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The problem is NP-hard even when all input integers are positive (and the target-sum T is a part of the input). This can be proved by a direct reduction from 3SAT. [2] It can also be proved by reduction from 3-dimensional matching (3DM): [3] We are given an instance of 3DM, where the vertex sets are W, X, Y. Each set has n vertices.
A minimum spanning tree of a weighted planar graph.Finding a minimum spanning tree is a common problem involving combinatorial optimization. Combinatorial optimization is a subfield of mathematical optimization that consists of finding an optimal object from a finite set of objects, [1] where the set of feasible solutions is discrete or can be reduced to a discrete set.
Set covering is equivalent to the hitting set problem. That is seen by observing that an instance of set covering can be viewed as an arbitrary bipartite graph, with the universe represented by vertices on the left, the sets represented by vertices on the right, and edges representing the membership of elements to sets. The task is then to find ...
If all the hard constraints are linear and some are inequalities, but the objective function is quadratic, the problem is a quadratic programming problem. It is one type of nonlinear programming. It can still be solved in polynomial time by the ellipsoid method if the objective function is convex; otherwise the problem may be NP hard.
MATLAB allows matrix manipulations, plotting of functions and data, implementation of algorithms, creation of user interfaces, and interfacing with programs written in other languages. Although MATLAB is intended primarily for numeric computing, an optional toolbox uses the MuPAD symbolic engine allowing access to symbolic computing abilities.
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 ...
Of particular use is the property that for any fixed set of ~ values, the optimal result to the Lagrangian relaxation problem will be no smaller than the optimal result to the original problem. To see this, let x ^ {\displaystyle {\hat {x}}} be the optimal solution to the original problem, and let x ¯ {\displaystyle {\bar {x}}} be the optimal ...
PROOF: Build a directed version of G by directing each edge to the node with the higher degree (breaking ties arbitrarily). So for every bad node, the number of out-going edges is more than 2 times the number of in-coming edges. So every bad edge, that enters a node v, can be matched to a distinct set of two edges that exit the node v.