原始数据

$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times10}} &A &B &C &D &E &F &G &H &I &J\\ \hline 初始概率P &0.5 &0.3 &0.6 &0.5 &0.4 &0.3 &0.6 &0.2 &0.1 &0.6\\ \hline A &0 &0.45 &0 &0.4 &0.45 &0.75 &0 &0.45 &0.45 &0\\ \hline B &0.25 &0 &0.28 &0.35 &0 &0.2 &0.38 &0.35 &0.35 &0.34\\ \hline C &0.55 &0.65 &0 &0.65 &0.7 &0.5 &0 &0.65 &0.65 &0.66\\ \hline D &0.4 &0.6 &0.51 &0 &0.55 &0.3 &0.53 &0.55 &0.55 &0.53\\ \hline E &0.3 &0.5 &0.39 &0.5 &0 &0.35 &0.47 &0 &0 &0.43\\ \hline F &0.4 &0.25 &0 &0.1 &0.25 &0 &0.27 &0.25 &0.25 &0.27\\ \hline G &0.55 &0.75 &0 &0.7 &0.75 &0.55 &0 &0 &0 &0.66\\ \hline H &0.1 &0.15 &0 &0.25 &0.25 &0.1 &0 &0 &0.3 &0.24\\ \hline I &0.05 &0 &0 &0.15 &0 &0.05 &0.15 &0.2 &0 &0.15\\ \hline J &0.55 &0.75 &0.59 &0.7 &0.75 &0.5 &0.66 &0 &0 &0\\ \hline \end{array} $$

第一行代表初始概率

初始概率

$$\begin{array}{c|c|c|c|c|c|c}{M_{10 \times1}} &初始概率 P\\ \hline A &0.5\\ \hline B &0.3\\ \hline C &0.6\\ \hline D &0.5\\ \hline E &0.4\\ \hline F &0.3\\ \hline G &0.6\\ \hline H &0.2\\ \hline I &0.1\\ \hline J &0.6\\ \hline \end{array} $$

概率关系矩阵

$$R=\begin{array}{c|c|c|c|c|c|c}{M_{10 \times10}} &A &B &C &D &E &F &G &H &I &J\\ \hline A &0 &0.45 &0 &0.4 &0.45 &0.75 &0 &0.45 &0.45 &0\\ \hline B &0.25 &0 &0.28 &0.35 &0 &0.2 &0.38 &0.35 &0.35 &0.34\\ \hline C &0.55 &0.65 &0 &0.65 &0.7 &0.5 &0 &0.65 &0.65 &0.66\\ \hline D &0.4 &0.6 &0.51 &0 &0.55 &0.3 &0.53 &0.55 &0.55 &0.53\\ \hline E &0.3 &0.5 &0.39 &0.5 &0 &0.35 &0.47 &0 &0 &0.43\\ \hline F &0.4 &0.25 &0 &0.1 &0.25 &0 &0.27 &0.25 &0.25 &0.27\\ \hline G &0.55 &0.75 &0 &0.7 &0.75 &0.55 &0 &0 &0 &0.66\\ \hline H &0.1 &0.15 &0 &0.25 &0.25 &0.1 &0 &0 &0.3 &0.24\\ \hline I &0.05 &0 &0 &0.15 &0 &0.05 &0.15 &0.2 &0 &0.15\\ \hline J &0.55 &0.75 &0.59 &0.7 &0.75 &0.5 &0.66 &0 &0 &0\\ \hline \end{array} $$

交叉影响矩阵的求解

$$ C_{ij}= \frac {1}{1-P_j}[ln( \frac {R_{ij}}{1-R_{ij}} ) - ln(\frac {P_i}{1-P_i} )] $$

$$CIA=\begin{array}{c|c|c|c|c|c|c}{M_{10 \times10}} &A &B &C &D &E &F &G &H &I &J\\ \hline A &0 &-0.287 &0 &-0.811 &-0.334 &1.569 &0 &-0.251 &-0.223 &0\\ \hline B &-0.503 &0 &-0.243 &0.457 &0 &-0.77 &0.894 &0.285 &0.254 &0.46\\ \hline C &-0.41 &0.305 &0 &0.427 &0.736 &-0.579 &0 &0.267 &0.237 &0.645\\ \hline D &-0.811 &0.579 &0.1 &0 &0.334 &-1.21 &0.3 &0.251 &0.223 &0.3\\ \hline E &-0.884 &0.579 &-0.105 &0.811 &0 &-0.305 &0.713 &0 &0 &0.309\\ \hline F &0.884 &-0.359 &0 &-2.7 &-0.419 &0 &-0.368 &-0.314 &-0.279 &-0.368\\ \hline G &-0.41 &0.99 &0 &0.884 &1.155 &-0.293 &0 &0 &0 &0.645\\ \hline H &-1.622 &-0.498 &0 &0.575 &0.479 &-1.158 &0 &0 &0.599 &0.584\\ \hline I &-1.494 &0 &0 &0.925 &0 &-1.067 &1.157 &1.014 &0 &1.157\\ \hline J &-0.41 &0.99 &-0.104 &0.884 &1.155 &-0.579 &0.645 &0 &0 &0\\ \hline \end{array} $$

截距选取并获得手性矩阵

输入截距λ
请输入截距值,该值为正整数。
由于定义的原因请务必选择原始矩阵是否需要转置

未取截距前手性对称矩阵如下

$$Ori=\begin{array}{c|c|c|c|c|c|c}{M_{10 \times10}} &A &B &C &D &E &F &G &H &I &J\\ \hline A &0 &-0.50263 &-0.40959 &-0.81093 &-0.88367 &0.88367 &-0.40959 &-1.62186 &-1.49443 &-0.40959\\ \hline B &-0.28667 &0 &0.30511 &0.57924 &0.57924 &-0.35902 &0.99021 &-0.49758 &0 &0.99021\\ \hline C &0 &-0.24291 &0 &0.10001 &-0.10462 &0 &0 &0 &0 &-0.10375\\ \hline D &-0.81093 &0.45652 &0.42715 &0 &0.81093 &-2.69985 &0.88367 &0.57536 &0.92525 &0.88367\\ \hline E &-0.33445 &0 &0.73639 &0.33445 &0 &-0.41886 &1.15525 &0.47947 &0 &1.15525\\ \hline F &1.56945 &-0.77 &-0.57924 &-1.21043 &-0.30511 &0 &-0.29256 &-1.15847 &-1.06745 &-0.57924\\ \hline G &0 &0.89437 &0 &0.30036 &0.7133 &-0.36831 &0 &0 &1.15656 &0.64457\\ \hline H &-0.25084 &0.28532 &0.26697 &0.25084 &0 &-0.31414 &0 &0 &1.01366 &0\\ \hline I &-0.22297 &0.25362 &0.2373 &0.22297 &0 &-0.27924 &0 &0.59889 &0 &0\\ \hline J &0 &0.46001 &0.64457 &0.30036 &0.30903 &-0.36831 &0.64457 &0.58404 &1.15656 &0\\ \hline \end{array} $$
镜像平移后的矩阵如下
$$\begin{array} {c|cccccccccc|cccccccccc}{M_{20 \times20}} &+A &+B &+C &+D &+E &+F &+G &+H &+I &+J &-A &-B &-C &-D &-E &-F &-G &-H &-I &-J\\ \hline +A &0 &0 &0 &0 &0 &\color{blue}{0.884} &0 &0 &0 &0 &0 &\color{red}{0.503} &\color{red}{0.41} &\color{red}{0.811} &\color{red}{0.884} &0 &\color{red}{0.41} &\color{red}{1.622} &\color{red}{1.494} &\color{red}{0.41}\\ +B &0 &0 &\color{blue}{0.305} &\color{blue}{0.579} &\color{blue}{0.579} &0 &\color{blue}{0.99} &0 &0 &\color{blue}{0.99} &\color{red}{0.287} &0 &0 &0 &0 &\color{red}{0.359} &0 &\color{red}{0.498} &0 &0\\ +C &0 &0 &0 &\color{blue}{0.1} &0 &0 &0 &0 &0 &0 &0 &\color{red}{0.243} &0 &0 &\color{red}{0.105} &0 &0 &0 &0 &\color{red}{0.104}\\ +D &0 &\color{blue}{0.457} &\color{blue}{0.427} &0 &\color{blue}{0.811} &0 &\color{blue}{0.884} &\color{blue}{0.575} &\color{blue}{0.925} &\color{blue}{0.884} &\color{red}{0.811} &0 &0 &0 &0 &\color{red}{2.7} &0 &0 &0 &0\\ +E &0 &0 &\color{blue}{0.736} &\color{blue}{0.334} &0 &0 &\color{blue}{1.155} &\color{blue}{0.479} &0 &\color{blue}{1.155} &\color{red}{0.334} &0 &0 &0 &0 &\color{red}{0.419} &0 &0 &0 &0\\ +F &\color{blue}{1.569} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &\color{red}{0.77} &\color{red}{0.579} &\color{red}{1.21} &\color{red}{0.305} &0 &\color{red}{0.293} &\color{red}{1.158} &\color{red}{1.067} &\color{red}{0.579}\\ +G &0 &\color{blue}{0.894} &0 &\color{blue}{0.3} &\color{blue}{0.713} &0 &0 &0 &\color{blue}{1.157} &\color{blue}{0.645} &0 &0 &0 &0 &0 &\color{red}{0.368} &0 &0 &0 &0\\ +H &0 &\color{blue}{0.285} &\color{blue}{0.267} &\color{blue}{0.251} &0 &0 &0 &0 &\color{blue}{1.014} &0 &\color{red}{0.251} &0 &0 &0 &0 &\color{red}{0.314} &0 &0 &0 &0\\ +I &0 &\color{blue}{0.254} &\color{blue}{0.237} &\color{blue}{0.223} &0 &0 &0 &\color{blue}{0.599} &0 &0 &\color{red}{0.223} &0 &0 &0 &0 &\color{red}{0.279} &0 &0 &0 &0\\ +J &0 &\color{blue}{0.46} &\color{blue}{0.645} &\color{blue}{0.3} &\color{blue}{0.309} &0 &\color{blue}{0.645} &\color{blue}{0.584} &\color{blue}{1.157} &0 &0 &0 &0 &0 &0 &\color{red}{0.368} &0 &0 &0 &0\\ \hline -A &0 &\color{red}{0.503} &\color{red}{0.41} &\color{red}{0.811} &\color{red}{0.884} &0 &\color{red}{0.41} &\color{red}{1.622} &\color{red}{1.494} &\color{red}{0.41} &0 &0 &0 &0 &0 &\color{blue}{0.884} &0 &0 &0 &0\\ -B &\color{red}{0.287} &0 &0 &0 &0 &\color{red}{0.359} &0 &\color{red}{0.498} &0 &0 &0 &0 &\color{blue}{0.305} &\color{blue}{0.579} &\color{blue}{0.579} &0 &\color{blue}{0.99} &0 &0 &\color{blue}{0.99}\\ -C &0 &\color{red}{0.243} &0 &0 &\color{red}{0.105} &0 &0 &0 &0 &\color{red}{0.104} &0 &0 &0 &\color{blue}{0.1} &0 &0 &0 &0 &0 &0\\ -D &\color{red}{0.811} &0 &0 &0 &0 &\color{red}{2.7} &0 &0 &0 &0 &0 &\color{blue}{0.457} &\color{blue}{0.427} &0 &\color{blue}{0.811} &0 &\color{blue}{0.884} &\color{blue}{0.575} &\color{blue}{0.925} &\color{blue}{0.884}\\ -E &\color{red}{0.334} &0 &0 &0 &0 &\color{red}{0.419} &0 &0 &0 &0 &0 &0 &\color{blue}{0.736} &\color{blue}{0.334} &0 &0 &\color{blue}{1.155} &\color{blue}{0.479} &0 &\color{blue}{1.155}\\ -F &0 &\color{red}{0.77} &\color{red}{0.579} &\color{red}{1.21} &\color{red}{0.305} &0 &\color{red}{0.293} &\color{red}{1.158} &\color{red}{1.067} &\color{red}{0.579} &\color{blue}{1.569} &0 &0 &0 &0 &0 &0 &0 &0 &0\\ -G &0 &0 &0 &0 &0 &\color{red}{0.368} &0 &0 &0 &0 &0 &\color{blue}{0.894} &0 &\color{blue}{0.3} &\color{blue}{0.713} &0 &0 &0 &\color{blue}{1.157} &\color{blue}{0.645}\\ -H &\color{red}{0.251} &0 &0 &0 &0 &\color{red}{0.314} &0 &0 &0 &0 &0 &\color{blue}{0.285} &\color{blue}{0.267} &\color{blue}{0.251} &0 &0 &0 &0 &\color{blue}{1.014} &0\\ -I &\color{red}{0.223} &0 &0 &0 &0 &\color{red}{0.279} &0 &0 &0 &0 &0 &\color{blue}{0.254} &\color{blue}{0.237} &\color{blue}{0.223} &0 &0 &0 &\color{blue}{0.599} &0 &0\\ -J &0 &0 &0 &0 &0 &\color{red}{0.368} &0 &0 &0 &0 &0 &\color{blue}{0.46} &\color{blue}{0.645} &\color{blue}{0.3} &\color{blue}{0.309} &0 &\color{blue}{0.645} &\color{blue}{0.584} &\color{blue}{1.157} &0\\ \hline \end{array} $$

取截距后对称矩阵如下

截距为:0.8$$对称矩阵hand=\begin{vmatrix}0&0&0&0&0&1&0&0&0&0&0&0&0&1&1&0&0&1&1&0\\ 0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&1&0&1&1&1&0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&1&1&0\\ 0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&1&1&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&1\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&1&1\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&1\\ 0&0&0&1&0&0&0&1&1&0&1&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\end{vmatrix} $$

获得最大连通域

$$最大连通域矩阵=\begin{array} {c|c|c|c|c|c|c|c}{M_{9 \times9}} &+A &+F &-B &-D &-E &-G &-H &-I &-J\\ \hline +A &0 &1 &0 &1 &1 &0 &1 &1 &0\\ \hline +F &1 &0 &0 &1 &0 &0 &1 &1 &0\\ \hline -B &0 &0 &0 &0 &0 &1 &0 &0 &1\\ \hline -D &1 &1 &0 &0 &1 &1 &0 &1 &1\\ \hline -E &0 &0 &0 &0 &0 &1 &0 &0 &1\\ \hline -G &0 &0 &1 &0 &0 &0 &0 &1 &0\\ \hline -H &0 &0 &0 &0 &0 &0 &0 &1 &0\\ \hline -I &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline -J &0 &0 &0 &0 &0 &0 &0 &1 &0\\ \hline \end{array} $$

求解最大连通域的对抗层级拓扑图

由最大连通域矩阵出发,求解其可达矩阵

相乘矩阵$ B= A+I$

$$B=\begin{array} {c|c|c|c|c|c|c|c}{M_{9 \times9}} &+A &+F &-B &-D &-E &-G &-H &-I &-J\\ \hline +A &1 &1 & &1 &1 & &1 &1 & \\ \hline +F &1 &1 & &1 & & &1 &1 & \\ \hline -B & & &1 & & &1 & & &1\\ \hline -D &1 &1 & &1 &1 &1 & &1 &1\\ \hline -E & & & & &1 &1 & & &1\\ \hline -G & & &1 & & &1 & &1 & \\ \hline -H & & & & & & &1 &1 & \\ \hline -I & & & & & & & &1 & \\ \hline -J & & & & & & & &1 &1\\ \hline \end{array} $$

可达矩阵$ \tilde R$ 由相乘矩阵一直乘下去直到不变。

$$R=\begin{array} {c|c|c|c|c|c|c|c}{M_{9 \times9}} &+A &+F &-B &-D &-E &-G &-H &-I &-J\\ \hline +A &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline +F &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline -B & & &1 & & &1 & &1 &1\\ \hline -D &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline -E & & &1 & &1 &1 & &1 &1\\ \hline -G & & &1 & & &1 & &1 &1\\ \hline -H & & & & & & &1 &1 & \\ \hline -I & & & & & & & &1 & \\ \hline -J & & & & & & & &1 &1\\ \hline \end{array} $$

对抗的层级抽取过程

两种层级抽取规则:

抽取的过程如下


结果优先——UP型抽取过程 原因优先——DOWN型抽取过程
$$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline +A&+A,+F,-B,-D,-E,-G,-H,-I,-J&+A,+F,-D \\\hline +F&+A,+F,-B,-D,-E,-G,-H,-I,-J&+A,+F,-D \\\hline -B&-B,-G,-I,-J&-B,-G \\\hline -D&+A,+F,-B,-D,-E,-G,-H,-I,-J&+A,+F,-D \\\hline -E&-B,-E,-G,-I,-J&-E \\\hline -G&-B,-G,-I,-J&-B,-G \\\hline -H&-H,-I&-H \\\hline -I&\color{red}{\fbox{-I}}&\color{red}{\fbox{-I}} \\\hline -J&-I,-J&-J \\\hline \end{array} $$ $$\begin{array} {c|c|c|c|c|c|c|c}{} &Q_{e} & T_{e} \\\hline +A&\color{blue}{\fbox{+A,+F,-D}}&\color{blue}{\fbox{+A,+F,-D}} \\\hline +F&\color{blue}{\fbox{+A,+F,-D}}&\color{blue}{\fbox{+A,+F,-D}} \\\hline -B&+A,+F,-B,-D,-E,-G&-B,-G \\\hline -D&\color{blue}{\fbox{+A,+F,-D}}&\color{blue}{\fbox{+A,+F,-D}} \\\hline -E&+A,+F,-D,-E&-E \\\hline -G&+A,+F,-B,-D,-E,-G&-B,-G \\\hline -H&+A,+F,-D,-H&-H \\\hline -I&+A,+F,-B,-D,-E,-G,-H,-I,-J&-I \\\hline -J&+A,+F,-B,-D,-E,-G,-J&-J \\\hline \end{array} $$
抽取出-I放置上层,删除后剩余的情况如下 抽取出+A,+F,-D放置下层,删除后剩余的情况如下
$$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline +A&+A,+F,-B,-D,-E,-G,-H,-J&+A,+F,-D \\\hline +F&+A,+F,-B,-D,-E,-G,-H,-J&+A,+F,-D \\\hline -B&-B,-G,-J&-B,-G \\\hline -D&+A,+F,-B,-D,-E,-G,-H,-J&+A,+F,-D \\\hline -E&-B,-E,-G,-J&-E \\\hline -G&-B,-G,-J&-B,-G \\\hline -H&\color{red}{\fbox{-H}}&\color{red}{\fbox{-H}} \\\hline -J&\color{red}{\fbox{-J}}&\color{red}{\fbox{-J}} \\\hline \end{array} $$ $$\begin{array} {c|c|c|c|c|c|c|c}{} &Q_{e} & T_{e} \\\hline -B&-B,-E,-G&-B,-G \\\hline -E&\color{blue}{\fbox{-E}}&\color{blue}{\fbox{-E}} \\\hline -G&-B,-E,-G&-B,-G \\\hline -H&\color{blue}{\fbox{-H}}&\color{blue}{\fbox{-H}} \\\hline -I&-B,-E,-G,-H,-I,-J&-I \\\hline -J&-B,-E,-G,-J&-J \\\hline \end{array} $$
抽取出-H、-J放置上层,删除后剩余的情况如下 抽取出-E,-H放置下层,删除后剩余的情况如下
$$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline +A&+A,+F,-B,-D,-E,-G&+A,+F,-D \\\hline +F&+A,+F,-B,-D,-E,-G&+A,+F,-D \\\hline -B&\color{red}{\fbox{-B,-G}}&\color{red}{\fbox{-B,-G}} \\\hline -D&+A,+F,-B,-D,-E,-G&+A,+F,-D \\\hline -E&-B,-E,-G&-E \\\hline -G&\color{red}{\fbox{-B,-G}}&\color{red}{\fbox{-B,-G}} \\\hline \end{array} $$ $$\begin{array} {c|c|c|c|c|c|c|c}{} &Q_{e} & T_{e} \\\hline -B&\color{blue}{\fbox{-B,-G}}&\color{blue}{\fbox{-B,-G}} \\\hline -G&\color{blue}{\fbox{-B,-G}}&\color{blue}{\fbox{-B,-G}} \\\hline -I&-B,-G,-I,-J&-I \\\hline -J&-B,-G,-J&-J \\\hline \end{array} $$
抽取出-B、-G放置上层,删除后剩余的情况如下 抽取出-B,-G放置下层,删除后剩余的情况如下
$$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline +A&+A,+F,-D,-E&+A,+F,-D \\\hline +F&+A,+F,-D,-E&+A,+F,-D \\\hline -D&+A,+F,-D,-E&+A,+F,-D \\\hline -E&\color{red}{\fbox{-E}}&\color{red}{\fbox{-E}} \\\hline \end{array} $$ $$\begin{array} {c|c|c|c|c|c|c|c}{} &Q_{e} & T_{e} \\\hline -I&-I,-J&-I \\\hline -J&\color{blue}{\fbox{-J}}&\color{blue}{\fbox{-J}} \\\hline \end{array} $$
抽取出-E放置上层,删除后剩余的情况如下 抽取出-J放置下层,删除后剩余的情况如下
$$\begin{array} {c|c|c|c|c|c|c|c}{} & R_{e} & T_{e} \\\hline +A&\color{red}{\fbox{+A,+F,-D}}&\color{red}{\fbox{+A,+F,-D}} \\\hline +F&\color{red}{\fbox{+A,+F,-D}}&\color{red}{\fbox{+A,+F,-D}} \\\hline -D&\color{red}{\fbox{+A,+F,-D}}&\color{red}{\fbox{+A,+F,-D}} \\\hline \end{array} $$ $$\begin{array} {c|c|c|c|c|c|c|c}{} &Q_{e} & T_{e} \\\hline -I&\color{blue}{\fbox{-I}}&\color{blue}{\fbox{-I}} \\\hline \end{array} $$
抽取出+A、+F、-D放置上层,删除后剩余的情况如下 抽取出-I放置下层,删除后剩余的情况如下

抽取方式的结果如下


层级 结果优先——UP型 原因优先——DOWN型
0 -I -I
1 -H,-J -J
2 -B,-G -B,-G
3 -E -E,-H
4 +A,+F,-D +A,+F,-D

一般性骨架矩阵求解

$$S=\begin{array} {c|c|c|c|c|c|c|c}{M_{9 \times9}} &+A &+F &-B &-D &-E &-G &-H &-I &-J\\ \hline +A & &1 & & &1 & &1 & & \\ \hline +F & & & &1 & & & & & \\ \hline -B & & & & & &1 & & &1\\ \hline -D &1 & & & & & & & & \\ \hline -E & & & & & &1 & & & \\ \hline -G & & &1 & & & & & & \\ \hline -H & & & & & & & &1 & \\ \hline -I & & & & & & & & & \\ \hline -J & & & & & & & &1 & \\ \hline \end{array} $$

带综合影响值的对抗拓扑层级图



计算对称中心,确定是否需要拆分,并进行拆分


手性中心计算

没有手性中心,无需拆分!!!!