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They are some of the very few NP problems not known to be in P or to be NP-complete. The graph isomorphism problem is the computational problem of determining whether two finite graphs are isomorphic. An important unsolved problem in complexity theory is whether the graph isomorphism problem is in P, NP-complete, or NP-intermediate.
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.
A simple example of an NP-hard problem is the subset sum problem. Informally, if H is NP-hard, then it is at least as difficult to solve as the problems in NP. However, the opposite direction is not true: some problems are undecidable, and therefore even more difficult to solve than all problems in NP, but they are probably not NP-hard (unless ...
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 ...
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.
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, Π 2 {\displaystyle \Pi _{2}} , to another problem, Π 1 {\displaystyle \Pi _{1}} , would indicate that ...
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).