The Complexity of Synthesizing $\textsf {nop}$-Equipped Boolean Petri Nets from g-Bounded Inputs

Abstract

Boolean Petri nets equipped with $\textsf {nop}$ allow places and transitions to be independent by being related by $\textsf {nop}$. We characterize for any fixed $g\in \mathbb {N}$ the computational complexity of synthesizing $\textsf {nop}$-equipped Boolean Petri nets from labeled directed graphs whose states have at most g incoming and at most g outgoing arcs.

1 Introduction

Boolean Petri nets are a basic model for the description of distributed and concurrent systems. These nets allow at most one token on each place p in every reachable marking. Therefore, p is considered a Boolean condition that is true if p is marked and false otherwise. A place p and a transition t of a Boolean Petri net N are related by one of the following Boolean interactions: no operation ($\textsf {nop}$), input ($\textsf {inp}$), output ($\textsf {out}$), unconditionally set to true ($\textsf {set}$), unconditionally reset to false ($\textsf {res}$), inverting ($\textsf {swap}$), test if true ($\textsf {used}$), and test if false ($\textsf {free}$). The relation between p and t determines which conditions p must satisfy to allow t’s firing and which impact has the firing of t on p: The interaction $\textsf {inp}$ ($\textsf {out}$) defines that p must be true (false) first and false (true) after t has fired. If p and t are related by $\textsf {free}$ ($\textsf {used}$) then t’s firing proves that p is false (true). The interaction $\textsf {nop}$ says that p and t are independent, that is, neither need p to fulfill any condition nor has the firing of t any impact on p. If p and t are related by $\textsf {res}$ ($\textsf {set}$) then p can be both false or true but after t’s firing it is false (true). Also, the interaction $\textsf {swap}$ does not require that p satisfies any particular condition to enable t. Here, the firing of t inverts p’s Boolean value.

Boolean Petri nets are classified by the interactions of I that they use to relate places and transitions. More exactly, a subset $\tau \subseteq I$ is called a type of net and a net N is of type $\tau$ (a $\tau$-net) if it applies at most the interactions of $\tau$. So far, research has explicitly discussed seven Boolean Petri net types: Elementary net systems $\{\textsf {nop}, \textsf {inp}, \textsf {out}\}$ [9], Contextual nets $\{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {used}, \textsf {free}\}$ [6], event/condition nets $\{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {used}\}$ [2], inhibitor nets $\{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {free}\}$ [8], set nets $\{\textsf {nop}, \textsf {inp}, \textsf {set}, \textsf {used}\}$ [5], trace nets $\{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {set}, \textsf {res}, \textsf {used}, \textsf {free}\}$ [3], and flip flop nets $\{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {swap}\}$ [10]. However, since we have eight interactions to choose from, there are actually a total of 256 different types.

This paper addresses the computational complexity of the $\tau$-synthesis problem. It consists in deciding whether a given directed labeled graph A, also called transition system, is isomorphic to the reachability graph of a $\tau$-net N and in constructing N if it exists. It has been shown that $\tau$-synthesis is NP-complete if $\tau =\{\textsf {nop},\textsf {inp},\textsf {out}\}$  [1], even if the inputs are strongly restricted [14, 17]. On the contrary, in [10], it has been shown that it becomes polynomial if $\tau =\{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {swap}\}$. These opposing results motivate the question which interactions of I make the synthesis problem hard and which make it tractable. In our previous work of [13, 15, 16], we answer this question partly and reveal the computational complexity of 120 of the 128 types that allow $\textsf {nop}$.

In this paper, we investigate for fixed $g\in \mathbb {N}$ the computational complexity of $\tau$-synthesis restricted to g-bounded inputs, that is, every state of A has at most g incoming and at most g outgoing arcs. On the one hand, inputs of practical applications tend to have a low bound g such as benchmarks of digital hardware design [4]. On the other hand, considering restricted inputs hopefully gives us a better understanding of the problem’s hardness. Thus, g-bounded inputs are interesting from both the practical and the theoretical point of view. In this paper, we completely characterize the complexity of $\tau$-synthesis restricted to g-bounded inputs for all types that allow places and transitions to be independent, that is, which contain $\textsf {nop}$. Figure 1 summarizes our findings: For the types of §1 and §2, we showed hardness of synthesis without restriction in [15]. In this paper, we strengthen these results to 2- and 3-bounded inputs, respectively, and show that these bounds are tight. The hardness result of the types of §3 originates from [16]. This paper shows that a bound less than 2 makes synthesis tractable. Hardness for the types of §4 to §8 has been shown for 2-bounded inputs in [16]. In this paper, we strengthen this results to 1-bounded inputs. The hardness part for the types of §9 origin from [13]. In this paper, we argue that the bound 2 is tight. Finally, while the results of §10 are new, the ones of §11 have been found in [15].

For all considered types $\tau$, the corresponding hardness results are based on a reduction of the so-called cubic monotone one-in-three 3SAT problem [7]. All reductions follow a common approach that represents clauses by directed labeled paths. Thus, this paper also contributes a very general way to prove NP-completeness of synthesis of Boolean types of nets.

Fig. 1.

The computational complexity of Boolean net synthesis from g-bounded TS for all types that contain $\textsf {nop}$.

2 Preliminaries

Transition Systems. A transition system (TS) $A=(S,E, \delta )$ is a directed labeled graph with states S, events E and partial transition function $\delta : S\times E \longrightarrow S$, where $\delta (s,e)=s'$ is interpreted as . For we say s is a source and $s'$ is a sink of e, respectively. An event e occurs at a state s, denoted by , if $\delta (s,e)$ is defined. An initialized TS $A=(S,E,\delta , s_0)$ is a TS with a distinct state $s_0\in S$ where every state $s\in S$ is reachable from $s_0$ by a directed labeled path. TSs in this paper are deterministic by design as their state transition behavior is given by a (partial) function. Let $g\in \mathbb {N}$. An initialized TS A is called g-bounded if for all $s\in S(A)$ the number of incoming and outgoing arcs at s is restricted by g: and .

Boolean Types of Nets [2]. The following notion of Boolean types of nets serves as vehicle to capture many Boolean Petri nets in a uniform way. A Boolean type of net $\tau =(\{0,1\},E_\tau ,\delta _\tau )$ is a TS such that $E_\tau$ is a subset of the Boolean interactions: $E_\tau \subseteq I = \{\textsf {nop}, \textsf {inp}, \textsf {out}, \textsf {set}, \textsf {res}, \textsf {swap}, \textsf {used}, \textsf {free}\}$. The interactions $i \in I$ are binary partial functions $i: \{0,1\} \rightarrow \{0,1\}$ as defined in Fig. 2. For all $x\in \{0,1\}$ and all $i\in E_\tau$ the transition function of $\tau$ is defined by $\delta _\tau (x,i)=i(x)$. Notice that I contains all binary partial functions $\{0,1\} \rightarrow \{0,1\}$ except for the entirely undefined function $\bot$. Even if a type $\tau$ includes $\bot$, this event can never occur, so it would be useless. Thus, I is complete for deterministic Boolean types of nets, and that means there are a total of 256 of them. By definition, a Boolean type $\tau$ is completely determined by its event set $E_\tau$. Hence, in the following we identify $\tau$ with $E_\tau$, cf. Fig. 3. Moreover, for readability, we group interactions by $\textsf {enter}= \{\textsf {out},\textsf {set},\textsf {swap}\}$, $\textsf {exit}= \{\textsf {inp},\textsf {res},\textsf {swap}\}$, $\textsf {keep}^+= \{\textsf {nop},\textsf {set},\textsf {used}\}$, and $\textsf {keep}^-= \{\textsf {nop},\textsf {res},\textsf {free}\}$.

Fig. 2.

All interactions in I. An empty cell means that the column’s function is undefined on the respective x. The entirely undefined function is missing in I.

Fig. 3.

Left: $\tau =\{\textsf {nop}, \textsf {out}, \textsf {res}, \textsf {swap}, \textsf {free}\}$. Right: $\tilde{\tau }=\{\textsf {nop}, \textsf {inp}, \textsf {set}, \textsf {swap}, \textsf {used}\}$. $\tau$ and $\tilde{\tau }$ are isomorphic. The isomorphism $\phi : \tau \rightarrow \tilde{\tau }$ is given by $\phi (s)=1-s$ for $s\in \{0,1\}$, $\phi (i)=i$ for $i\in \{\textsf {nop},\textsf {swap}\}$, $\phi (\textsf {out})=\textsf {inp}$, $\phi (\textsf {res})=\textsf {set}$ and $\phi (\textsf {free})=\textsf {used}$.

$\tau$-Nets. Let $\tau \subseteq I$. A Boolean Petri net $N = (P, T, H_0, f)$ of type $\tau$, ($\tau$-net) is given by finite and disjoint sets P of places and T of transitions, an initial marking $H_0: P\longrightarrow \{0,1\}$, and a (total) flow function $f: P \times T \rightarrow \tau$. A $\tau$-net realizes a certain behavior by firing sequences of transitions: A transition $t \in T$ can fire in a marking $M: P\longrightarrow \{0,1\}$ if $\delta _\tau (M(p), f(p,t))$ is defined for all $p\in P$. By firing, t produces the next marking $M' : P\longrightarrow \{0,1\}$ where $M'(p)=\delta _\tau (M(p), f(p,t))$ for all $p\in P$. This is denoted by . Given a $\tau$-net $N=(P, T, H_0, f)$, its behavior is captured by a transition system $A_N$, called the reachability graph of N. The state set of $A_N$ consists of all markings that, starting from initial state $H_0$, can be reached by firing a sequence of transitions. For every reachable marking M and transition $t \in T$ with the state transition function $\delta$ of A is defined as $\delta (M,t) = M'$.

$\tau$-Regions. Let $\tau \subseteq I$. If an input A of $\tau$-synthesis allows a positive decision then we want to construct a corresponding $\tau$-net N purely from A. Since A and $A_N$ are isomorphic, N’s transitions correspond to A’s events. However, the notion of a place is unknown for TSs. So-called regions mimic places of nets: A $\tau$-region of a given $A=(S, E, \delta , s_0)$ is a pair (sup, sig) of support $sup: S \rightarrow S_\tau = \{0,1\}$ and signature $sig: E\rightarrow E_\tau = \tau$ where every transition of A leads to a transition of $\tau$. While a region divides S into the two sets $sup^{-1}(b) = \{s \in S \mid sup(s) = b\}$ for $b \in \{0,1\}$, the events are cumulated by $sig^{-1}(i) = \{e \in E \mid sig(e) = i\}$ for all available interactions $i \in \tau$. We also use $sig^{-1}(\tau ') = \{e \in E \mid sig(e) \in \tau '\}$ for $\tau ' \subseteq \tau$. A region (sup, sig) models a place p and the corresponding part of the flow function f. In particular, sig(e) models f(p, e) and sup(s) models M(p) in the marking $M\in RS(N)$ corresponding to $s\in S(A)$. Every set $\mathcal {R}$ of $\tau$-regions of A defines the synthesized $\tau$-net $N^{\mathcal {R}}_A=(\mathcal {R}, E, f, H_0)$ with flow function $f((sup, sig),e)=sig(e)$ and initial marking $H_0((sup, sig))=sup(s_{0})$ for all $(sup, sig)\in \mathcal {R}, e\in E$. It is well known that $A_{N^{\mathcal {R}}_A }$ and A are isomorphic if and only if $\mathcal {R}$’s regions solve certain separation atoms [2], to be introduced next. A pair $(s, s')$ of distinct states of A defines a state separation atom (SSP atom). A $\tau$-region $R=(sup, sig)$ solves $(s,s')$ if $sup(s)\not =sup(s')$. The meaning of R is to ensure that $N^{\mathcal {R}}_A$ contains at least one place R such that $M(R)\not =M'(R)$ for the markings M and $M'$ corresponding to s and $s'$, respectively. If there is a $\tau$-region that solves $(s,s')$ then s and $s'$ are called $\tau$-solvable. If every SSP atom of A is $\tau$-solvable then A has the $\tau$-state separation property ($\tau$-SSP). A pair (e, s) of event $e\in E$ and state $s\in S$ where e does not occur at s, that is , defines an event state separation atom (ESSP atom). A $\tau$-region $R=(sup, sig)$ solves (e, s) if sig(e) is not defined on sup(s) in $\tau$, that is, $\lnot \delta _\tau (sup(s), sig(e))$. The meaning of R is to ensure that there is at least one place R in $N^{\mathcal {R}}_A$ such that for the marking M corresponding to s. If there is a $\tau$-region that solves (e, s) then e and s are called $\tau$-solvable. If every ESSP atom of A is $\tau$-solvable then A has the $\tau$-event state separation property ($\tau$-ESSP). A set $\mathcal {R}$ of $\tau$-regions of A is called $\tau$-admissible if for every of A’s (E)SSP atoms there is a $\tau$-region R in $\mathcal {R}$ that solves it. The following lemma, borrowed from [2, p.163], summarizes the already implied connection between the existence of $\tau$-admissible sets of A and (the solvability of) $\tau$-synthesis:

Lemma 1

([2]). A TS A is isomorphic to the reachability graph of a $\tau$-net N if and only if there is a $\tau$-admissible set $\mathcal {R}$ of A such that $N=N^{\mathcal {R}}_A$.

We say a $\tau$-net N $\tau$-solves A if $A_N$ and A are isomorphic. By Lemma 1, deciding if A is $\tau$-solvable reduces to deciding whether it has the $\tau$-(E)SSP. Moreover, it is easy to see that if $\tau$ and $\tilde{\tau }$ are isomorphic then deciding the $\tau$-(E)SSP reduces to deciding the $\tilde{\tau }$-(E)SSP:

Lemma 2

(Without proof). If $\tau$ and $\tilde{\tau }$ are isomorphic types of nets then a TS A has the $\tau$-(E)SSP if and only if A has the $\tilde{\tau }$-(E)SSP.

In particular, we benefit from the isomorphisms that map $\textsf {nop}$ to $\textsf {nop}$, $\textsf {swap}$ to $\textsf {swap}$, $\textsf {inp}$ to $\textsf {out}$, $\textsf {set}$ to $\textsf {res}$, $\textsf {used}$ to $\textsf {free}$, and vice versa.

Fig. 4.

Let $\tau =\{\textsf {nop},\textsf {set}, \textsf {swap}, \textsf {free}\}$. The TSs $A_1,\dots , A_4$ give examples for the presence and absence of the $\tau$-(E)SSP: TS $A_1$ has the $\tau$-ESSP as a occurs at every state. It has also the $\tau$-SSP: The region $R=(sup, sig)$ where $sup(s_0)=sup(s_2)=1$, $sup(s_1)=0$ and $sig(a)=\textsf {swap}$ separates the pairs $s_0,s_1$ and $s_2, s_1$. Moreover, the region $R'=(sup', sig')$ where $sup'(s_0)=0$ and $sup'(s_1)=sup'(s_2)=1$ and $sig'(a)=\textsf {set}$ separates $s_0$ and $s_1$. Notice that R and $R'$ can be translated into $\tilde{\tau }$-regions, where $\tilde{\tau }=\{\textsf {nop}, \textsf {res}, \textsf {swap}, \textsf {used}\}$, via the isomorphism of Fig. 3. For example, if $s\in S(A_1)$ and $e\in E(A_1)$ and $sup''(s)=\phi (sup(s))$ and $sig''(e)=\phi (sig(e))$ then the resulting $\tilde{\tau }$-region $R''=(sup'', sig'')$ separates $s_0, s_1$ and $s_2,s_1$. Thus, $A_1$ has also $\tilde{\tau }$-(E)SSP. TS $A_2$ has the $\tau$-SSP but not the $\tau$-ESSP as event a is not inhibitable at the state $s_2$. TS $A_3$ has the $\tau$-ESSP (a occurs at every state) but not the $\tau$-SSP as $s_1$ and $s_2$ are not separable. TS $A_4$ has neither the $\tau$-ESSP nor the $\tau$-SSP.

3 Hardness Results

In this section, for several types of nets $\tau \subseteq I$ and fixed $g\in \mathbb {N}$, we show that $\tau$-synthesis is NP-complete even if the input TS A is g-bounded, cf. Fig. 1. Since $\tau$-synthesis is known to be in NP for all $\tau \subseteq I$ [16], we restrict ourselves to the hardness part. All proofs are based on a reduction of the problem cubic monotone one-in-three 3-SAT which has been shown to be NP-complete in [7]. The input for this problem is a Boolean expression $\varphi =\{\zeta _0,\dots , \zeta _{m-1}\}$ of m negation-free three-clauses $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ such that every variable $X\in V(\varphi )=\bigcup _{i=0}^{m-1}\zeta _i$ occurs in exactly three clauses. Notice that the latter implies $\vert V(\varphi )\vert =m$. Moreover, we assume without loss of generality that if $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ then $i_0< i_1 < i_2$. The question to answer is whether there is a subset $M\subseteq V(\varphi )$ with $\vert M\cap \zeta _i\vert =1$ for all $i\in \{0,\dots , m-1\}$. For all considered types of nets $\tau$ and corresponding bounds g, we reduce a given instance $\varphi$ to a g-bounded TS $A^\tau _\varphi$ such that the following two conditions are true: Firstly, the TS $A^\tau _\varphi$ has an ESSP atom $\alpha$ which is $\tau$-solvable if and only if there is a one-in-three model M of $\varphi$. Secondly, if the ESSP atom $\alpha$ is $\tau$-solvable then all ESSP and SSP atoms of $A^\tau _\varphi$ are also $\tau$-solvable. A reduction that satisfies these conditions proves the NP-hardness of $\tau$-synthesis as follows: If $\varphi$ has a one-three-model then the conditions ensure that the TS $A^\tau _\varphi$ has the $\tau$-(E)SSP and thus is $\tau$-solvable. Conversely, if $A^\tau _\varphi$ is $\tau$-solvable then, by definition, it has the $\tau$-ESSP. In particular, there is a $\tau$-region that solves $\alpha$ which, by the first condition, implies that $\varphi$ has a one-in-three model. Consequently, $A^\tau _\varphi$ is $\tau$-solvable if and only if $\varphi$ has a one-in-three model. Due to space restrictions, we omit for all considered types the proof that $A^\tau _\varphi$ satisfies the second condition, that is, that the solvability of $\alpha$ implies the (E)SSP. However, the corresponding proofs can be found in the technical report [11].

A key idea, applied by all reductions in one way or another, is the representation of every clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$, $i\in \{0,\dots , m-1\}$, by a directed labeled path of $A^\tau _\varphi$ on which the variables of $\zeta _i$ occur as events:

The reductions ensure that if a $\tau$-region (sup, sig) solves the atom $\alpha$ then $sup(s_{i,0})\not =sup(s_{i,n})$. This makes the image of this path under (sup, sig) a directed path from 0 to 1 or from 1 to 0 in $\tau$. Thus, there has to be an event e on the path that causes the state change from $sup(s_{i,0})$ to $sup(s_{i,n})$ by sig(e). We exploit this property and ensure that this state change is unambiguously done by (the signature of) exactly one variable event per clause. As a result, the corresponding variable events define a searched model of $\varphi$ via their signature. The proof of the following theorem gives a first example of this approach, and Fig. 5 shows a full example reduction.

Theorem 1

For any fixed $g\ge 2$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau =\{\textsf {nop},\textsf {inp},\textsf {free}\}$, $\tau =\{\textsf {nop},\textsf {inp},\textsf {used}, \textsf {free}\}$, $\tau =\{\textsf {nop},\textsf {out},\textsf {used}\}$ and $\tau =\{\textsf {nop},\textsf {out},\textsf {used}, \textsf {free}\}$.

Fig. 5.

The TS $A^\tau _\varphi$ for $\varphi =\{\zeta _0,\dots , \zeta _{5}\}$ with clauses $\zeta _0=\{X_0,X_1,X_2\},\ \zeta _1= \{X_0,X_2,X_3\},\ \zeta _2= \{X_0,X_1,X_3\},\ \zeta _3= \{X_2,X_4,X_5\},\ \zeta _4=\{X_1,X_4,X_5\},\ \zeta _5= \{X_3,X_4,X_5\}$ . The red colored area sketc.hes the $\tau$-region (sup, sig) that solves $(k_1,h_0)$ and corresponds to the one-in-three model $M=\{X_0,X_4\}$. (Color figure online)

Proof

We argue for $\tau \in \{\{\textsf {nop},\textsf {inp},\textsf {free}\}, \{\textsf {nop},\textsf {inp},\textsf {used}, \textsf {free}\} \}$, which by Lemma 2 proves the claim for the other types, too.

Firstly, the TS $A^\tau _\varphi$ has the following gadget H (left hand side) that provides the events $k_0, k_1$ and the atom $\alpha =(k_1, h_0)$. Secondly, it has for every clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ the following gadget $T_i$ (right hand side) that applies $k_0$, $k_1$ and $\zeta _i's$ variables as events.

Finally, $A^\tau _\varphi$ uses the states $\bot _0,\dots , \bot _m$ and events $\ominus _1,\dots \ominus _m$ and $\oplus _0,\dots ,\oplus _m$ to join the gadgets $T_0,\dots , T_{m-1}$ and H by and , for all $i\in \{0,\dots , m-1\}$, and , cf. Fig. 5.

The gadget H ensures that if (sup, sig) is a region that solves $\alpha$ then $sup(h_0)=1$ and $sig(k_1)=\textsf {free}$ which implies $sup(h_1)=0$ and $sig(k_0)=\textsf {inp}$. This is because $sig(k_1)\in \{\textsf {inp},\textsf {used}\}$ and $sup(h_0)=0$ implies $sig(k_0)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$, which is impossible. Consequently, and imply $sup(s)=1$ and $sup(s')=0$, respectively. The TS $A^\tau _\varphi$ uses these properties to ensure via $T_0,\dots , T_{m-1}$ that the region (sup, sig) implies a one-in-three model of $\varphi$.

More exactly, if $i\in \{0,\dots , m-1\}$ then for $T_i$ we have by and that $sup(t_{i,0})=1$ and $sup(t_{i,3})=0$. Thus, there is an event $X_{i_j}$, where $j\in \{0,1,2\}$, such that $sig(X_{i_j})=\textsf {inp}$. Clearly, if $sig(X_{i_j})=\textsf {inp}$ then $sig(X_{i_\ell })\not =\textsf {inp}$ for all $j < \ell \in \{0,1,2\}$ as $X_{i_\ell }$’s sources have a 0-support. Consequently, there is exactly one variable event $X\in \zeta _i$ such that $sig(X)=\textsf {inp}$. Since i was arbitrary, this is simultaneously true for all clauses $\zeta _0,\dots , \zeta _{m-1}$. Thus, the set $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {inp}\}$ is a one-in-three model of $\varphi$.

Conversely, if $\varphi$ is one-in-three satisfiable then there is a $\tau$-region (sup, sig) of $A^\tau _\varphi$ that solves $\alpha$. In particular, if M is a one-in-three model of $\varphi$ then we first define $sup(\bot _0)=1$. Secondly, for all $e\in E(A^\tau _\varphi )$ we define $sig(e)=\textsf {free}$ if $e=k_1$, $sig(e)=\textsf {inp}$ if $e\in \{k_0\}\cup M$ and else $sig(e)=\textsf {nop}$. Since $A^\tau _\varphi$ is reachable, by inductively defining $sup(s_{i+1})=\delta _\tau (sup(s_i), sig(e_i))$ for all paths , this defines a fitting region (sup, sig), cf. Fig. 5.

This proves that $\alpha$ is $\tau$-solvable if and only if $\varphi$ is one-in-three satisfiable.

In the remainder of this section, we present the remaining hardness results in accordance to Fig. 1 and the corresponding reductions that prove them.

Theorem 2

For any fixed $g\ge 3$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau =\{\textsf {nop},\textsf {set},\textsf {res}\}\cup \omega$ and $\emptyset \not =\omega \subseteq \{\textsf {used},\textsf {free}\}$.

Proof

The TS $A^\tau _\varphi$ has the following gadgets $H_0, H_1$ and $H_2$ (in this order):

The gadget $H_0$ provides $\alpha =(k_0,h_{0,2})$. By symmetry, $A^\tau _I$ is $\{\textsf {nop},\textsf {set},\textsf {res},\textsf {used}\}$-solvable if and only if it is $\{\textsf {nop},\textsf {set},\textsf {res},\textsf {free}\}$- or $\{\textsf {nop},\textsf {set},\textsf {res},\textsf {free}, \textsf {used}\}$-solvable. Thus, in the following we assume $\tau =\{\textsf {nop},\textsf {set},\textsf {res},\textsf {used}\}$, $sig(k_0)=\textsf {used}$ and $sup(h_{0,2})=0$ if (sup, sig) $\tau$-solves $\alpha$. As a result, by $sig(k_0)=\textsf {used}$, implying $sup(h_{0,1})=1$, and $sup(h_{0,2})=0$ we have $sig(k_1)=\textsf {res}$. Especially, if then $sup(s)=1$ and if then $sup(s)=0$. Thus, $sup(h_{1,0})=sup(h_{2,1})=1$ and $sup(h_{1,1})=sup(h_{2,0})=0$ which implies $sig(k_2)=\textsf {res}$ and $sig(k_3)=\textsf {set}$.

The construction uses $k_2$ and $k_3$ to produce some neutral events. More exactly, the TS $A^\tau _\varphi$ implements for all $j\in \{0,\dots , 16m-1\}$ the following gadget $F_{j}$ that uses $k_2$ and $k_3$ to ensure that the events $z_j$ are neutral:

By $sig(k_2)=\textsf {res}$ and $sig(k_3)=\textsf {set}$ we have $sup(f_{j,1})=0$ and $sup(f_{j,4})=1$. This implies and and thus $sig(z_j)=\textsf {nop}$.

Finally, for every $i\in \{0, \dots , m-1\}$ and clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$, the TS $A^\tau _\varphi$ has the following four gadgets $T_{i,0}, T_{i,1}T_{i,2}$ and $T_{i,3}$ (in this order):

$T_{i,0},\dots , T_{i,4}$ ensure that there is exactly one $X\in \zeta _i$ with $sig(X)=\textsf {res}$: By $sig(k_0)=\textsf {used}$ and $sig(k_1)=\textsf {res}$ we get $sup(t_{i,0,0})=sup(t_{i,1,0})=sup(t_{i,2,0})=sup(t_{i,3,7})=1$ and $sup(t_{i,0,7})=sup(t_{i,1,7})=sup(t_{i,2,7})=sup(t_{i,3,0})=0$. Since $z_{16i}, \dots , z_{16i+11}$ are neutral, this implies $sup(t_{i,0,6})=sup(t_{i,1,6})=sup(t_{i,2,6})=0$ and that there is a variable event with a $\textsf {res}$-signature. Moreover, by $sup(t_{i,3,0})=0$ and $sup(t_{i,3,7})=1$ and the neutrality of $z_{16i+12}, \dots , z_{16i+15}$ there is an event of $y_{3i}, y_{3i+1}, y_{3i+2}$ with a $\textsf {set}$-signature. We argue that there is exactly one variable event with a $\textsf {res}$-signature: By $sup(t_{i,0,6})=sup(t_{i,1,6})=sup(t_{i,2,6})=0$, we have $sig(X)\not \in \{\textsf {set},\textsf {used}\}$ for all $X\in \{X_{i_0}, X_{i_1}, X_{i_2}\}$. Hence, if $sig(X_{i_0})=\textsf {res}$ then $sup(t_{i,0,2})=\dots =sup(t_{i,0,6})=0$ which implies $sig(y_{3i+1})\not =\textsf {set}$ and $sig(y_{3i+2})\not =\textsf {set}$ and thus $sig(y_{3i})=\textsf {set}$. By $sig(y_{3i})=\textsf {set}$ we have $sup(t_{i,1,2})=sup(t_{i,1,4})=1$ which implies $sig(X_{i_1})\not =\textsf {res}$ and $sig(X_{i_2})\not =\textsf {res}$.

If $sig(X_{i_1})=\textsf {res}$, then $sup(t_{i,0,4})=sup(t_{i,1,2})=0$ which implies $sig(y_{3i})\not =\textsf {set}$ and $sig(y_{3i+2})\not =\textsf {set}$ and thus $sig(y_{3i+1})=\textsf {set}$. By $sig(y_{3i+1})=\textsf {set}$ we have $sup(t_{i,0,2})=sup(t_{i,2,2})=1$ which implies $sig(X_{i_0})\not =\textsf {res}$ and $sig(X_{i_2})\not =\textsf {res}$.

Since $sig(X_{i_0})=\textsf {res}$ or $sig(X_{i_1})=\textsf {res}$ implies $sig(X_{i_2})\not =\textsf {res}$, we conclude that $sig(X_{i_2})=\textsf {res}$ implies $sig(X_{i_0})\not =\textsf {res}$ and $sig(X_{i_1})\not =\textsf {res}$. Thus, there is exactly one variable of the i-th clause with a signature $\textsf {res}$. Hence, the set $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {res}\}$ is a one-in-three model of $\varphi$.

To finally build $A^\tau _\varphi$, we use the states $\bot =\{\bot _0,\dots , \bot _{20m+2}\}$ and the events $\oplus =\{\oplus _{0},\dots , \oplus _{20m+2}\}$ and $\ominus =\{\ominus _{1},\dots , \ominus _{20m+2}\}$. The states of $\bot$ are connected by and for $j\in \{0,\dots , 20m+1\}$. Let $x=16m+3$ and $y=19m+3$. For all $i\in \{0,1,2\}$, for all $\ell \in \{0,\dots , 16m-1\}$ and for all $j\in \{0,\dots , m\}$ we add the following edges that connect the gadgets $H_0,H_1,H_2$ and $F_0,\dots , F_{16m-1}$ and $T_{0,0},T_{0,1},T_{0,2},\dots , T_{m-1,0},T_{m-1,1}T_{m-1,2}$ and

If M is a one-in-three model of $\varphi$ then $\alpha$ is $\tau$-solvable by a $\tau$-region (sup, sig): If $s\in \{h_{0,0}, h_{1,0}, h_{2,1}\}$ or $\{f_{j,0} \mid j\in \{0,\dots , 16m-1\}\}$ then $sup(s)=1$. The support values of the states of $T_{i,0},\dots , T_{i,3}$, where $i\in \{0,\dots , m-1\}$, are defined in accordance to which event of $X_{i_0}, X_{i_1}, X_{i_2}$ belongs to M. The red colored area above sketches $X_{i_0}\in M$. Moreover, we define $sup(s)=0$ for all $s\in \bot$. Let $e\in E(A^\tau _\varphi )\setminus \oplus$. We define $sig(e)=\textsf {used}$ if $e=k_0$ and $sig(e)=\textsf {res}$ if $e\in \{k_1\}\cup M$. For all $i\in \{0,\dots , m-1\}$ and clauses $\{X_{i_0}, X_{i_1}, X_{i_2}\}$ and all $j\in \{0,1,2\}$ we set $sig(e)=\textsf {set}$ if $e=y_{3i+j}$ and $X_{i_j}\in M$. Otherwise, we define $sig(e)=\textsf {nop}$. For all events $e\in \oplus$ and edges of A we define $sig(e)=\textsf {set}$ if $sup(s')=1$ and, otherwise, $sig(e)=\textsf {nop}$. The resulting $\tau$-region (sup, sig) of $A^\tau _\varphi$ solves $\alpha$.    $\square$

Theorem 3

For any fixed $g\ge 2$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if (1) $\tau =\{\textsf {nop},\textsf {inp},\textsf {set}\}$ or $\tau =\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {used}\}$ or $\tau =\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}\cup \omega$ and $\omega \subseteq \{\textsf {out},\textsf {used},\textsf {free}\}$ or if (2) $\tau =\{\textsf {nop},\textsf {out},\textsf {res}\}$ or $\tau =\{\textsf {nop},\textsf {out},\textsf {res},\textsf {free}\}$ or $\tau =\{\textsf {nop},\textsf {out},\textsf {res},\textsf {set}\}\cup \omega$ and $\omega \subseteq \{\textsf {inp},\textsf {used},\textsf {free}\}$.

Proof

We present a reduction for the types of (1). By Lemma 2, this proves the claim also for the types of (2). The TS $A^\tau _\varphi$ has the following gadget H:

The intention of the gadget H is to provide the atom $\alpha =(k, h_{0,6})$ and the events of $Z=\{z_0,\dots , z_{3m-1}\}$, $V=\{v_0,\dots , v_{3m-1}\}$ and $W=\{w_0,\dots , w_{3m-1}\}$.

Moreover, the TS $A^\tau _\varphi$ has the following two gadgets $F_0$ and $F_1$ and for all $i\in \{0,\dots , 6m-2\}$ the following gadget $G_i$ (in this order):

Finally, the TS $A^\tau _\varphi$ has for every clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$, $i\in \{0,\dots , m-1\}$, the following gadgets $T_{i,0}, T_{i,1}$ and $T_{i,2}$ (in this order):

In the following, we argue that $H,F_0,F_1$ and $G_0,\dots , G_{m-2}$ collaborate like this: If (sup, sig) is a $\tau$-region solving $\alpha$ then either $sig(k)=\textsf {inp}$, $V\subseteq sig^{-1}(\textsf {enter})$ and $W\subseteq sig^{-1}(\textsf {keep}^-)$ or $sig(k)=\textsf {out}$ and $V\subseteq sig^{-1}(\textsf {exit})$ and $W\subseteq sig^{-1}(\textsf {keep}^+)$. Moreover, we prove that this implies by the functionality of $T_{0,0},\dots , T_{m-1,2}$ that $M=\{X\in V(\varphi ) \mid sig(X)\not =\textsf {nop}\}$ is a one-in-three model of $\varphi$.

Let (sup, sig) be a $\tau$-region that solves $\alpha$. Since the interactions $\textsf {res},\textsf {set}$ and $\textsf {nop}$ are defined on both 0 and 1, this implies $sig(k)\in \{\textsf {inp},\textsf {out},\textsf {used},\textsf {free}\}$. If $sig(k)=\textsf {used}$ then $sup(s)=sup(s')=1$ for every transition . Hence, we have $sup(f_{0,3})=sup(f_{1,1})=sup(h_{0,4})=1$. By definition of $\textsf {inp}, \textsf {res}$ we have that and $sig(e)\in \{\textsf {inp},\textsf {res}\}$ implies $sup(s)=0$. Consequently, by and we get $sig(z_0), sig(q_0)\in \textsf {keep}^+$ and thus $sup(h_{0,4})=sup(h_{0,5})=sup(h_{0,6})=1$ which contradicts . Hence, $sig(k)\not =\textsf {used}$. Similarly, $sig(k) =\textsf {free}$ implies $sup(h_{0,6})=0$, which is a contradiction. Thus, we have that $sig(k)=\textsf {inp}$ and $sup(h_{0,6})=0$ or $sig(k)=\textsf {out}$ and $sup(h_{0,6})=1$.

As a next step, we show that $sig(k)=\textsf {inp}$ and $sup(h_{0,6})=0$ together imply $sig(v_0)\in \textsf {enter}$ and $sig(z_0)\in \textsf {keep}^-$. By $sig(k)=\textsf {inp}$ and and we get $sup(h_{0,1})=0$ and $sup(h_{0,3})=1$. Moreover, by and $sup(h_{0,6})=0$ we obtain $sig(z_0)\in \textsf {keep}^-$, which by $sup(h_{0,1})=0$ implies $sup(h_{0,2})=0$. Finally, $sup(h_{0,2})=0$ and $sup(h_{0,3})=1$ imply $sig(v_0)\in \textsf {enter}$. Notice that this reasoning purely bases on $sig(k)=\textsf {inp}$ and $sup(h_{0,6})=0$. Moreover, $A^\tau _\varphi$ uses for every $j\in \{0,\dots , 6m-2\}$ the TS $G_j$ to ensure $sup(h_{0,6})=sup(h_{1,6})=\dots =sup(h_{6m-1,6})$. This transfers $z_0\in \textsf {keep}^-$ and $v_0\in \textsf {enter}$ to $V\subseteq \textsf {enter}$ and $W\subseteq \textsf {keep}^-$. In particular, by $sig(k)=\textsf {inp}$ we have $sup(g_{i,0})=sup(g_{i,1})=1$ and $sup(g_{i,2})=sup(g_{i,3})=0$, that is, $sig(c_i)=\textsf {nop}$. Hence, if $sig(k)=\textsf {inp}$ and $sup(h_{0,6})=0$ then $sup(h_{i,6})=0$ for all $i\in \{0,\dots , 6m-1\}$. Perfectly similar to the discussion for $z_0$ and $v_0$ we obtain that $V\subseteq sig^{-1}(\textsf {enter})$ and $W\subseteq sig^{-1}(\textsf {keep}^-)$, respectively. Similarly, $sig(k)=\textsf {out}$ and $sup(h_{0,6})=1$ imply $V\subseteq sig^{-1}(\textsf {exit})$ and $W\subseteq sig^{-1}(\textsf {keep}^+)$.

We now argue that $T_{i,0},\dots , T_{m-1,2}$ ensure that $M=\{X\in V(\varphi )\mid sig(X)\not =\textsf {nop}\}$ is a one-in-three model of $\varphi$. Let $i\in \{0,\dots , m-1\}$ and $sig(k)=\textsf {inp}$ and $sup(h_{0,6})=0$ implying $V\subseteq sig^{-1}(\textsf {enter})$ and $W\subseteq sig^{-1}(\textsf {keep}^-)$. By $sig(k)=\textsf {inp}$ and $V\subseteq sig^{-1}(\textsf {enter})$ and $W\subseteq sig^{-1}(\textsf {keep}^-)$ we have that $sup(t_{i,0,2})=sup(t_{i,1,2})=sup(t_{i,2,2})=1$ and $sup(t_{i,0,5})=sup(t_{i,1,5})=sup(t_{i,2,5})=0$. As a result, every event $e\in \{X_{i_0}, X_{i_1}, X_{i_2}\}$ has a 0-sink, which implies $sig(e)\in \{\textsf {nop},\textsf {inp},\textsf {res}\}$, and every event $e\in \{x_{i_0}, x_{i_1}, x_{i_2}\}$ has a 1-sink, which implies $sig(e)\in \{\textsf {nop},\textsf {out},\textsf {set}\}$. By $sup(t_{i,0,2})=1$ and $sup(t_{i,0,5})=0$ there is a $X\in \{X_{i_0}, X_{i_1}, X_{i_2}\}$ such that $sig(X)\in \{\textsf {inp},\textsf {res}\}$. We argue that $sig(Y)=\textsf {nop}$ for $Y\in \{X_{i_0}, X_{i_1}, X_{i_2}\}\setminus \{X\}$. If $sig(X_{i_0})\in \{\textsf {inp},\textsf {res}\}$ then $sup(t_{i,0,3})=0$ which implies $sig(x_{i_0})\in \{\textsf {out},\textsf {set}\}$ and, therefore, $sup(t_{i,1,4})=1$. Since $sig(X_{i_1}), sig(X_{i_2})\not \in \{\textsf {out}, \textsf {set}\}$ and $sig(x_{i_1}), sig(x_{i_2})\not \in \{\textsf {inp}, \textsf {res}\}$, it holds $sup(t_{i,0,3})=sup(t_{i,0,4})=0$ and $sup(t_{i,1,3})=sup(t_{i,1,4})=1$, respectively. Thus, for all $e\in \{X_{i_1}, X_{i_2}\}$, there are edges and such that $sup(s)=0$ and $sup(s')=1$. This implies $sig(e)=\textsf {nop}$. Similarly, if $sig(X_{i_1})\in \{\textsf {inp},\textsf {res}\}$, then $sig(X_{i_0})=sig(X_{i_2})=\textsf {nop}$, and if $sig(X_{i_2})\in \{\textsf {inp},\textsf {res}\}$, then $sig(X_{i_0})=sig(X_{i_1})=\textsf {nop}$. Hence, every clause $\zeta _i$ has exactly one variable event with a signature different from $\textsf {nop}$. This makes $M=\{X\in V(\varphi )\mid sig(X)\not =\textsf {nop}\}$ a one-in-three model of $\varphi$. Similarly, if $sig(k)=\textsf {out}$ and $sup(h_{0,6})=1$, then M is also a one-in-three model of $\varphi$.

To join the gadgets and finally build $A^\tau _\varphi$, we use the states $\bot =\{\bot _0,\dots , \bot _{9m+1}\}$ and the events $\oplus =\{\oplus _{0},\dots , \oplus _{9m+1}\}$ and $\ominus =\{\ominus _{1},\dots , \ominus _{9m+1}\}$. The states of $\bot$ are connected by for $j\in \{0,\dots , 9m+1\}$. Let $x=6m+2$. For all $i\in \{0,\dots , 6m-2\}$, for all $j\in \{0,\dots , m-1\}$ and for all $\ell \in \{0,1,2\}$ we add the following edges that connect the gadgets $H_0, F_0,F_1, G_0,\dots , G_{6m-2}$ and $T_{0,0},T_{0,1},T_{0,2}$ up to $T_{m-1,0},T_{m-1,1},T_{m-1,2}$ to $A^\tau _\varphi$:

If M is a one-in-three model of $\varphi$ then there is a $\tau$-region (sup, sig) of $A^\tau _\varphi$ that solves $\alpha$. The red colored area of the figures introducing the gadgets indicates already a positive support of some states. In particular, if $s\in \{h_{j,0}, h_{j,3} \mid j\in \{0,\dots , 6m-1\}\}$ or $s\in \{f_{0,0}, f_{0,2}, f_{0,3}, f_{1,0},f_{1,1} \}$ $s\in \{g_{j,0}, g_{j,1} \mid j\in \{0,\dots , 6m-2\}\}$ then $sup(s)=1$. The support values of the states of $T_{i,0},\dots , T_{i,2}$, where $i\in \{0,\dots , m-1\}$, are defined in accordance to which of the events $X_{i_0}, X_{i_1}, X_{i_2}$ belongs to M. The red colored area above sketches the situation where $X_{i_0}\in M$. Moreover, for all $s\in \bot$, we define $sup(s)=0$. Let $e\in E(A^\tau _\varphi )\setminus \oplus$. We define $sig(e)=\textsf {inp}$ if $e\in \{k\}\cup M$. For all $i\in \{0,\dots , m-1\}$ and clauses $\{X_{i_0}, X_{i_1}, X_{i_2}\}$ and all $j\in \{0,1,2\}$ we set $sig(e)=\textsf {set}$ if $e=n$ or $e\in \{v_j,p_j\mid j\in \{0,\dots , 3m-1\}\}$ or $e=x_{i_j}$ and $X_{i_j}\in M$. Otherwise, we define $sig(e)=\textsf {nop}$. Finally, for all events $e\in \oplus$ and edges of A we define $sig(e)=\textsf {set}$ if $sup(s')=1$ and, otherwise, $sig(e)=\textsf {nop}$.

Joining of 1-Bounded Gadgets. In the following, we consider types $\tau$ where $\tau$-synthesis from 1-bounded inputs is NP-complete. All gadgets $A_0,\dots , A_n$ of the reductions are directed paths on pairwise distinct states $s^i_0,\dots , s^i_n$. For all types, the joining is the concatenation

with fresh states $\bot _1,\dots ,\bot _n$ and events $\ominus _1,\dots \ominus _n,\oplus _1,\dots \oplus _n$.

Theorem 4

For any fixed $g\ge 1$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau =\{\textsf {nop}, \textsf {inp},\textsf {out},\textsf {set}\}\cup \omega$ or $\tau =\{\textsf {nop}, \textsf {inp},\textsf {out},\textsf {res}\} \cup \omega$ and $\omega \subseteq \{\textsf {used}, \textsf {free}\}$.

Proof

Our construction proves the claim for $\tau =\{\textsf {nop}, \textsf {inp},\textsf {set},\textsf {out}\}\cup \omega$ with $\omega \subseteq \{\textsf {used}, \textsf {free}\}$. By Lemma 2, this proves the claim also for the other types.

The TS $A^\tau _\varphi$ has the following gadgets $H_0, H_1,H_2$ and $H_3$ (in this order):

If $\textsf {used}\in \tau$ then $A^\tau _\varphi$ has the following gadget $H_4$:

For all $i\in \{0,\dots , m-1\}$, the TS $A^\tau _\varphi$ has for the clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ and the variable $X_i\in V(\varphi )$ the following gadgets $T_i$ and $B_i$, respectively:

The gadget $H_0$ provides the atom $\alpha =(k_0,h_{0,6})$. Moreover, the gadgets $H_0,\dots , H_4$ ensure that if (sup, sig) is a $\tau$-region solving $\alpha$ then $sig(k_0)=\textsf {out}$ and $sig(k_1)\in \{\textsf {out}, \textsf {set}\}$. In particular, $H_4$ prevents the solvability of $\alpha$ by $\textsf {used}$. As a result, such a region implies $sup(t_{i,1})=1$, $sup(t_{i,4})=0$ and $sup(b_{i,1})=0$ for all $i\in \{0,\dots , m-1\}$. On the one hand, by $sup(b_{i,1})=0$ for all $i\in \{0,\dots , m-1\}$ we have $sig(X)\not \in \{\textsf {out}, \textsf {set}\}$ for all $X\in V(\varphi )$. On the other hand, by $sup(t_{i,1})=1$ and $sup(t_{i,4})=0$ there is an event $X\in \{X_{i_0}, X_{i_1},X_{i_2}\}$ such that $sig(X)=\textsf {inp}$. Since no variable event has an incoming signature we obtain immediately $sig(Y)\not =\textsf {inp}$ for $Y\in \{X_{i_0}, X_{i_1},X_{i_2}\}\setminus \{X\}$. Thus, $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {inp}\}$ is a one-in-three model of $\varphi$.

We argue that $H_0,\dots , H_4$ behave as announced. Let (sup, sig) be a region that solves $(k_0,h_{0,6})$. If $sig(k_0)=\textsf {inp}$ then $sup(h_{0,6})=0$ and $sig(h_{0,7})=1$, implying $sig(o)\in \{\textsf {out},\textsf {set}\}$ and $sup(h_{0,3})=1$. Thus, there is an event $e\in \{k_1, z_0,z_1\}$ with $sig(e)=\textsf {inp}$. By $sig(k_0)=\textsf {inp}$ we have $sup(h_{1,1})=sup(h_{2,1})=1$ and $sup(h_{3,1})=0$ implying $sig(e)\not =\textsf {inp}$ for all $e\in \{k_1, z_0,z_1\}$, a contradiction.

If $sig(k_0)=\textsf {free}$ then $sup(h_{0,6})=1$ and $sup(h_{0,1})=sup(h_{0,7})=sup(h_{1,1})=0$ which implies $sig(o)=\textsf {inp}$ and $sup(h_{0,2})=1$. By $sup(h_{0,1})=0$ and $sup(h_{0,2})=1$ we have $sig(z_0)\in \{\textsf {out}, \textsf {set}\}$ which by $sup(h_{1,1})=0$ is a contradiction.

If $sig(k_0)=\textsf {used}$ then $sup(h_{0,6})=0$ and $sup(h_{0,1})=sup(h_{0,7})=sup(h_{1,1})=sup(h_{2,1})=1$. This implies $sig(o)\in \{\textsf {out},\textsf {set}\}$ and $sup(h_{0,3})=1$. Thus, by $sup(h_{0,6})=0$ there is an event $e\in \{k_1,z_0,z_1\}$ with $sig(e)=\textsf {inp}$. By $sup(h_{1,1})=sup(h_{2,1})=1$, we have $e\not \in \{z_0,z_1\}$. If $sig(k_1)=\textsf {inp}$ then $sup(h_{4,1})=0$ and $sup(h_{4,2})=1$, implying $sig(z_0)\in \{\textsf {out},\textsf {set}\}$ and $sup(h_{0,6})=1$. This is a contradiction. Altogether, this proves $sig(k_0)\not \in \{\textsf {inp},\textsf {used},\textsf {free}\}$.

Consequently, we obtain $sig(k_0)=\textsf {out}$ and $sup(h_{0,6})=1$ which implies $sig(o)=\textsf {inp}$ and $sup(h_{0,3})=0$. By $sup(h_{0,6})=1$, this implies that there is an event $e\in \{k_1,z_0,z_1\}$ with $sig(e)\in \{\textsf {out}, \textsf {set}\}$. Again by $sig(k_0)=\textsf {out}$, we have $sup(h_{1,1})=sup(h_{2,1})=0$, which implies $e=k_1$. The signatures $sig(k_0)=\textsf {out}$ and $sig(k_1)\in \{\textsf {out},\textsf {set}\}$ and the construction of $T_0,\dots , T_{m-1}$ and $B_0,\dots , B_{m-1}$ ensure that $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {inp}\}$ is a one-in-three model of $\varphi$: By $sig(k_0)=\textsf {out}$ and $sig(k_1)\in \{\textsf {out},\textsf {set}\}$ we have $sup(t_{i,1})=1$ and $sup(t_{i,4})=sup(b_{i,1})=0$ for all $i\in \{0,\dots , m-1\}$. By $sup(t_{i,1})=1$ and $sup(t_{i,4})=0$, there is an event $X\in \zeta _i$ such that $sig(X)=\textsf {inp}$. Moreover, by $sup(b_{i,1})=0$, we get $sig(X_i)\not \in \textsf {enter}$ for all $i\in \{0,\dots , m-1\}$. Thus, X is unambiguous and thus M is a searched model.

Conversely, if M is a one-in-three model of $\varphi$ then there is a $\tau$-region (sup, sig) that solves $\alpha$. The red colored area above sketches states with a positive support. Which states of $T_{i}$, besides of $t_{i,0}, t_{i,1}$ and $t_{i,5}$, get a positive support depends for all $i\in \{0,\dots , m-1\}$ on which of $X_{i_0}, X_{i_1},X_{i_2}$ belongs to M. The red colored area above sketches the case $X_{i_0}\in M$. Moreover, we define $sup(s)=1$ if $s=b_{i,0}$ and $X_i\in M$ or if $s\in \bot$. The signature is defined as follows: $sig(k_1)=\textsf {set}$; for all $e\in E(A^\tau _\varphi )\setminus \{k_1\}$ and all , if $sup(s')>sup(s)$, then $sig(e)=\textsf {out}$; if $sup(s)>sup(s')$, then $sig(e)=\textsf {inp}$; else $sig(e)=\textsf {nop}$.    $\square$

Theorem 5

For any $g\ge 1$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau =\{\textsf {nop}, \textsf {inp}, \textsf {set}, \textsf {free}\}$ or $\tau =\{\textsf {nop}, \textsf {inp}, \textsf {set}, \textsf {used}, \textsf {free}\}$ or $\tau =\{\textsf {nop}, \textsf {out}, \textsf {res}, \textsf {used}\}$ or $\tau =\{\textsf {nop}, \textsf {out}, \textsf {res}, \textsf {used}, \textsf {free}\}$.

Proof

Our reduction proves the claim for $\tau =\{\textsf {nop}, \textsf {inp}, \textsf {set}, \textsf {free}\}$ and $\tau =\{\textsf {nop}, \textsf {inp}, \textsf {set}, \textsf {used}, \textsf {free}\}$ and thus by Lemma 2, for the other types, too.

The TS $A^\tau _\varphi$ has the following gadgets $H_0$ and $H_1$ providing the atom $(k_0,h_{0,3})$:

For all $i\in \{0,\dots , m-1\}$, the $A^\tau _\varphi$ for the clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ and the variable $X_i\in V(\varphi )$ the gadgets $T_i$ and $B_i$ as previously defined for Theorem 4. The gadgets $H_0$ and $H_1$ ensure that a $\tau$-region (sup, sig) solving $(k_0,h_{0,3})$ satisfies $sig(k_0)=\textsf {free}$ and $sig(k_1)=\textsf {set}$. This implies $sup(t_{i,1})=1$ and $sup(t_{i,4})=sup(b_{i,2})=0$ for all $i\in \{0,\dots , m-1\}$. By $sup(t_{i,1})=1$ and $sup(t_{i,4})=0$, there is an event $X\in \zeta _i$ such that $sig(X)=\textsf {inp}$ and, by $sup(b_{i,2})=0$ for all $i\in \{0,\dots , m-1\}$, we have $sig(X)\not =\textsf {set}$ for all $X\in V(\varphi )$. Thus, the event $X\in \zeta _i$ is unique and $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {inp}\}$ is a one-in-three model.

We briefly argue that $H_0$ and $H_1$ perform as announced: Let (sup, sig) be a $\tau$-region that solves $\alpha$. If $sig(k_0)=\textsf {inp}$ then $sup(h_{1,1})=0$ and $sup(h_{1,2})=1$ which implies $sig(z_0)=\textsf {set}$ and thus $sup(h_{0,3})=1$, a contradiction. Hence, $sig(k_{0})\not =\textsf {inp}$. If $sig(k_0)=\textsf {used}$ then $sup(h_{0,1})=sup(h_{1,2})=1$ and $sup(h_{0,3})=0$. Consequently, $sig(z_0)=\textsf {inp}$ or $sig(k_1)=\textsf {inp}$ but this contradicts $sup(h_{1,2})=1$ and $sup(h_{0,3})=0$. Thus, $sig(k_0)\not =\textsf {used}$. Thus, we have $sig(k_0)=\textsf {free}$ and $sup(h_{0,3})=1$, which implies that one of $k_1,z_0$ has a $\textsf {set}$-signature. By $sig(k_0)=\textsf {free}$, we get $sup(h_{1,3})=0$ and thus $sig(k_1)=\textsf {set}$.

If M is a one-in-three model of $\varphi$ then we can define an $\alpha$ solving region similar to the one of Theorem 4, where we replace $sig(k_0)=\textsf {inp}$ by $sig(k_0)=\textsf {free}$.

Theorem 6

For any fixed $g\ge 1$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau =\{\textsf {nop}, \textsf {inp},\textsf {res},\textsf {swap}\}\cup \omega$ or $\tau =\{\textsf {nop}, \textsf {out},\textsf {set},\textsf {swap}\}\cup \omega$ and $\omega \subseteq \{\textsf {used},\textsf {free}\}$.

Proof

The TS $A^\tau _\varphi$ has the following gadgets $H_0, H_1, H_2$ and $H_3$:

The gadgets $H_0,\dots , H_3$ provide the atom $\alpha =(k,h_{0,2})$ and ensure that a $\tau$-region (sup, sig) solving $\alpha$ satisfies $sig(k)=\textsf {inp}$ and $sup(h_{0,2})=0$. The TS $A^\tau _\varphi$ has the following gadgets $F_0, F_1$ and for all $j\in \{0,\dots , 10\}$ the gadget $G_j$:

For all $j\in \{0,\dots , 10\}$, the gadgets $F_0,F_1, G_j$ ensure $sig(u_j)=\textsf {swap}$ for any $\tau$-region (sup, sig) solving $\alpha$.

For all $i\in \{0,\dots , m-1\}$, the TS $A^\tau _\varphi$ has for the clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ some gadgets $T_{i,0},\dots , T_{i,6}$ and $B_i$. The purpose of these gadgets is to make the one-and-three satisfiability of $\varphi$ and the solvability of $\alpha$ the same. In particular, the TS $T_{i,0}$ is defined by:

The gadgets $T_{i,1}, T_{i,2}$ and $T_{i,3}$ are defined (in this order) as follows:

Moreover, the gadgets $T_{i,4}, T_{i,5}$ and $T_{i,6}$ are defined like this:

Finally, the gadget $B_i$ is defined as follows:

Let (sup, sig) be a $\tau$-region solving $\alpha$. We first argue that the gadgets $H_0,\dots ,H_3$ and $F_0,F_1$ and $G_0,\dots , G_{10}$ ensure that a $\tau$-region (sup, sig) solving $\alpha$ satisfies $sig(k)=\textsf {inp}$, $sup(h_{0,2})=0$ and $sig(u_0)=\dots =sig(u_{10})=\textsf {swap}$.

If $sig(k)=\textsf {free}$ and $sup(h_{0,2})=1$ then implies $sup(s)=sup(s')=0$. Especially, by $sup(h_{0,1})=0$ and $sup(h_{0,2})=1$ we have $sig(y_0)=\textsf {swap}$. Moreover, by $sup(h_{2,1})=sup(h_{2,4})=0$ and $sig(y_0)=\textsf {swap}$ we have that $sup(h_{2,2})=sup(h_{2,3})=1$. This implies $sig(y_1)\in \{\textsf {nop},\textsf {used}\}$. By $sup(h_{1,1})=0$ and this implies $sig(y_1)=\textsf {nop}$ and thus $sup(h_{1,2})=0$. Furthermore, by $sup(h_{1,2})=sup(h_{1,3})=0$ and this implies $sig(y_0)\not =\textsf {swap}$, a contradiction. Thus, we have $sig(k)\not =\textsf {free}$.

If $sig(k)=\textsf {used}$ and $sup(h_{0,2})=0$ then implies $sup(s)=sup(s')=1$. Thus, we get $sup(h_{0,1})=sup(h_{0,3})=sup(h_{1,3})=1$ which with $sup(h_{0,2})=0$ implies $sig(y_0)=sig(v)=\textsf {swap}$. Moreover, $sup(h_{1,3})=1$ and $sig(y_0)=\textsf {swap}$ imply $sup(h_{1,2})=0$. By $sup(h_{1,1})=1$, this implies $sig(y_1)\in \{\textsf {inp},\textsf {res}\}$. Finally, $sup(h_{3,3})=1$ and $sig(v)=sig(y_0)=\textsf {swap}$ imply $sup(h_{3,1})=1$. This contradicts $sig(y_1)\in \{\textsf {inp},\textsf {res}\}$. Thus, $sig(k)\not =\textsf {used}$. Altogether, this shows that $sig(k)=\textsf {inp}$ and $sup(h_{0,2})=0$, which implies $sig(v)=\textsf {swap}$.

By $sig(k)=\textsf {inp}$ we have $sup(f_{0,1})=sup(f_{1,1})=sup(g_{j,1})=0$ and $sup(f_{0,3})=sup(f_{1,3})=sup(g_{j,4})=1$. By $sig(v)=\textsf {swap}$, this implies $sup(f_{0,2})=sup(f_{1,2})=0$ and thus $sig(z_0), sig(z_1)\in \{\textsf {nop},\textsf {res},\textsf {free}\}$. Moreover, $sup(g_{j,1})=0$, $sup(g_{j,4})=1$ and $sig(z_0), sig(z_1)\in \{\textsf {nop},\textsf {res},\textsf {free}\}$ imply $sup(g_{j,2})=0$ and $sup(g_{j,3})=1$ and thus $sig(u_j)=\textsf {swap}$.

Let $i\in \{0,\dots ,m-1\}$. We now show that $T_{i,0},\dots , T_{i,6}$ and $B_i$ collaborate as announced. By $sig(k)=\textsf {inp}$ and $sig(u_{9})=sig(u_{10})=\textsf {swap}$, we have $sup(b_{i,1})=1$ for all $i\in \{0,\dots , m-1\}$. Since for all $i\in \{0,\dots , m-1\}$, the gadget $B_i$ ensures for all $X\in V(\varphi )$ that and $sup(s)\not =sup(s')$ imply $sig(X)=\textsf {swap}$.

The gadget $T_{i,0}$ works like this: By $sig(k)=\textsf {inp}$ we get that $sup(t_{i,0,1})=0$ and $sup(t_{i,0,8})=1$. Consequently, the image of the path performs an odd number of state changes between 0 to 1 in $\tau$. Since $sig(u_0)=\dots =sig(u_3)=\textsf {swap}$, the events $u_0,\dots , u_3$ perform an even number of state changes. Thus, either all of $X_{i_0}, X_{i_1}, X_{i_2}$ are mapped to $\textsf {swap}$ or exactly one of them. The construction of $T_{i,1},\dots , T_{i,6}$ guarantees that there is exactly one variable event mapped to $\textsf {swap}$.

In particular, the gadgets $T_{i,4}, T_{i,5}$ and $T_{i,6}$ ensure that if $e\in \{w_{3i}, w_{3i+1} ,w_{3i+2}\}$ then $sig(e)\not \in \{\textsf {nop},\textsf {used}\}$. We argue for $w_{3i}$: By $sig(k)=\textsf {inp}$ we get $sup(t_{i,4,1})=0$ and $sup(t_{i,4,4})=1$ which, by $sig(u_7)=sig(u_8)=\textsf {swap}$, implies $sup(t_{i,4,2})=1$ and $sup(t_{i,4,3})=0$. Clearly, this implies $sig(w_{3i})\not \in \{\textsf {nop},\textsf {used}\}$. Similarly, we obtain that $sig(w_{3i+1})\not \in \{\textsf {nop},\textsf {used}\}$ and $sig(w_{3i+2})\not \in \{\textsf {nop},\textsf {used}\}$.

Finally, the gadgets $T_{i,1}, T_{i,2}$ and $T_{i,3}$ ensure that no two variable events of the same clause can have a swap signature: By $sig(k)=\textsf {inp}$ we get that $sup(t_{i,1,1})=0$ and $sup(t_{i,1,7})=1$ which with $sig(u_4)=sig(u_5)=sig(u_6)=\textsf {swap}$ implies $sup(t_{i,1,3})=0$ and $sup(t_{i,1,6})=0$. Thus, if $sig(X_{i_0})=sig(X_{i_1})=\textsf {swap}$ then $sup(t_{i,1,4})=sup(t_{i,1,5})=1$ which implies $sig(w_{3i})\in \{\textsf {nop}, \textsf {used}\}$, a contradiction. Similarly, one uses $T_{i,2}$ and $T_{i,3}$ to show that neither $X_{i_0}$ and $X_{i_2}$ nor $X_{i_1}$ and $X_{i_2}$ can simultaneously be mapped to $\textsf {swap}$. As i was arbitrary, there is exactly one variable per clause that is mapped to $\textsf {swap}$. Thus, $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {swap}\}$ is a one-in-three model of $\varphi$.

Conversely, a one-in-three model M of $\varphi$ allows a $\tau$-region (sup, sig) that solves $\alpha$: The red colored area above indicates which states of $H_0,\dots , H_3$, $F_0,F_1$, $G_0,\dots , G_{10}$ and $T_{0,4},T_{0,5},T_{0,6},\dots , T_{m-1,4},T_{m-1,5},T_{m-1,6}$ have positive support. Moreover, we define $sup(s)=1$ for all $s\in \bot$. Which states of $T_{i,0}, \dots , T_{i,3}$, where $i\in \{0,\dots , m-1\}$, besides of k’s sources get a positive support depends on which of $X_{i_0}, X_{i_1}, X_{i_2}$ belongs to M. The red colored area sketches the situation for $X_{i_0}\in M$. It is easy to see that there is for all $e\in E(A^\tau _\varphi )$ a fitting sig-value making (sup, sig) a (solving) $\tau$-region where $sig(k)=\textsf {inp}$ and $sup(h_{0,2})=0$.    $\square$

Theorem 7

For any fixed $g\ge 1$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau =\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {swap}\}\cup \omega$ and $\omega \subseteq \{\textsf {out},\textsf {res},\textsf {used},\textsf {free}\}$ or if $\tau =\{\textsf {nop},\textsf {out},\textsf {res},\textsf {swap}\}\cup \omega$ and $\omega \subseteq \{\textsf {inp},\textsf {set},\textsf {used},\textsf {free}\}$.

Proof

We present the reduction for the types built by $\tau =\{\textsf {nop}, \textsf {inp},\textsf {set},\textsf {swap}\}\cup \omega$ where $\omega \subseteq \{\textsf {out}, \textsf {res}, \textsf {used}, \textsf {free}\}$. Again, the other types are covered by Lemma 2.

The TS $A^\tau _\varphi$ has the following gadgets $H_0,H_1,H_2$ and $H_3$:

If $\tau \cap \{\textsf {used},\textsf {free}\}\not =\emptyset$ then $A^\tau _\varphi$ has also the following gadgets $H_4,\dots , H_{12}$:

The gadgets $H_0,\dots , H_3$ ($H_4,\dots , H_{12}$, if added) provide $\alpha =(k,h_{3,3})$. They ensure that if (sup, sig) $\tau$-solves $\alpha$, then $sig(k)\in \{\textsf {inp},\textsf {out}\}$. The TS $A^\tau _\varphi$ adds the following gadgets $F_0,F_1,F_2$ and, for all $i\in \{0,\dots , 13\}$, the gadgets $G_i, N_i$:

The gadgets $F_0,F_1,F_2$ and $G_0,N_0,\dots , G_{13}, N_{13}$ guarantee that if (sup, sig) solves $\alpha$ then $sig(u_i)=\textsf {swap}$. Similarly to the reduction of Theorem 6, the TS $A^\tau _\varphi$ has for every $i\in \{0,\dots , m-1\}$ gadgets $T_{i,0},\dots , T_{i,6}$ and $B_i$ to make the one-in-three satisfiability of $\varphi$ and the $\tau$-solvability of $\alpha$ the same. These gadgets and the ones for Theorem 6 have basically the same intention. However, since the current types have different interactions, the peculiarity of these gadgets is slightly different. In particular, $A^\tau _\varphi$ has for each clause $\zeta _i=\{X_{i_0}, X_{i_1}, X_{i_2}\}$ the following gadget $T_{i,0}$:

Moreover, the gadgets $T_{i,1}, T_{i,2}$ and $T_{i,3}$ are defined as follows:

Furthermore, the gadgets $T_{i,4}, T_{i,5}$ and $T_{i,6}$ are defined by

Finally, the TS $A^\tau _\varphi$ has for all $i\in \{0,\dots , m-1\}$ the following gadget $B_i$:

We briefly argue for the announced functionality of the gadgets. Let (sup, sig) be a $\tau$-region solving $\alpha$. If $sig(k)=\textsf {free}$ then $sup(h_{3,3})=1$ and implies $sup(s)=sup(s')=0$. Since $sup(h_{3,1})=0$ and $sup(h_{3,3})=1$, there is an event $e\in \{v_0,v_1\}$ such that $sig(e)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$. If $sig(v_0)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$, then, by $sup(h_{1,1})=0$, we get $sig(v_{0})=\textsf {swap}$. Moreover, if $sig(v_1)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$, which implies $sig(h_{3,2})=1$, then, by $sup(h_{2,3})=0$, we get $sig(v_1)=\textsf {swap}$. By $sig(v_1)=\textsf {swap}$ and $sup(h_{2,3})=0$, we get $sup(h_{2,2})=1$. By $sup(h_{1,1})$, this implies $sig(v_0)=\textsf {swap}$. Thus, in any case we get $sig(v_0)=\textsf {swap}$. By $sig(v_0)=\textsf {swap}$ and $sup(h_{4,3})=sup(h_{5,1})=0$ we obtain $sup(h_{4,2})=sup(h_{5,2})=1$ which implies $sig(x)=\textsf {swap}$. Using this and $sup(s)=sup(s')=0$ if , we have that $sup(h_{j,2})=1$ for all $j\in \{6,\dots , 11\}$. This implies $sig(y_0)=sig(y_1)=sig(y_2)=\textsf {swap}$. By $sup(h_{12,1})=sup(h_{12,4})=0$, the image of is a path from 0 to 0 in $\tau$. The number of state changes between 0 and 1 on such a path is even. This contradicts $sig(y_0)=sig(y_1)=sig(y_2)=\textsf {swap}$. Thus, $sig(k)\not =\textsf {free}$. The assumption that $sig(k)=\textsf {used}$ and $sup(h_{3,3})=0$ yields a contradiction, too.

We conclude that $sig(k)=\textsf {inp}$ and $sup(h_{3,3})=0$. This implies $sig(v_0)\not \in \{\textsf {out},\textsf {set}\}$ and if , then $sup(s)=1$ and $sup(s')=0$. Thus, by $sup(h_{2,1})=0$ and $sup(h_{2,3})=1$ there is an event $e\in \{v_0,v_1\}$ such that $sig(e)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$. If $e=v_0$ then $sig(v_0)=\textsf {swap}$. Moreover, if $e=v_1$ then $sup(h_{3,2})=1$ which with $sup(h_{3,3})=0$ and $sup(h_{1,1})=1$ implies $sig(v_0)=\textsf {swap}$. Consequently, any case implies $sig(v_0)=\textsf {swap}$. This results in $sig(u_j)=\textsf {swap}$ for all $j\in \{0,\dots , 13\}$ as follows. By $sup(f_{0,3})=sup(f_{1,3})=1$ and $sig(v)=\textsf {swap}$ we obtain $sup(f_{0,2})=sup(f_{1,2})=0$ which with $sup(f_{0,1})=sup(f_{1,1})=0$ implies $sig(z_0), sig(z_1)\in \{\textsf {nop},\textsf {res},\textsf {free}\}$. Moreover, by $sig(z_0), sig(z_1)\in \{\textsf {nop},\textsf {res},\textsf {free}\}$ and $sup(f_{2,1})=0$ we get $sup(f_{2,3})=0$ which with $sup(f_{2,4})=1$ implies $sig(z_2)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$. By $sig(z_0)\in \{\textsf {nop},\textsf {res},\textsf {free}\}$ and $sup(g_{i,1})=0$, we get $sup(g_{i,2})=0$. Furthermore, $sig(z_1)\in \{\textsf {nop},\textsf {res},\textsf {free}\}$ and $sup(g_{i,4})=1$ yields $sig(z_1)=\textsf {nop}$ and $sup(g_{i,3})=1$. This implies $sig(u_i)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$. Finally, by $sup(n_{i,1})=0$ and $sig(z_2)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$, we get $sup(n_{i,1})=1$ and, by $sup(n_{i,4})=1$ and $sig(v_0)=\textsf {swap}$, we have $sup(n_{i,3})=0$. Since $sig(u_i)\in \{\textsf {out},\textsf {set},\textsf {swap}\}$, this yields $sig(u_i)=\textsf {swap}$ for all $i\in \{0,\dots , 13\}$.

The gadgets $T_{i,0},\dots , T_{i,6}$, where $i\in \{0,\dots , m-1\}$, use $sig(k)=\textsf {inp}$ and $sig(u_i)=\textsf {swap}$ for all $i\in \{0,\dots , 13\}$ similarly to the ones of Theorem 6 to ensure that $M=\{X\in V(\varphi ) \mid sig(X)=\textsf {swap}\}$ is a one-in-three model of $\varphi$: By $sup(t_{i,4,6})=sup(t_{i,5,6})=sup(t_{i,6,6})=1$ and $sig(u_{5})=sig(u_{6})=\textsf {swap}$ we have $sup(t_{i,4,4})=sup(t_{i,5,4})=sup(t_{i,6,4})=1$ for all $i\in \{0,\dots , m-1\}$. Thus, if $X\in V(\varphi )$, and $sup(s)\not =sup(s')$ then $sig(X)=\textsf {swap}$. Using this, one argues in a manner quite similar to that of the proof of Theorem 6 that $T_{i,0},\dots , T_{i,6}$ collaborate in such a way that there is exactly one variable event $X\in \{X_{i_0}, X_{i_1}, X_{i_2}\}$ such that $sig(X)=\textsf {swap}$. Thus, M is a corresponding model. Moreover, if $sig(k)=\textsf {out}$ and $sup(h_{3,3})=1$ then we obtain again that $sig(u_i)=\textsf {swap}$ for all $i\in \{0,\dots , 13\}$ which also guarantees that M is a searched model.

Conversely, if M is a one-in-three model of $\varphi$ then we can define analogously to Theorem 6 a $\tau$-region solving $\alpha$.    $\square$

Theorem 8

([12]). For any fixed $g\ge 1$, deciding if a g-bounded TS A is $\tau$-solvable is NP-complete if $\tau \in \{\textsf {nop},\textsf {inp},\textsf {out}\}\cup \{\textsf {used},\textsf {free}\}$.

Proof

The claim follows directly from our result of [12]. There we use 1-bounded cycle free gadgets to prove that synthesis of (pure) b-bounded Petri nets is NP-complete. The joining of [12] yields a 2-bounded TS. However, it is easy to see that the 1-bounded joining of this paper fits, too. The (pure) 1-bounded Petri net type is isomorphic to $\{\textsf {nop},\textsf {inp},\textsf {out},\textsf {used}\}$ ($\{\textsf {nop},\textsf {inp},\textsf {out}\}$). By symmetry, $\tau$-solving ESSP atoms by $\textsf {used}$ is equivalent to solving them by $\textsf {free}$.    $\square$

4 Polynomial Time Results

Theorem 9

For any fixed $g < 2$, one can decide in polynomial time if a g-bounded TS A is $\tau$-solvable if $\tau =\{\textsf {nop},\textsf {inp},\textsf {set}\}$ or $\tau =\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {used}\}$ or $\tau =\{\textsf {nop},\textsf {out},\textsf {res}\}$ or $\tau =\{\textsf {nop},\textsf {out},\textsf {res},\textsf {free}\}$ or $\tau =\{\textsf {nop},\textsf {set},\textsf {res}\}\cup \omega$ with non-empty $\omega \subseteq \{\textsf {inp},\textsf {out},\textsf {used},\textsf {free}\}$.

Proof

If A is $\tau$-solvable then no event e of A occurs twice in a row. Otherwise, the SSP atom $(s',s'')$ of a sequence is not $\tau$-solvable. Thus, in what follows, we assume that A has no event occurring twice in a row. Moreover, it is easy to see that a 1-bounded TS is a simple directed path on pairwise distinct states $s_0,\dots , s_m$ or a directed cycle, that is, all states $s_0,\dots , s_m$ except $s_0$ and $s_m$ are pairwise distinct. This proof proceeds as follows. First, we assume that $\tau =\{\textsf {nop},\textsf {inp},\textsf {set}\}$ and that A is a directed cycle and argue that the $\tau$-solvability of a given ESSP atom (k, s) or a SSP atom $(s,s')$ of A is decidable in polynomial time. Secondly, we argue that the presented algorithmic approach is applicable to directed paths, too. Thirdly, we show that the procedure introduced for $\{\textsf {nop},\textsf {inp},\textsf {set}\}$ can be extended to $\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {used}\}$. By Lemma 2, this proves the claim for $\{\textsf {nop},\textsf {out},\textsf {res}\}$ and $\{\textsf {nop},\textsf {out},\textsf {res},\textsf {free}\}$, too. After that we investigate the case where $\tau =\{\textsf {nop},\textsf {set},\textsf {res}\}\cup \omega$ with non-empty $\omega \subseteq \{\textsf {inp},\textsf {out},\textsf {used},\textsf {free}\}$. We argue that it is sufficient to decide the $\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}$- and $\{\textsf {nop},\textsf {res},\textsf {set},\textsf {used}\}$-solvability of A and that this is doable in polynomial time. The corresponding procedures again modify those introduced for $\{\textsf {nop},\textsf {inp},\textsf {set}\}$.

Fig. 6.

A sketch of a cyclic 1-bounded input A with ESSP atom $\alpha =(k,s^i_j)$.

Let $\tau =\{\textsf {nop},\textsf {inp},\textsf {set}\}$ and A be 1-bounded (cycle) TS with event $k\in E(A)$ that occurs m times. Since A is a cycle, we can assume that k occurs at A’s initial state: . Moreover, since k does not occur twice in a row, its occurrences partition A into m k-free subsequences $I_0,\dots , I_{m-1}$ such that , $i\in \{0,\dots , m-1\}$, and $s^{m-1}_{n_{m-1}}=\iota$, cf. Fig. 6.

Obviously, defining $sup(\iota )=1$, $sig(k)=\textsf {inp}$ and $sig(e)=\textsf {set}$ for all $e\in E(A)\setminus \{k\}$ inductively yields a region (sup, sig) solving the ESSP atoms (k, s) where . Thus, it remains to consider the case . Since , there is an $i\in \{0,\dots , m-1\}$ such that s is a state of the i-th subsequence $I_i$. In particular, there is a $j\in \{1,\dots ,n_i-1 \}$ such that $s=s^i_j$. The state $s^i_j$ divides $I_i$ into the sequences and , cf. Fig. 6.

If (sup, sig) is a region that solves $\alpha$ then $sig(k)=\textsf {inp}$ and $sup(s^i_j)=0$ is true. This implies for all $\ell \in \{0,\dots , m-1\}$ that $sup(s^\ell _0)=0$ and $sup(s^\ell _{n_\ell })=1$. Thus, it remains to define the signature of the events of $\bigcup _{\ell =0}^{m-1}E(I_\ell )$ such that , for all $\ell \in \{0,\dots , m-1\}\setminus \{i\}$, and and .

If there is for all $\ell \in \{0,\dots ,m-1\}\setminus \{i\}$ an event $e_\ell \in E(I_\ell )$ such that $e_\ell \not \in E(I^0_i)$ and if there is an event $e_i\in E(I^1_i)$ so that $e_i\not \in E(I^0_i)$ then $sup(\iota )=1$, $sig(k)=\textsf {inp}$, $sig(e_\ell )=\textsf {set}$ for all $\ell \in \{0,\dots , m-1\}$, and $sig(e)=\textsf {nop}$ for all $e\in E(A)\setminus \{k,e_0,\dots ,e_\ell \}$ yields a $\tau$-region (sup, sig) of A that solves $\alpha$. Clearly, whether A satisfies this property is decidable in polynomial time.

Otherwise, there is a sequence $I\in \{I_0,\dots , I_{i-1}, I^1_i,I_{i+1},\dots , I_{m-1}\}$ so that $E(I)\subseteq E(I^0_i)$. Thus, if (sup, sig) is a $\tau$-region that solves $\alpha$ then there is a $\ell \in \{1,\dots , j-1\}$ such that $sig(y^i_{\ell })=\textsf {set}$. Consequently, there has to be a $\ell ' \in \{\ell +1,\dots , j\}$ such that $sig(y^i_{\ell '})=\textsf {inp}$ and, in particular, $sig(y^i_{\ell ''})=\textsf {nop}$ for all $\ell ''\in \{\ell '+1,\dots ,j\}$. Using this, one finds that (sup, sig) implies a region $(sup',sig')$ that solves $\alpha$ and gets along with at most two $\textsf {inp}$-events. More exactly, defining $sup'(\iota )=1$, $sig'(k)=sig'(y^i_{\ell '})=\textsf {inp}$, $sig'(e)=\textsf {nop}$ for all $e\in \{y^i_{\ell '+1}, \dots , y^i_{j}\}$ and $sig'(e)=\textsf {set}$ for all $e\in E(A)\setminus (\{k,y^i_{\ell '},\dots , y^i_j\})$ yields a valid $\tau$-region $(sup', sig')$ that solves $\alpha$. Since (sup, sig) was arbitrary, these deliberations show that in the second case the atom $\alpha$ is $\tau$-solvable if and only if there is a corresponding region $(sup',sig')$. This yields the following polynomial procedure that decides whether $\alpha$ is $\tau$-solvable: For $\ell$ from j to 2 test if $(sup_\ell , sig_\ell )$ (inductively) defined by $sup_\ell (\iota )=1$, $sig_\ell (y^i_\ell )=\textsf {inp}$, $sig_\ell (y^i_{\ell '})=\textsf {nop}$ for all $\ell '\in \{\ell +1,\dots , j\}$ and $sig_\ell (e)=\textsf {set}$ for all $e\in E(A)\setminus (\{k,y^i_{\ell },\dots , y^i_j\})$ yields a $\tau$-region of A. If the test succeeds for any iteration then return yes, otherwise return no.

We can modify this approach to test the $\tau$-solvability of an SSP atom $\beta =(s,s')$ as follows. Since is a cycle we can assume without loss of generality that $s=\iota$ and $s'=s_i$ for some $i\in \{1,\dots , m-1\}$. The states $\iota$ and $s_i$ partition A into two subsequences and . If $\beta$ is solvable by a region $(sup', sig')$ such that $sup'(\iota )=1$ and $sup'(s_i)=0$ then there is an event $e\in I_0$ such that $sig(e)=\textsf {inp}$. In particular, there is a region (sup, sig) as follows: $sup(\iota )=1$, $sig(e_j)=\textsf {inp}$ and $j\in \{1,\dots , i\}$, $sig(e_{\ell })=\textsf {nop}$ for all $\ell \in \{j+1,\dots , i\}$ and $sig(e)=\textsf {set}$ for all $e\in E(A)\setminus \{e_j,\dots , e_i\}$. Similar to the approach for $\alpha$, we can check if such a region exists in polynomial time. Moreover, the case where $sup(\iota )=0$ and $sup(s_i)=1$ works symmetrically.

So far we have shown that the $\tau$-solvability of (E)SSP atoms of A are decidable in polynomial time if A is a cycle. If is a directed path then its cycle extension $A_c$ has a fresh event $\oplus \not \in E(A)$ and is defined by . The event $\oplus$ is unique thus an (E)SSP atom of A is solvable by a $\tau$-region of A if and only if it is solvable by a $\tau$-region of $A_c$. Thus, we can decide the solvability of atoms of A via $A_c$. Altogether, this proves that the $\tau$-solvability of (E)SSP atoms of 1-bounded inputs is decidable in polynomial time. Since we have at most $\vert S\vert ^2+\vert E\vert \cdot \vert S\vert$ atoms to solve, the decidability of the $\{\textsf {nop},\textsf {inp},\textsf {set}\}$-solvability for 1-bounded TS is polynomial.

Similar to the discussion for $\tau =\{\textsf {nop},\textsf {inp},\textsf {set}\}$, one argues that the following assertion is true: If $\tau =\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {used}\}$ then there is a $\tau$-region $(sup', sig')$ with $sig'(k)=\textsf {used}$ and $sup(s^i_j)=0$ if and only if there is a $\tau$-region (sup, sig) and an number $\ell \in \{1,\dots , j\}$ such that $sup(\iota )=1$, $sig(k)=\textsf {used}$, $sig(y^i_\ell )=\textsf {inp}$, $sig(y^i_{\ell '})=\textsf {nop}$ for all $\ell '\in \{\ell +1,\dots ,j\}$ and $sig(e)=\textsf {set}$ for all $e\in E(A)\setminus \{k,y^i_\ell ,\dots , y^i_j\}$. Clearly, the procedure introduced for $\{\textsf {nop},\textsf {inp},\textsf {set}\}$ can be extended appropriately to a procedure that works for $\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {used}\}$.

It remains to investigate the case where $\tau =\{\textsf {nop},\textsf {res},\textsf {set}\}\cup \omega$ with non-empty $\omega \subseteq \{\textsf {inp},\textsf {out},\textsf {used},\textsf {free}\}$. For a start, let’s argue that deciding the $\tau$-solvability is equivalent to deciding the $\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}$-solvability or the $\{\textsf {nop},\textsf {res},\textsf {set},\textsf {used}\}$-solvability of A. This can be seen as follows: If (sup, sig) is a region that solves an ESSP atom $\alpha =(k,s)$ such that $sig(k)=\textsf {inp}$ then there is a $\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}$-region $(sup', sig')$ that solves (k, s), too. The region $(sup',sig')$ originates from (sup, sig) by $sup'=sup$, $sig'(k)=\textsf {inp}$ and for all $e\in E(A)\setminus \{k\}$ by $sig'(e)=\textsf {nop}$ if $sig(e)\in \{\textsf {nop},\textsf {used},\textsf {free}\}$, $sig'(e)=\textsf {res}$ if $sig(e)\in \{\textsf {inp},\textsf {res}\}$ and, finally, $sig'(e)=\textsf {set}$ if $sig(e)\in \{\textsf {out},\textsf {set}\}$. Similarly, one argues that $\alpha$ is $\tau$-solvable such that $sig(k)=\textsf {out}$ if and only if it is $\{\textsf {nop},\textsf {out},\textsf {res},\textsf {set}\}$-solvable. Moreover, $\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}$ and $\{\textsf {nop},\textsf {out},\textsf {res},\textsf {set}\}$ are isomorphic thus $\tau$-solvability with $\textsf {inp}$ or $\textsf {out}$ reduces to $\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}$-solvability. Similarly, the $\tau$-solvability with $\textsf {used}$ or $\textsf {free}$ reduces to $\{\textsf {nop},\textsf {res},\textsf {set}, \textsf {used}\}$-solvability. It is easy to see that the procedure introduced for $\{\textsf {nop},\textsf {inp},\textsf {set}\}$ can be extended to the types $\{\textsf {nop},\textsf {inp},\textsf {res},\textsf {set}\}$ and $\{\textsf {nop},\textsf {res},\textsf {set}, \textsf {used}\}$. The only difference is that we now search for an event $y^i_\ell$ such that $sig(y^i_\ell )=\textsf {res}$ instead of $sig(y^i_\ell )=\textsf {inp}$.

Finally, we observe that a SSP atom $\beta =(s,s')$ is $\tau$-solvable if and only if it is $\{\textsf {nop},\textsf {res},\textsf {set}\}$-solvable. The states s and $s'$ induce again a partition $I_0$ and $I_1$ of A and we can adapt the approach above to $\{\textsf {nop},\textsf {res},\textsf {set}\}$.    $\square$

Theorem 10

For any fixed $g\in \mathbb {N}$, deciding whether a g-bounded TS A is $\tau$-solvable is polynomial if one of the following conditions is true:

• 1. $\tau =\{\textsf {nop},\textsf {inp},\textsf {free}\}$ or $\tau =\{\textsf {nop},\textsf {inp},\textsf {used}, \textsf {free}\}$ or $\tau =\{\textsf {nop},\textsf {out},\textsf {used}\}$ or $\tau =\{\textsf {nop},\textsf {out},\textsf {used}, \textsf {free}\}$ and $g < 2$.

• 2. $\tau =\{\textsf {nop}, \textsf {set}, \textsf {res}\} \cup \omega$ and $\emptyset \not =\omega \subseteq \{\textsf {used}, \textsf {free}\}$ and $g < 3$.

• 3. $\tau =\tau '\cup \omega$ and $\tau '\in \{\{\textsf {nop},\textsf {set},\textsf {swap}\}, \{\textsf {nop},\textsf {res},\textsf {swap}\}, \{\textsf {nop},\textsf {res},\textsf {set}, \textsf {swap}\} \}$ and $\emptyset \not =\omega \subseteq \{\textsf {used},\textsf {free}\}$ and $g < 2$.

• 4. $\tau \in \{ \{\textsf {nop}, \textsf {inp}\}, \{\textsf {nop},\textsf {inp},\textsf {used}\}, \{\textsf {nop},\textsf {out}\}, \{\textsf {nop},\textsf {out},\textsf {free}\} \}$ or $\tau \in \mathcal {T}=\{\{\textsf {nop},\textsf {set},\textsf {swap}\}, \{\textsf {nop},\textsf {res},\textsf {swap}\}, \{\textsf {nop},\textsf {set},\textsf {res}\}, \{\textsf {nop},\textsf {set},\textsf {res},\textsf {swap}\}\}$,

Proof

• (1):   It is easy to see that A is a loop, or that A is cycle free, since there is an unsolvable SSP atom otherwise. Moreover, if an event e occurs twice consecutively, , then $(s,s')$ is not $\tau$-solvable. Thus, for every $e\in E(A)$ there is a $s\in S(A)$ such that (e, s) has to be solved by $sig(e)=\textsf {inp}$ ($sig(e)=\textsf {out}$) and $sup(s)=0$ ($sup(s)=1$). If e occurs twice on the directed path A then such a region does not exist. On the other hand, A is $\tau$-solvable if every event occurs exactly once. Consequently, A is $\tau$-solvable if and only if it is 1-bounded and every event occurs exactly once.

• (2): Since ESSP atoms of a $\tau$-solvable input A are only solvable by $\textsf {used}$ and $\textsf {free}$, we have that if then . If $s=s''\not =s'$ or if $s,s',s''$ are pairwise distinct then $(s,s')$ is not $\tau$-solvable. This implies . As a result, $\tau$-solvable inputs have the shape

  

  Thus, if the loop erasement $A'$ of A originates from A by erasing all loops , that is, , then deciding the $\tau$-solvability of A reduces to deciding if $A'$ has the $\tau$-SSP and if all ESSP atoms (e, s) with of $A'$ are $\tau$-solvable. This is doable in polynomial time by the approach of Theorem 9.

• (3): Since ESSP atoms of an input A are only solvable by $\textsf {used}$ and $\textsf {free}$, if and $s\not =s'$ then . If and $s,s',s'', s'''$ are pairwise different, then the SSP atom $(s',s''')$ is not $\tau$-solvable. As a consequence, $\tau$-solvable inputs can have at most 3 different states.

• (4): Let $\tau \in \{\{\textsf {nop}, \textsf {inp}\}, \{\textsf {nop},\textsf {inp},\textsf {used}\}\}$. If A is $\tau$-solvable, then for all $e\in E(A)$ holds . Otherwise, $(e,\iota )$ is not $\tau$-solvable. Similarly, if $\tau \in \mathcal {T}$, then ESSP atoms are not $\tau$-solvable thus, every event occurs at $\iota$. A is g-bounded. This implies $\vert E(A)\vert \le g$. Thus, A has at most $2\cdot \vert \tau \vert ^g$ $\tau$-regions. Since g is fixed, $\tau$-synthesis is polynomial by brut-force. By Lemma 2, the claim follows.

   $\square$

5 Conclusion

In this paper, we fully characterize the computational complexity of $\textsf {nop}$-equipped Boolean Petri nets from g-bounded TS for any fixed $g\in \mathbb {N}$. Our results show that if $\tau$-synthesis is hard then it remains hard even for low bounds g. Moreover, they also show that when g becomes very small, sometimes it makes the difference between hardness and tractability, cf. Fig. 1 §1–§3 and §9, but sometimes it does not, cf. Fig. 1 §4–§7. In this sense, the parameter g helps to recognize interactions that contribute to the power of a type. By Theorem 3 and Theorem 9, $\{\textsf {nop}, \textsf {inp},\textsf {set}\}$-synthesis is hard if $g\ge 2$ and tractable if $g < 2$, respectively. By Theorem 5, $\{\textsf {nop},\textsf {inp},\textsf {set},\textsf {free}\}$-synthesis remains hard for all $g \ge 1$. Thus, if restricted to 1-bounded inputs then the test interaction $\textsf {free}$ makes the difference between hardness and tractability of synthesis. Surprisingly enough, by Theorem 9, replacing $\textsf {free}$ by $\textsf {used}$ makes synthesis from 1-bounded TS tractable again. It remains future work, to characterize the computational complexity of synthesis for the remaining 128 types which do not contain $\textsf {nop}$. Moreover, since $\tau$-synthesis generally remains hard even for (small) fixed g, the bound of a TS is ruled out for FPT-algorithms. Future work might be concerned with parameterizing $\tau$-synthesis by the dependence number of the searched $\tau$-net: If $N=(P, T, f, M_0)$ is a Boolean net, $p\in P$ and if the dependence number $d_p$ of p is defined by $d_p=\vert \{ t\in T\mid f(p,t)\not =\textsf {nop}\} \vert$ then the dependence number d of N is defined by $d=\text {max}\{ d_p \mid p\in P\}$. At first glance, d appears to be a promising parameter for FPT-approaches because this parameterization puts the problem into the complexity class XP: Since a $\tau$-region of $A=(S,E,\delta , \iota )$ is determined by $sup(\iota )$ and sig, for each (E)SSP atom $\alpha$ there are at most $2\cdot \vert \tau \vert ^d\cdot \sum _{i=0}^d \left( {\begin{array}{c}\vert E\vert \\ i\end{array}}\right)$ fitting $\tau$-regions solving $\alpha$. Thus, by $\vert \tau \vert \le 8$, $\tau$-synthesis parameterized by d is decidable in $\mathcal {O}(\vert E\vert ^d \cdot \vert S\vert \cdot \text {max}\{\vert S\vert ,\vert E\vert \} )$.