Finding an Optimal Label-Splitting to Make a Transition System Petri Net Implementable: a Complete Complexity Characterization

Transition System Petri Net Implementable: a Complete Complexity Characterization

Ronny Tredup

Universit¨at Rostock, Institut f¨ur Informatik, Theoretische Informatik, Albert-Einstein-Straße 22, 18059, Rostock,ronny.tredup@uni-rostock.de

Abstract. Petri netsynthesisis the task of finding a Petri netN for a given transition system (TS)Athat implements A, i.e.,N is an (exact net)realization, alanguage-simulationor anembedding ofAaccording to N’s degree of accuracy. Regardless of the sought implementation, there is not always a solution.Label-splitting converts a non-implementable TS Ainto an implementable one by giving edges with the same label now different labels. Since increasing the number of labels increases the complexity of the net synthesized, it is desired to keep the number of split labels small. This means that label-splitting can be considered as decision problem that asks for a given TSAand natural numberκwhether there is an implementable TSB with at mostκlabels that is derived fromA by splitting labels. Recently, Schlachter and Wimmel (2020) [18] showed that this problem is NP-complete if an embedding is sought. In this paper, we show that this remains true ifAis 2-bounded, i.e., every state has at most two incoming and two outgoing edges, and that this bound is tight.

Schlachter and Wimmel also alleged that label-splitting aiming at exact realization is NP-complete. In this paper, we prove this conjecture and show that label-splitting aiming at language-simulation or realization is NP-complete even ifAis 1-bounded.

1 Introduction

Petri net synthesis [2, 3, 5, 6] is the task of finding a Petri net N for a given transition system (TS)Athat implements A, i.e.,N is an (exact net)realization, alanguage-simulation or anembedding ofAaccording toN’s degree of accuracy.

Synthesis of Petri nets has applications in many areas like extracting concur- rency from sequential specifications like TS and languages [4], process discovery [1], supervisory control [14] or the synthesis of speed independent circuits [12].

However, regardless of the implementation sought, there is not always an implementing Petri net. In this case, Label-splitting [3, 11, 17] is an option, i.e, converting a non-implementable TS A into an implementable one by giving edges with the same label now different labels. Label-splitting has been originally

?Copyright c 2020 for this paper by its authors. Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0).

introduced in [10] for the synthesis of 1-bounded Petri nets and was extended to k-bounded Petri nets by the same authors in [9]. The corresponding heuristics have been implemented in the synthesis toolPetrify [12]. In [8], a novel view of the label-splitting techniques of [9, 10] has been shown, relating them to some NP-complete problems likechromatic number andweighted set cover. Inprocess mining, label-splitting is used to handle imprecise labels of event-logsto allow better fitting models instead of strongly over-approximating ones [15, 16]. Other applications of label-splitting can be found in [19], where it supports synthesis to get concise representations of service compositions, exhibiting concurrency.

Label-splitting is a powerful transformation that always guarantees an imple- mentable TS. In fact, an exact implementation is possible at the latest when every edge of the resulting TS is labeled differently. However, increasing the number of labels split, increases the complexity of the net derived and yields an over-fitting model. For applications in process-mining, this prevents users from getting a better insight about the behavior of the process [16]. Thus, it is desired to keep the number of labels split as small as possible. This means that label-splitting can be considered as an optimization problem that consist of convertingA into an implementable TS that has as few labels as possible. Equivalently [18], in this paper, we understandlabel-splittingas decision problem that asks for a given TS Aand natural numberκwhether there is an implementable TSB with at most κlabels that is derived fromAby splitting labels.

In [18], it has been shown that label-splitting is NP-complete if an embedding is sought. The presented reduction can be modified, such that the resulting TS is 3-bounded, i.e., every state has at most three incoming and three outgoing edges.

In this paper, we strengthen this result and show that label-splitting aiming at embedding remains NP-complete if Ais 2-bounded, and that this bound is tight.

In [18], it is also conjectured that label-splitting for realization is intractable.

In this paper, we prove this conjecture and show also that label-splitting for language-simulation and (exact) realization is NP-complete even ifAis 1-bounded.

We obtain all of our NP-completeness results by reductions from the vertex- cover problem on cubic graphs, which will be introduced in Section 3.

This paper is organized as follows. Section 2 introduces necessary definitions and supports them with examples. Section 3 and Section 4 present the complexity of label-splitting aiming at exact realization, language-simulation and embedding, respectively. Finally, Section 5 briefly closes the paper.

2 Preliminaries

This section introduces all necessary definitions and supports them with examples.

Transition Systems. A (deterministic)initialized transition system (TS, for short) A = (S, E, δ, ι) is a directed labeled graph with the set of nodes S (calledstates), the set of labelsE (calledevents), the partialtransition function δ : S ×E −→ S and the initial state ι ∈ S, where every state s ∈ S is reachable from ι by a directed labeled path. The set of arcs of A is defined by ∆A = {(s, e, s⁰) | s, s⁰ ∈ Se ∈ E : δ(s, e) = s⁰}. Event e occurs at state

s, denoted by s ^e , ifδ(s, e) is defined. We abridgeδ(s, e) =s⁰ by s ^e s⁰. If w=e1. . . en ∈E^∗, thens ^w denotes that there are states s=s0, . . . , sn∈S such thats_i^ei+1 s_i+1∈Afor alli∈ {0, . . . , n−1}. ThelanguageofAis defined by L(A) ={w∈E⁺|ι ^w }∪{ε}. Letb∈N.Ais calledb-bounded if, for every state s∈S, there are at most bincoming or outgoing arcs, i.e.,|{e∈E| ^e s}| ≤b and|{e∈E|s ^e }| ≤b. A cycle-free 1-bounded TSA is calledlinear. IfA is not explicitly defined, then we refer to its components byS(A) (states), E(A) (events)δ_A (function),ι_A (initial state).

Simulations.LetAandB be TS with the same set of events E. We say B simulatesA, if there is a mappingϕ:S(A)→S(B) such that ϕ(ιA) =ιB and s ^e s⁰ ∈A impliesϕ(s) ^e ϕ(s⁰)∈ B; such a mapping is called a simulation (betweenAandB).ϕis anembedding, denoted byA ,→B, if it is injective;ϕ is alanguage-simulation, denoted byA . B, ifϕ(s) ^e impliess ^e , implying L(A) =L(B) [3, p. 67];ϕis anisomorphism, denoted byA∼=B, if it is bijective andδ_A(s, e) =s⁰ if and only ifδ_B(ϕ(s), e) =ϕ(s⁰) for all s, s⁰∈S(A), e∈E(A).

Label-splitting.Let A= (S, E, δ, ι) be a TS andE={e1, . . . , en} ⊆ E a set of events. Thelabel-splitting ofe1, . . . , en into the (pairwise distinct) events e¹₁, . . . , e^m₁¹, . . . , e¹_n, . . . , e^m_nⁿ, where mi ≥ 2 for all i ∈ {1, . . . , n}, yields the event set E⁰ = (E\E)∪Sn

i=1{e¹_i, . . . , e^m_i ⁱ}. A TSB = (S, E⁰, δ⁰, ι) is an E⁰- label-splitting of A if there is a bijective mapping ψ : ∆_B → ∆_A such that ψ((s, e, s⁰)) = (s, e_i, s⁰) if e = e^j_i for some i ∈ {1, . . . , n}, j ∈ {1, . . . , mi} and, otherwise, ψ((s, e, s⁰)) = (s, e, s⁰) for all (s, e, s⁰)∈∆_B. We sayE isthe set of events ofAthat occur split inB.

Petri nets.APetri netN = (P, T, f, M0) consists of finite and disjoint sets of placesPandtransitionsT, a (total)flow functionf : ((P×T)∪(P×T))→Nand aninitial markingM0:P→N. A transitiont∈T canfire oroccur in a marking M :P→N, denoted byM ^t , ifM(p)≥f(p, t) for all placesp∈P. The firing of tin markingM leads to the markingM⁰(p) =M(p)−f(p, t)+f(t, p) for allp∈P, denoted byM ^t M⁰. Again, this notation extends to sequencesw∈T^∗ and the reachability setRS(N) ={M | ∃w∈T^∗:M₀ ^w M}contains all ofN’s reachable markings. Thereachability graphofN is the TSA_N = (RS(N), T, δ, M₀), where for every reachable marking M of N and transitiont ∈T with M ^t M⁰ the transition functionδofAN is defined byδ(M, t) =M⁰.

The following definitions relate TS and Petri nets via the reachability graph.

Implementations.A Petri netN is arealization, respectively alanguage- simulation, respectively anembedding of a TSAif there is a simulation such that A∼=AN, respectivelyA . AN, respectivelyA ,→AN.

Regions.If a TS A is implementable by a Petri net N, then we want to construct N purely from A. TS represents the behavior of a modeled system by means ofglobal states(states of TS) and transitions between them (events).

Dealing with a Petri net, we operate withlocal states(places) and their changing (transitions), while the global states of a net are markings, i.e., combinations of local states. SinceAN has to simulateA,N’s transitions correspond toA’s

events. The connection between global states in TS and local states in the sought net is given by regions of TS that mimic places: A region R = (sup, con, pro) of A= (S, E, δ, ι) consists of the mappings sup:S →Nandcon, pro:E→N such that for edge s ^e s⁰ of A it holds that con(e) ≤ sup(s) and sup(s⁰) = sup(s)−con(e) +pro(e). Notice that R is implicitly completely defined by sup(ι), con and pro: Since A is reachable, for every state s ∈ S, there is a pathι ^e1 . . . ^en sn such thats=sn. Thus, we inductively obtainsup(si+1) by sup(si+1) =sup(si)−con(ei+1) +pro(ei+1) for all i∈ {0, . . . , n−1} ands0=ι.

Since we can compute sup and, thus, R purely from sup(ι), con andpro, we often present regions only implicitly. A regionR= (sup, con, pro) models a place pand the corresponding part of the flow functionf, i.e.,con(e) modelsf(p, e) (the number of tokens that e consumes from p), pro(e, p) models f(e, p) (the number of tokens thateproduces onp) andsup(s) models (the number of tokens) M(p) (that are onp) in the markingM ∈RS(N) that corresponds to s∈S(A) via the simulation betweenAandA_N. Every setRof regions ofAdefines the synthesized net N_A^R= (R, E, f, M₀) withf(R, e) =con(e),f(e, R) =pro(e) and M0(R) =sup(ι) for allR= (sup, con, pro)∈ Rand alle∈E.

State and Event State Separation.To ensure that the input behavior is captured by the synthesized net, we have to distinguish global states, and prevent the firings of transitions when their corresponding events are not present in TS.

This is stated as so calledseparation atoms andseparation properties. A pair (s, s⁰) of distinct states ofAdefines astates separation atom (SSP atom). A region R= (sup, con, pro)solves (s, s⁰) ifsup(s)6=sup(s⁰). If every SSP atom ofA is solvable thenAhas thestate separation property (SSP, for short). A pair (e, s) of evente∈E and states∈S whereedoes not occur, that is¬s ^e , defines an event/state separation atom (ESSP atom). A region R= (sup, con, pro)solves (e, s) ifsup(s)< con(e). If every ESSP atom ofAis solvable thenAhas theevent state separation property (ESSP, for short). A setRof regions ofAis called a witness forA’s SSP, respectively ESSP, if for each SSP atom, respectively ESSP atom, there is a regionR inRthat solves it.

The next lemma ([3, p. 162], Proposition 5.10) establishes the connection between the existence of witnesses and the existence of an implementing netN in dependence of the testified property. Notice that Petri nets correspond to the type of netsτP T in [3, p. 130].

Lemma 1 ([3]). Let A be a TS. There is a Petri net N such that A ,→ AN, respectively A . AN, respectively A ∼= AN if and only if there is a witness R that testifiesA’s SSP, respectively ESSP, respectively both SSP and ESSP, and N =N_A^R.

By Lemma 1, deciding the existence of an implementing net is equivalent to deciding if the input TS has the property that corresponds to the implementation.

Moreover, there is an implementing Petri net for a given TS A if and only if there is a witnessRof regions that testifiesA’s SSP or ESSP or both according to the sought-for implementation.

A1=s0 a s1 b s2 b s3 a s4 a s5

A2=q0 q1

a a

A3=q0 q1

a a⁰

Fig. 1: The TSsA1 (Example 1),A2(Example 2) andA3 (Example 3). All of them are 1-bounded, but onlyA1 is linear.

a

R

1 ¹ b 0 1 2 3 4 5 . . .

a a a a a

b b b b b

Fig. 2: Left: The netN=N_A^R₁ (Example 1), where zero-valued flow arcs are omitted.

Right: A sketch of the (infinite) reachability graphAN, which embedsA1.

a 0 a R1

1 a

a⁰

1

R2 1

1

Fig. 3: Left:N1=N_A^R₂ (Example 2). Middle: The reachability graphAN₁. Right: The NetN2=N_A^R₃ (Example 3), whose reachability graph is isomorphic toA3.

Example 1 (Embedding). The TSA1 of Figure 1 has the SSP, since the following regionR= (sup, con, pro) solves all SSP atoms on one blow:sup(s_i) =ifor all i∈ {0, . . . ,5}andcon(a) =con(b) = 0 andpro(a) =pro(b) = 1. In particular, the setR={R}is a witness ofA₁’s SSP. The (infinite) reachability graph of the net N_A^R

1 = (R,{a, b}, f, M₀) synthesized fromR, wheref(R, a) =f(R, b) =con(a) = con(b) = 0 andf(a, R) =f(b, R) =pro(a) =pro(b) = 1 andM₀(R) = 0, embeds A₁, cf. Figure 2. An embeddingϕis given byϕ(si) =ifor alli∈ {0, . . . ,5}.

The TSA₁does not have the ESSP, since the atomα= (a, s₂) is not solvable.

This has already been proven in [7, p. 42], for self-containment we provide the proof: IfR= (sup, con, pro) is a region that solvesα, then (1)con(a)≤sup(s₀), sinceaoccurs ats₀; (2)sup(s₂)< con(a), implyingsup(s₀)−(con(a) +con(b)) + (pro(a) +pro(b)) < con(a), since R solves α; (3) con(a) ≤ sup(s4), implying

con(a)≤sup(s0)−2(con(a) +con(b)) + 2(pro(a) +pro(b)), since aoccurs ats4. If we combine (1) and (2), then we get−(con(a) +con(b)) + (pro(a) +pro(b))<0;

if we combine (2) and (3), then we get 0≤ −(con(a) +con(b)) + (pro(a) +pro(b)).

Since this is a contradiction,Rcannot exist and thus αis not solvable.

Example 2 (Language-Simulation). The TS A₂ of Figure 1 does not have any ESSP atom, since the only event a occur at all states. Hence, A₂ has the ESSP. The set R=∅ is a witness ofA₂’s ESSP. The reachability graph AN₁

of the synthesized netN1=N_A^R

2 = (∅,{a}, f, M0) simulatesA2 up to language equivalence, cf. Figure 3. A language-simulationϕis defined byϕ(q0) =ϕ(q1) = 0.

On the other hand,A₂does not have the SSP, since the SSP atomα= (s₀, s₁) is not solvable. This can be seen as follows: IfR= (sup, con, pro) is a region that solves α, then (1) sup(s₀)6=sup(s₁); (2) sup(s₀) = sup(s₁)−con(a) +con(a);

(3)sup(s₁) =sup(s₀)−con(a) +con(a). One easily verifies that the combination of (2) and (3) implies that sup(s0) =sup(s1), which contradicts (1). Hence,R cannot exist and αis not solvable.

Example 3 (Realization of anE⁰-Label Splitting).LetA2andA3be in accordance to Figure 1. The TSA2has the event setE={a}. The label-splitting of the event ainto the eventsaanda⁰ yields the event setE⁰= (E\ {a})∪ {a, a⁰}={a, a⁰}.

The TSA₃ is anE⁰-label-splitting ofA₂: Forq₀ ^a q₁∈A₂ there is exactly one x∈ {a, a⁰}, namelyx=a, such thatq0 x q1∈A3 and, forq1 a q0∈A2, there is exactly oney∈ {a, a⁰}, namelyy=a⁰, such thatq1 y q0∈A3. The set{a}

is the set of events of A2 that occur split inA3.

The TSA₃ has both the SSP and the ESSP. More exactly,A₃ has the SSP atom α₁ = (q₀, q₁) and the ESSP atoms α₂ = (a, q₁) and α₃ = (a⁰, q₀). The region R1 = (sup1, con1, pro1), where sup1(q0) = 1, sup1(q1) = 0, con1(a) = pro1(a⁰) = 1 andpro1(a) =con1(a⁰) = 0 solvesα1 andα2. Moreover, the region R2= (sup2, con2, pro2), wheresup2(q0) = 0,sup2(q1) = 1,con2(a⁰) =pro2(a) = 1 andpro2(a⁰) =con2(a) = 0 solvesα3. Thus,R={R1, R2}is a witness forA3’s SSP and ESSP. Figure 3 depicts the synthesized net N2=N_A^R

3 and it is easy to see that its reachability graphAN₂ is isomorphic toA3.

3 Label Splitting for Language-Simulation and Exact Realization

The following theorem presents our main result and states that label-splitting aiming at language-simulation or realization is NP-complete.

Theorem 1. Let A be a linear TS and κ∈ N. Deciding whether there is an E⁰-label-splitting B of A that satisfies |E⁰| ≤κand allows a Petri net N such that B . AN orB∼=AN is NP-complete.

The proof of Theorem 1 bases on a polynomial-time reduction of the vertex cover (VC) problem on cubic graphs, which is known to be NP-complete from [13]:

Cubic Vertex Cover (CVC)

Input: a GraphG= (V, Σ) with set of vertices V and edgesΣ such that everyv∈V is a member of exactly three distinctσ0, σ1, σ2∈Σ, a number λ∈N.

Decide: whether there is a (λ-VC)M ⊆V satisfying|M| ≤λandM∩σ6=∅ for allσ∈Σ.

Example 4 (CVC). The instance (G,3), where G = (V, Σ) such that V = {v0, v1, v2, v3} and Σ = {{v0, v1},{v0, v2},{v0, v3},{v1, v2},{v1, v3},{v2, v3}}, allows a positive decision, sinceM ={v0, v1, v2} is a 3-VC ofG.

In the remainder of this paper, if not explicitly stated otherwise, let (G, λ) be an arbitrary but fixed input of CVC, where G = (V, Σ) has n vertices V ={v₀, . . . , v_n−1}andmedgesΣ={σ₀, . . . , σ_m−1}such thatσ_i={v_i0, v_i1}for alli∈ {0, . . . , m−1}. For technical reasons, we assume without loss of generality that i0< i1 for the verticesvi₀, vi₁ of the edgeσi for alli∈ {0, . . . , m−1}.

For the proof of Theorem 1, we reduce (G, λ) to a pair (A, κ) of linear TS A and natural number κ as follows. If there is an E⁰-label-splitting B of A satisfying|E⁰| ≤κthat has the ESSP, then Ghas aλ-VC. Hence, ifB allows a language-simulation (implying its ESSP) or a realization (implying its SSP and its ESSP), then G has a λ-VC. Conversely, if G has aλ-VC, then there is an E⁰-label-splittingB of A satisfying|E⁰| ≤κthat has a set Rof regions that witnesses both B’s ESSP and SSP, which, by Lemma 1, implies both a language-simulation and a realization forB. Thus, (G, λ) is a yes-instance if and only if (A, κ) is a yes-instance, according to the implementation sought.

For a start, we defineκ=n+ 2(m−1) +λ, wheren+ 2(m−1) is the number of events of the announced TSA. Hence,λcorresponds to the maximum number of events ofAthat could possibly be split for a sought E⁰-label-splittingB. For everyi∈ {0, . . . , m−1}, the TSAhas the following directed pathT_i that uses the vertices of the edgeσ_i={v_i0, v_i1} as events:

T_i=^ti,0 ti,1 ti,2 ti,3 ti,4 ti,5

vi0 vi1 vi1 vi0 vi0

We use the states⊥1, . . . ,⊥_m−1and eventsy1, . . . , y_m−1,z1, . . . , z_m−1and, for all i∈ {1, . . . , m−1}, the edgest_i−1,5 ^yi ⊥iand⊥i zi ti,0to connectT₀, . . . , T_m−1 to finally buildA. The resulting linear TS can be sketched as follows:

A= T0 ^y1 ⊥1 ^z1 T1 ^y2 ⊥2 ^z2 . . . ^ym−1 ⊥_m−1 ^zm−1 T_m−1 We summarize events and states byY ={y1, . . . , y_m−1}, Z={z1, . . . , z_m−1}and

⊥={⊥₁, . . . ,⊥_m−1}. Notice that Ahas|V ∪Y ∪Z|=n+ 2·(m−1) events.

Lemma 2. If there is an E⁰-label-splitting B of A that satisfies |E⁰| ≤κand has the ESSP, then Ghas a λ-vertex cover.

Proof. Let B be anE⁰-label-splitting ofA that satisfies|E⁰| ≤ κand has the ESSP. LetE⊆E⁰ be the set of events ofAthat occur split inB. By definition of label-splitting, for everye∈E, there are at least two distinct eventse1, e2in E⁰ that originate from the splitting ofeand do not correspond to the splitting of another evente⁰∈E\ {e}. By|E|=n+ 2·(m−1) and|E⁰| ≤n+ 2·(m−1) +λ, this implies|E| ≤λ.

Leti∈ {0, . . . , m−1}. Notice that the TSA₁ of Figure 1 and the pathT_i (when considered as a TS) are isomorphic, that is, with the exception of names for states and events, they are structurally the same. Hence, just like in Example 1, one argues that if a TS has the pathTi, then it cannot have the ESSP, since the atom (vi₀, ti,2) is not solvable. Consequently, sinceB has the ESSP, the pathTi

t0,0 t0,1 t0,2 t0,3 t0,4 t0,5

v0 v1 v1 v0 v0

t1,0 t1,1 t1,2 t1,3 t1,4 t1,5

v0 v2 v2 v0 v0

t2,0 t2,1 t2,2 t2,3 t2,4 t2,5

v0 v3 v3 v0 v0

t3,0 t3,1 t3,2 t3,3 t3,4 t3,5

v1 v2 v2 v1 v1

t4,0 t4,1 t4,2 t4,3 t4,4 t4,5

v1 v3 v3 v1 v1

t5,0 t5,1 t5,2 t5,3 t5,4 t5,5

v2 v3 v3 v2 v2

Fig. 4: The pathsT0, . . . , T5 of TSAfor Theorem 1 that originates from Example 4.

t0,0 t0,1 t0,2 t0,3 t0,4 t0,5

v0⁰ v1 v1⁰ v0 v0

t1,0 t1,1 t1,2 t1,3 t1,4 t1,5

v0⁰ v2 v2⁰ v0 v0

t2,0 t2,1 t2,2 t2,3 t2,4 t2,5

v0 v3 v3 v⁰0 v0

t3,0 t3,1 t3,2 t3,3 t3,4 t3,5

v1⁰ v2 v2⁰ v1 v1

t4,0 t4,1 t4,2 t4,3 t4,4 t4,5

v1 v3 v3 v⁰1 v1

t5,0 t5,1 t5,2 t5,3 t5,4 t5,5

v2 v3 v3 v⁰2 v2

Fig. 5: According to the reduction for Theorem 1, the pathsT0⁰, . . . , T5⁰ of theE⁰-label- splittingBofAthat originates from the VCM ={v0, v1, v2}of Example 4.

(or an isomorphic version of it) cannot be present in B. SinceB is anE⁰-label- splitting ofA, this implies that there is an evente∈ {v_i0, v_i1}such thate∈E.

Sincei was arbitrary, this is simultaneously true for allT₀, . . . , T_m−1. Hence, if M =V ∩E, thenM∩E(Ti)6=∅, implyingM∩σi6=∅, for alli∈ {0, . . . , m−1}.

Finally, by|M|=|V ∩E| ≤ |E| ≤λ, we get thatM defines aλ-VC ofG.

Conversely, we have to show that ifGhas aλ-VC, then there is a searched E⁰-label-splitting forA. LetM ={vj0, . . . , v_jλ−1} ⊆V be aλ-VC ofG. For every i∈ {0, . . . , λ−1}, we split the eventv_ji into the two eventsv_ji andv⁰_j

i. This yieldsE⁰= (E\M)∪Sλ−1

i=0{vj_i, v_j⁰_i}. To define the announced E⁰-label-splitting B= (S, E⁰, δ⁰, t0,0) ofA, it is sufficient to defineδ⁰ on the states ofT0, . . . , Tm−1. For alli∈ {0, . . . , m−1},δ⁰ restricted toS(T_i) yields the pathT_i⁰ as follows:

– ifvi₀∈M andvi₁ 6∈M, thenT_i⁰=ti,0 vi0

ti,1 vi1

ti,2, ^vi1 ti,3 v⁰_i0

ti,4 vi0

ti,5; – ifv_i0, v_i1 ∈M, then T_i⁰ =t_i,0 ^v

0

i0 t_i,1 ^vi1 t_i,2, ^v

0

i1 t_i,3 ^vi0 t_i,4 ^vi0 t_i,5; – ifvi₀6∈M andvi₁ ∈M, thenT_i⁰=ti,0

vi0

ti,1 vi1

ti,2, ^v

0 i1 ti,3

vi0

ti,4 vi0

ti,5. By the following lemma, mappingssupandcon, prodefined on the states and events ofT₀⁰, . . . , T_m−1⁰ that, for alli∈ {0, . . . , m−1}, behave like a region ofT_i⁰ (when restricted toT_i⁰) can be extended to a regionR= (sup⁰, con⁰, pro⁰) ofB:

Lemma 3. Letsup:S\ ⊥ →Nand con, pro:E⁰\(Y ∪Z)→N be mappings such that if e∈ {vi₀, v⁰_i

0, vi₁, v_i⁰

1} and ti,j e ti,j+1∈B, thensup(ti,j)≤con(e) and sup(ti,j+1) =sup(ti,j)−con(e) +pro(e) for alli∈ {0, . . . , m−1} and j∈ {0, . . . ,4}. Ifsup⁰, con⁰, pro⁰ are defined as follows, thenR= (sup⁰, con⁰, pro⁰)is

a region ofB: for alls∈S, ifs∈ ⊥, thensup⁰(s) = 0, otherwisesup⁰(s) =sup(s);

for all e∈E⁰ and alli ∈ {0, . . . , m−1}, if e=y_i, then con⁰(e) = sup(t_i−1,4) and pro⁰(e) = 0; if e=z_i, then con⁰(e) = 0 and pro⁰(e) =sup(t_i,0); otherwise con⁰(e) =con(e)andpro⁰(e) =pro(e).

Proof. By the assumption aboutsup, conandpro, it is easy to see thats ^e s⁰∈ B impliessup⁰(s)≤con⁰(e) andsup⁰(s⁰) =sup⁰(s)−con⁰(e) +pro⁰(e).

The next lemma states that linear TSs and thus especiallyB have the SSP:

Lemma 4. IfA=s0 e1 . . . s_i−1 ^ei si ei+1 . . . ^en sn is a linear TS, thenAhas the SSP. Moreover, if the eventei occurs exactly once, then the atom(ei, s)is solvable for all s∈ {s0, . . . , sn} \ {s_i−1}.

Proof. The following regionR= (sup, con, pro) solves all SSP atoms on one blow:

sup(s0) = 0; for allj∈ {1, . . . , n},con(ej) = 0 and pro(ej) = 1.

The following regionR= (sup, con, pro) solve (e_i, s) for alls∈ {s0, . . . , s_i−2}:

sup(s₀) = 0; for all j ∈ {1, . . . , n}, ifj =i, then con(e_j) =i andpro(e_j) = 0;

otherwisecon(e_j) = 0 and pro(e_j) = 1.

The following regionR= (sup, con, pro) solve (ei, s) for alls∈ {si, . . . , sn}:

sup(s₀) = 1; for allj ∈ {1, . . . , n}, ifj =i, then con(e_j) = 1 and pro(e_j) = 0;

otherwisecon(e_j) =pro(e_j) = 0.

Lemma 5. There is a set Rof regions witnessing the ESSP and the SSP ofB. Proof. By Lemma 4,Bhas the SSP and the atom (e, s) is solvable for alle∈Y∪Z and (relevant)s∈S.

Letx∈E⁰\(Y∪Z) be arbitrary but fixed. SinceGis cubic, there are exactly three indices i < j < k ∈ {0, . . . , m−1}, such thatx∈T<sub>`</sub><sup>0</sup> for all` ∈ {i, j, k}.

Letno`(x) denote the number ofx’s occurrences inT<sub>`</sub>⁰ for all `∈ {i, j, k}. The following region solves (x, s) for alls ∈B\(S(T<sub>i</sub><sup>0</sup>)∪S(T<sub>j</sub><sup>0</sup>)∪S(T<sub>k</sub><sup>0</sup>)): If i = 0, thensup(t<sub>0,0</sub>) =no<sub>0</sub>(e), otherwisesup(t<sub>0,0</sub>) = 0. For alle∈E<sup>0</sup> and`∈ {i, j, k}, ife=z_{`</sub>, then con(e) = 0 and pro(e) =no<sub>`}(x); ife=x, then con(e) = 1 and pro(e) = 0; otherwisecon(e) =pro(e) = 0.

It remains to argue that (x, s) is also solvable when bothxandsbelong to the same gadget T_i⁰, i ∈ {0, . . . , m−1}. To do so, we proceed as follows. Let i∈ {0, . . . , m−1}; We argue, for all the (possible) casesv_i0∈M, v_i1 6∈M and v_i06∈M, v_i1 ∈Mandv_i0, v_i1 ∈M, that (x, s) is solvable for all relevant eventsx∈ {vi0, v_i1, v_i⁰

0, v⁰_i

1}and statess∈S(T_i⁰). By the arbitrariness ofi, this finally proves the ESSP forB. By Lemma 3, it suffices to define mappings for T₀⁰, . . . , T_m−1⁰ . Let S = {t_0,0, . . . , t_m−1,0} and E = E⁰ \(Y ∪Z). Let’s start with the case vi₀∈M, vi₁ 6∈M, which impliesT_i⁰ =ti,0

vi0

ti,1 vi1

ti,2, ^vi1 ti,3 v⁰_i0

ti,4 vi0

ti,5. (v_i0 andv_i⁰

0): Letj6=k∈ {0, . . . , m−1} \ {i}be such thatv_i0∈σ_j∩σ_k. The following regionR= (sup, con, pro) solves (vi₀, s) for all s∈ {ti,1, ti,2, ti,3, ti,5}:

For all s ∈S, if s =ti,0, then sup(s) = 1, ifs ∈ {tj,0, tk,0}, thensup(s) = 2;

otherwisesup(s) = 0. For alle∈E, if e=vi₀, thencon(e) = 1 andpro(e) = 0

(notice thatno_j(e), no_`(e)≤2, sincev_i0 ∈M); ife=v_i⁰

0, thencon(e) = 0 and pro(e) = 1; otherwise,con(e) =pro(e) = 0.

The following region R = (sup, con, pro) solves the SSP atom (v_i⁰₀, s) for all s∈ {ti,0, ti,1, ti,2, ti,4, ti,5}: For alls∈S, ifs∈ {tj,0, tk,0}, thensup(s) = 2;

otherwisesup(s) = 0. For alle∈E, if e=v_i⁰₀, thencon(e) = 2 andpro(e) = 0 (notice that noj(e), no`(e) ≤ 1); if e = vi1, then con(e) = 0 and pro(e) = 1;

otherwise,con(e) =pro(e) = 0.

(vi₁): Letj 6=k∈ {0, . . . , m−1}\{i}be such thatvi₁ ∈σj∩σk. The following regionR= (sup, con, pro) solves (vi₁, s) for alls∈ {ti,0, ti,3, ti,4}: For alls∈S, ifs∈ {tj,0, tk,0}, then sup(s) = 3; otherwisesup(s) = 0. For all e∈E, ife=vi₁, thencon(e) = 1 andpro(e) = 0 (notice that noj(e), no`(e)≤3); ife=vi₀, then con(e) = 0 andpro(e) = 2; otherwise,con(e) =pro(e) = 0.

The following region R = (sup, con, pro) solves (vi₁, ti,5): For all s ∈ S, if s = ti,0, then sup(s) = 2; if s ∈ {tj,0, tk,0}, then sup(s) = 3; otherwise sup(s) = 0. For alle∈E, ife=vi₁, then con(e) = 1 andpro(e) = 0; otherwise, con(e) =pro(e) = 0.

Similarly, one finds solving regions for all relevant ESSP atoms (x, s) ofT_i⁰ for the remaining casesv_i0 6∈M, v_i1 ∈M andv_i0, v_i1∈M. By the arbitrariness ofi, this proves the lemma.

4 Label Splitting for Embedding

In [18], it has been shown that label-splitting aiming at embedding is NP-complete.

The sixstrandsof the reduction presented in [18] can be consecutively arranged such that the resulting TS is 3-bounded. Thus, the problem is also NP-complete for 3-bounded inputs. In this section, we strengthen this result and show that label-splitting aiming at embedding remains NP-complete even for 2-bounded inputs and that this bound is tight.

Theorem 2. LetA be a b-bounded TS and κ∈N. Deciding if there is anE⁰- label-splitting B of Athat satisfies |E⁰| ≤κand allows a Petri netN such that A ,→A_N is NP-complete if b >1, otherwise it is polynomial.

The following lemma paves us the way for being able to prove easily the polynomial-time statement of Theorem 2:

Lemma 6. Let A=s0 e1 . . . ^en sn be a directed cycle TS, i.e., s0 =sn and si6=sj for alli6=j∈ {0, . . . , n−1}. If there is an i∈ {0, . . . , n−1} such that

¬sj ei for allj ∈ {0, . . . , n} \ {i−1}, that is,ei occurs exactly once in A, then A has a setRof regions that witnesses its SSP.

Proof. We argue that any SSP atom (s, s⁰) ofAis solvable. To do so, we assume without loss of generality thats₀=sands_j=s⁰ andi > j, that is,

A=s0 e1 . . . ^ej sj ej+1 . . . ^ej+k s_i−1 ^ei siei+1 . . . ^ei+`sn

where i−1 = j+k and n = j+k+`+ 1. This assumption is no essential restriction, sinceA is a cycle and thus can be transformed in the desired form by a simple renaming of its states. The following region R = (sup, con, pro) solves (s<sub>0</sub>, sj): sup(s<sub>0</sub>) = `; for all e ∈ E(A), if e 6= ei, then con(e) = 0 and pro(e) = 1; if e = ei, then con(e) = j+k+` and pro(e) = 0. This implies sup(s0) =`6=`+j=sup(sj), since` >0. Thus,Rseparates (s, s⁰).

IfA= (S, E, δ, ι) is a directed cycle that does not have the SSP, implying that every event occurs at least twice by Lemma 6, then we obtain an embeddable E⁰-label-splittingB ofA as follows: Letq ^x q⁰ ∈Abe arbitrary but fixed and E⁰= (E\ {x})∪ {x, x⁰}. The E⁰-label-splittingB= (S, E⁰, δ, ι) that originates fromAby relabeling the edge q ^x q⁰ byq ^x0 q⁰ (and nothing else), obviously, satisfies|E⁰|=|E|+ 1. Moreover, sincex⁰ occurs only once,B has the SSP by Lemma 6. Hence, ifA is a 1-bounded TS that has the SSP, which is particularly implied if it is linear, then (A, κ) is a yes-instance if and only if|E| ≤κ. Otherwise, (A, κ) is a yes-instance if and only if|E|+ 1≤κ.

Consequently, to complete the proof of Theorem 2, it remains to consider the case where Ais 2-bounded. To do so, we reduce an input (G, λ) of CVCto a pair (A, κ) of 2-bounded TSA= (S, E, δ, t_0,0) and natural numberκsuch that Ghas aλ-VC if and only if there is anE⁰-label-splitting B ofA that satisfies

|E⁰| ≤κand has a setRof regions that witnesses the SSP ofB.

Let’s introduce (A, κ). For a start, we defineκ=n+m−1+λ, wheren+m−1 is the number of events of A. Hence,λcorresponds to the maximum number of events ofAthat could possibly be split for a soughtE⁰-label-splitting B ofA.

For everyi∈ {0, . . . , m−1}, the TSA has the following gadgetTi that uses the vertices of the edgeσi={vi₀, vi₁} ofGas events:

Ti= ti,0

t_i,1

ti,2

ti,3

vi0 vi0

vi1 vi1

We use the eventsy1, . . . , ym−1and, for alli∈ {1, . . . , m−1}, the edgeti−1,2 yi ti,0

to connectT₀, . . . , T_m−1and buildA, which can be sketched as follows:

A= T₀ ^y1 T₁ ^y2 . . . ^ym−2 _Tm−2 ^ym−1 T_m−1

The initial state ofAist_0,0. LetY ={y₁, . . . , y_m−1}.Notice thatAis 2-bounded and has exactly|V ∪Y|=n+m−1 events.

If a TS has any ofT0, . . . , T_m−1, then it does not have the SSP:

Lemma 7. If A is a TS that has the paths P₁ = p₀ ^a p₁ ^a p₂ and P₂ = p0 b p3 b p2 with the same starting state p0 and final state p2, then A does not have the SSP, since the atom (p1, p3)is not solvable.