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Coding and decoding of MDS Fq-linear codes based onsuperregular matrices

Sara D. Cardell1 Joan-Josep Climent1 Verónica Requena2

1 Departament d’Estadística i Investigació OperativaUniversitat d’Alacant

2 Departamento de Estadística, Matemáticas e InformáticaUnivesidad Miguel Hernández de Elche

Ghent 2013

S.D. Cardell, J.-J. Climent, V. Requena MDS Fq -linear codes

Preliminaries Construction Decoding

Contents

1 Preliminaries

2 Construction

3 Decoding

Contents

1 Preliminaries

2 Construction

3 Decoding

Codes over extension alphabets

Let Fq be the Galois field of q elements and assume that CFq denotes a linearcode over Fq .

Definition

Let b be a positive integer. We say that CFbq

is an Fq-linear code of length n

over Fbq if CFq is a linear code of length nb over Fq .

Note that both CFq and CFbq

refer to the same set of codewords, but

considering the alphabets Fq and Fbq , respectively.

Definition

These codes appear in the literature called as Fq-linear codes, codes overextension alphabets or array codes.

References

These codes can be used in storage systems and communications, andprovide a good trade-off between error control power and complexity ofdecoding.

Codes designed to correct burst error patterns can be both MDS and simplerto decode than other equivalent codes.

References

Parameters

Let [N,K ,D] be the parameters of the linear code CFq , i.e.

N, K = dim CFq , D = d(CFq

The parameters of the Fq-linear code CFbq

are then [n, k , d ] with

n = N/b, k = K/b, d = d(CFb

k is called the normalized dimension of CFbq

over Fbq .

Remark

In general, d ≤ D.We cannot compare these two codes, since both alphabets are different.

Parameters

over Fbq .

Remark

Parameters

over Fbq .

Remark

Parameters

over Fbq .

Remark

Parameters

over Fbq .

Remark

MDS Fq-linear codes

For an Fq-linear code CFbq

with parameters [n, k , d ], the Singleton bound alsoholds; i.e.

d ≤ n − k + 1.

The code CFbq

is called maximum distance separable (MDS) if this bound isattained.

MDS Fq-linear codes

For an Fq-linear code CFbq

with parameters [n, k , d ], the Singleton bound alsoholds; i.e.

d ≤ n − k + 1.

The code CFbq

is called maximum distance separable (MDS) if this bound isattained.

MDS Fq-linear codes

Example

Consider the code CF2 whose (systematic) generator matrix is

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

.Here [N,K ,D] = [8, 4, 3] and CF2 is not an MDS code.

Now, the code CF22

has parameters [n, k , d ] = [4, 2, 3]. Therefore, it is an MDSF2-linear code.

We can consider the block matrix

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

.as its generator matrix.

MDS Fq-linear codes

Example

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

Now, the code CF22

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

MDS Fq-linear codes

Example

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

Now, the code CF22

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

MDS Fq-linear codes

The code CFbq

can be specified by either its generator matrix G of size kb× nbor its parity-check matrix H of size (n − k)b × nb, both over Fq .

M. BLAUM and R. M. ROTH.On lowest density MDS codes.IEEE Transactions on Information Theory, 45(1): 46–59 (1999).

Theorem

Let H =[A I(n−k)b

]be an (n − k)b × nb systematic parity-check matrix of

an Fq-linear code CFbq

with parameters [n, k ]. Assume that

A =[Aij]∈ Mat(n−k)b×kb

)where each Aij is a b × b matrix. Then CFb

MDS if and only if every square submatrix of A consisting of full blockssubmatrices Aij is nonsingular.

MDS Fq-linear codes

The code CFbq

can be specified by either its generator matrix G of size kb× nbor its parity-check matrix H of size (n − k)b × nb, both over Fq .

Theorem

Let H =[A I(n−k)b

with parameters [n, k ]. Assume that

A =[Aij]∈ Mat(n−k)b×kb

)where each Aij is a b × b matrix. Then CFb

MDS if and only if every square submatrix of A consisting of full blockssubmatrices Aij is nonsingular.

MDS Fq-linear codes

Definition

A matrix A ∈ Matm×t (Fq) is said to be a superregular matrix if every squaresubmatrix of A is nonsigular over Fq .

Definition

A matrix A ∈ Matbm×bt (Fq) is said to be a superregular b-block matrix if everysquare submatrix of A consisting of full blocks matrices of size b × b isnonsigular over Fq .

MDS Fq-linear codes

Definition

A matrix A ∈ Matm×t (Fq) is said to be a superregular matrix if every squaresubmatrix of A is nonsigular over Fq .

Definition

A matrix A ∈ Matbm×bt (Fq) is said to be a superregular b-block matrix if everysquare submatrix of A consisting of full blocks matrices of size b × b isnonsigular over Fq .

MDS Fq-linear codes

Theorem

Let H =[A I(n−k)b

with parameters [n, k ]. Then CFbq

is MDS if and only if Ais a superregular b-block matrix.

MDS Fq-linear codes

Example

Consider the code CF22

whose (systematic) parity-check matrix is

1 1 1 1 1 0 0 01 0 1 0 0 1 0 01 1 0 1 0 0 1 00 1 1 0 0 0 0 1

.Since the block matrices

[1 11 0

], A12 =

[1 11 0

], A21 =

[1 10 1

], A22 =

[0 11 0

]are nonsingular and matrix A is nonsingular, A is a superregular 2-blockmatrix.The F2-linear code CF2

2is MDS.

Note that the code CF2 is not MDS.

MDS Fq-linear codes

Example

1 1 1 1 1 0 0 01 0 1 0 0 1 0 01 1 0 1 0 0 1 00 1 1 0 0 0 0 1

[1 11 0

], A12 =

[1 11 0

], A21 =

[1 10 1

], A22 =

[0 11 0

2is MDS.

MDS Fq-linear codes

Example

1 1 1 1 1 0 0 01 0 1 0 0 1 0 01 1 0 1 0 0 1 00 1 1 0 0 0 0 1

[1 11 0

], A12 =

[1 11 0

], A21 =

[1 10 1

], A22 =

[0 11 0

2is MDS.

MDS Fq-linear codes

Example

1 1 1 1 1 0 0 01 0 1 0 0 1 0 01 1 0 1 0 0 1 00 1 1 0 0 0 0 1

[1 11 0

], A12 =

[1 11 0

], A21 =

[1 10 1

], A22 =

[0 11 0

2is MDS.

Contents

1 Preliminaries

2 Construction

3 Decoding

Construction

Let C be the companion matrix of a primitive polynomial

p(x) = p0 + p1x + p2x2 + pb−1xb−1 + xb ∈ Fq[x ]

0 0 · · · 0 −p0

1 0 · · · 0 −p1

0 1 · · · 0 −p2...

......

...0 0 · · · 0 −pb−2

0 0 · · · 1 −pb−1

It is well known thatFqb ≈ Fq[C].

Construction

0 0 · · · 0 −p0

1 0 · · · 0 −p1

0 1 · · · 0 −p2...

......

...0 0 · · · 0 −pb−2

0 0 · · · 1 −pb−1

Construction

0 0 · · · 0 −p0

1 0 · · · 0 −p1

0 1 · · · 0 −p2...

......

...0 0 · · · 0 −pb−2

0 0 · · · 1 −pb−1

Construction

Theorem

Let C be the companion matrix of a primitive polynomial p(x) ∈ Fq[x ] ofdegree b. If α ∈ Fqb is a primitive element, then the map ψ : Fqb −→ Fq[C]such that ψ(α) = C is a field isomorphism.

Theorem

Let C be the companion matrix of a primitive polynomial p(x) ∈ Fq[x ] ofdegree b and consider the field isomorphism ψ : Fqb −→ Fq[C] such thatψ(α) = C where α ∈ Fqb is a primitive element.Then, the map Ψ : Matm×t (Fqb ) −→ Matm×t (Fq[C]) given by

Ψ([αij])

=[ψ(αij )

is a ring isomorphism.

Construction

Theorem

Let C be the companion matrix of a primitive polynomial p(x) ∈ Fq[x ] ofdegree b. If α ∈ Fqb is a primitive element, then the map ψ : Fqb −→ Fq[C]such that ψ(α) = C is a field isomorphism.

Theorem

Let C be the companion matrix of a primitive polynomial p(x) ∈ Fq[x ] ofdegree b and consider the field isomorphism ψ : Fqb −→ Fq[C] such thatψ(α) = C where α ∈ Fqb is a primitive element.Then, the map Ψ : Matm×t (Fqb ) −→ Matm×t (Fq[C]) given by

Ψ([αij])

=[ψ(αij )

is a ring isomorphism.

Construction

Theorem

If A ∈ Matm×t (Fqb ) is a superrregular matrix, then H =[Ψ(A) Imb

], is the

parity check-matrix of an MDS Fq-linear code CFbq

with parameters[m + t , t ,m + 1].

Proof.

Since A ∈ Matm×t (Fqb ) is a superrregular matrix, Ψ(A) ∈ Matm×t (Fq[C]) is asuperregular b-block matrix.

Construction

Theorem

If A ∈ Matm×t (Fqb ) is a superrregular matrix, then H =[Ψ(A) Imb

], is the

parity check-matrix of an MDS Fq-linear code CFbq

with parameters[m + t , t ,m + 1].

Proof.

Since A ∈ Matm×t (Fqb ) is a superrregular matrix, Ψ(A) ∈ Matm×t (Fq[C]) is asuperregular b-block matrix.

Construction

Example

Consider the primitive polynomial p(x) = 1 + x2 + x3 ∈ F2[x ]. Then, its companionmatrix is

[0 0 11 0 00 1 1

Let α ∈ F23 be a primitive element and consider the superregular matrix

[α 11 α2

]∈ Mat2×2(F23 ).

Then Ψ(A) =

[C II C2

]∈ Mat2×2(F2[C]) is a superregular 2-block matrix.

Consequently,

is the parity-check matrix of an F2-linear code CF32

with parameters [4, 2, 3].

Construction

Example

[0 0 11 0 00 1 1

[α 11 α2

]∈ Mat2×2(F23 ).

Then Ψ(A) =

[C II C2

Consequently,

Construction

Example

[0 0 11 0 00 1 1

[α 11 α2

]∈ Mat2×2(F23 ).

Then Ψ(A) =

[C II C2

Consequently,

Construction

Example

[0 0 11 0 00 1 1

[α 11 α2

]∈ Mat2×2(F23 ).

Then Ψ(A) =

[C II C2

Consequently,

I6I2 C2

2with parameters [4, 2, 3].

Construction

Example

[0 0 11 0 00 1 1

[α 11 α2

]∈ Mat2×2(F23 ).

Then Ψ(A) =

[C II C2

Consequently,

0 0 1 1 0 0 1 0 0 0 0 01 0 0 0 1 0 0 1 0 0 0 00 1 1 0 0 1 0 0 1 0 0 01 0 0 0 1 1 0 0 0 1 0 00 1 0 0 0 1 0 0 0 0 1 00 0 1 1 1 1 0 0 0 0 0 1

Construction

Example

[0 0 11 0 00 1 1

[α 11 α2

]∈ Mat2×2(F23 ).

Then Ψ(A) =

[C II C2

Consequently,

0 0 1 1 0 0 1 0 0 0 0 01 0 0 0 1 0 0 1 0 0 0 00 1 1 0 0 1 0 0 1 0 0 01 0 0 0 1 1 0 0 0 1 0 00 1 0 0 0 1 0 0 0 0 1 00 0 1 1 1 1 0 0 0 0 0 1

Contents

1 Preliminaries

2 Construction

3 Decoding

Decoding

Let C be the companion matrix of a primitive polynomial p(x) ∈ F2[x ] ofdegree b. Assume that

A11 A12 · · · A1t

ImbA21 A22 · · · A2t

......

...Am1 Am2 · · · Amt

,where Aj` = C i(j,`), is the parity-check matrix of an MDS F2-linear code CFb

with parameters [m + t , t ,m + 1].

Decoding

Let C be the companion matrix of a primitive polynomial p(x) ∈ F2[x ] ofdegree b. Assume that

A11 A12 · · · A1t

ImbA21 A22 · · · A2t

......

...Am1 Am2 · · · Amt

,where Aj` = C i(j,`), is the parity-check matrix of an MDS F2-linear code CFb

with parameters [m + t , t ,m + 1].

Decoding

If we receive the word v =[

v1 v2 · · · v t v t+1 · · · v t+m], with

v` ∈ Fb2 for ` = 1, 2, . . . , t , t + 1, . . . , t + m, then the syndrome

s1 s2 · · · sm]

can be computed as

where e = v − c is the error vector and c is the codeword.

Decoding

v1 v2 · · · v t v t+1 · · · v t+m], with

s1 s2 · · · sm]

can be computed as

t∑`=1

Aj`vT` + vT

t+j , for j = 1, 2, . . . ,m.

Decoding

v1 v2 · · · v t v t+1 · · · v t+m], with

s1 s2 · · · sm]

can be computed as

t∑`=1

Aj`eT` + eT

t+j , for j = 1, 2, . . . ,m.

Decoding

v1 v2 · · · v t v t+1 · · · v t+m], with

s1 s2 · · · sm]

can be computed as

t∑`=1

Aj`eT` + eT

t+j , for j = 1, 2, . . . ,m.

Correcting one symbol in error

For F2-linear code with parameters [2 + t , t , 3] we can correct one error.

In this case,

[A11 A12 · · · A1t I2bA21 A22 · · · A2t

]and the syndrome s =

[s1 s2

]is given by

In this case,

[A11 A12 · · · A1t I2bA21 A22 · · · A2t

[s1 s2

]is given by

In this case,

[A11 A12 · · · A1t I2bA21 A22 · · · A2t

[s1 s2

]is given by

In this case,

[A11 A12 · · · A1t I2bA21 A22 · · · A2t

[s1 s2

]is given by

t∑`=1

A1`vT` + vT

t+1, sT2 =

t∑`=1

A2`vT` + vT

In this case,

[A11 A12 · · · A1t I2bA21 A22 · · · A2t

[s1 s2

]is given by

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

Assume that e =[

0 0 · · · 0 ej 0 · · · 0 0 0], then

sT1 = A1jeT

j , sT2 = A2jeT

and consequently, sT2 = A2jA−1

1j sT1 = C i(2,j)−i(1,j)sT

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

Assume that e =[

0 0 · · · 0 ej 0 · · · 0 0 0], then

sT1 = A1jeT

j , sT2 = A2jeT

1j sT1 = C i(2,j)−i(1,j)sT

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

Assume that e =[

0 0 · · · 0 ej 0 · · · 0 0 0], then

sT1 = A1jeT

j , sT2 = A2jeT

1j sT1 = C i(2,j)−i(1,j)sT

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

Assume that e =[

0 0 · · · 0 ej 0 · · · 0 0 0], then

sT1 = A1jeT

j , sT2 = A2jeT

1j sT1 = C i(2,j)−i(1,j)sT

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

Assume that e =[

0 0 · · · 0 ej 0 · · · 0 0 0], then

sT1 = A1jeT

j , sT2 = A2jeT

1j sT1 = C i(2,j)−i(1,j)sT

The location in error is given by the integer j satisfying the above expressionand can be computed as eT

j = C−i(1,j)sT1 = C−i(2,j)sT

t∑`=1

A1`eT` + eT

t+1, sT2 =

t∑`=1

A2`eT` + eT

Assume that e =[

0 0 · · · 0 ej 0 · · · 0 0 0], then

sT1 = A1jeT

j , sT2 = A2jeT

1j sT1 = C i(2,j)−i(1,j)sT

If no such j exists and one of the block syndromes is nonzero, then there is anerror in the corresponding parity-check block: et+1 = s1 or et+2 = s2.

Example

Consider the primitive polynomial p(x) = 1 + x + x2 ∈ F2[x ]. Then, its companionmatrix is

[0 11 1

[α 11 α

]∈ Mat2×2(F22 ).

Then Ψ(A) =

[C II C

Consequently,

0 1 1 0 1 0 0 01 1 0 1 0 1 0 01 0 0 1 0 0 1 00 1 1 1 0 0 0 1

Example

[0 11 1

[α 11 α

]∈ Mat2×2(F22 ).

Then Ψ(A) =

[C II C

Consequently,

0 1 1 0 1 0 0 01 1 0 1 0 1 0 01 0 0 1 0 0 1 00 1 1 1 0 0 0 1

Example

[0 11 1

[α 11 α

]∈ Mat2×2(F22 ).

Then Ψ(A) =

[C II C

Consequently,

0 1 1 0 1 0 0 01 1 0 1 0 1 0 01 0 0 1 0 0 1 00 1 1 1 0 0 0 1

Example

[0 11 1

[α 11 α

]∈ Mat2×2(F22 ).

Then Ψ(A) =

[C II C

Consequently,

0 1 1 0 1 0 0 01 1 0 1 0 1 0 01 0 0 1 0 0 1 00 1 1 1 0 0 0 1

Example

[0 11 1

[α 11 α

]∈ Mat2×2(F22 ).

Then Ψ(A) =

[C II C

Consequently,

0 1 1 0 1 0 0 01 1 0 1 0 1 0 01 0 0 1 0 0 1 00 1 1 1 0 0 0 1

Example (cont.)

Assume the received word is v =[

1 0 0 0 1 1 1 1].

Then the syndrome vector s =[

s1 s2]

is given by

]and sT

SinceC0−1sT

]6= sT

2 and C1−0sT1 =

There is one error in position 2 given by e2 = I−1sT1 =

The correct codeword is then

c = v − e

1 0 0 0 1 1 1 1]−[

0 0 1 0 0 0 0 0]

1 0 1 0 1 1 1 1]

Example (cont.)

1 0 0 0 1 1 1 1].

s1 s2]

is given by

]and sT

SinceC0−1sT

]6= sT

2 and C1−0sT1 =

c = v − e

1 0 0 0 1 1 1 1]−[

0 0 1 0 0 0 0 0]

1 0 1 0 1 1 1 1]

Example (cont.)

1 0 0 0 1 1 1 1].

s1 s2]

is given by

]and sT

SinceC0−1sT

]6= sT

2 and C1−0sT1 =

c = v − e

1 0 0 0 1 1 1 1]−[

0 0 1 0 0 0 0 0]

1 0 1 0 1 1 1 1]

Example (cont.)

1 0 0 0 1 1 1 1].

s1 s2]

is given by

]and sT

SinceC0−1sT

]6= sT

2 and C1−0sT1 =

c = v − e

1 0 0 0 1 1 1 1]−[

0 0 1 0 0 0 0 0]

1 0 1 0 1 1 1 1]

Example (cont.)

1 0 0 0 1 1 1 1].

s1 s2]

is given by

]and sT

SinceC0−1sT

]6= sT

2 and C1−0sT1 =

c = v − e

1 0 0 0 1 1 1 1]−[

0 0 1 0 0 0 0 0]

1 0 1 0 1 1 1 1]

Correcting two symbols in error

For F2-linear codes with parameters [4 + t , t , 5] we can correct two errors.

In this case,

A11 A12 · · · A1t

I4bA21 A22 · · · A2t

A31 A32 · · · A3t

A41 A42 · · · A4t

and the syndrome s =

[s1 s2 s3 s4

]is given by

In this case,

A11 A12 · · · A1t

I4bA21 A22 · · · A2t

A31 A32 · · · A3t

A41 A42 · · · A4t

[s1 s2 s3 s4

]is given by

In this case,

A11 A12 · · · A1t

I4bA21 A22 · · · A2t

A31 A32 · · · A3t

A41 A42 · · · A4t

[s1 s2 s3 s4

]is given by

t∑`=1

Aj`vT` + vT

t+j , for j = 1, 2, 3, 4.

In this case,

A11 A12 · · · A1t

I4bA21 A22 · · · A2t

A31 A32 · · · A3t

A41 A42 · · · A4t

[s1 s2 s3 s4

]is given by

t∑`=1

Aj`eT` + eT

t+j , for j = 1, 2, 3, 4.

Algorithm

1 If at least two of the syndromes s1, s2, s3, s4 are zero, then there are noerrors in the information symbols and the algorithm stops. Otherwhiseset `1 = 0.

2 Set `1 = `1 + 1. If `1 = t , then the algorithm stops and we declare thereare more than two errors. Otherwise, go to next step.

3 Compute the following vectors

yT1 = sT

1 + A1`1 A−14`1

sT4 , yT

2 = sT2 + A2`1 A−1

yT3 = sT

3 + A3`1 A−12`1

sT2 , yT

4 = sT4 + A4`1 A−1

4 If (y1, y2) = (0, 0), or (y2, y3) = (0, 0), or (y3, y4) = (0, 0), or(y4, y1) = (0, 0), then there is one single error in the informationsymbols in the position `1 given by eT

`1= A−1

k with k = 2, 3, 4, 1, andthe algorithm stops. Otherwise, go to next step.

Algorithm

yT1 = sT

1 + A1`1 A−14`1

sT4 , yT

2 = sT2 + A2`1 A−1

yT3 = sT

3 + A3`1 A−12`1

sT2 , yT

4 = sT4 + A4`1 A−1

`1= A−1

Algorithm

yT1 = sT

1 + A1`1 A−14`1

sT4 , yT

2 = sT2 + A2`1 A−1

yT3 = sT

3 + A3`1 A−12`1

sT2 , yT

4 = sT4 + A4`1 A−1

`1= A−1

Algorithm

yT1 = sT

1 + A1`1 A−14`1

sT4 , yT

2 = sT2 + A2`1 A−1

yT3 = sT

3 + A3`1 A−12`1

sT2 , yT

4 = sT4 + A4`1 A−1

`1= A−1

Algorithm

yT1 = sT

1 + A1`1 A−14`1

sT4 , yT

2 = sT2 + A2`1 A−1

yT3 = sT

3 + A3`1 A−12`1

sT2 , yT

4 = sT4 + A4`1 A−1

`1= A−1

Algorithm

yT1 = sT

1 + A1`1 A−14`1

sT4 , yT

2 = sT2 + A2`1 A−1

yT3 = sT

3 + A3`1 A−12`1

sT2 , yT

4 = sT4 + A4`1 A−1

`1= A−1

Algorithm (cont.)

5 If yT3 = Ck1 yT

2 with

k1 = i(3, `2)− i(2, `2) + Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

− Z (i(2, `1)− i(2, `2)− i(1, `1) + i(1, `2))

for some `2 ∈ {`1 + 1, `1 + 2, . . . , t}, go to next step. Otherwise, go to step 2.

6 If yT4 = Ck2 yT

3 with

k2 = i(4, `2)− i(3, `2) + Z (i(4, `1)− i(4, `2)− i(3, `1) + i(3, `2))

− Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

we declare there are errors in positions `1 and `2. The algorithm stops. In order toobtain the errors e`1 and e`2 , we solve the linear system

A1`1 eT`1

+ A1`2 eT`2

A2`1 eT`1

+ A2`2 eT`2

}Otherwise, go to step 2.

Algorithm (cont.)

5 If yT3 = Ck1 yT

2 with

k1 = i(3, `2)− i(2, `2) + Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

− Z (i(2, `1)− i(2, `2)− i(1, `1) + i(1, `2))

for some `2 ∈ {`1 + 1, `1 + 2, . . . , t}, go to next step. Otherwise, go to step 2.

6 If yT4 = Ck2 yT

3 with

k2 = i(4, `2)− i(3, `2) + Z (i(4, `1)− i(4, `2)− i(3, `1) + i(3, `2))

− Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

we declare there are errors in positions `1 and `2. The algorithm stops. In order toobtain the errors e`1 and e`2 , we solve the linear system

A1`1 eT`1

+ A1`2 eT`2

A2`1 eT`1

+ A2`2 eT`2

}Otherwise, go to step 2.

Let α ∈ Fqb be a primitive element. Then

Fqb ={

0, 1, α, α2, . . . , αqb−3, αqb−2}

where the multiplication and addition are given by

αi · αj = α(i+j) mod (qb−1),

αi + αj = αi (1 + αj−i ) = αi · αZ (j−i).

The Zech logarithm of k ∈ {0, 1, 2, . . . , qb − 3, qb − 2} in the basis α is theinteger Z (k) such that

αZ (k) = 1 + αk .

If C is the companion matrix of a primitive polynomial of degree b withcoefficients in Fq , then C is a primitive element of

Fq[C] ≈ Fqb

and therefore, we can apply the Zech logarithm.

Fqb ={

0, 1, α, α2, . . . , αqb−3, αqb−2}

αZ (k) = 1 + αk .

Fq[C] ≈ Fqb

Fqb ={

0, 1, α, α2, . . . , αqb−3, αqb−2}

αZ (k) = 1 + αk .

Fq[C] ≈ Fqb

Example

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

.Let α ∈ F24 be a primitive element and consider the superregular matrix

α14 1 α5 α8

α5 α13 α14 α4

α2 α4 α12 α13

α6 α α3 α11

∈ Mat4×4(F24 ).

Then Ψ(A) =

C14 I4 C5 C8

C5 C13 C14 C4

C2 C4 C12 C13

C6 C C3 C11

∈ Mat4×4(F2[C]) is a superregular

4-block matrix and consequently, H =[

Ψ(A) I16]

is the parity-check matrix of anF2-linear code CF4

Example

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

α14 1 α5 α8

α5 α13 α14 α4

α2 α4 α12 α13

α6 α α3 α11

∈ Mat4×4(F24 ).

Then Ψ(A) =

C14 I4 C5 C8

C5 C13 C14 C4

C2 C4 C12 C13

C6 C C3 C11

Ψ(A) I16]

Example

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

α14 1 α5 α8

α5 α13 α14 α4

α2 α4 α12 α13

α6 α α3 α11

∈ Mat4×4(F24 ).

Then Ψ(A) =

C14 I4 C5 C8

C5 C13 C14 C4

C2 C4 C12 C13

C6 C C3 C11

Ψ(A) I16]

Example

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

α14 1 α5 α8

α5 α13 α14 α4

α2 α4 α12 α13

α6 α α3 α11

∈ Mat4×4(F24 ).

Then Ψ(A) =

C14 I4 C5 C8

C5 C13 C14 C4

C2 C4 C12 C13

C6 C C3 C11

Ψ(A) I16]

Example (cont.)

Let v =[

0001 0000 0000 0000 0001 0101 0011 1110].

The syndrome vector s =[

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

.All are different from zero. So, `1 = 1 and

yT1 = sT

1 + A11A−141 sT

4 = 0T , yT2 = sT

2 + A21A−111 sT

1 = 0T ,

yT3 = sT

3 + A31A−121 sT

2 = 0T , yT4 = sT

4 + A41A−131 sT

3 = 0T .

We have an error in position `1 = 1 given by

eT1 = A−1

11 sT1 =

.S.D. Cardell, J.-J. Climent, V. Requena MDS Fq -linear codes

Example (cont.)

Let v =[

0001 0000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

4 = 0T , yT2 = sT

2 + A21A−111 sT

1 = 0T ,

yT3 = sT

3 + A31A−121 sT

2 = 0T , yT4 = sT

4 + A41A−131 sT

3 = 0T .

eT1 = A−1

11 sT1 =

Example (cont.)

Let v =[

0001 0000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

4 = 0T , yT2 = sT

2 + A21A−111 sT

1 = 0T ,

yT3 = sT

3 + A31A−121 sT

2 = 0T , yT4 = sT

4 + A41A−131 sT

3 = 0T .

eT1 = A−1

11 sT1 =

Example (cont.)

Let v =[

0001 0000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

4 = 0T , yT2 = sT

2 + A21A−111 sT

1 = 0T ,

yT3 = sT

3 + A31A−121 sT

2 = 0T , yT4 = sT

4 + A41A−131 sT

3 = 0T .

eT1 = A−1

11 sT1 =

Example (cont.)

c = v − e

0001 0000 0000 0000 0001 0101 0011 1110]

1101 0000 0000 0000 0000 0000 0000 0000]

1100 0000 0000 0000 0001 0101 0011 1110]

Example (cont.)

Let v =[

0000 1000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

, yT2 = sT

2 + A21A−111 sT

3 = sT3 + A31A−1

21 sT2 =

, yT4 = sT

4 + A41A−131 sT

.None of them are zero, that means we could have an error in `1 = 1.

Example (cont.)

Let v =[

0000 1000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

, yT2 = sT

2 + A21A−111 sT

3 = sT3 + A31A−1

21 sT2 =

, yT4 = sT

4 + A41A−131 sT

Example (cont.)

Let v =[

0000 1000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

, yT2 = sT

2 + A21A−111 sT

3 = sT3 + A31A−1

21 sT2 =

, yT4 = sT

4 + A41A−131 sT

Example (cont.)

Let v =[

0000 1000 0000 0000 0001 0101 0011 1110].

s1 s2 s3 s4]

is given by

, sT2 =

, sT3 =

and sT4 =

yT1 = sT

1 + A11A−141 sT

, yT2 = sT

2 + A21A−111 sT

3 = sT3 + A31A−1

21 sT2 =

, yT4 = sT

4 + A41A−131 sT

Example (cont.)

Let `2 = 2 and compute

k1 = i(3, `2)− i(2, `2) + Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

− Z (i(2, `1)− i(2, `2)− i(1, `1) + i(1, `2))

= 4− 13 + Z (6)− Z (−22) = 2.k2 = i(4, `2)− i(3, `2) + Z (i(4, `1)− i(4, `2)− i(3, `1) + i(3, `2))

− Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

= 1− 4 + Z (7)− Z (6) = −7.

Ck1 yT2 =

= yT3 and Ck2 yT

= yT4 .

Therefore, there are errors in positions `1 = 1 and `2 = 2.

Example (cont.)

k1 = i(3, `2)− i(2, `2) + Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

− Z (i(2, `1)− i(2, `2)− i(1, `1) + i(1, `2))

= 4− 13 + Z (6)− Z (−22) = 2.k2 = i(4, `2)− i(3, `2) + Z (i(4, `1)− i(4, `2)− i(3, `1) + i(3, `2))

− Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

= 1− 4 + Z (7)− Z (6) = −7.

Ck1 yT2 =

= yT3 and Ck2 yT

= yT4 .

Example (cont.)

k1 = i(3, `2)− i(2, `2) + Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

− Z (i(2, `1)− i(2, `2)− i(1, `1) + i(1, `2))

= 4− 13 + Z (6)− Z (−22) = 2.k2 = i(4, `2)− i(3, `2) + Z (i(4, `1)− i(4, `2)− i(3, `1) + i(3, `2))

− Z (i(3, `1)− i(3, `2)− i(2, `1) + i(2, `2))

= 1− 4 + Z (7)− Z (6) = −7.

Ck1 yT2 =

= yT3 and Ck2 yT

= yT4 .

Example (cont.)

In order to obtain the errors e`1 and e`2 , we solve the linear system

A1`1 eT`1

+ A1`2 eT`2

A2`1 eT`1

+ A2`2 eT`2

}and we obtain the errors

and eT2 =

.The correct codeword is then

c = v − e

0000 1000 0000 0000 0001 0101 0011 1110]

1100 1000 0000 0000 0000 0000 0000 0000]

1100 0000 0000 0000 0001 0101 0011 1110]

Example (cont.)

In order to obtain the errors e`1 and e`2 , we solve the linear system

A1`1 eT`1

+ A1`2 eT`2

A2`1 eT`1

+ A2`2 eT`2

}and we obtain the errors

and eT2 =

.The correct codeword is then

c = v − e

0000 1000 0000 0000 0001 0101 0011 1110]

1100 1000 0000 0000 0000 0000 0000 0000]

1100 0000 0000 0000 0001 0101 0011 1110]

Coding and decoding of MDS Fq-linear codes based onsuperregular matrices

Sara D. Cardell1 Joan-Josep Climent1 Verónica Requena2

1 Departament d’Estadística i Investigació OperativaUniversitat d’Alacant

2 Departamento de Estadística, Matemáticas e InformáticaUnivesidad Miguel Hernández de Elche

Ghent 2013

Cauchy matrices

A matrix A =[aij]∈ Matm×t (Fq) is a Cauchy matrix if

aij = (xi − xj )−1

with xi , xj ∈ Fq satisfaying the following conditions for i ∈ {0, 1, 2, . . . ,m − 1}and j ∈ {0, 1, 2, . . . , t − 1}

xi 6= yj ,

xi 6= xr , for r ∈ {1, 2, . . . ,m} \ {i},yj 6= ys, for s ∈ {1, 2, . . . , t} \ {j}.

Cauchy matrices are superregular matrices over Fq .

Cauchy matrices

A matrix A =[aij]∈ Matm×t (Fq) is a Cauchy matrix if

aij = (xi − xj )−1

with xi , xj ∈ Fq satisfaying the following conditions for i ∈ {0, 1, 2, . . . ,m − 1}and j ∈ {0, 1, 2, . . . , t − 1}

xi 6= yj ,

xi 6= xr , for r ∈ {1, 2, . . . ,m} \ {i},yj 6= ys, for s ∈ {1, 2, . . . , t} \ {j}.

Cauchy matrices are superregular matrices over Fq .

Cauchy matrices

Let {u1, u2, . . . , um}, {v1, v2, . . . , vt} ⊆ {0, 1, 2, . . . , qb − 2} satisfaying thefollowing conditions for i ∈ {0, 1, 2, . . . ,m − 1} and j ∈ {0, 1, 2, . . . , t − 1}

ui 6= vj ,

ui 6= ur , for r ∈ {0, 1, 2, . . . ,m − 1} \ {i},vj 6= vs, for s ∈ {0, 1, 2, . . . , t − 1} \ {j}.

If C is the companion matrix of a primitive polynomial of degree b withcoefficients in Fq , then

[Aij]∈ Matmb×tb(Fq[C]), with

Aij = (Cui − Cvj )−1,

is a superregular b-block matrix.

Cauchy matrices

Let {u1, u2, . . . , um}, {v1, v2, . . . , vt} ⊆ {0, 1, 2, . . . , qb − 2} satisfaying thefollowing conditions for i ∈ {0, 1, 2, . . . ,m − 1} and j ∈ {0, 1, 2, . . . , t − 1}

ui 6= vj ,

ui 6= ur , for r ∈ {0, 1, 2, . . . ,m − 1} \ {i},vj 6= vs, for s ∈ {0, 1, 2, . . . , t − 1} \ {j}.

If C is the companion matrix of a primitive polynomial of degree b withcoefficients in Fq , then

[Aij]∈ Matmb×tb(Fq[C]), with

Aij = (Cui − Cvj )−1,

is a superregular b-block matrix.

Addition and multiplication in F22

+ 0 1 2 30 0 1 2 31 1 0 3 22 2 3 0 13 3 2 1 0

· 0 1 2 30 0 0 0 01 0 1 2 32 0 2 3 13 0 3 1 2

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References

R. M. ROTH and G. SEROUSSI.On generator matrices of MDS codes.IEEE Transactions on Information Theory, 31(6): 826–830 (1985).

R. M. ROTH and A. LEMPEL.On MDS codes via Cauchy matrices.IEEE Transactions on Information Theory, 35(6): 1314–1319 (1989).

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References

R. M. ROTH and G. SEROUSSI.On generator matrices of MDS codes.IEEE Transactions on Information Theory, 31(6): 826–830 (1985).

R. M. ROTH and A. LEMPEL.On MDS codes via Cauchy matrices.IEEE Transactions on Information Theory, 35(6): 1314–1319 (1989).

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M. BLAUM, J. BRUCK, and A. VARDY.MDS array codes with independent parity symbols.IEEE Transactions on Information Theory, 42(2): 529–542 (1996).

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M. BLAUM, P. G. FARRELL, and H. C. A. VAN TILBORG.Array codes.In V. S. PLESS and W. C. HUFFMAN, editors, Handbook of Coding Theory, pages1855–1909. Elsevier, North-Holland, 1998.

M. BLAUM, J. L. FAN, and L. XU.Soft decoding of several classes of array codes.In Proceedings of the 2002 IEEE International Symposium on Information Theory (ISIT2002), page 368, Lausanne, Switzerland, July 2002.

E. LOUIDOR and R. M. ROTH.Lowest density MDS codes over extension alphabets.IEEE Transactions on Information Theory, 52(7): 46–59 (2006).

L. XU and J. BRUCK.X-code: MDS array codes with optimal encoding.IEEE Transactions on Information Theory, 45(1): 272–276 (1999).

L. XU, V. BOHOSSIAN, J. BRUCK, and D. G. WAGNER.Low-density MD codes and factors of complete graphs.IEEE Transactions on Information Theory, 45(6): 1817–1826 (1999).

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