Homework

1. Homework 2. Assume that the polynomial q(x) = x4 + x3 + 1 is used to construct a BCH code that corrects double errors with plaintext polynomials of the form a(x) GF2[x]. A table of powers of in GF16 = GF2[x]/(q(x)) is given on the next slide. As always, x has been replaced with in the polynomial elements of the field. Then the generator polyomial is g(x) = x8 + x4 + x2 + x + 1. Suppose we receive a message that we know has exactly two errors and the received polynomial is r(x) = x12 + x10 + x4 + x3 + 1. After calculation with Maple, we find that r() = 4, r(2) = 8, r(3) = 4, and r(a4) = a. Correct the two errors to find the codeword c(x) that was sent:

2. Correcting Two Errors • We have derived the following information j + k= 42j + 2k= 8 3j + 3k= 44j + 4k =  • Now for the algebraic “insight” when computing mod 2: (j + k) j+k + (2j + 2k) (j + k) = 3j + 3k Since we already know the value of j +  , we want to solve for j+k . • Substituting in the first equation, we get 4 j+k + 4 8= 4 • Thus, j+k + 8 = 1, so j+k= 1 + 8= 1 + 3+ 2+  = 6

3. Table of Powers for GF16

4. Correcting Two Errors • We have j + k = 4and j+k = 6 • We know that j+k = 6 or j+k = 21. So, first we check the possible values of j,k assuming j+k = 6: • Therefore, we have j = 1 and k = 5 and thus c(x) = r(x) + x5 + x = x12 + x10 + x4 + x3 + 1 + x5 + x = x12 + x10 + x5 + x4 + x3+x +1 • Dividing c(x) by the generator x8 + x4 + x2 + x + 1 yields the plaintext polynomial a(x) = x4+ x2+ 1