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Answer:

Answer is B, see explanations.

Explanation:

Answer is (B) NP−complete∩P=ϕ

Since, P≠NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say X. Since, P≠NP,X∉P .

Now, by definition, NP−complete problems are the hardest problems in NP and so X problem is in NP−complete. And being in NP, X can be reduced to all problems in NP−complete, making any other NP−complete problem as hard as X. So, since X∉P, none of the other NP−complete problems also cannot be in P.

MrRoyal

Answer:

B. NP-complete \cap P = \Phi

Explanation:

The is because no NP-Complete problem can be solved in polynomial time.

Assuming one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

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