# Edge exploration of temporal graphs

We introduce a natural temporal analogue of Eulerian circuits and prove that, in contrast with the static case, it is NP-hard to determine whether a given temporal graph is temporally Eulerian even if strong restrictions are placed on the structure of the underlying graph and each edge is active at only three times. However, we do obtain an FPT-algorithm with respect to a new parameter called interval-membership-width which restricts the times assigned to different edges; we believe that this parameter will be of independent interest for other temporal graph problems. Our techniques also allow us to resolve two open question of Akrida, Mertzios and Spirakis [CIAC 2019] concerning a related problem of exploring temporal stars.

## Authors

• 3 publications
• 12 publications
• ### Feedback Edge Sets in Temporal Graphs

The classical, linear-time solvable Feedback Edge Set problem is concern...
03/30/2020 ∙ by Roman Haag, et al. ∙ 0

• ### Changing times to optimise reachability in temporal graphs

Temporal graphs (in which edges are active only at specified time steps)...
02/16/2018 ∙ by Jessica Enright, et al. ∙ 0

• ### On Exploring Temporal Graphs of Small Pathwidth

We show that the Temporal Graph Exploration Problem is NP-complete, even...
07/31/2018 ∙ by Hans L. Bodlaender, et al. ∙ 0

• ### The Complexity of Transitively Orienting Temporal Graphs

In a temporal network with discrete time-labels on its edges, entities a...
02/12/2021 ∙ by George B. Mertzios, et al. ∙ 0

• ### Temporal Graph Classes: A View Through Temporal Separators

We investigate the computational complexity of separating two distinct v...
03/01/2018 ∙ by Till Fluschnik, et al. ∙ 0

• ### Sliding Window Temporal Graph Coloring

Graph coloring is one of the most famous computational problems with app...
11/12/2018 ∙ by George B. Mertzios, et al. ∙ 0

• ### How fast can we reach a target vertex in stochastic temporal graphs?

Temporal graphs are used to abstractly model real-life networks that are...
03/08/2019 ∙ by Eleni C. Akrida, et al. ∙ 0

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## 1 Introduction

Many real-world problems can be formulated and modeled in the language of graph theory. However, real-world networks are often not static. They change over time and their edges may appear or disappear (for instance friendships may change over time in a social network). Such networks are called dynamic or evolving or temporal and their structural and algorithmic properties have been the subject of active study in recent years [Akrida2017, ArnaudtempSurvey, Himmel2017, HOLME201297, OthonTempSurvey]. Some of the most natural and most studied topics in the theory of temporal graphs are temporal walks (in which consecutive edges appear at increasing times), paths and corresponding notions of temporal reachability [AKRIDA202065, axiotis2016size, temp_conn_cpts, temporalFeedbackVertexSet, mertzios2019temporal, KEMPE2002820, temp_path_comput, xuan2003computing]. Related to these notions is the study of explorability of a temporal graph which asks whether it is possible to visit all vertices or edges of a temporal graph via some temporal walk.

Temporal vertex-exploration problems (such as temporal variants of the Travelling Salesman problem) have already been thoroughly studied [starexp, OnTemporalGraphExploration, OthonTSP]. In contrast, here we focus on temporal edge-exploration and specifically we study temporally Eulerian graphs. Informally, these are temporal graphs admitting a temporal circuit that visits every edge at exactly one time (i.e. a temporal circuit that yields an Euler circuit in the underlying static graph).

Deciding whether a static graph is Eulerian is a prototypical example of a polynomial time solvable problem. In fact this follows from Euler’s characterization of Eulerian graphs dating back to the 18 century [eulerBridges]. In contrast, here we show that, unless , a characterization of this kind cannot exist for temporal graphs. In particular we show that deciding whether a temporal graph is temporally Eulerian is -complete even if strong restrictions are placed on the structure of the underlying graph and each edge is active at only three times.

The existence of problems that are tractable on static graphs, but -complete on temporal graphs is well-known [starexp, ArnaudtempSurvey, mertzios2019computing, OthonTempSurvey]. In fact there are examples of problems whose temporal analogues remain hard even on trees [starexp, mertzios2019computing]. Thus the need for parameters that take into account the temporal structure of the input is clear. Some measures of this kind (such as temporal variants of feedback vertex number and tree-width) have already been studied [temporalFeedbackVertexSet, temp_tw]. Unfortunately we shall see that these parameters will be of no use to us since the problems we consider here remain -complete even when these measures are bounded by constants on the underlying static graph. To overcome these difficulties, we introduce a new purely-temporal parameter called interval-membership-width. Parameterizing by this measure we find that the problem of determining whether a temporal graph is temporally Eulerian is in .

Temporal graphs of low interval-membership-width are ‘temporally sparse’ in the sense that only few edges are allowed to appear both before and after any given time. We point out that this parameter does not depend on the structure of the underlying static graph, but it is instead influenced only by the temporal structure. We believe that interval-membership-width will be a parameter of independent interest for other temporal graph problems in the future.

It turns out that our study of temporally Eulerian graphs is closely related to a temporal variant of the Travelling Salesman Problem concerning the exploration of temporal stars via a temporal circuit which starts at the center of the star and which visits all leaves. This problem was introduced and proven to be -complete by Akrida, Mertzios and Spirakis on temporal stars in which every edge has at most appearances times for all [starexp]. Although they also showed that the problem is polynomial-time solvable whenever each edge of the input temporal star has at most appearances, they left open the question of determining the hardness of the problem when each edge has at most or appearances. We resolve this open problem in the course of proving our results about temporally Eulerian graphs. Combined with Akrida, Mertzios and Spirakis’ results, this gives a complete dichotomy: their temporal star-exploration problem is in if each edge has at most appearances and is -complete otherwise.

As a potential ‘island of tractability’, Akrida, Mertzios and Spirakis proposed to restrict the input to their temporal star-exploration problem by requiring consecutive appearances of the edges to be evenly spaced (by some globally defined spacing). Using our new notion of interval-membership-width we are able to show that this restriction does indeed yield tractability parameterized by the maximum number of times per edge (thus partially resolving their open problem). Furthermore, we show that a slightly weaker result also holds for the problem of determining whether a temporal graph is temporally Eulerian in the setting with evenly-spaced edge-times.

Outline. We fix notation and provide background definitions in Section 2. We prove our hardness results in Section 3. Section 4 contains the definition of interval-membership-width as well as our algorithms parameterized by this measure. In Section 5 we show that Akrida, Mertzios and Spirakis’ temporal star-exploration problem is in parameterized by the maximum number of appearances of any edge in the input whenever the input temporal star has evenly-spaced times on all edges. We also show a similar result for our temporally Eulerian problem. Finally we provide concluding remarks and open problems in Section 6.

## 2 Background and notation

For any graph-theoretic notation not defined here, we refer the reader to Diestel’s textbook [Diestel2010GraphTheory]; similarly, for any terminology in parameterized complexity, we refer the reader to the textbook by Cygan et al. [cygan2015parameterized].

The formalism for the notion of dynamic or time-evolving graphs originated from the work of Kempe, Kleinberg, and Kumar [KEMPE2002820]. Formally, if is a function mapping edges of a graph to sets of integers, then we call the pair a temporal graph. We shall assume all temporal graphs to be finite and simple in this paper.

For any edge in , we call the set the time-set of . For any time we say that is active at time and we call the pair a time-edge. The set of all edges active at any given time is denoted . The latest time for which is non-empty is called the lifetime of a temporal graph (or equivalently ). Here we will only consider temporal graphs with finite lifetime.

In a temporal graph there are two natural notions of walk: one is the familiar notion of a walk in static graphs and the other is a truly temporal notion where we require consecutive edges in walks to appear at non-decreasing times. Formally, given vertices and in a temporal graph , a temporal -walk is a sequence of time-edges such that is a walk in starting at and ending at and such that . If , we denote by the temporal walk . We call a temporal -walk closed if and we call it a strict temporal walk if the times of the walk form a strictly increasing sequence. Hereafter we will assume all temporal walks to be strict.

Recall that an Euler circuit in a static graph is a circuit which traverses every edge of exactly once. In this paper we are interested in the natural temporal analogue of this notion.

###### Definition 2.1.

A temporal Eulerian circuit in a temporal graph is a closed temporal walk such that is an Euler circuit in the underlying static graph . If there exists a temporal Eulerian circuit in , then we call temporally Eulerian.

Note that if is a temporal graph in which every edge appears at exactly one time, then we can determine whether is temporally Euelrian in time linear in . To see this, note that, since every edge is active at precisely one time, there is only one candidate ordering of the edges (which may or may not give rise to an Eulerian circuit). Thus it is clear that the number of times per edge is relevant to the complexity of the associated decision problem – which we state as follows.

Input: A temporal graph where for every edge in the graph . Question: Is temporally Eulerian?

As we mentioned in Section 1, here we will show that is related to an analogue of the Travelling Salesman problem on temporal stars [starexp]. This problem (denoted as ) was introduced by Akrida, Mertzios and Spirakis [starexp]. It asks whether a given temporal star (where denotes the -leaf star) with at most times on each edge admits a closed temporal walk starting at the center of the star and which visits every leaf of . We call such a walk an exploration of . A temporal star that admits an exploration is called explorable. Formally we have the following decision problem.

Input: A temporal star where for every edge in the star . Question: Is explorable?

## 3 Hardness of temporal edge exploration

In this section we will show that is -complete for all at least (Corollary 3.4) and that is -complete for all (Corollary 3.2). This last result resolves an open problem of Akrida, Mertzios and Spirakis which asked to determine the complexity of and [starexp].

We begin by showing that is -hard. We will do so via a reduction from the -Coloringproblem (see for instance Garey and Johnson [GareyJohnson] for a proof of NP-completeness) which asks whether an input graph is -colorable.

-Coloring Input: A finite simple graph . Question: Does admit a proper -coloring?

Throughout, for an edge of a temporal star , we call any pair of times with a visit of . We say that is visited at in a temporal walk if the walk proceeds from the center of the star along at time and then back to the center at time . We say that two visits and of two edges and are in conflict with one another (or that ‘there is a conflict between them’) if there exists some time with and . Note that a complete set of visits (one visit for each edge of the star) which has no pairwise conflicts is in fact an exploration.

is -hard.

###### Proof.

Take any -Coloringinstance with vertices . We will construct a instance (where ) from .

Defining . The star is defined as follows: for each vertex in , we make one edge in while, for each edge with in , we make three edges , and in .

Defining . For and , let be the integer

 tiξ:=2in2+2ξ(n+1) (1)

and take any edge in with . Using the times defined in Equation (1), we then define and as

 τ(ei) :={ti0,ti1,ti2,ti3}% and (2) τ(eξjk) :={tjξ+2k−1,tjξ+2k,tkξ+2j−1,tkξ+2j}. (3)

Note that the elements of these sets are written in increasing order (see Figure 1).

Intuitively, the times associated to each edge corresponding to a vertex (Equation (2)) encode the possible colorings of via the three possible starting times of a visit of . The three edges , and corresponding to some are instead used to ‘force the colorings to be proper’ in . That is to say that, for a color , the times associated with the edge (Equation (3)) will prohibit us from entering at its -th appearance and also entering at its -th appearance (i.e. ‘coloring and the same color’).

Observe that the first two times in lie within an interval given by consecutive times in and that the same holds for the last two times in with respect to (see Figure 1). More precisely, it is immediate that for and , we have:

 tjξ

Given this set-up, we will now show that is a yes instance if and only if is.

Suppose is explorable. Define the coloring (to be shown proper) taking each vertex to the color whenever is entered at time within the exploration of (note that these are the only possible times at which can be entered, since every edge appears at exactly times). We claim that is a proper coloring. To see this, suppose on the contrary that there is a monochromatic edge with of color in . Then, this means that was entered at time and exited at time at least and similarly was entered at time and exited at time at least . But then, since all times in are contained either in the open interval or the open interval we know that cannot be explored (by (4)). This contradicts the assumption that is explorable, hence must be a proper coloring.

Conversely, suppose admits a proper -coloring . We define the following exploration of (see Figure 1):

• for every vertex in , if , then visit at

• for every edge in with and every color , define the visit of as follows: if , then visit at ; otherwise visit at .

Our aim now is to show that the visits we have just defined in terms of the coloring are disjoint (and thus witness the explorability of ).

Take any and any . By our definition of and , we must have . Thus we note that there are no conflicts between the visit of and the visit of .

Note that, for all and all pairs of edges and in with and , the visit of is in conflict with the visit of only if . To see this, observe that the visits of and

both consist of two consecutive times where the first time is odd. Thus we would only have a conflict if

which can be easily checked to happen only if and .

Finally we claim that there are no conflicts between the visit of and the visits of either or . To show this, we will only argue for the lack of conflicts between the visits of and since the same ideas suffice for the -case as well. Suppose , then we visit at and then since and since is a proper coloring. Similarly, if , then we visit at . As we observed in Inequality (4), we have . Thus, if is the visit of , then either or . In other words, no conflicts arise.

This concludes the proof since we have shown that the visits we assigned to the edges of constitute an exploration of . ∎

Observe that this hardness result can easily be extended to any by simply adding a new edge with times all prior to the times that are already in the star. This, together with the fact that Akrida, Mertzios and Spirakis [starexp] showed that is in for all , allows us to conclude the following corollary.

###### Corollary 3.2.

For all at least , is complete.

Next we shall reduce to . We point out that, for our purposes within this section, only the first point of the statement of the following result is needed. However, later (in the proof of Corollary 4.3) we shall make use of the properties stated in the second point of Lemma 3.3 (this is also why we allow any times per edge rather than just considering the case ). Thus we include full details here.

###### Lemma 3.3.

For all there is a polynomial-time-computable mapping taking every instance to a instance such that

1. is a yes instance for if and only if is a yes instance for and

2. is a graph obtained by identifying -copies of a cycle on three vertices along one center vertex and such that

 maxt∈N|{ e∈E(Dn):min(σ(e))≤t≤max(σ(e))}| ≤3maxt∈N|{e∈E(Sn):min(τ(e))≤t≤max(τ(e))}|.
###### Proof.

Note that we can assume without loss of generality that: (1) every edge in has exactly -times on each edge and (2) that all times are multiples of . This follows from the fact that we can construct from any -instance another -instance so that : is explorable if and only if also is and every time in satisfies conditions (1) and (2).

Now we will show how to construct a -instance from such that is temporally Eulerian if and only if is explorable (see Figure 2). Throughout, denote the vertices of the -th -cycle of as and let its edges be , and . For every with where , define the map as:

 σ(fi,1) :={t1,…,tk−1}, σ(fi,2) :={t1+1,…,tk−1+1}, σ(fi,3) :={t2,…,tk}.

Note that . Now suppose is a yes-instance witnessed by the sequence of visits of the edges of and observe that for all . We claim that the sequence of time-edges

 (f1,1,x1),(f1,2,x1+1),(f1,3,y1),…,(fn,1,xn),(fn,2,xn+1),(fn,3,yn)

is a temporal Eulerian circuit in . To see this, recall that (for ) and note that:

1. by definition is an Eulerian circuit in the underlying static graph (i.e. we walk along each -cycle in turn) and

2. for all since we assumed that .

Conversely, suppose is a yes-instance and let this fact be witnessed by the temporal Eulerian circuit . Since every Eulerian circuit in must run through each -cycle, we know that must consist – up to relabelling of the edges – of a sequence of time-edges of the form

 K:=(f1,1,x1,1),(f1,2,x1,2),(f1,3,y1,3) ,(f2,1,x2,1),(f2,2,x2,2),(f2,3,y2,3), … ,(fn,1,xn,1),(fn,2,xn,2),(fn,3,yn,3).

It follows immediately from the definition of that visiting each edge in at constitutes an exploration of , as desired. ∎

Since is clearly in (where the circuit acts as a certificate), our desired -completeness result follows immediately from Lemma 3.3 and Corollary 3.2.

###### Corollary 3.4.

is -complete for all at least .

As we noted earlier, is trivially solvable in time linear in the number of edges of the underlying static graph. Thus, towards obtaining a complexity dichotomy for , the only case remaining open is when .

Observe that the reduction in Lemma 3.3 rules out algorithms with respect to many standard parameters describing the structure of the underlying graph (for instance the path-width is and feedback vertex number is ). In fact we can strengthen these intractability results even further by showing that is hard even for instances whose underlying static graph has vertex-cover number . This motivates our search in Section 4 for parameters that describe the structure of the times assigned to edges rather than just the underlying static structure.

Notice that this time we will reduce from to (rather than from as in Lemma 3.3), so, in contrast to our previous reduction (Lemma 3.3), the proof of the following result cannot be used to show hardness of .

###### Theorem 3.5.

For all , the problem is -complete even on temporal graphs whose underlying static graph has vertex-cover number .

###### Proof.

Take any instance and assume that is even (if not, then simply add a dummy edge with all appearances strictly after the lifetime of the graph). Denoting by the center of and by its leaves, let be the double star constructed from by splitting into two twin centers; stating this formally, we define as

 Sc1,c2n=({c1,c2,x1,…,xn},{cixj:i∈[2] and j∈[n]}).

Notice that, since is even, is Eulerian and notice that the set is a vertex cover of .

Defining for all and as , we claim that the temporal graph is temporally Eulerian if and only if is explorable.

Suppose that is explorable and let this be witnessed by the sequence of visits . Then it follows immediately by the definition of that the following sequence of time-edges is a temporal circuit in :

 (c1x1,s1),(x1c2,t1),(c2x2,s2)(x2c1,t2),(c1x3,s3), (x3c2,t3), …, (c2xn,sn)(xnc1,tn).

To see this, note that this clearly induces an Eulerian circuit in the underlying static graph ; furthermore, since is an exploration in , it follows that , as desired.

Suppose now that is temporally Eulerian and that this fact is witnessed (without loss of generality – up to relabeling of vertices) by the temporal Eulerian circuit

 (c1x1,s1),(x1c2,t1),(c2x2,s2)(x2c1,t2),(c1x3,s3), (x3c2,t3), …, (c2xn,sn)(xnc1,tn).

Then, by the definition of in terms of and by similar arguments to the previous case, we have that is an exploration of . ∎

## 4 Interval-membership-width

As we saw in the previous section, both and are -complete for all even on instances whose underlying static graphs are very sparse (for instance even on graphs with vertex cover number ). Clearly this means that any useful paramterization must take into account the temporal structure of the input. Other authors have already proposed measures of this kind such as the temporal feedback vertex number [temporalFeedbackVertexSet] or temporal analogues of tree-width [temp_tw]. However these measures are all bounded on temporal graphs for which the underlying static graph has bounded feedback vertex number and tree-width respectively. Our reductions therefore show that is para--complete with respect to these parameters. Thus we do indeed need some new measure of temporal structure. To that end, here we introduce such a parameter called interval-membership-width which depends only on temporal structure and not on the structure of the underlying static graph. Parameterizing by this measure, we will show that both and lie in .

To first convey some intuition, consider again . As we noted earlier, this is trivially solvable in time linear in . The same is true for any -instance in which every edge is assigned a ‘private’ interval of times: that is to say that, for all distinct edges and in , either or . This holds because, on instances of this kind, there is only one possible relative ordering of edges available for an edge-exploration. It is thus natural to expect that, for graphs whose edges have intervals that are ‘almost private’ (defined formally below), we should be able to deduce similar tractability results.

Towards a formalization of this intuition, suppose that we are given a temporal graph which has precisely two edges and such that there is a time with and . It is easy to see that the problem is still tractable on graphs such as since there are only two possible relative edge-orderings for an edge exploration of (depending on whether we choose to explore before or before ). These observations lead to the following definition of interval-membership-width of a temporal graph.

###### Definition 4.1.

The interval membership sequence of a temporal graph is the sequence of edge-subsets of where and is the lifetime of . The interval-membership-width of is the integer .

Note that a temporal graph has unit interval-membership-width if and only if every edge is active at times spanning a ‘private interval’. Furthermore, we point out that the interval membership sequence of a temporal graph is not the same as the sequence . In fact, although , there exist classes of temporal graphs with unbounded interval-membership-width but such that every temporal graph in satisfies the property that at most one edge is active at any given time. To see this consider any graph with edges and let be the temporal graph defined by . Clearly , but we have .

Note that the interval membership sequence of a temporal graph can be computed in time by iterating over the edges of .

Armed with the notion of interval-membership-width, we will now show that both and are in when parameterized by this measure. We will do so first for (Theorem 4.2) and then we will leverage the reduction of Lemma 3.3 to deduce the fixed-parameter-tractability of as well (Corollary 4.3).

###### Theorem 4.2.

There is an algorithm that decides whether any temporal graph with vertices and lifetime is a yes-instance of in time where is the interval-membership-width of .

###### Proof.

Let be the interval membership sequence of and suppose without loss of generality that is not empty.

We will now describe an algorithm that proceeds by dynamic programming over the sequence to determine whether is temporally Eulerian. For each set we will compute a set consisting of triples of the form where and are vertices in and is a function mapping each edge in to an element of . Intuitively each entry of corresponds to the existence of a temporal walk starting at and ending at at time at most and such that, for any edge , we will have if and only if was traversed during this walk.

We will now define the entries recursively starting from the dummy set where is the function mapping every element in to