Bounded choice-free Petri net synthesis: algorithmic issues

Abstract

This paper describes a synthesis procedure dedicated to the construction of choice-free Petri nets from finite persistent transition systems, whenever possible. Taking advantage of the properties of choice-free Petri nets, a two-step approach is proposed. A pre-synthesis step checks necessary structural properties of the transition system and constructs some data structures needed for the second step. Then, a minimised set of simplified systems of linear inequalities is distilled from a general region-theoretic approach. This leads to a substantial narrowing of the sets of states for which linear inequalities must be solved, and allows an early detection of failures, supported by constructive error messages. The performance of the resulting algorithm is measured and compared numerically with existing synthesis tools.

Appendices

Keller’s theorem

The notion of a residual sequence, defined next, captures the subtraction of one sequence from another one “as much as possible”.

Definition 16

(Residues) Suppose that $\tau ,\sigma \in T^*$. The (left) residue of $\tau$ with respect to $\sigma$, denoted by $\tau {\mathop {-}\limits ^{\bullet }}\sigma$, arises from cancelling successively in $\tau$ the leftmost occurrences of all symbols from $\sigma$, read from left to right. Inductively: $\tau {\mathop {-}\limits ^{\bullet }}\varepsilon =\tau$; $\tau {\mathop {-}\limits ^{\bullet }}t=\tau$ if $t\notin supp (\tau )$; $\tau {\mathop {-}\limits ^{\bullet }}t=$ the sequence obtained by erasing the leftmost t in $\tau$ if $t\in supp (\tau )$; and $\tau {\mathop {-}\limits ^{\bullet }}(t\sigma )=(\tau {\mathop {-}\limits ^{\bullet }}t){\mathop {-}\limits ^{\bullet }}\sigma$.

For example, $acbcacbc{\mathop {-}\limits ^{\bullet }}abbcb=cacc$ and $abbcb{\mathop {-}\limits ^{\bullet }}acbcacbc=b$. In terms of Parikh vectors, Definition 16 directly yields the following consequence (where function $\min$ is extended componentwise to vectors):

Corollary 6

(Parikh vectors of residues) For $\tau ,\sigma \in T^*$, $\varPsi (\sigma )-\varPsi (\sigma {\mathop {-}\limits ^{\bullet }}\tau )=\varPsi (\tau )-\varPsi (\tau {\mathop {-}\limits ^{\bullet }}\sigma )=\min (\varPsi (\sigma ),\varPsi (\tau ))$.

E.g., $\varPsi (acbcacbc){-}\varPsi (acbcacbc{\mathop {-}\limits ^{\bullet }}abbcb){=}\varPsi (abbcb){-}\varPsi (abbcb{\mathop {-}\limits ^{\bullet }}acbcacbc){=}\varPsi (abbc)$.

Theorem 9

(Keller [32]) Let $(S,T,\rightarrow ,\imath )$ be a deterministic, persistent lts. Let $\tau$ and $\sigma$ be two label sequences enabled at some state s. Then $\tau (\sigma {\mathop {-}\limits ^{\bullet }}\tau )$ and $\sigma (\tau {\mathop {-}\limits ^{\bullet }}\sigma )$ are also enabled at s. Furthermore, the state reached after $\tau (\sigma {\mathop {-}\limits ^{\bullet }}\tau )$ equals the state reached after $\sigma (\tau {\mathop {-}\limits ^{\bullet }}\sigma )$.

Two nice little applications of this result are the following proofs.

Proof

(of Proposition 1) Let $s,s',s''\in S$ and $s[\alpha \rangle s'\wedge s[\alpha '\rangle s''\wedge \varPsi (\alpha ){=}\varPsi (\alpha ')$; we need to prove $s'=s''$ (see Definition 5). Theorem 9 (applicable by determinism and persistence) yields $s[\alpha \rangle s'[\alpha '{\mathop {-}\limits ^{\bullet }}\alpha \rangle q\wedge s[\alpha '\rangle s''[\alpha {\mathop {-}\limits ^{\bullet }}\alpha '\rangle q$. By $\varPsi (\alpha ){=}\varPsi (\alpha ')$, both $\alpha {\mathop {-}\limits ^{\bullet }}\alpha '$ and $\alpha '{\mathop {-}\limits ^{\bullet }}\alpha$ equal the empty sequence $\varepsilon$, entailing $s'=q=s''$. $\square$

Proof

(of Lemma 1) By the premise, we have $r[\kappa \alpha \rangle s$ and $r[\alpha \rangle s$. Theorem 9 yields both $s[\alpha {\mathop {-}\limits ^{\bullet }}(\kappa \alpha )\rangle q$ and $s[(\kappa \alpha ){\mathop {-}\limits ^{\bullet }}\alpha \rangle q$, for some state q. By $\varPsi (\alpha )\le \varPsi (\kappa \alpha )$, the sequence $\alpha {\mathop {-}\limits ^{\bullet }}(\kappa \alpha )$ is empty, and thus, $s=q$. Therefore, defining $\kappa '=(\kappa \alpha ){\mathop {-}\limits ^{\bullet }}\alpha$, we get $s[\kappa '\rangle s$. Clearly, $\varPsi (\kappa ')=\varPsi (\kappa )$, ending the proof. $\square$

Proof of Proposition 4

($\Rightarrow$): Obvious since the distance-path property requires that we have distance-paths for any pair of states $s'\in [s\rangle$, hence also for $s'\in [\imath \rangle$.

($\Leftarrow$): We assume that for all $s\in S$, there is a path $\imath [\alpha \rangle s$ such that $\varPsi (\alpha )=\Delta _{s}$, and we need to show that we have distance-paths for any pair of states $s'\in [s\rangle$, and not only when $s=\imath$.

Let ${\mathcal P}=\{\varUpsilon _1,\ldots ,\varUpsilon _n\}$. We shall proceed by contradiction and assume that there are two states $s,s'\in S$ such that $s'\in [s\rangle$ and the distance between them is not realised. By $s'\in [s\rangle$, there is a short path $s[\gamma \rangle s'$, and because this path does not realise the distance, $\varPsi (\gamma )\ge \varTheta$ for at least one Parikh vector $\varTheta \in {\mathcal P}$. We consider such a counterexample with the smallest possible $\gamma$ over the entire lts (see Fig. 17).

Since a subpath of a short path is itself short, without loss of generality we may assume that for some labels $u,v\in T$, $\gamma =u\delta v$ with $\varPsi (\gamma )(u)=\varTheta (u)$, $\varPsi (\gamma )(v)=\varTheta (v)$ and, for no $\varTheta '\in {\mathcal P}$, $\varPsi (\gamma )\ge \varTheta +\varTheta '$ (otherwise, we can find a smaller counterexample by suppressing a prefix and/or a suffix).

We may assume that $\imath [\alpha \rangle s$ with $\Delta _s=\varPsi (\alpha )$ and $\imath [\beta \rangle s'$ with $\Delta _{s'}=\varPsi (\beta )$, and we know that $\varPhi \not \le \Delta _s$ and $\varPhi \not \le \Delta _{s'}$ for all $\varPhi \in {\mathcal P}$, as well as that

$$\begin{aligned} \Delta _{s'}=\varPsi (\alpha \gamma )\,{\text {mod}}\,{\mathcal P}=\Delta _s+\varPsi (\gamma )-\varTheta -\sum _{\varPhi \in {\mathcal P}} k_\varPhi \cdot \varPhi \end{aligned}$$

(12)

for some natural numbers $k_\varPhi$. By $\Delta _s=\varPsi (\alpha )$ and $\Delta _{s'}=\varPsi (\beta )$, and by Corollary 6, this yields $\varPsi (\beta {\mathop {-}\limits ^{\bullet }}\alpha )=\varPsi (\alpha {\mathop {-}\limits ^{\bullet }}\beta )+\varPsi (\gamma )-\varTheta -\sum _{\varPhi \in {\mathcal P}} k_\varPhi \cdot \varPhi$.

If $\Delta _{s'}\not \le \Delta _{s}$, by Keller’s theorem we have $s[\beta {\mathop {-}\limits ^{\bullet }}\alpha \rangle \tilde{s}$ for some state $\tilde{s}\in S$, $\beta {\mathop {-}\limits ^{\bullet }}\alpha$ being nonempty. Since $\varPsi (\beta )\le \varPsi (\alpha )+\varPsi (\gamma )$, we have that $\varPsi (\beta {\mathop {-}\limits ^{\bullet }}\alpha )\le \varPsi (\gamma )$ and, by Keller’s theorem again, $\tilde{s}[\gamma {\mathop {-}\limits ^{\bullet }}(\beta {\mathop {-}\limits ^{\bullet }}\alpha )\rangle s'$, where $\varPsi ((\beta {\mathop {-}\limits ^{\bullet }}\alpha )(\gamma {\mathop {-}\limits ^{\bullet }}(\beta {\mathop {-}\limits ^{\bullet }}\alpha )))=\varPsi (\gamma )$, so that the path $s[\beta {\mathop {-}\limits ^{\bullet }}\alpha \rangle \tilde{s}[\gamma {\mathop {-}\limits ^{\bullet }}(\beta {\mathop {-}\limits ^{\bullet }}\alpha )\rangle s'$ is as short as $\gamma$, but it starts with labels from $\beta$ not needed to get $\varTheta$, since $\varPsi (\gamma {\mathop {-}\limits ^{\bullet }}(\beta {\mathop {-}\limits ^{\bullet }}\alpha ))=\varPsi (\gamma )-\varPsi (\beta {\mathop {-}\limits ^{\bullet }}\alpha ) \ge \varTheta$. As a consequence, we can construct a shorter counterexample by dropping $\beta {\mathop {-}\limits ^{\bullet }}\alpha$ in front of $\gamma {\mathop {-}\limits ^{\bullet }}(\beta {\mathop {-}\limits ^{\bullet }}\alpha )$ [see Fig. 17(l.h.s.)].

Fig. 17

We may have neither $\Delta _{s'}\not \le \Delta _{s}$ (l.h.s.) nor $\Delta _{s'}\le \Delta _{s}$ (r.h.s.)

Thus we may assume that $\Delta _{s'}\le \Delta _{s}$, and still by Keller’s theorem we have that $s'[\alpha {\mathop {-}\limits ^{\bullet }}\beta \rangle s$. Now, let us consider the path $s[u\delta \rangle s''[v\rangle s'$. There is a path $\imath [\tau \rangle s''$ with

$$\begin{aligned} \varPsi (\tau )=\Delta _{s''}=\varPsi (\alpha u \delta )\,{\text {mod}}\,{\mathcal P}=\Delta _{s'}+\varTheta -\varPsi (v)\ge \Delta _s \end{aligned}$$

The first and second equality come from the fact that $\tau$ realises the distance between $\imath$ and $s''$. The third equality is justified by first observing [from (12) by subtracting $\varPsi (v)$ on both sides] that

$$\begin{aligned} \varPsi (\alpha u\delta )=\varPsi (\beta )+\varTheta +\sum _{\varPhi \in {\mathcal P}} k_\varPhi \cdot \varPhi -\varPsi (v) \end{aligned}$$

and then taking the modulo on both sides. Moreover, $\Delta _{s'}=\varPsi (\tau v)\,{\text {mod}}\,{\mathcal P}\le \Delta _{s''}+\varPsi (v)$. Hence, with Keller again, there is a path $s'[\nu \rangle s''$ with $\varPsi (\nu )=\varTheta -\varPsi (v)$, so that there is a loop around $s'$ with Parikh vector $\varTheta$. Now, since loops can be transported Parikh-equivalently by Lemma 1, there is a loop around $s[\mu \rangle s$ with Parikh vector $\varTheta$ too. And again by Keller’s theorem, there is a path from $s[\gamma {\mathop {-}\limits ^{\bullet }}\mu \rangle s'$ from s to $s'$ with Parikh vector $\varPsi (\gamma )-\varTheta$, contradicting the shortness of $\gamma$ [see Fig. 17(r.h.s.)].

We can thus conclude that there is no counterexample to our claim, that each short path is a distance-path and that the distance-path property is valid for any pair of reachable states.