By Seibt P.

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

**Example text**

Let us change our outlook: We shall take the words of length n (n ≥ 1) as our new production units: x = aj1 aj2 · · · ajn will be the notation for these “big letters”. ,jn ≤N −1 (there are N n words of length n on an alphabet of N elements). Recall H(p(n) ) = n · H(p) (this was a previous exercise). 32 1 Data Compaction Proposition A memoryless source, producing the N letters a0 , a1 , . . , aN −1 , according to the probability distribution p = (p0 , p1 , . . , pN −1 ). Let us pass to an encoding of blocks in n letters.

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). Suppose, furthermore, that only the letter a was produced in the sequel ( 2493 4096 remains then the left end point A5 = A6 = A7 , etc. 1. 94). We obtain this way three source words s1 s2 s3 s4 s5 s6 s7 , s1 s2 s3 s4 s5 s6 s7 s8 , s1 s2 s3 s4 s5 s6 s7 s8 s9 , which produce the same code word 100110111. Back to the question: why An = 2,493 4,096 ?

PN −1 ). We shall always suppose p0 ≥ p1 ≥ · · · ≥ pN −1 . The arithmetic encoder will associate with a stream of source symbols aj1 aj2 · · · ajn · · · (which could be theoretically unlimited), a bitstream α1 α2 α3 · · · αl · · · (which would then also be unlimited). But let us stop after n encoding steps: The code word α1 α2 α3 · · · αl of l bits associated with the n ﬁrst source symbols aj1 aj2 · · · ajn will be the code word c(aj1 aj2 · · · ajn ) of a Shannon block encoding formally adapted to recursiveness according to the device: “every step yields a tree-antecedent to the next step”.