Why is NP not co-NP?

This arises because of the asymmetry in definition of a non-deterministic Turing machine. An NDTM accepts a language L, if at least one of the computation paths results in a “yes”. It could so happen that there are other computation paths which result in a “no”.

Is co-NP harder than NP?

Being coNP hard just means that you can reduce in polynomial time from every other problem in coNP. So if a problem is NP-hard, its complement is coNP hard.

Is co-NP NP-hard?

In complexity theory, computational problems that are co-NP-complete are those that are the hardest problems in co-NP, in the sense that any problem in co-NP can be reformulated as a special case of any co-NP-complete problem with only polynomial overhead.

Why is P co-NP?

The complement of L is also in P, so the complement of L is therefore in NP (because P ⊆ NP). Therefore, L is in co-NP. Consequently, P ⊆ co-NP. Hope this helps!

What’s the difference between NP and co-NP complete?

Co-* (Co-NP, Co-NP-complete) focuses on answering “no” to the complemented decision problem. For example, the complemented decision problem of Set Cover would be “For every combination of X subsets, is it impossible to cover all elements?”. Answering “no” to this question requires you to provide a counter-example.

When is a decision problem a member of co-NP?

A decision problem X is a member of co-NP if and only if its complement X is in the complexity class NP. The class can be defined as follows: a decision problem is in co-NP precisely if only no-instances have a polynomial-length ” certificate ” and there is a polynomial-time algorithm that can be used to verify any purported certificate.

Is the complexity class co-NP an open problem?

No. It is another open problem and certainly related, but different. The complexity class co- NP is the set of languages whose complements are in NP; that is, the set of decision problems for which a “no” answer has a deterministic polynomial-time verifier.

How to list the prime factors in co-NP?

Membership in co-NP is also straightforward: one can just list the prime factors of m, all greater or equal to n, which the verifier can confirm to be valid by multiplication and the AKS primality test.