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Finding a longest common subsequence between a run-length-encoded string and an uncompressed string

机译:在游程长度编码的字符串和未压缩的字符串之间找到最长的公共子序列

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摘要

In this paper, we propose an O(mn{mN, Mn}) time algorithm for finding a longest common subsequence of strings X and Y with lengths M and N, respectively, and run-length-encoded lengths m and n, respectively. We propose a new recursive formula for finding a longest common subsequence of Y and X which is in the run-length-encoded format. That is, Y = y_1Y_2…y_n and X=r_1~1 r_2~(l2)… r~(lm)_m, where r_1 is the repeated character of run i and l_i is the number of its repetitions. There are three cases in the proposed recursive formula in which two cases are for r_i matching y_j. The third case is for r-i mismatching y_j. We will look specifically at the prior two cases that r-i matches y_j. To determine which case will be used when r_i matches y_j, we have to find a specific value which can be obtained by using another of our proposed recursive formulas.
机译:在本文中,我们提出了一种O(mn {mN,Mn})时间算法,以找到长度分别为M和N以及游程长度编码长度为m和n的字符串X和Y的最长公共子序列。我们提出了一个新的递归公式,用于找到游程长度编码格式的Y和X的最长公共子序列。也就是说,Y = y_1Y_2…y_n且X = r_1〜1 r_2〜(l2)... r〜(lm)_m,其中r_1是运行i的重复字符,l_i是其重复次数。提议的递归公式中存在三种情况,其中两种情况用于r_i匹配y_j。第三种情况是r-i与y_j不匹配。我们将专门研究r-i与y_j匹配的前两种情况。为了确定在r_i与y_j匹配时将使用哪种情况,我们必须找到一个特定的值,该值可以通过使用我们提出的另一个递归公式获得。

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