[SOLVED] Cs3610 project 2

Huffman Coding
Assume that I want to store the message “opossum” on my computer. This string
of characters will be represented on my hard drive as a series of bits (1’s and 0’s).
Typically, an ASCII character takes up 8 bits of space on a computer. This means that
“opossum” will cost 56 bits of storage. For the sake of argument, let us assume that
this is a significant number.
One obvious way to reduce the number of bits in our message is to represent each
character in the message with fewer bits. Of course, using smaller bit sequences will
limit the number of unique characters we can encode. For example, a two bit sequence
is only capable of representing 4 unique characters. Consequently, our characters will
vary in length. This was expected though. So what are the real issues preventing us
from using fewer bits? Well, I have two for you right here:
1. Let us assume that the characters in “opossum”, ‘p’, ‘m’, ‘u’, ‘o’, and ‘s’, are
represented as 0, 1, 00, 01, and 10 respectively. What problems do you see with
this? Well, it just so happens that our most common characters ‘o’ and ‘s’ are
also some of our largest characters. We could have saved ourselves a little more
space if we instead represented ‘o’ and ‘s’ with the single bits 0 and 1, which were
naively used to denote ‘p’ and ‘m’. So the question arises, how do we efficiently
delegate bit sequences such that the most frequently occurring characters in our
messages take up the least amount of space?
2. Let us assume that the characters in “opossum”, ‘p’, ‘m’, ‘u’, ‘o’, and ‘s’, are
represented as 0, 1, 00, 01, and 10 respectively. Let us also assume that we have
used these characters to encode a new message “0001” on our hard drive. I now
ask you, what message did we just encode? Well, unfortunately, it is ambiguous.
One possible decoding could be “pppm”. Another could be “uo”. So the question
arises, how do we generate variable-length character codes such that our encoded
messages will not be ambiguous? In other words, how do we develop a prefix
coding system?
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One way to resolve these issues is by constructing a Huffman tree, just like the one
displayed right here:
A Huffman tree is a binary tree that encodes the characters of a message into a
reduced bit representation by implicitly storing the component bits along the tree’s
edges. Typically, left edges denote 0 while all the right edges denote 1. Thus, to find
the encoding of a specific character, we simply concatenate the bits found along the
edges of the path between the root node and the character’s node.
In the tree above, you can see that ‘o’ is represented as 11 because its path from the
root first follows a right edge and then another right edge. Take note that all character
nodes are leaf nodes and each path produces a unique string of bits. Also, you should
be aware that the most common characters, such as ‘o’ and ‘s’, appear closer to the
top of the tree, which gives them the shortened encoding that we desire. Conversely,
the lowest frequency characters reside near the bottom of the tree.
Using the Huffman tree above, “opossum” can now be written as
1100111010011010, which is just 16 bits as compared to the original 56. Better yet,
no characters share the same prefix, which means this binary message can be decoded
unambiguously. You can try it out for yourself by reading the bits left to right while
following the recorded bit path in the tree starting from the root. As soon as you hit a
leaf node, grab its character and continue reading the remaining bits starting from the
root again. So this is wonderful, I now have a compressed message that was easy to
encode and decode. However, the question remains, how do we construct the Huffman
tree? Well here is the algorithm:
1. Identify the frequency of each character in a given message. The frequency of
each character in “opossum” is listed alphabetically as follows:
m = 1 o = 2 p = 1 s = 2 u = 1
2. Package each character-frequency pair into a binary leaf node and insert each
node into a min heap using the frequency as the key.
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3. Extract the two smallest elements from the min heap and respectively attach
them to the left and right pointers of a new internal node. Set the key value of
the new internal node to be equal with the sum of the frequencies of the extracted
nodes. Insert this new internal node into the min heap.
(a) Result after extracting 1st min element (b) Result after extracting 2nd min element
(c) New node from extracted mins (d) New node inserted into heap
4. Repeat step 3 until the min heap has one element remaining.
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(a) New node from extracted mins (b) New node inserted into heap
(a) New node from extracted mins (b) New node inserted into heap
(a) New node from extracted mins (b) New node inserted into heap
5. Draw out the newly constructed Huffman tree
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Task
For this project, you have been provided an implementation of the Huffman tree construction algorithm (DO NOT modify this code. It is possible to construct multiple
Huffman trees with the same message if numerous characters in the message share the
same frequency value. The construct function that I set up ensures that your Huffman tree will produce results consistent with the test cases used for grading). This
algorithm utilizes a templated min heap class that you must implement. You must
also implement a function that prints the Huffman encoding of message using the
Huffman tree constructed from the same message. Specifically, you must implement
the following functions:
void MinHeap::insert(const T data, const int key) :
Insert the provided user data into the min heap using the provided key to make
comparisons with the other elements in the min heap.
To ensure that your min heap produces consistent results, stop bubbling up a
child node if it shares the same key value as its parent node.
T MinHeap::extract min() :
Remove from the min heap the element with the smallest key value and return
its data.
If you come across two sibling nodes that share the same key value while sifting
down a parent node with a larger key value, then you should swap the parent
node with the left child to ensure that your min heap produces consistent results.
T MinHeap::peek() const :
Retrieve the minimum element in the min heap and return its data to the user.
Do not remove this element from the min heap in this function.
void MinHeap::size() const :
Return the size of the min heap
void HuffmanTree::print() const :
Print the Huffman encoding of the member variable message assigned in the
construct function.
To ensure that you always produce a consistent output, DO NOT modify the completed code in the HuffmanTree class. You may however add print helper functions
if you feel it necessary.
Input
Input is read from the keyboard. The first line of the input will be an integer t > 0
that indicates the number of test cases. Each test case will contain a message on a
single line to be processed by the Huffman tree construct function. Each message will
contain at least 2 characters.
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Output
For each test case, print Test Case: followed by the test case number on one line. On
another line, print the Huffman encoding of the input message. Separate the individual
character encodings by a space.
Sample Test Cases
Use input redirection to redirect commands written in a file to the standard input, e.g.
$ ./a.out < input1.dat.
Input 1
3
opossum
hello world
message
Output 1
Test Case: 1
11 00 11 10 10 011 010
Test Case: 2
010 011 10 10 00 1100 1101 00 1110 10 1111
Test Case: 3
011 10 11 11 010 00 10
Timing Analysis
At the top of your main file, in comments, write the time complexity of constructing
a Huffman tree with a min heap in terms of the number of characters in the input
message, which you can denote as n. Also consider the time complexity of constructing
a Huffman tree without a min heap. Specifically, what running time can you expect
if you use a linear search to find minimum frequencies. Write these time complexities
using Big-O notation.
Turn In
Submit your source code to the TA through blackboard. If you have multiple files,
package them into a zip file.
Grading:
Total: 100 pts.
• 10/100 – Code style, commenting, general readability.
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• 05/100 – Compiles.
• 05/100 – Follows provided input and output format.
• 75/100 – Successful implementation of the min heap and Huffman print functions.
• 05/100 – Correct timing analysis.
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