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.252 17 — Communication over Constrained Noiseless Channels0❤ ✒✲M 1 = 21❤101❤0❤ ✒✲M 1 = 21 11❤1❤❅❅❅❘❤ ✒✲ ❤01M 2 = 3020❤ ✒✲M 1 = 21 1 21❤1❤1❤❅❅❅❘❤ ✒❅❅❅❘✲ ❤ ✒✲ ❤01M 2 = 302M 3 = 503Figure 17.4. Counting the numberof paths in the trellis of channelA. The counts next to the nodesare accumulated by passing fromleft to right across the trellises.(a) Channel AM 1 = 21❤1❅❅❅❘M 2 = 31❤1❅❅❅❘M 3 = 52❤1❅❅❅❘M 4 = 83❤1❅❅❅❘M 5 = 135❤1❅❅❅❘M 6 = 218❤1❅❅❅❘M 7 = 3413❤1❅❅❅❘M 8 = 5521❤10❤ ✒✲ ❤ ✒0 ✲ ❤ ✒0 ✲ ❤ ✒0 ✲ ❤ ✒0 ✲ ❤ ✒0 ✲ ❤ ✒0 ✲ ❤ ✒0 ✲ 0❤1 2 3 5 8 13 21 34M 1 = 2M 2 = 3M 3 = 5M 4 = 8M 5 = 13M 6 = 21M 7 = 34M 8 = 55(b) Channel B11❤1✲ ❤112✲ ❤113✲ ❤114✲ ❤116✲ ❤1110✲ ❤1117✲ ❤111❤ ✒ 1❤ ✒ 1❤ ✒ 1❤ ✒ 2❤ ✒ 4❤ ✒ 7❤ ✒ 11❤10❤❆❆❆❆❆❆❯1❅❅❅❘0❤❆❆❆❆❆❆❯1❅❅❅❘1❤0❆❆❆❆❆❆❯1❅❅❅❘2❤0❆❆❆❆❆❆❯1❅❅❅❘3❤0❆❆❆❆❆❆❯1❅❅❅❘4❤0❆❆❆❆❆❆❯1❅❅❅❘6❤0❆❆❆❆❆❆❯1❅❅❅❘10❤000❤✁ ✁✁✁✁✁✕✲ ❤✁ ✁✁✁✁✁✕00 ✲ ❤✁ ✁✁✁✁✁✕00 ✲ ❤✁ ✁✁✁✁✁✕00 ✲ ❤✁ ✁✁✁✁✁✕00 ✲ ❤✁ ✁✁✁✁✁✕00 ✲ ❤✁ ✁✁✁✁✁✕00 ✲ ❤✁ ✁✁✁✁✁✕00 ✲ 00❤1 1 1 2 4 7 11 17M 1 = 1M 2 = 2M 3 = 3M 4 = 5M 5 = 8M 6 = 13M 7 = 21M 8 = 34(c) Channel C11❤1❤1111❤1❤112❤112❤114❤117❤11❤ 00❆1 ❆❆❆❆❆❯❤ ✒ ❆❆❆❆❆❆❯1 ❤ ✒ ❆1 ❆❆❆❆❆❯1 ❤ ✒ ❆2 ❆❆❆❆❆❯1 ❤ ✒ ❆2 ❆❆❆❆❆❯1 ❤ ✒ ❆4 ❆❆❆❆❆❯1 ❤ ✒ ❆7 ❆❆❆❆❆❯1 ❤ ✒ 1011❤✁✕ ❅ ✁✕ ❅❅❘0❤ ✒❅ ✁✕1 ❅❅❘0❤ ✒❅ ✁✕1 ❅❅❘0❤ ✒❅ ✁✕1 ❅❅❘0❤ ✒❅ ✁✕3 ❅❅❘❤ ✒❅ ✁✕4 ❅❅❘0 ❤ ✒❅ ✁✕6 ❅❅❘0 ❤ ✒ 1100❤✁ ✁✁✁✁ ❅❅❤✁ ✁✁✁✁ ❅❅❅❘ ❤✁ ✁✁✁✁ ❅❅❅❘ ❤✁ ✁✁✁✁ ❅❅❅❘ ❤✁ ✁✁✁✁ ❅❅❅❘ ❤✁ ✁✁✁✁ ❅❅❅❘ ❤✁ ✁✁✁✁ ❅❅❅❘ ❤✁ ✁✁✁✁ ❅❘ ❤0000001001001003004006Figure 17.5. Counting the number of paths in the trellises of channels A, B, and C. We assume that atthe start the first bit is preceded by 00, so that for channels A and B, any initial characteris permitted, but for channel C, the first character must be a 1.
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.17.3: Counting the number of possible messages 253n M n M n /M n−1 log 2 M n1n log 2 M n1 2 1.0 1.002 3 1.500 1.6 0.793 5 1.667 2.3 0.774 8 1.600 3.0 0.755 13 1.625 3.7 0.746 21 1.615 4.4 0.737 34 1.619 5.1 0.738 55 1.618 5.8 0.729 89 1.618 6.5 0.7210 144 1.618 7.2 0.7211 233 1.618 7.9 0.7112 377 1.618 8.6 0.71100 9×10 20 1.618 69.7 0.70200 7×10 41 1.618 139.1 0.70300 6×10 62 1.618 208.5 0.70400 5×10 83 1.618 277.9 0.69Figure 17.6. Counting the numberof paths in the trellis of channel A.valid sequences is the number of paths. For the purpose of counting how manypaths there are through the trellis, we can ignore the labels on the edges andsummarize the trellis section by the connection matrix A, in which A ss ′ = 1if there is an edge from state s to s ′ , and A ss ′ = 0 otherwise (figure 17.2d).Figure 17.3 shows the state diagrams, trellis sections and connection matricesfor channels B and C.Let’s count the number of paths for channel A by message-passing in itstrellis. Figure 17.4 shows the first few steps of this counting process, andfigure 17.5a shows the number of paths ending in each state after n steps forn = 1, . . . , 8. The total number of paths of length n, M n , is shown along thetop. We recognize M n as the Fibonacci series.⊲ Exercise 17.6. [1 ] Show that the ratio of successive terms in the Fibonacci seriestends to the golden ratio,γ ≡ 1 + √ 52= 1.618. (17.11)Thus, to within a constant factor, M N scales as M N ∼ γ N as N → ∞, so thecapacity of channel A isC = lim 1 N log 2[constant · γN ] = log 2 γ = log 2 1.618 = 0.694. (17.12)How can we describe what we just did? The count of the number of pathsis a vector c (n) ; we can obtain c (n+1) from c (n) using:Soc (n+1) = Ac (n) . (17.13)c (N) = A N c (0) , (17.14)where c (0) is the state count before any symbols are transmitted. In figure 17.5we assumed c (0) = [0, 1] T , i.e., that either of the two symbols is permitted atthe outset. The total number of paths is M n = ∑ s c(n) s = c (n) · n. In the limit,c (N) becomes dominated by the principal right-eigenvector of A.c (N) → constant · λ N 1 e(0) R . (17.15)
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