Synthesis and reengineering of persistent systems

Abstract

In formal verification, a structural object, such as a program or a Petri net, is given, and questions are asked about its behaviour. In system synthesis, conversely, a behavioural object, such as a transition system, is given, and questions are asked about the existence of a structural object realising this behaviour. In system reengineering, one wishes to transform a given system into another one, with similar behaviour and other properties not enjoyed by the original system. This paper addresses synthesis and reengineering problems in the specific framework of finite-state labelled transition systems, place/transition Petri nets, and behaviour isomorphisms. Since algorithms solving these problems are prohibitively time-consuming in general, it is interesting to know whether they can be improved in restricted circumstances, and whether direct correspondences can be found between classes of behavioural and classes of structural objects. This paper is concerned with persistent systems, which occur in hardware design and in various other applications. We shall derive exact conditions for a finite persistent transition system to be isomorphically implementable by a bounded Petri net exhibiting persistence in a structural way, and derive an efficient algorithm to find such a net if one exists. For the class of marked graph Petri nets, this leads to an exact characterisation of their state spaces.

Appendix: Some properties of lts satisfying b, r, p, and \(\mathbf{P}{\varUpsilon }\)

Let ${ TS}=(S,\rightarrow ,T,s_0)$ be an lts satisfying properties b, r, p, and $\mathbf{P}{\varUpsilon }$.

First, we briefly recapitulate Keller’s theorem [22]. For sequences $\sigma ,\tau \in T^*, \tau {\mathop {{-}}\limits ^{\bullet }}\sigma$ denotes the residue of $\tau$ with respect to $\sigma$, i.e., the sequence left after cancelling successively in $\tau$ the leftmost occurrences of all symbols from $\sigma$, read from left to right. Formally and inductively: for $t\in T, \tau {\mathop {{-}}\limits ^{\bullet }}t=\tau$ if $\varPsi (\tau )(t)=0$; $\tau {\mathop {{-}}\limits ^{\bullet }}t=$the sequence obtained by erasing the leftmost $t$ in $\tau$ if $\varPsi (\tau )(t)\ne 0$; $\tau {\mathop {{-}}\limits ^{\bullet }}\varepsilon =\tau$; and $\tau {\mathop {{-}}\limits ^{\bullet }}(t\sigma )=(\tau {\mathop {{-}}\limits ^{\bullet }}t){\mathop {{-}}\limits ^{\bullet }}\sigma$.

Theorem 8

Keller’s theorem

If $s[\tau \rangle$ and $s[\sigma \rangle$ for some $s\in [s_0\rangle$, then $s[\tau (\sigma {\mathop {{-}}\limits ^{\bullet }}\tau )\rangle s'$ and $s[\sigma (\tau {\mathop {{-}}\limits ^{\bullet }}\sigma )\rangle s''$,

as well as $\varPsi (\tau (\sigma {\mathop {{-}}\limits ^{\bullet }}\tau ))=\varPsi (\sigma (\tau {\mathop {{-}}\limits ^{\bullet }}\sigma ))$ and $s'=s''$.

Lemma 6

Characterisation of small cycles

A cycle $s[\kappa \rangle s$ is small if and only if $\varPsi (\kappa )={\varUpsilon }$.

Proof

($\Rightarrow$): Immediate by $\mathbf{P}{\varUpsilon }$.

($\Leftarrow$): Suppose that $s[\kappa \rangle s$ is a cycle with $\varPsi (\kappa )={\varUpsilon }$. Assume that $\kappa$ is not small; we derive a contradiction. By the definition of (non-)smallness, there is some state $q\in [s_0\rangle$ and some cycle $q[\gamma \rangle q$ which is small, such that $\varPsi (\gamma )\lneqq \varPsi (\kappa )$. Also, by $\mathbf{P}{\varUpsilon }$, $\varPsi (\gamma )={\varUpsilon }=\varPsi (\kappa )$, a contradiction. Hence $s[\kappa \rangle s$ is small, proving ($\Leftarrow$). $\square$

Lemma 7

Paths with Parikh vector ${\varUpsilon }$ are cyclic

Let $s[\beta \rangle s'$ with $\varPsi (\beta )={\varUpsilon }$. Then $s=s'$.

Proof

First, we show that there exists some small cycle around state $s$. To this end, let $q\in [s_0\rangle$ be any state around which there is a small cycle $q[\gamma \rangle q$ with $\varPsi (\gamma )={\varUpsilon }$. Such a state exists, because by reversibility, we can choose some arbitrary cycle, and if the chosen one is not already small, the definition of (non-)smallness allows us to choose another one which is small. By strong connectedness (following from reversibility and total reachability), there is a sequence $q[\delta \rangle s$. By Keller’s theorem,

$$\begin{aligned} q[\gamma \delta \rangle s[\delta {\mathop {{-}}\limits ^{\bullet }}(\gamma \delta )\rangle \widehat{s}\;\;\text { and }\;\; q[\delta \rangle s[(\gamma \delta ){\mathop {{-}}\limits ^{\bullet }}\delta \rangle \widehat{s} \\ \end{aligned}$$

Because $\varPsi (\delta )\le \varPsi (\gamma \delta )$, the first conjunct yields $\delta {\mathop {{-}}\limits ^{\bullet }}(\gamma \delta )=\varepsilon$ and hence $\widehat{s}=s$. Then the second conjunct yields $s[(\gamma \delta ){\mathop {{-}}\limits ^{\bullet }}\delta \rangle s$. Moreover, $\varPsi ((\gamma \delta ){\mathop {{-}}\limits ^{\bullet }}\delta )=\varPsi (\gamma )={\varUpsilon }$. Thus, putting $\alpha =(\gamma \delta ){\mathop {{-}}\limits ^{\bullet }}\delta$, we have found a cycle $s[\alpha \rangle s$ with $\varPsi (\alpha )={\varUpsilon }$, which is moreover small by Lemma 6.

Now consider the sequence $s[\beta \rangle s'$ with $\varPsi (\beta )={\varUpsilon }$. By Keller’s theorem, using also $s[\alpha \rangle s$,

$$\begin{aligned} s[\alpha \rangle s[\beta {\mathop {{-}}\limits ^{\bullet }}\alpha \rangle \widehat{s}\;\;\text { and }\;\;s[\beta \rangle s'[\alpha {\mathop {{-}}\limits ^{\bullet }}\beta \rangle \widehat{s} \end{aligned}$$

Using $\varPsi (\alpha )=\varPsi (\beta )={\varUpsilon }$, we get $\alpha {\mathop {{-}}\limits ^{\bullet }}\beta =\beta {\mathop {{-}}\limits ^{\bullet }}\alpha =\varepsilon$. Hence $s=\widehat{s}=s'$. $\square$

Lemma 8

Cyclic extensions

Suppose $s[\alpha \rangle$ with $\alpha \in T^*$ and $\varPsi (\alpha )\le {\varUpsilon }$. Then there is a small cycle $s[\kappa \rangle s$ such that $\alpha$ is a prefix of $\kappa$.

Proof

As in the first part of the previous proof, we find a small cycle $s[\widetilde{\alpha }\rangle s$. By $\mathbf{P}{\varUpsilon }$, $\varPsi (\widetilde{\alpha })={\varUpsilon }$.

Suppose $s[\alpha \rangle s'$. By Keller’s theorem, $s[\alpha \rangle s'[\widetilde{\alpha }{\mathop {{-}}\limits ^{\bullet }}\alpha \rangle s''$. By $\varPsi (\alpha )\le {\varUpsilon }=\varPsi (\widetilde{\alpha }), \varPsi (\widetilde{\alpha })=\varPsi (\alpha (\widetilde{\alpha }{\mathop {{-}}\limits ^{\bullet }}\alpha ))$. By the cyclicity of $\widetilde{\alpha }, s''=s$. Choosing $\kappa =\alpha (\widetilde{\alpha }{\mathop {{-}}\limits ^{\bullet }}\alpha )$ finishes the proof, using $\varPsi (\kappa )=\varPsi (\widetilde{\alpha })={\varUpsilon }$ and Lemma 6($\Leftarrow$). $\square$

Lemma 9

Characterisation of short paths (cf. Definition 12)

Suppose that $s[\tau \rangle s'$. Then $s[\tau \rangle s'$ is short iff $\lnot ({\varUpsilon }\le \varPsi (\tau ))$.

Proof

($\Rightarrow$): By contraposition. Suppose that $s[\tau \rangle s'$ and that ${\varUpsilon }\le \varPsi (\tau )$. There is some cycle $s[\kappa \rangle s$ with $\varPsi (\kappa )={\varUpsilon }$. By Keller’s theorem, $s[\kappa \rangle s[\tau {\mathop {{-}}\limits ^{\bullet }}\kappa \rangle s''$. By $\varPsi (\kappa )={\varUpsilon }\le \varPsi (\tau ), \varPsi (\tau )=\varPsi (\kappa (\tau {\mathop {{-}}\limits ^{\bullet }}\kappa ))$, and therefore, $s'=s''$ (by determinacy, as implied by b). Since $\kappa$ contains every transition at least once and ${\varUpsilon }\le \varPsi (\tau ), |\tau {\mathop {{-}}\limits ^{\bullet }}\kappa |<|\tau |$. Hence $s[\tau \rangle s'$ is not short.

($\Leftarrow$): Suppose that $s[\tau \rangle s'$ and $\lnot ({\varUpsilon }\le \varPsi (\tau ))$. Consider any other path $s[\tau '\rangle s'$ from $s$ to $s'$. By reversibility, there is some path $\rho$ from $s'$ to $s$. Hence both $s'[\rho \tau \rangle s'$ and $s'[\rho \tau '\rangle s'$ are cycles at $s'$. By Keller’s theorem, $s'[\rho \tau '\rangle s'[(\rho \tau ){\mathop {{-}}\limits ^{\bullet }}(\rho \tau ')\rangle s'$. Hence $s'[\tau {\mathop {{-}}\limits ^{\bullet }}\tau '\rangle s'$, and since this is a cycle, $\varPsi (\tau {\mathop {{-}}\limits ^{\bullet }}\tau ')$ is a multiple of ${\varUpsilon }$. In view of $\lnot ({\varUpsilon }\le \varPsi (\tau ))$ and $1\le {\varUpsilon }$, this can only be the case if $\varPsi (\tau {\mathop {{-}}\limits ^{\bullet }}\tau ')=0$, i.e., $\tau {\mathop {{-}}\limits ^{\bullet }}\tau '=\varepsilon$. This implies, in particular, that $\varPsi (\tau )\le \varPsi (\tau ')$ and that $|\tau |\le |\tau '|$, and therefore, $s[\tau \rangle s'$ is short. $\square$

Lemma 10

Uniqueness of short Parikh vectors

Suppose that $s[\tau \rangle s'$ and $s[\tau '\rangle s'$ are both short. Then $\varPsi (\tau )=\varPsi (\tau ')$.

Proof

By Lemma 9, both $\lnot ({\varUpsilon }\le \varPsi (\tau ))$ and $\lnot ({\varUpsilon }\le \varPsi (\tau '))$. As in the second part of the previous proof, we may conclude, using some suitable (in fact any) path $s'[\rho \rangle s$, both $s'[\tau {\mathop {{-}}\limits ^{\bullet }}\tau '\rangle s'$ and $s'[\tau '{\mathop {{-}}\limits ^{\bullet }}\tau \rangle s'$. Therefore, both $\varPsi (\tau )\le \varPsi (\tau ')$ and $\varPsi (\tau ')\le \varPsi (\tau )$, implying $\varPsi (\tau )=\varPsi (\tau ')$. $\square$

Lemma 11

Characterisation of Parikh vectors of paths

Suppose that $s[\tau \rangle s'$. Then $\varPsi (\tau )=\varPsi (\tau ')+\ell {\cdot }{\varUpsilon }$, with some number $\ell \in {\mathbb {N}}$, where $s[\tau '\rangle s'$ is any short path from $s$ to $s'$.

Proof

Assume that $s[\tau \rangle s'$. Let $\ell$ be the maximal number in ${\mathbb {N}}$ such that $\varPsi (\ell {\cdot }{\varUpsilon })\le \varPsi (\tau )$. Let $s[\kappa \rangle s$ be some cycle with $\varPsi (\kappa )={\varUpsilon }$. Then also $s[\kappa ^\ell \rangle s$, with $\varPsi (\kappa ^\ell )=\ell {\cdot }{\varUpsilon }$. By Keller’s theorem, $s[\kappa ^\ell \rangle s[\tau '\rangle s'$, with $\tau '=\tau {\mathop {{-}}\limits ^{\bullet }}\kappa ^\ell$. By the maximality of $\ell , s[\tau '\rangle s'$ is short, and by $\varPsi (\kappa ^\ell )\le \varPsi (\tau ), \varPsi (\tau )$ can be written as $\varPsi (\tau )=\varPsi (\tau ')+\varPsi (\kappa ^\ell )$. By Lemma 10, the choice of $\tau '$ is arbitrary. $\square$

Lemma 12

Existence of short paths

Suppose that $s,s'$ are states. There is a short path from $s$ to $s'$.

Proof

By strong connectedness, $s[\tau \rangle s'$ for some sequence $\tau$. Just take the path $s[\tau '\rangle s'$ from the proof of Lemma 11. $\square$

So far, only $\mathbf{P}{\varUpsilon }$ was needed, but the remaining Lemmata depend on $\mathbf{P}1$.

Lemma 13

Every label has a unique singular enabling state

Under $\mathbf{P}1$, for every $x\in T$ there is a unique state $s_x$ at which only $x$ is enabled.

Proof

There must be at least one such state, because otherwise one can create a cycle without any $x$, by bypassing every outgoing edge labelled $x$ on every state and using the finiteness of the lts, eventually contradicting property $\mathbf{P}1$.

Suppose $s_x$ and $s_x'$ are two such states and let $s_x[x\rangle s$. By $\mathbf{P}1$, we can find $s[\alpha x\rangle s$, without any $x$ in $\alpha$. Let $s[\beta \rangle s_x'$ be a short path, which exists by Lemma 12. By Keller’s theorem, $s_x'[\alpha x{\mathop {{-}}\limits ^{\bullet }}\beta \rangle$. Hence all of $\alpha$ are wiped out by $\beta$ because $s_x'$ enables only $x$. Therefore, and because $\beta$ is short, every label except $x$ occurs at least once in $\beta$. Similarly, if $s_x'[x\rangle s'[\beta '\rangle s_x$ (with a short $\beta '$), then every label except $x$ occurs at least once in $\beta '$. Now consider the cycle $s_x[x\rangle s[\beta \rangle s_x'[x\rangle s'[\beta '\rangle s_x$. It has every label exactly twice, because $x$ occurs exactly twice in it and because by Lemma 11 and $\mathbf{P}1$, the Parikh vector of every cycle is constant. Therefore, $\beta$ has every label except $x$ exactly once, which implies $s_x=s_x'$ by Lemma 7. $\square$

Lemma 14

Labels on short paths into $s_x$

Under $\mathbf{P}1$, on any short path into $s_x$, there is no label $x$.

Proof

Assume that $r[x\alpha \rangle s_x$ is a short path such that $\alpha$ has no label $x$. (Other short paths into $s_x$ containing $x$ can be reduced to this case by taking suffixes.) Also, let $r[x\delta \rangle r$ be a cycle where $\delta$ contains no $x$ but every other label once. By Keller’s theorem, $s_x[x\delta {\mathop {{-}}\limits ^{\bullet }}x\alpha \rangle$ which cannot be empty (because otherwise $r[x\alpha \rangle s_x$ is not short) but also does not start with an $x$; contradiction. $\square$