Classification of self-orthogonal F q + u F q -codes Classification - - PowerPoint PPT Presentation

Self-orthogonal Codes over Fq + uFq

Rowena Alma Betty, Fidel Nemenzo and Lucky Erap Galvez University of the Philippines-Diliman LAWCI 2018

Classification of self-orthogonal Fq + uFq-codes

Classification of self-orthogonal Fq + uFq-codes

Classification of codes means finding complete set of representatives of equivalence classes.

Classification of self-orthogonal Fq + uFq-codes

Classification of codes means finding complete set of representatives of equivalence classes. For self-orthogonal codes, establishing the mass formula means finding a formula for

• [C]

|En| | Aut(C)| =

• C

1.

C: runs through self-orthogonal codes of length n over Fq + uFq [C]: runs through the equivalence classes of self-orthogonal codes of length n over Fq + uFq En is the full group of transformations allowed in defining the equivalence Aut(C): automorphism group of C

Survey: Mass formulas for Codes over Some Rings

Survey: Mass formulas for Codes over Some Rings

Number of self-dual codes over Z4 and self-dual codes over Fq + uFq, u2 = 0 (Gaborit)

Survey: Mass formulas for Codes over Some Rings

Number of self-dual codes over Z4 and self-dual codes over Fq + uFq, u2 = 0 (Gaborit)

p odd Zp2 s.d. BBN s.o. BM Z4 Fq + uFq s.d. G G s.o. BM ?

G Gaborit BBN Balmaceda-Betty-Nemenzo BM Betty-Munemasa

The ring Fq + uFq

Let R be the ring R = Fq[u]/(u2) = Fq + uFq where u commutes with the elements of Fq.

The ring Fq + uFq

Let R be the ring R = Fq[u]/(u2) = Fq + uFq where u commutes with the elements of Fq. The element u is a nilpotent element with nilpotency index 2. This finite chain ring is a local ring with unique maximal ideal (u) and Fq + uFq/(u) = Fq is the residue field.

The ring Fq + uFq

Let x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) ∈ Rn. The Euclidean inner product x · y on Rn is defined as x · y =

n

• i=1

xiyi. Let x = a + ub ∈ R and define x = a − ub. The Hermitian inner product x · y on Rn is defined as x · y =

n

• i=1

xiyi.

Codes over R

Definition A linear code C of length n over R is an R-submodule of the Rn module.

Codes over R

Definition A linear code C of length n over R is an R-submodule of the Rn module. Definition Two codes C and C′ over R are said to be equivalent if there exists a monomial matrix P such that C′ = CP = {c · P|c ∈ C}

Codes over R

Every code C of length n over R is permutation equivalent to a code with generator matrix Ik0 A + uB uD

• where A, B ∈ Mk0×(n−k0)(Fq) and D ∈ Mk1×(n−k0)(Fq). Such

code is said to be of type {k0, k1}.

Codes over R

Every code C of length n over R is permutation equivalent to a code with generator matrix Ik0 A + uB uD

• where A, B ∈ Mk0×(n−k0)(Fq) and D ∈ Mk1×(n−k0)(Fq). Such

code is said to be of type {k0, k1}. Definition Let x = (x1, x2, .., xn) ∈ Rn. If C is a code over R, the dual of C, denoted C⊥ is the set C⊥ = {x ∈ Rn|x · c = 0 ∀c ∈ C} .

Codes over R

Every code C of length n over R is permutation equivalent to a code with generator matrix Ik0 A + uB uD

• where A, B ∈ Mk0×(n−k0)(Fq) and D ∈ Mk1×(n−k0)(Fq). Such

code is said to be of type {k0, k1}. Definition Let x = (x1, x2, .., xn) ∈ Rn. If C is a code over R, the dual of C, denoted C⊥ is the set C⊥ = {x ∈ Rn|x · c = 0 ∀c ∈ C} . If C ⊆ C⊥, C is said to be (Euclidean/Hermitian) self-orthogonal.

Codes over R

Every code C of length n over R is permutation equivalent to a code with generator matrix Ik0 A + uB uD

• where A, B ∈ Mk0×(n−k0)(Fq) and D ∈ Mk1×(n−k0)(Fq). Such

code is said to be of type {k0, k1}. Definition Let x = (x1, x2, .., xn) ∈ Rn. If C is a code over R, the dual of C, denoted C⊥ is the set C⊥ = {x ∈ Rn|x · c = 0 ∀c ∈ C} . If C ⊆ C⊥, C is said to be (Euclidean/Hermitian) self-orthogonal. If C = C⊥, C is said to be (Euclidean/Hermitian) self-dual.

Codes over R

Definition For every code C over R, we associate the codes residue code of C: res(C) =

• v ∈ Fn

q |v ∈ C

• torsion code of C: tor(C) =
• v ∈ Fn

q |uv ∈ C

Codes over R

Definition For every code C over R, we associate the codes residue code of C: res(C) =

• v ∈ Fn

q |v ∈ C

• torsion code of C: tor(C) =
• v ∈ Fn

q |uv ∈ C

• If C has generator matrix

Ik0 A + uB uD

• , then

1 res(C) has generator matrix

• Ik0

A

• ;

Codes over R

Definition For every code C over R, we associate the codes residue code of C: res(C) =

• v ∈ Fn

q |v ∈ C

• torsion code of C: tor(C) =
• v ∈ Fn

q |uv ∈ C

• If C has generator matrix

Ik0 A + uB uD

• , then

1 res(C) has generator matrix

• Ik0

A

• ;

2 tor(C) has generator matrix

Ik0 A D

• where D is of full

row rank k1.

Self-orthogonal Codes over Fq + uFq

If the code C is self orthogonal over R, then

Self-orthogonal Codes over Fq + uFq

If the code C is self orthogonal over R, then res(C) ⊆ tor(C) ⊆ res(C)⊥.

Self-orthogonal Codes over Fq + uFq

If the code C is self orthogonal over R, then res(C) ⊆ tor(C) ⊆ res(C)⊥. In particular, if C is self-dual, then tor(C) = res(C)⊥.

Codes over R with prescribed residue and torsion

From now on, we assume the following:

1 C1 is a code of length n over Fq with dimension k0 and

generator matrix G1 =

• Ik0 A
• 2 C2 is a code of length n over Fq of dimension k0 + k1 and

generator matrix G2 = Ik0 A D

• where A, B ∈ Mk0×(n−k0)(Fq) and D ∈ Mk1×(n−k0)(Fq), of

full row rank.

Codes over R with prescribed residue and torsion

Lemma If C is a code of length n over R with residue code C1 and torsion code C2, then there exists a matrix N ∈ Mk0×(n−k0)(Fq) such that the matrix Ik0 A + uN uD

• is a generator matrix for C. Such matrix N is unique if C is a

free code, i.e. k1 = 0 and C1 = C2.

Codes over R with prescribed residue and torsion

Lemma If C is a code of length n over R with residue code C1 and torsion code C2, then there exists a matrix N ∈ Mk0×(n−k0)(Fq) such that the matrix Ik0 A + uN uD

• is a generator matrix for C. Such matrix N is unique if C is a

free code, i.e. k1 = 0 and C1 = C2. Now, we assume that C1 is self-orthogonal and C1 ⊆ C2 ⊆ C⊥

1 .

Then Ik0 + AAT ≡ (u) (1) DAT ≡ (u) (2)

Codes over R with prescribed residue and torsion

Lemma The number of free Euclidean self-orthogonal codes C over R with given residue (and torsion) code C1 is qk0(2n−3k0+ǫ)/2 where ǫ = −1 if char Fq = 2 and ǫ = 1 if char Fq = 2.

Codes over R with prescribed residue and torsion

Lemma The number of free Euclidean self-orthogonal codes C over R with given residue (and torsion) code C1 is qk0(2n−3k0+ǫ)/2 where ǫ = −1 if char Fq = 2 and ǫ = 1 if char Fq = 2.

• Proof. C has generator matrix [Ik0 A + uN] and since C is

self-orthogonal, Ik0 + AAT + u(ANT + NAT ) ≡ 0 (u2) ⇒ ANT + NAT ≡ 0 (u) by (1)

Codes over R with prescribed residue and torsion

Lemma The number of free Euclidean self-orthogonal codes C over R with given residue (and torsion) code C1 is qk0(2n−3k0+ǫ)/2 where ǫ = −1 if char Fq = 2 and ǫ = 1 if char Fq = 2.

• Proof. C has generator matrix [Ik0 A + uN] and since C is

self-orthogonal, Ik0 + AAT + u(ANT + NAT ) ≡ 0 (u2) ⇒ ANT + NAT ≡ 0 (u) by (1) Define the map ΨA : Mk0×(n−k0)(Fq) → Mk0(Fq) by ΨA(N) = ANT + NAT .

Codes over R with prescribed residue and torsion

Lemma The number of free Euclidean self-orthogonal codes C over R with given residue (and torsion) code C1 is qk0(2n−3k0+ǫ)/2 where ǫ = −1 if char Fq = 2 and ǫ = 1 if char Fq = 2.

• Proof. C has generator matrix [Ik0 A + uN] and since C is

self-orthogonal, Ik0 + AAT + u(ANT + NAT ) ≡ 0 (u2) ⇒ ANT + NAT ≡ 0 (u) by (1) Define the map ΨA : Mk0×(n−k0)(Fq) → Mk0(Fq) by ΨA(N) = ANT + NAT . So,

• N|ANT + NAT ≡ 0 (u)
• = |ker ΨA|

ΨA : Mk0×(n−k0)(Fq) → Mk0(Fq), ΨA(N) = AN T + NAT

Moreover, Im(ΨA) = Symk0(Fq), char Fq = 2 Altk0(Fq), char Fq = 2 . Since

• Symk0(Fq)
• = q

k0(k0+1) 2

and |Altk0(Fq)| = q

k0(k0−1) 2

, |ker ΨA| =

• qk0(n−k0)− k0(k0+1)

2

= q

k0(2n−3k0−1) 2

, char Fq = 2 qk0(n−k0)− k0(k0−1)

2

= q

k0(2n−3k0+1) 2

, char Fq = 2

Codes over R with prescribed residue and torsion

Lemma The number of free Hermitian self-orthogonal codes over Fq + uFq with given residue (and torsion) code C1 is qk0(2n−3k0+1)/2

Codes over R with prescribed residue and torsion

Lemma The number of free Hermitian self-orthogonal codes over Fq + uFq with given residue (and torsion) code C1 is qk0(2n−3k0+1)/2

• Proof. As in the proof of the previous lemma, C has generator

matrix [Ik0 A + uN] for some unique N ∈ Mk0×(n−k0)(Fq) and C is Hermitian self-orthogonal if and only if Ik0 + AAT + u(ANT − NAT ) ≡ 0 (u2).

Codes over R with prescribed residue and torsion

Lemma The number of free Hermitian self-orthogonal codes over Fq + uFq with given residue (and torsion) code C1 is qk0(2n−3k0+1)/2

• Proof. As in the proof of the previous lemma, C has generator

matrix [Ik0 A + uN] for some unique N ∈ Mk0×(n−k0)(Fq) and C is Hermitian self-orthogonal if and only if Ik0 + AAT + u(ANT − NAT ) ≡ 0 (u2). Since A is of full-row rank,

• AMT : M ∈ Mk0×(n−k0)(Fq)
• = Mk0(Fq).

Codes over R with prescribed residue and torsion

Define the map ΦA : Mk0×(n−k0)(Fq) − → Mk0(Fq) N − → ANT − NAT

Codes over R with prescribed residue and torsion

Define the map ΦA : Mk0×(n−k0)(Fq) − → Mk0(Fq) N − → ANT − NAT ΦA(Mk0×(n−k0)(Fq)) =

• ANT − NAT : N ∈ Mk0×(n−k0)(Fq)
• =
• S − ST : S ∈ Mk0(Fq)
• =

Skewk0(Fq)

Codes over R with prescribed residue and torsion

Define the map ΦA : Mk0×(n−k0)(Fq) − → Mk0(Fq) N − → ANT − NAT ΦA(Mk0×(n−k0)(Fq)) =

• ANT − NAT : N ∈ Mk0×(n−k0)(Fq)
• =
• S − ST : S ∈ Mk0(Fq)
• =

Skewk0(Fq) Hence, the number of free Hermitian self-orthogonal codes C with residue code C1 is

• N ∈ Mk0×(n−k0)(Fq)|ANT − NAT ≡ 0 (u)
• =

|ker ΦA| = q

k0(2n−3k0+1) 2

.

Codes over R with prescribed residue and torsion

Define the sets X =

• C | C of type {k0, 0}, C ⊆ C⊥, res(C) = C1
• ,

X′ =

• C′ | C′ of type {k0, k1}, C′ ⊆ C′⊥, res(C′) = C1, tor(C′) = C2
• ,

where self-orthogonality is either in the Euclidean or Hermitian sense.

Codes over R with prescribed residue and torsion

Define the sets X =

• C | C of type {k0, 0}, C ⊆ C⊥, res(C) = C1
• ,

X′ =

• C′ | C′ of type {k0, k1}, C′ ⊆ C′⊥, res(C′) = C1, tor(C′) = C2
• ,

where self-orthogonality is either in the Euclidean or Hermitian sense. Lemma If C ∈ X, there exist a unique code C′ ∈ X′ such that C ⊆ C′.

Codes over R with prescribed residue and torsion

Define the sets X =

• C | C of type {k0, 0}, C ⊆ C⊥, res(C) = C1
• ,

X′ =

• C′ | C′ of type {k0, k1}, C′ ⊆ C′⊥, res(C′) = C1, tor(C′) = C2
• ,

where self-orthogonality is either in the Euclidean or Hermitian sense. Lemma If C ∈ X, there exist a unique code C′ ∈ X′ such that C ⊆ C′. Lemma Let C′ ∈ X′. Then |{C ∈ X|C ⊆ C′}| = qk0k1.

Codes over R with prescribed residue and torsion

Theorem Let C1 and C2 be codes of length n over Fq where C1 ⊆ C2 ⊆ C⊥

1 . If dim C1 = k0 and dim C2 = k0 + k1, then

• i. the number of Euclidean self-orthogonal codes C of length n
• ver Fq + uFq with res(C) = C1 and tor(C) = C2 is

qk0(2n−3k0−2k1+ǫ)/2 where ǫ = −1 if char Fq = 2 and ǫ = 1 if char Fq = 2.

• ii. the number of Hermitian self-orthogonal codes C of length

n over Fq + uFq with res(C) = C1 and tor(C) = C2 is qk0(2n−3k0−2k1+1)/2

C1 ⊆ C2 ⊆ C⊥

1 , dim C1 = k0, dim C2 = k0 + k1

• Proof. WLOG, we may assume that C1 and C2 are codes with

generator matrices G1 =

• Ik0 A
• and G2 =

Ik0 A D

• ,

respectively.

C1 ⊆ C2 ⊆ C⊥

1 , dim C1 = k0, dim C2 = k0 + k1

• Proof. WLOG, we may assume that C1 and C2 are codes with

generator matrices G1 =

• Ik0 A
• and G2 =

Ik0 A D

• ,

respectively. qk0k1 X′

• =
• C′∈X′
• C ∈ X|C ⊆ C′

C1 ⊆ C2 ⊆ C⊥

1 , dim C1 = k0, dim C2 = k0 + k1

• Proof. WLOG, we may assume that C1 and C2 are codes with

generator matrices G1 =

• Ik0 A
• and G2 =

Ik0 A D

• ,

respectively. qk0k1 X′

• =
• C′∈X′
• C ∈ X|C ⊆ C′
• =
• C∈X
• C′ ∈ X′|C ⊆ C′

C1 ⊆ C2 ⊆ C⊥

1 , dim C1 = k0, dim C2 = k0 + k1

• Proof. WLOG, we may assume that C1 and C2 are codes with

generator matrices G1 =

• Ik0 A
• and G2 =

Ik0 A D

• ,

respectively. qk0k1 X′

• =
• C′∈X′
• C ∈ X|C ⊆ C′
• =
• C∈X
• C′ ∈ X′|C ⊆ C′
• =
• C∈X

1

C1 ⊆ C2 ⊆ C⊥

1 , dim C1 = k0, dim C2 = k0 + k1

• Proof. WLOG, we may assume that C1 and C2 are codes with

generator matrices G1 =

• Ik0 A
• and G2 =

Ik0 A D

• ,

respectively. qk0k1 X′

• =
• C′∈X′
• C ∈ X|C ⊆ C′
• =
• C∈X
• C′ ∈ X′|C ⊆ C′
• =
• C∈X

1 = |X| .

Main Result

Remarks.

1 Let σq(n, k) denote the number of distinct self-orthogonal

codes over Fq of length n and dimension k.

Main Result

Remarks.

1 Let σq(n, k) denote the number of distinct self-orthogonal

codes over Fq of length n and dimension k.

2 For k ≤ n, the Gaussian coefficient

n k

• q

is defined as n k

• q

= (qn − 1)(qn − q) · · · (qn − qk−1) (qk − 1)(qk − q) · · · (qk − qk−1) .

Main Result

Theorem The number of distinct Euclidean self-orthogonal codes over Fq + uFq of length n and type {k0, k1} is Mq(n, k0, k1)E = σq(n, k0) n − 2k0 k1

• q

qk0(2n−3k0−2k1+ǫ)/2 where ǫ = −1 if char Fq = 2 and ǫ = 1 if char Fq = 2 and the number of distinct Hermitian self-orthogonal codes over Fq + uFq of length n and type {k0, k1} is Mq(n, k0, k1)H = σq(n, k0) n − 2k0 k1

• q

qk0(2n−3k0−2k1+1)/2.

C ⊆ C⊥ ⇒ res(C) ⊆ tor(C) ⊆ res(C)⊥

• Proof. If C is a self-orthogonal code of length n over Fq + uFq of

type {k0, k1}, then we set C1 = res(C) and C2 = tor(C).

C ⊆ C⊥ ⇒ res(C) ⊆ tor(C) ⊆ res(C)⊥

• Proof. If C is a self-orthogonal code of length n over Fq + uFq of

type {k0, k1}, then we set C1 = res(C) and C2 = tor(C). There are σq(n, k0) self-orthogonal codes C1 of length n over Fq.

C ⊆ C⊥ ⇒ res(C) ⊆ tor(C) ⊆ res(C)⊥

• Proof. If C is a self-orthogonal code of length n over Fq + uFq of

type {k0, k1}, then we set C1 = res(C) and C2 = tor(C). There are σq(n, k0) self-orthogonal codes C1 of length n over Fq. Given C1, there are n − 2k0 k1

• q

codes C2 such that C1 ⊆ C2 ⊆ C⊥

1 .

Then the result follows from the previous theorem.

Main Results

Corollary The number of distinct Euclidean self-dual codes of length n

• ver Fq + uFq is given by
• 0≤k0≤⌊ n

2 ⌋

Mq(n, k0, n − 2k0)E (3) and the number of distinct Hermitian self-dual codes of length n

• ver Fq + uFq is given by
• 0≤k0≤⌊ n

2 ⌋

Mq(n, k0, n − 2k0)H. (4)

Main Results

Corollary The number of distinct Euclidean self-dual codes of length n

• ver Fq + uFq is given by
• 0≤k0≤⌊ n

2 ⌋

Mq(n, k0, n − 2k0)E (3) and the number of distinct Hermitian self-dual codes of length n

• ver Fq + uFq is given by
• 0≤k0≤⌊ n

2 ⌋

Mq(n, k0, n − 2k0)H. (4)

• Proof. Let C be a code over R of type {k0, k1}.

C is self-dual ⇒ C2 = C⊥

1 . Hence, k1 = n − 2k0.

Example

Consider self-orthogonal codes over F2 + uF2, n = 4, k0 = 1, k1 = 2.

Example

Consider self-orthogonal codes over F2 + uF2, n = 4, k0 = 1, k1 = 2. M2(4, 1, 2) = σ2(4, 1) 4 − 2 2

• 2

2(8−3−4+1)/2 = 7(1)(2) = 14.

Example

Consider self-orthogonal codes over F2 + uF2, n = 4, k0 = 1, k1 = 2. M2(4, 1, 2) = σ2(4, 1) 4 − 2 2

• 2

2(8−3−4+1)/2 = 7(1)(2) = 14. Let C1 and C2 be codes over F2 + uF2 with generator matrices   1 1 u u   ,   1 1 1 1 u u u u   , and |Aut(C1)| = 32, |Aut(C2)| = 192.

Example

Consider self-orthogonal codes over F2 + uF2, n = 4, k0 = 1, k1 = 2. M2(4, 1, 2) = σ2(4, 1) 4 − 2 2

• 2

2(8−3−4+1)/2 = 7(1)(2) = 14. Let C1 and C2 be codes over F2 + uF2 with generator matrices   1 1 u u   ,   1 1 1 1 u u u u   , and |Aut(C1)| = 32, |Aut(C2)| = 192. Hence,

2

• j=1

|En| |Aut(Cj)| = 244! 32 + 244! 192 = 12 + 2 = 14 ∴ there are 2 self-orthogonal codes over F2 + uF2 of length 4, type {1, 2}, up to equivalence.

Classification Results for Codes over Fq + uFq

The number of inequivalent Euclidean and Hermitian self-orthogonal codes of lengths 2 ≤ n ≤ 6 over F2 + uF2 and F3 + uF3

Classification Results for Codes over Fq + uFq

The number of inequivalent Euclidean and Hermitian self-orthogonal codes of lengths 2 ≤ n ≤ 6 over F2 + uF2 and F3 + uF3 n F2 + uF2 F3 + uF3 F3 + uF3 Euclidean Hermitian 2 4 3 3 3 10 10 14 4 28 29 46 5 70 77 135 6 206 283 647