| 结果优先——UP型抽取过程 |
| $$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline t1&\color{red}{\fbox{t1,t10}}&\color{red}{\fbox{t1,t10}} \\\hline t2&t1,t2,t10&t2 \\\hline t3&t1,t3,t10,t12&t3 \\\hline t4&t1,t4,t10,t11,t12,t13&t4 \\\hline t5&t1,t5,t10,t12&t5 \\\hline t6&t1,t6,t7,t8,t10,t13&t6 \\\hline t7&t1,t7,t10,t13&t7 \\\hline t8&t1,t7,t8,t10,t13&t8 \\\hline t9&t1,t9,t10,t12&t9 \\\hline t10&\color{red}{\fbox{t1,t10}}&\color{red}{\fbox{t1,t10}} \\\hline t11&t1,t10,t11,t12,t13&t11 \\\hline t12&t1,t10,t12&t12 \\\hline t13&t1,t10,t13&t13 \\\hline \end{array} $$ |
| 抽取出t1、t10 剩余的情况如下 |
| $$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline t2&\color{red}{\fbox{t2}}&\color{red}{\fbox{t2}} \\\hline t3&t3,t12&t3 \\\hline t4&t4,t11,t12,t13&t4 \\\hline t5&t5,t12&t5 \\\hline t6&t6,t7,t8,t13&t6 \\\hline t7&t7,t13&t7 \\\hline t8&t7,t8,t13&t8 \\\hline t9&t9,t12&t9 \\\hline t11&t11,t12,t13&t11 \\\hline t12&\color{red}{\fbox{t12}}&\color{red}{\fbox{t12}} \\\hline t13&\color{red}{\fbox{t13}}&\color{red}{\fbox{t13}} \\\hline \end{array} $$ |
| 抽取出t2、t12、t13 剩余的情况如下 |
| $$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline t3&\color{red}{\fbox{t3}}&\color{red}{\fbox{t3}} \\\hline t4&t4,t11&t4 \\\hline t5&\color{red}{\fbox{t5}}&\color{red}{\fbox{t5}} \\\hline t6&t6,t7,t8&t6 \\\hline t7&\color{red}{\fbox{t7}}&\color{red}{\fbox{t7}} \\\hline t8&t7,t8&t8 \\\hline t9&\color{red}{\fbox{t9}}&\color{red}{\fbox{t9}} \\\hline t11&\color{red}{\fbox{t11}}&\color{red}{\fbox{t11}} \\\hline \end{array} $$ |
| 抽取出t3、t5、t7、t9、t11 剩余的情况如下 |
| $$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline t4&\color{red}{\fbox{t4}}&\color{red}{\fbox{t4}} \\\hline t6&t6,t8&t6 \\\hline t8&\color{red}{\fbox{t8}}&\color{red}{\fbox{t8}} \\\hline \end{array} $$ |
| 抽取出t4、t8 剩余的情况如下 |
| $$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline t6&\color{red}{\fbox{t6}}&\color{red}{\fbox{t6}} \\\hline \end{array} $$ |
| 抽取出t6 剩余的情况如下 |
| 层级 | 结果优先——UP型 |
| 第0层 | t1,t10 |
| 第1层 | t2,t12,t13 |
| 第2层 | t3,t5,t7,t9,t11 |
| 第3层 | t4,t8 |
| 第4层 | t6 |
求解过程如链接所示:缩点、缩边,再把回路要素替代回去。这步是最难的,本处用的算法那人得了计算机界的诺奖-图领奖,算法为trajan算法的组合。现在的论文都忽略了这步。
可达矩阵 R的缩点矩阵 R'
$$R'=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &t1+t10 &t2 &t3 &t4 &t5 &t6 &t7 &t8 &t9 &t11 &t12 &t13\\ \hline t1+t10 &1 & & & & & & & & & & & \\ \hline t2 &1 &1 & & & & & & & & & & \\ \hline t3 &1 & &1 & & & & & & & &1 & \\ \hline t4 &1 & & &1 & & & & & &1 &1 &1\\ \hline t5 &1 & & & &1 & & & & & &1 & \\ \hline t6 &1 & & & & &1 &1 &1 & & & &1\\ \hline t7 &1 & & & & & &1 & & & & &1\\ \hline t8 &1 & & & & & &1 &1 & & & &1\\ \hline t9 &1 & & & & & & & &1 & &1 & \\ \hline t11 &1 & & & & & & & & &1 &1 &1\\ \hline t12 &1 & & & & & & & & & &1 & \\ \hline t13 &1 & & & & & & & & & & &1\\ \hline \end{array} $$缩点矩阵 R'的缩边矩阵 S' 公式:$ S'=R'-(R'-I)^2-I$
$$S'=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &t1+t10 &t2 &t3 &t4 &t5 &t6 &t7 &t8 &t9 &t11 &t12 &t13\\ \hline t1+t10 & & & & & & & & & & & & \\ \hline t2 &1 & & & & & & & & & & & \\ \hline t3 & & & & & & & & & & &1 & \\ \hline t4 & & & & & & & & & &1 & & \\ \hline t5 & & & & & & & & & & &1 & \\ \hline t6 & & & & & & & &1 & & & & \\ \hline t7 & & & & & & & & & & & &1\\ \hline t8 & & & & & & &1 & & & & & \\ \hline t9 & & & & & & & & & & &1 & \\ \hline t11 & & & & & & & & & & &1 &1\\ \hline t12 &1 & & & & & & & & & & & \\ \hline t13 &1 & & & & & & & & & & & \\ \hline \end{array} $$以最简菊花链表示回路代入回去,即为一般性骨架矩阵 $S$
$$S=\begin{array} {c|c|c|c|c|c|c|c}{M_{13 \times13}} &t1 &t2 &t3 &t4 &t5 &t6 &t7 &t8 &t9 &t10 &t11 &t12 &t13\\ \hline t1 & & & & & & & & & &1 & & & \\ \hline t2 &1 & & & & & & & & & & & & \\ \hline t3 & & & & & & & & & & & &1 & \\ \hline t4 & & & & & & & & & & &1 & & \\ \hline t5 & & & & & & & & & & & &1 & \\ \hline t6 & & & & & & & &1 & & & & & \\ \hline t7 & & & & & & & & & & & & &1\\ \hline t8 & & & & & & &1 & & & & & & \\ \hline t9 & & & & & & & & & & & &1 & \\ \hline t10 &1 & & & & & & & & & & & & \\ \hline t11 & & & & & & & & & & & &1 &1\\ \hline t12 & & & & & & & & & &1 & & & \\ \hline t13 & & & & & & & & & &1 & & & \\ \hline \end{array} $$