By Jacobus H. van Lint (auth.)

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Example text

Sk-l k Hence we can solve the equations. Now let hex) be a divisor of xn - 1, (n,q) 1. The linear recurring sequence 58 then is periodic vith period n. •. • '~-1) runs through all points of R(k) this is an (n,k) cyclic code and by the definition we see that lh(x -1) 1n tile check polynomial. tion at c The presenta- t k r iel c i $1 gives us another vay of 1nterpreting the code vords. We describe yet another representation of the code words in a cyclic code. =I Once again let (n,q) ~~d let V be a cyclic code of length n over GF(q).

E. 03 ~ is practically completely decoding failure. lculate A(z) for the binary Hamming codes. Ii' we take 1 - 1 25 columns of the parity check matrix H then there are 3 possibiUties: (i) the sum of these columns can be 0, (ii) the sum of these columns can be one of the chosen co1umns, (lii) the sum of these columns can be equal to one of the remaining columns. n The total number of possible chOices is (i-I)' Possibility (i) can occur in Ai _1 ways, possibility (ii) in (n-i+2)A i _2 ways and possibility (iii) in tA i ways.

A basis of ~(n). •• , ~-IYm' Yl~Y3' "', 29 Example: Let m a 4, n ~ 16. , 0000111 v, ..... 1~ ~ 0 ceo 0 001 0 0 001 0000 0 0 1 1 I 1 0 0 C 1 Co Co 0 0 0 0 1 0 0 0 COO C 0 1 0 a a a a 0 0 0 0 0 o COO o 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 100 0 0 001 0 0 0 000 0 1 100 o0 a 0 0 a 0 a a 0 0 0 1 1 o0 0 0 000 1 0 0 0 0 0 0 0 o0 a o COO 0 0 000 1 000 1 0 COO 000 0 0 0 001 0 o0 0 0 0 0 0 a a 000 0 0 000 0 000 000 0 0 1 0 0 0 We car. lear c::>des which were discovered by D. E. Muller and f::>r which I. S. Reed found a very nice decoding method.

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