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By formulating MAX-2-SAT as a problem of finding a cut (that is, a partition of the vertices into two subsets) maximizing the number of edges that have one endpoint in the first subset and one endpoint in the second, in a graph related to the implication graph, and applying semidefinite programming methods to this cut problem, it is possible to ...
The program is solvable in polynomial time if the graph has all undirected or all directed edges. Variants include the rural postman problem. [3]: ND25, ND27 Clique cover problem [2] [3]: GT17 Clique problem [2] [3]: GT19 Complete coloring, a.k.a. achromatic number [3]: GT5 Cycle rank; Degree-constrained spanning tree [3]: ND1
A polynomial-time problem can be very difficult to solve in practice if the polynomial's degree or constants are large enough. In addition, information-theoretic security provides cryptographic methods that cannot be broken even with unlimited computing power. "A large-scale quantum computer would be able to efficiently solve NP-complete problems."
[3] [4] At each iteration, the domain is partitioned into two parts, and the algorithm decides - based on a small number of function evaluations - which of these two parts must contain a root. In one dimension, the criterion for decision is that the function has opposite signs.
The main computer algebra systems (Maple, Mathematica, SageMath, PARI/GP) have each a variant of this method as the default algorithm for the real roots of a polynomial. The class of methods is based on converting the problem of finding polynomial roots to the problem of finding eigenvalues of the companion matrix of the polynomial, [1] in ...
Although this problem seems easier, Valiant and Vazirani have shown [25] that if there is a practical (i.e. randomized polynomial-time) algorithm to solve it, then all problems in NP can be solved just as easily. MAX-SAT, the maximum satisfiability problem, is an FNP generalization of SAT. It asks for the maximum number of clauses which can be ...
This polynomial is further reduced to = + + which is shown in blue and yields a zero of −5. The final root of the original polynomial may be found by either using the final zero as an initial guess for Newton's method, or by reducing () and solving the linear equation. As can be seen, the expected roots of −8, −5, −3, 2, 3, and 7 were ...
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. For instance, the Boolean satisfiability problem is NP-complete by the Cook–Levin theorem, so any instance of any ...