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The CLIQUE problem-- problem of finding the maximum clique in a graph is NP-complete. That is, CLIQUE is

a.) in NP and b.) there is an NP complete problem, 3-SAT for one, that reduces to CLIQUE in polynomial time.

Part (b) above is fine-- all over in every resource and very well explained. For Part (a), from what i know, we need to have the following: Given a specific solution instance, we need to show that it can be verified, in polynomial time, that that solution is an answer to this problem. So for instance, given a specific graph and a subgraph of it, we should be able to check whether that subgraph is a clique of maximum size in that graph.

The resources I've read so far are phrasing this Part (a) here as "easy, straightforward, etc" or "it can be shown in O(n^2) time that the given subgraph is a clique/not". However, the verification here is not just whether it's a clique, but also is whether it is a maximum clique in the graph. How can this be decided in polynomial time?

What am i missing here?

Roam
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1 Answers1

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You are confusing the optimization version of the problem with the decision version of the problem.

The decision version of clique asks if the graph has a clique of size k. Given a candidate solution, you can test its feasibility in polynomial time. Focus on the decision versions for NP-completeness proofs.

Certificates of optimality for the optimization problems are indeed harder to come by.

Tom Swifty
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  • clique enumeration is NP-hard-- no polynomial time decision on that one. the Q here is-- WHAT IS THE CERTIFICATE OF THE PROBLEM< HOW DOES ONE DETERMINE THAT. whats the difference between the certificates, if any, of CLIQUE, HALF-CLIQUE, MAXIMUM-CLIQUE-CONTAINING-A-SPECIFIC MEMBER. these are stuff you see errors on in literature. i dont need the distinction between decision & optimization. – Roam Oct 05 '13 at 19:35