Burrows - Wheeler Data Transform Algorithm - GeeksforGeeks

08 Aug, 2019
What is the Burrows-Wheeler Transform?The BWT is a data transformation algorithm that restructures data in such a way that the transformed message is more compressible. Technically, it is a lexicographical reversible permutation of the characters of a string. It is first of the three steps to be performed in succession while implementing the Burrows-Wheeler Data Compression algorithm that forms the basis of the Unix compression utility bzip2.
Why BWT? The main idea behind it.The most important application of BWT is found in biological sciences where genomes(long strings written in A, C, T, G alphabets) don’t have many runs but they do have many repeats.The idea of the BWT is to build an array whose rows are all cyclic shifts of the input string in dictionary order and return the last column of the array that tends to have long runs of identical characters. The benefit of this is that once the characters have been clustered together, they effectively have an ordering, which can make our string more compressible for other algorithms like run-length encoding and Huffman Coding.The remarkable thing about BWT is that this particular transform is reversible with minimal data overhead.
Steps involved in BWT algorithmLet’s take the word “banana$” as an example.
Step 1: Form all cyclic rotations of the given text.                                     banana$ 
       $    b                        $banana 
    a           a                    a$banan
   Cyclic rotations    ---------->   na$bana
    n         n                      ana$ban 
          a                          nana$ba
                                     anana$b

                                     banana$ 
       $    b                        $banana 
    a           a                    a$banan
   Cyclic rotations    ---------->   na$bana
    n         n                      ana$ban 
          a                          nana$ba
                                     anana$b

Step 2: The next step is to sort the rotations lexicographically. The ‘$’ sign is viewed as first letter lexicographically, even before ‘a’.banana$                    $banana
$banana                    a$banan
a$banan       Sorting      ana$ban
na$bana      ---------->   anana$b 
ana$ban    alphabetically  banana$
nana$ba                    na$bana
anana$b                    nana$ba

banana$                    $banana
$banana                    a$banan
a$banan       Sorting      ana$ban
na$bana      ---------->   anana$b 
ana$ban    alphabetically  banana$
nana$ba                    na$bana
anana$b                    nana$ba

Step 3: The last column is what we output as BWT.BWT(banana$) = annb$aa

BWT(banana$) = annb$aa

Examples:
Input: text = “banana$”Output: Burrows-Wheeler Transform = “annb$aa”
Input: text = “abracadabra$”Output: Burrows-Wheeler Transform = “ard$rcaaaabb”
Why last column is considered BWT?
The last column has a better symbol clustering than any other columns.If we only have BWT of our string, we can recover the rest of the cyclic rotations entirely. The rest of the columns don’t possess this characteristic which is highly important while computing the inverse of BWT.
The last column has a better symbol clustering than any other columns.
If we only have BWT of our string, we can recover the rest of the cyclic rotations entirely. The rest of the columns don’t possess this characteristic which is highly important while computing the inverse of BWT.
Why ‘$’ sign is embedded in the text?We can compute BWT even if our text is not concatenated with any EOF character (‘$’ here). The implication of ‘$’ sign comes while computing the inverse of BWT.
Way of implementation
Let’s instantiate “banana$” as our input_text and instantiate character array bwt_arr for our output.Let’s get all the suffixes of “banana$” and compute it’s suffix_arr to store index of each suffix.0 banana$                6 $   
1 anana$                 5 a$
2 nana$      Sorting     3 ana$
3 ana$     ---------->   1 anana$
4 na$     alphabetically 0 banana$
5 a$                     4 na$
6 $                      2 nana$
Iterating over the suffix_arr, let’s now add to our output array bwt_arr, the last character of each rotation.The last character of each rotation of input_text starting at the position denoted by the current value in the suffix array can be calculated with input_text[(suffix_arr[i] – 1 + n ) % n], where n is the number of elements in the suffix_arr.bwt_arr[0] 
  = input_text[(suffix_arr[0] - 1 + 7) % 7] 
  = input_text[5] 
  = a
bwt_arr[1] 
  = input_text[(suffix_arr[1] - 1 + 7) % 7] 
  = input_text[4] 
  = n

Let’s instantiate “banana$” as our input_text and instantiate character array bwt_arr for our output.
Let’s get all the suffixes of “banana$” and compute it’s suffix_arr to store index of each suffix.0 banana$                6 $   
1 anana$                 5 a$
2 nana$      Sorting     3 ana$
3 ana$     ---------->   1 anana$
4 na$     alphabetically 0 banana$
5 a$                     4 na$
6 $                      2 nana$

0 banana$                6 $   
1 anana$                 5 a$
2 nana$      Sorting     3 ana$
3 ana$     ---------->   1 anana$
4 na$     alphabetically 0 banana$
5 a$                     4 na$
6 $                      2 nana$

Iterating over the suffix_arr, let’s now add to our output array bwt_arr, the last character of each rotation.
The last character of each rotation of input_text starting at the position denoted by the current value in the suffix array can be calculated with input_text[(suffix_arr[i] – 1 + n ) % n], where n is the number of elements in the suffix_arr.bwt_arr[0] 
  = input_text[(suffix_arr[0] - 1 + 7) % 7] 
  = input_text[5] 
  = a
bwt_arr[1] 
  = input_text[(suffix_arr[1] - 1 + 7) % 7] 
  = input_text[4] 
  = n

bwt_arr[0] 
  = input_text[(suffix_arr[0] - 1 + 7) % 7] 
  = input_text[5] 
  = a
bwt_arr[1] 
  = input_text[(suffix_arr[1] - 1 + 7) % 7] 
  = input_text[4] 
  = n

Following is the code for the way of implementation explained above
// C program to find Burrows Wheeler transform// of a given text  #include <stdio.h>#include <stdlib.h>#include <string.h>  // Structure to store data of a rotationstruct rotation {    int index;    char* suffix;};  // Compares the rotations and// sorts the rotations alphabeticallyint cmpfunc(const void* x, const void* y){    struct rotation* rx = (struct rotation*)x;    struct rotation* ry = (struct rotation*)y;    return strcmp(rx->suffix, ry->suffix);}  // Takes text to be transformed and its length as// arguments and returns the corresponding suffix arrayint* computeSuffixArray(char* input_text, int len_text){    // Array of structures to store rotations and    // their indexes    struct rotation suff[len_text];      // Structure is needed to maintain old indexes of    // rotations after sorting them    for (int i = 0; i < len_text; i++) {        suff[i].index = i;        suff[i].suffix = (input_text + i);    }      // Sorts rotations using comparison    // function defined above    qsort(suff, len_text, sizeof(struct rotation),          cmpfunc);      // Stores the indexes of sorted rotations    int* suffix_arr        = (int*)malloc(len_text * sizeof(int));    for (int i = 0; i < len_text; i++)        suffix_arr[i] = suff[i].index;      // Returns the computed suffix array    return suffix_arr;}  // Takes suffix array and its size// as arguments and returns the// Burrows - Wheeler Transform of given textchar* findLastChar(char* input_text,                   int* suffix_arr, int n){    // Iterates over the suffix array to find    // the last char of each cyclic rotation    char* bwt_arr = (char*)malloc(n * sizeof(char));    int i;    for (i = 0; i < n; i++) {        // Computes the last char which is given by        // input_text[(suffix_arr[i] + n - 1) % n]        int j = suffix_arr[i] - 1;        if (j < 0)            j = j + n;          bwt_arr[i] = input_text[j];    }      bwt_arr[i] = '\0';      // Returns the computed Burrows - Wheeler Transform    return bwt_arr;}  // Driver program to test functions aboveint main(){    char input_text[] = "banana$";    int len_text = strlen(input_text);      // Computes the suffix array of our text    int* suffix_arr        = computeSuffixArray(input_text, len_text);      // Adds to the output array the last char    // of each rotation    char* bwt_arr        = findLastChar(input_text, suffix_arr, len_text);      printf("Input text : %s\n", input_text);    printf("Burrows - Wheeler Transform : %s\n",           bwt_arr);    return 0;}
Input text : banana$
Burrows - Wheeler Transform : annb$aa

Time Complexity: O(Logn). This is because of the method used above to build suffix array which has O(Logn) time complexity, due to O(n) time for strings comparisons in O(nLogn) sorting algorithm.
Exercise:
Compute suffix array in O(nLogn) time and then implement BWT.Implement Inverse of Burrows-Wheeler Transform.
Compute suffix array in O(nLogn) time and then implement BWT.
Implement Inverse of Burrows-Wheeler Transform.
This article is contributed by Anureet Kaur. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Suffix-Array
Advanced Data Structure
Strings
Strings
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Decision Tree Introduction with example
AVL Tree | Set 2 (Deletion)
Ordered Set and GNU C++ PBDS
Red-Black Tree | Set 2 (Insert)
Disjoint Set Data Structures
Write a program to reverse an array or string
Reverse a string in Java
Write a program to print all permutations of a given string
C++ Data Types
Longest Common Subsequence | DP-4