handout #28
CSE143—Computer Programming II
Programming Assignment #8
due: Friday,
This assignment is worth a total
of 30 points. It is divided into two
parts, each worth approximately half of the points. Please
note that no solutions will be accepted after
This program will give you
practice with binary trees and priority queues.
In this program you will explore how text files can be compressed by
using a coding scheme based on the frequency of characters. We will use a coding scheme called Huffman
coding. The basic idea is to abandon the
way that text files are usually stored.
Instead of using the usual seven or eight bits per character, Huffman's
method uses only a few bits for characters that are used often, more bits for
those that are rarely used.
You will solve this problem using
a structure known as a priority queue.
In a priority queue each value inserted into the queue has a priority
that determines when it will be removed.
There are many ways to specify the priorities. For this program you will construct objects
that implement the Comparable interface, with objects that are “less” being
given a higher priority (to be removed first).
The first step is to compute the
frequency of each character in the file you wish to encode. This allows you to determine which characters
should have the fewest bits, etc. The
next step is to build a “coding tree” from the bottom up according to the
frequencies. An example will help make
this clear. To make the example easier,
suppose we only want to encode the five letters (a, b, c, d, e)
and they have frequencies 3, 3, 1, 1, and 2, respectively.
We first create a leaf node for
each character/frequency pair and put them into a priority queue, so that the
characters with lower frequencies appear first:
+----+ +----+
+----+ +----+ +----+
| 1 | | 1
| |
2 | | 3 |
| 3 |
+----+ +----+ +----+
+----+ +----+
'c' 'd'
'e' 'a' 'b'
Now we pick the two nodes with the
smallest frequencies (the two at the front of the priority queue) and create a
new node with those two nodes as children (the first value from the queue
becomes the left, the second value from the queue becomes the right). We assign this new branch node a frequency that
is the sum of the frequencies of the two children. This new node is then put back into the
priority queue:
+----+ +----+
+----+ +----+
| 2 | | 2
| |
3 | | 3 |
+----+ +----+ +----+
+----+
'e'
/ \ 'a'
'b'
+----+ +----+
|
1 | | 1 |
+----+ +----+
'c' 'd'
Continuing in this way, we build
up larger and larger subtrees.
Here are the rest of the steps:
+----+ +----+
+----+
| 3 | | 3
| |
4 |
+----+ +----+ +----+
'a' 'b'
/ \
+----+ +----+
| 2 |
| 2 |
+----+ +----+
'e' /
\
+----+ +----+
| 1
| |
1 |
+----+ +----+
'c' 'd'
+----+ +----+
| 4 | | 6 |
+----+ +----+
/ \ / \
+----+ +----+
+----+ +----+
|
2 | | 2 |
| 3 | | 3
|
+----+ +----+
+----+ +----+
'e' /
\ 'a' 'b'
+----+ +----+
| 1 |
| 1 |
+----+ +----+
'c' 'd'
+----+
| 10 |
+----+
/ \
/ \
+----+ +----+
| 4 | | 6 |
+----+ +----+
/ \ / \
+----+ +----+
+----+ +----+
|
2 | | 2 |
| 3 | | 3
|
+----+ +----+
+----+ +----+
'e' /
\ 'a' 'b'
+----+ +----+
| 1 |
| 1 |
+----+ +----+
'c' 'd'
Note that the nodes with low
frequencies end up far down in the tree, and nodes with high frequencies end up
near the root of the tree. It turns out
that this structural description is exactly what is needed to create an
efficient encoding. The Huffman code is
derived from this coding tree simply by assigning a zero to each left branch
and a one to each right branch. The code
can be read directly from the tree. The
code for a is 10, the code for b is 11, the code for c
is 010, the code for d is 011 and the code for e is 00.
An interesting feature of the
Huffman code is that delimiters between characters are not stored, even though
different characters may be coded with different numbers of bits. The key is that a code created by this method
exhibits what is known as the prefix
property, which means that no code for a character is the prefix of the
code of any other character. Thus, to
decode a message we need only traverse our tree. When we reach a leaf, we know that we have
decoded one character, and can now start decoding the next character.
Part 1: Making a Code
For our purposes, we will encode what are known as “bytes”
(8 bits). This will allow us to encode
standard text files that use ASCII and binary files as well. From the point of view of your Huffman code,
you can think about the individual bytes as simple integers in the range of 0
to 255, each representing the ASCII value of a particular character. In part 1, you are working with a program
called MakeCode.
It prompts the user for a file to examine and it computes the frequency
of each character in the file. These
counts are passed as an array to your HuffmanTree
constructor.
The array passed to your
constructor will have exactly 256 values in it, but your program should not
depend on this. Instead, you can use the
length field of the array to know how many there are. In your constructor, you should use a
priority queue to build up the tree as described above. First you will add a leaf node for each
character that has a frequency greater than 0 (we don’t include the other
characters in our tree). These should be
added in increasing character order (character 0, character 1,
and so on).
Then you build the tree. Initially you have a bunch of leaf
nodes. Your goal is to get a single
tree. While you haven’t gotten down to a
single tree, you remove two values from the priority queue and combine them to
make a new branch node which you put back into the queue, as described
above. You continue combining subtrees until you get down to one tree. This becomes your Huffman tree.
You are to define a class called HuffmanTree
with the following public methods (more methods will be added in part 2 of this
assignment):
|
Method |
Description |
|
HuffmanTree(int[] count) |
Constructs a Huffman tree using the given array of frequencies where count[i] is the number of occurrences of the character with ASCII value i. |
|
void write(PrintStream output) |
Writes the current tree to the given output stream in standard format. |
In defining your class, you will
also define a node class called HuffmanNode. You may decide what data fields to include
with this class.
In writing the tree, you should
produce two lines of output for each character.
The first line should have the ASCII value of the character. The second line should have the code (0's and
1's) for the character with this ASCII value.
For the simple example above, the output would be (the letter “a” has
ASCII value 97):
101
00
99
010
100
011
97
10
98
11
The codes should be written in
“traversal order.” In other words, they
should be written in the order that any standard traversal of the tree would
visit them.
It turns out that Huffman coding
works best if one character is designated as “end of file,” meaning that every
file is guaranteed to end with such a character and it will be used for no
other purpose. Some operating systems
have such a character, but if we want to write a general-purpose program, we
have to do something that is not specific to any one operating system. So in addition to encoding the actual
characters that appear in the file, we will create a code for a fictitious
end-of-file character that will be used only by the Huffman encoding and
decoding programs. That means that in
addition to all of the legal characters, you are also going to introduce a
special character that will be used to signal end-of-file. We will refer to this as the “pseudo-eof” character. Its
value will be one higher than the value of the highest character in the
frequency array passed to the constructor.
It will always have a frequency of 1 because it appears exactly once at
the end of each file to be encoded. You
will have to manually add this character to your priority queue because it will
not be included as part of the frequency array.
The output listed above does not
include the pseudo-eof character. When you include the pseudo-eof character with a frequency of 1, the output becomes:
101
00
256
010
99
0110
100
0111
97
10
98
11
The java.util
package includes a PriorityQueue<E> class that
implements the Queue<E> interface.
You must use these to build your Huffman tree. You can read the Java api
description of them, but basically, they work like the Queue<E> interface
we have seen before except for the fact that the “enqueue”
method is called “offer” and the “dequeue” method is
called “remove.” The interface includes
the isEmpty and size methods we are used to
having. The only difference between a
priority queue and a standard queue is that it uses the natural ordering of the
objects to decide which object to dequeue first, with
objects considered “less” returned first.
You are going to be putting subtrees into your
priority queue, which means you’ll be adding values of type HuffmanNode. This means that your HuffmanNode
class will have to implement the Comparable<E> interface. It should use the frequency of the subtree to determine its ordering relative to other subtrees, with lower frequencies considered “less” than
higher frequencies. If two frequencies
are equal, the nodes should be considered equal.
The Queue<E> interface in
the java.util package is incompatible with the one we
used earlier in the quarter, so be sure not to include the Queue.java
and Queue.class files from the earlier assignment in
the same directory as your Huffman solution.
The Huffman solution is not
unique. You can obtain any one of
several different equivalent trees depending upon how certain decisions are
made. But if you implement it as we have
specified, then you should get exactly the same tree. Make sure that you use the built-in PriorityQueue class and that when you are combining pairs
of values taken from the priority queue, you make the first value removed from
the queue the left subtree and you make the second
value removed the right subtree.
You are being given two data files for this assignment called short.txt and hamlet.txt. The file short.txt is a short input file suitable for preliminary testing. The file hamlet.txt contains the full text of Shakespeare’s play Hamlet.
In terms of correctness, your class must provide
all of the functionality described above.
In terms of style, we will be grading on your use of comments, good
variable names, consistent indentation and good coding style to implement these
operations. Remember that you will lose
points if you declare variables as data fields that can instead be declared as
local variables. You should also avoid
extraneous cases (e.g., don’t make something into a special case if it doesn’t
have to be).
You should name your files HuffmanNode.java and HuffmanTree.java
and you should turn it in electronically from the “assignments” link on the
class web page. A collection of files
needed for the assignment is included on the web page as ass8.zip. You will need to have MakeCode.java
in the same directory as your files in order to run MakeCode. The zip file will also include the sample
data files (short.txt and hamlet.txt) along with their code files (short.code and hamlet.code).
Part 2: Encoding
and Decoding a File
There are two new main programs
that are used in this part of the assignment.
Recall that MakeCode.java examined an input
file and produced a code file for compressing it. The program Encode.java
uses this code and the original file to produce a binary file that is the
compressed version of the original. The
program Decode.java uses the code and the binary file
from Encode to reconstruct the original file.
Encode is a complete program that doesn’t need the Huffman tree. Decode depends on your HuffmanTree
class to do most of the work.
In
particular, you will have to add two new methods to your HuffmanTree
class:
|
Method |
Description |
|
HuffmanTree(Scanner input) |
Constructs a Huffman tree from the Scanner. Assumes the Scanner contains a tree description in standard format. |
|
void decode(BitInputStream input, PrintStream output, int eof) |
Reads bits from the given input stream and writes the corresponding characters to the output. Stops reading when it encounters a character with value equal to eof. This is a pseudo-eof character, so it should not be written to the output file. Assumes the input stream contains a legal encoding of characters for this tree’s Huffman code. |
The first of these methods is an
alternative constructor. In part 1 you wrote
a constructor that took an array of frequencies and constructed an appropriate
tree given those frequencies. In this
part you are given a Scanner that contains the file produced by your write
method from part 1. In other words, the
input for this part is the output you produced in part 1. You are using your own output to recreate the
tree. For this second part, the
frequencies are irrelevant because the tree has already been constructed once,
but you are using the same node class as before. You can set all of the frequencies to some
standard value like 0 or -1 for this part.
Remember that the standard format
was a series of pairs of lines where the first line has an integer representing
the character’s ASCII value and the second line has the code to use for that
character. You might be tempted to call nextInt()
to read the integer and nextLine() to read the code,
but remember that mixing token-based reading and line-based reading is not
simple. Here is an alternative that uses
a method called parseInt in the Integer class that
allows you to use two successive calls on nextLine():
int n = Integer.parseInt(input.nextLine());
String code = input.nextLine();
For the decoding part, you have to
read a BitInputStream. This is a special class that Stuart wrote
that works in conjunction with another class called BitOutputStream. They each have a very simple interface. They allow you to write and read compact
sequences of bits.
The only method you’ll use for BitInputStream is the following method which returns the
next bit from the file:
public int readBit()
Your method is doing the reverse
of the encoding process. It is reading
sequences of bits that represent encoded characters and it is figuring out what
the original characters must have been.
Your method should start at the top of your tree and should read bits
from the input stream, going left or right depending upon whether you get a 0
or 1 from the stream. When you hit a
leaf node, you know you’ve found the end of an encoded sequence. At that point, you should write the integer
code for that character to the output file.
In doing so, call this method from the PrintStream
class:
public void write(int b)
You don’t need to cast to
char. You just write the integer for
this particular character (the value between 0 and 255 that is stored in this
leaf). Once you’ve written this
character’s integer, you go back to the top of your tree and start over,
reading more bits and descending the tree until you hit a leaf again. At that point you write again, go back to the
top of the tree, read more bits and descend until you hit a leaf, then write
the leaf, go back to the top of the tree, and so on.
Recognizing the end of the input
will be tricky. Remember that we
introduced a pseudo-eof character with a special
value (256). The Encode method will
write exactly one occurrence of this character at the end of the file. At some point your decoding method will come
across this eof character. At that point it should stop decoding. It should not write this integer to the PrintStream because it isn’t actually part of the original
file. The eof
value will be 256 for this particular program, but your code isn’t supposed to
depend on this specific value, which is why it is passed to your decode method
as the third parameter.
If you fail to recognize the
pseudo-eof character, you might end up accidentally
reading past the end of the bit stream.
When that happens, the readBit method returns
a value of -1. So if you see a value of
-1 appearing, it’s because you’ve read too far in the bit stream.
You will have to be careful how
you use recursion in your decode method.
Java has a limit on the stack depth you can use. For a file like hamlet.txt, there are
hundreds of thousands of characters to decode.
That means it would not be appropriate to write code that requires a
stack that is hundreds of thousands of levels deep. You might be forced to write some parts of
this using a loop to make sure that you don’t exceed the stack depth.
In terms of correctness, your class must provide
all of the functionality described above.
In terms of style, we will be grading on your use of comments, good
variable names, consistent indentation and good coding style to implement these
operations. Remember that you will lose
points if you declare variables as data fields that can instead be declared as
local variables. You should also avoid
extraneous cases (e.g., don’t make something into a special case if it doesn’t
have to be).
There is one turn-in for both parts of
this assignment. You have to turn in HuffmanNode.java and HuffmanTree.java. The files needed for this part of the
assignment are included on the web page in ass8.zip. You should include Encode.java,
Decode.java, BitInputStream.java
and BitOutputStream.java in the same directory as
your other program files. The zip file
includes encoded versions of the sample files called short.short
and hamlet.short.
Your Decode program should be able to take one of these files and the
corresponding code file to reconstruct the original file.
If you would like to read more
about Huffman’s algorithm, there are many sources on the web including, for
example, http://www.maths.abdn.ac.uk/~igc/tch/mx4002/notes/node59.html.