Huffman Coding Algorithm

Huffman coding is a lossless data compression algorithm. In this algorithm, a variable-length code is assigned to input different characters. The code length is related to how frequently characters are used. Most frequent characters have the smallest codes and longer codes for least frequent characters.

There are mainly two parts. First one to create a Huffman tree, and another one to traverse the tree to find codes.

For an example, consider some strings “YYYZXXYYX”, the frequency of character Y is larger than X and the character Z has the least frequency. So the length of the code for Y is smaller than X, and code for X will be smaller than Z.

Complexity for assigning the code for each character according to their frequency is O(n log n)

Input and Output

Input:
A string with different characters, say “ACCEBFFFFAAXXBLKE”
Output:
Code for different characters:
Data: K, Frequency: 1, Code: 0000
Data: L, Frequency: 1, Code: 0001
Data: E, Frequency: 2, Code: 001
Data: F, Frequency: 4, Code: 01
Data: B, Frequency: 2, Code: 100
Data: C, Frequency: 2, Code: 101
Data: X, Frequency: 2, Code: 110
Data: A, Frequency: 3, Code: 111

Algorithm

huffmanCoding(string)

Input: A string with different characters.

Output: The codes for each individual characters.

Begin
define a node with character, frequency, left and right child of the node for Huffman tree.
create a list ‘freq’ to store frequency of each character, initially, all are 0
for each character c in the string do
increase the frequency for character ch in freq list.
done

for all type of character ch do
if the frequency of ch is non zero then
add ch and its frequency as a node of priority queue Q.
done

while Q is not empty do
remove item from Q and assign it to left child of node
remove item from Q and assign to the right child of node
traverse the node to find the assigned code
done
End

traverseNode(n: node, code)

Input: The node n of the Huffman tree, and the code assigned from the previous call

Output: Code assigned with each character

if a left child of node n ≠φ then
traverseNode(leftChild(n), code+’0’)     //traverse through the left child
traverseNode(rightChild(n), code+’1’)    //traverse through the right child
else
display the character and data of current node.

Example

#include
#include
#include
using namespace std;

struct node {
int freq;
char data;
const node *child0, *child1;

node(char d, int f = -1) { //assign values in the node
data = d;
freq = f;
child0 = NULL;
child1 = NULL;
}

node(const node *c0, const node *c1) {
data = 0;
freq = c0->freq + c1->freq;
child0=c0;
child1=c1;
}

bool operator<( const node &a ) const { //< operator performs to find priority in queue
return freq >a.freq;
}

void traverse(string code = "")const {
if(child0!=NULL) {
child0->traverse(code+'0'); //add 0 with the code as left child
child1->traverse(code+'1'); //add 1 with the code as right child
}else {
cout << "Data: " << data<< ", Frequency: "< qu;
int frequency[256];

for(int i = 0; i<256; i++)
frequency[i] = 0; //clear all frequency

for(int i = 0; i1) {
node *c0 = new node(qu.top()); //get left child and remove from queue
qu.pop();
node *c1 = new node(qu.top()); //get right child and remove from queue
qu.pop();
qu.push(node(c0, c1)); //add freq of two child and add again in the queue
}
cout << "The Huffman Code: "<


Output

The Huffman Code:
Data: K, Frequency: 1, Code: 0000
Data: L, Frequency: 1, Code: 0001
Data: E, Frequency: 2, Code: 001
Data: F, Frequency: 4, Code: 01
Data: B, Frequency: 2, Code: 100
Data: C, Frequency: 2, Code: 101
Data: X, Frequency: 2, Code: 110
Data: A, Frequency: 3, Code: 111
karthikeya Boyini

I love programming (: That's all I know