首页> 外文会议>International Symposium on Algorithms and Computation(ISAAC 2005); 20051219-21; Sanya(CN) >Efficient Algorithms for Finding a Longest Common Increasing Subsequence
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Efficient Algorithms for Finding a Longest Common Increasing Subsequence

机译:寻找最长公共递增子序列的高效算法

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We study the problem of finding a longest common increasing subsequence (LCIS) of multiple sequences of numbers. The LCIS problem is a fundamental issue in various application areas, including the whole genome alignment and pattern recognition. In this paper we give an efficient algorithm to find the LCIS of two sequences in O(min(r log l, nl + r) log log n + n log n) time where n is the length of each sequence and r is the total number of ordered pairs of positions at which the two sequences match and l is the length of the LCIS. For m sequences where m ≥ 3, we find the LCIS in O(min(mr~2, mr log l log~m r) + mn log n) time where r is the total number of m-tuples of positions at which the m sequences match. The previous results find the LCIS of two sequences in O(n~2) and O(nl log n) time. Our algorithm is faster when r is relatively small, e.g., for r < min(n~2/log log n,nl).
机译:我们研究寻找多个数字序列的最长公共递增子序列(LCIS)的问题。 LCIS问题是各个应用领域中的一个基本问题,包括整个基因组比对和模式识别。在本文中,我们提供了一种有效的算法,可以在O(min(r log l,nl + r)log log n + n log n)的时间内找到两个序列的LCIS,其中n是每个序列的长度,r是总数两个序列匹配的有序位置对的数量,l是LCIS的长度。对于m≥3的m个序列,我们在O(min(mr〜2,mr log l log〜mr)+ mn log n)时间找到LCIS,其中r是m处的m个元组的总数序列匹配。先前的结果在O(n〜2)和O(nl log n)时间中找到了两个序列的LCIS。当r较小时,例如r

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