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Results tagged with np
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user 41
Questions about decision problems that can be solved on nondeterministic Turing machines in time polynomial in the length of the input.
74
votes
Accepted
Are there subexponential-time algorithms for NP-complete problems?
It is closed under polynomials so no $\mathsf{NP}$-hard problem can be solved in this time without violating ETH.
III. … Take an $\mathsf{NP}$-complete problem like SAT. It has a brute-force algorithm that runs in time $2^{O(n)}$. …
- 237
2
votes
Is $NP$ "minimal", i.e. does $\Pi\notin NP$ imply $\Pi$ is $NP$-hard?
E.g. the fact that a problem cannot be solved in polynomial nondeterministic time does not imply that it is NP-complete (i.e. universal for NP). … For NP:
if P=NP all problems except trivial ones will be complete for NP (under Karp reductions). So assume P is a proper subset of NP (or alternatively use a weaker notion of reduction like AC0). …
- 22.5k
9
votes
Are there are problems in NP that have been shown to be not NP-complete but it is still not ...
In other words if you show some problem is in $\mathsf{NP}$ but is not $\mathsf{NP}$ complete would imply that $\mathsf{P}$ is not equal to $\mathsf{NP}$. … In short, there is no problem in $\mathsf{NP}$ that we know it is not $\mathsf{NP}$-complete. …
CommunityBot
- 1
7
votes
How to show that problems are in NP?
The easiest way to prove some problem is in $\mathsf{NP}$ is using the certificate definiiton of $\mathsf{NP}$ mentioned in other answers. … The nondeterministic definition of $\mathsf{NP}$ is usually not very useful for showing a problem belongs to $\mathsf{NP}$. …
- 22.5k
5
votes
NP $\subsetneq$ EXP?
So what you are saying is
$$\mathsf{NP} \subsetneq \mathsf{ExpTime} \implies \mathsf{NP} \subsetneq \mathsf{ExpTime}$$
which is trivially true. … If you are asking if $\mathsf{NP}\subsetneq \mathsf{ExpTime} $ then the answer is: it is unknown. …
- 22.8k
8
votes
Accepted
Problems that are Cook-reducible to a problem in NP $\cap$ co-NP
Stated in another way: $\mathsf{NP}$ is not closed under Cook reductions (assuming $\mathsf{P}\neq \mathsf{NP}$).
How about $\mathsf{NP}\cap\mathsf{coNP}$? … Is the only reason that $\mathsf{NP}$ is not closed under Cook reductions is because it is not closed under complement so if we take those $\mathsf{NP}$ problems whose complement is also $\mathsf{NP}$ …
- 22.5k