Boyer-Moore文本匹配算法(联合使用KMP和Horspool算法)
Boyer-Moore除了考虑Horspool算法(参考笔者的另一篇专门介绍Horspool算法的文章)的坏字符之外,还将模式串中已经匹配成功的后缀(叫做好后缀, good suffix)考虑进来,从而得到全部已经知道的启发信息(heuristic)。因此从理论上来说,BM算法应该是性能最佳的一个算法,实践中也证明了这一点。 这也是为什么BM算法经常用作精确匹配算法里面的性能测试基准算法。例如,在通过下面的图示就可以看出, KMP算法由于没有考虑进来bad character信息,比较次数比BM算法稍多:
(图一)
上面在i=4,j=4时出现mismatch,在KMP算法中的做法是找出j-1右边界位置的失败函数值作为下一个j值,这里f(j-1)=1, 因此j'=1(i值不变,即下次仍然将P[j']跟T[i]=C比较)。这里KMP算法的第二步只移动了3个字符位置。而在BM算法中的做法是首先找出bad character(=C)的位移值,这里C在pattern中未出现,用BM算法的case 1(参考前面另一篇文章中的BM简化版-Horspool算法的介绍),将模式串沿着文本向右移动m个位置,即向右移5;然后找出good suffix(长度等于0)的位移值,由于good suffix长度等于0,位移值为1。最后取good suffix shift和bad character shift的最大位移值,等于5,所以得出上面BM算法中第二步的位置。
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BM算法也是分两大步骤:1)预处理计算出bad character shift表(这一步跟Horspool算法中的做法一模一样,详细情况参考笔者的另一篇文章)和good suffix shift表;2)匹配。
匹配过程比较简单,即取bad character shift和good suffix shift值的最大值即可。从逻辑上来说,因为两个位移值都保证了不会遗漏能够成功匹配的子串,如果取最大值就可以更大幅度地将模式串向右移,从而减少比较次数。
设文本T长度为n,模式串P长度为m。T的当前位置指针为i(0<=i<n),P的当前位置指针为j(0<=j<m)。设 good suffix的长度为k(0<=k<m,即模式串中长度为k的后缀已经跟文本相应字符匹配成功),即好后缀为P[m-k...m-1]。好后 缀位移值是goodSuffixShiftTable[k]。
如果单考虑good suffix位移,可以分为以下几种情况(另外参考Horspool算法那篇文章介绍的bad character的cases):
case 1) k=0,即模式串中最后一个字符P[m-1]不匹配,做法是将模式串沿文本向右移动一个字符,即goodSuffixShiftTable[0]=1。
例如图二中case 1,这里X表示下一次模式串采用good suffix的位移位置。如果采用bad character位移,就是Y位置,从图中可见Y的移动幅度更大,在BM算法中应该取Y的位置。
case 2) k>0,但是在模式串中可以找到一个最右边的子串(非后缀)跟good suffix - P[m-k...m-1]完全相等。做法是将模式串沿文本向右移动goodSuffixShiftTable[k]个字符。例如在图二中case 2中,当j=2时,T[i]=C, P[j]=A,k=3,可以找到最右边的非后缀子串P[1...3]跟后缀BAB完全相等。这时需要将模式串沿文本向右移动2个字符,即 goodSuffixShiftTable[3]=2。
(图二)
case 3) k>0,但是不能找到像case 2中的最右边的非后缀子串跟好后缀(good suffix)完全相等。可以分成两个小cases:
case 3a) 模式串中没有任何前缀(prefix)跟模式串的good suffix的一个后缀(而且是一个proper后缀)完全相等,做法是将模式串向右移动m个字符,即 goodSuffixShiftTable[k]=m,例如图二中case 3a,这里j=1,T[i]=a,P[j]=z,k=2,显然没有任何prefix跟bc的唯一一个proper后缀“c”完全相等,于是得出X为模式串 的下一次匹配位置。
case 3b) 模式串中存在一个最长的前缀(prefix)跟模式串的good suffix的一个后缀(而且是一个proper后缀)完全相等,做法是将模式串向右移动goodSuffixShiftTable[k]个字符。例如图 二中case 3b,这里j=2,T[i]=z,P[j]=c,k=3,存在最长的前缀ab跟好后缀的后缀ab完全相等。 goodSuffixShiftTable[3]=4。
在Horspool算法的基础上,坏字符位移值(bad character shift)可以根据下面的公式得出:
badCharShift = j+1-min(j,1+q)
好后缀位移值就是goodSuffixShiftTable[k]。
下一个i指针位置向右移动量为m - j + goodSuffixShiftTable[k] - 1,即下一个i += m - j + goodSuffixShiftTable[k] - 1; 下一个j值永远是模式串尾部,即m-1,这跟Horspool算法是一样的。
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下面介绍一下如何计算good suffix位移表:
这里利用KMP算法中计算失败函数的做法来计算BM算法中的good suffix位移表,复杂度为O(m)。这样做的原因是KMP算法中的failure function从本质上等价于BM算法中的good suffix shift table,因为二者都是根据已经匹配成功的字符序列计算出的结果,只是形成序列的扫描方向相反。
首先将模式串转成逆序R,然后对逆序模式串计算出它的KMP失败函数f(j)。对f(j)从右往左循环一遍,计算上面case 2(即最右边子串等于长度为k的后缀)中的good suffix shift值,循环结束时有这样的结果:
1)还有一部分k(k>=0)对应的good suffix shift值为0(即数组初始值),这时对应case 1,case 3a和case 3b(见下面的说明)。
2)f(j)>0,如果这样的f(j)有多个,循环最后的那个f(j)就相当于原始模式串最右边的子串长度(该子串等于原始模式串中长度为k=f(j)的后缀),所以这个f(j)就是k值。位移值采用如下公式计算:
(m - f[j]) - (m-j-1)
接下来设置goodSuffixShiftTable[0]=1,这是k=0(即case 1)时的一个特定值。
最后考虑某些k的good suffix shift仍然为0的情况。因为f(m-1)就是case 3中匹配到的最长前缀。见下图中蓝色部分区域。如果f(m-1)=0,就是case 3a,位移值设置为m。
如果f(m-1)>0,对应case 3b,这时goodSuffixShiftTable[k]应修正为m-f(m-1),即将模式串首字符跟长度为f(m-1)的后缀的首字符对齐。需要说 明的是,在这种情况下(即f(m-1)>0), 这样的k必定满足条件k>f(m-1),用反证法证明:到目前为止goodSuffixShiftTable[k]=0的那些k值如果满足0<k<=f(m-1)的话,在上面2)中得到的结果会有goodSuffixShiftTable[k]>0, 理由是k<=f(m-1)说明一定存在一个长度为k子串与某一个长度为k的后缀完全相等,即满足case 2,显然对于case 2的情况必有goodSuffixShiftTable[k]>0,与goodSuffixShiftTable[k]=0前提矛盾。这也是为什么 下面实现代码中用"if(goodSuffixShift[k]==0)"代替"if(goodSuffixShift[k]==0 && prefixMaxLen<k)"的原因。
下图中的S就是good suffix shift值列表。
(图三)
实现:
import java.util.Arrays; /** * * Boyer-Moore Algorithm * * Copyright (c) 2011 ljs (http://blog.csdn.net/ljsspace/) * Licensed under GPL (http://www.opensource.org/licenses/gpl-license.php) * * @author ljs * 2011-06-21 * */ public class BM { private static final int CHARSET_SIZE = 256; //prepare the shift table: find the rightmost position //in the pattern private int[] makeBadCharTable(String pattern){ int[] shiftTable = new int[CHARSET_SIZE]; //init to -1 for(int i=0;i<CHARSET_SIZE;i++) shiftTable[i]=-1; //set the pattern chars for(int i=0;i<pattern.length();i++){ char c = pattern.charAt(i); //OK: the rightmost position i may overwrite the left positions shiftTable[c] = i; } return shiftTable; } //caculate KMP algorithm's failure function: f[0], f[1..m-1], m is the length of pattern private int[] calFailureF(String pattern){ int m = pattern.length(); int[] f = new int[m]; //i is the right border position int i = 1; int j = 0; f[0] = 0; //by definition while(i<m){ if(pattern.charAt(i)==pattern.charAt(j)){ //j is index from 0, f[i] is the length of suffix/prefix //so we need to add 1 f[i] = j + 1; i++; j++; }else if(j==0){ //find no valid prefix f[i] = 0; //move i only, j is still 0 i++; }else{ //move j only, i doesn't change position, thus f[i]'s value is not determined yet. //reuse the KMP algorithm: we already know f[j-1]'s value j = f[j-1]; } } return f; } //using KMP algorithm's failure function to caculate good-suffix shift table private int[] makeGoodSuffixShiftTable(String pattern){ String reverse = new StringBuffer(pattern).reverse().toString(); int[] f = calFailureF(reverse); int m = pattern.length(); int[] goodSuffixShift = new int[m]; //set goodSuffixShift by failure function for(int j=m-1;j>=0;j--){ //caculate d2 int index = m - j - 1; //the original(not reversed) index // e.g....BCD...BCD, the first BCD's start index = m - j - 1 // the length of BCD is f[j]; m - f[j] is the last BCD's start index // so m-f[j]-index is the shift distance from first B to the second B // if f[j] = 0, the goodSuffixShift[0] is always set with value 1 after // this loop int d2 = m - f[j] - index; //case 2 (f[j]>0) goodSuffixShift[f[j]] = d2; if(f[j]>0) { //only the last displayed k (rightmost) is valid System.out.format("k=%d: %d - > %d%n",f[j],index,m - f[j]); } } //case 1 goodSuffixShift[0] = 1; //j = 0; //case 3a and 3b (i.e. no substring fully overlapped suffix-k) int prefixMaxLen = f[m-1]; for(int k=1;k<=m-1;k++){ // goodSuffixShift[k]==0 is default value for int array //ie. no substring fully overlapped suffix-k //if(goodSuffixShift[k]==0 && prefixMaxLen<k){ if(goodSuffixShift[k]==0){ //if prefixMaxLen = 0, we hit case 3a; otherwise case 3b goodSuffixShift[k] = m - prefixMaxLen; System.out.format("k=%d: %d - > %d%n",k,0,m - prefixMaxLen); } } BM.printGoodSuffixShift(pattern,reverse,f,goodSuffixShift); return goodSuffixShift; } //return the first matched substring's position; //return -1 if no match public int match(String text,String pattern){ int n = text.length(); int m = pattern.length(); if(m>n) return -1; int[] badCharTable = makeBadCharTable(pattern); int[] goodSuffixShiftTable = makeGoodSuffixShiftTable(pattern); /****BEGIN TEST: the following code snippet can be commented out****/ StringBuilder sb = new StringBuilder(); for(int i=0;i<text.length();i++){ if(i%5==0){ sb.insert(i, String.valueOf(i)); } else{ sb.append(" "); } } System.out.format("%s%n",sb.toString()); System.out.format("%s%n",text); System.out.format("%s%n",pattern); /****END TEST: the above code snippet can be commented out****/ int i = m -1; int j = m -1; do{ int c = text.charAt(i); if(c == pattern.charAt(j)){ if(j==0){ //find a match return i; }else{ //BM algorithm: move from right to left i--; j--; } }else{ //use bad character shift int i_temp_badChar = i; //determine the i and j for next match attempt int p = badCharTable[c] + 1; int badCharShift = 1; if(j<=p){ i_temp_badChar += m - j; }else{ i_temp_badChar += m - p; badCharShift += j - p; } //or caculate the shift for bad character this way: //Note the fact: i_temp_badChar>i, j>j_last /* int badCharShift = (i_temp_badChar - i) - ((m-1) - j); if(badCharShift < 0) badCharShift = -badCharShift; */ //use good suffix shift //the length of good suffix int i_temp_goodSuffix = i; int k = m - j - 1; int goodSuffShift = goodSuffixShiftTable[k]; i_temp_goodSuffix += m - j + goodSuffShift - 1; //use the max shift between good-suffix and bad-character if(goodSuffShift > badCharShift) i = i_temp_goodSuffix; else i = i_temp_badChar; //BM algorithm: move j to the end of pattern j = m - 1; /****BEGIN TEST: the following code snippet can be commented out****/ int dotsCount = i - j; byte dot[] = new byte[dotsCount]; Arrays.fill(dot, (byte)'.'); System.out.format("%s%s%n",new String(dot),pattern); /****END TEST: the above code snippet can be commented out****/ } }while(i<=n-1); return -1; } //for test purpose only public static void printGoodSuffixShift(String pattern,String revPattern,int[] failureFunc,int[] goodSuffixShift){ //pattern index positions System.out.print("i:"); for(int i=0;i<pattern.length();i++){ System.out.format(" %2s", i); } System.out.println(); //the original pattern System.out.print("P:"); for(int i=0;i<pattern.length();i++){ System.out.format(" %2s", pattern.charAt(i)); } System.out.println(); //reversed pattern System.out.print("R:"); for(int i=0;i<revPattern.length();i++){ System.out.format(" %2s", revPattern.charAt(i)); } System.out.println(); //failure function output for reversed pattern System.out.print("f:"); for(int i=0;i<failureFunc.length;i++){ System.out.format(" %2d", failureFunc[i]); } System.out.println(); //good suffix shift System.out.println(); System.out.print("S:"); for(int k=0;k<goodSuffixShift.length;k++){ System.out.format(" %2d", goodSuffixShift[k]); } System.out.println(); System.out.println(); } //for test purpose only public static void findMatch(BM solver,String text,String pattern){ int index = solver.match(text, pattern); if(index>=0){ System.out.format("Found at position %d%n",index); }else{ System.out.format("No match%n"); } } public static void main(String[] args) { BM bm = new BM(); String pattern = "ABCBAB"; bm.makeGoodSuffixShiftTable(pattern); System.out.println("**********************"); pattern = "zzbc"; bm.makeGoodSuffixShiftTable(pattern); System.out.println("**********************"); pattern = "CBABAB"; bm.makeGoodSuffixShiftTable(pattern); System.out.println("**********************"); pattern = "ABABAB"; bm.makeGoodSuffixShiftTable(pattern); System.out.println("**********************"); String text = "A SLOW TURTLE"; pattern = "NEEDLE"; BM.findMatch(bm,text,pattern); System.out.println("**********************"); text = "After a long text, here's a needle ZZZZZ"; pattern = "ZZZZZ"; BM.findMatch(bm,text,pattern); System.out.println("**********************"); text = "The quick brown fox jumps over the lazy dog."; pattern = "lazy"; BM.findMatch(bm,text,pattern); System.out.println("**********************"); text = "Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna..."; pattern = "tempor"; BM.findMatch(bm,text,pattern); System.out.println("**********************"); text = "GGGGGGGGGGGGCGCAAAAGCGAGCAGAGAGAAAAAAAAAAAAAAAAAAAAAA"; pattern = "GCAGAGAG"; bm.makeGoodSuffixShiftTable(pattern); BM.findMatch(bm,text, pattern); } }
测试输出:
k=2: 0 - > 4 k=1: 1 - > 5 k=1: 3 - > 5 k=3: 0 - > 4 k=4: 0 - > 4 k=5: 0 - > 4 i: 0 1 2 3 4 5 P: A B C B A B R: B A B C B A f: 0 0 1 0 1 2 S: 1 2 4 4 4 4 ********************** k=1: 0 - > 4 k=2: 0 - > 4 k=3: 0 - > 4 i: 0 1 2 3 P: z z b c R: c b z z f: 0 0 0 0 S: 1 4 4 4 ********************** k=3: 1 - > 3 k=2: 2 - > 4 k=1: 3 - > 5 k=4: 0 - > 6 k=5: 0 - > 6 i: 0 1 2 3 4 5 P: C B A B A B R: B A B A B C f: 0 0 1 2 3 0 S: 1 2 2 2 6 6 ********************** k=4: 0 - > 2 k=3: 1 - > 3 k=2: 2 - > 4 k=1: 3 - > 5 k=5: 0 - > 2 i: 0 1 2 3 4 5 P: A B A B A B R: B A B A B A f: 0 0 1 2 3 4 S: 1 2 2 2 2 2 ********************** k=1: 1 - > 5 k=1: 2 - > 5 k=2: 0 - > 6 k=3: 0 - > 6 k=4: 0 - > 6 k=5: 0 - > 6 i: 0 1 2 3 4 5 P: N E E D L E R: E L D E E N f: 0 0 0 1 1 0 S: 1 3 6 6 6 6 0 5 10 A SLOW TURTLE NEEDLE ......NEEDLE .......NEEDLE .............NEEDLE No match ********************** k=4: 0 - > 1 k=3: 1 - > 2 k=2: 2 - > 3 k=1: 3 - > 4 i: 0 1 2 3 4 P: Z Z Z Z Z R: Z Z Z Z Z f: 0 1 2 3 4 S: 1 1 1 1 1 0 5 10 15 20 25 30 35 After a long text, here's a needle ZZZZZ ZZZZZ .....ZZZZZ ..........ZZZZZ ...............ZZZZZ ....................ZZZZZ .........................ZZZZZ ..............................ZZZZZ ...................................ZZZZZ Found at position 35 ********************** k=1: 0 - > 4 k=2: 0 - > 4 k=3: 0 - > 4 i: 0 1 2 3 P: l a z y R: y z a l f: 0 0 0 0 S: 1 4 4 4 0 5 10 15 20 25 30 35 40 The quick brown fox jumps over the lazy dog. lazy ....lazy ........lazy ............lazy ................lazy ....................lazy ........................lazy ............................lazy ................................lazy ...................................lazy Found at position 35 ********************** k=1: 0 - > 6 k=2: 0 - > 6 k=3: 0 - > 6 k=4: 0 - > 6 k=5: 0 - > 6 i: 0 1 2 3 4 5 P: t e m p o r R: r o p m e t f: 0 0 0 0 0 0 S: 1 6 6 6 6 6 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna... tempor ......tempor ............tempor ..................tempor .....................tempor ...........................tempor ...............................tempor ....................................tempor ..........................................tempor ................................................tempor ......................................................tempor ..........................................................tempor ...........................................................tempor .................................................................tempor ..................................................................tempor ........................................................................tempor .........................................................................tempor Found at position 73 ********************** k=1: 0 - > 7 k=4: 2 - > 4 k=3: 3 - > 5 k=2: 4 - > 6 k=1: 5 - > 7 k=5: 0 - > 7 k=6: 0 - > 7 k=7: 0 - > 7 i: 0 1 2 3 4 5 6 7 P: G C A G A G A G R: G A G A G A C G f: 0 0 1 2 3 4 0 1 S: 1 2 2 2 2 7 7 7 k=1: 0 - > 7 k=4: 2 - > 4 k=3: 3 - > 5 k=2: 4 - > 6 k=1: 5 - > 7 k=5: 0 - > 7 k=6: 0 - > 7 k=7: 0 - > 7 i: 0 1 2 3 4 5 6 7 P: G C A G A G A G R: G A G A G A C G f: 0 0 1 2 3 4 0 1 S: 1 2 2 2 2 7 7 7 0 5 10 15 20 25 30 35 40 45 50 GGGGGGGGGGGGCGCAAAAGCGAGCAGAGAGAAAAAAAAAAAAAAAAAAAAAA GCAGAGAG ..GCAGAGAG ....GCAGAGAG ......GCAGAGAG ...........GCAGAGAG ............GCAGAGAG ..............GCAGAGAG ...................GCAGAGAG .......................GCAGAGAG Found at position 23
BM文本匹配算法
作者:ljs
2011-06-21
(转载请注明出处,谢谢!)