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

Why use BinaryStdIn.readChar(8) instead of BinaryStdIn.readByte() to read in 8 bits? The primitive type char is a bit simpler to use because it is an unsigned type, whereas byte is a signed type. So, for example, you can't directly use a byte as an index into an array because it might be negative. You also need to be careful when converting from byte to char. For example, the statement char c = (char) b; first promotes the byte b to a 32-bit signed integer, then narrows it to a 16-bit signed integer, which is rarely what you want. Typically, you would want to use char c = (char) (b & 0xff); instead.

How do I convert between characters and integers? Java’s char type is a 16-bit signed integer. So, for example you can iterate over the characters in the alphabet with a for loop:

int R = 256;
for (char c = 0; c < R; c++) {
When you apply binary arithmetic or comparison operators to two char values, Java automatically promotes them both to type int. As a result, you can compare two char values with the binary comparison operators. Similarly, you can use a char as an index into an array.

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, 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.

About how many character compares will I need to make in order to compare two circular suffixes from a string of length n? It depends on the string. If the string has lots of redundancy (e.g., all As), you will need to make n character compares. If the string is “typical English text”, you can assume that number is about \(\Theta(\log n)\).

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 \(\Theta(R \, n)\) time in the best case? Is that okay. No. It’s fine to take \(\Theta(R \, n)\) time in the worst case. However, they should take \(O(n + R)\) time on inputs that arise when compressing typical English text.

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.

My transform/inverseTranform pipeline produces incorrect results for periodic string such as abababab and aaa. What could be the cause? A string is periodic if it consists of two or more repeated copies of another string. The circular suffixes of a periodic string are not all distinct (e.g., the string abababab has 8 circular suffices but only two are distinct: abababab and babababa). For periodic strings, there is ambiguity in the definition of the next[] array because each circular suffix appears in multiple positions. Your implementation is free to resolve the ambiguity in any way that you choose: the sequence s[next[first]], s[next[next[first]]], s[next[next[next[first]]]], ... is always the same. However, you must take care to write the correct number of characters.

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 construct a new String object (e.g., by calling substring()) 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.


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

Debugging. Debugging MoveToFront and BurrowsWheeler present extra challenges because they produce binary output (instead of text output) on standard output.

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 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 -kf 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 -kf 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.