# Boyer Moore Algorithm | Good Suffix heuristic

We have already discussed Bad character heuristic variation of Boyer Moore algorithm. In this article we will discuss **Good Suffix** heuristic for pattern searching. Just like bad character heuristic, a preprocessing table is generated for good suffix heuristic.

**Good Suffix Heuristic**

Let **t** be substring of text **T** which is matched with substring of pattern **P**. Now we shift pattern until :

1) Another occurrence of t in P matched with t in T.

2) A prefix of P, which matches with suffix of t

3) P moves past t

**Case 1: Another occurrence of t in P matched with t in T**

Pattern P might contain few more occurrences of **t**. In such case, we will try to shift the pattern to align that occurrence with t in text T. For example-

**Explanation:** In the above example, we have got a substring t of text T matched with pattern P (in green) before mismatch at index 2. Now we will search for occurrence of t (“AB”) in P. We have found an occurrence starting at position 1 (in yellow background) so we will right shift the pattern 2 times to align t in P with t in T. This is weak rule of original Boyer Moore and not much effective, we will discuss a **Strong Good Suffix rule** shortly.

**Case 2: A prefix of P, which matches with suffix of t in T**

It is not always likely that we will find the occurrence of t in P. Sometimes there is no occurrence at all, in such cases sometimes we can search for some **suffix of t** matching with some **prefix of P** and try to align them by shifting P. For example –

**Explanation:** In above example, we have got t (“BAB”) matched with P (in green) at index 2-4 before mismatch . But because there exists no occurrence of t in P we will search for some prefix of P which matches with some suffix of t. We have found prefix “AB” (in the yellow background) starting at index 0 which matches not with whole t but the suffix of t “AB” starting at index 3. So now we will shift pattern 3 times to align prefix with the suffix.

**Case 3: P moves past t**

If the above two cases are not satisfied, we will shift the pattern past the t. For example –

**Explanation:** If above example, there exist no occurrence of t (“AB”) in P and also there is no prefix in P which matches with the suffix of t. So, in that case, we can never find any perfect match before index 4, so we will shift the P past the t ie. to index 5.

**Strong Good suffix Heuristic**

Suppose substring **q = P[i to n]** got matched with **t** in T and **c = P[i-1]** is the mismatching character. Now unlike case 1 we will search for t in P which is not preceded by character **c**. The closest such occurrence is then aligned with t in T by shifting pattern P. For example –

**Explanation:** In above example, **q = P[7 to 8]** got matched with t in T. The mismatching character **c** is “C” at position P[6]. Now if we start searching t in P we will get the first occurrence of t starting at position 4. But this occurrence is preceded by “C” which is equal to c, so we will skip this and carry on searching. At position 1 we got another occurrence of t (in the yellow background). This occurrence is preceded by “A” (in blue) which is not equivalent to c. So we will shift pattern P 6 times to align this occurrence with t in T.We are doing this because we already know that character **c = “C”** causes the mismatch. So any occurrence of t preceded by c will again cause mismatch when aligned with t, so that’s why it is better to skip this.

**Preprocessing for Good suffix heuristic**

As a part of preprocessing, an array **shift** is created. Each entry **shift[i]** contain the distance pattern will shift if mismatch occur at position **i-1**. That is, the suffix of pattern starting at position **i** is matched and a mismatch occur at position **i-1**. Preprocessing is done separately for strong good suffix and case 2 discussed above.

**1) Preprocessing for Strong Good Suffix**

Before discussing preprocessing, let us first discuss the idea of border. A **border** is a substring which is both proper suffix and proper prefix. For example, in string **“ccacc”**, **“c”** is a border, **“cc”** is a border because it appears in both end of string but **“cca”** is not a border.

As a part of preprocessing an array **bpos** (border position) is calculated. Each entry **bpos[i]** contains the starting index of border for suffix starting at index i in given pattern P.

The suffix **φ** beginning at position m has no border, so **bpos[m]** is set to **m+1** where **m** is the length of the pattern.

The shift position is obtained by the borders which cannot be extended to the left. Following is the code for preprocessing –

void preprocess_strong_suffix(int *shift, int *bpos, char *pat, int m) { int i = m, j = m+1; bpos[i] = j; while(i > 0) { while(j <= m && pat[i-1] != pat[j-1]) { if (shift[j] == 0) shift[j] = j-i; j = bpos[j]; } i--; j--; bpos[i] = j; } }

**Explanation:** Consider pattern **P = “ABBABAB”, m = 7**.

- The widest border of suffix “AB” beginning at position i = 5 is φ(nothing) starting at position 7 so bpos[5] = 7.
- At position i = 2 the suffix is “BABAB”. The widest border for this suffix is “BAB” starting at position 4, so j = bpos[2] = 4.

We can understand **bpos[i] = j** using following example –

If character **#** Which is at position **i-1** is equivalent to character **?** at position **j-1**, we know that border will be **? + border of suffix at position i** which is starting at position **j** which is equivalent to saying that **border of suffix at i-1 begin at j-1 ** or **bpos[ i-1 ] = j-1** or in the code –

i--; j--; bpos[ i ] = j

But if character **#** at position **i-1** do not match with character **?** at position **j-1** then we continue our search to the right. Now we know that –

- Border width will be smaller than the border starting at position
**j**ie. smaller than**x…φ** - Border has to begin with
**#**and end with**φ**or could be empty (no border exist).

With above two facts we will continue our search in sub string **x…φ** from position **j to m**. The next border should be at **j = bpos[j]**. After updating **j**, we again compare character at position **j-1 (?)** with # and if they are equal then we got our border otherwise we continue our search to right **until j>m**. This process is shown by code –

while(j <= m && pat[i-1] != pat[j-1]) { j = bpos[j]; } i--; j--; bpos[i]=j;

In above code look at these conditions –

pat[i-1] != pat[j-1]

This is the condition which we discussed in case 2. When the character preceding the occurrence of t in pattern P is different than mismatching character in P, we stop skipping the occurrences and shift the pattern. So here **P[i] == P[j]** but **P[i-1] != p[j-1]** so we shift pattern from **i to j**. So **shift[j] = j-i** is recorder for **j**. So whenever any mismatch occur at position **j** we will shift the pattern **shift[j+1]** positions to the right.

In above code the following condition is very important –

if (shift[j] == 0 )

This condition prevent modification of **shift[j]** value from suffix having same border. For example, Consider pattern **P = “addbddcdd”**, when we calculate bpos[ i-1 ] for i = 4 then j = 7 in this case. we will be eventually setting value of shift[ 7 ] = 3. Now if we calculate bpos[ i-1 ] for i = 1 then j = 7 and we will be setting value shift[ 7 ] = 6 again if there is no test shift[ j ] == 0. This mean if we have a mismatch at position 6 we will shift pattern P 3 positions to right not 6 position.

**2) Preprocessing for Case 2**

In the preprocessing for case 2, for each suffix the **widest border of the whole pattern** that is contained in that suffix is determined.

The starting position of the widest border of the pattern at all is stored in **bpos[0]**

In the following preprocessing algorithm, this value bpos[0] is stored initially in all free entries of array shift. But when the suffix of the pattern becomes shorter than bpos[0], the algorithm continues with the next-wider border of the pattern, i.e. with bpos[j].

Following is the C implementation of the search algorithm –

`/* C program for Boyer Moore Algorithm with ` ` ` `Good Suffix heuristic to find pattern in ` ` ` `given text string */` ` ` `#include <stdio.h> ` `#include <string.h> ` ` ` `// preprocessing for strong good suffix rule ` `void` `preprocess_strong_suffix(` `int` `*shift, ` `int` `*bpos, ` ` ` `char` `*pat, ` `int` `m) ` `{ ` ` ` `// m is the length of pattern ` ` ` `int` `i=m, j=m+1; ` ` ` `bpos[i]=j; ` ` ` ` ` `while` `(i>0) ` ` ` `{ ` ` ` `/*if character at position i-1 is not equivalent to ` ` ` `character at j-1, then continue searching to right ` ` ` `of the pattern for border */` ` ` `while` `(j<=m && pat[i-1] != pat[j-1]) ` ` ` `{ ` ` ` `/* the character preceding the occurrence of t in ` ` ` `pattern P is different than the mismatching character in P, ` ` ` `we stop skipping the occurrences and shift the pattern ` ` ` `from i to j */` ` ` `if` `(shift[j]==0) ` ` ` `shift[j] = j-i; ` ` ` ` ` `//Update the position of next border ` ` ` `j = bpos[j]; ` ` ` `} ` ` ` `/* p[i-1] matched with p[j-1], border is found. ` ` ` `store the beginning position of border */` ` ` `i--;j--; ` ` ` `bpos[i] = j; ` ` ` `} ` `} ` ` ` `//Preprocessing for case 2 ` `void` `preprocess_case2(` `int` `*shift, ` `int` `*bpos, ` ` ` `char` `*pat, ` `int` `m) ` `{ ` ` ` `int` `i, j; ` ` ` `j = bpos[0]; ` ` ` `for` `(i=0; i<=m; i++) ` ` ` `{ ` ` ` `/* set the border position of the first character of the pattern ` ` ` `to all indices in array shift having shift[i] = 0 */` ` ` `if` `(shift[i]==0) ` ` ` `shift[i] = j; ` ` ` ` ` `/* suffix becomes shorter than bpos[0], use the position of ` ` ` `next widest border as value of j */` ` ` `if` `(i==j) ` ` ` `j = bpos[j]; ` ` ` `} ` `} ` ` ` `/*Search for a pattern in given text using ` ` ` `Boyer Moore algorithm with Good suffix rule */` `void` `search(` `char` `*text, ` `char` `*pat) ` `{ ` ` ` `// s is shift of the pattern with respect to text ` ` ` `int` `s=0, j; ` ` ` `int` `m = ` `strlen` `(pat); ` ` ` `int` `n = ` `strlen` `(text); ` ` ` ` ` `int` `bpos[m+1], shift[m+1]; ` ` ` ` ` `//initialize all occurrence of shift to 0 ` ` ` `for` `(` `int` `i=0;i<m+1;i++) shift[i]=0; ` ` ` ` ` `//do preprocessing ` ` ` `preprocess_strong_suffix(shift, bpos, pat, m); ` ` ` `preprocess_case2(shift, bpos, pat, m); ` ` ` ` ` `while` `(s <= n-m) ` ` ` `{ ` ` ` ` ` `j = m-1; ` ` ` ` ` `/* Keep reducing index j of pattern while characters of ` ` ` `pattern and text are matching at this shift s*/` ` ` `while` `(j >= 0 && pat[j] == text[s+j]) ` ` ` `j--; ` ` ` ` ` `/* If the pattern is present at the current shift, then index j ` ` ` `will become -1 after the above loop */` ` ` `if` `(j<0) ` ` ` `{ ` ` ` `printf` `(` `"pattern occurs at shift = %d\n"` `, s); ` ` ` `s += shift[0]; ` ` ` `} ` ` ` `else` ` ` `/*pat[i] != pat[s+j] so shift the pattern ` ` ` `shift[j+1] times */` ` ` `s += shift[j+1]; ` ` ` `} ` ` ` `} ` ` ` `//Driver ` `int` `main() ` `{ ` ` ` `char` `text[] = ` `"ABAAAABAACD"` `; ` ` ` `char` `pat[] = ` `"ABA"` `; ` ` ` `search(text, pat); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

pattern occurs at shift = 0 pattern occurs at shift = 5

**References**

This article is contributed by **Atul Kumar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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