Assuming P != NP, which of the following is true ?

(A) NP-complete = NP

(B) NP-complete P =

(C) NP-hard = NP

(D) P = NP-complete

**(A)** A

**(B)** B

**(C)** C

**(D)** D

**Answer:** **(B)** **Explanation:** The answer is B (no NP-Complete problem can be solved in polynomial time). Because, if 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.

Quiz of this Question

## Recommended Posts:

- Algorithms | NP Complete | Question 2
- Algorithms | NP Complete | Question 3
- Algorithms | NP Complete | Question 4
- Algorithms | NP Complete | Question 5
- Algorithms | NP Complete | Question 6
- Proof that Hamiltonian Path is NP-Complete
- Proof that vertex cover is NP complete
- Difference between NP hard and NP complete problem
- Proof that Hamiltonian Cycle is NP-Complete
- Proof that Independent Set in Graph theory is NP Complete
- Proof that Clique Decision problem is NP-Complete | Set 2
- Proof that Dominant Set of a Graph is NP-Complete
- Proof that Subgraph Isomorphism problem is NP-Complete
- Proof that Clique Decision problem is NP-Complete
- 3-coloring is NP Complete
- Set partition is NP complete
- Hitting Set problem is NP Complete
- Set cover is NP Complete
- Optimized Longest Path is NP Complete
- Subset Equality is NP Complete