# 六类纠缠辅助量子MDS码的构造Constructions of Six Classes of Entanglement-Assisted Quantum MDS Codes

DOI: 10.12677/PM.2019.95087, PDF, HTML, XML, 下载: 377  浏览: 1,483

Abstract: Entanglement-assisted quantum error-correcting codes make use of preexisting entanglement between the sender and receiver to improve information rate. In this paper, we obtain six classes of entanglement-assisted quantum MDS codes by means of characterizing the relationship between the number of entangled states and the size of defining set of constacyclic codes.

1. 引言

2. 预备知识

2.1. 常循环码

$\left(n,q\right)=1$。对于 $\eta \in {F}_{{q}^{2}}^{\text{*}}$，如果 $\left({c}_{0},{c}_{1},\cdots ,{c}_{n-1}\right)\in C$，有 $\left(\eta {c}_{n-1},{c}_{0},\cdots ,{c}_{n-2}\right)\in C$，称线性码C为 $\eta$ -常循环码。一个码字 $c=\left({c}_{0},{c}_{1},\cdots ,{c}_{n-1}\right)$ 也可以用一个多项式 $c\left(x\right)={c}_{0}+{c}_{1}+\cdots +{c}_{n-1}{x}^{n-1}$ 表示。因此每个 $\eta$ -常循环码都是商环 ${F}_{{q}^{2}}\left[x\right]/〈{x}^{n}－\eta 〉$ 的一个主理想，即 $C=〈g\left(x\right)〉$，其中 $g\left(x\right)|{x}^{n}-\eta$，且 $g\left(x\right)$ 是首1多项式。假设 $\omega \in {F}_{{q}^{2}}^{\text{*}}$ 是rn次本原单位根且 ${\omega }^{n}=\eta$，令 $\zeta ={\omega }^{r}$，则 $\zeta$ 是一个n次本原单位根。那么就有

${x}^{n}-\eta ={x}^{n}-{\omega }^{n}={\prod }_{j=0}^{n-1}\left(x-{\omega }^{1+jr}\right).$

$\Omega =\left\{1+jr|0\le j\le n-1\right\}$。对于 $\forall j\in \Omega$，设C是 $\eta$ -常循环码，且 $C=〈g\left(x\right)〉$，称 $Z=\left\{j\in \Omega |g\left({\omega }^{j}\right)=0\right\}$ 为C的定义集，因而 ${Z}^{\perp H}=\left\{j\in \Omega |-qj\left(\mathrm{mod}\text{\hspace{0.17em}}rn\right)\notin Z\right\}$${C}^{\perp H}$ 的定义集。称 ${C}_{j}=\left\{j{q}^{2i}\left(\mathrm{mod}rn\right)|i\in Z\right\}$ 为j模rn的 ${q}^{2}$ -分圆陪集。

2.2. 纠缠辅助量子码

3. 主要结果

3.1. 构造长为 $n=\frac{{q}^{2}-1}{r}$ 的EAQMDS码

$\begin{array}{c}c=|Z\cap \left(-qZ\right)|=|\left\{\left(j,k\right)|1+rj\equiv -q\left(1+rk\right)\left(\mathrm{mod}\text{\hspace{0.17em}}rn\right),0\le j,k\le \delta \right\}|\\ =|\left\{\left(j,k\right)|j+qk\equiv -\frac{q+1}{r}\left(\mathrm{mod}\text{\hspace{0.17em}}n\right),0\le j,k\le \delta \right\}|.\end{array}$

$j+qk\equiv -\frac{q+1}{r}\left(\mathrm{mod}n\right)$ 有整数解 $\left(j,k\right)$

$⇔$$l\in Z$ 来说， $nl-\frac{q+1}{r}=j+qk$ 有整数解 $\left(j,k\right)$

$\text{0}\le j,k\le n-1$，当 $r\ne 1$ 时，可得 $1\le l\le q$。在取定l后把 $nl-\frac{q+1}{r}$ 表示成 $j+qk$ 这种形式，其中 $\left(j,k\right)$ 的取法是多样的，例如：

$j+qk=nl-\frac{q+1}{r}=\frac{l\left({q}^{2}-1\right)}{r}-\frac{q+1}{r}=\left(\frac{l\left(q+1\right)}{r}-x\right)q+xq-\frac{\left(l+1\right)\left(q+1\right)}{r}.$

${b}_{l,x}=\mathrm{max}\left(j,k\right)=\mathrm{max}\left(xq-\frac{\left(l+1\right)\left(q+1\right)}{r},\frac{l\left(q+1\right)}{r}-x\right)$，得到所有的 ${b}_{l,x}$，只要给定r，就可以把 ${b}_{l,x}$ 从小到大排列。依次记为 ${b}_{1},{b}_{2},{b}_{3},\cdots$。若 ${b}_{i}$ 对应的 ${b}_{l,x}$${c}_{i}$ 个，则当 $0\le \delta \le {b}_{\text{1}}-1$ 时， $c\left(\delta \right)=0$ ；当 ${b}_{1}\le \delta \le {b}_{\text{2}}-1$ 时， $c\left(\delta \right)={c}_{1}$$\cdots$ ；当 ${b}_{i}\le \delta \le {b}_{i+1}-1$ 时， $c\left(\delta \right)={c}_{\text{1}}+{c}_{2}+\cdots +{c}_{i}$。接下来计算所有的 ${b}_{l,x}$

$l=t\left(1\le t\le q\right)$ 时，有 $t\frac{{q}^{2}-1}{r}-\frac{q+1}{r}=\left(\frac{t\left(q+1\right)}{r}-x\right)q+xq-\frac{\left(t+1\right)\left(q+1\right)}{r}$，因为 $\text{0}\le j,k\le n-1$，则

$0\le \frac{t\left(q+1\right)}{r}-x\le n-1$$0\le xq-\frac{\left(t+1\right)\left(q+1\right)}{r}\le n-1$

$⌈\frac{\left(t+1\right)\left(q+1\right)}{rq}⌉\le x\le ⌊\frac{{q}^{2}-1-r+\left(t+1\right)\left(q+1\right)}{rq}⌋.$

$xq-\frac{\left(t+1\right)\left(q+1\right)}{r}\ge \frac{t\left(q+1\right)}{r}-x$$x\ge ⌈\frac{2t+1}{r}⌉$ 时， ${b}_{t,x}=xq-\frac{\left(t+1\right)\left(q+1\right)}{r}$，否则 ${b}_{t,x}=\frac{t\left(q+1\right)}{r}-x$。得到所有的 ${b}_{t,x}$ 如下：

${b}_{t,⌈\frac{\left(t+1\right)\left(q+1\right)}{rq}⌉}=\frac{t\left(q+1\right)}{r}-⌈\frac{\left(t+1\right)\left(q+1\right)}{rq}⌉,\cdots ,{b}_{t,⌈\frac{2t+1}{r}⌉-1}=\frac{t\left(q+1\right)}{r}-\left(⌈\frac{2t+1}{r}⌉-1\right)$

${b}_{t,⌈\frac{2t+1}{r}⌉}=⌈\frac{2t+1}{r}⌉q-\frac{\left(t+1\right)\left(q+1\right)}{r},\cdots ,{b}_{t,⌊\frac{{q}^{2}-1-r+\left(t+1\right)\left(q+1\right)}{rq}⌋}=⌊\frac{{q}^{2}-1-r+\left(t+1\right)\left(q+1\right)}{rq}⌋q-\frac{\left(t+1\right)\left(q+1\right)}{r}.$

$l=\text{1}$，得到所有的 ${b}_{\text{1},x}$

${b}_{1,⌈\frac{3}{r}⌉}=⌈\frac{3}{r}⌉q-\frac{2\left(q+1\right)}{r},{b}_{1,⌈\frac{3}{r}⌉+1}=\left(⌈\frac{3}{r}⌉+1\right)q-\frac{2\left(q+1\right)}{r},\cdots ,{b}_{1,\frac{q+1}{r}}=\frac{q+1}{r}q-\frac{2\left(q+1\right)}{r}.$

$l=\text{2}$，得到所有的 ${b}_{\text{2},x}$

${b}_{2,⌈\frac{3\left(q+1\right)}{r}⌉}=\frac{2\left(q+1\right)}{r}-⌈\frac{3\left(q+1\right)}{r}⌉,\cdots ,{b}_{2,⌈\frac{5}{r}⌉-1}=\frac{2\left(q+1\right)}{r}-\left(⌈\frac{5}{r}⌉-1\right)$

${b}_{2,⌈\frac{5}{r}⌉}=\left(⌈\frac{5}{r}⌉\right)q-\frac{3\left(q+1\right)}{r},\cdots ,{b}_{2,⌊\frac{{q}^{2}-1-r+3\left(q+1\right)}{rq}⌋}=⌊\frac{{q}^{2}-1-r+3\left(q+1\right)}{rq}⌋q-\frac{3\left(q+1\right)}{r}.$

$⋮$

$l=q$，得到所有的 ${b}_{q,x}$

${b}_{q,\frac{q+1}{r}+1}=\frac{q\left(q+1\right)}{r}-\left(\frac{q+1}{r}+1\right)=n-1,\cdots ,{b}_{q,\frac{2\left(q+1\right)}{r}-1}=\frac{q\left(q+1\right)}{r}-\left(\frac{2\left(q+1\right)}{r}-1\right).$

2) 若C是一个在 ${F}_{{q}^{2}}$ 上，长度为 $n=\frac{{q}^{2}-1}{2}$$\eta$ -常循环码，它的定义集为 $Z={\cup }_{j=0}^{\delta }{C}_{1+\text{2}j}$，则存在参数为 ${〚n,n-2\left(\delta +1\right)+c\left(\delta \right),\delta +\text{2};c\left(\delta \right)〛}_{q}$ EAQMDS码。当 $0\le \delta \le q-\text{2}$ 时， $c\left(\delta \right)=0$，得 ${C}^{\perp H}\subseteq C$ ；当 $q-\text{1}\le \delta \le \frac{\text{3}q-5}{2}$ 时， $c\left(\delta \right)=\text{2}$ ；当 $\frac{\text{3}q-3}{2}\le \delta \le \text{2}q-3$ 时， $c\left(\delta \right)=4$ ；当 $\delta =\text{2}q-\text{2}$ 时， $c\left(\delta \right)=6$ ；当 $\text{2}q-1\le \delta \le \frac{5q-7}{2}$ 时， $c\left(\delta \right)=8$

$l=1$ 时， $q-2,2q-2,3q-2,4q-2,5q-2,\cdots ,\left(q-\text{1}\right)q-2.$

$l=\text{2}$ 时， $\text{2}q-2,2q-\text{3},3q-\text{3},4q-\text{3},5q-\text{3},\cdots ,\left(q-\text{1}\right)q-\text{3}\text{.}$

$l=\text{3}$ 时， $\text{3}q-2,\text{3}q-\text{3},3q-\text{4},4q-\text{4},5q-\text{4},\cdots ,\left(q-\text{1}\right)q-\text{4}\text{.}$

$l=\text{4}$ 时， $\text{4}q-2,\text{4}q-\text{3},\text{4}q-\text{4},4q-\text{5},5q-\text{5},\cdots ,\left(q-\text{1}\right)q-\text{5}\text{.}$

$⋮$

$l=t$ 时， $tq-2,tq-3,tq-4,tq-5,\cdots ,tq-t,tq-\left(t+1\right),\left(t+1\right)q-\left(t+1\right),\cdots ,\left(q-1\right)q-\left(t+1\right).$

$⋮$

$l=q+1$ 时， $\left(q+1\right)q-\left(q-2\right)=n-1.$

$\begin{array}{l}{b}_{1}=q-2,{c}_{1}=1;{b}_{2}=2q-3,{c}_{2}=1;{b}_{3}=2q-2,{c}_{3}=2;\\ {b}_{4}=3q-4,{c}_{4}=1;{b}_{5}=3q-3,{c}_{5}=2;{b}_{6}=3q-2,{c}_{6}=2.\end{array}$

2) 若C是一个在 ${F}_{{q}^{2}}$ 上，长度为 $n=\frac{{q}^{2}-1}{r}$$\eta$ -常循环码，其中 $r|q+1$$r\ge \text{4}$，r是一个偶数。令 $r=\text{2}m$$\lambda =\frac{q+1}{r}$。C的定义集为 $Z={\cup }_{j=0}^{\delta }{C}_{1+rj}$，则存在参数为 ${〚n,n-2\left(\delta +1\right)+c\left(\delta \right),\delta +\text{2};c\left(\delta \right)〛}_{q}$ 的EAQMDS码。当 $0\le \delta \le m\lambda -\text{2}$ 时， $c\left(\delta \right)=0$，得 ${C}^{\perp H}\subseteq C$ ；当 $m\lambda -\text{1}\le \delta \le \left(m+1\right)\lambda -\text{2}$ 时， $c\left(\delta \right)=2$ ；当 $\left(m+1\right)\lambda -1\le \delta \le \left(m+2\right)\lambda -2$ 时， $c\left(\delta \right)=4$ ；当 $\left(m+2\right)\lambda -1\le \delta \le \left(m+3\right)\lambda -2$ 时， $c\left(\delta \right)=6$$\cdots$ ；当 $\left(m+t\right)\lambda -1\le \delta \le \left(m+t+1\right)\lambda -2$ 时， $c\left(\delta \right)=2\left(t+1\right)$$\cdots$ ；当 $\left(2m-\text{2}\right)\lambda -1\le \delta \le q-2$ 时， $c\left(\delta \right)=2m-2$

$\delta =q-1$ 时， $c\left(\delta \right)=2m$

$l\le 2m-1$ 时，有 $l\frac{{q}^{2}-1}{r}-\frac{q+1}{r}=\left(\frac{l\left(q+1\right)}{r}-\text{1}\right)q+q-\frac{\left(l+\text{1}\right)\left(q+1\right)}{r}$

$l=2m$ 时，有 $2m\frac{{q}^{2}-1}{r}-\frac{q+1}{r}=\left(\frac{2m\left(q+1\right)}{r}-\text{2}\right)q+q-1$

$l=2m+1$ 时，有 $\left(2m+\text{1}\right)\frac{{q}^{2}-1}{r}-\frac{q+1}{r}=\left(q-1\right)q+q-\frac{q+1}{r}-1.$

$l=m$ 时，有

$q-\frac{\left(m+1\right)\left(q+1\right)}{r}=\frac{m\left(q+1\right)}{r}-1$${b}_{m,1}=m\frac{q+1}{r}-1$

$1\le l 时，有

$q-\frac{\left(l+\text{1}\right)\left(q+1\right)}{r}>\frac{l\left(q+1\right)}{r}-1$${b}_{l,1}=q-\frac{\left(l+\text{1}\right)\left(q+1\right)}{r}$

$m 时，有

$q-\frac{\left(l+\text{1}\right)\left(q+1\right)}{r}<\frac{l\left(q+1\right)}{r}-1$${b}_{l,1}=\frac{l\left(q+1\right)}{r}-1.$

$\frac{\left(m+t\right)\left(q+1\right)}{r}-1=q-\frac{\left(m-t+1\right)\left(q+1\right)}{r}$

$\begin{array}{l}{b}_{m,1}=m\frac{q+1}{r}-1,{b}_{m-1,1}={b}_{m+1,1}=\frac{\left(m+t\right)\left(q+1\right)}{r}-1,\cdots ,{b}_{m-t,1}={b}_{m+t,1}=\frac{\left(m+t\right)\left(q+1\right)}{r}-1\\ ,\cdots ,{b}_{1,1}={b}_{2m-1,1}=\frac{\left(2m-1\right)\left(q+1\right)}{r}-1,{b}_{2m,1}=q-1,{b}_{2m+1,1}=q-1.\end{array}$

$\begin{array}{l}{b}_{1}=m\lambda -1,{c}_{1}=1;{b}_{2}=\left(m+1\right)\lambda -1,{c}_{2}=2;{b}_{3}=\left(m+2\right)\lambda -1,{c}_{3}=2;\cdots ;\\ {b}_{t+1}=\left(m+t\right)\lambda -1,{c}_{t+1}=2;\cdots ;{b}_{m}=\left(2m-1\right)\lambda -1,{c}_{m}=2;{b}_{m+1}=q-1,{c}_{m+1}=2.\end{array}$

3.2. 构造长为 $n=\frac{{q}^{2}+1}{10}$，其中 $q\equiv 7\left(mod10\right)$ 与长为 $n=\frac{{q}^{2}+1}{34}$，其中 $q\equiv 21\left(mod34\right)$ 的 EAQMDS码

$\left(j,k\right)$$s-rj\equiv -q\left(s-rk\right)\left(\mathrm{mod}rn\right)$ 的解 $⇔$ $\left(k,j\right)$$s-rj\equiv -q\left(s+rk\right)\left(\mathrm{mod}rn\right)$ 的解。

$-q\left(j+qk\right)=-qj-{q}^{2}k=-qj+k-\left({q}^{2}+1\right)k=k-qj\equiv \text{0}\left(\mathrm{mod}n\right)$

$k-qj\equiv \text{0}\left(\mathrm{mod}n\right)$

$\left(k,j\right)$$j-qk\equiv 0\left(\mathrm{mod}n\right)$ 的解。同理可证充分性。

$-q\left(j+qk-\frac{q+1}{2}\right)=-qj-{q}^{2}k+\frac{{q}^{2}+q}{2}=-qj+k-\left({q}^{2}+1\right)k+\frac{{q}^{2}+1+q-1}{2}=k-qj+\frac{q-1}{2}\equiv \text{0}\left(\mathrm{mod}n\right)$

$k-qj+\frac{q-1}{2}\equiv \text{0}\left(\mathrm{mod}n\right)$

$\left(k,j\right)$$j-qk+\frac{q-1}{\text{2}}\equiv 0\left(\mathrm{mod}n\right)$ 的解。同理可证充分性。

$0\le \delta \le \frac{3q-\text{1}}{10}-\text{1}$ 时， $c\left(\delta \right)=\text{1}$ ；当 $\frac{3q-1}{10}\le \delta \le \frac{4q+2}{10}-1$ 时， $c\left(\delta \right)=5$

$\frac{4q+2}{10}\le \delta \le \frac{5q+5}{10}-1$ 时， $c\left(\delta \right)=\text{9}$ ；当 $\frac{5q+5}{10}\le \delta \le \frac{6q-\text{2}}{10}-1$ 时， $c\left(\delta \right)=\text{13}$

$\frac{6q-2}{10}\le \delta \le \frac{7q+1}{10}-1$ 时， $c\left(\delta \right)=\text{17}$ ；当 $\frac{7q+1}{10}\le \delta \le \frac{\text{8}q-6}{10}-1$ 时， $c\left(\delta \right)=\text{21}$

$\frac{\text{8}q-\text{6}}{10}\le \delta \le \frac{8q+4}{10}-1$ 时， $c\left(\delta \right)=\text{25}$ ；当 $\frac{8q+4}{10}\le \delta \le \frac{9q-3}{10}$ 时， $c\left(\delta \right)=\text{29}$

$\frac{9q-3}{10}\le \delta \le \frac{9q+7}{10}-1$ 时， $c\left(\delta \right)=\text{33}$

$\begin{array}{c}c=|Z\cap \left(-qZ\right)|\\ =|\left\{\left(j,k\right)|s±rj\equiv -q\left(s±rk\right)\left(\mathrm{mod}rn\right),1\le j,k\le q-1\right\}|+1\\ =2|\left\{\left(j,k\right)|j+qk\equiv 0\left(\mathrm{mod}n\right),1\le j,k\le q-1\right\}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+2|\left\{\left(j,k\right)|j-qk\equiv 0\left(\mathrm{mod}n\right),1\le j,k\le q-1\right\}|+1.\end{array}$

$j+qk\equiv \text{0}\left(\mathrm{mod}n\right)$ 有整数解 $\left(j,k\right)$，其中 $\text{1}\le j,k\le q-1,$

$⇔$$1\le l\le \text{9},l\in Z$ 来说， $nl=j+qk$ 有整数解 $\left(j,k\right)$，其中 $\text{1}\le j,k\le q-1$

$j-qk\equiv \text{0}\left(\mathrm{mod}n\right)$ 有整数解 $\left(j,k\right)$，其中 $\text{1}\le j,k\le q-1,$

$⇔$$-9\le l\le -1,l\in Z$ 来说， $nl=j-qk$ 有整数解 $\left(j,k\right)$，其中 $\text{1}\le j,k\le q-1$

$l=\text{1}$ 时，有 $\frac{{q}^{2}+1}{\text{10}}=\left(\frac{q-7}{10}\right)q+\frac{7q+1}{10}$${b}_{\text{1},1}=\frac{7q+1}{\text{10}}$

$l=\text{2}$ 时，有 $2\frac{{q}^{2}+1}{\text{10}}=\left(\frac{2\left(q-7\right)}{10}+1\right)q+\frac{4q+2}{10}$${b}_{\text{2},1}=\frac{\text{4}q+\text{2}}{\text{10}}$

$l=\text{3}$ 时，有 $\text{3}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{3}\left(q-7\right)}{10}+\text{2}\right)q+\frac{q+\text{3}}{10}$${b}_{3,1}=\frac{3q-1}{10}$

$l=\text{4}$ 时，有 $\text{4}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{4}\left(q-7\right)}{10}+\text{2}\right)q+\frac{\text{8}q+\text{4}}{10}$${b}_{4,1}=\frac{8q+4}{10}$

$l=\text{5}$ 时，有 $\text{5}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{5}\left(q-7\right)}{10}+\text{3}\right)q+\frac{\text{5}q+\text{5}}{10}$${b}_{5,1}=\frac{5q+5}{10}$

$l=\text{6}$ 时，有 $\text{6}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{6}\left(q-7\right)}{10}+\text{4}\right)q+\frac{\text{2}q+\text{6}}{10}$${b}_{6,1}=\frac{6q-2}{10}$

$l=\text{7}$ 时，有 $\text{7}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{7}\left(q-7\right)}{10}+\text{4}\right)q+\frac{\text{9}q+\text{7}}{10}$${b}_{7,1}=\frac{9q+7}{10}$

$l=\text{8}$ 时，有 $\text{8}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{8}\left(q-7\right)}{10}+\text{5}\right)q+\frac{\text{6}q+\text{8}}{10}$${b}_{8,1}=\frac{8q-6}{10}$

$l=\text{9}$ 时，有 $\text{9}\frac{{q}^{2}+1}{\text{10}}=\left(\frac{\text{9}\left(q-7\right)}{10}+\text{6}\right)q+\frac{\text{3}q+\text{9}}{10}$${b}_{9,1}=\frac{9q-3}{10}$

$\text{1}\le l\le 9,l\in Z$，将 ${b}_{l,1}=\mathrm{max}\left(j,k\right)$ 从小到大排列，依次记为 ${b}_{1},{b}_{2},{b}_{3},\cdots ,{b}_{9}$。得

${b}_{1}=\frac{3q-1}{10}$${b}_{2}=\frac{4q+2}{10}$${b}_{\text{3}}=\frac{5q+5}{10}$${b}_{\text{4}}=\frac{6q-\text{2}}{10}$${b}_{\text{5}}=\frac{7q+1}{10}$

${b}_{\text{6}}=\frac{\text{8}q-6}{10}$${b}_{7}=\frac{8q+4}{10}$${b}_{8}=\frac{9q-3}{10}$${b}_{9}=\frac{9q+7}{10}.$

$0\le \delta \le \frac{\text{2}q+\text{6}}{10}-\text{1}$ 时， $c\left(\delta \right)=\text{0}$ ；当 $\frac{2q+6}{10}\le \delta \le \frac{4q+2}{10}-1$ 时， $c\left(\delta \right)=\text{4}$

$\frac{4q+2}{10}\le \delta \le \frac{5q+5}{10}-1$ 时， $c\left(\delta \right)=8$ ；当 $\frac{5q+5}{10}\le \delta \le \frac{6q+8}{10}-1$ 时， $c\left(\delta \right)=\text{16}$

$\frac{6q+\text{8}}{10}\le \delta \le \frac{7q+1}{10}-1$ 时， $c\left(\delta \right)=\text{20}$ ；当 $\frac{7q+1}{10}\le \delta \le \frac{\text{7}q+\text{11}}{10}-1$ 时， $c\left(\delta \right)=\text{24}$

$\frac{7q+\text{1}1}{10}\le \delta \le \frac{\text{8}q+\text{4}}{10}-1$ 时， $c\left(\delta \right)=\text{28}$ ；当 $\frac{\text{8}q+\text{4}}{10}\le \delta \le \frac{\text{9}q-3}{10}-1$ 时， $c\left(\delta \right)=\text{32}$

$\frac{\text{9}q-3}{10}\le \delta \le \frac{9q+7}{10}-1$ 时， $c\left(\delta \right)=\text{36}$

$\begin{array}{c}c=|Z\cap \left(-qZ\right)|\\ =|\left\{\left(j,k\right)|s±r\left(\frac{n-1}{2}-j\right)\equiv -q\left(s±r\left(\frac{n-1}{2}-k\right)\left(\mathrm{mod}rn\right),0\le j,k\le q-1\right\}|\\ =2|\left\{\left(j,k\right)|j+qk-\frac{q+1}{2}\equiv 0\left(\mathrm{mod}n\right),0\le j,k\le q-1\right\}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+2|\left\{\left(j,k\right)|j-qk+\frac{q-1}{2}\equiv 0\left(\mathrm{mod}n\right),0\le j,k\le q-1\right\}|.\end{array}$

$j+qk-\frac{q+1}{2}\equiv \text{0}\left(\mathrm{mod}n\right)$ 有整数解 $\left(j,k\right)$，其中 $\text{0}\le j,k\le q-1,$

$⇔$$\text{0}\le l\le \text{9},l\in Z$ 来说， $nl+\frac{q+1}{2}=j+qk$ 有整数解 $\left(j,k\right)$，其中 $\text{0}\le j,k\le q-1$

$j-qk+\frac{q-1}{2}\equiv \text{0}\left(\mathrm{mod}n\right)$ 有整数解 $\left(j,k\right)$，其中 $\text{0}\le j,k\le q-1,$

$⇔$$-9\le l\le \text{0},l\in Z$ 来说， $nl-\frac{q-1}{2}=j-qk$ 有整数解 $\left(j,k\right)$，其中 $\text{0}\le j,k\le q-1$

${b}_{0}=\frac{2q+6}{10}$${b}_{1}=\frac{4q+2}{10}$${b}_{\text{2}}=\frac{\text{5}q+\text{5}}{10}$${b}_{\text{3}}=\frac{5q+5}{10}$${b}_{\text{4}}=\frac{6q+\text{8}}{10}$

${b}_{5}=\frac{7q+1}{10}$${b}_{\text{6}}=\frac{\text{7}q+\text{11}}{10}$${b}_{7}=\frac{8q+4}{10}$${b}_{8}=\frac{9q-\text{3}}{10}$${b}_{9}=\frac{9q+7}{10}$

${b}_{0}=0$${b}_{1}=\frac{5q-\text{3}}{34}$${b}_{\text{2}}=\frac{\text{8}q+\text{2}}{34}$${b}_{3}=\frac{10q-6}{34}$${b}_{\text{4}}=\frac{\text{11}q+\text{7}}{34}$

${b}_{\text{5}}=\frac{\text{13}q-\text{1}}{34}$${b}_{\text{6}}=\frac{\text{14}q+12}{34}$${b}_{\text{7}}=\frac{\text{15}q-\text{9}}{34}$${b}_{\text{8}}=\frac{\text{16}q+\text{4}}{34}$${b}_{\text{9}}=\frac{\text{17}q+\text{17}}{34}$

${b}_{\text{10}}=\frac{\text{18}q-\text{4}}{34}$${b}_{\text{11}}=\frac{\text{19}q+\text{9}}{34}$${b}_{\text{12}}=\frac{\text{20}q-\text{12}}{34}$${b}_{\text{13}}=\frac{\text{21}q+\text{1}}{34}$${b}_{\text{14}}=\frac{\text{22}q-20}{34}$

${b}_{\text{15}}=\frac{\text{22}q+\text{14}}{34}$${b}_{\text{16}}=\frac{\text{23}q-\text{7}}{34}$${b}_{\text{17}}=\frac{\text{24}q+\text{6}}{34}$${b}_{\text{18}}=\frac{\text{25}q-\text{15}}{34}$${b}_{\text{19}}=\frac{\text{25}q+\text{19}}{34}$

${b}_{\text{20}}=\frac{\text{26}q-2}{34}$${b}_{\text{21}}=\frac{\text{27}q-23}{34}$${b}_{\text{22}}=\frac{\text{27}q+11}{34}$${b}_{\text{23}}=\frac{\text{28}q-\text{10}}{34}$${b}_{\text{24}}=\frac{\text{28}q+24}{34}$

${b}_{\text{25}}=\frac{\text{29}q+\text{3}}{34}$${b}_{\text{26}}=\frac{\text{30}q-\text{18}}{34}$${b}_{\text{27}}=\frac{\text{30}q+16}{34}$${b}_{\text{28}}=\frac{\text{31}q-\text{5}}{34}$${b}_{\text{29}}=\frac{\text{31}q+\text{29}}{34}$

${b}_{\text{30}}=\frac{\text{32}q-26}{34}$${b}_{\text{31}}=\frac{\text{32}q+\text{8}}{34}$${b}_{\text{32}}=\frac{\text{33}q-13}{34}$${b}_{\text{33}}=\frac{\text{33}q+\text{21}}{34}$${b}_{\text{34}}=q$

$0\le i\le 33,i\in Z$，当 ${b}_{i}\le \delta \le {b}_{i+1}-1$ 时， $c\left(\delta \right)=4i+1$

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