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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 ...
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.
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 ...
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 ...
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.
The Boolean satisfiability problem on conjunctive normal form formulas is NP-complete. By the duality principle, so is the falsifiability problem on DNF formulas. Therefore, it is co-NP-hard to decide if a DNF formula is a tautology. Conversely, a DNF formula is satisfiable if, and only if, one of its conjunctions is satisfiable.
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 ...