The right answer is to eat the square to the right of the poisoned square on the first turn.You lose the game but you eat all the chocolate. Who's the loser now ?
I was about to give up, backwards induction wasn't helpful but there is a "mirror" strategy. First eat only 1 block up in right corner. If opponent choose to eat from upper edge, you eat equally many in lower edge (leaving a "hole" in the top right slot of remains) and if he eat lower one (can only eat exactly 1) you eat exactly 1 to continue that "hole" up in the right corner. That will continue until there is either 2x2 block left, a 1x2 block vertically stacked or a Yx1 block horizontally stacked, in later 2 cases you just eat everything except the poisoned. In the 2x2 block you eat the one in top right corner, they eat either of the 2 possible and then you eat the 2nd last leaving the poisoned one.
Is this solved for nxn sized chocolate? I am looking for a semester long project to wrap up my undergrad math degree. Was thinking this might be cool to study
So I have a question? How does this strategy scale if there are more than two rows say a chocolate bar that has thirty pieces arranged in a 5x6 rectangle?
This strategy cannot generalise. For larger rectangles, the strategy quickly gets more complicated. We only know that the first player has a winning strategy for rectangles of any size, but can't find an explicit strategy that generalises.
