2-CLUB is NP-hard for distance to 2-club cluster graphs

TL;DR

This paper proves that determining the 2-CLUB problem remains NP-hard even when restricted to graphs close to 2-club cluster graphs, highlighting computational complexity challenges.

Contribution

It establishes the NP-hardness of 2-CLUB for graphs with small distance to 2-club cluster graphs, a previously unknown complexity result.

Findings

2-CLUB is NP-hard for graphs close to 2-club cluster graphs

Complexity persists even with structural graph restrictions

Highlights computational difficulty in specialized graph classes

Abstract

We show that 2-CLUB is NP-hard for distance to 2-club cluster graphs.

Equations

$$V(G):=V(H)\cup A\cup \{a,b,u\}\cup X_1\cup X_2$$

$$E(G)=E(H)\bigcup a\times \{\{b\}\cup X_1\cup A\}\bigcup b\times \{X_1\cup V(H)\cup X_2\}\bigcup V(H)\times X_2\bigcup u\times \{A\cup V(H)\cup X_2\}\bigcup _{\forall i\in [n]}{v}_i\times V_i$$

Full text

1 Introduction

A complete graph or clique is a graph that contains an edge for every pair of distinct vertices. Diameter of a graph is the length of a longest shortest path in the graph. Any clique has diameter $1$. A generalization of this notion is $s$-club, a graph of diameter $s$. In general graphs, finding a set of vertices that induces a subgraph of diameter $s$ is NP-hard. For $s=2$, Hartung et al [HKN13] have studied the problem with many structral restrictions on the input graph. This paper answers one of the open problems mentioned in [HKN13].

Given a class of graphs with some property $\Pi$, we can define another class of graphs by the parameter distance to $\Pi$, namely the number of vertices that needs to be deleted from the graph to make the resultant graph have property $\Pi$. For example, distance $2$ to bipartiteness defines a class of graphs that become bipartite after deleting at most $2$ vectices. A graph where each connected component is an $s$-club is called $s$-club cluster graph. In this paper, we show that finding $2$-club in distance $d$ to $2$-club cluster graphs is NP-hard for $d\geq 2$.

2 Constant Distance to 2-club cluster

We define the $2$-CLUB problem as follows: Given an undirected graph $G=(V,E)$ and $k\in\mathbb{N}$, is there a vertex set $S\subseteq V$ of size at least $k$ such that $G[S]$ has diameter at most $2$?

Theorem 2.1

2-CLUB is NP-hard even on graphs with distance two to $2$-club cluster.

Proof

We reduce from the NP-hard CLIQUE problem: Given a positive integer $k$ and a graph $H$, the question is whether there is a clique of size at least $k$.

Given an instance $(H,k)$ of CLIQUE, we construct an undirected graph $G=(V,E)$.

Let $|V(H)|=n$. Define the vertex set

[TABLE]

where $a,b,u$ are vertices and $A,B,X_{1},X_{2}$ are sets of vertices with sizes $|X_{1}|=n^{3},|X_{2}|=n^{2}-n$ and $|A|=n^{2}$. For every vertex $v_{i}\in V(H)$, we lebel $n$ vertices of $A$ as $V_{i}=\{v_{i,1},...,v_{i,n}\}$. The edge set $E(G)$ is defined as

[TABLE]

Note that all the edges are undirected. See Figure below.

Claim

$H$ has a clique of size $k$ if and only if $G$ has a 2-club of size $n^{3}+n^{2}+(k-1)n+k+2$.

Proof

Let $S$ be a clique of size $k$ in $H$. Then, $X_{1}\cup X_{2}\cup S\cup(N(S)\cap A)\cup\{a,b\}$ is a 2-club of size $n^{3}+n^{2}+(k-1)n+k+2$.

Let $Y$ be a $2$-club of size $n^{3}+n^{2}+(k-1)n+k+2$ in $G$.

By size consideration $X_{1}\subset Y$. If $b\notin Y$, then none of $B$ and $X_{2}$ can be in $Y$. Consequently, the size of any $2$-club in $G$ can be $n^{3}+n^{2}+1$. Hence, we must have that $b\in Y$. By similar reasoning, we have that $a\in Y$.

If $A\cap Y=\emptyset$, then the size of the largest 2-club can be at most $n^{3}+n^{2}+2$ implying that $Y$ must intersect with $A$. Moreover, $|A\cap Y|$ must be a multiple of $n$ as for $v_{i}\in A$ contained in $Y$, the whole subset $V_{i}\subset A$ can be included in $Y$ preserving the $2$-club property. If $|A\cap Y|<(k-1)n$, then size of the maximum 2-club can be at most $n^{3}+n^{2}+(k-1)n+2$, the size of $X_{1}\cup X_{2}\cup V(H)\cup\{a,b\}\cup(A\cap Y)$ which is less than $n^{3}+n^{2}+(k-1)n+k+2$. Hence at least $k$ vertices in $V(H)\cap Y$ have neighbors in $A\cap Y$. This also implies that $V(H)\cap Y$ forms a clique in $H$. If $\{x,y\}\in V(H)\cap Y$ are not adjacent and have neighbors $\{x^{\prime},y^{\prime}\}\in A\cap Y$. Then, there is no path of length $\leq 2$ between $x$ and $y^{\prime}$. Hence, $H$ has a clique of size $k$.

Acknowledgements

I am grateful for the fruitful discussions with Daniel Lokshtanov and Markus Dregi.