Lecture 3 : Algorithms for source coding. September 30, 2016

Size: px

Start display at page:

Download "Lecture 3 : Algorithms for source coding. September 30, 2016"

Benedict Lindsey
6 years ago
Views:

1 Lecture 3 : Algorithms for source coding September 30, 2016

2 Outline 1. Huffman code ; proof of optimality ; 2. Coding with intervals : Shannon-Fano-Elias code and Shannon code ; 3. Arithmetic coding. 1/39

3 1. Coding and decoding with a prefix code Coding : Using a table indexed by the letters. Decoding : Using the associated binary tree Start at the root. For each bit turn left or right. When a leaf is attained, the corresponding letter is output and go back to the root. 2/39

4 Optimal Code Corollary of Kraft and Mac Millan theorems : Corollary 1. If there exists a uniquely decodable code with K words of length n 1, n 2,..., n K then there exists a prefix code with the same lengths. K k=1 1 2 n k 1. Definition A uniquely decodable code of a source X is optimal if there exists no other uniquely decodable code with a smaller average length. Proposition 1. For every source there exists an optimal prefix code. 3/39

5 Huffman Code Let X be a discrete source with alphabet X = {a 1,..., a K 2, a K 1, a K } with prob. distr. p. W.l.o.g. we can assume that p(a 1 )... p(a K 1 ) p(a K ) > 0. Let Y be the source with alphabet Y = {a 1,..., a K 2, b K 1 } and prob. distr. q(a k ) = p(a k ), k = 1..., K 2 q(b K 1 ) = p(a K 1 ) + p(a K ) Algorithm (Huffman) We compute recursively a prefix code of X. When K = 2, the words are given by ϕ K (a 1 ) = 0 and ϕ K (a 2 ) = 1. If K > 2, let ϕ K 1 be a Huffman code for Y, ϕ K (a k ) = ϕ K 1 (a k ) for k = 1..., K 2, ϕ K (a K 1 ) = ϕ K 1 (b K 1 ) 0, ϕ K (a K ) = ϕ K 1 (b K 1 ) 1, 4/39

6 Huffman Code : example 1 0,57 a 0,43 b 0,17 0,15 c 0,32 0,25 e 0,14 d 0,09 f 0,05 0,11 x P (x) log 2 (P (x)) ϕ(x) n x a b c d e f H = n = 2.28 E = 98.6% 5/39

7 The Huffman code is optimal Lemma 1. For every source, there exists an optimal prefix code such that 1. if p(x j ) > p(x k ) then φ(x j ) φ(x k ). 2. The two longest codewords have the same length and correspond to the least likely symbols. 3. These two codewords differ only in their last bit. 6/39

8 Proof of 1 1. If p(x j ) > p(x k ) then φ(x j ) φ(x k ). Let φ be an optimal code and assume that φ(x j ) > φ(x k ). φ is obtained by swapping the codewords of x j and x k. Then φ φ = p(x) φ (x) p(x) φ(x) = p(x j ) φ(x k ) + p(x k ) φ(x j ) p(x j ) φ(x j ) p(x k ) φ(x k ) = (p(x j ) p(x k ))( φ(x k ) φ(x j ) ) < 0 This contradicts the optimality of φ. 7/39

9 Proof of 2 2. The two longest codewords have the same length. If they do not have the same length then a bit can be removed from the longest one- the resulting code is still a prefix code. = a shorter prefix code is obtained in this way. By the previous point we know that they correspond to the least likely symbols. 8/39

10 Proof of The two longest codewords differ only in their last bit. Every φ(x) of maximal length has a brother (otherwise it would be possible to shorten the code). Among those, there are two which correspond to the least likely symbols. By swapping the leaves we can arrange these two symbols so that they become brothers. 9/39

11 The Huffman code is optimal Proof (Induction on k the alphabet size) Let f be optimal for X with p(a 1 ) p(a k 1 ) p(a k ), and f(a k 1 ) = m 0 et f(a k ) = m 1. We derive from f a code g on Y = (a 1,..., a k 2, b k 1 ) : { g(aj ) = f(a j ), j k 2 g(b k 1 ) = m f g = p(a k 1 ) + p(a k ) ϕ k ϕ k 1 = p(a k 1 ) + p(a k ) Hence f ϕ k = g φ k 1. Optimality of ϕ k 1 g φ k 1 0 f ϕ k 0. Optimality of f and f ϕ k 0 f ϕ k = 0 optimality of ϕ k. 10/39

12 Drawbacks of the Huffman code Need to know beforehand the source statistics. If the wrong probability distribution q instead of the true distribution p, instead of compressing with a rate of H(p) we compress with a rate of H(p) + D(p q). This can be dealt with by using an adaptive algorithm that computes the statistics on the fly and updates them during the coding process. Based on a memoryless source model. It is better to group letters by blocks of size l, for l large enough : (1) memoryless model sufficiently more realistic (2) coding efficiency 1. But space complexity = O( X l ). 11/39

13 Huffman code in practice Almost everywhere in the compression algorithms gzip, pkzip, winzip, bzip2 2. compressed images such as jpeg, png 3. compressed audio files such as mp3 Even the rather recent Google compression algorithm Brotli uses it... 12/39

14 2. Interval coding Gives a scheme for which we have in a natural way x i log p(x i ) bits and therefore φ H(X). Idea of arithmetic coding. 13/39

15 2-adic numbers In the interval [0, 1], they are of the form d i 2 i, where d i {0, 1} We write 0.d 1 d 2 d 3... i=1 For instance / Certain numbers have several different 2-adic representation, for instance and In the last case ; we choose the smallest representation. 14/39

16 Shannon-Fano-Elias Code Let X be a memoryless source with alphabet X, and probability distribution p. S(x) = x <x p(x ) p(x 1 ) + + p(x n 1 ) p(x 1 ) + + p(x i ) p(x 1 ) + p(x 2 ) p(x 1 ) The interval width is p(x i ). Each x i is encoded by its interval. 15/39

17 Shannon-Fano-Elias Coding Consider the middle of the interval S(x) def = 1 p(x) + S(x). 2 Let ϕ(x) for all x X be defined as the log p(x) + 1 first bits of the 2-adic representation of S(x). Proposition 2. The code ϕ is prefix and its average length ϕ verifies H(X) ϕ < H(X) /39

18 A simple explanation Lemma 2. For all reals u and v in [0, 1[, and for every integer l > 0,if u v 2 l then the l first bits of the 2-adic representation of u and v are distinct. 17/39

19 The Shannon-Fano-Elias code is prefix Proof W.l.o.g., we may assume y > x S(y) S(x) = p(y) 2 + p(x ) p(x) 2 p(x ) x <y x <x = p(y) 2 p(x) 2 + x x <y = p(y) 2 + p(x) 2 + > max x<x <y p(x ) p(x ) > max (2 log p(x) 1 log, 2 p(y) 1) ( p(y) 2, p(x) ) 2 Therefore they differ in their min( log p(x) + 1, log p(y) + 1) first bits : φ(x) et φ(y) can not be prefix of each other. 18/39

20 Shannon-Fano-Elias Code : example x p(x) log p(x) + 1 S(x) ϕ(x) Huffman a b c d e f /39

21 Shannon-Fano-Elias code : another example x p(x) log p(x) + 1 S(x) ϕ(x) Huffman a b c d x p(x) log p(x) + 1 S(x) ϕ(x) Huffman b a c d If the last bit of ϕ is deleted the code is not prefix anymore. If the last bit of ϕ is deleted the code is still a prefix code. 20/39

22 Shannon Code The Shannon code is defined in the same way as the Shannon-Fano-Elias code, with the exception of : The symbols are ordered by decreasing probability, The codeword of x is formed by the log p(x) first bits of S(x). (In particular, the codeword corresponding to the most likely letter is formed by log p(x) 0 ). Proposition 3. The Shannon code is prefix and its average length ϕ verifies H(X) ϕ < H(X) /39

23 The Shannon code is prefix Proof For all x, y X such that x < y (therefore log p(y) log p(x) ) S(y) S(x) = p(z) p(z) z<y z<x = p(z) x z<y = p(x) + x<z<y log p(x) p(z) p(x) 2 22/39

24 Proof for the lengths Proposition 4. The Shannon-Fano-Elias code ϕ verifies H(X) ϕ < H(X) + 2 Proof ϕ(x) = log p(x) + 1 < log 2 p(x) + 2. The conclusion follows by taking the average. Proposition 5. The Shannon code ϕ verifies H(X) ϕ < H(X) + 1 ϕ(x) = log p(x) < log 2 p(x) + 1. The conclusion follows by taking the average. 23/39

25 The Shannon code is optimal for a dyadic distribution In the dyadic case, the probabilities verify p(x) = 2 log p(x). In this case the lengths of encoding satisfy φ(x) = log p(x) = log 2 p(x) By taking the average p(x) φ(x) = p(x) log2 p(x) = H(X). As for the Huffman code : average coding length = entropy. 24/39

26 Non optimality of Shannon s code For the following source 1. output 1 with probability ; 2. output 0 with probability The Shannon code encodes 1 with a codeword of length log( ) = = 1 2. The Shannon code encodes 0 with a codeword of length log 2 10 = 10 = 10 The Huffman code uses 1 and 0, instead which is of course optimal. 25/39

27 3. Arithmetic coding One of the main drawbacks of the Huffman code is the space complexity when packets of length l are formed for the alphabet letters (space complexity C = O( X l )). r = redundancy def = 1 efficiency. Huffman code on the blocks of size l : r O ( 1 l ) O ( 1 log C Arithmetic coding allows to work on packets of arbitrary sizes with acceptable algorithmic cost. Here ( ) 1 r O, l but the dependency of the space complexity in l is much better. ). 26/39

28 The fundamental idea Instead of encoding the symbols of X, work directly on X l where l is the length of the word which is to encoded.. To encode x 1,..., x l 1. compute the interval [S(x 1,..., x l ), S(x 1,..., x l ) + p(x 1,..., x l )], with S(x 1,..., x l ) def = (y 1,...,y l )<(x 1,...,x l ) p(y 1,..., y l ), 2. encode (x 1,..., x l ) with an element of the interval whose 2-adic representation length is log(p(x 1,..., x l )) + 1. This identifies the interval deduce x 1... x l at the decoding step. 27/39

29 Advantage of arithmetic encoding codelength(x 1,..., x l ) = log p(x 1,..., x l ) + O(1). at most O(1) bits are used to encode l symbols and not O(l) as before! Very interesting in the previous example where 0 s are generated with probability The average coding length becomes h(2 10 )l + O(1) 0.011l instead of l for the Huffman code! 28/39

30 The major problem for implementing arithmetic coding Need to compute S(x 1,..., x l ) with a precision of log p(x 1,..., x l ) + 1 bits. perform computations with arbitrary large precision. 29/39

31 The key idea which allows to work with packets of size l Principle : efficient computation of p(x 1,..., x l ) and S(x 1,..., x l ). Computation of p(x 1,..., x l ) p(x 1,..., x l ) = p(x 1 )... p(x l ) iterative computation of S(x 1,..., x l ) : lexicographic order on X l. S(x 1,..., x l ) def = (y 1,...,y l )<(x 1,...,x l ) p(y 1,..., y l ). Proposition 6. S(x 1,..., x l ) = S(x 1,..., x l 1 ) + p(x 1,..., x l 1 )S(x l ). 30/39

32 Proof Let m be the smallest element in the alphabet S(x 1,..., x l ) = p(y 1, y 2,..., y l ) (y 1,...,y l )<(x 1,...,x l ) = A + B where A B def = def = p(y 1, y 2,..., y l ) (y 1,...,y l )<(x 1,...,x l 1,m) (x 1,...,x l 1,m) (y 1,...,y l )<(x 1,...,x l ) p(y 1, y 2,..., y l ) 31/39

33 Proof (cont d) A = p(y 1, y 2,..., y l ) = = = (y 1,...,y l )<(x 1,...,x l 1,m) y 1,...,y l :(y 1,...,y l 1 )<(x 1,...,x l 1 ) (y 1,...,y l 1 )<(x 1,...,x l 1 ) (y 1,...,y l 1 )<(x 1,...,x l 1 ) = S(x 1,..., x l 1 ). p(y 1, y 2,..., y l 1 )p(y l ) p(y 1, y 2,..., y l 1 ) p(y l ) y l p(y 1, y 2,..., y l 1 ) 32/39

34 Proof (end) B = p(y 1, y 2,..., y l ) = = (x 1,...,x l 1,m) (y 1,...,y l )<(x 1,...,x l ) p(x 1,..., x l 1, y l ) m y l <x l p(x 1,..., x l 1 )p(y l ) m y l <x l = p(x 1,..., x l 1 )S(x l ) 33/39

35 fractals... Example : Prob(a) = 1 2 ; Prob(b) = 1 6, Prob(c) = a b c 1 34/39

36 fractals Example : Prob(a) = 1 2 ; Prob(b) = 1 6, Prob(c) = a b c 1 aa ab ac ba bbbc ca cb cc 35/39

37 coding S(x 1,..., x l ) = S(x 1,..., x l 1 ) }{{} beginning of the previous interval + p(x 1,..., x l 1 ) }{{} S(x l ) length of. previous interval Therefore if the current interval is [a, b[,and if x i is read, then a new a + (b a)s(x i ) b new a new + (b a)p(x i ) The width of the interval becomes (b a)p(x i ) = p(x i )... p(x i 1 )p(x i ) = p(x 1,..., x i ). 36/39

38 Zooming Let [a, b[ be the current interval If b 1/2, [a, b[ [2a, 2b[, output ( 0 ) ; If a 1/2, [a, b[ [2a 1, 2b 1[, output ( 1 ) If 1/4 a < b 3/4, [a, b[ [2a 1/2, 2b 1/2[, no output (underflow expansion), but keep this in mind (with the help of a counter). 37/39

39 Decoding Let r = ϕ(x 1,..., x l ) be the real number whose 2-adic representation encodes (x 1,..., x l ) X l. The letters x 1,..., x l are the only ones for which S(x 1 ) < r < S(x 1 ) + p(x 1 ) S(x 1, x 2 ) < r. < S(x 1, x 2 ) + p(x 1, x 2 ) S(x 1,..., x l ) < r < S(x 1,..., x l ) + p(x 1,..., x l ) 38/39

40 Procedure S(x 1 ) < r < S(x 1 ) + p(x 1 ) Suppose that x 1 was found. We have S(x 1, x 2 ) < r < S(x 1, x 2 ) + p(x 1, x 2 ) S(x 1 ) + p(x 1 )S(x 2 ) r < S(x 1 ) + p(x 1 )S(x 2 ) + p(x 1 )p(x 2 ) p(x 1 )S(x 2 ) r S(x 1 ) < p(x 1 )(S(x 2 ) + p(x 2 )) S(x 2 ) r S(x 1 ) p(x 1 ) < S(x 2 ) + p(x 2 ) We have therefore to find out in which interval lies r 1 = r S(x 1) p(x 1 ). 39/39

Coding of memoryless sources 1/35

Coding of memoryless sources 1/35 Outline 1. Morse coding ; 2. Definitions : encoding, encoding efficiency ; 3. fixed length codes, encoding integers ; 4. prefix condition ; 5. Kraft and Mac Millan theorems