Abstract A hardness result and new algorithm for the longest common palindromic subsequence problem
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A hardness result and new algorithm for the longest common palindromic subsequence problem

机译:最长共同回文子序列问题的硬度结果和新算法

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AbstractThe 2-LCPS problem, first introduced by Chowdhury et al. (2014) , asks one to compute (the length of) a longest common palindromic subsequence between two given stringsAandB. We show that the 2-LCPS problem is at least as hard as the well-studied longest common subsequence problem for four strings. Then, we present a new algorithm which solves the 2-LCPS problem inO(σM2+n)time, wherendenotes the length ofAandB,Mdenotes the number of matching positions betweenAandB, andσdenotes the number of distinct characters occurring in bothAandB. Our new algorithm is faster than Chowdhury et al.'s sparse algorithm whenσ=o(log2nloglogn).HighlightsThe 2-LCPS problem is to compute a longest palindromic common subsequence between two strings.We show that the 2-LCPS problem is at least as hard as the longest common subsequence problem for four strings.We present a new efficient algorithm which solves the 2-LCPS problem.
机译: 摘要 第2个 -LCPS问题,由Chowdhury等人首先提出。 (2014),要求一个人计算两个给定字符串 A B 之间最长的常见回文子序列(的长度)。我们证明了2-LCPS问题至少与四个字符串最长的最长公共子序列问题研究得一样困难。然后,我们提出一种新算法,用于解决 O σ < mml:msup> M 2 + n time,其中 n 表示 A B M 表示 A B σ之间的匹配位置数表示同时出现在 A B 中的不同字符的数量。当 σ = o log < mml:mn> 2 n 日志 日志 ⁡< / mml:mo> n < / ce:abstract-sec> 突出显示 2-LCPS问题是计算两个字符串之间最长的回文通用子序列。 我们显示2-LCPS问题至少与四个字符串的最长公共子序列问题一样困难。 我们提出了一种解决2-LCPS问题的高效算法。

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