Numerical solution of fuzzy max–min systems

Numerical solution of fuzzy max–min systems

Applied Mathematics and Computation 174 (2006) 1321–1328 www.elsevier.com/locate/amc Numerical solution of fuzzy max–min systems S. Abbasbandy a,b,...

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Applied Mathematics and Computation 174 (2006) 1321–1328

www.elsevier.com/locate/amc

Numerical solution of fuzzy max–min systems S. Abbasbandy

a,b,* ,

E. Babolian c, M. Allame

a

a

Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran 14778, Iran b Department of Mathematics, Imam Khomeini International University, Ghazvin 34194, Iran c Faculty of Mathematical Sciences and Computer Engineering, Teacher Training University, Tehran 15618, Iran

Abstract For solvable systems of linear equations of the form A  x = b over the max–min algebra B ¼ f½0; 1;  ¼ max;  ¼ ming, we propose efficient methods for finding fuzzy determinant and fuzzy solution. We formulate a condition for linear systems of equations over B to have not a unique solution by using fuzzy determinant.  2005 Elsevier Inc. All rights reserved. Keywords: Fuzzy membership matrix; Fuzzy determinant; Max–min algebra

1. Introduction In many applications, e.g. in fuzzy control systems, in discrete dynamic systems, or in knowledge-based system of engineering, fuzzy relation equations

* Corresponding author. Address: Department of Mathematics, Imam Khomeini International University, P.O. Box 288, Ghazvin 34194, Iran. E-mail address: [email protected] (S. Abbasbandy).

0096-3003/$ - see front matter  2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.05.043

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play important role [4,5]. The solvability of fuzzy equations in Go¨del algebra was first studied in [14], later in [10–12]. Cechla´rova´ studied the unique solvability of linear systems of equations over the max–min fuzzy algebra by a necessary condition. Recently, she has described an efficient method for finding a Chebychev-best approximation for unsolvable systems [2,3]. Gavalec [6] has considered some necessary and sufficient conditions for the unique solvability of the max–min fuzzy linear equations. The question of unique solvability of such system is closely connected with the strong regularity of max–min matrices. The problem of the strong regularity for square matrices over a general max–min algebra was considered in [7–9]. Also, Thomason [15] defined fuzzy determinant of fuzzy square matrix and some properties of a square fuzzy matrix was considered in [13]. An important special case of max–min algebra is Go¨del algebra, in which the underlying set is the close unite interval with natural ordering of real numbers. This max–min algebra will be denoted by BG . An implication operator in BG is defined by uG(x, y) := 1, for x 6 y, and uG(x, y) := y for x > y. The aim of this paper is to present a condition for the unique solvability of a given system of linear equations in the max–min fuzzy algebra by using fuzzy determinant. Further, we describe efficient methods for finding fuzzy determinant and fuzzy solution of solvable systems.

2. Definitions and notations By a max–min fuzzy algebra B we mean a linear ordered set ([0, 1], 6, , ) with binary operations of  = maximum and  = minimum. For any natural n > 0, BðnÞ denotes set of all n-dimensional column vector over B, and Bðm; nÞ denotes the set of all matrices of type m · n over B. For x; y 2 BðnÞ, we write x 6 y, if xi 6 yi holds for all i, and we write x < y, if x 6 y and x 5 y. In this paper, we consider a system of linear equations of the form A  x ¼ b; where the matrix A 2 Bðm; nÞ and the vector b 2 BðmÞ are given, and the vector x 2 BðnÞ is unknown. The matrix operations use the operations  and  instead of the addition and multiplication in formally the same way as is done over a field. Definition 2.1 [15]. Fuzzy determinant of a square matrix A 2 Bðn; nÞ is defined by F detðAÞ ¼

max fminfa1h1 ; a2h2 ; . . . ; anhn gg;

ðh1 ;h2 ;...;hn Þ

where max is for all permutations (h1, h2, . . . , hn) of indices {1, 2, . . . , n}.

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In the rest of the paper, we use the notation N = {1, 2, . . . , n} and M = {1, 2, . . . , m}. Further, we denote the solution set SðA; bÞ :¼ fx 2 BðnÞ; A  x ¼ bg.

3. Algorithms Now, we are ready to state an algorithm for a given matrix A 2 Bðn; nÞ for evaluating fuzzy determinant. Step 1: Set MiR := maxj2N{aij} and MjC := maxi2N{aij} for i,j 2 N and so fd := min{mink{MkR}, mink{MkC}}.  by ½A  :¼ minf½A ; fdg. Step 2: Reduce A to A ij ij  then F det(A) := fd Step 3: If ‘‘fd’’ is occurred in every row and column of A, and stop.  which is smaller than fd, say Step 4: (Else) Choose maximum element of A  , and set fd :¼ ½A  and ½A  :¼ minf½A  ; fdg and return to step ½A pq pq ij ij 3. The execution of step 3, is as the same Hungarian algorithm in the assignment problem [1]. Example 1. Let us take the matrix A 2 Bð3; 3Þ, as an input matrix 2

0.3

6 A ¼ 4 0.7 1

0.9 0.4 1

1

3

7 1 5. 0.3

So we have M1R = 1, M2R = 1, M3R = 1, and M1C = 1, M2C = 1, M3C = 1, hence fd := 1 and 2 3 0.3 0.9 1 7 ¼A¼6 A 4 0.7 0.4 1 5. 1

1

0.3

 ¼ ½A  ¼ 0.9, hence fd := 0.9 and Again we have ½A pq 12 2 3 0.3 0.9 0.9 7 ¼6 A 4 0.7 0.4 0.9 5; 0.9

0.9

0.3

therefore F det(A) = 0.9 = min{a12, a23, a31}.

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Now, we are ready to state the algorithm for solving A  x = b for a given matrix A 2 Bðm; nÞ and a vector b 2 BðmÞ, also have N = {1, 2, . . . , n} and M = {1, 2, . . . , m}. Step 1: Step 2:

Step 3:

First arrange the equations by b1 6 b2 6    6 bm, and set x1 = x2 =    = xn = 1, N1 = N. For i = 1 to m If maxj2{1,2,. . .,n}[min(aij, xj)] < bi then A  x = b has not any solution and stop. Else set xj = bi for all index j that aij > bi and j 2 N1, and set N1 = N1  {j}. x = (x1, x2, . . . , xn)t is a solution of A  x = b and stop.

Lemma 3.1. For a given matrix A 2 Bðm; nÞ and b 2 BðmÞ, the above algorithm computed x, the solution of A  x = b, or say have not any solution. Proof. We consider the notations in [2]. For j 2 N = {1, 2, . . . , n}, suppose M j ðA; bÞ ¼ fi 2 Mjaij > bi g; N 0 ðA; bÞ ¼ fj 2 N jM j ðA; bÞ 6¼ £g; N 00 ðA; bÞ ¼ fj 2 N jM j ðA; bÞ ¼ £g; e j ðA; bÞ ¼ fi 2 Mjaij ¼ bi g. M For j 2 N 0 (A, b), let xj ¼ minfbi ji 2 M j ðA; bÞg; I j ðA; bÞ ¼ fi 2 M j ðA; bÞjbi ¼ xj g; e j ðA; bÞjaij ¼ bi 6 xj g; K j ðA; bÞ ¼ fi 2 M Lj ðA; bÞ ¼ I j ðA; bÞ [ K j ðA; bÞ; and for j 2 N00 (A, b), let xj ¼ 1; e j ðA; bÞ. Lj ðA; bÞ ¼ M We note that, x is changed in step 2 for i = i0, if    max min ai0 j ; xj < bi0 ; j2f1;2;...;ng

hence i0 62 ¨j2N Lj(A, b) and from Theorem 4 in [2], S(A, b) = B. But if for any index i in M, the above inequality does not hold, then 8i 2 M; 9j 2 N 3 ðaij  xj P bi Þ;

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and from Lemma 3 in [2], i 2 Lj and ¨j2NLj = M. Hence the solution exists by Theorem 4 in [2]. h Example 2. 2 0.3 6 0.5 6 4 0.6 0.4

Solve 0.9 0.8 0.1 0.3

2 3 3 0.3 1.0 2 3 x 1 6 7 0.7 7 74 x2 5 ¼ 6 0.7 7. 4 5 0.7 5 0.9 x3 0.8 0.7

By step 1, x1 = x2 = x3 = 1 and N = {1, 2, 3}. In step 2, for i = 1, x2 = x3 = 0.3 and N = {1}, and for i = 2 max ½minða2j ; xj Þ ¼ 0.5 < 0.7 ¼ b2 ;

j2f1;2;3g

hence we have not any solution. Example 3. 2 1.0 6 0.7 6 6 4 0.4 0.5

Solve 0.2 0.8 0.9 0.7

2 3 3 0.3 0.3 2 3 x 1 6 7 0.5 7 76 7 6 0.7 7 7. 74 x2 5 ¼ 6 4 0.7 5 0.7 5 x3 0.8 0.9

By step 1, x1 = x2 = x3 = 1 and N = {1, 2, 3}. In step 2, for i = 1, x1 = 0.3 and N = {2, 3}, and for i = 2, x2 = 0.7 and N = {3}. For i = 3, x is not changed, and for i = 4, x3 = 0.8. Finally the solution is x = (0.3, 0.7, 0.8)t.

4. Application of fuzzy determinant Theorem 4.1. For a given matrix A 2 Bðn; nÞ and b 2 BðnÞ, if F det (A) < mini2Nbi, then A  x = b has not unique solution. Proof. Let A  x = b has a unique solution, then from Theorem 9 of [2] there is a permutation p 2 Pn such that X aipðiÞ P bi > aipðjÞ  bj ; j2N fig

and hence F detðAÞ P min aipðiÞ P min bi ¼ bk ; i2N

i2N

for some indexes k 2 N, which is a contradiction. h

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Example 4. Let 2 0.9 0.2 6 4 0.3 0.8 0.5

0.7

32

x1

3

2

0.7

3

7 76 7 6 0.6 54 x2 5 ¼ 4 0.7 5.

0.8

0.3

0.7

x3

Here F det(A) = 0.6 and we have M 1 ¼ f1g;

M 2 ¼ f2; 3g;

M 3 ¼ f g.

f1 ¼ M f2 ¼ f g; M f3 ¼ f1g. Also Then x1 = 0.7, x2 = 0.7, x3 = 1 and M I 1 ¼ f1g; K 1 ¼ f g;

I 2 ¼ f2; 3g; I 3 ¼ f g; K 2 ¼ f g; K 3 ¼ f1g;

L1 ¼ f1g;

L2 ¼ f2; 3g;

L3 ¼ f1g.

Here {L1, L2, L3} is not minimal covering of M, but {L1, L2} is minimal covering of M. This means that N1(A, b) = {1}, N2(A, b) = {2, 3}, N3(A, b) = { }. Therefore all the solutions are of the form x = (0.7, 0.7, a)t where a 2 [0, 1]. Example 5. Let 2 0.3 0 6 4 0 0.3 0

0

32

0

x1

3

2

0.7

3

7 76 7 6 0 54 x2 5 ¼ 4 0.7 5. 0.3

0.7

x3

In this example, F det(A) = 0.3, and we have not any solution. In the rest of paper, we present three examples with maxi bi < F det(A), and with different cases. So in this case have not any relationship between F det(A) and maxibi. Example 6. Consider 2

32

3

2

3

0.9

0.2

0.7

6 4 0.3

0.4

76 7 6 7 0.6 54 x2 5 ¼ 4 0.3 5;

0.5

0.8

0.3

x1 x3

0.5 0.4

which F det(A) = 0.6, but we have no any solution. Because M 1 ¼ f1; 3g;

M 2 ¼ f2; 3g;

M 3 ¼ f1; 2g; P and x1 ¼ 0.4; x2 ¼ 0.3; x3 ¼ 0.3, also xj < b1 . Hence we have j2f1;2;3g a1j   not any solution, [6].

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Example 7. Consider 3 2 32 3 2 0.5 0.9 0.2 0.7 x1 7 6 76 7 6 4 0.3 0.4 0.6 54 x2 5 ¼ 4 0.5 5; 0.5

0.8

0.3

0.5

x3

which F det(A) = 0.6, but have many solutions. Because M 1 ¼ f1g;

M 2 ¼ f3g;

M 3 ¼ f1; 2g;

and x1 ¼ 0.5; x2 ¼ 0.5; x3 ¼ 0.5, also e 1 ¼ f3g; M

e 2 ¼ £; M

e 3 ¼ £; M

I 1 ¼ f1g;

I 2 ¼ f3g;

I 3 ¼ f2g;

K 1 ¼ f3g;

K 2 ¼ £;

K 3 ¼ £;

L1 ¼ f1; 3g;

L2 ¼ f3g;

L3 ¼ f2g.

Hence x is a solution. Also {L1, L2, L3} is not minimal covering of M = {1, 2, 3}, but {L1, L3} is minimal covering. Also N1 = {1}, N2 = { }, N3 = {2}. Therefore all the solutions are of the form x = (0.5, a, 0.5)t where a 2 [0, 0.5] [6]. Example 8. Consider 2

0.5

6 4 0.2 0.1

0.1 0.4 0.1

0.2

32

x1

3

2

0.3

3

76 7 6 7 0.1 54 x2 5 ¼ 4 0.3 5; 0.6 0.3 x3

which F det(A) = 0.4, but we have a unique solution. Because M 1 ¼ f1g;

M 2 ¼ f2g;

M 3 ¼ f3g;

and x1 ¼ 0.3; x2 ¼ 0.3; x3 ¼ 0.3, also e 1 ¼ £; M I 1 ¼ f1g;

e 2 ¼ £; M e 3 ¼ £; M I 2 ¼ f2g; I 3 ¼ f3g;

K 1 ¼ £; K 2 ¼ £; K 3 ¼ £; L1 ¼ f1g; L2 ¼ f2g; L3 ¼ f3g. Hence x is a solution and {L1, L2, L3} is minimal covering of M = {1, 2, 3}. Also N1 = {1}, N2 = {2}, P N3 = {3}, therefore x1 = 0.3, x2 = 0.3, x3 = 0.3. We note that here aii P bi > j6¼i aij  bj .

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5. Conclusion In this paper, we present an algorithm for calculate fuzzy determinant and an algorithm for solve A  x = b over the max–min algebra B ¼ f½0; 1;  ¼ max;  ¼ ming. It has been proved that if F det(A) < mini2Nbi then A  x = b has not a unique solution.

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