form:
Anne and Bob are playing a game on an $m \times n$ board. They alternatively chose a piece made out of $x$ $1 \times 1$ squares joined edge to edge. The loser is the one that cannot place a piece on the board. Knowing that Anne begins, does she have a winning strategy?
I was courious if there exists a general result for choices of $m,n,x$.
I searched the internet and found that when $x=2$ the game is called Cram, the strategy for even-even and even-odd boards being trivially known, while for odd-odd boards only in specific cases with smaller values of $m,n$, where the Sprague–Grundy value can be calculated.
I'm not sure if there are other generalizations of this :(.