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Informally, an NP-complete problem is an NP problem that is at least as "tough" as any other problem in NP. NP-hard problems are those at least as hard as NP problems; i.e., all NP problems can be reduced (in polynomial time) to them. NP-hard problems need not be in NP; i.e., they need not have solutions verifiable in polynomial time.
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.
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 ...
Therefore, the longest path problem is NP-hard. The question "does there exist a simple path in a given graph with at least k edges" is NP-complete. [2] In weighted complete graphs with non-negative edge weights, the weighted longest path problem is the same as the Travelling salesman path problem, because the longest path always includes all ...
Testing whether a graph is 1-tough is co-NP-complete. That is, the decision problem whose answer is "yes" for a graph that is not 1-tough, and "no" for a graph that is 1-tough, is NP-complete. The same is true for any fixed positive rational number q: testing whether a graph is q-tough is co-NP-complete (Bauer, Hakimi & Schmeichel 1990).
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 formula resulting from transforming all clauses is at most 3 times as long as its original; that is, the length growth is polynomial. [10] 3-SAT is one of Karp's 21 NP-complete problems, and it is used as a starting point for proving that other problems are also NP-hard. [b] This is done by polynomial-time reduction from 3-SAT to the other ...
Graph coloring is computationally hard. It is NP-complete to decide if a given graph admits a k-coloring for a given k except for the cases k ∈ {0,1,2}. In particular, it is NP-hard to compute the chromatic number. [33] The 3-coloring problem remains NP-complete even on 4-regular planar graphs. [34]