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  <front>
    <journal-meta>
      <journal-title-group>
        <journal-title>Corresponding author.
" fabrizio.montecchiani@unipg.it (F. Montecchiani); giacomo.ortali@unipg.it (G. Ortali);
tommaso.piselli@studenti.unipg.it (T. Piselli); alessandra.tappini@unipg.it (A. Tappini)</journal-title>
      </journal-title-group>
    </journal-meta>
    <article-meta>
      <title-group>
        <article-title>On the Parameterized Complexity of the -Club Cluster Edge Deletion Problem (Short Paper)</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Fabrizio Montecchiani</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Giacomo Ortali</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Tommaso Piselli</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Alessandra Tappini</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Università degli Studi di Perugia</institution>
          ,
          <country country="IT">Italy</country>
        </aff>
      </contrib-group>
      <pub-date>
        <year>2022</year>
      </pub-date>
      <volume>000</volume>
      <fpage>0</fpage>
      <lpage>0002</lpage>
      <abstract>
        <p>We study the parameterized complexity of the -Club Cluster Edge Deletion problem: Given a graph  and two integers  ≥ 2 and  ≥ 1, is it possible to remove at most  edges from  such that each connected component of the resulting graph has diameter at most ? This problem is known to be NP-hard already when  = 2. We prove that it admits a fixed-parameter tractable algorithm when parameterized by  and the treewidth of the input graph.</p>
      </abstract>
      <kwd-group>
        <kwd>eol&gt;-Club Cluster Edge Deletion</kwd>
        <kwd>Parameterized Complexity</kwd>
        <kwd>-Club</kwd>
        <kwd>Treewidth</kwd>
      </kwd-group>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>1. Introduction</title>
      <p>
        which each pair of vertices is at distance at most  ≥ 2 in the cluster. (Note that a 1-club is
in fact a clique.) We remark that defining clusters as -clubs proved to be efective in several
application scenarios such as social network analysis and bioinformatics [
        <xref ref-type="bibr" rid="ref12 ref13 ref14 ref15">12, 13, 14, 15, 16</xref>
        ]. The
-Club Cluster Edge Deletion problem can be stated analogously as Cluster Edge Deletion
by replacing cliques with -clubs (formal definitions are given later). Unfortunately, -Club
Cluster Edge Deletion is NP-complete already for  = 2 [17]. Also, 2-Club Cluster Edge
Deletion belongs to FPT parameterized by  [18, 17], and it admits no subexponential-time
parameterized algorithm in  [19]. More in general, for any  ≥ 2, -Club Cluster Edge
Deletion cannot be solved in time 2()(1) unless ETH fails [19].
      </p>
      <p>Based on the above discussion, we know that it is unlikely that -Club Cluster Edge
Deletion lies is FPT when parameterized by , whereas the complexity of the problem parameterized
by  +  is open to the best of our knowledge. In this paper, we instead focus on those scenarios
in which the solution size (measured by ) is large, and we still aim for tractable problems
based on alternative parameterizations. In this respect, treewidth is a central parameter in
the parameterized complexity analysis (see [20, 21]). We prove that -Club Cluster Edge
Deletion lies in FPT when parameterized by  + tw, where tw is an upper bound for the
treewidth of the input graph.</p>
      <p>Theorem 1 Let  be an -vertex graph of treewidth at most tw. There is an algorithm that solves
the -Club Cluster Edge Deletion problem on  in (22(tw2 log ) · ) time.</p>
      <p>From the technical viewpoint, the main crux of our approach lies in the definition of
sufifciently small records that allow to keep track of the distances between pairs of vertices in
a (partial) -club. With such records at hand, we then apply a standard DP algorithm over a
tree decomposition of the input graph, which still requires a nontrivial amount of technicalities
in order to update the records. Our records have similarities, but also several key diferences,
with those used in a technique presented by Dondi and Lafond in [22, Thm. 14], which solves a
related problem for the restricted case  = 2.</p>
      <p>For space constraints many technicalities are omitted and we only sketch the proof of
Theorem 1. See [23] for the full version of the paper.</p>
      <p>Preliminaries and notation. For any  ∈ Z+, we use [] as shorthand for the set {1, 2, . . . , }.
Let  = (, ) be a graph. For any  ⊂  , we denote by [ ] the subgraph of  induced
by the vertices of  . The neighborhood of a vertex  of  is defined as () = { :  ∈ }.
Given two vertices ,  ∈  , the distance in  between  and , denoted by (, ), is
the number of edges in any shortest path between  and  in . The diameter of  is the
maximum distance in  between any two of its vertices. An -club of , with  ≥ 1, is a subset
 ⊆  such that the diameter of [ ] is at most . A partition of  is a collection of subsets
 = {}∈[] such that: (a) ⋃︀</p>
      <p>=1  =  , and (b)  ∩  = ∅ for each ,  ∈ [] with  ̸= .</p>
      <p>We denote by  the set of all edges  of  such that ,  ∈ , for some  ∈ []. We study
the following problem.
-Club Cluster Edge Deletion
Input:  = (, ),  ≥ 1,  ≥ 2.</p>
      <p>Output: A partition  = {}∈[] of  such that  is an -club for each  ∈ [], and
| ∖ | ≤ .</p>
      <p>In what follows, for a graph  = (, ), the pair ( ,  ) denotes a nice tree-decomposition,
such that  = {}∈[ℓ] is a collection of subsets of vertices of  , called bags, and  is a tree
whose nodes are in one-to-one correspondence with the elements of  . We point the reader
to [24, 25] for the required background.</p>
    </sec>
    <sec id="sec-2">
      <title>2. Sketch of the Proof of Theorem 1</title>
      <p>The proof is based on a DP algorithm over a nice tree-decomposition. We first describe the
records to be stored at each bag, and we then sketch the algorithm.</p>
      <p>Definition of the records. Let  = (, ) be an -vertex graph and let ( ,  ) be a nice
tree-decomposition of  of width tw. For each  ∈ [ℓ], let  be the subtree of  rooted at the
bag  ∈  and let  = (, ) be the subgraph of  induced by the vertices that belong to
at least one bag of . A subset of vertices  ⊆  is a potential -club, and we let  =  ∩ 
and int() =  ∖ .</p>
      <p>The first item of the record is a table that stores the pairwise distances of the vertices in .
Namely, let () be a table having one row and one column for each vertex in , and such
that:</p>
      <p>⎧⎪0,
()[, ] = ⎨</p>
      <p>if  = 
⎪
⎩∞,
[](, ), if 1 ≤ [](, ) ≤ 
otherwise.</p>
      <p>The second item is a table that stores the distance between pairs of vertices such that one is
in  and the other is in int(). Two vertices , ′ in int() are equivalent with respect to ,
if for each vertex  ∈ , then either 1 ≤ [](, ) = [](′, ) ≤ , or [](, ) &gt; 
and [](′, ) &gt; . Namely, let () be a table having one column for each vertex  ∈ ,
and one row for each equivalence class with respect to , denoted by [] . We have:
()[, ] =</p>
      <p>∞,
{︃[](, ), if 1 ≤ [](, ) ≤ 
otherwise.</p>
      <p>The third (and last) item of the record represents the key diference to extend the result in [ 22]
to  ≥ 2. Suppose that  is a subset of an -club ′ of  and that there exist two vertices
, ′ ∈ int() whose distance in [] is larger than . Then, any path between  and ′ not
containing two vertices in  has length larger than . Hence, since  is a separator of ,
we have to consider paths in  between  and ′ going through some pair of vertices in .
We formalize this observation. Let ,  ∈ int() be two vertices such that [](, ) &gt; . A
request for , denoted by , is a table having one row and one column for each vertex in
. Namely, for each ,  ∈ , if there exists 2 ≤  ≤  − 2 such that connecting  and  with
a path  of length  makes the distance between  and  to be at most , then [, ] =  ,
while [, ] = ⋆ otherwise. Observe that if there exist two requests  and ′′ such
that [, ] = ′′ [, ] for each pair ,  ∈ , then  and ′ are equivalent with respect
to  (i.e., , ′ ∈ [] ), and the same holds for  and ′. Thus, we avoid storing duplicated
requests and we denote by () the set containing all distinct requests for .</p>
      <p>If a potential -club  is such that  = ∅ (recall that  ⊆ ), then we call it complete.
Consider a partitioning  of  into potential -clubs and let  = {, |  ∈ []} be the
potential -clubs in  that are not complete. Let  = {, |  ∈ []},  = {(,) |
 ∈ []}, ℋ = {(,) |  ∈ []}, and  = {(,) |  ∈ []}. A solution of  is a
tdhuiespntlicenectif=≤ ⟨.̸=T, wo,sℋ,oolru,tio,n̸=s⟩.H,=eorre⟨ℋi,̸=saℋn,iℋn,toe,rger,,c̸=al⟩leadned.dOgbe-sce=oruvn⟨ettehr,aet,qiufa,lℋtoan|,d∖,ar(e⟩naro)e|t,
distinct but  &lt; , then it sufices to consider only .</p>
      <p>Lemma 1 For a bag , there exist (22(tw2 log ) ) distinct solutions.</p>
      <p>Sketch of the algorithm. Let  be the current bag visited by the algorithm. We compute the
set of solutions for  based on the solutions computed for its child or children. If the resulting
set of solutions is empty, the algorithm halts and returns a negative answer. The running time
of the algorithm follows from Lemma 1. We only describe the case in which  is an introduce
bag. The cases in which  is a leaf, a forget, or a join bag are omitted in this extended abstract.
 is an introduce bag. Let  =  ∖ {} be the child of . The algorithm exhaustively
extends each solution  of  as follows. It first generates at most  new partitions by placing
 in each ′ ∈  . Also, it generates a partition in which  forms a new potential -club
 =  = {}. Consider one of the new partitions generated by the algorithm. In order to
build the corresponding new solution for , we distinguish the following two cases.
Case A ( = {}). () is trivially defined, () and () are empty.
Case B ( = ′ ∪ {}). The next observation immediately follows from the fact that
 = ′ ∪ {} and int() = int(′).</p>
      <p>Observation 1 Suppose that there exist ,  ∈ ′ such that [′](, ) &gt; [](, ), then
any shortest path between  and  in [] contains vertex .
– Computing () from (′).</p>
      <p>1. We add a new row and a new column for vertex .
2. For each vertex  ∈ ′, let   = min∈[]() (′)[, ], and note that   = 0 if
edge  belongs to []. Clearly, it holds that
()[, ] =
{︃∞, if   ∈ {, ∞}</p>
      <p>1 +  , otherwise.
3. By Observation 1, for each pair ,  ∈ ′, the corresponding value of () can be
updated as follows:</p>
      <p>()[, ] = min{(′)[, ], ()[, ] + ()[, ]}.
– Computing () from (′).</p>
      <p>1. We add a new column for vertex .
2. For each equivalence class []′ , let   = min∈[]() (′)[, ]. Since there is
no edge  such that  ∈ int(), it follows that
()[, ] =
{︃∞, if   ∈ {, ∞}</p>
      <p>1 +  , otherwise.
3. By Observation 1, for each pair of vertices  ∈ int(′) and  ∈ ′, the corresponding
value of () can be updated as follows:</p>
      <p>()[, ] = min{(′)[, ], ()[, ] + ()[, ]}.
– Computing () from (′). Note that the addition of  cannot lead to new requests but
it may actually yield the update of some request in (′).</p>
      <p>1. For each request  in (′), we verify whether, as a consequence of the introduction
of , there exists a cell [, ] such that ()[, ] ≤ [, ]. If such a cell exists,
we say that  is fulfilled . We add  to () if and only if  is not fulfilled.
2. If  is not fulfilled, before adding it to (), we update it as follows:
a) We add a row and a column for .
b) For each pair ,  ∈ ′, we compute</p>
      <p>= min{(()[, ] + ()[, ], ()[, ] + ()[, ]}.</p>
      <p>Observe that   + ()[, ] &gt; , otherwise the request would have been fulfilled
before.
c) By definition of request, we have [, ] =  −  , if   &lt;  − 1, and [, ] =
⋆, otherwise.</p>
      <p>Finally, in both Case A and Case B, we observe that, in order to obtain the edge-counter of
the new solution,  needs to be increased by the number of edges incident to  whose other
end-vertex is in  but not in . If the resulting edge-counter is greater than , the solution is
discarded.</p>
    </sec>
    <sec id="sec-3">
      <title>3. Discussion and Open Problems</title>
      <p>We have shown that the -Club Cluster Edge Deletion problem parameterized by  + tw
(where tw bounds the treewidth of the input graph) belongs to FPT. On the other hand, we
know that the problem parameterized by  alone is paraNP-hard. It remains open the complexity
of -Club Cluster Edge Deletion parameterized by tw alone. We conclude by remarking
that our approach can be slightly modified to solve a related problem, namely -Club Cluster
Vertex Deletion, in which we seek for  vertices whose removal yields a set of disjoint -clubs.
[16] R. Mokken, S. Laan, Close communication and 2-clubs in corporate networks: Europe 2010,</p>
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[20] R. G. Downey, M. R. Fellows, Parameterized Complexity, Monographs in Computer Science,</p>
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    </sec>
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