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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 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 ...
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 ...
Notice that the 3SAT formula is equivalent to the circuit designed above, hence their output is same for same input. Hence, If the 3SAT formula has a satisfying assignment, then the corresponding circuit will output 1, and vice versa. So, this is a valid reduction, and Circuit SAT is NP-hard. This completes the proof that Circuit SAT is NP ...
The PCP theorem implies that there exists an ε > 0 such that (1-ε)-approximation of MAX-3SAT is NP-hard. Proof: Any NP-complete problem (( ()), ()) by the PCP theorem. For x ∈ L, a 3-CNF formula Ψ x is constructed so that x ∈ L ⇒ Ψ x is satisfiable
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.
In computational complexity, an NP-complete (or NP-hard) problem is weakly NP-complete (or weakly NP-hard) if there is an algorithm for the problem whose running time is polynomial in the dimension of the problem and the magnitudes of the data involved (provided these are given as integers), rather than the base-two logarithms of their magnitudes.
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 ...