构造一系列新的最优循环填充及其相应的光正交码
Constructions of a New Infinite Class of Optimal Cyclic Packing and Their Related OOCs

1. 引言

光正交码一般分为码重单一的常重量光正交码 [1] 和多种码重的变重量光正交码 [2]。由于其在移动无线电通信、跳频扩频通信、雷达等光码分多址网络中有着广泛的应用 [3]，因而对光正交码的构造近年来备受关注。1998年，Yin为了构造最优常重量光正交码引入了区组大小单一的循环填充，并证明循环填充存在性等价于常重量光正交码的存在性 [4]。后来，为了能够构造最优变重量光正交码，2010年，Wu等人给出区组大小多样的循环填充相关定义并证明其存在性等价于变重量光正交码的存在性 [5]。

目前，利用循环填充构造常重量光正交码已取得较多的结果 [6] - [12]。同样地，利用循环填充构造双重量光正交码取得的结果主要是码重 $W\in \left\{\left\{3,4\right\},\left\{3,5\right\},\left\{3,6\right\},\left\{4,5\right\},\left\{4,6\right\}\right\}$ 时的情况 [5] [13] - [18]。对于含有码重大于7的双重量光正交码一系列直接具体的构造结果甚少。

根据文献 [2] 知，码重大的光正交码抗其他干扰的性能较好。因此，本文直接具体构造出一系列新的区组大小为3和7的最优循环填充及其相应的光正交码。即

定理：对于任何素数 $p\equiv 3\left(\mathrm{mod}4\right)$ 且 $p\ge 7$，存在最优的循环填充 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$ 和最优的变重量光正交码 $\left(27p,\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\}\right)-\text{OOC}$。

注：根据文献 [5] 知，若 $W=\left\{w_1,w_2,\cdots ,w_r\right\}$， $Q=\left\{q_1,q_2,\cdots ,q_r\right\}$ 是一个r-元组的正有理数且 ${\displaystyle \underset{i=1}{\overset{r}{\sum }}{q}_i}=1$，则符号 $\left(n,W,1,Q\right)-\text{OOC}$ 表示相关系数为1码字长度为n码重为 $w_i\in W$ 码字的个数占总码字个数的比例为 $q_i\in Q$ (其中 $i=1,2,\cdots ,r$ )的变重量光正交码。循环填充的符号 $2-CP\left(W,1,Q;n\right)$ 参见下文。

2. 预备知识

设G是一个交换群， $\mathcal{B}=\left\{B:B\subseteq G\right\}$。记符号 $\Delta B=\left\{x-y:x,y\in B,x\ne y\right\},B\subseteq G$。更进一步，记符号 $\Delta \mathcal{B}={\displaystyle {\cup }_{B\in \mathcal{B}}\Delta B}$。其中 $\Delta B$ 和 $\Delta \mathcal{B}$ 都是多重集合。

设 $W=\left\{w_1,w_2,\cdots ,w_r\right\}$ 是r个大于1的有序整数组， $Z_n$ 表示模n的剩余类环， $\mathcal{B}=\left\{B_j:B_j\subseteq Z_n,1\le j\le t\right\}$。称一个设计 $\left(Z_n,\mathcal{B}\right)$ 为循环填充 $2-CP\left(W,1;n\right)$ 若以下条件满足：

1) $\left|B_j\right|\in W,1\le j\le t$；2) $\Delta B$ 覆盖 $Z_n\setminus \left\{0\right\}$ 中的每个元素至多一次。

称 $B_j\in \mathcal{B},1\le j\le t$ 为循环填充的区组。若 $Q=\left\{q_1,q_2,\cdots ,q_r\right\}$ 是一个r-元组的正有理数且 ${\displaystyle \underset{i=1}{\overset{r}{\sum }}{q}_i}=1$。则用 $2-CP\left(W,1,Q;n\right)$ 表示区组大小等于 $w_i$ 的区组个数为 $q_i\left|\mathcal{B}\right|$ 的 $2-CP\left(W,1;n\right)$，其中 $q_i\in Q,1\le i\le r$。

特别的， $Z_{gv}$ 中的一个 $2-CP\left(W,1;gv\right)$，若 $\Delta \mathcal{B}={\displaystyle \underset{i=1}{\overset{t}{\cup }}\Delta }{B}_j$ 覆盖 $Z_{gv}\setminus v{Z}_{gv}$ 每个元素恰好一次，而不覆盖 $v{Z}_{gv}$ 的任何一个元素，则称为 $\text{g}$ -规则的。

关于最优循环填充和最优变重量光正交码之间的等价关系可用以下两个引理描述 [13]。

引理 1.1一个最优 $2-CP\left(W,1,Q;n\right)$ 等价于一个最优 $\left(n,W,1,Q\right)-\text{OOC}$。

引理 1.2设 $w={\displaystyle \underset{i=1}{\overset{r}{\sum }}{a}_i}{w}_i\left(w_i-1\right)$，其中 $w_i\in W$。若 $1\le g\le w$，则g-规则的 $2-CP\left(W,1,Q;gv\right)$ 是最优的。

设f是正整数， $p=2f+1$ 是奇素数，元素 $\theta$ 是 $Z_p^{\ast }=Z_p\setminus \left\{0\right\}$ 的一个生成元。则称 $C_0^2=\left\{{\theta }^{2i}:0\le i\le \frac{p-1}{2}\right\}$ 为二次剩余，称 $C_1^2=\theta C_0^2$ 为二次非剩余。

设 $A=\left\{\left(a_1,j_1\right),\left(a_2,j_2\right),\cdots ,\left(a_k,j_k\right)\right\}$ 是 $Z_p\times Z_m$ 的一个k元子集。再设K是一些正整数的集合，且每个元素都大于1。

若 $\mathcal{F}=\left\{A:A=\left\{\left(a_1,j_1\right),\left(a_2,j_2\right),\cdots ,\left(a_k,j_k\right)\right\}\subset Z_p\times Z_m,k\in K\right\}$。定义

1) $x\cdot A=\left\{\left(x{a}_1,j_1\right),\left(x{a}_2,j_2\right),\cdots ,\left(x{a}_k,j_k\right)\right\},x\in Z_p$；

2) $B\cdot A=\left\{b•A|b\in B\right\},B\subseteq Z_p$。

记符号 $L_i=\left\{a_l-a_s:\left\{\left(a_l,j_l\right),\left(a_s,j_s\right)\right\}\subset A\in \mathcal{F},i\equiv j_l-j_s\left(\mathrm{mod}m\right),1\le l,s\le k,k\in K\right\}$， $i=0,\cdots ,m-1$。

由文献 [16] 的构造(Construction I)知，当 $p\equiv 3\left(\mathrm{mod}4\right),p\nmid m$ 时，若 $\left|L_i\cap C_k^2\right|=1$， $0\le i\le m-1$， $k=0,1$ 且 $\mathcal{A}=\left\{C_0^2•A:A\in \mathcal{F}\right\}$。则 $\mathcal{A}$ 形成一个m-规则的 $2-CP\left(W,1,Q;mp\right)$。

为了方便，现将素数与二次剩余(二次非剩余)之间关系用以下引理 [19] 表示。

引理 1.3若 $p\equiv 3\left(\mathrm{mod}4\right)$ 是素数。则

1) $2\in C_0^2,3\in C_0^2,5\in C_0^2,7\in C_0^2\iff p\equiv 71,191,239,359,431,599\left(\mathrm{mod}840\right)$；

2) $2\in C_0^2,3\in C_0^2,5\in C_0^2,7\in C_1^2\iff p\equiv 311,479,551,671,719,839\left(\mathrm{mod}840\right)$；

3) $2\in C_1^2,3\in C_1^2,5\in C_1^2,7\in C_0^2\iff p\equiv 43,67,163,403,547,667\left(\mathrm{mod}840\right)$；

4) $2\in C_1^2,3\in C_1^2,5\in C_1^2,7\in C_1^2\iff p\equiv 187,283,307,523,643,787\left(\mathrm{mod}840\right)$；

5) $2\in C_0^2,3\in C_0^2,5\in C_1^2,7\in C_0^2\iff p\equiv 23,263,407,527,743,767\left(\mathrm{mod}840\right)$；

6) $2\in C_0^2,3\in C_0^2,5\in C_1^2,7\in C_1^2\iff p\equiv 47,143,167,383,503,647\left(\mathrm{mod}840\right)$；

7) $2\in C_1^2,3\in C_1^2,5\in C_0^2,7\in C_0^2\iff p\equiv 211,331,379,499,571,739\left(\mathrm{mod}840\right)$；

8) $2\in C_1^2,3\in C_1^2,5\in C_0^2,7\in C_1^2\iff p\equiv 19,139,451,619,691,811\left(\mathrm{mod}840\right)$；

9) $2\in C_0^2,3\in C_1^2,5\in C_0^2,7\in C_0^2\iff p\equiv 79,151,319,631,751,799\left(\mathrm{mod}840\right)$；

10) $2\in C_0^2,3\in C_1^2,5\in C_0^2,7\in C_1^2\iff p\equiv 31,199,271,391,439,559\left(\mathrm{mod}840\right)$；

11) $2\in C_1^2,3\in C_0^2,5\in C_1^2,7\in C_0^2\iff p\equiv 107,323,347,443,683,827\left(\mathrm{mod}840\right)$；

12) $2\in C_1^2,3\in C_0^2,5\in C_1^2,7\in C_1^2\iff p\equiv 83,227,467,563,587,803\left(\mathrm{mod}840\right)$；

13) $2\in C_1^2,3\in C_0^2,5\in C_0^2,7\in C_0^2\iff p\equiv 11,179,491,611,659,779\left(\mathrm{mod}840\right)$；

14) $2\in C_1^2,3\in C_0^2,5\in C_0^2,7\in C_1^2\iff p\equiv 59,131,251,299,419,731\left(\mathrm{mod}840\right)$；

15) $2\in C_0^2,3\in C_1^2,5\in C_1^2,7\in C_0^2\iff p\equiv 127,247,463,487,583,823\left(\mathrm{mod}840\right)$；

16) $2\in C_0^2,3\in C_1^2,5\in C_1^2,7\in C_1^2\iff p\equiv 103,223,367,607,703,727\left(\mathrm{mod}840\right)$。

3. 定理的证明

对于每一个素数 $p\equiv 3\left(\mathrm{mod}4\right)$，且 $p>7$，由于 $\gcd \left(27,p\right)=1$，因此根据中国剩余定理知 $Z_{27p}$ 同构于 $Z_p\times Z_{27}$。下面我们对 $p\equiv 3\left(\mathrm{mod}4\right)$ 分13种情形以引理的形式进行讨论。

引理 2.1若 $p\equiv 71,191,239,359,431,599,311,479,551,671,719,839\left(\mathrm{mod}840\right)$ 是素数， $\xi =\min \left\{x|x\in C_1\right\}$，则 $\mathcal{A}=\left\{C_0^2\cdot A_{1,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{1,1}=\left\{\left(0,0\right),\left(5,0\right),\left(1,1\right),\left(\xi ,3\right),\left(3,5\right),\left(2,11\right),\left(4,18\right)\right\}$ ,

$A_{1,2}=\left\{\left(0,0\right),\left(-5,4\right),\left(-1,12\right)\right\}$ , $A_{1,3}=\left\{\left(0,0\right),\left(1,6\right),\left(-1,13\right)\right\}$ .

证明：根据 $A_{1,1},A_{1,2},A_{1,3}$ 计算 $L_i,0\le i\le 26$。

易见 $L_s=-L_{27-s},14\le s\le 26$。因此，只需计算 $L_i,0\le i\le 13$。

$L_0=\left\{5,-5\right\}$， $L_1=L_9=\left\{1,-4\right\}$， $L_2=\left\{\xi -1,3-\xi \right\}$， $L_3=\left\{\xi ,\xi -5\right\}$， $L_4=\left\{2,-5\right\}$， $L_5=-L_{11}=\left\{3,-2\right\}$， $L_6=L_{13}=\left\{1,-1\right\}$， $L_7=\left\{2,-2\right\}$， $L_8=\left\{4,2-\xi \right\}$， $L_{10}=\left\{3,-1\right\}$， $L_{12}=\left\{\xi -4,-1\right\}$。

由引理1.3的(1)和(2)知， $2\in C_0^2,3\in C_0^2,5\in C_0^2$，且 $\xi -l\in C_0,1\le l<\xi$。不难验证 $\left|L_i\cap C_k^2\right|=1$， $0\le i\le 26$， $k=0,1$。由前述知， $\mathcal{A}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$。

利用引理1.3相应的条件，同理可证引理2.2~2.13成立。

引理 2.2 若 $p\equiv 43,67,163,403,547,667,187,283,307,523,643,787\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{2,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{2,1}=\left\{\left(1,0\right),\left(2,0\right),\left(3,1\right),\left(-3,3\right),\left(6,5\right),\left(0,11\right),\left(-2,18\right)\right\}$ ,

$A_{2,2}=\left\{\left(0,0\right),\left(4,4\right),\left(1,12\right)\right\}$ , $A_{2,3}=\left\{\left(0,0\right),\left(1,6\right),\left(3,13\right)\right\}$ .

引理 2.3若 $p\equiv 23,263,407,527,743,767,47,143,167,383,503,647\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{3,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{3,1}=\left\{\left(9,0\right),\left(5,0\right),\left(6,1\right),\left(4,3\right),\left(8,5\right),\left(7,11\right),\left(10,18\right)\right\}$ ,

$A_{3,2}=\left\{\left(0,0\right),\left(-1,4\right),\left(4,12\right)\right\}$ , $A_{3,3}=\left\{\left(0,0\right),\left(-5,6\right),\left(5,13\right)\right\}$ .

引理 2.4若 $p\equiv 19,139,451,619,691,811\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{4,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{4,1}=\left\{\left(0,0\right),\left(2,0\right),\left(1,1\right),\left(4,3\right),\left(5,5\right),\left(3,11\right),\left(-3,18\right)\right\}$ ,

$A_{4,2}=\left\{\left(0,0\right),\left(2,4\right),\left(6,12\right)\right\}$ , $A_{4,3}=\left\{\left(0,0\right),\left(2,6\right),\left(3,13\right)\right\}$ .

引理 2.5若 $p\equiv 211,331,379,499,571,739\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{5,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{5,1}=\left\{\left(0,0\right),\left(2,0\right),\left(1,1\right),\left(4,3\right),\left(5,5\right),\left(3,11\right),\left(-3,18\right)\right\}$ ,

$A_{5,2}=\left\{\left(0,0\right),\left(2,4\right),\left(3,12\right)\right\}$ , $A_{5,3}=\left\{\left(0,0\right),\left(2,6\right),\left(3,13\right)\right\}$ .

引理 2.6若 $p\equiv 31,199,271,391,439,559\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{6,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{6,1}=\left\{\left(2,0\right),\left(0,0\right),\left(-3,1\right),\left(6,3\right),\left(5,5\right),\left(3,11\right),\left(-4,18\right)\right\}$ ,

$A_{6,2}=\left\{\left(0,0\right),\left(3,4\right),\left(-1,12\right)\right\}$ , $A_{6,3}=\left\{\left(0,0\right),\left(2,6\right),\left(5,13\right)\right\}$ .

引理 2.7若 $p\equiv 79,151,319,631,751,799\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{7,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{7,1}=\left\{\left(2,0\right),\left(0,0\right),\left(-3,1\right),\left(6,3\right),\left(5,5\right),\left(3,11\right),\left(-4,18\right)\right\}$ ,

$A_{7,2}=\left\{\left(0,0\right),\left(3,4\right),\left(-1,12\right)\right\}$ , $A_{7,3}=\left\{\left(0,0\right),\left(5,6\right),\left(7,13\right)\right\}$ .

引理 2.8若 $p\equiv 83,227,467,563,587,803\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{8,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{8,1}=\left\{\left(9,0\right),\left(5,0\right),\left(7,1\right),\left(0,3\right),\left(8,5\right),\left(6,11\right),\left(4,18\right)\right\}$ ,

$A_{8,2}=\left\{\left(0,0\right),\left(2,4\right),\left(3,12\right)\right\}$ , $A_{8,3}=\left\{\left(0,0\right),\left(2,6\right),\left(1,13\right)\right\}$ .

引理 2.9若 $p\equiv 107,323,347,443,683,827\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{9,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{9,1}=\left\{\left(9,0\right),\left(5,0\right),\left(7,1\right),\left(0,3\right),\left(8,5\right),\left(6,11\right),\left(4,18\right)\right\}$ ,

$A_{9,2}=\left\{\left(0,0\right),\left(2,4\right),\left(3,12\right)\right\}$ , $A_{9,3}=\left\{\left(0,0\right),\left(2,6\right),\left(1,13\right)\right\}$ .

引理 2.10若 $p\equiv 59,131,251,299,419,731\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{10,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{10,1}=\left\{\left(4,0\right),\left(9,0\right),\left(5,1\right),\left(1,3\right),\left(10,5\right),\left(3,11\right),\left(8,18\right)\right\}$ ,

$A_{10,2}=\left\{\left(0,0\right),\left(-1,4\right),\left(2,12\right)\right\}$ , $A_{10,3}=\left\{\left(0,0\right),\left(2,6\right),\left(-1,13\right)\right\}$ .

引理 2.11若 $p\equiv 11,179,491,611,659,779\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{11,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{11,1}=\left\{\left(4,0\right),\left(9,0\right),\left(5,1\right),\left(1,3\right),\left(10,5\right),\left(3,11\right),\left(8,18\right)\right\}$ ,

$A_{11,2}=\left\{\left(0,0\right),\left(-1,4\right),\left(2,12\right)\right\}$ , $A_{11,3}=\left\{\left(0,0\right),\left(2,6\right),\left(-1,13\right)\right\}$ .

引理 2.12若 $p\equiv 103,223,367,607,703,727\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{12,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{12,1}=\left\{\left(5,0\right),\left(2,0\right),\left(7,1\right),\left(0,3\right),\left(3,5\right),\left(4,11\right),\left(9,18\right)\right\}$ ,

$A_{12,2}=\left\{\left(0,0\right),\left(4,4\right),\left(2,12\right)\right\}$ , $A_{12,3}=\left\{\left(0,0\right),\left(-1,6\right),\left(-7,13\right)\right\}$ .

引理 2.13若 $p\equiv 127,247,463,487,583,823\left(\mathrm{mod}840\right)$ 是素数，则 $\mathcal{A}=\left\{C_0^2\cdot A_{13,j}:j=1,2,3\right\}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$，其中

$A_{13,1}=\left\{\left(7,0\right),\left(4,0\right),\left(2,1\right),\left(9,3\right),\left(5,5\right),\left(6,11\right),\left(-1,18\right)\right\}$ ,

$A_{13,2}=\left\{\left(0,0\right),\left(4,4\right),\left(1,12\right)\right\}$ , $A_{13,3}=\left\{\left(0,0\right),\left(2,6\right),\left(-1,13\right)\right\}$ .

定理的证明：对于每个素数 $p\equiv 3\left(\mathrm{mod}4\right)$ 且 $p>7$，由引理2.1~2.13可得到一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$。对于 $p=7$，设

$\begin{array}{c}\mathcal{A}=\{\left\{0,1,3,7,12,20,44\right\},\left\{0,14,35,50,66,88,114\right\},\left\{0,10,28,55,95,118,160\right\},\\ \left\{0,25,58\right\},\left\{0,30,76\right\},\left\{0,34,83\right\},\left\{0,47,98\right\},\left\{0,54,116\right\},\left\{0,56,117\right\}\}\end{array}$

则 $\mathcal{A}$ 形成一个 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$。

因此，由引理1.1和1.2知，对于任意的素数 $p\equiv 3\left(\mathrm{mod}4\right)$ 且 $p\ge 7$，可得到一个最优的 $2-CP\left(\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\};27p\right)$ 和最优的变重量光正交码 $\left(27p,\left\{3,7\right\},1,\left\{\frac{2}{3},\frac{1}{3}\right\}\right)-\text{OOC}$。

基金项目

广西自然科学基金项目(2018GXNSFAA281259)。

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