Download Solutions to selected problems from Chapter 2

Error Control Coding — week 1: Binary field algebra and linear block codes
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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
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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
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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.