Cyclic Codes II - 上海交通大学数学系math.sjtu.edu.cn/upload/teachers/10473/8.pdfCyclic...

Cyclic Codes II

Jiyou [email protected]

Department of Mathematics, Shanghai Jiao Tong University

Tuesday 5th November, 2013

Li Jiyou Linear Codes II

Cyclic codes

C = (g(x)) ⊆ Fq[x ]/(xn − 1) and g(x)h(x) = xn − 1, withg(x) = g0 + g1x + · · ·+ gn−kxn−k ,h(x) = h0 + h1x + · · ·+ hkxk .

C has a basis g(x), xg(x), . . . , xk−1g(x).


Cyclic codes

C = (g(x)) ⊆ Fq[x ]/(xn − 1) and g(x)h(x) = xn − 1, withg(x) = g0 + g1x + · · ·+ gn−kxn−k ,h(x) = h0 + h1x + · · ·+ hkxk .C has a basis g(x), xg(x), . . . , xk−1g(x).


The generator matrix

G =

g0 g1 . . . gn−k 0 0 00 g0 g1 . . . gn−k 0 0

. . . . . . . . . . . .0 0 0 g0 g1 . . . gn−k

k×n

.

The parity check matrix

H =

hk hk−1 . . . h0 0 0 00 hk hk−1 . . . h0 0 0

. . . . . . . . . . . .0 0 0 hk hk−1 . . . h0

(n−k)×n

.


The generator matrix

G =

g0 g1 . . . gn−k 0 0 00 g0 g1 . . . gn−k 0 0

. . . . . . . . . . . .0 0 0 g0 g1 . . . gn−k

k×n

.

The parity check matrix

H =

hk hk−1 . . . h0 0 0 00 hk hk−1 . . . h0 0 0

. . . . . . . . . . . .0 0 0 hk hk−1 . . . h0

(n−k)×n

.


Example

Let Fq = F3 and n = 8, x8 − 1 = (x + 1)(x − 1)(x2 + 1)(x4 + 1).Take g(x) = x5 + 2x4 + x + 2, we then get a [8, 3, 4]3 code.

G =

2 1 0 0 2 1 0 00 2 1 0 0 2 1 00 0 2 1 0 0 2 1

,

H =

1 1 1 1 0 0 0 00 1 1 1 1 0 0 00 0 1 1 1 1 0 00 0 0 1 1 1 1 00 0 0 0 1 1 1 1

.


ExampleLet Fq = F2, n = 2m − 1 and g(x) be a primitive polynomial ofdegree m over F2[x ]. We then have a [2m − 1, 2m − 1−m, 3]2code. In fact it is the Hamming code Hm (whyº).


ProblemConstruct good cyclic codes using tools from algebra.


Trace representation of cyclic codes

TheoremSuppose the parity check polynomial h(x) is an irreduciblepolynomial of degree k and α is one of its root in someextension field of Fq (Thus Fq(α) = Fqk ). Then for anyc = (c0, c1, . . . , cn−1) ∈ C, there is a unique element βc ∈ Fqk

such thatci = Trace(βcα

−i).


Trace representation of cyclic codes

TheoremSuppose that the generator polynomial h(x) is a product ofirreducible polynomials of pi(x), 1 ≤ i ≤ s with deg pi(x) = diand αi are roots of pi(x) respectively in some extension field ofFq, i.e., Fq(αi) = Fqdi .Then for any c = (c0, c1, . . . , cn−1) ∈ C, there is a uniqueelement βc ∈ Fqk such that

ci =s∑

j=1

Trace(βcα−ij ).


ExampleOver Fq let h(x) be a primitive polynomial of degree k in Fq[x ].We then have a [n = qk − 1, k ]q cyclic code C.

Let α be a root of h(x). Applying the above theorem, for anyc = (c0, c1, . . . , cn−1) ∈ C, there is a unique element βc ∈ Fqk


−i).

This proves that C is a [qk − 1, k , qk − qk−1]q cyclic code.


ExampleOver Fq let h(x) be a primitive polynomial of degree k in Fq[x ].We then have a [n = qk − 1, k ]q cyclic code C.Let α be a root of h(x). Applying the above theorem, for anyc = (c0, c1, . . . , cn−1) ∈ C, there is a unique element βc ∈ Fqk


−i).



ExampleOver Fq let h(x) be a primitive polynomial of degree k in Fq[x ].We then have a [n = qk − 1, k ]q cyclic code C.Let α be a root of h(x). Applying the above theorem, for anyc = (c0, c1, . . . , cn−1) ∈ C, there is a unique element βc ∈ Fqk


−i).



Example

Over F2 take h(x) = x5 + x3 + x2 + x + 1 to be a primitivepolynomial of degree 5 in Fq[x ]. We then get a [31, 5, 16]2cyclic code.


Roots of cyclic codes

DefinitionSuppose the generator polynomial g(x) is a product ofirreducible polynomials of pi(x), 1 ≤ i ≤ s with deg pi(x) = diand αi are roots of pi(x) respectively in some extension field ofFq, i.e., Fq(αi) = Fqdi .Then

C = {c(x) ∈ Fq[x ]/(xn − 1), c(αi) = 0,∀1 ≤ i ≤ s}.


A new parity check matrix

The parity check matrix H is given by

H =

1 α1 α2

1 . . . αn−11

1 α2 α22 . . . αn−1

2. . . . . . . . . . . . . . .1 αs α2

s . . . αn−1s

(n−k)×n

.


Example

Let Fq = F2 and n = 7, Take g(x) = (x + 1)(x3 + x + 1). Onechecks that x3 + x + 1 is irreducible and is of order 7. Let α bea root of x3 + x + 1 and thus F2(α) = F8.Consider

C = {c(x) ∈ F2[x ]/(x7 − 1), c(1) = c(α) = 0}.

H =

(1 1 1 . . . 11 α α2 . . . α6

).

Over F2, we get a [7, 4, 4]2 code:

H =

1 1 1 1 1 1 11 0 0 1 0 1 10 1 0 1 1 1 00 0 1 0 1 1 1

.


Bose-Chaudhuri-Hocquenghem Codes

DefinitionLet (n, q) = 1 and Fqn = Fq(β). l and 2 ≤ δ ≤ n − 1 areintegers. Then q-ary cyclic code of roots β l , β l+1, . . . , β i+δ−2 isdefined as

C = {c(x) ∈ Fq[x ], c(β l) = c(β l+1) = · · · = c(β l+δ−2) = 0}

is called a BCH code of designed distance δ.


Bose-Chaudhuri-Hocquenghem Codes

TheoremThe minimum distance of a BCH code of designed distance δ isat least δ.


ExampleLet Fq = F2 and n = 2m − 1, Take g1(x) to be a primitivepolynomial and α is its root in some extension. One can showthe minimal polynomial g3(x) of α3 has degree m. Takeg(x) = g1(x)g3(x).Consider

C = {c(x) ∈ F2[x ]/(xn − 1), c(α) = c(α3) = 0}.

H =

(1 α α2 . . . αn−1

1 α3 α6 . . . α3(n−1)

).

We thus have a [n, n − 2m,≥ 5]2 code.


An example of Decoding BCH codes

Suppose u(x) = u0 + u1x + · · ·+ un−1xn−1 is the receivedpolynomial and u = (u0, u1, . . . , un−1) is the vector ofcoefficients;

Let u(x) = c(x) + ε(x) where c(x) is the sending messageand ε(x) is the error polynomial and ε is the error vector;

Compute HuT =( u(α)

u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);

If ε(x) = 0, no errors happened and output c(x) = u(x);If ε(x) = x i , 1 error happened and output c(x) = u(x) + x i ;If ε(x) = x i + x j , 2 errors happened and outputc(x) = u(x) + x i + x j .How to compute i and j?



Suppose u(x) = u0 + u1x + · · ·+ un−1xn−1 is the receivedpolynomial and u = (u0, u1, . . . , un−1) is the vector ofcoefficients;Let u(x) = c(x) + ε(x) where c(x) is the sending messageand ε(x) is the error polynomial and ε is the error vector;


u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);






u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);






u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);

If ε(x) = 0, no errors happened and output c(x) = u(x);

If ε(x) = x i , 1 error happened and output c(x) = u(x) + x i ;If ε(x) = x i + x j , 2 errors happened and outputc(x) = u(x) + x i + x j .How to compute i and j?





u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);

If ε(x) = 0, no errors happened and output c(x) = u(x);If ε(x) = x i , 1 error happened and output c(x) = u(x) + x i ;

If ε(x) = x i + x j , 2 errors happened and outputc(x) = u(x) + x i + x j .How to compute i and j?





u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);

If ε(x) = 0, no errors happened and output c(x) = u(x);If ε(x) = x i , 1 error happened and output c(x) = u(x) + x i ;If ε(x) = x i + x j , 2 errors happened and outputc(x) = u(x) + x i + x j .

How to compute i and j?





u(α3)

)= H(cT + εT ) = HεT =

( ε(α)ε(α3)

);



ExampleTake m = 4, and α is a root of primitive polynomialg1(x) = x4 + x + 1. One computes the minimal polynomial ofα3 is g2(x) = x4 + x3 + x2 + x + 1.Thus

C = {c(x) ∈ F2[x ], c(α) = c(α3) = 0}

is a [15, 7,≥ 5]2 cyclic code with generator polynomialg(x) = g1(x)g2(x).Over F2, we get a [15, 7,≥ 5]2 code. Supposec(x) = x + x2 + x5 + x6 + x7 + x10 and ε(x) = x5 + x7.Received u(x) = x + x2 + x6 + x10. Compute HuT =

( ε(α)ε(α3)

).

Solve this equation we get i = 5 and j = 7.


ExampleTaking l = 1, generalized Reed-solomon codes are cycliccodes with

H =

1 α α2 . . . αn−1

1 α2 α4 . . . α2(n−1)

. . . . . . . . . . . . . . .1 αk α2k . . . α(n−k)(n−1)

(n−k)×n

.


ExampleBinary Golay codes are cyclic codes.


Homework

Textbook Page 75: 1, 2, 3.


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