2013年6月11日 · From Wikipedia on Co-NP: A decision problem X is a member of co-NP if and only if its complement X is in the complexity class NP So I think, yes, the same verifier could be used to solve both problems. If you change NP to Co-NP and take the complement of the problem, it is essentially the same problem. I think Co-NP is useful to express the ...

2013年10月8日 · A language L is in co-NP iff its complement is in NP. So consider any language L ∈ P. The complement of L is also in P, so the complement of L is therefore in NP (because P ⊆ NP). Therefore, L is in co-NP. Consequently, P ⊆ co-NP. Hope this helps!

The polynomial hierarchy is a hierarchy of complexity classes that generalizes the classes ${\sf P}$, ${\sf NP}$ and ${\sf co-NP}$ to oracle machines and use as a scale to measure complexity of problems. It is known that If ${\sf NP}={\sf co-NP}$ or ${\sf P}={\sf NP}$ then the polynomial hierarchy collapses to its first level.

2020年4月18日 · Unsurprisingly, co-NP is exactly the set of languages accepted by some co-nondeterministic Turing machine, and the fact that NP=co-NP is open means exactly that it is not currently known whether nondeterministic and co-nondeterministic machines can compute the same things in polynomial time.

2015年12月19日 · It's because P = Co-P: if we can find a solution for a problem in Polynomial time, we can also decide in Polynomial time that it does NOT have a solution. Now, since P ⊆ NP we have that the complement of this problem is in Co-NP. But the complement of any problem in P is in Co-P and therefore any complement of any P problem is in Co-NP.

Of course if you take a problem in NP and just switch the answer from "YES" to "NO" and vice versa, you get a problem in co-NP, but that is rather boring. For example "is a number N composite" is in NP (just start with a non-trivial factor to show it is composite), and the opposite "is a number N prime or equal to 1" is in co-NP (just start ...

2015年4月17日 · Do we know anything about how BPP, RP, co-RP or ZPP intersect NP or co-NP? (apart from the fact that $(P \subset (ZPP = RP \cap co-RP)) \subset BPP$ and hence whatever intersection with NP and co-NP follows just from this)

2016年1月16日 · It's unknown whether NP = co-NP right now, and many people suspect the answer is no. Just as CLIQUE is NP-complete, the complement of CLIQUE is co-NP-complete. (More generally, the complement of any NP-complete problem is co-NP-complete). There's a theorem that if any co-NP-complete problem is in NP, then co-NP = NP,which would be a huge ...

2019年9月21日 · If I could reduce a NP-complete problem to a coNP-complete problem every problem in NP can be reduced to every coNP-complete problem and it would show that NP is a subset of coNP. Because every problem A in coNP has a complement co_A in NP. A in coNP => co_A in NP => co_A in coNP => A in NP => every Problem in coNP is in NP => NP = coNP

2020年7月9日 · But that doesn't prove that either is in NP or co-NP: To be in NP, you'd need a Turing machine that doesn't just solve SAT in exponential time. You'd need one that starts by writing a fantastically lucky hint on the tape, and then using the hint and the SAT problem, finds out that the SAT instance can be solved, and all that in polynomial time ...