2 ms·
Cperciva is right, but I think you're mistaken when you say "many problems harder than NP are NP complete", NP complete problems are both NP and NP hard. NP har
by sonoffett 16y ago
Cperciva is right, but I think you're mistaken when you say "many problems harder than NP are NP complete", NP complete problems are both NP and NP hard. NP hard is a class of problems such that every problem in NP can be reduced to any problem in NP hard. Meaning if P=NP then every problem in NP(including the NP complete problems) can be solved in polynomial time; all it takes is one algorithm that solves an NP complete problem in polynomial time (note that this does not show that P=NP hard, a much stronger claim).
- bdr 16y agoYou're right. I've got the brain hiccups.