Multistage Vertex Cover

Covering all edges of a graph by a minimum number of vertices, this is the NP-hard Vertex Cover problem, is among the most fundamental algorithmic tasks. Following a recent trend in studying dynamic and temporal graphs, we initiate the study of Multistage Vertex Cover. Herein, having a series of graphs with same vertex set but over time changing edge sets (known as temporal graph consisting of various layers), the goal is to find for each layer of the temporal graph a small vertex cover and to guarantee that the two vertex cover sets between two subsequent layers differ not too much (specified by a given parameter). We show that, different from classic Vertex Cover and some other dynamic or temporal variants of it, Multistage Vertex Cover is computationally hard even in fairly restricted settings. On the positive side, however, we also spot several fixed-parameter tractability results based on some of the most natural parameterizations.


Introduction
Vertex Cover (VC) 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.VC is NP-hard and it is a formative problem of algorithmics and combinatorial optimization.We study a timedependent, "multistage" version, namely a variant of VC 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 ⊆ V 2 × {1, . . ., τ }.Equivalently, a temporal graph G can be seen as a vector (G 1 , . . ., G τ ) of static graphs (layers), where each graph is defined over the same vertex set V .Then, our specific goal is to find a small vertex cover S i for each layer G i such that the sizes of the symmetric differences S i △ S i+1 between the vertex covers S i and S i+1 of every two consecutive layers G i and G i+1 are small.Formally, we thus introduce and study the following problem.
Throughout this paper we assume that 0 < k < |V | because otherwise we have a trivial instance.In our model, we follow the recently proposed multistage [4,5,14] 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 changing edge sets in the temporal graph) and a corresponding adaptation of the current solution comes with a cost.In this spirit, the parameter ℓ in the definition of MSVC allows to model that only moderate changes concerning the solution vertex set may be wanted when moving from one layer to the subsequent one.Indeed, in this sense ℓ can be interpreted as a parameter measuring the degree of (non-)conservation [1,15].
It is immediate that MSVC is NP-hard as it generalizes Vertex Cover (τ = 1).We will study its parameterized complexity regarding the problem-specific parameters k, τ , ℓ, and some of their combinations, as well as restrictions to temporal graph classes [11].
Related Work.The literature on vertex covering is extremely rich, even when focusing on parameterized complexity studies.Indeed, Vertex Cover (VC) can be seen as "drosophila" of parameterized algorithmics.Thus, we only consider on VC studies closely related to our setting.First, we mention in passing that VC is studied in dynamic graphs [3,17] and graph stream models [6].More importantly for us, Akrida et al. [2] studied a variant of VC on temporal graphs.Their model significantly differs from ours: They want an edge to be covered at least once over every time window of some given size ∆.That is, they define a temporal vertex cover as a set S ⊆ V × {1, . . ., τ } such that for every time window of size ∆ and for each edge e = {v, w} appearing in a layer contained in the time window it holds that (v, t) ∈ S or (w, t) ∈ S for some t in the time window with (e, t) ∈ E. For their model, they ask whether such an S of small cardinality exists.Note that if ∆ > 1, then for some t ∈ {1, . . ., τ }, the set S t ∶= {v | (v, t) ∈ S} is not necessarily a 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 difference between each S t and S t+1 may strongly vary.They provide several hardness results and algorithms (mostly referring to approximation or exact algorithms, but not to parameterized complexity studies).
A second related line of research, not directly referring to temporal graphs though, studies reconfiguration problems which arise when we wish to find a step-by-step transformation between two feasible solutions of a problem such that all intermediate results are also feasible solutions [13,16].Mouawad et al. [19,20] studied, among other reconfiguration problems, Vertex Cover Reconfiguration which takes as input a graph G, two vertex covers S and T of size at most k each, and an integer τ .The goal is to determine whether there is a sequence (S = S 1 , . . ., S τ = T ) such that each S t is a vertex cover of size at most k.The essential difference to our model is that from one "sequence element" to the next only one vertex may be changed and that the input graph does not change over time.Indeed, there is an easy reduction of this model to ours while the opposite direction is unlikely to hold.This is substantiated by the fact that Mouawad et al. [20] showed that Vertex Cover Reconfiguration is fixed-parameter tractable when parameterized by vertex cover size k while we show W [1]-hardness for the corresponding case of MSVC.
Finally, there is also a close relation to the research on dynamic parameterized problems [1,18].Krithika et al. [18] studied Dynamic Vertex Cover where one is given two graphs on the same vertex set and a vertex cover for one of them together with the guarantee that the cardinality of the symmetric difference between the two edge sets is upper-bounded by a parameter d.The task Table 1: Overview on our results.The column headings describe the restrictions on the input and each row corresponds to a parameter.p-NP-hard, PK, and NoPK, abbreviate para-NP-hard, polynomial problem kernel, and no problem kernel of polynomial size unless coNP ⊆ NP/ poly.† (Obs.2.5) general layers tree layers one-edge layers NP-hard (Thm.k + τ FPT, PK (Thm.5.9) FPT, PK (Thm.5.9) FPT, PK (Thm.5.9) then is to find a vertex cover for the second graph that is "close enough" (measured by a second parameter) to the vertex cover of the first graph.They show fixed-parameter tractability and a linear kernel with respect to d.
Our Contributions.Our results, focusing on the three perhaps most natural parameters, are summarized in Table 1.We highlight a few specific results.Multistage Vertex Cover remains NP-hard even if every layer consists of only one edge; clearly, the corresponding hardness reduction then 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 Multistage Vertex Cover already becomes NP-hard.MSVC parameterized by solution size k is fixed-parameter tractable if ℓ ≥ 2k, but becomes W [1]-hard if ℓ < 2k.Considering the tractability results for Dynamic Vertex Cover [18] and Vertex Cover Reconfiguration [20], this seems to be surprising and is our most technical result.Furthermore, in the former case (MSVC parameterized by k with ℓ ≥ 2k) does not admit a problem kernel of polynomial size unless coNP ⊆ NP/ poly.If one studies the combined parameter k + τ , however, then besides fixed-parameter tractability in all cases we also obtain polynomial-sized kernels.

Preliminaries
We denote by N and N 0 the natural numbers excluding and including zero, respectively.We denote by A △ B ∶= (A \ B) ∪ (B \ A) the symmetric difference of two sets A and B.
Temporal Graphs.A temporal graph G is a tuple (V, E, τ ) consisting of the set of vertices V , the set of temporal edges E, and a discrete time-horizon τ .A temporal edge e is an element in V 2 × {1, . . ., τ }.Equivalently, a temporal graph G is a vector of static graphs (G 1 , . . ., 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 for some computable function f only depending on k.A W [1]-hard parameterized problem is fixed-parameter intractable unless FPT=W [1].
Given a parameterized problem L, a kernelization is an algorithm that maps any instance for some computable function f (the size of the problem kernel) only depending on k.
We refer to Cygan et al. [7] for more material on parameterized complexity.
General Observations on MSVC.We state some simple but useful observations on Multistage Vertex Cover and its relation to Vertex Cover.
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 cover for each layer.The vector (S 1 , . . ., S τ ) with S i = S for all i ∈ {1, . . ., τ } is a solution for every ℓ ≥ 0. Proof.We first show that there is a solution S = (S 1 , . . ., S τ ) such that |S 1 | = k.Towards a contradiction assume that such a solution does not exist and that S = (S 1 , . . ., S τ ) is a solution such that |S 1 | is maximized.Let i ∈ {1, . . ., τ } be the maximum layer such that S j ⊆ S j−1 , for all j ∈ {2, . . ., i}.If i = τ , then we have that |S j | ≤ |S 1 | < k for all j ∈ {1, . . ., τ }.Hence, we can find subset X of V \ S 1 such that (S 1 ∪ X, . . ., S τ ∪ X) is a solution.This contradicts |S 1 | being maximized.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 ∈ {1, . . ., i}.This contradicts |S 1 | being maximized.Hence, there is a solution S = (S 1 , . . ., S τ ) such that |S 1 | = k.
Let Σ be the set of solutions such that the first vertex cover is of size k.Assume towards a contradiction that all solutions in Σ contain a vertex cover smaller than k − 1.Let Σ i ⊆ Σ be the set of solutions such that for (S 1 , . . ., S τ ) (a): Assume that there is a p ∈ {i + 1, . . ., τ } such that there is a w ∈ S p \ S p−1 and S j ⊆ S j−1 for all j ∈ {i + 1, . . ., p − 1}.The idea now is to keep v and add w in layer i and then remove v in layer p.We can achieve that by simply setting S q ∶= S q ∪ {v, w} for all q ∈ {i, . . ., p − 1}.Note that this is a solution which either contradicts that |S i | is maximized or that i is maximized.

Observation 2.3.
There is an algorithm that maps any instance (G, k) of Vertex Cover in τ ⋅

|V (G)|
O (1) time to an equivalent instance (G, k, ℓ) of MSVC with ℓ = 0, where G is a sequence of any τ subgraphs of G such that the underlying graph is G.
Proof.Construct the temporal graph G with sequence G 1 , . . ., G τ , where with S i ∶= S is solution to I ′ since G i is a subgraph of G for each i ∈ {1, . . ., τ }. (⇐) Let (S 1 , . . ., S τ ) with S i = S j be a solution to I ′ .We claim that S ∶= S 1 is a solution to I.
Suppose not, that is, there is an edge e ∈ E(G) such that S ∩ e = ∅.Since ⋃ τ i=1 E i = E, there is an i ∈ {1, . . ., τ } such that e ∈ E i .Since S = S 1 = S i , it follows that S i is not a vertex cover for G i , contradicting the fact that (S 1 , . . ., S τ ) is a solution to I ′ .Thus, S is a solution to I.
Observation 2.4.There is a polynomial-time algorithm that maps any instance (G, k, ℓ) of MSVC with ℓ = 0 to an equivalent instance (G ↓ (G), k) of Vertex Cover.
Proof.Let (G = (V, E, τ ), k, 0) be an arbitrary instance of MSVC.Construct the instance Clearly, since ℓ = 0, we have that S i = S j for all i, j ∈ {1, . . ., τ }.It is not difficult to see that S ∶= S 1 is a vertex cover for G ↓ , and hence the claim follows.
Proof.For each of the τ layers G i , we can construct an instance of VC of the form (G i , k).Since ℓ is large enough, we can solve each instance independent of the size of their symmetric differences.

Hardness On Restricted Inputs
Multistage Vertex Cover is NP-hard as it generalizes Vertex Cover (τ = 1).In this section we prove that MSVC remains NP-hard on inputs with only two layers one consisting of a path and the other consisting of a tree, and on inputs where every layer consists only of one edge.
and the first layer is a path and the second layer is a tree, or (ii) every layer contains only one edge and ℓ = 1.
Remark 3.2.Theorem 3.1(i) is tight regarding τ since Vertex Cover (i.e.MSVC with τ = 1) on trees is solvable in polynomial time.Theorem 3.1(ii) is tight regarding ℓ, because in the case of ℓ / = 1 either Observation 2.3 or Observation 2.5 is applicable.
Vertex Cover remains NP-complete on cubic Hamiltonian graphs when a Hamiltonian cycle is additionally given in the input [10]-we refer to this problem as Hamiltonian Cubic Vertec Cover (HCVC).To prove Theorem 3.1(i), we give a polynomial-time many-one reduction from HCVC to MSVC with two layers, one being a path, the other being a tree.

Proposition 3.3. There is an algorithm that maps any instance
) of MSVC with τ = 2 and the first layer G 1 being a path and second layer G 2 being a tree.
Proof.Let e ∈ E(C) be some edge of C, and let P = C − {e} be the Hamiltonian path obtained from C when removing e.Let E 1 = E(P ), and and G 2 = (V, E 2 ).Note that G 1 is a path and G 2 is the disjoint union of |V |/2 − 2 paths of length one and one path of length three.Add two special vertices z, z ′ , and connect z in G 1 with one endpoint of P and with z ′ and in G 2 with exactly one vertex of each connected component and In order to prove Theorem 3.1(ii), we give a polynomial-time many-one reduction from Vertex Cover to MSVC.

Proposition 3.4. There is an algorithm that maps any instance
) of MSVC where ℓ ′ = 1 and every layer G i contains only one edge.
Proof.Let the edges of G be enumerated

Parameter Vertex Cover Size
In this section, we study the parameter size k of the vertex cover of each layer for MSVC.Vertex Cover and Vertex Cover Reconfiguration [20] when parameterized by the vertex cover size are fixed-parameter tractable.We prove that this is no longer true for MSVC (unless FPT = W [1]).We first show the XP-algorithm (Section 4.2) and then prove W[1]-hardness (Section 4.8).

An XP-Algorithm
In this section, we prove the following.
In a nutshell, we first consider for each layer all subsets of vertices of size at most k that form a vertex cover.Second, we find a sequence of vertex covers for all layers such that the sizes of the symmetric differences for every two consecutive solutions is at most ℓ.We show that the second step can be solved via computing a directed source-sink path in a helper graph that we call configuration graph.
Note, that a similar idea leads to fixed-parameter tractability of Vertex Cover Reconfiguration [20].However, in the multistage setting we show an XP-algorithm regarding k to construct the configuration graph.This is due to the fact that we check for each subset of V (G) of size k or k − 1 whether it is a vertex cover in some layer.
This changes if we consider Minimal Multistage Vertex Cover where we additionally demand 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 (and hence all candidates for the i-th set of the solution) with the folklore search-tree algorithm for vertex cover (see e.g.[8]).This leads to O( 2k τ (G)) many vertices in the (k, ℓ)-configuration graph (for Minimal Multistage Vertex Cover) and thus to fixed-parameter tractability of Minimal Multistage Vertex Cover parameterized by the vertex cover size k.
However, it is not likely (unless FPT=W [1]) that one can improve substantially the algorithm behind Proposition 4.3 as we show next.

Fixed-parameter Intractability
In this section we show that MSVC is W[1]-hard when parameterized by k.This hardness result is established by the following parameterized reduction from the W[1]-complete [8] Clique problem, where, given an undirected graph G and a positive integer k, the question is whether G contains a clique of size k (that is, k vertices that are pairwise adjacent).Proposition 4.9.There is an algorithm that maps any instance (G, k) of Clique in polynomial time to an equivalent instance and each layer of G being a tree.
The proof of Proposition 4.9 is deferred to the end of this section.We construct an instance of Multistage Vertex Cover from an instance of Clique as follows (see Figure 1 for an illustrative example).Dashed vertical lines separate layers.
We construct a temporal graph G = (V ′ , E, τ ) as follows.Let V ′ be initially V ∪ E (note that E simultaneously describes the edge set of G and a vertex subset of G).We add the following vertex sets Let E be initially empty.We extend the set V ′ and define E through the τ = 2mκ + 1 layers we 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.
( This finishes the construction of G. ⬩ The construction essentially repeats the same gadget (which we call phase) κ times where the layer 2m ⋅ i + 1 is simultaneously last layer of phase i and the first layer of phase i + 1.In the beginning of phase i, a solution must contain the vertices of U i .The idea now is that 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 4.10 can be computed in polynomial time.Hence, it remains to prove the equivalence stated in Proposition 4.9.We prepare the proofs of the forward and backward direction in the following Sections 4.11.1 and 4.12.1, respectively.

Forward direction
The forward direction of Proposition 4.9 is-in a nutshell-as follows:  Observe that the clique G K covers in each phase K even layers.Hence, we replace, during the phase t ∈ {1, . . ., κ} (that is from layer (t − 1)2m + 1 to t2m + 1), the vertices U t with the vertices U t+1 .This also implies that the symmetric difference of two consecutive sets in S is exactly 2 = ℓ.
It follows that S is a solution for (G, k ′ , ℓ).

Backward direction
In this section we prepare the proof of the backward direction for the proof of Proposition 4.9.We first show that if an instance of Multistage Vertex Cover computed by Construction 4.10 is a yes-instance, then it is safe to assume that neither two vertices are deleted from nor added to a vertex cover in a consecutive step (we refer to these solutions as smooth, see Definition 4.14).Moreover, a vertex from C is only exchanged with another vertex from C and, at any time, there is exactly one vertex from C contained in the solution (similarly to the constructed solution in Lemma 4.12).We call these solutions one-centered (Definition 4.16).We then prove that there must be a phase t for any one-centered solution that is deleting at least k 2 times a vertex from "past" sets U t ′ , t ′ ≤ t.This at hand, we prove that such a phase witnesses a clique of size k.
That a solution needs to contain at least one vertex from C at any time follows immediately from the fact that there is either an edge between two vertices in C or there is a vertex in C which is the center of a star with k ′ + 1 leaves.
In the remainder of this section we denote which vertices are exchanged from set S i−1 to next set S i in a solution S by In case of S i−1 \ S i or S i \ S i−1 being of size at most one we will omit the set braces.
Proof.By Observation 2.1, we know that there is a solution S = (S 1 , . . ., S τ ) such that ) be initially S. Suppose there is j ∈ {1, . . ., i − 1} such that S j S j+1 = (c i , a), and let j ′ be the smallest among them.Then, set S ′ q ∶= S q \ {c i } for all q ∈ {1, . . ., j ′ − 1} to get a feasible solution contradicting the Hence, suppose there is no such j.If S i \ {c i } is a vertex cover of layer G i , then setting S ′ q ∶= S q \ {c i }, for all q ∈ {1, . . ., p} with c i ∈ S q for all q ∈ {1, . . ., p} and p being maximal, yields a feasible solution contradicting the minimality of S regarding |S 1 ∩ C|.
Hence, suppose there is no such j and c i ∈ S i .Let S i−1 S i = (a, b) for some a, b (each being possibly the empty set).Then for all q ∈ {1, . . ., i − 1}: Next, we show that there are solutions such that whenever we remove a vertex in C from the vertex cover then we simultaneously add another vertex from C to the vertex cover.Formally, we prove the following.Proof.Suppose not, that is, for every smooth solution (S 1 , . . ., S τ ) there is an i with S i−1 S i = (a, c) and c ∈ C and a / ∈ C. Let Σ be the non-empty (due to Lemma 4.17) set of smooth solutions (S 1 , . . ., S τ ) with |S 1 ∩ C| = 1.Let Σ ′ ⊂ Σ be the set of smooth solutions that maximizes i with S i−1 S i = (a, c q ) with c q ∈ C and a / ∈ C.Among those solutions, consider S = (S 1 , . . ., S τ ) ∈ Σ ′ to be the one with q being maximal.Note that due to Observation 4.13, Let S ′ i ∶= S i for all i ∈ {1, . . ., τ }.Case: i > 1 is odd.Since c i is the center of a star in layer i, c i has to be in S i .Subcase: ′ with c q ∈ S j for all i ≤ j ≤ q ′ and q ′ being maximal.It follows that (S ′ 1 , . . ., S ′ τ ) is again a feasible smooth solution contradicting i being maximal. Subcase: ) for all j ≤ p ≤ q with a ∈ S p for all j ≤ p ≤ q and q being maximal.Suppose that in between i and j, there are j 1 and j 2 such that S j 1 −1 S j 1 = (y, a) and ) is again a feasible smooth solution contradicting i being maximal.
Suppose that d = c p with p < q, then set S ′ j ∶= S j \ {c q } (i.e. ) for all i ≤ j < q, and S ′ q = (S q \ {d}) ∪ {c q } (i.e. S ′ q−1 S ′ q = (b, c q )) and S ′ j = (S j \ {d}) ∪ {c q } for all q < j ≤ q ′ with d ∈ S j for all q ≤ j ≤ q ′ and q ′ being maximal.
Suppose that d / ∈ C or if d = c p , then p > q.Then set S ′ j ∶= (S j \{c q })∪{d} (i.e. b, ∅)) for all q ≤ j ≤ q ′ with c q ∈ S j for all q ≤ j ≤ q ′ and q ′ being maximal.
′ with c i−1 ∈ S j for all i ≤ j ≤ q ′ and q ′ being maximal.
) is a feasible solution contradicting i being maximal.
Combining Observation 4.13 and Lemma 4.18, we can assume that given a yes-instance, there is a solution which is one-centered.In the remainder of this section, for each t ∈ {1, . . ., κ + 1} let the union of all U i be denoted by We introduce further notation regarding a one-centered solution S ∶= (S ′ , ℓ).Here, S t i is the i-th set of the phase t and thus the (2m(t − 1) + i)-th set of S. The set is the set of vertices e j from E in S t i such that the corresponding layer for e j in phase t is not before the layer i in phase t.The set is the set of layers from G in phase t where a vertex from Ūt is not carried over to the next layer's vertex cover.We now show that there is a phase t where |F  (G, k ′ , ℓ) from Construction 4.10 being a yes-instance.Then, there is a t ∈ {1, . . ., κ} such that contradicting S being a solution.
In the remainder of this section the value describes the number of vertices in Ūκ+1 which we could remove from S t i such that S t i is still a vertex cover for G 2m(t−1)+i (the i-th layer of phase t).Observe, that f t i ≥ 0 for all i ∈ {1, . . ., 2m + 1} and t ∈ {1, . . ., κ}, because we need in each layer exactly K vertices from Ūκ+1 in the vertex cover.

2
) then finishes the proof.

On Efficient Data Reduction
In this section, we study the possibility of effective data reduction for MSVC 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, MSVC admits no problem kernel of size polynomial in k (Section 5.1).Yet, when combining k and τ , we prove a problem kernel of size O(k 2 τ ) (Section 5.8).Moreover, we prove a problem kernel of size 5τ when each layer consists of only one edge (Section 5.12).Recall that MSVC is para-NP-hard regarding τ even if each layer is a tree.

No problem kernel of size polynomial in k
We prove that if (i) each layer consists only of one edge and ℓ = 1, and (ii) if each layer is planar and ℓ ≥ 2k, MSVC admits no kernel of size polynomial in k unless coNP ⊆ NP/ poly.Recall that fixed-parameter tractability of MSVC regarding k holds true in case of (i) (see Observation 2.5), while we left open whether it also holds true in case (ii).
Theorem 5.2.Unless coNP ⊆ NP/ poly, MSVC admits no polynomial kernel when parameterized by k, even if (i) each layer consists of one edge and ℓ = 1, or (ii) each layer is planar and ℓ ≥ 2k.
We prove Theorem 5.2 using AND-compositions.The following is the crucial connection to polynomial kernelization.
Note that coNP ⊆ NP/ poly implies a collapse of the polynomial-time hierarchy to its third level [21].
In the proof of Theorem 5.2(i), we use the following.Proof.We AND-cross-compose VC on planar graphs into MSVC with ℓ ≥ 2k.Let (G 1 , k), . . ., (G p , k) be p-instances of VC.Construct a temporal graph G with layers (G 1 , . . ., G p ). Set ℓ = 2k.This finishes the construction.It is not difficult to see that (G, k, ℓ) is a yes-instance of MSVC if and only if (G i , k) is a yes-instance of VC for all i ∈ {1, . . ., p}.
Proof of Theorem 5.2.Together with Theorem 5.4, Propositions 5.6 and 5.7 prove Theorem 5.2(i) and (ii), respectively.Recall that MSVC where each layer consists of one edge (Theorem 3.1) and Vertex Cover on planar graphs [12] are NP-hard.

A problem kernel of size O(k 2 τ )
MSVC remains NP-hard for τ = 2, even if each layer is a tree (Theorem 3.1).Moreover, MSVC does not admit a problem kernel of size polynomial in k, even if each layer consists of one edge (Theorem 5.2).Yet, when combining both parameters we obtain a problem kernel of cubic size.
Theorem 5.9.There is an algorithm that maps any instance of MSVC with at most 2k 2 τ vertices and k 2 τ temporal edges.
In the proof of Theorem 5.9, we apply three polynomial-time data reduction rules.These reduction rules can be understood as the temporal variants of the folklore reduction rules for Vertex Cover.Our first reduction rule is immediate.Reduction Rule 1 (Isolated vertices).If there is some vertex v ∈ V such that e ∩ v = ∅ for all e ∈ E(G ↓ ), then delete v.
For Vertex Cover when asking for a vertex cover of size q, there is the well-known reduction 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 MSVC a high-degree vertex can only appear in some layers, and hence deleting this vertex is in general not correct.However, there is a temporal variant of the high-degree rule as follows.
Reduction Rule 2 (High degree).If there exists a vertex v that there is an inclusion-maximal subset J ⊆ {1, . . ., 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 }. (Running time) For each vertex, we count the number of edges in each layer.If there are more than k edges in one layer, then we remember the index of the layer.For each layer, we compute for each vertex the degree and make the modification.Once for some v vertex w v is introduced, we add a pointer from v to w v , and add the edge {v, w v } in subsequent layers when needed.Hence, in each layer we touch each edge at most twice, yielding O(|V | Similarly as in the reduction rules for Vertex Cover, we now count the number of edges in each layer: If more than k 2 edges are contained in one layer, then no set of k vertices each of degree at most k can cover more than k 2 edges. Reduction Rule 3 (no-instance).If none of Reduction Rules 1 and 2 is applicable and there is a layer with more than k 2 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, the instance is either a no-instance or has "few" vertices and temporal edges.Lemma 5.11.Let (G, k, ℓ) be an instance of MSVC such that none of Reduction Rules 1 to 3 is applicable.Then G consists of at most 2k 2 τ (G) vertices and k 2 τ (G) temporal edges.
Proof.Let none of Reduction Rules 1 and 2 be applicable For each layer, it holds true that there is no isolated vertex and no vertex of degree larger than k.Due to Reduction Rule 3, either we deal with a trivial no-instance or each layer consists of at most k 2 edges, accounting to at most k 2 τ temporal edges in G. Consequently, due to Reduction Rule 1, there are at most 2k 2 τ vertices in G.
We are ready to prove the main result of this section.

A problem kernel of size 5τ
MSVC when each layer is a tree does not admit a problem kernel of any size in τ unless P = NP.Yet, when each layer consists of only one edge, then MSVC admits a problem kernel of size linear in τ .Proof.Observe that we can immediately output a trivial yes-instance if k ≥ τ (Observation 2.1) or ℓ ≥ 2 (Observation 2.5).Hence, assume that k ≤ τ − 1 and ℓ ≤ 1. Apply Reduction Rule 1 exhaustively (G, k, ℓ) to obtain (G ′ , k, ℓ).Since there are τ edges in G, there are at most 2τ vertices in G ′ .It follows that the encoding length of (G ′ , k, ℓ) is at most 5τ .

Conclusion
We introduced Multistage Vertex Cover, proved it to be NP-hard even on restricted inputs, and studied its parameterized complexity regarding the natural parameters k, ℓ, and τ (each given as input).We leave open whether MSVC parameterized by k is fixed-parameter tractable when each layer consists of only one edge (see Table 1).Moreover, it is open whether MSVC remains NP-hard on two layers each being a path (that is, strengthening Theorem 3.1(i)).
follows that there are at most k vertices covering all edges in the layers G 2i−1 , that is G. Theorem 3.1 now follows from Propositions 3.3 and 3.4.

Theorem 4 . 1 .
Multistage Vertex Cover parameterized by k is in XP and W[1]-hard.

Lemma 4 . 5 .
The (k, ℓ)-configuration graph of a temporal graph G with n vertices and time horizon τ (i) can be constructed in O(τ ⋅ n 2k+1 ) time, and (ii) contains at most τ ⋅ 2n k + 2 vertices and (τ − This finishes the proof.We are ready to prove Proposition 4.3.Proof of Proposition 4.3.First, compute the configuration graph D of the instance (G = (V, E, τ ), k, ℓ) of Multistage Vertex Cover in O(τ ⋅ |V | 2k+1 ) time (Lemma 4.5(i)).Then, find an s-t path in D with a breath-first search in O(τ ⋅ |V | 2k ) time (Lemma 4.5(ii)).If an s-t path is found, then return yes, otherwise return no (Lemma 4.6).Remark 4.7.The reason why the algorithm behind Proposition 4.3 is an XP-algorithm and not an FPT-algorithm regarding k for Multistage Vertex Cover is because we do not have a better upper bound on the number of vertices in the (k, ℓ)-configuration graph for G than O(τ (G)⋅|V (G)| k ).

Construction 4 . 10 .Figure 1 :
Figure 1: Illustration to Construction 4.10 on an example graph (left-hand side) and the first seven layers of the obtained graph (right-hand side).Star-shapes illustrate star graphs with k ′ + 1 leaves.

Remark 4 . 11 . 1 .
We can turn the instance (G, k ′ , ℓ) computed by Construction 4.10 into an equivalent instance (G ′ , k ′′ , ℓ) where each layer is a tree as follows.Set k ′′ = k ′ + Add a vertex x to G.In each layer of G, make x a star with k ′′ + 1 vertices and connect x with exactly one vertex of each connected component.Note that in every solution, x is contained in a vertex cover for each layer in G ′ .

Lemma 4 . 18 .
Let (G, k ′ , ℓ) from Construction 4.10 be a yes-instance.Then there is a smooth solution (S 1 , . . ., S τ ) such that S 1 ∩ C = {c 1 } and for all i with S i−1 S i = (a, c) and c ∈ C holds a ∈ C.

Corollary 4 . 19 .
Let (G, k ′ , ℓ) from Construction 4.10 be a yes-instance.Then, there is a solution S which is one-centered.

1 |
the clique of size k in G. Since |F t ≥ K, we know that, by Construction 4.10, at least K = k 2 layers are covered by vertices in V ∪ E ∪ Ūκ+1 ∪ {c t 2j+1 | j ∈ {1, . . ., m}} in phase t.Note that each of these layers correspond 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 ⋃ 2m+1 i=1 S t i ∩ V , these vertices induce a clique of size k in G.

Definition 5 . 3 .
An AND-composition for a parameterized problem L is an algorithm that given p instances (x 1 , k), . . .,(x p , k) of L, computes in time polynomial in ∑ p i=1 |x i | an instance (y, k ′ ) of L such that (i) (y, k ′ ) ∈ L if and only if (x i , k) ∈ L for all i ∈ {1, . . .,p}, and (ii) k′ is polynomially upper-bounded in k.

1 ).Proposition 5 . 6 . 2 = T i 1 ∪ 1 ( 1 \ T i 2 (
This finishes the construction.Note that τ = 2k(p − 1) + ∑ p i=1 τ i .⬩ Construction 5.5 gives an AND-composition used in the proof of Theorem 5.2(i).MSVC where each layer consists of one edge and ℓ = 1 AND-composes into itself parameterized by k.Proof.We AND-compose MSVC where each layer consists of one edge.Let I 1 = (G 1 = (V, E 1 , τ 1 ), k, ℓ), . . ., I p = (G p = (V, E p , τ p ), k, ℓ) be p instances of MSVC with ℓ = 1 where each layer consists of one edge.Apply Construction 5.5 to obtain instanceI = (G = (V, G, τ ), k, ℓ) of MSVC.We claim that I is a yes-instance if and only if I i is a yes-instance for all i ∈ {1, . . ., p}.(⇒)If I is a yes-instance, then for each i ∈ {1, . . ., p}, the subsequence of the solution restricted to the layers (G i j ) 1≤j≤τ i forms a solution to I i .(⇐) Let (S i 1 , . . ., S i τ i ) be a solution to I i for each i ∈ {1, . . ., p}.Clearly, (S i 1 , . . ., S i τ i ) forms a solution to the layers (G i j ) 1≤j≤τ i .For H i 1 , let T i 1 = S i τ i \ {v} for some v such that the unique edge of H i 1 is still covered.Next, set T i {w}, where w ∈ S i+1 1 with w being incident with the unique edge of H i 2 .Now, over the next 2k − 2 layers, transform T i 2 into S i+1 1 by first removing layer by layer the vertices in T i 2 \ S i+1 at most k − 1 many vertices), and then layer by layer add the vertices in S i+1 again, at most k − 1 vertices).This forms a solution to I. Turning a set of input instances of Vertex Cover on planar graphs (being equivalent to MSVC with one layer being a planar graph) into a sequence gives an AND-composition used in the proof of Theorem 5.2(ii).Proposition 5.7.MSVC with one layer being a planar graph AND-composes into MSVC with ℓ ≥ 2k and each layer being planar parameterized by k.

Figure 2 :
Figure 2: Illustration to Reduction Rule 2, exemplified for two vertices u, v and k = 5.The vertices w v , w u (gray squares) are introduced by the application of Reduction Rule 2.

Proof of Theorem 5 . 9 .
Apply Reduction Rules 1 to 3 exhaustively in O(|V | 2 τ ) time to obtain an equivalent instance (G ′ , k, ℓ).Due to Lemma 5.11, G ′ consists of at most 2k 2 τ vertices and at most k 2 τ temporal edges.
are subgraphs of G, and hence all edges are covered by S ′ .Moreover, all the edges in G i − E i , i ∈ {1, 2}, are incident with z and hence covered by S ′ .(⇐) Let (S 1 , S 2 ) be a minimal solution to (G, k ′ , ℓ ′ ) with S ′ ∶= S 1 = S 2 and |S ′ | ≤ k ′ .We can assume that z ∈ S ′ , since the edge {z, z ′ } is present in both G 1 and G 2 , 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 1 ∪ E 2 and hence, S forms a vertex cover of G of size at most k to the vertex set and edge set of a clique of size k, then there are K layers in each phase covered by V′ ∪ E ′ .Hence, having K layers where no vertices from C have to be exchanged, in each phase t we can exchange all vertices from U {1, . . ., κ}, we iteratively construct vertex covers for the layers (t − 1)2m + 2 until t2m in the following way.Let T ∶= (t − 1) ⋅ 2m.Let i ∈ {1, . . ., 2m − 1}, and assume the set S t i is already constructed and is a vertex cover for G T +i (this is possible due to the definition of S t 1 ) and thus, S is a smooth solution.Our goal is to prove the existence of the following type of solutions.
Definition 4.16.A smooth solution S = (S 1 , . . ., S τ ) for (G, k ′ , ℓ) from Construction 4.10 is onecentered if (i) for all i ∈ {1, . . ., τ } we have |S i ∩ C| = 1, and (ii) for all i ∈ {2, . . ., τ } and S i−1 S i = (a, b) we have that a ∈ C ⇔ b ∈ C. We now show that there are solutions where c 1 is the only vertex from C in the first set of the solution.Lemma 4.17.Let (G, k ′ , ℓ) from Construction 4.10 be a yes-instance.Then there is a smooth solution (S 1 , . . ., S τ ) such that S 1 ∩ C = {c 1 }.Proof.Suppose not, that is, due to Observation 4.13 in every smooth solution, the first vertex cover S 1 contains at least two vertices from C. Let Σ be the set of smooth solutions with |S 1 ∩ C| being minimal, where S 1 is the first vertex cover.Let S = (S 1 , . . ., S τ ) ∈ Σ be a smooth solution such that the smallest c i ∈ S 1 \ {c 1 } is maximal regarding i.Let S ′ = (S ′ 1 , . . ., S ′ τ