Some results on the resolution of fuzzy relation equations

Fuzzy Sets and Systems 60 (1993) 83-88 North-Holland

83

Some results on the resolution of fuzzy

relation equations G.I. Adamopoulos and C.P. Pappis Department of Industrial Management, University of Piraeus, Greece Received January 1993 Revised March 1993

Abstract: In this paper, two special cases of the solution of the Inverse Problem are investigated. The necessary and sufficient conditions for the existence of a solution such that the upper bound is the unit vector and for a unique solution of the Inverse Problem equal to the unit vector are presented. An algorithmic approach which derives the output vector and the state-matrix in each one of these cases is also proposed.

Keywords: Fuzzy relational equation; inverse problem.

1. Introduction The inverse problem of fuzzy systems can be stated as follows: Given an m x n fuzzy state matrix R and an n-dimensional fuzzy output vector b, determine an m-dimensional fuzzy input vector a such that aoR=b

where o denotes max-min composition and the elements of a, R and b represent grades of membership to corresponding fuzzy sets. Fuzzy relational equations, by means of which fuzzy systems may be modeled, have been investigated by many researchers. Sanchez [7] proved the necessary and sufficient conditions for the existence of a solution. Pappis and Sugeno [6] gave an analytical solution. Miyakoshi and Shimbo [3] presented a method to determine lower solutions, while Di Nola and Sessa [1] and Sessa [9] gave the sufficient and

Correspondence to: Professor C.P. Pappis, University of Piraeus, Department of Industrial Management, 80 Karaoli & Dimitriou Str., 18534 Piraeus, Greece.

necessary conditions for the existence of a unique solution. The number of the lower bounds have been investigated by Wang et al. [11]. Wang and Hsu [10] presented a solution procedure, by which the solution set and the type of solution (full set, unique minimal and unique solution) can be determined. Sessa [8] gave some theoretical aspects of fuzzy relation equations. Pappis and Adamopoulos [4, 5] presented two software routines to solve the Inverse Problem. Di Nola, Sessa, Pedrycz and Sanchez [2] gave the fundamental aspects of the theory of fuzzy relation equations. The necessary and sufficient conditions for the existence of a solution such that the upper bound is the unit vector and for a unique solution of the Inverse Problem equal to the unit vector are presented in this paper. Four algorithms are also presented. Algorithms 1 and 3 derive the set of all the possible state-matrices for a given output vector, which are such that the input vector and the upper bound of the inputs vectors are the unit vector, respectively. Algorithms 2 and 4 derive the output vector for a given state-matrix which is such that the input vector and the upper bound of the inputs vectors are the unit vector, respectively.

2. Notations and definitions In the sequel, small letters x, y etc. are used to denote scalars and when printed bold face, like a, b, etc., to denote vectors. Capital letters R, S etc., are used to denote fuzzy subsets and when printed bold face, like R, S, etc., to denote matrices. Any scalar and any elements of vectors or matrices are assumed to have their values in the interval [0, 1]. Let k, l be scalars, a = (a~, a2 . . . . . am), b = (bl, b2 . . . . . bin), be row vectors and R = [rii], S = [sij] be m × n matrices. The following

0165-0114/93/$06.00 (~) 1993---Elsevier Science Publishers B.V. All rights reserved

G.I. Adamopoulos, C.P. Pappis / Fuzzy relation equations

84 notations will be used:

the scalar

x/3y=

a i ~ b i , Vi, a < ~ b : ai<~bi, Vi, aAb: (a, A b j , a z A b 2 . . . . . a m ^ b i n ) , a v b : (al v bl, a2 v b2 . . . . . a,, v bm), a = 0 : ai = O, Vi (zero vector), a>~b:

a = 1 : a~ = 1, Vi (unit vector),

aT Rv /~R

: transpose of a, : transpose of R, : rlAr2A "''Arm,

VR

: rtvr2v...Vrm,

Y

Given a vector a = ( a l , a 2 , . . . , am) and a scalar x, the fl-composition of a with x, d e n o t e d by a / 3 x , is defined by the vector

a / 3 x = ( a l fix, a2/3x . . . . .

a,,, fix).

Given an m × n matrix R - - [ r q ] and a row vector a = (a~, a 2. . . . . an), the /3-composition of R with a, d e n o t e d by R / 3 a, is defined by the m × n matrix

rq >i sq, Vi, Vj, R<-S: rq<-sq, Vi, ¥j, R >1 S:

R /3 a = [rq /3 a/].

where ~ is the i-th row vector of R. o-composition. T h e o-composition of a vector a=(al,aE,...,am) with a c o l u m n vector b = ( b l , b2 . . . . , bm) T, d e n o t e d by aob, is defined by the scalar a ob = 7

ifx ~Y •

0

k A l : min(k, l), k v I : max(k, l),

6-composition. Given an m × n matrix R = , bn), the 6-composition of R with b, d e n o t e d by R 6 b, is defined by the rn × n matrix

[ri/] and a row vector b = (bl, b2 . . . .

R 6 b = [s,j],

so =

(ai A bi).

(rik a b k )

/3 (ri/flb/),

l

re, j = 1 . . . .

i=1 ..... The o-composition of a row vector with an m × n matrix R = [ r q ] , d e n o t e d by a oR, is defined by the row vector

,n.

Note that R 6 b can /~ (R a b ) T and R /3 b.

be

obtained

from

•-sets.

aoR=(ao~,ao~

.....

ao~)

where r/is the j-th c o l u m n vector of R. a - c o m p o s i t i o n . T h e a - c o m p o s i t i o n of a scalar x with a scalar y, d e n o t e d by x a y , is defined by the scalar

xay=

Given a column vector a-( a l , a 2 , . . . , a m ) v , such that ai=maxi(ai) or 0, i = 1 . . . . . m, the set @(a) of c o l u m n vectors q)(a) is defined as follows: • ( a ) = {~p(a)},

where

1

ifx<-y,

q,(a) = ( ~ , , ~2 . . . . .

y

ifx >y.

~Pi=0ormax(ai),

~m) ~, i= l,...,m,

i

Given a vector a = ( a ~ , a 2 . . . . . am) and a scalar x, the a - c o m p o s i t i o n of a with x, d e n o t e d by a a x , is defined by the vector a a x = (a~ a x , a2 a x . . . . .

am

aX).

Given an m × n matrix R = [ri/] and a row vector a = ( a l , a 2 . . . . . an), the a - c o m p o s i t i o n of R with a, d e n o t e d by R a a , is defined by the m × n matrix g a a = [rq a a , ] . fl-composition. T h e /3-composition of a scalar x with a scalar y, d e n o t e d by x / 3 y , is defined by

cPi = max (ai). i--I

i

Thus, if there are k n o n z e r o elements in a, there are k vectors in q~(a). N o t e that q}(a) is defined iff ai = maxi(ai) or 0, Vi.

3. Special cases o f the Inverse P r o b l e m

Let R be the matrix c o r r e s p o n d i n g to the fuzzy relation R, and a, b be the vectors corresponding to the fuzzy sets A, B,

85

G.I. Adamopoulos, C.P. Pappis / Fuzzy relation equations

respectively. Given the fuzzy relation R and the fuzzy subset B, all fuzzy subsets A such that a oR = b are given by V (¢p(R 6 b)) T ~ a <~/~ (R

V (R 6 b)

o{b) T,

6 b)

provided that there exists at least one A [6]. Let a' be the upper bound of the set L = {a} of the solutions of the Inverse Problem a oR = b. Theorem 1. Let ILl 4=0. T h e n a ' = 1 iff b / = max{riy}, Vj.

Proof. (a) Since ai /3 rii =

0

ifai
ro

if a, >~ rii

and because a, /3 rey = r,. (b) Since ai

/3 s d

of

a~-l>~r,j,

Vi,

we

have

0 ifa~ @

and because of a i = 1 1 > @ , Vi, we have ai/3 s u = &j. Therefore R = S, which is impossible because we have supposed that R =~S.

Proof. Since

1 riy °{ b =

by

Lemma 2. G i v e n three scalars x, y a n d z such that x, y ~ [0, 1] a n d z ~ (0, 1]. Then we h a v e (a) x c ~ y = l a n d x / 3 y = O imply x
if r u g b y , if r~j > b/

and because of b / = max{r/v} we have riy o:by = 1, V]. Therefore

x=y=0, (b) x o c y = l a n d x [ 3 y = z i m p l y x = y = z .

A (R o b)F: 1. For the opposite proposition, since a o R = V ( a i A rij ) i

and because of a ' = 1 we have (ai ^ rij)=riy. Therefore aoR=~/R=b

or

bj=max{rij},

Vj.

Example. Let

R =

0.3 0.7 0.8 0.3 I 0.4 0.6 0.2 0.7 0.9 0.4 0.5 0.8

b = ( 0 . 9 0.7 0.8 0.8). We have

R o~b=

1

1

1

1

I

1

1

1

1

1

1

1

and the upper bound vector is A(Ro~b)T=(1

1 1).

Lemma 1. G i v e n a vector a = (a¢, az . . . . . am) such that a = 1 a n d two m × n matrices R a n d S such that R 4=S. T h e n we h a v e (a) a / 3 R = R , (b) there does not exist S: a / 3 S = R .

Theorem 2. Let JL I 4=0. T h e n ILJ = 1 a n d a = 1 iff (a) by = max{r,y}, Vj, (b) there exists at least one j: riy = 1, Vi, (c) if b j ~ O there exists one a n d only one i: r,/ = by. Proof. R cv b = 1 and /~ (R c~b) v = 1 (Theorem 1). We have

r,yflbj = {0 if'0 < G r~y if riy >~ b/.

From (a) if by =0¢¢,r~y=0, Vi and r u / 3 b j = O . From (c) if by ~: 0 then one i is such that r~j = bj and ri/ /3 b i = by. For all others we have ru < b/ and r o/3 by = 0. Therefore there exists one and only one r~y/3by = r~y4: 0, Vi.

(1) Because of (1) 13bj = 1 and from (b) we have there exists at least one j: rv [3 b~ -- 1,

Vi.

(2)

Because o f / ~ (R olb) T= 1, from Lemma l(a) we obtain R6b=A(Ro~b)T/3(Rflb)=R/3b.

(3)

Because of (1) and (3), we have that the q~-set

G.I. Adamopoulos, C.P. Pappis / Fuzzy relation equations

86

has a unique element such that

this case we have

qb(n b b ) = R

there exists one and only one rq = bj 4=O,

6b.

Because of (2) and (3) we have that V (q~(R 6 b)) T = 1. From this result and Theorem 1 we have that ]LI = 1 and such that a=l. For the opposite proposition we have (a) if a = l then a o R = V ( a i ^ r q ) = V R . Therefore by = max{rii}, Vj. (c) if ILl = 1 then the lower bound is unique and equal to the upper bound. Therefore the q~-set has a unique element such that

~(R 6 b) = R 6 b.

(11)

Example. Let 0.3

R--

1 0.8

1 0.6 0.9 0.4

b=(1

0.3

0.7 , 0.8

0.2 1

1 1 0.8).

Obviously conditions (a), (b) Theorem 1 are satisfied. We have

(4)

1

1

1

R o:b=

there exists at least one f//= 1,

Vi,

A(Ro~b)T=(1

there exists at most one f/j :/= 0,

Vj.

It is easily seen that A ( R there exists a: a oR = b,

there exists at least one hi/= 1,

Vi,

(5)

there exists at most one hij 4=O,

Vj.

(6)

Rflb=

1 1).

0 0 0

R6b=

Vi.

or

rij=bj=Oifhq=O,

rq = b] = hi/ if hi] :/: O.

Vj.

O,

1

O.

0.8

bb,

V (~p(R 6 b)) T = V (R 6 b) T = (1 1 1). (8) (9)

(6) shows that each column vector of matrix R flb has at most one nonzero element. If it has one nonzero element, from (9) there exists a rq such that ri/= b~ = hi/ and from (8), all other rq are such that rq < by (first case). If for a column vector of the matrix R fib there exist nonzero elements (second case), then from (8) we have that rij = bi = 0, Vi (it cannot be that rq < b], otherwise /~ (R a~b)VoR = V R :/:b, therefore ILl = 0). In both cases we have

by = max{ru},

0I 0.8

We observe that the q~-set has a unique element:

c~(R 6 b ) = R

(b) From Lemma 2 also we have

rij
0 0

0 (7)

1

o:b)T°R =b. Thus

li 1° °I

From Lemma 2 and (5) we obtain there exists at least one rq = 1,

of

1 1 1 1 , 1 1 1 1

R f l b = R 6b. Therefore, for the matrix R flb = [hiE] we have

(c)

1

From the definition of ~-set and (4) the matrix R 6 b = [f/j] must be such that

From Theorem 1 and L e m m a l(b) we have

and

(10)

Only in the first case b] is such that b] 4= 0. In

It is obvious that

V(Rbb)T=A(Ro~b)T=(1

1 1).

4. An algorithmic approach From Theorem 1 we see that there exists an infinite number of pairs of matrices R and vectors b such that a ' = 1. Two algorithms may be derived from this theorem. Algorithm 1 begins with a given vector b and derives all the possible matrices R which are such that a ' = 1. Algorithm 2 begins with a given matrix R and derives the unique vector b which is such that a ' = 1.

G.I. Adamopoulos, C.P. Pappis / Fuzzy relation equations

Algorithm 1. Step 1: Define the dimensions m, n. Step 2: Define the elements of vector b arbitrarily.

Step 3: D e t e r m i n e the elements of matrix R as follows:

O<<-r,j<-bj Vi.

Example. Step 1: m = 2 a n d n = 4 . Step 2: b = (0.3 0.7 0.9 0.2). Step 3: O<~ri~ ~<0.3, Vi, 0 ~< ri2 ~< 0.7, Vi, 0 ~< r/3 ~< 0.9, Vi, 0 ~< ri4 ~< 0.2, Vi.

Algorithm 2. Step I: Define the dimensions m, n. Step 2: Define the elements of matrix R. Step 3: D e t e r m i n e the elements of vector b as follows:

bj = max{r/j},

V].

Example. Step 1: m = 4 a n d n = 3 . Step 2:

R=

0.9

0.8

0.4

0.2

0.5

0.7

1

0.3

0

0.1

0.4

0.7

Step 3: b = (1 0.8 0.7). T h e o r e m 2 states the necessary and sufficient conditions for the existence of a unique solution of the Inverse P r o b l e m which is such that a = 1. T w o m o r e algorithms m a y be derived f r o m T h e o r e m 2, which deal with the following subjects: (a) A l g o r i t h m 3. Given the vector b, find all the matrices R which are such that I L ) = 1 and a=l. (b) A l g o r i t h m 4. Given a matrix R such that: m~n, there exists at least one rij = 1, Vi, if bj 4=0 then max{rij} is unique, Vj, find the vector b which is such that ILl = 1 and a=l.

87

Note that for m > n, it is obvious that there exists at least o n e j: rij = rkj = 1. This is contrary to the third condition of T h e o r e m 2. T h e r e f o r e , it is obvious that if m > n there are no matrix R and vector b such that ILl = 1 and a = 1 (for p r o o f see also [1]).

Algorithm 3. Step 1: Define the elements of vector b. Step 2: C o m p u t e the n u m b e r k of the elements of vector b which are equal to 1. Step 3: If k = 0 there is no solution. G o to Step 7. Step 4: Put m = k . Step 5: D e t e r m i n e the elements of each c o l u m n vector of matrix R as follows: if bj = 0 then r;i = 0, Vi, else m - 1 elements are such that 0 ~< r~j < bj, one element is such that rgj = bj. Step 6: D e t e r m i n e the positions of these elements such that there exists at least one rij = 1,

Vi.

Step 7: Stop. N o t e that, in Step 6, there are m c o l u m n vectors such that max{r~j} = 1. Let p be the n u m b e r of elements of vector b, which are such that bj = 0. T h e n p = 0, 1 . . . . . n - m, therefore we m a y derive m! (m) " - m - p possible groups of matrices R.

Example. Step 1: b = ( 0 . 2 1 0.7 1 1). Step2: k = 3 . Step 3: k 4: 0. Step 4: m = k = 3 . Step 5: E a c h column vector has 3 elements which are given in Table 1 (not necessarily in this order). Table 1 Column vector

Element 1

Element 2

1

0.2

0~,<0.2

0~,<0.2

2 3 4 5

1 0.7 1 1

0~2<1

0~2<1 0~3<0.7 0~4<1

0~3<0.7 0~4<1 0~5<1

Element 3

0~5<1

G.I. Adamopoulos, C.P. Pappis / Fuzzy relation equations

88

Step 6: All the possible combinations of the positions of the elements of the above matrix, which satisfy the condition of Step 5 are (3!)(3) 5-3 = 5 4 (p = 0 ) and each one derives a group of matrices R. Each group contains an infinite number of matrices R because n(m - 1) elements of matrix R may take a value in an interval. Step 7: Stop.

Algorithm 4. Step 1: Define the matrix R. Step 2: If m > n then there is no solution. Go to Step 7. Step 3: If there does not exist at least one rij = 1, Vi, then there is no solution. Go to Step 7. Step 4: If there exists a column vector such that max{rij} is nonzero and not unique then there is no solution. Go to Step 7. Step 5: If there exists a column vector such that max{rii} = 0 and there exists k such that rki ~ 0 then there is no solution. G o to Step 7. Step 6: Determine the elements of vector b as follows:

by = max{rii},

Vj.

Step 7: Stop.

Example.

Step 1:

R =

Step Step Step Step

0.2

1

0.6

1

0.5

0.6

0

2: 3: 4: 5:

0.9

0.3

0.4

0.3

1

0.8

0.8

0.9

0

1

0.4

m-3
max{r 0} is unique, Vj. There does not exist column vector such that max{r 0} = 0. Step 6: Set b = (1 1 0.9 1 1 0.8). Step 7. Stop.

5. Conclusions In this paper we presented the necessary and sufficient conditions for the existence of a solution of the Inverse Problem such that the solutions' upper bound is the unit vector. We also presented the conditions for a unique solution such that a = 1. We considered two special cases of the Inverse Problem, in which the set of input vectors and either the state-matrix or the output vectors are given. Finally we proposed algorithmic methods to find the set of the state matrices and the output vectors which give this solution.

References [1] A. Di Nola and S. Sessa, Finite fuzzy relation equations with a unique solution in linear lattices, J. Math. Anal. Appl. 132 (1988) 39-49. [2] A. Di Nola, S. Sessa, W. Pedrycz and E. Sanchez,

Fuzzy Relation Equations and their Application to Knowledge Engineering (Kluwer Academic Publishers, Dordrecht, 1989). [3] M. Miyakoshi and M. Shimbo, Lower solutions of systems of fuzzy equations, Fuzzy Sets and Systems 19 (1986) 37-46. [4] C.P. Pappis and G.I. Adamopoulos, A computer algorithm for the solution of the inverse problem of fuzzy systems, Fuzzy Sets and Systems 39 (1991) 279-290. [5] C.P. Pappis and G.I. Adamopoulos, A software routine to solve the generalized inverse problem of fuzzy systems, Fuzzy Sets and Systems 47 (1992) 319-322. [6] C.P. Pappis and M. Sugeno, Fuzzy relational equations and the inverse problem, Fuzzy Sets and Systems 15 (1985) 79-90. [7] E. Sanchez, Resolution of composite fuzzy relation equations, Inf. Control 30 (1976) 38-48. [8] S. Sessa, Some results in the setting of fuzzy relation equations theory, Fuzzy Sets and Systems 14 (1984) 281-297. [9] S. Sessa, Finite fuzzy relation equations with unique solution in complete Brouwerian lattices, Fuzzy Sets and Systems 29 (1989) 103-113. [10] H.F. Wang and H.M. Hsu, An alternative approach to the resolution of fuzzy relations equations, Fuzzy Sets and Systems 45 (1992) 203-213. [11] P.Z. Wang, S. Sessa, A. Di Nola and W. Pedrycz, How many lower solutions does a fuzzy relation equation have?, BUSEFAL 18 (1984) 67-74.