Feedback Edge Sets in Temporal Graphs
aa r X i v : . [ c s . D M ] M a r Feedback Edge Sets in Temporal Graphs ∗ Roman Haag, Hendrik Molter, Rolf Niedermeier, and Malte RenkenAlgorithmics and Computational Complexity, Faculty IV, TU Berlin, Germany { h.molter,rolf.niedermeier,m.renken } @tu-berlin.de Abstract
The classical, linear-time solvable
Feedback Edge Set problem is concerned with findinga minimum number of edges intersecting all cycles in a (static, unweighted) graph. We providea first study of this problem in the setting of temporal graphs , where edges are present only atcertain points in time. We find that there are four natural generalizations of
Feedback EdgeSet , all of which turn out to be NP-hard. We also study the tractability of these problemswith respect to several parameters (solution size, lifetime, and number of graph vertices, amongothers) and obtain some parameterized hardness but also fixed-parameter tractability results.
A temporal graph G = ( V, E , τ ) has a fixed vertex set V and each time edge in E has a discretetime-label t ∈ { , , . . . , τ } , where τ denotes the lifetime of the temporal graph G . A temporal cycle in a temporal graph is a time-respecting sequence of edges starting and ending at the same vertex.We study the computational complexity of searching for small feedback edge sets , i.e., edge setswhose removal from the temporal graph destroys all temporal cycles. We distinguish between thefollowing two variants of feedback edge set problems.1. Temporal feedback edge sets, which consist of time-edges, that is, connections between twospecific vertices at a specific point in time.2.
Temporal feedback connection sets, which consist of vertex pairs { v, w } causing that all time-edges between v and w will be removed.Defining feedback edge set problems in temporal graphs is not straight-forward because fortemporal graphs the notions of paths and cycles are more involved than for static graphs. First, weconsider two different, established models of temporal paths. Temporal paths are time-respectingpaths in a temporal graph. Strict temporal paths have strictly increasing time-labels on consecutivetime-edges.
Non-strict temporal paths have non-decreasing time-labels on consecutive time-edges.Non-strictness can be used whenever the traversal time per edge is very short compared to the scaleof the time dimension. ∗ Supported by the DFG, project MATE (NI 369/17).
1e focus on finding temporal feedback edge sets and temporal feedback connection sets (formal-ized in Section 2) of small cardinality in unweighted temporal graphs, each time using both the strictand non-strict temporal cycle model. We call the corresponding problems (Strict) TemporalFeedback Edge Set and (Strict) Temporal Feedback Connection Set , respectively. (Strict) Temporal Feedback Edge Set ((S)TFES)
Input:
A temporal graph G = ( V, E , τ ) and k ∈ N . Question:
Is there a (strict) temporal feedback edge set E ′ ⊆ E of G with |E ′ | ≤ k ? (Strict) Temporal Feedback Connection Set ((S)TFCS) Input:
A temporal graph G = ( V, E , τ ) and k ∈ N . Question:
Is there a (strict) temporal feedback connection set C ′ of G with | C ′ | ≤ k ? Related Work.
In static connected graphs, removing a minimum-cardinality feedback edge setresults in a spanning tree. This can be done in linear time via depth-first or breadth-first search.Thus, it is natural to compare temporal feedback edge sets to the temporal analogue of a spanningtree. This analogue is known as the minimum temporally connected (sub)graph , which is a graphcontaining a time-respecting path from each vertex to every other vertex. The concept was firstintroduced by Kempe et al. [17], and Axiotis and Fotakis [4] showed that in an n -vertex graph sucha minimum temporally connected subgraph can have Ω( n ) edges while Casteigts et al. [10] showedthat complete temporal graphs admit sparse temporally connected subgraphs. Additionally, Akridaet al. [2] and Axiotis and Fotakis [4] proved that computing a minimum temporally connected sub-graph is APX-hard. Considering weighted temporal graphs, there is also (partially empirical) workon computing minimum spanning trees, mostly focusing on polynomial-time approximability [16].While feedback edge sets in temporal graphs seemingly have not been studied before, Agrawalet al. [1] investigated the related problem α − Simultaneous Feedback Edge Set , where theedge set of a graph is partitioned into α color classes and one wants to find a set of at most k edgesintersecting all monochromatic cycles. They show that this is NP-hard for α ≥ O ( kα ) poly( n )-time algorithm.Another related problem is finding s - t -separators in temporal graphs; this was studied byBerman [6], Kempe et al. [17], and Zschoche et al. [23]. Already here some differences were foundbetween the strict and the non-strict setting, a distinction that also matters for our results. Our Contributions.
Based on a reduction from 3-SAT, we show NP-hardness for all four prob-lem variants. The properties of the corresponding construction yield more insights concerningspecial cases. More specifically, the constructed graph uses τ = 8 distinct time-labels for thestrict variants and τ = 3 labels for the non-strict variants. Similarly, we observe that our con-structed graph has at most one time edge between any pair of vertices (i.e., is simple ), implyingthat the problems remain NP-hard when restricted to simple temporal graphs. Based on the Ex-ponential Time Hypothesis, we can derive the additional result stating that there is presumablyno subexponential-time algorithm solving (S)TFES or (S)TFCS. Moreover, we show that all fourproblem variants are W[1]-hard when parameterized by the solution size k , using a parameterizedreduction from the W[1]-hard problem Multicut in DAGs [18].On the positive side, based on a simple search tree, we first observe that all problem vari-ants are fixed-parameter tractable with respect to the combined parameter k + L , where L is the2able 1: Overview of our results for (Strict) Temporal Feedback Edge Set (marked with*) and (Strict) Temporal Feedback Connection Set (marked with **). Unmarked resultsapply to both variants. The parameter k denotes the solution size, τ the lifetime of the temporalgraph, L the maximum length of a minimal temporal cycle, and tw ↓ the treewidth of the underlyinggraph. Param. ComplexityStrict variant Non-strict variantnone NP-hard[Thm. 10*/Thm. 11**] NP-hard [Thm. 12] k W[1]-hard [Thm. 14] W[1]-hard [Thm. 14] τ τ ≥
8: NP-h.[Thm. 10*/Thm. 11**] τ ≥
3: NP-h. [Thm. 12] k + L O ( L k · | V | · |E| ) [Thm. 6] O ( L k · | V | · |E| ) [Thm. 6] k + τ O ( τ k · | V | · |E| ) [Thm. 7] open | V | O (2 | V | · | V | · τ )* [Thm. 17*] O (2 ( | V | −| V | ) · | V | · |E| )** O (2 | V | · | V | · τ )*[Thm. 17*] O (2 ( | V | −| V | ) · | V | · |E| )**tw ↓ + τ FPT [Thm. 23] FPT [Thm. 23]maximum length of a minimal temporal cycle. For the strict problem variants, this also impliesfixed-parameter tractability for the combined parameter τ + k . Our main result is to prove fixed-parameter tractability for (S)TFES with respect to the number of vertices | V | . (For (S)TFCS,the corresponding result is straightforward as there are ( | V | − | V | ) vertex pairs to consider.)Finally, studying the combined parameter τ plus treewidth of the underlying graph, we show fixed-parameter tractability based on an MSO formulation.Our results are summarized in Table 1. For the combined parameter k + τ , the non-strict caseremains open as we cannot use τ to upper-bound the length of temporal cycles. Comparing feedbackedge sets to feedback connection sets, we note a disparity for the parameter number of vertices | V | .While both (S)TFES and (S)TFCS are fixed-parameter tractable with respect to | V | , proving thisresult for (S)TFES was much more involved, because the maximum number of time-edges dependson τ and | V | whereas the number of “connections” only depends on | V | . We assume familiarity with standard notion from graph theory and from (parameterized) complex-ity theory. We denote the set of positive integers with N . For a ∈ N , we set [ a ] := { , . . . , a } .3 emporal Graphs. We use the following definition in which the vertex set does not change withtime and each time-edge has a discrete time-label [14, 15, 20].
Definition 1 (Temporal Graph, Underlying Graph) . A (undirected) temporal graph G = ( V, E , τ )is an ordered triple consisting of a set of vertices V , a set E ⊆ (cid:0) V (cid:1) × [ τ ] of (undirected) time-edges ,and a lifetime τ ∈ N .The underlying graph G ↓ is the static graph obtained by removing all time-labels from G andkeeping only one edge from every multi-edge. We call a temporal graph simple if each vertex pairis connected by at most one time-edge.Let G = ( V, E , τ ) be a temporal graph. For i ∈ [ τ ], let E i ( G ) := {{ v, w } | ( { v, w } , i ) ∈ E} be theset of edges with time-label i . We call the static graph G i ( G ) = ( V, E i ( G )) layer i of G . For t ∈ [ τ ],we denote the temporal subgraph consisting of the first t layers of G by G [ t ] ( G ) := ( V, { ( e, i ) | i ∈ [ t ] ∧ e ∈ E i ( G ) } , t ). We omit the function parameter G if it is clear from the context. For some E ′ ⊆ (cid:0) V (cid:1) × [ τ ], we denote G − E ′ := ( V, E \ E ′ , τ ). Definition 2 (Temporal Path, Temporal Cycle) . Given a temporal graph G = ( V, E , τ ), a temporalpath of length ℓ in G is a sequence P = ( e , e , . . . , e ℓ ) of time-edges e i = ( { v i , v i +1 } , t i ) ∈ E where v i = v j for all i, j ∈ [ ℓ ] and t i ≤ t i +1 for all i ∈ [ ℓ − strict if t i < t i +1 for all i ∈ [ ℓ − (Strict) Temporal Feedback Edge Set and (Strict) TemporalFeedback Connection Set (see Section 1) are based on the following two sets (problem and setnames are identical). Definition 3 ((Strict) Temporal Feedback Edge Set) . Let G = ( V, E , τ ) be a temporal graph. Atime-edge set E ′ ⊆ E is called a (strict) temporal feedback edge set of G if G ′ = ( V, E \ E ′ , τ ) doesnot contain a (strict) temporal cycle. Definition 4 ((Strict) Temporal Feedback Connection Set) . Let G = ( V, E , τ ) be a temporal graphwith underlying graph G ↓ = ( V, E ↓ ). An edge set C ′ ⊆ E ↓ is a (strict) temporal feedback connectionset of G if G ′ = ( V, E ′ , τ ) with E ′ = { ( { v, u } , t ) ∈ E | { v, u } / ∈ C ′ } does not contain a (strict)temporal cycle.The elements in a feedback connection set are known as underlying edges (edges of G ↓ ). Simple Observations.
We can compute shortest temporal paths from any given vertex to allother vertices in O ( |E| log( |E| )) time [22], respectively O ( |E| ) time for strict temporal paths [23].Thus, by searching for each time-edge ( { v, w } , t ) for a shortest temporal path from w to v whichstarts at time t and avoids the edge { v, w } , we can record the following observation. Observation 5. In O ( |E| log( |E| )) time, we can find a shortest temporal cycle or confirm thatnone exists. (For the strict case O ( |E| ) time suffices.) Given a shortest temporal cycle of length L , any temporal feedback edge or connection set mustcontain an edge or connection used by that cycle. By repeatedly searching for a shortest temporalcycle and then branching over all of its edges or connections, we obtain the following (again thelog-factor is only required in the non-strict case).4 bservation 6. Let G = ( V, E , τ ) be a temporal graph where each temporal cycle has length atmost L ∈ N . Then, (S)TFES and (S)TFCS can be solved in O ( L k · |E| log |E| ) time.Proof. We can construct a simple search tree based on the fact that at least one edge from eachcycle has to be in the solution. According to Theorem 5, we can confirm that G is cycle-freeor find some shortest cycle C in O ( |E| · | V | ) time. If we find a cycle C , then we branch overall of its | C | ≤ L time-edges and recursively solve the instance I ′ remaining after removing thistime-edge (underlying edge for (S)TFCS) and lowering k by one. Clearly, removing any time-edgecannot create a new temporal cycle and, thus, L is also an upper-bound for the length of a minimaltemporal cycle in I ′ . The size of the resulting search tree is upper-bounded by L k .Clearly, a strict temporal cycle cannot be longer than the lifetime τ . Thus, Theorem 6 imme-diately gives the following result. Corollary 7.
STFES and STFCS can be solved in O ( τ k · |E| ) time. Alternatively, we can also upper-bound L in terms of the length of any cycle of the underlyinggraph G ↓ , which in turn can be upper-bounded by 2 O (td ↓ ) [21], where td ↓ is the treedepth of theunderlying graph. Corollary 8.
Let G be a temporal graph and td ↓ be the treedepth of G ↓ . Then, (S)TFES and(S)TFCS can be solved in O ( td ↓ · k ) · |E| log( |E| ) time. In contrast to static graphs, | V | is to be considered as a useful parameter for temporal graphsbecause the maximum number of time-edges |E| can be arbitrarily much larger than | V | . However,the number of underlying edges is at most ( | V | − | V | ) which yields the following fixed-parametertractability result for (S)TFCS. Observation 9. (S)TFCS can be solved in O (2 ( | V | −| V | ) · |E| log |E| ) time.Proof. Let G be a temporal graph with underlying graph G ↓ = ( V, E ↓ ). As G ↓ is a static graph, wehave | E ↓ | ≤ ( | V | − | V | ). Thus, there are 2 | E ↓ | ≤ ( | V | −| V | ) possible feedback connection sets,each of which can be tested in O ( |E| log |E| ) time (Theorem 5). We now show that all four problem variants, (S)TFES and (S)TFCS, are NP-hard on simpletemporal graphs with constant lifetime. The proofs work by reduction from the classic 3-SATproblem.
Theorem 10.
STFES is NP-hard for simple temporal graphs with τ = 8 .Proof. We show NP-hardness via a polynomial-time many-one reduction from . For a booleanformula Φ in conjunctive normal form (CNF) with at most three variables per clause, asksif there is a satisfying truth assignment for Φ. Let Φ be such a formula with variables x , x , . . . , x n and clauses c , c , . . . , c m of the form c j = ( ℓ j ∨ ℓ j ∨ ℓ j ). We construct an STFES instance withtemporal graph G (Φ) and k = n + 2 m as follows.For each variable x i , we introduce a variable gadget (see Figure 1) with vertices v i , v Ti , and v Fi and edges e Ti := ( { v i , v Ti } , e Fi := ( { v i , v Fi } , e hi := ( { v Ti , v Fi } , i v Ti v Fi e Ti , e Fi , e hi , w j w j w j w j f j , f j , f j , f aj , f bj , s x T F12 3 x T F12 3 x T F12 3 x T F12 37 6 51 24 4 4 ( x ∨ ¬ x ∨ x ) ( ¬ x ∨ ¬ x ∨ x )Figure 2: Example: Reduction from 3-SAT to STFES/STFCS.the temporal cycle ( e hi , e Ti , e Fi ), any solution for STFES must contain at least one of them. Foreach clause c j , we introduce a clause gadget with four vertices, w j , w j , w j , and w j , and the edges f aj = ( { w j , w j } , f bj = ( { w j , w j } , f j := ( { c j , w j } , f j := ( { c j , w j } , f j := ( { c j , w j } , c j = ( ℓ j ∨ ℓ j ∨ ℓ j ) be a clause of Φ. If ℓ j = x i , then we add the edge ( { w j , v Ti } ,
4) and, if ℓ j = ¬ x i , we add( { w j , x Fi } ,
4) (edges for ℓ j and ℓ j analogously). Further, we connect a new vertex s to all variablegadgets by ( { s, v i } ,
1) for all i ∈ [ n ] and to all clause gadgets by ( { s, w j } ,
8) for all j ∈ [ m ].This creates three additional cycles per clause, each starting and ending in s . More precisely,if x i ( ¬ x i is handled analogously) is the z -th literal of clause c j , then G (Φ) contains the cycle C zij = (( { s, v i } , , ( { v i , v Ti } , , ( { v Ti , w zj } , , ( { w zj , w j } , − z ) , ( { w j , s } , G (Φ) can be computed in polynomial time. The general idea of thisreduction is to use the solution size constraint to ensure that exactly one edge from each variablegadget and exactly two edges from each clause gadget are taken. Thus, out of the three cyclesstarting in s and going through the clause gadget of c j , only two can be disconnected by pickingtwo edges from { f j , f j , f j } . The remaining cycle has to be disconnected inside its variable gadgetby picking either e Ti or e Fi which “selects” the variable that will satisfy the clause and gives usits truth assignment. Now we show that ( G (Φ) , k ) is a yes-instance of STFES if and only if Φ is6atisfiable.( ⇒ ) : Let E ′ be a solution to the constructed STFES instance. Due to the size constraint k ≤ n + 2 m and the cycles existing inside the gadgets, E ′ contains exactly one edge from eachvariable gadget and none of the edges adjacent to s or connecting variable and clause gadgets. Weobtain the solution for the 3-SAT instance by setting x i to true if e Ti ∈ E ′ and to false if e Fi ∈ E ′ or e hi ∈ E ′ . Assume towards contradiction that there is a clause c j = ( ℓ j ∨ ℓ j ∨ ℓ j ) which is notsatisfied. Then, in all three variable gadgets connected to w j , the edge needed to go from s to thecorresponding literal vertex of c j is present in G (Φ) − E ′ . As E ′ contains only two of the edges fromthe clause gadget, the path of one of the three literals can be extended to the vertex w j and fromthere back to s , contradicting that G (Φ) − E ′ is cycle-free.( ⇐ ) : For the other direction, suppose we have a satisfying truth assignment for Φ. We obtaina solution E ′ = E Var ∪ E Cl for the STFES instance ( G (Φ) = ( V, E , τ = 8), k = n + 2 m ) as follows.For the variable gadgets, we use the variable assignment to add the feedback edges E Var = { e Ti | i ∈ [ n ] , x i = true } ∪ { e Fi | i ∈ [ n ] , x i = false } . For each clause c j = ( ℓ j ∨ ℓ j ∨ ℓ j ), let z j ∈ [3] be the number of one of the literals satisfying theclause, i.e., ℓ z j j = true. We add the edges between w j and the other two literal vertices to thefeedback edge set: E Cl = { f zj | j ∈ [ m ] , z ∈ [3] , z = z j } . Note that this breaks all cycles inside the variable and clause gadgets and that |E ′ | = |E Var | + |E Cl | = n + 2 m . Cycles going through multiple gadgets but not starting in s are impossible as they woulduse at least two edges with time-label 4. It remains to show that G − E ′ does not contain any cyclestarting and ending in s . Assume towards contradiction that there is such a cycle going through thevariable gadget of x i and the clause gadget of c j . Further, assume that x i was set to true (the othercase is handled analogously) and that, therefore, ( { v i , v Fi } , ∈ E \ E ′ . Then, the cycle begins with( { s, v i } , , ( { v i , v Fi } , , ( { v Fi , w yj } ,
4) for some y ∈ [3]. Note that the edges e hi , f aj , and f bj cannot beused due to the time-labels. By construction of G (Φ), we know that if ( { v Fi , w yj } ,
4) exists, then ℓ yj isone of the literals satisfying the clause if x i = false. Since we assumed that x i = true, it holds that y = z j and, thus, f yj ∈ E Cl . It follows that there is no edge which can be appended to the temporalpath and, in particular, no possibility of reaching s , thus contradicting the assumption.The temporal graph constructed in the proof for Theorem 10 does not contain any pair ofvertices which is connected by more than one time-edge. Hence, each underlying edge correspondsto a single time-edge and thus the reduction implies the following corollary. Corollary 11.
STFCS is NP-hard even for simple temporal graphs with τ ≥ . A very similar reduction can also be used for the following.
Corollary 12.
TFES and TFCS are both NP-hard even for simple temporal graphs with τ ≥ .Proof sketch. The changes that need to be made to the reduction described in the proof of Theo-rem 10 are shown in Figure 3. Here, we have to subdivide the edges between variable and clausegadgets in order to avoid cycles which go through multiple gadgets but not through s . In turn,only three different time-labels are needed to create the required cycles.7 x T F12 2 x T F12 2 x T F12 2 x T F12 23 3 31 12 3 23 23 ( x ∨ ¬ x ∨ x ) ( ¬ x ∨ ¬ x ∨ x )Figure 3: Example: Reduction from 3-SAT to TFES/TFCS.We can also observe that the strict problem variants are NP-hard even if all edges are presentat all times. This problem is essentially equivalent to selecting a set of edges of the underlyinggraph that intersects all cycles of length at most τ , which is known to be NP-hard [9]. Observation 13.
STFES and
STFCS are NP-hard even on temporal graphs where all time edgesare present at all times, with τ = 3 , planar G ↓ , and ∆( G ↓ ) = 7 .Proof. Let G = ( V, E ,
3) be a temporal graph with three layers E = E = E . Then an edge set C ′ ⊆ E ↓ is a strict temporal feedback connection set if and only if it intersects all triangles in G ↓ .Similarly, it is easy to see that a strict temporal feedback edge set E ′ ⊆ E must contain at leastthree time-edges from every triangle in G ↓ and exactly three time-edges from each triangle suffice.Since it is NP-hard to determine whether there exists a set of k edges intersecting all trianglesin a planar graph with maximum degree 7 [9], the claim follows.We next show that our problems are W[1]-hard when parameterized by the solution size k witha parameterized reduction from Multicut in DAGs [18]. The idea here is that we can simulatea DAG D by an undirected temporal graph by first subdividing all edges of D and then assigningtime-labels according to a topological ordering. This ensures that each path in D corresponds toa path in the resulting temporal graph and vice versa. By adding a reverse edge from t to s foreach terminal pair ( s, t ) of the Multicut instance, an s - t -path in D produces a temporal cycleinvolving s , t and vice versa. Theorem 14. (S)TFES and (S)TFCS , parameterized by the solution size k , are W[1]-hard. We prove W[1]-hardness with a parameterized reduction from
Multicut in DAGs parameter-ized by the solution size.
Multicut in DAGs
Input:
A DAG D = ( V, A ), a set of terminal pairs T = { ( s i , t i ) | i ∈ [ r ] and s i , t i ∈ V } ,and an integer k . Question:
Is there a cut-set Z ⊆ V of at most k nonterminal vertices of G such that for all i ∈ [ r ] the terminal t i is not reachable from s i in G − Z ?8 b := t t + 2 a bv ab v ab ... v k +1 ab w ab w ab ... w k +1 ab t tt t + 1 t + 1 t + 1 t + 2 t + 2 t + 2 Figure 4: Heavy time-edge h( a, b, t ).This problem was shown to be W[1]-hard when parameterized by k by Kratsch et al. [18] whoalso provided the following lemma which will simplify our proof by further restricting the inputinstance. Lemma 15 (Kratsch et al. [18, Lemma 2.1]) . There exists a polynomial-time algorithm that,given a
Multicut in DAGs instance ( D, T , k ) with D = ( V, A ) , computes an equivalent instance ( D ′ , T ′ , k ′ ) with D ′ = ( V ′ , A ′ ) such that1. |T | = |T ′ | and k = k ′ ;2. T ′ = { ( s ′ i , t ′ i ) | i ∈ [ r ] } and all terminals s ′ i and t ′ i are pairwise distinct; and3. for each v ∈ V and i ∈ [ r ] we have ( v, s ′ i ) / ∈ A ′ and ( t ′ i , v ) / ∈ A ′ . We will from now on assume that we have an instance with these properties. The goal of ourreduction will be to create one temporal cycle for each terminal pair. Since there is a temporalpath from s i to t i (the pair can be ignored otherwise), we can create a cycle by adding a back-edge from t i to s i . To preserve the direction of the arcs in the (undirected) temporal graph, we willsubdivide each arc into small paths with ascending time-labels. As Multicut in DAGs asks for avertex set, we also need to subdivide each nonterminal vertex v into two new vertices v in and v out which are connected by one edge. Then, the vertex v is in the cut-set of the original problem if theedge between v in and v out is in the solution edge set of the STFES instance.Before stating our reduction, we introduce two auxiliary concepts. First, when we want toexclude edges from (S)TFES/(S)TFCS solutions, we will employ a gadget we call heavy time-edge which connects two vertices using k + 1 parallel paths. Definition 16 (Heavy time-edge) . Let I = ( G = ( V, E , τ ) , k ) be an instance of (S)TFES or(S)TFCS. For a, b ∈ V and t ≤ τ −
2, a heavy time-edge of G is a subgraph h( a, b, t ) := ( V h , E h , τ h = t + 2) connecting vertex a to vertex b with V h = { a, b } ∪ { v iab , w iab | i ∈ [ k + 1] } and E h = { ( { a, v iab } , t ) , ( { v iab , w iab } , t + 1) , ( { w iab , b } , t + 2) | i ∈ [ k + 1] } . b c d e f g h j c b a e f j d g h Figure 5: A DAG with values for π ( v ) (derived from an acyclic ordering) for each vertex v (top)and with the vertices aligned on a line according to π (bottom).The construction is shown in Figure 4. Let e h := h( a, b, t ) be a heavy time-edge. Due to thetime-labels, the gadget only connects a to b (and not b to a ) which we will use to model directedarcs with (undirected) time-edges. For (S)TFES/(S)TFCS solutions, it is easy to see that if thereis a temporal path from b to a , then the k + 1 cycles going through e h cannot be disconnectedby removing edges inside the gadget. Thus, without loss of generality we can assume that a givensolution contains no edges from e h . We also note that, for both temporal path models (i.e., strictor non-strict), it is possible to design a smaller gadget with identical properties, but we opted touse one which works for both models simultaneously.Second, in order to assign time-labels while preserving all paths of the input graph D , we willuse an acyclic ordering (also known as topological ordering) of D . For a directed graph D = ( V, A ),an acyclic ordering < is a linear ordering of the vertices with the property ( v, w ) ∈ A ⇒ v < w .In other words, if we place the vertices on a line in the order given by < , then all arcs point inone direction (see Figure 5 for an example). If D is a DAG, then such an ordering exists and canbe computed in linear time [5, Theorem 4.2.1]. For convenience, we represent this ordering as afunction π : V → N which maps each vertex to its position in the ordering. We now have all theingredients to prove the theorem. Proof of Theorem 14.
Let I = ( D, T , k ) be an instance of Multicut in DAGs with terminalvertices V T := { s i , t i | ( s i , t i ) ∈ T } . We construct an instance I ′ = ( G , k ′ = k ) of (S)TFES asfollows.1. We compute an acyclic ordering of the vertices V := V ( D ) and store it as a function π : V → N (see Figure 5). We use π to transform D into an equivalent temporal graph G = ( V ′ , E , τ =4 | V | ) by replacing each arc ( v, w ) ∈ A with the heavy time-edge e vw := h( v, w, π ( v ) + 1). Itis easy to verify that, for two vertices s, t ∈ V , the graph D contains a path from vertex s tovertex t if and only if G contains a temporal path from s to t . Note that starting each heavy10 a b c → Ga b in b out c Figure 6: Example: Reduction from
Multicut in DAGs with input digraph D and one terminalpair ( a, c ) to TFES. Double lines represent heavy time-edges (Theorem 16).time-edge with time-label 4 π ( v ) + 1 leaves layer 4 π ( v ) empty which we will use in the nextstep.2. In G , we replace each nonterminal vertex v ∈ V \ V T with two new vertices v in and v out connected by time-edge e v = ( { v in , v out } , π ( v )) and update the edges adjacent to v as follows.For each (incoming) edge of the form e uv := ( { u, v } , t ) with t < π ( v ), replace e uv with( { u, v in } , t ). For each (outgoing) edge of the form e vw := ( { v, w } , π ( v ) + 1), replace e vw with( { v out , w } , π ( v ) + 1). Let G denote the resulting graph. Clearly, for two vertices s, t ∈ V ,removing v in G disconnects all ( s, t )-paths if and only if removing e v in G disconnects all( s, t )-paths.3. We obtain G = G by adding a back-edge h( t i , s i , τ be ) with τ be = 4 | V + 1 | for each terminalpair ( s i , t i ). Since there is a temporal path from s i to t i , this creates at least one cycle foreach terminal pair.Figure 6 shows a small example. It is easy to see that the construction can be done in polynomialtime. Now we show that I = ( D, T , k ) is a yes-instance of Multicut in DAGs if and only if I ′ = ( G , k ′ = k ) is a yes-instance of (S)TFES.( ⇒ ) : Let Z be a solution of I . We claim that E ′ = { ( { v in , v out } , π ( v )) | v ∈ Z } is a solutionof I ′ . We first show that, for any terminal pair ( s i , t i ), the graph G − E ′ contains no temporal pathfrom s i to t i . For G , this is easily verified as D − Z contains no ( s i , t i )-path. In G = G , thisclaim holds if no temporal path from s i to t i contains a back-edge e t j s j = h( t j , s j , τ be ) added instep 3. Due to the starting time-label of the back-edges, no temporal path can contain more thanone back-edge and, if it does, then this back-edge must be at its end. Clearly, the temporal pathcannot end at both t i and s j , unless t i = s j which we excluded by applying Theorem 15. Now,assume towards contradiction that G − E ′ contains a cycle C . Since G was cycle-free, C must usesome back edge e t i s i introduced in step 3 and, as reasoned above, this back edge must be the lastedge of C . However, there is no temporal path from s i to t i in G − E ′ and, thus, C cannot be atemporal cycle.( ⇐ ) : Let E ′ be a solution of I ′ , i.e., G − E ′ contains no cycles. Recall that V T := { s i , t i | ( s i , t i ) ∈ T } is the set of terminal vertices of I . As observed above, E ′ does not contain any edgesfrom heavy time-edges, thus we have E ′ ⊆ { e v | v ∈ V \ V T } and define the solution for I as Z := { v | e v ∈ E ′ } . Assume towards a contradiction that D − Z contains an ( s i , t i )-path for someterminal pair ( s i , t i ). This path induces a temporal path from s i to t i in G − E ′ which we can extendback to s i by appending the back edge e t i s i to obtain a cycle in G − E ′ and, thus, a contradiction.For both directions, we have | Z | = |E ′ | ≤ k = k ′ meeting the requirements for the solution size.As the constructed temporal graph G contains no pair of vertices connected by more than onetime-edge, we can easily transform a minimal feedback edge set of G into a minimal feedback11onnection set. Thus, the arguments presented in this proof also hold for (S)TFCS. After having shown computational hardness for the single parameters solution size k and life-time τ in Section 3, we now consider larger and combined parameters, and present fixed-parametertractability results. As shown in Theorem 9, (S)TFCS is trivially fixed-parameter tractable with respect to the numberof vertices | V | . For (S)TFES, however, the same result is much more difficult to show as the sizeof the search space is only upper-bounded by 2 τ ( | V | −| V | ) . Here, the dependence on τ prevents usfrom using the (brute-force) approach that worked for (S)TFCS. Theorem 17.
STFES can be solved in O (2 | V | · | V | · τ ) time and TFES can be solved in O (2 | V | · | V | · τ ) time, both requiring O (2 | V | ) space. We prove Theorem 17 using a dynamic program which computes the minimum number of time-edges which have to be removed to achieve a specified connectivity at a specified point in time.The key idea is that in order to compute optimal solutions for time t , we only need to know whichtemporal paths were possible at time t −
1. In particular, we will only need the start and end pointsof these temporal paths and, thus, avoid storing the time-edges that were removed before t whichwould cause problems when aiming at a fixed-parameter tractability result for the parameter | V | .With the dynamic programming table available, we can solve (S)TFES by looking up the entry attime τ for a connectivity specification that excludes temporal cycles.Before formally describing the dynamic program, we need to introduce some notations andintermediate results. Let G = ( V, E , τ ) be a temporal graph with V = { v , v , . . . , v n } . The firstdimension of the dynamic programming table will be the connectivity between the vertices of G .We will store this as a connectivity matrix A ∈ { , ? } n × n encoding the following connectivityrelationships: a ij = 0 ⇒ there is no temporal path from vertex v i to v j (resp. no temporal cycle if i = j ) and a ij = ? ⇒ there might be a temporal path (resp. cycle) from v i to v j . Next, we define two functions, srd(
G, B, A ) ( s trict r equired d eletions) and nrd( G, B, A ) ( n on-strict r equired d eletions), which return the solution to the following subproblem. Given connectivity B ( b efore) at time t −
1, what is the minimum number of edge deletions required in G t to ensureconnectivity A ( a fter) at time t ? Figure 7 illustrates this problem for two vertices v i and v j . If a ij = 0 and there is some vertex v k which might be reachable from v i (i.e., b ik = ? representedby the dotted path), then we must remove the edge between v k and v j . In order to guaranteecorrectness, we have to assume that every “?” in B represents an existing path. Additionally, if A and B encode incompatible connectivity, then the function value is defined as ∞ .12ertices time t − tv i ... v k ... v j v i v k v j v i v k v j “ b ik = ?” e Figure 7: Illustration of the subproblem solved by srd/nrd(
G, B, A ). If we want to make sure thatno temporal path from v i to v j exists at time t (that is, we have a ij = 0), then the time-edge e has to be removed from the graph because there might be a temporal path from v i to v k ( b ik = ?,dotted line). Definition 18.
Let G = ( V, E ) be a static graph with | V | = n and let A, B ∈ { , ? } n × n be twoconnectivity matrices. Function srd( G, B, A ) is as follows.If ∃ i, j ∈ [ n ] : b ij = ? ∧ a ij = 0, then srd( G, B, A ) := ∞ . Otherwise, srd( G, B, A ) := |{{ v k , v j } ∈ E | ∃ i ∈ [ n ] : a ij = 0 ∧ ( b ik = ? ∨ i = k ) }| .Note the clause b ik = ? ∨ i = k in the formulation of Theorem 18. This is due to the factthat a vertex v k is always reachable from itself by a trivial temporal path, regardless of whether atemporal cycle at v k exists.Next, we show that srd( G, B, A ) can be computed in polynomial time.
Lemma 19.
Algorithm 1 computes function srd(
G, B, A ) in O ( | V | ) time.Proof. In order to show correctness, we first point out that lines 4 and 5 check for the case wheresrd(
G, B, A ) = ∞ in a straightforward manner. For the other case of Theorem 18, it can be easilyverified that the loops and conditions in lines 3 and 6 to 9 will count an edge { v k , v j } ∈ E if ∃ i, j, k ∈ [ n ] : b ij = 0 ∧ a ij = 0 ∧ b ik = ?. The running time is determined by the loops in line 3( O ( | V | )) and line 7 ( O ( | V | ) as we have |{{ v k , v j } ∈ E ( G t ) }| ≤ | V | for any fixed vertex v j ). Usinga hash-set as data structure for E ′ , we can add edges in constant time for an overall running timeof O ( | V | ).Since, in the non-strict case, a temporal path can successively use multiple edges from G t , itis not possible to consider each entry a ij = 0 separately (a single edge might be part of multipleunwanted temporal walks). Instead, we have to find an optimal edge-cut disconnecting all “prob-lematic” pairs ( v k , v j ) in G t where ∃ i ∈ [ n ] : a ij = 0 ∧ ( b ik = ? ∨ i = k ) } . This problem is known asthe Multicut problem. We will use
Multicut to define the second function, nrd(
G, B, A ).13 lgorithm 1
Algorithm computing srd(
G, B, A ) (for strict temporal paths)
Parameters G : static graph A, B : connectivity matrices
Output srd(
G, B, A ) function StrictRequiredDeletions ( G, B, A ) E ′ ← {} for i, j ∈ [ n ] do if b ij = ? ∧ a ij = 0 then return ∞ else if b ij = 0 ∧ a ij = 0 then for all e := { v k , v j } ∈ E ( G ) with b ik = ? do E ′ ← E ′ ∪ { e } end for end if end for return | E ′ | end function Multicut (Optimization variant)
Input:
An undirected, static graph G = ( V, E ) and a set of r terminal pairs T = { ( s i , t i ) | i ∈ [ r ] and s i , t i ∈ V } . Output:
A minimum-cardinality edge set E ′ ⊆ E whose removal disconnects all terminalpairs in T . Definition 20.
Let G = ( V, E ) be a static graph with | V | = n and let A, B ∈ { , ? } n × n be twoconnectivity matrices. Function nrd( G, B, A ) is as follows.If ∃ i, j ∈ [ n ] : b ij = ? ∧ a ij = 0, then nrd( G, B, A ) := ∞ .Otherwise, let E ′ be a solution to Multicut ( G, T ) with T = { ( v k , v j ) | ∃ i ∈ [ n ] : a ij =0 ∧ ( b ik = ? ∨ i = k ) } . Then, nrd( G, B, A ) := | E ′ | .In order to compute nrd( G, B, A ), we have to solve
Multicut which was shown to be APX-hard [12]. While there exist FPT algorithms [8, 19] for the parameter solution size, our best upperbound for the solution size is | V | − | V | and thus using these algorithms would result in a worserunning time than the brute-force approach we will use to prove the next lemma. Lemma 21.
Function nrd(
G, B, A ) can be computed in O (2 | V | · | V | ) time.Proof. The number of subsets of E is at most 2 | V | −| V | and we can verify if a subset is a Multicut solution by performing a breadth-first search in O ( | V | + | E | ) = O ( | V | ) time. Thus, nrd( G, B, A )can be computed in O (2 | V | · | V | ) time.We can now define the dynamic program which we will use to prove Theorem 17. Let A ∈{ , ? } n × n be a connectivity matrix. The table entry T ( A, t ) ∈ N contains the minimum number of14ime-edges which have to be removed from G [ t ] in order to achieve the connectivity specified by A .We define T as follows. T ( A,
0) := 0 ∀ A ∈ { , ? } n × n (1)strict paths: T ( A, t ) := min B ∈{ , ? } n × n T ( B, t −
1) + srd( G t , B, A ) (2a)non-strict paths: T ( A, t ) := min B ∈{ , ? } n × n T ( B, t −
1) + nrd( G t , B, A ) (2b) Lemma 22.
Let G = ( V, E , τ ) be a temporal graph with | V | = n and let A ∈ { , ? } n × n be aconnectivity matrix. Then, there exists no E ′ ⊆ E with |E ′ | < T ( A, t ) for which ( G − E ′ ) [ t ] possessesthe connectivity specified by A .Proof. We prove the lemma via induction over t . Recall that the connectivity matrix A can onlyencode that certain temporal paths must not exist. Thus, the correctness of the initialization T ( A,
0) = 0 is easy to see since no temporal paths exist at time t = 0. For the correctness of theupdate step (Equations (2a) and (2b)), we note that, by minimizing over all possible B ∈ { , ? } n × n ,we always find the optimal state B for the time t −
1. With some B fixed, it remains to show that T ( B, t −
1) + srd/nrd( G t , B, A ) is minimal for achieving both connectivity B at time t − A at time t . By induction hypothesis, we know that T ( B, t −
1) is minimal. To showcorrectness and minimality of srd( G t , B, A ) and nrd( G t , B, A ), we analyze how the time-edges oflayer G t influence the possible temporal paths up to time t and which changes (i.e., time-edgedeletions) are required to achieve connectivity A . For any two vertices v i , v j ∈ V , we compare theconnectivity for time t − b ij to the target connectivity given by a ij and identify thefollowing four cases.(Case 1) b ij = ? ∧ a ij = ?: Here, we do not care if v j is reachable from v i and, thus, we do notneed to remove any edges.(Case 2) b ij = ? ∧ a ij = 0: We cannot disconnect a temporal path that already exists at time t − t and, therefore, cannot guarantee a ij = 0. In this case, there is nosolution for the input parameters. Both functions are defined (see Theorems 18 and 20) to return ∞ in this case.(Case 3) b ij = 0 ∧ a ij = ?: Identical to Case 1.(Case 4) b ij = 0 ∧ a ij = 0: We have to ensure that v j is not reachable from v i in G [ t ] . Bothsrd( G t , B, A ) and nrd( G t , B, A ) are defined based on the following concept. If there is a vertex v k which was reachable from v i in the past, then we cannot keep an edge (strict case) or any path(non-strict case) connecting v k and v j in layer t . Assuming that Case 2 is already excluded for allvertex pairs, it can be easily verified that Theorem 18 uses exactly the set of such edges { v k , v j } .For the non-strict case, the set T of terminal pairs in Theorem 20 is defined to exactly contain allsuch pairs ( v k , v j ). The minimality and correctness of nrd( G t , B, A ) follows from the definition of Multicut .Finally, we must show that assuming “ b ij = ? ⇒ there is a temporal path from v i to v j ” attime t − G t , B, A ) did not result in unnecessary time-edge deletions. Tothis end, assume b ij = ? and that there is no temporal path from v i to v j at time t −
1. Let B ′ be aconnectivity matrix identical to B except for b ′ ij = 0. As the path from v i to v j does not exist, wehave T ( B, t −
1) = T ( B ′ , t − b ij = ? resulted in an unnecessary time-edge deletion,15.e., srd/nrd( G t , B, A ) > srd/nrd( G t , B ′ , A ), then the minimum function in Equations (2a) and (2b)will not chose the value computed using B .Since we are only interested in specifying that certain paths (cycles) must not exist, we choseto use “?”-entries in the connectivity matrices to represent entries we do not care about. Theadvantage is evident in Cases 1 and 3 of the previous proof. We now have all required ingredientsto prove Theorem 17. Proof of Theorem 17.
Let ( G , k ) be an instance of (S)TFES. Further, let A ∗ be an n × n connectivitymatrix with a ∗ ij = 0 if i = j and a ∗ ij = ? otherwise. As “?”-entries cannot require more time-edge deletions than “0”-entries, A ∗ is the cheapest connectivity specification that does not allowany temporal cycle to exist. Thus, it follows from Theorem 22 that ( G , k ) is a yes-instance if T ( A ∗ , τ ) ≤ k , and a no-instance otherwise.For the running time, we first note that a connectivity matrix has size | V | with two possiblechoices for each entry resulting in 2 | V | possible connectivity matrices. Thus, the table size of thedynamic program is 2 | V | · τ . To compute each table entry, we have to compute srd/nrd( G t , B, A )for each of the 2 | V | possible choices for B . Together with Theorems 19 and 21, we obtain therunning times stated in the theorem. The computation requires O (2 | V | ) space as we only need thetable entries for time t − t . Thus, it is not necessary tostore more than two columns of the table, each of size 2 | V | .We note that our dynamic program indeed solves the optimization variant of (S)TFES. That is,given a temporal graph G , it finds the smallest k for which ( G , k ) is a yes-instance of the decisionvariant stated defined in Section 1. As shown in the previous proof, we can easily use the result tosolve any instance ( G , k ′ ) of the decision variant by comparing k ′ to k .For the ease of presentation, we did not store the actual solution, that is, the feedback edgeset of size T ( A, t ). However, the functions srd( G t , B, A ) and nrd( G t , B, A ) can easily be changedto return the solution edge sets for each layer t . Using linked lists, which can be concatenated inconstant time, it is possible to include the solutions sets in the dynamic programming table withoutchanging the asymptotic running time. Finally, we show that all our problem variants are fixed-parameter tractable when parameterized bythe combination of the treewidth of the underlying graph and the lifetime. To this end we employan optimization variant of Courcelle’s famous theorem on graph properties expressible in monadicsecond-order (MSO) logic [3, 11].
Theorem 23. (S)TFES and (S)TFCS are fixed-parameter tractable when parameterized by thecombination of the treewidth of the underlying graph and the lifetime.
To prove this result we require an auxiliary (static) graph S whose vertex set is the disjointunion of • the set V of vertices of G , • the set E ( G ↓ ) of underlying edges, • the set [ τ ] of points in time, and 16 the set E of time-edges.Its edges are given by the (disjoint union of) the following binary relations, where we write R ( e, v )as a shortcut for ( e, v ) ∈ R : • the incidence relation inc ⊆ E × V where inc( e, v ) ⇐⇒ v ∈ e , • the time relation time ⊆ E × [ τ ] where time(( e, t ) , t ′ ) ⇐⇒ t = t ′ , • the edge relation edge ⊆ E × E where edge(( e, t ) , e ′ ) ⇐⇒ e = e ′ , and • the presence relation pres ⊆ E × [ τ ] where pres( e, t ) ⇐⇒ ( e, t ) ∈ E .A monadic second-order (MSO) formula over S is a formula that uses • the above relations, • the logical operators ∧ , ∨ , ¬ , =, and parentheses, • a finite set of variables, each of which is either taken as an element or a subset of V ( S ), and • the quantifiers ∀ and ∃ .Additionally we will use some folklore shortcuts such as =, ⊆ , ∈ , and ⊕ (exclusive or) whichcan themselves be replaced by MSO formulas.By the following theorem, for any property that can be expressed by an MSO formula, aminimum subset that satisfies it can be computed in linear time. Theorem 24 (Arnborg et al. [3, Thm 5.6]) . There exists an algorithm that, given • an MSO formula φ with free variables X , . . . , X r , • an affine function α ( x , . . . , x r ) , and • a graph G together with a tree decomposition of width w ,finds the minimum of α ( | X | , . . . , | X r | ) over all X , . . . , X r ⊆ V ( G ) for which formula φ is satisfiedon G . The running time is f ( | φ | , w ) · | G | , where | φ | is the length of φ .Proof sketch of Theorem 23. Let ( G = ( V, E , τ ) , k ) be a problem instance of one of our problemvariants. We will show that a solution can be found by applying Theorem 24 to the auxiliary graph S defined above and a suitable MSO formula.First we observe that the treewidth of S is bounded in terms of tw( G ↓ ) + τ . To this end let( T, { B t } t ∈ V ( T ) ) be an optimal tree decomposition of G ↓ where T is a tree and B t ⊆ V ( G ) for t ∈ V ( T ). Then ( T, { B ′ t } t ∈ V ( T ) ) with B ′ t := B t ∪ [ τ ] ∪ ( E ∩ B t ) ∪ { ( e, t ) ∈ E | e ∈ E ∩ B t } is a treedecomposition of S of width at most O ( τ · tw( G ↓ ) ). Note that a suitable tree decomposition of S can be computed in f ( τ + tw( G ↓ )) · |S| for some function f [7].Now we construct MSO formulas to express our problem variants. We begin by giving severalauxiliary subformulas of constant size to ease readability. • Adjacency of two vertices v, w at some time t :tadj( v, w, t ) := ∃ e ∈ E : inc( e, v ) ∧ inc( e, w ) ∧ pres( e, t )17 Connectivity of two vertices v, w using only edges from some edge set E ′ ⊆ E :conn( v, w, E ′ ) := ∃ X ⊆ V ∀∅ 6 = Y ⊂ X ∃ x ∈ X \ Y ∃ y ∈ Y ∃ e ∈ E ′ : v ∈ X ∧ w ∈ X ∧ inc( e, x ) ∧ inc( e, y ) • Connectivity of two vertices v, w using only edges from { e | ( e, t ) ∈ E ′ } for some given t and E ′ ⊆ E :ttconn( v, w, t, E ′ ) := ∃ E ′ ⊆ E ∀ e ∈ E ′ ∃ ε ∈ E ′ : edge( ε, e ) ∧ time( ε, t ) ∧ conn( v, w, E ′ ) • Connectivity of two vertices v, w using only edges from E ′ ∩ E t :tconn( v, w, t, E ′ ) := ∃E ′ ⊆ E∀ ε ∈ E ′ ∃ e ∈ E ′ : edge( ε, e ) ∧ ttconn( v, w, t, E ′ ) • Testing whether ( { v, w } , t ) ∈ E ′ for some given E ′ ⊆ E :teelem( v, w, t, E ′ ) := ∃ ε ∈ E ′ ∃ e ∈ E : edge( ε, e ) ∧ inc( v, e ) ∧ inc( w, e ) ∧ time( ε, t )The formulas expressing the existence of a temporal cycle for our four problem variants area thenthe following. • Existence of a strict cycle using only time edges whose underlying edges are contained in E ′ ⊆ E :cycle SC ( E ′ ) := ∃ v , v , . . . , v τ ∈ V : τ − _ t ∗ =1 tadj( v t ∗ , v τ , t ∗ ) ∧ τ ^ t = t ∗ +1 ( v t = v t − ∨ ( ∗ )) ! where ( ∗ ) := tadj( v t , v t − , t ) ∧ { v t , v t − } ∈ E ′ ∧ { v t , v t − } 6 = { v t ∗ , v τ } Note that the sequence of vertices v , v , . . . , v τ might not be a temporal cycle, but then itcontains a subsequence which is. • Existence of a non-strict cycle using only time edges whose underlying edges are contained in E ′ ⊆ E :cycle C ( E ′ ) := cycle SC ( E ′ ) , but with ( ∗ ) replaced by ( ∗∗ ), where( ∗∗ ) := ∃ E ′′ ⊆ E ∀ e ∈ E ′ : ( e ∈ E ′′ ⊕ e = { v t ∗ , v τ } ) ∧ tconn( v t , v t − , t, E ′′ ) • Existence of a strict cycle using only time edges in E ′ ⊆ E :cycle SE ( E ′ ) := cycle SC ( E ′ ) , but with ( ∗ ) replaced by ( ∗∗∗ ), where( ∗∗∗ ) := tadj( v t , v t − , t ) ∧ teelem( v t , v t − , t, E ′ ) ∧ { v t , v t − } 6 = { v t ∗ , v τ }• Existence of a non-strict cycle using only time edges in E ′ ⊆ E :cycle E ( E ′ ) := cycle SC ( E ′ ) , but with ( ∗ ) replaced by ( ∗∗∗∗ ), where( ∗∗∗∗ ) := ∃E ′′ ⊆ E∀ ε ∈ E ′ : ( ε ∈ E ′′ ⊕ edge( ε, { v t ∗ , v τ } )) ∧ ttconn( v t , v t − , t, E ′′ )18t is easy to check that the sizes of the formulas are in O ( τ ). We can now build a formula identifyinga feedback set for each of the problem variants as follows. For (S)TFES : φ (S)TFES = ∃E ′ : ¬ cycle (S)E ( E \ E ′ ) , and for (S)TFCS : φ (S)TFCS = ∃ E ′ : ¬ cycle (S)C ( E \ E ′ ) . The result now follows from Theorem 24 (for α ( x ) = x ) since |S| ∈ O ( τ ·|G| ), tw( S ) ∈ O ( τ · tw( G ↓ ) ),and | φ (S)TFES | , | φ (S)TFCS | ∈ O ( τ ). To summarize, we investigated the parameterized computational complexity of the problem ofremoving edges from a temporal graph to destroy all temporal cycles. We showed NP-hardnesseven for temporal graphs with constant lifetime and W[1]-hardness for the solution size parameter.On the positive side our main results are fixed-parameter tractability for the parameter “numberof vertices” and the treewidth of the underlying graph combined with the lifetime.We conclude with some challenges for future research. For the parameter lifetime τ , it remainsopen whether there exists a polynomial-time algorithm for instances with 3 ≤ τ ≤ τ = 2 in the non-strict case. We believe that, for the strict case, our 3-SAT reductioncan be modified to use only seven time-labels. Similarly to the work of Zschoche et al. [23] in thecontext of temporal separators, we could not resolve the question whether the non-strict variantsare fixed-parameter tractable for the combined parameter τ + k , whereas for the strict case, this isalmost trivial. Additionally, it seems natural to study (S)TFES and (S)TFCS variants restrictedto specific temporal graph classes (e.g., see Fluschnik et al. [13]).Finally, we remark that we focused on finding feedback edge sets, ignoring the presumablyharder vertex variant; however, one can observe that our W[1]-hardness result also transfers to theproblem of finding feedback vertex sets in temporal graphs. References [1] Akanksha Agrawal, Fahad Panolan, Saket Saurabh, and Meirav Zehavi. 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