在另一个序列中找到最相似的子序列

Question

我需要编写一个算法，在 N log(N) 中找到 S1 中与另一个字符串 S2 最相似的子字符串（S1 中与 S2 具有最小汉明距离的子字符串），其中 N = len( S1) + len(S2), len(S2)

例如：
S1 = AGTCAGTC
S2 = GTC
答案：GTC（距离0）

S1 = AAGGTTCC
S2 = TCAA
答：TTCC（距离3）

时间复杂度不得超过 O(N Log(N))。空间复杂度无关紧要。

LCS（最长公共子序列）在我的情况下不起作用。例如：

S1 = GAATCCAGTCTGTCT S2 = AATATAT LCS 答案：AATCCAGTC GAATCCAGTCTGTCT ||| 阿塔塔特 正确答案：AGCTGTCT GAATCCAGTCTGTCT | | | | | 阿塔塔特

Answer 1

我认为您正在尝试解决longest common subsequence problem。此问题涉及尝试找到将一个字符串转换为另一个字符串所需的最少修改量。

您说您正在尝试编写一个算法来执行此操作。看看 LCS 问题，如果您想使用自己的算法，请尝试在谷歌上搜索它，或者您可以利用命令行实用程序 diff。

就像最长的公共子序列问题一样，diff 需要两个文件并找到一个添加和删除序列，从而将 file1 转换为 file2 的修改最少。Diff 非常有效，我想它对于您的目的来说足够快.我很确定大多数差异的空间和时间复杂度为 O(Nlog(N)) 或更小，但您可能想自己验证这一点。更多关于差异在这里http://en.wikipedia.org/wiki/Diff_utility

我写了一个小 Python 程序，它使用 diff 来查找最长的连续公共子序列。这适用于 unix，但我确信您使用的任何平台都有一个 diff 实用程序。

文件每行应该有一个字符。您可能必须编写一个程序来对文件执行该转换。

import sys
import subprocess
import os
import re

def getLinesFromShellCommand(command):
        p = subprocess.Popen(command, shell=True, stdout=subprocess.PIPE, stderr=subprocess.STDOUT, cwd=os.getcwd())
        lines = []
        for curLine in p.stdout.readlines():
                lines.append(curLine)
        retVal = p.wait()
        return (retVal, lines)

#We need to find the longest sequence of lines that start with a space( a space meant the line was in common).
#We could use some kind of while loop to detect the start and end of a group of lines that start with spaces. 
#However there is a much simpler method. Create a string by concatenating the first char from each line and use regex
#to find all the subsequences that start with spaces. After that just take the longest subsequence.
def findLongestCommonSubsequence(diffOutput):
    outputFirstCharStr = reduce(lambda x, y: x+y[:1], diffOutput, "")
    commonSubsequences = [(m.start(0), m.end(0)) for m in re.finditer(" +", outputFirstCharStr)]
    longestCommonSubsequence = max(commonSubsequences, key=lambda (start,end) : end - start)
    return longestCommonSubsequence

def main():
    if len(sys.argv) != 3:
        sys.stderr.write("usage: python LongestCommonSubsequence.py <file1> <file2>\n")
        sys.exit(1)
    commandOutput = getLinesFromShellCommand("diff -u {0} {1}".format(sys.argv[1], sys.argv[2]))
    if commandOutput[0] != 1: # apparently diff returns 1 if its successful
        sys.stderr.write("diff failed with input files.\n")
        sys.exit(1)
    diffOutput = commandOutput[1]
    diffOutput = diffOutput[3:] # the first three lines are not needed
    longestCommonSubsequence = findLongestCommonSubsequence(diffOutput)
    print "Indices:",longestCommonSubsequence
    for i in range(longestCommonSubsequence[0], longestCommonSubsequence[1]):
        print diffOutput[i],

if __name__ == "__main__":
    main()

用法

python LongestCommonSubsequence.py f1.txt f2.txt

如果 f1.txt 和 f2.txt 是您给出的第二个示例，则输出：

Indices: (5, 7)
 T
 C

编辑：我看到您的 cmets 关于为什么上述方法不起作用。不过，您可能对这篇文章感兴趣：https://cs.stackexchange.com/questions/2519/efficiently-calculating-minimum-edit-distance-of-a-smaller-string-at-each-positi