Huffman Coding.ppt
- 2. Encoding messages Encode a message composed of a string of characters Codes used by computer systems ASCII • uses 8 bits per character • can encode 256 characters Unicode • 16 bits per character • can encode 65536 characters • includes all characters encoded by ASCII ASCII and Unicode are fixed-length codes all characters represented by same number of bits
- 3. Problems Suppose that we want to encode a message constructed from the symbols A, B, C, D, and E using a fixed-length code How many bits are required to encode each symbol? at least 3 bits are required 2 bits are not enough (can only encode four symbols) How many bits are required to encode the message DEAACAAAAABA? there are twelve symbols, each requires 3 bits 12*3 = 36 bits are required
- 4. Drawbacks of fixed-length codes Wasted space Unicode uses twice as much space as ASCII • inefficient for plain-text messages containing only ASCII characters Same number of bits used to represent all characters ‘a’ and ‘e’ occur more frequently than ‘q’ and ‘z’ Potential solution: use variable-length codes variable number of bits to represent characters when frequency of occurrence is known short codes for characters that occur frequently
- 5. Advantages of variable-length codes The advantage of variable-length codes over fixed- length is short codes can be given to characters that occur frequently on average, the length of the encoded message is less than fixed-length encoding Potential problem: how do we know where one character ends and another begins? • not a problem if number of bits is fixed! A = 00 B = 01 C = 10 D = 11 0010110111001111111111 A C D B A D D D D D
- 6. Prefix property A code has the prefix property if no character code is the prefix (start of the code) for another character Example: 000 is not a prefix of 11, 01, 001, or 10 11 is not a prefix of 000, 01, 001, or 10 … Symbol Code P 000 Q 11 R 01 S 001 T 10 01001101100010 R S T Q P T
- 7. Code without prefix property The following code does not have prefix property The pattern 1110 can be decoded as QQQP, QTP, QQS, or TS Symbol Code P 0 Q 1 R 01 S 10 T 11
- 8. Problem Design a variable-length prefix-free code such that the message DEAACAAAAABA can be encoded using 22 bits Possible solution: A occurs eight times while B, C, D, and E each occur once represent A with a one bit code, say 0 • remaining codes cannot start with 0 represent B with the two bit code 10 • remaining codes cannot start with 0 or 10 represent C with 110 represent D with 1110 represent E with 11110
- 9. Encoded message Symbol Code A 0 B 10 C 110 D 1110 E 11110 DEAACAAAAABA 1110111100011000000100 22 bits
- 10. Another possible code Symbol Code A 0 B 100 C 101 D 1101 E 1111 DEAACAAAAABA 1101111100101000001000 22 bits
- 11. Better code Symbol Code A 0 B 100 C 101 D 110 E 111 DEAACAAAAABA 11011100101000001000 20 bits
- 12. What code to use? Question: Is there a variable-length code that makes the most efficient use of space? Answer: Yes!
- 13. Huffman coding tree Binary tree each leaf contains symbol (character) label edge from node to left child with 0 label edge from node to right child with 1 Code for any symbol obtained by following path from root to the leaf containing symbol Code has prefix property leaf node cannot appear on path to another leaf note: fixed-length codes are represented by a complete Huffman tree and clearly have the prefix property
- 14. Building a Huffman tree Find frequencies of each symbol occurring in message Begin with a forest of single node trees each contain symbol and its frequency Do recursively select two trees with smallest frequency at the root produce a new binary tree with the selected trees as children and store the sum of their frequencies in the root Recursion ends when there is one tree this is the Huffman coding tree
- 15. Example Build the Huffman coding tree for the message This is his message Character frequencies Begin with forest of single trees A G M T E H _ I S 1 1 1 1 2 2 3 3 5 1 1 3 1 2 1 2 3 5 A G I S M T E H _
- 16. Step 1 1 1 3 1 2 1 2 3 5 A G I S M T E H _ 2
- 17. Step 2 1 1 3 1 2 1 2 3 5 A G I S M T E H _ 2 2
- 18. Step 3 1 1 3 1 1 3 5 A G I S M T _ 2 2 2 2 E H 4
- 19. Step 4 1 1 3 1 1 3 5 A G I S M T _ 2 2 2 2 E H 4 4
- 20. Step 5 1 1 3 1 1 3 5 A G I S M T _ 2 2 2 2 E H 4 4 6
- 21. Step 6 3 3 5 I S _ 2 2 E H 4 1 1 1 1 A G M T 2 2 4 6 8
- 22. Step 7 3 3 5 I S _ 2 2 E H 4 1 1 1 1 A G M T 2 2 4 6 8 11
- 23. Step 8 3 3 5 I S _ 2 2 E H 4 1 1 1 1 A G M T 2 2 4 6 8 11 19
- 24. Label edges 3 3 5 I S _ 2 2 E H 4 1 1 1 1 A G M T 2 2 4 6 8 11 19 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
- 25. Huffman code & encoded message S 11 E 010 H 011 _ 100 I 101 A 0000 G 0001 M 0010 T 0011 This is his message 00110111011110010111100011101111000010010111100000001010
- 26. Average Code Length The average code length of the Huffman tree can be determined by using the formula given below: Average Code Length = ∑ ( frequency × code length ) / ∑ ( frequency ) This is his message Symbol Frequency (F) Code Code Length (CL) Total (F*CL) S 5 11 2 10 E 2 010 3 6 H 2 011 3 6 _ 3 100 3 9 I 3 101 3 9 A 1 0000 4 4 G 1 0001 4 4 M 1 0010 4 4 T 1 0011 4 4 Total_Freq 19 56 Average = 56/19 = 2.94737 bits Length of the string??