# Arithmetic Coding. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20062 How we can do better than Huffman? - I As we have seen, the.

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Arithmetic Coding

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20062 How we can do better than Huffman? - I As we have seen, the main drawback of Huffman scheme is that it has problems when there is a symbol with very high probability Remember static Huffman redundancy bound where is the probability of the most likely simbol

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20063 How we can do better than Huffman? - II The only way to overcome this limitation is to use, as symbols, “blocks” of several characters. In this way the per-symbol inefficiency is spread over the whole block However, the use of blocks is difficult to implement as there must be a block for every possible combination of symbols, so block number increases exponentially with their length

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20064 How we can do better than Huffman? - III Huffman Coding is optimal in its framework static model one symbol, one word adaptive Huffman blocking arithmetic coding

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20065 The key idea Arithmetic coding completely bypasses the idea of replacing an input symbol with a specific code. Instead, it takes a stream of input symbols and replaces it with a single floating point number in The longer and more complex the message, the more bits are needed to represents the output number

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20066 The key idea - II The output of an arithmetic coding is, as usual, a stream of bits However we can think that there is a prefix 0, and the stream represents a fractional binary number between 0 and 1 In order to explain the algorithm, numbers will be shown as decimal, but obviously they are always binary

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20067 An example - I String bccb from the alphabet {a,b,c} Zero-frequency problem solved initializing at 1 all character counters When the first b is to be coded all symbols have a 33% probability (why?) The arithmetic coder maintains two numbers, low and high, which represent a subinterval [low,high) of the range [0,1) Initially low=0 and high=1

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-20068 An example - II The range between low and high is divided between the symbols of the alphabet, according to their probabilities low high 0 1 0.3333 0.6667 a b c (P[c]=1/3) (P[b]=1/3) (P[a]=1/3)

9 An example - III low high 0 1 0.3333 0.6667 a b c b low = 0.3333 high = 0.6667  P[a]=1/4  P[b]=2/4  P[c]=1/4 new probabilities

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200610 An example - IV new probabilities P[a]=1/5 P[b]=2/5 P[c]=2/5 low high 0.3333 0.6667 0.4167 0.5834 a b c c low = 0.5834 high = 0.6667 (P[c]=1/4) (P[b]=2/4) (P[a]=1/4)

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200611 An example - V new probabilities P[a]=1/6 P[b]=2/6 P[c]=3/6 low high 0.5834 0.6667 0.6001 0.6334 a b c c low = 0.6334 high = 0.6667 (P[c]=2/5) (P[b]=2/5) (P[a]=1/5)

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200612 An example - VI Final interval [0.6390,0.6501) we can send 0.64 low high 0.6334 0.6667 0.6390 0.6501 a b c low = 0.6390 high = 0.6501 b (P[c]=3/6) (P[b]=2/6) (P[a]=1/6)

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200613 An example - summary Starting from the range between 0 and 1 we restrict ourself each time to the subinterval that codify the given symbol At the end the whole sequence can be codified by any of the numbers in the final range (but mind the brackets...)

14 An example - summary 0 1 0.3333 0.6667 a b c 0.3333 1/3 0.4167 0.5834 1/4 2/4 1/4 a b c 0. 5834 0. 6667 2/5 1/5 0.6001 0.6334 a b c 0. 6667 0.6334 a b c 0.6390 0.6501 3/6 2/6 1/6 [0.6390, 0.6501)0.64

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200615 Another example - I Consider encoding the name BILL GATES Again, we need the frequency of all the characters in the text. chrfreq. space0.1 A0.1 B0.1 E0.1 G0.1 I0.1 L0.2 S0.1 T0.1

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200616 Another example - II characterprobabilityrange space0.1[0.00, 0.10) A0.1[0.10, 0.20) B0.1[0.20, 0.30) E0.1[0.30, 0.40) G0.1[0.40, 0.50) I0.1[0.50, 0.60) L0.2[0.60, 0.80) S0.1[0.80, 0.90) T0.1[0.90, 1.00)

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200617 Another example - III chrlowhigh 0.01.0 B0.20.3 I0.250.26 L0.2560.258 L0.25720.2576 Space0.257200.25724 G0.2572160.257220 A0.25721640.2572168 T0.257216760.2572168 E0.2572167720.257216776 S0.25721677520.2572167756 The final low value, 0.2572167752 will uniquely encode the name BILL GATES

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200618 Decoding - I Suppose we have to decode 0.64 The decoder needs symbol probabilities, as it simulates what the encoder must have been doing It starts with low=0 and high=1 and divides the interval exactly in the same manner as the encoder (a in [0, 1/3), b in [1/3, 2/3), c in [2/3, 1)

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200619 Decoding - II The trasmitted number falls in the interval corresponding to b, so b must have been the first symbol encoded Then the decoder evaluates the new values for low (0.3333) and for high (0.6667), updates symbol probabilities and divides the range from low to high according to these new probabilities Decoding proceeds until the full string has been reconstructed

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200620 Decoding - III 0.64 in [0.3333, 0.6667) b 0.64 in [0.5834, 0.6667) c... and so on...

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200621 Why does it works? More bits are necessary to express a number in a smaller interval High-probability events do not decrease very much interval range, while low probability events result a much smaller next interval The number of digits needed is proportional to the negative logarithm of the size of the interval

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200622 Why does it works? The size of the final interval is the product of the probabilities of the symbols coded, so the logarithm of this product is the sum of the logarithm of each term So a symbol s with probability Pr[s] contributes bits to the output, that is equal to symbol probability content (uncertainty)!!

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200623 Why does it works? For this reason arithmetic coding is nearly optimum as number of output bits, and it is capable to code very high probability events in just a fraction of bit In practice, the algorithm is not exactly optimal because of the use of limited precision arithmetic, and because trasmission requires to send a whole number of bits

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200624 A trick - I As the algorithm was described until now, the whole output is available only when encoding are finished In practice, it is possible to output bits during the encoding, which avoids the need for higher and higher arithmetic precision in the encoding The trick is to observe that when low and high are close they could share a common prefix

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200625 A trick - II This prefix will remain forever in the two values, so we can transmit it and remove from low and high For example, during the encoding of “bccb”, it has happened that after the encoding of the third character the range is low=0.6334, high=0.6667 We can remove the common prefix, sending 6 to the output and transforming low and high into 0.334 and 0,667

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200626 The encoding step To code symbol s, where symbols are numbered from 1 to n and symbol i has probability Pr[i] low_bound = high_bound = range = high - low low = low + range * low_bound high = low + range * high_bound

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200627 The decoding step The symbols are numbered from 1 to n and value is the arithmetic code to be processed Find s such that Return symbol s Perform the same range-narrowing step of the encoding step

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200628 Implementing arithmetic coding As mentioned early, arithmetic coding uses binary fractional number with unlimited arithmetic precision Working with finite precision (16 or 32 bits) causes compression be a little worser than entropy bound It is possible also to build coders based on integer arithmetic, with another little degradation of compression

29 Arithmetic coding vs. Huffman coding In tipical English text, the space character is the most common, with a probability of about 18%, so Huffman redundancy is quite small. Moreover this is an upper bound On the contrary, in black and white images, arithmetic coding is much better than Huffman coding, unless a blocking technique is used A A. coding requires less memory, as symbol representation is calculated on the fly A A. coding is more suitable for high performance models, where there are confident predictions

Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-200630 Arithmetic coding vs. Huffman coding HH. decoding is generally faster than a. decoding HIn a. coding it is not easy to start decoding in the middle of the stream, while in H. coding we can use “starting points”  In large collections of text and images, Huffman coding is likely to be used for the text, and arithmeting coding for the images

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