As an aside, I hadn’t considered the nD problem when I first posted this. It seems to be a pretty straight-forward generalisation of the 1D case though. A bit disappointing 😦

]]>Actually, more generally, if there are equal numbers of squares on either side of the chocolate, make your friend start first. Then however many he breaks off, break off the same number from the opposite end. That way your friend always gets the bar with equal numbers of squares on each side of the poison, and eventually it will reach 0 and he will have to eat the poison.

If there is a different number of squares on one side than the other, start first and break off the difference on the longer side, so that your friend gets the bar with equal numbers of squares and it reduces to the case above.

Then find a new friend and repeat. Eventually you will kill all your time and all your friends.

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