On Binary Words Being Petri Net Solvable

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On Binary Words Being Petri Net Solvable

Kamila Barylska^1,2?†, Eike Best^1∗,??, Evgeny Erofeev^{1? ? ?}, Lukasz Mikulski^2†, Marcin Pia_,tkowski^2†

1 Department of Comp. Sci., Carl von Ossietzky Univ. Oldenburg, Germany {eike.best,evgeny.erofeev}@informatik.uni-oldenburg.de

2 Faculty of Math. and Comp. Sci., Nicolaus Copernicus University Toru´n, Poland {kamila.barylska,lukasz.mikulski,marcin.piatkowski}@mat.umk.pl

Abstract. A finite word is called Petri net solvable if it is isomorphic to the reachability graph of some unlabelled Petri net. In this paper, the class of two-letter Petri net solvable words is studied.

Keywords:Binary words, labelled transition systems, Petri nets, synthesis

1 Introduction

Region theory [1] provides a polynomial algorithm, based on solving linear inequations, that checks whether a given finite labelled transition system is the reachability graph of some place/transition Petri net [5], and if it is, synthesises one of them. Due to the size of a transition system, this algorithm may be very time-consuming. Moreover, it may produce one out of a class of different nets, and there may not be a unique simplest one. For some applications, only certain types of labelled transition systems are relevant. This leads to the idea of investigating properties of labelled transition systems before synthesising them, in the hope of obtaining more efficient and possibly also more deterministic synthesis algorithms.

It may even be possible to find exact structural characterisations, based solely on graph-theoretical properties, such as for the class of finite labelled transition systems which correspond to T-systems [2].

This paper reports progress on a similar effort about characterising the set of finite words over an alphabet{a, b} which are Petri net solvable, i.e., for which a place/transition net with an isomorphic reachability graph exists. We shall put forward two conjectures, and describe some progress in analysing them. This

?Supported by a research and a visiting fellowship within the project ”Enhancing Educational Potential of Nicolaus Copernicus University in the Disciplines of Mathe- matical and Natural Sciences” (project no.POKL.04.01.01-00-081/10)

?? Supported by the German Research Foundation through the DFG/RFBR cooperation projectCAVER, BE 1267/14-1

? ? ?

Supported by the German Research Foundation through the Research Training Group (DFG GRK 1765) SCARE–http://www.uni-oldenburg.de/en/scare/

†Supported by the National Science Center under grantno. 2013/09/D/ST6/03928

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work could well be of interest in a wider context, as it might entail a nontrivial necessary condition for the solvability of an arbitrary labelled transition system.

If the latter is solvable, then finding a PN-unsolvable word as a path in it may have a strong impact on its structure / shape.

2 Basic notations and conventions used in this paper

Afinite labelled transition systemwith initial state is a tupleT S= (S,→, T, s0) with nodesS (a finite set of states), edge labelsT, edges→⊆(S×T×S), and an initial states0∈S. A labeltis enabled ats∈S, written formally ass[ti, if

∃s⁰ ∈S: (s, t, s⁰)∈→. A state s⁰ is reachable fromsthrough the execution of σ∈T^∗, denoted by s[σis⁰, if there is a directed path fromstos⁰ whose edges are labelled consecutively by σ. The set of states reachable froms is denoted by [si. A (firing) sequence σ∈ T^∗ is allowed from a state s, denoted by s[σi, if there is some state s⁰ such that s[σis⁰. Two ltsT S₁ = (S₁,→₁, T, s₀₁) and T S2 = (S2,→2, T, s02) are isomorphic if there is a bijection ζ:S1 →S2 with ζ(s01) =s02 and (s, t, s⁰)∈→1⇔(ζ(s), t, ζ(s⁰))∈→2, for alls, s⁰∈S1.

A word over T is a sequence w ∈ T^∗, and it is binary if |T| = 2. A word t₁t₂. . . t_n of length n ∈ N uniquely corresponds to a finite transition system ({0, . . . , n},{(i−1, t_i, i)|0< i≤n∧t_i∈T}, T,0).

Aninitially marked Petri netis denoted asN= (P, T, F, M0) whereP is a finite set of places,T is a finite set of transitions,F is the flow functionF: ((P×T)∪ (T×P))→Nspecifying the arc weights, andM0 is the initial marking (where a marking is a mapping M:P →N, indicating the number of tokens in each place). A side-place is a placepwithp^•∩^•p6=∅, wherep^•={t∈T |F(p, t)>0}

and ^•p = {t ∈ T | F(t, p)>0}. N is pure or side-place free if it has no side- places. A transition t ∈ T is enabled at a marking M, denoted by M[ti, if

∀p∈P:M(p)≥F(p, t). The firing oftleads fromM toM⁰, denoted byM[tiM⁰, ifM[tiandM⁰(p) =M(p)−F(p, t) +F(t, p). This can be extended, as usual, to M[σiM⁰ for sequencesσ∈T^∗, and [Midenotes the set of markings reachable fromM. The reachability graphRG(N) of a bounded (such that the number of tokens in each place does not exceed a certain finite number) Petri netN is the labelled transition system with the set of vertices [M0i, initial stateM0, label setT, and set of edges{(M, t, M⁰)|M, M⁰ ∈[M0i ∧M[tiM⁰}. If an ltsT S is isomorphic to the reachability graph of a Petri net N, we say thatN solvesT S.

3 Separation problems, and an example

In region theory, a labelled transition system (S,→, T, s₀) is assumed to be given as an input. In order to synthesise (if possible) a Petri net with isomorphic reachability graph,T is used as the set of transitions, and for the places, ¹₂·(|S|·(|S|−1)) state separation problems and up to|S|·|T|event/state separation problems have

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to be solved. A state separation problem consists of a set of states {s, s⁰}with s6=s⁰, and for each such set, one needs a place that distinguishes them. Such problems are always solvable for words; for instance, we might introduce a counting place which simply hasjtokens in statej. An event/state separation problem consists of a pair (s, t)∈S×T with¬(s[ti). For every such problem, we need a placepsuch that M(p)< F(p, t) for the markingM corresponding to state s, whereF refers to the arcs of the hoped-for net.

0 a 1 a 2 b 3

T S1 w=aab

a b

q p

2 N1

0

1

2 3

4 5 a

b b

a a T S2

w=abbaa

Fig. 1.N1 solvesT S1. No solution ofT S2 exists.

For example, in figure 1, the labelled transition systemsT S1andT S2correspond to the words aabandabbaa, respectively. The former is PN-solvable, since the reachability graph of N1 is isomorphic to T S1. T S2 contains an unsolvable event/state separation problem. The states= 2 just between the twob’s satisfies

¬(s[ai). We need a placeqwhose number of tokens in (the marking corresponding to) state 2 is less than needed for transitionato be enabled. Such a placeqhas the general form shown on the right-hand side of figure 2. It is useful to speak of theeffectE(τ) of a sequenceτ∈T^∗ on placeq. For the lettera, the effect is defined asE(a) = (a+−a−), and this can be generalised easily. Thus, for instance, the effect E(abbaa) isE(abbaa) = 3·(a+−a₋) + 2·(b+−b₋). If qpreventsaat state 2 inabbaa, then it must satisfy the following inequalities, amongst others:

a₋≤m, sinceais enabled initially; a₋≤m+E(abba), since state 4 enables a;

andm+E(ab)< a₋, or equivalently, 0≤ −m−E(ab) +a₋−1, expressing the fact that q solves the event/state separation problem¬(2[ai). Later, we show that this set of inequalities cannot be solved in the natural numbers.

m p

a b

a−

a+

b−

b+

m q

a b

a−

a+

b−

b+

Fig. 2.Placesp/qwith four arc weightsa−, a+, b−, b+ and initial markingm.

They are similar, butpwill be used for preventingb, andq for preventinga.

In a word of lengthn, the equation system for a single event/state separation problem comprises n+ 1 inequations. In binary words, we have n+ 2 such problems, one for every state 0, . . . , n−1 and two for the last state. Thus, a wordwof length nis PN-solvable if and only if all thosen+ 2 systems, each havingn+ 1 inequalities and five unknownsa₋, a₊, b₋, b₊, m, are solvable inN. The question dealt with in this paper is whether the set of binary words that are PN-(un)solvable can be characterised equivalently, in a more structural way. We shall assume, from now on, thatT ={a, b}.

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4 Minimal unsolvable binary words, and some conjectures

Let a wordw⁰ ∈T^∗be called asubword(orfactor) ofw∈T^∗ if∃(u1, u2∈T^∗) : w=u1w⁰u2, and let #t(w) denote the number of times the letter toccurs inw.

Observe that ifwis PN-solvable then all its subwords are, too. To see this, let the Petri net solvingwbe executed up to the state beforew⁰, take this as the new initial marking, and add a pre-place with #a(w⁰) tokens toaand a pre-place with #b(w⁰) tokens tob. Thus, if a subword ofwis unsolvable, thenwis. For this reason, the notion of aminimal unsolvable wordis well-defined (namely, as an unsolvable word all of whose subwords are solvable). A complete list of minimal unsolvable words up to length 110 can be found in [6]. As a consequence of the next proposition, any minimal unsolvable word either starts and ends withaor starts and ends withb.

Proposition 1. Solvability of aw andwbimplies solvability of awb If both awandwbare solvable, thenawb is also solvable.

Proof:Assume thatawandwbare PN-solvable words over{a, b}. Ifw=b^k (for k∈N) then awb=ab^k+1 is obviuosly solvable, hence we assume thatbcontains at least onea. LetN1= (P1,{a, b}, F1, M01) and N2= (P2,{a, b}, F2, M02) be Petri nets such that N1 solvesaw andN2 solves wb. We can assume thatN1

andN2are disjoint, except for their transitionsaandb. Forming the union of N₁ andN₂ gives a net which is synchronised at aandb, and which allows all (and only) sequences allowed by bothN₁ andN₂. We modifyN₁ andN₂before

forming their union, as follows:

(i) Modify N1 by adding, to each place p in^•b∩P1, another F1(p, b) tokens.

This allows an additionalb.

(ii) Modify N2 by adding, to each placeq in ^•a∩P2, another F2(q, a) tokens.

This allows an additionala. Further, for each placepina^•∩P2∩^•b, add the quantityF₂(a, p) both toF₂(p, b) and toF₂(b, p). The new arc weights lead to the same effect ofb onp, but prevent premature occurrences ofbwhich could have been allowed by adding the tokens in front ofb in step (i).

Define N as the union of the two nets thus modified, and see figure 3 for an example. (The added tokens are drawn as hollow circles.) In general,N solves awb in the following way: The initialais allowed inN1 by definition and inN2

by the additional tokens. The subsequentwis allowed in both nets, and hence in their synchronisation. The finalbis allowed inN2by definition and inN1by the additional tokens. No prematureb is allowed by the arc weight increase, and no final additionalais allowed becauseN1 does not allow it. 1

¿From the list [6], it can be observed that all minimal unsolvable words starting and ending withaare of the following general form:

s0[ (abα)b^∗is[ (baα)⁺ir[ai whereα∈T^∗ ands0,s,rare states (1)

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a b

◦

•

•• N1

a b

p•

◦

•◦

2

N2

a b

p•

•

••

2 3

2

N Fig. 3.N1(black tokens) solvesaw=abab.N2 (black tokens) solveswb=babb.

N (redundant places omitted) solvesawb=ababb.

withanot being separated ats. For example,abbaasatisfies (1) withα=ε, the star^∗being repeated zero times, and the plus⁺being repeated just once. Indeed, it is easy to prove that such words are generally PN-unsolvable:

Proposition 2. Sufficient condition for the unsolvability of a word If a word over {a, b}has a subword of the form (1), then it is not PN-solvable.

Proof:For a wordwof the form (1), we prove thatwcannot be solved (implying the proposition in the context of the considerations above). Becausebaαoccurs at least once after states,bis enabled at s, andais not enabled ats. Suppose that some place q as in figure 2 (r.h.s.) exists which separates a ats. Let E beE(abα), i.e., the effect of abα onq, and letE_b =E(b). E(b) ≥1 because q separatesaatsbut not ats+ 1. Forw, we derive the following inequalities:

(0) a₋ ≤m

(s+1) a₋ ≤m+E+k·E_b+E_b for some fixed k≥0

(r) a₋ ≤m+E+k·Eb+`·E for the samek and some fixed` >0 (sep) 0 ≤ −m−E−k·Eb+a₋−1 for the samek

(0) is true because at s₀,a is enabled. (s+1) is true becausea is enabled one state after s. (r) is true becauseais enabled at r; and` >0 because of the⁺. Finally, (sep) is true becauseq disablesaat states. Adding (s+1)+(sep) gives 1≤ Eb. Adding (0)+(sep) gives 1≤ −E−k·Eb, and using also 1≤Eb gives 1≤ −E−k·Eb ≤ −E. Adding (r)+(sep) gives 1≤`·E, contradicting 1≤ −E.

The system cannot be solved, and no placeqseparating aatsexists. 2 Conjecture 1.A converse of proposition 2

Suppose a word over{a, b}is non-PN-solvable and minimal with that property.

Then it is (modulo swappinga andb) of the form given in (1).Strengthened conjecture: It is either of the formab b^j

|{z}

α

b^kba b^j

|{z}

α

awith j≥0, k≥1 or of the

formabα(baα)^`awith`≥1 Conj. 1

Proposition 3. Another sufficient condition for unsolvability Let wbe of the form s0[αis[βir[aisuch that αstarts withaandβ starts withb.

If #a(β)·#b(α) ≥ #a(α)·#b(β) (2)

thenw is unsolvable.

For instance, forw=abbaa,α=ab andβ=ba, and (2) holds true.

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Proof:If a placeq separates aats and has markingm ats₀, then forE_α= E(α) = #_a(α)·E_a+ #_b(α)·E_b andE_β=E(β) = #_a(β)·E_a+ #_b(β)·E_b we have:

(0) a₋≤m (sinceαstarts witha) (r) a₋≤m+E_α+E_β

(sep) 0 ≤ −m−Eα+a₋−1 (since¬s[ai)

Adding (0)+(sep) yields 1≤ −Eα, hence (A):−(#a(α)Ea+ #b(α)Eb)≥1, where Ea andEb denote the effects ofa andb onq, respectively. As before, Eb ≥1.

Adding (r)+(sep) yields 1≤Eβ, hence (B): (#a(β)Ea+ #b(β)Eb)≥1. Then,

−#a(β) ≥ #a(β)#a(α)Ea+ #a(β)#b(α)Eb (algebra, and by (A))

≥ #_a(β)#_a(α)E_a+ #_a(α)#_b(β)E_b (using (2) andE_b≥1)

≥ #a(α) (algebra, and by (B))

However,−#a(β)≥#_a(α) implies #_a(β) = #_a(α) = 0, and this is a contradiction sinceαcontains at least onea. Thus, such a placeqdoes not exist. 3 Words in which the lettersaandb strictly alternate are easy to solve. Therefore, it stands to reason to investigate cases in which a letter occurs twice in a row.

Proposition 4. Solvable words starting witha can be prefixed bya If a word av is PN-solvable thenaav is, too.

Proof:LetN = (P,{a, b}, F, M0) be a net solvingav. We shall construct a net which solvesaav. The idea is to obtain such a net by “unfiring”aonce from the initial marking ofN. Since this may lead to a non-semipositive marking which we would like to avoid, we will first normalise and modify the netN, obtaining another solutionN⁰ ofav, and then construct a solutionN⁰⁰ foraav(cf. Fig. 4).

For normalisation, we assume that there are two placesp_bandq_a; the first prevents bexplicitly in the initial phase, and the second preventsaafter the last occurrence of a. They are defined byM0(pb) = 1,F(a, pb) = 1,F(b, pb) =`+1 =F(pb, b), where`is the number ofabefore the firstbinav, andM0(qa) =k,F(qa, a) = 1, wherekis the number ofain av. (All otherF values = 0.)

LetNUF(a) ={p∈a^•|M₀(p)< F(a, p)}be the set of places which do not allow the “unfiring” ofaatM₀. Note that neitherp_b nor q_a are inNUF(a). Note also that for every p∈NUF(a), F(p, a)≤M0(p)< F(a, p) – the first becausea is initially enabled, the second byp∈NUF(a). That is,ahas a positive effect onp.

Without loss of generality,b has a negative effect onp(otherwise, thanks to the normalising placepb,pcould be deleted without changing the behaviour ofN).

For everyp∈NUF(a) we add the quantityF(a, p) uniformly toM₀(p), toF(p, b), and toF(b, p), eventually obtainingN⁰= (P⁰,{a, b}, F⁰, M₀⁰), and we show that N⁰ also solvesav. First, bothM0[ai ∧ ¬M0[biand M₀⁰[ai ∧ ¬M₀⁰[bi(the former by definition, the latter by construction). For an inductive proof, suppose that M0[aiM1[τiM andM₀⁰[aiM₁⁰[τiM⁰. We have M[biiffM⁰[biby construction. If

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M[ai, then alsoM⁰[ai, sinceM ≤M⁰. Next, suppose that¬M[ai; then there is some placeqsuch thatM(q)< F(q, a). We show that, without loss of generality, q /∈NUF(a), so thatqalso disablesaatM⁰ inN⁰. IfM disables the lastainav, we can takeq=qa∈/ NUF(a). IfM disables someawhich is not the last one in av, thenqcannot be inNUF(a), sinceb acts negatively on such places.

Now, we construct a netN⁰⁰= (P⁰,{a, b}, F⁰, M₀⁰⁰) fromN⁰ by definingM₀⁰⁰(p) = M₀⁰(p)−F⁰(a, p) +F⁰(p, a) for every placep. By construction, aav is a firing sequence ofN⁰⁰. Furthermore,M₀⁰⁰ does not enablebbecause of p_b. 4

a b

p

•

qa

•• pb

• 2

2 N

a b

• p

•

qa

•• pb

• 2

2 2

N⁰

a b

p

• •

qa

••• pb

2 2

2

N⁰⁰ Fig. 4.N is normalised and solvesabab.N⁰ solvesababas well.N⁰⁰solvesaabab.

As a consequence, ifav is minimally PN-unsolvable, thenv starts with ab.

Proposition 5. Noaa andbbinside a minimal unsolvable word

If a minimal non-PN-solvable word is of the form u=aαa, then eitherαdoes not contain the factor aaorαdoes not contain the factor bb.

Proof:By contraposition. Assume thatαcontains a factoraaand a factorbb.

Two cases are possible:

Case 1:There is a group ofa’s which goes after a group ofb’s. Leta^mandbⁿ be such groups, assume thata^mgoes afterbⁿ and that there are no groups ofa or ofb between them. Thenuis of the following form

s₀[. . .iq[abⁿ(ab)^ka^mir[. . .i

wheren, m≥2, k≥0. Recombine the letters inuto the following form:

s0[. . .iq[ (ab)bⁿ⁻²(ba)^k+1aa^m−2ir[. . .i

Sinceuends witha, (ab)bⁿ⁻²(ba)^k+1ais a proper subword ofu. But it has the form (abw)b^∗(baw)⁺a, with w=ε, which implies its unsolvability by proposition 2, contradicting the minimality ofu.

Case 2:All groups ofaprecede all groups ofb. In this caseuis of the form aa^x⁰ba^x¹. . . ba^xⁿb^y⁰ab^y¹ab^y². . . ab^y^ma

where at least one ofxi and one of yj is greater than 1. Consider`= max{i| xi > 1}. If ` = 0, we get a contradiction to proposition 4. Hence, ` >0. Let t= min{j|yj>1}. Thenuhas the form

s0[a . . .iq[ba^x^`(ba)^n−`(ba)^tb^y^t ir[. . . ai

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Recombine the letters inuto the form

s0[a . . .iq[ (ba)a^x^`⁻²(ab)^n−`+t+1bb^y^t⁻²ir[. . . ai

Hence, u has a proper subword (ba)a^x^`⁻²(ab)^n−`+t+1b, which is of the form (baw)a^∗(abw)⁺bwith w=ε, implying its non-P N-solvability, due to proposition 2 with invertedaandb. This again contradicts the minimality of u. 5 For these reasons, we are particularly interested in words of the following form:

either ab^x¹a . . . ab^xⁿa or b^x¹a . . . ab^xⁿ wherex_i≥1 andn >1 (3) In the first form, there are no factorsaa. If factorsbbare excluded and the word starts and ends with ana, then we get words that are of the second form, except for swappingaandb. This swapping is useful in order to understand how words of the two forms are interrelated.

Conjecture 2.A converse of proposition 3

If a word is of the formw=αβawhereαstarts withaandβ starts withb, and if wis minimal non-PN-solvable and also of the form given in (3), then inequation (2) holds. A stronger variant of this conjecture: Ifw=αβais of the form

w = [ab^x¹a . . . ab^x^k⁻¹

| {z }

α

is[ba . . . ab^xⁿ

| {z }

β

ai withn≥3 andxi≥1

thenais not separated at statesiff #_a(β)·#_b(α) = #_a(α)·#_b(β). Conj. 2

5 Some results about words of the form b

^x¹

a . . . ab

^xⁿ

5.1 Side-places in words of the form b^x¹a . . . ab^xⁿ

If a wordw=b^x¹a . . . ab^xⁿcan be solved at all, then side-places may be necessary to do it. However, we will show that in the worst case, only some side-place q arounda, preventingaat some state, are necessary.

Lemma 1. side-place-freeness aroundb

If w=b^x¹a . . . ab^xⁿ is solvable, thenw is solvable without side-place aroundb.

Proof:Letwand its intermediate states be of the form¹

w = s₀[b^x¹is₁[ab^x²is₂[ai. . . s_n−1[ab^xⁿis_n (4) Suppose some placeppreventsbat some state sk, for 1≤k≤n−1. (The only other state at whichbmust be prevented is statesn, but that can clearly be done

1 A note on convention: in the following, we use the lettersto denote states at which bhas to be prevented, andpfor places doing this. Similarly, we use the letterrto denote states at whichahas to be prevented, andqfor places doing this.

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by a non-side-place, e.g. by an incoming place of transitionbthat has initially Pn

i=1x_i tokens.) Note thatb₋> b₊, because place pallowsb to be enabled at the state precedings_k but not ats_k. Similarly,a₋ < a₊, becausebis not enabled at statesk but at the immediately following state, which is reached after firinga.

¿From the form (4) ofw, we have b₊ ≤m+x₁(b₊−b₋)

b₊ ≤m+ (x₁+x₂)(b₊−b₋) + (a₊−a₋)

· · ·

b+ ≤m+ (x1+. . .+xn)(b+−b₋) + (n−1)(a+−a₋)

0 ≤ −m−(x1+. . .+xk)(b+−b₋)−(k−1)(a+−a₋) +b₋−1

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The firstninequations assert the semipositivity of the marking of place p(more precisely, its boundedness from below byb+, sincepmay be a side-place) at the n states s1, . . . , sn. In our context, if these inequalities are fullfilled, then the marking is ≥b₊ atallstates, as a consequence of b₋ ≥b₊,a₋ ≤a₊, and the special form of the word. The last inequality comes from¬(sk[bi).

We certainly have 0≤b+ ≤b₋ ≤m, because ofb₋ ≥b+ as noted above, and becausebis initially enabled. Ifb+= 0, thenpis not a side-place around b, and there is nothing more to prove (forp). Ifb+≥1, we consider the transformation

b⁰₊=b+−1 andb⁰₋ =b₋−1 andm⁰=m−1

The relation 0≤b⁰₊≤b⁰₋≤m⁰still holds for the new values. Also, all inequalities in (5) remain true for the new values: in the firstnlines, 1 is subtracted on each side, and on the last line, the increase in−mis offset by the decrease inb₋. Thus, we get a solution preventingb with a ‘smaller’ side-place, and we can continue until eventuallyb₊ becomes zero.

A side-place aroundb might, however, still be necessary to prevent aat some state. We show next that such side-places are also unnecessary.

Letwand its intermediate states be of the form

[b^x¹⁻¹ir₁[b a b^x²⁻¹ir₂[b ai. . .[a b^x^k⁻¹ir_k[bis_k[ai. . .[b^xⁿ⁻¹⁻¹ir_n−1[b a b^xⁿi Suppose some placeqas on the right-hand side of figure 2 preventsaat staterk, for 1≤k≤n−1. Symmetrically to the previous case, we haveb+> b−. This is true because, whileqdoes not have enough tokens to enable aat staterk, it must have enough tokens to enableaat the directly following state (which we may continue to calls_k). But we also have (w.l.o.g.)a₊< a₋. Fork≥2, this follows from the fact that if the previousa(enabled at the states_k−1just after r_k−1) acts positively onq, thenqalso has sufficiently many tokens to enableaat stater_k. Fork= 1, it is possible to argue thata₊≤a₋ is valid without loss of generality. For suppose thatqdisablesaonly atr₁ and nowhere else. (This is no loss of generality because for the other statesrk,k≥2, copies ofqcan be used.) Then we may considerq⁰which is an exact copy ofq, except thata+=a₋−1 for q⁰. This placeq⁰ also disablesaat stater1(because it has the same marking as

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q). Moreover, it does not disableaat any other state afterr₁ because it always has≥a₋−1 tokens, and after the nextb,≥a₋ tokens, sinceb₊> b₋.

Because of b+ ≥b− and a+ ≤a−, q also preventsa at all prior states in the same group ofb’s. Moreover, in the last (i.e.n’th) group ofb’s,acan easily be prevented side-place-freely. For placeq with initial markingm, we have

a+ ≤m+x1(b+−b₋) + (a+−a₋)

a+ ≤m+ (x1+x2)(b+−b₋) + 2(a+−a₋)

· · ·

a+ ≤m+ (x1+. . .+xn−1)(b+−b−) + (n−1)(a+−a−)

0 ≤ −m−(x1+. . .+xk−1)(b+−b−)−(k−1)(a+−a−) +a−−1 (6)

The firstn−1 inequations assert the semipositivity of the marking of place q (more precisely, its boundedness from below bya₊, sinceqmay be a side-place ofa) at then−1 statesr₁, . . . , r_n−1. If these inequalities are fullfilled, then the marking is≥a+ atallstates after the firsta, as a consequence ofb+≥b₋ and the special form of the word. The last inequality asserts that placeqprevents transitionaat staterk, hence effects the event/state separation ofaatrk. Ifb₋ is already zero, placeqis not a side-place ofb. Otherwise, we may perform the transformation

b⁰₊=b+−1 andb⁰₋ =b−−1 andm⁰=m

because of b+ ≥ b₋ as noted above. The left-hand sides of the first n−1 inequalities in (6) do not decrease, and neither do the right-hand sides. The same

is true for the last inequality. 1

Lemma 2. Side-place-freeness around a, preventingb

Suppose w=b^x¹ab^x²a . . . ab^xⁿ. Ifw is solvable by a net in which some placep separatesb, then we may w.l.o.g. assume thatpis not a side-place around a.

Proof:The equation system (5) is invariant under the transformation a⁰₊ =a₊−1 anda⁰₋=a₋−1 andm⁰=m

as neither the left-hand sides nor the right-hand sides change their values. 2 If some placeqprevents transitiona, then a side-placeqconnected to amay be present. It may not always be possible to remove such a side-place. Consider, for instance, the wordw=bbbabab. It is of the form (4), and any net solvingbbbabab necessarily contains a side-place around transitiona.²The word bbabbabababcan also not be solved without a side-place (butbbabbababcan). So far, no tight (weak) sufficient conditions for solvability, or solvability without side-places arounda, are known. However, the next lemma shows that the presence of a side-place around amay be due to there being “many” initialb’s. That is, ifx1is “small enough”, then such a side-place is not necessary.

2 The reader is invited to use [7] in order to verify this claim; we will not include a proof in this paper for lack of space.

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Lemma 3. Side-place-freeness around a, preventinga

Suppose w=b^x¹ab^x²a . . . ab^xⁿ. If x1 ≤min{x2, . . . , xn−1} and if w is solvable by a net in which some placeq prevents transitionaat staterk with 1≤k≤n, then we may w.l.o.g. assume that qis not a side-place around a.

Proof:For preventing aat state rn, we only need a place with no input and outputa(weight 1) which hasn−1 tokens initially.

Supposeqpreventsaat stater_k, with 1≤k≤n−1. ¿From previous considerations, we know a₊ < a₋ andb₊> b₋, and we may assume, w.l.o.g., thatq is not a side-place aroundb, i.e., thatb₋= 0. The initial markingm ofqand the remaining arc weightsa₊, a₋, b₊ satisfy the system of inequations (6), except that it is simplified byb₋= 0. Ifa+= 0, thenq is already of the required form.

Fora+>0, we distinguish two cases.

Case 1:m >0 anda₊>0. Then consider the transformation m⁰=m−1 anda⁰₊=a+−1 anda⁰₋=a₋−1

Bym >0 anda− ≥a+>0, we get new valuesm⁰, a⁰₊, a⁰₋ ≥0. Moreover, (6) remains invariant under this transformation. So,q⁰ serves the same purpose asq, and it has one incoming arc from aless thanq. By repeating this procedure, we either get a place which serves the same purpose asq, or we hit Case 2.

Case 2:m= 0 anda+>0. In this case, we consider the transformation m⁰=m= 0 anda⁰₊= 0 and a⁰₋=a₋

Such a transformation also guaranteesm⁰, a⁰₊, a⁰₋≥0. Also, the last line of (6) is clearly satisfied with these new values, since the value of its right-hand stays the same (fork= 1) or increases (for k >1). To see that the firstn−1 lines of (6) are also true with the new values (and withb₋= 0), and that we can, therefore, replace qbyq⁰, we may argue as follows. At any markingme reached along the execution ofw, we have the following:

m(q)e ≥ m(qe ⁰) ≥ 0 (7)

These inequalities imply that the new placeq⁰ preventsaatrk, whenever the old one,q, does, and that, moreover, no occurrences ofaare excluded by the place q⁰ where they should not be prohibited.

The first of the inequalities (7) holds because it holds initially (whenme =m, then m(q) =e m=m⁰=m(qe ⁰)), and because the effect ofabefore the transformation is (a+−a₋), and after the transformation, it is (−a₋). In other words,areduces the token count onq⁰ more than it does so onq, while bhas the same effect on q⁰ as onq. To see the second inequality in (7), letx= min{x2, . . . , x_n−1}. Then

a₋ ≤ x₁·b₊ ≤ x·b₊

The first inequality follows becausem= 0 andq has enough tokens after the first x1 occurrences ofbin order to enablea. The second inequality follows from

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x₁≤x. But then, sinceaonly removesa₋ tokens fromq⁰ and the subsequent block ofb’s puts at least x·b₊ tokens back on q⁰, the marking onq⁰ is always

≥0, up to and including the last block ofb’s. 3

Corollary 1. Side-place-free solvability with few initial b’s

If w = b^x¹ab^x²a . . . ab^xⁿ is solvable, then side-places are necessary, at worst, betweenaand q, where q is some place preventing aat one of the states rk with 1≤k < n−1. Ifw=b^x¹ab^x²a . . . ab^xⁿ is solvable and x1≤min{x2, . . . , x_n−1},

thenw is solvable side-place-freely. 1

5.2 Solving words aw from words of the form w=b^x¹a . . . ab^xⁿ Solving a word of the formw=b^x¹a . . . ab^xⁿ side-place-freely allows us to draw some conclusion about prepending a letterato it, as follows.

Lemma 4. Preventing ain aw

Suppose w=b^x¹ab^x²a . . . ab^xⁿ is solvable side-place-freely. Then in aw, all occurrences ofacan be separated side-place-freely.

Proof:In order to preventainw side-place-freely at any staterk, the system (6) has a solution with a+ = 0 andb₋ = 0 for any fixed 1 ≤k ≤n−1. This refers to a pure input placeqofa, which may or may not be an output place of b. In order to preventainawside-place-freely, we need to consider the statesrk

as before (but shifted to the right by one index position, still just before the last b of thek’th group ofb’s) and a correspondingly modified system as follows:

0≤m⁰+ (x₁+. . .+x_i)·(b⁰₊) + (i+ 1)·(−a⁰₋) for all 0≤i≤n−1

0≤ −m⁰−(x1+. . .+xk−1)·(b⁰₊)−k·(−a⁰₋) +a⁰₋−1 (8) wherem⁰,b⁰₊ anda⁰₋ refer to a new pure placeq⁰ preventing aat stater_k inaw.

The line withi= 0 was added becausem⁰ is required to be bounded from below andamust be enabled initially. (In (6), nonnegativity ofmfollows from the line withi= 1 andb being the first transition ofw, which is no longer true inaw.) Consider the transformation

m⁰=m+a₋ andb⁰₊=b+ anda⁰₋=a₋

These values satisfy (8), provided m, b₊ and a₋ (together with a₊ = 0 and b₋ = 0) satisfy (6). The line with i = 0 follows from m⁰ =m+a₋ ≥0. The other lines corresponding toi≥1 reduce to the corresponding lines in (6), since the additional (−a₋) at the end of each line is offset by the additional (+a₋) at the beginning of the line. The last line (which belongs to stater_k at which ais separated) corresponds to the last line of (6), because the decrease bya₋ at the beginning of the line is offset by an increase bya₋ in the termk·(−a⁰₋)

(compared with (k−1)·(−a₋) as in (6)). 4

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Note 1:In order to disableaat r_k,qcould be replaced by a placeq⁰ obtained by duplicatingqand changing the initial markingmtom⁰=m+a₋. Intuitively, this means thatm⁰ is computed frommby “unfiring”aonce.

Note 2:Placeqshould not be removed as soon asq⁰ is added, becauseqcould also be preventing a at some other r_k. In that case, a new place q⁰⁰ must be computed fromqfor this different value ofk. We may forget aboutqonly after all the relevant indiceskhave been processed.

Next, consider an input placepofbin a side-place-free solution ofwand suppose thatppreventsbat statesk. Suppose that we want to solveaw. Ifpis not also an output place ofa, then it can simply be retained unchanged, and with the same marking, preventbat corresponding states in awand inw. However, ifp is also an output place of a, there may be a problem, because “unfiring” ain the initial marking may lead to negative tokens onp. This is illustrated by the example babbabbwhich has a side-place-free solution, as shown on the left-hand side of figure 5.

a b

•p

• q1

q2

• • 2

2

solves babbabb

a b

−1 p⁰

••• q⁰1

••• q⁰2

2 2

“solves”

ababbabb

a b

p⁰

••• q⁰1

••• q2⁰

5 3

2

solves ababbabb Fig. 5.Solvingbabbabb(l.h.s.), (almost)ababbabb(middle), andababbabb(r.h.s.).

The places q₁, q₂ can be treated as in the above proof (that is, by changing their markings by “unfiring”a) and yield new places q₁⁰, q₂⁰ with marking {(q₁⁰,3),(q₂⁰,3)}. If we allowed negative markings, then a new placep⁰ with initial marking (p⁰,−1) (and otherwise duplicatingp) would do the job of solving ababbabb(as in the middle of the figure). However, we shall need a more refined argument in order to avoid negative markings.

Letp⁰ be a general new place which is supposed to preventbat state s_k inaw.

In order to check the general solvability ofawifwis side-place-freely solvable, we consider a general transformation

m⁰ =m+µ , b⁰₊=b++β+ , b⁰₋=b₋+β₋ , a⁰₊=a++α+ , a⁰₋=a₋+α₋ whereµ≥ −m,β₊≥ −b+,β₋≥ −b−, α₊ ≥ −a+ andα₋ ≥ −a−, as well as a new inequation system:

b⁰₊≤m⁰+ (x1+. . .+xi)·(b⁰₊−b⁰₋) +i·(a⁰₊−a⁰₋) for 1≤i≤n 0 ≤ −m⁰−(x₁+. . .+x_k)·(b⁰₊−b⁰₋)−k·(a⁰₊−a⁰₋) +b⁰₋−1

This system has to be compared with a restricted form of (5) (settingb+=a₋= 0, since the solution ofwis pure). Doing this by line-wise comparison, we get the

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following inequation system for the new value differences:

µ ≥ −m, β+ ≥ −b+, β₋ ≥ −b−, α₊ ≥ −a+, α₋ ≥ −a−

β+ ≤ µ+ (x1+. . .+xi)·(β+−β−) +i·(α+−α−) +a+

0 ≤ −µ−(x1+. . .+xk)·(β+−β₋)−k·(α+−α₋)−a++β₋

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The lines withimust be solved simultaneously for every 1≤i≤nwhile the line withk must be solved individually for every 1≤k ≤n−1, in order to get a place preventingbat states_k. This leads to the following lemma.

Lemma 5. Solving aw from w

Supposew=b^x¹ab^x²a . . . ab^xⁿ is solvable side-place-freely. Thenawis solvable.

Proof:Suppose that a pure place pwith parameters b₋ (arc intob), a+ (arc froma) andm(initial marking) is given and suppose it separatesbfromsk inw.

This place solves (5) for that particulark. We distinguish two cases:

Case 1:a₊≤m. In this case, the placepcan essentially be re-used for the same purpose in the solution (that we construct in this way) foraw, since (9) is solved by putting

µ=−a₊, β₊=β₋= 0, α₊=α₋= 0

Hence, a placep⁰ which differs from ponly by its initial marking (m⁰=m−a₊ instead ofm) separatesbats_k inaw.

Case 2:a+> m. In this case, (9) can be solved by

µ=−m , β+=β₋=a+−m , α+=α₋= 0

That is, we may replacepby a placep⁰ with zero initial marking and adding uniformly the valuea+−mto the incoming and outgoing arcs of b, creating a

side-place aroundb. 5

For instance, in the solution of babbabb shown on the left-hand side of figure 5, the placepfromatob satisfiesm=1,b₋=1, b+=0, a₋=0 anda+=2. (9) is solved byµ=−1,β₋= 2,β+= 0,α₋= 0 andα+= 3. Hence withm⁰=m−1, b⁰₋=b₋+ 2, b⁰₊=b+, a⁰₋=a₋ anda⁰₊=a++ 3, the net shown on the right-hand side of figure 5 is a pure solution ofababbabb. (Placep⁰ preventsb not only in statess1 ands2 but also in the initial state and in the final state.) There exist words such asw1=bbbababorw2=bbabbababab, however, which can be solved but for whichawis not solvable. We have a converse of Lemma 5:

Lemma 6. Solving w side-place-freely fromaw If awhas a solution, thenw has a side-place-free solution.

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Proof:Suppose thatawhas a solution in which some placeq⁰, preventinga, has a side-place around a. Becauseq⁰ preventsa, a⁰₋ > a⁰₊ (unless it is the first a, but then we don’t needq⁰ in solvingw). Becauseais enabled initially,m⁰≥a⁰₋. But then, the transformation a⁰⁰₋ = a⁰₋−a⁰₊, a⁰⁰₊ = 0, m⁰⁰ = m⁰−a⁰₊ yields another placeq⁰⁰ which is not a side-place aroundabut serves the same purpose asq⁰. The rest of the proof follows because the above transformations (removing side-places aroundb, or side-places around awhich prevent b) do not introduce

any new side-places arounda. 6

Corollary 2. Side-place-free solvability of b^x¹ab^x²a . . . ab^xⁿ

w=b^x¹ab^x²a . . . ab^xⁿ is solvable side-place-freely iff awis solvable. 2

6 Concluding remarks

If the characterisations of minimal Petri net solvable binary words proposed in this paper are valid, then the usual linear solver for detecting them can be replaced by a pattern-matching algorithm based on conjecture 1, or by a letter- counting algorithm based on conjecture 2. We have described a variety of results providing some insight into this class of words. There are several other facts about them which we did not have space to describe. For example, if a word is solvable side-place-freely, then so is the reverse word. Also, if a word is solvable, then it is solvable by places having exactly one outgoing transition. (This property is not shared by words with three or more letters, a counterexample beingabcbaa.) Moreover, PN-solvable words are balanced in the following sense. Referring to w=b^x¹ab^x²a . . . ab^xⁿ, callwbalanced if there is somexsuch thatx_i∈ {x, x+ 1}

for all 2≤i≤n−1. We can prove that ifw=b^x¹ab^x²a . . . ab^xⁿ is PN-solvable, then w is balanced, and moreover, x_n ≤ x+ 1. Presenting these, and other, properties of PN-solvability must however be left to future publications.

Acknowledgements: We would like to thank Harro Wimmel and Uli Schlachter for valuable comments, and Uli for finding weird words (amongst them,abcbaaas mentioned in the conclusion) using [7, 3]. We thank also the anonymous reviewers for their remarks which allowed to improve the presentation of the paper.

References

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Lectures on Petri Nets I: Basic Models. LNCS Vol. 1491, 529–586 (1998).

2. E. Best, R. Devillers: State Space Axioms for T-Systems. G. L¨uttgen, F. Corradini (eds): Special volume on the occasion of Walter Vogler’s 60th birthday. Acta Informatica, Vol. 52(2-3), 133-152 (2015).

3. E. Best, U. Schlachter: Analysis of Petri Nets and Transition Systems. ICE’2015.

4. B. Caillaud:http://www.irisa.fr/s4/tools/synet/

5. T. Murata: Petri Nets: Properties, Analysis and Applications. Proc. of the IEEE, Vol. 77(4), 541-580 (1989).

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