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An overview on error correcting codes Linear codes Cyclic codes Encoding and decoding with cyclic codes

An introduction to cyclic codes

Emanuele Betti1, Emmanuela Orsini2

[email protected]

Department of Mathematics, University of Florence, Italy

[email protected]

Department of Mathematics, University of Milan, Italy

May 1, 2006

1 An overview on error correcting codes

2 Linear codesDefinitionsAn exampleHamming distanceParity-check and decoding

3 Cyclic codesDefinitionsAlgebraic structureRoots of unity and defining setAn exampleParity-check matrixBCH codes

4 Encoding and decoding with cyclic codesPolynomial encodingDecodingReferences

A communication schema

SOURCE DESTINATIONINFORMATION

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SOURCE DESTINATIONTRASMITTERINFORMATION

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SOURCE DESTINATIONRECEIVERTRASMITTER

SIGNAL RECEIVEDSIGNAL

INFORMATION

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SOURCE DESTINATIONRECEIVER

NOISESOURCE

TRASMITTER

INFORMATION

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SOURCE DESTINATIONRECEIVER

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PROCEDURECODING

PROCEDUREDECODING

Definitions

A Definition for Linear Codes

Let q ∈ N such that q = pk , p prime. We denote with Fq

the field with q elements.

(Fq)n forms an n−dimensional vector space over Fq.

Definition

A linear code C is a k-dimensional vector sub-space of (Fq)n ,

(0 ≤ k ≤ n). The elements of a code are called words. n is calledthe length and k is called the dimension .

C⊥ = {v ∈ (Fq)n | 〈v , c〉 = 0 ∀ c ∈ C}

Is called the dual code of C .

Definitions

Definition

C⊥ = {v ∈ (Fq)n | 〈v , c〉 = 0 ∀ c ∈ C}

Definitions

Definition

C⊥ = {v ∈ (Fq)n | 〈v , c〉 = 0 ∀ c ∈ C}

Definitions

Definition

C⊥ = {v ∈ (Fq)n | 〈v , c〉 = 0 ∀ c ∈ C}

Definitions

The generator-matrix of a linear code

Let C be a linear code over Fq. Since C is a k−dimensional vectorspace, there exist k vectors c1, . . . , ck in (Fq)

n that generate C .Let G be the k × n matrix in (Fq)

c2...ck

then G is a generator matrix for C .

v ∈ (Fq)k ⇒ vG ∈ C .

An example

We first associate a vector in (F2)2 at each

color:→ (0, 0) = v1 → (0, 1) = v2

→ (1, 0) = v3 → (1, 1) = v4

Let C be the linear code of length n = 6 anddimension k = 2 generated by:

(0 0 0 1 1 11 1 1 0 0 0

C = {(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1), (0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)}

Trasmit viG ∈ C ⊂ (F2)6 instead of vi ∈ (F2)

An example

color:→ (0, 0) = v1 → (0, 1) = v2

→ (1, 0) = v3 → (1, 1) = v4

(0 0 0 1 1 11 1 1 0 0 0

C = {(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1), (0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)}

An example

color:→ (0, 0) = v1 → (0, 1) = v2

→ (1, 0) = v3 → (1, 1) = v4

(0 0 0 1 1 11 1 1 0 0 0

C = {(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1), (0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)}

An example

color:→ (0, 0) = v1 → (0, 1) = v2

→ (1, 0) = v3 → (1, 1) = v4

(0 0 0 1 1 11 1 1 0 0 0

C = {(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1), (0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)}

An example

color:→ (0, 0) = v1 → (0, 1) = v2

→ (1, 0) = v3 → (1, 1) = v4

(0 0 0 1 1 11 1 1 0 0 0

C = {(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1), (0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)}

An example

A noised communication

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 1, 1) ∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 1) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 0) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 0, 0, 0) ∈ C

Error Correction Capability Error Detection Capability

An example

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 1, 1) ∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 1) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 0) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 0, 0, 0) ∈ C

An example

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 1, 1) ∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 1) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 0) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 0, 0, 0) ∈ C

An example

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 1, 1) ∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 1) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 0) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 0, 0, 0) ∈ C

An example

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 1, 1) ∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 1) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 0) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 0, 0, 0) ∈ C

An example

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 1, 1) ∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 1) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 1, 0, 0) /∈ C

→ (0, 1)G−→ (0, 0, 0, 1, 1, 1) −→ (0, 0, 0, 0, 0, 0) ∈ C

Hamming distance

The Hamming distance

Definition (Hamming Distance)

dH : (Fq)n × (Fq)

n −→ N

dH(u, v) = |{i s.t. ui 6= vi , 1 ≤ i ≤ n}|, u = (ui ), v = (vi ).

The weight of u is w(u) = dH(u, 0)

Definition (Hamming Distance of a Code)

dH(C ) = min{dH(ci , cj) s.t. ci , cj ∈ C , ci 6= cj}.

Theorem

If C is linear: dH(C ) = min{w(c) s.t. c ∈ C , c 6= 0} .

Hamming distance

dH : (Fq)n × (Fq)

n −→ N

Theorem

Hamming distance

dH : (Fq)n × (Fq)

n −→ N

Theorem

Hamming distance

dH : (Fq)n × (Fq)

n −→ N

Theorem

Hamming distance

dH : (Fq)n × (Fq)

n −→ N

Theorem

Hamming distance

Theorem

Let C be a Fq[n, k, d ] code.

C can correct t = bd−12 c errors.

C can detect e = d − 1 errors.

Singleton bound: d ≤ n − k + 1.

Example

{(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1),(0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)

}has distance 3, then t = 1 and e = 2.

Hamming distance

Theorem

Example

{(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1),(0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)

Hamming distance

Theorem

Example

{(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1),(0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)

Hamming distance

Theorem

Example

{(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1),(0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)

Hamming distance

Theorem

Example

{(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1),(0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)

Hamming distance

Theorem

Example

{(0, 0, 0, 0, 0, 0), (1, 1, 1, 1, 1, 1),(0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 0, 0)

Parity-check and decoding

Parity–check matrix and syndromes

It is easy to see that C may be expressed as the null space of amatrix H, which is called a parity–check matrix for C .

∀x ∈ (Fq)n, Hx = 0⇔ x ∈ C

Observe that H is an (n − k)× n matrix, its coefficients maylie in an extension field of Fq and it is full rank over Fq.Obviously, H is not unique

HT is a generator matrix for C⊥

Definition

The elements in (Fq)n−k , s = Hx are called syndromes. We say

that s is the syndrome corresponding to x .

∀x ∈ (Fq)n, Hx = 0⇔ x ∈ C

Definition

∀x ∈ (Fq)n, Hx = 0⇔ x ∈ C

Definition

∀x ∈ (Fq)n, Hx = 0⇔ x ∈ C

Definition

A decoding procedure

Let C be as in the previous example.A parity check matrix for C is:

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

Let c be a word of a linear code and let x = c + e be the receivedvector after the trasmission. Then:

s = Hc ′ = H(c + e) = Hc + He = He

Theorem (Correctable syndrome)

If no more then t errors are occurred, there exists only one error ecorresponding to a syndrome s (i.e. s = He).

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

0 0 0 1 0 10 0 0 0 1 11 0 1 0 0 01 1 0 0 0 0

(000111) −→c ′

(000101)H−→

(0100) ↔e

(000010) ⇒ c = c ′ − e

Remark

s = Hc ′ = H(c + e) = Hc + He = He

Definitions

A first definition for cyclic codes

Definition

A Fq[n, k, d ] code (linear) C is called cyclic if any cyclic shift of acodeword is a codeword, i.e.

(c0, . . . , cn−1) ∈ C ⇒ (cn−1, c0, . . . , cn−2) ∈ C

Example

1 C = {(0000), (1111)} is cyclic

2 C = {(000), (110), (101), (011)} is cyclic

3 C = {(000), (100), (011), (111)} is not cyclic

Definitions

Definition

(c0, . . . , cn−1) ∈ C ⇒ (cn−1, c0, . . . , cn−2) ∈ C

Example

1 C = {(0000), (1111)} is cyclic

2 C = {(000), (110), (101), (011)} is cyclic

3 C = {(000), (100), (011), (111)} is not cyclic

Definitions

Definition

(c0, . . . , cn−1) ∈ C ⇒ (cn−1, c0, . . . , cn−2) ∈ C

Example

1 C = {(0000), (1111)} is cyclic

2 C = {(000), (110), (101), (011)} is cyclic

3 C = {(000), (100), (011), (111)} is not cyclic

Definitions

Definition

(c0, . . . , cn−1) ∈ C ⇒ (cn−1, c0, . . . , cn−2) ∈ C

Example

1 C = {(0000), (1111)} is cyclic

2 C = {(000), (110), (101), (011)} is cyclic

3 C = {(000), (100), (011), (111)} is not cyclic

Definitions

Definition

(c0, . . . , cn−1) ∈ C ⇒ (cn−1, c0, . . . , cn−2) ∈ C

Example

1 C = {(0000), (1111)} is cyclic

2 C = {(000), (110), (101), (011)} is cyclic

3 C = {(000), (100), (011), (111)} is not cyclic

Algebraic structure

Cyclic codes: an algebraic correspondence

Let R be the ring

R =Fq[x ]

xn − 1

There is a bijective correspondence between the vectors of (Fq)n

and residue class of polynomials in R.

v = (v0, . . . , vn−1)↔ v0 + v1x + · · ·+ vn−1xn−1

Algebraic structure

Let C be a Fq[n, k, d ] code

C is a cyclic code iff C is an ideal of R

Multiplying by x modulo xn − 1 corresponds to cyclic shift:

(c0, c1, . . . , cn−1)→ (cn−1, c0, . . . , cn−2)

xc(c0 + c1x + · · ·+ cn−1xn−1 = cn−1 + c0x + · · ·+ cn−2x

Algebraic structure

Let C be a Fq[n, k, d ] code

C is a cyclic code iff C is an ideal of R

Multiplying by x modulo xn − 1 corresponds to cyclic shift:

(c0, c1, . . . , cn−1)→ (cn−1, c0, . . . , cn−2)

xc(c0 + c1x + · · ·+ cn−1xn−1 = cn−1 + c0x + · · ·+ cn−2x

Algebraic structure

Properties of the ring R

R is a principal ideals ring.An ideal C of R is generated by a monic polynomial (residue class)

g that divides xn − 1.

C = 〈g〉

If C has length n and dimension k, the degree of g is n − kA generator matrix for C is given by:

gxg...

g0 g1 . . . gn−k 0 . . . 00 g0 . . . gn−k−1 an−k 0 . . ....

. . ....

0 . . . 0 g0 g1 . . . gn−k

n−k∑i=0

gixi , gn−k = 1.

Algebraic structure

C = 〈g〉

gxg...

g0 g1 . . . gn−k 0 . . . 00 g0 . . . gn−k−1 an−k 0 . . ....

. . ....

0 . . . 0 g0 g1 . . . gn−k

n−k∑i=0

gixi , gn−k = 1.

Algebraic structure

C = 〈g〉

gxg...

g0 g1 . . . gn−k 0 . . . 00 g0 . . . gn−k−1 an−k 0 . . ....

. . ....

0 . . . 0 g0 g1 . . . gn−k

n−k∑i=0

gixi , gn−k = 1.

Algebraic structure

C = 〈g〉

gxg...

g0 g1 . . . gn−k 0 . . . 00 g0 . . . gn−k−1 an−k 0 . . ....

. . ....

0 . . . 0 g0 g1 . . . gn−k

n−k∑i=0

gixi , gn−k = 1.

Algebraic structure

C = 〈g〉

gxg...

g0 g1 . . . gn−k 0 . . . 00 g0 . . . gn−k−1 an−k 0 . . ....

. . ....

0 . . . 0 g0 g1 . . . gn−k

n−k∑i=0

gixi , gn−k = 1.

Algebraic structure

Parity check polynomial

Let C generated by g , deg(g) = n − k.g divides xn − 1⇒ xn − 1 = hgh ∈ Fq[x ] unique, deg(h) = k.

h = h0 + h1 + · · ·+ hkxk

is called the parity checkpolynomial of C .

Theorem

Let H ∈ (Fq)(n−k)×n,

0 0 . . . 0 hk . . . h1 h0

0 0 . . . hk . . . h1 h0 0...

...hk . . . h1 h0 0 . . . 0 0

H is a parity check matrix for C .

Algebraic structure

h = h0 + h1 + · · ·+ hkxk

Theorem

0 0 . . . 0 hk . . . h1 h0

0 0 . . . hk . . . h1 h0 0...

...hk . . . h1 h0 0 . . . 0 0

Algebraic structure

h = h0 + h1 + · · ·+ hkxk

Theorem

0 0 . . . 0 hk . . . h1 h0

0 0 . . . hk . . . h1 h0 0...

...hk . . . h1 h0 0 . . . 0 0

Algebraic structure

h = h0 + h1 + · · ·+ hkxk

Theorem

0 0 . . . 0 hk . . . h1 h0

0 0 . . . hk . . . h1 h0 0...

...hk . . . h1 h0 0 . . . 0 0

Roots of unity and defining set

Primitive root of unity

Cyclic codes of length n over Fq are gen. by divisors of xn − 1. Let

xn − 1 =r∏

fj , fj irreducible over Fq

Cyclic codes of length n over Fq ←→ subsets of {fj}rj=1

If GCD(n, q) = 1 , there exists α ∈ Fq s.t.:

xn − 1 =n−1∏i=0

x − αi .

α is a n−th primitive root of unity over Fq.

xn − 1 =r∏

xn − 1 =n−1∏i=0

x − αi .

xn − 1 =r∏

xn − 1 =n−1∏i=0

x − αi .

xn − 1 =r∏

xn − 1 =n−1∏i=0

x − αi .

Families of good cyclic codes

Proposition (Castagnoli/Massey/Schoeller, 1991)

If there exists a family of “good” codes {Cm}m over Fq of lengthsm with GCD(m, q) 6= 1, there exists a family {C ′

n}n withGCD(n, q) = 1 with the same properties.

Cyclic codes of length coprime with q enjoy a useful decriptiongiven by the existence n−th primitive roots of unity

Families of good cyclic codes

Proposition (Castagnoli/Massey/Schoeller, 1991)

If there exists a family of “good” codes {Cm}m over Fq of lengthsm with GCD(m, q) 6= 1, there exists a family {C ′

n}n withGCD(n, q) = 1 with the same properties.

Cyclic codes of length coprime with q enjoy a useful decriptiongiven by the existence n−th primitive roots of unity

Defining set

The generator polynomial gC of a cyclic code divides xn − 1. Theroots of gC are powers of any n−th primitive root of unity.

Definition

Let α be a n−th primitive root of unity over Fq and let C be aFq[n, k, d ] cyclic code with generator polynomial gC . The set:

SC ,α = SC = {0 ≤ i ≤ n − 1 | gC (αi ) = 0}

Is called the complete defining set of C .

#(SC ) = deg(gC ) = n − k

Defining set depends by the choice of α:

For any α n−th primitive root of unity SC has a different form.

Defining set

Definition

SC ,α = SC = {0 ≤ i ≤ n − 1 | gC (αi ) = 0}

#(SC ) = deg(gC ) = n − k

Defining set

Definition

SC ,α = SC = {0 ≤ i ≤ n − 1 | gC (αi ) = 0}

#(SC ) = deg(gC ) = n − k

Defining set

Definition

SC ,α = SC = {0 ≤ i ≤ n − 1 | gC (αi ) = 0}

#(SC ) = deg(gC ) = n − k

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

Consider C the cyclic code generated by g = f0 · f1

SC = {0, 1, 2, 4}

Respect to a n−th primitive root of unity α s.t. f1(α) = 0.

n = 7 k = 3 d =?

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

SC = {0, 1, 2, 4}

n = 7 k = 3 d =?

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

SC = {0, 1, 2, 4}

n = 7 k = 3 d =?

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

SC = {0, 1, 2, 4}

n = 7 k = 3 d =?

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

SC = {0, 1, 2, 4}

n = 7 k = 3 d =?

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

SC = {0, 1, 2, 4}

n = 7 k = 3 d =?

An example

x7 − 1 over F2:

f0 f1 f3q q q

x7 − 1 = (x + 1) (x3 + x2 + 1) (x3 + x + 1)

SC = {0, 1, 2, 4}

n = 7 k = 3 d =?

Parity-check matrix

Let C be an Fq[n, k, d ] cyclic code with (n, q) = 1. LetSC = {i1, . . . , in−k} be its defining set and α be a primitive n–rootof unity in the splitting field of xn − 1 on Fq. Then

1 αi1 α2i1 · · · α(n−1)i1

1 αi2 α2i2 · · · α(n−1)i2

......

.... . .

1 αin−k α2in−k · · · α(n−1)in−k

is a parity-check matrix for C .

Let c ∈ C then cHT = 0 is equivalent to sayn−1∑τ=0

cτατ is = 0, ∀is ∈ Sc ,

that is c(αis ) = 0, for any is ∈ SC .

Parity-check matrix

Let C be an Fq[n, k, d ] cyclic code with (n, q) = 1. LetSC = {i1, . . . , in−k} be its defining set and α be a primitive n–rootof unity in the splitting field of xn − 1 on Fq. Then

1 αi1 α2i1 · · · α(n−1)i1

1 αi2 α2i2 · · · α(n−1)i2

......

.... . .

1 αin−k α2in−k · · · α(n−1)in−k

is a parity-check matrix for C .

Let c ∈ C then cHT = 0 is equivalent to sayn−1∑τ=0

cτατ is = 0, ∀is ∈ Sc ,

that is c(αis ) = 0, for any is ∈ SC .

BCH codes

BCH bound–BCH codes

Theorem (BCH bound)

Let C be an Fq[n, k, d ] cyclic code with defining setSC = {i1, . . . , in−k} and let (n, q) = 1. Suppose there are δ − 1consecutive numbers in SC , say{m0,m0 + 1, . . . m0 + δ − 2} ⊂ SC . Then

d ≥ δ

Definition

Let S = (m0,m0 + 1, . . . ,m0 + δ − 2) be such that

0 ≤ m0 ≤ · · · ≤ m0 + δ − 2 ≤ n − 1

If C is the Fq[n, k, d ] cyclic code with defining set S , we say thatC is a BCH code of designed distance δ

BCH codes

BCH bound–BCH codes

Theorem (BCH bound)

Let C be an Fq[n, k, d ] cyclic code with defining setSC = {i1, . . . , in−k} and let (n, q) = 1. Suppose there are δ − 1consecutive numbers in SC , say{m0,m0 + 1, . . . m0 + δ − 2} ⊂ SC . Then

d ≥ δ

Definition

Let S = (m0,m0 + 1, . . . ,m0 + δ − 2) be such that

0 ≤ m0 ≤ · · · ≤ m0 + δ − 2 ≤ n − 1

If C is the Fq[n, k, d ] cyclic code with defining set S , we say thatC is a BCH code of designed distance δ

BCH codes

An example

The F2[7, 3, d ] code of previous example with SC = {0, 1, 2, 4} is aBCH code of designed distance δ = 4.

d ≥ 4 (BCH ).

A minimum weight word of the code is given by the generatorpolynomial

gC = x4 + x2 + x + 1

BCH codes

An example

d ≥ 4 (BCH ).

gC = x4 + x2 + x + 1

BCH codes

An example

d ≥ 4 (BCH ).

gC = x4 + x2 + x + 1

BCH codes

An example

d ≥ 4 (BCH ).

gC = x4 + x2 + x + 1

BCH codes

An example

d ≥ 4 (BCH ).

gC = x4 + x2 + x + 1

Polynomial encoding

Polynomial encoding (1)

Let C be an [n, k, d ] cyclic code over Fq, then C is capable ofencoding q–ary messages of length k and requires n − kredundancy symbols.

Non–systematic method

given a message w = (w0, . . . , wk−1), we consider its

polynomial representation w(x) ∈ Fq[x]xn−1

multiply w(x) by the generator polynomial g(x)

c(x) = w(x)g(x) ∈ C is the encoded word

Polynomial encoding

Example (Systematic method)

Let C be an F2[7, 4] code, g(x) = x3 + x2 + 1 and w= (1, 0, 1, 0).

find the associated polynomial: w(x) = x2 + 1

multiply it by xn−k : (x2 + 1) ∗ x3 = x5 + x3

divide this results by g(x) and take the remainder r(x):

x5 + x3 = (x2 + x) ∗ (x3 + x2 + 1) + (x2 ∗ x)

r(x) as an (n − k)–vector: r = (0, 1, 1)

encode using r , preceded by the message (1, 0, 1, 0):

(1, 0, 1, 0)→ (1, 0, 1, 0, 0, 1, 1)

Polynomial encoding

x5 + x3 = (x2 + x) ∗ (x3 + x2 + 1) + (x2 ∗ x)

(1, 0, 1, 0)→ (1, 0, 1, 0, 0, 1, 1)

Polynomial encoding

x5 + x3 = (x2 + x) ∗ (x3 + x2 + 1) + (x2 ∗ x)

(1, 0, 1, 0)→ (1, 0, 1, 0, 0, 1, 1)

Polynomial encoding

x5 + x3 = (x2 + x) ∗ (x3 + x2 + 1) + (x2 ∗ x)

(1, 0, 1, 0)→ (1, 0, 1, 0, 0, 1, 1)

Polynomial encoding

x5 + x3 = (x2 + x) ∗ (x3 + x2 + 1) + (x2 ∗ x)

(1, 0, 1, 0)→ (1, 0, 1, 0, 0, 1, 1)

Polynomial encoding

x5 + x3 = (x2 + x) ∗ (x3 + x2 + 1) + (x2 ∗ x)

(1, 0, 1, 0)→ (1, 0, 1, 0, 0, 1, 1)

Decoding

Polynomial decoding

Suppose that the decoder receive exactly the sent word (but itdoes not know it!)

v= (1, 0, 1, 0, 0, 1, 1)→ v(x) = x6 + x5 + x2 + 1

the decoder divides v(x) by g(x):

x6 + x5 + x2 + 1 = (x3 + x2 + 1) ∗ (x3 + 1) + 0

(probably) no errors have occurred

the message is formed by the first 4 components of thereceived word, i.e. w= (1, 0, 1, 0)

Decoding

Polynomial decoding

v= (1, 0, 1, 0, 0, 1, 1)→ v(x) = x6 + x5 + x2 + 1

x6 + x5 + x2 + 1 = (x3 + x2 + 1) ∗ (x3 + 1) + 0

Decoding

Polynomial decoding

v= (1, 0, 1, 0, 0, 1, 1)→ v(x) = x6 + x5 + x2 + 1

x6 + x5 + x2 + 1 = (x3 + x2 + 1) ∗ (x3 + 1) + 0

Decoding

Polynomial decoding

v= (1, 0, 1, 0, 0, 1, 1)→ v(x) = x6 + x5 + x2 + 1

x6 + x5 + x2 + 1 = (x3 + x2 + 1) ∗ (x3 + 1) + 0

Decoding

Polynomial decoding

v= (1, 0, 1, 0, 0, 1, 1)→ v(x) = x6 + x5 + x2 + 1

x6 + x5 + x2 + 1 = (x3 + x2 + 1) ∗ (x3 + 1) + 0

References

C. E. Shannon, “A mathematical theory of communication”, BellSystems Technical Journal,, vol. 27, p. 379–423, p. 623–656, 1948.

F.J. MacWilliams, N.J.A. Sloane, The Theory of Error-CorrectingCodes, North Holland, 1977.

E. Prange, “Cyclic Error-Correcting Codes in Two Symbols”, AirForce Cambridge Research Center, Cambridge, MA, Tech. Rep.AFCRC-TN-57-103, 1957.

G. Castagnoli, J. L. Massey, P. A. Schoeller, and N. von Seeman,“On repeated-root cyclic codes”, IEEE Trans.Inf. Theory, vol. 37,no.2, pp. 337342, Mar. 1991.

R. C. Bose and D. K. Ray-Chaudhuri, “On a class of error correctingbinary group codes,” Inform. Control, vol. 3, pp. 68-79, 1960.

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