Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
My understanding is that Ladner’s theorem is basically this:
P != NP implies that there exists a set NPI where NPI is not in P and
NPI is not NP-complete
What happens to this theorem if we assume that P = NP rather than P != NP? We know that if NP Intermediate doesn’t exist, then P = NP. But can NP Intermediate exist if P = NP?
NPI must imply that it is in NP, but that it is not NP-complete.
If P = NP, then all problems in P and NP will be NP-complete, because any problem will be reducible to another one in polynomial time (∅ and Σ* cannot be NP-complete, because we can’t map an arbitrary problem to either of them – we won’t have anything to map to for the positive/negative case. However, since they are in P, we don’t care about them for the purpose of this question.)
Since all problems in NP are NP-complete, NPI cannot exist.