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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.
The longest increasing subsequence problem is closely related to the longest common subsequence problem, which has a quadratic time dynamic programming solution: the longest increasing subsequence of a sequence is the longest common subsequence of and , where is the result of sorting.
In computer programming and computer science, "maximal munch" or "longest match" is the principle that when creating some construct, as much of the available input as possible should be consumed. The earliest known use of this term is by R.G.G. Cattell in his PhD thesis [ 1 ] on automatic derivation of code generators for compilers .
Comparison of two revisions of an example file, based on their longest common subsequence (black) A longest common subsequence (LCS) is the longest subsequence common to all sequences in a set of sequences (often just two sequences).
The most common problem being solved is the 0-1 knapsack problem, which restricts the number of copies of each kind of item to zero or one. Given a set of items numbered from 1 up to , each with a weight and a value , along with a maximum weight capacity ,
OPAL — an SIMD C/C++ library for massive optimal sequence alignment; diagonalsw — an open-source C/C++ implementation with SIMD instruction sets (notably SSE4.1) under the MIT license; SSW — an open-source C++ library providing an API to an SIMD implementation of the Smith–Waterman algorithm under the MIT license
With that knowledge, everything after the "c" looks like the reflection of everything before the "c". The "a" after the "c" has the same longest palindrome as the "a" before the "c". Similarly, the "b" after the "c" has a longest palindrome that is at least the length of the longest palindrome centered on the "b" before the "c". There are some ...
The maximum sum is 1, attained by giving one agent the item with value 1 and the other agent nothing. But the max-min allocation gives each agent value at least e, so the sum must be at most 3e. Therefore the POF is 1/(3e), which is unbounded. Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The ...