Nieminen, U. J.

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Institute of Mathematics AS CR, v. v. i.

KybernetikaU. J. NieminenA graphical way to solve the Boolean matrix equations AX = B and XA = BKybernetika, Vol. 10 (1974), No. 1, (61)--65Persistent URL: http://dml.cz/dmlcz/125747Terms of use:© Institute of Information Theory and Automation AS CR, 1974Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access to digitizeddocuments strictly for personal use. Each copy of any part of this document must contain theseTerms of use.This paper has been digitized, optimized for electronic delivery and stamped withdigital signature within the project DML-CZ: The Czech Digital Mathematics Libraryhttp://project.dml.czKYBERNETIKA — VOLUME 10 (1974), NUMBER 1 A Graphical Way to Solve the Boolean Matrix Equations AX=B and XA = B U . J. NlEMINEN A graphical way to find all the solutions of the Boolean matrix equations AX = B and XA = B is proposed and an example is given. 1. INTRODUCTION AND BASIC CONCEPTS As shown by Ledley in [2, pp. 448-484] and in [3, 479-494] , the determination of the solutions for the Boolean matrix equations AX = BandXA = B has important applications to switching theory and logical problems. A way to find all the solutions is given in the books cited above. Recently, Rudeanu [4] has derived a complete solution to the equations AX = B and XA = B in parametric form. In this paper we apply a well known graphtheoretic representation of a Boolean matrix to find a graphical way to determine the complete solution to the equations AX = B and XA = B. We assume that the reader is familiar with the basic concepts in graph theory. By a Boolean matrix Q = [a,/J we shall mean in this paper a (0, l)-matrix. The join of two Boolean n x m matrices A and B is the matrix [au u fr;/J, and the product of the matrices C and D of orders n x p and p x m, respectively, is an n x m matrix CD = [\jcisdsj~\. Further, AT is the transpose of A and A' the com-plement of A, i.e. AT =s [oji] and A' = \a'is\- A ^ -B if and only if a;j ^ bti for any index pair ij. It is well known that with every m x n Boolean matrix Q one can naturally as-sociate a bipartite graph Gb(Q) as follows (see e.g. Hedetniemi [ l ] ) : The set of vertices V(Gb(Q)) of Gb(Q) consists of two disjoint subsets {«; | / = 1, . . . , m} and {vj\j = 1, •••,«} which correspond to the rows and columns of Q, respectively. An edge («,-, Vj) joining «,- and vp belongs to the edge set E(Gb(Q)) only if a,7 = 1 in Q. Conversely, every bipartite graph Gb can be translated into a Boolean matrix according to the rules above. In the following we shall concentrate on the equation AX = B. As known, the solution of XA = B is analogous to that of AX = B. 2. THE BOOLEAN MATRIX EQUATION AX = B Consider the product of two Boolean matrices A and B, and let the vertex sets of the bipartite graphs Gb{A) and Gb{B) be V(G„(A)) = {uAi \ i = 1,. . . , . . . , m} u u {vAs | s = 1,..., k} and V{Gb{B)) = {wBs | S = 1,.... k] u {vBj\j = 1,..., n}. Let us draw the bipartite graphs Gfe(A) and Gb{B) such that the vertices in the sets {vAs} and {uBs} are common, and denote the graph thus obtained by Gj,(A) Gb{B). Then, according to the formula AB = ~XJaisbsj~\, in the bipartite graph Gb{AB) a vertex uABi is connected by an edge to a vertex vABJ if and only if there is a path of length two from uAi to vBj in the graph G6(A) Bb{B). As an illustration, see the graphs of Fig. 1. This graphical form of the product of two Boolean matrices can be applied to the determination of a complete solution to AX = B. A = Г , B = 1 0 0 l l , AB - [° 1] 1 0 ,0 11 U A B l Ч ч O Д B 1 \ UAB2° *> AB2 GjAB) Fig. 1. As shown in the literature, the equation AX = B has a solution if and only if the matrix {ATB')' is a solution to AX = B, i.e. A{ArB')' = B. Moreover, the solutions of AX = B form a join semilattice, denoted by LU{X), where (ATB')' is the greatest element. Hence, if AQ = B, g u (ATB')' = (ATB')'. Thus, in order to obtain the complete set of solutions, one needs to determine the greatest element and the minimum elements of the semilattice LU{X), if such exist. First we consider a direct way to determine the graph Gb{{ArB')'), and the matrix (AT5')' as well, and then we show an obvious way to find all the solutions of AX = B. Assume that the equation AX = B has a solution. Now clearly a bipartite graph Gj,(X0) corresponds to the greatest solution of AX = B, if in the graph Gft(A) Gb(-X'o) every vertex uAi, corresponding to uBi in Gb{B), is connected by a path of length two to every vertex vXo-j, corresponding to vBj in Gb(B), for which (uBi, vBj) $ E(Gb(B)), i.e. (uBi, vBj) e E(Gb(B')). Thus the following simple rule can be obtained to find the graph Gb(X'0): Rule 1. Connect in Gb(X'0) the vertices TuAi = {vAil,..., vAir} = {uXo,it,..., uXtJ.ir}, uAieV(Gb(A)), to all the vertices vXol for which (uBi, vBJ) e E(Gb(B')). It should be noted that the matrix X0 determined by the rule above does not give any indication of the non-consistency of the equation AX = B. As an illuminating example, consider the following consistent Boolean matrix equation (0 [iiMU The graphs Gb(A) and Gb(B) are given in Fig. 2, and the graph Gb(X'0) can be seen in the graph Gb(A) Gb(X'0) determined by Rule 1. Hence, X0 = 1 0 1 1 1 0 - . • X M G K ( B . ) Fig. 2. Consider now a way to find all the solutions of AX = B. We construct a solution matrix base, denoted by Zx, Z2,..., Zt, where every Zw, w = 1, ..., t, is a Boolean matrix of the order of X and corresponds to an edge, say (uBi, vB}), of Gb(B) such that Gb(Zw) contains any edge which gives in Gb(A) Gb(Zw) a path of length two from UAI t 0 vz„j { = vB]) and no edges such that there would be a path of length two in Gb(A) Gb(Zw) determining an edge of Gb(B'). Since the matrix product is distributive with respect to the join operation and AZW ^ B, A(ZX u Z2 u ... u Z,) = B according to the definition of the matrices Zw, if Zw > 0 for any w, w = 1, . . . , t. Furthermore, as every G6(ZW) contains all the edges giving in Gb(A) Gb(Zw) the edge of Gb(B) which determines Gb(Zw), Zx u ... u Z, = (ATB')' = X0, the greatest element of the solution join semilattice LU(X). According to the definition of Zw, the matrix equation AX = B is consistent if and only if Zw > 0, i.e. E(Gb(Zw)) #= 0, for any w, w = 1, . . . , t. A matrix Q is a solution of AX = B, if Q n Z w > 0 for every w, and g 0 is a mini­mum element of LU(X) if and only if the equation Q00 n Z w > 0 does not hold for any matrix <200 < Q0, w = 1,..., t. For the determination of a matrix Z w corresponding to an edge (uB;, vBj) e E(Gb(B)) we obtain the following simple rule: Rule 2. Connect in Gb(Zw) the vertices of TuAi = {vAh,..., vAir} = {wZwll,... • • •> "z w r r }> UAI e ^(o»(^))> t 0 p z„j a n t ! remove then the edges which belong to Gb(X'0). VAЗ G ъ ( A ) G b ( Z l ) -вio- 1 0 0 0 1 0 uB l o 1 0 o 1 0 1 o _ O 0 j 0 1 O 0 j Fig. 3. Consider as an example the matrix equation in (1). Fig. 3 shows the determi­nations of the basis matrices Zx, Z2, and Z 3 corresponding to the edges (uB1, vB1), (uB2, vB1), and (uB2, vB2), respectively. The dotted lines in Fig. 3 mean the edges of Gb(X'0). Since Z l 5 Z 2 , Z 3 > 0, the equation in (1) is consistent. [0 0 As one can readily check, the minimum elements of LU(X) are X, = 0 1 1 0 and * 2 = 1 0 1 1 0 0 . Z^ u Z 2 u Z 3 = "l o" 1 1 1 0 = x0 = (ATB')'. The other solutions to AX = = B, which are between X1 and X0 in LU(X), are X3 = "1 0" 0 1 1 0 and X, = "0 o" 1 1 1 0 There is an other way to construct a solution matrix base. After determining the matrices Z , , . . . , Z , defined above, we substitute the matrix Zw, w= \,...,t, by a set {YlH„ Y2w,..., Yj„w} of matrices, where Ylw u ... u YSwW = Z„„ Ykw > 0 and Ykw contains a single one for any k, k = 1, ..., sw. Every solution to AX = B is obtained by forming all possible joins (U^) of the matrices in the sets {Yiw,..., YSwW} such that ({JY) n Zw > 0 for any value of w. In the example considered before, "1 0" 0 0 0 0 r 1 2 = "0 0" 0 0 1 0 ү2l = 0 o 1 0 0 0 Y„ = 0 0 "0 0 0 0 , and Y31 = 0 1 1 0 0 0 Thus X i X, = Y12 u Y22 u Y31 = Y12 u Y31 = Y22 u y 3 1, X2 = YuuY21uY31, 3 = Ylt u Y22 u Y31 = Yu u Y12 u Y22 u Y31, XA = Y2 u Y2l u y 3 1 = y l 2 u u y 2 1 u Y22 u Y31, and X0 = Yn u Y12 u Y21 u Y31 = Yn u Y21 u Y22 u Y31 = = YUUY12UY21UY22UY31. In the case of the equation XA = B, Rule 1 and Rule 2 can be expressed as follows: Rule 1'. Connect in Gb(X'0) the vertices TvAi = {uAir ..., uAir} = {vXo-h, ..., vXo,ir}, vAieV(Gb(A)), to all the vertices uXo,jfor which (uBj, uBi)eE(Gb(B')). Rule 2'. Connect in Gb(Zw) the vertices of FvAJ = {uAjl,..., uAJr} = {vZyvjl, ... ..., vZnJr}, vAJ e V(Gb(A)), to uZwi and remove then the edges which belong to Gb(X'0). (Received August 6, 1973.) REFERENCES [1] S. T. Hedetniemi: Graphs of (0, l)-matrices. In: Recent Trends in Graph Theory (ed. by M. Capobianco, J. B. Frechen, and M. Krolik). Springer-Verlag, Berlin—Heidelberg— New York 1971, 157-171. [2] R. S. Ledley: Digital computer and control engineering. McGraw-Hill, New Y o r k — T o r o n t o -London 1960. [3] R. S. Ledley: Programming and utilizing digital computers. McGraw-Hill, New Y o r k -Toronto—London 1962. [4] S. Rudeanu: On Boolean matrix equations. Rev. Roum. Math. Pures et Appl. AT//(1972), 7, 1075-1090. U. J. Nieminen, Research assistant, Finnish Academy, Department of Technical Sciences, Lauttasaarentie 1, 00200 Helsinki 20, Finland. 

A graphical way to solve the Boolean matrix equations $AX=B$ and $XA=B$

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A graphical way to solve the Boolean matrix equations $AX=B$ and $XA=B$

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A graphical way to solve the Boolean matrix equations AX=BAX=BAX=B and XA=BXA=BXA=B

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Institute of Mathematics AS CR, v. v. i.

A graphical way to solve the Boolean matrix equations $AX=B$ and $XA=B$