﻿ 模糊关系的分块合成运算

# 模糊关系的分块合成运算Block Composition of Fuzzy Relation

Abstract: Compositional operation of fuzzy relations is an important operation, which is widely used in pattern recognition, machine learning, and data mining. In this paper, block operation of fuzzy relations is studied, and the complexity of block synthesis is analyzed. The results show that block operations can simplify the form of fuzzy relation synthesis operations, but they do not increase the complexity of the operation.

1. 引言

2. 模糊关系的分块合成运算

$A\circ B=\left[\begin{array}{cc}\left(0.4\wedge 1\right)\vee \left(0.7\wedge 0.4\right)\vee \left(0\wedge 0\right)& \left(0.4\wedge 0.7\right)\vee \left(0.7\wedge 0.6\right)\vee \left(0\wedge 0.3\right)\\ \left(1\wedge 1\right)\vee \left(0.8\wedge 0.4\right)\vee \left(0.5\wedge 0\right)& \left(1\wedge 0.7\right)\vee \left(0.8\wedge 0.6\right)\vee \left(0.5\wedge 0.3\right)\end{array}\right]=\left[\begin{array}{cc}0.4& 0.6\\ 1& 0.7\end{array}\right]$

% command window

>> A=[0.4 0.7 0;1 0.8 0.5];

>> B=[1 0.7;0.4 0.6;0 0.3];

>> C=Max_Min(A,B)

C=

0.4000 0.6000

1.0000 0.7000

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccccc}& {n}_{1}& \text{\hspace{0.17em}}\text{ }{n}_{2}& \cdots & \text{\hspace{0.17em}}\text{\hspace{0.17em}}{n}_{l}\end{array}\\ A=\begin{array}{c}{s}_{1}\\ {s}_{2}\\ ⋮\\ {s}_{t}\end{array}\left[\begin{array}{cccc}{A}_{11}& {A}_{12}& \cdots & {A}_{1l}\\ {A}_{21}& {A}_{22}& \cdots & {A}_{2l}\\ ⋮& ⋮& & ⋮\\ {A}_{t1}& {A}_{t2}& \cdots & {A}_{tl}\end{array}\right]\end{array}$$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccccc}& {m}_{1}& \text{ }{m}_{2}& \cdots & \text{\hspace{0.17em}}{m}_{r}\end{array}\\ B=\begin{array}{c}{n}_{1}\\ {n}_{2}\\ ⋮\\ {n}_{l}\end{array}\left[\begin{array}{cccc}{B}_{11}& {B}_{12}& \cdots & {B}_{1r}\\ {B}_{21}& {B}_{22}& \cdots & {B}_{2r}\\ ⋮& ⋮& & ⋮\\ {B}_{l1}& {B}_{l2}& \cdots & {B}_{lr}\end{array}\right]\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccccc}& {m}_{1}& \text{ }{m}_{2}& \cdots & \text{\hspace{0.17em}}{m}_{r}\end{array}\\ C=A\circ B=\begin{array}{c}{s}_{1}\\ {s}_{2}\\ ⋮\\ {s}_{t}\end{array}\left[\begin{array}{cccc}{C}_{11}& {C}_{12}& \cdots & {C}_{1r}\\ {C}_{21}& {C}_{22}& \cdots & {C}_{2r}\\ ⋮& ⋮& & ⋮\\ {C}_{t1}& {C}_{t2}& \cdots & {C}_{tr}\end{array}\right]\end{array}$

${C}_{pq}={A}_{p1}\circ {B}_{1q}\cup {A}_{p2}\circ {B}_{2q}\cup \cdots \cup {A}_{pl}\circ {B}_{lq}$$\left(p=1,2,\cdots ,t;q=1,2,\cdots ,r\right)$

${s}_{p}{n}_{1}{m}_{q}+{s}_{p}{n}_{2}{m}_{q}+\cdots +{s}_{p}{n}_{l}{m}_{q}={s}_{p}{m}_{q}n$ .

${s}_{i}{m}_{1}n+{s}_{i}{m}_{2}n+\cdots +{s}_{i}{m}_{r}n={s}_{i}mn$ .

2.1. 列分块合成运算

$A=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right]=\left({\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}\right)$$B=\left[\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2n}\\ ⋮& ⋮& & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}\right]$

$\begin{array}{c}A\circ B=\left({\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}\right)\circ \left[\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2n}\\ ⋮& ⋮& & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}\right]\\ =\left({\alpha }_{1}\circ {b}_{11}\cup {\alpha }_{2}\circ {b}_{21}\cup \cdots \cup {\alpha }_{n}\circ {b}_{n1},{\alpha }_{1}\circ {b}_{12}\cup {\alpha }_{2}\circ {b}_{22}\cup \cdots \cup {\alpha }_{n}\circ {b}_{n2},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\cdots ,{\alpha }_{1}\circ {b}_{1n}\cup {\alpha }_{2}\circ {b}_{2n}\cup \cdots \cup {\alpha }_{n}\circ {b}_{nn}\right)\\ =\left({\beta }_{1},{\beta }_{2},\cdots ,{\beta }_{n}\right)\end{array}$

$\begin{array}{c}A\circ B=\left({\alpha }_{1},{\alpha }_{2},{\alpha }_{3}\right)\circ \left[\begin{array}{cc}1& 0.7\\ 0.4& 0.6\\ 0& 0.3\end{array}\right]=\left({\alpha }_{1}\circ 1\cup {\alpha }_{2}\circ 0.4\cup {\alpha }_{3}\circ 0,{\alpha }_{1}\circ 0.7\cup {\alpha }_{2}\circ 0.6\cup {\alpha }_{3}\circ 0.3\right)\\ =\left[\left[\begin{array}{c}0.4\\ 1\end{array}\right]\circ \left[1\right]\cup \left[\begin{array}{c}0.7\\ 0.8\end{array}\right]\circ \left[0.4\right]\cup \left[\begin{array}{c}0\\ 0.5\end{array}\right]\circ \left[0\right],\left[\begin{array}{c}0.4\\ 1\end{array}\right]\circ \left[0.7\right]\cup \left[\begin{array}{c}0.7\\ 0.8\end{array}\right]\circ \left[0.6\right]\cup \left[\begin{array}{c}0\\ 0.5\end{array}\right]\circ \left[0.3\right]\right]\\ =\left[\left[\begin{array}{c}0.4\\ 1\end{array}\right]\cup \left[\begin{array}{c}0.4\\ 0.4\end{array}\right]\cup \left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{c}0.4\\ 0.7\end{array}\right]\cup \left[\begin{array}{c}0.6\\ 0.6\end{array}\right]\cup \left[\begin{array}{c}0\\ 0.3\end{array}\right]\right]=\left[\begin{array}{cc}0.4& 0.6\\ 1& 0.7\end{array}\right]\end{array}$

%command window

>> A=[0.4 0.7 0;1 0.8 0.5];

>> B=[1 0.7;0.4 0.6;0 0.3];

>> C=leftrightblock(A,B)

C=

0.4000 0.6000

1.0000 0.7000

2.2. 行分块合成运算

$A=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right]$$B=\left[\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2n}\\ ⋮& ⋮& & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}\right]=\left[\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\\ ⋮\\ {\beta }_{n}\end{array}\right]$

$A\circ B=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right]\circ \left[\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\\ ⋮\\ {\beta }_{n}\end{array}\right]=\left[\begin{array}{c}{a}_{11}\circ {\beta }_{1}\cup {a}_{12}\circ {\beta }_{2}\cup \cdots \cup {a}_{1n}\circ {\beta }_{n}\\ {a}_{21}\circ {\beta }_{1}\cup {a}_{22}\circ {\beta }_{2}\cup \cdots \cup {a}_{2n}\circ {\beta }_{n}\\ ⋮\\ {a}_{n1}\circ {\beta }_{1}\cup {a}_{n2}\circ {\beta }_{2}\cup \cdots \cup {a}_{nn}\circ {\beta }_{n}\end{array}\right]$ .

$\begin{array}{c}A\circ B=\left[\begin{array}{ccc}0.4& 0.7& 0\\ 1& 0.8& 0.5\end{array}\right]\circ \left[\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\\ {\beta }_{3}\end{array}\right]=\left[\begin{array}{c}0.4\circ {\beta }_{1}\cup 0.7\circ {\beta }_{2}\cup 0\circ {\beta }_{3}\\ 1\circ {\beta }_{1}\cup 0.8\circ {\beta }_{2}\cup 0.5\circ {\beta }_{3}\end{array}\right]\\ =\left[\begin{array}{c}\left[0.4\right]\circ \left[\begin{array}{cc}1& 0.7\end{array}\right]\cup \left[0.7\right]\circ \left[\begin{array}{cc}0.4& 0.6\end{array}\right]\cup \left[0\right]\circ \left[\begin{array}{cc}0& 0.3\end{array}\right]\\ \left[1\right]\circ \left[\begin{array}{cc}1& 0.7\end{array}\right]\cup \left[0.8\right]\circ \left[\begin{array}{cc}0.4& 0.6\end{array}\right]\cup \left[0.5\right]\circ \left[\begin{array}{cc}0& 0.3\end{array}\right]\end{array}\right]\\ =\left[\begin{array}{c}\left[\begin{array}{cc}0.4& 0.4\end{array}\right]\cup \left[\begin{array}{cc}0.4& 0.6\end{array}\right]\cup \left[\begin{array}{cc}0& 0\end{array}\right]\\ \left[\begin{array}{cc}1& 0.7\end{array}\right]\cup \left[\begin{array}{cc}0.4& 0.6\end{array}\right]\cup \left[\begin{array}{cc}0& 0.3\end{array}\right]\end{array}\right]=\left[\begin{array}{cc}0.4& 0.6\\ 1& 0.7\end{array}\right]\end{array}$

%command window

>> A=[0.4 0.7 0;1 0.8 0.5];

>> B=[1 0.7;0.4 0.6;0 0.3];

>> C=rightblock(A,B)

C=

0.4000 0.6000

1.0000 0.7000

2.3. 行列分块合成运算

$A=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right]=\left({\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}\right)$$B=\left[\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2n}\\ ⋮& ⋮& & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}\right]=\left[\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\\ ⋮\\ {\beta }_{n}\end{array}\right]$

$A\circ B=\left[\begin{array}{cccc}{\alpha }_{1}& {\alpha }_{2}& \cdots & {\alpha }_{n}\end{array}\right]\circ \left[\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\\ ⋮\\ {\beta }_{n}\end{array}\right]={\alpha }_{1}\circ {\beta }_{1}\cup {\alpha }_{2}\circ {\beta }_{2}\cup \cdots \cup {\alpha }_{n}\circ {\beta }_{n}$ .

$\begin{array}{c}A\circ B=\left[\begin{array}{cc}\begin{array}{c}{\alpha }_{1}\end{array}& \begin{array}{cc}{\alpha }_{2}& \begin{array}{c}{\alpha }_{3}\end{array}\end{array}\end{array}\right]\circ \left[\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\\ {\beta }_{n}\end{array}\right]={\alpha }_{1}\circ {\beta }_{1}\cup {\alpha }_{2}\circ {\beta }_{2}\cup {\alpha }_{3}\circ {\beta }_{3}\\ =\left[\begin{array}{c}0.4\\ 1\end{array}\right]\circ \left[\begin{array}{cc}1& 0.7\end{array}\right]\cup \left[\begin{array}{c}0.7\\ 0.8\end{array}\right]\circ \left[\begin{array}{cc}0.4& 0.6\end{array}\right]\cup \left[\begin{array}{c}0\\ 0.5\end{array}\right]\circ \left[\begin{array}{cc}0& 0.3\end{array}\right]\\ =\left[\begin{array}{cc}0.4& 0.4\\ 1& 0.7\end{array}\right]\cup \left[\begin{array}{cc}0.4& 0.6\\ 0.4& 0.6\end{array}\right]\cup \left[\begin{array}{cc}0& 0\\ 0& 0.3\end{array}\right]=\left[\begin{array}{cc}0.4& 0.6\\ 1& 0.7\end{array}\right]\end{array}$

%command window

>> A=[0.4 0.7 0;1 0.8 0.5];

>> B=[1 0.7;0.4 0.6;0 0.3];

>> C=leftright(A,B)

C=

0.4000 0.6000

1.0000 0.7000

3. 结论

1) 合成运算的MATLAB程序代码为：

function[C]=Max_Min(A,B)

C=[];

if (s1~=s)

return

end

for (i=1:m)

for (j=1:n)

C(i,j)=0;

for (k=1:s)

x=0;

if (A(i,k)

x=A(i,k);

else

x=B(k,j);

end

if (C(i,j)

C(i,j)=x;

end

end

end

end

2) 列分块合成运算的MATLAB程序代码为：

function [C,time1] = leftblock(A,B)

C = zeros(m,n);

if (s1 ~= s)

return

end

C = zeros(m,n);

for j = 1:n

for k = 1:s

for i = 1:m

if A(i,k) > B(k,j);

x = B(k,j);

else

x = A(i,k);

end

if C(i,j) < x

C(i,j) = x;

end

end

end

End

3) 合成运算的MATLAB程序代码为：

function [C,time1] = rightblock(A,B)

C = zeros(m,n);

if (s1 ~= s)

return

end

C = zeros(m,n);

for i = 1:m

for k = 1:s

for j = 1:n

if A(i,k) > B(k,j);

x = B(k,j);

else

x = A(i,k);

end

if C(i,j) < x

C(i,j) = x;

end

end

end

End

4) 行列合成运算的MATLAB程序代码为：

function [C]=leftright(A,B)

C=zeros(m,n);

if (s1~=s)

return

end

for i=1:s

Ai=A(:,i);

Bi=B(i,:);

Ci=Max_Min(Ai,Bi);

for j=1:m

for k=1:n

if (Ci(j,k)>C(j,k))

C(j,k)=Ci(j,k);

end

end

end

end

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