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# s5 - #21 ECE 253a Digital Image Processing Pamela Cosman...

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#21 ECE 253a Digital Image Processing Pamela Cosman 11/30/10 Solutions to homework 5 These solutions are written by Dawei Wang. Problem 1: Huffman Coding a) One possible codeword assignment is: Symbol Probability Codeword Length Codeword A 0.5 1 0 B 0.25 2 10 C 0.1 3 110 D 0.05 4 1110 E 0.05 5 11110 F 0.05 5 11111 Other correct answers are also acceptable, and the average length is 2 bits. The entropy of this R.V. is H ( 1 2 , 1 4 , 1 10 , 1 20 , 1 20 , 1 20 ) 1 . 98 bits. b) For quaternary Huffman coding, we merge 4 nodes with smallest probability to one node in every iteration. For this particular example, one possible code can be: Symbol Probability Codeword Length Codeword A 0.5 1 a B 0.25 1 b C 0.1 2 ca D 0.05 2 cb E 0.05 2 cc F 0.05 2 cd and the average length is 1.25 quaternary units. The quaternary alphabet symbol d is not used in the first position of the codeword, and this encoding method is not efficient. In general, if the cardinality of input symbols is not in the form of 3 N + 1, N Q + for quaternary code (( M - 1) N + 1, N Q + for M -ary code), we would waste some codeword length in the final step of the iteration. To solve this problem, a zero-probability symbol padding is used. Basically, we introduce some dummy symbols with zero probability to make the starting number of input symbols be of the form 3 N + 1, N Q + . For this particular example: Symbol Probability Codeword Length Codeword A 0.5 1 a B 0.25 1 b C 0.1 1 c D 0.05 2 da E 0.05 2 db F 0.05 2 dc G(dummy) 0 2 dd 1

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and the average length is 1.15 quaternary units. Because the use of dummy symbols was not explained to you, both answers to this problem are acceptable. c) If we generate a binary code from the first quaternary code of part b), the codewords are: Symbol Probability Codeword Length Codeword A 0.5 2 00 B 0.25 2 01 C 0.1 4 1000 D 0.05 4 1001 E 0.05 4 1010 F 0.05 4 1011 and the average length of this code is 2.5 bits, which is longer than 2 bits in part a).
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