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Chapter 5

Chapter 5. Markov processes Run length coding Gray code. Markov Processes. Transition Graph. Transition Matrix. Weather Example : Let j = 1. Think: a means “fair” b means “rain” c means “snow”. ½. next symbol. b. ⅓. ⅓. c u r r e n t s t a t e. ¼. ⅓ ⅓ ⅓

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Chapter 5

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  1. Chapter 5 Markov processes Run length coding Gray code

  2. Markov Processes Transition Graph Transition Matrix Weather Example: Let j = 1. Think: a means “fair” b means “rain” c means “snow” ½ next symbol b ⅓ ⅓ c u r r e n t s t a t e ¼ ⅓ ⅓ ⅓ ¼ ½ ¼ ¼ ¼ ½ Let S = {s1, …, sq} be a set of symbols. A jth-order Markov process has probabilities p(si | si1 … sij) associated with it, the conditional probability of seeing siafter having seen si1… sij. This is said to be a j-memory source, and there are qjstates in the Markov process. a ¼ ¼ = M ¼ c ⅓ ½ p(si | sj) p(si | sj) i = column, j = row sj si transition probability ∑ outgoing edges = 1 5.2

  3. Ergodic Equilibriums Definition: A first-order Markov process M is said to be ergodic if • From any state we can eventually get to any other state. • The system reaches a limiting distribution. 5.2

  4. Predictive Coding Assume a prediction algorithm for a binary source which given all prior bits predicts the next. input stream prediction s1 ….. sn1pnen = pnsn error What is transmitted is the error, ei. By knowing just the error, the predictor also knows the original symbols. en en   sn sn destination channel source pn pn predictor predictor must assume that both predictors are identical, and start in the same state 5.7

  5. Accuracy: The probability of the predictor being correct is p = 1  q; constant (over time) and independent of other prediction errors. Let the probability of a run of exactly n 0’s, (0n1), be p(n) = pn ∙ q. The probability of runs of any length n = 0, 1, 2, … is: Note: alternate method for calculating f(p), look at 5.8

  6. Coding of Run Lengths Send a k-digit binary number to represent a run of zeroes whose length is between 0 and 2k 2. (small runs are in binary) For run lengths larger than 2k 2, send 2k 1 (k ones) followed by another k-digit binary number, etc. (large runs are in unary) Let n = run length. Fix k = block length. Use division to get: n = i ∙ m + j 0 ≤ j < m = 2k 1 (like “reading” the “matrix” with m cells and ∞ many rows) 5.9

  7. Expected length of run length code Let p(n) = the probability of a run of exactly n 0’s: 0n1. The expected code length is: But every n can be written uniquely as i∙m + j where i ≥ 0, 0 ≤ j < m = 2k 1. 5.9

  8. Gray Code Consider an analog-to-digital “flash” converter consisting of a rotating wheel: 0 0 The maximum error in the scheme is ± ⅛ rotation because … 1 0 imagine “brushes” contacting the wheel in each of the three circles 0 0 0 1 1 1 0 0 0 0 0 1 1 0 1 1 1 1 0 1 1 The Hamming Distance between adjacent positions is 1. In ordinary binary, the maximum distance is 3 (the max. possible). 5.15-17

  9. Binary ↔ Gray Inductively: Computationally: 5.15-17

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