## Passing time with Toblerone

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You and a friend have time to kill and lots and lots of Toblerone (thanks Santa!) so you have decided to pass the time by playing a lethal game involving chocolate.

The game involves one bar of Toblerone chocolate and some poison. Your friend will paint a single square of chocolate in a Toblerone bar with poison. Because the poison is bright green you will both know which square has been poisoned. For example:

The point of the game is to not eat the poisoned square of chocolate. This means that your friend will have to eat it, but c’est la vie, right?

Each turn you must break off 1 or more squares from one end (either end) of the bar, eat them then pass what’s left to your friend so that he can play his turn. If you are passed the poison square then you have to eat it and you lose. And you die. You can choose to go first or to make your friend go first. What do you choose and how do you break the chocolate so that you don’t die?

Coming up: generalisations to 2D and nD blocks of chocolate…

If the poison is at one end, break off all the other chocolate and you win. If there is exactly one square on either side of the poison, make your friend start first. If there is one square between the poison and the end, break off all but one square on the other side (so there is one square left on either side of the poison), and you win. If there are exactly two squares on either side of the poison, make your friend start first.

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.

qwandorJanuary 5, 02014 at 22:48

Nice, I like the generalisation to n friends as well.

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 😦

Christo FogelbergJanuary 8, 02014 at 18:41

My brain hurts.

Claire FlemingJanuary 22, 02014 at 23:39