Burrows–Wheeler Assignment

Implement the Burrows–Wheeler data compression algorithm. This revolutionary algorithm outcompresses gzip and PKZIP, is relatively easy to implement, and is not protected by any patents. It forms the basis of the Unix compression utility bzip2.

The Burrows–Wheeler data compression algorithm consists of three algorithmic components, which are applied in succession:

Burrows–Wheeler transform. Given a typical English text file, transform it into a text file in which sequences of the same character occur near each other many times.
Move-to-front encoding. Given a text file in which sequences of the same character occur near each other many times, convert it into a text file in which certain characters appear much more frequently than others.
Huffman compression. Given a text file in which certain characters appear much more frequently than others, compress it by encoding frequently occurring characters with short codewords and infrequently occurring characters with long codewords.

Step 3 is the only one that compresses the message: it is particularly effective because Steps 1 and 2 produce a text file in which certain characters appear much more frequently than others. To expand a message, apply the inverse operations in reverse order: first apply the Huffman expansion, then the move-to-front decoding, and finally the inverse Burrows–Wheeler transform. Your task is to implement the Burrows–Wheeler and move-to-front components.

Binary input and binary output. To enable your programs to work with binary data, use BinaryStdIn and BinaryStdOut, which are described in Algorithms, 4th edition. You can use HexDump, to display the binary output when debugging: it takes a command-line argument n; reads bytes from standard input; and writes them to standard output in hexadecimal, n per line.

~/Desktop/burrows> more abra.txt
ABRACADABRA!

~/Desktop/burrows> java-algs4 edu.princeton.cs.algs4.HexDump < abra.txt
41 42 52 41 43 41 44 41 42 52 41 21
96 bits

Note that in ASCII, 'A' is 41 (hex), 'B' is 42 (hex), and '!' is 21 (hex).

Huffman compression and expansion. Huffman (Program 5.10 in Algorithms, 4th edition) implements the classic Huffman compression and expansion algorithms.

~/Desktop/burrows> java-algs4 edu.princeton.cs.algs4.Huffman - < abra.txt | java-algs4 edu.princeton.cs.algs4.HexDump 16
50 4a 22 43 43 54 a8 40 00 00 01 8f 96 8f 94
120 bits

~/Desktop/burrows> java-algs4 edu.princeton.cs.algs4.Huffman - < abra.txt | java-algs4 edu.princeton.cs.algs4.Huffman +
ABRACADABRA!

Do not write any code for this step.

Move-to-front encoding and decoding. The main idea of move-to-front encoding is to maintain an ordered sequence of the characters in the alphabet by repeatedly reading a character from the input message; printing the position in the sequence in which that character appears; and moving that character to the front of the sequence. As a simple example, if the initial ordering over a 6-character alphabet is A B C D E F, and we want to encode the input CAAABCCCACCF, then we would update the move-to-front sequence as follows:

move-to-front    in   out
-------------    ---  ---
 A B C D E F      C    2 
 C A B D E F      A    1
 A C B D E F      A    0
 A C B D E F      A    0
 A C B D E F      B    2
 B A C D E F      C    2
 C B A D E F      C    0
 C B A D E F      C    0
 C B A D E F      A    2
 A C B D E F      C    1
 C A B D E F      C    0
 C A B D E F      F    5
 F C A B D E

If equal characters occur near one another other many times in the input, then many of the output values will be small integers (such as 0, 1, and 2). The resulting high frequency of certain characters (0s, 1s, and 2s) provides exactly the kind of input for which Huffman coding achieves favorable compression ratios.

Move-to-front encoding. Your task is to maintain an ordered sequence of the 256 extended ASCII characters. Initialize the sequence by making the ith character in the sequence equal to the ith extended ASCII character. Now, read each 8-bit character c from standard input, one at a time; output the 8-bit index in the sequence where c appears; and move c to the front.
```
~/Desktop/burrows> java-algs4 MoveToFront - < abra.txt | java-algs4 edu.princeton.cs.algs4.HexDump 16
41 42 52 02 44 01 45 01 04 04 02 26
96 bits
```
Move-to-front decoding. Initialize an ordered sequence of 256 characters, where extended ASCII character i appears ith in the sequence. Now, read each 8-bit character i (but treat it as an integer between 0 and 255) from standard input one at a time; write the ith character in the sequence; and move that character to the front. Check that the decoder recovers any encoded message.
```
~/Desktop/burrows> java-algs4 MoveToFront - < abra.txt | java-algs4 MoveToFront +
ABRACADABRA!
```

Name your program MoveToFront.java and organize it using the following API:

public class MoveToFront {

    // apply move-to-front encoding, reading from stdin and writing to stdout
    public static void encode()

    // apply move-to-front decoding, reading from stdin and writing to stdout
    public static void decode()

    // if args[0] is "-", apply move-to-front encoding
    // if args[0] is "+", apply move-to-front decoding
    public static void main(String[] args)

}

Performance requirements. The running time of both move-to-front encoding and decoding must be proportional to n R (or better) in the worst case and proportional to n + R (or better) on inputs that arise when compressing typical English text, where n is the number of characters in the input and R is the alphabet size. The amount of memory used by both move-to-front encoding and decoding must be proportional to n + R (or better) in the worst case.

Circular suffix array. To efficiently implement the key component in the Burrows–Wheeler transform, you will use a fundamental data structure known as the circular suffix array, which describes the abstraction of a sorted array of the n circular suffixes of a string of length n. As an example, consider the string "ABRACADABRA!" of length 12. The table below shows its 12 circular suffixes and the result of sorting them.

 i       Original Suffixes           Sorted Suffixes         index[i]
--    -----------------------     -----------------------    --------
 0    A B R A C A D A B R A !     ! A B R A C A D A B R A    11
 1    B R A C A D A B R A ! A     A ! A B R A C A D A B R    10
 2    R A C A D A B R A ! A B     A B R A ! A B R A C A D    7
 3    A C A D A B R A ! A B R     A B R A C A D A B R A !    0
 4    C A D A B R A ! A B R A     A C A D A B R A ! A B R    3
 5    A D A B R A ! A B R A C     A D A B R A ! A B R A C    5
 6    D A B R A ! A B R A C A     B R A ! A B R A C A D A    8
 7    A B R A ! A B R A C A D     B R A C A D A B R A ! A    1
 8    B R A ! A B R A C A D A     C A D A B R A ! A B R A    4
 9    R A ! A B R A C A D A B     D A B R A ! A B R A C A    6
10    A ! A B R A C A D A B R     R A ! A B R A C A D A B    9
11    ! A B R A C A D A B R A     R A C A D A B R A ! A B    2

We define index[i] to be the index of the original suffix that appears ith in the sorted array. For example, index[11] = 2 means that the 2nd original suffix appears 11th in the sorted order (i.e., last alphabetically).

Your job is to implement the following circular suffix array API, which provides the client access to the index[] values:

public class CircularSuffixArray {

    // circular suffix array of s
    public CircularSuffixArray(String s)

    // length of s
    public int length()

    // returns index of ith sorted suffix
    public int index(int i)

    // unit testing (required)
    public static void main(String[] args)

}

Corner cases. Throw an IllegalArgumentException in the constructor if the argument is null. Throw an IllegalArgumentException in the method index() if i is outside its prescribed range (between 0 and n − 1).

Unit testing. Your main() method must call each public method directly and help verify that they work as prescribed (e.g., by printing results to standard output).

Performance requirements. On typical English text, your data type must use space proportional to n + R (or better) and the constructor must take subquadratic time (better than n²). The methods length() and index() must take constant time in the worst case.

Burrows–Wheeler transform. The goal of the Burrows–Wheeler transform is not to compress a message, but rather to transform it into a form that is more amenable for compression. The Burrows–Wheeler transform rearranges the characters in the input so that there are lots of clusters with repeated characters, but in such a way that it is still possible to recover the original input. It relies on the following intuition: if you see the letters hen in English text, then, most of the time, the letter preceding it is either t or w. If you could somehow group all such preceding letters together (mostly t’s and some w’s), then you would have a propitious opportunity for data compression.

Burrows–Wheeler transform. The Burrows–Wheeler transform of a string s of length n is defined as follows: Consider the result of sorting the n circular suffixes of s. The Burrows–Wheeler transform is the last column in the sorted suffixes array t[], preceded by the row number first in which the original string ends up. Continuing with the "ABRACADABRA!" example above, we highlight the two components of the Burrows–Wheeler transform in the table below.

 i     Original Suffixes          Sorted Suffixes       t    index[i]
--    -----------------------     -----------------------    --------
 0    A B R A C A D A B R A !     ! A B R A C A D A B R A    11
 1    B R A C A D A B R A ! A     A ! A B R A C A D A B R    10
 2    R A C A D A B R A ! A B     A B R A ! A B R A C A D    7
*3    A C A D A B R A ! A B R     A B R A C A D A B R A !   *0
 4    C A D A B R A ! A B R A     A C A D A B R A ! A B R    3
 5    A D A B R A ! A B R A C     A D A B R A ! A B R A C    5
 6    D A B R A ! A B R A C A     B R A ! A B R A C A D A    8
 7    A B R A ! A B R A C A D     B R A C A D A B R A ! A    1
 8    B R A ! A B R A C A D A     C A D A B R A ! A B R A    4
 9    R A ! A B R A C A D A B     D A B R A ! A B R A C A    6
10    A ! A B R A C A D A B R     R A ! A B R A C A D A B    9
11    ! A B R A C A D A B R A     R A C A D A B R A ! A B    2

Since the original string ABRACADABRA! ends up in row 3, we have first = 3. Thus, the Burrows–Wheeler transform is

3
ARD!RCAAAABB

Notice how there are 4 consecutive As and 2 consecutive Bs—these clusters make the message easier to compress.

~/Desktop/burrows> java-algs4 BurrowsWheeler - < abra.txt | java-algs4 edu.princeton.cs.algs4.HexDump 16
00 00 00 03 41 52 44 21 52 43 41 41 41 41 42 42
128 bits

Also, note that the integer 3 is represented using 4 bytes (00 00 00 03). The character 'A' is represented by hex 41, the character 'R' by 52, and so forth.

Burrows–Wheeler inverse transform. Now, we describe how to invert the Burrows–Wheeler transform and recover the original input string. If the jth original suffix (original string, shifted j characters to the left) is the ith row in the sorted order, we define next[i] to be the row in the sorted order where the (j + 1)st original suffix appears. For example, if first is the row in which the original input string appears, then next[first] is the row in the sorted order where the 1st original suffix (the original string left-shifted by 1) appears; next[next[first]] is the row in the sorted order where the 2nd original suffix appears; next[next[next[first]]] is the row where the 3rd original suffix appears; and so forth.
- Inverting the message given t[], first, and the next[] array. The input to the Burrows–Wheeler decoder is the last column t[] of the sorted suffixes along with first. From t[], we can deduce the first column of the sorted suffixes because it consists of precisely the same characters, but in sorted order.
```
 i      Sorted Suffixes     t      next[i]
--    -----------------------      -------
 0    ! ? ? ? ? ? ? ? ? ? ? A        3
 1    A ? ? ? ? ? ? ? ? ? ? R        0
 2    A ? ? ? ? ? ? ? ? ? ? D        6
*3    A ? ? ? ? ? ? ? ? ? ? !        7
 4    A ? ? ? ? ? ? ? ? ? ? R        8
 5    A ? ? ? ? ? ? ? ? ? ? C        9
 6    B ? ? ? ? ? ? ? ? ? ? A       10
 7    B ? ? ? ? ? ? ? ? ? ? A       11
 8    C ? ? ? ? ? ? ? ? ? ? A        5
 9    D ? ? ? ? ? ? ? ? ? ? A        2
10    R ? ? ? ? ? ? ? ? ? ? B        1
11    R ? ? ? ? ? ? ? ? ? ? B        4
```
  Now, given the next[] array and first, we can reconstruct the original input string because the first character of the ith original suffix is the ith character in the input string. In the example above, since first = 3, we know that the original input string appears in row 3; thus, the original input string starts with 'A' (and ends with '!'). Since next[first] = 7, the next original suffix appears in row 7; thus, the next character in the original input string is 'B'. Since next[next[first]] = 11, the next original suffix appears in row 11; thus, the next character in the original input string is 'R'.
- Constructing the next[] array from t[] and first. Amazingly, the information contained in the Burrows–Wheeler transform suffices to reconstruct the next[] array, and, hence, the original message! Here’s how. It is easy to deduce a next[] value for a character that appears exactly once in the input string. For example, consider the suffix that starts with 'C'. By inspecting the first column, it appears 8th in the sorted order. The next original suffix after this one will have the character 'C' as its last character. By inspecting the last column, the next original appears 5th in the sorted order. Thus, next[8] = 5. Similarly, 'D' and '!' each occur only once, so we can deduce that next[9] = 2 and next[0] = 3.
```
 i      Sorted Suffixes     t      next[i]
--    -----------------------      -------
 0    ! ? ? ? ? ? ? ? ? ? ? A        3
 1    A ? ? ? ? ? ? ? ? ? ? R
 2    A ? ? ? ? ? ? ? ? ? ? D
*3    A ? ? ? ? ? ? ? ? ? ? !
 4    A ? ? ? ? ? ? ? ? ? ? R 
 5    A ? ? ? ? ? ? ? ? ? ? C
 6    B ? ? ? ? ? ? ? ? ? ? A
 7    B ? ? ? ? ? ? ? ? ? ? A
 8    C ? ? ? ? ? ? ? ? ? ? A        5
 9    D ? ? ? ? ? ? ? ? ? ? A        2
10    R ? ? ? ? ? ? ? ? ? ? B
11    R ? ? ? ? ? ? ? ? ? ? B
```
  However, since 'R' appears twice, it may seem ambiguous whether next[10] = 1 and next[11] = 4, or whether next[10] = 4 and next[11] = 1. Here’s the key rule that resolves the apparent ambiguity:
  If sorted row i and j both start with the same character and i < j, then next[i] < next[j].
  This rule implies next[10] = 1 and next[11] = 4. Why is this rule valid? The rows are sorted, so row 10 is lexicographically less than row 11. Thus, the ten unknown characters in row 10 must be less than the ten unknown characters in row 11 (since both start with 'R'). We also know that between the two rows that end with 'R', row 1 is less than row 4. But, the ten unknown characters in row 10 and 11 are precisely the first ten characters in rows 1 and 4. Thus, next[10] = 1 and next[11] = 4 or this would contradict the fact that the suffixes are sorted.
Check that the inverse transform recovers any transformed message.
```
~/Desktop/burrows> java-algs4 BurrowsWheeler - < abra.txt | java-algs4 BurrowsWheeler +
ABRACADABRA!
```

Name your program BurrowsWheeler.java and organize it using the following API:

public class BurrowsWheeler {

    // apply Burrows-Wheeler transform,
    // reading from standard input and writing to standard output 
    public static void transform()

    // apply Burrows-Wheeler inverse transform,
    // reading from standard input and writing to standard output
    public static void inverseTransform()

    // if args[0] is "-", apply Burrows-Wheeler transform
    // if args[0] is "+", apply Burrows-Wheeler inverse transform
    public static void main(String[] args)

}

Performance requirements. The running time of your Burrows–Wheeler transform must be proportional to n + R (or better) in the worst case, excluding the time to construct the circular suffix array. The running time of your Burrows–Wheeler inverse transform must be proportional to n + R (or better) in the worst case. The amount of memory used by both the Burrows–Wheeler transform and inverse transform must be proportional to n + R (or better) in the worst case.

Analysis. Once you have MoveToFront.java and BurrowsWheeler.java working, compress some text files. Then, test it on some binary files. Calculate the compression ratio achieved for each file and report the time to compress and expand each file. (Here, compression and expansion consists of applying BurrowsWheeler, MoveToFront, and Huffman in succession.) Finally, determine the order of growth of the running time of each of your methods, both in the worst case and on typical English text inputs.

Deliverables. Submit MoveToFront.java, BurrowsWheeler.java, and CircularSuffixArray.java. Also submit a readme.txt and answer all questions.

Grading.

file points

MoveToFront.java 10

CircularSuffixArray.java 10

BurrowsWheeler.java 14

readme.txt 6

40

file	points
`MoveToFront.java`	10
`CircularSuffixArray.java`	10
`BurrowsWheeler.java`	14
`readme.txt`	6
	40

Reminder: You can lose up to 4 points for poor style and up to 4 points for inadequate unit testing.