Multistage Vertex Cover

Abstract

The NP-complete Vertex Cover problem asks to cover all edges of a graph by a small (given) number of vertices. It is among the most prominent graph-algorithmic problems. Following a recent trend in studying temporal graphs (a sequence of graphs, so-called layers, over the same vertex set but, over time, changing edge sets), we initiate the study of Multistage Vertex Cover. Herein, given a temporal graph, the goal is to find for each layer of the temporal graph a small vertex cover and to guarantee that two vertex cover sets of every two consecutive 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 themost natural parameterizations.

Full text

Theoryof ComputingSystems ( 2022) 6 6:454 483
https://doi.org/10.1007/s00224-022-10069-w
Multistage Vertex Cover
Till Fluschnik1·Rolf Niedermeier1·Valentin Rohm1·Philipp Zschoche1
Accepted: 2 January 2022
©The Author(s) 2022, corrected publication 2022
Abstract
The NP-complete VERTEX COVER problem asks to cover all edges of a graph by a
small (given) number of vertices. It is among the most prominent graph-algorithmic
problems. Following a recent trend in studying temporal graphs (a sequence of
graphs, so-called layers, over the same vertex set but, over time, changing edge sets),
we initiate the study of MULTISTAGE VERTEX COVER. Herein, given a temporal
graph, the goal is to find for each layer of the temporal graph a small vertex cover and
to guarantee that two vertex cover sets of every two consecutive 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.
Keywords Parameterized algorithmics ·NP-completeness ·Temporal graphs ·Data
reduction
An extended abstract of this paper appeared in Proceedings of the 14th International Symposium on
Parameterized and Exact Computation (IPEC’19), LIPIcs, Schloss Dagstuhl - Leibniz-Zentrum f ¨ur
Informatik, 2019, 148(14):1–14 [1]. This paper contains all details and proofs missing in the
extended abstract. Till Fluschnik acknowledges support by the DFG, project TORE (NI 369/18).
Till Fluschnik
till.fluschnik@tu-berlin.de
Philipp Zschoche
zschoche@tu-berlin.de
Rolf Niedermeier
rolf.niedermeier@tu-berlin.de
Valentin Rohm
valentinl.rohm@campus.tu-berlin.de
1Algorithmics and Computational Complexity, Technische Universit¨
at Berlin, Berlin, Germany
–
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1 Introduction
VERTEX COVER asks, given an undirected graph Gandanintegerk≥0, whether
at most kvertices can be deleted from Gsuch that the remaining graph contains no
edge. VERTEX COVER is NP-complete and it is a formative problem of algorithmics
and combinatorial optimization. We study a time-dependent,“multistage” version,
namely a variant of VERTEX COVER on temporal graphs. A temporal graph Gis
a tuple (V , E,τ) consisting of a set Vof 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 (G1,...,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 Sifor each layer Gisuch that the size of the symmetric difference SiSi+1=
(Si\Si+1)∪(Si+1\Si)of the vertex covers Siand Si+1of every two consecutive
layers Giand Gi+1is small. Formally, we thus introduce and study the following
problem (see Fig. 1for an illustrative example).
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 [2–14]
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 cur-
rent solution comes with a cost. In this spirit, the parameter in the definition of
MULTISTAGE VERTEX COVER allows to model that only moderate changes con-
cerning the solution vertex set may be wanted when moving from one layer to the
subsequent one. Indeed, in this sense can be interpreted as a parameter measuring
the degree of (non-)conservation [15,16].
It is immediate that MULTISTAGE VERTEX COVER is NP-hard as it generalizes
VERTEX COVER (τ=1). We will study its parameterized complexity regarding
the problem-specific parameters k,τ,, and some of their combinations, as well as
restrictions to temporal graph classes [17,18].
Fig. 1 An illustrative example with temporal graph G=(G1,G
2,G
3)over the vertex set V=
{v1,...,v
4}. A solution S=({v2,v
3},{v3},{v1,v
3})fork=2and=1 is highlighted
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Related Work The literature on vertex covering is extremely rich, even when focus-
ing on parameterized complexity studies. Indeed, VERTEX COVER can be seen as
“drosophila” of parameterized algorithmics. 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 [19,20] and graph stream models [21]. More
importantly for our work, Akrida et al. [22] studied a variant of VERTEX COVER
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 .Thatis,
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 con-
tained in the time window, it holds that (v, t) ∈Sor (w, t ) ∈Sfor some tin the
time window with (e, t) ∈E. For their model, Akrida et al. ask whether such an S
of small cardinality exists. Note that if >1, then for some t∈{1,...,τ}the
set St:={v|(v, t ) ∈S}is not necessarily a vertex cover of layer Gt.For=1,
each Stmust be a vertex cover of Gt. However, in Akrida et al.’s model the size of
each Stas well as the size of the symmetric difference between each Stand St+1may
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 interme-
diate results are feasible solutions as well [23,24]. Among other reconfiguration
problems, Mouawad et al. [25,26] studied VERTEX COVER RECONFIGURATION:
given a graph G, two vertex covers Sand Teach of size at most k, and an inte-
ger τ, the question is whether there is a sequence (S =S1,...,S
τ=T) such
that each St,1≤t≤τ, is a vertex cover of size at most k. The essential differ-
ence 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. [25] showed that VER-
TEX COVER RECONFIGURATION is fixed-parameter tractable when parameterized
by vertex cover size kwhile we show W[1]-hardness for the corresponding case of
MULTISTAGE VERTEX COVER.
Finally, there is also a close relation to the research on dynamic parameterized
problems [16,27]. Krithika et al. [27] studied DYNAMIC VERTEX COVE R 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 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 Tab l e 1summarized our results, focusing on the three perhaps
most natural parameters. We highlight a few specific results. MULTISTAGE VER-
TEX COVER remains NP-hard even if every layer consists of only one edge; not
surprisingly, the corresponding hardness reduction exploits an unbounded number τ
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Table 1 Overview of 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-size problem
kernel, and no problem kernel of polynomial size unless coNP ⊆NP/poly
general layers tree layers one-edge layers
0≤<2k≥2k0≤<2k=1
NP-hard NP-hard NP-hard NP-hard
(Theorem 4.1(i)) (Theorem 4.1(ii))
τp-NP-hard p-NP-hard p-NP-hard FPT, PK
(Theorem 4.1) (Theorem 4.1) (Theorem 4.1) (Observation 6.1)
kXP, W[1]-h., FPT†, NoPK XP, W[1]-h. open,NoPK
(Theorem 5.1) (Observation 3.5, (Theorem 5.1, (Theorem 6.1)
Theorem 6.1) Corollary 5.3)
k+τFPT, PK FPT, PK FPT, PK FPT, PK
(Theorem 6.2) (Theorem 6.2) (Theorem 6.2) (Theorem 6.2)
of time layers. If there are one two layers, however, one of them being a tree and
the other being a path, then again MULTISTAGE VERTEX COV ER already becomes
NP-hard. MULTISTAGE VERTEX COVER parameterized by solution size kis fixed-
parameter tractable if ≥2k, but becomes W[1]-hard if <2k. Considering
the tractability results for DYNAMIC VERTEX COVE R [27]andVERTEX COVER
RECONFIGURATION [25], this hardness is surprising; it is our most technical result.
Furthermore, MULTISTAGE VERTEX COVER parameterized by kwith ≥2kdoes
not admit a problem kernel of polynomial size unless coNP ⊆NP/poly. Finally, for
the combined parameter k+τwe obtain polynomial-sized problem kernels (and thus
fixed-parameter tractability) in all cases without any further constraints.
Outline In Section 2, we provide some preliminaries. For MULTISTAGE VERTEX
COVER, we provide some first and general observations in Section 3, study the
parameterized complexity regarding kin Section 5, and discuss the possibilities for
efficient data reduction in Section 6. We conclude in Section 7.
2 Preliminaries
We denote by Nand N0the natural numbers excluding and including zero, respec-
tively. For two sets Aand B,wedenotebyAB:=(A \B) ∪(B \A) =
(A ∪B) \(A ∩B) the symmetric difference of Aand B, and by ABthe disjoint
union of Aand B. For static graphs, we use standard notations [28].
Temporal Graphs A temporal graph Gis a tuple (V , E,τ)consisting of the set Vof
vertices, the set Eof temporal edges, and a discrete time-horizon τ. A temporal edge e
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is an element in V
2×{1,...,τ}. Equivalently, a temporal graph Gcan be defined
as a vector of static graphs (G1,...,G
τ), where each graph is defined over the same
vertex set V. We also denote by V(G),E(G),andτ(G)the set of vertices, the set
of temporal edges, and the discrete (and finite) time-horizon of G, respectively. The
underlying graph G↓=G↓(G)of a temporal graph Gis the static graph with vertex
set V(G)and edge set {e|∃t∈{1,...,τ(G)}:(e, t ) ∈E(G)}.
Parameterized Complexity Theory Let be a finite alphabet. A parameterized prob-
lem Lis a subset L⊆{(x, k) ∈∗×N0}. An instance (x , k) ∈∗×N0is
ayes-instance of Lif 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 Lis fixed-
parameter tractable (FPT) if for every input (x, k) one can decide whether (x, k ) ∈
Lin f(k) ·|x|O(1)time, where fis some computable function only depending
on k. A parameterized problem Lis in XP if for every instance (x, k) one can
decide whether (x, k ) ∈Lin time |x|f(k) for some computable function fonly
depending on k. A W[1]-hard parameterized problem is fixed-parameter intractable
unless FPT=W[1].
Given a parameterized problem L,akernelization is an algorithm that maps any
instance (x, k ) of Lin time polynomial in |x|+kto 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 computable function f(the size of the problem
kernel) only depending on k.
3 Basic Observations
In this section, we state some preliminary simple-but-useful observations on MULTI-
STAGE VERTEX COVER and its relation to VERTEX COVER.
Observation 3.1 Every instance (G,k,) of MULTISTAGE VERTEX COVER
with k≥τ(G)
i=1|E(Gi)|is a yes-instance.
Proof Clearly, a graph with medges always admits a vertex cover of size m. Hence,
there is a vertex cover S⊆Vof size kof G↓(G), and hence, Sis a vertex cover for
each layer. The vector (S1,...,S
τ)with Si=Sfor all i∈{1,...,τ}is a solution
for every ≥0.
Next, we state that if we are facing a yes-instance, then we can assume that there
exists a solution where each layer’s vertex cover is either of size kor k−1.
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=(S1,...,S
τ)such that |S1|=
kand k−1≤|Si|≤kfor all i∈{1,...,τ}.
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Proof We first show that there is a solution S=(S1,...,S
τ)for I:=(G,k,)
such that |S1|=k. (Recall that we assume k<|V(G)|.) Towards a contradiction
assume that such a solution does not exist. Let S=(S1,...,S
τ)be a solution such
that |S1|is maximal over all solutions for I.Leti∈{1,...,τ}be the maximum
index such that Sj⊆Sj−1,forallj∈{2,...,i}.Ifi=τ,thenwehavethat
|Sj|≤|S1|<kfor all j∈{1,...,τ}. Hence, we can find a subset X⊆V\S1such
that (S1∪X,...,S
τ∪X) is a solution. This contradicts |S1|being maximal. Now
let i<τ. Hence, there is a vertex v∈Si+1\Si. Now we can adjust the solution by
adding vto Sjfor all j∈{1,...,i}. This contradicts |S1|being maximal. Hence,
there is a solution S=(S1,...,S
τ)such that |S1|=k.
Let Ψbe the set of solutions such that the first vertex cover is of size k. Assume
towards a contradiction that all solutions in Ψcontain a vertex cover smaller
than k−1. Let Ψi⊆Ψbe the set of solutions such that for each (S1,...,S
τ)∈Ψiwe
have that |Si|<k−1and|Sj|≥k−1forallj∈{1,...,i −1}.Leti∈{1,...,τ}
be maximal such that Ψi=∅. Furthermore, let S=(S1,...,S
τ)∈Ψisuch that |Si|
is maximal over all solutions in Ψi. Hence, there is a vertex v∈Si−1\Si.We
distinguish two cases.
(a): Assume that there is a p∈{i+1,...,τ}such that there is a w∈Sp\Sp−1
and Sj⊆Sj−1for all j∈{i+1,...,p−1}. The idea now is to keep vand add w
in the i-th layer and then remove vin the p-th layer. We can achieve this by simply
setting Sq:=Sq∪{v, w}for all q∈{i,...,p−1}.
(b): Assume that Sj⊆Sj−1for all j∈{i+1,...,τ}. In this case we take an
arbitrary vertex w∈V\Siand set Sq:=Sq∪{v, w}for all q∈{i,...,τ}.
In either of the cases (a) or (b), the obtained solution either contradicts that |Si|is
maximal, or that iis maximal, or that every solution in Ψcontains a vertex cover of
size smaller than k−1.
With the next two observations, we show that the special case of MULTISTAGE
VERTEX COVER with =0 is equivalent to VERTEX COVER under polynomial-time
many-one reductions.
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
VERTEX COVER where =0 and every layer Gicontains only one edge.
Proof Let the edges E={e1,...,e
m}of Gbe ordered in an arbitrary way. Set τ=
mand =0. Set Gi=(V , {ei})for each i∈{1,...,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 MULTISTAGE VERTEX COVER.
(⇒)Let Sbe a vertex cover of Gof size at most k.SetSi:=Sfor all i∈
{1,...,τ}. Clearly, Siis a vertex cover of Gifor all i∈{1,...,τ}of size at
most k. Moreover, by construction, SiSi+1=0foralli∈{1,...,τ −1}. Hence,
(S1,...,S
τ)forms a solution to (G,k,).
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(⇐)Let S=(S1,...,S
τ)be a solution to (G,k,). Observe that |iSi|≤k.It
follows that there are at most kvertices covering all edges of the layers Gi,thatis,
E=τ
i=1E(Gi), and hence they cover all edges of G.
Observation 3.4 There is a polynomial-time algorithm that maps any instance (G,
k, ) of MULTISTAGE VERTEX COVE R with =0 to an equivalent instance (G, k)
of VERTEX COVER.
Proof Now let (G=(V, E,τ),k,0)be an arbitrary instance of MULTISTAGE VER-
TEX COVER. Construct the instance (G↓,k) of VERTEX COVER. We claim that
(G,k,0)is a yes-instance if and only if (G↓,k)is a yes-instance.
(⇐)Let S⊆Vbeavertexcoverofsizeatmostk.SinceSis a vertex cover
for G↓,Scovers each layer of G. Hence, Si:=Sfor all i∈{1,...,τ}forms a
solution to (G,k,0).
(⇒)Let (S1,...,S
τ)be a solution to (G,k,0). Clearly, since =0, we have
that Si=Sjfor all i, j ∈{1,...,τ}. Thus, S:=S1is a vertex cover for G↓,and
hence the claim follows.
Finally, the special case of MULTISTAGE VERTEX COVER with ≥2k(that is,
where vertex covers of any two consecutive layers can be even disjoint) is Turing-
reducible to VERTEX COVER.
Observation 3.5 Any instance (G,k,)of MULTISTAGE VERTEX COVER with ≥
2kand G=(G1,...,G
τ)can be solved by deciding each instance of the
set {(Gi,k) |1≤i≤τ}of VERTEX COVER-instances.
Proof For each of the layers Gi,i∈{1,...,τ}, we can construct an instance of
VERTEX COVER of the form (Gi,k). We can solve each instance independently, since
the symmetric difference of any two size-at-most-ksolutions is at most 2k≤.
4 Hardness for Restricted Input Instances
MULTISTAGE VERTEX COVER is NP-hard as it generalizes VERTEX COVER (τ=1).
In this section, we prove that MULTISTAGE VERTEX COVE R 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 contains only one edge.
Theorem 4.1 MULTISTAGE VERTEX COVE R 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 ≤1.
Remark 4.1 Theorem 4.1(i) is tight regarding τsince VERTEX COVER (i.e., MUL-
TISTAGE VERTEX COVER with τ=1) on trees is solvable in linear time. Theorem
4.1(ii) is tight regarding because if >1, then Observation 3.5 is applicable.
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It is known that VERTEX COVER remains NP-complete on cubic Hamiltonian
graphs when a Hamiltonian cycle is additionally given as part of the input [29]: 1
Proposition 4.1 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 VER-
TEX COVER with τ=2and the first layer G1being a path and the second layer G2
beingatree.
Proof Let e∈E(C) be some edge of C,andletP=C−ebe the Hamiltonian
path obtained from Cwhen removing e.LetE1:=E(P ),andE2:=E\E(P).Set
initially G1=(V , E1)and G2=(V , E2). Note that G1is a path. Moreover, observe
that G2is the disjoint union of |V|/2−2 paths of length one and one path of length
three: the graph G−E(C) is a matching of size |V|/2. This is because each vertex is
of degree three in Gand each vertex is adjacent to two vertices in C. Thus, all vertices
in G−E(C) have degree one. Since G−E(C) =G2−e, edge econnects two paths
of length one to one path of length three in G2. Add two special vertices z, zto V.
In G1, connect zwith zand with one endpoint of P.InG2, connect zwith zand
with exactly one vertex of each connected component. Set k=k+1and=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beavertexcoverofGof size at most k. We claim that S:=S∪{z}is
a vertex cover for both G1and G2. Observe that G1[E1]and G2[E2]are subgraphs
of G, and hence all edges are covered by S. Moreover, all edges in Gi−Ei,i∈{1,2},
are incident with zand hence covered by S.
(⇐)Let (S1,S
2)be a minimal solution to (G,k
,
)with S:=S1=S2and |S|≤
k. We can assume that z∈Ssince the edge {z, z}is present in both G1and G2,
and exchanging zin zdoes not cover less edges. Moreover, we can assume that
not both zand zare in Sdue to the minimality of S.LetS:=S\{z}.Observe
that Scovers all edges in E1∪E2and, hence, Sforms a vertex cover of Gof size at
most k=k−1.
Note that with Observation 3.3, we already proved that MULTISTAGE VERTEX
COVER is NP-hard even if =0 and each layer contains only one edge. In order
to prove Theorem 4.1(ii) (with =1), we adjust the polynomial-time many-one
reduction behind Observation 3.3.
1A graph is cubic if each vertex is of degree exactly three; a graph is Hamiltonian if it contains a subgraph
being a Hamiltonian cycle, that is, a cycle that visits each vertex in the graph exactly once.
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Proposition 4.2 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
VERTEX COVER where =1and every layer Gicontains only one edge.
Proof Let the edges E={e1,...,e
m}of Gbe arbitrarily ordered. Set τ=2m.
Set V=V∪W,whereW={w1,...,w
2m}.SetG2i−1=(V ,{ei})and G2i=
(V ,{wi,w
i+m})for each i∈{1,...,m}.Setk=k+1and=1. 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 MULTISTAGE VERTEX COVER.
(⇒)Let Sbe a vertex cover of Gof size at most k.SetS2i−1:=S,andS2i:=
S∪{wi}for all i∈{1,...,m}. Clearly, Siis a vertex cover of Gifor all i∈{1,...,τ}
of size at most k=k+1. Moreover, by construction, SiSi+1≤1foralli∈
{1,...,2τ−1}. Hence, (S1,...,S
τ)forms a solution to (G,k
,
).
(⇐)Let S=(S1,...,S
τ)be a solution to (G,k
,
). Observe that |iSi|≤
k+τ. We know that |W∩iSi|≥τ. It follows that there are at most kvertices
covering all edges of the layers G2i−1, that is, covering E=m
i=1E(G2i−1), and,
hence, covering all edges of G.
Theorem 4.1 directly follows from Proposition 4.1 and 4.2.
5 Parameter Vertex Cover Size
In this section, we study the parameter size kof the vertex cover of each layer for
MULTISTAGE VERTEX COVER.VERTEX COVE R and VERTEX COVER RECONFIG-
URATION [25], when parameterized by the vertex cover size, are fixed-parameter
tractable. We prove that this is no longer true for MULTISTAGE VERTEX COVER
(unless FPT =W[1]).
Theorem 5.1 MULTISTAGE VERTEX COVE R parameterized by kis in XP and W[1]-
hard.
We first present the XP-algorithm (Section 5.1), and then prove the W[1]-hardness
(Section 5.2) and discuss its implications.
5.1 XP-Algorithm
Here, we prove the following.
Proposition 5.1 Every instance (G,k,) of MULTISTAGE VERTEX COVER can be
decided in O(τ(G)·|V(G)|2k+1)time.
In a nutshell, to prove Proposition 5.1 we first consider for each layer all vertex
subsets of size at most kthat 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 . We show that the second step can be solved via
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computing a source-sink path in an auxiliary directed graph that we call configuration
graph (see Fig. 2for an illustrative example).
Definition 5.1 GivenaninstanceI=(G,k,) of MULTISTAGE VERTEX COVER,
the configuration graph of Iis the directed graph D=(V,A,γ) with V=V1
··· Vτ{s, t }, being equipped with a function γ:V→{V⊆V(G)||V|≤k}
such that
(i) for every i∈{1,...,τ(G)}, it holds true that Sis a vertex cover of Giof size
exactly k−1orkif and only if there is a vertex v∈Viwith γ(v) =S,
(ii) there is an arc from v∈Vto w∈Vif and only if v∈Vi,w∈Vi+1,
and γ(v)γ(w)≤,and
(iii) there is an arc (s, v) for all v∈V1and an arc (v , t) for all v∈Vτ.
Note that Mouawad et al. [25] used a similar configuration graph to show fixed-
parameter tractability of VERTEX COVER RECONFIGURATION parameterized by
the vertex cover size k. In the multistage setting, the configuration graph is too
large for showing fixed-parameter tractability regarding k.However,weshowan
XP-algorithm regarding kto construct the configuration graph.
Lemma 5.1 The configuration graph of an instance (G,k,) of MULTISTAGE
VERTEX COVER,whereGhas nvertices and time horizon τ,
(i) can be constructed in O(τ ·n2k+1)time, and
Fig. 2 Illustrative example of a configuration graph. (a) Temporal graph instance I=(G,k,)from Fig. 1
with G=(G1,G
2,G
3),k=2, and =1. (b) Configuration graph of Ifrom (a); a directed s-tpath is
highlighted corresponding to the solution depicted in Fig. 1
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(ii) contains at most τ·2nk+2vertices and (τ −1)n2k+4nkarcs.
Proof Compute the set S={V⊆V(G)|k−1≤|V|≤k}in O(nk)time. For
each layer Giand each set S∈S, check in O(|E(Gi)|)time whether Sis a vertex
cover for Gi.LetSi⊆Sdenote the set of vertex covers of size k−1orkof layer Gi.
For each S∈Si,addavertexvto Viand set γ(v) =S. Lastly, add the vertices s
and t. Hence, we can construct the vertex set Vof the configuration graph Dof
size τ·2nk+2inO(nk+2·τ) time. For every i∈{1,...,τ −1},andeveryv∈Vi
and w∈Vi+1, check whether γ(v)γ(w)≤in O(k) time. If this is the case, then
add the arc (v, w). The latter steps can be done in O(n2k+1·(τ −1)) time, because
there are at most n2karcs from Vito Vi+1,foralli∈{1,...τ −1}. Finally, add
the arc (s, v) for each v∈V1and the arc (v, t ) for each v∈Vτin O(nk)time,
because |V1|,|Vτ|≤nk. This finishes the construction of D=(V =V1···Vτ
{s, t},A,γ). Note that we added at most (τ −1)n2k+4nkarcs to D.
The crucial observation is that we can decide any instance by checking for an s-t
path in its configuration graph.
Lemma 5.2 AMULTISTAGE VERTEX COVE R-instance I=(G,k,) is a
yes
-
instance if and only if there is an s-tpath in the configuration graph Dof I.
Proof Let D=(V =V1··· Vτ{s, t },A,γ).
(⇒)Let (S1,...,S
τ)be a solution to (G,k,). By Observation 3.2, we can
assume without loss of generality that k−1≤|Si|≤k,foralli∈
{1,...,τ}. Hence for each Si, there is a vi∈Visuch that γ(v
i)=Si,for
all i∈{1,...,τ}. Note that the arc (vi,v
i+1)is contained in Afor each i∈
{1,...,τ −1}since γ(v
i)γ(v
i+1)=SiSi+1≤. Hence, P=({v1,...,v
τ}∪
{s, t},{(s, v1), (vτ,t)}∪τ−1
i=1{(vi,v
i+1)})is an s-tpath in D.
(⇐)Let P=({v1,...,v
τ}∪{s, t },{(s, v1), (vτ,t)}∪τ−1
i=1{(vi,v
i+1)})be an s-
tpath in D. We claim that (γ (vi))i∈{1,...,τ }forms a solution to (G,k,). First, note
that for all i∈{1,...,τ},γ(v
i)is a vertex cover for Giof size at most k. Moreover,
for all i∈{1,...,τ−1},γ(v
i)γ(v
i+1)≤since the arc (vi,v
i+1)is present in D.
This finishes the proof.
Eventually, we are ready to prove Proposition 5.1.
Proof of Proposition 5.1 First, compute the configuration graph Dof the instance
(G=(V , E,τ),k,) of MULTISTAGE VERTEX COVER in O(τ ·|V|2k+1)time
(Lemma 5.1(i)). Then, find an s-tpath in Dwith a breadth-first search in O(τ·|V|2k)
time (Lemma 5.1 (ii)). If an s-tpath is found, then return yes, otherwise return no
(Lemma 5.2).
Remark 5.1 The reason why the algorithm behind Proposition 5.1 is only an XP-
algorithm and not an FPT-algorithm regarding kstems from the fact that we do not
have a better upper bound on the number of vertices in the configuration graph for
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(G,k,) than O(τ(G)·|V(G)|k). This is because we check for each subset of V(G)
of size kor 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 Gi. Here, we can enumerate for each layer Giall minimal vertex cov-
ers 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. This leads to O(2kτ(G)) many
vertices in the configuration graph (for MINIMAL MULTISTAGE VERTEX COV ER)
and thus to fixed-parameter tractability of MINIMAL MULTISTAGE VERTEX COVER
parameterized by the vertex cover size k.
It is unlikely (unless FPT=W[1]), however, that one can substantially improve the
algorithm behind Proposition 5.1, as we show next.
5.2 Presumable Fixed-Parameter Intractability
Here, we show that MULTISTAGE VERTEX COVER is W[1]-hard when parameterized
by k. This hardness result is established by the following parameterized reduction
from the W[1]-complete CLIQUE problem [30], where, given an undirected graph G
and a positive integer k, the question is whether Gcontains a clique of size k(that is,
kvertices that are pairwise adjacent and kis the parameter).
Proposition 5.2 There is an algorithm that maps any instance (G, k) of CLIQUE
in polynomial time to an equivalent instance (G,k
,) of MULTISTAGE VERTEX
COVER with k=2k
2+k+1,=2, and each layer of Gbeing a forest with O(k4)
edges.
In the remainder of this section, we prove Proposition 5.2. We next give the con-
struction of the MULTISTAGE VERTEX COVER instance, then prove the forward
(Section 5.2.1) and backward (Section 5.2.2) direction of the equivalence, and finally
(in Section 5.2.3) put the pieces together and derive two corollaries.
We construct an instance of MULTISTAGE VERTEX COVE R from an instance of
CLIQUE as follows (see Fig. 3for an illustrative example).
Construction 5.1 Let (G =(V , E), k) be an instance of CLIQUE with m:=|E|and
E={e1,...,e
m}.Let
K:=k
2,k
:=2K+k+1,and κ:=K+k+3.
We construct a MULTISTAGE VERTEX COVE R instance (G,k
,) with :=2, where
we construct the temporal graph G=(V ,E,τ) as follows. Let Vbe initialized
to V∪E(note that Esimultaneously describes the edge set of Gand a vertex subset
of G). We add the following vertex sets
Ut:={ut
j|j∈{1,...,K}} for every t∈{1,...,κ +1},and
C:={c1,...,c
2mκ+1}(we refer to Cas the set of center vertices).
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Fig. 3 Illustration of Construction 5.1 on an example graph (left-hand side) and the first seven layers of
the obtained graph (right-hand side). Dashed vertical lines separate layers, and for each layer all present
edges (but only their incident vertices) are depicted. Star-shapes illustrate star graphs with k+1leaves.
Vertices in a solution (layers’ vertex covers) are highlighted
Let Ebe initially empty. We extend the set Vand define Ethrough the τ:=2mκ +1
layers we construct in the following.
1. In each layer Giwith ibeing odd, make cithe center of a star with k+1leaves.
2
2. In each layer G2mj +1,j∈{0,...,κ}, make each vertex in Uj+1the center of
a star with k+1leaves.
3. For each j∈{0,...,κ −1}, in each layer G2mj +iwith i∈{1,...,2m+1},
make uj+1
xadjacent to uj+2
xfor each x∈{1,...,K}.
4. For each even i, add the edge {ci,c
i+1}to Giand to Gi+1.
5. For each j∈{0,...,κ−1}, for each i∈{1,...,m},inG2mj +2i,makecj2m+2i
adjacent to ei={v, w},v,andw.
This finishes the construction of G.
The construction essentially repeats the same gadget (which we call phase)
κtimes, where the layer 2m·i+1 is simultaneously the last layer of phase iand the
first layer of phase i+1. In the beginning of phase i, a solution has to contain the
vertices of Ui. The idea now is that during phase ione has to exchange the vertices
of Uiwith the vertices of Ui+1.
It is not difficult to see that the instance in Construction 5.1 can be computed in
polynomial time. Hence, it remains to prove the equivalence stated in Proposition
5.2. We prove the forward and the backward direction in Sections 5.2.1 and 5.2.2,
respectively, and finally prove Proposition 5.2 in Section 5.2.3.
2A star (graph) is a tree where at most one vertex (the so-called center) has degree larger than one.
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5.2.1 Forward Direction
The forward direction of Proposition 5.2 is—in a nutshell—as follows: If V∪E
with V⊆Vand E⊆Ecorresponds to the vertex set and edge set of a clique
of size k, then there are Klayers in each phase covered by V∪E. Hence, hav-
ing Klayers where no vertices from Chave to be exchanged, in each phase twe can
exchange all vertices from Utto Ut+1. Starting with set S1:= U1∪V∪E∪{c1}
then yields a solution.
Lemma 5.3 Let (G, k) be an instance of CLIQUE and (G,k
,) be the instance
of MULTISTAGE VERTEX COVER resulting from Construction 5.1. If (G, k) is a
yes
-instance, then (G,k
,) is a
yes
-instance.
Proof Let G=(V ,E
)be the clique of size kin G. We construct a solution S=
(S1
1,...,S1
2m,S1
2m+1=S2
1,...,Sκ
2m+1=Sκ+1
1)for (G,k
,) in the following way.
For each t∈{1,...,κ +1}we set St
1:= V∪E∪Ut∪{c(t −1)2m+1},whichisa
vertex cover of size kfor G(t −1)2m+1.
Now, for each t∈{1,...,κ}, we iteratively construct vertex covers for the layers
(t −1)2m+2 until t2min the following way. Let T:=(t −1)·2m.Leti∈
{1,...,2m−1}, and assume that the set St
iis already constructed and is a vertex
cover for GT+i(this is possible due to the definition of St
1). We distinguish two cases.
Case 1: iis odd. We know that cT+i∈St
i.If(St
i\{cT+i})∪{cT+i+2}is a vertex
cover for GT+i+1,thenwesetSt
i+1:= (St
i\{cT+i})∪{cT+i+2}. Otherwise we set
St
i+1:= (St
i\{cT+i})∪{cT+i+1}. In both cases St
i+1is a vertex cover for GT+i+1
and either St
i+1∩C={cT+i+1}or St
i+1∩C={cT+i+2}.
Case 2: iis even. We know that cT+ior cT+i+1is in St
i.IfcT+i∈St
i,thenwe
set St
i+1=(St
i\{cT+i})∪{cT+i+1}, which is a vertex cover for GT+i+1.If
cT+i+1∈St
i,thenSt
iis already a vertex cover for GT+i+1and the vertices in
V∪Ecover all edges incident with cT+iin the graph GT+i. In this case we say
that Gcovers the layer T+iand set St
i+1=(St
i\{ut
j})∪{ut+1
j},whereut
jis an
arbitrary vertex in St
i∩Ut.
Observe that the clique Gcovers Keven-numbered layers in each phase. Hence,
we replace, during phase t∈{1,...,κ}(that is, from layer (t −1)2m+1tot2m+1),
the vertices Utwith the vertices Ut+1. This also implies that the symmetric differ-
ence of two consecutive sets in Sis exactly 2 =. It follows that Sis a solution
for (G,k
,).
5.2.2 Backward Direction
In this section, we prove the backward direction for the proof of Proposition 5.2.
We first show that if an instance of MULTISTAGE VERTEX COVER computed by
Construction 5.1 is a yes-instance, then it is safe to assume that two vertices are
neither deleted from nor added to a vertex cover in a consecutive step (we refer to
these solutions as smooth, see Definition 5.2). Moreover, a vertex from the vertex
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set Cis only exchanged with another vertex from Cand, at any time, there is exactly
one vertex from Ccontained in the solution (similarly to the constructed solution in
Lemma 5.3. We call these (smooth) solutions one-centered (Definition 5.3). We then
prove that there must be a phase tfor any one-centered solution where at least k
2
times a vertex from “past” sets Ut,t≤t, is deleted. This at hand, we prove that
such a phase witnesses a clique of size k.
The fact that a solution needs to contain at least one vertex from Cat any time
immediately follows from the fact that there is either an edge between two vertices
in Cor there is a vertex in Cwhich is the center of a star with k+1leaves.
Observation 5.2 Let (G,k
,) from Construction 5.8 be a yes-instance. Then for
each solution (S1,...,S
τ)it holds true that |Si∩C|≥1foralli∈{1,...,τ(G)}.
In the remainder of this section, we denote the vertices which are removed from
the set Si−1and added to the next set Siin a solution S=(...,S
i−1,S
i,...)
by
Si−1Si:=(Si−1\Si,S
i\Si−1).
If Si−1\Sior Si\Si−1have size one, then we will omit the brackets of the singleton.
Definition 5.2 A solution S=(S1,...,S
τ)for (G,k
,) from Construction 5.8 is
smooth if for all i∈{2,...,τ}we have |Si−1\Si|≤1and|Si\Si−1|≤1.
In fact, if there is a solution, then there is also a smooth solution.
Observation 5.3 Let (G,k
, =2)from Construction 5.1 be a yes-instance. Then
there is a smooth solution (S1,...,S
τ).
Proof By Observation 3.1, we know that there is a solution S=(S1,...,S
τ)such
that |S1|=kand k−1≤|Si|≤kfor all i∈{1,...,τ}. Hence, for all i∈
{2,...,τ}it holds true that 
|Si|−|Si−1|
≤1. By Construction 5.8, we have that
|SiSi−1|≤=2, for all i∈{2,...,τ}. It follows that |Si−1\Si|≤1and
|Si\Si−1|≤1, and thus, Sis a smooth solution.
Our next goal is to prove the existence of the following type of solutions.
Definition 5.3 A smooth solution S=(S1,...,S
τ)for (G,k
,) from Construc-
tion 5.1 is one-centered if
1. for all i∈{1,...,τ}it holds true that |Si∩C|=1, and
2. for all i∈{2,...,τ}and Si−1Si=(α, β) it holds true that α∈C⇐⇒ β∈
C.
With the next two lemmata, we prove that in case of a yes-instance, a one-
centered solution exists. We first show that if the output instance of Construction 5.1
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is a yes-instance, then there is a solution where c1∈Cis the only vertex from Cin
the first set of the solution.
Lemma 5.4 Let (G,k
,) from Construction 5.1 be a
yes
-instance. Then there is a
smooth solution (S1,...,S
τ)for (G,k
,) such that S1∩C={c1}.
Proof Suppose towards a contradiction that such a smooth solution does not exist.
Since we know from Observation 5.2 that there is at least one smooth solution, our
assumption means that, in every smooth solution the first vertex cover S1contains
at least two vertices from C(due to Observation 5.1, S1must contain at least one).
Let Ψbe the set of smooth solutions with |S1∩C|being minimal, where S1is the
first vertex cover. Let S=(S1,...,S
τ)∈Ψbe a smooth solution such that the
value i:=min{j∈{1,...,τ}|cj∈S1\{c1}} is maximal. Let S=(S
1,...,S
τ)
be initially S.
Suppose that ciwas moved out of the solution before the i-th layer. That is, there is
aj∈{1,...,i−1}such that SjSj+1=(ci,α).Letj:=min{j∈{1,...,i−1}|
SjSj+1=(ci,α)}be the smallest among them. Then, set S
q:=Sq\{ci}for all q∈
{1,...,j−1}to get a feasible solution (note that S
j−1S
j=(∅,α) is feasible
since |S
j−1|≤k−1). This contradicts the minimality of Sregarding |S1∩C|.
Hence, suppose that there is no such j,thatis,thereisnoj∈{1,...,i −1}such
that SjSj+1=(ci,α).IfSi∗\{ci}is a vertex cover of layer Gi∗for all i∗≤i,then
setting S
q:=Sq\{ci},forallq∈{1,...,p}with p:=max{p∈{1,...,τ}|∀q∈
{1,...,p
}: ci∈Sq}, yields a feasible solution. This contradicts the minimality
of Sregarding |S1∩C|.
Finally, suppose that there is no j∈{1,...,i −1}such that SjSj+1=(ci,α)
(and hence ci∈Si)andSi∗\{ci}is no vertex cover of layer Gi∗, with i∗≤ismallest
possible. Note that i∗∈{i−1,i}, and we distinguish the two cases.
Case 1: i∗=i.LetSi−1Si=(α, β) for some α, β (each being possibly the
empty set). Note that β= ci. Then for all q∈{1,...,i −1}do the following (we
distinguish two cases):
Case 1.1: β=crwith r<i. We remove cifrom the first i−1 vertex covers
and βfrom all vertex covers after icontaining β. Formally, set S
q:=Sq\
{ci}and S
q:=Sq\{β}(i.e. S
i−1S
i=(α, ci))forallq∈{i,...,p}
with p:=max{p∈{1,...,τ}|∀p ∈{i,...,p
}: β∈Sp }. Recall that ci
is dispensable from the first i−1 vertex covers, Si\{β}is a vertex cover of Gi
(since, if iis odd, then ciis the center of a star, or if ieven, then βis isolated),
and βis isolated in all layers after the i-th one. Moreover, from the (i −1)-st
to i-th vertex cover, we exchange αby ciinstead of αby β. Hence, the obtained
sequence is a solution. This contradicts the minimality of Sregarding |S1∩C|.
Case 1.2: β=crwith r>i,orβ∈ C. We replace ciby βin the first i−1
vertex covers. Formally, set S
q:=(Sq\{ci})∪{β}(note that S
i=Si
and hence S
i−1S
i=(α, ci)). Note that if there is a p∈{2,...,i −1}
with Sp−1Sp=(β, x ) or Sp−1Sp=(x, β ),thenwegetS
p−1S
p=(∅,x)
and S
p−1S
p=(x, ∅), respectively. Recall that ciis dispensable from the
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first i−1 vertex covers, and βis still contained in each vertex cover it con-
tained before the modification. Moreover, from the (i −1)-st to the i-th vertex
cover, we exchange αby ciinstead of αby β. Hence, the obtained sequence is
a solution. In the case of β=crwith r>i, this contradicts the fact that ciis
maximal regarding i. In the case of β∈ C, this contradicts the minimality of S
regarding |S1∩C|.
Case 2: i∗=i−1. It follows that iis odd. Let Si−2Si−1=(α, β) for some α, β
(each being possibly the empty set). Note that β= ci.SinceSi−1\{ci}is not a
vertex cover of Gi−1,wehavethatci−1∈ Si−1. It follows that β= ci−1.Then
for all q∈{1,...,i −2}do the following (we distinguish two cases):
Case 2.1: β=crwith r<i−1. We remove cifrom the first i−2 vertex covers
and βfrom all vertex covers after (i −2)-nd containing β.Formally,setS
q:=
Sq\{ci}and S
q:=Sq\{β}(i.e. S
i−2S
i−1=(α, ci))forallq∈{i−1,...,p}
with p:=max{p∈{1,...,τ}|∀p ∈{i−1,...,p
}: β∈Sp }. Recall
that ciis not part of any of the first i−2vertexcoversandβis isolated in all
layers after the (i −2)-nd one. Moreover, from the (i −2)-nd to i-th vertex
cover, we exchange αby ciinstead of αby β. Hence, the obtained sequence is
a solution. This contradicts the minimality of Sregarding |S1∩C|.
Case 2.2: β=crwith r>i,orβ∈ C. We replace ciby βin the first i−2
vertex covers. Formally, set S
q:=(Sq\{ci})∪{β}(note that S
i−1=Si−1
and hence S
i−2S
i−1=(α, ci)). Note that if there is a p∈{2,...,i −1}
with Sp−1Sp=(β, x ) or Sp−1Sp=(x, β ),thenwegetS
p−1S
p=(∅,x)
and S
p−1S
p=(x, ∅), respectively. Note that ciis isolated in the first i−2
layers, and βis still contained in each vertex cover it was contained in before
the modification. Moreover, from the (i−2)-ndtothe(i −1)-st vertex cover, we
exchange αby ciinstead of αby β. Hence, the obtained sequence is a solution.
If β=crwith r>i, then this contradicts the fact that ciis maximal regarding i.
If β∈ C, then this contradicts the minimality of Sregarding |S1∩C|.
In every case, we obtain a contradiction, concluding the proof.
Next, we show that there are solutions such that whenever we remove a vertex in
the set of center vertices Cfrom the vertex cover, then we simultaneously add another
vertex from Cto the vertex cover. Formally, we prove the following.
Lemma 5.5 Let (G,k
,) from Construction 5.1 be a
yes
-instance. Then there is a
smooth solution (S1,...,S
τ)with S1∩C={c1}such that for all i∈{1,...,τ}with
Si−1Si=(α, c) and c∈Cwe also have α∈C.
Proof Suppose towards a contradiction the contrary. That is, let for every smooth
solution (S1,...,S
τ)exist an i∈{1,...,τ}with Si−1Si=(α, c) and c∈C
and α∈ C.LetΨbe the non-empty (due to Lemma 5.4) set of smooth solutions
(S1,...,S
τ)with |S1∩C|=1. Let Ψ⊆Ψbe the set of smooth solutions that
maximizes the first index iwith Si−1Si=(α, cq)with cq∈Cand α∈ C. Among
those solutions, consider S=(S1,...,S
τ)∈Ψto be the one with qbeing maximal.
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Note that due to Observation 5.10, we have that |Si−1∩C|≥1. Let S
j:=Sjfor
all j∈{1,...,τ}.
Case 1: i>1is odd.Sinceciis the center of a star in layer i,cihas to be in Si.
We distinguish three subcases regarding the relation of qand i, that is, the cases
of qbeing smaller than, equal to, or larger than i.
Case 1.1: q<i. We remove cqfrom the longest sequence of vertex covers con-
taining cqstarting from the i-th vertex cover. Formally, set S
j:= (Sj\{cq})
(i.e., S
i−1S
i=(α, ∅))forallj∈{i,...,q}with q:=max{q ∈{i,...,τ}|
∀j∈{i,...,q }: cq∈Sj}. Note that cqwith q<iis isolated in all layers
after the i-th one and Si\{cq}is a vertex cover of layer Gisince ci∈Si.Itfol-
lows that (S
1,...,S
τ)is again a feasible smooth solution contradicting ibeing
maximal.
Case 1.2: q=i.Wehaveci∈ Si−1. Since the edge {ci−1,c
i}must be covered
in layer Gi−1, it follows that ci−1∈Si−1. Moreover, ci−1∈Si(and possibly
more subsequent vertex covers) since α∈ C.Asci−1is isolated in all layers
after the i-th and ci∈Si, we remove ci−1and add αin the i-th and subsequent
layers (i.e., instead of exchanging αwith cqfrom the (i −1)-st to the i-th
layer, we exchange ci−1with cq). Formally, set S
p:= (Sp\{ci−1})∪{α}
(i.e., S
i−1S
i=(ci−1,c
q))forallp∈{i,...,j},wherej>iis minimal
such that Sj−1Sj=(ci−1,x),orτif such a jdoes not exist. If there is a
minimal j>isuch that Sj−1Sj=(ci−1,x),thensetS
p:= (Sp\{α})(i.e.,
S
j−1S
j=(α, x))forallp∈{j,...,q}with q:=max{q ∈{i,...,τ}|
∀p∈{i,...,q}: α∈Sp}. Suppose that between iand j, there are j1and j2
such that Sj1−1Sj1=(y, α ) and Sj2−1Sj2=(α, y). Note that S
j1−1S
j1=
(y, ∅)and S
j1−1S
j1=(∅,y). It follows that (S
1,...,S
τ)is again a feasible
smooth solution, contradicting ibeing maximal.
Case 1.3: q>i.Thenci∈Si−1.Letq∗≤qbe the smallest index with Sq∗\
{cq}beingnovertexcoverofGq∗. Note that q∗∈{q−1,q}. Observe that if
no such q∗exists, then we can exclude cqfrom all vertex cover starting from
the i-th one. Formally, set S
j:= Sj\{cq}(i.e., S
i−1S
i=(α, ∅))forallj∈
{i,...,q}where q:= max{q ∈{i,...,τ}|∀j∈{i,...,q }: cq∈Sj},
contradicting the fact that iis maximal. We distinguish two cases:
Case 1.3.1: q∗=q.LetSq−1Sq=(β, d ). Note that if d=cq,thenwe
remove cqfrom all vertex covers before the q-th one, yielding a contradiction
to ibeing maximal. We distinguish two cases regarding d.
Case 1.3.1.1: d=cpwith p<q.Instead of adding cqto the i-th vertex
cover, we add cqinstead of dto the q-th vertex cover (recall that cqis
dispensable from all vertex covers before the q-th). Note that if qis even,
then dis isolated in all layers after the (q −1)-st one, and if qis odd,
then Sq\{d}is a vertex cover of Gq(since cqis added to Sq)anddis
isolated in all layers after the q-th one. Formally, set S
j:= Sj\{cq}
(i.e., S
i−1S
i=(α, ∅))forallj∈{i,...,q −1}. Moreover, set S
j:=
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(Sj\{d})∪{cq}(i.e., S
q−1S
q=(β, cq))forallj∈{q,...,q}with q:=
max{q ∈{q,...,τ}|∀j∈{q,...,q}: d∈Sj}. The obtained
sequence is a solution.
Case 1.3.1.2: d∈ Cor if d=cp,thenp>q.Instead of introducing cqin
the i-th and din the q-th vertex cover, we swap their timings and intro-
duce din the i-th vertex cover and cqin the q-th vertex cover. Set S
j=
(Sj\{cq})∪{d}(i.e., S
i−1S
i=(α, d))forallj∈{i,...,q−1}.More-
over, set S
j=Sj∪{cq}(i.e., S
q−1S
q=(β, cq)or S
q−1S
q=(β, ∅))for
all j∈{q,...,q}with q:=max{q ∈{q,...,τ}|∀j∈{q,...,q}:
cq∈Sj}. Recall that cqis dispensable from all vertex covers before
the q-th one. Hence, the obtained sequence is a solution.
In either case, we have that (S
1,...,S
τ)is a feasible solution contradict-
ing either ibeing maximal (d∈ C,ord=cpwith p<q)orqbeing
maximal (d=cpwith p>q).
Case 1.3.2: q∗=q−1. It follows that qmust be odd and that cq−1∈ Sq−1.
Let Sq−2Sq−1=(β, d ). Note that d= cq−1and that if d=cq,then
we remove cqfrom all vertex covers before the (q −1)-st one, yielding a
contradiction to ibeing maximal. We distinguish two cases regarding d.
Case 1.3.2.1: d=cpwith p<q−1.Instead of adding cqto the i-th ver-
tex cover, we add cqinstead of dto the (q −1)-st vertex cover. Formally,
set S
j=Sj\{cq}(i.e., S
i−1S
i=(α, ∅))forallj∈{i,...,q −2}.
Moreover, set S
j=(Sj\{d})∪{cq}(i.e., S
q−2S
q−1=(β, cq))for
all j∈{q−1,...,q}with q:=max{q ∈{q−1,...,τ}|∀j∈
{q−1,...,q }: d∈Sj}.Sinced=cpfor p<q−1andqis odd, dis
isolated in all layers after the (q −2)-nd. Hence, the obtained sequence is
a solution.
Case 1.3.2.2: d∈ Cor if d=cp,thenp>q.Instead of introducing cqin
the i-th and din the (q −1)-st vertex cover, we swap their timings and
introduce din the i-th vertex cover and cqin the (q −1)-st vertex cover.
Set S
j:= (Sj\{cq})∪{d}(i.e., S
i−1S
i=(α, d))forallj∈{i,...,q−
2}. Moreover, set S
j:= Sj∪{cq}(i.e., S
q−2S
q−1=(β, cq)or S
q−2
S
q−1=(β, ∅))forallj∈{q−1,...,q}with q:=max{q ∈{q−
1,...,τ}|∀j∈{q−1,...,q}:cq∈Sj}. Note that cqis isolated before
the (q −1)-st layer, and dappears in a superset of vertex covers after the
modification. Hence, the obtained sequence is a solution.
In either case, we have that (S
1,...,S
τ)is a feasible solution contradict-
ing either ibeing maximal (d∈ C,ord=cpwith p<q)orqbeing
maximal (d=cpwith p>q).
Case 2: i>1is even.Thenci−1∈Si−1and cq∈{ci,c
i+1}.Sinceci−1is isolated
in all layers after the (i −1)-st, we can exchange ci−1instead of αby cq. Formally,
set S
j:=(Sj\{ci−1})∪{α}(i.e., S
i−1S
i=(ci−1,c
q))forallj∈{i,...,q}
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with q:=max{q ∈{i,...,τ}|∀j∈{i,...,q}: ci−1∈Sj}.Then
(S
1,...,S
τ)is a feasible solution contradicting ibeing maximal.
Combining Observation 5.1 and Lemma 5.5, we can assume that for every given
yes-instance, there is a solution which is one-centered.
Corollary 5.1 Let (G,k
,) from Construction 5.1 be a
yes
-instance. Then there is
a solution Swhich is one-centered.
In the remainder of this section, for each t∈{1,...,κ +1}let the union of all Ui
be denoted by

Ut:=t
i=1Ui.
We introduce further notation regarding a one-centered solution S:=(S1
1,...,
S1
2m+1=S2
1,...,...,Sκ
1,...,Sκ
2m+1)for (G,k
,). Here, St
iis the i-th set of phase t
and thus the (2m(t −1)+i)-th set of S.Theset
Yt
i:={ej∈St
i∩E|2j≥i}(1)
is the set of vertices ejfrom Ein St
isuch that the corresponding layer for ejin phase
tis not before the layer iin phase t.Theset
Ft
i:={j>i|St
j−1St
j=(u, β) with u∈
Ut}(2)
is the set of layers from Gin phase twhere a vertex from 
Utis not carried over to the
next layer’s vertex cover. We now show that there is a phase twhere |Ft
1|≥K.
Lemma 5.6 Let S=(S1
1,...,S1
2m+1=S2
1,...,...,Sκ
1,...,Sκ
2m+1)be a one-
centered solution to (G,k
,) from Construction 5.1. Then, there is a t∈{1,...,κ}
such that |Ft
1|≥K.
Proof Suppose towards a contradiction that for all t∈{1,...,κ}it holds true
that |Ft
1|<K. Then, for each i∈{2,...,κ +1},wehavethat|Si
1∩
Ui−1|≥i−1.
Since Sis a solution, we know that Uκ+1⊆Sκ+1
1and hence |Sκ+1
1∩Uκ+1|=K.
Thus, we have that
|Sκ+1
1|≥|Sκ+1
1∩Uκ+1|+|Sκ+1
1∩
Uκ|≥K+κ−1=2K+k+2>k
,
contradicting Sbeing a solution.
In the remainder of this section, the value
ft
i:=|St
i∩
Uκ+1|−K(3)
describes the number of vertices in 
Uκ+1which we could remove from St
isuch that
St
iis still a vertex cover for G2m(t−1)+i(the i-th layer of phase t). Observe that
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ft
i≥0forallt∈{1,...,κ}and all i∈{1,...,2m+1}, because we need in each
layer exactly Kvertices from 
Uκ+1in the vertex cover.
We now derive an invariant which must be true in each phase.
Lemma 5.7 Let S=(S1
1,...,S1
2m+1=S2
1,...,...,Sκ
1,...,Sκ
2m+1)be a one-
centered solution to (G,k
,) from Construction 5.1. Then, for all t∈{1,...,κ}and
all i∈{1,...,2m+1}, it holds true that |Ft
i|−|Yt
i|≤ft
i.
Proof Let t∈{1,...,κ}be arbitrary but fixed. For all i∈{1,...,2m+1}let
εi:=|Ft
i|−|Yt
i|−ft
i.
We claim that εi−εi−1≥0foralli∈{1,...,2m+1}.SinceSis one-centered,
in Table 2all relevant tuples for St
i−1St
iare shown. As each relevant tuple results
in εi−εi−1∈{0,1,2}, the claim follows.
We want to prove that εi≤0foralli∈{1,...,2m+1}. So, assume towards
a contradiction that there is a j∈{1,...,2m+1}such that εj>0. Since εi−
εi−1≥0foralli∈{1,...,2m+1},wehavethatε2m+1>0, which is equivalent
to |Ft
2m+1|−|Yt
2m+1|>f
t
2m+1. By definition, we have that |Yt
2m+1|=0(see(1))
and |Ft
2m+1|=0(see(2)). Moreover, since Sis a solution and each vertex cover
needs at least Kvertices from 
Uτ,wehavethatft
2m+1≥0. It follows that 0 =
|Ft
2m+1|−|Yt
2m+1|>ft
2m+1≥0, yielding a contradiction.
Next, we prove that in a phase twith |Ft
1|≥K, there are at most kvertices from V
contained in the union of the vertex covers of phase t.
Lemma 5.8 Let S=(S1
1,...,S1
2m+1=S2
1,...,...,Sκ
1,...,Sκ
2m+1)be a one-
centered solution to (G,k
,) from Construction 5.1, and let t∈{1,...,κ}be such
that |Ft
1|≥K. Then, |2m+1
i=1St
i∩V|≤k.
Table 2 Overview of all tuples of St
i−1St
irelevant in the proof of Lemma 5.17 and their possible values
of εi−εi−1=|Ft
i|−|Ft
i−1|−(|Yt
i|−|Yt
i−1|)−(f t
i−ft
i−1)
St
i−1St
i|Ft
i|−|Ft
i−1|−(|Yt
i|−|Yt
i−1|)−(f t
i−ft
i−1)ε
i−εi−1
(u, β) β ∈E∈{−1,0}∈{0,1}1∈{0,1,2}
β∈
Uκ+1∈{−1,0}10∈{0,1}
β∈V,β=∅ ∈{−1,0}11∈{1,2}
(α, u) α ∈E0∈{1,2}-1 ∈{0,1}
α∈V,α=∅ 0 1 -1 0
(α, v) α ∈E0∈{1,2}0∈{1,2}
α∈V,α=∅ 01 0 1
(α, e) α ∈V01 0 1
α∈E,α=∅ 0∈{0,1}0∈{0,1}
In the tuples, u,v,anderepresent some vertices from 
Uκ+1,V,andE, respectively
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Proof From Lemma 5.7, we know that |Yt
1|≥K−ft
1.Let
|Yt
1|=K−ft
1+λ
for some λ∈N0,andletεi=|Ft
i|−|Yt
i|−ft
i,foralli∈{1,...,2m+1}.
We now show that there are at most λlayers where we exchange a vertex currently
in the vertex cover with a vertex in V.Leti∈{2,...,2m+1}such that St
i−1St
i=
(α, v) with v∈V.FromTable2(recall that one-centered solutions are smooth), we
know that εi≥εi−1+1.
Assume towards a contradiction that there are λ+1 many of these exchanges.
Then, there is a j∈{1,...,2m+1}such that
εj≥ε1+λ+1=|Ft
1|−|Yt
1|−ft
1+λ+1
≥K−(K −ft
1+λ) −ft
1+λ+1≥1⇐⇒ | Ft
j|−|Yt
j|>ft
j.
This contradicts the invariant of Lemma 5.7.
In the beginning of phase t,wehaveatmostk−λvertices from Vin the vertex
cover because
|St
1∩V|≤K+k−|Yt
1|−ft
1=K+k−(K −ft
1+λ) −ft
1=k−λ.
Since there are at most λmany exchanges St
i−1St
i=(α, v) where v∈Vand i∈
{2,...,2m+1}, we know that the vertex set 2m+1
i=1St
i∩Vis of size at most k.
We are set to prove the backward direction of Proposition 5.2.
Lemma 5.9 Let (G, k) be an instance of CLIQUE and (G,k
,) be the instance
of MULTISTAGE VERTEX COVER resulting from Construction 5.1. If (G,k
,) is a
yes
-instance, then (G, k) is a
yes
-instance.
Proof Let (G,k
,) be a yes-instance. From Corollary 5.16 it follows that there
is a one-centered solution S=(S1
1,...,S1
2m+1=S2
1,...,...,Sκ
1,...,Sκ
2m+1)for
(G,k
,). By Lemma 5.6, there is a t∈{1,...,κ}such that |Ft
1|≥K=k
2.By
Lemma 5.8, we know that |2m+1
i=1St
i∩V|≤k. Now we identify the clique of size
kin G.Since|Ft
1|≥K, we know that, by Construction 5.1, at least K=k
2layers
are covered by vertices in V∪E∪
Uκ+1∪{ct
2j+1|j∈{1,...,m}} in phase t.Note
that each of these layers corresponds to an edge e={v, w}in Gand that we need in
particular the vertices vand win the vertex cover. Since we have at most kvertices
in 2m+1
i=1St
i∩V, these vertices induce a clique of size kin G.
5.2.3 Proof of Proposition 5.2 and two corollaries
We proved the forward and backward direction of Proposition 5.2 in Sections 5.2.1
and 5.2.2, respectively. It remains to put everything together.
Proof of Proposition 5.2 Let (G, k) be an instance of CLIQUE and (G,k
,) be the
instance of MULTISTAGE VERTEX COVER resulting from Construction 5.1. Observe
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that Construction 5.1 runs in polynomial time, and that each layer of Gis a forest with
O(k2)edges. We know that if (G, k) is a yes-instance of CLIQUE,then(G,k
,)
is a yes-instance of MULTISTAGE VERTEX COVER (Lemma 5.3), and vice versa
(Lemma 5.9). Finally, the W[1]-hardness of CLIQUE [30] regarding kand the fact
that k∈O(k2)then finish the proof.
From a motivation point of view, it is natural to assume that the change over
time modeled by the temporal graph is rather of evolutionary character, meaning
that the difference of a layer to its predecessor is limited. However, Proposition 5.2
gives a bound (in terms of the desired vertex cover size in input instance) on the
number of edges of each layer. In particular, we also obtain the following W[1]-
hardness.
Corollary 5.2 MULTISTAGE VERTEX COVE R parameterized by the maximum num-
ber maxi∈{1,...,τ }|E(Gi)|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 when parameterized for example by the combination of kand the maximum
size of symmetric difference between two consecutive layers.
Furthermore, we can turn the instance (G,k
,) resulting from Construction 5.1
into an equivalent instance (G,k
,)where each layer is a tree as follows. Set k :=
k+1. Add a vertex xto G. In each layer of G,makexthe center of a star with k
(new) leaf vertices, and connect xwith exactly one vertex of each connected com-
ponent. Note that, in every solution, xis contained in a vertex cover for each layer
in G.
Corollary 5.3 MULTISTAGE VERTEX COVER parameterized by kis W[1]-hard,
even if each layer is a tree.
Note, however, that in Corollary 5.3, maxi∈τ|E(Gi)|is unbounded and we cannot
hope to strengthen the reduction in this sense because if each layer is a tree, then we
have exactly |V|−1 edges in each layer. This would contradict Proposition 5.1.
6 On Efficient Data Reduction
In this section, we study the possibility of efficient and effective data reduction for
MULTISTAGE VERTEX COVER when parameterized by k,τ,andk+τ,thatis,
the possible existence of problem kernels of polynomial size. We prove that unless
coNP ⊆NP/poly, MULTISTAGE VERTEX COVER admits no problem kernel of size
polynomial in keven on quite restricted inputs (Section 6.1). Yet, when combining k
and τ, we prove a problem kernel of size O(k2τ) (Section 6.2). Moreover, we prove
a problem kernel of size 5τwhen each layer consists of only one edge (Section 6.3).
Recall (from Theorem 4.1) that MULTISTAGE VERTEX COVER is para-NP-hard
regarding τeven if each layer is a tree.
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6.1 No Problem Kernel of Size Polynomial in
k
for Restricted Input Instances
In this section, we prove the following.
Theorem 6.1 Unless coNP ⊆NP/poly, MULTISTAGE VERTEX COVE R admits no
polynomial kernel when parameterized by k, even
(i) if each layer consists of one edge and =1,or
(ii) if each layer is planar3and ≥2k.
Recall that MULTISTAGE VERTEX COVER parameterized by kis fixed-parameter
tractable in 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 [31].
Definition 6.1 An AND-composition for a parameterized problem Lis an algo-
rithm that, given pinstances (x1,k),...,(x
p,k)of L, computes in time polynomial
in p
i=1|xi|an instance (y, k )of Lsuch that
1. (y, k )∈Lif and only if (xi,k) ∈Lfor all i∈{1,...,p},and
2. kis polynomially upper-bounded in k.
The following is the crucial connection to polynomial kernelization: If a param-
eterized problem whose unparameterized version is NP-hard admits an AND-
composition, then coNP ⊆NP/poly [32]. Note that coNP ⊆NP/poly implies a
collapse of the polynomial-time hierarchy to its third level [33].
In the proof of Theorem 6.1(i), we use an AND-composition. The idea is to take
pinstances of MULTISTAGE VERTEX COVER on the same vertex set with =1and
identical k, and stack all these instances one after the another in the time dimension.
Here, we connect the i-th instance with (i +1)-st instance by just repeating the first
layer of the (i +1)-st instance so often such that there is enough time to transfer from
a solution of the i-th instance to a solution of the (i +1)-st instance without violating
the upper bound on the symmetric difference between two consecutive vertex covers.
Formally, we use the following construction.
Construction 6.1 Let (G1,k,),...,(Gp,k,) be pinstances of MULTISTAGE
VERTEX COVER where =1 and each layer of each Gq=(V , Eq,τ
q),q∈
{1,...,p}, consists of one edge. We construct an instance (G=(V , E,τ),k,)
of MULTISTAGE VERTEX COVER as follows. Denote by (Gi
1,...,G
i
τi)the
sequence of layers of Gi. Initially, let Gbe the temporal graph with layer
sequence ((Gi
j)1≤j≤τi)1≤i≤p. Next, for each i∈{1,...,p−1}, insert between Gi
τi
and Gi+1
1the sequence (H i
1,Hi
2,...,Hi
2k):=(Gi
τi,G
i+1
1,...,G
i+1
1). This finishes
the construction. Note that τ:=2k(p −1)+p
i=1τi.
3A graph is planar if it can be drawn on the plane such that no two edges cross each other.
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In the next two propositions, we prove that Construction 6.1 forms AND-
compositions, then used in the proof of Theorem 6.1(i).
Proposition 6.1 MULTISTAGE VERTEX COVE R where each layer consists of one
edge and =1admits an AND-composition when parameterized by k.
Proof We AND-compose MULTISTAGE VERTEX COVER where each layer consists
of one edge. Let I1=(G1=(V , E1,τ
1),k,),...,I
p=(Gp=(V , Ep,τ
p), k, )
be pinstances of MULTISTAGE VERTEX COVER with =1 where each layer
consists of one edge. Apply Construction 6.1 to obtain instance I=(G=
(V , G,τ),k,)of MULTISTAGE VERTEX COV ER. We claim that Iis a yes-instance
if and only if Iiis a yes-instance for all i∈{1,...,p}.
(⇒)If Iis a yes-instance, then for each i∈{1,...,p}, the subsequence of the
solution restricted to the layers (Gi
j)1≤j≤τiforms a solution to Ii.
(⇐)Let (Si
1,...,Si
τi)be a solution to Iifor each i∈{1,...,p}.
Clearly, (Si
1,...,Si
τi)forms a solution to the layers (Gi
j)1≤j≤τi.ForHi
1,letTi
1=
Si
τi\{v}for some vsuch that the unique edge of Hi
1is still covered. Next, set Ti
2=
Ti
1∪{w},wherew∈Si+1
1with wbeing incident with the unique edge of Hi
2.Now,
over the next 2k−2 layers, transform Ti
2into Si+1
1by first removing layer by layer
the vertices in Ti
2\Si+1
1(at most k−1 many vertices), and then layer by layer add the
vertices in Si+1
1\Ti
2(again, at most k−1 vertices). This forms a solution to I.
Turning a set of input instances of MULTISTAGE VERTEX COVE R with only one
layer (τ=1) additionally being a planar graph into a sequence gives an AND-
composition to be used in the proof of Theorem 6.1(ii).
Proposition 6.2 MULTISTAGE VERTEX COVE R where each layer is planar and ≥
2kadmits an AND-composition when parameterized by k.
Proof We AND-compose MULTISTAGE VERTEX COVER where the temporal graph
has only one layer which is planar (and ≥2k)intoM
ULTISTAGE VERTEX COVER
with ≥2k.Let(G1,k,
),...,(G
p,k,
)be pinstances of MULTISTAGE VER-
TEX COVER with one layer being a planar graph. Construct a temporal graph Gwith
layers (G1,...,G
p).Set:= 2k. This finishes the construction. It is not difficult to
see that (G,k,) is a yes-instance of MULTISTAGE VERTEX COVER if and only if
(Gi,k)is a yes-instance of VERTEX COVER for all i∈{1,...,p}.
Proposition 6.1 and 6.2 at hand, we are set to prove this section’s main result.
Proof of Theorem 6.1 Using Drucker’s result [32] for AND-compositions, Proposi-
tion 6.1 and 6.2 prove Theorem 6.1(i) and (ii), respectively. Recall that MULTISTAGE
VERTEX COVER where each layer consists of one edge (Theorem 4.1) and MUL-
TISTAGE VERTEX COVER on one layer being a planar graph (basically, VERTEX
COVER on planar graphs) [34] are NP-hard.
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6.2 A Problem Kernel of Size
O
(
k
2τ)
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 of size polynomial in k, even if each layer consists of only one edge
(Theorem 6.1). Yet, when combining both parameters, we obtain a problem kernel of
cubic size.
Theorem 6.2 There is an algorithm that maps any instance (G,k,) of MULTI-
STAGE VERTEX COVER in O(|V(G)|2τ) time to an instance (G,k,) of MULTI-
STAGE VERTEX COVER with at most 2k2τ(G)vertices and at most k2τ(G)temporal
edges.
To prove Theorem 6.2, we apply three polynomial-time data reduction rules. These
reduction rules can be understood as temporal variants of the folklore reduction rules
for VERTEX COVER. Our first reduction rule is immediate.
Reduction Rule 6.1 (Isolated vertices) If there is some vertex v∈Vsuch 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 vof
degree larger than q, then delete vand its incident edges and decrease qby one.
For MULTISTAGE VERTEX COVER a high-degree vertex can only appear in some
layers, and hence deleting this vertex is in general not correct. However, the following
is a temporal variant of the high-degree rule (see Fig. 4for an illustration).
Reduction Rule 6.2 (High degree) If there exists a vertex vsuch that there is an
inclusion-maximal subset ∅ = J⊆{1,...,τ}such that degGi(v) > k for all i∈J,
then add a vertex wvto Vand for each i∈J, remove all edges incident to vin Gi,
and add the edge {v, wv}.
Fig. 4 Illustration of Reduction Rule 6.2, exemplified for two vertices u, v and k=5. Each ellipse for a
graph Giand G
i, respectively, represents Gi−{u, v}and G
i−{u, v, wu,w
v}. The vertices wv,w
u(gray
squares) are introduced by the application of Reduction Rule 6.8. Note that u(v)hasahighdegreeinG1
(G2)andG4
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We now show how Reduction Rule 6.2 can be applied and that it does not turn a
yes-instance into a no-instance or vice versa.
Lemma 6.1 Reduction Rule 6.2 is correct and exhaustively applicable in O(|V|2
τ) time.
Proof (Correctness)LetI=(G,k,) be an instance with G=(G1,...,G
τ),and
let I=(G,k,) be the instance with G=(G
1,...,G

τ)obtained from Iapplying
Reduction Rule 6.2 with vertex vandindexsetJ. We prove that Iis a yes-instance
if and only if Iis a yes-instance.
(⇒)Let (S1,...,S
τ)be a solution to I. Observe that for all i∈J,deg
Gi(v) > k
and hence v∈Si. It follows that (S1,...,S
τ)is a solution to I.
(⇐)Let (S
1,...,S
τ)be a solution to I. Observe that for each i∈J,S
i∩
{v, wv} =∅.SetSi:= (S 
i\{wv})∪{v}for all i∈J. Note that Siis a vertex
cover for Gisince vcovers all its incident edges and Si\{v}is a vertex cover
for Gi−{v}=G
i−{v, wv}. For each i∈{1,...,τ}\J,setSi:= S
iif wv∈ S
i,
and Si:= (S
i\{wv})∪{v}otherwise. Note that Siis a vertex cover of Gi=G
i−{wv}.
Finally, observe that |Si|≤|S
i|for all i∈{1,...,τ},andthat|SiSi+1|≤for
all i∈{1,...,τ −1}. It follows that (S1,...,S
τ)is a solution to I.
(Running time) For each vertex, we count the number of edges in each layer. If
there are more than kedges 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 vvertex wvis introduced, we add a pointer from vto wv, and add the
edge {v, wv}in subsequent layers when needed. Hence, in each layer we touch each
edge at most twice, yielding O(|V(
G)|2)time per layer.
Similarly as in the reduction rules for VERTEX COVER, we now count the number
of edges in each layer: if more than k2edges are contained in one layer, then no set
of kvertices, each of degree at most k, can cover more than k2edges.
Reduction Rule 6.3 (no-instance) If neither Reduction Rule 6.1 nor Reduction Rule
6.2 is applicable and there is a layer with more than k2edges, then output a trivial
no
-instance.
We are ready to prove that when none of the Reduction Rules 6.1, 6.2 and 6.3 can
be applied, then the instance contains “few” vertices and temporal edges.
Lemma 6.2 Let (G,k,) be an instance of MULTISTAGE VERTEX COV ER such
that none of Reduction Rules 6.1, 6.2 and 6.3 is applicable. Then Gconsists of at
most 2k2τ(G)vertices and k2τ(G)temporal edges.
Proof Since none of Reduction Rules 6.1 and 6.2 is applicable, for each layer it
holds true that there is no isolated vertex and no vertex of degree larger than k.Since
Reduction Rule 6.3 is not applicable, each layer consists of at most k2edges. Hence,
there are at most k2τtemporal edges in G. Consequently, due to Reduction Rule 6.1,
there are at most 2k2τvertices in G.
480 Theory of Computing Systems (2022) 66:454 483
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We are ready to prove the main result of this section.
Proof of Theorem 6.2 Given an instance I=(G,k,) of MULTISTAGE VERTEX
COVER, apply Reduction Rules 6.1, 6.2 and 6.3 exhaustively in O(|V(G)|2τ(G))
time either to decide that Iis a trivial no-instance or to obtain an instance (G,k,)
equivalent to I. Due to Lemma 6.2, Gconsists of at most 2k2τ(G)vertices and at
most k2τ(G)temporal edges.
6.3 A Problem Kernel of Size 5τ
MULTISTAGE VERTEX COVER, even when each layer is a tree, does not admit a
problem kernel of any size in τunless P =NP. Yet, when each layer consists of only
one edge, then each instance of MULTISTAGE VERTEX COVER contains at most τ
edges and, hence, at most 2τnon-isolated vertices. Thus, MULTISTAGE VERTEX
COVER admits a straight-forward problem kernel of size linear in τ.
Observation 6.3 Let (G,k,) be an instance of MULTISTAGE VERTEX COVER
where each layer consists of one edge. Then we can compute in O(|V(G)|·τ) time
an instance (G,k,) of size at most 5τ(G).
Proof Let (G,k,) be an instance of MULTISTAGE VERTEX COVER where each
layer of G=(V , E,τ)consists of one edge. Observe that we can immediately output
atrivialyes-instance if k≥τ(Observation 3.1) or ≥2 (Observation 3.5). Hence,
assume that k≤τ−1and≤1. Apply Reduction Rule 6.1 exhaustively on (G,k,)
to obtain (G,k,). Since there are τedges in G, there are at most 2τvertices in G.
It follows that the size of (G,k,) is at most 5τ.
7 Conclusion
We introduced MULTISTAGE VERTEX COVE R, proved its NP-hardness even on very
restricted input instances, and studied its parameterized complexity regarding the
natural parameters k,,andτ(all given as part of the input). The technical high-
light is the W[1]-hardness described in Section 5.2 which, because it holds on very
restricted instances of MULTISTAGE VERTEX COVER, may turn out to be useful to
provide W[1]-hardness results for other problems in the multistage setting. We leave
open whether MULTISTAGE VERTEX COVER parameterized by the vertex cover size
bound kis fixed-parameter tractable when each layer consists of only one edge
(see Table 1). Moreover, it is open whether MULTISTAGE VERTEX COV ER remains
NP-hard on two layers each being a path (which would strengthen Theorem 4.1(i)).
Funding Open Access funding enabled and organized by Projekt DEAL.
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