Information Theory, Inference, and Learning ... - Inference Group

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Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.108 5 — Symbol Codescomplete q-ary tree if A mod (q −1) is equal to 1. Otherwise, we know thatwhatever prefix code we make, it must be an incomplete tree with a numberof missing leaves equal, modulo (q−1), to A mod (q−1) − 1. For example, ifa ternary tree is built for eight symbols, then there will unavoidably be onemissing leaf in the tree.The optimal q-ary code is made by putting these extra leaves in the longestbranch of the tree. This can be achieved by adding the appropriate numberof symbols to the original source symbol set, all of these extra symbols havingprobability zero. The total number of leaves is then equal to r(q−1) + 1, forsome integer r. The symbols are then repeatedly combined by taking the qsymbols with smallest probability and replacing them by a single symbol, asin the binary Huffman coding algorithm.Solution to exercise 5.33 (p.104). We wish to show that a greedy metacode,which picks the code which gives the shortest encoding, is actually suboptimal,because it violates the Kraft inequality.We’ll assume that each symbol x is assigned lengths l k (x) by each of thecandidate codes C k . Let us assume there are K alternative codes and that wecan encode which code is being used with a header of length log K bits. Thenthe metacode assigns lengths l ′ (x) that are given byl ′ (x) = log 2 K + min l k (x). (5.39)kWe compute the Kraft sum:S = ∑ 2 −l′ (x) = 1 ∑2 − min k l k (x) . (5.40)KxLet’s divide the set A X into non-overlapping subsets {A k } K k=1such that subsetA k contains all the symbols x that the metacode sends via code k. ThenS = 1 ∑ ∑2 −lk(x) . (5.41)Kx∈A kkNow if one sub-code k satisfies the Kraft equality ∑ x∈A X2 −lk(x) = 1, then itmust be the case that ∑2 −lk(x) ≤ 1, (5.42)x∈A kwith equality only if all the symbols x are in A k , which would mean that weare only using one of the K codes. SoS ≤ 1 K∑1 = 1, (5.43)Kk=1with equality only if equation (5.42) is an equality for all codes k. But it’simpossible for all the symbols to be in all the non-overlapping subsets {A k } K k=1 ,so we can’t have equality (5.42) holding for all k. So S < 1.Another way of seeing that a mixture code is suboptimal is to considerthe binary tree that it defines. Think of the special case of two codes. Thefirst bit we send identifies which code we are using. Now, in a complete code,any subsequent binary string is a valid string. But once we know that weare using, say, code A, we know that what follows can only be a codewordcorresponding to a symbol x whose encoding is shorter under code A thancode B. So some strings are invalid continuations, and the mixture code isincomplete and suboptimal.For further discussion of this issue and its relationship to probabilisticmodelling read about ‘bits back coding’ in section 28.3 and in Frey (1998).x
Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links.About Chapter 6Before reading Chapter 6, you should have read the previous chapter andworked on most of the exercises in it.We’ll also make use of some Bayesian modelling ideas that arrived in thevicinity of exercise 2.8 (p.30).109
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