Monday, September 08, 2003


Okay then, here's the solution to the puzzle I set on Friday:

For the purposes of the solution there are n prisoners in total, X is the prisoner who can set them all free and Y is the group of all the other prisoners.

Statistically, the longer prisoner X waits before making his pronouncement, the more likely he is to be correct but he cannot be absolutely certain about getting it right by relying on the time elapsed alone; it's possible that one of the prisoners has been very unlucky and not been exercised. So, how can he be sure about what he's saying?

Because the prisoners are all transported on the same bus and are aware of what will happen while they are in the prison, they collectively decide to do the following:

Y prisoners are only allowed to move the lever from A to B and they are each only allowed to move it once. Prisoner X only moves the lever from B to A and can move it as often as he likes.

1. Whenever a Y prisoner is exercised he checks the lever. If it is in position A and he has not moved the lever before then he moves it to B. Otherwise he leaves it alone.
2. Whenever X is exercised he checks the lever and if it is in position B he moves the lever back to A. Otherwise he does nothing.

If X keeps a tally of how many times he has moved the lever, once he has moved it n-1 times he can say with absolute confidence that everyone has been out in the exercise yard at least once.

For large numbers of n, this process could take a very long time, with most of the prisoners having been exercised many times, but it is the best way for X to be completely certain in his timing. After all, if he gets it wrong, they'll be incarcerated forever.

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