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In computational complexity theory, co-NP is a complexity class. A problem $mathcal\{X\}$ is a member of co-NP if and only if its complement $overline\{mathcal\{X\}\}$ is in complexity class NP. In simple terms, co-NP is the class of problems for which efficiently verifiable proofs of no instances, sometimes called counterexamples, exist.## References

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An example of an NP-complete problem is the subset sum problem: given a finite set of integers is there a non-empty subset which sums to zero? The complementary problem is in co-NP and asks: "given a finite set of integers, does every non-empty subset have a nonzero sum?" To give a proof of a "no" instance one must specify a non-empty subset which does sum to zero. This proof is then easy to verify.

P, the class of polynomial time solvable problems, is a subset of both NP and co-NP. P is thought to be a strict subset in both cases (and demonstrably cannot be strict in one case but not the other). NP and co-NP are also thought to be unequal. If so, then no NP-complete problem can be in co-NP and no co-NP-complete problem can be in NP.

This can be shown as follows. Assume that there is an NP-complete problem that is in co-NP. Since all problems in NP can be reduced to this problem it follows that for all problems in NP we can construct a non-deterministic Turing machine that decides the complement of the problem in polynomial time, i.e., NP is a subset of co-NP. From this it follows that the set of complements of the problems in NP is a subset of the set of complements of the problems in co-NP, i.e., co-NP is a subset of NP. Since we already knew that NP is a subset of co-NP it follows that they are the same. The proof for the fact that no co-NP-complete problem can be in NP is symmetrical.

If a problem can be shown to be in both NP and co-NP, that is generally accepted as strong evidence that the problem is probably not NP-complete (since otherwise NP = co-NP).

An example of a problem which is known to be in NP and in co-NP is integer factorization: given positive integers m and n determine if m has a factor less than n and greater than one. Membership in NP is clear; if m does have such a factor then the factor itself is a certificate. Membership in co-NP is more subtle; one must list the prime factors of m and provide a primality certificate for each one.

Integer factorization is often confused with the closely related primality problem. Both primality testing and factorization have long been known to be NP and co-NP problems. The AKS primality test, published in 2002, proves that primality testing also lies in P, while factorization may or may not have a polynomial-time algorithm.

- Complexity Zoo: coNP

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Last updated on Wednesday June 04, 2008 at 11:43:13 PDT (GMT -0700)

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Last updated on Wednesday June 04, 2008 at 11:43:13 PDT (GMT -0700)

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