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Max-sum MSSP is a special case of MKP in which the value of each item equals its weight. The knapsack problem is a special case of MKP in which m=1. The subset-sum problem is a special case of MKP in which both the value of each item equals its weight, and m=1. The MKP has a Polynomial-time approximation scheme. [6]
Let A be the sum of the negative values and B the sum of the positive values; the number of different possible sums is at most B-A, so the total runtime is in (()). For example, if all input values are positive and bounded by some constant C , then B is at most N C , so the time required is O ( N 2 C ) {\displaystyle O(N^{2}C)} .
Equal-cardinality partition is a variant in which both parts should have an equal number of items, in addition to having an equal sum. This variant is NP-hard too. [5]: SP12 Proof. Given a standard Partition instance with some n numbers, construct an Equal-Cardinality-Partition instance by adding n zeros. Clearly, the new instance has an equal ...
In the th step, it computes the subarray with the largest sum ending at ; this sum is maintained in variable current_sum. [ note 3 ] Moreover, it computes the subarray with the largest sum anywhere in A [ 1 … j ] {\displaystyle A[1\ldots j]} , maintained in variable best_sum , [ note 4 ] and easily obtained as the maximum of all values of ...
Minimum maximal independent set a.k.a. minimum independent dominating set [4] NP-complete special cases include the minimum maximal matching problem, [3]: GT10 which is essentially equal to the edge dominating set problem (see above). Metric dimension of a graph [3]: GT61 Metric k-center; Minimum degree spanning tree; Minimum k-cut
The algorithm performs summation with two accumulators: sum holds the sum, and c accumulates the parts not assimilated into sum, to nudge the low-order part of sum the next time around. Thus the summation proceeds with "guard digits" in c , which is better than not having any, but is not as good as performing the calculations with double the ...
In this variant of the problem, which allows for interesting applications in several contexts, it is possible to devise an optimal selection procedure that, given a random sample of size as input, will generate an increasing sequence with maximal expected length of size approximately . [11] The length of the increasing subsequence selected by ...
(x, y, and z represent expressions made from the functions S, K, and I, and set values): I returns its argument: Ix = x. K, when applied to any argument x, yields a one-argument constant function Kx, which, when applied to any argument y, returns x: Kxy = x. S is a substitution operator. It takes three arguments and then returns the first ...