By San Ling

Interested by effectively transmitting facts via a loud channel, coding thought might be utilized to digital engineering and communications. in line with the authors' broad educating adventure, this article offers a very sleek and available path at the topic. It comprises sections on linear programming and interpreting tools crucial for modern arithmetic. quite a few examples and routines make the amount perfect for college students and teachers.

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**Extra resources for Coding Theory: A First Course**

**Sample text**

59 of ref. [11]), then gcd( f (x), g(x)) = p1 (x)min{e1 ,d1 } · · · pn (x)min{en ,dn } and lcm( f (x), g(x)) = p1 (x)max{e1 ,d1 } · · · pn (x)max{en ,dn } . (ii) Let f (x), g(x) ∈ F[x] be two nonzero polynomials. Then there exist two polynomials u(x), v(x) with deg(u(x)) < deg(g(x)) and deg(v(x)) < deg( f (x)) such that gcd( f (x), g(x)) = u(x) f (x) + v(x)g(x). (iii) From (ii), it is easily shown that gcd( f (x)h(x), g(x)) = gcd( f (x), g(x)) if gcd(h(x), g(x)) = 1. There are many analogies between the integral ring Z and a polynomial ring F[x].

0 or 1 would be a root of g(x), but g(0) = g(1) = 1 ∈ Z2 . (iii) Using the same arguments as in (ii), we can show that both 1 + x + x 3 and 1 + x 2 + x 3 are irreducible over Z2 as they have no linear factors. 3 Let f (x) ∈ F[x] be a polynomial of degree n ≥ 1. Then, for any polynomial g(x) ∈ F[x], there exists a unique pair (s(x), r (x)) of polynomials with deg(r (x)) < deg( f (x)) or r (x) = 0 such that g(x) = s(x) f (x) + r (x). The polynomial r (x) is called the (principal) remainder of g(x) divided by f (x), denoted by (g(x) (mod f (x))).

Since 21 is a divisor of 26 − 1, we consider the field F64 . Let α be a root of 1 + x + x 6 . It can be verified that α is a primitive element of F64 (check that α 3 = 1, α 7 = 1, α 9 = 1 and α 21 = 1). We list the cyclotomic cosets of 2 modulo 63 containing multiples of 3: C0 = {0}, C9 = {9, 18, 36}, C21 = {21, 42}, C3 = {3, 6, 12, 24, 48, 33}, C15 = {15, 30, 60, 57, 51, 39}, C27 = {27, 54, 45}. Hence, we obtain M (0) (x) = 1 + x, M (3) (x) = (x − α j ) = 1 + x + x 2 + x 4 + x 6 , j∈C3 M (x) = (x − α j ) = 1 + x 2 + x 3 , (9) j∈C9 M (15) (x) = (x − α j ) = 1 + x 2 + x 4 + x 5 + x 6 , j∈C15 M (21) (x) = (x − α j ) = 1 + x + x 2 , j∈C21 M (27) (x) = (x − α j ) = 1 + x + x 3 .