By Seibt P.

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**Extra info for Algorithmic Information Theory. Mathematics of Digital Information Processing**

**Example text**

10101, D1 < An < D2 =⇒ [An , Bn [⊂ [D1 , D2 [ ≡ the interval of ab. =⇒ s1 s2 = ab. 9 , B2 = 21 Thus, A2 = 16 32 . 1011101, An < D1 =⇒ [An , Bn [⊂ [A2 , D1 [ ≡ the interval of aba. =⇒ s1 s2 s3 = aba. 9 81 , B3 = 128 . 1001111111, D2 = 1024 An < D1 =⇒ [An , Bn [⊂ [A3 , D1 [ ≡ the interval of abaa. =⇒ s1 s2 s3 s4 = abaa. 9 , B4 = 315 Thus, A4 = 16 512 . 100110111101, An and D2 have the same binary notation – until the masked part of An . Question How shall we continue? e. [An , Bn [⊂ [D2 , B4 [). 26).

2 Universal Codes: The Example LZW 43 The decoder will recover sn+1 , since it knows p(n) – the actual probability distribution after the production of the nth character – due to the histogram established with the information of anterior decoding. 2 Universal Codes: The Example LZW The algorithms for data compaction which we shall treat now are “universal” in the following sense: the idea of a memoryless source – which is a nice but very rare object – will be sacriﬁced. We shall joyfully accept source productions of a high statistical complexity; this means that explicit statistical evaluations of the source production will disappear.

PN −1 ) with p0 ≥ p1 ≥ · · · ≥ pN −1 > 0. (2) A (memoryless) binary source p = (p0 , p1 ) with p0 = 34 , p1 = 14 . Decode 1101010. (3) The binary source of exercise (2). The decoder receives a code stream beginning with 1001. Find the ﬁrst three bits of the source stream. (4) A memoryless source which produces the four letters a, b, c, d according to the probability distribution p given by p(a) = 34 , p(b) = 18 , p(c) = 1 . Decode 11101111000101010000. p(d) = 16 (5) A (binary) arithmetic encoder produces zeros and ones; hence, it is nothing but a binary source.