/** * Implementation of the Boyer-Moore Algorithm for pattern matching. * @author V.Boutchkova */ public class BoyerMoore { public static final int ALPHABET_SIZE = Character.MAX_VALUE + 1; private String text; private String pattern; private int[] last; private int[] match; private int[] suffix; public BoyerMoore(String pattern, String text) { this.text = text; this.pattern = pattern; last = new int[ALPHABET_SIZE]; match = new int[pattern.length()]; suffix = new int[pattern.length()]; } /** * Searches the pattern in the text. * Returns the position of the first occurrence, if found and -1 otherwise. */ public int match() { // Preprocessing computeLast(); computeMatch(); // Searching int i = pattern.length() - 1; int j = pattern.length() - 1; while (i < text.length()) { if (pattern.charAt(j) == text.charAt(i)) { if (j == 0) { //the left-most match is found return i; } j--; i--; } else { //a difference i += pattern.length() - j - 1 + Math.max(j - last[text.charAt(i)], match[j]); j = pattern.length() - 1; } } return -1; } /** * Computes the function last and stores its values in the array last. * The function is defined as follows: *
   * last(Char ch) = the index of the right-most occurrence of the character ch
   *                                                           in the pattern; 
   *                 -1 if ch does not occur in the pattern.
   * 
* The running time is O(pattern.length() + |Alphabet|). */ private void computeLast() { for (int k = 0; k < last.length; k++) { last[k] = -1; } for (int j = pattern.length()-1; j >= 0; j--) { if (last[pattern.charAt(j)] < 0) { last[pattern.charAt(j)] = j; } } } /** * Computes the function match and stores its values in the array match. * The function is defined as follows: *
   * match(j) = min{ s | 0 < s <= j && p[j-s]!=p[j]
   *                            && p[j-s+1]..p[m-s-1] is suffix of p[j+1]..p[m-1] }, 
   *                                                         if such s exists, else
   *            min{ s | j+1 <= s <= m 
   *                            && p[0]..p[m-s-1] is suffix of p[j+1]..p[m-1] }, 
   *                                                         if such s exists,
   *            m, otherwise,
   * where m is the pattern's length and p is the pattern.
   * 
* The running time is O(pattern.length()). */ private void computeMatch() { /* Phase 1 */ for (int j = 0; j < match.length; j++) { match[j] = match.length; } //O(m) computeSuffix(); //O(m) /* Phase 2 */ //Uses an auxiliary array, backwards version of the KMP failure function. //suffix[i] = the smallest j > i s.t. p[j..m-1] is a prefix of p[i..m-1], //if there is no such j, suffix[i] = m //Compute the smallest shift s, such that 0 < s <= j and //p[j-s]!=p[j] and p[j-s+1..m-s-1] is suffix of p[j+1..m-1] or j == m-1}, // if such s exists, for (int i = 0; i < match.length - 1; i++) { int j = suffix[i + 1] - 1; // suffix[i+1] <= suffix[i] + 1 if (suffix[i] > j) { // therefore pattern[i] != pattern[j] match[j] = j - i; } else {// j == suffix[i] match[j] = Math.min(j - i + match[i], match[j]); } } //End of Phase 2 /* Phase 3 */ //Uses the suffix array to compute each shift s such that //p[0..m-s-1] is a suffix of p[j+1..m-1] with j < s < m //and stores the minimum of this shift and the previously computed one. if (suffix[0] < pattern.length()) { for (int j = suffix[0] - 1; j >= 0; j--) { if (suffix[0] < match[j]) { match[j] = suffix[0]; } } int j = suffix[0]; for (int k = suffix[j]; k < pattern.length(); k = suffix[k]) { while (j < k) { if (match[j] > k) match[j] = k; j++; } } }//endif } /** * Computes the values of suffix, which is an auxiliary array, * backwards version of the KMP failure function. *
* suffix[i] = the smallest j > i s.t. p[j..m-1] is a prefix of p[i..m-1], * if there is no such j, suffix[i] = m, i.e.
* p[suffix[i]..m-1] is the longest prefix of p[i..m-1], if suffix[i] < m. *
* The running time for computing the suffix is O(m). */ private void computeSuffix() { suffix[suffix.length-1] = suffix.length; int j = suffix.length - 1; //suffix[i] = m - the length of the longest prefix of p[i..m-1] for (int i = suffix.length - 2; i >= 0; i--) { while (j < suffix.length - 1 && pattern.charAt(j) != pattern.charAt(i)) { j = suffix[j + 1] - 1; } if (pattern.charAt(j) == pattern.charAt(i)) { j--; } suffix[i] = j + 1; } } }