Boyer Moore Algorithm for Pattern Searching
Last Updated :
11 Mar, 2024
Pattern searching is an important problem in computer science. When we do search for a string in a notepad/word file, browser, or database, pattern searching algorithms are used to show the search results.
A typical problem statement would be-
” Given a text txt[0..n-1] and a pattern pat[0..m-1] where n is the length of the text and m is the length of the pattern, write a function search(char pat[], char txt[]) that prints all occurrences of pat[] in txt[]. You may assume that n > m. “
Examples:
Input: txt[] = “THIS IS A TEST TEXT”
pat[] = “TEST”
Output: Pattern found at index 10
Input: txt[] = “AABAACAADAABAABA”
pat[] = “AABA”
Output: Pattern found at index 0
Pattern found at index 9
Pattern found at index 12
In this post, we will discuss the Boyer Moore pattern searching algorithm. Like KMP and Finite Automata algorithms, Boyer Moore algorithm also preprocesses the pattern.
Boyer Moore is a combination of the following two approaches.
- Bad Character Heuristic
- Good Suffix Heuristic
Both of the above heuristics can also be used independently to search a pattern in a text. Let us first understand how two independent approaches work together in the Boyer Moore algorithm.
If we take a look at the Naive algorithm, it slides the pattern over the text one by one. KMP algorithm does preprocessing over the pattern so that the pattern can be shifted by more than one. The Boyer Moore algorithm does preprocessing for the same reason. It processes the pattern and creates different arrays for each of the two heuristics. At every step, it slides the pattern by the max of the slides suggested by each of the two heuristics. So, it uses greatest offset suggested by the two heuristics at every step.
Unlike the previous pattern searching algorithms, the Boyer Moore algorithm starts matching from the last character of the pattern.
In this post, we will discuss the bad character heuristic and the Good Suffix heuristic in the next post.
Bad Character Heuristic
The idea of bad character heuristic is simple. The character of the text which doesn’t match with the current character of the pattern is called the Bad Character. Upon mismatch, we shift the pattern until –
- The mismatch becomes a match.
- Pattern P moves past the mismatched character.
Case 1 – Mismatch become match
We will lookup the position of the last occurrence of the mismatched character in the pattern, and if the mismatched character exists in the pattern, then we’ll shift the pattern such that it becomes aligned to the mismatched character in the text T.
Explanation:
In the above example, we got a mismatch at position 3.
Here our mismatching character is “A”. Now we will search for last occurrence of “A” in pattern. We got “A” at position 1 in pattern (displayed in Blue) and this is the last occurrence of it. Now we will shift pattern 2 times so that “A” in pattern get aligned with “A” in text.
Case 2 – Pattern move past the mismatch character
We’ll lookup the position of last occurrence of mismatching character in pattern and if character does not exist we will shift pattern past the mismatching character.
Explanation:
Here we have a mismatch at position 7.
The mismatching character “C” does not exist in pattern before position 7 so we’ll shift pattern past to the position 7 and eventually in above example we have got a perfect match of pattern (displayed in Green). We are doing this because “C” does not exist in the pattern so at every shift before position 7 we will get mismatch and our search will be fruitless.
Implementation:
In the following implementation, we pre-process the pattern and store the last occurrence of every possible character in an array of size equal to alphabet size. If the character is not present at all, then it may result in a shift by m (length of pattern). Therefore, the bad character heuristic takes O(n/m) time in the best case.
Below is the implementation of the above idea:
C++
#include <bits/stdc++.h>
using namespace std;
#define NO_OF_CHARS 256
void badCharHeuristic(string str, int size,
int badchar[NO_OF_CHARS])
{
int i;
for (i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
for (i = 0; i < size; i++)
badchar[(int)str[i]] = i;
}
void search(string txt, string pat)
{
int m = pat.size();
int n = txt.size();
int badchar[NO_OF_CHARS];
badCharHeuristic(pat, m, badchar);
int s = 0;
while (s <= (n - m)) {
int j = m - 1;
while (j >= 0 && pat[j] == txt[s + j])
j--;
if (j < 0) {
cout << "pattern occurs at shift = " << s
<< endl;
s += (s + m < n) ? m - badchar[txt[s + m]] : 1;
}
else
s += max(1, j - badchar[txt[s + j]]);
}
}
int main()
{
string txt = "ABAAABCD";
string pat = "ABC";
search(txt, pat);
return 0;
}
|
C
#include <limits.h>
#include <stdio.h>
#include <string.h>
#define NO_OF_CHARS 256
int max(int a, int b) { return (a > b) ? a : b; }
void badCharHeuristic(char* str, int size,
int badchar[NO_OF_CHARS])
{
int i;
for (i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
for (i = 0; i < size; i++)
badchar[(int)str[i]] = i;
}
void search(char* txt, char* pat)
{
int m = strlen(pat);
int n = strlen(txt);
int badchar[NO_OF_CHARS];
badCharHeuristic(pat, m, badchar);
int s = 0;
while (s <= (n - m)) {
int j = m - 1;
while (j >= 0 && pat[j] == txt[s + j])
j--;
if (j < 0) {
printf("\n pattern occurs at shift = %d", s);
s += (s + m < n) ? m - badchar[txt[s + m]] : 1;
}
else
s += max(1, j - badchar[txt[s + j]]);
}
}
int main()
{
char txt[] = "ABAAABCD";
char pat[] = "ABC";
search(txt, pat);
return 0;
}
|
Java
class AWQ {
static int NO_OF_CHARS = 256;
static int max(int a, int b) { return (a > b) ? a : b; }
static void badCharHeuristic(char[] str, int size,
int badchar[])
{
for (int i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
for (int i = 0; i < size; i++)
badchar[(int)str[i]] = i;
}
static void search(char txt[], char pat[])
{
int m = pat.length;
int n = txt.length;
int badchar[] = new int[NO_OF_CHARS];
badCharHeuristic(pat, m, badchar);
int s = 0;
while (s <= (n - m)) {
int j = m - 1;
while (j >= 0 && pat[j] == txt[s + j])
j--;
if (j < 0) {
System.out.println(
"Patterns occur at shift = " + s);
s += (s + m < n) ? m - badchar[txt[s + m]]
: 1;
}
else
s += max(1, j - badchar[txt[s + j]]);
}
}
public static void main(String[] args)
{
char txt[] = "ABAAABCD".toCharArray();
char pat[] = "ABC".toCharArray();
search(txt, pat);
}
}
|
C#
using System;
public class AWQ {
static int NO_OF_CHARS = 256;
static int max(int a, int b) { return (a > b) ? a : b; }
static void badCharHeuristic(char[] str, int size,
int[] badchar)
{
int i;
for (i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
for (i = 0; i < size; i++)
badchar[(int)str[i]] = i;
}
static void search(char[] txt, char[] pat)
{
int m = pat.Length;
int n = txt.Length;
int[] badchar = new int[NO_OF_CHARS];
badCharHeuristic(pat, m, badchar);
int s = 0;
while (s <= (n - m)) {
int j = m - 1;
while (j >= 0 && pat[j] == txt[s + j])
j--;
if (j < 0) {
Console.WriteLine(
"Patterns occur at shift = " + s);
s += (s + m < n) ? m - badchar[txt[s + m]]
: 1;
}
else
s += max(1, j - badchar[txt[s + j]]);
}
}
public static void Main()
{
char[] txt = "ABAAABCD".ToCharArray();
char[] pat = "ABC".ToCharArray();
search(txt, pat);
}
}
|
Javascript
<script>
let NO_OF_CHARS = 256;
function max (a,b)
{
return (a > b)? a: b;
}
function badCharHeuristic(str,size,badchar)
{
for (let i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
for (i = 0; i < size; i++)
badchar[ str[i].charCodeAt(0)] = i;
}
function search(txt,pat)
{
let m = pat.length;
let n = txt.length;
let badchar = new Array(NO_OF_CHARS);
badCharHeuristic(pat, m, badchar);
let s = 0;
while(s <= (n - m))
{
let j = m-1;
while(j >= 0 && pat[j] == txt[s+j])
j--;
if (j < 0)
{
document.write("Patterns occur at shift = " + s);
s += (s+m < n)? m-badchar[txt[s+m].charCodeAt(0)] : 1;
}
else
s += max(1, j - badchar[txt[s+j].charCodeAt(0)]);
}
}
let txt="ABAAABCD".split("");
let pat = "ABC".split("");
search(txt, pat);
</script>
|
Python3
NO_OF_CHARS = 256
def badCharHeuristic(string, size):
badChar = [-1]*NO_OF_CHARS
for i in range(size):
badChar[ord(string[i])] = i
return badChar
def search(txt, pat):
m = len(pat)
n = len(txt)
badChar = badCharHeuristic(pat, m)
s = 0
while(s <= n-m):
j = m-1
while j >= 0 and pat[j] == txt[s+j]:
j -= 1
if j < 0:
print("Pattern occur at shift = {}".format(s))
s += (m-badChar[ord(txt[s+m])] if s+m < n else 1)
else:
s += max(1, j-badChar[ord(txt[s+j])])
def main():
txt = "ABAAABCD"
pat = "ABC"
search(txt, pat)
if __name__ == '__main__':
main()
|
Output
pattern occurs at shift = 4
Time Complexity : O(m*n)
Auxiliary Space: O(1)
The Bad Character Heuristic may take O(m*n) time in worst case. The worst case occurs when all characters of the text and pattern are same. For example, txt[] = “AAAAAAAAAAAAAAAAAA” and pat[] = “AAAAA”. The Bad Character Heuristic may take O(n/m) in the best case. The best case occurs when all the characters of the text and pattern are different.
Boyer Moore Algorithm | Good Suffix heuristic
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