Huffman tree
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The document explains Huffman coding, a file compression technique developed by Dr. David A. Huffman in 1952, which is often used in data transmission. It describes the process of creating a binary tree and priority queue to build a coding scheme based on character frequency, ultimately leading to significant data size reduction. The example demonstrates encoding a sample text, resulting in a more efficient representation compared to standard ASCII encoding.
Huffman tree
- 1. Al-amin Hossain 123 1173 642 North South University
- 2. An application of Binary Tree and priority Queue. Huffman coding is a technique used to compress files for transmission. Developed by Dr. David A. Huffman in 1952 “A Method for the Construction of Minimum Redundancy Codes.” Application: Common application is data transmission in Fax Machines Encoding-Decoding text.
- 3. Leaf nodes are character. Left Branches are labeled with 0. Right Branches are labeled with 1. Following a path from root to leaf You can get the encoding of the Character in the leaf. Like: 101= ‘ I ’
- 4. Scan text to be compressed and tally occurrence of all characters. Sort or prioritize characters based on number of occurrences in text. Build Huffman code tree based on prioritized list. Perform a traversal of tree to determine all code words. Scan text again and create new file using the Huffman codes.
- 5. Lets consider this text, “Eerie eyes seen near lake.” Count up the occurrences of all characters.
- 6. What characters are present ? E e r i space y s n a r l k . Now the Frequency of each character in the text. Char Freq. Char Freq. Char Freq. E 1 y 1 k 1 e 8 s 2 . 1 r 2 n 2 i 1 a 2 space 4 l 1
- 7. Create Binary tree nodes with character and frequency of each character. Place nodes in a priority queue The lower the occurrence (frequency), the higher the priority in the queue.
- 8. • After inserting all nodes in the queue E 1 i 1 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2 sp 4 e 8
- 9. While priority queue contains two or more nodes Create new node Dequeue node and make it left subtree Dequeue next node and make it right subtree Frequency of new node equals sum of frequency of left and right children Enqueue new node back into queue
- 10. • Dequeue 2 lowest frequency nodes • Lowest will be left child • Other will be right child • Sum of the two frequencies will be the parent. E 1 i 1 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2 sp 4 e 8 2
- 11. • Enqeue the new created node to the queue E 1 i 1 y 1 l 1 k 1 . 1 r 2 s 2 n 2 a 2 sp 4 e 8 2
- 12. • Following the same procedure two lowest Dequeue creating new Tree and Enqueue E 1 i 1 k 1 . 1 r 2 s 2 n 2 a 2 sp 4 e 8 2 y 1 l 1 2
- 14. • Again following the same procedure Two lowest Dequeue -> New Tree-> Enqueue E 1 i 1 r 2 s 2 n 2 a 2 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2
- 16. Two lowest Dequeue -> New Tree-> Enqueue E 1 i 1 r 2 s 2 n 2 a 2 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2
- 18. Two lowest Dequeue -> New Tree-> Enqueue E 1 i 1 n 2 a 2 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4
- 20. Two lowest Dequeue -> New Tree -> Enqueue i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 E 1
- 22. Two lowest Dequeue -> New Tree -> Enqueue E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4
- 24. Two lowest Dequeue -> New Tree -> Enqueue E 1 i 1 sp 4 e 82 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6
- 26. Two lowest Dequeue -> New Tree -> Enqueue E 1 i 1 sp 4 e 82 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8
- 28. Two lowest Dequeue -> New Tree -> Enqueue E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10
- 30. Two lowest Dequeue ->New Tree -> Enqueue E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16
- 32. Two lowest Dequeue ->New Tree -> Enqueue E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16 26
- 33. After Enqueing this node there is only one node left in priority queue. E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16 26
- 34. And the last Dequeue We got a Tree This tree contains the new code words for each character. Freq of root = Num of char in text Root=26 E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16 26 Eerie eyes seen near lake. 26 characters
- 36. • By traversing tree we will get new code words for each character • Going left is a 0 • Going Right is a 1 • Code word is only completed when a leaf node is reached E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16 26
- 37. E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16 26 Char Code E 0000 i 0001 y 0010 l 0011 k 0100 . 0101 Space 011 e 10 r 1100 s 1101 n 1110 a 1111
- 38. Char Code E 0000 i 0001 y 0010 l 0011 k 0100 . 0101 Space 011 e 10 r 1100 s 1101 n 1110 a 1111 Rescan text and encode file using new code words. Eerie eyes seen near lake. 0000101100000110011 1000101011011010011 1110101111110001100 1111110100100101
- 39. Now the question is, have we made things any better? Answer is YES! If we use ASCII, it would take 8*26 = 208 bits. But now after encoding its just 73 bits.
- 41. Read the Frequency table and rebuild the tree. Read one bit at a time when you read a bit 1, go to the Right child when you read a bit 0, go to the Left child Return to the root and continue reading bits The tree allows us to easily overcome the challenge of determining the character boundaries!
- 42. • If got 0000 0 0 0 0 Char Code E 0000 i 0001 y 0010 l 0011 k 0100 . 0101 Space 011 e 10 r 1100 s 1101 n 1110 a 1111 E 1 0000101100000110011 1000101011011010011 1110101111110001100 1111110100100101
- 43. Final Tree is looks like that. E 1 i 1 sp 4 e 8 2 y 1 l 1 2 k 1 . 1 2 r 2 s 2 4 n 2 a 2 4 4 6 8 10 16 26