Introduction to Coding Theory - en:group [Algo]

15 Pages·2011·0.12 MB·English

E´COLE POLYTECHNIQUE FE´DE´RALE DE LAUSANNE
Final Exam June 30, 2011
Introduction to Coding Theory
June 30, 2011
• Any document or material is forbidden, except a hand-written recto verso A4 formula
sheet.
• Use a separate sheet of paper for every problem you are working on, write your name
on and number additionnal sheets.
• Within the same problem, you can use the answer of a question to solve the following
ones.
• There are a total of 90 points to obtain.
• You have exactly three hours. Good luck!
Name:
Problem Problem Problem Problem Problem Problem
1 2 3 4 5 6
/ 10 points / 10 points / 15 points / 20 points / 25 points / 10 points
Total
Introduction to Coding Theory - Spring 2010
CONVENTIONS & REMINDERS
• F stands for the finite field with q elements.
q
• If C is a code, puncturing C at position i means that you delete the ith coordinate
of all codewords ; shortening C at position i means that you puncture at position i
the subcode of C formed by all codewords with a zero at the ith coordinate.
• If C is a code over F for some t ≥ 1, C = C∩Fn stands for the subfield subcode.
qt |Fq q
• The MacWilliams identity is between a q-ary code C and its dual C⊥ is
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W (x,y) = W (y −x,y +(q −1)x).
C⊥ |C| C
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Introduction to Coding Theory - Spring 2010
Problem 1 [10 points]. Let G be the generating matrix of a [n,k,d] -code. Show that
q
every k×(n−d+1) submatrix of G has rank k and that furthermore, d is the largest number
with that property.
Solution :
We can assume without loss of generality that we are considering the s × k submatrix B
formed by the last columns and write G = [A|B]. By permutation we can always bring the
positions under consideration to the end. If the rank of B is less than k, there exists a non-
zero linear combination u ∈ Fk(cid:114){0} such that uB = 0. Then c = uG is a codeword of weight
q
≤ n−s. The word c cannot be zero since G is of rank k and u (cid:54)= 0. So d ≤ wgt(c) ≤ n−s,
i.e. s ≤ n−d. The result follows.
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Introduction to Coding Theory - Spring 2010
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Introduction to Coding Theory - Spring 2010
Problem 2 [10 points]. Let C be a [n,k] -code and (A ) its weight distribution.
4 i 0≤i≤n
n
1 (cid:88)
1. Show that C⊥ contains A 3n−i(−1)i codewords of weight n.
4k i
i=0
2. Show that if C has only even weight codewords, then C⊥ has a vector of weight n.
Solution :
1. We apply Mac Williams identity :
n
1 1 (cid:88)
W = W (y−x,y+(q−1)x) = A (y+3x)n−i(y−x)i.
C⊥ |C| C 4k i
i=1
n
1 (cid:88)
The coefficient of xn is A 3n−i(−1)i
4k i
i=0
n/2
2. Ifthecodehasonlyevenweightwords, thereare 1 (cid:88)A 3n−2i(cid:48) codewordsofweightn.
4k i
i(cid:48)=0
This number is clearly strictly positive.
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Introduction to Coding Theory - Spring 2010
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Introduction to Coding Theory - Spring 2010
Problem 3 [15 points]. Let t ≥ 1 be an integer.
1. Let C be a cyclic code over F . Show that the subfield subcode C is also a cyclic code
qt |Fq
over F .
q
2. Let α be a root of x2 −x+2 ∈ F [x] and F be defined as F [α]. The element α is
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primitive in F . Show that the code C given by the following check matrix is cyclic
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 
1 1 1 1 1 1
H = 1 α8 α16 1 α8 α16.
1 α20 α16 −1 α8 α4
3. Find a check matrix as well as its generating polynomial for the subfield subcode C
|F5
of the code defined in the previous question.
Solution :
1. We need to check that C(cid:48) = C is stable under cyclic shift. Let c = (c ,...,c ) be
|Fq 0 n−1
a code word of C(cid:48). By definition, c ∈ C, so by cyclicity of C, c(cid:48) = (c ,c ,...,c )
n−1 0 n−2
belongs to C. But c(cid:48) ∈ Fn, so c(cid:48) is in C(cid:48) and thus C(cid:48) is cyclic.
q
2. We observe that each row of H contains the successive powers of 1, α8 and α20 succes-
sively, which are 6-th roots of unity. If we interpret a codeword c = (c ,...,c ) as
0 n−1
a polynomial c(x) = c +c x+···+c xn−1, that imposes on c that c(1) = c(α8) =
0 1 n−1
c(α20) = 0. In other word, c(x) is a multiple of g(x) = (x−1)(x−α8)(x−α20).
3. A check relation over F can be brought to 2 linear relation over F by projection of
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the coefficient on the basis (1,α). Now, α4 = 2 + 2α, α8 = 1 + 2α, α16 = α12α4 =
−α4 = 3+3α = and α20 = −α8 = 4+3α. We obtain the following equations :
 
1 1 1 1 1 1
0 0 0 0 0 0
 
1 1 3 1 1 3
 .
0 2 3 0 2 3
 
1 4 3 4 1 2
0 3 3 0 2 2
We can eliminate the second line, to adjust the rank and the size of the matrix to get
 
1 1 1 1 1 1
1 1 3 1 1 3
 
H(cid:101) = 0 2 3 0 2 3.
 
1 4 3 4 1 2
0 3 3 0 2 2
as a a check matrix. To find the generator polynomial, we can note that if β is a root of
the g(x), all the conjugates by the Frobenius now must be roots of g(x), the generating
(cid:101)
polynomial of C(cid:48). But x6 −1 = (x−1)(x+1)(x2 +x+1)(x2 −x+1). The minimal
polynomial of α8 and its conjugates is x2 +x+1, the minimal polynomial of α20 and
its conjugates is x2−x+1. So g(x) = (x−1)(x2+x+1)(x2−x+1).
(cid:101)
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Introduction to Coding Theory - Spring 2010
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Introduction to Coding Theory - Spring 2010
Problem 4 [20 points]. The two parts of this problem are independent.
Part A
Let p(x) = x3+x+1 ∈ F [x], q(x) = x3+x2+1 ∈ F [x], and α be a root of p. Let E be the
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elliptic curve defined by the equation f(x,y) = x3+x+1+y+y2 = 0.
1. Show that p and q are irreducible. Identify K = F [α]. Express the roots of p and q
2
in K.
2. Show that for any root β of q, we have p(β) = β+β2.
3. Count the number of points of E on F and on K.
2
4. Describe the space Γ of polynomial functions on E over K and give its dimension.
<m
5. Construct a [12,9,3] -code. Compare its parameters with the Singleton bound.
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Part B (not part of the exam)
We denote by M (n,k) the largest minimal distance of a [n,k] -code. In the sequel, C stands
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for a [n,k,d]-code and C(cid:48) for a [n(cid:48),k(cid:48),d(cid:48)]-code with d,d(cid:48) ≥ 2.
1. Suppose that C(cid:48) obtained from C by shortening. Compare the parameters of C and C(cid:48).
Deduce that M (n+1,k+1) ≤ M (n,k).
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2. Assume that C(cid:48) obtained from C by puncturing. Compare the parameters of C and C(cid:48).
Deduce that M (n+1,k) ≤ M (n,k)+1.
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3. Let H be the check matrix of a [10,8] -code. Show that there must be two linearly
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dependent columns in H. Deduce that M (10,8) ≤ 2.
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4. Prove that a [12,9,3] -code has an optimal minimum distance.
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Solution :
Part A
1. As p and q have degree 3, they are decomposable if and only if they are divisible by
a linear factor, i.e. if they have a zero on the ground field F . This is not the case as
2
p(0) = p(1) = q(0) = q(1) = 1. Thus α defines a field extension of degree 3. So K is
the field F = {0,1,α,...,α6}. The roots of p are the conjugates of α : α, α2 and α4.
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Since q is irreducible of degree 3, the roots of q are the remaining elements : α3,α5 and
α6. (On can check that they are conjugate).
2. If q(β) = β3+β2+1 = 0, one clearly has that p(β) = β3+β2+1−β2+β = β2+β.
3. We need to solve the equation y2 +y+p(x) = 0 in F for various values of x ∈ F . If
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x = 0 or x = 1, we get y2+y+1 = 0, so y should be a third root of unity which does
not exist in F (we would need to be in an extension of F , i.e. in F with k even). If
8 4 2k
x is a root of p, y is 0 or 1. If x = β is a root of q, we have an obvious solution y = β
from the question before and thus a second solution (which by the way is y = β +1).
In total that give us 3·2+3·2 = 12 points on E over F and no points over F .
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Introduction to Coding Theory - Spring 2010
4. We have
Γ = F [y] ⊕xF [y] ⊕x2F [y]
<m 8 <m 8 <m−1 8 <m−2
and dimension is 3m−3.
5. We take the AG code C(E,m = 4). We have directly a [12,3m−3,15−3m] -code, i.e.
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a [12,9,3] -code. The singleton bound would give 12 = k+d ≤ n+1 = 13. We have a
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defect of 1.
Part B
1. Wehaven(cid:48) = n−1, k(cid:48) = k−1(exceptifthereareonlyzerosattheithposition), d = d(cid:48).
Now if C is a code that achieves M (n+1,k +1), C(cid:48) is a [n,k] code with minimum
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distance d so d ≤ M (n,k), i.e. M (n+1,k+1) ≤ M (n,k).
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2. We have n(cid:48) = n−1, k(cid:48) = k (because d > 1 and so the ith axe is not a line contained
in C) and d(cid:48) = d or d − 1 depending wether there exists a minimal codeword with
support containing the position that is punctured. Now if C is a code that achieves
d = M (n+1,k), we can construct a code C(cid:48) such that d(cid:48) = d or d−1 and by definition
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d(cid:48) ≤ M (n,k). So min(d,d−1) ≤ M (n,k). Thus M (n+1,k) ≤ M (n,k)+1.
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3. Either there is a zero column in H. Or, if not, we observe that there are 9 lines in the
plane F2 (directed by the vectors (1,α), α ∈ F or (0,1)). So any 2×10 matrix over F
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has two columns that span the same line, thus two linearly dependant columns. This
shows that a [10,8] code has surely a codeword of weight 2 (deduced from the linear
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relation on H). So M (10,8) ≤ 2.
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4. Wehave, bycombiningthepreviousquestions, M (12,9) ≤ M (11,8) ≤ M (10,8)+1 ≤
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3. This proves optimality.
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