Normal and Sinkless Petri Nets. Hsu-Chun Yen. National Taiwan University. Taipei, Taiwan ROC. November 24, Abstract

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1 Normal and Sinkless Petri Nets Rodney R. Howell Dept. of Computing and Information Sciences Kansas State University Manhattan, KS USA Hsu-Chun Yen Dept. of Electrical Engineering National Taiwan University Taipei, Taiwan ROC Louis E. Rosier y Dept. of Computer Sciences The University of Texas at Austin Austin, TX 7872 USA November 24, 992 Abstract We examine both the modeling power of normal and sinkless Petri nets and the computational complexities of various classical decision problems with respect to these two classes. We argue that although neither normal nor sinkless Petri nets are strictly more powerful than persistent Petri nets, they nonetheless are both capable of modeling a more interesting class of problems. On the other hand, we give strong evidence that normal and sinkless Petri nets are easier to analyze than persistent Petri nets. In so doing, we apply techniques originally developed for conict-free Petri nets a class dened solely in terms of the structure of the the net to sinkless Petri nets a class dened in terms of the behavior of the net. As a result, we give the rst comprehensive complexity analysis of a class of potentially unbounded Petri nets dened in terms of their behavior. Introduction Many aspects of the fundamental nature of computation are often studied via formal models, such as Turing machines, nite-state machines, and push-down automata (see, e.g., [HU79]). One formalism that has been used to model parallel computations is the Petri net (PN) [Pet8, Rei85]. As a means of gaining a better understanding of the PN model, the decidability and computational complexity of typical automata theoretic problems concerning PNs have been examined. These problems include boundedness, reachability, containment, and equivalence. Lipton [Lip76] and Racko [Rac78] have shown exponential space lower and upper bounds, respectively, for the boundedness problem. Also, Rabin [Bak73] and Hack [Hac76] have shown the containment and equivalence problems, respectively, to be undecidable. No tight bounds, however, have This work was supported in part by National Science Foundation Grant No. CCR A preliminary version was presented at the 7th International Conference on Fundamentals of Computation Theory, Szeged, Hungary, August 989. y Louis Rosier passed away on May 6, 99, while this paper was under review.

2 yet been established for the complexity of the reachability problem. The best lower bound known for this problem is exponential space [Lip76], but the only known algorithm is nonprimitive recursive [May84]. (See also [Kos82, Lam87].) Even the decidability of this problem was an open question for many years. Early eorts to show the reachability problem to be decidable included the study of various restricted subclasses of PNs [CLM76, CRM75, GY80, Gra80, HP79, LR78, May8, MM8, MM82, Mul8, VVN8]. The only classes for which completeness results have been given concerning all four of the problems mentioned above are -conservative PNs [JLL77], -conservative free choice PNs [JLL77], symmetric PNs [CLM76, MM82, Huy85], and conict-free PNs [JLL77, HRY87, HR88] (the proofs in [JLL77] also apply to -bounded PNs and elementary nets). Of these classes only symmetric and conict-free PNs are potentially unbounded, and both symmetric and conict-free PNs are dened only in terms of their structure, not in terms of their behavior. Thus, until now there has been no class of potentially unbounded PNs dened in terms of their behavior for which a comprehensive complexity analysis has been given. Of all of the PN classes for which completeness results have been shown concerning all four of the problems mentioned above, the class for which the decision procedures are most ecient is that of conict-free PNs. In particular, the boundedness problem is complete for polynomial time [HRY87], the reachability problem is NP-complete [JLL77, HR88], and the containment and equivalence problems are P 2 -complete [HR88], where P 2 is the set of all languages whose complements are in the second level of the polynomial-time hierarchy [Sto77]. However, since conict-free PNs comprise such a simple class, their modeling power is very limited. In particular, as we show in this paper, conict-free PNs cannot model the producer/consumer problem if more than one consumer is involved and the actions of each consumer are to be modeled by separate transitions. Furthermore, the obvious generalization of conict-free PNs to persistent PNs (see [LR78]) is not very helpful: not only is it impossible to model the above problem with persistent PNs, but the complexities of the various problems regarding persistent PNs appear to be worse than for conict-free PNs. In particular, all four problems are PSPACE-hard [JLL77]. The known upper bounds are much worse: the best upper bound known for boundedness is exponential space [Rac78], and no primitive recursive algorithms are known for the other three problems, though they are known to be decidable [Gra80, May8, Mul8]. Even the problem of recognizing a persistent PN, though known to be decidable [Gra80, May8, Mul8], is not known to be primitive recursive, and is PSPACE-hard [JLL77]. More recently, Yamasaki [Yam84] has dened two other generalizations of conict-free PNs, normal PNs and sinkless PNs. The relationship of normal PNs to sinkless PNs is analogous to the relationship of conict-free PNs to persistent PNs; i.e., normal PNs are those PNs that are sinkless for every initial marking [Yam84], just as conict-free PNs are those PNs that are persistent for every initial marking [LR78]. In addition, normal PNs, like conict-free PNs, are dened in terms of the structure of the net, whereas sinkless PNs, like persistent PNs, are dened in terms of the behavior of the net. However, both normal and sinkless PNs are incomparable to the class of persistent PNs; i.e., there are persistent PNs that are not sinkless, and normal PNs that are not persistent. In this paper, we examine both the modeling power of normal and sinkless PNs and the computational complexities of the four problems mentioned above with respect to these two classes. We show that both in terms of modeling power and ease of analysis, normal and sinkless PNs compare very favorably to conict-free and persistent PNs. 2

3 Concerning the modeling power, we rst show that the producer/consumer problem mentioned above, which cannot be modeled by persistent PNs, can be modeled by normal PNs. We then examine the mutual exclusion problem. We show that although this problem cannot be modeled by sinkless PNs, a version in which a bounded number of exclusions take place can be modeled by normal PNs (persistent PNs cannot even model one exclusion). We therefore conclude that although not all persistent PNs are sinkless, the class of problems that can be modeled by sinkless (or even normal) PNs is somewhat more interesting than that class modeled by persistent PNs. We then examine whether the more \useful" nature of sinkless PNs causes a corresponding increase in the complexities of the classical problems (as compared to persistent PNs). We show that this is not the case; i.e., we show that for both normal and sinkless PNs, the boundedness problem is co-np-complete, the reachability problem is NP-complete, and the containment and equivalence problems are P 2 -complete. Note that with the exception of the boundedness problem, the complexities of these problems are identical to those for conict-free PNs an extremely simple class. In fact, the techniques used in deriving these results for normal and sinkless PNs are simply more sophisticated applications of the techniques developed in [HR88] for conict-free PNs. Thus, techniques originally developed for analyzing problems involving conict-free PNs a class dened solely in terms of the structure of the net have been generalized to apply not only to normal PNs, but also to sinkless PNs a class dened in terms of the behavior of the net. Furthermore, these results represent the rst comprehensive complexity analysis of the classical problems concerning a class of potentially unbounded PNs dened in terms of their behavior. We also examine the question of how much easier it is to recognize a normal PN than to recognize a sinkless PN. Recall that the problem of recognizing a persistent PN is not known to be primitive recursive, and is at least PSPACE-hard [JLL77], whereas a conict-free PN can easily be recognized in polynomial time. The main problem in deciding persistence is that some sort of reachability analysis is necessary due to the fact that persistence is a behavioral property. Since sinkless PNs are likewise dened in terms of their behavior, whereas normal PNs are dened solely in terms of their structure, one might suppose that normal PNs would be easier to recognize than sinkless PNs. However, in order to determine whether a PN is normal, a rather complex property of the graphical representation of the PN must be tested. The end result is that both the problem of determining whether a PN is normal and the problem of determining whether a PN is sinkless are co-np-complete. We therefore conclude that in most applications, one might as well consider the entire class of sinkless PNs rather than the more restricted class of normal PNs. The remainder of the paper is organized as follows. In Section 2, we formally dene the concepts used throughout the paper. In Section 3, we compare the modeling power of normal and sinkless PNs with that of persistent PNs. Finally, in Section 4, we examine the complexities of the various problems regarding normal and sinkless PNs. 3

4 2 Denitions Let N denote the set of nonnegative integers, R the set of rational numbers, N k (R k ) the set of vectors of k nonnegative integers (rational numbers, respectively), and N km (R km ) the set of k m matrices of nonnegative integers (rational numbers, respectively). For a k-dimensional vector v, let v(i), i k, denote the ith component of v. For a k m matrix A, let A(i,j), i k, j m, denote the element in the ith row and the jth column of A, and let a j denote the jth column of A. For a given value of k, let 0 denote the vector of k zeros (i.e., 0(i) = 0 for i =,...,k). Given k-dimensional vectors u, v, and w, we say: v = w i v(i) = w(i) for i =,...,k; v w i v(i) w(i) for i =,...,k; v > w i v w and v 6= w; and u = v + w i u(i) = v(i) + w(i) for i =,...,k. A Petri Net (PN, for short) is a tuple (P,T,', 0 ), where P is a nite set of places, T is a nite set of transitions, ' is a ow function ' : (P T) [ (T P)! N, and 0 is the initial marking 0 : P! N (in this paper, we only consider PNs for which the range of ' is f0,g). A marking is a mapping : P! N. A transition t 2 T is enabled at a marking i for every p 2 P, '(p,t) (p). A transition t may re at a marking if t is enabled at. We then write! t 0, where 0 (p) = (p) { '(p,t) + '(t,p) for all p 2 P. A sequence of transitions = t...t n is a ring sequence from 0 (or a ring sequence of (P,T,'; 0 )) i t! t 0 2! t! n n for some sequence of markings,..., n. We also write 0! n. For ; 0 2 T*, 0 = t...t n, let. 0 be inductively dened as follows. Let 0 be. If t i 2 i?, let i be i? with the last occurrence of t i deleted; otherwise, let i = i?. Finally, let. 0 = n. We only consider Petri nets (P,T,'; 0 ) such that the range of ' is f0,g; hence, we may view them from two dierent perspectives. The rst perspective is graph-theoretical: P [ T forms the set of vertices of a directed graph, and (u,v) is an edge i '(u,v) =. Since ' is a function on (P T) [ (T P), the graph representation of the PN is bipartite. This graph theoretic perspective yields a natural pictorial representation for PNs. In such pictures, we adopt the convention of denoting a place by a circle and a transition by a bar; the marking is represented by (p) dots (or tokens) in the circle denoting each place p. The other perspective is algebraic. Suppose P contains k elements and T contains m elements. By establishing an ordering on the elements of P and T (i.e., P = fp,...,p k g and T = ft,...,t m g), we dene the k m addition matrix T of (P,T,'; 0 ) so that T(i,j) = '(t j,p i ) { '(p i,t j ). Thus, if we view a marking as a k-dimensional column vector in which the ith component is (p i ), each column t j of T is then a k-dimensional vector such that if! tj 0, then 0 = + t j. (Note that by this convention, the notations (p i ) and (i) are interchangeable.) For a given alphabet, let : *! (! N) be the Parikh mapping so that for 2 *, a 2 ; ()(a) is the number of occurrences of a in. For = T, we can view () as an m-dimensional column vector in which the jth component is ()(t j ). Then if 0!, 0 + T () = (note that the converse does not necessarily hold). 4

5 Let P = (P,T,'; 0 ) be a PN. The reachability set of P is the set R(P) = f j 0! for some g. Let c = u,u 2,...,u n,u be a circuit in the graph of P, and let be a marking of P. For convenience, we will always assume u is a place. Let pl(c) = fu,u 3,...,u n? g denote the set of places in c, and let tr(c) = u 2 u 4...u n denote the sequence of transitions in c. We dene (c) = P p i2pl(c) (i). We say c is token-free in i (c) = 0. c is said to be minimal i pl(c) does not properly include the set of places in any other circuit. (Note that the transitions in c are ignored in this denition.) c is said to have a sink i for some 2 R(P) and some and 0 such that! 0, (c) > 0, but 0 (c) = 0. c is said to be sinkless i it does not have a sink. P is said to be sinkless i each minimal circuit of P is sinkless. P is said to be normal i for every minimal circuit c and each transition t j, P p i2pl(c) T(i,j) 0; i.e., no transition can decrease the token count of a minimal circuit by ring at any marking. We say P is persistent if for every 2 R(P), when any pair of distinct transitions t and t 2 are both enabled at, t t 2 is a ring sequence from ; i.e., no enabled transition can ever be disabled by ring some other transition. P is said to be conict-free i for every place p for which there are two or more transitions t such that '(p,t) =, for each such transition, '(t,p) =. (This denition of conict-freedom was given in [LR78] for PNs whose ow function has a range of f0,g; see [CRM75, HRY87, HR88, HR89] for somewhat more general denitions.) As was shown in [Yam84], the relationship of normal PNs to sinkless PNs is analogous to the relationship of conict-free PNs to persistent PNs: normal PNs are those PNs that are sinkless for every initial marking [Yam84], while conict-free PNs are those PNs that are persistent for every initial marking [LR78]. Part of this paper is dedicated to comparing the modeling power of sinkless PNs with that of other classes of PNs. In order to formalize the notion of a PN modeling a particular problem, such as the producer/consumer problem or the mutual exclusion problem, we adopt a language-theoretic approach (see also [Hac75]). A modeling problem Q is given by an action language L(Q) over a nite alphabet of actions such that if 2 L(Q) and 0 is a prex of, then 0 2 L(Q). Thus, the action language gives all possible nite sequences of actions to be modeled. Let and 2 be two nite alphabets, and let h : *! 2 *. If for any strings ; 2 *, h() = h()h(), then we say that h is a homomorphism. A labeled Petri net is a tuple (P,T,'; 0 ;,h), where (P,T,'; 0 ) is a PN, is a nite alphabet, and h is a homomorphism h : T*! * such that for any t 2 T, the length of h(t) is at most ; we call h the labeling function. We dene the language of a labeled PN, L(P,T,'; 0 ;,h) = fh() j is a ring sequence of (P,T,'; 0 )g. In order to formally dene what it means for a PN to model a problem, we would like to identify the language of a labeled PN with the action language of the problem. However, we must also ensure that for any ring sequence that models a sequence of actions that can be extended, can also be extended in the same manner. Therefore, we say a PN P models a modeling problem Q with action alphabet i there is a labeled PN P 0 = (P;,h) such that L(P 0 ) = L(Q) and for any string 2 2 L(Q), if is a ring sequence of P such that h( ) =, then there is a ring sequence 2 of P such that h( 2 ) = 2. Aside from examining the modeling power of various classes of PNs, we also examine the computational complexities of a number of problems concerning normal and sinkless PNs. Given a marking of a given PN P, the reachability problem (RP) is to determine whether 2 R(P). The boundedness problem (BP) is to determine whether R(P) is nite. The sink-detection problem is the problem of determining whether there is a minimal circuit of P with a sink. Given two PNs P and P 0, the containment and equivalence problems (CP 5

6 and EP, respectively) are to determine whether R(P) R(P 0 ) and whether R(P) = R(P 0 ), respectively. In examining the latter two problems, we use concepts from linear algebra and the theory of semilinear sets. For any vector v 0 2 N k and any nite set V = fv,...,v m g N k, the set L(v 0,V) = fx j 9c,...,c m 2 N such that x = v 0 + P m i= c iv i g is called the linear set with base v 0 over the set of periods V. A nite union of linear sets is called a semilinear set (SLS for short). If x = P m i= a iv i for some a,...,a m 2 R, then x is a linear combination of the vectors in V. If a i 0 for all i, then x is a nonnegative linear combination of the vectors in V. If in addition for some i, a i > 0, then x is a positive linear combination of the vectors in V. 3 Modeling Power In this section, we examine the modeling power of normal and sinkless PNs, comparing it with that of conictfree and persistent PNs. It has already been shown that the class of conict-free PNs is properly contained in both the classes of persistent PNs [LR78] and sinkless PNs [Yam84]. Furthermore, it is clear from the denitions that the class of sinkless PNs properly includes the class of normal PNs (see also [Yam84]). On the other hand, it is not hard to see that the class of persistent PNs is incomparable to both the class of normal PNs and the class of sinkless PNs; i.e., there is a persistent PN that is not sinkless (Figure 3-), and there is a normal PN that is not persistent (Figure 3-2; see also [Yam84]). These relationships are summarized by the Venn diagram in Figure 3-3. Figure 3-: A persistent PN that is not sinkless. Figure 3-2: A normal PN that is not persistent. One of the shortcomings of conict-free and persistent PNs is that their modeling power is severely limited. By the denition of persistence, only a very limited type of nondeterminism is allowed: if more than 6

7 PNs Sinkless Normal Conflict- Free Persistent Figure 3-3: Relationships between Petri net classes. one transition is enabled, the next transition to re may be nondeterministically chosen, but the ring of this transition cannot disable any others. Although there are modeling problems, such as a simple version the producer/consumer problem, that can be modeled by conict-free PNs, the severely restricted version of nondeterminism prohibits even persistent PNs from modeling many \interesting" problems. In particular, we show that neither a more general producer/consumer problem nor the mutual exclusion problem can be modeled by persistent PNs. Given these limitations of persistent PNs and the fact that there are normal PNs that are not persistent, it is natural to ask whether normal PNs can model some of the classical modeling problems that persistent PNs cannot. In this section, we show that although the mutual exclusion problem cannot be modeled even by sinkless PNs, the generalized producer/consumer problem and a version of the mutual exclusion problem in which the total number of exclusions in any computation is bounded by a xed constant both can be modeled by normal PNs (and, hence, by sinkless PNs). On the other hand, we show that persistent PNs cannot even model one exclusion. In the next section, we will present evidence that sinkless PNs may be signicantly easier to analyze than persistent PNs, but not signicantly more dicult to analyze than normal PNs. These results suggest that sinkless PNs may be a more useful class of PNs than many of the classes that have been studied in the past. We rst introduce a simple version of the well-known producer/consumer problem PC ;. Informally, the problem involves two processes, the producer and the consumer, and an unbounded buer. The producer iterates a loop consisting of a sequence of two actions, produce (denoted p) followed by send (denoted s). Thus, in any computation, the number of produces is never less than the number sends and never exceeds the number of sends by more than one. Meanwhile, the consumer iterates a loop consisting of a sequence of two actions, receive (denoted r) followed by consume (denoted c). Finally, the number of receives can never exceed the number of sends. Thus, the action alphabet of PC ; is ; = fp,s,r,cg. Recalling that ()(a) denotes the number of occurrences of a in, we then dene L(PC ; ) as the set of all strings in 7

8 ; * such that for any prex, ()(s) ()(p) ()(s) + ; ()(c) ()(r) ()(c) + ; and ()(r) ()(s). The PN shown in Figure 3-4 is easily seen to model PC ; and to be conict-free. <s> <r> p p p p p <p> <c> Figure 3-4: A conict-free PN to model PC ;. Suppose we wish to generalize the above problem to m producers and n consumers. In order to be able to dierentiate between actions of individual producers and individual consumers, we must dene the action alphabet as m;n = fp i,s i,r j,c j j i m, j ng. We then dene L(PC m;n ) as the set of all strings in m;n * such that for any prex, ()(s i ) ()(p i ) ()(s i ) +, for i m; ()(c j ) ()(r j ) ()(c j ) +, for j n; and P n j= ()(r j) P m i= ()(s i). We will now show that there is no persistent PN that models PC ;2. We rst reproduce the following easily shown lemma from [LR78]. Lemma 3. (from [LR78]): If and 0 are ring sequences of some persistent PN, then ( 0. ) is also a ring sequence of that PN. Theorem 3.: There is no persistent PN that models PC ;2. Proof. Suppose some persistent PN P models PC ;2 with labeling function h. Since p s r and p s r 2 are both in L(PC ;2 ), there must be ring sequences and 2 of P such that h() = p s, h( ) = 8

9 r, and h( 2 ) = r 2. Since P is persistent, from Lemma 3., ( 2. ) must also be a ring sequence of P. Since there is no transition labeled r 2 in, h( ( 2. )) = p s r r 2, which is not in L(PC ;2 ) a contradiction. 2 The above theorem clearly demonstrates the limitations of persistent PNs as modeling tools. We will now show that these limitations are, at least in part, eliminated by normal PNs. Consider the PN in Figure 3-5. It is not hard to see that this PN models PC m;n. Furthermore, it is obvious by inspection of the graph that no transition can decrease the token count of any minimal circuit, so the PN must be normal. We therefore have the following theorem. Theorem 3.2: For each positive m and n, there is a normal PN to model PC m;n. <s > <r > <p > <c >.... <s > m <r > n <p > m <c > n Figure 3-5: A normal PN to model PC m;n. We have therefore seen an example of a well-known modeling problem that can be modeled by a normal PN, but not by a persistent PN. One of the fundamental building blocks for many modeling problems (e.g., readers/writers, dining philosophers) is the mutual exclusion problem. We will now examine the question of whether normal or sinkless PNs can model mutual exclusion. Informally, the n-process mutual exclusion 9

10 problem ME n consists of n processes, each of which has a critical region. Each process P i iterates a loop consisting of a sequence of two actions: entering its critical region (denoted en i ) followed by exiting its critical region (denoted ex i ). Furthermore, no two processes may be in their respective critical regions simultaneously. Thus, the action alphabet of ME n is n = fen i, ex i j i ng. We then dene L(ME n ) to be the set of all strings in n * such that for any prex, ()(ex i ) ()(en i ) ()(ex i ) +, for i n; and P n i= ( ()(en i)? ()(ex i )). Figure 3-6 shows a PN that models ME n for any n 2. It can be shown that this PN is neither persistent nor sinkless. In fact, the following theorem can be shown in a manner similar to Theorem 3.. Theorem 3.3: There is no persistent PN that models ME 2. <en > <ex > <en n >... <ex n > Figure 3-6: A PN to model ME n. We now show that there is also no sinkless PN to model ME 2. We rst reproduce the following lemma from [Yam84]. Lemma 3.2 (from [Yam84]): Let P = (P,T,'; 0 ) be a PN with k places and m transitions. For each w 2 N m, there is some ring sequence of P with () = w if. 0 + Tw 0, and 2. for each ring sequence 0 of P and each circuit c, if ( 0 tr(c)) w, then (c) > 0, where 0 0!. Theorem 3.4: There is no sinkless PN that models ME 2. Proof. Let P = (P,T,'; 0 ) be any PN that models ME 2 with labeling function h. Since (en ex )* L(ME 2 ), for each positive k there must exist a k such that h( k ) = en ex and ::: k is a ring sequence of P. Consider the innite sequence of markings ; 2 ; :::, where k! k? k. As was shown in [KM69], it follows from Konig's Innity Lemma [Kon36] that there exist positive i and j such that i < j and i j. Let = ::: i ; 2 = i+ ::: j ; and w = ( 2 ). Then Tw 0. Since (en ex ) j en 2 2 L(ME 2 ), there must 0

11 be a 3 such that h( 3 ) = en 2 and 2 3 0! 0 for some 0. Since h( 2 3 ) 2 (en ex )*en 2 and h( 2 ) 2 (en ex ) +, there is no rable at 0 such that () = w. Therefore, either condition () or condition (2) of Lemma 3.2 must fail for (P,T,'; 0 ) and w. Since Tw 0, condition () must hold. Thus, there must be a 0 and a circuit c such that ( 0 tr(c)) w and (c) = 0, where 0! 0. Consider the ring sequence = of P. Clearly, 0!. Since (c) = 0, some subset of pl(c) is the set of places on a minimal circuit with a token count of 0 in. However, since (tr(c)) w = ( 2 ), and since 2 has already red in, each of the places in pl(c) must have had a positive token count at one time in. Thus, decreases the token count of some minimal circuit to 0, and P is not sinkless. 2 We conclude this section by noting that by using a proof similar to that of Theorem 3., it can be shown that there is no persistent PN to model the problem whose action language is all prexes of en ex en 2 ex 2. In contrast, there is a normal PN to model this problem. In fact, Figure 3-7 shows a normal PN to model the problem ME k n, for any positive k and n, dened so that L(ME k n) = f j 2 L(ME n ) and P n i= ()(en i) kg. Thus, normal PNs can be used to model mutual exclusion if the total number of exclusions is bounded by a xed constant. We therefore have the following theorems. Theorem 3.5: There is no persistent PN to model ME 2. Theorem 3.6: For each positive k and n, there is a normal PN to model ME k n. The results of this section, summarized in Table 3-, serve to demonstrate that even though the modeling power of sinkless PNs is still quite limited, even normal PNs provide a more usable version of nondeterminism than persistent PNs. and Problem Modeled by persistent? Modeled by normal/sinkless? PC ;2 no yes PC m;n no yes ME 2 no no ME 2 no yes ME k n no yes Table 3-4 Complexity Results In this section, we examine the complexities of various problems involving normal and sinkless PNs. In [HR88], we developed several techniques for analyzing conict-free PNs. These techniques relied heavily upon the structural property of conict-freedom. In this section, we show that these same techniques also apply, albeit in a more sophisticated manner, to sinkless PNs, a class dened not in terms of structure, but in terms of behavior. We begin by developing an important lemma (Lemma 4.3) which will be used in deriving most of the upper bounds in this section. In [HR88], we showed that for any conict-free PN P and any marking of P,

12 <en >... <en n >... <ex >... <ex n > <en >... <en n >... <ex >... <ex > n Figure 3-7: A normal PN to model ME k n. there is an instance of integer linear programming that has a solution i is reachable in P. Furthermore, we showed that this instance of integer linear programming can be guessed in polynomial time. It therefore followed that the reachability problem for conict-free PNs is in NP. We will show in Lemma 4.3 that a similar fact holds for sinkless PNs. In so doing, we make use of the following lemma which follows immediately from Lemma 3.2 and was rst stated in [Yam84]. Lemma 4. (from [Yam84]): If a PN P = (P,T,'; 0 ) has no token-free circuits in every reachable marking, then R(P) = f j = 0 + Tx 0 for some x 2 N m g, where m is the number of transitions in T. The above lemma gives an instance of integer linear programming whose solution set gives the reachability set of a PN. The only requirement is that no reachable marking can have a token-free circuit. In order to enable us to work with PNs whose reachability sets contain no markings with token-free circuits, we give the following lemma. 2

13 Lemma 4.2: Let P = (P,T,'; 0 ) be a sinkless PN, and let P 0 = (P,T 0 ; ' 0 ; ) be such that 0! in P for some, T 0 T such that each t 2 T 0 is enabled at some point in the ring of from 0, and ' 0 is the restriction of ' to (P T 0 ) [ (T 0 P). Then P 0 has no token-free circuits in every reachable marking. Proof. Suppose P 0 has some reachable marking 0 with a token-free circuit c 0. Since 0 is reachable in P 0, it must also be reachable from in P. Let 0 be a ring sequence of P such that 0!! 0 0. Also, since c 0 is a circuit in P 0, it must also be a circuit in P. Thus, there must be a minimal circuit c of P such that pl(c) pl(c 0 ). Clearly, 0 (c) = 0. Let p be any place in c, and let t be the transition following p in c 0. Since t 2 T 0, t must have been enabled at some marking 00 reached in the ring of from 0 in P. Thus, 00 (p) > 0, so 00 (c) > 0. Since 0 (c) = 0 and 0 is reachable from 00, c has a sink a contradiction. 2 Given the above two lemmas, we can now outline our strategy for showing the RP to be in NP, and the BP to be in co-np. This strategy will then be the basis for most of the other upper bounds shown in this paper. Let P = (P,T,'; 0 ) be a sinkless PN, and consider the sequence P,...,P n, where each P h = (P,T h ; ' h ; h? ) such that T 0 = ;, and for h n, T h = T h? [ ft jh g for some t jh 62 T h? enabled at h?, for h n; ' h is the restriction of ' to (P T h ) [ (T h P), for 0 h n; and h? h?! h for some h? 2 T h *, h n. From Lemma 4.2, no P h contains a token-free circuit for any marking in R(P h ). Thus, from Lemma 4., there is an instance of integer linear programming S h whose solution set describes the reachability set of P h. Furthermore, S h can clearly be constructed from P h in polynomial time. Also note that the initial marking of P h is given by a solution to S h?. Thus, a portion of R(P) can be given by the solution set of a system S of linear inequalities over the integers, where S is constructed in polynomial time from P and a sequence of distinct transitions of P. To formalize these ideas, let P = fp,...,p k g, T = ft,...,t m g, and let = t j...t jn be any sequence of distinct transitions from T. We dene the characteristic system of inequalities for P and as S(P; ) = S 0 [ :::[ S n, where S 0 = fx 0 = 0 g, S h = fx h? (i) '(p i,t jh ), x h = x h? + A h y h j i kg, and A h is the k h matrix whose columns are t j,...,t jh for h n. The variables in S are the components of the k-dimensional column vectors x 0,...,x n and the h-dimensional column vectors y h, h n. We will now show that for any marking, 2 R(P) i there is a sequence of distinct transitions = t j...t jn such that S(P; ) has a nonnegative integer solution in which x n =. It is precisely this fact that allows us to apply the techniques of [HR88], developed specically for the structurally-dened conict-free PNs, to the behaviorally-dened sinkless PNs. Lemma 4.3: Let P = (P,T,'; 0 ) be a sinkless PN, and let be any marking of P. Then there is some 2 T* such that 0! i there is some sequence = t j...t jn of distinct transitions in T such that S(P; ) has a nonnegative integer solution in which x n =. Furthermore, and can be chosen such that = ::: n, where 0 = x 0! 2 x! n! xn =, t jh is enabled at x h?, h 2 ft j,...,t jh g*, and y h (h 0 ) gives the number of times t jh 0 occurs in h, for h 0 h n. Proof. ): Let 2 R(P). Then for some 2 T*, 0!. Let = ::: n such that for h n, h begins with some transition t jh that does not occur in ::: h?, and h contains only transitions from the 3

14 set ft j,...,t jh g. Let = t j...t jn. By letting y h (h 0 ) be the number of times t jh 0 occurs in h, h 0 h n, it is easily seen that S(P; ) has a nonnegative integer solution for which 0 = x 0! 2 x! n! xn =. (: We will show by induction on n that for any sequence n = t j...t jn of n distinct transitions from T and any marking, if S(P; n ) has a nonnegative integer solution for which x n =, then 0 = x 0! x 2! n! xn = such that t jh is enabled at x h? ; h 2 ft j,...,t jh g*, and t jh 0 occurs y h (h 0 ) times in h for h 0 h n. The lemma will then follow. Base: Let n = 0. Then S(P; 0 ) = fx 0 = 0 g. Clearly, the only solution to S is 0, and 0! 0. Induction Hypothesis: Let n be some positive integer, and assume that for any sequence of distinct transitions n? = t j...t jn? from T and any marking, if S(P; n? ) has a nonnegative integer solution in which x n? =, then 0 = x 0! 2 n? x!! xn? = such that t jh is enabled at x h?, h 2 ft j,...,t jh g*, and t jh 0 occurs y h (h 0 ) times in h for h 0 h n {. Induction Step: Let n = t j...t jn be a sequence of distinct transitions in T, be any marking of P, n? = t j...t jn?, and T 0 = ft j,...,t jn g. Then S(P; n ) = S(P; n? ) [ fx n? (i) '(p i,t jn ), x n = x n? + T 0 y n j i kg. Suppose S(P; n ) has a nonnegative integer solution in which the value of x n is. Since any solution to S(P; n ) is clearly a solution to S(P; n? ), from the induction hypothesis, 0 = x 0! x 2 n?!! xn? such that t jh is enabled at x h?, h 2 ft j,...,t jh g*, and t jh 0 occurs y h (h 0 ) times in h for h 0 h n {. Let 0 = ::: n?. Since x n? (i) '(p i,t jn ) for i k, t jn is enabled at x n?. Thus, each t 2 T 0 is enabled at some point in the ring of 0 from 0, so from Lemma 4.2, (P,T 0 ; ' 0,x n? ) has no token-free circuits in every reachable marking, where ' 0 is the restriction of ' to (P T 0 ) [ (T 0 P). Since = x n = x n? + T 0 y n for some y n 2 N n, from Lemma 3.2, x n? n! xn = for some n 2 T 0 * containing y n (h 0 ) occurrences of t jh 0 for h 0 n. Thus, 0 = x 0! 2 x! n! xn = such that t jh is enabled at x h?, h 2 ft j,...,t jh g*, and t jh 0 occurs y h (h 0 ) times in h for h 0 h n. 2 The following corollory follows from Lemmas 4. and 4.2 and the proof of Lemma 4.3. Corollary 4.: Let P = (P,T,'; 0 ) be a sinkless PN, be a sequence of n distinct transitions from T, T 0 be the set of transitions in, ' 0 be the restriction of ' to (P T 0 ) [ (T 0 P), and be any marking of P such that for some nonnegative integer solution of S(P; ), x n =. Then R(P,T 0 ; ' 0 ; ) = f 0 j 0 = + T 0 x for some x 2 N n g. Lemma 4.3 can now be coupled with the fact that integer linear programming is in NP [BT76] to show that the RP for sinkless PNs is in NP. Since the RP is NP-hard for conict-free PNs [JLL77], it will then follow that the RP for sinkless (normal) PNs is NP-complete. We therefore have the following theorem. Theorem 4.: The RP for sinkless (normal) PNs is NP-complete. Proof. From [JLL77], the RP for conict-free PNs is NP-hard. Since any conict-free PN is normal [Yam84], it follows that the RP for normal PNs is NP-hard. Since any normal PN is sinkless, we need only show that the RP for sinkless PNs is in NP. We use the following nondeterministic algorithm to decide the RP for any given PN P = (P,T,'; 0 ) and any marking of P. First, guess a sequence of n distinct transitions from T. Then construct S(P; ) in polynomial time. Next, construct S = S(P; ) [ fx n = g. Since integer 4

15 linear programming is in NP [BT76], we can guess a solution to S and verify it in polynomial time. Clearly, S has a nonnegative integer solution i S(P; ) has a nonnegative integer solution in which x n =. From Lemma 4.3, there is a such that S(P; ) has a nonnegative integer solution in which x n = i 2 R(P). Therefore, the RP for sinkless (normal) PNs is NP-complete. 2 In [HRY87], we showed the BP to be PTIME-complete for conict-free PNs. However, we will now show the problem to be co-np-complete for both normal and sinkless PNs. Theorem 4.2: The BP for sinkless (normal) PNs is co-np-complete. Proof. We rst show the BP for sinkless PNs to be in co-np. We use the following nondeterministic algorithm to decide, for any given PN P = (P,T,'; 0 ), whether P is unbounded. First, guess a sequence of n distinct transitions from T. Then construct S(P; ) in polynomial time. Next, construct S = S(P; ) [ f T 0 z > 0g, where T 0 is the set of rules in. As in Theorem 4., we can guess a solution to S and verify it in polynomial time. Suppose S has a nonnegative integer solution, and let be the value of x n and value of z in that solution. From Lemma 4.3, 2 R(P), and from Corollary 4., + T 0 be the is reachable from. Thus, since T 0 > 0, P must be unbounded. On the other hand, suppose P is unbounded. From [KM69], there is a ring sequence 0 of P such that T ( 0 ) > 0. Let 0 0!. From Lemma 4.3, there is a choice of for which there is a nonnegative integer solution of S(P; ) in which x n has the value, and for which all transitions in 0 appear in. Thus, we can clearly set z to a value that gives us a nonnegative integer solution to S. Therefore, there is a choice of for which S has a nonnegative solution i P is unbounded. It follows that the BP is in co-np. We now show that the BP for normal PNs is co-np-hard. We show this by reducing 3SAT to the complement of BP. Let C = fc,...,c m g be the set of clauses and V = fx,...,x n g be the set of variables in an arbitrary instance of 3SAT, where C i = f i ; i2 ; i3 g and ij 2 fx,x j x 2 Vg for i m, j 3. We construct a normal PN P = (P,T,'; 0 ) (shown in Figure 4-) such that P is unbounded i C is satisable (i.e., if there is an assignment of truth values to V such that V m i= W 3 j= ij is true). For each variable x j, let a j be a place for which 0 (a j ) =. We then dene transitions t j and f j for j n. t j and f j both have a j as an input place; thus, since a j will not be the output place of any transition, t j and f j cannot both re in any ring sequence. For each clause C i, let c i be a place such that 0 (c i ) = 0. For i m and j n, t j will have c i as an output place i x j 2 C i, and f j will have c i as an output place i x j 2 C i. Since no c i will be an output place for any other transitions, all of the c i s can become simultaneously positive i C is satisable. Finally, we dene the transitions s and s 2 and the places p and p 2 as shown in Figure 4-. The resulting PN is clearly normal, and is unbounded i C is satiable. Therefore, the BP is co-np-complete for sinkless (normal) PNs. 2 We can also use Lemma 4.3 to show the sink detection problem for PNs to be NP-complete. Theorem 4.3: The sink detection problem for PNs is NP-complete. Proof. Our nondeterministic algorithm for this problem is similar to those for the RP and the BP. Given P = (P,T,'; 0 ), we guess = t j...t jn and a minimal circuit c. We can clearly verify in polynomial time that c is a circuit. Furthermore, we can verify that c is minimal by verifying for each pair of places p and p 0 5

16 a a n... t f t n f n c c m s s 2 p 2 p Figure 4-: The lower bound for BP. in pl(c) and each transition t in T that '(p,t) = '(t,p 0 ) = only if p,t,p 0 is a segment of c; this verication can clearly be done in polynomial time. We then guess an integer m, m n, and a marking such that (c) = and (p) = 0 for all p 62 pl(c). We then construct S = S(P; ) [ fx n (c) = 0, y y m, x m? + A m y g, where A m is the matrix whose columns are t j,...,t jm. We claim that there is some choice of, c, m, and for which S has a nonnegative integer solution i P has a sink. First, suppose that for some choice of, c, m, and, S has a nonnegative integer solution. In order derive a contradiction, assume P is sinkless. From Lemma 4.3, there is a nonnegative integer solution of S such that 0 = x 0! 2 x! n! xn and each t jh is enabled at x h, for some ::: n 2 T*. Consider P 0 = (P,T 0 ; ' 0,x m? ), where T 0 = ft j,...,t jm g and ' 0 is the restriction of ' to (P T 0 ) [ (T 0 P). Since each transition in T 0 is enabled at some point in the ring of ::: m? from 0, from Lemma 4.2, P 0 has no token-free circuits in every reachable marking. Thus, from Lemma 4., there is some 2 T 0 * such that 6

17 x m?! xm? + T 0 y = 0 for some 0. Since T 0 = A m, 0, and hence, 0 (c) > 0. Since y y m, Lemma 4.2 also guarantees that there is some 0 2 T 0 * such that 0! T 0 (y m { y) = x m. Thus, ::: m? 0! 0 0 m+ ::: n! x n, where x n (c) = 0. Thus, P has a sink a contradiction. Therefore, if there is some choice of, c, m, and such that S has a nonnegative solution, P must have a sink. Now suppose P has a sink. Then there exist 0 and 00 and a minimal circuit c such that 0 (c) > 0, 00 (c) = 0, and 0! 0! 0 00 for some 0 2 T*. From Lemma 4.3, there is a of m transitions, m 0, such that S(P; ) has a nonnegative integer solution in which x m = 0. We may assume without loss of generality that m, since at least one transition t must be enabled at 0 (because 0 6= 00 ). Since 0 (c) > 0, x m (c) for some such that (c) = and (p) = 0 for all p 62 pl(c). By proceeding as in Lemma 4.3, it is not hard to see that there is a 2 of n { m transitions, n, such that S(P; 2 ) has a nonnegative solution in which x n = 00 and x m? + A m y = 0, where A m is the addition matrix formed from the transitions in and y y m. Thus, S = S(P; 2 ) [ fx n (c) = 0, y y m, x m? + A m y g has a nonnegative integer solution. Therefore, there is some choice of, c, m, and for which S has a nonnegative integer solution i P has a sink. We will now show the sink detection problem to be NP-hard. We again use a reduction from 3SAT. Let C = fc,...,c m g be the set of clauses and V = fx,...,x n g be the set of variables in an arbitrary instance of 3SAT. Consider again the PN P constructed in the proof of Theorem 4.2 (Figure 4-). We construct P 0 from P by adding a transition s 3 with input place p and no output places. Thus P 0 has a sink i p can become positive i C is satisable. Therefore, the sink detection problem is NP-hard. 2 Theorem 4.4: The problem of determining whether a PN is normal is co-np-complete. Proof. In order to show that a given PN P = (P,T,'; 0 ) is not normal, we need to nd a minimal circuit c and a transition t j such that P p i2pl(c) T(i,j) < 0. The proof of Theorem 4.3 shows how we can guess a minimal circuit c and verify that it is minimal in polynomial time. Clearly, we can also guess a transition t j and verify that P p i2pl(c) T(i,j) < 0 in polynomial time. Thus, the problem is in co-np. We will now show the problem to be co-np-hard. We do this via a reduction from 3UNSAT (the complement of 3SAT). Let C = fc,...,c m g be the set of clauses and V = fx,...,x n g be the set of variables in an arbitrary instance of 3UNSAT, where C i = f i ; i2 ; i3 g and ij 2 fx,x j x 2 Vg for i m, j 3. We construct a PN P = (P,T,'; 0 ) (shown in Figure 4-2) that is normal i C is not satisable. In considering whether a PN is normal or not, it is convenient to view the PN as simply a directed bipartite graph; places and transitions are dierentiated only in connection with their respective roles regarding minimal circuits. The initial marking is therefore irrelevant, and will be chosen to be the zero marking. The graph will contain a special place z such that P will be normal i z is not on a minimal circuit. Our objective will then be to show that z is on a minimal circuit i C is satisable. The graph will consist of three components Q, R, and S, each of which will be acyclic. Furthermore, the three components will be interconnected in such a manner that any cycle in the graph must contain vertices from each of the three components. We will rst describe each of the three components, then describe how they are interconnected. 7

18 Q s S R a m a 2 m a 3 m p t p f cm xt x f qt q f. y.. c 2 y n- a a2 a3 pn t pn f c xt n xn f s2 z qt n yn qn f Figure 4-2: The lower bound for Theorem

19 Q is the simplest of the three components, consisting of only a single transition s. Component R is dened in terms of the set V. For each x j 2 V, R contains the places p t j, pf j, qt j, and qf j, and the transitions x t j, xf j, and y j. These places and transitions are interconnected as shown in Figure 4-2 so that R is acyclic, and any path within R contains at most one of the two transitions x t j and xf j for each j. Component S is dened in terms of the set C. For each C i 2 C, S contains the places a i, a2 i, and a3 i, plus the transition c i. In addition, S contains the place z and the transition s 2. The places and transitions are interconnected as shown in Figure 4-2 so that S is acyclic, and any path within S contains at most one of the places a i, a2 i, and a 3 i for each i. We now add edges to the graph to interconnect the three components. We rst add the edges (s,p t ), (s,p f ), (y n,z), (a m,s ), (a 2 m,s ), and (a 3 m,s ). These edges have the eect of creating a number of circuits, each of which contains s, z, exactly one of x t j or xf j for each variable x j, and exactly one of a i, a2 i, or a3 i for each clause C i. Next, we add additional edges from R to S, each of which \short-circuits" some of the previously created circuits. For each literal ik in each clause in C, we add the edge (x f j,ak i ) if ik = x j, or the edge (x t j,ak i ) if ik = x j. Thus, none of these newly created circuits contains z, but all contain s. We will now show that the PN P constructed above is normal i z is not on a minimal circuit. If P is normal, z clearly must not be on any minimal circuit c, since transition s 2 would then have an input place in pl(c) but no output place in pl(c). On the other hand, suppose z is not on a minimal circuit. In order to derive a contradiction, assume there is some minimal circuit c and some transition t such that t has an input place in pl(c) but no output place in pl(c). Notice that z is the only place in P that is an input place to more than one transition. Since z 62 pl(c), this implies that t is in the circuit c. Therefore, t must have an output place in pl(c) a contradiction. Thus, P is normal i z is not on a minimal circuit. We conclude the proof by showing that z is on a minimal circuit i C is satisable; it will then follow that P is normal i C is not satisable. First suppose z is on a minimal circuit c. Since any circuit containing z must contain exactly one of x t j or x f j for each variable x j, consider the truth assignment such that x j is true i x t j is in c. Let c i be an arbitrary clause in C. Since any circuit containing z contains exactly one of a i, a2 i, or a3 i, let ak i be on c. Suppose ik = x j (the case in which ik = x j is symmetric). Then there is an edge from x f j to ak i. Since c is a minimal circuit, xt j must be on c, so ik has been assigned a value of true. Thus, each clause evaluates to true, so C is satisable. Conversely, suppose C is satisable. Then for any satisfying assignment, there must be some circuit containing x t j if x j is true in that assignment, x f j if x j is false in that assignment, some place a k i such that ik is true with that assignment for each clause C i, and the place z. In order to derive a contradiction, assume c is not minimal. Then there must be some edge from a transition in R, say (without loss of generality) x t j, to some place in S, say ak i, such that pt j (and hence x t j ) and ak i are both on c. Since xt j and ak i are both on c, x j and i must both evaluate to true. However, since there is an edge from x t j to ak i, i = x j a contradiction. Therefore, z is on a minimal circuit i C is satisable, and hence, P is normal i C is not satisable. 2 We conclude this section by showing the CP and EP for normal and sinkless PNs to be P 2 -complete, where P 2 is the class of languages whose complements are in the second level of the polynomial-time hierarchy (see, e.g., [Sto77]). The strategy we use is again similar to that developed in [HR88] (see also 9

20 [HRHY86, Huy86]). Recall that sinkless PNs have eectively computable semilinear sets [Yam84]. We use Lemma 4.3 to give an upper bound on the size of the SLS representation of the reachability set of a given sinkless PN. Although the CP and EP for SLSs are known to be P 2 -complete [Huy82], the bound on the size of the SLS representation of the reachability set must be at least exponential in the sizes of the PNs even for conict-free PNs (see [HR88]). However, as is the case with conict-free PNs [HR88], we can show that for sinkless PNs the SLS representation can be chosen to have a high degree of symmetry. Proceeding as in [HR88], we use this symmetry to show the CP (and, hence, the EP) to be in P 2. Since the EP is known to be P -hard for conict-free PNs [HR88], the CP and EP will have then been shown to be 2 P -complete 2 for sinkless and normal PNs. The following lemma gives an SLS representation of the reachability set of a sinkless PN. The strategy follows that developed in [HR88]. Lemma 4.4: Let P = (P,T,'; 0 ) be a sinkless PN with k places and m transitions such that no component of 0 is larger than n. Then there exist constants c, c 2, d, and d 2, independent of k, m, and n, such that R(P) = S 2 L(; ), where is the set of all reachable markings with no component larger than (c kmn) c2km, and is the set of all 2 N k such that:. for some 2 T*,! + ; 2. has no component larger than (d kmn) d2km ; 3. if (i) = 0, then (i) = 0, for i k; and 4. 6= 0. Proof. Clearly, S 2 L(; ) R(P). We therefore only need to show that R(P) S 2 L(; ). Let be an arbitrary marking in R(P). We dene 0 so that 0 (i) = 0 if (i) = 0, 0 (i) = otherwise. From Lemma 4.3, there is some sequence of m 0 m distinct transitions from T such that S(P; ) has a solution in which x m 0 =. Let ^x denote this solution, and let T 0 be the set of transitions in. Clearly, the system S = S(P; ) [ fx m 0 0 g must have ^x in its solution set. Let be the value of x m 0 in some minimal solution ^y ^x. Also, let 2 T 0 * be the ring sequence given by ^x according to Lemma 4.3, and let 0 2 T 0 * be the ring sequence likewise given by ^y. Thus, 0 0!, each transition in T 0 is enabled at some point in 0, and ( 0 ) () (in this proof, is dened in terms of T 0, not T). Furthermore, since 0 ; (i) = 0 i (i) = 0, i k. From results in [Huy82, VzGS78] involving integer linear programming, there exist constants c and c 2 such that no component of is larger than (c kmn) c2km. Note that since n, c and c 2 are independent of 0, and hence of. If we now assign the values of c and c 2 to the constants (of the same name) in the denition of given in the statement of the lemma, then 2. We will now show that 2 L( ; ); i.e., we will exhibit a series of vectors ; :::; j such that each j 2 and + P j i= i =. Since each i 2 i, there will be some ring sequence i from such that T 0 ( i ) = i. Furthermore, since each i > 0, it will be the case that 0 0! j!! j+ =. We already have that 0! and 0 0!, where ( 0 ) () and. It will therefore be sucient to show that for any i, if 0 0! i?!! i <, where i and P i? h=0 ( h) < (), then there is a i 2 and a i in T 0 * such that i i! i + i = i+ and P i h=0 ( h) (). We use 20