1 Introduction
In a graph modification problem, we are usually interested in modifying a given graph , via a small number of operations, into some other graph that has a certain desired property. This property often describes a certain graph class to which must belong. Such graph modification problems allow to capture a variety of classical graphtheoretic problems. Indeed, if for instance only vertex deletions are allowed and must be a stable set or a clique, we obtain the Stable Set or Clique problem, respectively.
Now, instead of specifying a graph class to which should belong, we may ask for a specific graph parameter to decrease. In other words, given a graph , a set of one or more graph operations and an integer , the question is whether can be transformed into a graph by using at most operations from such that for some threshold . Such problems are called blocker problems as the set of vertices or edges involved can be viewed as “blocking” the parameter . Notice that identifying such sets may provide important information relative to the structure of the graph .
Blocker problems have been well studied in the literature (see for instance [1, 2, 3, 5, 9, 10, 15, 16, 17, 18, 19]) and relations to other wellknown graph problems have been presented (see for instance [9, 16]). So far, the literature mainly focused on the following graph parameters: the chromatic number, the independence number, the clique number, the matching number and the vertex cover number. Furthermore, the set consisted of a single graph operation, namely either vertex deletion, edge contraction, edge deletion or edge addition. Since these blocker problems are usually hard in general graphs, a particular attention has been paid to their computational complexity when restricted to special graph classes.
In this paper, we focus on another parameter, namely the domination number , and we restrict to a single graph operation, the edge contraction. More specifically, let be a graph. The contraction of an edge removes vertices and from and replaces them by a new vertex that is made adjacent to precisely those vertices that were adjacent to or in (without introducing selfloops nor multiple edges). We say that a graph can be contracted into a graph , if can be transformed into by a sequence of at most edge contractions, for an integer . We will be interested in the following problem, where is a fixed integer.
.99 Edge Contraction()
Instance:  A connected graph 

Question:  Can be edge contracted into a graph such that ? 
In other words, we are interested in a blocker problem with parameter , graph operations set edge contraction and threshold . Notice that if that is, contains a dominating vertex, then is always a Noinstance for Edge Contraction(). Reducing the domination number using edge contractions was first considered in [14]; given a graph , the authors denote by the minimum number of edge contractions required to transform into a graph such that and prove that for a connected graph such that , we have . It follows that a graph with is always a Yesinstance of Edge Contraction(), if . The authors [14] further give necessary and sufficient conditions for to be equal to 1, respectively 2.
Theorem 1.1 ([14])
For a connected graph , the following holds.

if and only if there exists a minimum dominating set in that is not a stable set.

if and only if every minimum dominating set in is a stable set and there exists a dominating set in of size such that contains at least two edges.
To the best of our knowledge, a systematic study of the complexity of Edge Contraction() has not yet been attempted in the literature. We here initiate such a study as it has been done for other parameters and several graph operations. Our paper is organised as follows^{1}^{1}1Missing proofs will be marked by and are in the appendix for reviewing purposes.. In Section 2, we present definitions and notations that are used throughout the paper. In Section 3, we prove the ()hardness of Edge Contraction() for . We further show that Edge Contraction() is [1]hard parameterized by the size of a minimum dominating set plus the mimwidth of the input graph, and that it remains hard when restricted to free graphs, bipartite graphs and free graphs for any . Finally, we present in Section 4 some positive results; in particular, we show that for any , Edge Contraction() is polynomialtime solvable for free graphs and that it can be solved in time and time when parameterized by treewidth and mimwidth, respectively.
2 Preliminaries
Throughout the paper, we only consider finite, undirected, connected graphs that have no selfloops nor multiple edges. We refer the reader to [8] for any terminology and notation not defined here and to [6] for basic definitions and terminology regarding parameterized complexity.
Let be a graph and let . We denote by , or simply if it is clear from the context, the set of vertices that are adjacent to i.e., the neighbors of , and let . Two vertices are said to be true twins (resp. false twins), if (resp. if ).
For a family of graphs, is said to be free if has no induced subgraph isomorphic to a graph in ; if we may write free instead of free. For a subset , we let denote the subgraph of induced by , which has vertex set and edge set .
We denote by , or simply if it is clear from the context, the length of a shortest path from to in . Similarly, for any subset , we denote by , or simply if it is clear from the context, the minimum length of a shortest path from to some vertex in i.e., .
For a vertex , we write and for a subset we write . For an edge , we denote by the graph obtained from by contracting the edge . The subdivision of an edge consists in replacing it by a path , where are new vertices.
For , the path and cycle on vertices are denoted by and respectively. A graph is bipartite if every cycle contains an even number of vertices.
A subset is called an stable set of if any two vertices in are nonadjacent; we may also say that is stable. A subset is called a dominating set, if every vertex in is adjacent to at least one vertex in ; the domination number is the number of vertices in a minimum dominating set. For any and , is said to dominate (in particular, dominates itself); furthermore, is a private neighbor of with respect to if has no neighbor in . We say that contains an edge (or more) if the graph contains an edge (or more). The Dominating Set problem is to test whether a given graph has a dominating set of size at most , for some given integer .
3 Hardness results
In this section, we present hardness results for the Edge Contraction() problem. Recall that for , the problem is trivial; we show that for , it becomes ()hard. To this end, we introduce the following problem.
.99 Contraction Number(,)
[2pt] Instance: A connected graph . Question: Is ?
Theorem 3.1
Contraction Number(,) is hard.
Proof
We reduce from 1in3 Positive 3Sat, where each variable occurs only positively, each clause contains exactly three positive literals, and we want a truth assignment such that each clause contains exactly one true variable. This problem is known to be complete [12]. Given an instance of this problem, with variable set and clause set , we construct an equivalent instance of Contraction Number(,) as follows. For any variable , we introduce a copy of , which we denote by , with two distinguished truth vertices and (see Fig. 1); in the following, the third vertex of is denoted by . For any clause containing variables and , we introduce the gadget depicted in Fig. 1 (where it is connected to the corresponding variable gadgets). The vertex set of the clique corresponds to the set of subsets of size 1 of (hence the notation); for any , the vertex (resp. ) is connected to every vertex such that (resp. ). Finally, for , we add an edge between (resp. ) and the truth vertex (resp. ). Our goal now is to show that is satisfiable if and only if . In the remainder of the proof, given a clause , we denote by , and the variables occuring in and thus assume that (resp. ) is adjacent to (resp. ) for . Let us first start with some easy observations.
Observation 1
Let be a dominating set of . Then for any , and for any , . In particular, .
Clearly, for any , since must be dominated. Also, in order to dominate vertices and in some gadget , we need at least 4 distinct vertices, since their neighborhoods are pairwise disjoint and so, , for any .
Observation 2
Let be a dominating set of . For any clause gadget and , .
This immediately follows from the fact that every vertex needs to be dominated and its neighbors are and for .
Observation 3
Let be a dominating set of . For any clause gadget , if , then and , for any .
If for some , then it follows from Observation 2 that for any . This implies that at least two vertices among and belong to for otherwise there would exist such that is not dominated. In particular, there must exist such that ; but then, is not dominated. Similarly, if for some , it follows from Observation 2 that for any . But then, in order to dominate the vertices of , either but then is not dominated; or and with , is not dominated.
Now suppose that for some . Then by Observation 2, we conclude that for and . This implies that for otherwise we would have . But then, since , must contain at least two vertices among and in order to dominate the vertices of ; in particular, there exists such that and so, is not dominated.
Observation 4
Let be a minimum dominating set of and suppose that . Then for any vertices , we have .
Indeed, if are at distance at most 2, we conclude by Theorem 1.1(ii) that , a contradiction.
Observation 5
Let be a minimum dominating set of and suppose that . Then for any clause gadget and , if and only if .
This readily follows from Observation 4. Further note that we may assume that for any , if and only if ; is equivalent to and we may always replace by .
Observation 6
Let be a minimum dominating set of and suppose that . Then for any clause gadget , .
If it weren’t the case then, by Observation 4, no or () would belong to . But since and must be dominated, it follows that and by Observation 5, we conclude that contains two vertices at distance two (namely, and for some ).
Observation 7
Let be a minimum dominating set of and suppose that . Then for any clause gadget , .
Indeed, if we assume, without loss of generality, that , then by Observation 4, . Furthermore, if then since must be dominated and (it would otherwise be within distance at most 2 from a vertex in belonging to ); and, if , then by Observation 4. Thus, we conclude by Observation 5 that if and only if . It then follows from Observation 4 that and so, for otherwise at least one vertex in would not be dominated. But then, is not dominated seeing that by Observation 4, since , and as shown previously.
Claim 1 ()
if and only if .
Claim 2
if and only if is satisfiable.
Assume first that and consider a minimum dominating set of . We construct a truth assignment from satisfying as follows. For any , if , set to true; otherwise, set to false. We claim that each clause has exactly one true variable. Indeed, it follows from Observation 1 that for any , and from Claim 1 that . But then, by Observation 3, for any , if and only if ( would otherwise not be dominated). It then follows from Observations 6 and 7 that and for any ; but by Observation 5 we conclude that if and only if , which proves our claim.
Conversely, assume that is satisfiable and consider a truth assignment satisfying . We construct a dominating set of as follows. If variable is set to true, we add to ; otherwise, we add to . For any clause and , if , then add to ; otherwise, add to . Since every clause has exactly one true variable, it follows that and ; finally add to where . Now clearly and every vertex in is dominated. Thus, and so by Observation 1, , which concludes this proof.
By observing that for any graph , is a Yesinstance for Contraction Number(,) if and only if is a Noinstance for Edge Contraction(), we deduce the following corollary from Theorem 3.1.
Corollary 1
Edge Contraction() is hard.
It is thus hard to decide whether for a graph ; and in fact, it is hard to decide whether equality holds, as stated in the following.
Theorem 3.2 ()
Contraction Number(,) is hard.
We finally consider the case .
Theorem 3.3
Edge Contraction() is hard even when restricted to free graphs, with .
Proof
We reduce from Dominating Set: given an instance of this problem, we construct an equivalent instance of Edge Contraction() as follows. We denote by the vertex set of . The graph consists of copies of , denoted by , connected in such a way that for any and , the copies and of a vertex of are true twins in the subgraph of induced by ; and for any and , the copies and of a vertex of are false twins in the subgraph of induced by . Next, we add pairwise nonadjacent vertices , which are made adjacent to every vertex in ; is further made adjacent to every vertex in , for all . Finally, we add a vertex adjacent to only (see Fig. 2). Note that the fact that for all and , and (resp. and ) are false (resp. true) twins within the graph induced by (resp. ) is not made explicit on Fig. 2 for the sake of readability. In the following, we denote by and . We now claim the following.
Claim 3
.
It is clear that is a dominating set of ; thus, . If and is a minimum dominating set of , it is easily seen that is a dominating set of . Thus, and so, . Now, suppose to the contrary that and consider a minimum dominating set of . We first make the following simple observation.
Observation 8
For any minimum dominating set of , .
Now, since , there exists such that (otherwise, and combined with Observation 8, would be of size at least ). But then, must dominate every vertex in , and so . Since (recall that ), we then have , a contradiction. Thus, .
We now show that is a Yesinstance for Dominating Set if and only if is a Yesinstance for 1Edge Contraction().
First assume that . Then, by the previous claim, and if is a minimum dominating set of , then is a minimum dominating set of which is not stable. Hence, by Theorem 1.1(i), is a Yesinstance for 1Edge Contraction().
Conversely, assume that is a Yesinstance for 1Edge Contraction() i.e., there exists a minimum dominating set of which is not stable (see Theorem 1.1(i)). Then, Observation 8 implies that there exists such that ; indeed, if it weren’t the case, then by Claim 3 we would have and thus, would consist of and either or . In both cases, would be stable, a contradiction. It follows that must dominate every vertex in and thus, . But (recall that ) and so by Claim 3, that is, is a Yesinstance for Dominating Set.
Finally, we can prove that if is free then is free. However, due to lack of space, this proof has been placed in Section 0.C of the Appendix. ∎
Given the hardness of 1Edge Contraction() and its close relation to Dominating Set, it is natural to consider the complexity of the problem when parameterized by the size of a minimum dominating set of the input graph. In the following, we denote by the mimwidth parameter, and show that 1Edge Contraction() is hard when parameterized by . We first state two simple facts regarding the mimwidth parameter.
Observation 9
Let be a graph and be two vertices that are true (resp. false) twins in . Then .
Observation 10
Let be a graph and . Then .
Theorem 3.4
1Edge Contraction() is hard parameterized by .
Proof
We give a parameterized reduction from Dominating Set parameterized by solution size plus mimwidth, which is a problem that was recently shown to be hard by Fomin et al. [11]. Given an instance of Dominating Set, the construction of the equivalent instance for 1Edge Contraction() is the same as the one introduced in the proof of Theorem 3.3; and it is there shown that is a Yesinstance for Dominating Set if and only if is a Yesinstance for 1Edge Contraction(). Now, note that can be obtained from by the addition of true twins (the set ), the addition of false twins (the sets ), and the addition of vertices (). By Observation 9, the addition of true (resp. false) twins does not increase the mimwidth of a graph and, by Observation 10, the addition of a vertex can only increase the mimwidth of by one; thus, and since by Claim 3, we conclude that . ∎
In order to obtain complexity results for further graph classes, let us now consider subdivisions of edges.
Lemma 1 ()
Let be a graph and let be the graph obtained by 3subdividing every edge of . Then is a Yesinstance for 1Edge Contraction() if and only if is a Yesinstance for 1Edge Contraction().
By 3subdividing every edge of a graph sufficiently many times, we deduce the following two corollaries from Lemma 1.
Corollary 2
1Edge Contraction() is hard when restricted to bipartite graphs.
Corollary 3
For any , 1Edge Contraction() is hard when restricted to free graphs.
We finally observe that, even if an edge is given, deciding whether contracting this particular edge decreases the domination number is unlikely to be solvable in polynomial time as shown in the following result.
Theorem 3.5 ()
There exists no polynomialtime algorithm deciding whether contracting a given edge decreases the domination number, unless .
4 Algorithms
We now deal with cases in which Edge Contraction() is tractable, for . A first simple approach to the problem, from which Proposition 1 readily follows, is based on brute force.
Proposition 1 ()
For , Edge Contraction() can be solved in polynomial time for a graph class , if either

is closed under edge contractions and Dominating Set can be solved in polynomial time for ; or

for every , , where is some fixed contant; or

is the class of free graphs, where is a fixed constant and Edge Contraction() is polynomialtime solvable on free graphs.
Proposition 1(b) provides an algorithm for 1Edge Contraction() parameterized by the size of a minimum dominating set of the input graph running in time. Note that this result is optimal as 1Edge Contraction() is [1]hard with such parameterization from Theorem 3.4.
We further show that even though simple, this brute force method provides polynomialtime algorithms for a number of relevant classes of graphs, such as graphs of bounded treewidth and graphs of bounded mimwidth. We first state the following result and observation.
Theorem 4.1
[20] Given a graph and a decomposition of width , Dominating Set can be solved in time when parameterized by treewidth, and in time when parameterized by mimwidth.
Observation 11 ()
.
Proposition 2
Given a decomposition of width , Edge Contraction() can be solved in time in graphs of treewidth at most and in time in graphs of mimwidth at most , for .
Proof
We use the abovementioned brute force approach and Theorem 4.1. That is, for , the algorithm first computes and then computes for every . For , the algorithm proceeds similarly for every pair of edges. We next show that the width parameters increase by a constant when contracting at most two edges. It is a wellknown fact that and so, . By Observation 11, which implies that . Also note that, given a tree (resp. mim) decomposition of width for , we can construct in polynomial time decompositions of width (resp. at most ) for and . This implies that and can also be computed in time if is a graph of treewidth at most , and in time if is a graph of mimwidth at most . ∎
Proposition 2 provides an algorithm for 1Edge Contraction() parameterized by mimwidth running in time; this result is optimal as 1Edge Contraction() is [1]hard parameterized by mimwidth from Theorem 3.4.
Since Dominating Set is polynomialtime solvable in free graphs (see [13]), it follows from Proposition 1(a) that Edge Contraction() can also be solved efficiently in this graph class. However, Dominating Set is complete for free graphs (see [4]) and thus, it is natural to examine the complexity of Edge Contraction() for this graph class. As we next show, Edge Contraction() is in fact polynomialtime solvable on free graphs, for .
Lemma 2
If is a free graph with , then .
Proof
Let be a free graph and be a minimum dominating set of . Suppose that is a stable set and consider such that . Since is free, and, since is stable, . We distinguish two cases depending on this distance.
Case 1. . Let (resp. ) be the neighbor of (resp. ) on a shortest path from to . Then, ; indeed, if is a neighbor of , then is nonadjacent to (recall that ) and thus, is adjacent to either or for otherwise and would induce a in . The same holds for any neighbor of . Consequently, is a minimum dominating set of which is not stable; the result then follows from Theorem 1.1(i).
Case 2. . Since is stable and , it follows that every is at distance two from both and . Let (resp. ) be the vertex on a shortest path from (resp. ) to some vertex .
Suppose first that . If every private neighbor of with respect to is adjacent to then is a minimum dominating set of which is not stable; the result then follows from Theorem 1.1(i). We conclude similarly if every private neighbor of or with respect to is adjacent to . Thus, we may assume that (resp. ; ) has a private neighbor (resp. ; ) with respect to which is nonadjacent to . Since is free, it then follows that , and are pairwise adjacent. But then, and induce a , a contradiction.
Finally, suppose that (we may also assume that as we otherwise fall back in the previous case). Then, for and would otherwise induce a . Now, if is a private neighbor of with respect to then is adjacent to either or ( and otherwise induce a ); we conclude similarly that any private neihbor of with respect to is adjacent to either or . If is adjacent to both and but not , then it is adjacent to (and ) as and ( and ) would otherwise induce a . But then, is a minimum dominating set of which is not stable; thus, by Theorem 1.1(i), which concludes the proof. ∎
Theorem 4.2
Edge Contraction() is polynomialtime solvable on free graphs, for .
Proof
If has a dominating vertex, then is clearly a Noinstance for both . Now, for every , we check whether is a dominating set. If it is the case, then by Theorem 1.1(i), is a Yesinstance for Edge Contraction() for . If no edge of is dominating, we consider all the pairs of nonadjacent vertices of . If there exists such a pair dominating and then by Theorem 1.1(i), we have a Noinstance for 1Edge Contraction() since this implies that every minimum dominating set of is stable. For the case , if has two nonadjacent vertices dominating , we then consider all triples of vertices of to check whether there exists one which is dominating and contains at least two edges (see Theorem 1.1(ii)). Finally, both for and , if has no dominating set of size at most two, then by Lemma 2, is a Yesinstance for Edge Contraction(). ∎
5 Conclusion
In this paper, we studied the Edge Contraction() problem and provided the first complexity results. In particular, we showed that Edge Contraction() is hard for free graphs, , but polynomialtime solvable for free graphs; it would be interesting to determine the complexity status for free graphs, for . Similarly, the complexity of Edge Contraction() for free graphs, with , remains an interesting open problem.
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Missing Proofs
Appendix 0.A The proof of Claim 1 in Theorem 3.1
Assume that and consider a minimum dominating set of . We first show that is a stable set which would imply that (see Theorem 1.1(i)). First note that Observation 1 implies that and , for any variable and any clause . It then follows from Observation 3 that no truth vertex is dominated by some vertex or in some clause gadget with ; in particular, this implies that there can exist no edge in having one endvertex in some gadget () and the other in some gadget (). Hence, it is enough to show that for any , is a stable set.
Now consider a clause gadget . It follows from Observation 3 that if there exists such that then since must be dominated (also note that by Observation 3, if then ). Hence, for any , exactly one of and belongs to . But then, by Observation 3 and since , we immediately conclude that is a stable set and so, is a stable set.
Now, suppose to the contrary that i.e., there exists a dominating set of of size containing two edges and (see Theorem 1.1(ii)). First assume that there exists such that . Then, for any , ; and for any , which by Observation 3 implies that for any . Since as shown previously, is then a stable set, it follows that contains at most one egde, a contradiction.
Thus, there must exist some such that . We then claim that . Indeed, since
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