Self-dual and LCD double circulant and double negacirculant codes over $\mathbb{F}_q+u\mathbb{F}_q+v\mathbb{F}_q$

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Self-dual and LCD double circulant and double negacirculant codes over F _q + uF _q + vF _q

Shikha Yadav, Habibul Islam, Om Prakash, Patrick Solé

To cite this version:

Shikha Yadav, Habibul Islam, Om Prakash, Patrick Solé. Self-dual and LCD double circulant and dou-

ble negacirculant codes over

_q

_q

_q. Journal of Applied Mathematics and Computing,

Springer, 2021. �hal-03112296�

(will be inserted by the editor)

Self-dual and LCD double circulant and double negacirculant codes over F

+ u F

+ v F

Shikha Yadav · Habibul Islam · Om Prakash · Patrick Sol´e

Received: date / Accepted: date

Abstract Let qbe an odd prime power, and denote by F^q the finite field withq elements. In this paper, we consider the ringR=F^q+uF^q+vF^q, whereu²=u, v²=v, uv=vu= 0 and study double circulant and double negacirculant codes over this ring. We first obtain the necessary and sufficient conditions for a double circulant code to be self-dual (resp. LCD). Then we enumerate self-dual and LCD double circulant and double negacirculant codes overR. Last but not the least, we show that the family of Gray images of self-dual and LCD double circulant codes over R are good. Several numerical examples of self-dual and LCD codes overF⁵as the Gray images of these codes overRare given in short lengths.

Keywords Double circulant code·Self-dual code·LCD code·Gray map Mathematics Subject Classification (2000) 94B05·94B15·94B35·94B60

1 Introduction

Cyclic codes are one of the oldest family of block codes. They have received very intensive attention during the last six decades [6]. In that period, several studies have shown their important uses in and out of mathematics. Many times, they have appeared through their generalized classes [4, 14, 15, 25] and have produced lots of good codes. Along with some other classes, namely, constacyclic, skew cyclic etc., quasi-cyclic codes have led to record-breaking codes [8, 9]. Recall that a linear code is said to be a quasi-cyclic code of indexl, if it is invariant underT^l, whereT denotes the cyclic shift operator. In particular, a quasi-cyclic code of index 1 is indeed a cyclic code. In 2001, Ling and Sol´e [17] presented a new approach to study quasi-cyclic codes over finite fields. They regarded quasi-cyclic codes over a finite fieldF as linear codes of lengthlover the polynomial ringR(F, m) =F[x]/(x^m−1) wherem= ⁿ_l.Essential to that approach was the decomposition ofR(F, m) into local rings via the Chinese Remainder Theorem for polynomials. Later, in 2003, they extended their study to the case whenF is itself a chain ring [18]. In 2016, Guneri et al. [7] have shown that quasi-cyclic codes include families of good LCD codes. Double circulant codes are particular types of quasi-cyclic codes having index 2. In 2018, Alahmadi et al. [2] have shown the self-dual double circulant codes of odd dimension to be dihedral or constadihedral depending upon the characteristic of the field. Meanwhile, self-dual negacirculant codes overfinite fields were studiedin [1]. To generalize the concept over finite rings, recently, Shi et al. [21] considered the finite commutative semi-local non-chain ringF^q+uF^q,u²=u and studied double circulant self-dual or LCD codes. They first enumerated these codes for self-dual and LCD codes, respectively, and later obtained distance bounds on them. A similar work has been reported over a semi-local non-chain ring F^q+uF^q +u²F^q, where u³ = u in [28]. On the other

Shikha Yadav Habibul Islam Om Prakash

Department of Mathematics

Indian Institute of Technology Patna, Patna 801 106, India

E-mail: 1821ma10@iitp.ac.in, habibul.pma17@iitp.ac.in,om@iitp.ac.in Patrick Sol´e

I2M, (CNRS, Aix-Marseille University, Centrale Marseille) Marseilles, France

E-mail: sole@enst.fr

side, double circulant LCD codes overZ⁴ in [22],Zp² in [10] and Galois ring in [23] were studied, respectively. Further, for some related studies on these topics, interested readers can see [20, 24, 26, 27, 29]. Therefore, because of available works on non-chain rings [21, 28], it is logical to investigate these codes over other semi-local non-chain rings. Motivated by the above studies, here we consider the finite commutative semi-local non-chain ringR=Fq+uFq+vFq, whereu²=u, v²=v, uv=vu= 0.

In fact, we first determine the necessary and sufficient conditions (Lemma 1) of double circulant codes to be self-dual and LCD. Then weenumerate self-dual and LCD double circulant codes of length 2nover R when nis odd and double negacirculant codes when nis even, respectively. Finally, by taking Gray images of such codes we show that both families are good in terms of distance bounds (Theorem 2). It is worth mentioning that the ringRhas an important interest and several classes of codes were considered over it [3, 12, 13] in the literature.

The paper isorganized as follows: Section 2 contains some basic definitions and Gray maps. In Section 3, we study the structure of double circulant and double negacirculant codes overR and enumerate self-dual and LCD double circulant and double negacirculant codes. Section 4 provides the distance bounds andestablishes that the families of Gray images of self-dual double circulant codes, and LCD double circulant codes overRare good. In Section 5, we present several non-trivial examples of self-dual and LCD codes overF⁵from the Gray images of these codes overR. Section 6 concludes the paper.

2 Preliminary

Letqbe an odd prime power such that there existsω∈F^q with ω²=−1 (for the existence of such element see [18]). Throughout, we fixR=F^q+uF^q+vF^q,u²=u, v²=v, uv=vu= 0. Now, following [3, 12], we recall thatRis a semi-local non-chain ring with three maximal idealsh1−ui,h1−viand hu+vi. Again, by applying the Chinese Remainder Theorem (CRT) decomposition, we can write R ∼= (1−u−v)Fq+uFq+vFq. Hence, an arbitrary element r ∈ R has a unique representation r= (1−u−v)r1+ur2+vr3, wherer1, r2, r3∈Fq. Further,Rhas (q−1)³units andq³−(q−1)³non- unit elements where the set of units is calculated by the fact thatris a unit inRif and only ifr1, r2, r3

are non-zero inF^q. Now, we define a Gray mapφ1 :R→F³q by φ1(a+ub+vc) = (−b,2a+b, c), for all a, b, c ∈ F^q. In addition, we consider another Gray map φ2 : R → F³q defined in [12] by φ2(a+ub+vc) = (a, a+b, a+c), for alla, b, c∈Fq. It is evident to check that φi is anFq-linear bijective map and can be naturally extended over Rⁿ. In later portion, we show that these Gray maps preserve the orthogonality of a linear code, and hence carry Euclidean LCD and self-dual codes fromRtoF^q.

A linear code C of length n over R is an R-submodule of Rⁿ. The Hamming weight wH(c) of a vector c = (c0, c1, . . . , cn−1) ∈ Rⁿ is the number of non-zero coordinates while the minimum Hamming distance of the codeC is

dH(C) = min{wH(c) : 06=c∈C}.

Now, we define the Lee weightwL(c) of a vectorc= (c0, c1, . . . , cn−1)∈Rⁿ aswL(c) =wH(φi(c)) while the minimum Lee distance ofC is given by

dL(C) = min{w_L(c) : 06=c∈C}.

Therefore, it is checked thatφi is a linear isometric map from (Rⁿ, dL) to (F³ⁿq , dH) for i = 1,2.

For any two elementss= (s0, s1, . . . , sn−1) and t= (t0, t1, . . . , tn−1) in Rⁿ, theirEuclidean (resp.

Hermitian)inner product is defined by

s·t=

n−1

X

i=0

siti

and

hs, ti_H=

n−1

X

i=0

sit¯i,

respectively, where forx+uy+vz∈Rits conjugate is defined byx+uy+vz=x

√q+uy

√q+vz

√q. In this way, the Euclidean (resp. Hermitian) dual of a linear codeC is denoted byC^⊥ (resp.C^⊥H) and defined by

C^⊥={a∈Rⁿ:a·c= 0 for allcinC}

and

C^⊥H ={a∈Rⁿ:ha, ciH= 0 for allcinC},

respectively. A linear codeC is said to be a Euclidean (resp. Hermitian)LCD code if and only if C∩C^⊥={0} (resp.C∩C^⊥H ={0}). Also,C is said to be aEuclidean(resp.Hermitian)self-dual code ifC =C^⊥ (resp.C=C^⊥H). Now, the next result shows thatφi preserve the orthogonality of a linear code.

Theorem 1 LetC be a linear code overR. ThenC is a Euclidean LCD (resp. self-dual) code if and only ifφi(C)is a Euclidean LCD (resp. self-dual) code overF^q fori= 1,2.

Proof The proof depends on the main factφi(C^⊥) = (φi(C))^⊥, which can be verified by using the similar procedure of [[21], Theorem 2.2].

For solving the nonlinear equations in Theorem 3, we use the concept ofnorm functions which is defined asNorm:F^qn →F^q given by

N orm(x) =x

qn−1

q−1 , for x∈F^qn.

ThenN ormis a multiplicative surjective function andN orm(0) = 0. Further, each element inF^∗q is a norm of exactly ^q_q−1ⁿ⁻¹ elements in F^∗qⁿ (see [16], Theorem 2.28). Now, we recall that a linear code is said to beadouble circulant (resp.negacirculant)code, if its generator matrix is of the form

G= (I, A)

whereAis a circulant (resp. negacirculant) matrix, i.e., the matrix whose rows can be obtained by successive circular shifts (resp. negashifts) of the first row. LetC<n> be a family of codes having parameters [n, kn, dn] overF^q. Then therate ρandrelative distance δ are defined asρ= lim sup

n→∞

kn n

and δ = lim sup

n→∞

dn

n. This family is said to be good , if ρδ 6= 0. To derive the main result which proposes that the Gray images of a subfamily of double circulant (LCD or self dual) codes overR are good(Theorem 2), we will use the entropy function [11] defined by

Hq(x) =

(0, ifx= 0

xlog_q(q−1)−xlog_q(x)−(1−x) log_q(1−x), if 0< x≤1−¹_q . Now, we state one of the main result of this paper, and prove it at the end of Section 4.

Theorem 2 Let q be an odd prime power, and δ > 0 be given. Then there are families of double circulant self-dual (resp. LCD) codes of length 2n over R, with code rate ¹₂, and with Gray images of relative distanceδ as long as Hq(δ)< ₁₂¹ (resp.Hq(δ)< ¹₆). Moreover, we conclude that both of these families of codes are good.

Remark:This result shows that for all >0 arbitrarily small there are families of the said types of relative distanceδ0−,with Hq(δ0) = ₁₂¹ (resp.Hq(δ0) = ¹₆). Unfortunately, the method does not allow us to make= 0.

3 Double circulant and double negacirculant codes

In this section, we enumerate self-dual and LCD double circulant codes overR. For this, we first obtain the necessary and sufficient conditions of double circulant codes to be self-dual or LCD. Further, the enumeration of self-dual and LCD double negacirculant codes is provided.

3.1 Enumeration of double circulant codes whennis odd

We assume thatnis an odd positiveinteger and the factorization ofxⁿ−1 into distinct irreducible polynomials overRis as follows:

xⁿ−1 =a(x−1)

s

Y

i=2

gi(x)

t

Y

j=1

hj(x)h^∗_j(x), where

– a∈R^∗ whereR^∗ denotes the set of all units inR,

– gi(x) (2≤i≤s) are self-reciprocal polynomials of even degree 2ei,respectively, – h^∗_j(x) (1≤j≤t) are reciprocal polynomials ofhj(x) with degreedj,respectively.

By the Chinese Remainder Theorem (CRT), we have

R[x]

hxⁿ−1i ∼= R[x]

hx−1i⊕

⊕^s_i=2 R[x]

hg_i(x)i

⊕ ⊕^t_j=1

R[x]

hh_j(x)i

⊕

R[x]

hh^∗_j(x)i

!

∼=R⊕(⊕^s_i=2R2ei)⊕ ⊕^t_j=1Rdj⊕Rdj

,

whereRr=F^qr+uF^qr+vF^qr,u²=u, v²=v, uv=vu= 0 forr= 2eiordj. The above decomposition can be naturally extended as

R[x]

hxⁿ−1i 2

∼=R²⊕

⊕^s_i=2(R2ei)²

⊕

⊕^t_j=1(Rdj)²⊕(Rdj)² .

Now, using this decomposition,any linear codeC of length 2 over _hx^R[x]n−1i can be decomposed as C∼=C1⊕(⊕^s_i=2Ci)⊕

⊕^t_j=1(C_j⁰⊕C_j⁰⁰)

(1) whereC1 is a linear code overR,Ci is a linear code overR2ei, for 2≤i≤sandC_j⁰, C_j⁰⁰ are linear codes overRdj, for 1≤j≤t.

Lemma 1 Let C be a double circulant code over R given in the CRT decomposition (1) and α1 = (1, ce1), αi= (1, cei), α⁰j= (1, c⁰_dj), α⁰⁰j = (1, c⁰⁰_dj)be generators of the constituent codesC1, Ci, Cj⁰, Cj⁰⁰

overR, R2ei,Rdj andRdj, respectively, for2≤i≤s, 1≤j≤t. Then

(1) C is a self-dual code if and only if1 +c²e1 = 0, 1 +c^1+qei ^ei = 0and1 +c⁰_djc⁰⁰_dj = 0.

(2) C is a Euclidean LCD code if and only if1 +c²e1 ∈R^∗,1 +c^1+qei ^ei ∈R^∗2ei and 1 +c⁰_djc⁰⁰_dj ∈R^∗_dj. Proof Let C be a double circulant code overR given by the CRT decomposition (1). By following the same procedure of[[18], Theorem 4.2], we get thatC is a self-dual code if and only ifC1, Ci are self-dual codes with respect to the Euclidean, Hermitian inner product, respectively for 2≤ i ≤ s andC_j⁰⁰ is the dual code ofC_j⁰ with respect to Euclidean inner product, for 1≤j≤t. In our case, it further implies thatCis a self-dual code if and only if 1 +c²e1= 0, 1 +c^1+qei ^ei = 0 and 1 +c⁰_djc⁰⁰_dj = 0.

Now, following the same method of [[7], Theorem 3.1], we get C is an LCD code if and only ifC1, Ci are LCD codes with respect to the Euclidean, Hermitian inner product, respectively for 2≤i ≤sand (C_j⁰⁰)^⊥∩C_j⁰ = {0}, C_j⁰⁰ ∩(C_j⁰)^⊥ ={0}, for 1≤ j≤ t. In our case, it further implies thatC is a Euclidean LCD code if and only if 1 +c²_e1 ∈R^∗, 1 +c^1+q_ei ^ei ∈R^∗_2ei and 1 +c⁰_djc⁰⁰_dj ∈R^∗_dj. Using the necessary and sufficient conditions given in Lemma 1, we now find the total number of self-dual or LCD double circulant codes overRin the following results.

Theorem 3 Assume that for an odd integern, the factorization ofxⁿ−1overR is xⁿ−1 =a(x−1)

s

Y

i=2

gi(x)

t

Y

j=1

hj(x)h^∗j(x),

wherea∈R^∗ and n= 1 +Σi=2^s 2ei+ 2Σj=1^t dj. Then the total number of self-dual double circulant codes overRis

8

s

Y

i=2

(q^ei+ 1)³

t

Y

j=1

(q^dj −1)³.

Proof We can obtain the total number of self-dual double circulant codes by just counting the con- stituent codes. There are 8 choices forC1which have the generator polynomials as (1, ω),(1,−ω),(1, ω(1−

2v)),(1, ω(2v−1)),(1, ω(1−2u)),(1, ω(2u−1)),(1, ω(1−2u−2v)) and (1, ω(−1+2u+2v)), respectively, whereω²=−1.

For the second constituent codes, to count self-dual codes with respect to the Hermitian inner product, we need to find the number of solutions of the equation 1 +ceic^qe^eii = 0. Letcei =x(1−u− v) +yu+zv, for somex, y, z∈Fq^2ei. Then

1 + (x(1−u−v) +yu+zv)(x(1−u−v) +yu+zv)^qei = 0

if and only if

xx^qei =−1, yy^qei =xx^qei =−1 andzz^qei =xx^qei =−1

i.e., N orm(x) = −1, N orm(y) = −1 and N orm(z) = −1. There are q^ei+ 1 solutions for each N orm(x) =−1, N orm(y) =−1 and N orm(z) =−1, respectively. Therefore, the total number of solutions for the above system is (q^ei+ 1)³.

Now, we count the dual pairs (w.r.t. Euclidean inner product) of codes. For this, we need to find the number of solutions of the equation 1 +c⁰_djc⁰⁰_dj = 0. We have the following possibilities:

– Ifc⁰_dj ∈R^∗_dj,thenc⁰⁰_dj =− ¹

c⁰ dj

and we have|R^∗_dj|= (q^dj −1)³ choices for the pair{c⁰_dj, c⁰⁰_dj}. – Ifc⁰_dj ∈Rdj \R^∗_dj, then c⁰_dj =x(1−u−v) +yu+zv, for somex, y, z∈F_qdj and at least one of

x, yorzis 0. Forc⁰⁰_dj =β1(1−u−v) +β2u+β3v∈Rdj, whereβ1, β2, β3∈Fq^dj, we have 1 +c⁰djc⁰⁰dj = (1 +xβ1) + (yβ2−xβ1)u+ (zβ3−xβ1)v= 0,

which implies thatxβ1=−1, yβ2=−1, zβ3=−1. If any one ofx, yorzis 0, then we get−1 = 0, a contradiction. Therefore, these cases doesn’t occur.

Hence, we have (q^dj −1)³ choices for the dual pairs. Combining above all cases we get the desired result.

With the same notations used in the above theorem, we now count the total number of LCD double circulant codes overR.

Theorem 4 The total number of LCD double circulant codes overRis

(q−2)³

s

Y

i=2

(q^2ei−q^ei−1)³

t

Y

j=1

(q^6dj−3q^5dj + 6q^4dj−7q^3dj+ 6q^2dj −3q^dj+ 1).

Proof We can obtain the total number of LCD double circulant codes by just counting the constituent codes as done for the self-dual codes. IfC1is an LCD code with respect to the Euclidean inner product, then 1 +c²_e1∈R^∗ and we have the following possibilities:

– Ifc1= 0, then 1 +c²₁= 1∈R^∗.

– If 06=c1∈ h1−u−viandc1=x(1−u−v), for somex∈F^∗q, then 1+c²1= 1+x²(1−u−v)∈R^∗if and only ifx6=±ω, whereω²=−1. Therefore, we haveq−3choices. Similarly, when 06=c1∈ hui orhvi, we haveq−3 choices for each.

– If 0 6= c1 ∈ h1−u−v, ui and c1 = x(1−u−v) +yu, for some x, y ∈ F^∗q, then 1 +c²₁ = 1 +x²(1−u−v) +y²u∈R^∗ if and only ifx6=±ω andy6=±ω, whereω²=−1. Therefore, we have (q−3)² choices. Similarly, when 06=c1∈ hu, viorh1−u−v, vi, we have (q−3)²choices for each.

– If 0 6= c1 ∈ h1−u−v, u, vi and c1 = x(1−u−v) +yu+zv, for some x, y, z ∈ F^∗q, then 1 +c²₁= 1 +x²(1−u−v) +y²u+z²v∈R^∗ if and only ifx, y, z6=±ω, whereω²=−1. Therefore, we have (q−3)³choices.

So, we have 1 + 3(q−3) + 3(q−3)²+ (q−3)³= (q−2)³choices forC1.

Now, we find the choices for Cei such that it is an LCD code over R2ei with respect to the Hermitian inner product. The linear codeCei is Hermitian LCD if 1 +c^1+q_ei ^ei ∈R^∗_2ei and we have the following possibilities:

– Ifcei= 0, then 1 +c^1+q_ei ^ei = 1∈R^∗_2ei.

– If 06=cei∈ h1−u−viandcei=x(1−u−v), for somex∈F^∗q²ei, then 1 +c^1+q_ei ^ei = 1 +x^1+qei(1− u−v) ∈R^∗_2ei if and only if x^1+qei 6= −1. Therefore, we haveq^2ei−q^ei−2 choices. Similarly, when 06=cei∈ huiorhvi, we haveq^2ei−q^ei−2 choices for each.

– If 06=cei ∈ h1−u−v, uiandcei =x(1−u−v) +yu, for somex, y ∈F^∗q²ei, then 1 +c^1+q_ei ^ei = 1 +x^1+qei(1−u−v) +y^1+qeiu∈R^∗_2ei if and only ifx^1+qei 6=−1 andy^1+qei 6=−1. Therefore, we have (q^2ei−q^ei−2)² choices. Similarly, when 0 6= cei ∈ hu, vi or h1−u−v, vi, we have (q^2ei−q^ei−2)²choices for each.

– If 0 6= cei ∈ h1−u−v, u, viand cei = x(1−u−v) +yu+zv, for some x, y, z ∈F^∗q²ei, then 1+c^1+qei ^ei = 1+x^1+qei(1−u−v)+y^1+qeiu+z^1+qeiv∈R^∗2eiif and only ifx^1+qei 6=−1, y^1+qei 6=−1 andz^1+qei 6=−1. Therefore, we have (q^2ei−q^ei−2)³ choices.

So, in this case, we have 1 + 3(q^2ei−q^ei−2) + 3(q^2ei−q^ei−2)²+ (q^2ei−q^ei−2)³= (q^2ei−q^ei−1)³ choices forCi, where 2≤i≤s.

Now, for the last case we need to find choices for the pairs{c⁰_d

j, c⁰⁰_dj}such that 1 +c⁰_djc⁰⁰_dj ∈R^∗_dj and we have the following possibilities:

– Ifc⁰_dj = 0, then 1 +c⁰_djc⁰⁰_dj ∈R^∗_dj for anyc⁰⁰_dj ∈R^∗_dj. So, we haveq^3dj choices forc⁰⁰_dj. – Ifc⁰_dj ∈R^∗_dj then c⁰⁰_dj ∈R^∗_dj − ¹

c⁰_dj and|R^∗_d

j − ¹

c⁰_dj|=|R^∗_d

j|. We have|R^∗_d

j|² = (q^dj −1)⁶ choices for the pairs{c⁰_dj, c⁰⁰_dj}.

– If 0 6= c⁰_dj ∈ h1−u −vi and c⁰_dj = x(1−u−v), for some x ∈ F^∗_qdj. Assume that c⁰⁰_dj = β1(1−u−v) +β2u+β3v, for someβ1, β2, β3∈F_qdj. Then 1 +c⁰_djc⁰⁰_dj = 1 +xβ1(1−u−v)∈R^∗_dj if and only ifxβ16=−1. Therefore, we have (q^dj−1)²q^2dj choices. Similarly, when 06=cei ∈ hui orhvi, we have (q^dj−1)²q^2dj choices for each.

– If 06=c⁰_dj ∈ h1−u−v, uiandc⁰_dj =x(1−u−v) +yu, for somex, y∈F^∗_qdj. Assume thatc⁰⁰_dj = β1(1−u−v)+β2u+β3v, for someβ1, β2, β3∈F_qdj Then 1+c⁰_djc⁰⁰_dj = 1+xβ1(1−u−v)+yβ2u∈R^∗_dj if and only ifxβ16=−1 andyβ26=−1. Therefore, we have (q^dj−1)⁴q^dj choices. Similarly, when 06=c⁰_dj ∈ hu, viorh1−u−v, vi, we have (q^dj −1)⁴q^dj choices for each.

In this case, we haveq^3dj+ (q^dj −1)⁶+ 3(q^dj −1)²q^2dj+ 3(q^dj−1)⁴q^dj = (q^6dj −3q^5dj + 6q^4dj − 7q^3dj + 6q^2dj −3q^dj + 1) choices for the pairs{C_j⁰, C_j⁰⁰}, where 1≤ j ≤t. Now, summing all these above choices we get the required result.

3.2 Enumeration of double negacirculant codes whennis even

The present subsection deals with the enumeration of self-dual or LCD double negacirculant codes overR. Here, we take nto be an even positive integer such thatgcd(n, q) = 1. We assume that the factorization ofxⁿ+ 1 into distinct irreducible polynomials overRis as follows:

xⁿ+ 1 =a

s

Y

i=1

gi(x)

t

Y

j=1

hj(x)h^∗_j(x),

wherea∈R^∗,gi(x) (1 ≤ i≤s) are self-reciprocal polynomials of even degree 2ei andh^∗_j(x) (1≤ j≤t) are reciprocal polynomials ofhj(x) with degreedj, respectively. Using the arguments similar to double circulant codes, we get

R[x]

hxⁿ+ 1i

∼= (⊕^s_i=1R2ei)⊕ ⊕^t_j=1Rdj⊕Rdj

,

whereRr=Fq^r+uFq^r+vFq^r,u²=u, v²=v, uv=vu= 0 forr= 2ei ordj. Also, any linear code C of length 2 can be written as

C∼= (⊕^s_i=1Ci)⊕

⊕^t_j=1(Cj⁰ ⊕Cj⁰⁰)

, (2)

whereCiis a linear code overR2ei, for 1≤i≤sandCj⁰, Cj⁰⁰ are linear codes overRdj, for 1≤j≤t.

To enumerate the self-dual and LCD double negacirculant codes, we need the following result which can be proved using the same procedure of Lemma 1.

Lemma 2 LetC be a double negacirculant code over Randαi= (1, cei), α⁰_j= (1, c⁰_dj), α⁰⁰_j = (1, c⁰⁰_dj) be generators of the constituent codesCi, Cj⁰, Cj⁰⁰ overR2ei,Rdj andRdj, respectively, for1≤i≤s, 1≤j≤t. Then

(1) C is a self-dual code if and only if1 +c^1+q_ei ^ei = 0and1 +c⁰_djc⁰⁰_dj = 0.

(2) C is a Euclidean LCD code if and only if1 +c^1+qei ^ei ∈R_2e^∗_i and1 +c⁰_djc⁰⁰_dj ∈R^∗_dj.

Using this lemma, we now enumerate self-dual and LCD double negacirculant codes overR.

Theorem 5 Assume that for an even integern, the factorization ofxⁿ+ 1overRis

xⁿ+ 1 =a

s

Y

i=1

gi(x)

t

Y

j=1

hj(x)h^∗j(x),

where a∈ R^∗ and n =

s

P

i=1

2ei+ 2

t

P

j=1

dj. The total number of self-dual double negacirculant codes overRis

s

Y

i=1

(q^ei+ 1)³

t

Y

j=1

(q^dj −1)³.

Proof We enumerate self-dual double negacirculant codes by counting the constituent codes. To count the choices forCi, we need to find the number of solutions of the equation 1 +ceic^q_e^ei_i = 0. Let cei=x(1−u−v) +yu+zv, for somex, y, z∈Fq²ei. Then

1 + (x(1−u−v) +yu+zv)(x(1−u−v) +yu+zv)^qei = 0 if and only if

xx^qei =−1, yy^qei =xx^qei =−1 andzz^qei =xx^qei =−1

i.e., N orm(x) = −1, N orm(y) = −1 and N orm(z) = −1. There are q^ei+ 1 solutions for each N orm(x) =−1, N orm(y) =−1 and N orm(z) =−1, respectively. Therefore, the total number of solutions for the above system is (q^ei+ 1)³.

Now, to count the choices for the dual pairs{C_j⁰, C_j⁰⁰}, we need to find the number of solutions of the equation 1 +c⁰_djc⁰⁰_dj = 0. There are the following possibilities:

– Ifc⁰_dj ∈R^∗_dj,then c⁰⁰_dj =− ¹

c⁰ dj

, i.e., a unique choice forc⁰⁰_dj corresponding to eachc⁰_dj.Therefore, there are|R^∗_d

j|= (q^dj−1)³choices forc⁰_dj and hence for the pair{c⁰_d

j, c⁰⁰_dj}.

– Ifc⁰_dj ∈Rdj \R_d^∗_j, thenc⁰_dj =x(1−u−v) +yu+zv, for somex, y, z∈F_qdj, where not all x, y orzare non-zero. Forc⁰⁰_dj =β1(1−u−v) +β2u+β3v∈Rdj, whereβ1, β2, β3∈Fq^dj, we have

1 +c⁰djc⁰⁰dj = (1 +xβ1) + (yβ2−xβ1)u+ (zβ3−xβ1)v= 0,

which implies thatxβ1=−1, yβ2=−1, zβ3=−1. Any one ofx, yorzequal to 0, yields−1 = 0, a contradiction. Therefore, these cases are not possible.

From all the above cases, we conclude that there areQs

i=1(q^ei+ 1)³Qt

j=1(q^dj−1)³self-dual double negacirculant codes overR.

With the same assumptions and notations of Theorem 5, the following result provides the enu- meration of LCD double negacirculant codes overR.

Theorem 6 The total number of LCD double negacirculant codes overRis

s

Y

i=1

(q^2ei−q^ei−1)³

t

Y

j=1

(q^6dj −3q^5dj+ 6q^4dj−7q^3dj + 6q^2dj−3q^dj+ 1).

Proof The total number of LCD double negacirculant codes over R can be obtained by counting the constituent codes. To count the choices forCi, we need to find the number of solutions for the equation 1 +c^1+q_ei ^ei ∈R^∗_2ei. The following cases arise:

– Ifcei= 0, then clearly 1 +c^1+qei ^ei ∈R^∗2ei.

– Letcei∈ h1−u−viandcei=x(1−u−v),for somex∈F^∗q^2ei. Then 1 +c^1+qei ^ei = 1 +x^1+qei(1− u−v)∈R^∗2ei if and only ifx^1+qei 6=−1. Therefore, we haveq^2ei−q^ei−2 choices forxand hence forcei. Similarly, there areq^2ei−q^ei−2 choices for each case 06=cei∈ huiorhvi.

– Let cei ∈ h1−u−v, ui andcei =x(1−u−v) +yu, for somex, y ∈F^∗q²ei. Then 1 +c^1+qei ^ei = 1 +x^1+qei(1−u−v) +y^1+qeiu∈R^∗2ei if and only ifx^1+qei 6=−1 andy^1+qei 6=−1. Therefore, we haveq^2ei−q^ei−2 choices for eachx, y and hence (q^2ei−q^ei−2)² choices forcei. Similarly, there are (q^2ei−q^ei−2)² choices for 06=cei ∈ hu, viorh1−u−v, vieach.

– Let cei ∈ h1−u−v, u, vi and cei = x(1−u−v) +yu+zv, for some x, y, z ∈ F^∗q²ei. Then 1+c^1+q_ei ^ei = 1+x^1+qei(1−u−v)+y^1+qeiu+z^1+qeiv∈R^∗_2eiif and only ifx^1+qei 6=−1, y^1+qei 6=−1 andz^1+qei 6=−1. Therefore, we haveq^2ei−q^ei−2 choices for eachx, y, zand hence (q^2ei−q^ei−2)³ choices forcei.

Hence, there are 1 + 3(q^2ei−q^ei−2) + 3(q^2ei−q^ei−2)²+ (q^2ei−q^ei−2)³= (q^2ei−q^ei−1)³choices forCi, where 1≤i≤s.

Now, to count the choices for the pairs {C_j⁰, C_j⁰⁰}, we need to find the number of solution pairs {c⁰_dj, c⁰⁰_dj}for the equation 1 +c⁰_djc⁰⁰_dj ∈R^∗_dj. Here the following cases arise:

– Ifcdj = 0, then clearly 1 +c⁰_djc⁰⁰_dj ∈R_d^∗_j for allc^”_dj ∈R^∗_dj. Thus, there areq^3dj choices forc^”_dj. – Ifc⁰_dj ∈R^∗_dj,thenc⁰⁰_dj ∈R^∗_dj− ¹

c⁰_dj and we have|R^∗_d

j − ¹

c⁰_dj|=|R^∗_d

j|choices forc⁰⁰_dj corresponding to eachc⁰_dj. Therefore, there are|R^∗_dj|²= (q^dj −1)⁶ choices for the pair{c⁰_dj, c⁰⁰_dj}.

– If c⁰_dj ∈ h1−u−vi \ {0}, then c⁰_dj = x(1−u−v), for some x ∈ F^∗_qdj. Assume that c⁰⁰_dj = β1(1−u−v) +β2u+β3v, for someβ1, β2, β3∈F_qdj. Then 1 +c⁰_djc⁰⁰_dj = 1 +xβ1(1−u−v)∈R^∗_dj if and only ifxβ16=−1. Oncec⁰_dj is fixed, there areq^dj−1 choices forβ1andq^dj choices for each β2, β3. Therefore, there are (q^dj−1)²q^2dj choices for the pair{c⁰_d

j, c⁰⁰_dj}. Similarly, if 06=cei ∈ hui orhvi, there are (q^dj −1)²q^2dj choices for each case.

– If c⁰_dj ∈ h1−u−v, uiand c⁰_dj = x(1−u−v) +yu, for some x, y ∈F^∗_qdj. Assume that c⁰⁰_dj = β1(1−u−v)+β2u+β3v, for someβ1, β2, β3∈Fq^dj Then 1+c⁰_djc⁰⁰_dj = 1+xβ1(1−u−v)+yβ2u∈R^∗_dj if and only ifxβ16=−1 andyβ26=−1. Once c⁰_dj is fixed, there areq^dj−1 choices for eachβ1, β2

andq^dj choices forβ3. Therefore, there are (q^dj−1)⁴q^dj choices for the pair{c⁰_dj, c⁰⁰_dj}. Similarly, if 06=c⁰_dj ∈ hu, viorh1−u−v, vi, we have (q^dj−1)⁴q^dj choices for each pair{c⁰_dj, c⁰⁰_dj}.

Summing all these cases, we get that there areq^3dj+ (q^dj−1)⁶+ 3(q^dj−1)²q^2dj+ 3(q^dj−1)⁴q^dj = (q^6dj−3q^5dj+ 6q^4dj−7q^3dj+ 6q^2dj−3q^dj+ 1) choices for the pairs{C_j⁰, Cj^”}, where 1≤j≤t. Now, combining all the above choices forCi and the pairs {C_j⁰, C_j⁰⁰}, we get the desired result.

4Distance boundsfor double circulant codes

In this section, we provide distance bounds for self-dual or LCD double circulant codes and show that the families of Gray images of self-dual or LCD double circulant codes are good (which is one of the main results of this paper).

Letnbe an odd prime andqbe a primitive rootmodulo n. We assume that the factorization of xⁿ−1 into distinct irreducible polynomials overRis

xⁿ−1 = (x−1)(1 +x+· · ·+xⁿ⁻¹) = (x−1)h(x), (3) where h(x) = 1 +x+· · ·+xⁿ⁻¹ is an irreducible polynomial over R. By the Chinese Remainder Theorem(CRT), we have

R[x]

hxⁿ−1i ∼= R[x]

hx−1i⊕ R[x]

hh(x)i

∼=R⊕R1,

whereR1=Fqⁿ⁻¹+uFqⁿ⁻¹+vFqⁿ⁻¹,u²=u, v²=v, uv=vu= 0. We denoteR= _hh(x)i^R[x] .

Definition 1 LetC be a cyclic code overRof odd length nandh(x) be a polynomial given in the above discussion. Then a non-zero codeword inC is said to be a constant vector if it is generated by h(x).

Now, we provide two lemmas which will be used to prove the main result.

Lemma 3 Letz= (e, f)∈R²ⁿbe a non-zero vector such thateis not a constant vector. Then there are at mostq²ⁿ⁺¹ double circulant codes Ca= (1, a)overR such thatz∈Ca, wherea∈ R.

Proof The vector z can be written as z = (e, f) ∼= (e1, f1)⊕(e2, f2) by the CRT decomposition.

Asz∈Ca, we have f =ea, f1 = e1a1 andf2 = e2a2, wheree1, f1, a1 ∈R ande2, f2, a2 ∈ R. Let a1 = r1(1−u−v) +us1+vt1 and a2 = r2(1−u−v) +us2+vt2, for some r1, s1, t1 ∈ F^q and r2, s2, t2∈Fqⁿ⁻¹. Now, we discuss the first constituent of the code Cathroughe1.

– Ife1= 0, then we haveq³ choices fora1.

– If 06=e1∈ h1−u−vi, thene1= (1−u−v)x1andf1= (1−u−v)x⁰₁, for somex1∈F^∗q, x⁰₁∈Fq. Now,

f1= (1−u−v)x⁰₁= (1−u−v)x1a1= (1−u−v)x1r1. This implies thatr1= ^x

0 1

x1 and we haveq²choices fora1. Similarly, when 06=e1∈ huiorhvi, we haveq²choices for each case.

– If 0 6= e1 ∈ h1−u−v, ui and e1 = x1(1−u−v) +y1u, for some x1, y1 ∈ F^∗q, then f1 = x⁰₁(1−u−v) +y₁⁰u, for somex⁰₁, y₁⁰ ∈F^q. Now,

f1= (1−u−v)x⁰1+uy⁰1= ((1−u−v)x1+uy1)a1= (1−u−v)x1r1+uy1s1. This implies thatr1= ^x

0 1 x1, s1 = ^y

0 1

y1 and we have qchoices fora1. Similarly, when 06=c1∈ hu, vi orh1−u−v, vi, we haveqchoices for each case.

– Ife1∈R^∗, then we have a unique choice fora1= ^f_e¹

1. Therefore, for each case we have at mostq³choices fora1.

For the second constituent code ofCa, we discuss choices fora2 throughe2.

– If e2 = 0, then e is a constant vector, i.e.,e ≡ 0 (modh(x)), which is a contradiction to the choice ofe.

– If 06=e2∈ h1−u−vi, thene2= (1−u−v)x2andf2= (1−u−v)x⁰2, for somex2∈F^∗qⁿ⁻¹, x⁰2∈ Fqⁿ⁻¹. Now,

f2= (1−u−v)x⁰2= (1−u−v)x2a2= (1−u−v)x2r2. This implies thatr2= ^x

0 2

x2 and we haveq²ⁿ⁻² choices fora2. Similarly, when 06=e2∈ huiorhvi, we haveq²ⁿ⁻² choices for each case.

– If 0 6= e2 ∈ h1−u−v, ui and e2 = x2(1−u−v) +y2u, for some x2, y2 ∈ F^∗qⁿ⁻¹, then f2 = x⁰2(1−u−v) +y2⁰u, for somex⁰2, y2⁰ ∈Fqⁿ⁻¹. Now,

f2= (1−u−v)x⁰2+uy⁰2= ((1−u−v)x2+uy2)a2= (1−u−v)x2r2+uy2s2. This implies thatr2= ^x

0 2 x2, s2= ^y

0 2

y2 and we haveqⁿ⁻¹choices fora2. Similarly, when 06=c2∈ hu, vi orh1−u−v, vi, we haveqⁿ⁻¹choices for each case.

– Ife2∈ R^∗, then we have a unique choice fora2= ^f_e²

2.

From the above cases, we conclude that the number of choices fora2 is at most q²ⁿ⁻². Therefore, there are at mostq²ⁿ⁺¹ choices forasuch thatz∈Ca.

Keeping the same notations, we have the following result.

Lemma 4 Letz= (e, f)∈R²ⁿbe a non-zero vector such thateis not a constant vector. Then there are at most8(1 +qⁿ⁻¹² )² self-dual codesCa= (1, a)such thatz∈Ca, wherea∈ R.

Proof Intially, we discuss the first constituent of the codeCa. From Theorem 3, there are at most 8 choices forC1, a self-dual double circulant code overR.

For the second constituent code of Ca, we discuss choices fora2 through e2. Let a2 = (1−u− v)r2+us2+vt2, for somer2, s2, t2∈Fqⁿ⁻¹.

– If e2 = 0, then e is a constant vector, i.e.,e ≡ 0 (modh(x)), which is a contradiction to the choice ofe.

– If 06=e2∈ h1−u−vi, thene2= (1−u−v)x2andf2= (1−u−v)x⁰₂, for somex2∈F^∗qⁿ⁻¹, x⁰₂∈ Fqⁿ⁻¹. Now,

f2= (1−u−v)x⁰2= (1−u−v)x2a2= (1−u−v)x2r2. This implies thatr2= ^x

0 2

x2. Further, asCais self-dual, 1 +a2¯a2= 1 +a2a^q

n−1 2

2 = 0 which implies thatr2r^q

n−1 2

2 =−1, s2s^q

n−1 2

2 =−1 andt2t^q

n−1 2

2 =−1, i.e.,N orm(r2) =−1, N orm(s2) =−1 and N orm(t2) =−1. Therefore, we have (1 +qⁿ⁻¹² )² choices for a2. Similarly, when 06=e2∈ huior hvi, we have (1 +qⁿ⁻¹² )²choices for each case.

– If 0 6= e2 ∈ h1−u−v, ui and e2 = x2(1−u−v) +y2u, for some x2, y2 ∈ F^∗qⁿ⁻¹, then f2 = x⁰2(1−u−v) +y2⁰u, for somex⁰2, y2⁰ ∈Fqⁿ⁻¹. Now,

f2= (1−u−v)x⁰2+uy⁰2= ((1−u−v)x2+uy2)a2= (1−u−v)x2r2+uy2s2. This implies thatr2= ^x

0 2 x2, s2= ^y

0 2

y2. Further, asCais self-dual, 1 +a2¯a2= 1 +a2a^q

n−1 2

2 = 0, which implies thatr2r^q

n−1 2

2 =−1, s₂s^q

n−1 2

2 =−1 andt2t^q

n−1 2

2 =−1, i.e.,N orm(r2) =−1, N orm(s₂) =

−1 and N orm(t2) =−1. Therefore, we have at most (1 +qⁿ⁻¹² ) choices for a2. Similarly, when 06=c2∈ hu, viorh1−u−v, vi, we have at most (1 +qⁿ⁻¹² ) choices for each case.