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By Henk C. A. van Tilborg (auth.)

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D. D. codes for S. 12 both codes, C and C* , satisfy properties PI-PS. -l and f... in C differ only in their last coordinate, as do f~-l and f~ in C* . Now apply one step of the reduction process to C and C*. One obtains a Huffman code D and a prefix code D* , with expected lengths M resp. M* . By the induction hypothesis M ~ M* . The proof now follows from 53 Huffman Codes L - L* " • pil = (~li i=1 " l~ • p;) (~ i=1 = " li • Pi - P,,-1 - p,,) - (~l~ " • Pi - P,,-1 - p,,) (~ i=1 i=1 11-2 11-2 i=1 i=1 = ( ~ li· Pi + (III - 1)(PIl_l + PII) - ( ~ l~ • Pi + (l~ - 1) (P"-1 + p,,)} = =M-M*sO.

O , O~ i ~ n - 1, are clearly independent So the dimension of il (f) is at least n. Let f be a polynomial of degree n, say f (x) = defined by L" Cixi. ~o S (x) .. • C,,-IX +C". 12) i~ Instead of writing a (f), we shall also use the notation S (x) E a (f). (Si }i~O E is uniquely detennined by explicit. • , S,,_I a (f). (Si) i~O E and We know that S (x) We shall now make this dependency more f~). 9). J&. 14) . *(x) .. , = (~ SA:xA:) (~C"-l Xl) = .. , =~ ( ~ C.. -l Sj-l " = 't(x) + ~ (~Ci j~" )x j SU-II)+i = )x j = 't(x) .

So in this case L k+1 = k + 1 = k + 1 - Lk . This proves the fust induction step. We now proceed by induction. So we assume that for Lk :s; j :s; k - 1, Lj;-l L j=O c/k ) Sj-Lj;+j = Sj. 19) = k, then Lk+l = Lb c(k+1) (x) = C(k) (x) and there remains nothing ta prove. 28 L k+1 ~ max {Lk , k + 1 - Lk }. We shall prove that equality holds. 21) Because we already proved the fust induction step. this number m is well detined. 22) j =m. = m + 1- Lm. Detine L = max {Lbk + 1 - LI;}. Shift Register Sequences C 33 L-L (x) = X 1 C (L) A L-(k+I-L) (X) + X will be a suitable choice for C (k+l) (x).

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