MMultistage Vertex Cover
Till Fluschnik
Technische Universität Berlin, Algorithmics and Computational Complexity, Germanytill.fl[email protected]
Rolf Niedermeier
Technische Universität Berlin, Algorithmics and Computational Complexity, [email protected]
Valentin Rohm
Technische Universität Berlin, Algorithmics and Computational Complexity, [email protected]
Philipp Zschoche
Technische Universität Berlin, Algorithmics and Computational Complexity, [email protected]
Abstract
Covering all edges of a graph by a small number of vertices, this is the NP-complete
Vertex Cover problem. It is among the most fundamental graph-algorithmic problems. Following a recent trend instudying temporal graphs (a sequence of graphs, so-called layers, over the same vertex set but, overtime, changing edge sets), we initiate the study of
Multistage Vertex Cover . Herein, given atemporal graph, the goal is to find for each layer of the temporal graph a small vertex cover and toguarantee that two vertex cover sets of every two consecutive layers differ not too much (specified bya given parameter). We show that, different from classic
Vertex Cover and some other dynamicor temporal variants of it,
Multistage Vertex Cover is computationally hard even in fairlyrestricted settings. On the positive side, however, we also spot several fixed-parameter tractabilityresults based on some of the most natural parameterizations.
Mathematics of computing → Graph algorithms
Keywords and phrases parameterized algorithmics, NP-completeness, temporal graphs, data reduc-tion
Funding
Till Fluschnik : Supported by the DFG, project TORE (NI 369/18).
Vertex Cover asks, given an undirected graph G and an integer k ≥
0, whether at most k vertices can be deleted from G such that the remaining graph contains no edge. VertexCover is NP-complete and it is a formative problem of algorithmics and combinatorialoptimization. We study a time-dependent , “ multistage ” version, namely a variant of
VertexCover on temporal graphs. A temporal graph G is a tuple ( V, E , τ ) consisting of a set V of vertices, a discrete time-horizon τ , and a set of temporal edges E ⊆ (cid:0) V (cid:1) × { , . . . , τ } .Equivalently, a temporal graph G can be seen as a vector ( G , . . . , G τ ) of static graphs( layers ), where each graph is defined over the same vertex set V . Then, our specific goalis to find a small vertex cover S i for each layer G i such that the size of the symmetricdifference S i S i +1 = ( S i \ S i +1 ) ∪ ( S i +1 \ S i ) of the vertex covers S i and S i +1 of every twoconsecutive layers G i and G i +1 is small. Formally, we thus introduce and study the followingproblem (see Figure 1 for an illustrative example). a r X i v : . [ c s . CC ] A p r Multistage Vertex Cover
Multistage Vertex Cover
Input:
A temporal graph G = ( V, E , τ ) and two integers k ∈ N , ‘ ∈ N . Question:
Is there a sequence S = ( S , . . . , S τ ) such that(i) for all i ∈ { , . . . , τ } , it holds true that S i ⊆ V is a size-at-most- k vertex coverfor layer G i , and(ii) for all i ∈ { , . . . , τ − } , it holds true that | S i S i +1 | ≤ ‘ ?Throughout this paper we assume that 0 < k < | V | because otherwise we have a trivialinstance. In our model, we follow the recently proposed multistage [4–6, 9, 15, 19, 22, 24]view on classical optimization problems on temporal graphs.In general, the motivation behind a multistage variant of a classical problem such as Vertex cover is that the environment changes over time (here reflected by the changingedge sets in the temporal graph) and a corresponding adaptation of the current solutioncomes with a cost. In this spirit, the parameter ‘ in the definition of Multistage VertexCover allows to model that only moderate changes concerning the solution vertex set maybe wanted when moving from one layer to the subsequent one. Indeed, in this sense ‘ can beinterpreted as a parameter measuring the degree of (non-)conservation [1, 23].It is immediate that Multistage Vertex Cover is NP-hard as it generalizes
VertexCover ( τ = 1). We will study its parameterized complexity regarding the problem-specificparameters k , τ , ‘ , and some of their combinations, as well as restrictions to temporal graphclasses [8, 18]. G v v v v G v v v v G v v v v Figure 1
An illustrative example with temporal graph G = ( G , G , G ) over the vertex set V = { v , . . . , v } . A solution S = ( { v , v } , { v } , { v , v } ) for k = 2 and ‘ = 1 is highlighted. The literature on vertex covering is extremely rich, even when focusing on parameterizedcomplexity studies. Indeed,
Vertex Cover can be seen as “drosophila” of parameterizedalgorithmics. Thus, we only consider
Vertex Cover studies closely related to our setting.First, we mention in passing that
Vertex Cover is studied in dynamic graphs [3, 26] andgraph stream models [10]. More importantly for our work, Akrida et al. [2] studied a variantof
Vertex Cover on temporal graphs. Their model significantly differs from ours: theywant an edge to be covered at least once over every time window of some given size ∆. Thatis, they define a temporal vertex cover as a set S ⊆ V × { , . . . , τ } such that, for every timewindow of size ∆ and for each edge e = { v, w } appearing in a layer contained in the timewindow, it holds that ( v, t ) ∈ S or ( w, t ) ∈ S for some t in the time window with ( e, t ) ∈ E .For their model, Akrida et al. ask whether such an S of small cardinality exists. Notethat if ∆ >
1, then for some t ∈ { , . . . , τ } the set S t := { v | ( v, t ) ∈ S } is not necessarilya vertex cover of layer G t . For ∆ = 1, each S t must be a vertex cover of G t . However,in Akrida et al.’s model the size of each S t as well as the size of the symmetric differencebetween each S t and S t +1 may strongly vary. They provide several hardness results andalgorithms (mostly referring to approximation or exact algorithms, but not to parameterizedcomplexity studies). . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 3 Table 1
Overview of our results. The column headings describe the restrictions on the inputand each row corresponds to a parameter. p-NP-hard, PK, and NoPK abbreviate para-NP-hard,polynomial-size problem kernel, and no problem kernel of polynomial size unless coNP ⊆ NP / poly.general layers tree layers one-edge layers0 ≤ ‘ < k ‘ ≥ k ≤ ‘ < k ≤ ‘ < τ p-NP-hard p-NP-hard p-NP-hard FPT, PK (Thm. 4.1) (Thm. 4.1) (Thm. 4.1) (Obs. 6.9) k XP, W[1]-h., FPT † , NoPK XP, W[1]-h. open , NoPK (Thm. 5.1) (Obs. 3.5, Thm. 6.1) (Thm. 5.1, Cor. 5.21) (Thm. 6.1) k + τ FPT, PK FPT, PK FPT, PK FPT, PK (Thm. 6.6) (Thm. 6.6) (Thm. 6.6) (Thm. 6.6)
A second related line of research, not directly referring to temporal graphs though, studiesreconfiguration problems which arise when we wish to find a step-by-step transformationbetween two feasible solutions of a problem such that all intermediate results are feasiblesolutions as well [21, 25]. Among other reconfiguration problems, Mouawad et al. [28, 29]studied
Vertex Cover Reconfiguration : given a graph G , two vertex covers S and T each of size at most k , and an integer τ , the question is whether there is a sequence ( S = S , . . . , S τ = T ) such that each S t is a vertex cover of size at most k . The essential differenceto our model is that from one “sequence element” to the next only one vertex may be changedand that the input graph does not change over time. Indeed, there is an easy reduction of thismodel to ours while the opposite direction is unlikely to hold. This is substantiated by the factthat Mouawad et al. [29] showed that Vertex Cover Reconfiguration is fixed-parametertractable when parameterized by vertex cover size k while we show W[1]-hardness for thecorresponding case of Multistage Vertex Cover .Finally, there is also a close relation to the research on dynamic parameterized problems [1,27]. Krithika et al. [27] studied
Dynamic Vertex Cover where one is given two graphson the same vertex set and a vertex cover for one of them together with the guarantee thatthe cardinality of the symmetric difference between the two edge sets is upper-bounded bya parameter d . The task then is to find a vertex cover for the second graph that is “closeenough” (measured by a second parameter) to the vertex cover of the first graph. They showfixed-parameter tractability and a linear kernel with respect to d . Our results, focusing on the three perhaps most natural parameters, are summarized inTable 1. We highlight a few specific results.
Multistage Vertex Cover remains NP-hardeven if every layer consists of only one edge; not surprisingly, the corresponding hardnessreduction exploits an unbounded number τ of time layers. If one only has two layers,however, one of them being a tree and the other being a path, then again MultistageVertex Cover already becomes NP-hard.
Multistage Vertex Cover parameterizedby solution size k is fixed-parameter tractable if ‘ ≥ k , but becomes W[1]-hard if ‘ < k .Considering the tractability results for Dynamic Vertex Cover [27] and
Vertex CoverReconfiguration [29], this hardness is surprising, and it is our most technical result.Furthermore,
Multistage Vertex Cover parameterized by k with ‘ ≥ k does not admit Multistage Vertex Cover a problem kernel of polynomial size unless coNP ⊆ NP / poly. Finally, for the combinedparameter k + τ we obtain polynomial-sized problem kernels (and thus fixed-parametertractability) in all cases without any further constraints. In Section 2, we provide some preliminaries. For
Multistage Vertex Cover , we give somefirst and general observations in Section 3, study the parameterized complexity regarding k in Section 5, and discuss the possibilities for efficient data reduction in Section 6. We concludein Section 7. We denote by N and N the natural numbers excluding and including zero, respectively.For two sets A and B , we denote by A B := ( A \ B ) ∪ ( B \ A ) = ( A ∪ B ) \ ( A ∩ B ) thesymmetric difference of A and B , and by A ] B the disjoint union of A and B . A temporal graph G is a tuple ( V, E , τ ) consisting of the set V of vertices, the set E of temporaledges, and a discrete time-horizon τ . A temporal edge e is an element in (cid:0) V (cid:1) × { , . . . , τ } .Equivalently, a temporal graph G can be defined as a vector of static graphs ( G , . . . , G τ ),where each graph is defined over the same vertex set V . We also denote by V ( G ), E ( G ),and τ ( G ) the set of vertices, the set of temporal edges, and the discrete time-horizon of G ,respectively. The underlying graph G ↓ = G ↓ ( G ) of a temporal graph G is the static graphwith vertex set V ( G ) and edge set { e | ∃ t ∈ { , . . . , τ ( G ) } : ( e, t ) ∈ E ( G ) } . Let Σ be a finite alphabet. A parameterized problem L is a subset L ⊆ { ( x, k ) ∈ Σ ∗ × N } .An instance ( x, k ) ∈ Σ ∗ × N is a yes -instance of L if and only if ( x, k ) ∈ L (otherwise, it is a no -instance). Two instances ( x, k ) and ( x , k ) of parameterized problems L, L are equivalent if ( x, k ) ∈ L ⇐⇒ ( x , k ) ∈ L . A parameterized problem L is fixed-parameter tractable(FPT) if for every input ( x, k ) one can decide whether ( x, k ) ∈ L in f ( k ) · | x | O (1) time,where f is some computable function only depending on k . A parameterized problem L is in XP if for every instance ( x, k ) one can decide whether ( x, k ) ∈ L in time | x | f ( k ) forsome computable function f only depending on k . A W[1]-hard parameterized problem isfixed-parameter intractable unless FPT=W[1].Given a parameterized problem L , a kernelization is an algorithm that maps any in-stance ( x, k ) of L in time polynomial in | x | + k to an instance ( x , k ) of L (the problem kernel )such that (i) ( x, k ) ∈ L ⇐⇒ ( x , k ) ∈ L , and (ii) | x | + k ≤ f ( k ) for some computablefunction f (the size of the problem kernel) only depending on k .We refer to Downey and Fellows [13] and Cygan et al. [11] for more material on parame-terized complexity. In this section, we state some preliminary simple-but-useful observations on
MultistageVertex Cover and its relation to
Vertex Cover . . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 5 (cid:66) Observation 3.1.
Every instance ( G , k, ‘ ) of Multistage Vertex Cover with k ≥ P τ ( G ) i =1 | E ( G i ) | is a yes -instance. Proof.
It is easy to see that a graph with m edges always admits a vertex cover of size m .Hence, there is a vertex cover S ⊆ V of size k of G ↓ ( G ), and hence, S is a vertex coverfor each layer. The vector ( S , . . . , S τ ) with S i = S for all i ∈ { , . . . , τ } is a solution forevery ‘ ≥ (cid:74) Next, we state that if we are facing a yes -instance, then we can assume that there exists asolution where each layer’s vertex cover is either of size k or k − (cid:66) Observation 3.2.
Let ( G , k, ‘ ) be an instance of Multistage Vertex Cover . If( G , k, ‘ ) is a yes -instance, then there is a solution S = ( S , . . . , S τ ) such that | S | = k and k − ≤ | S i | ≤ k for all i ∈ { , . . . , τ } . Proof.
We first show that there is a solution S = ( S , . . . , S τ ) for I := ( G , k, ‘ ) such that | S | = k . Towards a contradiction assume that such a solution does not exist. Let S =( S , . . . , S τ ) be a solution such that | S | is maximal over all solutions for I . Let i ∈ { , . . . , τ } be the maximum index such that S j ⊆ S j − , for all j ∈ { , . . . , i } . If i = τ , then we havethat | S j | ≤ | S | < k for all j ∈ { , . . . , τ } . Hence, we can find a subset X ⊆ V \ S suchthat ( S ∪ X, . . . , S τ ∪ X ) is a solution. This contradicts | S | being maximal. Now let i < τ .Hence, there is a vertex v ∈ S i +1 \ S i . Now we can adjust the solution by adding v to S j for all j ∈ { , . . . , i } . This contradicts | S | being maximal. Hence, there is a solution S = ( S , . . . , S τ ) such that | S | = k .Let Ψ be the set of solutions such that the first vertex cover is of size k . Assume towardsa contradiction that all solutions in Ψ contain a vertex cover smaller than k −
1. Let Ψ i ⊆ Ψbe the set of solutions such that for each ( S , . . . , S τ ) ∈ Ψ i we have that | S i | < k − | S j | ≥ k − j ∈ { , . . . , i − } . Let i ∈ { , . . . , τ } be maximal such that Ψ i = ∅ .Furthermore, let S = ( S , . . . , S τ ) ∈ Ψ i such that | S i | is maximal over all solutions in Ψ i .Hence, there is a vertex v ∈ S i − \ S i . We distinguish two cases. (a): Assume that there is a p ∈ { i +1 , . . . , τ } such that there is a w ∈ S p \ S p − and S j ⊆ S j − for all j ∈ { i + 1 , . . . , p − } . The idea now is to keep v and add w in the i -th layer andthen remove v in the p -th layer. We can achieve this by simply setting S q := S q ∪ { v, w } for all q ∈ { i, . . . , p − } . Note that this is a solution which either contradicts that | S i | ismaximal or that i is maximal. (b): Now assume that S j ⊆ S j − for all j ∈ { i + 1 , . . . , τ } . In this case we take an arbitraryvertex w ∈ V \ S i and set S q := { v, w } for all q ∈ { i, . . . , τ } . This contradicts i beingmaximal. (cid:74) With the next two observations, we show that the special case of
Multistage VertexCover where ‘ = 0 is equivalent to Vertex Cover under polynomial-time many-onereductions. (cid:66)
Observation 3.3.
There is a polynomial-time algorithm that maps any instance ( G =( V, E ) , k ) of Vertex Cover to an equivalent instance ( G , k, ‘ ) of Multistage VertexCover where ‘ = 0 and every layer G i contains only one edge. Proof.
Let the edges E = { e , . . . , e m } of G be ordered in an arbitrary way. Set τ = m and ‘ = 0. Set G i = ( V, { e i } ) for each i ∈ { , . . . , τ } . We claim that ( G = ( V, E ) , k ) is a yes -instance of Vertex Cover if and only if ( G , k, ‘ ) is a yes -instance of MultistageVertex Cover . Multistage Vertex Cover ( ⇒ ) Let S be a vertex cover of G of size at most k . Set S i := S for all i ∈ { , . . . , τ } .Clearly, S i is a vertex cover of G i for all i ∈ { , . . . , τ } of size at most k . Moreover, byconstruction, | S i S i +1 | ≤ i ∈ { , . . . , τ − } . Hence, ( S , . . . , S τ ) forms a solutionto ( G , k, ‘ ).( ⇐ ) Let S = ( S , . . . , S τ ) be a solution to ( G , k, ‘ ). Observe that | S i S i | ≤ k . It followsthat there are at most k vertices covering all edges of the layers G i , that is, E = S τi =1 E ( G i ),and hence they cover all edges of G . (cid:74)(cid:66) Observation 3.4.
There is a polynomial-time algorithm that maps any instance ( G , k, ‘ )of Multistage Vertex Cover with ‘ = 0 to an equivalent instance ( G, k ) of
VertexCover . Proof.
Now let ( G = ( V, E , τ ) , k,
0) be an arbitrary instance of
Multistage Vertex Cover .Construct the instance ( G ↓ , k ) of Vertex Cover . We claim that ( G , k,
0) is a yes -instanceif and only if ( G ↓ , k ) is a yes -instance.( ⇐ ) Let S ⊆ V be a vertex cover of size at most k . Since S is a vertex cover for G ↓ , S covers each layer of G . Hence, S i := S for all i ∈ { , . . . , τ } forms a solution to ( G , k, ⇒ ) Let ( S , . . . , S τ ) be a solution to ( G , k, ‘ = 0, we have that S i = S j for all i, j ∈ { , . . . , τ } . It is not difficult to see that S := S is a vertex cover for G ↓ , andhence the claim follows. (cid:74) Finally, the special case of
Multistage Vertex Cover with ‘ ≥ k (that is, where vertexcovers of any two consecutive layers can be even disjoint) is Turing-reducible to VertexCover . (cid:66) Observation 3.5.
Any instance ( G , k, ‘ ) of Multistage Vertex Cover with ‘ ≥ k and G = ( G , . . . , G τ ) can be decided by deciding each instance of the set { ( G i , k ) | ≤ i ≤ τ } of Vertex Cover -instances.
Proof.
For each of the layers G i , i ∈ { , . . . , τ } , we can construct an instance of VertexCover of the form ( G i , k ). We can solve each instance independently, since the symmetricdifference of any two size-at-most- k solutions is at most 2 k ≤ ‘ . (cid:74) Multistage Vertex Cover is NP-hard as it generalizes
Vertex Cover ( τ = 1). In thissection we prove that Multistage Vertex Cover remains NP-hard on inputs with onlytwo layers, one consisting of a path and the other consisting of a tree, and on inputs whereevery layer consists only of one edge. (cid:73)
Theorem 4.1.
Multistage Vertex Cover is NP -hard even if(i) τ = 2 , ‘ = 0 , and the first layer is a path and the second layer is a tree, or(ii) every layer contains only one edge and ‘ ≤ . (cid:73) Remark 4.2.
Theorem 4.1(i) is tight regarding τ since Vertex Cover (i.e.,
MultistageVertex Cover with τ = 1) on trees is solvable in linear time. Theorem 4.1(ii) is tightregarding ‘ , because in the case of ‘ > Vertex Cover remains NP-complete on cubic Hamiltonian graphs when a Hamiltoniancycle is additionally given in the input [16]: A graph is cubic if each vertex is of degree exactly three; A graph is Hamiltonian if it contains asubgraph being a Hamiltonian cycle, that is, a cycle that visits each vertex in the graph exactly once. . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 7
Hamiltonian Cubic Vertex Cover (HCVC)
Input:
An undirected, cubic, Hamiltonian graph G = ( V, E ), an integer k ∈ N , and a Hamilto-nian Cycle C = ( V, E ) of G . Question:
Is there a set S ⊆ V such that S is a size-at-most- k vertex cover for G ? To prove Theorem 4.1(i), we give a polynomial-time many-one reduction from HCVC to
Multistage Vertex Cover with two layers, one being a path, the other being a tree. (cid:73)
Proposition 4.3.
There is a polynomial-time algorithm that maps any instance ( G =( V, E ) , k, C ) of HCVC to an equivalent instance ( G , k , ‘ ) of Multistage Vertex Cover with τ = 2 and the first layer G being a path and second layer G being a tree. Proof.
Let e ∈ E ( C ) be some edge of C , and let P = C − e be the Hamiltonian path obtainedfrom C when removing e . Let E := E ( P ), and E := E \ E ( P ). Set initially G = ( V, E )and G = ( V, E ). Note that G is a path. Moreover, observe that G is the disjoint unionof | V | / − G − E ( C ) is adisjoint union of | V | / G , isadjacent to two vertices in C , and thus has degree one in G − E ( C ); Since G − E ( C ) = G − e ,edge e connects two paths of length one to one path of length three in G . Add two specialvertices z, z to V . In G , connect z with z and with one endpoint of P . In G , connect z with z and with exactly one vertex of each connected component. Set k = k + 1 and ‘ = 0.We claim that ( G = ( V, E ) , k, C ) is a yes -instance if and only if ( G , k , ‘ ) is a yes -instance.( ⇒ ) Let S be a vertex cover of G of size at most k . We claim that S := S ∪ { z } is avertex cover for both G and G . Observe that G [ E ] and G [ E ] are subgraphs of G , andhence all edges are covered by S . Moreover, all edges in G i − E i , i ∈ { , } , are incidentwith z and hence covered by S .( ⇐ ) Let ( S , S ) be a minimal solution to ( G , k , ‘ ) with S := S = S and | S | ≤ k . Wecan assume that z ∈ S since the edge { z, z } is present in both G and G , and exchanging z in z does not cover less edges. Moreover, we can assume that not both z and z are in S due to the minimality of S . Let S := S \ { z } . Observe that S covers all edges in E ∪ E and, hence, S forms a vertex cover of G of size at most k = k − (cid:74) Note that Theorem 4.1(ii) for ‘ = 0 is already shown by Observation 3.3. In order toprove Theorem 4.1(ii) for ‘ = 1, we adjust the polynomial-time many-one reduction behindObservation 3.3. (cid:73) Proposition 4.4.
There is a polynomial-time algorithm that maps any instance ( G =( V, E ) , k ) of Vertex Cover to an equivalent instance ( G , k , ‘ ) of Multistage VertexCover where ‘ = 1 and every layer G i contains only one edge. Proof.
Let the edges E = { e , . . . , e m } of G be arbitrarily ordered. Set τ = 2 m . Set V = V ∪ W , where W = { w , . . . , w τ } . Set G i − = ( V , { e i } ) and G i = ( V , { w i , w i + τ } ) foreach i ∈ { , . . . , τ } . Set k = k + 1 and ‘ = 1. We claim that ( G = ( V, E ) , k ) is a yes -instanceof Vertex Cover if and only if ( G , k , ‘ ) is a yes -instance of Multistage Vertex Cover .( ⇒ ) Let S be a vertex cover of G of size at most k . Set S i − := S , and S i := S ∪ { w i } for all i ∈ { , . . . , τ } . Clearly, S i is a vertex cover of G i for all i ∈ { , . . . , τ } of size atmost k = k + 1. Moreover, by construction, | S i S i +1 | ≤ i ∈ { , . . . , τ − } . Hence,( S , . . . , S τ ) forms a solution to ( G , k , ‘ ).( ⇐ ) Let S = ( S , . . . , S τ ) be a solution to ( G , k , ‘ ). Observe that | S i S i | ≤ k + τ . Weknow that | W ∩ S i S i | ≥ τ . It follows that there are at most k vertices covering all edges ofthe layers G i − , that is, E = S τi =1 E ( G i − ), and, hence, covering all edges of G . (cid:74) Theorem 4.1 now follows from Propositions 4.3 and 4.4.
Multistage Vertex Cover
In this section, we study the parameter size k of the vertex cover of each layer for MultistageVertex Cover . Vertex Cover and
Vertex Cover Reconfiguration [29] whenparameterized by the vertex cover size are fixed-parameter tractable. We prove that this isno longer true for
Multistage Vertex Cover (unless FPT = W[1]). (cid:73)
Theorem 5.1.
Multistage Vertex Cover parameterized by k is in XP and W[1] -hard.
We first show the XP-algorithm (Section 5.1), and then prove the W[1]-hardness (Section 5.2)and discuss its implications.
In this section, we prove the following. (cid:73)
Proposition 5.2.
Every instance ( G , k, ‘ ) of Multistage Vertex Cover can be decidedin O ( τ ( G ) · | V ( G ) | k +1 ) time. In a nutshell, to prove Proposition 5.2 we first consider for each layer all vertex subsetsof size at most k that form a vertex cover. Second, we find a sequence of vertex covers for alllayers such that the sizes of the symmetric differences for every two consecutive solutions isat most ‘ . We show that the second step can be solved via computing a source-sink path inan auxiliary directed graph that we call configuration graph (see Figure 2 for an illustrativeexample). (cid:73) Definition 5.3.
Given an instance I = ( G , k, ‘ ) of Multistage Vertex Cover , the configuration graph of I is the directed graph D = ( V, A, γ ) with V = V ] · · · ] V τ ] { s, t } ,being equipped with a function γ : V → { V ⊆ V ( G ) | | V | ≤ k } such that(i) for every i ∈ { , . . . , τ ( G ) } , it holds true that S is a vertex cover of G i of size exactly k − or k if and only if there is a vertex v ∈ V i with γ ( v ) = S ,(ii) there is an arc from v ∈ V to w ∈ V if and only if v ∈ V i , w ∈ V i +1 , and | γ ( v ) γ ( w ) | ≤ ‘ , and(iii) there is an arc ( s, v ) for all v ∈ V and an arc ( v, t ) for all v ∈ V τ . Note that Mouawad et al. [29] used a similar configuration graph to show fixed-parametertractability of
Vertex Cover Reconfiguration parameterized by the vertex cover size k .In the multistage setting the configuration graph is too large for fixed-parameter tractabilityregarding k . However, we show an XP-algorithm regarding k to construct the configurationgraph. (cid:73) Lemma 5.4.
The configuration graph of an instance ( G , k, ‘ ) of Multistage VertexCover , where temporal graph G has n vertices and time horizon τ ,(i) can be constructed in O ( τ · n k +1 ) time, and(ii) contains at most τ · n k + 2 vertices and ( τ − n k + 4 n k arcs. Proof.
Compute the set S = { V ⊆ V ( G ) | k − ≤ | V | ≤ k } in O ( n k ) time. For each layer G i and each set S ∈ S , check in O ( | E ( G i ) | ) time whether S is a vertex cover for G i . Let S i ⊆ S denote the set of vertex covers of size k − k of layer G i . For each S ∈ S i , add a vertex v to V i and set γ ( v ) = S . Lastly, add the vertices s and t . Hence, we can construct the vertex set V of the configuration graph D of size τ · n k +2 in O ( n k +2 · τ ) time. For every i ∈ { , . . . , τ − } ,and every v ∈ V i and w ∈ V i +1 , check whether | γ ( v ) γ ( w ) | ≤ ‘ in O ( k ) time. If this is . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 9 G v v v v G v v v v G v v v v (a) s t { v , v }{ v , v }{ v , v } V { v , v }{ v }{ v , v }{ v , v }{ v }{ v , v }{ v , v } V { v , v }{ v , v } V (b) Figure 2
Illustrative example of a configuration graph. (a) Temporal graph instance I = ( G , k, ‘ )from Figure 1 with G = ( G , G , G ), k = 2, and ‘ = 1. (b) Configuration graph of I from (a); adirected s - t path is highlighted corresponding to the solution depicted in Figure 1. the case, then add the arc ( v, w ). The latter steps can be done in O ( n k +1 · ( τ − s, v ) for each v ∈ V and the arc ( v, t ) for each v ∈ V τ in O ( n k ) time.The finishes the construction of D = ( V = V ] · · · ] V τ ] { s, t } , A, γ ). (cid:74) The crucial observation is that we can decide any instance by checking for an s - t path inits configuration graph. (cid:73) Lemma 5.5.
Multistage Vertex Cover -instance I = ( G , k, ‘ ) is a yes -instance if andonly if there is an s - t path in the configuration graph D of I . Proof.
Let D = ( V = V ] · · · ] V τ ] { s, t } , A, γ ).( ⇒ ) Let ( S , . . . , S τ ) be a solution to ( G , k, ‘ ). By Observation 3.2, we can assumewithout loss of generality that k − ≤ | S i | ≤ k , for all i ∈ { , . . . , τ } . Hence for each S i ,there is a v i ∈ V i such that γ ( v i ) = S i , for all i ∈ { , . . . , τ } . Note that the arc ( v i , v i +1 ) iscontained in A for each i ∈ { , . . . , τ − } since | γ ( v i ) γ ( v i +1 ) | = | S i S i +1 | ≤ ‘ . Hence, P =( { v , . . . , v τ } ∪ { s, t } , { ( s, v ) , ( v τ , t ) } ∪ S τ − i =1 { ( v i , v i +1 ) } ) is an s - t path in D .( ⇐ ) Let P = ( { v , . . . , v τ } ∪ { s, t } , { ( s, v ) , ( v τ , t ) } ∪ S τ − i =1 { ( v i , v i +1 ) } ) be an s - t pathin D . We claim that ( γ ( v i )) i ∈{ ,...,τ } forms a solution to ( G , k, ‘ ). First, note that for all i ∈{ , . . . , τ } , γ ( v i ) is a vertex cover for G i of size at most k . Moreover, for all i ∈ { , . . . , τ − } , | γ ( v i ) γ ( v i +1 ) | ≤ ‘ since the arc ( v i , v i +1 ) is present in D . This finishes the proof. (cid:74) We are ready to prove Proposition 5.2. of Proposition 5.2.
First, compute the configuration graph D of the instance ( G =( V, E , τ ) , k, ‘ ) of Multistage Vertex Cover in O ( τ · | V | k +1 ) time (Lemma 5.4(i)). Then,find an s - t path in D with a breadth-first search in O ( τ · | V | k ) time (Lemma 5.4(ii)). Ifan s - t path is found, then return yes , otherwise return no (Lemma 5.5). (cid:74)(cid:73) Remark 5.6.
The reason why the algorithm behind Proposition 5.2 is only an XP-algorithmand not an FPT-algorithm regarding k is because we do not have a better upper bound onthe number of vertices in the configuration graph for ( G , k, ‘ ) than O ( τ ( G ) · | V ( G ) | k ). This is u vwe e e (cid:32) c c c c c c c c c c c c c u u u u u u u u u u u u e u v e v w e u w Figure 3
Illustration of Construction 1 on an example graph (left-hand side) and the first sevenlayers of the obtained graph (right-hand side). Dashed vertical lines separate layers, and for eachlayer all present edges (but only their incident vertices) are depicted. Star-shapes illustrate stargraphs with k + 1 leaves. Vertices in a solution (layers’ vertex covers) are highlighted. due to the fact that we check for each subset of V ( G ) of size k or k − Minimal Multistage Vertex Cover where we additionallydemand the i -th set in the solution to be a minimal vertex cover for the layer G i . Here,we can enumerate for each layer G i all minimal vertex covers of size at most k (andhence all candidates for the i -th set of the solution) with the folklore search-tree algorithmfor vertex cover. This leads to O (2 k τ ( G )) many vertices in the configuration graph (for Minimal Multistage Vertex Cover ) and thus to fixed-parameter tractability of
MinimalMultistage Vertex Cover parameterized by the vertex cover size k .However, it is unlikely (unless FPT=W[1]) that one can substantially improve thealgorithm behind Proposition 5.2, as we show next. In this section we show that
Multistage Vertex Cover is W[1]-hard when parameterizedby k . This hardness result is established by the following parameterized reduction from theW[1]-complete [12] Clique problem, where, given an undirected graph G and a positiveinteger k , the question is whether G contains a clique of size k (that is, k vertices that arepairwise adjacent). (cid:73) Proposition 5.7.
There is an algorithm that maps any instance ( G, k ) of Clique inpolynomial time to an equivalent instance ( G , k , ‘ ) of Multistage Vertex Cover with k =2 (cid:0) k (cid:1) + k + 1 , ‘ = 2 , and each layer of G being a forest with O ( k ) edges. In the remainder of this section, we prove Proposition 5.7. We next give the constructionof the
Multistage Vertex Cover instance, then prove the forward (Section 5.2.1) andbackward (Section 5.2.2) direction of the equivalence, and finally (in Section 5.2.3) put thepieces together and derive two corollaries.We construct an instance of
Multistage Vertex Cover from an instance of
Clique as follows (see Figure 3 for an illustrative example). . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 11 (cid:73)
Construction 1.
Let ( G = ( V, E ) , k ) be an instance of Clique with m := | E | and E = { e , . . . , e m } . Let K := (cid:0) k (cid:1) , k := 2 K + k + 1 , and κ := K + k + 3 . We construct a temporal graph G = ( V , E , τ ) as follows. Let V be initialized to V ∪ E (notethat E simultaneously describes the edge set of G and a vertex subset of G ). We add thefollowing vertex sets U t := { u tj | j ∈ { , . . . , K }} for every t ∈ { , . . . , κ + 1 } , and C := { c , . . . , c mκ +1 } (we refer to C as the set of center vertices) . Let E be initially empty. We extend the set V and define E through the τ := 2 mκ + 1 layerswe construct in the following.(1) In each layer G i with i being odd, make c i the center of a star with k + 1 leaves. (2) In each layer G mj +1 , j ∈ { , . . . , κ } , make each vertex in U j +1 the center of a starwith k + 1 leaves.(3) For each j ∈ { , . . . , κ − } , in each layer G mj + i with i ∈ { , . . . , m + 1 } , make u j +1 x adjacent to u j +2 x for each x ∈ { , . . . , K } .(4) For each even i , add the edge { c i , c i +1 } to G i and to G i +1 .(5) For each j ∈ { , . . . , κ − } , for each i ∈ { , . . . , m } , in G mj +2 i , make c j m +2 i adjacentwith e i = { v, w } , v , and w .This finishes the construction of G . (cid:7) The construction essentially repeats the same gadget (which we call phase ) κ times, wherethe layer 2 m · i + 1 is simultaneously the last layer of phase i and the first layer of phase i + 1.In the beginning of phase i , a solution has to contain the vertices of U i . The idea now isthat during phase i one has to exchange the vertices of U i with the vertices of U i +1 .It is not difficult to see that the instance in Construction 1 can be computed in polynomialtime. Hence, it remains to prove the equivalence stated in Proposition 5.7. Recall that weprove the forward and the backward direction in Sections 5.2.1 and 5.2.2, respectively, andfinally prove Proposition 5.7 in Section 5.2.3. The forward direction of Proposition 5.7 is—in a nutshell—as follows: If V ∪ E with V ⊆ V and E ⊆ E corresponds to the vertex set and edge set of a clique of size k , then thereare K layers in each phase covered by V ∪ E . Hence, having K layers where no verticesfrom C have to be exchanged, in each phase t we can exchange all vertices from U t to U t +1 .Starting with set S = U ∪ V ∪ E ∪ { c } then yields a solution. (cid:73) Lemma 5.8.
Let ( G, k ) be an instance of Clique and ( G , k , ‘ ) be the instance of Mul-tistage Vertex Cover resulting from Construction 1. If ( G, k ) is a yes -instance, then ( G , k , ‘ ) is a yes -instance. Proof.
Let G = ( V , E ) be the clique of size k in G . We construct a solution S =( S , . . . , S m , S m +1 = S , . . . , S κ m +1 = S κ +11 ) for ( G , k , ‘ ) in the following way. For each t ∈{ , . . . , κ + 1 } we set S t = V ∪ E ∪ U t ∪ { c ( t − m +1 } , which is a vertex cover of size k for G ( t − m +1 . A star (graph) is a tree where at most one vertex (so-called center) is of degree larger than one.
Now, for each t ∈ { , . . . , κ } , we iteratively construct vertex covers for the layers ( t − m + 2 until t m in the following way. Let T := ( t − · m . Let i ∈ { , . . . , m − } , andassume that the set S ti is already constructed and is a vertex cover for G T + i (this is possibledue to the definition of S t ). We distinguish two cases. Case 1: i is odd . We know that c T + i ∈ S ti . If ( S ti \ { c T + i } ) ∪ { c T + i +2 } is a vertex coverfor G T + i +1 , then we set S ti +1 = ( S ti \ { c T + i } ) ∪ { c T + i +2 } . Otherwise we set S ti +1 =( S ti \ { c T + i } ) ∪ { c T + i +1 } . In both cases S ti +1 is a vertex cover for G T + i +1 and either S ti +1 ∩ C = { c T + i +1 } or S ti +1 ∩ C = { c T + i +2 } . Case 2: i is even . We know that c T + i or c T + i +1 is in S ti . If c T + i ∈ S ti , then we set S ti +1 = ( S ti \ { c T + i } ) ∪ { c T + i +1 } , which is a vertex cover for G T + i +1 . If c T + i +1 ∈ S ti ,then S ti is already a vertex cover for G T + i +1 and the vertices in V ∪ E cover all edgesincident with c T + i in the graph G T + i . In this case we say that G covers the layer T + i and set S ti +1 = ( S ti \ { u tj } ) ∪ { u t +1 j } , where u tj is an arbitrary vertex in S ti ∩ U t .Observe that the clique G covers K even-numbered layers in each phase. Hence, wereplace, during phase t ∈ { , . . . , κ } (that is, from layer ( t − m + 1 to t m + 1), thevertices U t with the vertices U t +1 . This also implies that the symmetric difference of twoconsecutive sets in S is exactly 2 = ‘ . It follows that S is a solution for ( G , k , ‘ ). (cid:74) In this section we prove the backward direction for the proof of Proposition 5.7. We firstshow that if an instance of
Multistage Vertex Cover computed by Construction 1 is a yes -instance, then it is safe to assume that two vertices are neither deleted from nor added toa vertex cover in a consecutive step (we refer to these solutions as smooth , see Definition 5.10).Moreover, a vertex from the vertex set C is only exchanged with another vertex from C and, at any time, there is exactly one vertex from C contained in the solution (similarlyto the constructed solution in Lemma 5.8). We call these (smooth) solutions one-centered (Definition 5.12). We then prove that there must be a phase t for any one-centered solutionwhere at least (cid:0) k (cid:1) times a vertex from “past” sets U t , t ≤ t is deleted. This at hand, weprove that such a phase witnesses a clique of size k .The fact that a solution needs to contain at least one vertex from C at any timeimmediately follows from the fact that there is either an edge between two vertices in C orthere is a vertex in C which is the center of a star with k + 1 leaves. (cid:66) Observation 5.9.
Let ( G , k , ‘ ) from Construction 1 be a yes -instance. Then for eachsolution ( S , . . . , S τ ) it holds true that | S i ∩ C | ≥ i ∈ { , . . . , τ ( G ) } .In the remainder of this section we denote the vertices which are removed from the set S i − and added to the next set S i in a solution S = ( . . . , S i − , S i , . . . ) by S i − (cid:5) S i := ( S i − \ S i , S i \ S i − ) . If S i − \ S i or S i \ S i − have size one, then we will omit the brackets of the singleton. (cid:73) Definition 5.10.
A solution S = ( S , . . . , S τ ) for ( G , k , ‘ ) from Construction 1 is smooth if for all i ∈ { , . . . , τ } we have | S i − \ S i | ≤ and | S i − \ S i | ≤ . In fact, if there is a solution, then there is also a smooth solution. (cid:66)
Observation 5.11.
Let ( G , k , ‘ ) from Construction 1 be a yes -instance. Then there is asmooth solution ( S , . . . , S τ ). . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 13 Proof.
By Observation 3.1, we know that there is a solution S = ( S , . . . , S τ ) such that | S | = k and k − ≤ | S i | ≤ k for all i ∈ { , . . . , τ } . Hence, for all i ∈ { , . . . , τ } it holdstrue that (cid:12)(cid:12) | S i | − | S i − | (cid:12)(cid:12) ≤
1. It follows that | S i − \ S i | ≤ | S i − \ S i | ≤
1, and thus, S isa smooth solution. (cid:74) Our next goal is to prove the existence of the following type of solutions. (cid:73)
Definition 5.12.
A smooth solution S = ( S , . . . , S τ ) for ( G , k , ‘ ) from Construction 1 is one-centered if(i) for all i ∈ { , . . . , τ } it holds true that | S i ∩ C | = 1 , and(ii) for all i ∈ { , . . . , τ } and S i − (cid:5) S i = ( α, β ) it holds true that α ∈ C ⇐⇒ β ∈ C . We now show that if the output instance of Construction 1 is a yes -instance, then there is asolution where c ∈ C is the only vertex from C in the first set of the solution. (cid:73) Lemma 5.13.
Let ( G , k , ‘ ) from Construction 1 be a yes -instance. Then there is a smoothsolution ( S , . . . , S τ ) for ( G , k , ‘ ) such that S ∩ C = { c } . Proof.
Suppose towards a contradiction that such a smooth solution does not exist. Thatis, in every smooth solution the first vertex cover S contains at least two vertices from C (due to Observation 5.9, S must contain at least one). Let Ψ be the set of smooth solutionswith | S ∩ C | being minimal, where S is the first vertex cover. Let S = ( S , . . . , S τ ) ∈ Ψ bea smooth solution such that the value i := min { j ∈ { , . . . , τ } | c j ∈ S \ { c }} is maximal.Let S = ( S , . . . , S τ ) be initially S .Suppose there is a j ∈ { , . . . , i − } such that S j (cid:5) S j +1 = ( c i , α ). Let j := min { j ∈{ , . . . , i − } | S j (cid:5) S j +1 = ( c i , α ) } be the smallest among them. Then, set S q := S q \ { c i } for all q ∈ { , . . . , j − } to get a feasible solution (note that S j − (cid:5) S j = ( ∅ , α ) is feasiblesince | S j − | ≤ k − S regarding | S ∩ C | .Hence, suppose that there is no such j , that is, there is no j ∈ { , . . . , i − } suchthat S j (cid:5) S j +1 = ( c i , α ). If S i \ { c i } is a vertex cover of layer G i , then setting S q := S q \ { c i } ,for all q ∈ { , . . . , p } with p := max { p ∈ { , . . . , τ } | ∀ q ∈ { , . . . , p } : c i ∈ S q } , yields afeasible solution. This contradicts the minimality of S regarding | S ∩ C | .Finally, suppose that there is no j ∈ { , . . . , i − } such that S j (cid:5) S j +1 = ( c i , α ) (andhence c i ∈ S i ) and S i \ { c i } is no vertex cover of layer G i . Let S i − (cid:5) S i = ( α, β ) for some α, β (each being possibly the empty set). Then for all q ∈ { , . . . , i − } do the following (wedistinguish two cases): Case 1: β = c r with r < i . Set S q := S q \ { c i } and S q := S q \ { β } (i.e. S i − (cid:5) S i = ( α, c i ))for all q ∈ { i, . . . , p } with p := max { p ∈ { , . . . , τ } | ∀ p ∈ { i, . . . , p } : β ∈ S p } . Thiscontradicts the minimality of S regarding | S ∩ C | . Case 2: β = c r with r > i , or β C . Set S q := ( S q \ { c i } ) ∪ { β } (note that S i = S i andhence S i − (cid:5) S i = ( α, c i )). Note that if there is a p ∈ { , . . . , i − } with S p − (cid:5) S p = ( β, x )or S p − (cid:5) S p = ( x, β ), then we get S p − (cid:5) S p = ( ∅ , x ) and S p − (cid:5) S p = ( x, ∅ ), respectively.In the case of β = c r with r > i , this contradicts the fact that c i is maximal regarding i .In the case of β C , this contradicts the minimality of S regarding | S ∩ C | .In every case, we obtain a contradiction, concluding the proof. (cid:74) Next we show that there are solutions such that whenever we remove a vertex in C fromthe vertex cover, then we simultaneously add another vertex from C to the vertex cover.Formally, we prove the following. (cid:73) Lemma 5.14.
Let ( G , k , ‘ ) from Construction 1 be a yes -instance. Then there is a smoothsolution ( S , . . . , S τ ) with S ∩ C = { c } such that for all i ∈ { , . . . , τ } with S i − (cid:5) S i = ( α, c ) and c ∈ C we also have α ∈ C . Proof.
Suppose towards a contradiction the contrary. That is, let for every smooth solution( S , . . . , S τ ) exist an i ∈ { , . . . , τ } with S i − (cid:5) S i = ( α, c ) and c ∈ C and α C . Let Ψ bethe non-empty (due to Lemma 5.13) set of smooth solutions ( S , . . . , S τ ) with | S ∩ C | = 1.Let Ψ ⊆ Ψ be the set of smooth solutions that maximizes the first index i with S i − (cid:5) S i =( α, c q ) with c q ∈ C and α C . Among those solutions, consider S = ( S , . . . , S τ ) ∈ Ψ to bethe one with q being maximal. Note that due to Observation 5.9, we have that | S i − ∩ C | ≥ S j := S j for all j ∈ { , . . . , τ } . Case 1: i > is odd . Since c i is the center of a star in layer i , c i has to be in S i . Wedistinguish three subcases regarding the relation of q and i , that is, the cases of q beingsmaller, equal, or larger than i . Case 1.1: q < i . Set S j = ( S j \ { c q } ) (i.e., S i − (cid:5) S i = ( α, ∅ )) for all j ∈ { i, . . . , q } with q := max { q ∈ { i, . . . , τ } | ∀ j ∈ { i, . . . , q } : c q ∈ S j } . It follows that ( S , . . . , S τ )is again a feasible smooth solution contradicting i being maximal. Case 1.2: q = i . Then c i S i − , and hence c i − ∈ S i − since the edge { c i − , c i } mustbe covered in layer G i − . Set S p = ( S p \ { c i − } ) ∪ { α } (i.e., S i − (cid:5) S i = ( c i − , c q ))for all p ∈ { i, . . . , j } , where j > i is minimal such that S j − (cid:5) S j = ( c i − , x ), or τ ifsuch a j does not exist. If there is a minimal j > i such that S j − (cid:5) S j = ( c i − , x ),then set S p = ( S p \ { α } ) (i.e., S j − (cid:5) S j = ( α, x )) for all p ∈ { j, . . . , q } with q :=max { q ∈ { i, . . . , τ } | ∀ p ∈ { i, . . . , q } : α ∈ S p } . Suppose that between i and j ,there are j and j such that S j − (cid:5) S j = ( y, α ) and S j − (cid:5) S j = ( α, y ). Notethat S j − (cid:5) S j = ( y, ∅ ) and S j − (cid:5) S j = ( ∅ , y ). It follows that ( S , . . . , S τ ) is againa feasible smooth solution, contradicting i being maximal. Case 1.3: q > i . Then c i ∈ S i − . Let S q − (cid:5) S q = ( β, d ). We distinguish into two casesregarding d . Case 1.3.1: d = c p with p < q . Set S j = S j \ { c q } (i.e., S i − (cid:5) S i = ( α, ∅ )) forall j ∈ { i, . . . , q − } . Moreover, set S j = ( S j \ { d } ) ∪ { c q } (i.e., S q − (cid:5) S q = ( β, c q ))for all j ∈ { q, . . . , q } with q := max { q ∈ { q, . . . , τ } | ∀ j ∈ { q, . . . , q } : d ∈ S j } . Case 1.3.2: d C or if d = c p , then p > q . Set S j = ( S j \ { c q } ) ∪ { d } (i.e., S i − (cid:5) S i = ( α, d )) for all j ∈ { i, . . . , q − } . Moreover, set S j = S j ∪ { c q } (i.e., S q − (cid:5) S q =( β, c q ) or S q − (cid:5) S q = ( β, ∅ )) for all j ∈ { q, . . . , q } with q := max { q ∈ { q, . . . , τ } |∀ j ∈ { q, . . . , q } : c q ∈ S j } .In either case, we have that ( S , . . . , S τ ) is a feasible solution contradicting either i being maximal ( d C , or d = c p with p < q ) or q being maximal ( d = c p with p > q ). Case 2: i > is even . Then c i − ∈ S i − and c q ∈ { c i , c i +1 } . Set S j := ( S j \ { c i − } ) ∪ { α } (i.e., S i − (cid:5) S i = ( c i − , c q )) for all j ∈ { i, . . . , q } with q := max { q ∈ { i, . . . , τ } | ∀ j ∈{ i, . . . , q } : c i − ∈ S j } . Then ( S , . . . , S τ ) is a feasible solution contradicting i beingmaximal. (cid:74) Combining Observation 5.9 and Lemma 5.14, we can assume that for every given yes -instance,there is a solution which is one-centered. (cid:73)
Corollary 5.15.
Let ( G , k , ‘ ) from Construction 1 be a yes -instance. Then there is asolution S which is one-centered. In the remainder of this section, for each t ∈ { , . . . , κ + 1 } let the union of all U i be denoted . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 15 by b U t := S ti =1 U i . We introduce further notation regarding a one-centered solution S := ( S , . . . , S m +1 = S , . . . , . . . , S κ , . . . , S κ m +1 ) for ( G , k , ‘ ). Here, S ti is the i -th set of phase t and thusthe (2 m ( t −
1) + i )-th set of S . The set Y ti := { e j ∈ S ti ∩ E | j ≥ i } (1)is the set of vertices e j from E in S ti such that the corresponding layer for e j in phase t isnot before the layer i in phase t . The set F ti := { j > i | S tj − (cid:5) S tj = ( u, β ) with u ∈ b U t } (2)is the set of layers from G in phase t where a vertex from b U t is not carried over to the nextlayer’s vertex cover. We now show that there is a phase t where | F t | ≥ K . (cid:73) Lemma 5.16.
Let S = ( S , . . . , S m +1 = S , . . . , . . . , S κ , . . . , S κ m +1 ) be a one-centeredsolution to ( G , k , ‘ ) from Construction 1. Then, there is a t ∈ { , . . . , κ } such that | F t | ≥ K . Proof.
Suppose towards a contradiction the contrary, that is, that for all t ∈ { , . . . , κ } itholds true that | F t | < K . Then, for each i ∈ { , . . . , κ + 1 } , we have that | S i ∩ b U i − | ≥ i − S is a solution, we know that U κ +1 ⊆ S κ +11 and hence | S κ +11 ∩ U κ +1 | = K . Thus, wehave that | S κ +11 | ≥ | S κ +11 ∩ U κ +1 | + | S κ +11 ∩ b U κ | ≥ K + κ − K + k + 2 > k , contradicting S being a solution. (cid:74) In the remainder of this section, the value f ti := | S ti ∩ b U κ +1 | − K (3)describes the number of vertices in b U κ +1 which we could remove from S ti such that S ti isstill a vertex cover for G m ( t − i (the i -th layer of phase t ). Observe that f ti ≥ t ∈ { , . . . , κ } and all i ∈ { , . . . , m + 1 } , because we need in each layer exactly K verticesfrom b U κ +1 in the vertex cover.We now derive an invariant which must be true in each phase. (cid:73) Lemma 5.17.
Let S = ( S , . . . , S m +1 = S , . . . , . . . , S κ , . . . , S κ m +1 ) be a one-centeredsolution to ( G , k , ‘ ) from Construction 1. Then, for all t ∈ { , . . . , κ } and all i ∈ { , . . . , m +1 } , it holds true that | F ti | − | Y ti | ≤ f ti . Proof.
Let t ∈ { , . . . , κ } be arbitrary but fixed. For all i ∈ { , . . . , m + 1 } let ε i := | F ti | − | Y ti | − f ti . We claim that ε i − ε i − ≥ i ∈ { , . . . , m + 1 } . Since S is one-centered, in Table 2 allrelevant tuples for S ti − (cid:5) S ti are shown. As each relevant tuple results in ε i − ε i − ∈ { , , } ,the claim follows.We want to prove that ε i ≤ i ∈ { , . . . , m + 1 } . So, assume towards acontradiction that there is a j ∈ { , . . . , m + 1 } such that ε j >
0. Since ε i − ε i − ≥ i ∈ { , . . . , m + 1 } , we have that ε m +1 >
0, which is equivalent to | F t m +1 | − | Y t m +1 | >f t m +1 . By definition, we have that | Y t m +1 | = 0 (see (1)) and | F t m +1 | = 0 (see (2)). Moreover,since S is a solution and each vertex cover needs at least K vertices from b U τ , we have that f t m +1 ≥
0. It follows that 0 = | F t m +1 | − | Y t m +1 | > f t m +1 ≥
0, yielding a contradiction. (cid:74)
Table 2
Overview of all tuples of S ti − (cid:5) S ti relevant in the proof of Lemma 5.17 and their possiblevalues of ε i − ε i − = | F ti | − | F ti − | − ( | Y ti | − | Y ti − | ) − ( f ti − f ti − ). In the tuples, u , v , and e representsome vertex from b U κ +1 , V , and E , respectively. S ti − (cid:5) S ti | F ti | − | F ti − | − ( | Y ti | − | Y ti − | ) − ( f ti − f ti − ) ε i − ε i − ( u, β ) β ∈ E ∈ {− , } ∈ { , } ∈ { , , } β ∈ b U κ +1 ∈ {− , } ∈ { , } β ∈ V , β = ∅ ∈ {− , } ∈ { , } ( α, u ) α ∈ E ∈ { , } -1 ∈ { , } α ∈ V , α = ∅ α, v ) α ∈ E ∈ { , } ∈ { , } α ∈ V , α = ∅ α, e ) α ∈ V α ∈ E , α = ∅ ∈ { , } ∈ { , } Next, we prove that in a phase t with | F t | ≥ K , there are at most k vertices from V contained in the union of the vertex covers of phase t . (cid:73) Lemma 5.18.
Let S = ( S , . . . , S m +1 = S , . . . , . . . , S κ , . . . , S κ m +1 ) be a one-centeredsolution to ( G , k , ‘ ) from Construction 1, and let t ∈ { , . . . , κ } be such that | F t | ≥ K . Then, | S m +1 i =1 S ti ∩ V | ≤ k . Proof.
From Lemma 5.17, we know that | Y t | ≥ K − f t . Let | Y t | = K − f t + λ for some λ ∈ N , and let ε i = | F ti | − | Y ti | − f ti , for all i ∈ { , . . . , m + 1 } .We now show that there are at most λ layers where we exchange a vertex currently in thevertex cover with a vertex in V . Let i ∈ { , . . . , m +1 } such that S ti − (cid:5) S ti = ( α, v ) with v ∈ V .From Table 2 (recall that one-centered solutions are smooth), we know that ε i ≥ ε i − + 1.Assume towards a contradiction that there are λ + 1 many of these exchanges. Then,there is a j ∈ { , . . . , m + 1 } such that ε j ≥ ε + λ + 1 = | F t | − | Y t | − f t + λ + 1 ≥ K − ( K − f t + λ ) − f t + λ + 1 ≥ ⇐⇒ | F tj | − | Y tj | > f tj . This contradicts the invariant of Lemma 5.17.In the beginning of phase t , we have at most k − λ vertices from V in the vertex cover,because | S t ∩ V | ≤ K + k − | Y t | − f t = K + k − ( K − f t + λ ) − f t = k − λ. Since there are at most λ many exchanges S ti − (cid:5) S ti = ( α, v ) where v ∈ V and i ∈{ , . . . , m + 1 } , we know that the vertex set S m +1 i =1 S ti ∩ V is of size at most k . (cid:74) We are set to prove the backward direction of Proposition 5.7. (cid:73)
Lemma 5.19.
Let ( G, k ) be an instance of Clique and ( G , k , ‘ ) be the instance of Multistage Vertex Cover resulting from Construction 1. If ( G , k , ‘ ) is a yes -instance,then ( G, k ) is a yes -instance. . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 17 Proof.
Let ( G , k , ‘ ) be a yes -instance. From Corollary 5.15 it follows that there is aone-centered solution S = ( S , . . . , S m +1 = S , . . . , . . . , S κ , . . . , S κ m +1 ) for ( G , k , ‘ ). ByLemma 5.16, there is a t ∈ { , . . . , κ } such that | F t | ≥ K = (cid:0) k (cid:1) . By Lemma 5.18, weknow that | S m +1 i =1 S ti ∩ V | ≤ k . Now we identify the clique of size k in G . Since | F t | ≥ K , we know that, by Construction 1, at least K = (cid:0) k (cid:1) layers are covered by verticesin V ∪ E ∪ b U κ +1 ∪ { c t j +1 | j ∈ { , . . . , m }} in phase t . Note that each of these layerscorresponds to an edge e = { v, w } in G and that we need in particular the vertices v and w in the vertex cover. Since we have at most k vertices in S m +1 i =1 S ti ∩ V , these vertices inducea clique of size k in G . (cid:74) We proved the forward and backward direction of Proposition 5.7 in Sections 5.2.1 and 5.2.2,respectively. It remains to put everything together. of Proposition 5.7.
Let (
G, k ) be an instance of
Clique and ( G , k , ‘ ) be the instanceof Multistage Vertex Cover resulting from Construction 1. Observe that Construction 1runs in polynomial time, and that each layer of G is a forest with O ( k ) edges. We knowthat if ( G, k ) is a yes -instance of
Clique , then ( G , k , ‘ ) is a yes -instance of MultistageVertex Cover (Lemma 5.8), and vice versa (Lemma 5.19). Finally, the W[1]-hardnessof
Clique [12] regarding k and the fact that k ∈ O ( k ) then finishes the proof. (cid:74) From a motivation point of view, it is natural to assume that the change over time modeledby the temporal graph is rather of evolutionary character, meaning that the difference of alayer to its predecessor is limited. However, Proposition 5.7 gives a bound (in terms of thedesired vertex cover size in input instance) on the number of edges of each layer. Hence, wealso have the following W[1]-hardness. (cid:73)
Corollary 5.20.
Multistage Vertex Cover parameterized by the maximum num-ber max i ∈{ ,...,τ } | E ( G i ) | of edges in a layer is W[1] -hard, even if each layer is a forest.
Thus, we cannot hope for fixed-parameter tractability of
Multistage Vertex Cover whenparameterized for example by the combination of k and the maximum size of symmetricdifference between two consecutive layers.Furthermore, we can turn the instance ( G , k , ‘ ) computed by Construction 1 into anequivalent instance ( G , k , ‘ ) where each layer is a tree as follows. Set k = k + 1. Add avertex x to G . In each layer of G , make x the center of a star with k + 1 (new) leaf verticesand connect x with exactly one vertex of each connected component. Note that in everysolution x is contained in a vertex cover for each layer in G . (cid:73) Corollary 5.21.
Multistage Vertex Cover parameterized by k is W[1] -hard, even ifeach layer is a tree.
However, in Corollary 5.21, max i ∈ τ | E ( G i ) | is unbounded and we cannot hope to strengthenthe reduction in this sense because if each layer is a tree, then we have exactly | V | − In this section, we study the possibility of efficient and effective data reduction for
Multi-stage Vertex Cover when parameterized by k , τ , and k + τ , that is, the possible existence of problem kernels of polynomial size. We prove that unless coNP ⊆ NP / poly, MultistageVertex Cover admits no problem kernel of size polynomial in k (Section 6.1). Yet, whencombining k and τ , we prove a problem kernel of size O ( k τ ) (Section 6.2). Moreover, weprove a problem kernel of size 5 τ when each layer consists of only one edge (Section 6.3).Recall that Multistage Vertex Cover is para-NP-hard regarding τ even if each layer isa tree. k for Restricted InputInstances In this section, we prove the following. (cid:73) Theorem 6.1.
Unless coNP ⊆ NP / poly , Multistage Vertex Cover admits no poly-nomial kernel when parameterized by k , even(i) if each layer consists of one edge and ‘ = 1 , or(ii) if each layer is planar and ‘ ≥ k . Recall that
Multistage Vertex Cover parameterized by k is fixed-parameter tractablein case of (ii) (see Observation 3.5), while we left open whether it also holds true in case (i).We prove Theorem 6.1 using AND-compositions [7]. (cid:73) Definition 6.2. An AND-composition for a parameterized problem L is an algorithmthat, given p instances ( x , k ) , . . . , ( x p , k ) of L , computes in time polynomial in P pi =1 | x i | aninstance ( y, k ) of L such that(i) ( y, k ) ∈ L if and only if ( x i , k ) ∈ L for all i ∈ { , . . . , p } , and(ii) k is polynomially upper-bounded in k . The following is the crucial connection to polynomial kernelization. (cid:73)
Theorem 6.3 (Drucker [14]) . If a parameterized problem whose unparameterized versionis NP -hard admits an AND-composition, then coNP ⊆ NP / poly . Note that coNP ⊆ NP / poly implies a collapse of the polynomial-time hierarchy to its thirdlevel [30].In the proof of Theorem 6.1(i), we use an AND-composition. The idea is to take p instances of Multistage Vertex Cover on the same vertex set with ‘ = 1 and identical k ,and stack all these instances one after the another in the time dimension. Here, we connectthe i -th instance with ( i + 1)-th instance by just repeating the first layer of the ( i + 1)-stinstance so often such that there is enough time to transfer from a solution of the i -thinstance to a solution of the ( i + 1)-th instance without violating the upper bound on thesymmetric difference between two consecutive vertex covers. Formally, we use the followingconstruction. (cid:73) Construction 2.
Let ( G , k, ‘ ) , . . . , ( G p , k, ‘ ) be p instances of Multistage Vertex Cover where ‘ = 1 and each layer of each G q = ( V, E q , τ q ), q ∈ { , . . . , p } , consists of one edge. Weconstruct an instance ( G = ( V, E , τ ) , k, ‘ ) of Multistage Vertex Cover as follows. Denoteby ( G i , . . . , G iτ i ) the sequence of layers of G i . Initially, let G be the temporal graph with layersequence (( G ij ) ≤ j ≤ τ i ) ≤ i ≤ p . Next, for each i ∈ { , . . . , p − } , insert between G iτ i and G i +11 the sequence ( H i , H i , . . . , H i k ) := ( G iτ i , G i +11 , . . . , G i +11 ). This finishes the construction.Note that τ := 2 k ( p −
1) + P pi =1 τ i . (cid:7) A graph is planar if it can be drawn on the plane such that no two edges cross each other. . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 19
In the next two propositions, we prove that Construction 2 forms AND-compositions, usedin the proof of Theorem 6.1(i). (cid:73)
Proposition 6.4.
Multistage Vertex Cover where each layer consists of one edgeand ‘ = 1 admits an AND-composition when parameterized by k . Proof.
We AND-compose
Multistage Vertex Cover where each layer consists of one edge.Let I = ( G = ( V, E , τ ) , k, ‘ ) , . . . , I p = ( G p = ( V, E p , τ p ) , k, ‘ ) be p instances of MultistageVertex Cover with ‘ = 1 where each layer consists of one edge. Apply Construction 2 toobtain instance I = ( G = ( V, G , τ ) , k, ‘ ) of Multistage Vertex Cover . We claim that I is a yes -instance if and only if I i is a yes -instance for all i ∈ { , . . . , p } .( ⇒ ) If I is a yes -instance, then for each i ∈ { , . . . , p } , the subsequence of the solutionrestricted to the layers ( G ij ) ≤ j ≤ τ i forms a solution to I i .( ⇐ ) Let ( S i , . . . , S iτ i ) be a solution to I i for each i ∈ { , . . . , p } . Clearly, ( S i , . . . , S iτ i )forms a solution to the layers ( G ij ) ≤ j ≤ τ i . For H i , let T i = S iτ i \ { v } for some v such thatthe unique edge of H i is still covered. Next, set T i = T i ∪ { w } , where w ∈ S i +11 with w being incident with the unique edge of H i . Now, over the next 2 k − T i into S i +11 by first removing layer by layer the vertices in T i \ S i +11 (at most k − S i +11 \ T i (again, at most k − I . (cid:74) Turning a set of input instances of
Multistage Vertex Cover with only one layer ( τ = 1)which additionally is a planar graph into a sequence gives an AND-composition used in theproof of Theorem 6.1(ii). (cid:73) Proposition 6.5.
Multistage Vertex Cover where each layer is planar and ‘ ≥ k admits an AND-composition when parameterized by k . Proof.
We AND-compose
Multistage Vertex Cover with one layer being a planar graph(and ‘ ≥ k ) into Multistage Vertex Cover with ‘ ≥ k . Let ( G , k, ‘ ) , . . . , ( G p , k, ‘ )be p -instances of Multistage Vertex Cover with one layer being a planar graph. Con-struct a temporal graph G with layers ( G , . . . , G p ). Set ‘ = 2 k . This finishes the construction.It is not difficult to see that ( G , k, ‘ ) is a yes -instance of Multistage Vertex Cover ifand only if ( G i , k ) is a yes -instance of Vertex Cover for all i ∈ { , . . . , p } . (cid:74) Propositions 6.4 and 6.5 at hand, we are set to prove this section’s main result. of Theorem 6.1.
Using Drucker’s result [14] for AND-compositions, Propositions 6.4 and 6.5prove Theorem 6.1(i) and (ii), respectively. Recall that
Multistage Vertex Cover whereeach layer consists of one edge (Theorem 4.1) and
Multistage Vertex Cover on one layerbeing a planar graph (basically,
Vertex Cover on planar graphs) [20] are NP-hard. (cid:74) O ( k τ ) Multistage Vertex Cover remains NP-hard for τ = 2, even if each layer is a tree(Theorem 4.1). Moreover, Multistage Vertex Cover does not admit a problem kernel ofsize polynomial in k , even if each layer consists of only one edge (Theorem 6.1). Yet, whencombining both parameters we obtain a problem kernel of cubic size. (cid:73) Theorem 6.6.
There is an algorithm that maps any instance ( G , k, ‘ ) of MultistageVertex Cover in O ( | V ( G ) | τ ) time to an instance ( G , k, ‘ ) of Multistage VertexCover with at most k τ ( G ) vertices and at most k τ ( G ) temporal edges. G u v G u v G u v G u vG u vw u w v (cid:32) G u vw u w v (cid:32) G u vw u w v (cid:32) G u vw u w v (cid:32) Figure 4
Illustration of Reduction Rule 2, exemplified for two vertices u, v and k = 5. Eachellipse for a graph G i and G i , respectively, represents G i − { u, v } and G i − { u, v, w u , w v } . Thevertices w v , w u (gray squares) are introduced by the application of Reduction Rule 2. Note that u ( v ) has a high degree in G ( G ) and G . To prove Theorem 6.6, we apply three polynomial-time data reduction rules. These reductionrules can be understood as temporal variants of the folklore reduction rules for
VertexCover . Our first reduction rule is immediate. (cid:66)
Reduction Rule 1 (Isolated vertices).
If there is some vertex v ∈ V such that e ∩ v = ∅ forall e ∈ E ( G ↓ ), then delete v .For Vertex Cover , when asking for a vertex cover of size q , there is the well-knownreduction rule dealing with high-degree vertices: If there is a vertex v of degree larger than q ,then delete v and its incident edges and decrease q by one. For Multistage Vertex Cover a high-degree vertex can only appear in some layers, and hence deleting this vertex is ingeneral not correct. However, the following is a temporal variant of the high-degree rule (seeFigure 4 for an illustration). (cid:66)
Reduction Rule 2 (High degree).
If there exists a vertex v such that there is an inclusion-maximal subset J ⊆ { , . . . , τ } such that deg G i ( v ) > k for all i ∈ J , then add a vertex w v to V and for each i ∈ J , remove all edges incident to v in G i , and add the edge { v, w v } .We now show how Reduction Rule 2 can be applied and that it does not turn a yes -instanceinto a no -instance or vice versa. (cid:73) Lemma 6.7.
Reduction Rule 2 is correct and exhaustively applicable in O ( | V | τ ) time. Proof. ( Correctness ) Let I = ( G , k, ‘ ) be an instance with G = ( G , . . . , G τ ), and let I =( G , k, ‘ ) be the instance with G = ( G , . . . , G τ ) obtained from I applying Reduction Rule 2with vertex v and index set J . We prove that I is a yes -instance if and only if I is a yes -instance.( ⇒ ) Let ( S , . . . , S τ ) be a solution to I . Observe that for all i ∈ J , deg G i ( v ) > k andhence v ∈ S i . It follows that ( S , . . . , S τ ) is a solution to I .( ⇐ ) Let ( S , . . . , S τ ) be a solution to I . Observe that for each i ∈ J , S i ∩ { v, w v } 6 = ∅ .Set S i = ( S i \ { w v } ) ∪ { v } for all i ∈ J . Note that S i is a vertex cover for G i since v covers all its incident edges and S i \ { v } is a vertex cover for G i − { v } = G i − { v, w v } .For each i ∈ { , . . . , τ } \ J , set S i = S i if w v S i , and S i = ( S i \ { w v } ) ∪ { v } otherwise.Note that S i is a vertex cover of G i = G i − { w v } . Finally, observe that | S i | ≤ | S i | forall i ∈ { , . . . , τ } , and that | S i S i +1 | ≤ ‘ for all i ∈ { , . . . , τ − } . It follows that ( S , . . . , S τ )is a solution to I .( Running time ) For each vertex, we count the number of edges in each layer. If there aremore than k edges in one layer, then we remember the index of the layer. For each layer, we . Fluschnik, R. Niedermeier, V. Rohm, P. Zschoche 21 compute for each vertex the degree and make the modification. Once for some v vertex w v isintroduced, we add a pointer from v to w v , and add the edge { v, w v } in subsequent layerswhen needed. Hence, in each layer we touch each edge at most twice, yielding O ( | V ( G ) | ) timeper layer. (cid:74) Similarly as in the reduction rules for
Vertex Cover , we now count the number of edges ineach layer: if more than k edges are contained in one layer, then no set of k vertices, eachof degree at most k , can cover more than k edges. (cid:66) Reduction Rule 3 ( no -instance). If neither Reduction Rule 1 nor Reduction Rule 2 isapplicable and there is a layer with more than k edges, then output a trivial no -instance.We are ready to prove that when none of the Reduction Rules 1 to 3 can be applied, thenthe instance contains “few” vertices and temporal edges. (cid:73) Lemma 6.8.
Let ( G , k, ‘ ) be an instance of Multistage Vertex Cover such thatnone of Reduction Rules 1 to 3 is applicable. Then G consists of at most k τ ( G ) verticesand k τ ( G ) temporal edges. Proof.
Since none of Reduction Rules 1 and 2 is applicable, for each layer it holds true thatthere is no isolated vertex and no vertex of degree larger than k . Since Reduction Rule 3is not applicable, each layer consists of at most k edges. Hence, there are at most k τ temporal edges in G . Consequently, due to Reduction Rule 1, there are at most 2 k τ verticesin G . (cid:74) We are ready to prove the main result of this section. of Theorem 6.6.
Given an instance I = ( G , k, ‘ ) of Multistage Vertex Cover , applyReduction Rules 1 to 3 exhaustively in O ( | V ( G ) | τ ( G )) time either to decide that I is atrivial no -instance or to obtain an instance ( G , k, ‘ ) equivalent to I . Due to Lemma 6.8, G consists of at most 2 k τ ( G ) vertices and at most k τ ( G ) temporal edges. (cid:74) τ Multistage Vertex Cover , even when each layer is a tree, does not admit a problemkernel of any size in τ unless P = NP. Yet, when each layer consists of only one edge, theneach instance of Multistage Vertex Cover contains at most τ edges and, hence, atmost 2 τ non-isolated vertices. Thus, Multistage Vertex Cover admits a straight-forwardproblem kernel of size linear in τ . (cid:66) Observation 6.9.
Let ( G , k, ‘ ) be an instance of Multistage Vertex Cover where eachlayer consists of one edge. Then we can compute in O ( | V ( G ) | · τ ) time an instance ( G , k, ‘ )of size at most 5 τ ( G ). Proof.
Let ( G , k, ‘ ) be an instance of Multistage Vertex Cover where each layerof G = ( V, E , τ ) consists of one edge. Observe that we can immediately output a trivial yes -instance if k ≥ τ (Observation 3.1) or ‘ ≥ k ≤ τ − ‘ ≤
1. Apply Reduction Rule 1 exhaustively on ( G , k, ‘ ) to obtain ( G , k, ‘ ). Since thereare τ edges in G , there are at most 2 τ vertices in G . It follows that the size of ( G , k, ‘ ) is atmost 5 τ . (cid:74) We introduced
Multistage Vertex Cover , proved it to be NP-hard even on very restrictedinput instances, and studied its parameterized complexity regarding the natural parameters k , ‘ , and τ (each given as input). A highlight is the W[1]-hardness described in Section 5.2 which,because it holds on very restricted instances of Multistage Vertex Cover , may turn outto be useful to provide W[1]-hardness results for other problems in the multistage setting.We leave open whether
Multistage Vertex Cover parameterized by k is fixed-parametertractable when each layer consists of only one edge (see Table 1). Moreover, it is openwhether Multistage Vertex Cover remains NP-hard on two layers each being a path(that is, strengthening Theorem 4.1(i)).
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