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grad
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Three prisoners (request solution)
« on: Sep 5th, 2006, 5:12am » |
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I found a riddle and I want to ask if anyone knows the solution. I am not sure if it is correctly presented, so, there is a chance it has no solution at all...
Well, here it is:
Three prisoners have a number written on the back of their heads. There are no mirrors and the prisoners are not allowed to speak to each other or make gestures etc. They know that one number is the sum of the other two but they do not know which one of them. The first one who will find HIS number will be set free.
The first day no one speaks. The second day no one speaks. The third day one of them says: "I have the number 50!" and he is set free. How did he find it?
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towr
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Re: Three prisoners (request solution)
« Reply #1 on: Sep 5th, 2006, 5:52am » |
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It'd be easier if there was a limit on how large the numbers might be.
The clue is that people's silence can also convey information. It helps you to know that others don't have enough information to figure something out.
The actual solution will take a bit of puzzling though. And the lack of an upper limit to the number a prisoner might have on his head is troubling me. (Otherwise you could consider all possible sums, and eliminate possibilities)
One thing you know at the start of the second day, is that you don't have a 0, otherwise the other two would know their number (as it'd be identical to the other) and they would have said so at the end of the first day.
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grad
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Re: Three prisoners (request solution)
« Reply #2 on: Sep 5th, 2006, 6:11am » |
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Oh, I forgot to say that the prisoners have only one chance per day to say the right number or say nothing at all and each prisoner has only one chance to guess the number.
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grad
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Re: Three prisoners (request solution)
« Reply #3 on: Sep 5th, 2006, 6:13am » |
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on Sep 5th, 2006, 5:52am, towr wrote:It'd be easier if there was a limit on how large the numbers might be.
The clue is that people's silence can also convey information. It helps you to know that others don't have enough information to figure something out.
The actual solution will take a bit of puzzling though. And the lack of an upper limit to the number a prisoner might have on his head is troubling me. (Otherwise you could consider all possible sums, and eliminate possibilities)
One thing you know at the start of the second day, is that you don't have a 0, otherwise the other two would know their number (as it'd be identical to the other) and they would have said so at the end of the first day.
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The prisoner of course has an upper limit in his mind. It is the sum of the other two numbers.
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Astrix
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Re: Three prisoners (request solution)
« Reply #4 on: Sep 5th, 2006, 6:27am » |
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No one can have a 0, or else 2 numbers would be the sum of the other two. And that's the key to a possible day by day elimination.
So if you saw 10,10, you'd have to be 20, and you'd answer on day one.
If you saw 10,20, and no one answers on day one, you'd know you've got 30.
and if you saw 20,30, and no one answers on day two, you know you've got 50.
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grad
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Re: Three prisoners (request solution)
« Reply #5 on: Sep 5th, 2006, 6:33am » |
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Astrix I cannot understand your solution.
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Astrix
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Re: Three prisoners (request solution)
« Reply #6 on: Sep 5th, 2006, 6:47am » |
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A little more verbosely
The rules say One person has the total of the others. If anyone has 0, then Two people would have the total of the others. If I'm wrong about this, then that would adjust the answer by one day, but a day three answer would then have to be divisible by 3, and 50 is not.
So assuming I'm reading correctly,
If I see two 10's, I'm either 0 or 20. I've ruled out 0, so I must be 20. And I can answer on day one.
If I see 10 and 20, I'm either 10 or 30. But if I was 10, then the guy who has 20 would have seen 10,10 and he would have answered on day one. His silence on day one tells me on day two that I'm not 10, so I must be 30.
If I see 20 and 30, I'm either 10 or 50. If I was 10, the guy with 30 would have figured out his number on day 2. He didn't, so I must be 50. And I will give my answer on day three.
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towr
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Re: Three prisoners (request solution)
« Reply #7 on: Sep 5th, 2006, 6:50am » |
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on Sep 5th, 2006, 6:13am, grad wrote:| The prisoner of course has an upper limit in his mind. It is the sum of the other two numbers. |
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Often common knowledge plays a role, and you'd need to know what limits the other have in mind, and what they think you have in mind etc, which you can't know without knowing your number.
But Astrix's solution seems to solve it, without needing extra info.. So nevermind.
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| « Last Edit: Sep 5th, 2006, 6:59am by towr » |
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Astrix
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Re: Three prisoners (request solution)
« Reply #8 on: Sep 5th, 2006, 7:38am » |
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Here's a follow-up riddle that's a little more challenging.
One of the prisoners announces, 899. What is the earliest day he could have figured that out?
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grad
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Re: Three prisoners (request solution)
« Reply #9 on: Sep 5th, 2006, 7:59am » |
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Well, I think I got it now! From your solution I understand that the only possibility for the prisoner to solve the riddle is that his co-prisoners have the numbers 20 and 30. That means that if they had any other number the problem cannot be solved in 3 days, right?
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towr
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Re: Three prisoners (request solution)
« Reply #10 on: Sep 6th, 2006, 2:00am » |
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on Sep 5th, 2006, 7:38am, Astrix wrote:| One of the prisoners announces, 899. What is the earliest day he could have figured that out? |
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I'll go with the 4th.
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Deedlit
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Re: Three prisoners (request solution)
« Reply #11 on: Sep 6th, 2006, 3:43pm » |
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7 days.
How about a simple proof that the number announced is the sum of the other two numbers?
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Grimbal
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Re: Three prisoners (request solution)
« Reply #12 on: Sep 6th, 2006, 4:14pm » |
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on Sep 5th, 2006, 7:38am, Astrix wrote:Here's a follow-up riddle that's a little more challenging.
One of the prisoners announces, 899. What is the earliest day he could have figured that out? |
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I find day 7. With 5 possible configurations.
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towr
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Re: Three prisoners (request solution)
« Reply #13 on: Sep 7th, 2006, 1:13am » |
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oops, I had mistakenly used 889 (7*127)
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