Construction of Hermitian Self-Orthogonal Codes and Application

Abstract

We introduce some methods for constructing quaternary Hermitian self-orthogonal (HSO) codes, and construct quaternary $[n, 5]$ HSO for $342\le n\le 492$. Furthermore, we present methods of constructing Hermitian linear complementary dual (HLCD) codes from known HSO codes, and obtain many HLCD codes with good parameters. As an application, 31 classes of entanglement-assisted quantum error correction codes (EAQECCs) with maximal entanglement can be obtained from these HLCD codes. These new EAQECCs have better parameters than those in the literature.

1. Introduction

Let $F_4=\left\{0, 1, \omega , {\omega }^2\right\}$ be the field with four elements, where ${\omega }^2=1+\omega$, and $F_4^n$ be the n-dimensional row space over $F_4$. For x$\in F_4$, its conjugate is $\overline{x}=x^2$. A quaternary $[n, k]$ code $\mathcal{C}$ is a k-dimensional subspace of $F_4^n$, vectors in $\mathcal{C}$ are called as codewords of $\mathcal{C}$. If the minimal Hamming weights of non-zero codewords in $\mathcal{C}$ is d, then $\mathcal{C}$ is denoted as $\mathcal{C}$$=[n, k, d]$. A linear code $[n, k, d]$ is optimal if there is no $[n, k, d+1]$ code; such a code is denoted as $\mathcal{C}$$=[n, k, d_{op}(n, k)]$. For $u=(u_1, u_2\cdots u_n)$, $v=(v_1, v_2\cdots v_n)\in F_4^n$, their Hermitian inner product is ${(u, v)}_h={\sum }_{i=1}^n{u}_i·\overline{v_i}={\sum }_{i=1}^n{u}_i·v_i^2$. The Hermitian dual code $\mathcal{C}$${\mathcal{C}}^{{\perp }_h}$ is defined as ${\mathcal{C}}^{{\perp }_h}=\{u\in F_4^n\mid {(u, v)}_h=0,\forall v\in \mathcal{C}\}$. If $\mathcal{C}\subseteq {\mathcal{C}}^{{\perp }_h}$, then $\mathcal{C}$ is called as an HSO code. In particular, if $\mathcal{C}={\mathcal{C}}^{{\perp }_h}$, then $\mathcal{C}$ is called a Hermitian self-dual code. If $\mathcal{C}\cap {\mathcal{C}}^{{\perp }_h}$$=\left\{0\right\}$, then $\mathcal{C}$ is called an HLCD code.

In the past 30 years, much work has been conducted concerning quaternary optimal linear codes for both theoretical and practical reasons (see, e.g., [1,2,3,4,5] and the references given therein). By 1996, the parameters of all optimal linear codes with $k\le 4$ were determined [2,3,4]. Since 1996, people have paid much attention to optimal linear codes with $k=5$; to date, there are yet 104 open cases on the parameters of optimal linear codes (see [3,4,5,6]). If $d_{op}(n, k)$ is not determined for a given $n, k$, a code $\mathcal{C}$ with the largest known distance $d_{bk}(n, k)$ is denoted as $\mathcal{C}$$=[n, k, d_{bk}(n, k)]$. Quaternary HSO codes and HLCD codes are special kinds of linear codes, these two kinds of codes have connections with many branches of mathematics and quantum information [7,8,9,10,11,12,13,14,15,16,17,18,19]. In recent years, there has been an increasing interest in optimal HSO and HLCD codes. A HSO (or HLCD) $[n, k, d]$ code is called optimal if there is no HSO (or HLCD) $[n, k, d^{\prime }]$ code for $d^{\prime }>d$. Entanglement-assisted (EA) stabilizer formalism was devised by Brun et al. in [20]. It has been proven that each $[n, k, d]$ quaternary HLCD code gave a maximal entanglement-assisted quantum code with a parameter ${\left[[n, k, d; n-k]\right]}_2$ by [14,19,21]. Under this EA stabilizer formalism, any quaternary HLCD code can be transformed into a maximal EAQECC if the shared entanglement is available between sender and receiver. Hence, it is important to study optimal quaternary HLCD codes for constructing ${\left[[n, k, d; n-k]\right]}_2$ EAQECCs.

Recall that, from 1978 to 1998, people paid much attention to special HSO codes—self-dual codes with short lengths (see [7,8,9,10]). In 1998, Calderbank et al. [11] set up connections among quantum codes, binary symplectic codes, and HSO codes; this inspired people to study HSO codes over $F_4$ for general length n. Bouyukliev et al. [12] discussed the classification of optimal HSO codes over $F_4$ for length $n\le 29$ and low dimensions. Ma et al. [13] determined the parameters of optimal HSO codes over $F_4$ for all n and $k\le 3$. Recently, Refs. [17,18] determined the parameters of optimal $[n, 4]$ HSO codes for all $n\ge 8$ and most $[n, 5]$ HSO codes for $n\le 341$. In [14], Lu et al. showed that an $[n, k, d]$ HLCD (also called zero radical) code can derive an ${\left[[n, k, d;n-k]\right]}_2$ EAQECC, they construct many good EAQECCs. According to [14,15,16], the parameters of optimal HLCD codes were determined for all n and $k\le 3$. Refs. [14,19] discussed the construction of $[n, 4]$ HLCD codes with $4\le n\le 85$ and $[n, 5]$ HLCD codes with $5\le n\le 341$, respectively. They have given some optimal HLCD codes and good low bounds on the distance of optimal HLCD codes. According to [3], the parameters of optimal $[n, 5]$ linear codes for $n\ge 492$ are known and can be constructed using a unified method of puncturing. Thus, the construction of $[n, 5]$ optimal HSO codes for $n\ge 492$ can be conducted as we conducted in [18] for constructing $[n, 4]$ HSO codes with $n\ge 124$. According to [6,14,18,19], when discussing $[n, 5]$ HSO codes and HLCD codes, we should consider codes with length n such that $342\le n\le 492$.

This paper is organized as follows. In Section 2, we prepare the definitions, notations, and basic results used in this paper. In Section 3, the construction of $[n, 5]$ HSO codes is presented. In Section 4, we derive $[n, 5]$ HLCD codes from known HSO codes and related EAQECCs. Finally, in Section 5, we conclude this paper.

2. Preliminaries

In this section, some notations, definitions, and basic results are given (for details, see Ref. [22]).

Throughout this paper, we use the following notations. We assume all codes are linear codes over $F_4$, and use 2 and 3 to represent $\omega$ and ${\omega }^2$ in $F_4$, respectively. Let ${\mathbf{1}}_{\mathbf{n}}$ = ${(1, 1, \dots , 1)}_{1\times n}$ and ${\mathbf{0}}_{\mathbf{n}}$ = ${(0, 0, \dots , 0)}_{1\times n}$ denote the all-one vector and the all-zero vector of length n, respectively. Let 0 denote a zero matrix of appropriate size and $I_k$ denote the identity matrix of order k. Let $A^T$ denote the transpose of a matrix A, and let $A^{†}$ denote the conjugate transpose of A.

Let $\mathcal{C}$ = $[n, k]$. A $k\times n$ matrix G whose rows form a basis of $\mathcal{C}$ is called a generator matrix. The weight enumerator of $\mathcal{C}$ is $w\left(z\right)={\sum }_{i=0}^n{A}_i{z}^i=A_0+A_{1}z+\cdots +A_n{z}^n$, where $A_i$ is the number of codewords in $\mathcal{C}$ with weight equal to i for $0\le i\le n$. We say that two $[n, k]$ codes ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ are equivalent, provided there is a monomial matrix M such that ${\mathcal{C}}_2$ = ${\mathcal{C}}_{1}M$. A code $\mathcal{C}$ is called an even code if all its codewords have even weights [22]. According to Ref. [11], $\mathcal{C}$ is a HSO code if and only if it is an even code. Using the generator matrix, one can give the following criterion for a code to be HSO or HLCD.

Proposition 1

([15]). Let G be a generator matrix of $\mathcal{C}$; then,

(1) 

$\mathcal{C}$ is an HSO code if and only if $G{G}^{†}=0$.

(2) 

$\mathcal{C}$ is a HLCD code if and only if $G{G}^{†}$ is nonsingular.

Definition 1.

If $A=A_{k, m}$ is a $k\times m$ matrix and the vectors formed by row linear combination of A have the largest weight δ, then A is called as an $(m, \delta )$ block. If $A{A}^{†}=0$, A is called an $(m, \delta )$ HSO block. If $A{A}^{†}$ is nonsingular, A is called an $(m, \delta )$ HLCD block.

We introduce some methods for constructing new codes from known ones. Lemmas 1 and 2 are directly obtained from [17] for linear codes. Lemma 3 and 4 can be derived from [18]. These four lemmas give the constructions of HSO codes by juxtaposition, pasting, puncturing, and shorting, respectively.

Lemma 1

([17]). Suppose ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ are $[n_1, k, d_1]$ and $[n_2, k, d_2]$ HSO codes, respectively. If ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ have generator matrices $G_1$ and $G_2$, respectively, then $(G_1\mid G_2)$ generates an $[n_1+n_2, k, d_1+d_2]$ HSO code.

Lemma 2

([17]). Suppose ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ are $[n_1, k, d_1]$ and $[n_2, k-1, d_2]$ HSO codes, respectively. If ${\mathcal{C}}_1$ contains a codeword of weight at least $d_1+d_2$, then there exists an $[n_1+n_2, k, d_1+d_2]$ HSO code.

Lemma 3.

Suppose $\mathcal{C}$$=[n, k, d]$ is an HSO code with generator matrix $G_{k,n}$ and $G_{k,n}$ has a $k\times m$ sub-matrix A. If A is an $(m, \delta )$ HSO (HLCD) block, then there is an $[n-m, k, d-\delta ]$ HSO (HLCD) code.

Proof. 

Let generator matrix $G_{k,n}=(A_{k,m}\mid B_{k n-m})$, then

$$G_{k,n}{G}_{k,n}^{†}=A_{k,m}{A}_{k,m}^{†}+B_{k,n-m}{B}_{k,n-m}^{+}$$

according to $\mathcal{C}$ is a HSO code.

Let $A_{k,m}$ generate an HSO code ${\mathcal{C}}_A$. $\alpha$ is a codeword in ${\mathcal{C}}_A$ with minimum Hamming weight $\delta$. Since HSO code $\mathcal{C}$$=[n, k, d]$, there is a codeword $\beta$ in $\mathcal{C}$ with minimum Hamming weight d. $B_{k,n-m}=(G_{k,n}\setminus A_{k,m})$. According to [22], $B_{k,n-m}$ generates an $[n-m,k,d_B]$ code with $d_B=d-\delta$.

(1)

If $A_{k,m}$ is an $(m,\delta )$ HSO block, $rank\left(G_{k,n}{G}_{k,n}^{†}\right)=rank\left(A_{k,m}{A}_{k,m}^{†}\right)=0$. Then, $rank(B_{k,n-m}$$B_{k,n-m}^{†})=0$ and $B_{k,n-m}$ generates an $[n-m,k,d-\delta ]$ HSO code.

(2)

If $A_{k,m}$ is an $(m,\delta )$ HLCD block, $rank\left(G_{k,n}{G}_{k,n}^{†}\right)=0$ and $rank\left(A_{k,m}{A}_{k,m}^{†}\right)=k$. Then, $rank\left(B_{k,n-m}{B}_{k,n-m}^{†}\right)=k$, $B_{k,n-m}$ generates an $[n-m,k,d-\delta ]$ HLCD code.

□

Lemma 4.

Suppose $n\ge 341+4$ and $\mathcal{C}$$=[n, 5, d]$ is an HSO code with $d\ge 6$. Then, there are $[n-i,4,d-2\lceil i/2\rceil ]$ HSO codes for $i=1,2,3,4$.

3. Constructing HSO Codes

In this section, we discuss the construction of $[n,5]$ HSO codes for $342\le n\le 492$; our results are given in two subsections.

3.1. $[n,5]$ HSO Codes for $342\le n\le 407$

In [3], the authors introduced a dual transform method for constructing new codes from known codes; they derived the existence of three codes with parameters $[364, 5, 272]$, $[386, 5, 288]$, and $[407, 5, 304]$ from three known codes with parameters $[27, 5, 16]$, $[38, 5, 24]$, and $[28, 5, 16]$, respectively. Using Magma [23], we can check that these three codes $[364, 5, 272]$, $[386, 5, 288]$, $[407, 5, 304]$ have generator matrices $G_{5,364}$, $G_{5,386}$, and $G_{5,407}$ (see Appendix A.1) and weight enumerators $1+942x^{272}+81x^{288}$, $1+213x^{288}+42x^{304}$, and $1+924x^{304}+99x^{320}$, respectively.

We try to find HSO blocks in $G_{5,364}$. It is not difficult to see that $A_{01a}$ and $A_{01b}$ have submatrices $S{A}_{01a}$ and $S{A}_{01a}^{\prime }$ and $S{A}_{01b}$ and $S{A}_{01b}^{\prime }$ as follows, where

$$S{A}_{01a}=\left(\begin{array}{c}0000\\ 1111\\ 0000\\ 0123\\ 1111\end{array}\right), S{A}_{01a}^{\prime }=\left(\begin{array}{c}0000\\ 1111\\ 1111\\ 3333\\ 0123\end{array}\right), S{A}_{01b}=\left(\begin{array}{c}0000\\ 1111\\ 2222\\ 0123\\ 3102\end{array}\right), S{A}_{01b}^{\prime }=\left(\begin{array}{c}0000\\ 1111\\ 3333\\ 0123\\ 1032\end{array}\right),$$

Columns ${\left(00010\right)}^T$, ${\left(00001\right)}^T$, ${\left(00012\right)}^T$, and ${\left(00011\right)}^T$ from $A_{00}$ are added to matrices $S{A}_{01a}$, $S{A}_{01a}^{\prime }$, $S{A}_{01b}$, and $S{A}_{01b}^{\prime }$, respectively. One can obtain four $5\times 5$ matrices, $X_1$, $X_2$, $X_3$, and $X_4$, of $G_{5,364}$, respectively, and they all satisfy $X_i·X_i^{†}=0$ for $i=1, 2, 3, 4$. It is easy to see that these $X_i$ are formed by different columns of $G_{5,364}$, and each is a $(5, 4)$ HSO block; thus, $G_{5,364}$ has $(5, 4)$, $(10, 8)$, $(15, 12)$, and $(20, 16)$ HSO blocks. According to Lemma 3, by removing these blocks, in turn, from $G_{5,364}$, HSO codes $[364-5i, 5, 272-4i]$ for $i=0, 1, 2, 3, 4$ can be constructed from $G_{5,364}$. From previous discussions, using Lemmas 3 and 4, we can achieve the following theorem.

Theorem 1.

Based on the $[364, 5, 272]$ HSO code, HSO codes with the following parameters can be constructed:

(1) 

$[364-5i, 5, 272-4i]$ for $i=0, 1, 2, 3, 4$;

(2) 

$[364-5i-j, 5, 272-4i-2\lceil \frac{j}{2}\rceil ]$ for $i=0, 1, 2, 3$ and $j=1, 2, 3, 4$.

Similar to the above discussion, we can show that $G_{5,386}$ has four $5\times 5$ submatrices, $Y_1$, $Y_2$, $Y_3$, and $Y_4$, and $G_{5,407}$ has four $5\times 5$ submatrices, $Z_1$, $Z_2$, $Z_3$, and $Z_4$, where

$$\begin{array}{c}{Y}_1=\left(\begin{array}{c}0000 0\\ 0000 0\\ 0111 1\\ 1012 3\\ 0333 3\end{array}\right), Y_2=\left(\begin{array}{c}0111 1\\ 0000 0\\ 0000 0\\ 1012 3\\ 1012 3\end{array}\right), Y_3=\left(\begin{array}{c}0111 1\\ 0000 0\\ 0222 2\\ 1012 3\\ 3031 2\end{array}\right), Y_4=\left(\begin{array}{c}0111 1\\ 0111 1\\ 1012 3\\ 0111 1\\ 0000 0\end{array}\right);\\ Z_1=\left(\begin{array}{c}0000 0\\ 0000 0\\ 0111 1\\ 0333 3\\ 1012 3\end{array}\right), Z_2=\left(\begin{array}{c}0000 0\\ 0111 1\\ 0000 0\\ 1012 3\\ 1230 1\end{array}\right), Z_3=\left(\begin{array}{c}0000 0\\ 0111 1\\ 0111 1\\ 1012 3\\ 2132 0\end{array}\right), Z_4=\left(\begin{array}{c}0000 0\\ 0111 1\\ 0222 2\\ 1012 3\\ 3120 3\end{array}\right).\end{array}$$

It is easy to see that these $Y_i$ are formed by different columns of $G_{5,386}$, with each being a $(5, 4)$ HSO block; these $Z_i$ are formed by different columns of $G_{5,407}$, with each being a $(5,4)$ HSO block for $i=1, 2, 3, 4$. Hence, both of $G_{5,386}$ and $G_{5,407}$ have $(5, 4)$, $(10, 8)$, $(15, 12)$, and $(20, 16)$ HSO blocks. According to Lemma 3, by removing these blocks, in turn, from $G_{5,386}$ and $G_{5,407}$, HSO codes $[386-5i, 5, 288-4i]$ and $[407-5i, 5, 304-4i]$ for $i=0, 1, 2, 3, 4$ can be constructed from $G_{5,386}$ and $G_{5,407}$, respectively. From previous discussions, by using Lemmas 3 and 4, we can achieve the following theorem.

Theorem 2.

Based on the $[386, 5, 288]$ and $[407, 5, 304]$ HSO codes, HSO codes with the following parameters can be constructed:

(1) 

$[386-5i, 5, 288-4i]$ for $i=0, 1, 2, 3, 4$;

(2) 

$[386-5i-j, 5, 288-4i-2\lceil \frac{j}{2}\rceil ]$ for $i=0, 1, 2, 3$ and $j=1, 2, 3, 4$;

(3) 

$[407-5i,5,304-4i]$ for $i=0, 1, 2, 3, 4$;

(4) 

$[407-5i-j, 5, 304-4i-2\lceil \frac{j}{2}\rceil ]$ for $i=0, 1, 2, 3$ and $j=1, 2, 3, 4$.

3.2. $[n, 5]$ HSO Codes for $408\le n\le 492$

In this subsection, we use the McDonald code $[256, 5, 192]$ and four known codes given in [4] to construct $[n, 5]$ HSO codes for $408\le n\le 492$.

In [4], four optimal codes $[172, 5, 128]$, $[194, 5, 144]$, $[215, 5, 160]$, and $[236, 5, 176]$ and their generator matrices are given. It is easy to see that these four codes are HSO codes. Using column permutation (special equivalent transform M), we obtain four equivalent HSO codes with generator matrices $G_{5,172}=(I_5\mid A_{5,167})$, $G_{5,194}=(I_5\mid A_{5,189})$, $G_{5,215}=(I_5\mid A_{5,210})$, and $G_{5,236}=(I_5\mid A_{5,231})$, respectively, all these matrices are given in Appendix A.2.

Lemma 5.

If $n=256+m$, $m\ge 170$ and there is an $[m, 5, d_{5,m}]$ HSO code, then there are HSO codes with the following parameters: $[n, 5, d]=[256+m, 5, 192+d_{5,m}]$, $[n-5i, 5, 192+d_{5,m}-4i]$ for $i=1, 2, 3, 4$ and $[n-5i-j, 5, 192+d_{5,m}-4i-2\lceil \frac{j}{2}\rceil ]$ for $i=0, 1, 2, 3, 4$ and $j=1, 2, 3, 4$.

Proof. 

Suppose $G_{5,n}=(M_5\mid G_{5,m})$, where $G_{5,m}=(I\mid A_{5,m-5})$ is a generator matrix of an $[m, 5, d_{5,m}]$ HSO code. Then, $G_{5,n}$ generates an $[n, 5, d]=[256+m, 5, 192+d_{5,m}]$ HSO code.

There are four submatrices in $G_{5,256}$:

$$G_a=\left(\begin{array}{c}0111\\ 0000\\ 1231\\ 0000\\ 1231\end{array}\right), G_b=\left(\begin{array}{c}1111\\ 0123\\ 0000\\ 0000\\ 1111\end{array}\right), G_c=\left(\begin{array}{c}1111\\ 0000\\ 0123\\ 1111\\ 1111\end{array}\right), G_d=\left(\begin{array}{c}0000\\ 0000\\ 0000\\ 0111\\ 1231\end{array}\right).$$

Adding column vectors ${(1, 0, 0, 0, 0)}^T$, ${(0, 1, 0, 0, 0)}^T$, ${(0, 0, 1, 0, 0)}^T$, and ${(0, 0, 0, 1, 0)}^T$ from $G_{5,m}$ to submatrices $G_a$, $G_b$, $G_c$, and $G_d$, respectively, we obtain four $5\times 5$ submatrices, $U_1$, $U_2$, $U_3$, and $U_4$, of $G_{5,n}$. It is obvious that these are formed by different columns of $G_{5,n}$, all $U_i$ satisfy $U_i{U}_i^{†}=0$ and are $(5,4)$ HSO blocks for $1\le i\le 4$. Hence, $G_{5,n}$ have $(5,4)$, $(10,8)$, $(15,12)$, and $(20,16)$ HSO blocks.

By removing $U_i(1\le i\le 4)$ from $G_{5,n}$, in turn, one can derive that there are $[n-5i,5,192+d_{5,m}-4i]$ HSO codes for $0\le i\le 4$. From $n-5i\ge 345$, we can obtain $[428-5i-j,5,320-4i-2\lceil \frac{j}{2}\rceil ]$ HSO codes for $(0\le i\le 4$, $1\le j\le 4)$. □

Since there are four HSO codes $[172, 5, 128]$, $[194, 5, 144]$, $[215, 5, 160]$, and $[236, 5, 176]$, we have the following corollary.

Corollary 1.

There are four groups of HSO codes:

(1) 

$[428-5i, 5, 320-4i]$ for $0\le i\le 4$, and $[428-5i-j, 5, 320-4i-2\lceil \frac{j}{2}\rceil ]$ for $0\le i\le 3$ and $1\le j\le 4$;

(2) 

$[450-5i, 5, 336-4i]$ for $0\le i\le 4$, and $[450-5i-j, 5, 336-4i-2\lceil \frac{j}{2}\rceil ]$ for $0\le i\le 3$ and $1\le j\le 4$;

(3) 

$[472-5i, 5, 352-4i]$ for $0\le i\le 4$, and $[472-5i-j, 5, 352-4i-2\lceil \frac{j}{2}\rceil ]$ for $0\le i\le 3$ and $1\le j\le 4$;

(4) 

$[492-5i, 5, 368-4i]$ for $0\le i\le 4$, and $[492-5i-j, 5, 368-4i-2\lceil \frac{j}{2}\rceil ]$ for $0\le i\le 3$ and $1\le j\le 4$.

Summarizing the above two subsections, we construct $[n, 5]$ HSO codes for each n with $342\le n\le 492$.

4. Construction of HLCD Codes

In this section, we focus on constructing HLCD codes from known HSO codes in the last section by puncturing some HLCD blocks.

Lemma 6.

Let $A_i$ be $(5, 4)$ HSO blocks for $1\le i\le 4$ and $A=(A_1, A_2, A_3, A_4)$. If $n\ge 341$, $G_{5,n}=(A\mid G_{5,n-20})$ is a generator matrix of an $[n, 5, d]$ HSO code and $G_{5,n-20}$ has $(j, j)$ HLCD blocks for $5\le j\le 9$. Then, there are $[n-5i-j, 5, d-4i-j]$ HLCD codes for $0\le i\le 3$ and $5\le j\le 9$.

Proof. 

Let $B_j$ be $(j, j)$ HLCD blocks of $G_{5,n-20}$ for $5\le j\le 9$. Let $D_{5i+j}=(A_1, \cdots ,A_i\mid B_j)$ for $1\le i\le 4$ and $5\le j\le 9$. Then, these $D_{5i+j}$ are $(5i+j, 4i+j)$ HLCD blocks of $G_{5,n}$ for $1\le i\le 4$ and $5\le j\le 9$. Puncturing these blocks form $G_{5,n}$; then, one can obtain the generator matrix of $[n-5i-j, 5, d-4i-j]$ HLCD codes for $0\le i\le 3$ and $5\le j\le 9$. □

According to Section 3.1, for $n=364, 386, 407$, let $d=272, 288, 304$, respectively; there are $[n, 5, d]$ HSO codes with generator matrices $G_{5,n}=(A\mid G_{5,n-20})$, where $A=(A_1, A_2, A_3, A_4)$, as shown in Section 3, and $G_{5,n-20}=(G_{5,n}\setminus A)$. Thus, if we can find that each $G_{5,n-20}$ has $(j, j)$ HLCD blocks $B_{m,j}$ for $5\le j\le 9$ and $m=n-20$, then we can obtain $[n-5i-j, 5, d-4i-j]$ HLCD codes. We check these facts in three cases.

Case 1.

Let $m=344$ and $G_{5,344}=(G_{5,364}\setminus (X_1, X_2, X_3, X_4)$. It is easy to check that $G_{5,344}$ has five $(j, j)$ HLCD blocks $B_{m,j}$ for $5\le j\le 9$, as follows:

$$\begin{array}{c}{B}_{m,5}=\left(\begin{array}{c}1000 0\\ 2100 0\\ 2011 1\\ 1010 3\\ 1333 0\end{array}\right), B_{m,6}=\left(\begin{array}{c}10000 0\\ 21000 0\\ 20111 1\\ 10110 3\\ 13233 0\end{array}\right), B_{m,7}=\left(\begin{array}{c}100000 0\\ 210000 0\\ 201111 1\\ 101310 3\\ 132133 0\end{array}\right), B_{m,8}=\left(\begin{array}{c}1000000 0\\ 2100000 0\\ 2011111 1\\ 1011310 3\\ 1302133 0\end{array}\right),\hfill \\ B_{m,9}=\left(\begin{array}{c}10000000 0\\ 21000000 0\\ 20111111 1\\ 10101310 3\\ 13022133 0\end{array}\right).\hfill \end{array}$$

Case 2.

Let $m=366$, $G_{5,366}=(G_{5,386}\setminus (Y_1, Y_2, Y_3, Y_4)$. It is easy to check that $G_{5,344}$ has five $(j,j)$ HLCD blocks $B_{m,j}$ for $5\le j\le 9$, as follows:

$$\begin{array}{c}{B}_{m,5}=\left(\begin{array}{c}00011\\ 00111\\ 11313\\ 23223\\ 13231\end{array}\right), B_{m,6}=\left(\begin{array}{c}000011\\ 100111\\ 211313\\ 323223\\ 313231\end{array}\right), B_{m,7}=\left(\begin{array}{c}0011011\\ 0011111\\ 1133313\\ 2333223\\ 1311231\end{array}\right), B_{m,8}=\left(\begin{array}{c}00011011\\ 10011111\\ 21133313\\ 32333223\\ 31311231\end{array}\right),\hfill \\ B_{m,9}=\left(\begin{array}{c}001111011\\ 001111111\\ 111313313\\ 232323223\\ 133131231\end{array}\right).\hfill \end{array}$$

Case 3.

Let $m=387$ and $G_{5,387}=(G_{5,407}\setminus (Z_1, Z_2, Z_3, Z_4)$. It is easy to check that $G_{5,387}$ has five $(j, j)$ HLCD blocks $B_{m,j}$ for $5\le j\le 9$ as follows:

$$\begin{array}{c}{B}_{m,5}=\left(\begin{array}{c}1000 0\\ 1110 0\\ 3201 1\\ 2121 2\\ 2002 2\end{array}\right), B_{m,6}=\left(\begin{array}{c}10000 0\\ 11110 0\\ 32201 1\\ 23121 2\\ 22002 2\end{array}\right), B_{m,7}=\left(\begin{array}{c}100000 0\\ 111110 0\\ 313201 1\\ 213121 2\\ 223002 2\end{array}\right), B_{m,8}=\left(\begin{array}{c}1000000 0\\ 1111110 0\\ 3313201 1\\ 2113121 2\\ 2223002 2\end{array}\right),\hfill \\ B_{m,9}=\left(\begin{array}{c}10000000 0\\ 11111110 0\\ 33113201 1\\ 22213121 2\\ 23223002 2\end{array}\right).\hfill \end{array}$$

According to Section 3, for $n=428, 450, 471, 492$, let $d=320, 336, 352, 368$, respectively; there are $[n, 5, d]$ HSO codes with generator matrices $G_{5,n}=(U\mid G_{5,241}\mid A_{5,n-261})$, where $U=(U_1, U_2, U_3, U_4)$ and $G_{5,241}=((G_{5,256}\mid I_5)\setminus U)$. Thus, if we can find that each $A_{5,n-261}$ has $(j, j)$ HLCD blocks $D_{m,j}$ for $5\le j\le 9$ and $m=n-261$, then we can obtain $[n-5i-j, 5, d-4i-j]$ HLCD codes. We check these facts in four cases.

Case 4.

Let $m=167$, and $A_{5,167}$ is given in Appendix A.2. It is easy to check that $A_{5,167}$ has five $(j, j)$ HLCD blocks $D_{m,j}$ for $5\le j\le 9$, as follows:

$$\begin{array}{c}{D}_{m,5}=\left(\begin{array}{c}0221 3\\ 2132 2\\ 2000 0\\ 1302 0\\ 1100 0\end{array}\right), D_{m,6}=\left(\begin{array}{c}01221 3\\ 22132 2\\ 20000 0\\ 13302 0\\ 13100 0\end{array}\right), D_{m,7}=\left(\begin{array}{c}022021 3\\ 211332 2\\ 200000 0\\ 123202 0\\ 101000 0\end{array}\right), D_{m,8}=\left(\begin{array}{c}0122021 3\\ 2211332 2\\ 2000000 0\\ 1323202 0\\ 1301000 0\end{array}\right),\hfill \\ D_{m,9}=\left(\begin{array}{c}00322021 3\\ 21011332 2\\ 20000000 0\\ 11223202 0\\ 13001000 0\end{array}\right).\hfill \end{array}$$

Case 5.

Let $m=189$, and $A_{5,189}$ is given in Appendix A.2. It is easy to check that $A_{5,189}$ has five $(j, j)$ HLCD blocks $D_{m,j}$ for $5\le j\le 9$, as follows:

$$\begin{array}{c}{D}_{m,5}=\left(\begin{array}{c}00313\\ 32202\\ 22200\\ 03311\\ 11111\end{array}\right), D_{m,6}=\left(\begin{array}{c}000313\\ 321202\\ 221200\\ 030311\\ 111111\end{array}\right), D_{m,7}=\left(\begin{array}{c}0011313\\ 3200202\\ 2212200\\ 0303311\\ 1111111\end{array}\right), D_{m,8}=\left(\begin{array}{c}02011313\\ 30200202\\ 23212200\\ 02303311\\ 11111111\end{array}\right),\hfill \\ D_{m,9}=\left(\begin{array}{c}010011313\\ 332100202\\ 232112200\\ 023003311\\ 111111111\end{array}\right).\hfill \end{array}$$

Case 6.

Let $m=210$, and $A_{5,210}$ is given in Appendix A.2. It is easy to check that $A_{5,210}$ has five $(j, j)$ HLCD blocks $D_{m,j}$ for $5\le j\le 9$, as follows:

$$\begin{array}{c}{D}_{m,5}=\left(\begin{array}{c}33131\\ 01211\\ 10210\\ 00200\\ 22031\end{array}\right), D_{m,6}=\left(\begin{array}{c}333131\\ 101211\\ 310210\\ 000200\\ 122031\end{array}\right), D_{m,7}=\left(\begin{array}{c}3133131\\ 1201211\\ 3110210\\ 0200200\\ 1322031\end{array}\right), D_{m,8}=\left(\begin{array}{c}32133131\\ 11201211\\ 30110210\\ 01200200\\ 12322031\end{array}\right),\hfill \\ D_{m,9}=\left(\begin{array}{c}132133131\\ 211201211\\ 030110210\\ 201200200\\ 212322031\end{array}\right).\hfill \end{array}$$

Case 7.

Let $m=231$, and $A_{5,231}$ is given in Appendix A.2. It is easy to check that $A_{5,231}$ has five $(j, j)$ HLCD blocks $D_{m,j}$ for $5\le j\le 9$, as follows:

$$\begin{array}{c}{D}_{m,5}=\left(\begin{array}{c}12210\\ 30012\\ 01021\\ 20131\\ 32332\end{array}\right), D_{m,6}=\left(\begin{array}{c}112210\\ 230012\\ 201021\\ 120131\\ 032332\end{array}\right), D_{m,7}=\left(\begin{array}{c}1112210\\ 2230012\\ 2201021\\ 1120131\\ 0032332\end{array}\right), D_{m,8}=\left(\begin{array}{c}11112210\\ 22230012\\ 02201021\\ 31120131\\ 20032332\end{array}\right),\hfill \\ D_{m,9}=\left(\begin{array}{c}111112210\\ 232230012\\ 032201021\\ 311120131\\ 200032332\end{array}\right).\hfill \end{array}$$

Summarizing previous discussions, from seven HSO codes (which are also optimal codes), $[364, 5, 272]$, $[386, 5, 288]$, $[407, 5, 304]$, $[428, 5, 320]$, $[450, 5, 336]$, $[471, 5, 352]$, and $[492, 4, 368]$, we can derive seven groups of HLCD codes, as follows.

Theorem 3.

Let $0\le i\le 4,5\le j\le 9$. There are seven groups of HLCD codes with lengths $342\le n\le 487$:

$[364-5i-j, 5, 272-4i-j]$, $[386-5i-j, 5, 288-4i-j]$, $[407-5i-j, 5, 304-4i-j]$, $[428-5i-j, 5, 320-4i-j]$, $[450-5i-j, 5, 336-4i-j]$, $[471-5i-j, 5, 352-4i-j]$, and $[492-5i-j, 4, 368-4i-j]$.

Comparing the parameters of the above new HLCD codes with those in [19], one can see that 31 of our HLCD codes have larger distances than those $[n, 5]$ of the same lengths in [19], and most of the others have the same distances as those in [19]. Table 1 shows our 31 HLCD codes and theirs.

Table 1. Comparison of HLCD codes.

No.	HLCD in [19]	Our HLCD Codes	No.	HLCD in [19]	Our HLCD Codes
1	$[359, 5, 266]$	$[359, 5, 267]$	17	$[466, 5, 346]$	$[466, 5, 347]$
2	$[397, 5, 294]$	$[397, 5, 295]$	18	$[472, 5, 350]$	$[472, 5, 351]$
3	$[400, 5, 296]$	$[400, 5, 297]$	19	$[475, 5, 352]$	$[475, 5, 353]$
4	$[401, 5, 297]$	$[401, 5, 298]$	20	$[476, 5, 353]$	$[476, 5, 354]$
5	$[402, 5, 298]$	$[402, 5, 299]$	21	$[477, 5, 354]$	$[477, 5, 355]$
6	$[417, 5, 309]$	$[417, 5, 310]$	22	$[478, 5, 354]$	$[478, 5, 355]$
7	$[418, 5, 310]$	$[418, 5, 311]$	23	$[479, 5, 355]$	$[479, 5, 356]$
8	$[420, 5, 311]$	$[420, 5, 312]$	24	$[480, 5, 356]$	$[480, 5, 357]$
9	$[421, 5, 312]$	$[421, 5, 313]$	25	$[481, 5, 357]$	$[481, 5, 358]$
10	$[422, 5, 313]$	$[422, 5, 314]$	26	$[482, 5, 357]$	$[482, 5, 359]$
11	$[423, 5, 314]$	$[423, 5, 315]$	27	$[483, 5, 358]$	$[483, 5, 359]$
12	$[440, 5, 326]$	$[440, 5, 327]$	28	$[484, 5, 359]$	$[484, 5, 360]$
13	$[460, 5, 341]$	$[460, 5, 342]$	29	$[485, 5, 360]$	$[485, 5, 361]$
14	$[461, 5, 342]$	$[461, 5, 343]$	30	$[486, 5, 361]$	$[486, 5, 362]$
15	$[464, 5, 344]$	$[464, 5, 345]$	31	$[487, 5, 361]$	$[487, 5, 363]$
16	$[465, 5, 345]$	$[465, 5, 346]$

For each of our $[m, 5, d]$ HLCD codes given in Table 1, we can derive $[341s+m, 5, 256s+d]$ HLCD codes for $s\ge 0$.

Theorem 4.

If $[m, 5, d]$ is one of our 31 HLCD codes given in Table 1, then there are $\left[\right[341s+m, 5, 256s+d; 341s+m-5{\left]\right]}_2$ EAQECCs for $s\ge 0$. Thus, we obtain 31 classes of EAQECCs better than those in [19] of the same lengths.

5. Conclusions

In this paper, we have studied the construction of HSO codes and HLCD codes with good minimum distances from known codes and further constructed EAQECCs with good parameters.

The largest minimum distance $d_{so}$ of HSO codes for $342\le n\le 492$ has been given above. If $d_{op}(n,5)$ is determined for a given n, for any optimal linear code $[n, 5, d_{op}(n, 5)]$, an HSO code with $d_{so}(n, 5)=2\lfloor {\displaystyle \frac{d_{op}(n, 5)}{2}}\rfloor$ could be constructed. If $d_{op}(n, 5)$ is not determined for given n, for any linear code $[n, 5, d_{bk}(n, 5)]$, an HSO code with $d_{so}(n, 5)=2\lfloor {\displaystyle \frac{d_{bk}(n, 5)}{2}}\rfloor$ could be constructed. The minimum distance has been optimized for all the above HSO codes.

Based on these HSO codes, we can further construct HLCD codes with lengths $342\le n\le 492$. The parameters of these HLCD codes are as follows: $[n-5i-j, 5, d_{so}(n, 5)-4i-j]$ for $n=364, 386, 407, 428, 450, 471, 492$, $0\le i\le 4$ and $5\le j\le 9$. By comparing with ones in the literature, it is easy to know that our 31 HLCD codes in Table 1 have better parameters. From these HLCD codes, we have obtained 31 classes of entanglement-assisted quantum codes with maximal entanglement.

Appendix A.1. Generator matrices G 5,386 and G 5,407 in Section 3.1

Let $G_{5,364}=(A_{00}, A_{01a}, A_{01b}, A_{10a}, A_{10b}, A_{11a}, A_{11b}, A_{12a}, A_{12b}, A_{13a}, A_{13b})$, where

$$\begin{array}{c}{A}_{00}=\left(\begin{array}{c}0000000000000000\\ 0000000000000000\\ 0000111111111111\\ 0111001111223333\\ 1012230123130122\end{array}\right), A_{01a}=\left(\begin{array}{c}0000000000000000000000000000\\ 1111111111111111111111111111\\ 0000000000001111111111111111\\ 0001112223330001112222233333\\ 0131231321230120231122301123\end{array}\right),\hfill \\ A_{01b}=\left(\begin{array}{c}00000000000000000000000000000000\\ 11111111111111111111111111111111\\ 22222222222222223333333333333333\\ 00011122222333330000111122223333\\ 12312301223012330123012301231233\end{array}\right), A_{10a}=\left(\begin{array}{c}111111111111111111111111111 1\\ 0000000000000000000000000000000 0\\ 000000000000111111111111111 1\\ 001111222333000111112222333 3\\ 010123123013012012230123012 3\end{array}\right),\hfill \\ A_{10b}=\left(\begin{array}{c}1111111111111111111111111111111 1\\ 0000000000000000000000000000000 0\\ 2222222222222222333333333333333 3\\ 0000111122222333000011112222233 3\\ 0123012301122123012301230122123 3\end{array}\right), A_{11a}=\left(\begin{array}{c}11111111111111111111111111111111111 1\\ 11111111111111111111111111111111111 1\\ 00000000000000001111111111111111111 1\\ 00011111222233330000011111222233333 3\\ 12301123012301230123300123012201223 3\end{array}\right),\hfill \\ A_{11b}=\left(\begin{array}{c}111111111111111111111111111111111111111 1\\ 111111111111111111111111111111111111111 1\\ 222222222222222222223333333333333333333 3\\ 000011111122222333330000011111222233333 3\\ 012300112301223011230112211223012300123 3\end{array}\right), A_{12a}=\left(\begin{array}{c}11111111111111111111111111111111111 1\\ 22222222222222222222222222222222222 2\\ 00000000000000001111111111111111111 1\\ 00001122222333330000011111222222333 3\\ 12330200123011230122301233011233012 2\end{array}\right),\hfill \\ A_{12b}=\left(\begin{array}{c}111111111111111111111111111111111111111 1\\ 222222222222222222222222222222222222222 2\\ 222222222222222222223333333333333333333 3\\ 000001111122223333330001111122222233333 3\\ 0112301223012300112221230123301123300112 2\end{array}\right), A_{13a}=\left(\begin{array}{c}11111111111111111111111111111111111 1\\ 33333333333333333333333333333333333 3\\ 00000000000000001111111111111111111 1\\ 00011111222233330001111122222233333 3\\ 01201123012301230230112301223300112 3\end{array}\right),\hfill \\ A_{13b}=\left(\begin{array}{c}111111111111111111111111111111111111111 1\\ 333333333333333333333333333333333333333 3\\ 222222222222222222223333333333333333333 3\\ 000111112222223333330000111111222223333 3\\ 012012230011230122330123011223011231223 3\end{array}\right).\hfill \end{array}$$

Let $G_{5,386}=(B_{00}, B_{01a}, B_{01b}, B_{10a}, B_{10b}, B_{11a}, B_{11b}, B_{12a}, B_{12b}, B_{13a}, B_{13b})$, where

$$\begin{array}{c}{B}_{00}=\left(\begin{array}{c}00000000000000000 0\\ 00000000000000000 0\\ 00001111111111111 1\\ 11110000111122233 3\\ 01130123012311323 3\end{array}\right), B_{01a}=\left(\begin{array}{c}000000000000000000000000000 0\\ 111111111111111111111111111 1\\ 000000000000001111111111111 1\\ 000111222333330001112222333 3\\ 123013123012331231230123012 3\end{array}\right),\hfill \\ B_{01b}=\left(\begin{array}{c}00000000000000000000000000000000000 0\\ 11111111111111111111111111111111111 1\\ 22222222222222222233333333333333333 3\\ 00001111122223333300001112222222333 3\\ 02230122301131223312330130012223022 3\end{array}\right), B_{10a}=\left(\begin{array}{c}1111111111111111111111111111111 1\\ 0000000000000000000000000000000 0\\ 0000000000000011111111111111111 1\\ 0000111222233300011112222333333 3\\ 0123012012301311201230122001122 3\end{array}\right),\hfill \\ B_{10b}=\left(\begin{array}{c}1111111111111111111111111111111 1\\ 0000000000000000000000000000000 0\\ 2222222222222222223333333333333 3\\ 0000111122222233330001112222333 3\\ 0123012301122301230121230123012 2\end{array}\right), B_{11a}=\left(\begin{array}{c}111111111111111111111111111111111111111 1\\ 111111111111111111111111111111111111111 1\\ 000000000000000000111111111111111111111 1\\ 000011111222233333000111122222222333333 3\\ 012301123011301223123012200122333001123 3\end{array}\right),\hfill \\ B_{11b}=\left(\begin{array}{c}111111111111111111111111111111111111111 1\\ 111111111111111111111111111111111111111 1\\ 222222222222222222333333333333333333333 3\\ 000111112222222333000011111111222333333 3\\ 013012330011223123001200112233123001112 3\end{array}\right), B_{12a}=\left(\begin{array}{c}11111111111111111111111111111111111 1\\ 22222222222222222222222222222222222 2\\ 00000000000000000011111111111111111 1\\ 00011112222222333300000111122223333 3\\ 01302330011223012301123122322330112 3\end{array}\right),\hfill \\ B_{12b}=\left(\begin{array}{c}1111111111111111111111111111111111111111111 1\\ 2222222222222222222222222222222222222222222 2\\ 2222222222222222222222333333333333333333333 3\\ 0001111111222222233333000000011111112222333 3\\ 0230011223001223301223001122301122330123001 2\end{array}\right), B_{13a}=\left(\begin{array}{c}11111111111111111111111111111111111 1\\ 33333333333333333333333333333333333 3\\ 00000000000000111111111111111111111 1\\ 00001112222333000011111112222222333 3\\ 01121230123123012301122330112233011 3\end{array}\right),\hfill \\ B_{13b}=\left(\begin{array}{c}1111111111111111111111111111111111111111111 1\\ 3333333333333333333333333333333333333333333 3\\ 2222222222222222222222333333333333333333333 3\\ 0000011111112223333333000001111122223333333 3\\ 0122300122331230011223011230122312230011223 3\end{array}\right).\hfill \end{array}$$

Let $G_{5,407}=(D_{00}, D_{01a}, D_{01b}, D_{10a}, D_{10b}, D_{11a}, D_{11b}, D_{12a}, D_{12b}, D_{13a}, D_{13b})$, where

$$\begin{array}{c}{D}_{00}=\left(\begin{array}{c}000000000000000000 0\\ 000000000000000000 0\\ 000011111111111111 1\\ 011100111112222333 3\\ 112323012231223012 3\end{array}\right), D_{01a}=\left(\begin{array}{c}00000000000000000000000000000 0\\ 11111111111111111111111111111 1\\ 00000000000000011111111111111 1\\ 00011111222223300111122222333 3\\ 12301223001231213122301223011 2\end{array}\right),\hfill \\ D_{01b}=\left(\begin{array}{c}0000000000000000000000000000000000000 0\\ 1111111111111111111111111111111111111 1\\ 2222222222222222222333333333333333333 3\\ 0001111112222233333000011111222223333 3\\ 1230012330122301223012301223012330123 3\end{array}\right), D_{10a}=\left(\begin{array}{c}11111111111111111111111111111 1\\ 00000000000000000000000000000 0\\ 00000000000111111111111111111 1\\ 00111223333000011111222223333 3\\ 23023230123123300123012330012 3\end{array}\right),\hfill \\ D_{10b}=\left(\begin{array}{c}1111111111111111111111111111111111111 1\\ 0000000000000000000000000000000000000 0\\ 2222222222222222222333333333333333333 3\\ 0000011122222333333000001111122223333 3\\ 0223312301233112233012230122301230122 3\end{array}\right), D_{11a}=\left(\begin{array}{c}1111111111111111111111111111111111111 1\\ 1111111111111111111111111111111111111 1\\ 0000000000000000000111111111111111111 1\\ 0000011122222233333000011112222223333 3\\ 0123302300123301123012301230112230122 3\end{array}\right),\hfill \\ D_{11b}=\left(\begin{array}{c}111111111111111111111111111111111111111111111 1\\ 111111111111111111111111111111111111111111111 1\\ 222222222222222222222223333333333333333333333 3\\ 000000111112222233333330000011112222222333333 3\\ 011223012231122301122330122302330112233001122 3\end{array}\right),\hfill \\ D_{12a}=\left(\begin{array}{c}11111111111111111111111111111111111111111 1\\ 22222222222222222222222222222222222222222 2\\ 00000000000000000001111111111111111111111 1\\ 00000111122222233330000001111122222233333 3\\ 01223012300223301220012330112301123300122 3\end{array}\right),\hfill \\ D_{12b}=\left(\begin{array}{c}11111111111111111111111111111111111111111 1\\ 22222222222222222222222222222222222222222 2\\ 22222222222222222223333333333333333333333 3\\ 00011111222222333330000001111112222223333 3\\ 02302233001223012230112230122330112330012 3\end{array}\right),\hfill \\ D_{13a}=\left(\begin{array}{c}11111111111111111111111111111111111111111 1\\ 33333333333333333333333333333333333333333 3\\ 00000000000000000001111111111111111111111 1\\ 00011111222223333330000011111222222333333 3\\ 12301233012330012230123300133112233001223 3\end{array}\right),\hfill \\ D_{13b}=\left(\begin{array}{c}11111111111111111111111111111111111111111 1\\ 33333333333333333333333333333333333333333 3\\ 22222222222222222222222333333333333333333 3\\ 00000111111222222233333000111122222233333 3\\ 01233001133001123301123023012300112300112 3\end{array}\right).\hfill \end{array}$$

Appendix A.2. Generator Matrices G_5,172, G_5,194, G_5,215, and G_5,236 in Section 3.2

Let $G_{5,172}=(E_1, E_2)$, where

$$\begin{array}{c}{E}_1=\left(\begin{array}{cc}10000\hspace{1em}& 02220313122330200013330311323322120001332022303103320202333210021033322320123101222331320 0\\ 01000\hspace{1em}& 02222232311012210101320131312031213223110033312013022113310120133330110002223310031302120 3\\ 00100\hspace{1em}& 00011310333333303301030000000000022022022222222222222222122121111011111000100303303333333 3\\ 00010\hspace{1em}& 11102221101100202303131331120202200311311322322223031022010200113213232233211002001211032 1\\ 00001\hspace{1em}& 00000101111111111110111113313133300300300003030003233222233132221232222222322320323233202 3\end{array}\right),\hfill \\ E_2=\left(\begin{array}{c}2000021213322132120031113002233333322010011012323223030113102223310210322021 3\\ 3221021313320001113202331220001032123122301100223320030012232100132021011332 2\\ 3333333333333333013313313113101111111111111112121112221212222222222000000000 0\\ 3013222332221311031031020230232201201132311110130101320202002010111231223202 0\\ 3223303020333000230030030320323233133311132113111112131111110111111333001000 0\end{array}\right).\hfill \end{array}$$

Let $G_{5,194}=(H_1, H_2)$, where

$$\begin{array}{c}{H}_1=\left(\begin{array}{cc}10000\hspace{1em}& 22213001230023323032031103312221301012120203203200121121130233112011233331010320322313221 0\\ 01000\hspace{1em}& 22213001103310010310213321133321301103031021021022312212203103112011233331023013011021003 2\\ 00100\hspace{1em}& 10333322111122330011000033222211123222103001122333322110000333333221111000033301112223332 2\\ 00010\hspace{1em}& 01222233000033221100111122333311123222103001122333322110000330000112222333300032221110001 1\\ 00001\hspace{1em}& 00000000000000000000000000000011111111111111111111111111111111111111111111111111111111111 1\end{array}\right),\hfill \\ H_2=\left(\begin{array}{c}20213202113302113022211300200221312033303222003331231231021100023303203200313123200130021113001131 3\\ 02031313002213002132211300200221303122212333112113013013212233310033203203020210133312203331210020 2\\ 21000330000111122222223333301111100000322211111222233001100003333220033221112333000111033322211220 0\\ 12333003333222211110001111123333322222100033333000011223322221111001122330003222111000122233300331 1\\ 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 1\end{array}\right).\hfill \end{array}$$

Let $G_{5,215}=(K_1, K_2, K_3)$, where

$$\begin{array}{c}{K}_1=\left(\begin{array}{cc}10000\hspace{1em}& 00030033303333030030000030001131131312222221210010001010303030333333230322220220211211111 2\\ 01000\hspace{1em}& 03210201321313232232322032200031010001103330311001112322130302023222233220023103031031311 0\\ 00100\hspace{1em}& 21011132323001221010232011111133203202201123310213221103130012220110133303331102122230003 2\\ 00010\hspace{1em}& 12211211121311111031010300032202222223230003032232221010303030301101101113311331303302032 0\\ 00001\hspace{1em}& 23321211010132211100233230033310001322312210033020110013121102302233303023311002121232031 1\end{array}\right),\hfill \\ K_2=\left(\begin{array}{c}1222222121112212110101022020033333333033301130110000202211233333322001113123223223303311001101201 2\\ 1100232120112213333323332320101010001100010001000111110132132333322321011101101002302222322232332 2\\ 3011310113222230030221232133202201103333300002220100331031220100321323102200233211332303202211333 1\\ 3002222121112212333333333331111002202002201123113333030011022222200330003013113113323300220023123 1\\ 0220211003333320302112022332230031122213311100333322013321121100312003003221112030303203122222100 2\end{array}\right),\hfill \\ K_3=\left(\begin{array}{c}222222222122133213313 1\\ 333231330231211120121 1\\ 032011103333033011021 0\\ 332221321221200120020 0\\ 033003110121211232203 1\end{array}\right).\hfill \end{array}$$

Let $G_{5,236}=(L_1, L_2, L_3)$, where

$$\begin{array}{c}{L}_1=\left(\begin{array}{cc}10000\hspace{1em}& 00120323121113210203300323222200330320323032233311211311113101033300000112201100223202002 1\\ 01000\hspace{1em}& 11231102012231233322322303111133211031213223211133222132223320322001101010021123012102013 0\\ 00100\hspace{1em}& 32310232333302330120201121100010113320123332001032301303220101201023302313212123130313213 2\\ 00010\hspace{1em}& 32211232221130003302132213232232221112023323000132210213221010201300021031031310203121130 2\\ 00001\hspace{1em}& 33010102311131203100333110332223123120102032122200222131001231001233312300023101200102313 2\end{array}\right),\hfill \\ L_2=\left(\begin{array}{c}1313311333332301122222220323300103330210202210213010310200031100301331031201112332110333303322022 1\\ 1233023011103111101232323212332123322303202213101221021203231130012301210130111230111101120311322 3\\ 0210003310122313202021021302021302202130201110112032200233110013110320312231231103112013320233201 1\\ 1300013200133303203302213130312021020213033301113022200333000011222112031113013320220330022233311 1\\ 1231231133203012330120033323002232311132310002311312021321201132312301022011333010002330012311000 2\end{array}\right),\hfill \\ L_3=\left(\begin{array}{c}022220202222320313132321013231001111111221 0\\ 022331303221202131230211302120223223223001 2\\ 123102313100322310130112302232332103220102 1\\ 123003213111323122302330131100010231112013 1\\ 133002203112311023323122012132101320003233 2\end{array}\right).\hfill \end{array}$$