《1 引言》

1 引言

《2 三角算子的定义》

2 三角算子的定义

$\begin{array}{l}Τ\left(x,y\right)=Τ\left(y,x\right)；\\ Τ\left[Τ\left(x,y\right),z\right]=Τ\left[x,Τ\left(y,z\right)\right]；\\ x\le u\phantom{\rule{0.25em}{0ex}}\text{a}\text{n}\text{d}\phantom{\rule{0.25em}{0ex}}y\le v\phantom{\rule{0.25em}{0ex}}⇒Τ\left(x,y\right)\le Τ\left(u,v\right)；\\ Τ\left(0,x\right)=0,Τ\left(1,x\right)=x。\end{array}$

$\begin{array}{l}S\left(x,y\right)=S\left(y,x\right)；\\ S\left[S\left(x,y\right),z\right]=S\left[x,S\left(y,z\right)\right]；\\ x\le u\phantom{\rule{0.25em}{0ex}}\text{a}\text{n}\text{d}\phantom{\rule{0.25em}{0ex}}y\le v\phantom{\rule{0.25em}{0ex}}⇒S\left(x,y\right)\le S\left(u,v\right)；\\ S\left(0,x\right)=x,S\left(1,x\right)=1。\end{array}$

《3 Hamacher算子的定义》

3 Hamacher算子的定义

1) $\forall u\in U,\phantom{\rule{0.25em}{0ex}}r\in \left[0,+\infty \right)\phantom{\rule{0.25em}{0ex}}$

1) A与B的Hamacher积记为C=A

《图1》

B, 定义为

《图2》

2) $\stackrel{˜}{A}$$\stackrel{˜}{B}$Hamacher和记为

《图3》

, $\stackrel{˜}{B}$ 定义为

$\begin{array}{l}\stackrel{˜}{C}\left(u\right)=\left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right)\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)+\left[r-2\right]\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}。\text{ }\text{ }\text{ }\left(2\right)\end{array}$

《图4》

) 被称为Hamacher算子。[1]

《图5》

) 如下:

1) $\stackrel{˜}{A}$$\stackrel{˜}{B}$Hamacher积记为

《图6》

, 定义为

$\begin{array}{l}\\ \left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)\left(u\right)=\left\{\begin{array}{cc}0\hfill & r=0\text{且}\stackrel{˜}{A}\left(u\right)=\stackrel{˜}{B}\left(u\right)=0‚\hfill \\ \frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{r+\left[1-r\right]\left[\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right]}\hfill & \text{其}\text{他}。\hfill \end{array}\text{ }\text{ }\text{ }\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left(3\right)\end{array}$

2) $\stackrel{˜}{A}$$\stackrel{˜}{B}$Hamacher和记为,

《图7》

$\begin{array}{l}\stackrel{˜}{C}\left(u\right)=\left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right)\left(u\right)=\left\{\begin{array}{cc}1& r=0\text{且}\stackrel{˜}{A}\left(u\right)=\stackrel{˜}{B}\left(u\right)=1‚\\ \frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)+\left[r-2\right]\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}& \text{其}\text{他}。\end{array}\text{ }\text{ }\text{ }\left(4\right)\\ \end{array}$

《4 Hamacher算子的一个新重要性质》

4 Hamacher算子的一个新重要性质

$\begin{array}{l}\\ \frac{\text{d}Y{}_{1}\left(r\right)}{\text{d}r}=\frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\left[\left(1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right)-\left(\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-2\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right)\right]}{\left[1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right]{}^{2}}=\\ \frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\left[1-\left(\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right)\right]}{\left[1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right]{}^{2}}\ge 0\end{array}$

$Y{}_{1}\left(r\right)=\stackrel{˜}{C}\left(u\right)↑\square$

$\begin{array}{l}\frac{\text{d}Y{}_{2}\left(r\right)}{\text{d}r}=\frac{-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\left[1-\left(\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right)\right]}{\left[r\left(1-\stackrel{˜}{A}\left(u\right)-\stackrel{˜}{B}\left(u\right)+\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right)+\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)+\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right]{}^{2}}\le 0‚\\ \end{array}$

$Y{}_{2}\left(r\right)=\stackrel{˜}{C}\left(u\right)↓\square$

《5 连续的Fuzzy算子和其他常用的三角算子的关系》

5 连续的Fuzzy算子和其他常用的三角算子的关系

$\begin{array}{l}\stackrel{˜}{A}\otimes \stackrel{˜}{B}\subseteq \stackrel{˜}{A}\stackrel{·}{\epsilon }\stackrel{˜}{B}\subseteq \stackrel{˜}{A}\stackrel{^}{·}\stackrel{˜}{B}\subseteq \stackrel{˜}{A}\cap \stackrel{˜}{B}\subseteq \\ \stackrel{˜}{A}\cup \stackrel{˜}{B}\subseteq \stackrel{˜}{A}\stackrel{^}{+}\stackrel{˜}{B}\subseteq \stackrel{˜}{A}\stackrel{+}{\epsilon }\stackrel{˜}{B}\subseteq \stackrel{˜}{A}\otimes \stackrel{˜}{B}\end{array}$

《图8》

《图9》

1) 对于定义4中的连续的Fuzzy算子, 当参数变量r=1时, ∀uU, 分别记定义4中的$\stackrel{˜}{C}$ (u) 为 (

《图10》

) 1与 (

《图11》

) 1, 则Hamacher算子 (

《图12》

) 就变成了概率积与概率和算子 (

《图13》

) :

$\begin{array}{l}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{1}\left(u\right)=\left(\stackrel{˜}{A}\stackrel{^}{·}\stackrel{˜}{B}\right)\left(u\right)=\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)‚\\ \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{1}\left(u\right)=\left(\stackrel{˜}{A}\stackrel{^}{+}\stackrel{˜}{B}\right)\left(u\right)=\\ \stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)。\end{array}$

2) 对于定义4中的连续的Fuzzy算子, 当参数变量r=2时, ∀uU, 分别记定义4中的

《图14》

《图15》

《图16》

) 2, 则Hamacher算子 (

《图17》

) 就变成了Einstain算子 (

《图18》

) :

$\begin{array}{l}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{2}\left(u\right)=\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{\epsilon }\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1+\left[1-\stackrel{˜}{A}\left(u\right)\right]\left[1-\stackrel{˜}{B}\left(u\right)\right]}‚\\ \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{2}\left(u\right)=\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{^}{\epsilon }\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)\left(u\right)=\frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)}{1+\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}。\end{array}$

《图19》

$\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{0}\subseteq \left(\stackrel{˜}{A}\cap \stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\cup \stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{0}。\text{ }\text{ }\text{ }\left(5\right)$

《图20》

) 0与 (

《图21》

) 0, 即:

$\begin{array}{l}\\ \stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{0}\left(u\right)=\left\{\begin{array}{cc}0& \stackrel{˜}{A}\left(u\right)=\stackrel{˜}{B}\left(u\right)=0‚\\ \frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}& \text{其}\text{他}；\end{array}\\ \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{0}\left(u\right)=\left\{\begin{array}{cc}1& \stackrel{˜}{A}\left(u\right)=\stackrel{˜}{B}\left(u\right)=1,\\ \frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-2\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}& \text{其}\text{他}。\end{array}\\ \end{array}$

$\begin{array}{l}\left(\stackrel{˜}{A}\cap \stackrel{˜}{B}\right)\left(u\right)=\stackrel{˜}{A}\left(u\right)\wedge \stackrel{˜}{B}\left(u\right)‚\\ \left(\stackrel{˜}{A}\cup \stackrel{˜}{B}\right)\left(u\right)=\stackrel{˜}{A}\left(u\right)\vee \stackrel{˜}{B}\left(u\right)。\end{array}$

《图22》

《图23》

1) 设$\stackrel{˜}{A}\left(u\right)\ge \stackrel{˜}{B}\left(u\right)$, 则有

$\begin{array}{l}\left(\stackrel{˜}{A}\stackrel{·}{r}\stackrel{˜}{B}\right){}_{0}\left(u\right)-\left(\stackrel{˜}{A}\cap \stackrel{˜}{B}\right)\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}-\stackrel{˜}{B}\left(u\right)=\\ \stackrel{˜}{B}\left(u\right)\frac{\stackrel{˜}{B}\left(u\right)\left[\stackrel{˜}{A}\left(u\right)-1\right]}{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}\le 0‚\\ \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{0}\left(u\right)-\left(\stackrel{˜}{A}\cup \stackrel{˜}{B}\right)\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-2\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}-\stackrel{˜}{A}\left(u\right)=\\ \frac{\stackrel{˜}{B}\left(u\right)\left[1-\stackrel{˜}{A}\left(u\right)\right]{}^{2}}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}\ge 0。\end{array}$

2) 设$\stackrel{˜}{A}\left(u\right)<\stackrel{˜}{B}\left(u\right)$, 则有

$\begin{array}{l}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{0}\left(u\right)-\left(\stackrel{˜}{A}\cap \stackrel{˜}{B}\right)\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}-\stackrel{˜}{A}\left(u\right)=\\ \stackrel{˜}{A}\left(u\right)\frac{\stackrel{˜}{A}\left(u\right)\left[\stackrel{˜}{B}\left(u\right)-1\right]}{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}\le 0‚\\ \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{0}\left(u\right)-\left(\stackrel{˜}{A}\cup \stackrel{˜}{B}\right)\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-2\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}-\stackrel{˜}{B}\left(u\right)=\\ \frac{\stackrel{˜}{A}\left(u\right)\left[1-\stackrel{˜}{B}\left(u\right)\right]{}^{2}}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}\ge 0。\end{array}$

《图24》

) 与有界积与有界和算子 (⨂, ⊕) 之间的关系为

$\begin{array}{l}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{+\infty }\subseteq \\ \left(\stackrel{˜}{A}\otimes \stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\oplus \stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{+\infty }。\text{ }\text{ }\text{ }\left(6\right)\end{array}$

《图25》

) +∞$\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{+}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{+\infty }$, 即

$\begin{array}{l}\underset{r\to +\infty }{\mathrm{lim}}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)=\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{+\infty }‚\\ \underset{r\to +\infty }{\mathrm{lim}}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{+}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)=\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{+}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{+\infty }。\end{array}$

$\begin{array}{l}\left(\stackrel{˜}{A}\otimes \stackrel{˜}{B}\right)\left(u\right)=\mathrm{max}\phantom{\rule{0.25em}{0ex}}\left[0‚\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-1\right]‚\\ \left(\stackrel{˜}{A}\oplus \stackrel{˜}{B}\right)\left(u\right)=\mathrm{min}\phantom{\rule{0.25em}{0ex}}\left[\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)‚1\right]。\end{array}$

1) 如果$0<\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)$, 则有

$\begin{array}{l}\underset{r\to +\infty }{\mathrm{lim}}\left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right)\left(u\right)=\underset{r\to +\infty }{\mathrm{lim}}\frac{\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-2\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{1-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)+r\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}=\\ 1\ge \left(\stackrel{˜}{A}\oplus \stackrel{˜}{B}\right)\left(u\right)；\end{array}$

$\underset{r\to +\infty }{\mathrm{lim}}\left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right)\left(u\right)=\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)=\left(\stackrel{˜}{A}\oplus \stackrel{˜}{B}\right)\left(u\right)$

2) 如果$\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\ne 0$$\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)=1$, 则有

$\begin{array}{l}\underset{r\to +\infty }{\mathrm{lim}}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)\left(u\right)=\underset{r\to +\infty }{\mathrm{lim}}\frac{\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}{r\left[1-\stackrel{˜}{A}\left(u\right)-\stackrel{˜}{B}\left(u\right)+\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)\right]+\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)}=\\ \stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(u\right)=\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-1=\left(\stackrel{˜}{A}\otimes B\right)\left(u\right)；\\ \end{array}$

$\stackrel{˜}{A}\left(u\right)+\stackrel{˜}{B}\left(u\right)-\stackrel{˜}{A}\left(u\right)\stackrel{˜}{B}\left(U\right)\ne 1$, 则有

$\begin{array}{l}\underset{r\to +\infty }{\mathrm{lim}}\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)\left(u\right)=0\le \mathrm{max}\left[0‚\stackrel{˜}{A}\left(u\right)+\\ \stackrel{˜}{B}\left(u\right)-1\right]=\left(\stackrel{˜}{A}\otimes \stackrel{˜}{B}\right)\left(u\right)；\end{array}$

《图26》

$\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\cap \stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\cup \stackrel{˜}{B}\right)\subseteq \left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right)。\text{ }\text{ }\text{ }\left(7\right)$

《6 一个新的统一的Fuzzy综合决策准则》

6 一个新的统一的Fuzzy综合决策准则

$\stackrel{˜}{\mathbit{R}}=\left[\begin{array}{cccc}0.7& 0.2& 0.1& 0\\ 0.5& 0.4& 0.1& 0\\ 0.4& 0.4& 0.1& 0.1\\ 0.3& 0.5& 0& 0.2\\ 0.4& 0.3& 0.2& 0.1\end{array}\right]‚$

$\left(\cap ‚\cup \right)⇒\stackrel{˜}{B}=\left[0.3\text{ }0.3\text{ }0.2\text{ }0.2\right]$,

《图27》

《图28》

) , 当参数变量r=0时,

$\begin{array}{l}\left(\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}‚\stackrel{+}{r}\right){}_{0}=\left[\left(\stackrel{˜}{A}\phantom{\rule{0.25em}{0ex}}\stackrel{·}{r}\phantom{\rule{0.25em}{0ex}}\stackrel{˜}{B}\right){}_{0}‚\left(\stackrel{˜}{A}\stackrel{+}{r}\stackrel{˜}{B}\right){}_{0}\right]⇒\\ \stackrel{˜}{B}=\left[0.632\text{ }0.456\text{ }0.257\text{ }0.224\right]‚\end{array}$

《图29》

《图30》

《图31》

, $⇒\stackrel{˜}{\text{B}}=\left[0.314\text{ }0.240\text{ }0.060\text{ }0.054\right]$

《图32》

《图33》

《图34》

) +∞亦不能作出评判。由定理1和定理2可知, 此时连续Fuzzy算子 (

《图35》

) +∞中的三角范式

《图36》

《图37》

) 0, (

《图38》

) 1, 和 (

《图39》

) 2逐渐增加评判结果的精度。根据定理1和定理2可知, 随着参数变量r的增加, 三角范式将逐渐被弱化, 而反三角范式将逐渐被强化;当参数变量r从最小值r=0开始, 随着参数变量r的增加, 即随着连续Fuzzy算子的变化, 模糊综合评判的结果将逐渐从模糊到清晰。但是, 随着参数变量r的进一步增加, 模糊综合评判的结果又将逐渐从清晰到模糊。

《图40》

) 或 (∩,

《图41》

) , (∩, ⊕) , (⨂, ∪) 均不能作出评判。只有用算子对 (

《图42》

, ∪) 或 (

《图43》

, ∪) 可以作出评判。换言之, 当Zadeh算子 (∩, ∪) 失效时, 应该选择那对算子组合来进行决策, 亦是一个困难的问题。

《7 结语》

7 结语

1) 直接使用未转换的数据;

2) 避免丢失原始信息;

3) 避免增加干扰信息;

4) 对于不同文献中的相关结论, 用此标准容易判断出谁的结论最好。