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If both the source and code alphabets are binary the rate is just N / K output symbols per input, and the symbol compression function of the coder is clear. 1-2) that depends on the source distribution as well as the code, so we cannot define a rate for them. n A simple block code example will help fix ideas. 1. 05. 2). 1-1 has binary-symbol code words of length 3 that represent sequences of 5 outcomes. If any other sequence appears besides the eight shown, it is encoded as 111, which, of course, is an encoding error.

3-9). Assume a bounded per-letter distortion measure p(z, w) and a discrete memoryless source defined by pz(z). 3-10a) and rate 1 K IOg2 m < R(D) + t. 3-8). From these words make up several sets: i) SD(Z) = {w : (l/K)p(z, w) ~ D + t/2}, the set of reproducer words close enough to z; ii) SR(Z) = {w : (l/K)log{q(wlz)/q(w)} ~ R(D) + t/2}, the set of reproducer words with small enough information about Zj iii) SDR(Z) = SD(Z)nSR(Z), the set ofreproducer words satisfying both (i) and (ii). The proof consists of three parts that show that with certainty as K grows this m-word code has average distortion D + t.

We can view the scheme now as variable-to-block. The following example illustrates these points. 2-2. 2-1. The coder output integer 0, represented by 000, means the next source symbol is a 0; integer 6, or 110, means a run of 6 is terminated by a 0; integer 7 means a run of 7 15 that mayor may not be followed by a o. 50 bits/source symbol. The code tree is Fig. 1-1(c). 1-1, but also it is ellSier to use and hIlS zero failure probability. 2-1), but it is much ellSier to use than the zero-failure Huffman version.

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