Binary construction of pure additive quantum codes with distance five or six

Abstract

This paper discusses binary quantum stabilizer codes with distance five or six constructed from binary self-orthogonal codes using Steane construction. First, nineteen special binary one-generator quasi-cyclic self-orthogonal $[pk,k]$ codes with dual distance five or six for $12 \le k \le 16$ are built. Second, a feasible algorithm for searching subcodes of linear codes and an extension strategy for pairs of nested self-orthogonal codes are proposed, then thirty-eight code pairs are designed from obtained quasi-cyclic self-orthogonal codes. Third, thirty-two good binary quantum stabilizer codes are constructed from the code pairs obtained through Steane construction. Thirty of them are previously known codes. In particular, two codes $[[52,31,6]]$ and $[[56,34,6]]$ have improved codes $[[52,31,5]]$ and $[[56,34,5]]$ constructed by quaternary construction, and thus, they are record breaking ones.

Appendices

Appendix 1: Proof for Lemma 3

Proof

For binary self-orthogonal code $C = [n,k]$, the equations $GG^T=\mathbf{0}_{k \times k}$ and $G\mathbf{1}^T_n = \mathbf{0}_{k \times 1}$ hold. Because $d(C^{\perp }) = d^{\perp }$, any $d^{\perp } - 1$ columns in $G$ are linearly independent and there exist $d^{\perp }$ linearly dependent columns in $G$. Suppose $G = \left( \begin{array}{cccc}\alpha _1&\alpha _2&\cdots&\alpha _n\end{array} \right)$, then

$$\begin{aligned} G_1 = \left( \begin{array}{cccc} \alpha _1&{}\alpha _2&{}\cdots &{} \alpha _n\\ 1&{}1&{}\cdots &{} 1 \end{array} \right) , \quad G_{2} = \left( \begin{array}{c|cccc} 0&{}\alpha _1&{}\alpha _2&{}\cdots &{}\alpha _n\\ 1&{}1&{}1&{}\cdots &{} 1 \end{array} \right) =(L \mid R), \end{aligned}$$

where $\alpha _j$ is the binary $k$-dimensional column vector for $j = 1, 2, \cdots , n$. Let $\alpha _{j_1}, \alpha _{j_2}, \cdots , \alpha _{j_{d^{\perp }}}$ be the linearly dependent columns in $G$ such that

$$\begin{aligned} \sum \limits _{i=1}^{d^{\perp }} {\alpha _{j_i}} = \mathbf{0}_{k \times 1}. \end{aligned}$$

In order to prove the conclusions for $C_i$, the linear dependence of all rows in $G_i$, the self-orthogonality of $C_i$, and the number of linear independent or dependent columns in $G_i$ must be verified step by step.

(1). Firstly, all rows in $G_1$ are linearly independent because $\mathbf{1}_n \notin C$. Secondly, the fact that $\mathbf{1}_n\mathbf{1}^T_n = 0$ for even $n$ can deduce the equation

$$\begin{aligned} G_1G^T_1 =\left( \begin{array}{c} G\\ \mathbf{1}_n\\ \end{array} \right) \left( \begin{array}{cc} G^T &{} \mathbf{1}^T_n\\ \end{array} \right) =\left( \begin{array}{cc} GG^T&{}G\mathbf{1}^T_n \\ \mathbf{1}_nG^T&{}\mathbf{1}_n\mathbf{1}^T_n \end{array} \right) =\mathbf{0}_{(k+1) \times (k+1)}. \end{aligned}$$

They mean that $G_1$ can produce a binary self-orthogonal code $C_1 = [n, k+1]$.

Finally, we determine $d(C^{\perp }_1)$ under the assumptions of even $d^{\perp }$ and odd $d^{\perp }$, respectively.

Case I: $d^{\perp }$ is even. The conclusion that any $d^{\perp } - 1$ columns in $G$ are linearly independent implies that any $d^{\perp } - 1$ columns in $G_1$ are linearly independent. Furthermore, the equation $\sum \nolimits _{i=1}^{d^{\perp }} {\left( \begin{array}{c}\alpha _{j_i}\\ 1\end{array} \right) } = \mathbf{0}_{(k+1) \times 1}$ holds because $d^{\perp }$ is even, i.e., $G_1$ has a set of $d^{\perp }$ linearly dependent columns. It means that $d(C^{\perp }_1)$ equals $d^{\perp }$ if $d^{\perp }$ is even.

Case II: $d^{\perp }$ is odd. Let $\sum \nolimits _{i=1}^{d^{\perp }} {k_i\left( \begin{array}{c}\alpha _{l_i}\\ 1\end{array} \right) } = \mathbf{0}_{(k+1) \times 1}$, where $k_i \in \mathbb {F}_2$ and $\left( \begin{array}{c}\alpha _{l_i}\\ 1\end{array} \right)$ is the column in $G_1$ for $i = 1, 2, \cdots , d^{\perp }$. The equation is equivalent to a equation system $\sum \nolimits _{i=1}^{d^{\perp }} {k_i\alpha _{l_i}} = \mathbf{0}_{k \times 1}$ and $\sum \nolimits _{i=1}^{d^{\perp }} {k_i} = 0$.

If there exist some $k_i$’s such that $k_i = 1$, then the number of such $k_i$’s must be even and less than or equal to $d^{\perp } - 1$ because $d^{\perp }$ is odd. We can conclude that there are at most $d^{\perp } - 1$ linearly dependent columns in $G$. It is in contradiction with $d(C^{\perp }) = d^{\perp }$. That is, all $k_i = 0$ for $i = 1, 2, \cdots , d^{\perp }$, which implies any $d^{\perp }$ columns in $G_1$ are linearly independent in this case. It further states that $d(C^{\perp }_1)$ is greater than $d^{\perp }$ if $d^{\perp }$ is odd.

(2). It is obvious that all rows in $G_2$ are linearly independent and the following equation holds because $\mathbf{1}_n\mathbf{1}^T_n = 1$ for odd $n$.

$$\begin{aligned} G_2G^T_2 = \left( \begin{array}{c@{\quad }c} 0 &{} G\\ 1 &{}\mathbf{1}_n\\ \end{array} \right) \left( \begin{array}{c@{\quad }c} 0 &{} 1\\ G^T &{}\mathbf{1}^T_n\\ \end{array} \right) = \left( \begin{array}{c@{\quad }c} GG^T &{} G\mathbf{1}^T_n\\ \mathbf{1}^T_nG &{}1 + \mathbf{1}_n\mathbf{1}^T_n\\ \end{array} \right) = \mathbf{0}_{(k + 1) \times (k + 1)}. \end{aligned}$$

Thus, $G_2$ produces a binary self-orthogonal code $C_2 = [n + 1, k + 1]$. We subsequently determine $d(C^{\perp }_1)$ as below.

Case I: $d^{\perp }$ is even.

There are two types for any $d^{\perp } - 1$ columns in $G_2$.

Subcase I.1: All $d^{\perp } - 1$ columns come from the submatrix $R$ in $G_2$. In this subcase, the fact that any $d^{\perp } - 1$ columns in $G$ are linearly independent implies that any $d^{\perp } - 1$ columns in $R$ are linearly independent.

Subcase I.2: The one comes from the submatrix $L$ and the rest $d^{\perp } - 2$ columns come from the submatrix $R$. In this subcase, let

$$\begin{aligned} k_0\left( \begin{array}{c}0\\ 1\end{array} \right) + \sum \limits _{i=1}^{d^{\perp } - 2} {k_i\left( \begin{array}{c}\alpha _{l_i}\\ 1\end{array} \right) } = \mathbf{0}_{(k+1) \times 1}, \end{aligned}$$

where $k_0, k_i \in \mathbb {F}_2$ and $\left( \begin{array}{c}\alpha _{l_i}\\ 1\end{array} \right)$ is the column in $R$ for $i = 1, 2, \cdots , d^{\perp } - 2$. The equation is equivalent to a equation system $\sum \nolimits _{i=1}^{d^{\perp } - 2} {k_i\alpha _{l_i}} = \mathbf{0}_{k \times 1}$ and $k_0 + \sum \nolimits _{i=1}^{d^{\perp } - 2} {k_i} = 0$. The equation $\sum \nolimits _{i=1}^{d^{\perp } - 2} {k_i\alpha _{l_i}} = \mathbf{0}_{k \times 1}$ implies $k_i = 0$ for $i = 1, 2, \cdots , d^{\perp } - 2$ because $d(C^{\perp }) = d^{\perp }$. Furthermore, the equation $k_0 + \sum \nolimits _{i=1}^{d^{\perp } - 2} {k_i} = 0$ and the conclusion $k_i = 0$ for $i = 1, 2, \cdots , d^{\perp } - 2$ lead to that $k_0 = 0$.

It shows that any $d^{\perp } - 1$ columns in $G_2$ are linearly independent in both subcases.

The equation $\sum \nolimits _{i=1}^{d^{\perp }} {\alpha _{j_i}} = \mathbf{0}_{k \times 1}$ indicates the conclusion $\sum \nolimits _{i=1}^{d^{\perp }} {\left( \begin{array}{c}\alpha _{l_i}\\ 1\end{array} \right) } = \mathbf{0}_{(k+1) \times 1}$ because $d^{\perp }$ is even.

So, the conclusion of $d(C^{\perp }_2) = d^{\perp }$ holds in this case.

Case II: $d^{\perp }$ is odd. There are two types for any $d^{\perp }$ columns in $G_2$.

Subcase II.1: All $d^{\perp }$ columns come from the submatrix $R$ in $G_2$. The proof procedure for linearly independence of these columns is similar to Case II in (1).

Subcase II.2: The one comes from the submatrix $L$ and the rest $d^{\perp } - 1$ columns come from the submatrix $R$. The discussion for this subcase is similar to subcase I.2 in (2).

The fact that $\left( \begin{array}{c} 0\\ 1 \end{array} \right) + \sum \nolimits _{i=1}^{d^{\perp }} {\left( \begin{array}{c}\alpha _{l_i}\\ 1\end{array} \right) } = \mathbf{0}_{(k+1) \times 1}$ for odd $d^{\perp }$ declares the existence of $d^{\perp } + 1$ linearly dependent columns in $G_2$.

So, the conclusion that $d(C^{\perp }_2) = d^{\perp } + 1$ holds if $d^{\perp }$ is odd.

Appendix 2: Binary quasi-cyclic self-orthogonal codes

See Table 2.

Table 2 Binary quasi-cyclic self-orthogonal codes with dual distance five or six

Appendix 3: Dimensionality reduction matrices

• 1. $k=12$

  • $n=36,48$

    $$\begin{aligned} T^{(36)}_{6 \times 12}= \begin{pmatrix} 100000001100\\ 010010010111\\ 001000011110\\ 000110010110\\ 000001001101\\ 000000110001 \end{pmatrix}, \quad T^{(48)}_{6 \times 12}= \begin{pmatrix} 100000111011\\ 010000100110\\ 001000010011\\ 000100110010\\ 000010011001\\ 000001110111 \end{pmatrix}. \end{aligned}$$

    (10)

• 2. $k=13$

  • $n=39$

    $$\begin{aligned} T^{(39)}_{12 \times 13}= \begin{pmatrix} 1000000000001\\ 0100000000001\\ 0010000000001\\ 0001000000001\\ 0000100000001\\ 0000010000001\\ 0000001000001\\ 0000000100001\\ 0000000010001\\ 0000000001001\\ 0000000000101\\ 0000000000011 \end{pmatrix}, \quad T^{(39)}_{6 \times 12}= \begin{pmatrix} 100000100111\\ 010000010010\\ 001000111001\\ 000100011101\\ 000010111101\\ 000001000010 \end{pmatrix}. \end{aligned}$$

    (11)

  • $n=52$

    $$\begin{aligned} T^{(52)}_{12 \times 13}= \begin{pmatrix} 1000000000001\\ 0100000000001\\ 0010000000001\\ 0001000000001\\ 0000100000001\\ 0000010000001\\ 0000001000001\\ 0000000100001\\ 0000000010001\\ 0000000001001\\ 0000000000101\\ 0000000000011 \end{pmatrix}, \quad T^{(52)}_{7 \times 12}= \begin{pmatrix} 100000000111\\ 010000100101\\ 001000100001\\ 000100001110\\ 000010001001\\ 000001000011\\ 000000010101 \end{pmatrix}. \end{aligned}$$

    (12)

  • $n=65$

    $$\begin{aligned} T^{(65)}_{8 \times 13}= \begin{pmatrix} 1000000000001\\ 0100000000100\\ 0010000011100\\ 0001000011111\\ 0000100000010\\ 0000010001000\\ 0000001000010\\ 0000000100100 \end{pmatrix}. \end{aligned}$$

    (13)

• 3. $k=14$

  • $n=42$

    $$\begin{aligned} T^{(42)}_{13 \times 14}= \begin{pmatrix} 10000000000001\\ 01000000000000\\ 00100000000001\\ 00010000000000\\ 00001000000001\\ 00000100000000\\ 00000010000000\\ 00000001000000\\ 00000000100001\\ 00000000010000\\ 00000000001001\\ 00000000000100\\ 00000000000011 \end{pmatrix}, \quad T^{(42)}_{6 \times 13}= \begin{pmatrix} 1000001101110\\ 0100000101110\\ 0010000011011\\ 0001001110110\\ 0000101011101\\ 0000010011101 \end{pmatrix}. \end{aligned}$$

    (14)

  • $n=56$

    $$\begin{aligned} T^{(56)}_{13 \times 14}= \begin{pmatrix} 10000000000000\\ 01000000000010\\ 00100000000010\\ 00010000000010\\ 00001000000010\\ 00000100000000\\ 00000010000000\\ 00000001000010\\ 00000000100010\\ 00000000010010\\ 00000000001000\\ 00000000000110\\ 00000000000001 \end{pmatrix}, \quad T^{(56)}_{7 \times 13}= \begin{pmatrix} 1000000001000\\ 0100000101101\\ 0010000010101\\ 0001000010011\\ 0000100100101\\ 0000010111110\\ 0000001001010\\ \end{pmatrix}. \end{aligned}$$

    (15)

  • $n=70$

    $$\begin{aligned} T^{(70)}_{8 \times 14}= \begin{pmatrix} 10000000111110\\ 01000000101101\\ 00100000111011\\ 00010000110111\\ 00001000110110\\ 00000100000010\\ 00000010010101\\ 00000001000001\\ \end{pmatrix}. \end{aligned}$$

    (16)

  • $n=84$

    $$\begin{aligned} T^{(84)}_{12 \times 14}= \begin{pmatrix} 10000000000010\\ 01000000000001\\ 00100000000010\\ 00010000000001\\ 00001000000010\\ 00000100000001\\ 00000010000010\\ 00000001000001\\ 00000000100010\\ 00000000010001\\ 00000000001010\\ 00000000000101\\ \end{pmatrix}, \quad T^{(84)}_{8 \times 12}= \begin{pmatrix} 100000001101\\ 010000000010\\ 001000001101\\ 000100001000\\ 000010000010\\ 000001001011\\ 000000101000\\ 000000010110\\ \end{pmatrix}. \end{aligned}$$

    (17)

• 4. $k=15$

  • $n=45$

    $$\begin{aligned} T^{(45)}_{14 \times 15}= \begin{pmatrix} 100000000000000\\ 010000000000000\\ 001000000000001\\ 000100000000001\\ 000010000000001\\ 000001000000000\\ 000000100000001\\ 000000010000001\\ 000000001000001\\ 000000000100001\\ 000000000010001\\ 000000000001001\\ 000000000000100\\ 000000000000010 \end{pmatrix}, \quad T^{(45)}_{7 \times 14}= \begin{pmatrix} 10000001100010\\ 01000000100000\\ 00100000110010\\ 00010000110110\\ 00001000111100\\ 00000101001011\\ 00000011000111 \end{pmatrix}. \end{aligned}$$

    (18)

  • $n=60$

    $$\begin{aligned} T^{(60)}_{14 \times 15}= \begin{pmatrix} 100000000000000\\ 010000000000001\\ 001000000000001\\ 000100000000001\\ 000010000000000\\ 000001000000001\\ 000000100000001\\ 000000010000001\\ 000000001000001\\ 000000000100000\\ 000000000010001\\ 000000000001001\\ 000000000000101\\ 000000000000010 \end{pmatrix}, \quad T^{(60)}_{7 \times 14}= \begin{pmatrix} 10000001100101\\ 01000001011001\\ 00100000100010\\ 00010000111011\\ 00001000011110\\ 00000100001100\\ 00000010001100\\ \end{pmatrix}. \qquad \end{aligned}$$

    (19)

  • $n=75, 90$

    $$\begin{aligned} T^{(75)}_{8 \times 15}= \begin{pmatrix} 100000001011010\\ 010000001011110\\ 001000001001111\\ 000100000000001\\ 000010000101001\\ 000001000100111\\ 000000101001010\\ 000000010010101\ \end{pmatrix}, \quad T^{(90)}_{9 \times 15}= \begin{pmatrix} 100000000111011\\ 010000001111000\\ 001000000111111\\ 000100001001110\\ 000010001111011\\ 000001000111000\\ 000000100111101\\ 000000011011100\\ \end{pmatrix}.\qquad \end{aligned}$$

    (20)

  • $n=105$

    $$\begin{aligned} T^{(105)}_{9 \times 15}= \begin{pmatrix} 100000000010000\\ 010000000101001\\ 001000000101111\\ 000100000010110\\ 000010000101010\\ 000001000111011\\ 000000100111000\\ 000000010101010\\ 000000001001101 \end{pmatrix}. \end{aligned}$$

    (21)

• 5. $k=16$

  • $n=64$

    $$\begin{aligned} T^{(64)}_{14 \times 16}= \begin{pmatrix} 1000000000000010\\ 0100000000000010\\ 0010000000000001\\ 0001000000000010\\ 0000100000000001\\ 0000010000000001\\ 0000001000000010\\ 0000000100000001\\ 0000000010000001\\ 0000000001000001\\ 0000000000100001\\ 0000000000010001\\ 0000000000001001\\ 0000000000000101 \end{pmatrix}, \quad T^{(64)}_{7 \times 14}= \begin{pmatrix} 1000000110110111\\ 0100000111100010\\ 0010000011100000\\ 0001000001010010\\ 0000100001011101\\ 0000010011000100\\ 0000001111001010 \end{pmatrix}.\qquad \end{aligned}$$

    (22)

  • $n=80$

    $$\begin{aligned} T^{(80)}_{8 \times 16}= \begin{pmatrix} 1000000001011010\\ 0100000000101011\\ 0010000000111100\\ 0001000001000000\\ 0000100010010100\\ 0000010010001100\\ 0000001001001011\\ 0000000101110011\\ \end{pmatrix}. \end{aligned}$$

    (23)

  • $n=96$

    $$\begin{aligned} T^{(96)}_{15 \times 16}= \begin{pmatrix} 1000000000000000\\ 0100000000000000\\ 0010000000000000\\ 0001000000000001\\ 0000100000000001\\ 0000010000000001\\ 0000001000000000\\ 0000000100000000\\ 0000000010000001\\ 0000000001000000\\ 0000000000100001\\ 0000000000010001\\ 0000000000001001\\ 0000000000000101\\ 0000000000000010 \end{pmatrix}, \quad T^{(96)}_{9 \times 15}= \begin{pmatrix} 100000000010111\\ 010000000110010\\ 001000000010100\\ 000100000011111\\ 000010000100011\\ 000001000100010\\ 000000100101010\\ 000000010110000\\ 000000001010100\\ \end{pmatrix}.\qquad \end{aligned}$$

    (24)

  • $n=112, 128$

    $$\begin{aligned} T^{(112)}_{9 \times 16}= \begin{pmatrix} 1000000000101011\\ 0100000000110011\\ 0010000000001000\\ 0001000000010100\\ 0000100001101100\\ 0000010000001110\\ 0000001000011100\\ 0000000101010011\\ 0000000010101111 \end{pmatrix}, \quad T^{(128)}_{9 \times 16}= \begin{pmatrix} 1000000000110101\\ 0100000001101100\\ 0010000001111001\\ 0001000001001111\\ 0000100001001000\\ 0000010000101110\\ 0000001001001000\\ 0000000100100111\\ 0000000010101010 \end{pmatrix}.\qquad \end{aligned}$$

    (25)