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Play on a classic: computers


Best Answer harey, 10 May 2014 - 08:27 PM

Spoiler for Surprisingly, the answer does not depend so much on the total number of computers, but mainly on the number of suspicious computers. And even more surprisingly, there is no "worst case" neither "best case". An even more general solution working for k suspicious computer and at least (k+1) good computers:
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#11 LVan Toren

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Posted 20 March 2014 - 11:08 PM

Spoiler for spoiler


Edited by bonanova, 21 March 2014 - 06:24 AM.
spoiler

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#12 harey

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Posted 10 May 2014 - 08:27 PM   Best Answer

Spoiler for Surprisingly, the answer does not depend so much on the total number of computers, but mainly on the number of suspicious computers. And even more surprisingly, there is no "worst case" neither "best case". An even more general solution working for k suspicious computer and at least (k+1) good computers:

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#13 m00li

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Posted 24 May 2014 - 09:44 AM

Spoiler for Surprisingly, the answer does not depend so much on the total number of computers, but mainly on the number of suspicious computers. And even more surprisingly, there is no "worst case" neither "best case". An even more general solution working for k suspicious computer and at least (k+1) good computers:

 

this answer assumes that we know the number of bad computers, but we don't. what is the actual answer?


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