# 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

DOI: 10.12677/PM.2019.99129, PDF, HTML, XML, 下载: 298  浏览: 2,060  国家自然科学基金支持

Abstract: Cyclic packing is one of efficient approaches for constructing optical orthogonal code (OOC). In this paper, a new infinite class of optimal variable-weight with length 27p and weights w={3,7} are obtained via constructing cyclic packing with blocks 3 and 7, for any prime p≡3(mod4) and p≥7.

1. 引言

2. 预备知识

$W=\left\{{w}_{1},{w}_{2},\cdots ,{w}_{r}\right\}$ 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) $|{B}_{j}|\in W,1\le j\le t$；2) $\Delta B$ 覆盖 ${Z}_{n}\setminus \left\{0\right\}$ 中的每个元素至多一次。

${B}_{j}\in \mathcal{B},1\le j\le t$ 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 $\underset{i=1}{\overset{r}{\sum }}{q}_{i}=1$。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}|\mathcal{B}|$$2-CP\left(W,1;n\right)$，其中 ${q}_{i}\in Q,1\le i\le r$

$A=\left\{\left({a}_{1},{j}_{1}\right),\left({a}_{2},{j}_{2}\right),\cdots ,\left({a}_{k},{j}_{k}\right)\right\}$${Z}_{p}×{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。

$\mathcal{F}=\left\{A:A=\left\{\left({a}_{1},{j}_{1}\right),\left({a}_{2},{j}_{2}\right),\cdots ,\left({a}_{k},{j}_{k}\right)\right\}\subset {Z}_{p}×{Z}_{m},k\in K\right\}$。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) $B\cdot A=\left\{b•A|b\in B\right\},B\subseteq {Z}_{p}$

1) $2\in {C}_{0}^{2},3\in {C}_{0}^{2},5\in {C}_{0}^{2},7\in {C}_{0}^{2}⇔p\equiv 71,191,239,359,431,599\left(\mathrm{mod}840\right)$

2) $2\in {C}_{0}^{2},3\in {C}_{0}^{2},5\in {C}_{0}^{2},7\in {C}_{1}^{2}⇔p\equiv 311,479,551,671,719,839\left(\mathrm{mod}840\right)$

3) $2\in {C}_{1}^{2},3\in {C}_{1}^{2},5\in {C}_{1}^{2},7\in {C}_{0}^{2}⇔p\equiv 43,67,163,403,547,667\left(\mathrm{mod}840\right)$

4) $2\in {C}_{1}^{2},3\in {C}_{1}^{2},5\in {C}_{1}^{2},7\in {C}_{1}^{2}⇔p\equiv 187,283,307,523,643,787\left(\mathrm{mod}840\right)$

5) $2\in {C}_{0}^{2},3\in {C}_{0}^{2},5\in {C}_{1}^{2},7\in {C}_{0}^{2}⇔p\equiv 23,263,407,527,743,767\left(\mathrm{mod}840\right)$

6) $2\in {C}_{0}^{2},3\in {C}_{0}^{2},5\in {C}_{1}^{2},7\in {C}_{1}^{2}⇔p\equiv 47,143,167,383,503,647\left(\mathrm{mod}840\right)$

7) $2\in {C}_{1}^{2},3\in {C}_{1}^{2},5\in {C}_{0}^{2},7\in {C}_{0}^{2}⇔p\equiv 211,331,379,499,571,739\left(\mathrm{mod}840\right)$

8) $2\in {C}_{1}^{2},3\in {C}_{1}^{2},5\in {C}_{0}^{2},7\in {C}_{1}^{2}⇔p\equiv 19,139,451,619,691,811\left(\mathrm{mod}840\right)$

9) $2\in {C}_{0}^{2},3\in {C}_{1}^{2},5\in {C}_{0}^{2},7\in {C}_{0}^{2}⇔p\equiv 79,151,319,631,751,799\left(\mathrm{mod}840\right)$

10) $2\in {C}_{0}^{2},3\in {C}_{1}^{2},5\in {C}_{0}^{2},7\in {C}_{1}^{2}⇔p\equiv 31,199,271,391,439,559\left(\mathrm{mod}840\right)$

11) $2\in {C}_{1}^{2},3\in {C}_{0}^{2},5\in {C}_{1}^{2},7\in {C}_{0}^{2}⇔p\equiv 107,323,347,443,683,827\left(\mathrm{mod}840\right)$

12) $2\in {C}_{1}^{2},3\in {C}_{0}^{2},5\in {C}_{1}^{2},7\in {C}_{1}^{2}⇔p\equiv 83,227,467,563,587,803\left(\mathrm{mod}840\right)$

13) $2\in {C}_{1}^{2},3\in {C}_{0}^{2},5\in {C}_{0}^{2},7\in {C}_{0}^{2}⇔p\equiv 11,179,491,611,659,779\left(\mathrm{mod}840\right)$

14) $2\in {C}_{1}^{2},3\in {C}_{0}^{2},5\in {C}_{0}^{2},7\in {C}_{1}^{2}⇔p\equiv 59,131,251,299,419,731\left(\mathrm{mod}840\right)$

15) $2\in {C}_{0}^{2},3\in {C}_{1}^{2},5\in {C}_{1}^{2},7\in {C}_{0}^{2}⇔p\equiv 127,247,463,487,583,823\left(\mathrm{mod}840\right)$

16) $2\in {C}_{0}^{2},3\in {C}_{1}^{2},5\in {C}_{1}^{2},7\in {C}_{1}^{2}⇔p\equiv 103,223,367,607,703,727\left(\mathrm{mod}840\right)$

3. 定理的证明

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

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

${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\}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\begin{array}{c}\mathcal{A}=\left\{\left\{0,1,3,7,12,20,44\right\},\left\{0,14,35,50,66,88,114\right\},\left\{0,10,28,55,95,118,160\right\},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{0,25,58\right\},\left\{0,30,76\right\},\left\{0,34,83\right\},\left\{0,47,98\right\},\left\{0,54,116\right\},\left\{0,56,117\right\}\right\}\end{array}$

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

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