Constructions of maximum distance separable symbol-pair codes using cyclic and constacyclic codes

Abstract

Symbol-pair code is a new coding framework which is proposed to correct errors in the symbol-pair read channel. In particular, maximum distance separable (MDS) symbol-pair codes are a kind of symbol-pair codes with the best possible error-correction capability. Employing cyclic and constacyclic codes, we construct three new classes of MDS symbol-pair codes with minimum pair-distance five or six. Moreover, we find a necessary and sufficient condition which ensures a class of cyclic codes to be MDS symbol-pair codes. This condition is related to certain property of a special kind of linear fractional transformations. A detailed analysis on these linear fractional transformations leads to an algorithm, which produces many MDS symbol-pair codes with minimum pair-distance seven.

Appendix

Let q be a prime power. For $u,v,w,z \in \mathbb {F}_q$ and $\delta \in \mathbb {F}_{q^2}$, define a linear fractional transformation from $\mathbb {F}_{q^2}$ to $\mathbb {F}_{q^2}$ by

$$\begin{aligned} f_{u,v,w,z}(\delta )=\frac{u+v\delta }{w+z\delta }, \end{aligned}$$

where $w+z\delta \ne 0$ and $uz-vw \ne 0$. We further assume that $z \ne 0$, since otherwise, $f_{u,v,w,z}$ degenerates into a linear function. Below, we will study this special kind of linear fractional transformation. In particular, suppose $\delta \in \mathbb {F}_{q^2}{\setminus } \mathbb {F}_q$, we will present a necessary and sufficient condition such that

$$\begin{aligned} \delta ^i=\frac{u+v\delta }{w+z\delta } \end{aligned}$$

for some integer i. This condition provides a criterion to determine whether the linear fractional transformation $f_{u,v,w,z}$ maps $\delta$ to an element belonging to the multiplicative cyclic group generated by $\delta$.

Proposition 17

Let $\delta \in \mathbb {F}_{q^2}{\setminus } \mathbb {F}_q$. Let $x^2-bx-c$ be the monic minimal polynomial of $\delta$ over $\mathbb {F}_q$. For an integer $i \ge 2$, define

$$\begin{aligned} a_0^{(i)}=\sum _{j=0}^{\lfloor \frac{i-2}{2} \rfloor } {i-2-j \atopwithdelims ()j}b^{i-2-2j}c^{j+1}, \quad a_1^{(i)}=\sum _{j=0}^{\lfloor \frac{i-1}{2} \rfloor } {i-1-j \atopwithdelims ()j}b^{i-1-2j}c^{j}. \end{aligned}$$

(4)

Then for $i \ge 0$, $\delta ^i=\frac{u+v\delta }{w+z\delta }$ if and only if one of the following holds:

• (1) If $i=0$, then $u=w$, $v=z$.

• (2) If $i=1$, then $b=\frac{v-w}{z}$ and $c=\frac{u}{z}$.

• (3) If $i \ge 2$, then

  $$\begin{aligned} a_1^{(i)}\ne 0, \quad b=-\frac{a_0^{(i)}}{a_1^{(i)}}+\frac{v}{za_1^{(i)}}-\frac{w}{z}, \quad c=-\frac{wa_0^{(i)}}{za_1^{(i)}}+\frac{u}{za_1^{(i)}}. \end{aligned}$$

Proof

(1) and (2) are trivial. We only consider (3) below. Since $\delta ^i=\frac{u+v\delta }{w+z\delta }$ and $z \ne 0$, we have $\delta ^{i+1}+\frac{w}{z}\delta ^{i}-\frac{v}{z}\delta -\frac{u}{z}=0$. Therefore, $\delta$ is a root of the polynomial $x^{i+1}+\frac{w}{z}x^{i}-\frac{v}{z}x-\frac{u}{z}$ and

$$\begin{aligned} x^{i+1}+\frac{w}{z}x^{i}-\frac{v}{z}x-\frac{u}{z} \equiv 0 \pmod {x^2-bx-c}. \end{aligned}$$

For an integer $i\ge 0$, we define a polynomial $T_i(x)=x^{i+1}+\frac{w}{z}x^{i}$. For any $i \ge 2$, we have the following recurrence relation:

$$\begin{aligned} T_i(x)&\equiv x^{i+1}+\frac{w}{z}x^{i} \\&\equiv bx^{i}+cx^{i-1}+\frac{w}{z}(bx^{i-1}+cx^{i-2}) \\&\equiv b(x^i+\frac{w}{z}x^{i-1})+c(x^{i-1}+\frac{w}{z}x^{i-2}) \\&\equiv bT_{i-1}(x)+cT_{i-2}(x) \pmod {x^2-bx-c}. \end{aligned}$$

By employing this recurrence relation repeatedly, we have

$$\begin{aligned} T_i(x)&\equiv d_2^{(i)}T_2(x)+d_1^{(i)}T_1(x) \\&\equiv e_1^{(i)}T_1(x)+e_0^{(i)}T_0(x) \pmod {x^2-bx-c}, \end{aligned}$$

where $d_1^{(i)},d_2^{(i)},e_0^{(i)},e_1^{(i)}\in \mathbb {F}_q$. Now, we aim to determine $e_0^{(i)}$ and $e_1^{(i)}$ explicitly. The recurrence relation implies that $T_0(x)$ necessarily originates from $T_2(x)$ by subtracting a proper multiple of $x^2-bx-c$. Since $T_2(x)\equiv bT_{1}(x)+cT_{0}(x) \pmod {x^2-bx-c}$, we have $e_0^{(i)}=c d_2^{(i)}$. Apparently, $d_2^{(i)}$ is a summation of monomials regarding of b and c. More precisely, suppose $i-2$ can be expressed as an ordered sum containing $i-2-2j$ ones and j twos. Then this ordered sum corresponds to a monomial $b^{i-2-2j}c^j$ in the summation of $d_2^{(i)}$. Recall that there are $\left( {\begin{array}{c}i-2-j\\ j\end{array}}\right)$ ways to decompose $i-2$ into distinct ordered sums containing $i-2-2j$ ones and j twos. Therefore, we have

$$\begin{aligned} d_2^{(i)}=\sum _{j=0}^{\lfloor \frac{i-2}{2} \rfloor } {i-2-j \atopwithdelims ()j}b^{i-2-2j}c^{j}, \end{aligned}$$

and

$$\begin{aligned} e_0^{(i)}=cd_2^{(i)}=\sum _{j=0}^{\lfloor \frac{i-2}{2} \rfloor } {i-2-j \atopwithdelims ()j}b^{i-2-2j}c^{j+1}=a_0^{(i)}. \end{aligned}$$

Similarly, by analyzing the decomposition of $i-1$ into ordered sums consisting of ones and twos, we have

$$\begin{aligned} e_1^{(i)}=\sum _{j=0}^{\lfloor \frac{i-1}{2} \rfloor } {i-1-j \atopwithdelims ()j}b^{i-1-2j}c^{j}=a_1^{(i)}. \end{aligned}$$

Consequently,

$$\begin{aligned} x^{i+1}+\frac{w}{z}x^{i}-\frac{v}{z}x-\frac{u}{z}&\equiv T_i(x)-\frac{v}{z}x-\frac{u}{z} \\&\equiv a_1^{(i)}T_1(x)+a_0^{(i)}T_0(x)-\frac{v}{z}x-\frac{u}{z} \\&\equiv a_1^{(i)}x^2+(a_0^{(i)}+\frac{wa_1^{(i)}}{z}-\frac{v}{z})x+\frac{wa_0^{(i)}}{z}-\frac{u}{z} \\&\equiv 0 \pmod {x^2-bx-c}. \end{aligned}$$

Hence, we must have $a_1^{(i)}\ne 0$ and $x^2+(\frac{a_0^{(i)}}{a_1^{(i)}}+\frac{w}{z}-\frac{v}{za_1^{(i)}})x+\frac{wa_0^{(i)}}{za_1^{(i)}}-\frac{u}{za_1^{(i)}}=x^2-bx-c$. The conclusion follows by comparing the coefficients. $\square$

Particularly, given $\delta \in \mathbb {F}_{q^2}{\setminus } \mathbb {F}_q$ and an integer $i\ge 2$, we have the following easy criterion to determine if $\delta ^i=\frac{1-\theta \delta }{-\theta +\delta }$ for some $\theta \in \mathbb {F}_q^*$.

Corollary 18

Let $\delta \in \mathbb {F}_{q^2}{\setminus } \mathbb {F}_q$. Let $x^2-bx-c$ be the monic minimal polynomial of $\delta$ over $\mathbb {F}_q$. For an integer $i \ge 2$, $\delta ^i=\frac{1-\theta \delta }{-\theta +\delta }$ for some $\theta \in \mathbb {F}_q^*$ if and only if $a_1^{(i)}\ne 0$ and one of the following condition holds

• 1) If $a_1^{(i)}=1$, then $a_0^{(i)}=-b$ and $c \ne 1$,

• 2) If $a_0^{(i)}=0$, then $a_1^{(i)}=\frac{1}{c}$ and $b \ne 0$,

• 3) If $a_0^{(i)}\ne 0$ and $a_1^{(i)}\ne 1$, then $a_1^{(i)}c\ne 1$ and $\frac{a_1^{(i)}b+a_0^{(i)}}{a_1^{(i)}-1}=\frac{a_1^{(i)}c-1}{a_0^{(i)}}$.

where $a_0^{(i)}$ and $a_1^{(i)}$ are defined in (4). Moreover, let $\mathbb {F}_r$ be a subfield of $\mathbb {F}_q$. If $b,c \in \mathbb {F}_r$, then $\delta ^i=\frac{1-\theta \delta }{-\theta +\delta }$ for some $i\ge 2$ only if $\theta \in \mathbb {F}_r$.

Proof

By setting $u=z=-1$ and $v=w=\theta$ in Proposition 17, we have $\delta ^i=\frac{1-\theta \delta }{-\theta +\delta }$ for some $\theta \in \mathbb {F}_q^*$ if and only if

$$\begin{aligned} b=\frac{(a_1^{(i)}-1)\theta -a_0^{(i)}}{a_1^{(i)}}, \quad c=\frac{a_0^{(i)}\theta +1}{a_1^{(i)}}. \end{aligned}$$

If $a_0^{(i)}=0$ and $a_1^{(i)}=1$, then we have $b=0$ and $c=1$, which is impossible since $x^2-1$ is reducible over $\mathbb {F}_q$. If either $a_1^{(i)}=1$ or $a_0^{(i)}=0$, then the Condition 1) or the Condition 2) holds. If $a_0^{(i)}\ne 0$ and $a_1^{(i)}\ne 1$, the Condition 3) is derived from the expressions of b and c. Suppose b and c belong to a subfield $\mathbb {F}_r$, then $a_0^{(i)},a_1^{(i)}\in \mathbb {F}_r$ by definition. Since we have either $a_0^{(i)}\ne 0$ or $a_1^{(i)}\ne 1$, it is easy to see that $\theta \in \mathbb {F}_r$. $\square$