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11-1Data Structures---HuffmanCoding
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The course is mainly about Huffman coding.Generally covered Huffman coding: An exercise in the use of priority queue;and how you could to encode symbols so that the binary file has the smallest size and so on.

1.Huffman coding An exercise in the use of priority queue Symbol Frequency Symbol Frequency Symbol Frequency space 186 b 47 g 15 e 103 d 32 p 15 t 80 l 32 b 13 a 64 u 23 v 8 o 63 c 22 k 5 i 57 f 21 j 1 n 57 m 20 q 1 s 51 w 18 x 1 r 48 y 16 z 1 Source: Donald Knuth, The Art of Computer Programming (Volume 3) p 441 This table shows the average occurrence of individual letters in every 1000 letters. Each letter can be encoded by a custom binary code. The problem. How will you encode these symbols so that the binary file has the smallest size?

2.Straight ASCII (that will use n bytes for coding n characters is not the optimal solution, when you know the frequencies of these characters. An efficient solution will use only a few bits for the frequently used characters, but may use more bits to encode less frequently used characters. Of course this will be a custom encoding scheme. [Application: saving the transmission bandwidth] A naïve solution is as follows. Suppose the frequencies of the following five letters satisfy the order e > a> c > b > d 0 1 e 0 1 a 0 1 c 0 1 b d Here are the codes: e = 0, a = 10, c = 110, b= 1110, d = 1111 This is ok, but may not be optimal when the frequencies are known.

3.A smaller scale example e r s t n l z x 34 22 24 28 15 10 9 8 Frequency in an average sample of size 150 letters Enqueue these in a priority queue Dequeue (the letter/subtree with smallest count) Dequeue (the letter/subtree with smallest count) Form a subtree by adding a common parent to the above two and enqueue into the priority queue again Repeat these steps till a binary tree is formed. The tree is shown in the next page. This leads to the following codes. z = 0000 n = 0110 x = 0001 l = 0111 r = 001 e = 10 s = 010 t = 11

4. 9+8 = 17 34 22 24 28 15 10 z x e r s t n l 9+8 = 17 34 22 24 28 15+10 = 25 z x e r s t n l 17+22 = 39 34 24 28 25 e s t n l r z x 39 34 24+25 = 49 28 e t r s z x n l 39 34+28 = 62 24+25 = 49 r e t s z x n l 39+49=88 62 e t r s n l z x 88+62=150 0 1 0 1 0 1 1 0 1 0 e t s 0 1 0 1 r n l z x

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