{"id":25081,"date":"2017-10-15T14:11:55","date_gmt":"2017-10-15T08:41:55","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=25081"},"modified":"2017-10-15T14:11:55","modified_gmt":"2017-10-15T08:41:55","slug":"efficient-huffman-coding-sorted-input","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/efficient-huffman-coding-sorted-input\/","title":{"rendered":"Efficient Huffman Coding for Sorted Input"},"content":{"rendered":"

We recommend to read following post as a prerequisite for this.<\/p>\n

Huffman Coding<\/p>\n

Time complexity of the algorithm discussed in above post is O(nLogn). <\/span>If we know that the given array is sorted (by non-decreasing order of frequency), we can generate Huffman codes in O(n) time. Following is a O(n) algorithm for sorted input.<\/p>\n

1.<\/strong> Create two empty queues.<\/p>\n

2.<\/strong> Create a leaf node for each unique character and Enqueue it to the first queue in non-decreasing order of frequency. Initially second queue is empty.<\/p>\n

3.<\/strong> Dequeue two nodes with the minimum frequency by examining the front of both queues. Repeat following steps two times
\n\u2026..a)<\/strong> If second queue is empty, dequeue from first queue.
\n\u2026..b)<\/strong> If first queue is empty, dequeue from second queue.
\n\u2026..c)<\/strong> Else, compare the front of two queues and dequeue the minimum.<\/p>\n

4.<\/strong> Create a new internal node with frequency equal to the sum of the two nodes frequencies. Make the first Dequeued node as its left child and the second Dequeued node as right child. Enqueue this node to second queue.<\/p>\n

5.<\/strong> Repeat steps#3 and #4 until there is more than one node in the queues. The remaining node is the root node and the tree is complete.<\/p>\n

c<\/span> <\/div> 
\/\/ C Program for Efficient Huffman Coding for Sorted input#include <stdio.h>#include <stdlib.h> \/\/ This constant can be avoided by explicitly calculating height of Huffman Tree#define MAX_TREE_HT 100 \/\/ A node of huffman treestruct QueueNode{    char data;    unsigned freq;    struct QueueNode *left, *right;}; \/\/ Structure for Queue: collection of Huffman Tree nodes (or QueueNodes)struct Queue{    int front, rear;    int capacity;    struct QueueNode **array;}; \/\/ A utility function to create a new Queuenodestruct QueueNode* newNode(char data, unsigned freq){    struct QueueNode* temp =       (struct QueueNode*) malloc(sizeof(struct QueueNode));    temp->left = temp->right = NULL;    temp->data = data;    temp->freq = freq;    return temp;} \/\/ A utility function to create a Queue of given capacitystruct Queue* createQueue(int capacity){    struct Queue* queue = (struct Queue*) malloc(sizeof(struct Queue));    queue->front = queue->rear = -1;    queue->capacity = capacity;    queue->array =      (struct QueueNode**) malloc(queue->capacity * sizeof(struct QueueNode*));    return queue;} \/\/ A utility function to check if size of given queue is 1int isSizeOne(struct Queue* queue){    return queue->front == queue->rear && queue->front != -1;} \/\/ A utility function to check if given queue is emptyint isEmpty(struct Queue* queue){    return queue->front == -1;} \/\/ A utility function to check if given queue is fullint isFull(struct Queue* queue){    return queue->rear == queue->capacity - 1;} \/\/ A utility function to add an item to queuevoid enQueue(struct Queue* queue, struct QueueNode* item){    if (isFull(queue))        return;    queue->array[++queue->rear] = item;    if (queue->front == -1)        ++queue->front;} \/\/ A utility function to remove an item from queuestruct QueueNode* deQueue(struct Queue* queue){    if (isEmpty(queue))        return NULL;    struct QueueNode* temp = queue->array[queue->front];    if (queue->front == queue->rear)  \/\/ If there is only one item in queue        queue->front = queue->rear = -1;    else        ++queue->front;    return temp;} \/\/ A utility function to get from of queuestruct QueueNode* getFront(struct Queue* queue){    if (isEmpty(queue))        return NULL;    return queue->array[queue->front];} \/* A function to get minimum item from two queues *\/struct QueueNode* findMin(struct Queue* firstQueue, struct Queue* secondQueue){    \/\/ Step 3.a: If second queue is empty, dequeue from first queue    if (isEmpty(firstQueue))        return deQueue(secondQueue);     \/\/ Step 3.b: If first queue is empty, dequeue from second queue    if (isEmpty(secondQueue))        return deQueue(firstQueue);     \/\/ Step 3.c:  Else, compare the front of two queues and dequeue minimum    if (getFront(firstQueue)->freq < getFront(secondQueue)->freq)        return deQueue(firstQueue);     return deQueue(secondQueue);} \/\/ Utility function to check if this node is leafint isLeaf(struct QueueNode* root){    return !(root->left) && !(root->right) ;} \/\/ A utility function to print an array of size nvoid printArr(int arr[], int n){    int i;    for (i = 0; i < n; ++i)        printf("%d", arr[i]);    printf("\\n");} \/\/ The main function that builds Huffman treestruct QueueNode* buildHuffmanTree(char data[], int freq[], int size){    struct QueueNode *left, *right, *top;     \/\/ Step 1: Create two empty queues    struct Queue* firstQueue  = createQueue(size);    struct Queue* secondQueue = createQueue(size);     \/\/ Step 2:Create a leaf node for each unique character and Enqueue it to    \/\/ the first queue in non-decreasing order of frequency. Initially second    \/\/ queue is empty    for (int i = 0; i < size; ++i)        enQueue(firstQueue, newNode(data[i], freq[i]));     \/\/ Run while Queues contain more than one node. Finally, first queue will    \/\/ be empty and second queue will contain only one node    while (!(isEmpty(firstQueue) && isSizeOne(secondQueue)))    {        \/\/ Step 3: Dequeue two nodes with the minimum frequency by examining        \/\/ the front of both queues        left = findMin(firstQueue, secondQueue);        right = findMin(firstQueue, secondQueue);         \/\/ Step 4: Create a new internal node with frequency equal to the sum        \/\/ of the two nodes frequencies. Enqueue this node to second queue.        top = newNode('$' , left->freq + right->freq);        top->left = left;        top->right = right;        enQueue(secondQueue, top);    }     return deQueue(secondQueue);} \/\/ Prints huffman codes from the root of Huffman Tree.  It uses arr[] to\/\/ store codesvoid printCodes(struct QueueNode* root, int arr[], int top){    \/\/ Assign 0 to left edge and recur    if (root->left)    {        arr[top] = 0;        printCodes(root->left, arr, top + 1);    }     \/\/ Assign 1 to right edge and recur    if (root->right)    {        arr[top] = 1;        printCodes(root->right, arr, top + 1);    }     \/\/ If this is a leaf node, then it contains one of the input    \/\/ characters, print the character and its code from arr[]    if (isLeaf(root))    {        printf("%c: ", root->data);        printArr(arr, top);    }} \/\/ The main function that builds a Huffman Tree and print codes by traversing\/\/ the built Huffman Treevoid HuffmanCodes(char data[], int freq[], int size){   \/\/  Construct Huffman Tree   struct QueueNode* root = buildHuffmanTree(data, freq, size);    \/\/ Print Huffman codes using the Huffman tree built above   int arr[MAX_TREE_HT], top = 0;   printCodes(root, arr, top);} \/\/ Driver program to test above functionsint main(){    char arr[] = {'a', 'b', 'c', 'd', 'e', 'f'};    int freq[] = {5, 9, 12, 13, 16, 45};    int size = sizeof(arr)\/sizeof(arr[0]);    HuffmanCodes(arr, freq, size);    return 0;}Run on IDE<\/code><\/pre> <\/div>\n
Output:<\/strong><\/h4>\n
f: 0\r\nc: 100\r\nd: 101\r\na: 1100\r\nb: 1101\r\ne: 111<\/pre>\n
Time complexity:<\/strong> O(n)<\/p>\n
If the input is not sorted, it need to be sorted first before it can be processed by the above algorithm. Sorting can be done using heap-sort or merge-sort both of which run in Theta(nlogn). So, the overall time complexity becomes O(nlogn) for unsorted input.<\/p>\n[ad type=”banner”]\n","protected":false},"excerpt":{"rendered":"
Efficient Huffman Coding for Sorted Input – Greedy Algorithm – Time complexity of the algorithm discussed in above post is O(nLogn). If we know that the given array is sorted (by non-decreasing order of frequency).<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,83476],"tags":[71215,71202,71082,71195,71197,71090,71067,71203,71084,71200,71208,71207,71199,71097,71113,71114,71068,71119,71209,70480,71079,71112,71116,71083,71085,71204,71103,71091,71108,71101,71201,71089,71100,71107,71213,71086,71206,71080,71088,71118,71214,71205,71212,71211,71115,71087,71099,71105,71109,71092,71117,71210,71193,71194,71111,71073,71064,71071,71192,71196,71198,71102],"class_list":["post-25081","post","type-post","status-publish","format-standard","hentry","category-coding","category-huffman-coding","tag-adaptive-huffman-coding-in-c","tag-adaptive-huffman-coding-tree-example","tag-application-of-huffman-coding-in-data-compression","tag-arithmetic-coding-for-data-compression-example","tag-binary-code-example","tag-c-program-to-implement-huffman-code","tag-codetree","tag-coding-algorithms","tag-encoding-and-decoding-huffman-code-in-java","tag-example-of-encoding","tag-example-of-huffman-coding","tag-examples-of-trees","tag-ffman-coding-in-c-huffman-program-in-c","tag-frequency-tree","tag-half-man-coding","tag-how-to-find-codeword-in-huffman-coding","tag-huffman-algorithm-in-data-structure","tag-huffman-algorithm-java","tag-huffman-code-tree-generator","tag-huffman-coding","tag-huffman-coding-algorithm-in-c","tag-huffman-coding-algorithm-in-matlab","tag-huffman-coding-animation","tag-huffman-coding-example","tag-huffman-coding-example-pdf","tag-huffman-coding-implementation","tag-huffman-coding-in-c-language","tag-huffman-coding-in-c-using-array","tag-huffman-coding-in-matlab-program","tag-huffman-coding-matlab-program","tag-huffman-coding-ppt-presentation","tag-huffman-coding-program-in-c","tag-huffman-coding-program-in-c-with-output","tag-huffman-coding-program-in-java","tag-huffman-coding-simple-program-in-c","tag-huffman-coding-steps","tag-huffman-coding-using-matlab","tag-huffman-coding-with-example","tag-huffman-compression-algorithm","tag-huffman-encoding-and-decoding-example-in-java","tag-huffman-encoding-and-decoding-in-c","tag-huffman-encoding-and-decoding-in-matlab","tag-huffman-encoding-in-c","tag-huffman-encoding-online","tag-huffman-encoding-tree","tag-huffman-tree","tag-huffman-tree-c","tag-implementation-of-huffman-coding-in-c","tag-online-huffman-encoder","tag-optimal-huffman-code","tag-program-for-huffman-coding-in-c","tag-shannon-coding-example","tag-shannon-fano-algorithm","tag-shannon-fano-coding-example","tag-simple-huffman-coding-c","tag-simple-huffman-coding-in-c","tag-source-code-for-huffman-coding-in-c","tag-tree-code","tag-tree-tutorial","tag-variance-symbol","tag-write-a-program-for-hu","tag-write-a-program-for-huffman-coding"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/25081","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=25081"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/25081\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=25081"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=25081"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=25081"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}