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Topic: Prisoners Challenge Once Again (Read 1688 times) |
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Barukh
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Prisoners Challenge Once Again
« on: Jul 8th, 2007, 11:24pm » |
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Sorry if this was posted before...
As usual, there are 100 prisoners who are going to be challenged by the management of the jail.
On a Sunday morning, every prisoner will be assigned a number from 1 to 100, which will be written at his forehead. Any number may be assigned to an arbitrary number of prisoners (repeatitions allowed!)
As usual, every prisoner will see the numbers of all his mates, but won't see his own number. They then guess their own numbers simultaneously.
If at least one prisoner guesses his number right, they are all released immediately. Otherwise, they will need to wait another year for another challenge.
On the evening before, they are allowed to gather together and device a strategy.
What's the best strategy they can come with and what are their chances to get free?
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towr
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Re: Prisoners Challenge Once Again
« Reply #1 on: Jul 9th, 2007, 1:26am » |
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Can they hear each other's guesses? Cause then, if they don't get out the first time, there's certainly a strategy that gets them out the second time.
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Barukh
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Re: Prisoners Challenge Once Again
« Reply #2 on: Jul 9th, 2007, 2:03am » |
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on Jul 9th, 2007, 1:26am, towr wrote:| Can they hear each other's guesses? |
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Are you referring to the next year's challenge? It's not relevant, since the task will be completely different.
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Grimbal
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Re: Prisoners Challenge Once Again
« Reply #3 on: Jul 9th, 2007, 5:15am » |
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I think they can guarantee a release on the first try.
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Barukh
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Re: Prisoners Challenge Once Again
« Reply #4 on: Jul 9th, 2007, 5:16am » |
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on Jul 9th, 2007, 5:15am, Grimbal wrote:| I think they can guarantee a release on the first try. |
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Good guess! But how?!
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SMQ
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Re: Prisoners Challenge Once Again
« Reply #5 on: Jul 9th, 2007, 5:32am » |
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We saw this before somewhere, although I'm not finding it at the moment... each prisoner assumes the sum of all assigned numbers will be a different 0 <= x <= 99 mod 100, sums the other 99 prisoner's numbers, and guesses their own number accordingly. Exactly one of them is guaranteed to be correct.
[edit]Aha, found it.[/edit]
--SMQ
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| « Last Edit: Jul 9th, 2007, 5:34am by SMQ » |
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--SMQ
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Grimbal
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Re: Prisoners Challenge Once Again
« Reply #6 on: Jul 9th, 2007, 5:32am » |
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Well, I don't want to spoil the fun for others, but
think modulo 100
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| « Last Edit: Jul 9th, 2007, 5:32am by Grimbal » |
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Barukh
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Re: Prisoners Challenge Once Again
« Reply #7 on: Jul 9th, 2007, 8:12am » |
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on Jul 9th, 2007, 5:32am, SMQ wrote:
Yes, that's it. Somehow I completely overlooked this thread.
The solution is really beautiful, isn't it?
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