Longest Common Subsequence: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
Line 32: | Line 32: | ||
[[File:Longest Common Subsequence - Space.png|1000px]] | [[File:Longest Common Subsequence - Space.png|1000px]] | ||
== | == Space-Time Tradeoff Improvements == | ||
[[File:Longest Common Subsequence - Pareto Frontier.png|1000px]] | [[File:Longest Common Subsequence - Pareto Frontier.png|1000px]] |
Revision as of 14:34, 15 February 2023
Description
The longest common subsequence (LCS) problem is the problem of finding the longest subsequence common to all sequences in a set of sequences (often just two sequences).
Related Problems
Subproblem: Longest Common Substring with don't cares
Parameters
$n$: length of the longer input string
$m$: length of the shorter input string
$r$: length of the LCS
$s$: size of the alphabet
$p$: the number of dominant matches (AKA number of minimal candidates), i.e. the total number of ordered pairs of positions at which the two sequences match
Table of Algorithms
Currently no algorithms in our database for the given problem.
Time Complexity Graph
Error creating thumbnail: Unable to save thumbnail to destination
Space Complexity Graph
Error creating thumbnail: Unable to save thumbnail to destination
Space-Time Tradeoff Improvements
Error creating thumbnail: Unable to save thumbnail to destination
Reductions FROM Problem
Problem | Implication | Year | Citation | Reduction |
---|---|---|---|---|
UOV | If: to-time: $O((nm)^{({1}-\epsilon)})$, where $|x| = O(nd)$ and $|y| = O(md)$ Then: from-time: $O((nm)^{({1}-\epsilon/{2})})$ |
2015 | https://arxiv.org/pdf/1502.01063.pdf | link |
References/Citation
https://link.springer.com/chapter/10.1007/978-3-662-43948-7_4