Home > Co-NP

P is a subset of both NP and co-NP. That subset is thought to be strict in both cases. 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. One example is integer factorization, the problem of finding the prime factors of a number. It is in both NP and co-NP, but is generally suspected to be outside P, outside NP-complete, and outside co-NP-complete.