Abstract
The sClub problem asks whether a given undirected graph G contains a vertex set S of size at least k such that G[S], the subgraph of G induced by S, has diameter at most s. We consider variants of sClub where one additionally demands that each vertex of G[S] is contained in at least \(\ell \) triangles in G[S], that each edge of G[S] is contained in at least \(\ell \) triangles in G[S], or that S contains a given set W of seed vertices. We show that in general these variants are W[1]hard when parameterized by the solution size k, making them significantly harder than the unconstrained sClub problem. On the positive side, we obtain some FPT algorithms for the case when \(\ell =1\) and for the case when G[W], the graph induced by the set of seed vertices, is a clique.
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1 Introduction
Finding cohesive subgroups in social or biological networks is a fundamental task in network analysis. A classic formulation of cohesiveness is based on the observation that cohesive groups have small diameter. This observation led to the sclub model originally proposed by Mokken [17]. An sclub in a graph \(G=(V,E)\) is a set of vertices S such that the subgraph of G induced by S has diameter at most s. The 1clubs are thus precisely the cliques and the larger the value of s, the more the cliquedefining constraint of having diameter one is relaxed. In the sClub problem we aim to decide whether G contains an sclub of size at least k.
A big drawback of sclubs is that the largest sclubs are often not very cohesive with respect to other cohesiveness measures such as density or minimum degree. This behavior is particularly pronounced for \(s=2\): the largest 2club in a graph is often the vertex v of maximum degree together with its neighbors [12]. To avoid these socalled hubandspoke structures, it has been proposed to augment the sclub definition with additional constraints [5, 16, 18, 20].
One of these augmented models, proposed by Carvalho and Almeida [5], asks that every vertex is part of a triangle. This property was later generalized to the vertex\(\ell \)triangle property, which asks that every vertex of S is in at least \(\ell \) triangles in G[S] [1]. For an example we refer to Fig. 1. This model is motivated by clique relaxations based on clustering coefficients, socalled local and global\(\alpha \) clusters, proposed by Ertem et al. [9]. The idea is that the vertex\(\ell \)triangle property leads to higher clustering coefficients.
Vertex Triangle sClub
Input: An undirected graph G = (V, E), and two integers \(k,\ell \ge 1\).
Question: Does G contain an sclub S of size at least k that fulfills the vertex\(\ell \)triangle property?
The vertex\(\ell \)triangle constraint entails some desirable properties for cohesive subgraphs. Let \(D_v\) be the set of vertices which are in a triangle with \(v\in V(G)\). Since \(D_v> \sqrt{2\ell }\) we obtain that the minimum degree of a vertex\(\ell \)triangle sclub is larger than \(\sqrt{2\ell }\). Moreover, vertex\(\ell \)triangle sclubs S are robust with respect to some specific edge deletions: Since \(D_v> \sqrt{2\ell }\) we may delete up to \(\sqrt{2\ell }\) edges with one endpoint being v and the other in \(D_v\) without disconnecting S. This property is not fulfilled, if we instead of the vertex\(\ell \)triangle constraint put the additional property of having minimum degree at least \(\sqrt{2\ell }+1\) on sclubs: In such a model there may exist one vertex v such that each deletion incident with v disconnects the graph, for example, a star with \(\sqrt{2\ell }+1\) leaves, where a clique of size \(\sqrt{2\ell }\) is attached to each leaf. However, some undesirable behavior of hubandspoke structures remains: vertex\(\ell \)triangle sclubs are not robust with respect to each possible edge deletion. For an example, we refer to Fig. 2.
To overcome this problem, we introduce a new model where we put triangle constraints on the edges of the sclub instead of the vertices. More precisely, we say that a vertex set S of a graph G fulfills the edge\(\ell \)triangle property if G[S] contains a spanning subgraph \(G':= (S,E')\) such that every edge in \(E(G')\) is in at least \(\ell \) triangles in \(G'\) and the diameter of \(G'\) is at most s. For an example we refer to Fig. 1.
Edge Triangle sClub
Input: An undirected graph \(G=(V,E)\), and two integers \(k,\ell \ge 1\).
Question: Does G contain a vertex set S of size at least k that fulfills the edge\(\ell \)triangle property?
Note that in this definition, the triangle and diameter constraints are imposed on a spanning subgraph of G[S]. In contrast, for Vertex Triangle sClub they are imposed directly on G[S]. The reason for this distinction is that we would like to have properties that are closed under edge insertions. Properties which are closed under edge insertions are also wellmotivated from an application point of view since adding a new connection within a group should not destroy this group. If we would impose the triangle constraint on the induced subgraph G[S] instead, then an edge\(\ell \)triangle sclub S would not be robust to edge insertions. For an example, we refer to Fig. 3. By searching for a subgraph we can overcome this issue, since this allows us to ignore such edges.
Observe that every set that fulfills the edge\(\ell \)triangle property also fulfills the vertex\(\ell \)triangle property. Also note that the converse is not true: A vertex\(\ell \)triangle sclub is not necessarily also an edge\(\ell \)triangle sclub. For example, consider a graph consisting of two cliques of size d in which one edge e with exactly one endpoint in each clique is added (see Fig. 2). This graph is a vertex\(\left( {\begin{array}{c}d1\\ 2\end{array}}\right) \)triangle3club. But after deleting e the graph is disconnected. Also note that in an edge\(\ell \)triangle sclub each vertex has degree at least \(\ell +1\): Consider an arbitrary edge uv. Since uv is in at least \(\ell \) triangles \(\{u,v,w_1\},\ldots ,\{u,v,w_\ell \}\), we thus conclude that u and v have degree at least \(\ell \). We can show an even stronger statement: an edge\(\ell \)triangle sclub S is robust against up to \( \ell \) edge deletions, as desired.
Proposition 1
Let \(G=(V,E)\) be a graph and let S be an edge\(\ell \)triangle sclub in G. Let \(G'\) be a spanning subgraph of G[S] such that every edge in \(G'\) is in at least \(\ell \) triangles and the diameter of \(G'\) is at most s. If \( \ell \) edges of \(G'\) are removed from G, then S is still an \( (s+\ell )\)club and a (2s)club in G.
Proof
We show that if \( \ell \) edges are removed from \( G' \), the diameter of the resulting graph \( \widetilde{G} \) increases by at most \( \ell \). Let \( P=(v_1,\ldots ,v_{s+1}) \) be a path of length s in \( G' \). Since S is an edge\(\ell \)triangle sclub, every edge \( v_i v_{i+1} \) of P is part of at least \( \ell \) triangles in \(G'\). Thus, for two vertices \( v_i\) and \(v_{i+1}\) in P there is a path of length at most two from \( v_i \) to \( v_{i+1} \) in \( G' \), either directly through the edge \( v_i v_{i+1}\) or via a vertex u that forms one of the \( \ell \) triangles with \( v_i \) and \( v_{i+1} \) in \( G' \). Thus, \( \text {dist}(v_{i},v_{i+1}) \) increases by at most 1 after one edge deletion and only if \(v_i v_{i+1}\) is removed. Since at most \( \ell \) of the edges in P are removed, we have \( \text {dist}(v_1,v_{s+1}) \le \text {dist}(v_1,v_2) + \ldots + \text {dist}(v_s,v_{s+1}) \le s + \ell \) in \( \widetilde{G} \) and thus also in G. By the same arguments, we also have \(\text {dist}(v_1,v_{s+1}) \le 2s\). Thus, after removing \( \ell \) edges of \(G'\) from G, S is an \( (s+\ell ) \)club and a (2s)club in G. \(\square \)
The following further variant of sClub is also practically motivated but not necessarily by concerns about the robustness of the sclub. Here, the difference to the standard problem is simply that we are given a set of seed vertices W and aim to find a large sclub that contains all seed vertices.
Seeded sClub
Input: An undirected graph G = (V, E), a subset \(W\subseteq V\), and an integer \(k\ge 1\).
Question: Does G contain an sclub S of size at least k such that \({W\subseteq S}\)?
This variant has applications in community detection, where we are often interested in finding communities containing some set of fixed vertices [14, 21].
In this work, we study the parameterized complexity of the three abovementioned problems with respect to the standard parameter solution size k. Our goal is to determine whether FPT results for sClub [6, 19] transfer to these practically motivated problem variants.
Known Results. The sClub problem is NPhard for all \(s\ge 1\) [4], even when the input graph has diameter \(s+1\) [2]. For \(s=1\), sClub is equivalent to Clique and thus W[1]hard with respect to k. In contrast, for \(s>1\), sClub is fixedparameter tractable (FPT) with respect to the solution size k [6, 19]. This fixedparameter tractability can be shown via a Turing kernel with \(\mathcal {O}(k^{2})\) vertices for even s and \(\mathcal {O}(k^3)\) vertices for odd s [6, 19]. In these Turing kernels n instances of sClub of size \(\mathcal {O}(k^2)\), or \(\mathcal {O}(k^3)\), respectively, are created. These n instances can then be solved independently.
The complexity of sClub has been also studied with respect to different classes of input graphs [10] and with respect to structural parameters such as the degeneracy of the input graph [13]. The sClub problem can be solved efficiently in practice, in particular for \(s=2\) [4, 6, 12].
Vertex Triangle \(s\)Club is NPhard for all \(s\ge 1\) and for all \(\ell \ge 1\) [1, 5]. We are not aware of any algorithmic studies of Edge Triangle \(s\)Club or Seeded \(s\)Club. NPhardness of Edge Triangle \(s\)Club for \(\ell =1\) can be shown via the reduction for Vertex Triangle \(s\)Club for \(\ell =1\) [1]. Also, the NPhardness of Seeded \(s\)Club follows directly from the fact that an algorithm for the case where \(W=1\) can be used as a black box to solve sClub.
Further robust models of sclubs, which are not considered in this work, include thereditary sclubs [18], trobust sclubs [20], and tconnected sclubs [16, 22]. For an overview on clique relaxation models and complexity issues for the corresponding subgraph problems we refer to the relevant surveys [15, 18].
Our Results. For Vertex Triangle \(s\)Club and Edge Triangle \(s\)Club, we provide a complexity dichotomy into cases that are FPT or W[1]hard with respect to k, respectively. For Seeded \(s\)Club, we provide a kernel with respect to k for clique seeds W and W[1]hardness with respect to k for some other cases. For \(s=2\), our results provide a dichotomy into FPT and W[1]hardness with respect to k. An overview of our results is given in Table 1. Our W[1]hardness reduction for Edge Triangle \(s\)Club for \(\ell \ge 2\) also shows the NPhardness of this case.
The FPT algorithms are obtained via adaptions of the Turing kernelization for sClub. Interestingly, Vertex Triangle \(s\)Club with \(\ell =1\) is FPT only for larger s, whereas Edge Triangle \(s\)Club is FPT for all s. Hence, the edge \(\ell \)triangle sclub model provides better robustness guarantees than the vertex \(\ell \)triangle sclub model. Moreover, the edge \(\ell \)triangle sclub model is also preferable from an algorithmic standpoint as it allows to employ Turing kernelization as a part of the solving procedure, at least for \(\ell =1\). For the variants with triangle constraints, we only obtain Turing kernels and it is easy to see that standard problem kernels are unlikely to exist: sclubs are necessarily connected and thus taking the disjoint union of graphs gives a trivial orcomposition; consequently, a polynomial problem kernel implies coNP \(\subseteq \) NP/poly [3].
All W[1]hardness results are shown via reduction from Clique. The idea is to use the triangle constraints to ensure that selecting some vertex \(v'\) in the (Vertex/Edge) Triangle sClub instance which corresponds to a vertex v in the Clique instance triggers a cascade which will imply that we need to select a full vertex gadget corresponding to v. We can then use the distance constraint to make sure that full vertex gadgets are chosen only if the corresponding vertices are adjacent. While this idea is very natural, using the triangle constraint without creating many vertices that are too close to each other turned out to be technically challenging.
For Seeded \(s\)Club we provide two reductions from Clique. One for the case \(s=2\) if the seed contains at least two nonadjacent vertices u and z, and one for the case \(s\ge 3\) if the seed contains at least two connected components U and Z. In both cases we add two copies \(X_1\) and \(X_2\) of the graph of the Clique instance to the new instance of Seeded \(s\)Club such that each vertex of \(X_1\) has distance at most s to u or U if its copy in \(X_2\) is also part of the solution. We show a similar property for \(X_2\) and z or Z. This feature will then ensure that the same clique has to be chosen from both copies.
Our W[1]hardness results, in particular those for Seeded \(s\)Club, show that the FPT results for sClub are quite brittle and that adding even simple further constraints makes finding smalldiameter subgraphs much harder.
Preliminaries. For integers p, q, we denote \([p,q]:= \{p,p+1, \ldots ,q\}\) and \([q]:= [1,q]\). For a graph G, we let V(G) denote its vertex set and E(G) its edge set. We let n and m denote the order of G and the number of edges in G, respectively. A path of length p is a sequence of pairwise distinct vertices \(v_1,\ldots , v_{p+1}\) such that \(v_iv_{i+1}\in E(G)\) for each \(i\in [p]\). We let \(C_p\) denote the cycle graph on p vertices. The distance dist\(_G(u,v)\) is the length of a shortest path between vertices u and v. Furthermore, we define dist\(_G(u,W):= \min _{w\in W}\text {dist}(u,w)\). We denote by \(\text {diam}_G(G):= \max _{u,v\in V(G)}\text {dist}_G(u,v)\) the diameter of G. Let \(S\subseteq V(G)\) be a vertex set. We denote by \(N_i(S):= \{u \in V \mid \text {dist}(u,S) = i \}\) the open ineighborhood of S and by \(N_i[S]:= \bigcup _{j\le i}N_i(S)\cup S\) the closed ineighborhood of S. For a vertex \(v\in V(G)\), we write \(N_i(v):=N_i(\{v\})\) and \(N_i[v]:=N_i[\{v\}]\). By \(G[S]:= (S,\{uv\in E(G)\mid u,v\in S\})\) we denote the subgraph induced by S. Furthermore, by \(GS:= G[V\setminus S]\) we denote the subgraph obtained after the deletion of the vertices in S. A complete subgraph is called a clique and a clique consisting of three vertices is referred to as a triangle.
Parameterized complexity is a twodimensional framework for describing the computational complexity of decision problems [7, 8]. An instance (x, k) of a parameterized problem \(\textrm{L} \subseteq \Sigma ^{*} \times \mathbb{I}\mathbb{N} \) consists of the input x of a decision problem and a parameter k. A parameterized problem L is called fixedparameter tractable, if there is a computable function f such that for every instance \( (x,k) \subseteq \Sigma ^{*} \times \mathbb{I}\mathbb{N}\) it can be decided in \( f(k) \cdot x^{\mathcal {O}(1)}\) time whether (x, k) is a yesinstance of L. The class W[1] is a basic class of presumed parameterized intractability.
A parameterized reduction is an algorithm that takes an instance \( I_1 = (x_1,k_1) \) of a problem \( L_1 \) and transforms it into an instance \( I_2 = (x_2,k_2) \) of a problem \( L_2 \) such that \( I_1 \in L_1 \) if and only if \( I_2 \in L_2 \) and \( k_2 \le g(k_1) \) with running time \( f(k_1) \cdot x_1^{\mathcal {O}(1)} \) for some computable functions f and g. Parameterized reductions can be used to show that a parameterized problem L is not likely to be fixedparameter tractable by reducing some W[1]hard parameterized problem to L.
Let L be a parameterized problem and let \(f:\mathbb{I}\mathbb{N} \rightarrow \mathbb{I}\mathbb{N}\) be a computable function. A Turing kernelization of size f for L is an algorithm that decides whether a given instance \((x, k) \in \Sigma ^* \times \mathbb{I}\mathbb{N}\) is a yesinstance of L in time \({(x+k)^{\mathcal {O}(1)}}\), when given access to an oracle that decides membership in L for any instance \((x', k')\) with \(x'+ k' \le f(k)\) in a single step. For more details on parameterized complexity theory, we refer to the standard monographs [7, 8].
All of our hardness results are shown by a parameterized reduction from Clique.
Clique
Input: An undirected graph \(G=(V,E)\) and an integer k.
Question: Does G contain a clique of size at least k?
2 Vertex Triangle sClub
In this section, we settle the parameterized complexity of Vertex Triangle sClub with respect to the solution size k. First, we show that this problem is fixedparameter tractable when \(\ell =1\) and \(s\ge 4\). Afterwards, we show W[1]hardness for all remaining cases, that is, for \(\ell \ge 2\) and \(s\ge 2\), and also for \(\ell =1\) and \(s\in \{2,3\}\).
2.1 FPTAlgorithms
The overall idea is based on the idea of the Turing kernel for 2Club, that is, bounding the size of \(N_s[v]\) for each vertex \(v\in V(G)\). The first step is to remove all vertices which are not in a triangle.
Reduction Rule 1
Let (G, k) be an instance of Vertex Triangle sClub. Delete all vertices from G which are not part of any triangle.
Clearly, Reduction Rule 1 is correct and can be exhaustively applied in polynomial time. The application of Reduction Rule 1 has the following effect: if some vertex v is close to many vertices, then (G, k) is a trivial yesinstance.
Lemma 1
Let (G, k) be an instance of Vertex Triangle sClub with \(\ell =1\) and \(s\ge 4\) to which Reduction Rule 1 is applied. Then, (G, k) is a yesinstance if \(N_{\lfloor s/2\rfloor 1}[v]\ge k\) for some vertex \(v\in V(G)\).
Proof
Let \(v\in V(G)\) be a vertex such that \(N_{\lfloor s/2\rfloor 1}[v]\ge k\). We construct a vertex1triangle sclub T of size at least \(N_{\lfloor s/2\rfloor 1}[v]\ge k\). Initially, we set \(T:= N_{\lfloor s/2\rfloor 1}[v]\). Now, for each vertex \(w\in N_{\lfloor s/2\rfloor 1}(v)\) we do the following: Since Reduction Rule 1 is applied, we conclude that there exist two vertices x and y such that \(G[\{ w,x,y\}]\) is a triangle. We add x and y to the set T. We call the set of vertices added in this step the Texpansion.
Next, we show that T is indeed a vertex1triangle sclub for \(s\ge 4\). Observe that each vertex in T is either in \(N_{\lfloor s/2\rfloor 1}[v]\) or a neighbor of a vertex in \(N_{\lfloor s/2\rfloor 1}(v)\). Hence, each vertex in T has distance at most \(\lfloor s/2\rfloor \) to vertex v. Thus, T is an sclub. It remains to show that each vertex of T is in a triangle. Observe that for each vertex \(w\in N_{\lfloor s/2\rfloor 2}[v]\) we have \(N(w)\subseteq N_{\lfloor s/2\rfloor 1}[v]\). Recall that since Reduction Rule 1 is applied, each vertex in G is contained in a triangle. Thus, each vertex of \(N_{\lfloor s/2\rfloor 2}[v]\) is contained in a triangle in T. Furthermore, all vertices in \(N_{\lfloor s/2\rfloor 1}(v)\cup (T\setminus N_{\lfloor s/2\rfloor 1}[v])\) are in a triangle because of the Texpansion. Since \(T\ge N_{\lfloor s/2\rfloor 1}[v]\ge k\), the statement follows. \(\square \)
Next, we show that Lemma 1 implies the existence of a Turing kernel for \(s\ge 4\). We do this by showing that \(N_s[v]\) is bounded for every vertex in the graph. This in turn implies that the problem is fixedparameter tractable. It is sufficient to bound the size of \(N_s[v]\) for each \(v\in V(G)\) since we then can query the oracle for an sclub of size k.
Theorem 2
Vertex Triangle sClub for \(\ell =1\) admits a \(k^4\)vertex Turing kernel if \(s=4\) or \(s=7\), a \(k^5\)vertex Turing kernel if \(s=5\), and a \(k^3\)vertex Turing kernel if \(s=6\) or \(s\ge 8\).
Proof
First, we apply Reduction Rule 1. Because of Lemma 1 we conclude that (G, k) is a trivial yesinstance if \(N_{\lfloor s/2\rfloor 1}[v]\ge k\) for any vertex \(v\in V(G)\). Thus, in the following we can assume that \(N_{\lfloor s/2\rfloor 1}[v]<k\) for each vertex \(v\in V(G)\). We use this fact to bound the size of \(N_s[v]\) in nontrivial instances.
Note that if \(s=4\) or \(s=5\) we have \(\lfloor s/2\rfloor 1=1\). In these cases from \(N_{\lfloor s/2\rfloor 1}[v]<k\) we obtain that the size of the neighborhood of each vertex is bounded. Thus, we obtain a \(k^4\)vertex Turing kernel for \(s=4\) and a \(k^5\)vertex Turing kernel for \(s=5\). Furthermore, if \(s=7\) we have \(\lfloor s/2\rfloor 1=2\). Thus, we obtain a \(k^4\)vertex Turing kernel for \(s=7\) since \(N_7[v]\subseteq N_8[v]=N_2[N_2[N_2[N_2[v]]]]\).
If \(s=6\) or \(s\ge 8\), then \(\lfloor s/2\rfloor 1\ge \lceil s/3\rceil \). Observe that \(N_s[v]\) is contained in \(N_{\lceil s/3\rceil }[N_{\lceil s/3\rceil }[N_{\lceil s/3\rceil }[v]]]\). Thus, we obtain a \(k^3\)vertex Turing kernel for \(s=6\) or \(s\ge 8\). \(\square \)
Note that \(s\ge 4\) is necessary to ensure \(\lfloor s/2\rfloor 1\ge 1\). In our arguments to obtain a Turing kernel \(\ell =1\) is necessary for the following reason: if \(\ell \ge 2\), then the remaining vertices of the other triangles of a vertex in the Texpansion may be contained in \(N_{\lfloor s/2\rfloor +1}[v]\) and, thus, adding them will not necessarily give an sclub. Also note that using \(N_t[v]\) for some \(t<\lfloor s/2\rfloor 1\) does not help: The remaining vertices of the other triangles of a vertex in the Texpansion may be contained in \(N_{t+1}[v]\). But now, another Texpansion for the vertices in \(N_{t+1}[v]\) is necessary. This may lead to a cascade of Texpansions where eventually we add vertices with distance at least \(s+1\) to v. Thus, the constructed set is no sclub anymore.
2.2 Parameterized Hardness
In the following, we prove W[1]hardness for Vertex Triangle sClub parameterized by the solution size k for all cases not covered by Theorem 2, that is, \(\ell \ge 2\) and \(s\ge 2\), and also for \(\ell =1\) and \(s\in \{2,3\}\).
Theorem 3
Vertex Triangle sClub is W[1]hard for parameter k if \(\ell \ge 2\), and if \(\ell =1\) and \(s\in \{2,3\}\).
For some combinations of s and \(\ell \) we provide hardness for restricted input graphs. More precisely, we prove that Vertex Triangle sClub is W[1]hard even if each vertex \(v\in V(G)\) is contained in exactly \(\ell \) triangles in the input graph. In other words, the hardness does not depend on the fact that we could choose different triangles. We provide this hardness for the case \(s\ge 3\) and arbitrary \(\ell \) and also for \(s=2\) when \(\ell =\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \) for some integer c.
We prove the theorem by considering four subcases. The proofs for the four cases all use a reduction from the W[1]hard Clique problem. In these constructions, each vertex v of the Clique instance is replaced by a vertex gadget \(T^v\) such that every vertex\(\ell \)triangle sclub S either contains \(T^v\) completely or contains no vertex of \(T^v\). This property is obtained since each vertex in \(T^v\) will be in exactly \(\ell \) triangles and each of these triangles is within \(T^v\). The idea is that if \(uv\notin E(G)\) then there exists a vertex \(x\in T^u\) and a vertex \(y\in T^v\) such that \(\text {dist}(x,y)\ge s+1\).
Vertex Triangle 2Club. First, we handle the case \(s=2\).
Construction 1
Let (G, k) be an instance of Clique and let c be the smallest number such that \(\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \ge \ell \). We construct an instance \((G',c(k+1),\ell )\) of Vertex Triangle 2Club as follows.

For each vertex \(v\in V(G)\), we add a clique \(T^v:= \{x_1^v, \ldots , x_c^v\}\) of size c to \(G'\).

For each edge \(vw\in E(G)\), we connect the cliques \(T^v\) and \(T^w\) by adding the edge \(x^v_{2i1}x^w_{2i}\) and \(x^w_{2i1}x^v_{2i}\) to \(G'\) for each \(i\in [\lfloor c/2\rfloor ]\).

Furthermore, we add a clique \(Y:= \{y_1, \ldots , y_c\}\) of size c to \(G'\).

We also add, for each \(i\in [c]\) and each \(v\in V(G)\), the edge \(x^v_iy_i\) to \(G'\).
Note that the clique size c ensures that each vertex \(x\in V(G')\) is contained in at least \(\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \ge \ell \) triangles in \(G'\). Furthermore, note that the clique Y is only necessary when c is odd to ensure that the vertices \(x^v_c\) and \(x^w_c\) have a common neighbor. We add the clique Y in both cases to unify the construction and the correctness proof. Next, we show that for each vertex gadget \(T^v\) the intersection with each vertex\(\ell \)triangle 2club is either empty or \(T^v\).
Lemma 2
Let S be a vertex\(\ell \)triangle 2club in \(G'\). Then,

a)
\(S\cap T^v\ne \emptyset \Leftrightarrow T^v\subseteq S\), and

b)
\(S':= S\cup Y\) is also a vertex\(\ell \)triangle 2club in \(G'\).
Proof
First, we show statement a). Assume that for a vertex \(z\in T^v\) for some \(v\in V(G)\) we have \(z\in S\) for some vertex\(\ell \)triangle 2club S. Note that \(T^v\) contains all vertices which form a triangle with vertex z. Since c is minimal such that \(\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \ge \ell \) and since \(T^v\) is a clique, we conclude that \(T^v\subseteq S\) to fulfill the property that vertex z is contained in at least \(\ell \) triangles in G[S]. Thus, \(T^v\subseteq S\).
Second, we show statement b). Since each vertex \(y\in Y\) forms only triangles with vertices in Y and Y has size \(\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \), we conclude that \(Y\subseteq S^*\Leftrightarrow S^*\cap Y\ne \emptyset \) for each vertex\(\ell \)triangle 2club \(S^*\). In the following, let S be a vertex\(\ell \)triangle 2club such that \(Y\cap S=\emptyset \). From statement a) we conclude that \(S:= \bigcup _{v\in P}T^v\) for some set \(P\subseteq V(G)\). Next, we show that \(S':= S\cup Y\) is also a vertex\(\ell \)triangle 2club. Since each vertex is contained in a clique of size c in S, it is in sufficiently many triangles. Thus, it remains to prove that \(S'\) is a 2club. Consider some vertex \(x^v_i\) and some vertex \(y_j\) for some \(i,j\in [c]\) and \(v\in P\). Then, \(y_i\) is a common neighbor of \(x^v_i\) and \(y_j\). Hence, \(S'\) is a vertex\(\ell \)triangle 2club and thus b) holds. \(\square \)
Now, we prove the correctness of Construction 1.
Lemma 3
For each \(\ell \in \mathbb{I}\mathbb{N}\), the Vertex Triangle 2Club problem parameterized by k is W[1]hard.
Proof
We prove that G contains a clique of size k if and only if \(G'\) contains a vertex\(\ell \)triangle 2club of size at least \(c(k+1)\).
Let C be a clique of size at least k in G. We argue that \(S:= Y\cup \bigcup _{v\in C}T^v\) is a vertex\(\ell \)triangle 2club of size \(c(k+1)\) in \(G'\). Note that for each vertex \(v\in T^v\) we have \(T^v=c\). Since \(T^v\) is a clique, we conclude that each vertex in \(T^v\) is contained in exactly \(\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \ge \ell \) triangles. The same is true for each vertex in Y. Hence, each vertex in S is contained in at least \(\ell \) triangles. Thus, it remains to show that S is a 2club. Consider the vertices \(x^v_i\) and \(x^w_j\) for \(v,w\in C\), \(i\in [c1]\), and \(j\in [c]\). If i is odd, then \(x^w_{i+1}\in N(x^v_i)\cap N(x^w_j)\). Otherwise, if i is even, \(x^w_{i1}\in N(x^v_i)\cap N(x^w_j)\). In both cases, we obtain \(\text {dist}(x^v_i,x^w_j)\le 2\). Next, consider two vertices \(x^v_c\) and \(x^w_c\) in S. Observe that \(y_c\in N(x^v_c)\cap N(x^w_c)\). Since Y is a clique, it remains to consider vertices \(x^v_i\) and \(y_j\) in S for \(i\in [c]\) and \(j\in [c]\). Observe that \(x^v_j\in N(y_j)\cap N[x^v_i]\). Thus, \(G'\) contains a vertex\(\ell \)triangle 2club of size at least \(c(k+1)\).
Conversely, suppose that \(G'\) contains a vertex\(\ell \)triangle 2club S of size at least \(c(k+1)\). By Lemma 2, we can assume that \(Y\subseteq S\) and for each vertex gadget \(T^v\in G'\) we either have \(T^v\subseteq S\) or \(T^v\cap S=\emptyset \). Hence, S contains at least k cliques of the form \(T^v\). Assume towards a contradiction that S contains two cliques \(T^v\) and \(T^w\) such that \(vw\notin E(G)\) and consider the two vertices \(x^v_1\in T^v\) and \(x^w_2\in T^w\). Note that these vertices always exist since \(c\ge 3\). Observe that \(N[x^v_1]=T^v\cup \{x^u_2\mid uv\in E(G)\}\cup \{y_1\}\) and \(N[x^w_2]=T^w\cup \{x^u_1\mid uw\in E(G)\}\cup \{y_2\}\). Thus, \(\text {dist}(x^v_1,x^w_2)\ge 3\), a contradiction. Hence, for each two distinct vertex gadgets \(T^v\) and \(T^w\) that are contained in S, we observe that \(vw\in E(G)\). Consequently, the set \(\{v \mid T^v\subseteq S\}\) is a clique of size at least k in G. \(\square \)
If \(\ell =\left( {\begin{array}{c}c1\\ 2\end{array}}\right) \) for some integer c, then Lemma 3 also holds for the restriction that each vertex is contained in exactly \(\ell \) triangles in the input graph \(G'\).
Vertex Triangle sClub for \(s=3\) and for \(s\ge 4\) and \(\ell \ge 2\). Now, we provide hardness for the remaining cases. We consider three subcases. Case 1 deals with odd s. Case 2 covers the case that s is even and \(\ell \ge 3\). Case 3 deals with the case that s is even and \(\ell =2\). All three cases use the same vertex gadget. Only the edges between these gadgets, called connector edges, differ. The idea is to construct the vertex gadgets \(T^v\) in such a way that there are pairs of vertices in \(T^v\) of distance \(2s^*\) which is almost s. Thus, in a vertex\(\ell \)triangle sclub, the distance between two different vertex gadgets must be small. The first part of the following construction describes the vertex gadget which is used in all three cases. For an illustration of this construction see Fig. 4.
Construction 2
We set \(s^*:= \lfloor (s1)/2\rfloor \). Let (G, k) be an instance of Clique. We construct an equivalent instance \((G',3\ell ks^*,\ell )\) of Vertex Triangle sClub. Recall that \(\ell \ge 1\) and \(s=3\), or \(\ell \ge 2\) and \(s\ge 4\). For each vertex \(v\in V(G)\) we construct a vertex gadget \(T^v\). This vertex gadget is used for each reduction of the three cases.

We add the vertices \(p^v_i\), and \(q^v_i\), and an edge \(p^v_iq^v_i\) for each \(i\in [\ell ]\) to \(G'\).

We add a vertex \(x^v_{j,i}\) for each \(i\in [\ell ]\) and each \(j\in [s^*]\) to \(G'\).

We add an edge \(y^v_{t,i}z^v_{t,i}\) for each \(i\in [\ell ]\) and each \(t\in [s^*1]\) to \(G'\). Note that these vertices only exist, if \(s\ge 5\).
The vertices \(x^v_{j,i}\), \(y^v_{t,i}\), and \(z^v_{t,i}\) are referred to as the cascading vertices. They are used to ensure that all vertices in \(T^v\) are in exactly \(\ell \) triangles and that there are vertex pairs of distance \(2s^*\) within \(T^v\). Note that since \(s\ge 3\) we create at least \(\ell \) many xvertices. We connect these vertices as follows:

We add the edges \(p^v_ix^v_{1,j}\) and \(q^v_ix^v_{1,j}\) for each \(i\in [\ell 1]\) and each \(j\in [\ell ]\) to \(G'\).

We add the edges \(p^v_\ell x^v_{s^*,j}\) and \(q^v_\ell x^v_{s^*,j}\) for each \(j\in [\ell ]\) to \(G'\).

We add the edges \(y^v_{t,i}x^v_{t,i}\) and \(z^v_{t,i}x^v_{t,i}\) for each \(i\in [\ell ]\), and each \(t\in [s^*1]\) to \(G'\).

We add the edges \(y^v_{t,i}x^v_{t+1,j}\) and \(z^v_{t,i}x^v_{t+1,j}\) for each \(i\in [\ell ]\), each \(j\in [\ell ]\setminus \{i\}\), and each \(t\in [s^*1]\) to \(G'\).
Note that if \(s=3\) or \(s=4\), the graph \(G'\) is a noninduced biclique where one partite set consists of the vertices \(\{x^v_{1,j}\mid j\in [\ell ]\}\) and the other partite set consists of the vertices \(\{p^v_i,q^v_i\mid i\in [\ell ]\}\). Furthermore, the additional edges are \(p^v_iq^v_i\) for each \(i\in [\ell ]\). Also, observe that each vertex gadget \(T^v\) consists of exactly \(3\ell s^*\) vertices.
From now on, the construction differs for the three cases. We now connect these vertex gadgets by introducing the connector edges: For each edge \(vw\in E(G)\) we add edges between the vertex gadgets \(T^v\) and \(T^w\) of the corresponding vertices. We distinguish between the three cases.
 Case I::

s is odd. We add the edges \(p^v_iq^w_i\) and \(q^v_ip^w_i\) for each \(i\in [\ell ]\) to \(G'\).
 Case II::

s is even and \(\ell \ge 3\). We add the edges \(p^v_1q^w_1\), \(q^v_1p^w_1\), \(p^v_\ell q^w_\ell \), and \(q^v_\ell p^w_\ell \) to \(G'\).
 Case III::

s is even and \(\ell =2\). We add the edges \(p^v_1x^w_{s^*,1}\), and \(p^w_1x^v_{s^*,1}\) to \(G'\).
We make the following observation about the connector edges between different vertex gadgets: If s is odd (Case I), or if s is even and \(\ell \ge 3\) (Case II), we have \(N(p^v_i)\setminus T^v\subseteq \{q^w_i\mid vw\in E(G)\}\) and also \(N(q^v_i)\setminus T^v\subseteq \{p^w_i\mid vw\in E(G)\}\) for each \(v\in V(G)\) and each \(i\in [\ell ]\). Otherwise, if s is even and \(\ell =2\) (Case III), observe that we have \(N(p^v_1)\setminus T^v=\{x^w_{s^*,1}\mid vw\in E(G)\}\) and also \(N(x^v_{s^*,1})\setminus T^v=\{p^w_{1}\mid v^w\in E(G)\}\) for each \(v\in V(G)\). Since only the connector edges have endpoints in different vertex gadgets, we thus observe the following.
Observation 4
All three endpoints of each triangle in \(G'\) are contained in exactly one vertex gadget.
Next, we show that each vertex in a vertex gadget \(T^v\) is contained in exactly \(\ell \) triangles in \(G'[T^v]\). Together with Observation 4 this implies that each vertex in \(G'\) is contained in exactly \(\ell \) triangles and also that for each vertex gadget \(T^v\) any vertex\(\ell \)triangle sclub S in \(G'\) either contains all vertices of \(T^v\) or none of them.
Lemma 4
Let \(T^v\) be a vertex gadget. Each vertex in \(T^v\) is contained in exactly \(\ell \) triangles in \(G'[T^v]\).
Proof
We make a case distinction, that is, for each vertex \(a\in T^v\) we present exactly \(\ell \) triangles containing vertex a. We distinguish between the different vertices of a vertex gadget.
Case 1: Consider vertex \(p^v_d\) for some \(d\in [\ell 1]\). Because of Observation 4 we only have to consider the neighbors of \(p^v_d\) in \(T^v\). By construction we have \(N(p^v_d)\cap T^v=\{q^v_d\}\cup \{x^v_{1,j}\mid j\in [\ell ]\}\). Observe that the only edges within \(N(p^v_d)\cap T^v\) are the edges \(q^v_dx^v_{1,j}\) for each \(j\in [\ell ]\). Thus, \(p^v_d\) is contained in the \(\ell \) triangles \(\{\{p^v_d,q^v_d,x^v_{1,i}\}\mid i\in [\ell ]\}\). By similar arguments the same is true for vertex \(q^{v}_{d}\).
Case 2: Consider vertex \(p^v_\ell \). By construction we have \(N(p^v_\ell )= \{q^v_\ell \}\cup \{x^v_{s^*,i}\mid i\in [\ell ]\}\cup \{q^w_\ell \mid vw\in E(G)\}\) if s is odd or s is even and \(\ell \ge 3\). If s is even and \(\ell =2\) we have \(N(p^v_\ell )=\{q^v_\ell \}\cup \{x^v_{s^*,i}\mid i\in [\ell ]\}\). Observe that the only edges within \(N(p^v_\ell )\) are the edges \(q^v_\ell x^v_{s^*,i}\) for each \(i\in [\ell ]\). Thus, \(p^v_\ell \) is contained in the \(\ell \) triangles \(\{\{p^v_\ell ,q^v_\ell ,x^v_{s^*,i}\}\mid i\in [\ell ]\}\). By similar arguments the same is true for vertex \(q^{v}_\ell \).
Case 3: Now, consider vertices \(x^{v}_{1,i}\) and \(x^v_{s^*,i}\) for some \(i\in [\ell ]\). Here we have to distinguish if \(s\in \{3,4\}\) or \(s\ge 5\) since \(y^v_{t,i}\) exists only in the second case.
First, consider the case \(s=3\) or \(s=4\). Note that we now have \(x^v_{1,i}=x^v_{s^*,i}\). By construction we have \(N(x^v_{1,i})=\{p^v_j\mid j\in [\ell ]\}\cup \{q^v_j\mid j\in [\ell ]\}\). Note that the only edges within \(N(x^v_{1,i})\) have the form \(p^v_jq^v_j\) for each \(j\in [\ell ]\). Thus, \(x^{v}_{1,i}\) is contained in the \(\ell \) triangles \(\{\{x^{v}_{1,i},p^v_j,q^{v}_{j}\}\mid j\in [\ell ]\}\).
Second, consider the case \(s\ge 5\). First, we investigate vertex \(x^v_{1,i}\). By construction we have \(N(x^v_{1,i})=\{p^{v}_{j}\mid j\in [\ell 1]\}\cup \{q^{v}_{j}\mid j\in [\ell 1]\}\cup \{y^{v}_{1,i},z^{v}_{1,i}\}\). The only edges within \(N(x^v_{1,i})\) are the edge \(p^v_jq^v_j\) for each \(j\in [\ell 1]\) and the edge \(y^v_{1,i}z^{v}_{1,i}\). Thus, \(x^{v}_{1,i}\) is contained in the \(\ell 1\) triangles \(\{\{x^v_{1,i},p^v_j,q^v_j\}\mid j\in [\ell 1]\}\), and in the triangle \(\{x^v_{1,i},y^v_{1,i},z^v_{1,i}\}\).
Now, consider vertex \(x^{v}_{s^*,i}\). Because of Observation 4 we only have to consider the neighbors of \(x^v_{s^*,i}\) in \(T^v\). By construction we observe that \(N(x^v_{s^*,i})\cap T^v= \{p^v_\ell ,q^v_\ell \}\cup \{y^v_{s^*1,j},z^v_{s^*1,j}\mid j\in [\ell ]\setminus \{i\}\}\). Note that the only edges within \(N(x^v_{s^*,i})\cap T^v\) are the edge \(p^v_\ell q^v_\ell \) and the edge \(y^v_{s^*1,j}z^v_{s^*1,j}\) for each \(j\in [\ell ]\setminus \{i\}\). Thus, \(x^v_{s^*,i}\) is contained in the triangle \(\{x^v_{s^*,i},p^v_\ell ,q^v_\ell \}\) and in the \(\ell 1\) triangles \(\{\{x^v_{s^*,i},y^v_{s^*1,j},z^v_{s^*1,j}\}\mid j\in [\ell ]\setminus \{i\}\}\).
Case 4: Consider vertex \(x^v_{r,i}\) for some \(i\in [\ell ]\) and some \(r\in [2,s^*1]\). Recall that these vertices only exist if \(s\ge 7\). By construction \(N(x^v_{r,i})=\{y^v_{r,i},z^v_{r,i}\}\cup \{y^v_{r1,j},z^v_{r1,j}\mid j\in [\ell ]\setminus \{i\}\}\). The only edges within \(N(x^v_{r,i})\) have the form \(y^v_{r,i}z^v_{r,i}\) and \(y^v_{r1,j}z^v_{r1,j}\) for each \(j\in [\ell ]\setminus \{i\}\). Thus, \(x^v_{r,i}\) is contained in the triangle \(\{x^v_{r,i},y^v_{r,i},z^v_{r,i}\}\) and in the \(\ell 1\) triangles \(\{\{x^v_{r,i},y^v_{r1,j},z^v_{r1,j}\}\mid j\in [\ell ]\setminus \{i\}\}\).
Case 5: Finally, consider vertex \(y^v_{t,i}\) for some \(i\in [\ell ]\) and some \(t\in [s^*1]\). Recall that these vertices only exist if \(s\ge 5\). By construction we have \(N(y^v_{t,i})=\{x^v_{t,i},z^v_{t,i}\}\cup \{x^v_{t+1,j}\mid j\in [\ell ]\setminus \{i\}\}\). The only edges within \(N(y^v_{t,i})\) are the edge \(x^v_{t,i}z^v_{t,i}\) and the edge \(x^v_{t+1,j}z^v_{t,i}\) for each \(j\in [\ell ]\setminus \{i\}\). Thus, \(y^{v}_{t,i}\) is contained in the triangle \(\{y^v_{t,i},z^v_{t,i},x^v_{t,i}\}\) and in the \(\ell 1\) triangles \(\{\{y^v_{t,i},z^v_{t,i}, x^v_{t+1,j}\}\mid j\in [\ell ]\setminus \{i\}\}\). By similar arguments the same holds for vertex \(z^v_{t,i}\).
Thus, each vertex in \(T^v\) is contained in exactly \(\ell \) triangles. \(\square \)
From Lemma 4 and Observation 4, we conclude the following.
Observation 5
Let S be a vertex\(\ell \)triangle sclub for \(s=3\) and \(\ell \ge 1\) or for \(s\ge 4\) and \(\ell \ge 2\) in \(G'\). Then, \(S\cap T^v\ne \emptyset \Leftrightarrow T^v\subseteq S\).
Now, we prove the correctness of Construction 2.
Lemma 5
For \(s=3\) and each \(\ell \ge 1\), and also for each \(s\ge 4\) and \(\ell \ge 2\) the Vertex Triangle sClub problem parameterized by k is W[1]hard, even if each vertex in the input graph is contained in exactly \(\ell \) triangles.
Proof
We show that G contains a clique of size at least k if and only if \(G'\) contains a vertex\(\ell \)triangle sclub of size at least \(3\ell ks^*\).
Let K be a clique of size at least k in G. We argue that \(S:= \bigcup _{v\in K}T^v\) is a vertex\(\ell \)triangle sclub of size at least \(3\ell ks^*\) in \(G'\). The size bound follows from the fact that each \(T^v\) consists of exactly \(3\ell s^*\) vertices. Furthermore, by Lemma 4 each vertex in \(T^v\) for some \(v\in V(G)\) is contained in exactly \(\ell \) triangles in \(G'[T^v]\). Hence, it remains to show that S is an sclub.
To do so, we first prove the following two claims. To formulate the claims, we need some further notation. We define \(T^v_0:= \{p^v_1, \ldots , p^v_{\ell 1}\}\cup \{q^v_1, \ldots , q^v_{\ell 1}\}\) and \(T^v_\ell := \{p^v_\ell ,q^v_\ell \}\). Recall that if \(\ell =1\), then \(T^v_0=\emptyset \). Otherwise, both sets are nonempty.
Claim 1
For \(\ell \ge 2\), we have \(\text {dist}_{G'}(u,a)+\text {dist}_{G'}(u,b)\le 2s^*\) for each vertex \(a\in T^v_0\), each vertex \(b\in T^v_\ell \), and for each vertex \(u\in T^v\setminus (T^v_0\cup T^v_\ell )\).
Proof of Claim
By \(X^v_1:= \{x^v_{1,j}\mid j\in [\ell ]\}\) we denote the neighbors of \(T^v_0\). We define the sets \(X^v_2, \ldots , X^v_{s^*}\) and \(Y^v_1, Z^v_1, \ldots , Y^v_{s^*1},Z^v_{s^*1}\) analogously. We first make some observations about the neighborhoods of these vertex sets, which then allows us to obtain an upper bound for \(\text {dist}_{G'}(u,a)+\text {dist}_{G'}(u,b)\). Note that here we only consider paths that are entirely contained in the vertex gadget \(T^v\).

For \(s\in \{3,4\}\): \(N(x^v_{1,i})\cap T^v= T^v_0\cup T^v_\ell \) for each \(i\in [\ell ]\). Hence \(N(X^v_1)\cap T^v=T^v_0\cup T^v_\ell \).

For \(s\ge 5\): \(N(x^v_{1,i}) = T^v_0\cup \{y^v_{1,i},z^v_{1,i}\}\) for each \(i\in [\ell ]\). Hence \(N(X^v_1)=T^v_0\cup Y^v_1\cup Z^v_1\).

For \(s\ge 5\): \(N(x^v_{s^*,i})\cap T^v = T^v_\ell \cup \{y^v_{s^*1,j},z^v_{s^*1,j}\mid j\in [\ell ]\setminus \{i\}\}\) for each \(i\in [\ell ]\). Hence, \(N(X^v_{s^*})\cap T^v=T^v_\ell \cup Y^v_{s^*1}\cup Z^v_{s^*1}\).

For \(s\ge 7\): \(N(x^v_{t,i})=\{y^v_{t,i},z^v_{t,i}\} \cup \{y^v_{t1,j},z^v_{t1,j}\mid j\in [\ell ]\setminus \{i\}\}\) for each \(t\in [2,s^*1]\) and each \(i\in [\ell ]\). Hence, \(N(X^v_t)=Y^v_t\cup Z^v_t\cup Y^v_{t1}\cup Z^v_{t1}\) for each \(t\in [2,s^*1]\).

For \(s\ge 5\): \(N(y^v_{t,i}) = \{x^v_{t,i},z^v_{t,i}\}\cup \{x^v_{t+1,j}\mid j\in [\ell ]\setminus \{i\}\} \) and also \(N(z^v_{t,i}) = \{x^v_{t,i},y^v_{t,i}\}\cup \{x^v_{t+1,j}\mid j\in [\ell ]\setminus \{i\}\} \) for each \(t\in [s^*1]\) and each \(i\in [\ell ]\). Hence, \(N(Y^v_t)=X^v_t\cup Z^v_t\cup X^v_{t+1}\) and \(N(Z^v_t)=X^v_t\cup Y^v_t\cup X^v_{t+1}\) for each \(t\in [s^*1]\).
From the above and for a vertex \(u\in X^v_z\) we obtain a path from u to a of length \(2z1\) by using subsequent vertices from \(Y^v_{z1},X^v_{z1},\ldots , Y^v_1,X^v_1\). A similar observation can be made for a path from u to b. The cases \(u\in Y^v_z\cup Z^v_z\) are treated similarly. This implies that \(\text {dist}_{G'}(u,a)+\text {dist}_{G'}(u,b)\le 2s^*\) for each vertex \(a\in T^v_0\), each vertex \(b\in T^v_\ell \), and for each vertex \(u\in T^v\setminus (T^v_0\cup T^v_\ell )\). \(\diamond \)
Now, we use Claim 1 to show that \(T^v\) is a \((2s^*)\)club.
Claim 2
If s is odd, then \(T^v\) is an \((s1)\)club and if s is even, then \(T^v\) is an \((s2)\)club for each vertex \(v\in V(G)\).
Proof of Claim
Recall that for \(s\in \{3,4\}\) the gadget \(T^v\) is a noninduced biclique. Hence, the statement is true in this case. In the following, we assume that \(s\ge 5\). Note that this implies that \(\ell \ge 2\). We only consider the case that s is odd. The case that s is even follows analogously.
By construction we have \(T^v_{0}\subseteq N(x^v_{1,1})\) and \(T^v_{\ell }\subseteq N(x^v_{s^*,1})\), so each pair of vertices in \(T^v_{0}\) has distance at most 2 to each other and the same is true for the vertices in \(T^v_{\ell }\).
Consider a pair of vertices \(u,\widetilde{u}\) with \(u \in T^v_0\) and \(\widetilde{u}\in T^v\setminus T^v_0\). From Claim 1 we get \(\text {dist}_{G'}(u,\widetilde{u})\le s1\). The case \(u \in T^v_{\ell }\) and \(\widetilde{u}\in T^v\setminus T^v_{\ell }\) follows analogously.
Now, consider a pair of vertices \(u,\widetilde{u}\) of \(T^v\setminus (T^v_0\cup T^v_\ell )\). To bound the distance of u and \(\widetilde{u}\), we consider one path via a vertex in \(T^v_0\) and one via a vertex in \(T^v_\ell \). We have \(\text {dist}_{G'}(u,\widetilde{u})\le \min (\text {dist}_{G'}(u,a)+\text {dist}_{G'}(a,\widetilde{u}),\text {dist}_{G'}(u,b)+\text {dist}_{G'}(b,\widetilde{u}))\) for each vertex \(a\in T^v_0\), and each vertex \(b\in T^v_\ell \). From Claim 1 we know that \(\text {dist}_{G'}(u,a)+\text {dist}_{G'}(u,b)\le 2s^*\) and that \(\text {dist}_{G'}(\widetilde{u},a)+\text {dist}_{G'}(\widetilde{u},b)\le 2s^*\). Hence, \(\text {dist}_{G'}(u,\widetilde{u})\le 2s^*\le s1\). Thus, \(T^v\) is indeed an \((s1)\)club. \(\diamond \)
Now, we show that for two vertices v and w in K, a vertex \(u\in T^v\), and a vertex \(\widetilde{u}\in T^w\) we have \(\text {dist}_{G'}(u,\widetilde{u})\le s\). We consider the three cases:
Case I: s is odd. Observe that since \(vw\in E(G)\), each vertex \(u\in (T^v_0\cup T^v_\ell )\) has one neighbor in \(T^w\). Since \(T^w\) is an \((s1)\)club by Claim 2, we obtain that for each vertex \(\widetilde{u}\in T^w\) we have \(\text {dist}_{G'}(u,\widetilde{u})\le s\) if \(u\in T^v_0\cup T^v_\ell \). Hence, it remains to consider the case that \(u\in T^v\setminus (T^v_0\cup T^v_\ell )\) and that \(\widetilde{u}\in T^w\setminus (T^w_0\cup T^w_\ell )\). For this, let \(u^v_1:= \text {dist}_{G'}(u,p^v_1), \widetilde{u}^w_1:= \text {dist}_{G'}(q^w_1,\widetilde{u}), u^v_\ell := \text {dist}_{G'}(u,p^v_\ell )\), and \(\widetilde{u}^w_\ell := \text {dist}_{G'}(q^w_\ell ,\widetilde{u})\). Note that
In this inequality, the ’\(+1\)’ is the result of the fact that we have to use an edge to get from gadget \(T^v\) to gadget \(T^w\). By Claim 1 we know that \(u^v_1+u^v_\ell \le 2s^*\) and that \(\widetilde{u}^w_1+\widetilde{u}^w_\ell \le 2s^*\). Since s is odd, we obtain that \(\text {dist}_{G'}(u,\widetilde{u})\le 1 + 2s^*=s\). Thus, S is a vertex\(\ell \)triangle sclub of size \(3\ell ks^*\).
Case II: s is even and \(\ell \ge 3\). For each vertex pair \(u\in T^v\setminus T^v_0\) and \(\widetilde{u}\in T^w\setminus T^w_0\) the proof of \(\text {dist}_{G'}(u,\widetilde{u})\le s\) is analogous to the proof in Case I handling odd values of s. Hence, it remains to show that each vertex \(u\in T^v_0\) has distance at most s to each vertex \(\widetilde{u}\in T^w\). Observe that \(\text {dist}_{G'}(u,p^v_1)\le 2\) since \(T^v_0\subseteq N(x^v_{1,1})\) and \(x_{1,1}^v\) is a neighbor of \(p^v_1\). Thus, \(\text {dist}_{G'}(u,q^w_1)\le 3\) since \(vw\in E(G)\). Furthermore, for each vertex \(\widetilde{u}\in T^w\setminus T^w_\ell \) we have \(\text {dist}_{G'}(\widetilde{u},q^w_1)\le s3\) by the proof of Claim 2. Hence, \(\text {dist}(u,\widetilde{u})\le s\). Thus, it remains to consider the case that \(u\in T^v_0\) and that \(\widetilde{u}\in T^w_\ell =\{p^w_\ell ,q^w_\ell \}\). Since \(vw\in E(G)\) we conclude that \(\widetilde{u}\) has a neighbor in \(T^v\). Since \(T^v\) is an \((s2)\)club by Claim 2 we conclude that \(\text {dist}_{G'}(u,\widetilde{u})=s1\). Hence, S is a vertex\(\ell \)triangle sclub of size \(3\ell ks^*\).
Case III: s is even and \(\ell =2\). Let \(x^w:= x_{s^*,1}^w\) and \(x^v:= x_{s^*,1}^v\). Furthermore, let \(u^v_1:= \text {dist}_{G'}(u,p^v_1), \widetilde{u}^w_x:= \text {dist}_{G'}(x^w,\widetilde{u}), u^v_x:= \text {dist}_{G'}(u,x^v)\), and \(\widetilde{u}^w_1:= \text {dist}_{G'}(p^w_1,\widetilde{u})\). We have
Again, the ’\(+1\)’ results from the fact that we have to use an edge to go from gadget \(T^v\) to \(T^w\). Consider the following claim.
Claim 3
For each vertex \(u\in T^v\) we have \(u^v_1+u^v_x\le s1\).
Claim 3 directly implies that
and thus \(\text {dist}_{G'}(u,\widetilde{u})\le s\). This in turn implies that S is a vertex\(\ell \)triangle sclub of size \(3\ell ks^*\). Thus, it remains to prove Claim 3.
Proof of Claim
We first consider the case that \(u\in T^v_{\ell }\). By construction we have \(T^v_{\ell }\subseteq N(x^v_{s^*,1})\) and thus \(u^v_x = 1\). Since \(T^v\) is an \((s2)\)club by Claim 2, we have \(u^v_1\le s2\) and hence Claim 3 is true in this case.
Next, we consider the case that \(u\in T^v_0\). Since \(T^v_0\subseteq N(x^v_{1,1})\) we conclude that \(u^v_1\le 2\). Furthermore, \(T^v_0\subseteq N(x^v_{1,i})\) for each \(i\in [\ell ]\), so there is a path of length \(2(s^*1)+1= s3\) from u to \(x^v_{s^*,1}\) and we conclude that \(u^v_x=s3\). Hence, Claim 3 is true in this case.
For all remaining cases, it is sufficient to prove the claim for each vertex \(u\in T^v\setminus (T^v_0\cup T^v_\ell )\). First, we consider the case that \(u:= x^v_{t,i}\) for some \(t\in [s^*]\) and some \(i\in [\ell ]\). Then we have \(\text {dist}_{G'}(x^v_{t,i},p^v_1)=2t1\) and \(\text {dist}_{G'}(x^v_{t,i},x^v_{s^*,1})\le 2(s^*t)+2\) and hence \(\text {dist}_{G'}(x^v_{t,i},p^v_1)+\text {dist}_{G'}(x^v_{t,i},x^v_{s^*,1})\le 2t1+2s^*2t+2=2s^*+1\le s1\).
Second, we consider the case that \(u:= y^v_{t,i}\) for some \(t\in [s^*1]\) and some \(i\in [\ell ]\). Then we have \(\text {dist}_{G'}(y^v_{t,i},p^v_1)=2t\) and \(\text {dist}_{G'}(y^v_{t,i},x^v_{s^*,1})\le 2(s^*t)+1\) and hence we obtain that \(\text {dist}_{G'}(y^v_{t,i},p^v_1)+\text {dist}_{G'}(y^v_{t,i},x^v_{s^*,1})\le 2t+2s^*2t+1=2s^*+1\le s1\). The case that \(u:= z^v_{t,i}\) follows by the same argumentation. \(\diamond \)
Conversely, suppose that \(G'\) contains a vertex\(\ell \)triangle sclub S of size at least \(3\ell ks^*\). Because of Observation 5 for each vertex gadget \(T^v\) we either have \(T^v\subseteq S\) or \(T^v\cap S=\emptyset \). Hence, S contains at least k vertex gadgets. We assume towards a contradiction that S contains two vertex gadgets \(T^v\) and \(T^w\) such that \(vw\notin E(G)\). In each case, we will determine a vertex \(u_v\in T^v\) and a vertex \(u_w\in T^w\) such that \(\text {dist}_{G'}(u_v,u_w)\ge s+1\). This contradiction to the sclub property allows us to conclude that the set \(\{v \mid T^v\subseteq S\}\) is a clique of size at least k in G.
Case I: s is odd. We define vertex \(u_v\) as follows. The vertex \(u_w\) is defined analogously. Recall that in this case we have \(s^*=(s1)/2\).

If \(s\equiv 3\mod 4\), we set \(u_v:= x^v_{(s+1)/4,1}\).

Otherwise, if \(s\equiv 1\mod 4\), we set \(u_v:= y^v_{(s1)/4,1}\).
Observe that for each vertex \(u\in T^v_0\cup T^v_\ell \) we have \(\text {dist}_{G'}(u_v,u)=(s1)/2\). Furthermore, recall that the vertices in \(T^v_0\cup T^v_\ell \) are the only vertices in \(T^v\) with neighbors in other vertex gadgets.
Similarly, for each vertex \(u'\in T^w_0\cup T^w_\ell \) we have \(\text {dist}_{G'}(u_w,u')=(s1)/2\). But since \(vw\notin E(G)\) there are no edges between \(T^v\) and \(T^w\) and we obtain for each choice of u and \(u'\) that \(\text {dist}_{G'}(u_v,u_w)\ge \text {dist}_{G'}(u_v,u)+2+\text {dist}_{G'}(u',u_w)=(s1)/2+2+(s1)/2=s+1\), a contradiction.
Case II: s is even and \(\ell \ge 3\). We set \(u_v:= p^v_2\) and \(u_w:= x^w_{s^*,1}\). Note that \(u_v\in T^v_0\) since \(\ell \ge 3\). By the construction we obtain that for each vertex \(u_0\in \{p^v_1,q^v_1\}\) we have \(\text {dist}_{G'}(u_v,u_0)=2\) since \(T^v_0\subseteq N(x^v_{1,1})\). Furthermore, for each vertex \(u_\ell \in \{p^v_\ell , q^v_\ell \}\) we have \(\text {dist}_{G'}(u_v,u_\ell )=s2\). Next, observe that for each vertex \(u'_0\in \{p^w_1, q^w_1\}\) we have \(\text {dist}_{G'}(u_w,u'_0)=s3\) and for each vertex \(u'_\ell \in \{p^w_\ell , q^w_\ell \}\) we have \(\text {dist}(u_w,u'_\ell )=1\).
Since \(vw\notin E(G)\), there are no edges between the gadgets \(T^v\) and \(T^w\). Hence, \(\text {dist}_{G'}(u_v,u_w)\ge \min (\text {dist}_{G'}(u_v, u_0)+2+\text {dist}_{G'}(u'_0,u_w),\text {dist}_{G'}(u_v, u_\ell )+2+\text {dist}_{G'}(u'_\ell ,u_w))\) for each vertex \(u_0\in \{p^v_1, q^v_1\}\), each \(u'_0\in \{p^w_1,q^w_1\}\), each \(u_\ell \in T^v_\ell \), and each \(u'_\ell \in T^w_\ell \). By the above argumentation we obtain \(\text {dist}_{G'}(u_v,u_w)\ge \min (2+2+s3,s2+2+1)=s+1\), a contradiction.
Case III: s is even and \(\ell =2\). We define the vertices \(u_v\) and \(u_w\) as follows:

If \(s=4\), we set \(u_v:= x^v_{1,2}\) and \(u_w:= p^w_2\).

If \(s\equiv 0\mod 4\) and \(s\ge 8\), we set \(u_v:= x^v_{s/4,1}\) and \(u_w:= y^w_{s/4,1}\).

If \(s\equiv 2\mod 8\), we set \(u_v:= y^v_{(s2)/4,2}\) and \(u_w:= x^w_{(s+2)/4,1}\).

If \(s\equiv 6\mod 8\), we set \(u_v:= y^v_{(s2)/4,1}\) and \(u_w:= x^w_{(s+2)/4,2}\).
From the definition of these vertices we obtain that \(\text {dist}_{G'}(u_v,p^v_1)=s/21\), \(\text {dist}_{G'}(u_v,x^v_{s^*,1})=s/2\), \(\text {dist}_{G'}(u_w,p^w_1)=s/2\), and that \(\text {dist}_{G'}(u_w,x^w_{s^*,1})=s/21\). Recall that the vertices \(p^u_1\) and \(x^u_{s^*,1}\) are the only vertices in \(T^u\) which have neighbors outside \(T^u\) for each \(u\in V(G)\). Furthermore, observe that all neighbors of \(p^u_1\) which are not contained in \(T^u\) are the vertices \(x^b_{s^*,1}\) where \(ub\in E(G)\). Similar, all neighbors of \(x^u_{s^*,1}\) which are not in \(T^u\) are of the form \(p^b_1\) where \(ub\in E(G)\). We conclude that \(\text {dist}_{G'}(u_v,u_w)\ge \text {dist}_{G'}(u_v,p^v_1)+3+\text {dist}_{G'}(x^w_{s^*,1},u_w)\). Here, the ’\(+3\)’ results from the fact that at least 3 edges to switch the vertex gadgets have to be used: one is not sufficient since \(uw\notin E(G)\) and also two are not sufficient since in two steps one can only reach a vertex \(p^c_1\) for \(c\in V(G)\) from \(p^v_1\) but no vertex \(x^d_{s^*,1}\) for \(d\in V(G)\). Hence, \(\text {dist}_{G'}(u_v,u_w)\ge s/21+3+s/21=s+1\), a contradiction. \(\square \)
3 Edge Triangle sClub
In this section we settle the parameterized complexity of Edge Triangle sClub with respect to the solution size k. Recall that a vertex set S is an edge\(\ell \)triangle sclub if G[S] contains a spanning subgraph \(G'=(S,E')\) such that each edge in \(E(G')\) is contained in at least \(\ell \) triangles within \(G'\) and the diameter of \(G'\) is at most s. First, we show that Edge Triangle sClub is FPT with respect to k when \(\ell =1\) irrespective of the value of s by providing a Turing kernel. To show this, it is sufficient to delete edges which are not part of a triangle. Afterwards, we prove W[1]hardness of Edge Triangle sClub with respect to k for all fixed \(\ell \ge 2\).
3.1 Edge Triangle sClub with \(\ell =1\)
Now, we prove that Edge Triangle sClub for \(\ell =1\) admits a Turing kernel with respect to k implying that the problem is FPT. To obtain the kernel we need the following reduction rule which removes edges which are in no triangle.
Reduction Rule 6
Let (G, k) be an instance of Edge Triangle sClub. Delete all edges from G which are not part of any triangle.
It is clear that Reduction Rule 6 is correct and can be applied in polynomial time. The idea is that after Reduction Rule 6 is applied, we can bound the size of the neighborhood of each vertex. Next, we prove that after the application of Reduction Rule 6 each edge with both endpoints in the closed neighborhood of a vertex w is contained in a triangle which is contained completely in N[w].
Lemma 6
Let (G, k) be an instance of Edge Triangle sClub with \(\ell =1\) to which Reduction Rule 6 is applied. For each vertex \(v\in V(G)\) each edge in G[N[v]] is contained in at least one triangle in G[N[v]].
Proof
First, we consider edges of the form uv where u is a neighbor of v. Since Reduction Rule 6 is applied, there exists another vertex \(w\in V(G)\) such that \(G[\{u,v,w\}]\) is a triangle. Observe that \(w\in N(v)\).
Second, each edge uw with \(u,w\in N(v)\) is in a triangle with vertex v. \(\square \)
Next, we show that any instance such that sufficiently many vertices are close to some vertex v is a yesinstance.
Lemma 7
Let (G, k) be an instance of Edge Triangle sClub with \(\ell =1\) to which Reduction Rule 6 is applied. Then, (G, k) is a yesinstance if \(N_{\lfloor s/2\rfloor }[v]\ge k\) for some vertex \(v\in V(G)\).
Proof
By Lemma 6 each edge in \(G[N_{\lfloor s/2\rfloor }[v]]\) is contained in at least one triangle in \(G[N_{\lfloor s/2\rfloor }[v]]\) since \(N_{\lfloor s/2\rfloor }[v]= \bigcup _{w\in N_{\lfloor s/2\rfloor 1}[v]}N[w]\) for each \(s\ge 4\). Furthermore, each vertex in \(N_{\lfloor s/2\rfloor }[v]\) has distance at most \(\lfloor s/2\rfloor \) to vertex v. Hence, \(N_{\lfloor s/2\rfloor }[v]\) is an sclub and by assumption \(N_{\lfloor s/2\rfloor }[v]\ge k\). \(\square \)
Lemma 7 implies a Turing kernel for k which implies that the problem is fixedparameter tractable.
Theorem 7
Edge Triangle sClub for \(\ell =1\) admits a \(k^2\)vertex Turing kernel if s is even and a \(k^3\)vertex Turing kernel if s is odd and \(s\ge 3\).
Proof
First, we apply Reduction Rule 6. Because of Lemma 7 we conclude that (G, k) is a trivial yesinstance, if \(N_{\lfloor s/2\rfloor }[v]\ge k\) for some \(v\in V(G)\). Hence, in the following we can assume that \(k> N_{\lfloor s/2\rfloor }[v]\) for each vertex \(v\in V(G)\).
First, we consider the case that s is even. Then \(\lfloor s/2\rfloor =s/2\) and we obtain that \(N_s[v]\subseteq N_{s/2}[N_{s/2}[v]]\) for each \(v\in V(G)\). Thus, \(N_s[v]\le k^2\).
Second, we consider the case that s is odd. Observe that we have \(N_s[v]\subseteq N_{\lfloor s/2\rfloor }[N_{\lfloor s/2\rfloor }[N_{\lfloor s/2\rfloor }[v]]]\) for each \(v\in V(G)\). Thus, \(N_s[v]\le k^3\). \(\square \)
3.2 Edge Triangle sClub for \(\ell \ge 2\)
Now we show W[1]hardness for the remaining cases.
Theorem 8
Edge Triangle sClub is W[1]hard for parameter k if \(\ell \ge 2\).
Next, we describe the construction of the reduction to prove Theorem 8. We reduce from Clique. The idea is to construct one vertex gadget for each vertex of the Clique instance and to add edges between two different vertex gadgets if and only if the two corresponding vertices are adjacent in such a way that all these edges are in exactly \(\ell \) triangles. For an illustration of this construction see Fig. 5.
Construction 3
Let (G, k) be an instance of Clique with \(k\ge 3\). We construct an equivalent instance \((G',k')\) of EdgeTriangle sClub for some fixed \(\ell \ge 2\) as follows. Let \(\ell ^*:= \lceil \ell /2\rceil \) and let \(x:= 6\cdot \ell ^* (s1)+\lfloor \ell /2\rfloor \). For each vertex \(v\in V(G)\), we construct the following vertex gadget \(T^v\). For better readability, all subindices of the vertices in \(T^v\) are considered modulo x. Our construction distinguishes between even and odd values of \(\ell \). First, we describe the part of the construction which both cases have in common.

1.
We add vertex sets \(A_v:= \{a^v_i\mid i\in [0,x]\}\) and \(B_v:= \{b^v_i\mid i\in [0,x]\}\) to \(G'\).

2.
We add the edges \(a^v_ia^v_{i+j}\), and \(b^v_ib^v_{i+j}\) for each \(i\in [0,x]\) and each \(j\in [3\ell ^*,3\ell ^*]\setminus \{0\}\) to \(G'\).

3.
We add the edge \(a^v_ib^v_{i+j}\) for each \(i\in [0,x]\) and each \(j\in [3\ell ^*,3\ell ^*]\) to \(G'\).
In other words, an edge \(a^v_i b^v_j\) is added if the indices differ by at most \(3\ell ^*\). For even \(\ell \), this completes the construction of \(T^v\). For odd \(\ell \), we extend \(T^v\) as follows:

01.
We add the vertex set \(C_v:= \{c^v_{i}\mid i\in [0,x] \text { and } i\equiv 0\mod \ell ^*\}\) to \(G'\). Note that \(C_v\) consists of exactly \(6s5\) vertices.

02.
We add the edges \(c^v_ia^v_{i+j}\) and \(c^v_ib^v_{i+j}\) for each \(i\in [0,x]\) such that \(i\equiv 0\mod \ell ^*\) and each \(j\in [3\ell ^*,3\ell ^*]\) to \(G'\).

03.
Also, we add the edge \(c^v_ic^v_{i+j}\) to \(G'\) for each \(i\in [0,x]\) such that \(i\equiv 0\mod \ell ^*\) and each \(j\in [3\ell ^*,3\ell ^*]\setminus \{0\}\) to \(G'\) if the corresponding vertex \(c^v_{i+j}\) exists.
In other words, an edge between \(c^v_i\) and \(a^v_j\), \(b^v_j\), or \(c^v_j\) is added if the indices differ by at most \(3\ell ^*\). Now, for each edge \(uv\in E(G)\), we add the following to \(G'\):

04.
We add the edges \(a^v_ib^u_{i+j}\) and \(a^u_ib^v_{i+j}\) for each \(i\in [0,x]\) and \(j\in [0,\lfloor \ell /2\rfloor ]\).

05.
If \(\ell \) is odd, we also add the edges \(c^v_ib^u_{i+j}\) and \(c^u_ib^v_{i+j}\) for each \(i\in [0,x]\) such that \(i\equiv 0\mod \ell ^*\) and each \(j\in [0,\lfloor \ell /2\rfloor ]\) to \(G'\). Observe that each vertex \(b^u_{i+j}\) is adjacent to exactly one vertex in \(C_v\).
In other words, an edge between \(a^v_i\) or \(c^v_i\) and \(b^v_j\) is added if j exceeds i by at most \(\lfloor \ell /2\rfloor \). Finally, if \(\ell \) is even, we set \(k':= 2(x+1)k=(\ell (6s5)+2)\cdot k\), and if \(\ell \) is odd, we set \(k':= (2(x+1)+6s5)k=(\ell +2)(6s5)\cdot k\).
Construction 3 has two key mechanisms: First, if \(uv\notin E(G)\) then for each vertex \(a\in A_v\) there is at least one vertex \(b\in B_u\) such that \(\text {dist}(a,b)>s\). Second, each edge with one endpoint in \(A_v\) and one endpoint in \(B_u\) is contained in exactly \(\ell \) triangles. Furthermore, if \(\ell \) is odd, then this also holds for each edge with one endpoint in \(C_v\) and one in \(B_u\). Consider an edge\(\ell \)triangle sclub S in G and let \(\widetilde{G}=(S,\widetilde{E})\) be a spanning subgraph of G[S] with the maximal number of edges, such that each edge of \(\widetilde{E}\) is contained in at least \(\ell \) triangles in \(\widetilde{G}\) and the diameter of \(\widetilde{G}\) is s. As we will show, the two mechanisms ensure that an edge with one endpoint in \(A_v\) (or \(C_v\)) and the other endpoint in \(B_u\) is contained in \(\widetilde{E}\) if and only if S contains all vertices of \(A_v\) (and \(C_v\)) and \(B_u\). We call this the enforcement property. Next, we formalize this property. To this end, we introduce the following notation. By \(E_{uv}\) we denote the set of all edges with one endpoint in \(A_v\) (or \(C_v\) if \(\ell \) is odd) and the other endpoint in \(B_u\).
Lemma 8
Let S be an edge\(\ell \)triangle sclub in the graph \(G'\) constructed in Construction 3. More precisely, let \(\widetilde{G}=(S,\widetilde{E})\) be a maximal subgraph of \(G'[S]\) such that each edge in \(E(\widetilde{G})\) is contained in at least \(\ell \) triangles within \(\widetilde{G}\) and the diameter of \(\widetilde{G}\) is at most s. Let \(e\in E_{uv}\). Then \(e\in E(\widetilde{G})\) if and only if \(A_v,B_u\subseteq S\) (and \(C_v\subseteq S\), if \(\ell \) is odd).
Proof
Before we show the two implications, we prove the following cascading property of edge\(\ell \)triangle sclubs which contain at least one edge of \(E_{uv}\).
Claim 4
If \(a^v_ib^u_j\in E(\widetilde{G})\) or \(c^v_ib^u_j\in E(\widetilde{G})\), then \(E_{uv}\subseteq E(\widetilde{G})\).
Proof of Claim
First, we consider even values of \(\ell \). Note that \(\ell ^*=\lfloor \ell /2\rfloor =\ell /2=\lceil \ell /2\rceil \). By construction we have:

\(N[a^v_i]= \{a^v_{i+i'}, b^v_{i+i'}\mid i'\in [3\ell /2,3\ell /2]\}\cup \{b^w_{i+i'}\mid i'\in [0,\ell /2] \text { and }vw\in E(G)\}\), and

\(N[b^u_j]=\{a^u_{j+i'},b^u_{j+i'}\mid i'\in [3\ell /2,3\ell /2]\}\cup \{a^w_{ji'}\mid i'\in [0,\ell /2] \text { and }uw\in E(G)\}\).
Since \(a^v_ib^u_j\in E(G')\) we obtain by Part 04. of Construction 3 that \(j=i+z\) for some \(z\in [0,\ell /2]\). Without loss of generality we assume that \(z> 0\). Let \(y:=\ell /2z\) and observe that \(N(a^v_i)\cap N(b^u_j)=\{a^v_{i+i'}\mid i'\in [y,z]\setminus \{0\}\}\cup \{b^u_{j+i'} \mid i'\in [z,y]\setminus {\{0\}}\}\). Thus, the edge \(a^v_ib^u_j\) is contained in exactly \(\ell \) triangles whose vertex sets are all contained in \(G'[A_v\cup B_u]\). Since \(a^v_ib^u_j\in E(\widetilde{G})\), we thus conclude that all vertices and edges which form these \(\ell \) triangles are contained in \(\widetilde{G}\). Since \(a^v_ib^u_j\in E(\widetilde{G})\) and we assume that \(z>0\), we obtain that \(a^v_{i+1}\in V(\widetilde{G})\) and \(a^v_{i+1}b^u_j\in E(\widetilde{G})\). Now, \(j=i+1+z'\) where \(z'=z1\) and thus \(z<\ell /2\). By similar arguments as before, we can show that \(a^v_{i+1}b^u_{j+1}\in E(\widetilde{G})\) and thus also that \(a^v_{i+1},b^u_{j+1}\in V(\widetilde{G})\). Now, since \(a^v_{i+1}b^u_{j+1}\in E(\widetilde{G})\) we can repeat the above argumentation for the edge \(a^v_{i+1}b^u_{j+1}\) and inductively for the edge \(a^v_{i+q}b^v_{i+q}\) for all \(q\in [x]\). We then have verified that \(A_v\cup B_u\subseteq V(\widetilde{G})\) and that each edge in \(E_{uv}\) is contained in \(E(\widetilde{G})\). Recall that \(A_v\) and \(B_u\) have size \(x+1\).
Second, we consider odd values of \(\ell \), that is \(\ell =2t+1\) for some integer t. Note that \(\lfloor \ell /2\rfloor =t\) and that \(\ell ^*=\lceil \ell /2\rceil =t+1\). Furthermore, observe that vertex \(b^u_j\) has exactly one neighbor \(c^v_{ji'}\) in \(C_v\) for some \(i'\in [0,t]\) such that \(ji'\mod (t+1)=0\).
Hence, by construction we have

\(N[a^v_i]=N[c^v_i]=\{a^v_{i+i'},b^v_{i+i'}\mid i'\in [3(t+1),3(t+1)]\}\cup \{c^v_{i+i'}\mid i'\in [3(t+1),3(t+1)] \text { and } (i+i')\mod (t+1)=0\}\cup \{b^w_{i+i'}\mid i'\in [0,t] \text { and }vw\in E(G)\}\), and

\(N[b^u_j]=\{a^u_{j+i'},b^u_{j+i'}\mid i'\in [3(t+1),3(t+1)]\}\cup \{c^u_{j+i'}\mid i'\in [3(t+1),3(t+1)] \text { and } (j+i')\mod (t+1)=0\}\cup \{a^w_{ji'}\mid i'\in [0,t] \text { and }uw\in E(G)\}\cup \{c^w_{ji'}\mid i'\in [0,t] \text { and } (ji')\mod (t+1)=0 \text { and }uw\in E(G)\}\).
Since \(a^v_ib^u_j\in E(G')\) we obtain by Part 04. of Construction 3 that \(j=i+z\) for some \(z\in [0,t]\). Without loss of generality we assume that \(z> 0\). Now, let \(y:= tz\) and let \(c^v_{i'}\) be the unique neighbor of \(b^u_j\) in \(C_v\). We conclude that \(N(a^v_i)\cap N(b^u_j)=\{a^v_{i+j'}\mid j'\in [y,z]\setminus \{0\}\}\cup \{b^u_{j+j'} \mid j'\in [z,y]\setminus {0}\}\cup \{c^v_{i'}\}\).
Hence, both vertices have exactly \(t+t+1=\ell \) common neighbors and thus \(N(a^v_i)\cap N(b^u_j)\subseteq S\). By similar arguments a similar statement can be shown for the edge \(c^v_ib^u_j\). Thus, the edges \(a^v_ib^u_j\) and \(c^v_ib^u_j\) are contained in exactly \(\ell \) triangles whose vertex sets are all contained in \(G'[A_v\cup C_v\cup B_u]\). Since \(a^v_ib^u_j\in E(\widetilde{G})\) we thus conclude that all vertices and edges which form these \(\ell \) triangles are contained in \(\widetilde{G}\).
Furthermore, since \(a^v_ib^u_j\in E(\widetilde{G})\) and we assume that \(z>0\), we obtain that \(a^v_{i+1}\in V(\widetilde{G})\) and \(a^v_{i+1}b^u_j\in E(\widetilde{G})\). Now, \(j=i+1+z'\) where \(z'=z1\) and thus \(z<\ell /2\). By similar arguments as before, we can show that \(a^v_{i+1}b^u_{j+1}\in E(\widetilde{G})\) and thus also that \(a^v_{i+1},b^u_{j+1}\in V(\widetilde{G})\). Now, since \(a^v_{i+1}b^u_{j+1}\in E(\widetilde{G})\) we can repeat the above argumentation for the edge \(a^v_{i+1}b^u_{j+1}\) and inductively for the edge \(a^v_{i+q}b^v_{i+q}\) for all \(q\in [x]\). We then have verified that \(A_v\cup B_u\cup C_v\subseteq V(\widetilde{G})\) and that each edge in \(E_{uv}\) is contained in \(E(\widetilde{G})\). \(\diamond \)
Now, we are ready to prove the two implications. The implication that if \(e\in E(\widetilde{G})\) then \(A_v,B_u\subseteq S\) (and \(C_v\subseteq S\), if \(\ell \) is odd) directly follows from Claim 4. It remains to show the other implication. From Claim 4 we conclude that either \(E_{uv}\subseteq E(\widetilde{G})\) or \(E_{uv}\cap E(\widetilde{G})=\emptyset \). If \(E_{uv}\subseteq E(\widetilde{G})\), then we are done, so we assume towards a contradiction that \(E_{uv}\cap E(\widetilde{G})=\emptyset \). Recall that \(A_v\cup B_u\subseteq S\) (and also \(C_v\subseteq S\) is \(\ell \) is odd). Furthermore, recall that \(\widetilde{G}\) is maximal, that is, there exists no spanning subgraph of \(G'[S]\) which has more edges than \(\widetilde{G}\). The proof of Claim 4 also shows that each edge in \(E_{uv}\) is contained in exactly \(\ell \) triangles. Thus, adding all edges in \(E_{uv}\) to \(\widetilde{G}\) is still an edge\(\ell \)triangle sclub, a contradiction to the assumption that \(\widetilde{G}\) contains the maximal number of possible edges. \(\square \)
Now, we prove the correctness of the reduction for Theorem 8.
Proof (of Theorem 8)
We show that G contains a clique of size at least k if and only if \(G'\) contains an edge\(\ell \)triangle sclub of size at least \(k'\).
Let K be a clique of size k in G. Recall that \(T^v\) is the gadget of vertex \(v\in V(G)\). We verify that \(S:=\{v\in V(T^v)\mid v\in K\}\) is an edge\(\ell \)triangle sclub of size at least \(k'\). More precisely, we show that S fulfills all properties of being an edge\(\ell \)triangle sclub. Since \(T^v=2\ell ^*(6s5)+2\) if \(\ell \) is even and \(T^v=(2\ell ^*+1)(6s5)\) if \(\ell \) is odd for each \(v\in K\) and since \(K\ge k\), we have \(S=V(\widetilde{G})\ge k'\). It remains to show that S is an edge\(\ell \)triangle sclub. For this, we show the following:
Claim 5
\(\widetilde{G}\) is an sclub.
Proof of Claim
First, we show that \(T^v\) is an sclub. For this, consider the vertex pair \(\{a^v_i,a^v_j\}\) for some \(v\in K\). Observe that \(P:= (a^v_i,a^v_{i+1},\ldots , a^v_{i+p})\) for \(i+p=j\) is a path of length p from \(a^v_i\) to \(a^v_j\) and that \(Q:= (a^v_i,a^v_{i1},\ldots , a^v_{iq})\) for \(iq=j\) is a path of length q from \(a^v_i\) to \(a^v_j\). Clearly, \(p+q=x+1\). Hence, \(\min (p,q)\le (x+1)/2\le 3\ell ^*(s1)+\lfloor \ell /2\rfloor \). Without loss of generality, assume that the minimum is achieved by path P and assume that \(p=\alpha \cdot (3\ell ^*)+\beta \) for some \(\alpha \in [s1]\) and some \(\beta <3\ell ^*\). Recall that by Part 2 of Construction 3, \(a^v_{i'}a^v_{j'}\in E(G')\) if and only if \(j'=i'+z\) for some \(z\in [3\ell ^*,3\ell ^*]\setminus \{0\}\). Hence, \((a^v_i,a^v_{i+1\cdot (3\ell ^*)}, \ldots , a^v_{i+\alpha \cdot (3\ell ^*)}, a^v_{i+\alpha \cdot (3\ell ^*)+\beta })\) is a path of length at most \((s1)+1=s\) from \(a^v_i\) to \(a^v_j\).
These arguments also apply symmetrically to the vertex pairs \(\{b^v_i,b^v_j\}\) and \(\{a^v_i,b^v_j\}\) for each \(v\in K\). Furthermore, if \(\ell \) is odd, observe that the above argumentation can also be used to show that the vertex pairs \(\{c^v_i,a^v_j\}\), \(\{c^v_i,b^v_j\}\), and \(\{c^v_i,c^v_j\}\) have distance at most s to each other.
Second, we show that \(a^v_i\) has distance at most s to \(b^u_j\). Note that by Part 04. of Construction 3, \(a^v_i\) has neighbors \(b^u_i, \ldots , b^u_{i+\lfloor \ell /2\rfloor }\) since \(uv\in E(G)\). In the following, we assume that \(j\ne i+z\) for all \(z\in [0,\lfloor \ell /2\rfloor ]\). Consider the paths \(P:= (a^v_i,b^u_{i+\lfloor \ell /2\rfloor },b^u_{i+\lfloor \ell /2\rfloor +1},\ldots , ,b^u_{i+\lfloor \ell /2\rfloor +p})\) for \(i+\lfloor \ell /2\rfloor +p= j\) of length \(p+1\) and \(Q:= (a^v_i,b^u_i,b^u_{i1}, \ldots , b^u_{iq})\) for \(iq= j\) of length \(q+1\). Observe that \((p+1)+(q+1)=(x+3)\lfloor \ell /2\rfloor \). Thus, \(p+q=(x+1)\lfloor \ell /2\rfloor =6\ell ^*(s1)+1\). Since p and q are integers we have \(\min (p,q)\le \lfloor ((x+1)\lfloor \ell /2\rfloor )/2\rfloor =3\ell ^*(s1)\).
Without loss of generality assume that the minimum is achieved by path P and assume that \(p=\alpha \cdot (3\ell ^*)+\beta \) for some \(\alpha \in [s2]\) and some \(\beta \in [3\ell ^*]\). Recall that by Part 2 of Construction 3, we have \(b^u_{i'}b^u_{j'}\in E(G')\) if and only if \(j'=i'+z\) for some \(z\in [3\ell ^*,3\ell ^*]\setminus \{0\}\). Now, observe that
is a path of length at most \(1+(s2)+1=s\) from \(a^v_i\) to \(b^u_j\).
Furthermore, if \(\ell \) is odd, observe that the above argumentation can also be used to show that the vertex pairs \(\{c^v_i,b^u_j\}\) have distance at most s to each other by replacing \(a^v_i\) with \(c^v_i\) in the paths P and Q.
The fact that vertices \(a^v_i\) and \(a^u_j\), and \(b^v_i\) and \(b^u_j\), respectively, have distance at most s to each other can be proven similarly as we showed that \(a^v_i\) and \(b^u_j\) have distance at most s by observing that \(a^v_i\) has distance 2 to each vertex \(a^u_{i+z}\) with \(z\in [3\ell ^*,\lfloor \ell /2\rfloor +3\ell ^*]\) since \(a^v_i\) has neighbors \(b^u_i,\ldots , b^u_{i+\lfloor \ell /2\rfloor }\) and since by Part 2 of Construction 3, we have \(b^u_{i'}a^u_{j'}\in E(G')\) if and only if \(j'=i'+z'\) for some \(z'\in [3\ell ^*,3\ell ^*]\).
Furthermore, if \(\ell \) is odd, observe that the above argumentation can also be used to show that the vertex pairs \(\{c^v_i,a^u_j\}\) and \(\{c^v_i,c^u_j\}\) have distance at most s to each other by replacing \(a^v_i\) with \(c^v_i\) and replacing \(a^u_j\) with \(c^u_j\), respectively, in the paths P and Q.
Hence, S is indeed an sclub. \(\diamond \)
Next, we show that also the triangle constraint is fulfilled.
Claim 6
Each edge in \(E(\widetilde{G})\) is contained in at least \(\ell \) triangles which are contained in \(\widetilde{G}\).
Proof of Claim
Consider the edge \(a^v_ia^v_{i+j}\) for some \(j\in [3\ell ^*,3\ell ^*]\setminus \{0\}\). Without loss of generality, assume that \(j>0\). By Part 2 of Construction 3, both vertices are adjacent to each vertex \(a^v_{i+i'}\) with \(i'\in [3\ell ^*]\setminus \{j\}\). Hence, both vertices are in at least \(3\ell ^*1\ge \ell \) triangles.
Furthermore, the statement can be shown analogously for the edges \(b^v_ib^v_{i+j}\) (and \(c^v_ic^v_{i+j}\) if \(\ell \) is odd) for some \(j\in [3\ell ^*,3\ell ^*]\setminus \{0\}\).
Also, the statement can be shown analogously for the edge \(a^v_ib^v_{i+j}\) for some \(j\in [3\ell ^*,3\ell ^*]\). If \(j\ge 0\), then \(a^v_{i+z}\) for each \(z\in [3\ell ^*]\) is a common neighbor of both vertices and thus the edge \(a^v_ib^v_{i+j}\) is contained in at least \(\ell \) triangles. The case \(j<0\) can be shown analogously.
For odd \(\ell \), the statement can be shown analogously for the edges \(a^v_ic^v_{i+j}\) and \(b^v_ic^v_{i+j}\) for some \(j\in [3\ell ^*,3\ell ^*]\), and for the edges \(c^v_ic^v_{i+j}\) for some \(j\in [3\ell ^*,3\ell ^*]\setminus \{0\}\).
The fact that the edges \(a^v_ib^u_{i+j}\) and \(c^v_ib^u_{i+j}\) are contained in exactly \(\ell \) triangles follows from the proof of Lemma 8.
We conclude that each edge in \(E(\widetilde{G})\) is contained in at least \(\ell \) triangles in \(\widetilde{G}\). \(\diamond \)
Thus, S is indeed an edge\(\ell \)triangle sclub of size \(k'\).
Conversely, let S be an edge\(\ell \)triangle sclub of size at least \(k'\) in \(G'\). More precisely, let \(\widetilde{G}\) be a maximal spanning subgraph of \(G'[S]\) which has diameter at most s and such that each edge in \(E(\widetilde{G})\) is contained in at least \(\ell \) triangles is \(\widetilde{G}\). We show that G contains a clique of size at least k.
First, we show that for each vertex \(x_i\in A_v\cup B_v\cup C_v\) there exists a vertex \(y_j\in A_u\cup B_u\cup C_u\) such that \(\text {dist}(x_i,y_{j})\ge s+1\) if \(uv\notin E(G)\). For this, recall that by construction each two vertices with subindices \(i'\) and \(j'\) are not adjacent if their difference (modulo x) is larger than \(3\ell ^*\). Also recall that the vertices \(C_v\) only exist if \(\ell \) is odd.
Claim 7
In \(G'\) we have \(\text {dist}(x_i,y_{j})\ge s+1\) for each \(i\in [0,x]\), \(j:= i+\lfloor \ell /2\rfloor +3\ell ^*(s1)\), \(x_i\in \{a^v_i,b^v_i,c^v_i\}\), and \(y_j\in \{a^u_j,b^u_j,c^u_j\}\) if \(uv\notin E(G)\).
Proof of Claim
There are two possible paths from \(x_i\) to \(y_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\) with respect to the indices. First, there is a subsequence of the indices which is increasing (\(i,i+1,\ldots ,i+\lfloor \ell /2\rfloor +3\ell ^*(s1)\)). This path has length \(\lfloor \ell /2\rfloor +3\ell ^*(s1)\). Second, there is a subsequence of the indices which is decreasing (\(i,i1,\ldots ,ii'\), where \(i':=\lfloor \ell /2\rfloor +3\ell ^*(s1)\)). This path has length \(3\ell ^*(s1)+1\).
Hence, each path from \(x_i\) to \(y_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\) has to overcome at least \(3\ell ^*(s1)+1\) indices. Observe that whenever an edge between \(A_p\) (or \(C_p\)) and \(B_q\) for \(p,q\in V(G)\) with \(pq\in E(G)\) is traversed, by construction the index can increase/decrease by at most \(\lfloor \ell /2\rfloor \). Now, we use the fact that \(uv\notin E(G)\): There are no edges between the vertex gadgets \(T^v\) and \(T^u\). Thus, at least two times such a traversal of at most \(\lfloor \ell /2\rfloor \) indices has to be done. Hence, the index i can increase or decrease by at most \(2\cdot \lfloor \ell /2\rfloor +3\ell ^*(s2)<3\ell ^*(s1)+1\) if at least 2 edge traversals between different vertex gadgets are necessary. Thus, the two vertices \(x_i\) and \(y_j\) have distance at least \(s+1\). \(\diamond \)
The following statement directly follows from Claim 7.
Claim 8
If \(A_v\subseteq S\) (or \(C_v\subseteq S\) if \(\ell \) is odd) and \(B_u\subseteq S\) then \(uv\in E(G)\).
We now use Claims 7 and 8 to show that G contains a clique of size at least k. We distinguish between the cases whether S contains only parts of one of the gadgets \(A_v\), \(B_v\), or \(C_v\) or whether S contains all vertices of the gadgets \(A_v\) (or \(B_v\), or \(C_v\)) completely.
First, assume that for some vertex \(v\in V(G)\) we have \(A_v\cap S\ne \emptyset \) and \(A_v\not \subseteq S\). In the following, we show that S only contains vertices of gadget \(T^v\) and from gadgets \(T^u\) such that \(uv\in E(G)\). Since \(A_v\not \subseteq S\), we conclude that in \(\widetilde{G}\) we have \(N_{\widetilde{G}}(A_v\cap S)\subseteq (B_v\cup C_v)\): Otherwise, since vertex \(a_v\in A_v\cap S\) has a neighbor \(b_u\in B_u\) and by Lemma 8 we would obtain \(A_v\subseteq S\), a contradiction to the assumption \(A_v\not \subseteq S\). If \(B_v\not \subseteq S\), then by Lemma 8 no vertex in \(B_v\) can have a neighbor \(a^w_i\) or \(c^w_i\) for some \(w\ne v\). Note that \(B_v\) may have neighbors in \(C_v\). Hence, \(S\cap T^v\) would be a connected component of size at most \(3(x+1)\), a contradiction to the size of S since \(k\ge 3\). Thus, we may assume that \(B_v\subseteq S\).
Observe that if \(a^w_i\in S\) or \(c^w_i\in S\) for some \(w\in V(G)\) such that \(vw\in E(\widetilde{G})\), that is, also \(vw\in E(G')\) and thus also \(yw\in E(G)\), then we have \(A_w\subseteq S\) and \(C_w\subseteq S\) by Lemma 8 since each vertex \(a^w_i\) and \(c^w_i\) has a neighbor in \(B_v\). Let \(W:= \{w_1,\ldots , w_t\}\) denote the set of vertices \(w_j\) such that \(vw_j\in E(G)\) and \(A_{w_j},C_{w_j}\subseteq S\). If \(w_xw_y\notin E(G)\) for some \(x,y\in [t]\) with \(x\ne y\), then \(a^{w_x}_0\) and \(a^{w_y}_{\lfloor \ell /2\rfloor +3\ell ^*(s1)}\) have distance at least \(s+1\) by Claim 7. Thus \(w_xw_y\in E(G)\) for each \(x,y\in [t]\) with \(x\ne y\).
Assume towards a contradiction that \(a^p_i\in S\) or \(c^p_i\in S\) for some \(p\in V(G)\setminus W\) with \(p\ne v\). Note that \(pv\notin E(G)\) since otherwise \(p\in W\) by the definition of W. Observe that since \(B_v\subseteq S\) we also have \(b^v_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\in S\). But since \(pv\notin E(G)\) we obtain from Claim 7 that \(\text {dist}(z_i,b^v_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)})\ge s+1\) for \(z_i=a^p_i\) or \(z^p_i=c_i\), a contradiction. We conclude that S does not contain any vertex \(a^p_i\) or \(c^p_i\) with \(p\ne v\) or \(p\ne w_j\) for \(j\in [t]\).
Next, assume towards a contradiction that \(b^p_i\in S\) for some \(p\in V(G)\setminus W\) with \(p\ne v\). If \(pv\notin E(G)\), then \(b^p_i\) and \(b^v_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\) have distance at least \(s+1\) again by Claim 7. Thus, we can assume that \(pv\in E(G)\). Recall that \(b^v_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\in S\). As defined by Claim 7, each shortest path from \(b^p_i\) to \(b^v_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\) can swap at most once between different vertex gadgets. In this case, there is exactly one swap from \(T^p\) to \(T^v\). From the above we know that \((A_p\cup C_p)\cap S=\emptyset \). Thus, each shortest path from \(b^p_i\) to \(b^v_{i+\lfloor \ell /2\rfloor +3\ell ^*(s1)}\) uses at least one vertex in \(A_v\cup C_v\). Since at least one edge with an endpoint in \(A_v\cup C_v\) is contained in \(E(\widetilde{G})\), we conclude from Lemma 8 that \(A_v\cup C_v\subseteq S\), a contradiction to the assumption \(A_v\not \subseteq S\).
Hence, there is no vertex \(p\ne v\) and \(p\notin W\) such that \(T^p\cap S\ne \emptyset \). In other words, S contains only vertices from the gadget \(T^v\) and from gadgets \(T^u\) with \(vu\in E(G)\). Thus, \(S\subseteq T^v\cup \bigcup _{j=1}^t T^{w_j}\). By definition of \(k'\), we have \(t\ge k1\) and we conclude that G contains a clique of size at least k. The case that we have \(B_v\cap S\ne \emptyset \) and \(B_v\not \subseteq S\) or \(C_v\cap S\ne \emptyset \) and \(C_v\not \subseteq S\) for some vertex \(v\in V(G)\) can be handled similarly.
Second, consider the case that for each set \(A_v\) with \(A_v\cap S\ne \emptyset \) we have \(A_v\subseteq S\), that for each set \(B_v\) with \(B_v\cap S\ne \emptyset \) we have \(B_v\subseteq S\) , and if \(\ell \) is odd, that for each set \(C_v\) with \(C_v\cap S\ne \emptyset \) we have \(C_v\subseteq S\). Let \(W_A:= \{w_A^j\in V(G)\mid A_{w_j}\subseteq S\}\), \(W_B:= \{w_B^j\in V(G)\mid B_{w_j}\subseteq S\}\), and \(W_C:= \{w_C^j\in V(G)\mid C_{w_j}\subseteq S\}\). If \(W_A=\emptyset \) or \(W_B=\emptyset \) or \(W_C=\emptyset \) (recall that these vertices only exist if \(\ell \) is odd), then each connected component in \(G'[S]\) has size at most \(2(x+1)<k'\). Thus, we may assume that \(W_A\ne \emptyset \), \(W_B\ne \emptyset \), and \(W_C\ne \emptyset \). By Claim 8, we have \(w^i_Aw^j_B\in E(G)\) for each \(w^i_A\in W_A\) and \(w^j_B\in W_B\) and, if \(\ell \) is odd, also \(w^i_Cw^j_B\in E(G)\) for each \(w^i_C\in W_C\) and \(w^j_B\in W_B\). Furthermore, by Claim 7, we have \(w^j_Bw^{j'}_B\in E(G)\) for distinct \(w^j_B,w^{j'}_B\in W_B\), \(w^i_Aw^{i'}_A\in E(G)\) for distinct \(w^i_A,w^{i'}_A\in W_A\), and if \(\ell \) is odd we also have \(w^i_Cw^{i'}_C\in E(G)\) for distinct \(w^i_C,w^{i'}_C\in W_C\) and \(w^i_Aw^{i'}_C\in E(G)\) for \(w^i_A\in W_A\) and \(w^{i'}_C\in W_C\).
Recall that \(k'=yk\) where \(y:= 2(x+1)\) if \(\ell \) is even and \(y:= (2(x+1)+6s5)\) if \(\ell \) is odd and that for each vertex \(v\in V(G)\) we have \(A_v+B_v+C_v=y\). Hence, we obtain that \(\max (W_A,W_B,W_C)\ge k\). Without loss of generality, we assume that \(W_A\ge k\). Then since \(w^i_Aw^{i'}_A\in E(G)\) for distinct \(w^i_A,w^{i'}_A\in W_A\) we conclude that G contains a clique of size at least k. \(\square \)
4 Seeded sClub
In this section we study the parameterized complexity of Seeded sClub with respect to the standard parameter solution size k. Recall that in this problem we aim to find an sclub containing a given seed of vertices. Here, we assume that \(W<k\) since otherwise the problem can be solved in polynomial time.
4.1 Tractable Cases
For clique seeds, we provide the following kernel. Note that here we present a kernel and not only a Turing kernel.
Theorem 9
Seeded sClub admits a kernel with \(\mathcal {O}(k^{2W+1})\) vertices if W induces a clique.
Note that this kernel has polynomial size if W has constant size. In the following, assume that G[W] is a clique. To prove the kernel, we first remove all vertices with distance at least \(s+1\) to any vertex in W. Second, we show that if the remaining graph, that is \(N_s[W]\), is sufficiently large, then (G, W, k) is a trivial yesinstance.
Reduction Rule 10
Let (G, W, k) be an instance of Seeded sClub. If G contains a vertex u such that \(\text {dist}(u,w)\ge s+1\) for some \(w\in W\), then remove u.
Clearly, Reduction Rule 10 is correct and can be applied in polynomial time. Next, we show that if the remaining graph is sufficiently large, then (G, W, k) is a yesinstance of Seeded sClub.
Lemma 9
An instance (G, W, k) of Seeded sClub with \(N_{s1}[W]\ge k^2\) is a yesinstance.
To prove Lemma 9 for \(s\ge 3\) we need the following technical lemmas.
Lemma 10
An instance (G, W, k) of Seeded sClub with \(s\ge 3\) is a yesinstance if \(N_{\lfloor (s+1)/2\rfloor 1}[W]\ge k\).
Proof
By definition \(W\subseteq N_{\lfloor (s+1)/2\rfloor 1}[W]\) and \(N_{\lfloor (s+1)/2\rfloor 1}[W]\ge k\). Thus, it remains to show that \(N_{\lfloor (s+1)/2\rfloor 1}[W]\) is an sclub. For this, consider a pair of vertices \(u, v\in N_{\lfloor (s+1)/2\rfloor 1}[W]\). Observe that by definition \(\text {dist}(u,W)\le \lfloor (s+1)/2\rfloor 1\) and \(\text {dist}(v,W)\le \lfloor (s+1)/2\rfloor 1\). Since W is a clique, we have \(\text {dist}(u,v)\le (\lfloor (s+1)/2\rfloor 1)+1+(\lfloor (s+1)/2\rfloor 1)\le s\). Hence, the lemma follows. \(\square \)
Note that the assumption \(s\ge 3\) in Lemma 10 is necessary to guarantee that \(\lfloor (s+1)/2\rfloor 1\ge 1\). Next, we show that if a vertex in \(N_{\lfloor (s+1)/2\rfloor 1}(W)\) has many vertices close to it, then (G, W, k) is a yesinstance.
Lemma 11
An instance (G, W, k) of Seeded sClub with \(s\ge 3\) and with \(N_{\lfloor s/2\rfloor }[v]\ge k\) for some vertex \(v\in N_{\lfloor (s+1)/2\rfloor 1}(W)\) is a yesinstance.
Proof
Let v be a vertex as specified in the lemma. By definition of v, there exists a path \(P:= (q_0,q_1,\ldots , q_{\lfloor (s+1)/2\rfloor 1})\) of length \(\lfloor (s+1)/2\rfloor 1\) in G such that \(q_{\lfloor (s+1)/2\rfloor 1}=v\), \(q_0\in W\), and \(q_i\in N_i(q_0)\). We show that \(S:= N_{\lfloor s/2\rfloor }[v]\cup W\cup P\) is an sclub of size k containing W. Clearly, \(W\subseteq S\) and \(S\ge k\). Thus, it remains to show that S is an sclub.
Consider a vertex \(w\in W\). Vertex w has distance at most \(i+1\) to vertex \(q_i\). In particular, \(\text {dist}(w,v)\le \lfloor (s+1)/2\rfloor \). Since each vertex \(u\in N_{\lfloor s/2\rfloor }[v]\) has distance at most \(\lfloor s/2\rfloor \) to v we obtain that \(\text {dist}(w,u)\le \lfloor (s+1)/2\rfloor +\lfloor s/2\rfloor =s\). By similar arguments we can also show that vertex \(q_i\) for \(i\in [\lfloor (s+1)/2\rfloor 1]\) has distance at most s to each vertex in S.
Finally, consider two vertices \(x,y\in N_{\lfloor s/2\rfloor }[v]\). Note that \(\text {dist}(x,v)\le \lfloor s/2\rfloor \) and also \(\text {dist}(y,v)\le \lfloor s/2\rfloor \) and thus \(\text {dist}(x,y)\le s\). Thus, S is indeed an sclub. \(\square \)
With those two lemmas we are now able to prove Lemma 9.
Proof (of Lemma 9)
First, we consider the case \(s=2\). Since \(W\le k\) and since \(N[W]\ge k^2\), it is sufficient to show that (G, W, k) is a yesinstance if \(N[w]\ge k\) for some \(w\in W\). Since all vertices in N(w) have the common neighbor w, we conclude that N[w] is a 2club. Also, since W is a clique, we have \(W\subseteq N[w]\). The size bound of \(N_{s1}[W]\) follows from \(N[w]\ge k\). Thus, (G, W, k) is a yesinstance.
Second, we consider the case \(s\ge 3\). Observe that
By Lemma 10, (G, W, k) is a yesinstance if \(N_{\lfloor (s+1)/2\rfloor 1}[W]\ge k\). Furthermore, by Lemma 11, (G, W, k) is a yesinstance if \(N_{\lfloor s/2\rfloor }[v]\ge k\) for some \(v\in N_{\lfloor (s+1)/2\rfloor 1}(W)\). Thus, by the above equality we conclude that (G, W, k) is a yesinstance. \(\square \)
Finally, we bound the size of \(N_s(W)\). There we assume that \(N_{s1}[W]<k^2\) by Lemma 9 and that Reduction Rule 10 is applied.
Lemma 12
An instance (G, W, k) of Seeded sClub such that \(N_s(W)\ge k^{2W+1}\) which is reduced with respect to Reduction Rule 10 is a yesinstance.
Proof
Since Reduction Rule 10 has been applied exhaustively, each vertex \(p\in N_s(W)\) has distance exactly s to each vertex in W. In other words, for each vertex \(w_\ell \in W\) there exists a vertex \(u^{\ell }_{s1}\in N_{s1}(w_\ell )\) such that \(pu^\ell _{s1}\in E(G)\). Note that \(N_{s1}(w_\ell )\subseteq N_{s1}[W]\). Moreover, by Lemma 9 we may assume that \(N_{s1}[W]<k^2\). In particular: \(N_{s1}(W)<k^2\). Since \(N_s(W)\ge k^{2W+1}\), by the pigeonhole principle there exists a set \(\{u^1_{s1},u^2_{s1},\ldots , u^{W}_{s1}\}\) with \(u^\ell _{s1}\in N_{s1}(w_\ell )\) for \(\ell \in [W]\) such that the set \(P:= N_s(W)\cap \bigcap _{\ell \in [W]}N(u^\ell _{s1})\) has size at least k. The size bound follows from the observation that each \(N_{s1}(w_\ell )\) has size at most \(k^2\) and we have exactly W many of these sets. By the definition of vertex \(u^\ell _{s1}\), there exists for each \(i\in [s2]\) a vertex \(u_i^\ell \in N_i(w_\ell )\) such that \(w_\ell , u_1^\ell , \ldots , u_{s1}^\ell \) is a path of length \(s1\) in G. We define the set \(U:= \{u_i^\ell \mid \ell \in [W], i\in [s1]\}\). Next, we show that \(Z:= P\cup W\cup U\) induces an sclub.
First, observe that all vertices in P have distance at most 2 to each other since they have the common neighbor \(u^1_{s1}\). Second, note that the vertices \(w_\ell \), \(u^\ell _1,\ldots , u^\ell _{s1}\), \(p,u^j_{s1},\ldots , u^j_1,w_j\) form a cycle with \(2s+1\) vertices, for each \(p\in P\) and each two indices \(j,\ell \in [W]\). Each vertex in this cycle has distance at most s to each other vertex in that cycle. Hence, Z is indeed an sclub. \(\square \)
Recall that Lemma 9 showed that the number of vertices with distance at most \(s1\) to W is bounded by \(k^2\). Together with Lemma 12 now Theorem 9 is proven.
4.2 Intractable Cases
Now, we show hardness for some of the remaining cases.
Theorem 11
Let H be a fixed graph. Seeded sClub is W[1]hard parameterized by k even if G[W] is isomorphic to H, when

\(s=2\) and H contains at least two nonadjacent vertices, or if

\(s\ge 3\) and H contains at least two connected components.
Hardness for s\(=2\). First, we prove hardness for \(s=2\) when H contains at least one nonedge. For an illustration of Construction 4 we refer to Fig. 6.
Construction 4
Let (G, k) be an instance of Clique. We construct an equivalent instance \((G',k')\) of Seeded sClub as follows. Initially, we add the set W to \(G'\), and add edges such that \(G'[W]\) is isomorphic to H. Since H is not a clique, there exist two vertices \(u,v\in V(H)\) such that \(uv\notin E(H)\). Let \(R:= W\setminus \{u,v\}\). Next, we add two copies \(G_u\) and \(G_v\) of G to \(G'\), make u adjacent to each vertex in \(V(G_u)\), and make v adjacent to each vertex in \(V(G_v)\). For \(x\in V(G)\) we denote with \(x_u\) and \(x_v\) the copies of x in \(G_u\) and \(G_v\), respectively. Next, we add the edge \(x_ux_v\) for each \(x\in V(G)\). Furthermore, we add a new vertex p and make it adjacent to each vertex in W. Next, we add a new vertex \(u^*\) adjacent to p, each vertex in \(V(G_u)\), and each vertex in R. Analogously, we add a new vertex \(v^*\) which is adjacent to p, each vertex in \(V(G_v)\), and each vertex in R. Finally, we set \(k':= 2k+W+3\).
Now, we prove the correctness of Construction 4.
Lemma 13
For any graph H which is not a clique, Seeded 2Club parameterized by k is W[1]hard if the subgraph induced by W is isomorphic to H.
Proof
We prove that G contains a clique of size k if and only if \(G'\) contains a 2club S containing W of size \(k'=2k+W+3\).
Let K be a clique of size k in G and let \(K_u\) and \(K_v\) be the copies of K in \(G_u\) and \(G_v\). We argue that \(S:= K_u\cup K_v\cup W\cup \{u^*,v^*,p\}\) is a 2club of size at least \(k'\) containing W. Clearly, \(S=k'\) and S contains W. Thus, it remains to show that S is a 2club in \(G'\).
First, we show that each vertex in R has distance at most 2 to each vertex in S: All vertices in R have the common neighbors p, \(u^*\), and \(v^*\). Since u and v are neighbors of p, each vertex in \(V(G_u)\) is a neighbor of \(u^*\), and since each vertex in \(V(G_v)\) is a neighbor of \(v^*\), we conclude that each vertex in R has distance at most 2 to any vertex in S.
Second, we show that each vertex in \(K_u\) has distance at most 2 to each vertex in \(S\setminus R\). Observe that \(\{u,u^*,x_v\}\cup K_u\subseteq N[x_u]\) for each vertex \(x_u\in K_u\). Hence, \(x_u\) has distance at most 2 to p via \(u^*\), u, v and to \(v^*\) via \(x_v\), and to each vertex in \(K_v\) via the corresponding vertex in \(K_u\). By symmetric arguments the statement also holds for each vertex in \(K_v\).
Finally, each pair of vertices of \(\{p,u,u^*,v,v^*\}\) has distance at most 2 to each other since \(u,u^*,v,v^*\in N(p)\). Thus, S is indeed a 2club.
Conversely, suppose that \(G'\) contains a 2club S of size at least \(2k+W+3\) which contains all vertices of W. Observe that we have \(N(x_u)\cap N(v)=\{x_v\}\) for each vertex \(x_u\in V(G_u)\), and symmetrically \(N(x_v)\cap N(u)=\{x_u\}\) for each vertex \(x_v\in V(G_v)\). Hence, \(x_u\in S\) if and only if \(x_v\in S\). Let \(K_u:= S\cap V(G_u)\). By definition of \(k'\) we obtain that \(K_u\ge k\). Assume towards a contradiction that \(K_u\) contains a pair of nonadjacent vertices \(x_u\) and \(y_u\). By the argumentation above we obtain \(y_v\in S\). Now, observe that \(N(x_u)=\{u,u^*,x_v\}\cup \bigcup \{z_u\mid xz\in E(G)\}\) and that \(N(y_v)=\{v,v^*,y_u\}\cup \bigcup \{z_v\mid yz\in E(G)\}\). Since \(xy\notin E(G)\) we thus obtain \(N(x_u)\cap N(y_v)=\emptyset \), a contradiction. Thus, G contains a clique of size k. \(\square \)
Hardness for seeds with at least two connected components and \(s\ge 3\). Now, we show W[1]hardness for the case \(s\ge 3\) when the seed contains at least two connected components. Fix a graph H with at least two connected components. We show W[1]hardness for \(s\ge 3\) even if G[W] is isomorphic to H.
Construction 5
Let (G, k) be an instance of Clique. We construct an equivalent instance \((G',k')\) of Seeded sClub as follows. Initially, we add the set W to \(G'\), and add edges such that \(G'[W]\) is isomorphic to H. Let \(D_1\) be one connected component of \(G'[W]\). By assumption, \(D_2:= W\setminus D_1\) is not empty. Next, we add two copies \(G_1\) and \(G_2\) of G to \(G'\). Then, we add edges to \(G'\) such that each vertex in \(D_1\) is adjacent to each vertex in \(V(G_1)\) and such that each vertex in \(D_2\) is adjacent to each vertex in \(V(G_2)\). Furthermore, we add a path \((p_1, \ldots , p_{s1})\) consisting of exactly \(s1\) new vertices to \(G'\), make \(p_1\) adjacent to each \(u\in D_1\), and make \(p_{s1}\) adjacent to each \(v\in D_2\). By \(P:= \{p_i\mid i\in [s1]\}\) we denote the set of these newly added vertices. Now, for each \(x\in V(G)\) we do the following. Consider the copies \(x_1\in V(G_1)\) and \(x_2\in V(G_2)\) of vertex \(x\in V(G)\). We add a path \((x_1, q_1^x, \ldots , q_{s2}^x, x_2)\) consisting of \(s2\) new vertices to \(G'\). By \(Q_x:= \{q^x_i\mid i\in [s2]\}\) we denote the set of the new internal path vertices. Finally, we set \(k':= sk+W+s1\).
Now, we prove the correctness of Construction 5.
Lemma 14
Let H be a fixed graph with at least two connected components. Seeded sClub for \(s\ge 3\) parameterized by k is W[1]hard even if G[W] is isomorphic to H.
Proof
We show that G contains a clique of size k if and only if \(G'\) contains an sclub containing W of size at least \(k'=sk+W+s1\).
Let K be a clique of size k in G. Furthermore, let \(K_1\) and \(K_2\) denote the copies of K in \(G_1\) and \(G_2\), respectively. We show that \(S:= W\cup P\cup K_1\cup K_2\cup \bigcup _{x\in K}Q_x\) is an sclub containing W of size at least \(k'\). Clearly, \(S=k'\) and S contains W. Thus, it remains to verify that S is an sclub in \(G'\). Note that since each vertex in \(V(G_1)\) is adjacent to each vertex in \(D_1\), each two vertices in \(D_1\) have distance at most 2, and similarly each two vertices in \(V(G_1)\) have distance at most 2. Analogously, we can show that each two vertices in \(D_2\) and each two vertices in \(V(G_2)\) have distance at most 2. Furthermore, the vertices \((x,p_1,p_2,\ldots ,p_{s1},y,u_2,q^u_{s2},\ldots , q^u_1,u_1)\) for each vertex \(x\in D_1\), each vertex \(y\in D_2\), and each vertex \(u\in K\) form a \(C_{2s+1}\), a cycle with \(2s+1\) vertices. Observe that also the vertices \((u_1,q^u_1,\ldots ,q^u_{s2},u_2,v_2,q^v_{s2},\ldots , q^v_1,v_1)\) form a \(C_{2s}\) for each two vertices \(u,v\in K\). Since all remaining distances are covered by these two kinds of cycles, we conclude that S is indeed an sclub.
Conversely, suppose that \(G'\) contains an sclub S containing W of size at least \(k'\). Let \(Q'_v:= \{v_1,q^v_1,\ldots , q^v_{s2},v_2\}\) for each \(v\in V(G)\). We show that \(Q'_v\cap S\ne \emptyset \) if and only if \(Q'_v\subseteq S\). Assume towards a contradiction, that \(Q'_v\cap S\ne \emptyset \) for some \(v\in V(G)\) such that \(Q'_v\not \subseteq S\). If \(v_1\notin S\), and also \(v_2\notin S\), then no vertex in \(S\cap Q'_v\) is connected to any vertex in \(S\setminus Q'_v\). Hence, we can assume without loss of generality that \(v_1\in S\). Note that \(N(D_2)=V(G_2)\cup \{p_{s1}\}\). Furthermore, observe that \(\text {dist}(v_1,p_{s1})=s\), that \(\text {dist}(v_1,q^v_{s2})=s2\), and that \(\text {dist}(v_1,q^u_{s2})\ge s1\) for each \(u\in V(G)\setminus \{v\}\). Thus, the unique path of length at most s from \(v_1\) to \(D_2\) contains all vertices in \(Q'_v\). Hence, \(Q'_v\cap S\ne \emptyset \) if and only if \(Q'_v\subseteq S\). By the definition of \(k'\) we may thus conclude that \(Q'_v\subseteq S\) for at least k vertices \(v\in V(G)\).
Now, assume towards a contradiction that \(Q'_u\subseteq S\) and \(Q'_v\subseteq S\) such that \(uv\notin E(G)\). We consider the vertices \(v_1\) and \(u_2\). Observe that by construction each path from \(v_1\) to \(u_2\) containing any vertex \(p_i\) has length at least \(s+2\). Hence, each shortest path from \(v_1\) to \(u_2\) contains the vertex set of \(Q'_w\) for some \(w\in V(G)\). Since the path induced by each \(Q'_w\) has length \(s1\), we conclude that \(w=u\) or \(w=v\). Assume without loss of generality that \(w=v\). Hence, the \((s1)\)th vertex on the shortest path from \(u_2\) to \(v_1\) is \(v_2\). Since \(uv\notin E(G)\) we have by construction that \(u_2v_2\notin E(G')\). Hence, \(\text {dist}(v_1,u_2)\ge s+1\), a contradiction to the fact that S is an sclub. Thus, \(\{v\mid Q'_v\subseteq S\}\) is a clique of size at least k in G. \(\square \)
5 Conclusion
For Seeded sClub we were not able to provide a parameterized complexity dichotomy for the parameter solution size k with respect to the seed structure. More precisely, we could not resolve the complexity for \(s\ge 3\) if the seed is connected and contains at least one nonedge; for all other cases we either presented an FPTalgorithm or showed W[1]hardness. To show hardness, it would thus be sufficient to provide a reduction if the seed contains at least one \(P_3\). For future work, it seems interesting to study the complexity of the considered variants of sClub with respect to further parameters, for example with respect to structural parameters of the input graph G such as the treewidth of G. Additionally, the parameterized complexity of further robust variants of sClub such as tHereditary sClub [16, 18] with respect to k remains open.
In followup work, we provided an efficient branchandbound solver for Vertex Triangle 2Club and Edge Triangle 2Club [11]. This solver can compute optimal solutions for graphs with up to \(1\, 000\, 000\) edges within 1 hour for arbitrary values of \(\ell \). For Vertex Triangle 2Club, this solver outperforms ILPs by Almeida and Brás [1], and for Edge Triangle 2Club it is the first solver. The experiments show that edge\(\ell \)triangle 2clubs can not only be found faster than vertex\(\ell \)triangle 2clubs but also that the local and global clustering coefficients of the solutions are usually higher in the edgevariant than in the vertexvariant [11]. Summarizing, the edge\(\ell \)triangle 2club model has better robustness guarantees, and appears to be preferable regarding the running time and the solution quality.
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Acknowledgements
Frank Sommer is supported by the Deutsche Forschungsgemeinschaft (DFG), project EAGR (KO 3669/61)
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An extended abstract of this work appeared in the Proceedings of the 33rd International Workshop on Combinatorial Algorithms (IWOCA ’22) held in Trier, Germany.
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Frank Sommer is supported by the Deutsche Forschungsgemeinschaft (DFG), project EAGR (KO 3669/61).
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Garvardt, J., Komusiewicz, C. & Sommer, F. The Parameterized Complexity of sClub with Triangle and Seed Constraints. Theory Comput Syst 67, 1050–1081 (2023). https://doi.org/10.1007/s0022402310135x
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DOI: https://doi.org/10.1007/s0022402310135x