By Hankerson, Darrel R.

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B) Prove recursively that deg (Ri(x)) â ¤iâ â (1 + Di)/2 , after first showing that the choice of qi(0) forces the coefficient of xi in Pi(x)q-1(x) to be zero. (c) Prove that since all the Dj are distinct, that at least one of the Dj, jâ ¤i, must be at least i; and hence that either Diâ ¥i or Dziâ ¥i. < previous page page_152 next page > < previous page page_153 next page > Page 153 (d) When D2t â ¥ 2t, show that deg(P2t(x))â ¤t, and that since deg(R2t(x)) â ¤ t note that at least t consecutive coefficients of P2t(x)q-1(x) are zero (meaning that at least t consecutive Newton's identities are satisfied).

2. 1 for finding the error locator polynomial when erasures are included. 6 (Berlekamp-Massey decoding with erasures) Suppose C is an RS(2r,Î´ code with generator g(x)=(Î²m + 1 + x)Â·Â·Â·(Î²m + Î´â 1 + x) and let w be a received word. Let Ï B(x)= B0 + B1x + ... + xâ be the erasure locator polynomial of w. 1 for finding the error locator polynomial of w 59 60 as follows: 1. Calculate sj=w(Î²j) for m + 1 â ¤ j â ¤ m + Î´ â 2. Calculate â . = B0sj + B1sj + 1 + Â·Â·Â· +Bâ 1. â 1 + sj + â for m + 1 â ¤ j â ¤ m + Î´ â 1â 3.

The algorithm is essentially the algorithm of Berlekamp and Massey for calculating the error locator polynomial Ï (x) given the syndromes sj = w(Î²j) for m + 1â ¤ j â ¤m + 2t. Let Ï R(x) = 1 + Ï tâ 1x + Ï tâ 2x2+Â·Â·Â·+ Ï 0xt; that is, Ï R(x) is the ''reverse" of the error locator polynomial Ï (x). Let s(x) = 1 + sm+1x + sm+ 2x2 +Â·Â·Â·+ sm+ 2tx2t be the syndrome polynomial. Using the division algorithm, we can write Ï R(x)s(x) = q(x)x2t + 1 + r(x) with deg(r(x))) â ¤ 2t. 4), in fact deg(r(x)) â ¤ t.