# 交换折叠交叉立方体的Hamilton分解及其性质The Hamiltonian Decomposition and Its Properties of Exchanged Folded Crossed Cube

DOI: 10.12677/CSA.2020.106116, PDF, 下载: 174  浏览: 254

Abstract: Exchanged folded crossed cube (EFCQ(s,t)) is a new interconnection network for parallel computation. In this article, author proved EFCQ(s,t) is Hamiltonian decomposition, when s=t=1;2. And EFCQ(s,t) can be decomposed into a Hamiltonian cycle and s perfect matching, when s=t=1;2;3. Finally, some properties of EFCQ(s,t) are proved.

1. 引言

2. 基本概念

V是图中顶点的集合，且 $V=\left\{{a}_{s-1}\cdots {a}_{0}{b}_{t-1}\cdots {b}_{0}c|{a}_{i},{b}_{j},c\in \left\{0,1\right\},i\in \left[0,s-1\right],j\in \left[0,t-1\right]\right\}$

E是图中边的集合，且E是由四种互不相交的边集 ${E}_{1}\cup {E}_{2}\cup {E}_{3}\cup {E}_{4}$ 组成，对任意两个顶点有 $u,v\in V,{E}_{1},{E}_{2},{E}_{3},{E}_{4}$，定义如下

1) $\left(u,v\right)\in {E}_{1}$，当且仅当 $u\left[0\right]\ne v\left[0\right],u\oplus v=1$ 其中 $\oplus$ 是异或操作。

2) $\left(u,v\right)\in {E}_{2}$，当且仅当 $u\left[t:1\right]=v\left[t:1\right],u\left[0\right]=v\left[0\right]=0$ 并且 $\left(u\left[s+t:t+1\right],v\left[s+t:t+1\right]\right)\in E\left(C{Q}_{S}\right)$

3) $\left(u,v\right)\in {E}_{3}$，当且仅当 $u\left[s+t:t+1\right]=v\left[s+t:t+1\right],u\left[0\right]=v\left[0\right]=1$ 并且 $\left(u\left[t:1\right],v\left[t:1\right]\right)\in E\left(C{Q}_{t}\right)$

4) $\left(u,v\right)\in {E}_{4}$，当且仅当 $u=\left\{{a}_{s-1}\cdots {a}_{0}{b}_{t-1}\cdots {b}_{0}c\right\},v=\left\{\stackrel{¯}{{a}_{s-1}}\cdots \stackrel{¯}{{a}_{0}}\stackrel{¯}{{b}_{t-1}}\cdots \stackrel{¯}{{b}_{0}}\stackrel{¯}{c}\right\}$$\stackrel{¯}{{a}_{i}}=1-{a}_{i},\stackrel{¯}{{b}_{j}}=1-{b}_{j},\stackrel{¯}{c}=1-c$

${a}_{i},{b}_{j},c\in \left\{0,1\right\},i\in \left[0,s-1\right],j\in \left[0,t-1\right]$

3. 主要结果

$EFCQ\left(s,t\right)$ 是非正则的，若 $s\ne t,s\ge 1,t\ge 1$

2) $EFCQ\left(2,2\right)$ 是Hamilton可分解的。

Figure 1. $EFCQ\left(1,1\right)$

$EFCQ\left(1,1\right)$ 中的Hamilton圈为：

100-101-111-110-010-011-001-000-100

$EFCQ\left(1,1\right)$ 中的完美对集为：

100-011, 101-010, 111-000, 110-001

$EFCQ\left(2,2\right)$ 中的Hamilton圈 ${H}_{1}$ 为：

01000-01001-01011-01010-11010-11011-11001-11000-10000-10001-10011-10010-00010-00011-00111-00110-01110-01111-01101-01100-11100-11101-11111-11110-10110-10111-10101-10100-00100-00101-00001-00000-01000

$EFCQ\left(2,2\right)$ 中的Hamilton圈 ${H}_{2}$ 为：

01000-11000-00111-00101-11010-10010-01101-01001-10110-00110-11001-11101-00010-01010-10101-10001-01110-11110-00001-00011-11100-10100-01011-01111-10000-00000-11111-11011-00100-01100-10011-10111-01000

Figure 2. $EFCQ\left(2,2\right)$

2) $EFCQ\left(3,3\right)$ 可以分解为32个4圈16个8圈和一个完美对集并。

3) $EFCQ\left(3,3\right)$ 可以分解为一个Hamilton圈22个4圈5个8圈和一个完美对集的并(如图3)。

a2001011-a2001010-a2011010-a2011011-a2011111-a2011110-a2001110-a2001111-a2001011；

a2001001-a2001000-a2011000-a2011001-a2011101-a2011100-a2001100-a2001101-a2001001；

a2000011-a2000010-a2010010-a2010011-a2010111-a2010110-a2000110-a2000111-a2000011；

a2000001-a2000000-a2010000-a2010001-a2010101-a2010100-a2000100-a2000101-a2000001；

a2101011-a2101010-a2111010-a2111011-a2111111-a2111110-a2101110-a2101111-a2101011；

a2101001-a2101000-a2111000-a2111001-a2111101-a2111100-a2101100-a2101101-a2101001；

a2100011-a2100010-a2110010-a2110011-a2110111-a2110110-a2100110-a2100111-a2100011；

a2100001-a2100000-a2110000-a2110001-a2110101-a2110100-a2100100-a2100101-a2100001；

${a}_{2}{a}_{1}{a}_{0}\in \left\{000,001,010,011,100,101,110,111\right\}$，24个8圈中剩余8个8圈可以表示为：

${a}_{2}{a}_{1}{a}_{0}1101\text{-}{a}_{2}{a}_{1}{a}_{0}1111\text{-}{a}_{2}{a}_{1}{a}_{0}0011\text{-}{a}_{2}{a}_{1}{a}_{0}0001\text{-}{a}_{2}{a}_{1}{a}_{0}1001\text{-}{a}_{2}{a}_{1}{a}_{0}1011$ $\text{-}{a}_{2}{a}_{1}{a}_{0}0111\text{-}{a}_{2}{a}_{1}{a}_{0}0101\text{-}{a}_{2}{a}_{1}{a}_{0}1101$

Figure 3. $ECQ\left(3,3\right)$

${b}_{2}{b}_{1}{b}_{0}c\in \left\{1000,1100,0000,0100\right\},{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\in \left\{1010,1110,0010,0110\right\}$

$EFCQ\left(3,3\right)$ 中有16个4圈可以表示为：

$\begin{array}{l}000{b}_{2}{b}_{1}{b}_{0}c\text{-}010{b}_{2}{b}_{1}{b}_{0}c\text{-}110{b}_{2}{b}_{1}{b}_{0}c\text{-}100{b}_{2}{b}_{1}{b}_{0}c\text{-}000{b}_{2}{b}_{1}{b}_{0}c\\ 001{b}_{2}{b}_{1}{b}_{0}c\text{-}011{b}_{2}{b}_{1}{b}_{0}c\text{-}101{b}_{2}{b}_{1}{b}_{0}c\text{-}111{b}_{2}{b}_{1}{b}_{0}c\text{-}001{b}_{2}{b}_{1}{b}_{0}c\\ 000{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}010{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}110{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}100{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}000{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\\ 001{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}011{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}101{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}111{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\text{-}001{{b}^{\prime }}_{2}{{b}^{\prime }}_{1}{{b}^{\prime }}_{0}{c}^{\prime }\end{array}$

$EFCQ\left(3,3\right)$ 中的完美对集为M：

0001000-1110111, 0001001-1110110, 0001011-1110100, 0001010-1110101, 0001100-1110011,

0001101-1110010, 0001111-1110000, 0001110-1110001, 0000000-1111111, 0000001-1111110,

0000011-1111100, 0000010-1111101, 0000100-1111011, 0000101-1111010, 0000111-1111000,

0000110-1111001, 0011000-1100111, 0011001-1100110, 0011011-1100100, 0011010-1100101,

0011100-1100011, 0011101-1100010, 0011111-1100000, 0011110-1100001, 0010000-1101111,

0010001-1101110, 0010011-1101100, 0010010-1101101, 0010100-1101011, 0010101-1101010,

0010111-1101000, 0010110-1101001, 0101000-1010111, 0101001-1010110, 0101011-1010100,

0101010-1010101, 0101100-1010011, 0101101-1010010, 0101111-1010000, 0101110-1010001,

0100000-1011111, 0100001-1011110, 0100011-1011100, 0100010-1011101, 0100100-1011011,

0100101-1011010, 0100111-1011000, 0100110-1011001, 0111000-1000111, 0111001-1000110,

0111011-1000100, 0111010-1000101, 0111100-1000011, 0111101-1000010, 0111111-1000000,

0111110-1000001, 0110000-1001111, 0110001-1001110, 0110011-1001100, 0110010-1001101,

0110100-1001011, 0110101-1001010, 0110111-1001000, 0110110-1001001.

${a}_{2}{a}_{1}{a}_{0}=000$ 时，同一个圈内 ${{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}=001$${a}_{2}{a}_{1}{a}_{0}=010$ 时，同一个圈内 ${{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}=011$ ；当 ${a}_{2}{a}_{1}{a}_{0}=100$ 时，同一个圈内 ${{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}=101$${a}_{2}{a}_{1}{a}_{0}=110$ 时，同一个圈内 ${{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}=111$$EFCQ\left(3,3\right)$ 中的16个8圈可以表示为：

${a}_{2}{a}_{1}{a}_{0}1000\text{-}{a}_{2}{a}_{1}{a}_{0}1001\text{-}{a}_{2}{a}_{1}{a}_{0}1011\text{-}{a}_{2}{a}_{1}{a}_{0}1010\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1010\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1011$ $\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1001\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1000\text{-}{a}_{2}{a}_{1}{a}_{0}1000$

${a}_{2}{a}_{1}{a}_{0}1100\text{-}{a}_{2}{a}_{1}{a}_{0}1101\text{-}{a}_{2}{a}_{1}{a}_{0}1111\text{-}{a}_{2}{a}_{1}{a}_{0}1110\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1110\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1111$ $\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1101\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}1100\text{-}{a}_{2}{a}_{1}{a}_{0}1100$

${a}_{2}{a}_{1}{a}_{0}0000\text{-}{a}_{2}{a}_{1}{a}_{0}0001\text{-}{a}_{2}{a}_{1}{a}_{0}0011\text{-}{a}_{2}{a}_{1}{a}_{0}0010\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0010\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0011$ $\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}\text{0001-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0000\text{-}{a}_{2}{a}_{1}{a}_{0}0000$

${a}_{2}{a}_{1}{a}_{0}0100\text{-}{a}_{2}{a}_{1}{a}_{0}0101\text{-}{a}_{2}{a}_{1}{a}_{0}0111\text{-}{a}_{2}{a}_{1}{a}_{0}0110\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0110\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0111$ $\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0101\text{-}{{a}^{\prime }}_{2}{{a}^{\prime }}_{1}{{a}^{\prime }}_{0}0100\text{-}{a}_{2}{a}_{1}{a}_{0}0100$

${a}_{2}{a}_{1}{a}_{0}\in \left\{000,001,010,011,100,101,110,111\right\}$$EFCQ\left(3,3\right)$ 中的32个4圈其中16个可以表示为：

${a}_{2}{a}_{1}{a}_{0}1001\text{-}{a}_{2}{a}_{1}{a}_{0}1101\text{-}{a}_{2}{a}_{1}{a}_{0}0101\text{-}{a}_{2}{a}_{1}{a}_{0}0001\text{-}{a}_{2}{a}_{1}{a}_{0}1001$

${a}_{2}{a}_{1}{a}_{0}1011\text{-}{a}_{2}{a}_{1}{a}_{0}0111\text{-}{a}_{2}{a}_{1}{a}_{0}0011\text{-}{a}_{2}{a}_{1}{a}_{0}1111\text{-}{a}_{2}{a}_{1}{a}_{0}1011$

$EFCQ\left(3,3\right)$ 中有32个4圈剩余的16个和定理4 1)的证明中所找到的16个4圈相同。

0011010-0011011-0010111-0010110-0000110-0000111-0000101-0000100-0010100-0010101-0010001-0010000-0000000-0000001-0000011-0000010-0010010-0010011-0011111-0011110-0001110-0001111-0001101-0001100-0011100-0011101-0011001-0011000-0001000-0001001-0001011-0001010-0101010-0101011-0101001-0101000-0111000-0111001-0111101-0111100-0101100-0101101-0101111-0101110-0111110-0111111-0110011-0110010-0100010-0100011-0100001-0100000-0110000-0110001-0110101-0110100-0100100-0100101-0100111-0100110-0110110-0110111-0111011-0111010-1011010-1011011-1010111-1010110-1000110-1000111-1000101-1000100-1010100-1010101-1010001-1010000-1000000-1000001-1000011-1000010-1010010-1010011-1011111-1011110-1001110-1001111-1001101-1001100-1011100-1011101-1011001-1011000-1001000-1001001-1001011-1001010-1101010-1101011-1101001-1101000-1111000-1111001-1111101-1111100-1101100-1101101-1101111-1101110-1111110-1111111-1110011-1110010-1100010-1100011-1100001-1100000-1110000-1110001-1110101-1110100-1100100-1100101-1100111-1100110-1110110-1110111-1111011-1111010-0011010

0011001-0011011-0011111-0011101-0010101-0010111-0010011-0010001-0011001；

0111001-0111011-0111111-0111101-0110101-0110111-0110011-0110001-0111001；

1011001-1011011-1011111-1011101-1010101-1010111-1010011-1010001-1011001；

1111001-1111011-1111111-1111101-1110101-1110111-1110011-1110001-1111001；

0001010-0011010-0111010-0101010-1101010-1111010-1011010-1001010-0001010。

$EFCQ\left(3,3\right)$ 中的22个4圈，其中14个4圈为定理4 1)证明中所找到的16个4圈中除去两个4圈0001010-0101010-1101010-1001010-0001010；0011010-0111010-1011010-1111010-0011010所得。

${a}_{2}{a}_{1}{a}_{0}\in \left\{000,010,100,110\right\}$，22个4圈中剩余的8个4圈则可以表示为：

${a}_{2}{a}_{1}{a}_{0}1001\text{-}{a}_{2}{a}_{1}{a}_{0}1101\text{-}{a}_{2}{a}_{1}{a}_{0}0101\text{-}{a}_{2}{a}_{1}{a}_{0}0001\text{-}{a}_{2}{a}_{1}{a}_{0}1001$

${a}_{2}{a}_{1}{a}_{0}1011\text{-}{a}_{2}{a}_{1}{a}_{0}0111\text{-}{a}_{2}{a}_{1}{a}_{0}0011\text{-}{a}_{2}{a}_{1}{a}_{0}1111\text{-}{a}_{2}{a}_{1}{a}_{0}1011$

2) $EFCQ\left(2,2\right)$ 可以分解为一个Hamilton圈和2个完美对集的并。

3) $EFCQ\left(3,3\right)$ 可以分解为一个Hamilton圈和3个完美对集的并。

M1: 01000-11000, 01100-00100, 00000-10000, 11100-10100, 01010-00010, 01110-11110, 00110-10110, 11010-10010, 01001-01101, 01011-01111, 00001-00011, 00101-00111, 11001-11101, 11011-11111, 10001-10101, 10011-10111.

M2: 01000-10111, 01001-10110, 01011-10100, 01010-10101, 01100-10011, 01101-10010, 01111-10000, 01110-10001, 00000-11111, 00001-11110, 00011-11100, 00010-11101, 00100-11011, 00101-11010, 00111-11000, 00110-11001.

Ea: 0001000-0101000, 0001100-0101100, 0000000-0100000, 0000100-0100100, 0011000-0111000, 0011100-0111100, 0010000-0110000, 0010100-0110100, 1001000-1101000, 1001100-1101100, 1000000-1100000, 1000100-1100100, 1011000-1111000, 1011100-1111100, 1010000-1110000, 1010100-1110100.

Eb: 0001010-0011010, 0001110-0101110, 0000010-0100010, 0000110-0100110, 0011110-0111110, 0010010-0110010, 0010110-0110110, 0101010-0111010, 1001010-1011010, 1001110-1101110, 1000010-1100010, 1000110-1100110, 1011110-1111110, 1010010-1110010, 1010110-1110110, 1101010-1111010.

Ec: 0001000-1001000, 0001100-1001100, 0000000-1000000, 0000100-1000100, 0011000-1111000, 0011100-1111100, 0010000-1110000, 0010100-1110100, 0101000-1101000, 0101100-1101100, 0100000-1100000, 0100100-1100100, 0111000-1011000, 0111100-1011100, 0110000-1010000, 0110100-1010100.

Ed: 0001010-1001010, 0001110-1001110, 0000010-1000010, 0000110-1000110, 0011010-0111010, 0011110-1111110, 0010010-1110010, 0010110-1110110, 0101010-1101010, 0101110-1101110, 0100010-1100010, 0100110-1100110, 0111110-1011110, 0110010-1010010, 0110110-1010110, 1011010-1111010.

Ee: 0001001-0001101, 0000001-0000101, 0011101-0010101, 0011001-0010001, 0011011-0011111, 0010011-0010111, 0101001-0101101, 0100001-0100101, 0111101-0110101, 0111001-0110001, 0111011-0111111, 0110011-0110111, 1001001-1001101, 1000001-1000101, 1011101-1010101, 1011001-1010001, 1011011-1011111, 1010011-1010111, 1101001-1101101, 1100001-1100101, 1111101-1110101, 1111001-1110001, 1111011-1111111, 1110011-1110111.

Ef: 0001101-0000101, 0001001-0000001, 0011001-0011011, 0011101-0011111, 0010001-0010011, 0010101-0010111, 0101101-0100101, 0101001-0100001, 0111001-0111011, 0111101-0111111, 0110001-0110011, 0110101-0110111, 1001101-1000101, 1001001-1000001, 1011001-1011011, 1011101-1011111, 1010001-1010011, 1010101-1010111, 1101101-1100101, 1101001-1100001, 1111001-1111011, 1111101-1111111, 1110001-1110011, 1110101-1110111.

Eg: 0001011-0001111, 0000011-0000111, 0100011-0100111, 0101011-0101111, 1001011-1001111, 1000011-1000111, 1101011-1101111, 1100011-1100111.

Eh: 0001111-0000011, 0001011-0000111, 0101111-0100011, 0101011-0100111, 1001111-1000011, 1001011-1000111, 1101111-1100011, 1101011-1100111.

${E}_{i}=E\left( M \right)$

$\begin{array}{l}{M}_{1},{M}_{2},{M}_{3},{M}_{4},{M}_{5},{M}_{6},{M}_{7},{M}_{8},{M}_{9},{M}_{10},{M}_{11},{M}_{12},{M}_{13},{M}_{14},{M}_{15},{M}_{16},M\\ {M}_{1}=\left({E}_{a}\cup {E}_{b}\cup {E}_{e}\cup {E}_{g}\right),{M}_{2}=\left({E}_{a}\cup {E}_{b}\cup {E}_{f}\cup {E}_{h}\right),{M}_{3}=\left({E}_{a}\cup {E}_{b}\cup {E}_{e}\cup {E}_{h}\right),\\ {M}_{4}=\left({E}_{a}\cup {E}_{b}\cup {E}_{f}\cup {E}_{g}\right),{M}_{5}=\left({E}_{c}\cup {E}_{d}\cup {E}_{f}\cup {E}_{h}\right),{M}_{6}=\left({E}_{c}\cup {E}_{d}\cup {E}_{e}\cup {E}_{g}\right)\\ {M}_{7}=\left({E}_{c}\cup {E}_{d}\cup {E}_{e}\cup {E}_{h}\right),{M}_{8}=\left({E}_{c}\cup {E}_{d}\cup {E}_{f}\cup {E}_{g}\right),{M}_{9}=\left({E}_{c}\cup {E}_{b}\cup {E}_{e}\cup {E}_{g}\right)\end{array}$

$\begin{array}{l}{M}_{10}=\left({E}_{c}\cup {E}_{b}\cup {E}_{f}\cup {E}_{h}\right),{M}_{11}=\left({E}_{c}\cup {E}_{b}\cup {E}_{e}\cup {E}_{h}\right),{M}_{12}=\left({E}_{c}\cup {E}_{b}\cup {E}_{f}\cup {E}_{g}\right)\\ {M}_{13}=\left({E}_{a}\cup {E}_{d}\cup {E}_{f}\cup {E}_{h}\right),{M}_{14}=\left({E}_{a}\cup {E}_{d}\cup {E}_{e}\cup {E}_{g}\right),{M}_{15}=\left({E}_{a}\cup {E}_{d}\cup {E}_{e}\cup {E}_{h}\right)\\ {M}_{16}=\left({E}_{a}\cup {E}_{d}\cup {E}_{f}\cup {E}_{g}\right)\end{array}$

$\begin{array}{l}E\left(EFCQ\left(3,3\right)\right)\\ =E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{1}\right)\cup E\left({M}_{5}\right)=E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{2}\right)\cup E\left({M}_{6}\right)\\ =E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{3}\right)\cup E\left({M}_{8}\right)=E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{4}\right)\cup E\left({M}_{7}\right)\\ =E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{9}\right)\cup E\left({M}_{13}\right)=E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{10}\right)\cup E\left({M}_{14}\right)\\ =E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{11}\right)\cup E\left({M}_{16}\right)=E\left(H\right)\cup E\left(M\right)\cup E\left({M}_{12}\right)\cup E\left( M 15 \right)\end{array}$

4. 结束语

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