An important problem in computational biology is finding the longest common subsequence (LCS) of two nucleotide sequences. This paper examines the correctness and performance of a recently proposed parallel LCS algorithm that uses successor tables and pruning rules to construct a list of sets from which an LCS can be easily reconstructed. Counterexamples are given for two pruning rules that were given with the original algorithm. Because of these errors, performance measurements originally reported cannot be validated. The work presented here shows that speedup can be reliably achieved by an implementation in Unified Parallel C that runs on an Infiniband cluster. This performance is partly facilitated by exploiting the software cache ...
AbstractThe problem of finding the longest common subsequence (lcs) of a given set of sequences over...
Finding the longest common subsequence (LCS) of multiple strings is an NP-hard problem, with many ...
现在几个最常用的解决最长公共子序列(LCS)问题的算法的时间复杂度分别是O(pn),O(n(m-p)).这里m、n为两个待比较字符串的长度,p是最长公共子串的长度.给出一种时间复杂度为O(p(m-p)...
An important problem in computational biology is finding the longest common subsequence (LCS) of two...
An important problem in computational biology is finding the longest common subsequence (LCS) of two...
Analysis and performance of a UPC implementation of a parallel longest common subsequence algorithm ...
Abstract Objective Finding the longest common subsequence (LCS) among sequences is NP-hard. This is ...
Information in various applications is often expressed as character sequences over a finite alphabet...
Background: The longest common subsequence (LCS) problem is a classical problem in computer science,...
Multiple longest common subsequence (MLCS) mining (a classical NP-hard problem) is an important tas...
The problem of computing the length of the maximal common subsequences of two strings is quite well...
As new genes are sequenced, it is necessary for molecular biologists to compare the new gene’s biose...
The multiple longest common subsequence (MLCS) problem involves finding all the longest common subse...
We present algorithms for finding a longest common increasing subsequence of two or more input seque...
The Longest Common Subsequence (LCS) is one of the most basic similarity measures and it captures im...
AbstractThe problem of finding the longest common subsequence (lcs) of a given set of sequences over...
Finding the longest common subsequence (LCS) of multiple strings is an NP-hard problem, with many ...
现在几个最常用的解决最长公共子序列(LCS)问题的算法的时间复杂度分别是O(pn),O(n(m-p)).这里m、n为两个待比较字符串的长度,p是最长公共子串的长度.给出一种时间复杂度为O(p(m-p)...
An important problem in computational biology is finding the longest common subsequence (LCS) of two...
An important problem in computational biology is finding the longest common subsequence (LCS) of two...
Analysis and performance of a UPC implementation of a parallel longest common subsequence algorithm ...
Abstract Objective Finding the longest common subsequence (LCS) among sequences is NP-hard. This is ...
Information in various applications is often expressed as character sequences over a finite alphabet...
Background: The longest common subsequence (LCS) problem is a classical problem in computer science,...
Multiple longest common subsequence (MLCS) mining (a classical NP-hard problem) is an important tas...
The problem of computing the length of the maximal common subsequences of two strings is quite well...
As new genes are sequenced, it is necessary for molecular biologists to compare the new gene’s biose...
The multiple longest common subsequence (MLCS) problem involves finding all the longest common subse...
We present algorithms for finding a longest common increasing subsequence of two or more input seque...
The Longest Common Subsequence (LCS) is one of the most basic similarity measures and it captures im...
AbstractThe problem of finding the longest common subsequence (lcs) of a given set of sequences over...
Finding the longest common subsequence (LCS) of multiple strings is an NP-hard problem, with many ...
现在几个最常用的解决最长公共子序列(LCS)问题的算法的时间复杂度分别是O(pn),O(n(m-p)).这里m、n为两个待比较字符串的长度,p是最长公共子串的长度.给出一种时间复杂度为O(p(m-p)...