Can a problem be both NP and co-NP?

Can a problem be both NP and co-NP?

There are some problems in both NP and co-NP, for example all problems in P or integer factorization. However, it is not known if the sets are equal, although inequality is thought more likely. See co-NP and NP-complete for more details.

Which of the following problems are in co-NP?

So co-NP contains problems such as non-3-colourability, Boolean unsatisfiability and so on.

How do you prove something is co-NP?

A problem is in NP if every instance with answer “yes” you has a proof that can be checked on polynomial time. A problem is in co-NP if every instance with answer “no” has a proof that can be checked in polynomial time.

Is factoring in co-NP?

FACTORING ∈ NP ∩ coNP. Proof. The problem is in NP because the prime factor p is a polynomial-length certificate for YES instances: Using the polynomial-time algorithm for PRIMES, we can verify in polynomial-time that p is a prime.

Is sat in CoNP?

The real answer is, nobody knows that SAT isn’t in CoNP. If we knew that, we would know that CoNP=NP, but this is still an open problem. So, nobody can point to a specific reason that you can’t build a machine that decides the complement of SAT it nondeterministic polynomial time.

Is NP same as NP?

Since all problems in NP can be reduced to X, it follows that for every problem in NP, we can construct a non-deterministic Turing machine that decides its complement in polynomial time; i.e., NP ⊆ co-NP. Thus co-NP = NP. The proof that no co-NP-complete problem can be in NP if NP ≠ co-NP is symmetrical.

Is CoNP harder than NP?

3 Answers. A language L is NP-hard if for every language R in NP there exists a function f computable in polynomial time such that for all x, x∈R iff f(x)∈L. A language L is coNP-hard if for every language R in coNP there exists a function f computable in polynomial time such that for all x, x∈R iff f(x)∈L.

Is NP equal to P?

The “P versus NP problem” asks whether these two classes are actually identical; that is, whether every NP problem is also a P problem. Practical experience overwhelmingly suggests that P does not equal NP. But until someone provides a sound mathematical proof, the validity of the assumption remains open to question.

Is sat in co-NP?

What is a co-NP problem?

That is, co-NP is the set of decision problems where there exists a polynomial p(n) and a polynomial-time bounded Turing machine M such that for every instance x, x is a no-instance if and only if: for some possible certificate c of length bounded by p(n), the Turing machine M accepts the pair (x, c). …

Why is sat not in P?

1 Answer. It’s actually not known whether the complement of SAT is in NP. If P = NP, then since all P languages are closed under complementation, then the complement of SAT must be in NP (since it’s in P).

Is 2 sat a NP?

#2SAT is the problem of counting the number of satisfying assignments to a given 2-CNF formula. This counting problem is #P-complete, which implies that it is not solvable in polynomial time unless P = NP.