# co-NP

(Redirected from CoNP)
 Unsolved problem in computer science: Is NP = co-NP ? (more unsolved problems in computer science)

In computational complexity theory, co-NP is a complexity class. A decision problem ${\displaystyle {\mathcal {X}}}$ is a member of co-NP if and only if its complement ${\displaystyle {\overline {\mathcal {X}}}}$ is in the complexity class NP. In simple terms, co-NP is the class of problems for which there is a polynomial-time algorithm that can verify no instances (sometimes called counterexamples) given the appropriate certificate. Equivalently, co-NP is the set of decision problems where the "no" instances can be accepted in polynomial time by a non-deterministic Turing machine.

An example of an NP-complete problem is the subset sum problem: given a finite set of integers, is there a non-empty subset that sums to zero? To give a proof of a "yes" instance, one must specify a non-empty subset that does sum to zero. The complementary problem is in co-NP and asks: "given a finite set of integers, does every non-empty subset have a non-zero sum?".

## Relationship to other classes

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 and not strict in the other). NP and co-NP are also thought to be unequal.[1] If so, then no NP-complete problem can be in co-NP and no co-NP-complete problem can be in NP.[2]

This can be shown as follows. Suppose there exists an NP-complete problem ${\displaystyle {\mathcal {X}}}$ that is in co-NP. Since all problems in NP can be reduced to ${\displaystyle {\mathcal {X}}}$, 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., NPco-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-NPNP. Thus co-NP = NP. The proof that no co-NP-complete problem can be in NP if NPco-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 that is known to belong to both NP and co-NP is Integer_factorization#Difficulty_and_complexity: 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 also straightforward: one can just list the prime factors of m, which the verifier can confirm to be valid by multiplication and the AKS primality test. It is presently not known whether there is a polynomial-time algorithm for factorization, equivalently that integer factorization is in P, and hence this example is interesting as one of the most natural problems known to be in NP and coNP but not known to be in P.[3]

## References

1. ^ Hopcroft, John E. (2000). Introduction to Automata Theory, Languages, and Computation (2nd Edition). Boston: Addison-Wesley. ISBN 0-201-44124-1. Chap. 11.
2. ^ Goldreich, Oded (2010). P, NP, and NP-completeness: The Basics of Computational Complexity. Cambridge University Press. p. 155. ISBN 9781139490092.
3. ^ http://mathoverflow.net/q/31821