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 &0.075 &0.079 &0.093 &0.077 &0.087 &0.09 &0.088 &0.094 &0.013 &0.015 &0.085 &0.094 &0.025 &0.075 &0.011\\ \hline C2 &0.087 &0 &0.054 &0.093 &0.094 &0.023 &0.031 &0.075 &0.094 &0.016 &0.054 &0.043 &0.053 &0.054 &0.043 &0.028\\ \hline C3 &0.078 &0.024 &0 &0.092 &0.032 &0.075 &0.011 &0.094 &0.09 &0.003 &0.003 &0.058 &0.07 &0.047 &0.068 &0.032\\ \hline C4 &0.074 &0.054 &0.021 &0 &0.092 &0.075 &0.026 &0.076 &0.086 &0.013 &0.022 &0.053 &0.09 &0.033 &0.064 &0.023\\ \hline C5 &0.045 &0.047 &0.042 &0.074 &0 &0.051 &0.05 &0.069 &0.092 &0.018 &0.022 &0.031 &0.092 &0.044 &0.063 &0.032\\ \hline E1 &0.022 &0.016 &0.01 &0.075 &0.026 &0 &0.065 &0.053 &0.092 &0.034 &0.047 &0.024 &0.092 &0.037 &0.09 &0.059\\ \hline E2 &0.021 &0.074 &0.011 &0.092 &0.034 &0.086 &0 &0.074 &0.088 &0.09 &0.084 &0.022 &0.088 &0.065 &0.072 &0.087\\ \hline E3 &0.074 &0.034 &0.026 &0.094 &0.045 &0.051 &0.055 &0 &0.064 &0.024 &0.065 &0.021 &0.09 &0.037 &0.044 &0.065\\ \hline F1 &0.09 &0.094 &0.068 &0.09 &0.09 &0.076 &0.086 &0.088 &0 &0.023 &0.021 &0.012 &0.021 &0.053 &0.021 &0.063\\ \hline F2 &0.011 &0.016 &0.001 &0.011 &0.012 &0.032 &0.037 &0.021 &0.016 &0 &0.082 &0.016 &0.016 &0.063 &0.026 &0.068\\ \hline F3 &0.047 &0.063 &0.011 &0.012 &0.026 &0.084 &0.084 &0.094 &0.026 &0.087 &0 &0.013 &0.015 &0.074 &0.081 &0.079\\ \hline U1 &0.074 &0.092 &0.074 &0.068 &0.016 &0.066 &0.034 &0.084 &0.048 &0.021 &0.042 &0 &0.088 &0.076 &0.058 &0.032\\ \hline U2 &0.074 &0.052 &0.064 &0.079 &0.053 &0.076 &0.074 &0.085 &0.068 &0.022 &0.021 &0.093 &0 &0.013 &0.007 &0.086\\ \hline D1 &0.021 &0.011 &0.037 &0.073 &0.068 &0.09 &0.073 &0.067 &0.024 &0.011 &0.086 &0.011 &0.068 &0 &0.08 &0.094\\ \hline D2 &0.064 &0.074 &0.012 &0.087 &0.074 &0.087 &0.042 &0.094 &0.091 &0.043 &0.035 &0.085 &0.053 &0.094 &0 &0.032\\ \hline D3 &0.013 &0.011 &0.01 &0.092 &0.022 &0.094 &0.087 &0.076 &0.008 &0.068 &0.075 &0.089 &0.034 &0.076 &0.094 &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 &0.325 &0.367 &0.287 &0.534 &0.387 &0.484 &0.413 &0.52 &0.486 &0.2 &0.268 &0.34 &0.476 &0.317 &0.405 &0.317\\ \hline C2 &0.355 &0.25 &0.232 &0.462 &0.356 &0.366 &0.313 &0.441 &0.42 &0.173 &0.264 &0.261 &0.376 &0.297 &0.327 &0.284\\ \hline C3 &0.33 &0.258 &0.169 &0.441 &0.283 &0.391 &0.276 &0.434 &0.397 &0.147 &0.203 &0.263 &0.374 &0.274 &0.33 &0.271\\ \hline C4 &0.331 &0.291 &0.195 &0.362 &0.342 &0.397 &0.298 &0.426 &0.401 &0.163 &0.227 &0.263 &0.398 &0.267 &0.331 &0.27\\ \hline C5 &0.294 &0.275 &0.204 &0.417 &0.247 &0.365 &0.308 &0.406 &0.392 &0.164 &0.221 &0.234 &0.386 &0.269 &0.32 &0.271\\ \hline E1 &0.259 &0.236 &0.163 &0.4 &0.259 &0.305 &0.313 &0.377 &0.375 &0.179 &0.239 &0.218 &0.368 &0.257 &0.335 &0.289\\ \hline E2 &0.318 &0.342 &0.203 &0.499 &0.327 &0.463 &0.319 &0.48 &0.445 &0.27 &0.329 &0.266 &0.438 &0.343 &0.388 &0.377\\ \hline E3 &0.32 &0.265 &0.19 &0.438 &0.291 &0.372 &0.32 &0.347 &0.37 &0.176 &0.265 &0.23 &0.388 &0.268 &0.311 &0.306\\ \hline F1 &0.366 &0.345 &0.25 &0.482 &0.365 &0.43 &0.376 &0.47 &0.353 &0.19 &0.251 &0.243 &0.369 &0.31 &0.327 &0.33\\ \hline F2 &0.14 &0.137 &0.085 &0.196 &0.141 &0.207 &0.182 &0.207 &0.175 &0.089 &0.195 &0.124 &0.174 &0.19 &0.175 &0.202\\ \hline F3 &0.281 &0.28 &0.163 &0.353 &0.262 &0.391 &0.339 &0.42 &0.322 &0.235 &0.211 &0.211 &0.309 &0.302 &0.341 &0.317\\ \hline U1 &0.35 &0.339 &0.254 &0.453 &0.291 &0.415 &0.322 &0.46 &0.389 &0.183 &0.264 &0.229 &0.418 &0.325 &0.35 &0.298\\ \hline U2 &0.345 &0.303 &0.242 &0.46 &0.317 &0.421 &0.356 &0.457 &0.403 &0.186 &0.243 &0.313 &0.336 &0.267 &0.303 &0.343\\ \hline D1 &0.273 &0.245 &0.196 &0.425 &0.313 &0.416 &0.341 &0.417 &0.338 &0.172 &0.293 &0.223 &0.374 &0.241 &0.354 &0.34\\ \hline D2 &0.363 &0.347 &0.212 &0.502 &0.368 &0.464 &0.358 &0.5 &0.455 &0.219 &0.28 &0.323 &0.417 &0.366 &0.321 &0.323\\ \hline D3 &0.268 &0.25 &0.173 &0.444 &0.273 &0.423 &0.355 &0.428 &0.326 &0.228 &0.292 &0.294 &0.349 &0.319 &0.371 &0.258\\ \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}= 0.31406679641503 $

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 = 0.089159864851877 $

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 = 0.089334517056183 $

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}= 0.092083912459046 $

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

$\sigma_{s}= 0.019629174775939 $

方差 $ {\sigma}^{2}= \sigma ^{2} $

$\sigma^{2}= 0.0079494815004049 $

选择的截距方式为:$\lambda= \bar{x} $

$\lambda=0.31406679641503 $

\begin{CD} T@>\lambda=0.31406679641503>> A \\ \end{CD}

$$ a_{ij}= \begin{cases} 1 , \text{ $e_i$}\rightarrow \text{$e_j$ 当: $ t_{ij} > \lambda=0.31406679641503 $} \\ 0, \text{ $e_i$}\rightarrow \text{$e_j$ 当: $ t_{ij} < \lambda=0.31406679641503 $} \end{cases} $$

$\lambda= 0.31406679641503$ 截取后的关系矩阵$ A$

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