Search results
Results From The WOW.Com Content Network
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 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 ...
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.
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 ...
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 ...
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 ...
3-SAT is NP-complete (like any other k-SAT problem with k>2) while 2-SAT is known to have solutions in polynomial time. As a consequence, [ f ] the task of converting a formula into a DNF , preserving satisfiability, is NP-hard ; dually , converting into CNF, preserving validity , is also NP-hard; hence equivalence-preserving conversion into ...
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).