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For example, for the array of values [−2, 1, −3, 4, −1, 2, 1, −5, 4], the contiguous subarray with the largest sum is [4, −1, 2, 1], with sum 6. Some properties of this problem are: If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array.
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)} .
Given such an instance, construct an instance of Partition in which the input set contains the original set plus two elements: z 1 and z 2, with z 1 = sum(S) and z 2 = 2T. The sum of this input set is sum(S) + z 1 + z 2 = 2 sum(S) + 2T, so the target sum for Partition is sum(S) + T. Suppose there exists a solution S′ to the SubsetSum instance
Legendre's three-square theorem states which numbers can be expressed as the sum of three squares; Jacobi's four-square theorem gives the number of ways that a number can be represented as the sum of four squares. For the number of representations of a positive integer as a sum of squares of k integers, see Sum of squares function.
Thus, in the equation relating the Bell numbers to the Stirling numbers, each partition counted on the left hand side of the equation is counted in exactly one of the terms of the sum on the right hand side, the one for which k is the number of sets in the partition. [8] Spivey 2008 has given a formula that combines both of these summations:
In number theory, zero-sum problems are certain kinds of combinatorial problems about the structure of a finite abelian group. Concretely, given a finite abelian group G and a positive integer n , one asks for the smallest value of k such that every sequence of elements of G of size k contains n terms that sum to 0 .
3-satisfiability can be generalized to k-satisfiability (k-SAT, also k-CNF-SAT), when formulas in CNF are considered with each clause containing up to k literals. [ citation needed ] However, since for any k ≥ 3, this problem can neither be easier than 3-SAT nor harder than SAT, and the latter two are NP-complete, so must be k-SAT.