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Quadratic programming (NP-hard in some cases, P if convex) Subset sum problem [3]: SP13 Variations on the Traveling salesman problem. The problem for graphs is NP-complete if the edge lengths are assumed integers. The problem for points on the plane is NP-complete with the discretized Euclidean metric and rectilinear metric.
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.
Clearly, a #P problem must be at least as hard as the corresponding NP problem, since a count of solutions immediately tells if at least one solution exists, if the count is greater than zero. Surprisingly, some #P problems that are believed to be difficult correspond to easy (for example linear-time) P problems. [ 18 ]
Solution of a travelling salesman problem: the black line shows the shortest possible loop that connects every red dot. In the theory of computational complexity, the travelling salesman problem (TSP) asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the ...
In computational complexity theory, a computational problem H is called NP-hard if, for every problem L which can be solved in non-deterministic polynomial-time, there is a polynomial-time reduction from L to H. That is, assuming a solution for H takes 1 unit time, H ' s solution can be used to solve L in polynomial time.
Computationally, the problem is NP-hard, and the corresponding decision problem, deciding if items can fit into a specified number of bins, is NP-complete. Despite its worst-case hardness, optimal solutions to very large instances of the problem can be produced with sophisticated algorithms. In addition, many approximation algorithms exist.
Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of rational numbers. Rational numbers can be written in the form p/q, where p and ...
The most common problem being solved is the 0-1 knapsack problem, which restricts the number of copies of each kind of item to zero or one. Given a set of items numbered from 1 up to , each with a weight and a value , along with a maximum weight capacity ,