The Burrows–Wheeler data compression algorithm consists of three algorithmic components, which are applied in succession:
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.
Note that in ASCII,~/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
'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
Do not write any code for this step.~/Desktop/burrows> java-algs4 edu.princeton.cs.algs4.Huffman - < abra.txt | java-algs4 edu.princeton.cs.algs4.Huffman + ABRACADABRA!
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:
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 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
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
~/Desktop/burrows> java-algs4 MoveToFront - < abra.txt | java-algs4 MoveToFront + ABRACADABRA!
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. Your implementation must achieve the following performance requirements, where n is the number of characters in the input and R is the alphabet size:
encode()
and decode()
must take \(O(nR)\) time in the worst case.
encode()
and decode()
must take \(O(n + R)\) time on inputs that arise when compressing typical English text.
encode()
and decode()
must use
\(O(n + R)\) memory 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.
We definei 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
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:
Corner cases. Throw anpublic 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) }
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. Your implementation must achieve the following performance requirements, where n is the number of characters in the input and R is the alphabet size:
length()
and index()
must take
\(\Theta(1)\) 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.
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.
Since the original stringi 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
ABRACADABRA!
ends up in row 3,
we have first
= 3
. Thus, the Burrows–Wheeler transform is
Notice how there are 4 consecutive3 ARD!RCAAAABB
A
s and 2 consecutive
B
s—these clusters make the message easier to compress.
Also, note that the integer 3 is represented using 4 bytes (~/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
00 00 00 03
).
The character 'A'
is represented by hex 41
, the character
'R'
by 52
, and so forth.
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.
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.
Now, given thei 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
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'
.
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
.
However, sincei 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
'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!
BurrowsWheeler.java
and organize it using the following API:
Performance requirements. Your implementation must achieve the following performance requirements, where n is the number of characters in the input and R is the alphabet size: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) }
transform()
method must take
\(O(n + R)\) time in the worst case, excluding the time to construct the
circular suffix array.
inverseTransform()
method must take \(O(n + R)\)
time in the worst case.
transform()
and inverseTransform()
methods
must use \(O(n + R)\) space 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.
Submission.
Submit MoveToFront.java
, BurrowsWheeler.java
,
and CircularSuffixArray.java
.
Finally, submit readme.txt
and acknowledgments.txt
files and answer the questions.
Grading.
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.