Contracting edges to destroy a pattern: A complexity study

TL;DR

This paper studies the computational complexity of contracting edges in a graph to eliminate a certain pattern, showing NP-completeness in most cases and analyzing fixed-parameter tractability for H-free contractions.

Contribution

It characterizes the complexity of H-free Contraction problems, proving NP-completeness for all but trivial cases and analyzing fixed-parameter tractability for tree patterns.

Findings

H-free Contraction is polynomial-time solvable only when H is a small complete graph.

Most H-free Contraction problems are NP-complete.

H-free Contraction is W[2]-hard for most trees, with a few exceptions.

Abstract

Given a graph G and an integer k, the objective of the $\Pi$-Contraction problem is to check whether there exists at most k edges in G such that contracting them in G results in a graph satisfying the property $\Pi$. We investigate the problem where $\Pi$ is `H-free' (without any induced copies of H). It is trivial that H-free Contraction is polynomial-time solvable if H is a complete graph of at most two vertices. We prove that, in all other cases, the problem is NP-complete. We then investigate the fixed-parameter tractability of these problems. We prove that whenever H is a tree, except for seven trees, H-free Contraction is W[2]-hard. This result along with the known results leaves behind three unknown cases among trees.

Tables

Table 1. Table 1 : Complexities of various graph modification problems where the target property is H 𝐻 H -free. The number of vertices, the number of edges, and the number of nonedges in H 𝐻 H are denoted by n , m , m ′ 𝑛 𝑚 superscript 𝑚 ′ n,m,m^{\prime} respectively.

Problem	P	NPC	FPT	W-hard
Edge Editing	$n\le 2$ [trivial]	otherwise [1]	For all $H$ [6]
Edge Deletion	$m\le 1$ [trivial]	otherwise [1]	For all $H$ [6]
Edge Completion	$m^{\prime }\le 1$ [trivial]	otherwise [1]	For all $H$ [6]
Vertex Deletion	$n\le 1$ [trivial]	otherwise [19]	For all $H$ [6]
Edge Contraction	$K_1,K_2$ [trivial]	otherwise [Theorem 3.1]	$K_t$ ($t\ge 3$) [15, 20], $P_3,P_4,K_2+K_1$( [8, 20], MSO1 expressibility)	W[2]-hard for 3-connected non-complete graphs, diamond [8, 15], $C_t$ ($t\ge 4$) [8, 20], all trees except 7 trees [Theorem 4.1]

Full text

1 Introduction

Let $\Pi$ be any graph property. Given a graph $G$ and an integer $k$, the objective of the $\Pi$-Contraction problem is to check whether $G$ contains at most $k$ edges so that contracting them results in a graph with property $\Pi$. This is a vertex partitioning problem in disguise: Find whether there is a partition $\mathcal{P}$ of the vertices of $G$ such that each set in $\mathcal{P}$ induces a connected subgraph of $G$, $G/\mathcal{P}$ (the graph obtained by contracting each set in $\mathcal{P}$ into a vertex) has property $\Pi$, and $n-|\mathcal{P}|\leq k$. These problems, for various graph properties $\Pi$, have been studied for the last four decades. Asano and Hirata [2] proved that the problem is NP-complete if $\Pi$ is any of the following classes - planar, series-parallel, outerplanar, chordal. When $\Pi$ is a singleton set $\{H\}$, then the problem is known as $H$-Contraction. Brouwer and Veldman [5] proved that $H$-Contraction is polynomial-time solvable if $H$ is a star, and NP-complete if $H$ is a connected triangle-free graph other than a star graph. Belmonte, Heggernes, and van ’t Hof [3] proved that it is polynomial-time solvable when $H$ is a split graph. Golovach, Kaminski, Paulusma, and Thilikos [13] studied the problem when $\Pi$ is ‘minimum degree at least $d$’ and proved that the problem is NP-complete even for $d=14$ and W[1]-hard when parameterized by $k$. Heggernes, van ’t Hof, Lokshtanov, and Paul [16] proved that the problem is fixed parameter tractable when $\Pi$ is the class of bipartite graphs. Guillemot and Marx [14] obtained a faster FPT algorithm for the problem. Cai, Guo [8], and Lokshtanov, Misra, and Saurabh [20] proved that the problem is W[2]-hard when $\Pi$ is the class of chordal graphs. Garey and Johnson [11] mentioned that, given two graphs $G$ and $H$, the problem of checking whether $H$ can be obtained from $G$ by edge contractions is NP-complete. Edge contraction has applications in Graph minor theory (see [21]), Hamiltonian graph theory [17], and geometric model simplification [12].

We consider the $H$-free Contraction problem: Given a graph $G$ and an integer $k$, find whether $G$ can be transformed, by at most $k$ edge contractions, into a graph without any induced copies of $H$. The parameter we consider is $k$. Unlike graph contraction problems, other major graph modification problems are well-understood for these target graph classes. In particular, P versus NP-complete dichotomies are known for $H$-free Edge Editing, $H$-free Edge Deletion, $H$-free Edge Completion [1], and $H$-free Vertex Deletion [19] (here, the allowed operations are edge editing, edge deletion, edge completion, and vertex deletion respectively). It is also known that all these problems are in FPT for every graph $H$ [6]. The picture is far from complete for $H$-free Contraction. See Table 1. It is trivial to note that $H$-free Contraction is polynomial-time solvable if $H$ is a complete graph of at most 2 vertices. Cai, Guo [8, 15], and Lokshtanov, Misra, and Saurabh [20] proved the following results for $H$-free Contraction.

• •

FPT when $H$ is a complete graph

• •

If $H$ is a path or a cycle, then the problem is FPT when $H$ has at most 3 edges, and W[2]-hard otherwise.

• •

W[2]-hard when $H$ is 3-connected but not complete, or a star graph on at least 5 vertices, or a diamond.

The W[2]-hardness results mentioned above also imply NP-completeness of the problems. Guo [15] proved that the problem is NP-complete when $H$ is a complete graph on $t$ vertices, for every $t\geq 3$. Eppstein [10] proved that the Hadwiger number problem (find whether the size of a largest clique minor of a graph is at least $k$) is NP-complete. This problem is essentially $2K_{1}$-free Contraction, if we ignore the parameter. This result implies that the problem is NP-complete when $H$ is a $P_{3}$. We build on these results and prove the following.

• •

$H$-free Contraction is NP-complete if $H$ is not a complete graph on at most 2 vertices.

• •

$H$-free Contraction is W[2]-hard if $H$ is a tree which is neither a star on at most 4 vertices ( , , , ) nor a bistar in $\{$ , , $\}$.

Our W[2]-hardness results, along with known positive results, leaves behind only three open cases among trees -
,
,
.

2 Preliminaries

Graphs.

All graphs considered in this paper are simple and undirected. A complete graph and a path on $t$ vertices are denoted by $K_{t}$ and $P_{t}$ respectively. A universal vertex of a graph is a vertex adjacent to every other vertex of the graph. An isolated vertex is a vertex with degree 0. For an integer $t\geq 0$, a star on $t+1$ vertices, denoted by $K_{1,t}$, is a tree with a single universal vertex and $t$ degree-1 vertices. The universal vertex in $K_{1,t}$ is also called the center of the star. For integers $t,t^{\prime}$ such that $t\geq t^{\prime}\geq 0$, a bistar on $t+t^{\prime}+2$ vertices, denoted by $T_{t,t^{\prime}}$, is a tree with two adjacent vertices $v$ and $v^{\prime}$, where $t$ degree-1 vertices are attached to $v$ and $t^{\prime}$ degree-1 vertices are attached to $v^{\prime}$. The bistar $T_{t,0}$ is the star $K_{1,t+1}$ and the bistar $T_{1,1}$ is $P_{4}$. We say that $vv^{\prime}$ is the central edge of the bistar. By $G_{1}+G_{2}$ we denote the disjoint union of the graphs $G_{1}$ and $G_{2}$. A graph is $H$-free if it does not contain any induced copies of $H$. In a graph $G$, replacing a vertex $v$ with a graph $H$ is the graph obtained from $G$ by removing $v$, introducing a copy of $H$, and adding edges between every vertex of the $H$ and every neighbor of $v$ in $G$. A separator $S$ of a connected graph $G$ is a subset of its vertices such that $G-S$ (the graph obtained from $G$ by removing the vertices in $S$) is disconnected. A separator is universal if every vertex of the separator is adjacent to every vertex outside the separator. We will be using the term ‘universal $K_{1}$ (resp. $K_{2}$) separator’ to denote a universal separator which induces a $K_{1}$ (resp. $K_{2}$). Let $V^{\prime}$ be a subset of vertices of a graph $H$. By $H[V^{\prime}]$ we denote the graph induced by $V^{\prime}$ in $H$. For a graph $G$ and two subsets $A$ and $B$ of vertices of $G$, by $E[A,B]$ we denote the set of edges in $G$, where each edge in the set is having one end point in $A$ and the other end point in $B$.

Contraction.

Contracting an edge $uv$ in a graph $G$ is the operation in which the vertices $u$ and $v$ are identified to be a new vertex $w$ such that $w$ is adjacent to every vertex adjacent to either $u$ or $v$. Given a graph $G$ and a subset $F$ of edges of $G$, the graph $G/F$ obtained by contracting the edges in $F$ does not depend on the order in which the edges are contracted. Every vertex $w$ in $G/F$ represents a subset $W$ of vertices (which are contracted to $w$) of $G$ such that $W$ induces a connected graph in $G$. Let $G_{F}$ be the subgraph of $G$ containing all vertices of $G$ and the edges in $F$. There is a partition $\mathcal{P}$ of vertices of $G$ implied by $F$: Every set in $\mathcal{P}$ corresponds to the vertices of a connected component in $G_{F}$. We note that many subsets of edges may imply the same partition - it does not matter which all edges of a connected subgraph are contracted to get a single vertex. The graph $G/F$ is nothing but the graph in which there is a vertex corresponding to every set in $\mathcal{P}$ and two vertices in $G/F$ are adjacent if and only if there is at least one edge in $G$ between the corresponding sets in $\mathcal{P}$. The graph $G/F$ is equivalently denoted by $G/\mathcal{P}$. Assume that $\mathcal{P}^{\prime}$ is a partition of a subset of vertices of $G$. Then by $G/\mathcal{P}^{\prime}$ we denote the graph obtained from $G$ by contracting each set in $\mathcal{P}^{\prime}$ into a single vertex. The cost of a set $P$ in $\mathcal{P}$ is the number $|P|-1$, which is equal to the minimum number of edges required to form the set $P$. The cost of $\mathcal{P}$, denoted by cost$(\mathcal{P})$, is the sum of costs of the sets in $\mathcal{P}$. We observe that $|\mathcal{P}|+$cost$(\mathcal{P})$ $=n$, where $n$ is the number of vertices of $G$. We say that $F$ touches a subset $W$ of vertices of $G$, if there is at least one edge $uv$ in $F$ such that either $u$ or $v$ is in $W$. Let $u,v$ be two non-adjacent vertices of $G$. Identifying $u$ and $v$ in $G$ is the operation of removing $u$ and $v$, adding a new vertex $w$, and making $w$ adjacent to every vertex adjacent to either $u$ or $v$.

Fixed parameter (in)tractability.

A parameterized problem is fixed-parameter tractable (FPT) if it can be solved in time $f(k)|I|^{O(1)}$-time, where $f$ is a computable function and $(I,k)$ is the input. Parameterized problems fall into different levels of complexities which are captured by the W-hierarchy. A parameterized reduction from a parameterized problem $Q^{\prime}$ to a parameterized problem $Q$ is an algorithm which takes as input an instance $(I^{\prime},k^{\prime})$ of $Q^{\prime}$ and outputs an instance $(I,k)$ of $Q$ such that the algorithm runs in time $f(k^{\prime})|I^{\prime}|^{O(1)}$ (where $f$ is a computable function), and $(I^{\prime},k^{\prime})$ is a yes-instance of $Q^{\prime}$ if and only if $(I,k)$ is a yes-instance of $Q$, and $k\leq g(k^{\prime})$ (for a computable function $g$). We use parameterized reductions to transfer fixed-parameter intractability. For more details on these topics, we refer to the textbook [9]. The problem that we deal with in this paper is defined as follows.

$H$-free Contraction: Given a graph $G$ and an integer $k$, can $G$ be modified into an $H$-free graph by at most $k$ edge contractions?

3 NP-completeness

In this section we prove that $H$-free Contraction is NP-complete whenever $H$ is not a complete graph of at most two vertices. In Section 3.1 we obtain reductions for the cases when $H$ is connected but does not have any universal $K_{1}$ separator and universal $K_{2}$ separator. In Section 3.2 we deal with non-star graphs with universal $K_{1}$ separator or universal $K_{2}$ separator. In Section 3.3 we resolve the stars. The cases when $H$ is a $2K_{2}$ or a $K_{2}+K_{1}$ are handled in Section 3.4. These come as base cases in the inductive proof of the main result of the section given in Section 3.5. We crucially use the following results.

Proposition 1 ([15, 10])

$H$-free Contraction* is NP-complete when $H$ is a $2K_{1}$, or a $P_{3}$, or a $K_{t}$, for any $t\geq 3$.*

3.1 A general reduction

A vertex cover of a graph $G$ is a subset $V^{\prime}$ of its vertices such that for every edge $uv$ of $G$, either $u$ or $v$ is in $V^{\prime}$. Vertex Cover is the decision problem in which the objective is to check whether the given graph has a vertex cover of size at most $k$. The problem is very well-known to be an NP-complete problem. Inspired by a reduction by Asano and Hirata [2], we introduce the following reduction from Vertex Cover  which handles connected graphs $H$ without any universal $K_{1}$ separator and universal $K_{2}$ separator.

Construction 1

Let $G^{\prime}$ be a graph without any isolated vertices, and $H$ be a connected non-complete graph. We obtain a graph $G$ from $G^{\prime}$ and $H$ as follows.

• •

Subdivide each edge of $G^{\prime}$ once, i.e., for every edge $uv$, introduce a new vertex and make it adjacent to both $u$ and $v$, and delete the edge $uv$.

• •

Replace each new vertex by a copy $H$.

• •

Let $w$ be a non-universal vertex of $H$ (the existence of $w$ is guaranteed as $H$ is not a complete graph). Identify $w$ of every copy of $H$ (introduced in the previous step) to be a single vertex named $w$.

Let the resultant graph be $G$. The vertices in $G$ copied from $G^{\prime}$ form the set $V^{\prime}$, which forms an independent set in $G$. For each edge $uv$ in $G^{\prime}$, the vertices, except $w$, of the copy of $H$ is denoted by $W_{uv}$. By $W$ we denote any such set. We note that $w$ is adjacent to every vertex in $V^{\prime}$, as $G^{\prime}$ does not have any isolated vertices. This completes the construction. An example is shown in Figure 1.

Let $H$ be a connected non-complete graph with $h$ vertices and without any universal $K_{1}$ or $K_{2}$ separator. Let $G^{\prime}$ be a graph without any isolated vertices. Let $G$ be obtained from $(G^{\prime},H)$ by Construction 1. Assume that $(G^{\prime},k)$ is a yes-instance of Vertex Cover and let $T$ be any vertex cover of size at most $k$ of $G^{\prime}$. Let $F=\{wu:u\in T\}$. Let $w$ itself denote the vertex obtained by contracting the edges in $F$. The following can be observed directly from the fact that $T$ is a vertex cover.

Observation 1

The vertex $w$ is a universal vertex in $G/F$.

Before, proving that $G/F$ is $H$-free, we prove the following lemma.

Lemma 1

Let $v$ be a vertex in $V^{\prime}\setminus T$. Let $u$ be any neighbor of $v$ in $G^{\prime}$. Then none of the subsets of $W_{uv}\cup\{v,w\}$ induces an $H$ in $G/F$.

Proof

For a contradiction, assume that a subset $U$ of $W_{uv}\cup\{v,w\}$ induces an $H$ in $G/F$. Since $w$ is not a universal vertex in $H$, and both $w$ (by Observation 1) and $v$ (by the construction) are universal to $W_{uv}$ in $G/F$, we obtain that neither $W_{uv}\cup\{v\}$ nor $W_{uv}\cup\{w\}$ induces $H$ in $G/F$. Therefore, $U=(W_{uv}\setminus\{x\})\cup\{v,w\}$, for some vertex $x$ in $W_{uv}$. Let $m$ be the number of edges in $H$. Let $m^{\prime}$ be the number of edges in the graph induced by $W_{uv}\setminus\{x\}$ in $H$. Clearly, $m=m^{\prime}+d_{x}+d_{w}-d_{wx}$, where $d_{x}$ and $d_{w}$ denote the degree of $x$ and degree of $w$ respectively in $H$, and $d_{wx}$ is 1 if there is an edge between $w$ and $x$ in $H$, and 0 otherwise. Since $U$ induces an $H$ in $G/F$, we obtain that $m=m^{\prime}+h-1+h-1-1=m^{\prime}+2h-3$ (note that both $w$ and $v$ are universal in the $H$ induced by $U$). Therefore, we obtain that $d_{x}+d_{w}-d_{xw}=2h-3$. Since $w$ is not universal in $H$, we obtain that $d_{w}\leq h-2$. Therefore, $d_{x}-d_{xw}\geq h-1$. If $x$ is universal in $H$, then we get that $d_{x}-d_{xw}=h-1-1=h-2$, and if $x$ is not universal in $H$, then we obtain that $d_{x}-d_{xw}\leq h-2-0=h-2$. Therefore, we get a contradiction.

Lemma 2

If $(G^{\prime},k)$ is a yes-instance of Vertex Cover, then $(G,k)$ is a yes-instance of $H$-free Contraction.

Proof

Let $T$ be a vertex cover of size at most $k$ of $G^{\prime}$ and $F=\{wu:u\in T\}$. We claim that $G/F$ is $H$-free. For a contradiction, assume that there is a copy of $H$ induced by a set $U$ in $G/F$. By Observation 1, $w$ is a universal vertex in $G/F$. Since the vertices in $V^{\prime}\setminus T$ along with $w$ induces a star graph in $G/F$, and $H$ is a not a subgraph of a star graph (a non-complete star graph has a universal $K_{1}$ separator), $U$ cannot be a subset of $(V^{\prime}\setminus T)\cup\{w\}$. Therefore, $U$ must include some vertices from at least one copy of $W$, say $W_{uv}$. Since $w$ is universal to $W_{uv}$ and $w$ is not a universal vertex in $H$, we obtain that $W_{uv}\cup\{w\}$ does not induce an $H$. Since $T$ is a vertex cover, either $u$ or $v$ is in $T$. Assume that both are in $T$. Then $U$ must contain at least one vertex not in $W_{uv}\cup\{w\}$ and then $w$ is a universal $K_{1}$ separator in the $H$ induced by $U$. Assume that $u\in T$ and $v\notin T$. By Lemma 1, none of the subsets of $W_{uv}\cup\{v,w\}$ induces an $H$. Let $W_{v}$ denote the union of all $W_{u^{\prime}v}$ (for every edge $u^{\prime}v$ in $G^{\prime}$). If $U$ contains at least one vertex not in $\{v,w\}\cup W_{v}$, then $w$ is a universal $K_{1}$ separator in the $H$ induced by $U$. If $U$ does not contain any such vertex, then $U$ must contain a vertex from $W_{u^{\prime}v}$ for some $u^{\prime}\neq u$. Then either $w$ or $v$ is a universal $K_{1}$ separator or $wv$ is a universal $K_{2}$ separator in the $H$ induced by $U$. Thus, in all these cases, we obtain contradictions.

Lemma 3

Let $(G,k)$ be a yes-instance of $H$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of Vertex Cover.

Proof

Let $F$ be a set of edges of $G$ such that $|F|\leq k$ and $G/F$ is $H$-free. We create a vertex cover $T$ of $G^{\prime}$ as follows.

• •

For every vertex $u$ of $G^{\prime}$, if $wu$ is in $F$, then include $u$ in $T$.

• •

For every edge $uv$ of $G^{\prime}$, if there is an edge $uv^{\prime}\in F$, where $v^{\prime}\in W_{uv}$, then include $u$ in $T$.

• •

For every edge $uv$ of $G^{\prime}$, if there is an edge $u^{\prime}v^{\prime}\in F$, or an edge $wu^{\prime}\in F$, where $u^{\prime},v^{\prime}\in W_{uv}$, include either $u$ or $v$ in $T$ arbitrarily.

Clearly, $|T|\leq|F|\leq k$. Now we prove that $T$ is a vertex cover of $G^{\prime}$. We note that $W_{uv}\cup\{w\}$ induces an $H$ in $G$ for every edge $uv$ in $G^{\prime}$. Therefore, a vertex in $W_{uv}\cup\{u,v\}$ must be touched by $F$ (the only neighbors of $W_{uv}$ are $u,v$, and $w$). Therefore, by the construction of $T$, either $u$ or $v$ is in $T$. Hence $T$ is a vertex cover of $G^{\prime}$.

NP-completeness of Vertex Cover and Lemmas 2, 3 prove Lemma 4.

Lemma 4

Let $H$ be a connected non-complete graph with neither a universal $K_{1}$ separator nor a universal $K_{2}$ separator. Then $H$-free Contraction is NP-complete.

3.2 Graphs with universal clique separators

In this section, we handle the graphs $H$ with either a universal $K_{1}$ separator (except stars) or with a universal $K_{2}$ separator. We note that $H$ cannot have both a universal $K_{1}$ separator and a universal $K_{2}$ separator. Further $H$ cannot have more than one such separator.

Construction 2

Let $G^{\prime},H$ be any graphs and let $V^{\prime}$ be any subset of vertices of $H$. Let $b,c,k$ be positive integers. We obtain a graph $G$ from $(G^{\prime},H,V^{\prime},b,c,k)$ as follows. For every set $S$ of vertices of $G^{\prime}$, where $S$ induces a clique on $b$ vertices in $G^{\prime}$, do the following: Introduce $k+c$ copies of $H[V^{\prime}]$ and make every vertex of the copies adjacent to every vertex of $S$. Let $W_{S}$ denote the set of new vertices introduced for $S$, and let $W$ be the set of all new vertices.

Let $H$ be a graph with a universal $K_{1}$ separator or a universal $K_{2}$ separator. Let the set of vertices of the separator be denoted by $K$. Assume that there exists at least two non-isomorphic components in $H-K$ (therefore, $H$ cannot be a star). Let $J$ be a component in $H-K$ with minimum number of vertices. Let $c$ be the number of times $J$ appears (as a component) in $H-K$. Let $V^{\prime}$ be the set of vertices of a copy of $J$. Let $H^{\prime}$ be the graph obtained from $H$ by removing the vertices of every component isomorphic to $J$ in $H-K$. Let $(G^{\prime},k)$ be an instance of $H^{\prime}$-free Contraction. Let $G$ be obtained from $(G^{\prime},H,V^{\prime},b=|K|,c,k)$ by Construction 2. An example of the construction is shown in Figure 2.

Lemma 5

Let $(G^{\prime},k)$ be a yes-instance of $H^{\prime}$-free Contraction. Then $(G,k)$ is a yes-instance of $H$-free Contraction.

Proof

Let $\mathcal{P}^{\prime}$ be a partition of vertices of $G^{\prime}$ such that each set induces a connected graph, cost(\mathcal{P}^{\prime})$$\leq k, and $G/\mathcal{P}^{\prime}$ is $H^{\prime}$-free. Let $\mathcal{P}$ be obtained from $\mathcal{P}^{\prime}$ by adding the singleton sets corresponding to vertices in $W$. Let $\mathcal{P}_{W}$ be the singleton sets corresponding to vertices in $W$, i.e., $\mathcal{P}=\mathcal{P}^{\prime}\cup\mathcal{P}_{W}$. Clearly, cost(\mathcal{P})$$=cost(\mathcal{P}^{\prime})$$\leq k. We claim that $G/\mathcal{P}$ is $H$-free. For a contradiction, assume that $\mathcal{U}\subseteq\mathcal{P}$ induces an $H$ in $G/\mathcal{P}$. Since there are at least two components in $H-K$, we obtain that degree of a vertex in $J$ is strictly less than that of a vertex in $K$ in $H$. Therefore, None of the sets in $\mathcal{P}_{W}$ can act as a universal vertex (a vertex in $K$) in the $H$ induced by $\mathcal{U}$. Assume that $|K|=1$. Let $P_{u}\in\mathcal{P^{\prime}}$ be the universal vertex of the induced $H$. Since $J$ is a component in $H-K$ with minimum number of vertices, if a singleton set corresponding to a vertex in $W_{\{u\}}$ (for any vertex $u\in P_{u}$) is in $\mathcal{U}$, then each of the singleton sets corresponding to vertices in $W_{\{u\}}$ (forming a copy of $J$) is in $\mathcal{U}$. Among the sets in $P_{W}$, the set $P_{u}$ is adjacent to only the singleton sets corresponding to vertices in $W_{\{u\}}$ (for $u\in P_{u}$). Therefore, the singleton sets corresponding to $W_{v}$, for $v\notin P_{u}$, are not in $\mathcal{U}$. Therefore, there must be an induced copy of $H^{\prime}$ in $G^{\prime}/\mathcal{P}^{\prime}$, which is a contradiction. The case when $|K|=2$ can be proved in a similar way.

Lemma 6

Let $(G,k)$ be a yes-instance of $H$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of $H^{\prime}$-free Contraction.

Proof

Let $F$ be a subset of edges of $G$ such that $G/F$ is $H$-free and $|F|\leq k$. Let $\mathcal{P}$ be the partition of vertices of $G$ corresponding to $F$. We create a partition $\mathcal{P}^{\prime}$ from $\mathcal{P}$ as follows. For every $P\in\mathcal{P}$, add the set $P\setminus W$ to $\mathcal{P}^{\prime}$. Clearly, $\mathcal{P}^{\prime}$ is a partition of vertices of $G^{\prime}$ and each set in $\mathcal{P}^{\prime}$ induces a connected subgraph in $G^{\prime}$. Further, cost$(\mathcal{P}^{\prime})$ $\leq$ cost$(\mathcal{P})$ $\leq k$. Now, it is sufficient to prove that $G^{\prime}/\mathcal{P}^{\prime}$ is $H^{\prime}$-free. For a contradiction, assume that $G^{\prime}/\mathcal{P}^{\prime}$ has an $H^{\prime}$ induced by $\mathcal{U}\subseteq\mathcal{P}^{\prime}$.

Let $|K|=1$. Let $P^{\prime}\in\mathcal{U}$ be a set corresponding to the universal $K_{1}$ separator (of $H$) which is part of the induced $H^{\prime}$. Consider any vertex $u\in P^{\prime}$. Since there are $k+c$ copies of $J$ attached to $u$, at least $c$ of them are not touched by $\mathcal{P}$ (i.e., each vertex of those $c$ copies of $J$ forms a singleton set in $\mathcal{P}$). Therefore, the singleton sets in $\mathcal{P}$ corresponding the $c$ copies of $J$ along with the sets in $\mathcal{P}$ containing sets in $\mathcal{U}$ induce an $H$ in $G/\mathcal{P}$, which is a contradiction.

With similar arguments we can handle the case of $|K|=2$. Let $|K|=2$. Let $P_{u}^{\prime},P_{v}^{\prime}\in\mathcal{U}$ be two sets corresponding to the two vertices in the universal $K_{2}$ separator (of $H$), which is part of the induced $H^{\prime}$. Since $P_{u}^{\prime}$ and $P_{v}^{\prime}$ are adjacent, there exists a vertex $u\in P_{u}^{\prime}$ and a vertex $v\in P_{v}^{\prime}$ such that $uv$ is an edge in $G^{\prime}$. Since there are $k+c$ copies of $J$ attached to $uv$, at least $c$ of them are not touched by $\mathcal{P}$ (i.e., each vertex of those $c$ copies of $J$ forms a singleton set in $\mathcal{P}$). Therefore, the singleton sets in $\mathcal{P}$ corresponding the $c$ copies of $J$ along with the sets in $\mathcal{P}$ containing sets in $\mathcal{U}$ induce an $H$ in $G/\mathcal{P}$, which is a contradiction.

We obtain Lemma 7 from Lemmas 5 and 6.

Lemma 7

Let $H$ be a graph with a universal $K_{1}$ separator or a universal $K_{2}$ separator, denoted by $K$. Assume that $H-K$ has at least two components which are not isomorphic. Let $J$ be a component in $H-K$ with minimum number of vertices. Let $H^{\prime}$ be obtained from $H$ by removing all components of $H-K$ isomorphic to $J$. Then there is a polynomial-time reduction from $H^{\prime}$-free Contraction to $H$-free Contraction.

What remains to handle is the case when $H$ has a universal $K_{1}$ separator or a universal $K_{2}$ separator $K$ such that $H-K$ is a disjoint union of a graph $J$. The diamond graph is an example. For this we need the concept of an enforcer - a structure to forbid contraction of certain edges. Enforcers are used widely in connection with proving hardness results for edge modification problems (see [7, 15, 22]).

We define an enforcer of a graph $H$ as a graph $H_{E}$ with two specified adjacent vertices $v,w$ such that the following hold true for every instance $(G,k)$ of $H$-free Contraction and for every edge $ab$ of $G$: $(\hat{G},k)$ is a yes-instance of $H$-free Contraction if and only if there is a partition $\mathcal{P}$ of vertices of $G$ (where each set in $\mathcal{P}$ induces a connected subgraph of $G$) such that $G/\mathcal{P}$ is $H$-free, cost(\mathcal{P})$$\leq k, and $a$ and $b$ are in different sets in $\mathcal{P}$. Here, $\hat{G}$ is the graph obtained from $G$ by attaching $k+1$ copies of $H_{E}$ by identifying $a$ with $v$, and $b$ with $w$ for each copy of $H_{E}$.

Now, we will handle the case in which $H-K$ is the disjoint union of $t\geq 2$ copies of a connected graph $J$. Assume that $H$ is not a star graph. Let $u$ be a vertex in $K$ and $v$ be a vertex in $H-K$. Clearly, $uv$ is an edge. Subdivide $uv$ by introducing a new vertex $w$, making it adjacent to both $u$ and $v$, and deleting the edge $uv$. We note that contracting the edge $vw$ in $H_{E}$ creates $H$. We will prove that $H_{E}$ along with $v,w$ is an enforcer for $H$.

Let $\hat{G}$ be obtained from $(G,a,b,H_{E},v,w,k)$ as described above. An example is shown in Figure 3. Let $S_{1},S_{2},\ldots,S_{k+1}$ denote the set of vertices of $k+1$ copies of $H_{E}$, except $\{v,w\}$. Let $S$ be the set of all new vertices, i.e., $S=\bigcup_{i=1}^{i=k+1}S_{i}$. Let the number of vertices in $H$ be $h$.

Lemma 8

Let $\mathcal{\hat{P}}$ be a partition of vertices of $\hat{G}$ (where each set in $\mathcal{\hat{P}}$ induces a connected subgraph in $\hat{G}$) such that cost(\mathcal{\hat{P}})$$\leq k, and $\hat{G}/\mathcal{\hat{P}}$ is $H$-free. Let $P_{a}$ and $P_{b}$ denote the sets in $\mathcal{\hat{P}}$ containing $a$ and $b$ respectively. Then $P_{a}\neq P_{b}$.

Proof

For a contradiction, assume that $P_{a}=P_{b}$. There are $k+1$ copies of $H$ in $\hat{G}/\{P_{a}\}$ where all the copies are attached to $P_{a}=P_{b}$. This implies that all of them cannot be destroyed by $\mathcal{\hat{P}}$. Then there exists a set $S_{i}$ such that every vertex in $S_{i}$ is a singleton set in $\mathcal{\hat{P}}$. Then those sets and $P_{a}$ induce an $H$ in $\hat{G}/\mathcal{\hat{P}}$, which is a contradiction.

Lemma 9

Let $\mathcal{P}$ be a partition of vertices of $G$ (where each set in $\mathcal{P}$ induces a connected subgraph in $G$) such that $a$ and $b$ are not in the same set in $\mathcal{P}$, and $G/\mathcal{P}$ is $H$-free. Let $\mathcal{\hat{P}}$ be obtained from $\mathcal{P}$ by adding singleton sets corresponding to vertices in $S$. Then $\hat{G}/\mathcal{\hat{P}}$ is $H$-free.

Proof

For a contradiction, assume that $\mathcal{U}\subseteq\mathcal{\hat{P}}$ induces an $H$ in $\hat{G}/\mathcal{\hat{P}}$. Clearly, none of the subsets of $\mathcal{P}$ induces an $H$ in $\hat{G}$. Hence at least one set in $\mathcal{\hat{P}}\setminus\mathcal{P}$ is part of $\mathcal{U}$, let it be from $S_{i}$. Let $\mathcal{P}_{i}$ be the subset of singleton sets in $\mathcal{\hat{P}}$ corresponding to $S_{i}$. Also assume that $P_{a}$ and $P_{b}$ be the sets in $\mathcal{\hat{P}}$ containing $a$ and $b$ respectively. Clearly, $\mathcal{P}_{i}\cup\{P_{a},P_{b}\}$ induces an $H_{E}$ and not an $H$. Therefore, $\mathcal{U}$ contains sets corresponding to $S_{j}$ ($j\neq i$) or from $\mathcal{P}\setminus\{P_{a},P_{b}\}$. Therefore, either $P_{a}$ or $P_{b}$ acts as a universal vertex in the $H$ induced by $\mathcal{U}$. Recall that $b$ is identified with $w$, a degree two vertex in $H_{E}$, whose neighbors are not adjacent to each other (one of them is $v$ which is identified with $a$). Therefore, if $P_{b}$ acts as a universal vertex, then $J$ can have only one vertex. This is a contradiction as $H$ is not a star graph. Recall that $a$ is identified with a vertex $v$ of $H_{E}$ which is adjacent to at most $|V(J)|$ vertices including $w$, where $w$ is not adjacent to any neighbors of $v$. Therefore, if $v$ acts as a universal vertex, $J$ can have only one vertex, which is a contradiction as $H$ is not a star graph.

Lemmas 8 and 9 imply that $(H_{E},v,w)$ is an enforcer for $H$.

Lemma 10

$(H_{E},v,w)$* is an enforcer for $H$.*

Proof

We need to prove that $(\hat{G},k)$ is a yes-instance of $H$-free Contraction if and only if there is a partition $\mathcal{P}$ of vertices of $G$ (where each set in $\mathcal{P}$ induces a connected subgraph of $G$) such that $G/\mathcal{P}$ is $H$-free, cost(\mathcal{P})$$\leq k, and $a$ and $b$ are in different sets in $\mathcal{P}$. Lemma 9 proves the backward direction, i.e., if there is a partition $\mathcal{P}$ as specified, then $(\hat{G},k)$ is a yes-instance. For the forward direction, assume that $(\hat{G},k)$ is a yes-instance. Let $\mathcal{\hat{P}}$ be a partition of vertices of $\hat{G}$ such that cost(\mathcal{\hat{P}})$$\leq k and $\hat{G}/\mathcal{\hat{P}}$ is $H$-free. Let $P_{a}$ and $P_{b}$ be the sets in $\mathcal{\hat{P}}$ containing $a$ and $b$ respectively. By Lemma 8, $P_{a}\neq P_{b}$. Create $\mathcal{P}$ from $\mathcal{\hat{P}}$ as follows: For every $\hat{P}\in\mathcal{\hat{P}}$, include $\hat{P}\setminus S$ in $\mathcal{P}$. It is straight-forward to verify that $G/\mathcal{P}$ is $H$-free. Clearly, cost(\mathcal{P})$$\leqcost(\mathcal{\hat{P}})$$\leq k. Therefore, $(H_{E},v,w)$ is an enforcer for $H$.

Now we will come up with a reduction from $H^{\prime}$-free Contraction to $H$-free Contraction, where $H^{\prime}$ is $H-K$, i.e., $H^{\prime}$ is isomorphic to $tJ$.

Construction 3

Let $G^{\prime},H_{E}$ be graphs and $k$ be an integer. Let $v,w$ be two vertices in $H_{E}$. We obtain a graph $G$ from $G^{\prime}$ by introducing two new adjacent vertices $y,z$ and making them adjacent to all vertices of $G^{\prime}$. Then we obtain a graph $\hat{G}$ from $G$ as follows:

• •

For every edge $xy$ (for $x\in V(G^{\prime})$), attach $k+1$ copies of $H_{E}$ by identifying $x$ with $v$ and $y$ with $w$.

• •

For every edge $xz$ (for $x\in V(G^{\prime})$), attach $k+1$ copies of $H_{E}$ by identifying $x$ with $v$ and $z$ with $w$.

• •

Attach $k+1$ copies of $H_{E}$ by identifying $y$ with $v$ and $z$ with $w$.

Recall that $H$ is a non-star graph with a universal $K_{1}$ separator or a universal $K_{2}$ separator, where the universal separator is denoted by $K$. Let $(G^{\prime},k)$ be an instance of $H^{\prime}$-free Contraction, where $H^{\prime}=H-K$. Let $(H_{E},v,w)$ be the enforcer we created for $H$ (Lemma 10). Let $G$ and $\hat{G}$ be the graphs obtained from $(G^{\prime},H_{E},v,w,k)$ by Construction 3.

Lemma 11

Let $(G^{\prime},k)$ be a yes-instance of $H^{\prime}$-free Contraction. Then $(\hat{G},k)$ is a yes-instance of $H$-free Contraction.

Proof

To prove that $(\hat{G},k)$ is a yes-instance of $H$-free Contraction, by Lemma 10, it is sufficient to prove that $(G,k)$ is a yes-instance with a solution $\mathcal{P}$ such that $x$ and $y$ are not in the same set in $\mathcal{P}$, $x$ and $z$ are not in the same set in $\mathcal{P}$ (for every $x\in V(G^{\prime})$), and $y$ and $z$ are not in the same set. Let $\mathcal{P}^{\prime}$ be a partition of vertices of $G^{\prime}$ such that $G^{\prime}/\mathcal{P}^{\prime}$ is $H^{\prime}$-free and cost(\mathcal{P}^{\prime})$$\leq k. We obtain $\mathcal{P}$, a partition of vertices of $G$, from $\mathcal{P}^{\prime}$ by including two singleton sets $P_{y}=\{y\}$ and $P_{z}=\{z\}$. We observe that if $G/\mathcal{P}$ has an induced $H$, then $P_{y}$ and $P_{z}$ can act only as universal vertices in the $H$ and hence $G^{\prime}/\mathcal{P}^{\prime}$ must contain an induced $H^{\prime}$, which is a contradiction. It is straight-forward to verify that $\mathcal{P}$ satisfies the rest of the required properties.

Lemma 12

Let $(\hat{G},k)$ be a yes-instance of $H$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of $H^{\prime}$-free Contraction.

Proof

Let $(\hat{G},k)$ be a yes-instance of $H$-free Contraction. Then Lemma 10 implies that there exists a partition $\mathcal{P}$ of vertices of $G$ such that $P_{x}\neq P_{y}$, $P_{x}\neq P_{z}$, and $P_{y}\neq P_{z}$, for every vertex $x\in V(G^{\prime})$, where $P_{x},P_{y},P_{z}$ denote the sets containing $x,y,z$ respectively. Further, $G/\mathcal{P}$ is $H$-free and cost(\mathcal{P})$$\leq k. This implies that $P_{y}=\{y\}$ and $P_{z}=\{z\}$. Let $\mathcal{P}^{\prime}=\mathcal{P}\setminus\{P_{y},P_{z}\}$. Clearly, cost(\mathcal{P}^{\prime})$$=cost(\mathcal{P})$$\leq k. We claim that $G^{\prime}/\mathcal{P}^{\prime}$ is $H$-free. For a contradiction, assume that $\mathcal{U}\subseteq\mathcal{P}^{\prime}$ induces an $H^{\prime}$ in $G^{\prime}/\mathcal{P}^{\prime}$. Then, we obtain that there is an $H$ induced in $G$ by $\mathcal{U}\cup\{P_{y}\}$, if $|K|=1$, or by $\mathcal{U}\cup\{P_{y},P_{z}\}$, if $|K|=2$. This gives us a contradiction.

Lemmas 11 and 12 prove Lemma 13.

Lemma 13

Let $H$ be a graph with a universal $K_{1}$ separator or a universal $K_{2}$ separator, denoted by $K$. Assume that $H-K$ is a disjoint union of $t$ copies of a graph $J$, for some $t\geq 2$. Let $H$ be not a star and let $H^{\prime}$ be $tJ$. Then there is a polynomial-time reduction from $H^{\prime}$-free Contraction to $H$-free Contraction.

3.3 Stars

Here, we prove that $H$-free Contraction is NP-complete whenever $H$ is a star graph of at least 4 vertices.

Construction 4

Let $G^{\prime}$ be a graph and $k,t$ be integers. We obtain a graph $G$ from $G^{\prime}$ as follows.

• •

Introduce four cliques $X,Y,X^{\prime},Y^{\prime}$ having $k+1$ vertices each. Make every vertex of $X\cup Y\cup X^{\prime}\cup Y^{\prime}$ adjacent to $V(G^{\prime})$. Similarly, make every vertex of $X$ adjacent to every vertex of $Y$, and every vertex of $X^{\prime}$ adjacent to every vertex of $Y^{\prime}$.

• •

*For every vertex $z\in X\cup Y\cup X^{\prime}\cup Y^{\prime}$, introduce $t-2$ cliques $Z_{1},Z_{2},\ldots,Z_{t-2}$ of $k+1$ vertices each, and make $z$ adjacent to every vertex of $Z_{z}=Z_{1}\cup Z_{2}\cup\ldots\cup Z_{t-2}$. Let $Z$ denote the union of all *$Z_{z}$s.

An example is given in Figure 4. For two integers $a,b\geq 1$, by $K_{1,a,b}$ we denote the graph obtained by introducing a vertex adjacent to all vertices of $a$ copies of $K_{b}$s. The following is a trivial observation.

Observation 2

Let $(G,k)$ be a yes-instance of $K_{1,t}$-free Contraction for any integer $t\geq 2$. Then $G$ does not have $K_{1,t,k+1}$ as an induced subgraph.

We give a reduction from $2K_{1}$-free Contraction to $K_{1,t}$-free Contraction, for every $t\geq 3$. For an instance $(G^{\prime},k)$ of $2K_{1}$-free Contraction, we obtain an instance $(G,k)$ of $K_{1,t}$-free Contraction by applying Construction 4 on $(G^{\prime},k,t)$.

Lemma 14

Let $(G^{\prime},k)$ be a yes-instance of $2K_{1}$-free Contraction. Then $(G,k)$ is a yes-instance of $K_{1,t}$-free Contraction.

Proof

Let $\mathcal{P}^{\prime}$ be a partition of vertices of $G^{\prime}$ such that cost(\mathcal{P}^{\prime})$$\leq k, $G^{\prime}/\mathcal{P}^{\prime}$ is a clique (i.e., $2K_{1}$-free), and every set in $\mathcal{P}^{\prime}$ induces a connected graph in $G^{\prime}$. Construct $\mathcal{P}$ from $\mathcal{P}^{\prime}$ by introducing singleton sets corresponding to the vertices $X\cup Y\cup X^{\prime}\cup Y^{\prime}\cup Z$. Clearly, cost(\mathcal{P})$$=cost(\mathcal{P}^{\prime})$$\leq k. It is straight-forward to verify that a maximum sized induced star graph in $G/\mathcal{P}$ is $K_{1,t-1}$. Therefore, $G/\mathcal{P}$ is $K_{1,t}$-free.

Lemma 15

Let $(G,k)$ be a yes-instance of $K_{1,t}$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of $2K_{1}$-free Contraction.

Proof

Let $\mathcal{P}$ be a partition of vertices of $G$ such that $G/\mathcal{P}$ is $K_{1,t}$-free and cost(\mathcal{P})$$\leq k. For a vertex $a$ in $G$, let $P_{a}$ denote the set containing $a$ in $\mathcal{P}$. First we claim that $P_{u}\neq P_{x}$, where $u\in V(G^{\prime}),x\in X$. If, on the contrary, $P_{u}=P_{x}$ for some $u\in V(G^{\prime})$, we observe that, in $G/P_{u}$, there is a $K_{1,t,k+1}$ induced by $P_{u}$ and the vertices in $Z_{x}\cup Y\cup X^{\prime}$. Therefore, by Observation 2, the claim is true. By symmetry, we obtain that $P_{u}\neq P_{y}$, $P_{u}\neq P_{x^{\prime}}$, and $P_{u}\neq P_{y^{\prime}}$ for vertices $y\in Y,x^{\prime}\in X^{\prime},y^{\prime}\in Y^{\prime}$. This implies that there is no set in $\mathcal{P}$ containing vertices from both $V(G^{\prime})$ and $X\cup Y\cup X^{\prime}\cup Y^{\prime}\cup Z$. Therefore, there is a subset $\mathcal{P}^{\prime}$ of $\mathcal{P}$ such that $\mathcal{P}^{\prime}$ is a partition of vertices of $G^{\prime}$ and every set in $\mathcal{P}^{\prime}$ induces a connected subgraph in $G^{\prime}$. Now, it is sufficient to prove that that $G^{\prime}/\mathcal{P}^{\prime}$ is a clique. For a contradiction assume that there is a $2K_{1}$ induced in $G^{\prime}/\mathcal{P}^{\prime}$ by $P_{u},P_{v}\in\mathcal{P}^{\prime}$. We observe that for every clique $K$ attached to $x\in X$, there is a partition $P$ in $\mathcal{P}$ such that $P\subseteq K$ (all the vertices in $K$ cannot be contracted to $x$ as the budget is only $k$ and $K$ has $k+1$ vertices). Therefore, for every vertex $x\in X$, $P_{x}$ is a center of a $K_{1,t-2}$ in $G/\mathcal{P}$, where the leaves are from the $t-2$ cliques attached to $x$. This along with the sets $P_{u},P_{v}$ induce a $K_{1,t}$ in $G/\mathcal{P}$, which is a contradiction.

Lemmas 14 and 15, and the NP-completeness of $2K_{1}$-free Contraction (Proposition 1) prove Lemma 16.

Lemma 16

For $t\geq 3$, $K_{1,t}$-free Contraction is NP-complete.

3.4 Small graphs

In this section, we prove that $H$-free Contraction is NP-complete when $H$ is either a $K_{2}+K_{1}$ or a $2K_{2}$. First we handle the case of $2K_{2}$ with a simple reduction from $2K_{1}$-free Contraction.

Construction 5

Let $G^{\prime}$ be a graph and $k$ be an integer. We obtain a graph $G$ by attaching $k+1$ pendant vertices, denoted by a set $Z_{u}$, to every vertex $u$ in $G^{\prime}$. Let $Z$ denote the set of all newly added vertices.

Let $(G^{\prime},k)$ be an instance of $2K_{1}$-free Contraction. We obtain an instance $(G,k)$ of $2K_{2}$-free Contraction by applying Construction 5 on $(G^{\prime},k)$.

Lemma 17

Let $(G^{\prime},k)$ be a yes-instance of $2K_{1}$-free Contraction. Then $(G,k)$ is a yes-instance of $2K_{2}$-free Contraction.

Proof

Let $\mathcal{P}^{\prime}$ be a partition of vertices of $G^{\prime}$ such that cost(\mathcal{P}^{\prime})$$\leq k and $G^{\prime}/\mathcal{P}^{\prime}$ is $H$-free. We obtain a partition $\mathcal{P}$ of vertices of $G$ from $\mathcal{P}^{\prime}$ by introducing singleton sets corresponding to the vertices in $Z$. Clearly, cost(\mathcal{P})$$=cost(\mathcal{P}^{\prime})$$\leq k. Since there is no edge induced by the sets corresponds to vertices in $Z$, we obtain that if $G/\mathcal{P}$ is not $2K_{2}$-free, then there is an induced $2K_{1}$ in $G^{\prime}/\mathcal{P}^{\prime}$, which is a contradiction.

Lemma 18

Let $(G,k)$ be a yes-instance of $2K_{2}$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of $2K_{1}$-free Contraction.

Proof

Let $\mathcal{P}$ be a partition of vertices of $G$ such that $G/\mathcal{P}$ is $2K_{2}$-free and cost(\mathcal{P})$$\leq k. We obtain a partition $\mathcal{P}^{\prime}$ of vertices of $G^{\prime}$ as follows: For every set $P\in\mathcal{P}$, include $P\setminus Z$ in $\mathcal{P}^{\prime}$. Since $P$ induces a connected graph in $G$, $P\setminus Z$ induces a connected graph in $G^{\prime}$. Assume that there is a $2K_{1}$ induced by $P^{\prime}_{u},P^{\prime}_{v}\in\mathcal{P}^{\prime}$. Let $u\in P^{\prime}_{u}$ and $v\in P^{\prime}_{v}$. Since there is a set $Z_{u}$ of $k+1$ pendant vertices attached to $u$ and a set $Z_{v}$ of $k+1$ pendant vertices attached to $v$, at least one vertex from $Z_{u}$ and at least one vertex from $Z_{v}$ form singleton sets in $\mathcal{P}$. Then, those two sets along with the sets containing $P^{\prime}_{u}$ and the set containing $P^{\prime}_{v}$ in $\mathcal{P}$ induces a $2K_{2}$ in $G/\mathcal{P}$, which is a contradiction.

Now, the NP-completeness of $2K_{2}$-free Contraction follows from that of $2K_{1}$-free Contraction (Proposition 1) and Lemmas 17 and 18.

Lemma 19

$2K_{2}$-free Contraction* is NP-complete.*

Next we prove the hardness for $K_{2}+K_{1}$-free Contraction by a reduction from Domatic Number. Domatic number of a graph is the size of a largest set of disjoint dominating sets of the graph, which partitions the vertices of the graph. For example, the domatic number of a complete graph is the number of vertices of it, and that of a star graph is 2. The Domatic Number problem is to find whether the domatic number of the input graph is at least $k$ or not. It is known that Domatic Number is NP-complete [11] even for various classes of graphs [4, 18]. Recall that $K_{2}+K_{1}$-free graphs are exactly the class of complete multipartite graphs. The reduction that we use is exactly the same as the reduction for the NP-completeness of Hadwiger number problem (which is equivalent to $2K_{1}$-free Contraction) described by Eppstein [10]. The proof requires some adaptation.

Let $(G^{\prime},d)$ be an instance of Domatic Number such that $G^{\prime}$ has $n\geq 2$ vertices. Let $\{v_{1},v_{2},\ldots,v_{n}\}$ be the vertices of $G^{\prime}$. We can safely assume that there are no universal vertices in $G^{\prime}$ (otherwise, remove it and decrease $d$ by 1 to obtain an equivalent instance). We obtain a graph $G$ as follows:

• •

Introduce a clique $U=\{u_{1},u_{2},\ldots,u_{d}\}$ of size $d$, called the upper vertices.

• •

Introduce an independent set $M=\{m_{1},m_{2},\ldots,m_{n}\}$ of size $n$, called the middle vertices.

• •

Introduce a clique $L=L_{1}\cup L_{2}\cup\ldots\cup L_{n}$ of $n(n+1)$ vertices, where $L_{i}=\{\ell_{i,1},\ell_{i,2},\ldots,\ell_{i,n+1}\}$ is a clique of $n+1$ vertices, for $1\leq i\leq n$. These vertices are called the lower vertices.

• •

Make every vertex in $U$ adjacent to every vertex in $M$.

• •

The vertex $m_{i}$ is adjacent to (every vertex of) $L_{j}$ if and only if $v_{i}$ dominates $v_{j}$, i.e., either $i=j$ or $v_{i}$ is adjacent to $v_{j}$ in $G^{\prime}$.

Let $N$ be the number of vertices in $G$, i.e., $N=n(n+1)+n+d$. Let $h=n(n+1)+d$. This completes the reduction. An example is given in Figure 5.

The idea behind Eppstein’s reduction is the following: If $G^{\prime}$ has $d$ disjoint dominating sets, then a partition of vertices of $G$ can be obtained by the subsets of $M$ corresponding to the dominating sets of $G^{\prime}$ (subsets of $M$ is an independent set, so take one vertex from $U$ to make sure that each set induces a connected subgraph). On the other hand, if $(G,h)$ is a yes-instance of $2K_{1}$-free Contraction, then there exists a partition $\mathcal{P}$ (which is a solution) such that each subset of $M$ contained in $P_{u}$ (the set containing a vertex $u\in U$ in the partition) implies a dominating set in $G^{\prime}$. It turns out that, the largest complete multipartite graph that can be obtained from $G$ by contractions is a complete graph.

Lemma 20

If the domatic number of $G^{\prime}$ is at least $d$, then there is a partition $\mathcal{P}$ of vertices of $G$ such that every set in $\mathcal{P}$ induces a connected graph in $G$ and $G/\mathcal{P}$ is a complete multipartite graph of at least $h$ vertices.

Proof

Assume that $\{D_{1},D_{2},\ldots,D_{d}\}$ is a set of $d$ mutually disjoint dominating sets of $G^{\prime}$. We define a partition $\mathcal{P}$ of vertices of $G$ as follows:

• •

Let $M_{i}$ be the subset of vertices of $M$ corresponding to $D_{i}$, i.e., $M_{i}=\{m_{j}:v_{j}\in D_{i}\}$. Form a set $M_{i}\cup\{u_{i}\}$, for $1\leq i\leq d$.

• •

Every vertex in $L$ forms a singleton set in $\mathcal{P}$.

It is straight-forward to verify that there are $h$ sets in $\mathcal{P}$, each set induces a connected graph in $G$, and $G/\mathcal{P}$ is a complete graph.

Let $\mathcal{P}$ be a partition of vertices of $G$ such that each set in $\mathcal{P}$ forms a connected subgraph and $G/\mathcal{P}$ is a complete multipartite graph. Further assume that $|\mathcal{P}|\geq h$. Observations 3, 4, 5, and Lemma 21 are essentially taken from [10]. Observation 3 follows from the fact that cost(\mathcal{P})$$+|\mathcal{P}|=N.

Observation 3

cost$(\mathcal{P})$ $\leq n$.

Observation 4

For each $1\leq i\leq n$, there is a set $L_{i}^{\prime}\in\mathcal{P}$ such that $L_{i}^{\prime}\subseteq L_{i}$.

Proof

Assume that for every vertex $\ell_{i,j}$ in $L_{i}$, the set in $\mathcal{P}$ containing $\ell_{i,j}$ contains at least one vertex from outside $L_{i}$. Then cost$(\mathcal{P})$ is at least $n+1$ as there are $n+1$ vertices in $L_{i}$, which contradicts Observation 3.

Observations 5, 6, and Lemma 21 assumes that $G/\mathcal{P}$ is a clique.

Observation 5

$\mathcal{P}$* cannot contain a singleton subset of $M$.*

Proof

Since there are no universal vertices in $G^{\prime}$, the degree of a vertex in $M$ is at most $(n-1)(n+1)+d=n(n+1)+d-(n+1)<h-1$. Since $G/\mathcal{P}$ is a clique of size at least $h$, we obtain that $\mathcal{P}$ cannot contain a singleton subset of $M$.

Observation 6 follows from Observation 4 and the assumption that $G/\mathcal{P}$ is a clique.

Observation 6

If $P\in\mathcal{P}$ contains a vertex in $U$, then it contains a vertex from $M$.

Lemma 21

$G^{\prime}$* has a set of at least $d$ mutually disjoint dominating sets.*

Proof

By Observation 3, cost$(\mathcal{P})$ $\leq n$. Let $\mathcal{P}_{M}$ be the subset of $\mathcal{P}$ such that each set in $\mathcal{P}_{M}$ contains at least one vertex from $M$. By Observation 5, each set $P\in\mathcal{P}_{M}$ contains at least two vertices. Since $M$ is an independent set and $P$ induces a connected subgraph of $G$, we obtain that $P$ is not a subset of $M$. Therefore, cost($\mathcal{P}_{M}$)$=n$. This implies that $P$ contains exactly one vertex outside $M$. By Observation 6, every set $P_{i}$ containing a vertex $u_{i}$ has at least one vertex from $M$. As a set containing a vertex from $M$ contains exactly one vertex outside $M$, we obtain that $P_{i}$ does not contain any vertex from $L$. Now we obtain $d$ mutually disjoint dominating sets of $G^{\prime}$ as follows. For every $P_{i}$, $D_{i}$ is the vertices in $G^{\prime}$ corresponding to $P_{i}\cap M$, i.e., $D_{i}=\{v_{j}|m_{j}\in P_{i}\}$. It is sufficient to prove that $D_{i}$ is a dominating set of $G^{\prime}$. Assume that a vertex $v_{\ell}$ is not dominated by $D_{i}$ in $G^{\prime}$. By Observation 4, there is a set $L_{i}^{\prime}\in\mathcal{P}$ such that $L_{i}^{\prime}\subseteq L_{i}$. Then, $P_{i}$ is not adjacent to that set, which is a contradiction.

Lemma 22

Assume that there is a partition $\mathcal{P}$ of vertices of $G$ into sets inducing connected subgraphs such that $G/\mathcal{P}$ is a complete multipartite graph $Q$ on at least $h$ vertices. Then $Q$ is a clique.

Proof

For a contradiction, assume that $Q$ is not a clique. Let $r,s$ be two nonadjacent vertices of $Q$. Let $R,S$ be the sets in $\mathcal{P}$ containing $r,s$ respectively.

Claim 1: $R$ cannot be a subset of $U$.

Proof of Claim 1: Assume that $R\subseteq U$. Clearly, $S$ cannot contain any vertex from $U\cup M$. Therefore, $S\subseteq L$. Without loss of generality, assume that $S$ contains at least one vertex, say $l_{i,j}$, from $L_{i}$. Then every other vertex in $L_{i}$ contributes towards the cost of $\mathcal{P}$. For example, if $l_{i,q}$ ($q\neq j$) is part of $S$, then it contributes toward the cost of $\mathcal{P}$. If $l_{i,q}$ is not part of $S$, then the set containing $l_{i,q}$ must contain some other vertices from outside $L_{i}$ to make sure that it is adjacent to $R$ (otherwise there is a $K_{1}+K_{2}$ induced by $R,S,$ and the set containing $l_{i,q}$). Since cost$(\mathcal{P})$ is at most $n$ (Observation 3), this implies that sets in $\mathcal{P}$ containing vertices from $L_{i}\setminus\{\ell_{i,j}\}$ are the only sets containing at least two vertices and all other sets are singleton sets. Further, each set $P\in\mathcal{P}$ containing at least one vertex from $L_{i}\setminus\{\ell_{i,j}\}$ contains exactly one vertex outside $L_{i}$ and that vertex must be from $M$ (to make sure that $P$ is adjacent to $R$ - otherwise there is a $K_{2}+K_{1}$ induced by $R,S$, and $P$). This implies that every vertex in $L\setminus L_{i}$ forms a singleton set in $\mathcal{P}$. Then any such set (recall that $n\geq 2$) along with $R$ and $S$ induces a $K_{2}+K_{1}$, which is a contradiction.

Claim 2: It cannot happen that $R$ contains a middle vertex and $S$ contains a lower vertex.

Proof of Claim 2: For a contradiction, assume that $R$ contains a middle vertex $m$ and $S$ contains a lower vertex $l_{i,j}\in L_{i}$. Then every other vertex in $L_{i}$ contributes toward the cost of $\mathcal{P}$ (as obtained in the proof of Claim 1). Therefore, $L_{i}$ contributes $n$ toward the cost$(\mathcal{P})$. This implies that the sets $P\in\mathcal{P}$ containing at least one vertex from $L_{i}\setminus\{\ell_{i,j}\}$ are the only sets containing at least two vertices and all other sets in $\mathcal{P}$ are singleton sets. Further, each such set $P$ can contain only one vertex outside $L_{i}$ - clearly, that extra vertex cannot be from $U$. Then $S$ is a singleton set and every vertex in $U$ forms a singleton set in $\mathcal{P}$. Then there is a $K_{2}+K_{1}$ induced in $Q$ by $R,S,$ and one singleton set formed by a vertex in $U$, which is a contradiction.

Claim 1 and Claim 2 implies that both $R$ and $S$ are subsets of $M$. Since both $R$ and $S$ must induce connected subgraph of $G$, we obtain that both $R$ and $S$ are singleton sets. Let $R=\{m\}$. Let $L_{i}$ be nonadjacent to $m$ (this is guaranteed as there are no universal vertices in $G^{\prime}$). By Claim 2, a set $P\in\mathcal{P}$ containing at least one vertex from $L_{i}$ must be adjacent to $m$. Then every vertex in $L_{i}$ contribute toward cost$(\mathcal{P})$. This is a contradiction as $|L_{i}|=n+1$ and cost$(\mathcal{P})$ $\leq n$.

Lemmas 20, 21, 22 imply that there is a polynomial-time reduction from Domatic Number to $K_{2}+K_{1}$-free Contraction.

Lemma 23

$K_{2}+K_{1}$-free Contraction* is NP-complete.*

3.5 Putting them together

Recall that the reduction from Vertex Cover in Section 3.1 does not handle disconnected graph. This is the main ingredient that remains to be added to obtain the main result of the section. This turns out to be easy. Guo [15] has a reduction for transferring the hardness of $H^{\prime}$-free Contraction to $H$-free Contraction, where $H^{\prime}$ is any component of $H$.

Proposition 2 ([15])

Let $H$ be a disconnected graph. Let $H^{\prime}$ be any component of it. Then there is a polynomial-time reduction from $H^{\prime}$-free Contraction to $H$-free Contraction.

Proposition 2 does not help us to prove the hardness when every component of $H$ is either a $K_{1}$ or a $K_{2}$. But there are simple reductions to handle them.

Lemma 24

Let $H$ be a disconnected graph with an isolated vertex $v$. There is a polynomial-time reduction from $(H-v)$-free Contraction to $H$-free Contraction.

Proof

It is straight-forward to verify that $(G^{\prime},k)$ is a yes-instance of $(H-v)$-free Contraction if and only if $(G^{\prime}+K_{1},k)$ is a yes-instance of $H$-free Contraction.

Repeated application of Lemma 24 implies that there is a polynomial-time reduction from $2K_{1}$-free Contraction to $tK_{1}$-free Contraction, for every $t\geq 3$. Then the NP-Completeness of $2K_{1}$-free Contraction (Proposition 1) implies Lemma 25.

Lemma 25

For every $t\geq 3$, $tK_{1}$-free Contraction is NP-complete.

Now, we handle the case when $H$ is a disjoint union of $t$ copies of $K_{2}$.

Lemma 26

Let $H=tK_{2}$, for any integer $t\geq 3$ and let $H^{\prime}$ be $(t-1)K_{2}$. There is a polynomial-time reduction from $H^{\prime}$-free Contraction to $H$-free Contraction.

Proof

It is straight-forward to verify that $(G^{\prime},k)$ is a yes-instance of $H^{\prime}$-free Contraction if and only if $(G^{\prime}+K_{k+2},k)$ is a yes-instance of $H$-free Contraction.

Repeated application of Lemma 26 and the NP-completeness of $2K_{2}$-free Contraction (Lemma 19) give us the following Lemma.

Lemma 27

For every $t\geq 3$, $tK_{2}$-free Contraction is NP-complete.

Now we are ready to prove the main result of this section.

Theorem 3.1

Let $H$ be any graph other than $K_{1}$ and $K_{2}$. Then $H$-free Contraction is NP-complete.

Proof

We prove this by induction on $n$, the number of vertices of $H$. The base cases are when $n=2$ and $n=3$, i.e., when $H$ is $2K_{1}$ or $P_{3}$ or triangle (Proposition 1), or $3K_{1}$ (Lemma 25), or $K_{2}+K_{1}$ (Lemma 23). Assume that $n\geq 4$.

Let $H$ be a disconnected graph. Assume that $H$ has a component $H^{\prime}$ with at least three vertices. By Proposition 2, there is a polynomial-time reduction from $H^{\prime}$-free Contraction to $H$-free Contraction. Then we are done by induction hypothesis. Assume that every component of $H$ is either a $K_{2}$ or a $K_{1}$. If $H$ has an isolated vertex, then we are done by Lemma 24. If there are no isolated vertex in $H$, then $H$ is isomorphic to $tK_{2}$, for $t\geq 2$. Then we are done by Lemma 27.

Let $H$ be a connected graph. If $H$ is complete, then Proposition 1 is sufficient. Assume that $H$ is non-complete. Therefore, there is a non-universal vertex in $H$. Assume that $H$ has neither a universal $K_{1}$ separator nor a universal $K_{2}$ separator. Then we are done by Lemma 4. Assume that $H$ has either a universal $K_{1}$ separator or a universal $K_{2}$ separator, denoted by $K$. Further assume that $H$ is not a star. Let $H-K$ has at least two non-isomorphic components. Let $J$ be any component in $H-K$ with least number of vertices. Let $H^{\prime}$ be obtained from $H$ by removing all copies of $J$ in $H-K$. Then by Lemma 7, there is a polynomial-time reduction from $H^{\prime}$-free Contraction to $H$-free Contraction, and we are done. Assume that $H-K$ is disjoint union of $t$ copies of a graph $J$, for some $t\geq 2$. Then Lemma 13 gives us a polynomial-time reduction from $tJ$-free Contraction to $H$-free Contraction. For the last case, assume that $H$ is a star graph. Then we are done by Lemma 16.

4 W[2]-hardness

In this section, we prove that $H$-free Contraction is W[2]-hard whenever $H$ is a tree, except for 7 trees. All our reductions are from Dominating Set, which is well-known to be a W[2]-hard problem. First we obtain a reduction for all trees which are neither stars nor bistars. Then we come up with a reduction for a subset of bistars, a corner case of the same proves the case of stars. Then we come up with a reduction which handles the remaining bistars.

Recall that a dominating set $D$ of a graph $G$ is a subset of vertices of $G$ such that every vertex of $G$ is either in $D$ or adjacent to a vertex in $D$. The objective of the Dominating Set problem is to check whether a graph has a dominating set of size at most $k$ or not.

4.1 A general reduction for trees

Here we handle all trees which are neither stars nor bistars. The reduction that we use is an adapted version of a reduction used in [8] (to handle 3-connected graphs) and a reduction used in [20] (to handle cycles).

Construction 6

Let $G^{\prime},H$ be graphs and $k$ be an integer. Let $\{v_{1},v_{2},\ldots,v_{n}\}$ be the set of vertices of $G^{\prime}$. We construct a graph $G$ from $(G^{\prime},H,k)$ as follows.

• •

Introduce a clique $X=\{x_{1},x_{2},\ldots,x_{n}\}$.

• •

Introduce $n$ copies of $H$ denoted by $H_{1},H_{2},\ldots,H_{n}$. Let $V_{i}$ denote the set of vertices of $H_{i}$, for $1\leq i\leq n$.

• •

Let $w$ be any vertex in $H$. Identify $w$’s of all copies of $H$. Let the vertex obtained so be denoted by $w$. Let the remaining vertices in each copy $H_{i}$ be denoted by $W_{i}$, i.e., $W_{i}\cup\{w\}$ induces $H$, for $1\leq i\leq n$.

• •

Make $w$ adjacent to every vertex in $X$.

• •

Make $x_{i}$ adjacent to every vertex in $W_{j}$ if and only if $i=j$ or $v_{i}$ is adjacent to $v_{j}$ in $G^{\prime}$.

This completes the construction. An example is shown in Figure 6.

Let $(G^{\prime},k)$ be an instance of Dominating Set. Let $H$ be a tree which is neither a star nor a bistar. We obtain a graph $G$ from $(G^{\prime},H,k)$ by applying Construction 6.

Lemma 28

Let $(G^{\prime},k)$ be a yes-instance of Dominating Set. Then $(G,k)$ is a yes-instance of $H$-free Contraction.

Proof

Let $D$ be a dominating set of size at most $k$ of $G^{\prime}$. Let $F=\{wx_{i}|v_{i}\in D\}$. Clearly, $|F|=|D|\leq k$. We claim that $G/F$ is $H$-free. Let $w$ itself denote the vertex obtained by contracting the edges in $F$. To get a contradiction, assume that there is an $H$ induced by a set $U$ of vertices of $G/F$. We observe that $w$ is a universal vertex in $G/F$, due to the fact that $D$ is a dominating set of $G^{\prime}$. Since $H$ does not contain a universal vertex (a tree has a universal vertex if and only if it is a star), $w$ cannot be in $U$. Since $H$ is triangle-free, $U$ can have at most two vertices from $X$. Assume that $U$ has no vertex in $X$. Then $U$ must be a subset of $W_{i}$, which is a contradiction (observe that $W_{i}$ and $W_{j}$ are nonadjacent in $G/F$ for $i\neq j$). Assume that $U$ has exactly one vertex, say $x_{i}$, from $X$. Then the rest of the vertices are from $W_{j}$s adjacent to $x_{i}$. Recall that $x_{i}$ is adjacent to either all or none of vertices in $W_{j}$. Since $H$ is triangle-free, the vertices in $U$ from $W_{j}$s adjacent to $x_{i}$ form an independent set. Then $H$ is a star, which is a contradiction. If $U$ has exactly two vertices from $X$, then with similar arguments, we obtain that $H$ is a bistar, which is a contradiction.

Lemma 29

Let $(G,k)$ be a yes-instance of $H$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of Dominating Set.

Proof

Let $F^{\prime}$ be a subset of edges of $G$ such that $G/F^{\prime}$ is $H$-free and $|F^{\prime}|\leq k$. We construct a new solution $F$ from $F^{\prime}$ as follows. If $F^{\prime}$ contains an edge touching $W_{j}$, then we replace that edge with the edge $wx_{j}$ in $F$. Clearly $|F|\leq|F^{\prime}|\leq k$. Since an edge touching $W_{j}$ kills only the $H$ induced by $W_{j}\cup\{w\}$, which is killed by $wx_{j}$, we obtain that $G/F$ is $H$-free. Let $\mathcal{P}$ be the partition of vertices of $G$ corresponds to $F$. Let $P_{w}$ be the set in $\mathcal{P}$ containing $w$. Let $D=P_{w}\cap X$. Clearly, $|D|\leq|$cost(\mathcal{P})$$|\leq k. We claim that $D^{\prime}=\{v_{i}|x_{i}\in D\}$ is a dominating set of $G^{\prime}$. Assume that there is a vertex $v_{j}$ in $G^{\prime}$ not dominated by $D^{\prime}$. Then $W_{j}\cup\{w\}$ induces an $H$ in $G/F$, which is a contradiction.

Lemmas 29 and 28 imply that there is a parameterized reduction from Dominating Set to $H$-free Contraction.

Lemma 30

Let $H$ be a tree which is neither a star nor a bistar. Then $H$-free Contraction is W[2]-hard.

4.2 Stars and Bistars

First we generalize a reduction given in [15] for $K_{1,4}$-free Contraction. This generalized reduction covers all bistars $T_{t,t^{\prime}}$ such that $t\geq 3$ and $t>t^{\prime}\geq 0$. As a boundary case (when $t^{\prime}=0$) we obtain hardness result for all stars of at least 5 vertices.

Construction 7

Let $G^{\prime}$ be a graph with a vertex set $V^{\prime}=\{v_{1},v_{2},\ldots,v_{n}\}$ and let $k,t,t^{\prime}$ be integers. We construct a graph $G$ from $(G^{\prime},k,t,t^{\prime})$ as follows.

• •

Create two cliques $X=\{x_{1},x_{2},\ldots,x_{n}\},Y=\{y_{1},y_{2},\ldots,y_{n}\}$.

• •

Make $x_{i}$ adjacent to $y_{j}$ if and only if $i=j$ or $v_{i}v_{j}$ is an edge in $G^{\prime}$.

• •

Create a vertex $w$ and two cliques $A$ and $B$ of $k+1$ vertices each.

• •

Make $w$ adjacent to all vertices of $X\cup A\cup B$. Make $A\cup B\cup Y$ a clique.

• •

Introduce $t-1$ cliques with $k+1$ vertices each and make them adjacent to $A$. Let $A^{\prime}$ denote the set of these vertices.

• •

Introduce $t-1$ cliques with $k+1$ vertices each and make them adjacent to $B$. Let $B^{\prime}$ denote the set of these vertices..

• •

Introduce $t-1$ cliques with $k+1$ vertices each and make them adjacent to $X$. Let $X^{\prime}$ denote the set of these vertices.

• •

Introduce a clique of $k+1$ vertices and make it adjacent to $Y$. Let $Y^{\prime}$ denote the set of these vertices.

• •

For every vertex in $A^{\prime}\cup B^{\prime}\cup X^{\prime}\cup Y^{\prime}$, attach $t^{\prime}$ degree-1 vertices.

Let $H$ be $T_{t,t^{\prime}}$ for $t>t^{\prime}\geq 0$ and $t\geq 3$. Let $(G^{\prime},k)$ be an instance of Dominating Set. We obtain $G$ from $(G^{\prime},t,t^{\prime},k)$ by Construction 7. An example is shown in Figure 7.

Lemma 31

Let $(G^{\prime},k)$ be a yes-instance of Dominating Set. Then $(G,k)$ is a yes-instance of $H$-free Contraction.

Proof

Let $D$ be a dominating set of size at most $k$ of $G^{\prime}$. Let $F=\{wx_{i}|v_{i}\in D\}$. Let $w$ itself denote the vertex obtained by contracting edges in $F$. Since $D$ is a dominating set, $w$ is adjacent to every vertex of $Y$ in $G/F$. We claim that $G/F$ is $H$-free. For a contradiction, assume that there is an induced $T_{t,t^{\prime}}$ in $G/F$ with the central edge being $vv^{\prime}$ (we do not distinguish between $v$ and $v^{\prime}$, i.e., we do not assume that $v$ is the vertex attached to $t$ degree-1 vertices and $v^{\prime}$ is the vertex attached to $t^{\prime}$ degree-1 vertices in the induced $T_{t,t^{\prime}}$). Let $v\in A$ and $v^{\prime}\in B$. Then the largest bistar graph rooted at $v$ and $v^{\prime}$ is $T_{t-1,t-1}$ which does not have a $T_{t,t^{\prime}}$ as an induced subgraph. The case when $v=w$ and $v^{\prime}\in A\cup B$ can be proved similarly. Let $v\in A\cup B\cup X$ and $v^{\prime}\in Y$. Then the largest bistar graph rooted at $v$ and $v^{\prime}$ is $T_{t-1,2}$ which does not have an induced $T_{t,t^{\prime}}$. Let $v=w$ and $v^{\prime}\in Y$. Then the largest bistar with the central edge $vv^{\prime}$ is $T_{t-1,1}$, which does not have a $T_{t,t^{\prime}}$ as an induced subgraph. For a pair of adjacent vertices $v,v^{\prime}$ in $G/F$ let $I(v,v^{\prime})$ denote the size of a maximum independent set in the graph induced by $(N(v)\setminus N[v^{\prime}])\cup(N(v^{\prime})\setminus N[v])$ in $G/F$. It is sufficient to prove that there exists no adjacent pair $(v,v^{\prime})$ such that $I(v,v^{\prime})\geq t+t^{\prime}$. Clearly, none of the degree-1 vertex can be in the pair we are looking for. The rest of the cases are handled below, each of them is straight-forward to verify.

• •

Let $v,v^{\prime}\in A^{\prime}$ or $v,v^{\prime}\in B^{\prime}$ or $v,v^{\prime}\in X^{\prime}$ or $v,v^{\prime}\in Y^{\prime}$. Then $I(v,v^{\prime})=t^{\prime}+t^{\prime}<t+t^{\prime}$.

• •

Let $v,v^{\prime}\in A$ or $v,v^{\prime}\in B$. Then $I(v,v^{\prime})=0<t+t^{\prime}$.

• •

Let $v,v^{\prime}\in X$ or $v,v^{\prime}\in Y$. Then $I(v,v^{\prime})\leq 1<t+t^{\prime}$.

• •

Let $v=w$ and $v^{\prime}\in X$. Then $I(v,v^{\prime})\leq 1<t+t^{\prime}$.

• •

Let $v=w$ and $v^{\prime}\in X^{\prime}$. Then $I(v,v^{\prime})=t^{\prime}+t-2+1=t+t^{\prime}-1<t+t^{\prime}$.

• •

Let $v\in A$ and $v^{\prime}\in A^{\prime}$, or $v\in B$ and $v^{\prime}\in B^{\prime}$, $v\in X$ and $v^{\prime}\in X^{\prime}$. Then $I(v^{\prime},v^{\prime})=t^{\prime}+t-2+1=t+t^{\prime}-1<t+t^{\prime}$.

• •

Let $v\in Y$ and $v^{\prime}\in Y^{\prime}$. Then $I(v,v^{\prime})=t^{\prime}+2<t+t^{\prime}$.

By extending the definition of $K_{1,a,b}$, we define $K_{1,a,b,c}$ as follows. Create a vertex $w$ and make it adjacent to $a+1$ copies of $K_{b}$s. Attach $c$ degree-1 vertices to each of the $b$ vertices of one of such $K_{b}$s. Let $w$ be known as the root of $K_{1,a,b,c}$. An example is shown in Figure 8. The following observation is the bistar counterpart of Observation 2.

Observation 7

Let $K$ denote a copy of $K_{1,t,k+1,t^{\prime}}$. Let $G$ has $K$ as an induced subgraph such that the degree-1 vertices in the $K$ are degree-1 vertices in $G$. Then $(G,k)$ is a no-instance of $T_{t,t^{\prime}}$-free Contraction.

Observation 8

Let $F$ be a subset of edges of $G$ such that $|F|\leq k$ and $G/F$ is $H$-free. Then $F$ does not contain any edge from $E[w,A]\cup E[w,B]\cup E[X,Y]$.

Proof

Assume that $wv\in F$ for some vertex $v\in A$. Then there is a $K_{1,t,k+1,t^{\prime}}$ in $G/\{wv\}$ induced by $w,A^{\prime},B,X$, and the degree-1 vertices attached to the vertices of any one of the $t-1$ cliques formed by $A^{\prime}$. Further, it satisfies the condition in Observation 8. Then we get a contradiction by Observation 7. By symmetry, we obtain that $F$ does not contain any edge from $E[w,B]$. Assume that $xy\in F$, where $x\in X,y\in Y$. Then there is a $K_{1,t,k+1,t^{\prime}}$ induced by $x,X^{\prime},Y^{\prime},A$, and the degree-1 vertices attached to the vertices of $Y^{\prime}$. Further, it satisfies the condition in Observation 8. This results in a contradiction.

Observation 9

Let $\mathcal{P}$ be a partition of vertices of $G$ such that $G/\mathcal{P}$ is $H$-free. Further assume that every set in $\mathcal{P}$ is either a singleton set, or a subset of $X\cup\{w\}$ or a subset of $Y$. Let $P_{w}$ denote the set in $\mathcal{P}$ containing $w$. Then, for every set $P_{y}\in\mathcal{P}$ such that $P_{y}\subseteq Y$, at least one vertex of $P_{w}\cap X$ is adjacent to at least one vertex of $P_{y}$.

Proof

For a contradiction, assume that none of the vertices in $P_{y}$ is adjacent to none of the vertices of $P_{w}\cap X$. Then there is a $T_{t,t^{\prime}}$ induced by $P_{w},P_{y}$ and singleton sets corresponding to a vertex in $A$ and $t-1$ vertices (one from each clique) corresponding to $A^{\prime}$, and $t^{\prime}$ vertices attached to one selected vertex from $A^{\prime}$. Therefore, $G/\mathcal{P}$ is not $T_{t,t^{\prime}}$-free, which is a contradiction.

Lemma 32

Let $(G,k)$ be a yes-instance of $H$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of Dominating Set.

Proof

Let $F$ be a minimal subset of edges of $G$ such that $G/F$ is $H$-free and $|F|\leq k$. By Observation 8, the edges from $E[w,A]\cup E[w,B]\cup E[X,Y]$ cannot be in $F$. It can be easily verified that, due to the limited budget and the minimality of $F$, $F$ does not contain any edges of $G$ other than those from $E[w,X]\cup E[X,X]\cup E[Y,Y]$. Let $\mathcal{P}$ be the partition of vertices of $G$ corresponding to $F$. Let $P_{w}$ denote the set containing $w$. We create a dominating set $D$ of size at most $k$ of $G^{\prime}$ as follows. Include the vertices corresponding to $P_{w}\cap X$ in $D$, i.e., include $\{v_{i}|x_{i}\in P_{w}\cap X\}$ in $D$. By Observation 9, for each $P_{y}\in\mathcal{P}$ such that $P_{y}\subseteq Y$, at least one vertex of $P_{y}$ is adjacent to at least one vertex of $P_{w}\cap X$. Let $y\in P_{y}$ is adjacent to a vertex in $P_{w}\cap X$. Then for every vertex $y_{i}\in P_{y}\setminus\{y\}$, include $v_{i}$ in $D$. We claim that $D$ is a dominating set in $G^{\prime}$. For a contradiction, assume that there is a vertex $v_{j}$ not adjacent to any vertex in $D$. Let $P_{y_{j}}\in\mathcal{P}$ be the set containing $y_{j}$. By the construction of $D$, for every vertex in $P_{y_{j}}$ either $v_{j}$ or a vertex adjacent to $v_{j}$ is included in $D$. This gives us a contradiction. Now, it is sufficient to prove that cost(\mathcal{P})$$\leq k. This follows from the fact that for $P_{w}$ we included $|P_{w}|-1$ vertices in $D$ and for every set $P_{y}\in\mathcal{P}$ such that $P_{y}\subseteq Y$, we included at most $|P_{y}|-1$ vertices in $D$.

Lemmas 31 and 32 imply that there is a parameterized reduction from Dominating Set to $T_{t,t^{\prime}}$-free Contraction.

Lemma 33

Let $t>t^{\prime}\geq 0$ and $t\geq 3$. Then $T_{t,t^{\prime}}$-free Contraction is W[2]-hard.

Now, we are left with the bistars $T_{t,t^{\prime}}$ where $t=t^{\prime}$. For this, we come up with a reduction that handles more than this case. The reduction is a parameterized reduction but not a polynomial-time reduction, as the main structure, that we define next, used in the reduction has size exponential in $k$. For integers $t,k\geq 1$, a $(t,k)$-canopy is a graph obtained as follows. The vertices are arranged in $k+1$ levels - level 0 to level $k$. The set of vertices in $i$th level is denoted by $L_{i}$. The set $L_{0}$ contains a single vertex, which is called the root. For $1\leq i\leq k$, $L_{i}$ contains $t^{\lfloor i/2\rfloor}(k+1)^{\lfloor(i-1)/2\rfloor}$ cliques, denoted by $L_{i,1},L_{i,2},\ldots$, each of size $k+1$. The single vertex in $L_{0}$ is adjacent to all vertices in the clique $L_{1}$. For $1\leq i\leq k-1$, the edges between cliques in $L_{i}$ and cliques in $L_{i+1}$ are as follows. If $i$ is odd, then for every clique $L_{i,j}$ there are $t$ unique cliques in $L_{i+1}$ such that edges of a perfect matching is added between $L_{i,j}$ and each such clique, i.e., for every vertex in $L_{i,j}$, there is a unique neighbor in each such clique . If $i$ is even, then every vertex in a clique $L_{i,j}$ is adjacent to all vertices of a unique clique in $L_{i+1}$. An example is shown in Figure 9.

Construction 8

Let $G^{\prime}$ be a graph with $n$ vertices $\{v_{1},v_{2},\ldots,v_{n}\}$ and let $k,t,t^{\prime}$ be integers. A graph $G$ is constructed from $(G^{\prime},k,t,t^{\prime})$ as follows.

• •

Introduce a set of $t$ cliques each of size $k+1$. These cliques are denoted by $A_{1},A_{2},\ldots,A_{t}$, and the set of all these vertices is denoted by $A$.

• •

Introduce a clique $X=\{x_{1},x_{2},\ldots,x_{n}\}$ of $n$ vertices.

• •

Introduce a vertex $w$ and make it adjacent to $A\cup X$.

• •

Introduce $t^{\prime}$ cliques each of size $n$. These cliques are denoted by $Y_{1},Y_{2},\ldots,Y_{t^{\prime}}$ and the set of all these vertices is denoted by $Y$. Let the vertices in $Y_{1}$ be $\{y_{1,1},y_{1,2},\ldots,y_{1,n}\}$.

• •

The vertices $x_{i}$ is adjacent to $y_{1,j}$ if and only if $i=j$ or $v_{i}$ and $v_{j}$ are adjacent in $G^{\prime}$.

• •

For $i=2$ to $t^{\prime}$, the edges of a perfect matching is added between $X$ and $Y_{i}$, i.e., each vertex in $X$ is made adjacent to a unique vertex in $Y_{i}$.

• •

For each vertex in $y\in Y$, introduce a $(t^{\prime},k)$-canopy and identify $y$ with the root of the canopy.

An example of the construction is given in Figure 10.

Let $(G^{\prime},k)$ be an instance of Dominating Set. Let $H$ be $T_{t,t^{\prime}}$, where $t\geq 3$ and $3\leq t^{\prime}\leq t$. Let $G$ be constructed from $(G^{\prime},k,t,t^{\prime})$ by Construction 8.

Lemma 34

Let $(G^{\prime},k)$ be a yes-instance of Dominating Set. Then $(G,k)$ is a yes-instance of $H$-free Contraction.

Proof

Let $D$ be a dominating set of size at most $k$ of $G^{\prime}$. Let $F=\{wx_{i}|v_{i}\in D\}$. Clearly $|F|=|D|\leq k$. We claim that $G/F$ is $H$-free. Let $w$ itself denote the vertex obtained by contracting the edges in $F$. Since $D$ is a dominating set of $G^{\prime}$, $w$ is adjacent to all vertices of $Y_{1}$ in $G/F$. For a contradiction assume that there is an induced $T_{t,t^{\prime}}$ in $G/F$ with the central edge $vv^{\prime}$. If $v=w$ and $v^{\prime}\in X$, then the largest bistar where $vv^{\prime}$ is the central edge is $T_{t+1,t^{\prime}-1}$ which does not have an induced $T_{t,t^{\prime}}$. If $v=w$ and $v^{\prime}\in Y$, then the largest bistar where $vv^{\prime}$ is the central edge is $T_{t+t^{\prime},1}$ which does not have an induced $T_{t,t^{\prime}}$. For a pair of adjacent vertices $v,v^{\prime}$ in $G/F$ let $I(v,v^{\prime})$ denote the size of a maximum independent set in the graph induced by $(N(v)\setminus N[v^{\prime}])\cup(N(v^{\prime})\setminus N[v])$ in $G/F$. It is sufficient to prove that there exists no adjacent pair $(v,v^{\prime})$ such that $I(v,v^{\prime})\geq t+t^{\prime}$. The remaining cases are handled below, each of them is easy to verify.

• •

Let $v,v^{\prime}\in K$ where $K$ is a clique in an odd-numbered level in any $(t^{\prime},k)$-canopy attached to the vertices of $Y$. Then $I(v,v^{\prime})\leq t^{\prime}<t+t^{\prime}$.

• •

Let $v,v^{\prime}\in K$ where $K$ is a clique in an $i$th level, for any even $i\geq 2$ in any $(t^{\prime},k)$-canopy attached to the vertices of $Y$. Then $I(v,v^{\prime})=3<t+t^{\prime}$.

• •

Let $v,v^{\prime}\in Y_{i}$, for $1\leq i\leq t^{\prime}$. Then $I(v,v^{\prime})=3<t+t^{\prime}$.

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Let $v,v^{\prime}\in X$. Then $I(v,v^{\prime})=t^{\prime}<t+t^{\prime}$.

• •

Let $v,v^{\prime}\in A$. Then $I(v,v^{\prime})=0$.

• •

Let $v\in L_{i}$ and $v^{\prime}\in L_{i+1}$, where $0\leq i\leq k-1$ for a $(t^{\prime},k)$-canopy attached to a vertex in $Y$. Then $I(v,v^{\prime})\leq t^{\prime}+2\leq t+t^{\prime}$.

• •

Let $v\in X$ and $v^{\prime}\in Y$. Then $I(v,v^{\prime})\leq 2+t^{\prime}<t+t^{\prime}$.

• •

Let $v=w,v^{\prime}\in A$. Then $I(v,v^{\prime})=0$.

The following observation is straight-forward to verify.

Observation 10

Let $\mathcal{P}$ be a partition of vertices of $G$ such that each set induces a connected graph and $G/\mathcal{P}$ is $H$-free and cost(\mathcal{P})$$\leq k. Then for every clique $K$ introduced while constructing $G$, there is a subset $K^{\prime}$ of $K$ such that $K^{\prime}\in\mathcal{P}$.

Observation 11

Let $\mathcal{P}$ be a partition of vertices of $G$ such that each set induces a connected graph, $G/\mathcal{P}$ is $H$-free, and cost(\mathcal{P})$$\leq k. Further assume that $\mathcal{P}$ is such a partition with least cost$(\mathcal{P})$. Let $P_{w}\in\mathcal{P}$ be the set containing $w$. Then $P_{w}$ does not contain any vertex in any of the $(t,k)$-canopies attached to vertices in $Y$.

Proof

For a contradiction, assume that $P_{w}$ contains a $(t^{\prime},k)$-canopy vertex and let $i$ be the largest index such that $P_{w}$ contains a vertex in $L_{i}$ among all $(t,k)$-canopies attached to vertices of $Y$. Since the budget is only $k$, we obtain that $i\leq k-2$. Let $v$ be any vertex in $P_{w}\cap L_{i}$. Assume that $i$ is an even number. Then there is a $T_{t,t^{\prime}}$ induced by $P_{w}$, $t$ sets in $\mathcal{P}$ subsets of $t$ cliques in $A$ (Observation 10), a set $P_{v^{\prime}}$ corresponding to a clique in $L_{i+1}$ adjacent to $v$, and $t^{\prime}$ sets corresponding to the $t^{\prime}$ vertices in $L_{i+2}$ adjacent to $P_{v^{\prime}}$. Let $i\geq 1$ be an odd number. Then there is a $T_{t,t^{\prime}}$ induced by $P_{w}$, $t$ sets in $\mathcal{P}$ subsets of $t$ cliques in $A$ (Observation 10), a set $P_{v^{\prime}}$ corresponding to the clique in $v$ is part of (such that $v\notin P_{v^{\prime}}$), and $t^{\prime}$ sets corresponding to the vertices in $L_{i+1}$ adjacent to $P_{v^{\prime}}$.

Lemma 35

Let $(G,k)$ be a yes-instance of $H$-free Contraction. Then $(G^{\prime},k)$ is a yes-instance of Dominating Set.

Proof

Let $\mathcal{P}$ be a partition of vertices of $G$ into connected subgraphs such that $G/\mathcal{P}$ is $H$-free and cost(\mathcal{P})$$\leq k. Further assume that $\mathcal{P}$ is having the least cost$(\mathcal{P})$. Let $P_{w}\in\mathcal{P}$ denote the set containing $w$. By Observation 9, $P_{w}$ does not contain any vertex in the $(t,k)$-canopies introduced during the construction of $G$. Now, as we have done in the proof of Lemma 32, we can construct a dominating set $D$ of $G^{\prime}$: Add the vertices in $G^{\prime}$ corresponding to $P_{w}\cap X$ in $D$. For every $P_{y}\in\mathcal{P}$ such that $y\in Y_{1}$, we know that at least one vertex $v$ of $P_{y}$ must be adjacent to $P_{w}\cap X$. For all other vertices $v_{i}$ in $P_{y}\cap Y_{1}$, add $v_{i}$ in $D$. It is straight-forward to verify that $D$ is a dominating set of size at most $k$ of $G^{\prime}$.

Lemmas 34 and 35 imply that there is a parameterized reduction from Dominating Set to $T_{t,t^{\prime}}$-free Contraction.

Lemma 36

Let $t\geq t^{\prime}\geq 3$. Then $T_{t,t^{\prime}}$-free Contraction is W[2]-hard.

Now, Lemmas 30, 33, and 36 imply the main result of this section.

Theorem 4.1

Let $T$ be a tree which is neither a star of at most 4 vertices ($\{K_{1},K_{2},P_{3},K_{1,3}\}$) nor a bistar in $\{T_{1,1},T_{2,1},T_{2,2}\}$. Then $T_{t,t^{\prime}}$-free Contraction is W[2]-hard.

We believe that our W[2]-hardness result on trees will be a stepping stone for an eventual parameterized complexity classification of $H$-free Contraction. The most challenging hurdle for such a complete classification can be the graphs $H$ where each component is of at most 2 vertices, and the case of claw, the usual trouble-maker for other graph modification problems to $H$-free graphs.

We conclude with some folklore observations. As noted in a version of [20], the property that “there exists at most $k$ edges contracting which results in an $H$-free graph” can be expressed in MSO1. The length of the corresponding MSO1 formula will be a function of $k$. Then, there exists FPT algorithms for $H$-free Contraction, whenever $H$-free graphs have bounded rankwidth (See Chapter 7 of the textbook [9]). This, in particular, implies that $K_{2}+K_{1}$-free Contraction can be solved in FPT time. It is known that every component of a paw-free graph is either triangle-free or complete multipartite [23], where where paw is the graph
. Then the existance of FPT algorithms for $K_{3}$-free Contraction and $K_{2}+K_{1}$-free Contraction imply that there exists an FPT algorithm for $paw$-free Contraction.