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Suppose a polynomial time algorithm is discovered that correctly computes the largest clique in a given graph. In this scenario, which one of the following represents the correct Venn diagram of the complexity classes P, NP and NP Complete (NPC)?

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http://cs.stackexchange.com/q/7453/15154

Shouldn't the answer be (C)  instead? Considering the nature of $\Sigma^*$ and $\emptyset$ which belong to class P but no other problem can be reduced to them.

Clique is in NPC. By definition of NPC, all NP problems can be reduced to Clique in polynomial time. So, if clique can be solved in polynomial time, any NP problem can also be solved in polynomial time making P=NP and hence P=NP=NPC.

http://gatecse.in/wiki/NP,_NP_Complete,_NP_Hard

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Sir, why not option C?
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If D is true, C can't be true :) IT says NPC is a subset of NP=P, which is not possible as if 1 NPC problem is reduced to a P problem, then all NPC problem can also be reduced to P.
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"a polynomial time algorithm is discovered that correctly computes the largest clique in a given graph.".this means this NPC problem is polynomial solvable hence all NP are polynomial solvable since all NP reduces to NPC so NP=P.. But Sir, does it mean this NPC problem(or P problem since NP=P) can be reduced to some another P problem.. Solving in Polynomial time  means it is reducible to another P problem?
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yes, and that is the "a polynomial time algorithm" from the beginning.
if there is a polynomial time algorithm is  discovered for clique , then there is polynomial time algo for all NP complete problems.

hence option D is correct

No problem can be reduced to either $\emptyset$ or $\Sigma^*$.

http://cs.stackexchange.com/q/7453/15154

edited