Departement Elektrotechniek ESAT-SISTA/TR Minimal state space realization of SISO systems in the max algebra 1.

Transcription

1 Katholieke Universiteit Leuven Departement Elektrotechniek ESAT-SISTA/TR 9- inimal state space realization of SISO systems in the max algebra 1 Bart De Schutter and Bart De oor October 199 A short version of this paper has appeared in the Proceedings of the 11th International Conference on Analysis and Optimization of Systems, Sophia-Antipolis, France, June 1-1, 199, pp. {8. 1 This paper is also available by anonymous ftp from ftp.esat.kuleuven.ac.be as le 9-.ps.Z in the directory pub/sista/deschutter/reports. ESAT/SISTA, K.U.Leuven, Kardinaal ercierlaan 9, B-001 Leuven (Heverlee), Belgium, tel , fax , bart.deschutter@esat.kuleuven.ac.be, bart.demoor@esat.kuleuven.ac.be. Bart De Schutter is a research assistant with the N.F.W.O. (Belgian National Fund for Scientic Research) and Bart De oor is a senior research associate with the N.F.W.O.

2 Abstract First we determine necessary and for some cases also sucient conditions for a polynomial to be the characteristic polynomial of a matrix with elements in R max. Then we indicate how to construct a matrix such that its characteristic polynomial is equal to a given monic polynomial in S max, the extension of R max. Next we use these results to develop a procedure to nd the minimal state space realization of a single input single output (SISO) discrete event system, given its arkov parameters.

3 1 Introduction 1.1 Overview There exists a wide range of frameworks to model and to analyze discrete event systems Petri nets, generalized semi-arkov processes, formal languages, perturbation analysis, computer simulation and so on. In this paper we concentrate on discrete event systems that can be described with the max algebra. We address the minimal state space realization problem for max-algebraic single input single output (SISO) systems. We show that the characteristic equation in the max algebra plays an important role in the solution of this problem. Therefore we rst make a study of the characteristic equation of a matrix in the max algebra. Next we use these results to propose a procedure to nd a minimal state space description of a max-linear time-invariant SISO discrete event system. In the rst section we introduce the notations and some of the denitions and properties that will be used throughout the remainder of this report. In the second section we give some necessary and sucient conditions for the coecients of a polynomial such that it is the characteristic polynomial of a matrix in the max algebra. Then we indicate how to construct a matrix for which the characteristic polynomial is equal to a given polynomial. These results will then be used to determine a lower bound for the minimal order of the state space description of a SISO system in the max algebra. This will enable us to nd the minimal realization of a SISO discrete event system, given its arkov parameters. Finally we shall illustrate this procedure with a few examples. 1. Notations One of the mathematical tools used in this report is the max algebra. In this introduction we only explain the notations we use to represent the max-algebraic operations. A complete introduction to the max algebra can be found in [1]. In this report we use the following notations a b = max(a; b) and a b = a + b. " =?1 is the neutral element for in R max = (R[ f"g; ; ). To avoid confusion we always write the sign explicitly. The inverse element of a = " for in R max is denoted by a?1. The division is dened as follows a b = a b?1 if b = " If A is an m by n matrix then the element on the i-th row and on the j-th column is denoted by a ij. E n is the n by n identity matrix in R max e ij = 0 if i = j and e ij = " if i = j. The operations and are extended to matrices in the usual way. A k = A A A. {z } k times We also use the extension of the max algebra S max that was introduced in [1, ]. S max is a kind of symmetrization of R max. We shall restrict ourselves to the most important features of S max. For a more formal derivation the interested reader is referred to []. There are three kinds of elements in S max the positive elements (S max, this corresponds to R max ), the negative elements (S max) and the balanced elements (S max). The positive and the negative elements are called signed (S _ max = S max [ S max as follows a b = a if a > b ; a b = b if a < b ; a a = a 1 ). The operation in S max is dened

4 If a S max then it can be written as a = a + a? where a + is the positive part of a and a? is the negative part of a; jaj = a + a? is the absolute value of a. There are three possible cases if a S max then a + = a and a? = ", if a S max then a + = " and a? = a and if a S max then a + = a? = jaj. Example 1.1 Let a = S max, then a+ =, a? = and jaj =. For b = S max we have b + = ", b? = and jbj =. This symmetrization allows us to 'solve' equations that have no solutions in R max. Unfortunately we then have to introduce balances (r) instead of equalities. Informally an sign in a balance means that the element should be at the other side so r means r. If both sides of a balance are signed (positive or negative) we can replace the balance by an equality. To select submatrices of a matrix we use the following notation A([i 1 ; i ; ; i k ]; [j 1 ; j ; ; j l ]) is the matrix resulting from A by eliminating all rows except for rows i 1 ; i ; ; i k and all columns except for columns j 1 ; j ; ; j l. A(i; ) is the i-th row of A and A(; j) is the j-th column of A. [1 n] stands for [1; ; ; n]. 1. Some denitions and theorems Denition 1. (Determinant) dened as no det A = sgn () P n i=1 a i(i) Consider a matrix A S nn max. The determinant of A is where P n is the set of all permutations of f1; ; ng, and sgn () = 0 if the permutation is even and sgn () = 0 if the permutation is odd. Theorem 1. Let A S nn max. The homogeneous linear balance A x r " has a non-trivial signed solution if and only if det A r ". Proof See []. The proof given there is constructive so it can be used to nd a solution. Denition 1. (Characteristic equation) Let A S nn max. The characteristic equation of A is dened as det(a E n ) r ". This leads to n n p=1 a p n?p r " which will be called a monic balance, since the coecient of n equals 0 (i.e. the identity element for ). If we dene p = a + p and p = a? p and if we move all terms with negative coecients to the right hand side we get n n i=1 i n?i r n j=1 j n?j with p ; p R max. In [8] Olsder denes a variant of this equation using the dominant instead of the determinant. This leads to signed coecients a Olsder p S _ max or Olsder p p Olsder = ".

5 Theorem 1. (Cayley-Hamilton) In S max every square matrix satises its characteristic equation. Proof See [] and [8]. The characteristic equation of a positive matrix A positive matrix is a matrix the elements of which lie in R max. In this section we derive necessary conditions for a polynomial in S max to be generated by a matrix with elements in R max..1 The characteristic equation In this subsection we derive a formula for the coecients of the characteristic equation. Property.1 Consider A S nn max and k S max, then det(k A) = k n det A no Proof det(k A) = sgn () (k a i(i) ) P n i=1 no = sgn () k n (k a i(i) ) P n i=1 no = k n sgn () (k a i(i) ) P n = k n det A i=1 We know that det(a E n ) r " represents the characteristic equation of A. But because of property.1 we have that det(a E n ) = ( 0) n det( E n A) and since a r " if and only if a r " the characteristic equation can also be represented by det(e n A) r ". Denition. (Principal submatrix) Let A S nn max and let fi 1 ; i ; ; i k g be a combination of k elements out of f1; ; ng. Then the matrix A([i 1 ; i ; ; i k ]; [i 1 ; i ; ; i k ]) is a k by k principal submatrix of A. It can be obtained from A by deleting n? k rows and columns.! n Every square n by n matrix A has principal submatrices of size k k. k We represent the max-algebraic sum of the determinants of all k by k submatrices of A as E k (A) E k (A) = det A([i 1 ; i ; ; i k ]; [i 1 ; i ; ; i k ]) 'C k n where C k n is the set of all combinations of k numbers out of f1; ; ng and ' = fi 1 ; i ; ; i k g. Property. If we represent the characteristic equation of A S nn max as n n?p r " then a p = ( 0) p E p (A).! n p=1 a p

6 Proof det( E n A) = det a 11 a 1 a 1n a 1 a a n a n1 a n a nn Since det(u 1 ; ; u i v i ; ; u n ) = det(u 1 ; ; u i ; ; u n ) det(u 1 ; ; v i ; ; u n ) we can split this determinant up as det det " " " " det " " " a 1n " a n " " a nn 'C p n with B ' (; i) = A(; i) if i ' ; B ' (; i) = E n (; i) if i ' If ' = fi 1 ; i ; ; i p g, we have that a 11 " " a 1 " det a n1 " det B '. det det B ' = n?p ( 0) p det A([i 1 ; i ; ; i p ]; [i 1 ; i ; ; i p ]) a 1 " " a " " a n a 11 a 1 a 1n a 1 a a n a n1 a n a nn since B ' has n? p diagonal entries that are equal to and p columns that contain elements of A. We have used property.1 to put the signs before the determinant. So we nd that det( E n A) = n p=0 ( 0) p E p (A) n?p with E p (A) the max-algebraic sum of the determinants of all possible p by p principal submatrices of A. This results in a 0 = 0 ; a 1 = tr (A) = Example. Consider A = n i=1 a n = ( 0) n det A a ii ; The characteristic equation of A is r "... Properties of the characteristic equation Proposition. In S max every monic n-th order linear balance is the characteristic equation of an n n matrix.

7 Proof Suppose that the linear balance has the following form n a 1 n?1 a n?1 a n r " We shall prove that this is the characteristic equation of the matrix A = " 0 " " " " 0 " " " " 0 a n a n?1 a n? a 1 We use the formula of property. to calculate the coecients of n?p in the characteristic equation of A. If we take the p by p principal submatrices of A we see that each of them has an "-column { and thus a determinant equal to " { except for B p = A([n?p+1 n]; [n?p+1 n]). So a p = ( 0) p E p (A) = ( 0) p det B p. B p is obtained by deleting the rst n? p rows and columns of A, so B p = " 0 " " " " 0 " " " " 0 a p a p?1 a p? a 1 If we develop the determinant of B p to the rst column we get det B p = ( 0) p?1 ( a p ) = ( 0) p a p so the coecient of n?p equals ( 0) p ( 0) p a p = ( 0) p a p = a p. Thus the linear balance is indeed the characteristic equation of A. In the next section we shall see that not every monic polynomial corresponds to the characteristic polynomial of a positive matrix.. Properties of the characteristic polynomial of positive matrices Property. If A R nn max then a 1 S max. n Proof We know that a 1 = tr (A) = i=1 a ii with a ii R max so a 1 S max. To prove the following property we rst need a lemma involving permutations. The parity of a permutation can be determined in various ways. We use Property. The parity of a permutation is equal to the parity of the number of its elementary cycles of even length. First consider a circular permutation c of n elements c (i 1 ) = i ; c (i ) = i ; ; c (i n?1 ) = i n ; c (i n ) = i 1 The graph of this permutation is

8 i1 q q in QQk Q Q Qq in?1 i -q QQQ i Q Qs q i + q This permutation has a cycle of length n. If n is even, then c P n is odd because there is 1 cycle of even length. If n is odd, then c P n is even because there are 0 cycles of even length. If a permutation of n numbers is not circular we can decompose it into r elementary cycles C i of length l i, with r > 1 and rx i=1 l i = n. Each cycle will be a circular permutation. Example.8 Let P be dened as (1) = ; () = ; () = 1 and () =. The graph of this permutation is 1 This permutation can be decomposed into three elementary cycles, each of which is a circular permutation of its vertices. There is one cycle of even length (1!! 1). So is an odd permutation. Lemma.9 If k;even (k > 0) is an even permutation of k elements, then it can be decomposed into two even permutations of an odd number of elements or two odd permutations of an even number of elements k;even = l+1;even [ k?l?1;even or m;odd [ k?m;odd If k+1;odd (k > 0) is an odd permutation of k + 1 elements, then it can be decomposed into an even permutation of an odd number of elements and an odd permutation of an even number of elements k+1;odd = p+1;even [ k?p;odd Proof First consider k;even. This is an even permutation of an even number of elements so it is not circular and it can be decomposed into elementary cycles. Suppose that there are c even cycles of even length each having n even;i elements and c odd cycles of odd length each having n odd;j elements. Let n tot;even = cx even i=1 n even;i and n tot;odd = cx odd j=1 n odd;j. Since the parity of k;even is even, c even should also be even. n tot;even is always even. The total number of elements

9 n tot = k is even, so we have that n tot;odd is also even and hence that c odd is even. There are two cases c even = 0 and c even = 0. If c even = 0 then c odd = 0 because k = 0. Take one cycle of odd length l + 1. This corresponds to an even permutation of l + 1 elements l+1;even. The other cycles form a permutation with 0 cycles of even length, so it is an even permutation of the remaining k? l? 1 elements k?l?1;even. If c even = 0 we take one cycle of even length m. This corresponds to m;odd. The remaining cycles constitute a permutation with an odd number (c even? 1) of cycles of even length k?m;odd. So we have proven that k;even can be decomposed as l+1;even [ k?l?1;even or m;odd [ k?m;odd. Now consider k+1;odd. This is an odd permutation of an odd number of elements so it is not circular and it can be decomposed into elementary cycles. Since the parity of k+1;odd is odd, c even should also be odd. n tot;even is always even, and since the total number of elements n tot = k + 1 is odd we have that n tot;odd is odd and hence that c odd is odd. This means that c odd = 0. So let us take one cycle of odd length p + 1. This corresponds to an even permutation of p + 1 elements p+1;even. The other cycles will then correspond to a permutation of with an odd number (c even ) of cycles of even length, so it is an odd permutation of k? p elements k?p;odd. So k+1;odd = p+1;even [ k?p;odd. Now we give some properties of a p = ( 0) p E p (A) = a + p a? p. First we suppose that we don't simplify. This means that for a = we have a + = and a? =. Later we shall see how we have to adapt the properties to take simplication into account, because then we shall have that a = results in a = or a + = " and a? =. Property.10 Let A R nn max and let a p = ( 0) p E p (A) = a + p a? p (without simplifying b p c ). Then 8p f; ; ng a + p a? r a? p?r ; where bxc stands for the largest integer number less than or equal to x. Proof We know that a p = ( 0) p E p (A) = ( 0) p 'C p n P p sgn () a i1 i (1) a i i () a ipi (p) with ' = fi 1 ; i ; ; i p g. If we extract the positive and the negative part of a p (without simplifying ), we nd for k > 0 a + k = a? k = 'C k n 'C k n P k;even a i1 i (1) a i i () a ik i (k) (1) P k;odd a i1 i (1) a i i () a ik i (k) ()

10 a + k+1 = a? k+1 = 'C k+1 n 'C k+1 n P k+1;odd a i1 i (1) a i i () a ik+1 i (k+1) () P k+1;even a i1 i (1) a i i () a ik+1 i (k+1) () Let us rst consider a + k. The terms of a+ k are generated by even permutations of k elements. According to lemma.9 such a permutation can be decomposed into two even permutations of odd lengths or two odd permutations of even lengths. So if we consider all possible concatenations of two even permutations of odd lengths (corresponding to a? l+1 a? ) or k?l?1 two odd permutations of even length (a? m a? k?m ), we are sure to have included all terms of a + k. In other words a+ k?1 k a? r a? k?r. Since (a? r a? k?r ) (a? k?r a? r ) = a? r a? k?r we nd a + k k a? r a? k?r. Now consider a + k+1, the terms of which are generated by odd permutations of k+1 elements. Lemma.9 also tells us that such a permutation can be decomposed into an odd permutation of an even number of elements and an even permutation of an odd number of elements. Using the same reasoning as for a + k, we nd that a+ k k+1 a? r a? k+1?r. Combining the two inequalities leads to a + p b p c a? r a? p?r. We don't have a similar expression for a? p because then some of the generating permutations are circular, and these cannot be decomposed into more than one elementary cycle. Normally we simplify ; by setting a? p = " if a? p < a+ p and a + p = " if a? p > a+ p. Therefore we shall from now on represent the characteristic equation of A R nn max as with 8 >< > n n i= i n?i r 1 n?1 p = a + p ; p = " if a + p > a? p ; p = "; p = a? p if a + p < a? p ; p = a + p ; p = a? p if a + p = a? p n j= j n?j So there are three possible cases p = ", p = " or p = p. We already have omitted 1 because property. leads to 1 = a + 1 = ". We have that p a + p, p a? p and ja pj = p p. Property.11 8i f; ; ng i b c i for the largest integer number less than or equal to x. ( r r ) ( i?r i?r ) ; where bxc stands 8

11 Proof Using the fact that a? i ja i j property.10 leads to a + i b c i b c i ja? r j ja? i?r j We also know that i a + i. So i b c i ( r r ) ( i?r i?r ) ( r r ) ( i?r i?r ) We even have a more stringent property Property.1 8i f; ; ng at least one of the following statements is true 1) i ) i < ) i < b c i b c i r= i?1 r= r i?r r i?r r i?r where bxc stands for the largest integer number less than or equal to x. Proof Take an arbitrary i f; ; ng. Then according to property.10 there exists an i s such that i a + i a? s a? i?s We have that either a? s = s or a? s < s and the same goes for a? i?s. This means that at least one of the following inequalities holds 1) i s i?s ) i < s i?s b c i b c i r= r i?r r i?r ) i < s i?s s i?s i?1 r i?r In the last two max-algebraic sums we can start from r = because 1 = ". r= Property.1 gives necessary conditions for the coecients of an S max polynomial such that it is the characteristic polynomial of a positive matrix. 9

12 Necessary and sucient conditions for a polynomial to be the characteristic polynomial of a positive matrix We now give some necessary and sucient conditions for the coecients of the characteristic equation of a positive matrix. These conditions will play an important role when one wants to determine the minimal order of a SISO system in the max algebra, as will be shown in section. We shall prove that the conditions are sucient by giving for each set of conditions a matrix the characteristic equation of which will satisfy the conditions. So if we have a monic polynomial in S max the results of this section will allow us to 1. check whether the given polynomial can be the characteristic polynomial of a positive matrix and. construct a matrix such that its characteristic polynomial is equal to the given polynomial. For the lower dimensional cases we can give an analytic description of the matrix we are looking for. For higher dimensional cases we shall rst state a conjecture and then develop a heuristic algorithm that will (in most cases) nd a solution. In this section we shall encounter matrices with the following structure A = 0;1 0; 0; 0;n?1 0;n 0 1; 1; 1;n?1 1;n " 0 ; ;n?1 ;n " " " n?;n?1 n?;n " " " 0 n?1;n The coecients of the characteristic equation of A (without simplication of ) are given by a + k = a? k = C p n ; p even C p n ; p odd where = fi 1 ; i ; ; i p g. P p (jr?ir )=k ; ir<jri r+1 P p (jr?ir)=k ; ir <jri r+1 Proof We know that a k = ( 0) k E k (A) with E k (A) = det A([i 1 ; i ; ; i k ]; [i 1 ; i ; ; i k ]) C k n and by reordering if necessary i 1 < i < < i k. Now let B(i 1 ; i ; ; i k ) = A([i 1 ; i ; ; i k ]; [i 1 ; i ; ; i k ]) = i1 ;j 1 i ;j ip;j p () i1 ;j 1 i ;j ip;j p () i1?1;i 1 i1?1;i i1?1;i i1?1;i k i ;i 1 i?1;i i?1;i i?1;i k " i ;i i?1;i i?1;i k " " " ik?1;i k 10

13 with ( ir+1 ;i r = 0 if i r+1 = i r + 1 ; = " if i r+1 = i r + 1 We shall prove by induction that det B(i 1 ; i ; ; i k ) = ( 0) K C p n ; p odd C p n ; p even P p (jr?ir )=K ; ir<jri r+1 P p (jr?ir)=k ; ir <jri r+1 with K the number of columns of B(i 1 ; i ; ; i k ). i1 ;j 1 ip;j p i1 ;j 1 ip;j p () K = 1 det B(i 1 ) = det[ i1?1;i 1 ] = i1?1;i 1 with i 1? (i 1? 1) = 1 = K and i 1? 1 < i 1 so equation () is satised. K = k Suppose that equation () holds for K = 1; ; k? 1. By developing the determinant to the rst column we get det B(i 1 ; i ; ; i k ) = i1?1;i 1 det B(i ; ; i k ) i ;i 1 f i1?1;i det B(i ; ; i k ) i ;i [ i1?1;i det B(i ; ; i k ) = i1?1;i 1 det B(i ; ; i k ) ]g i ;i 1 i1?1;i det B(i ; ; i k ) i ;i i ;i 1 i1?1;i det B(i ; ; i k ) ( 0) k? ik?1;i k? i ;i 1 i1?1;i k?1 det B(i k) Consider the l-th term ( 0) k?1 ik ;i k?1 i ;i 1 i1?1;i k ( 0) l?1 ik?1;i k? i ;i 1 i1?1;i l det B(i l+1 ; ; i k ) We have that K = k? (l + 1) + 1 = k? l for B(i l+1 ; ; i k ) so det B(i l+1 ; ; i k ) = ( 0) k?l C p n ; p odd P p C p n ; p even P p (jr?ir)=k?l ; ir<jri r+1 (jr?ir )=k?l ; ir<jri r+1 11 i1 ;j 1 ip;j p i1 ;j 1 ip;j p

14 Now ik?1;i k? i ;i 1 = 0 if i l = i l?1 + 1 So if i l = i 1 + l? 1 the l-th term becomes. i = i + 1 i = i = " otherwise 9 >= >; or i l = i 1 + l? 1 ; ( 0) l?1 i1?1;i 1 +l?1 ( 0) k?l p even P (jr?i r )=k?l po ir;j r P p odd (jr?i r )=k?l po ir;j r or ( 0) k?1 P p odd (jr?i r )=k po ir ;j r P p even (jr?i r )=k po ir ;j r So we nally nd that a k = ( 0) k = p even i 1 <<i k det B(i 1 ; ; i k ) P (jr?i r )=k po ir;j r p odd P (jr?i r )=k po ir;j r Extracting the positive and the negative parts leads to equations () and (). In the next subsections we shall case by case determine necessary and sucient conditions for n n i=1 i n?i r n j=1 j n?j (8) to be the characteristic equation of a positive matrix and indicate how such a matrix can be found. In all cases we have 1 = " as a necessary condition. j We also dene i;j = if i = " ; i = " if i = ".1 The 1 1 case The only necessary and also sucient condition is 1 = ". The matrix [ 1 ] has r 1 as its characteristic equation. 1

15 . The case ( The necessary and also sucient conditions are 1 = " 1 1 " # The matrix 1 has 0 r 1 as its characteristic equation. 1; Proof From property.10 we know that 1 1. This means that if 1 = " then also = ". First we prove that 1 1; = 1 1; = 1 1 = if 1 = " ; = " " = " if 1 = " and thus also = " We always have that 1; 1 because 1; = 1 if 1 = " ; 1 = " 1 if 1 = " Using formulas () and () we nd a 1 = 1 1; = 1 a = 1 1; =. The case 8 The necessary and also sucient conditions are >< > 1 = " or < 1 The matrix 1 0 1; 1; has r 1 as its " 0 " characteristic equation. Proof We already know that 1; 1 and that 1 1; =. Analogously we can prove that 1 1; = since if 1 = " we also have = ". Now we prove that 1; if > and that ;1 < if < If > the necessary condition for becomes 1. So 1; = if 1 = " ; 1 = " if 1 = " If < we have that < 1 so 1 = " and 1

16 1; = < 1 So we always have that 1; =. We nd a 1 = 1 1; = 1 a = 1 1; 1; = a = 1 1; =. The case First we distinguish three possible cases Case A 1 or < 1 Case B > 1 and > 1 and and ( 1 = " or = " or = ) Case C > 1 and > 1 and and = = " and = " If the coecients don't fall into (exactly) one of these three cases, they cannot correspond to a positive matrix. The necessary 8 and sucient conditions are 1 = " >< > or < 1 for Case A no extra conditions for Case B 1 or 1 < for Case C 1 = and 1 =..1 Extra conditions First we derive some extra conditions that automatically follow from the necessary and sucient conditions. Property.1 In Case B and Case C we have 1) = " ) = " ) ) 1 1 Proof The condition > 1 can only be fullled if = ". Since we then have that = " or equivalently. 1

17 Assume that < 1 1. This means that 1 = ". If we use this in the rst necessary and sucient condition for Case B we get 1 < 1 1 or < 1. But this is in contradiction with the fact that > 1 in Case B. The second necessary and sucient condition would lead to < 1 whereas > 1 in Case B. The necessary and sucient conditions for Case C would also lead to < 1, which is impossible since > 1 in Case C. So clearly our initial assumption was false and therefore we conclude that 1 1. Property. In Case C we have 1) 1 = " ) = = 1 1 ) = 1 = ( 1 ) ) = " ) = = 1 = ( 1 ) Proof We know that 1 1. Since = " we should have that 1 = ". From property.1 we know that 1 1. So 1 1. Since = this leads to = = 1 1. The condition 1 = then results in = 1 = 1 = = ( 1 ). The condition 1 = leads to = 1 = 1 ( 1 ) = ( 1 ) =. Since = 1 and > 1 we have > or equivalently = "... Necessary conditions Now we prove that the conditions for Case B and Case C are necessary. Remark The following properties are also valid for the coecients of an arbitrary n n matrix with n >. We shall need some expressions that can be derived from formulas (1) { () a? 1 = C 1 n a + = 'C n a? = 'C n a i1 i 1 a j1 j 1 a j j (10) a j1 j a j j 1 (11) (9) a + = C n a? = C n a + = C n a k1 k 1 a k k a k k (1) a k1 k 1 a k k a k k C n a l1 l 1 a l l a l l a l l C n 1 a k1 k a k k a k k 1 (1) a l1 l 1 a l l a l l a l l (1)

18 C n a? = C n with = fi 1 g ' = fj 1 ; j g = fk 1 ; k ; k g = fl 1 ; l ; l ; l g a l1 l a l l 1 a l l a l l a l1 l 1 a l l a l l a l l C n Property. If > 1 and > 1 then 1) = " a l1 l a l l a l l a l l 1 (1) ) = C n a l1 l a l l 1 a l l a l l with = fl 1 ; l ; l ; l g ) 1 1 ) ) Proof > 1 is only possible if = " and thus = a +. From formula (1) we know that = t 1 t t with t 1 = C n a l1 l 1 a l l a l l a l l t = C n t = C n a l1 l 1 a l l a l l a l l a l1 l a l l 1 a l l a l l We know that 1 = a? 1. If a+ > a? then we have > a? and if a+ a? = a?. So > 1 and > 1 means that > a? 1 a?. Using formulas (9) and (1) we nd that then we have a? 1 a? = t t with t = t = C 1 n ;C n C 1 n ;C n a i1 i 1 a k1 k 1 a k k a k k a i1 i 1 a k1 k a k k a k k 1 If we compare t 1 and t we see that t contains more terms than t 1 so we have that t 1 t. Analogously we nd t t. Combining these inequalities leads to t 1 t t t (1) 1

19 But we know that > a? 1 a? or equivalently t 1 t t > t t. Because of equation (1) we necessarily have that t 1 < t and t < t and thus = t = C n a l1 l a l l 1 a l l a l l (1) Suppose that the maximum of t is reached for = f 1 ; ; ; g and that a 1 a 1 > a a. We know that t <. First take k = and k =. Then we know that k 1 = and k 1 =. The inequality t < then reduces to or i 1 ;k 1 = ;k 1 = a i1 i 1 a k1 k 1 a a a 1 a 1 a a a i1 i 1 a k1 k 1 a 1 a 1 a? i 1 ;k 1 = ;k 1 = because a? = a j1 j a j j 1. j 1 =j Analogously we nd that i 1 ;k 1 = 1 ;k 1 = a i1 i 1 a k1 k 1 a? if we take k = 1 and k =. Combining the last two inequalities yields i 1 ;k 1 a i1 i 1 a k1 k 1 a? This leads to 1 1 = a? 1 a? 1 = p 1 ;q 1 a p1 p 1 a q1 q 1 a? From property.10 we know that a + a? 1 a? 1, so we have a+ a? or a? = and thus 1 1. a + a? also leads to. We already know that > a? 1 a? = a? a? a? =. so according to property.10 we should have that Now we prove that the conditions for Case B are necessary Property. If > 1 and > 1 then at least one of the following statements is true 1) 1 ) 1 < 1

20 Proof Using equation (1) and the fact that = a + if > 1, we nd a? 1 a+ = a i1 i 1 a l1 l a l l 1 a l l a l l C 1 n ; C n Combining formulas (11) and (1) leads to a? a+ = a j1 j a j j 1 a k1 k 1 a k k a k k 'C n ;C n Take an arbitrary term a i1 i 1 a l1 l a l l 1 a l l a l l of a? 1 a+. If i 1 = l 1 or if i 1 = l we know that i 1 = l and i 1 = l and then we see that the term a i1 i 1 a l1 l a l l 1 a l l a l l corresponds to the term of a? a+ with j 1 = l 1 ; j = l ; k 1 = i 1 ; k = l and k = l. Otherwise we have that i 1 = l 1 and that i 1 = l and then we can take the term of a? a+ with j 1 = l ; j = l ; k 1 = i 1 ; k = l 1 and k = l. So we have demonstrated that each term of a? 1 a+ also appears in a? a+ and thus that a? 1 a+ a? a+. We have that 1 = a? 1. From property. we know that a+ = and that a? =. If a + > a? we have = a + and thus 1. On the other hand if a + < a? we have > a + and this would lead to 1 <. The conditions for Case C are also necessary Property. If > 1 and > 1 and = = " and = " then 1) 1 = ) 1 = Proof First we prove that under the conditions > 1 and > 1 and = we should have that a + a + a? 1. We know that = a j1 j 1 a j j a + 'C n Suppose that the maximum of a + is reached for ' = f 1; g. Because = we have that a + = a? = a j1 j a j j 1 and thus 'C n a j1 j a j j 1 a 1 1 a ; 8j 1 ; j f1; ; ng We already know that a + = C n a l1 l a l l 1 a l l a l l Formulas (9) and (1) result in a? 1 a+ = a i1 i 1 a k1 k 1 a k k a k k C 1 n ;C n 18

21 Take an arbitrary term a l1 l a l l 1 a l l a l l of a +. If = l 1 or = l we know that = l and = l and then we see that a l1 l a l l 1 a l l a l l is less than or equal to the term of a? 1 a+ with i 1 = 1 ; k 1 = ; k = l and k = l. Otherwise we have that = l 1 and = l and then we take the term with i 1 = 1 ; k 1 = ; k = l 1 and k = l. So we have demonstrated that each term of a + also appears in a? 1 a+ and thus that a + a? 1 a+ (18) From property. we know that a + =. If a + < a? we have > a +. Then we get a+ = > 1 > a? 1 a+ but this is impossible because of equation (18). So we have to conclude that we always have that a + > a?. So = a + and thus 1. If we combine this with the condition > 1 we nd = 1 Because = = " we have = a + = a? =. In general we have that 1 1 but from property. we know that 1 1 =. This leads to 1 1 = = So 1 = 1 1 =. If = " then a? < a+ because we already know that = ". Assume that the maximum in equation (1) is reached for = f 1 ; ; ; g. Since = " we have that a 1 a 1 = " and that a a = ". Formula (1) then leads to a? = C n a l1 l 1 a l l a l l a l l < a 1 a 1 a a Considering the terms with l = and l = leads to l 1 =l ;l 1 ;l f 1 ; g a l1 l 1 a l l a a < a 1 a 1 a a and since a a = " we then have that l 1 =l ;l 1 ;l f 1 ; g a l1 l 1 a l l < a 1 a 1 a? = a+ Using an analogous reasoning with l = 1 and l = we nd l 1 =l ;l 1 ;l f ; g a l1 l 1 a l l < a + and combining this with the previous result and formula (10) we get a + = a 1 1 a a 1 1 a a a a a 19

22 because all other terms of the form a j1 j 1 a j j are less than a +. Since we haven't put any restrictions or conditions on the indices, we may assume without loss of generality that = a + = a 1 1 a. We already know that = 1 1. So we conclude that 1 = a 1 1 = a Because a 1 a 1 a + = a 1 1 a we have = a 1 a 1 a a a a 1 1 a a a? 1 a+ = 1 But since we know that = 1 the inequalities in the previous expression should be equalities. This leads to a 1 a 1 = a a 1 1 = 1 1 = Analogously we also nd that and thus = a a = Then we get 1 = = = Finally we have to demonstrate that we have considered all possible cases Property. The coecients of the characteristic equation of a positive n by n matrix with n > always fall into exactly one of the following cases Case A 1 or < 1 Case B > 1 and > 1 and and ( 1 = " or = " or = ) Case C > 1 and > 1 and and = = " and = " Proof According to property.1 we have that Case 1 1 or Case < 1 or Case or Case <. From property. we know that if we are neither in Case 1 nor in Case then we are in Case. So it is not necessary to consider Case. Case 1 and Case correspond to Case A. From now on we assume that Case 1 and Case are not true if we are in Case. Then we know from property.1 that = " so we have that either = or = ". If we have 1 = " or = " or =, then we are in Case B. Otherwise we know that 1 = ", = " and = ". Because of property.1 we have that = ". This means that = and thus we have Case C. So we have indeed considered all possible cases and since the three cases are mutually exclusive for the coecients of a positive matrix, we always have exactly one of the three cases. 0

23 Property. can be considered as an extra general necessary condition for an n-th order polynomial (n > ) to be the characteristic polynomial of a positive matrix... Sucient conditions Now we demonstrate that the conditions are also sucient. If we search a matrix such that its characteristic equation is we nd for Case A for Case C 1 0 1; 1; 1; " 0 " " " " 0 " 1 " " 0 " " " " 0 ; ; " " 0 " ; for Case B 1 0 1; 1; " " 0 " ; " " 0 " r 1 and Proof for Case A From the and the case we already know that 1; 1 1; or 1; < 1 1; = 1 1; = Using the same reasoning we nd for Case A 1; or 1; < 1 1; = This leads to a 1 = 1 1; = 1 a = 1 1; 1; = a = 1 1; 1; = a = 1 1; = Proof for Case B Because we have that ;. 1

24 Now we use the necessary and sucient conditions for Case B to prove that 1 ; if > and that 1 ; < if < If > and one of the necessary conditions is fullled we have that > 1. From property.1 we know that = ". This leads to > 1 = 1 ;. If < the necessary conditions result in > 1 and this leads to > 1 ;. So we always have that 1 ; =. Now we prove that under the conditions of Case B we have that 1 1; ; If 1 = " or = " then 1 1; ; = ". Otherwise we have = and =. So 1 1; ; = 1 1 = = We nd a 1 = 1 1; = 1 a = 1 1; 1; ; = a = 1 1; 1 ; 1; ; = 1 ; since 1; 1 = a = ; 1 1; ; = Proof for Case C We know that = " and from property. we also know that 1 = ". If 1 = then ; = = 1 and 1 ; = 1 =. Because we 1 know that ;. = " according to property.. We also have that ; = and that ; =. Finally 1 = leads to 1 ; = 1 =. We nd a 1 = 1 ; = 1 a = 1 ; ; = a = 1 ; ; = a = ; =

25 Example. Consider the monic polynomial 9 1 We have that = 1 > " = " = 1 = 1 > 11 = 9 = 1 = 1 > 1 = = Since the coecients don't belong to one of the three possible cases, the given polynomial cannot be the characteristic polynomial of a positive matrix. Example.8 Consider 9 r " We have that = 9 > " = " = 1 = 9 > 9 = = 1 = 9 1 = = = 9 = 9 = so we are in Case B. The necessary and sucient conditions are fullled 1 = " = = = 1 1 = 9 = = 1 1 = 9 = 1 1 = = The matrix A = " 9 0 " " 0 " " " 0 ". The general case has the given monic balance as its characteristic equation. Here we have not yet found sucient conditions, but we shall outline a heuristic algorithm that will (in most cases) result in a positive matrix for which the characteristic polynomial will be equal to a given polynomial. Extrapolating the results of the previous subsections and supported by many examples we state the following conjecture Conjecture.9 If n n i= i n?i r 1 n?1 n j= j n?j is the characteristic equation of a matrix A R nn max then it is also the characteristic equation of an upper Hessenberg matrix of the form K = k 0;1 k 0; k 0; k 0;n?1 k 0;n 0 k 1; k 1; k 1;n?1 k 1;n " 0 k ; k ;n?1 k ;n " " " 0 k n?1;n We shall use this conjecture in our heuristic algorithm to construct a matrix for which the characteristic polynomial will be equal to a given polynomial. However in [] we have presented a method to construct such a matrix that works even if Conjecture.9 would not be true. The major disadvantage of this method is its computational complexity. Therefore

26 we now present a heuristic algorithm that will on the average be much faster. If a result is returned, it is right. But it could be possible that sometimes no result is returned although there is a solution (in which case we have to fall back on the method of []). A heuristic algorithm First we check whether the coecients of the given polynomial satisfy the conditions of Property.1. Then we reconstruct the a? p 's by setting a? 1 = 1 and a? p = max( p? ; p ) for p = ; ; ; n with a small strictly positive real number. Consider K 1 = where i;j = a? 1 a? a? a? n 0 a? 1 a? a? n?1 " 0 a? 1 a? n? " " " a? 1 j a? i if a? i = " ; = " if a? i = " and K = " " " " " 1; 1; 1;n " " ; ;n " " " n?1;n We shall make a judicious choice out of the elements of K 1 and K to compose a matrix for which the characteristic equation will coincide with (8) We start with A = a? 1 a? a? a? n 0 " " " " 0 " " " " " ". Now we shall column by column transfer non- " elements of K to A (one element per column) such that the coecients of the characteristic equation of A are less than or equal to those of (8). If this doesn't lead to a valid result we shift a? 1 along its diagonal and repeat the procedure. We keep shifting a? 1 until it reaches the n-th column. If this still doesn't yield a result we put a? 1 back in the rst column and repeat the procedure but now with a?, and so on. Finally, if we have found A we remove redundant entries these are elements that can be removed without altering the characteristic equation. The results of this section will now be used to determine a minimal state space realization of a SISO discrete event system. inimal state space realization.1 Realization and minimal realization Suppose that we have a single input single output (SISO) discrete event system that can be described by an n-th order state space model x[k + 1] = A x[k] B u[k] (19) y[k] = C x[k] (0) with A R nn max ; B Rn1 max and C R1n max. u is the input, y is the output and x is the state vector. We dene the unit impulse e as e[k] = 0 if k = 0 ; = " otherwise

27 If we apply a unit impulse to the system and if we assume that the initial state x[0] satises x[0] = " or A x[0] B, we get the impulse response as the output of the system x[1] = B x[] = A B. x[k] = A k?1 B ) y[k] = C A k?1 B Let g k = C A k B. The g k 's are called the arkov parameters. Let us now reverse the process suppose that A; B and C are unknown, and that we only know the arkov parameters (e.g. from experiments { where we assume that the system is max-linear and time-invariant and that there is no noise present). How can we construct A; B and C from the g k 's? This process is called realization. If we make the dimension of A minimal, we have a minimal realization. Although there have been some attempts to solve this problem [,, 9], this problem has at present { to the authors' knowledge { not been solved entirely.. A lower bound for the minimal system order In this section we shall use the following property Property.1 Consider A S nn max ; B Sn1 max and C S 1n max. If A satises an equation of the form n p=0 a p A n?p r " (e.g. its characteristic equation) then the arkov parameters satisfy n p=0 a p g k+n?p r " for k = 0 ; 1; ; Proof We know that n p=0 multiplication by B we get nally get n p=0 a p A n?p r ". After left multiplication by C A k and right n p=0 a p g k+n?p r ". a p C A k+n?p B r " and since g k = C A k B we Suppose that we have a system that can be described by equations (19) and (0), with unknown system matrices. If we want to nd a minimal realization of this system the rst question that has to be answered is that of the minimal system order. Consider the semi-innite Hankel matrix H = g 0 g 1 g g 1 g g g g g......

28 As a direct consequence of theorem 1. and property.1 we have that the columns of H satisfy n p=0 a p H(; k + n? p) r " for k = 1; ; (1) where the coecients a p are the coecients of the characteristic equation of the system matrix A. Now we shall reverse this reasoning rst we construct a p by q Hankel matrix H p;q = g 0 g 1 g g q?1 g 1 g g g q g g g g q g p?1 g p g p+1 g p+q? with p and q large enough p; q n, where n is the real (but unknown) system order. Then we try to nd n and a 0 ; a 1 ; ; a n such that the columns of H p;q satisfy an equation of the form (1), which will lead to the characteristic equation of the unknown system matrix A. We propose the following procedure First we look for the largest square submatrix of H pq with consecutive column indices H sub;r = H p;q ([i 1 ; i ; ; i r ]; [j + 1; j + ; ; j + r])enspace; the determinant of which is not balanced det H sub;r r= ". If we add one row and the j + r + 1- st column we get an r + 1 by r + 1 matrix H sub;r+1 that has a balanced determinant. So according to theorem 1. the set of linear balances H sub;r+1 a r " has a signed solution a = h a r a r?1 a 0 i t. We now search a solution a that corresponds to the characteristic equation of a matrix with elements in R max (this should not necessarily be a signed solution { a signed solution would correspond to the a Olsder p 's). First of all we normalize a 0 to 0 and then we check if the necessary (and sucient) conditions of section are satised. If they are not satised we augment r and repeat the procedure. We continue until we get the following stable relation among the columns of H p;q H p;q (; k + r) a 1 H p;q (; k + r? 1) a r H p;q (; k) r " () for k f1; ; q? rg Since we assume that the system can be described by equations (19) and (0) and that p; q n, we can always nd such a stable relationship, by gradually augmenting r. The r that results from this procedure is a lower bound for the minimal system order.. Determination of the system matrices In [] we have described a method to nd all solutions of a set of multivariate polynomial (in)equalities in the max algebra. Now we can use this method to nd the A; B and C matrices of an r-th order SISO system with arkov parameters g 0 ; g 1 ; g ;. If the algorithm doesn't

29 nd any solutions, this means that the output behavior can't be described by an r-th order SISO system. In that case we have to augment our estimate of the system order and repeat the procedure. Since we assume that the system can be described by the state space model (19) { (0) we shall always get a minimal realization. However in many cases we can use the results of section to nd a minimal realization. Starting from the coecients a 1 ; a ; ; a r of equation () we search a matrix A with elements in R max such that its characteristic equation is n r p=1 a p r?p r " () Once we have found the A matrix, we have to nd a B and a C with elements in R max such that C A k B = g k for k = 0; 1; ; In practice it seems that we only have to take the transient behavior and the rst cycles of the steady-state behavior into account. So we may limit ourselves to the rst, say, N arkov parameters. Let's take a closer look at equations of the form C RB = s with C Rmax 1n ; R Rnn max R ; B n1 max and s R max. This equation can be rewritten as n n i=1 j=1 c i r ij b i = s So if we take the rst N arkov parameters into account, we get a set of N multivariate polynomial equations in the max algebra, with the elements of B and C as unknowns and R = A k?1 and s = g k?1 in the k-th equation. This problem can also be solved using the algorithm described in []. However one has to be careful since it is not always possible to nd a B and a C for every matrix that has equation () as its characteristic equation as will be shown in example.. In that case we have to search another A matrix or we could fall back on the method described in [, ], which nds all possible minimal realizations. The reason that it is not always possible to nd a B and C for A is that in S max all triples (A; B; C) that result in the same output behavior are connected by a kind of similarity transformation. We have to pick a triple ( ~ A; ~ B; ~ C) that is completely in Rmax. Examples We now illustrate the procedure of the preceding section with a few examples. Example.1 Here we reconsider the example of [, 9]. We start from a system with system matrices A = 1?1??1 0? 1 ; B = 0 " " h and C = 0 " " i

30 Now we are going to construct the system matrices from the impulse response of the system. This impulse response is given by fg k g = 0 ; 1 ; ; ; ; ; ; 8 ; 10 ; 1 ; 1 ; First we construct the Hankel matrix H 8;8 = " # 0 1 The determinant of H sub; = H 8;8 ([1; ]; [1; ]) = is not balanced. We add one row 8 and the third column and then we search a solution of the set of linear balances a a a 0 r " The solution a 0 = 0; a 1 = ; a = satises the necessary and sucient conditions for the by case since 1 = " and = = = 1 1. This solution also corresponds to a stable relation among the columns of H 8;8 H 8;8 (; k + ) H 8;8 (; k) = H 8;8 (; k + 1) for k f1; ; ; g ; or to the following characteristic equation r " This leads to a second order system with A = " " 0 1 whole set of solutions for B and C. One of the solutions is B = #. Using the technique of [] we get a " #? h i and C = " 0. 0 Apart from a permutation of the two state variables this result is the same as that of [9]. We now give another example that doesn't satisfy the assumptions of [9], where only impulse responses with a uniformly up-terrace behavior are considered. Example. We start from the system (A; B; C) with A = 1 0 " 0 " ; B = 0 1 h and C = 0 " " i 8

31 The impulse response of this system is 0 ; ; ; 9 ; 1 ; 1 ; 0 ; ; ;. Since there are two dierent alternating increments in steady state ( and ), we can't use the technique of [9]. First we construct the Hankel matrix H 8;8 = The determinant H sub; = H 8;8 ([1; ; ]; [1; ; ]) = The set of linear balances a a a 1 a 0 r " is not balanced. has a solution a 0 = 0; a 1 = ; a = ; a = 10 that satises the necessary and sucient 1 = " conditions of subsection. 8 >< > = " = = 1 1 = = = 1 This solution also corresponds to a stable relation among the columns of H 8;8 H 8;8 (; k + ) 10 H 8;8 (; k) = H 8;8 (; k + ) H 8;8 (; k + 1) for k f1; ; ; g, or to the following characteristic equation 10 r " This would lead to A 1 = " 0 " " 0 ". But it is impossible to nd positive vectors B 1 and C 1 such that (A 1 ; B 1 ; C 1 ) is a realization of the given impulse response. However the on the rst row andin the second column of A 1 is redundant, because if we remove it, we get A = " " 0 " which has the same characteristic equation as A 1, but " 0 " for A it is possible to nd a corresponding B and C B = 0? " h and C = 0 " i. 9

32 Conclusions and future research We have derived necessary and for some cases also sucient conditions for an S max polynomial to be the characteristic polynomial of an R max matrix. Then we have indicated how such a matrix can be constructed. The results were then applied to develop a procedure to nd a minimal state space realization of a SISO system, given its arkov parameters. This procedure is an alternative to the method of [], which nds all possible minimal realizations but which has one disadvantage its computational complexity. Since we allow an upper Hessenberg form for the matrix A, our method incorporates both the companion form of [] and the bidiagonal form of [9]. In the future we shall expand our theory and ll the gaps that are still left how many arkov parameters are necessary to nd the system matrices, how do we select a new matrix A if the matrix that resulted from the heuristic algorithm didn't lead to a realization of the given impulse response, etc.. We shall also try to improve the performance of our heuristic algorithm. Then we shall turn our attention to multiple input multiple output (IO) discrete event systems. References [1] F. Baccelli, G. Cohen, G.J. Olsder, and J.P. Quadrat, Synchronization and Linearity. New York John Wiley & Sons, 199. [] R.A. Cuninghame-Green, \Algebraic realization of discrete dynamic systems," in Proceedings of the 1991 IFAC Workshop on Discrete Event System Theory and Applications in anufacturing and Social Phenomena, Shenyang, China, pp. 11{1, June [] B. De Schutter and B. De oor, \A method to nd all solutions of a system of multivariate polynomial equalities and inequalities in the max algebra," Tech. rep. 9-1, ESAT/SISTA, K.U.Leuven, Leuven, Belgium, Dec Accepted for publication in Discrete Event Dynamic Systems Theory and Applications. [] B. De Schutter and B. De oor, \inimal realization in the max algebra is an extended linear complementarity problem," Systems & Control Letters, vol., no., pp. 10{111, ay 199. [] S. Gaubert, Theorie des Systemes Lineaires dans les Diodes. PhD thesis, Ecole Nationale Superieure des ines de Paris, France, July 199. [] P. oller, \Theoreme de Cayley-Hamilton dans les diodes et application a l'etude des systemes a evenements discrets," in Proceedings of the th International Conference on Analysis and Optimization of Systems, Antibes, France, vol. 8 of Lecture Notes in Control and Information Sciences, pp. 1{, Berlin, Germany Springer-Verlag, 198. [] G.J. Olsder and R.E. de Vries, \On an analogy of minimal realizations in conventional and discrete-event dynamic systems," in Algebres Exotiques et Systemes a Evenements Discrets, CNRS/CNET/INRIA Seminar, CNET, Issy-les-oulineaux, France, pp. 19{ 1, June 198. [8] G.J. Olsder and C. Roos, \Cramer and Cayley-Hamilton in the max algebra," Linear Algebra and Its Applications, vol. 101, pp. 8{108,

33 [9] L. Wang and X. Xu, \On minimal realization of SISO DEDS over max algebra," in Proceedings of the nd European Control Conference, Groningen, The Netherlands, pp. {0, June