Difference between revisions of "161-A2.1"

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Recall that Shannon's Noiseless Coding Theorem states that <math>H\left(S\right)\leq L = 1.75\,\mathrm{bits}</math>. Thus, codes 1 and 2 are not uniquely decodable since they have average code lengths less than <math>L</math>.
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For code 1, since there are two symbols sharing an encoded symbol, we can say that it is not a distinct code, and therefore not uniquely decodable. We can check code 2 by taking a sequence of symbols encoded using this code: <math>0011</math>. We can easily see that this message can be decoded in different ways, either as <math>0, 0, 1, 1</math>, or <math>00, 11</math>.

Revision as of 20:12, 17 September 2020

  • Activity: Source Coding
  • Instructions: In this activity, you are tasked to
    • Walk through the examples.
    • Write a short program to compress and decompress a redundant file.
  • Should you have any questions, clarifications, or issues, please contact your instructor as soon as possible.

Uniquely Decodable Codes

Let us try to encode a source with just four symbols in its alphabet, i.e. , with probability distribution . We can calculate the entropy of this source as:

 

 

 

 

(1)

Let us look at a few bit sequence assignments and see if they are uniquely decodable or not.

Symbol Probability Code 1 Code 2 Code 3 Code 4 Code 5
0.5 0 0 0 0 00
0.25 0 1 10 01 01
0.125 1 00 110 011 10
0.125 10 11 111 0111 11
Average Code Length: 1.125 1.25 1.75 1.875 2

Recall that Shannon's Noiseless Coding Theorem states that . Thus, codes 1 and 2 are not uniquely decodable since they have average code lengths less than .

For code 1, since there are two symbols sharing an encoded symbol, we can say that it is not a distinct code, and therefore not uniquely decodable. We can check code 2 by taking a sequence of symbols encoded using this code: . We can easily see that this message can be decoded in different ways, either as , or .