Abstract Algebra Theory and Applications - Computer Science ...

20.2 POLYNOMIAL CODES 355We see that g(t) = (1 + t + t 3 ) generates an ideal C in R 7 . This code is a(7, 4)-block code. As in Example 5, it is easy to calculate a generator matrixby examining what g(t) does to the polynomials 1, t, t 2 , and t 3 . A generatormatrix for C is⎛⎞1 0 0 01 1 0 00 1 1 0G =1 0 1 1.⎜ 0 1 0 1⎟⎝ 0 0 1 0 ⎠0 0 0 1In general, we can determine a generator matrix for an (n, k)-code C bythe manner in which the elements t k are encoded. Let x n − 1 = g(x)h(x) inZ 2 [x]. If g(x) = g 0 + g 1 x + · · · + g n−k x n−k and h(x) = h 0 + h 1 x + · · · + h k x k ,then the n × k matrix⎛⎞g 0 0 · · · 0g 1 g 0 · · · 0. . . .. .G =g n−k g n−k−1 · · · g 00 g n−k · · · g 1⎜⎝.. . ..⎟. ⎠0 0 · · · g n−kis a generator matrix for the code C with generator polynomial g(t). Theparity-check matrix for C is the (n − k) × n matrixH =⎛⎜⎝0 · · · 0 0 h k · · · h 00 · · · 0 h k · · · h 0 0· · · · · · · · · · · · · · · · · · · · ·h k · · · h 0 0 0 · · · 0⎞⎟⎠ .We will leave the details of the proof of the following proposition as anexercise.Proposition 20.11 Let C = 〈g(t)〉 be a cyclic code in R n and suppose thatx n − 1 = g(x)h(x). Then G and H are generator and parity-check matricesfor C, respectively. Furthermore, HG = 0.