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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.
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 can also find that solution quickly. Since the former describes the class of problems termed NP, while the latter describes P, the question is equivalent to asking whether all problems in NP are ...
It is in NP because (given an input) it is simple to check whether M accepts the input by simulating M; it is NP-complete because the verifier for any particular instance of a problem in NP can be encoded as a polynomial-time machine M that takes the solution to be verified as input. Then the question of whether the instance is a yes or no ...
The problem has been shown to be NP-hard (more precisely, it is complete for the complexity class FP NP; see function problem), and the decision problem version ("given the costs and a number x, decide whether there is a round-trip route cheaper than x") is NP-complete. The bottleneck travelling salesman problem is also NP-hard.
The class of problems with such verifiers for the "no"-answers is called co-NP. In fact, it is an open question whether all problems in NP also have verifiers for the "no"-answers and thus are in co-NP. In some literature the verifier is called the "certifier", and the witness the "certificate". [2]
A problem is NP-complete if it is both in NP and NP-hard. The NP-complete problems represent the hardest problems in NP. If some NP-complete problem has a polynomial time algorithm, all problems in NP do. The set of NP-complete problems is often denoted by NP-C or NPC.
There are decision problems that are NP-hard but not NP-complete such as the halting problem. That is the problem which asks "given a program and its input, will it run forever?" That is a yes/no question and so is a decision problem. It is easy to prove that the halting problem is NP-hard but not NP-complete.
The problem of finding a Hamiltonian cycle or path is in FNP; the analogous decision problem is to test whether a Hamiltonian cycle or path exists. The directed and undirected Hamiltonian cycle problems were two of Karp's 21 NP-complete problems. They remain NP-complete even for special kinds of graphs, such as: bipartite graphs, [12]
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