Ad
related to: p vs np problem explained in detail
Search results
Results From The WOW.Com Content Network
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 ...
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 ...
A problem p in NP is NP-complete if every other problem in NP can be transformed (or reduced) into p in polynomial time. [citation needed] It is not known whether every problem in NP can be quickly solved—this is called the P versus NP problem.
Whether these problems are not decidable in polynomial time is one of the greatest open questions in computer science (see P versus NP ("P = NP") problem for an in-depth discussion). An important notion in this context is the set of NP-complete decision problems, which is a subset of NP and might be informally described as the "hardest ...
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 .
Quadratic programming (NP-hard in some cases, P if convex) Subset sum problem [3]: SP13 Variations on the Traveling salesman problem. The problem for graphs is NP-complete if the edge lengths are assumed integers. The problem for points on the plane is NP-complete with the discretized Euclidean metric and rectilinear metric.
If #P=FP, then the functions that determine the number of certificates for problems in NP are efficiently solvable. 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 # ...
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).