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The P versus NP problem is a major unsolved problem in theoretical computer science. Informally, it asks whether every problem whose solution can be quickly verified ...
Euler diagram for P, NP, NP-complete, and NP-hard set of problems (excluding the empty language and its complement, which belong to P but are not NP-complete) Main article: P versus NP problem The question is whether or not, for all problems for which an algorithm can verify a given solution quickly (that is, in polynomial time ), an algorithm ...
If P and NP are different, then there exist decision problems in the region of NP that fall between P and the NP-complete problems. (If P and NP are the same class, then NP-intermediate problems do not exist because in this case every NP-complete problem would fall in P, and by definition, every problem in NP can be reduced to an NP-complete ...
Thus the class of NP-complete problems contains the most difficult problems in NP, in the sense that they are the ones most likely not to be in P. Because the problem P = NP is not solved, being able to reduce a known NP-complete problem, , to another problem, , would indicate that there is no known polynomial-time solution for .
Euler diagram for P, NP, NP-complete, and NP-hard set of problems. Under the assumption that P ≠ NP, the existence of problems within NP but outside both P and NP-complete was established by Ladner. [1] In computational complexity theory, NP (nondeterministic polynomial time) is a complexity class used to classify decision problems.
The Subgraph Isomorphism problem is NP-complete. The graph isomorphism problem is suspected to be neither in P nor NP-complete, though it is in NP. This is an example of a problem that is thought to be hard, but is not thought to be NP-complete. This class is called NP-Intermediate problems and exists if and only if P≠NP.
Assuming P ≠ NP, the following are true for computational problems on integers: [3] If a problem is weakly NP-hard, then it does not have a weakly polynomial time algorithm (polynomial in the number of integers and the number of bits in the largest integer), but it may have a pseudopolynomial time algorithm (polynomial in the number of integers and the magnitude of the largest integer).
And since computing the number of certificates is at least as hard as determining whether a certificate exists, it must follow that if #P=FP then P=NP (it is not known whether this holds in the reverse, i.e. whether P=NP implies #P=FP). [23] Just as FP is the function problem equivalent of P, FNP is the function problem equivalent of NP.