Burrows–Wheeler Checklist

Frequently Asked Questions (General)

How should I read and write the data? You must use BinaryStdIn and BinaryStdOut, which read and write sequences of bytes. Do not use either StdIn or StdOut, which read and write sequences of Unicode characters.

My programs don’t work properly with binary data. Why not? Be absolutely sure that you use only BinaryStdIn and BinaryStdOut for reading and writing data. Also, be sure to call either BinaryStdOut.flush() or BinaryStdOut.close() after you are done writing; for an example, see RunLength.java.

Why does BinaryStdIn return the 8-bits as a (16-bit unsigned) char instead of as an (8-bit signed) byte? The primitive type byte is annoying to use in Java. When you operate on a byte, it is typically promoted to an int and you must be careful because the type byte is signed. For example, to convert a byte b to a char c, you must write c = (char) (b & 0xff) instead of c = (char) b. By using char, we avoid the hassle.

I'm curious. Which compression algorithm is used in PKZIP? In gzip? In bzip2? PKZIP uses LZW compression followed by Shannon–Fano (an entropy encoder similar to Huffman). The Unix utility gzip uses a variation of LZ77 (similar to LZW) followed by Huffman coding. The program bzip2 combines the Burrows–Wheeler transform, Huffman coding, and a (fancier) move-to-front style rule.

How can I view the contents of a binary file and determine its size? Use HexDump.java, as in the assignment. The command-line argument specifies the number of bytes per line to print; if the argument is 0, all output except for the number of bits will be suppressed.

Frequently Asked Questions (CircularSuffixArray)

Must I use BinaryStdIn and BinaryStdOut with CircularSuffixArray? The constructor and methods neither write to standard output nor read from standard input, so there is no need to use either BinaryStdIn or BinaryStdOut. You are free to use StdOut.println() when testing in main().

Can I form the n circular suffixes using the substring() method from the String data type? No. Beginning with Java 7, Update 6, the substring() method takes time and space proportional to the length of the substring. So, explicitly forming the n circular suffixes in this way would take both quadratic time and space.

Frequently Asked Questions (MoveToFront)

How should I read the binary input in encode()? The input is a sequence of extended ASCII characters (0x00 to 0xFF). You should read them one character at a time using BinaryStdIn.readChar() until BinaryStdIn.isEmpty().

How should I read the binary input in decode()? Same as encode().

My encode and decode methods take time proportional to R n in the best case? Is that okay. No. It’s fine to take time proportional to R n in the worst case. However, they should take time proportional to n + R on inputs that arise when compressing typical English text.

How do I convert between characters and integers? Java’s char type is an integral type. So, no explicit conversion should be necessary.

Frequently Asked Questions (BurrowsWheeler)

How should I read the binary input in transform()? The input is a sequence of extended ASCII characters (0x00 to 0xFF). You can read it using BinaryStdIn.readString().

How should I read the binary input in inverseTransform()? The input is an integer, followed by a sequence of extended ASCII characters (0x00 to 0xFF). You can read it using BinaryStdIn.readInt(), followed by BinaryStdIn.readString().

My BurrowsWheeler works except for large files. What could be the cause? Be sure that you read and write the number first as a 32-bit int, not a char.

Can I assume that inverseTransform() receives only valid inputs (e.g., that correspond to the output of transform())? Yes.

For the Burrows–Wheeler transform, in which order do I use to sort the suffixes? Use lexicographic order to sort the suffixes, which is the natural order of the String data type.

For the Burrows–Wheeler inverse transform, does next[0] always equal first? No. This is just a coincidence with the input string "ABRACADABRA!". Consider any two input strings that are cyclic rotations of one another, e.g., "ABRACADABRA!" and "CADABRA!ABRA". They will have the same sorted suffixes and t[] array—their only difference will be in the index first.

How much memory can my program consume? The Burrows–Wheeler transform may use quite a bit, so you may need to use the -Xmx option when executing. You must use space linear in the input size n and alphabet size R. (Industrial strength Burrows–Wheeler compression algorithms typically use a fixed block size, and encode the message in these smaller chunks. This reduces the memory requirements, at the expense of some loss in compression ratio.) Therefore, depending on your operating system and configuration there may be some very large files for which your program will not have enough memory even with the -Xmx option.

I’m running out of memory in the transform() method in Burrows–Wheeler. Any ideas? Be sure not to create a new String object for each circular suffix created in CircularSuffixArray, It is fine to have multiple references to the same String object.

What is meant by “typical English text inputs”? Inputs such as Aesop’s Fables, Moby Dick, or your most recent essay. We do not mean inputs with very long repeated substrings (such as aesop-2copies.txt or an input will 1 million consecutive As) or random inputs.

Testing

Input. To fully test your programs, you should use not only text files but also binary files (such as .class or .jpg files).

Reference solutions. For reference, we have provided the output of compressing aesop.txt and us.gif. We have also provided the results of applying each of the three encoding algorithms in isolation. Note that the binary file us.gif is already compressed.

To compare the contents of two files, you can use the following Bash command:

~/Desktop/burrows> cmp aesop.txt us.gif
aesop.txt us.gif differ: byte 1, line 1

~/Desktop/burrows> cmp us.gif us.copy.gif

Compression ratio. You can use the ls command to determine the size of a file (in bytes).

~/Desktop/burrows> ls -l
total 60536
-rw-rw-r--@  1 wayne  staff       723 Jul 14 10:24 COS 226.iml
-rw-rw-r--@  1 wayne  staff     24567 May  4  2012 CS_bricks.jpg
-rw-rw-r--@  1 wayne  staff         1 May  4  2012 a.txt
-rw-rw-r--@  1 wayne  staff        12 Mar 22  2009 abra.txt
-rw-rw-r--@  1 wayne  staff        19 Mar 22  2009 abra.txt.bwt.mtf.huf
-rw-rw-r--@  1 wayne  staff    191943 Nov 10  2005 aesop.txt
-rw-rw-r--@  1 wayne  staff     66026 Mar 22  2009 aesop.txt.bwt.mtf.huf
...

For example, aesop.txt uses 191,943 bytes; after compression, it (aesop.txt.bwt.mtf.huf) uses only 66,026 bytes; the compression ratio is 66026/191943 = 0.344.

Timing your program. Use the following Bash commands for compression and expansion, respectively:

~/Desktop/burrows> time java-algs4 edu.princeton.cs.algs4.Huffman - < mobydick.txt - >| mobyDickOutputFileName1

~/Desktop/burrows> time java-algs4 edu.princeton.cs.algs4.Huffman + < mobyDickOutputFileName1 >| moby-copy.txt

~/Desktop/burrows> time java-algs4 BurrowsWheeler - < mobydick.txt | java-algs4 MoveToFront - | java-algs4 edu.princeton.cs.algs4.Huffman - >| mobyDickOutputFileName2
real	0m1.341s
user	0m1.447s
sys	0m0.477s

~/Desktop/burrows> time java-algs4 edu.princeton.cs.algs4.Huffman + < mobyDickOutputFileName2 | java-algs4 MoveToFront + | java-algs4 BurrowsWheeler + >| moby-copy.txt
real	0m0.419s
user	0m0.750s
sys	0m0.372s

The “real” value is the wall clock time; the “user” value is the amount of CPU time spent in user-mode; the “system” value is the amount of CPU time spent in kernel mode. The CPU time may exceed the real time if your computer is using multiple CPUs.

Note that the internal Bash command time measures the time of the pipeline. If you use an external version of the command time (or an internal version associated with a different shell), it may measure only the first command in the pipeline.

We use >| instead of > to redirect standard output to a file and force overwriting (in case a file with that name already exists).

Timing using gzip or bzip2. If you are using Bash, you should have access to the the following data compression and expansion commands: gzip, gunzip, bzip2, or bunzip2. You can time them using the following Bash commands:

~/Desktop/burrows> time gzip -kf mobydick.txt
real	0m0.218s
user	0m0.133s
sys	0m0.007s

~/Desktop/burrows> time gunzip mobydick.txt.gz
real	0m0.063s
user	0m0.026s
sys	0m0.008s

~/Desktop/burrows> time bzip2 -kf mobydick.txt
real	0m0.176s
user	0m0.145s
sys	0m0.007s

~/Desktop/burrows> time bunzip2 mobydick.txt.bz2
real	0m0.099s
user	0m0.057s
sys	0m0.008s

Possible Progress Steps

These are purely suggestions for how you might make progress. You do not have to follow these steps.

Download burrows.zip. It contains sample input files and reference solutions.
Implement the CircularSuffixArray. Be sure not to create new String objects when you sort the suffixes. That would take quadratic space. A natural approach is to define a nested class CircularSuffix that represents a circular suffix implicitly (via a reference to the input string and a pointer to the first character in the circular suffix). The constructor of CircularSuffix should take constant time and use constant space. You might also consider making CircularSuffix implement the Comparable<CircularSuffix> interface. Note, that while this is, perhaps, the cleanest solution, it is not the fastest.
Implement the Burrows–Wheeler transform, using the CircularSuffixArray class.
The Burrows–Wheeler decoding is the trickiest part conceptually, but it is very little code once you understand how it works. (Not including declarations and input, our solution is about 10 lines of code.) You should find the key-indexed counting algorithm from the string sorting lecture to be especially useful.
Implement the move-to-front encoding and decoding algorithms. Not including comments and declarations, our solutions are about 10 lines of code each. If yours is significantly longer, try to simplify it. Do not worry about optimizing the worst-case performance—the goal is good performance on typical English text inputs.

A video on compressing and another on expanding are provided for those wishing additional assistance. Be forewarned that video was made in early 2014 and is somewhat out of date. For example, the API has changed.