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)?
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
According to D, $P=NP=NPC$
3SAT is $NPC$. Can 3SAT be reduced to $\phi$ ? (which is in P)
The answer is no. Hence, C is the correct choice.
D is the correct choice iff we exclude $\phi$ and $\Sigma^*$ from $P$
Answer is option C.
No problem can be reduced to either $\emptyset $ or $\Sigma^*$.
NPC should be a subset of P=NP.
If P=NP=NPC then that means that all P and NP problems are NP Complete and thus NP Hard. This means that every P and NP problem is polynomial time reducible to every other problem. This is not true.
Hence C.
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