流程图


输入

直接影响矩阵

参数设置

第一、归一化方法的设置

第二、截距值的获得

输出结果

第一、一组成对的对抗层级拓扑图

第二、带综合影响值的MR的直角坐标几何分布图

选择规范化方式

选择截距方式

原始矩阵(直接影响矩阵)为


$$Ori=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0 &7.12 &7.52 &8.9 &7.36 &8.24 &8.56 &8.4 &8.92 &1.28 &1.4 &8.08 &8.92 &2.36 &7.12 &1.04\\ \hline C2 &8.32 &0 &5.12 &8.88 &8.92 &2.16 &2.96 &7.16 &8.92 &1.56 &5.16 &4.12 &5 &5.12 &4.12 &2.68\\ \hline C3 &7.44 &2.32 &0 &8.8 &3.08 &7.12 &1.04 &8.92 &8.56 &0.24 &0.26 &5.52 &6.64 &4.48 &6.52 &3.08\\ \hline C4 &7.04 &5.12 &2.04 &0 &8.76 &7.16 &2.48 &7.28 &8.16 &1.2 &2.12 &5.08 &8.6 &3.12 &6.08 &2.2\\ \hline C5 &4.24 &4.52 &3.96 &7.08 &0 &4.84 &4.76 &6.56 &8.72 &1.76 &2.12 &2.96 &8.8 &4.16 &6 &3.04\\ \hline E1 &2.12 &1.5 &0.92 &7.12 &2.48 &0 &6.2 &5.04 &8.76 &3.28 &4.52 &2.24 &8.72 &3.5 &8.6 &5.64\\ \hline E2 &2 &7.04 &1.08 &8.72 &3.24 &8.2 &0 &7.08 &8.4 &8.6 &8 &2.12 &8.4 &6.2 &6.84 &8.24\\ \hline E3 &7.02 &3.2 &2.5 &8.96 &4.24 &4.88 &5.24 &0 &6.12 &2.24 &6.2 &2 &8.6 &3.52 &4.22 &6.2\\ \hline F1 &8.56 &8.92 &6.52 &8.6 &8.6 &7.2 &8.2 &8.4 &0 &2.2 &2 &1.12 &2 &5.08 &2.04 &6\\ \hline F2 &1.04 &1.52 &0.096 &1.08 &1.12 &3.04 &3.52 &2.04 &1.56 &0 &7.8 &1.52 &1.5 &6 &2.52 &6.5\\ \hline F3 &4.5 &6.04 &1 &1.16 &2.52 &8 &8.04 &8.96 &2.5 &8.25 &0 &1.24 &1.4 &7.04 &7.72 &7.5\\ \hline U1 &7.08 &8.72 &7.08 &6.52 &1.5 &6.28 &3.2 &8 &4.56 &2.04 &4 &0 &8.4 &7.2 &5.56 &3\\ \hline U2 &7.04 &4.96 &6.08 &7.52 &5.02 &7.2 &7 &8.12 &6.5 &2.12 &2 &8.88 &0 &1.24 &0.64 &8.2\\ \hline D1 &2 &1.04 &3.56 &6.96 &6.52 &8.6 &6.96 &6.4 &2.28 &1.02 &8.2 &1.02 &6.5 &0 &7.64 &8.92\\ \hline D2 &6.08 &7.02 &1.12 &8.32 &7 &8.24 &4.04 &8.92 &8.68 &4.12 &3.32 &8.12 &5.08 &8.92 &0 &3.04\\ \hline D3 &1.2 &1 &0.92 &8.76 &2.12 &8.96 &8.24 &7.28 &0.8 &6.52 &7.16 &8.52 &3.2 &7.2 &8.92 &0\\ \hline \end{array} $$

规范直接关系矩阵求解过程 $$ \require{cancel} \require{AMScd} \begin{CD} O @>>>N \\ \end{CD} $$


  • $$\mathcal{N}=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0 &0.066 &0.069 &0.082 &0.068 &0.076 &0.079 &0.077 &0.082 &0.012 &0.013 &0.074 &0.082 &0.022 &0.066 &0.01\\ \hline C2 &0.077 &0 &0.047 &0.082 &0.082 &0.02 &0.027 &0.066 &0.082 &0.014 &0.048 &0.038 &0.046 &0.047 &0.038 &0.025\\ \hline C3 &0.069 &0.021 &0 &0.081 &0.028 &0.066 &0.01 &0.082 &0.079 &0.002 &0.002 &0.051 &0.061 &0.041 &0.06 &0.028\\ \hline C4 &0.065 &0.047 &0.019 &0 &0.081 &0.066 &0.023 &0.067 &0.075 &0.011 &0.02 &0.047 &0.079 &0.029 &0.056 &0.02\\ \hline C5 &0.039 &0.042 &0.036 &0.065 &0 &0.045 &0.044 &0.06 &0.08 &0.016 &0.02 &0.027 &0.081 &0.038 &0.055 &0.028\\ \hline E1 &0.02 &0.014 &0.008 &0.066 &0.023 &0 &0.057 &0.046 &0.081 &0.03 &0.042 &0.021 &0.08 &0.032 &0.079 &0.052\\ \hline E2 &0.018 &0.065 &0.01 &0.08 &0.03 &0.076 &0 &0.065 &0.077 &0.079 &0.074 &0.02 &0.077 &0.057 &0.063 &0.076\\ \hline E3 &0.065 &0.029 &0.023 &0.083 &0.039 &0.045 &0.048 &0 &0.056 &0.021 &0.057 &0.018 &0.079 &0.032 &0.039 &0.057\\ \hline F1 &0.079 &0.082 &0.06 &0.079 &0.079 &0.066 &0.076 &0.077 &0 &0.02 &0.018 &0.01 &0.018 &0.047 &0.019 &0.055\\ \hline F2 &0.01 &0.014 &0.001 &0.01 &0.01 &0.028 &0.032 &0.019 &0.014 &0 &0.072 &0.014 &0.014 &0.055 &0.023 &0.06\\ \hline F3 &0.041 &0.056 &0.009 &0.011 &0.023 &0.074 &0.074 &0.083 &0.023 &0.076 &0 &0.011 &0.013 &0.065 &0.071 &0.069\\ \hline U1 &0.065 &0.08 &0.065 &0.06 &0.014 &0.058 &0.029 &0.074 &0.042 &0.019 &0.037 &0 &0.077 &0.066 &0.051 &0.028\\ \hline U2 &0.065 &0.046 &0.056 &0.069 &0.046 &0.066 &0.064 &0.075 &0.06 &0.02 &0.018 &0.082 &0 &0.011 &0.006 &0.076\\ \hline D1 &0.018 &0.01 &0.033 &0.064 &0.06 &0.079 &0.064 &0.059 &0.021 &0.009 &0.076 &0.009 &0.06 &0 &0.07 &0.082\\ \hline D2 &0.056 &0.065 &0.01 &0.077 &0.064 &0.076 &0.037 &0.082 &0.08 &0.038 &0.031 &0.075 &0.047 &0.082 &0 &0.028\\ \hline D3 &0.011 &0.009 &0.008 &0.081 &0.02 &0.083 &0.076 &0.067 &0.007 &0.06 &0.066 &0.078 &0.029 &0.066 &0.082 &0\\ \hline \end{array} $$

综合影响矩阵求解过程 $$\begin{CD} N @>>>T \\ \end{CD} $$


  综合影响矩阵如下

$T=\mathcal{N}(I-\mathcal{N})^{-1}$

$$T=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0.157 &0.205 &0.168 &0.291 &0.215 &0.263 &0.229 &0.282 &0.27 &0.099 &0.132 &0.194 &0.264 &0.16 &0.22 &0.155\\ \hline C2 &0.203 &0.12 &0.132 &0.256 &0.206 &0.183 &0.161 &0.24 &0.236 &0.087 &0.145 &0.141 &0.2 &0.162 &0.172 &0.145\\ \hline C3 &0.188 &0.133 &0.082 &0.246 &0.148 &0.215 &0.136 &0.242 &0.224 &0.069 &0.096 &0.148 &0.206 &0.147 &0.183 &0.14\\ \hline C4 &0.187 &0.16 &0.103 &0.173 &0.198 &0.217 &0.152 &0.232 &0.225 &0.081 &0.115 &0.146 &0.225 &0.139 &0.181 &0.137\\ \hline C5 &0.158 &0.15 &0.114 &0.228 &0.118 &0.193 &0.166 &0.22 &0.222 &0.085 &0.113 &0.124 &0.219 &0.144 &0.175 &0.141\\ \hline E1 &0.132 &0.12 &0.081 &0.219 &0.134 &0.146 &0.175 &0.2 &0.213 &0.1 &0.132 &0.114 &0.209 &0.137 &0.193 &0.161\\ \hline E2 &0.16 &0.191 &0.1 &0.272 &0.169 &0.254 &0.153 &0.257 &0.244 &0.164 &0.191 &0.135 &0.241 &0.19 &0.212 &0.215\\ \hline E3 &0.181 &0.14 &0.101 &0.244 &0.156 &0.198 &0.174 &0.166 &0.201 &0.093 &0.15 &0.119 &0.218 &0.141 &0.165 &0.171\\ \hline F1 &0.208 &0.199 &0.145 &0.265 &0.209 &0.233 &0.211 &0.258 &0.17 &0.099 &0.128 &0.12 &0.186 &0.168 &0.165 &0.18\\ \hline F2 &0.069 &0.071 &0.039 &0.097 &0.07 &0.112 &0.102 &0.107 &0.087 &0.044 &0.126 &0.065 &0.088 &0.116 &0.095 &0.124\\ \hline F3 &0.15 &0.156 &0.08 &0.174 &0.134 &0.219 &0.195 &0.235 &0.163 &0.147 &0.104 &0.105 &0.153 &0.174 &0.195 &0.183\\ \hline U1 &0.196 &0.196 &0.15 &0.243 &0.146 &0.223 &0.166 &0.252 &0.204 &0.094 &0.141 &0.11 &0.233 &0.183 &0.189 &0.154\\ \hline U2 &0.193 &0.165 &0.141 &0.25 &0.17 &0.229 &0.197 &0.25 &0.218 &0.097 &0.124 &0.185 &0.161 &0.132 &0.148 &0.195\\ \hline D1 &0.137 &0.12 &0.106 &0.23 &0.174 &0.234 &0.191 &0.225 &0.17 &0.088 &0.173 &0.111 &0.204 &0.115 &0.201 &0.199\\ \hline D2 &0.198 &0.194 &0.107 &0.273 &0.204 &0.254 &0.187 &0.274 &0.252 &0.12 &0.148 &0.185 &0.22 &0.21 &0.153 &0.167\\ \hline D3 &0.132 &0.123 &0.084 &0.245 &0.137 &0.238 &0.202 &0.233 &0.158 &0.137 &0.17 &0.173 &0.18 &0.183 &0.214 &0.125\\ \hline \end{array} $$

区段截取的处理


$T$的相关统计数据求解

平均数,均值 $\bar{x}$ 

$\bar{x}= 0.16948736690616 $

总体标准差$\sigma=\sqrt { \frac {\sum \limits_{i=1}^{n^2} ({x_i-\bar{x}})^2 }{n^2} } $ ( $n$为要素的数目)

$\sigma = 0.05161062797326 $

样本标准差一:$S=\sqrt { \frac {\sum \limits_{i=1}^{n^2} ({x_i-\bar{x}})^2 }{n^2-1} }$ ( $n$为要素的数目)

$S = 0.051711726264022 $

样本标准差二:$ \bar {S}=\sqrt { \frac {\sum \limits_{i=1}^{n^2} ({x_i-\bar{x}})^2 }{n^2-n} } $ ( $n$为要素的数目)

$ \bar {S}= 0.053303227367398 $

标准误差 $\sigma_{s}= \frac {\sigma}{n }$ ( $n$为要素的数目)

$\sigma_{s}= 0.010592960431635 $

方差 $ {\sigma}^{2}= \sigma ^{2} $

$\sigma^{2}= 0.0026636569197942 $

选择的截距方式为:$\lambda= \bar{x}+ \sigma$

$\lambda=0.22109799487942 $

\begin{CD} T@>\lambda=0.22109799487942>> A \\ \end{CD}

$$ a_{ij}= \begin{cases} 1 , \text{ $e_i$}\rightarrow \text{$e_j$ 当: $ t_{ij} > \lambda=0.22109799487942 $} \\ 0, \text{ $e_i$}\rightarrow \text{$e_j$ 当: $ t_{ij} < \lambda=0.22109799487942 $} \end{cases} $$

$\lambda= 0.22109799487942$ 截取后的关系矩阵$ A$

$$ A=\begin{array}{c|c|c|c|c|c|c}{M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0 &0 &0 &1 &0 &1 &1 &1 &1 &0 &0 &0 &1 &0 &0 &0\\ \hline C2 &0 &0 &0 &1 &0 &0 &0 &1 &1 &0 &0 &0 &0 &0 &0 &0\\ \hline C3 &0 &0 &0 &1 &0 &0 &0 &1 &1 &0 &0 &0 &0 &0 &0 &0\\ \hline C4 &0 &0 &0 &0 &0 &0 &0 &1 &1 &0 &0 &0 &1 &0 &0 &0\\ \hline C5 &0 &0 &0 &1 &0 &0 &0 &0 &1 &0 &0 &0 &0 &0 &0 &0\\ \hline E1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline E2 &0 &0 &0 &1 &0 &1 &0 &1 &1 &0 &0 &0 &1 &0 &0 &0\\ \hline E3 &0 &0 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline F1 &0 &0 &0 &1 &0 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline F2 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline F3 &0 &0 &0 &0 &0 &0 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline U1 &0 &0 &0 &1 &0 &1 &0 &1 &0 &0 &0 &0 &1 &0 &0 &0\\ \hline U2 &0 &0 &0 &1 &0 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline D1 &0 &0 &0 &1 &0 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline D2 &0 &0 &0 &1 &0 &1 &0 &1 &1 &0 &0 &0 &0 &0 &0 &0\\ \hline D3 &0 &0 &0 &1 &0 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$