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The multiple subset sum problem is an optimization problem in computer science and operations research. It is a generalization of the subset sum problem . The input to the problem is a multiset S {\displaystyle S} of n integers and a positive integer m representing the number of subsets.
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)} .
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 ...
Summation by parts is frequently used to prove Abel's theorem and Dirichlet's test. One can also use this technique to prove Abel's test: If is a convergent series, and a bounded monotone sequence, then = = converges. Proof of Abel's test.
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 ...
For each triplet t = {w i,x j,y k} in E, the set A contains an element u t = 10r 4-kr 3-jr 2-ir. For each triplet t = {w i,x j,y k} in E, the set B contains w it, C contains x jt, and D contains y kt. So for each of w i, x j, y k, there may be many corresponding elements in B, C, D - one for each triplet in which they appear. We consider one of ...
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 ...
That is, for source code where the average line is 60 or more characters long, the hash or checksum for that line might be only 8 to 40 characters long. Additionally, the randomized nature of hashes and checksums would guarantee that comparisons would short-circuit faster, as lines of source code will rarely be changed at the beginning.