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# Download Solutions to selected problems from Chapter 2

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Error Control Coding — week 1: Binary field algebra and linear block codes 1 Solutions to selected problems from Chapter 2 2.9 Denote the equations with the numbers (1) – (4) starting from the top one. Then, summing the equations (3) and (4) (modulo 2) yields (3) + (4) ⇒ X +Y +Y +Z +Z +W +W =X =1 Similarly, summation of the equations (2) and (3) results in (2) + (3) ⇒ X +X +Y +Z +Z +W +W =Y =1 From equation (1) we obtain (recall that in GF (2) subtraction is equivalent to addition) (1) ⇒ W = 1 + X + Y = 1 Finally, equation (4) gives (4) ⇒ Z =1+Y +W =0 Thus, the solution is [X Y W Z] = [1 1 1 0]. 2.10 The polynomial a(x) = x5 + x3 + 1 is irreducible over GF (2) if it is not divisible by any polynomial p(x) over GF (2) of degree 1 ≤ deg(p(x)) ≤ 2 = bdeg(a(x))/2c. Thus, we have to check if there exists a polynomial of the form p(x) = p2 x2 + p1 x + p0 over GF (2), which divides a(x). (a) As a(0) = 1 we can conclude that x − 0 = x is not a factor of a(x). (b) As a(1) = 1 we can conclude that x − 1 = x + 1 is not a factor of a(x). If there exists a polynomial p(x) = p2 x2 + p1 x + p0 which divides a(x), we can conclude from (a) that p0 = 1 and from (b) that p2 = 1. Thus p(x) = x2 + p1 x + 1 and we have to check the two polynomials x2 + 1 and x2 + x + 1. Clearly, as x2 + 1 = (x + 1)2 and with x + 1 not being a factor of a(x), it follows directly that x2 + 1 cannot be a factor of a(x) either. It remains to test x2 + x + 1. After performing a division we see that x2 + x + 1 doesn’t divide a(x) and thus we can finally conclude that a(x) is irreducible. 2.11 a) The proof of if and only if -type of statements involves two proofs. The first part is the proof of the one-sided statement: f ∗ (X) is irreducible over GF (2) if f (X) is irreducible over GF (2). The second part is to prove the other side of the statement: f (X) is irreducible over GF (2) if f ∗ (X) is irreducible over GF (2). Error Control Coding — week 1: Binary field algebra and linear block codes 2 We will prove the first part this statement by assuming the opposite, and then reducing this assumption to a contradiction. To this end, assume that f (X) is irreducible, but its reciprocal f ∗ (X) is not. Then f ∗ (X) = a(X) · b(X), for some polynomials a(X) and b(X) whose degrees are k and m, respectively, k, m > 0. Clearly, k + m = n. Since f (X) is the reciprocal of f ∗ (X), we have that 1 1 1 1 n ∗ k m n ∗ =X f =X a X b . f (X) = X f X X X X This means that f (X) is not irreducible, which is the contradiction to the starting hypothesis. Hence, f ∗ (X) must be irreducible if f (X) is irreducible. Similarly, we prove that if f ∗ (X) is irreducible, then f (X) is also irreducible. This completes the proof. b) Suppose that f (X) is primitive but f ∗ (X) is not. Then there exists a positive integer k < 2n − 1 such that f ∗ (X) divides (X k + 1), i.e., X k + 1 = f ∗ (X) q(X) for some polynomial q(X) of nonzero degree. Taking the reciprocals of both sides of the above equality yields (X k + 1)∗ = f (X)q ∗ (X) Since (X k + 1)∗ = X k (X −k + 1) = X k + 1, we further obtain: X k + 1 = f (X) q ∗ (X) = f (X) X deg(q(X)) q 1 X which implies that f (X) divides (X k + 1) for k < 2n − 1. This contradicts the hypothesis that f (X) is primitive. Hence, we conclude that if f (X) is primitive, f ∗ (X) must also be primitive. Similarly, we can prove that if f ∗ (X) is primitive, then f (X) must also be primitive. Thus, we have proved that f ∗ (X) is primitive if and only if f (X) is primitive. 2.12 Irreducible polynomials of degree 5 over GF (2) are listed below. These polynomials are all primitive. x5 + x2 + 1 x5 + x3 + 1 x5 + x4 + x 3 + x2 + 1 x5 + x4 + x 3 + x + 1 x5 + x4 + x 2 + x + 1 x5 + x3 + x 2 + x + 1 Error Control Coding — week 1: Binary field algebra and linear block codes 3 2.22 According to Theorem 2.22, S is a subspace of Vn over a field F if (i) ∀ u, v ∈ S u + v ∈ S (ii) ∀ a ∈ F a·u∈S The condition (i) is already given, we only need to show that, for F = GF (2), the condition (ii) is already implied by the condition (i). Indeed, for any u ∈ S it follows from (i) that u + u = 0 ∈ S. Then, for any a ∈ GF (2), i.e., a ∈ {0, 1} we have ( 0, for a = 0 a·u= ∈ S. u, for a = 1 Hence, we conclude that S is a subspace. 2.23 Let V denote the set of all polynomials v(x) over GF (2) of degree deg(v(x)) ≤ n−1, and let + and · denote addition and multiplication. The polynomials are given by v(x) = vn−1 xn−1 +vn−2 xn−2 +...+v1 x+v0 , where binary coefficients vi , 0 ≤ i ≤ n − 1 form an n-tuple v = [vn−1 ... v1 v0 ] of length n. Multiplication of a polynomial v(x) with a scalar from GF (2), as well as addition of two polynomials are defined as usually. It is easy to verify that the following statements hold (i) ∀ v(x), u(x) ∈ V v(x) + u(x) = u(x) + v(x) (ii) ∀ a ∈ GF (2), ∀ v(x) ∈ V a · v(x) = avn−1 xn−1 + ... + av1 x + av0 ∈ V (iii) Distributive laws: ∀ a, b ∈ GF (2), ∀ v(x), u(x) ∈ V a(u(x) + v(x)) = au(x) + av(x) (a + b)v(x) = av(x) + bv(x) (iv) Associative law: ∀ a, b ∈ GF (2), ∀ v(x) ∈ V (v) ∀ v(x) ∈ V (a · b) · v(x) = a · (b · v(x)) 1 · v(x) = v(x) ∈ V Equivalent statements hold for vector representation v of the polynomial coefficients. According to Section 2.7, (i)-(v) are necessary and sufficient conditions for a set V to be a vector space over the binary field GF (2). 2.26 Null space of a matrix H is a subset of vectors g such that Hg T = 0, where g is assumed to be a 1 × N row vector. We can verify that the above relation holds for all rows of the matrix G, or, in matrix notation, HGT = 0, that is GH T = 0, holds. Hence, the row space of G is the null space of H, and vice versa.