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Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981 You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links. 114 6 — Stream Codes symbol given the string thus far. Let these probabilities be: Context (sequence thus far) Probability of next symbol P (a) = 0.425 P (b) = 0.425 P (✷) = 0.15 b P (a | b) = 0.28 P (b | b) = 0.57 P (✷ | b) = 0.15 bb P (a | bb) = 0.21 P (b | bb) = 0.64 P (✷ | bb) = 0.15 bbb P (a | bbb) = 0.17 P (b | bbb) = 0.68 P (✷ | bbb) = 0.15 bbba P (a | bbba) = 0.28 P (b | bbba) = 0.57 P (✷ | bbba) = 0.15 Figure 6.4 shows the corresponding intervals. The interval b is the middle 0.425 of [0, 1). The interval bb is the middle 0.567 of b, and so forth. a b ✷ ba bb b✷ bba bbb bbba bbbb bbb✷ bb✷ 00000 0000 00001 000 00010 0001 00011 00 00100 0010 00101 001 00110 0011 00111 0 01000 0100 01001 010 01010 0101 01011 01 01100 0110 01101 011 01110 0111 01111 10000 ✂ 1000 10001 ✂ 100 10010 ✂ 1001 ✂ 10011 10 10100 1010 ❇ 10101 ❇ 101 10110 ❇ 1011 10111 ❇ 1 11000 1100 11001 110 11010 1101 11011 11 11100 1110 11101 111 11110 1111 11111 bbba bbbaa bbbab bbba✷ 10010111 10011000 10011001 10011010 10011011 10011100 10011101 10011110 ❈❈❖ 10011111 ❈10100000 ❈ 100111101 When the first symbol ‘b’ is observed, the encoder knows that the encoded string will start ‘01’, ‘10’, or ‘11’, but does not know which. The encoder writes nothing for the time being, and examines the next symbol, which is ‘b’. The interval ‘bb’ lies wholly within interval ‘1’, so the encoder can write the first bit: ‘1’. The third symbol ‘b’ narrows down the interval a little, but not quite enough for it to lie wholly within interval ‘10’. Only when the next ‘a’ is read from the source can we transmit some more bits. Interval ‘bbba’ lies wholly within the interval ‘1001’, so the encoder adds ‘001’ to the ‘1’ it has written. Finally when the ‘✷’ arrives, we need a procedure for terminating the encoding. Magnifying the interval ‘bbba✷’ (figure 6.4, right) we note that the marked interval ‘100111101’ is wholly contained by bbba✷, so the encoding can be completed by appending ‘11101’. Figure 6.4. Illustration of the arithmetic coding process as the sequence bbba✷ is transmitted. 10011
Copyright Cambridge University Press 2003. On-screen viewing permitted. Printing not permitted. http://www.cambridge.org/0521642981 You can buy this book for 30 pounds or $50. See http://www.inference.phy.cam.ac.uk/mackay/itila/ for links. 6.2: Arithmetic codes 115 Exercise 6.1. [2, p.127] Show that the overhead required to terminate a message is never more than 2 bits, relative to the ideal message length given the probabilistic model H, h(x | H) = log[1/P (x | H)]. This is an important result. Arithmetic coding is very nearly optimal. The message length is always within two bits of the Shannon information content of the entire source string, so the expected message length is within two bits of the entropy of the entire message. Decoding The decoder receives the string ‘100111101’ and passes along it one symbol at a time. First, the probabilities P (a), P (b), P (✷) are computed using the identical program that the encoder used and the intervals ‘a’, ‘b’ and ‘✷’ are deduced. Once the first two bits ‘10’ have been examined, it is certain that the original string must have been started with a ‘b’, since the interval ‘10’ lies wholly within interval ‘b’. The decoder can then use the model to compute P (a | b), P (b | b), P (✷ | b) and deduce the boundaries of the intervals ‘ba’, ‘bb’ and ‘b✷’. Continuing, we decode the second b once we reach ‘1001’, the third b once we reach ‘100111’, and so forth, with the unambiguous identification of ‘bbba✷’ once the whole binary string has been read. With the convention that ‘✷’ denotes the end of the message, the decoder knows to stop decoding. Transmission of multiple files How might one use arithmetic coding to communicate several distinct files over the binary channel? Once the ✷ character has been transmitted, we imagine that the decoder is reset into its initial state. There is no transfer of the learnt statistics of the first file to the second file. If, however, we did believe that there is a relationship among the files that we are going to compress, we could define our alphabet differently, introducing a second end-of-file character that marks the end of the file but instructs the encoder and decoder to continue using the same probabilistic model. The big picture Notice that to communicate a string of N letters both the encoder and the decoder needed to compute only N|A| conditional probabilities – the probabilities of each possible letter in each context actually encountered – just as in the guessing game. This cost can be contrasted with the alternative of using a Huffman code with a large block size (in order to reduce the possible onebit-per-symbol overhead discussed in section 5.6), where all block sequences that could occur must be considered and their probabilities evaluated. Notice how flexible arithmetic coding is: it can be used with any source alphabet and any encoded alphabet. The size of the source alphabet and the encoded alphabet can change with time. Arithmetic coding can be used with any probability distribution, which can change utterly from context to context. Furthermore, if we would like the symbols of the encoding alphabet (say, 0 and 1) to be used with unequal frequency, that can easily be arranged by subdividing the right-hand interval in proportion to the required frequencies. How the probabilistic model might make its predictions The technique of arithmetic coding does not force one to produce the predictive probability in any particular way, but the predictive distributions might
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Information Theory, Inference, and Learning ... - MAELabs UCSD

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