arithExample2011

arithExample2011 - # 18 ECE 253a Digital Image Processing...

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ECE 253a Digital Image Processing Pamela Cosman 11/15/11 Example of Arithmetic Coding We have 3 symbols in the alphabet, with probabilities p (1) = 0 . 8 p (2) = 0 . 02 p (3) = 0 . 18 The cdf corresponding to this pdf has values: F (0) = 0 F (1) = 0 . 8 F (2) = 0 . 82 F (3) = 1 The input sequence we will code is X 1 , X 2 , X 3 , X 4 . . . = 1 , 3 , 2 , 1 . . . We initialize the lower and upper endpoints of the interval to be l (0) = 0 u (0) = 1 After reading input symbol X n , we will update the sub-interval endpoints as follows: l ( n ) = l ( n - 1) + [ u ( n - 1) - l ( n - 1) ] F ( X n - 1) u ( n ) = l ( n - 1) + [ u ( n - 1) - l ( n - 1) ] F ( X n ) We begin with the first symbol: X 1 = 1 . We apply the endpoint updating equations to obtain: l (1) = 0 + (1 - 0) × 0 = 0 u (1) = 0 + (1 - 0) × 0 . 8 = 0 . 8 This interval straddles 0.5, so there is no output of bits. So we proceed to look at the next symbol: X 2 = 3 . We find the new sub-interval by applying the endpoint updating equations: l (2) = 0 + (0 . 8 - 0) F (2) = 0 . 8 × 0 . 82 = 0 . 656 u (2) = 0 + (0 . 8 - 0) F (3) = 0 . 8 × 1 = 0 . 8 Again, we compare to 0.5, and we see this is entirely in the upper half of the unit interval. So we output the bit 1. Now we can rescale with the rescaling function
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arithExample2011 - # 18 ECE 253a Digital Image Processing...

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