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# Execution Table

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Posted 22 April 2013 - 03:58 PM

I believe this question is a classic but I am unable to find it in the forums so I will venture the wrath of the moderator gods by posting it anyways:

"Only the smartest one shall survive," said the executioner. "All of you prisoners will be seated around this round table. I will chop off the head of the prisoner in seat #1, skip seat #2, chop #3, etc. Beheaded bodies (and chairs) will be cleared away post haste. When I get to the end I will continue to chop, skip, chop, until there is but one prisoner remaining, who will then be freed."

a) If there are 13 prisoners, which is the lucky seat number?

b) In the morning, you will be told how many prisoners, n. You must figure out the best seat quickly, just knowing n. Which seat, k, should you 'head' for, in order to be freed?

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### #2 TimeSpaceLightForce

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Posted 22 April 2013 - 04:59 PM

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### #3 k-man

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Posted 22 April 2013 - 07:26 PM

Spoiler for looks like...

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### #4 plasmid

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Posted 23 April 2013 - 06:31 AM

I remember this one.

http://brainden.com/...p/topic/7688--/

Admittedly something that cannot really be found using a search... unless you know that that version had "sword" and "kill", and you're willing to look through a few pages of search results.

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### #5 bgm1961

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Posted 23 April 2013 - 02:31 PM

I agree with Time - the intuitive and easy part of the answer is that the smartest person will choose

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But calculating how to arrive at that answer vice using intuition will have to wait till after breakfast!

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### #6 bhramarraj

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Posted 23 April 2013 - 04:41 PM

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### #7 bhramarraj

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Posted 23 April 2013 - 04:59 PM

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Posted 23 April 2013 - 07:49 PM   Best Answer

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### #9 k-man

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Posted 23 April 2013 - 10:12 PM

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Respectfully disagree.

Let's test this solution for 5 prisoners. Execution order will be 1,3,5,4 with #2 surviving.

For 13 prisoners: 1,3,5,7,9,11,13,4,8,12,6,2 with #10 survivng.

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Posted 24 April 2013 - 12:32 AM

I think we are interpreting the OP differently. In your case, as I read your answer, the executioner never stops, cutting in a circle until they are done. The way I meant for the op to be interpreted is that after the first run of kills, he/she starts back over and kills the new first person, continue through and starting over once they come to the end with the 'next' first person.

so with 13 people:round 1,, 1,3,5,7,9,11,13...round 2: 2,6,10...round 3: 4, 12...8 lives

I like your solution too though.

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Respectfully disagree.

Let's test this solution for 5 prisoners. Execution order will be 1,3,5,4 with #2 surviving.
For 13 prisoners: 1,3,5,7,9,11,13,4,8,12,6,2 with #10 survivng.

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