114_1_image_proc_exam_solns_2010_Part2

114_1_image_proc_exam_solns_2010_Part2 - p +p 1 +p 2 +p 3...

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EE114 Introduction to Speech and Image Processing Winter Quarter, 2010 Page 5 of 6 Solution: 1 0 1 1 0 0 0 2 0 1 0 4. Huffman Coding (25 points) Consider a set of symbols s 0 , s 1 , s 2 , s 3 , and s 4 , with associated respective probabilities p 0 , p 1 , p 2 , p 3 , and p 4 . Of course, p 0 +p 1 +p 2 +p 3 +p 4 =1. Suppose that two different Huffman codes for this symbol set are designed, using normal Huffman code design procedures. In both cases, the same set of probabilities p i is used. The resulting codes are shown below, and are both valid Huffman codes for this symbol set. Code 1 Code 2 s 0 0 00 s 1 100 010 s 2 101 011 s 3 110 10 s 4 111 11 Given the codes listed above, what can be determined about the relative values of the probabilities p 0 , p 1 , p 2 , p 3 , and p 4 ? For full credit, your answer must provide all the information that can be determined about the relationships among these five probability values (other than
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Unformatted text preview: p +p 1 +p 2 +p 3 +p 4 =1, which was already given). Solution: First we draw the code trees for code 1 and 2 Code 1 EE114 Introduction to Speech and Image Processing Winter Quarter, 2010 Page 6 of 6 s s 1 s 2 s 3 s 4 1 1 1 1 Code 2 s s 1 s 2 s 3 s 4 1 1 1 Looking at the trees, we can see that 1 2 3 4 p p p p p + + = because of the differences between the two trees (the code designer has the ability to associate ( 1 2 s s + ) either with s ,or with ( 3 4 s s + )). Note that this also implies that 3 4 1 p p p p = + and 3 4 2 p p p p = + ....
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114_1_image_proc_exam_solns_2010_Part2 - p +p 1 +p 2 +p 3...

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