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Maximum subarray problems arise in many fields, such as genomic sequence analysis and computer vision.. Genomic sequence analysis employs maximum subarray algorithms to identify important biological segments of protein sequences that have unusual properties, by assigning scores to points within the sequence that are positive when a motif to be recognized is present, and negative when it is not ...
Max-sum MSSP: for each subset j in 1,...,m, there is a capacity C j. The goal is to make the sum of all subsets as large as possible, such that the sum in each subset j is at most C j. [1] Max-min MSSP (also called bottleneck MSSP or BMSSP): again each subset has a capacity, but now the goal is to make the smallest subset sum as large as ...
The Minkowski sum of two sets and of real numbers is the set + := {+:,} consisting of all possible arithmetic sums of pairs of numbers, one from each set. The infimum and supremum of the Minkowski sum satisfy, if A ≠ ∅ ≠ B {\displaystyle A\neq \varnothing \neq B} inf ( A + B ) = ( inf A ) + ( inf B ) {\displaystyle \inf(A+B)=(\inf A ...
Given the two sorted lists, the algorithm can check if an element of the first array and an element of the second array sum up to T in time (/). To do that, the algorithm passes through the first array in decreasing order (starting at the largest element) and the second array in increasing order (starting at the smallest element).
For data in which the maximum key size is significantly smaller than the number of data items, counting sort may be parallelized by splitting the input into subarrays of approximately equal size, processing each subarray in parallel to generate a separate count array for each subarray, and then merging the count arrays.
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
The picture shows two strings where the problem has multiple solutions. Although the substring occurrences always overlap, it is impossible to obtain a longer common substring by "uniting" them. The strings "ABABC", "BABCA" and "ABCBA" have only one longest common substring, viz. "ABC" of length 3.
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.