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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]
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 ...
Whenever the sum of the current element in the first array and the current element in the second array is more than T, the algorithm moves to the next element in the first array. If it is less than T, the algorithm moves to the next element in the second array. If two elements that sum to T are found, it stops. (The sub-problem for two elements ...
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.
Clearly, the new instance has an equal-cardinality equal-sum partition iff the original instance has an equal-sum partition. See also Balanced number partitioning. Product partition is the problem of partitioning a set of integers into two sets with the same product (rather than the same sum). This problem is strongly NP-hard. [14]
It may be used to prove Nicomachus's theorem that the sum of the first cubes equals the square of the sum of the first positive integers. [2] Summation by parts is frequently used to prove Abel's theorem and Dirichlet's test.
The k-way merge problem consists of merging k sorted arrays to produce a single sorted array with the same elements. Denote by n the total number of elements. n is equal to the size of the output array and the sum of the sizes of the k input arrays. For simplicity, we assume that none of the input arrays is empty.
The value of this limit, should it exist, is the (C, α) sum of the integral. Analogously to the case of the sum of a series, if α = 0, the result is convergence of the improper integral. In the case α = 1, (C, 1) convergence is equivalent to the existence of the limit