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An engineer has n supposedly identical integrated-circuit chips that in principle are capable of testing each other. The engineer test jig accommodates two chips at a time. When the jig is loaded, each chip tests the other and reports whether it is good or bad. A good chip always reports accurately whether the other chip is good or bad, but the engineer cannot trust the answer of a bad chip. Assume that thenumber of good chips is greater than the number of bad chips. Thenanswer the following question:

Is it possible to design an algorithm that finds all the good chips after at most O(n log n) pairwise tests?

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  • $\begingroup$ Is it possible to design an answer to the question of whether this is a school assignment? $\endgroup$ – Mark L. Stone May 30 at 20:50
  • $\begingroup$ I know two solutions one in O (n ^ 2) and one in O (n) but I would like to know if a solution is possible in O (n log n). $\endgroup$ – neider May 30 at 21:22
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    $\begingroup$ Hmm, isn't O(n) better than O(n logn), leaving constants aside? $\endgroup$ – Mark L. Stone May 30 at 22:15
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    $\begingroup$ @MarkL.Stone is correct. $O(n)$ is better. Even more, if we look at $O$ notation (and not $\Theta$ - tight notation), $O(n)$ would qualify for an $O(n\log n)$. $\endgroup$ – Anton Menshov♦ May 30 at 22:36
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The answer is yes. It is possible. What have you tried so far? Have you implemented any code to test different algorithms?

I also suspect this is homework, but I will give you some ways to think about this problem. What would be the -worst- way to test the chips? If you check any chip with every other one you would have n² checks to perform. Is there any way you could use the additional information, that there are more correct than incorrect readings, to improve this?

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  • $\begingroup$ The only way that I think possible is an algorithm with a recurrence T (n) = 2T (n / 2) + cn. But how to split the set in two ensuring that one part has more good chips than bad ones? $\endgroup$ – neider Jun 2 at 19:38
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    $\begingroup$ You are thinking of a divide-and-conquer algorithm. Maybe you don't even need that. Lets say you pick up one chip, and test all the other ones with this one (that takes N cheks). you sort them into two stacks wether they agree with the one that you are testing with and not.. $\endgroup$ – MPIchael Jun 4 at 9:01

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