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- 1. ENSC 424 - Multimedia Communications Engineering Topic 6: Arithmetic Coding 1 Jie Liang Engineering Science Simon Fraser University JieL@sfu.ca J. Liang: SFU ENSC 424 9/20/2005 1
- 2. Outline Introduction Basic Encoding and Decoding Scaling and Incremental Coding Integer Implementation Adaptive Arithmetic Coding Binary Arithmetic Coding Applications JBIG, H.264, JPEG 2000 J. Liang: SFU ENSC 424 9/20/2005 2
- 3. Huffman Coding: The Retired Champion Replacing an input symbol with a codeword Need a probability distribution Hard to adapt to changing statistics Need to store the codeword table Minimum codeword length is 1 bit Arithmetic Coding: The Rising Star Replace the entire input with a single floating-point number Does not need the probability distribution Adaptive coding is very easy No need to keep and send codeword table Fractional codeword length J. Liang: SFU ENSC 424 9/20/2005 3
- 4. History of Arithmetic Coding Claude Shannon: 1916-2001 A distant relative of Thomas Edison 1932: Went to University of Michigan. 1937: Master thesis at MIT became the foundation of digital circuit design: “The most important, and also the most famous, master's thesis of the century“ 1940: PhD, MIT 1940-1956: Bell Lab (back to MIT after that) 1948: The birth of Information Theory A mathematical theory of communication, Bell System Technical Journal. Earliest idea of arithmetic coding Robert Fano: 1917- Shannon-Fano code: proved to be sub-optimal by Huffman 1952: First Information Theory class. Students included: David Huffman: Huffman Coding Peter Elias: Recursive implementation of arithmetic coding Frederick Jelinek Also Fano’s student: PhD MIT 1962 (now at Johns Hopkins) 1968: Further development of arithmetic coding 1976: Rediscovered by Pasco and Rissanen Practical implementation: since 1980’s Bell Lab for Sale: http://www.spectrum.ieee.org/sep05/1683 J. Liang: SFU ENSC 424 9/20/2005 4
- 5. Introduction Recall table look-up decoding of Huffman code N: alphabet size 1 L: Max codeword length 00 Divide [0, 2^L] into N intervals One interval for one symbol 010 011 Interval size is roughly proportional to symbol prob. 000 010 011 100 Arithmetic coding applies this idea recursively Normalizes the range [0, 2^L] to [0, 1]. Map an input sequence to a unique tag in [0, 1). abcd….. dcba….. 0 1 J. Liang: SFU ENSC 424 9/20/2005 5
- 6. 0 1 Arithmetic Coding a b c Disjoint and complete partition of the range [0, 1) [0, 0.8), [0.8, 0.82), [0.82, 1) Each interval corresponds to one symbol Interval size is proportional to symbol probability The first symbol restricts the tag 0 1 position to be in one of the intervals The reduced interval is partitioned 0 1 recursively as more symbols are processed. 0 1 Observation: once the tag falls into an interval, it never gets out of it J. Liang: SFU ENSC 424 9/20/2005 6
- 7. Some Questions to think about: Why compression is achieved this way? How to implement it efficiently? How to decode the sequence? Why is it better than Huffman code? J. Liang: SFU ENSC 424 9/20/2005 7
- 8. Possible Ways to Terminate Encoding 1. Define an end of file (EOF) symbol in the alphabet. Assign a probability for it. 0 1 a b c EOF 2. Encode the lower end of the final range. 3. If number of symbols is known to the decoder, encode any nice number in the final range. J. Liang: SFU ENSC 424 9/20/2005 8
- 9. Example: 1 2 3 Symbol Prob. 1 0.8 0 0.8 0.82 1.0 2 0.02 Map to real line range [0, 1) 3 0.18 Order does not matter Decoder needs to use the same order Disjoint but complete partition: 1: [0, 0.8): 0, 0.799999…9 2: [0.8, 0.82): 0.8, 0.819999…9 3: [0.82, 1): 0.82, 0.999999…9 J. Liang: SFU ENSC 424 9/20/2005 9
- 10. Encoding Input sequence: “1321” 1 2 3 Range 1 0 0.8 0.82 1.0 1 2 3 Range 0.8 0 0.64 0.656 0.8 1 2 3 Range 0.144 0.656 0.7712 0.77408 0.8 1 2 3 Range 0.00288 0.7712 0.773504 0.7735616 0.77408 Termination: Encode the lower end (0.7712) to signal the end. Difficulties: 1. Shrinking of interval requires very high precision for long sequence. 2. No output is generated until the entire sequence has been processed. J. Liang: SFU ENSC 424 9/20/2005 10
- 11. Encoder Pseudo Code Probability Mass Function Cumulative Density Function (CDF) 0.4 For continuous distribution: x 0.2 0.2 0.2 FX ( x) = P ( X ≤ x) = −∞ ∫ p( x)dx 1 2 3 4 X For discrete distribution: i 1.0 FX (i ) = P( X ≤ i ) = ∑ P( X = k ) k = −∞ CDF 0.8 0.4 Properties: 0.2 Non-decreasing Piece-wise constant X 1 2 3 4 Each segment is closed at the lower end. J. Liang: SFU ENSC 424 9/20/2005 11
- 12. Encoder Pseudo Code low=0.0, high=1.0; Keep track of while (not EOF) { LOW, HIGH, RANGE n = ReadSymbol(); Any two are RANGE = HIGH - LOW; sufficient, e.g., HIGH = LOW + RANGE * CDF(n); LOW and RANGE. LOW = LOW + RANGE * CDF(n-1); } output LOW; Input HIGH LOW RANGE Initial 1.0 0.0 1.0 1 0.0+1.0*0.8=0.8 0.0+1.0*0 = 0.0 0.8 3 0.0 + 0.8*1=0.8 0.0 + 0.8*0.82=0.656 0.144 2 0.656+0.144*0.82=0.77408 0.656+0.144*0.8=0.7712 0.00288 1 0.7712+0.00288*0=0.7712 0.7712+0.00288*0.8=0.773504 0.002304 J. Liang: SFU ENSC 424 9/20/2005 12
- 13. Decoding Receive 0.7712 1 2 3 Decode 1 0 0.8 0.82 1.0 1 2 3 Decode 3 0 0.64 0.656 0.8 1 2 3 Decode 2 0.656 0.7712 0.77408 0.8 1 2 3 Decode 1 0.7712 0.773504 0.7735616 0.77408 Drawback: need to recalculate all thresholds each time. J. Liang: SFU ENSC 424 9/20/2005 13
- 14. Simplified Decoding x − low Normalize RANGE to [0, 1) each time x← range No need to recalculate the thresholds. Receive 0.7712 1 2 3 Decode 1 x =(0.7712-0) / 0.8 0 0.8 0.82 1.0 = 0.964 1 2 3 Decode 3 0 0.8 0.82 1.0 x =(0.964-0.82) / 0.18 = 0.8 1 2 3 Decode 2 x =(0.8-0.8) / 0.02 0 0.8 0.82 1.0 =0 Decode 1. 1 2 3 Stop. 0 0.8 0.82 1.0 J. Liang: SFU ENSC 424 9/20/2005 14
- 15. Decoder Pseudo Code Low = 0; high = 1; x = GetEncodedNumber(); While (x ≠ low) { n = DecodeOneSymbol(x); output symbol n; x = (x - CDF(n-1)) / (CDF(n) - CDF(n-1)); }; J. Liang: SFU ENSC 424 9/20/2005 15
- 16. Outline Introduction Basic Encoding and Decoding Scaling and Incremental Coding Integer Implementation Adaptive Arithmetic Coding Binary Arithmetic Coding Applications JBIG, H.264, JPEG 2000 J. Liang: SFU ENSC 424 9/20/2005 16
- 17. Scaling and Incremental Coding Problems of Previous examples: Need high precision No output is generated until the entire sequence is encoded Key Observation: As the RANGE reduces, many MSB’s of LOW and HIGH become identical: Example: Binary form of 0.7712 and 0.773504: 0.1100010.., 0.1100011.., We can output identical MSB’s and re-scale the rest: Incremental encoding This also allows us to achieve infinite precision with finite-precision integers. Three kinds of scaling: E1, E2, E3 J. Liang: SFU ENSC 424 9/20/2005 17
- 18. E1 and E2 Scaling E1: [LOW HIGH) in [0, 0.5) 0 0.5 1.0 LOW: 0.0xxxxxxx (binary), HIGH: 0.0xxxxxxx. 0 0.5 1.0 Output 0, then shift left by 1 bit [0, 0.5) [0, 1): E1(x) = 2 x E2: [LOW HIGH) in [0.5, 1) 0 0.5 1.0 LOW: 0.1xxxxxxx, HIGH: 0.1xxxxxxx. 0 0.5 1.0 Output 1, subtract 0.5, shift left by 1 bit [0.5, 1) [0, 1): E2(x) = 2(x - 0.5) J. Liang: SFU ENSC 424 9/20/2005 18
- 19. Encoding with E1 and E2 Symbol 1 Prob. 0.8 Input 1 2 0.02 0 0.8 1.0 3 0.18 Input 3 0 0.656 0.8 E2: Output 1 Input 2 2(x – 0.5) 0.312 0.5424 0.54816 0.6 E2: Output 1 0.0848 0.09632 E1: 2x, Output 0 0.1696 0.19264 E1: Output 0 0.3392 0.38528 E1: Output 0 0.6784 0.77056 E2: Output 1 Input 1 Encode any value 0.3568 0.54112 in the tag, e.g., 0.5 Output 1 0.3568 0.504256 All outputs: 1100011 J. Liang: SFU ENSC 424 9/20/2005 19
- 20. To verify LOW = 0.5424 (0.10001010... in binary), HIGH = 0.54816 (0.10001100... in binary). So we can send out 10001 (0.53125) Equivalent to E2 E1 E1 E1 E2 After left shift by 5 bits: LOW = (0.5424 – 0.53125) x 32 = 0.3568 HIGH = (0.54816 – 0.53125) x 32 = 0.54112 Same as the result in the last page. J. Liang: SFU ENSC 424 9/20/2005 20
- 21. Symbol Prob. Note: Complete all possible scaling before 1 0.8 encoding the next symbol 2 0.02 3 0.18 Comparison with Huffman Input Symbol 1 does not cause any output Input Symbol 3 generates 1 bit Input Symbol 2 generates 5 bits Symbols with larger probabilities generates less number of bits. Sometimes no bit is generated at all Advantage over Huffman coding Large probabilities are desired in arithmetic coding Can use context-adaptive method to create larger probability and to improve compression ratio. J. Liang: SFU ENSC 424 9/20/2005 21
- 22. Incremental Decoding Input 1100011 Decode 1: Need ≥ 5 bits (verify) 0 0.8 1.0 Read 6 bits: Tag: 110001, 0.765625 0 0.656 0.8 Decode 3, E2 scaling Tag: 100011 (0.546875) 0.312 0.5424 0.54816 0.6 Decode 2, E2 scaling Tag: 000110 (0.09375) 0.0848 0.09632 E1: Tag: 001100 (0.1875) 0.1696 0.19264 E1: Tag: 011000 (0.375) 0.3392 0.38528 E1: Tag: 110000 (0.75) 0.6784 0.77056 E2: Tag: 100000 (0.5) 0.3568 0.54112 Decode 1 Summary: Complete all possible scaling before further decoding Adjust LOW, HIGH and Tag together. J. Liang: SFU ENSC 424 9/20/2005 22
- 23. Summary Introduction Encoding and Decoding Scaling and Incremental Coding E1, E2 Next: Integer Implementation E3 scaling Adaptive Arithmetic Coding Binary Arithmetic Coding Applications JBIG, H.264, JPEG 2000 J. Liang: SFU ENSC 424 9/20/2005 23

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