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 algorithm
Let’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- 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
- Step 3: The last column is what we output as BWT.
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.
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
Following is the code for the way of implementation explained above
C++
#include <bits/stdc++.h>
using namespace std;
struct rotation {
int index;
char* suffix;
};
int 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);
}
int* computeSuffixArray(char* input_text, int len_text)
{
struct rotation suff[len_text];
for (int i = 0; i < len_text; i++) {
suff[i].index = i;
suff[i].suffix = (input_text + i);
}
qsort(suff, len_text, sizeof(struct rotation), cmpfunc);
int* suffix_arr = (int*)malloc(len_text * sizeof(int));
for (int i = 0; i < len_text; i++)
suffix_arr[i] = suff[i].index;
return suffix_arr;
}
char* findLastChar(char* input_text, int* suffix_arr, int n)
{
char* bwt_arr = (char*)malloc(n * sizeof(char));
int i;
for (i = 0; i < n; i++) {
int j = suffix_arr[i] - 1;
if (j < 0)
j = j + n;
bwt_arr[i] = input_text[j];
}
bwt_arr[i] = '\0';
return bwt_arr;
}
int main()
{
char input_text[] = "banana$";
int len_text = strlen(input_text);
int* suffix_arr
= computeSuffixArray(input_text, len_text);
char* bwt_arr
= findLastChar(input_text, suffix_arr, len_text);
cout << "Input text : " << input_text << endl;
cout << "Burrows - Wheeler Transform : " << bwt_arr
<< endl;
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct rotation {
int index;
char* suffix;
};
int cmpfunc(const void* x, const void* y)
{
struct rotation* rx = (struct rotation*)x;
struct rotation* ry = (struct rotation*)y;
return strcmp(rx->suffix, ry->suffix);
}
int* computeSuffixArray(char* input_text, int len_text)
{
struct rotation suff[len_text];
for (int i = 0; i < len_text; i++) {
suff[i].index = i;
suff[i].suffix = (input_text + i);
}
qsort(suff, len_text, sizeof(struct rotation),
cmpfunc);
int* suffix_arr
= (int*)malloc(len_text * sizeof(int));
for (int i = 0; i < len_text; i++)
suffix_arr[i] = suff[i].index;
return suffix_arr;
}
char* findLastChar(char* input_text,
int* suffix_arr, int n)
{
char* bwt_arr = (char*)malloc(n * sizeof(char));
int i;
for (i = 0; i < n; i++) {
int j = suffix_arr[i] - 1;
if (j < 0)
j = j + n;
bwt_arr[i] = input_text[j];
}
bwt_arr[i] = '\0';
return bwt_arr;
}
int main()
{
char input_text[] = "banana$";
int len_text = strlen(input_text);
int* suffix_arr
= computeSuffixArray(input_text, len_text);
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;
}
|
Java
import java.util.Arrays;
class Rotation implements Comparable<Rotation> {
int index;
String suffix;
public Rotation(int index, String suffix)
{
this.index = index;
this.suffix = suffix;
}
@Override public int compareTo(Rotation o)
{
return this.suffix.compareTo(o.suffix);
}
}
public class BurrowsWheelerTransform {
public static int[] computeSuffixArray(String inputText)
{
int lenText = inputText.length();
Rotation[] suff = new Rotation[lenText];
for (int i = 0; i < lenText; i++) {
suff[i]
= new Rotation(i, inputText.substring(i));
}
Arrays.sort(suff);
int[] suffixArr = new int[lenText];
for (int i = 0; i < lenText; i++) {
suffixArr[i] = suff[i].index;
}
return suffixArr;
}
public static String findLastChar(String inputText,
int[] suffixArr)
{
int n = inputText.length();
StringBuilder bwtArr = new StringBuilder();
for (int i = 0; i < n; i++) {
int j = suffixArr[i] - 1;
if (j < 0) {
j = j + n;
}
bwtArr.append(inputText.charAt(j));
}
return bwtArr.toString();
}
public static void main(String[] args)
{
String inputText = "banana$";
int[] suffixArr = computeSuffixArray(inputText);
String bwtArr = findLastChar(inputText, suffixArr);
System.out.println("Input text : " + inputText);
System.out.println("Burrows - Wheeler Transform : "
+ bwtArr);
}
}
|
Python3
def cmp_func(x, y):
return (x[1] > y[1]) - (x[1] < y[1])
def compute_suffix_array(input_text, len_text):
suff = [(i, input_text[i:]) for i in range(len_text)]
suff.sort(key=lambda x: x[1])
suffix_arr = [i for i, _ in suff]
return suffix_arr
def find_last_char(input_text, suffix_arr, n):
bwt_arr = ""
for i in range(n):
j = suffix_arr[i] - 1
if j < 0:
j = j + n
bwt_arr += input_text[j]
return bwt_arr
input_text = "banana$"
len_text = len(input_text)
suffix_arr = compute_suffix_array(input_text, len_text)
bwt_arr = find_last_char(input_text, suffix_arr, len_text)
print("Input text :", input_text)
print("Burrows - Wheeler Transform :", bwt_arr)
|
C#
using System;
using System.Collections.Generic;
public class BurrowsWheelerTransform
{
public struct Rotation
{
public int Index;
public string Suffix;
}
private static int CompareRotations(Rotation x, Rotation y)
{
return string.Compare(x.Suffix, y.Suffix);
}
private static int[] ComputeSuffixArray(string inputText)
{
int lenText = inputText.Length;
Rotation[] suff = new Rotation[lenText];
for (int i = 0; i < lenText; i++)
{
suff[i].Index = i;
suff[i].Suffix = inputText.Substring(i);
}
Array.Sort(suff, CompareRotations);
int[] suffixArr = new int[lenText];
for (int i = 0; i < lenText; i++)
{
suffixArr[i] = suff[i].Index;
}
return suffixArr;
}
private static string FindLastChar(string inputText, int[] suffixArr)
{
int n = suffixArr.Length;
char[] bwtArr = new char[n];
for (int i = 0; i < n; i++)
{
int j = suffixArr[i] - 1;
if (j < 0)
{
j += n;
}
bwtArr[i] = inputText[j];
}
return new string(bwtArr);
}
public static void Main(string[] args)
{
string inputText = "banana$";
int lenText = inputText.Length;
int[] suffixArr = ComputeSuffixArray(inputText);
string bwtArr = FindLastChar(inputText, suffixArr);
Console.WriteLine($"Input text : {inputText}");
Console.WriteLine($"Burrows-Wheeler Transform : {bwtArr}");
}
}
|
Javascript
function cmp_func(x, y) {
return (x[1] > y[1]) - (x[1] < y[1]);
}
function compute_suffix_array(input_text, len_text) {
let suff = [];
for (let i = 0; i < len_text; i++) {
suff.push([i, input_text.slice(i)]);
}
suff.sort(cmp_func);
let suffix_arr = suff.map(item => item[0]);
return suffix_arr;
}
function find_last_char(input_text, suffix_arr, n) {
let bwt_arr = "";
for (let i = 0; i < n; i++) {
let j = suffix_arr[i] - 1;
if (j < 0) {
j = j + n;
}
bwt_arr += input_text[j];
}
return bwt_arr;
}
let input_text = "banana$";
let len_text = input_text.length;
let suffix_arr = compute_suffix_array(input_text, len_text);
let bwt_arr = find_last_char(input_text, suffix_arr, len_text);
console.log("Input text: " + input_text);
console.log("Burrows-Wheeler Transform: " + bwt_arr);
|
OutputInput 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.
This article is contributed by Anureet Kaur. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.