1 Introduction
We consider graph modification problems, where the aim is to test if a given graph can be modified to have some desired property using a small number of graph operations. Many such problems can be formulated as a vertex transversal problem. That is, the modified graph, obtained after deleting a number of vertices of the original graph, must not contain any graph of a prescribed family . More formally, we are looking for a small transversal of a graph , where an transversal of is a subset such that is free. In other words, intersects every subset of that induces a subgraph isomorphic to a graph in . For instance, let denote the path on vertices. Then a set is a vertex cover if is a transversal. Note that is an independent set. A set is a feedback vertex set if is an transversal for the infinite family (where is the cycle on vertices). In this case, the graph is a forest. To give another example, a set is an odd cycle transversal if is an transversal for the infinite family . In this case, the graph is bipartite.
Vertex covers, feedback vertex sets, and odd cycle transversals are wellstudied concepts. The corresponding decision problems that ask whether a given graph has a vertex cover, feedback vertex set, or odd cycle transversal of size at most for some given integer are called Vertex Cover, Feedback Vertex Set and Odd Cycle Transversal, respectively. These problems are well known to be NPcomplete [14, 20]. In this paper we focus on the connected variants of these problems. We say that a vertex cover of a graph is connected if induces a connected subgraph of . The notions of a connected feedback vertex set and connected odd cycle transversal are defined in a similar way.
.99 Connected Vertex Cover
[2pt] Instance: a graph and an integer . Question: does have a connected vertex cover with ?
.99 Connected Feedback Vertex Set
[2pt] Instance: a graph and an integer . Question: does have a connected feedback vertex set with ?
.99 Connected Odd Cycle Transversal
[2pt] Instance: a graph and an integer . Question: does have a connected odd cycle transversal with ?
1.1 Related Work
The three connected problem variants are known to be NPcomplete. It is therefore a natural question whether there exist special graph classes for which they become polynomialtime solvable. Below we briefly survey known hard and tractable results for each of the three problems.
Already in the seventies, Garey and Johnson [15] proved that Connected Vertex Cover is NPcomplete even for planar graphs of maximum degree 4. Later, Priyadarsini and Hemalatha [25] strengthened this result to 2connected planar graphs of maximum degree 4, and Fernau and Manlove [13] strengthened it to planar bipartite graphs of maximum degree 4. Wanatabe, Kajita, and Onaga [32] proved NPcompleteness of Connected Vertex Cover for 3connected graphs. More recently, Munaro [23] established NPcompleteness of Connected Vertex Cover for line graphs of planar cubic bipartite graphs and for planar bipartite graphs of arbitrarily large girth. Ueno, Kajitani, and Gotoh [31] proved that Connected Vertex Cover can be solved in polynomial time for graphs of maximum degree at most 3 and for trees. Escoffier, Gourvès, and Monnot [12] improved the latter result by showing that Connected Vertex Cover is polynomialtime solvable for chordal graphs. It is known that Vertex Cover is polynomialtime solvable for chordal graphs as well. The same authors proposed to study the complexity of Connected Vertex Cover for other graph classes for which Vertex Cover is polynomialtime solvable.
Grigoriev and Sitters [17] proved that Connected Feedback Vertex Set is NPcomplete for planar graphs with maximum degree 9. Besides this result not much is known on the computational complexity of this problem except that it is fixed parameter tractable when parameterized by [21]. The Connected Odd Cycle Transversal has been mainly studied from parameterized point of view [9, 10]. It is implicit in these works that Connected Odd Cycle Transversal is NPcomplete, though we were not able to find any proof of this (simple) result. For the sake of completeness, we note now that it is implied by the stronger results we present in Theorems 2.2 and 2.4.
1.2 Our Results.
We continue the investigations on the complexity of Connected Vertex Cover and also initiate a complexity study of Connected Feedback Vertex Set and Connected Odd Cycle Transversal. Our paper consists of two parts. In the first part (Section 2) we prove new hardness results and in the second part (Section 3) new tractable results. As we explain below we try to do so in a systematic way. We refer to Table 1 for a survey of our results together with some related known results. This table leads to some natural open problems, which we discuss in Section 4, together with some other directions for future work.
The girth of a graph is the length of a shortest cycle in it. We prove that Connected Feedback Vertex Set and Connected Odd Cycle Transversal are NPcomplete on graphs of girth at least for any fixed constant . In order to obtain this result for Connected Odd Cycle Transversal we first prove NPcompleteness of Odd Cycle Transversal for graphs of girth at least for any fixed constant (analogous results were already known for Vertex Cover and Feedback Vertex Set). In the same section we also show that Connected Feedback Vertex Set and Connected Odd Cycle Transversal are NPcomplete for line graphs.
The above hardness results imply that Connected Feedback Vertex Set and Connected Odd Cycle Transversal are NPcomplete on free graphs whenever is a graph that contains a cycle or a claw. Due to the results of Munaro [23] the same holds for Connected Vertex Cover. Moreover, the same holds for the original variants Feedback Vertex Set and Odd Cycle Transversal but not for Vertex Cover, which is polynomialtime solvable for clawfree graphs. Due to the hardness results we can restrict ourselves to free graphs where is a linear forest, that is, the disjoint union of one or more paths.
We prove that Connected Vertex Cover, Connected Feedback Vertex Set, and Connected Odd Cycle Transversal are polynomialtime solvable for free graphs for every constant . Since Vertex Cover is polynomialtime solvable for free graphs for any ,^{1}^{1}1This follows from combining results of [1, 30], as explained in more detail in Section 3. our result for Connected Vertex Cover restricted to free graphs can be seen as a new step in the aforementioned research direction of Escoffier, Gourvès, and Monnot [12].
girth  line graphs  free  

Con. Vertex Cover  NPc [23]  NPc [23]  P 
Con. Feedback Vertex Set  NPc  NPc  P 
Con. Odd Cycle Transversal  NPc  NPc  P 
Vertex Cover  NPc [24]  P [27]  P [1, 30] 
Feedback Vertex Set  NPc [24]  NPc [29]  P 
Odd Cycle Transversal  NPc  NPc  P 
To prove our tractability results we first focus on the original problems Vertex Cover, Feedback Vertex Set, and Odd Cycle Transversal. In particular we will use the enumeration algorithms of Tsukiyama, Ide, Ariyoshi and Shirakawa [30] and Schwikowski and Speckenmeyer [28] for listing all maximal independent sets and all maximal feedback vertex sets, respectively. We will also use the fact that the number of maximal independent sets is polynomially bounded for free graphs [1]. However, we need to prove analogous results for feedback vertex sets and odd cycle transversals. This leads to polynomialtime enumeration algorithms for minimal feedback vertex sets of an free graph and minimal odd cycle transversals of an free graph. What remains to do is to relate these results to the connected problem variants and below we describe how to do this.
We need to consider the effect of adding the connectivity constraint on the minimum size of an transversal for a graph family . This effect is in fact measured by a known concept called the price of connectivity. This concept was coined by Cardinal and Levy [8] for vertex cover, but can be defined for any graph property for which a connected variant is meaningful: for a class of graphs and graph property , the price of connectivity is the worstcase ratio over all connected graphs , where and denote the smallest subset and smallest connected subset, respectively, of the vertices of satisfying . Apart from further results on vertex cover [4, 5, 6], the price of connectivity has been studied for dominating set [4, 5, 7, 11, 33], face hitting set [17, 26], and feedback vertex set [2], until in [18] known results for vertex cover and feedback vertex set for free graphs were extended in a larger framework of transversals. In particular, it is known that for free graphs demanding connectivity results only in an additive constant with respect to the size of a smallest vertex cover, feedback vertex set, or odd cycle transversal [2, 18]. This is exactly what we need to finalize our proofs.
2 Hardness Results
In this section we prove our hardness results (see also Table 1). We first need to introduce some terminology. The line graph of a graph has the edges of as its vertices, and two vertices and of are adjacent if and only if and have an endvertex in common in . The claw is the graph with vertices , , , and edges , , . It is wellknown and readily seen that every line graph is clawfree.
Our proofs for the Odd Cycle Transversal and Connected Odd Cycle Transversal problems can be found in Section 2.1 and for the Connected Feedback Vertex Set problem in Section 2.2.
2.1 Odd Cycle Transversal and Connected Odd Cycle Transversal
We will first prove that Odd Cycle Transversal and Connected Odd Cycle Transversal are NPcomplete for line graphs.
A 2factor of a graph is a set of cycles such that every vertex of belongs to exactly one of them. Let Even Cycle Factor denote the problem of deciding if a graph has an even cycle factor, that is, a 2factor in which each cycle is even.
We need the following lemmas due to Hell, Kirkpatrick, Kratchovíl, and Kříž.
Lemma 1 ([19])
Even Cycle Factor is NPcomplete.
We also need the next lemma. Here, we let denote the subgraph of a graph induced by a subset .
Lemma 2
Let be a graph on vertices and edges. If has an odd cycle transversal of size at most , then has an even cycle factor.
Proof
Let be the vertices of that do not belong to the odd cycle transversal. As has vertices, contains at least vertices. By definition, is bipartite which means in particular that the neighbours of each vertex are independent. Thus, as is a line graph and so clawfree, no vertex in has degree greater than 2. Hence is the disjoint union of a set of paths and even cycles. Each path on vertices in corresponds to a path on vertices in , and each cycle on vertices in corresponds to a cycle on vertices in . Thus, if of the components of are paths, then the number of vertices of incident with is . As has at least vertices and has exactly vertices, we must have , and so is the disjoint union of even cycles that corresponds to an even cycle factor of . ∎
Theorem 2.1
Odd Cycle Transversal is NPcomplete for line graphs.
Proof
We reduce from Even Cycle Factor, which is NPcomplete by Lemma 1. Let be an instance of this problem with vertices and edges. We claim that has an even cycle factor if and only if its line graph has an odd cycle transversal of size at most .
If has an even cycle factor, then contains a set of vertices that induce a disjoint union of cycles. Hence is an odd cycle transversal of that has size . To complete the proof, we apply Lemma 2. ∎
Theorem 2.2
Connected Odd Cycle Transversal is NPcomplete for line graphs.
Proof
As in the proof of Theorem 2.1 we reduce from Even Cycle Factor. Let be an instance of this problem with vertices and edges. We add a new vertex to and an edge between and each vertex of . We also add three vertices with edges , , and . We call the resulting graph . We observe that has an even cycle factor if and only if has an even cycle factor. Note that has vertices and edges. We construct the line graph of and claim that has an even cycle factor if and only if has a connected odd cycle transversal of size at most .
First suppose that has an even cycle factor . By construction, contains the cycle on vertices , which we denote by . As a consequence, no edge for belongs to a cycle of . Then contains a set of vertices that induce a disjoint union of cycles that do not contain any vertex for . Hence is an odd cycle transversal of that has size and that contains all vertices of the form . We need to show that induces a connected subgraph of . This follows from the fact that every vertex is adjacent to (and ), which belongs to , and the fact that the set form a clique in .
Now suppose that has a connected odd cycle transversal of size at most . Then we can apply Lemma 2 (noting again that has vertices and edges) to find that has an even cycle factor. ∎
For our next hardness result on Connected Odd Cycle Transversal we first need to show the same result for Odd Cycle Transversal. For that we need the following lemma.
Lemma 3
Let be a graph obtained from a graph after subdividing an edge of twice. Then the size of a minimum odd cycle transversal of is equal to the size of a minimum odd cycle transversal of .
Proof
Let be the edge of that is subdivided twice. Let be the two new vertices obtained after the subdividing so that contains the path .
First suppose that is a minimum odd cycle transversal of . If one or two of are in , then is an odd cycle transversal of (as we can safely place in ). The same is true if , as any even cycle in is still even after subdividing twice.
Now suppose that is a minimum odd cycle transversal of . If one or two of are in , then are not in by minimality of , meaning that is also an odd cycle transversal of . Suppose that are not in . If or is in , then we can safely replace or by or , respectively, to obtain a new odd cycle transversal of with the same size as . If none of is in , then, as contracting the path back to does not result in an odd cycle, we again find that is an odd cycle transversal of . ∎
Repeatedly applying Lemma 3 in combination with the fact that Odd Cycle Transversal is NPcomplete (due to Theorem 2.1) implies the following theorem.
Theorem 2.3
For every constant , Odd Cycle Transversal is NPcomplete for graphs of girth at least .
We now show that we also have hardness when the problem asks for a connected transversal.
Theorem 2.4
For every constant , Connected Odd Cycle Transversal is NPcomplete for graphs of girth at least .
Proof
We reduce from Connected Vertex Cover restricted to graphs of girth at least . Recall that Munaro [23] proved that this problem is NPcomplete. Let be a graph of girth at least . We may assume without loss of generality that has at least two edges. Between each pair of adjacent vertices of add a path with edges. Let the graph obtained be and note that it also has girth at least . We will show that has a connected vertex cover of size at most if and only if has a connected odd cycle transversal of size at most .
Let be a connected vertex cover of of at most . As every cycle in contains two vertices that are adjacent in , the set is also an odd cycle transversal of , and it is also connected in as contains all the edges of .
Let be a connected odd cycle transversal of of at most . We can assume that all vertices of belong to (if this is not the case, then we can move all vertices from to without creating an odd cycle in or disconnecting ). As each pair of adjacent vertices in belong to the odd cycle of that contains them and the path on edges added between, at least one of them must belong to . Thus is a connected vertex cover of (of size at most ). ∎
Corollary 1
Let be a graph. Then Connected Odd Cycle Transversal is NPcomplete for free graphs if contains a cycle or a claw.
2.2 Connected Feedback Vertex Set
We will now consider the Connected Feedback Vertex Set problem. We first prove the following result.
Theorem 2.5
Connected Feedback Vertex Set is NPcomplete for line graphs.
Proof
Let be an instance of Hamiltonian Path with vertices and edges. We will describe a reduction to Connected Feedback Vertex Set. We add two new vertices and to and an edge between and each vertex of and an edge between and each vertex of . We also add two vertices and and edges and . We call the resulting graph . We observe that has a Hamiltonian path if and only if has a Hamiltonian path from to . Note that has vertices and edges. We claim that has a Hamiltonian path if and only if its line graph has a connected feedback vertex set of size at most .
First suppose that has a Hamiltonian path (on vertices). Let and be the two edges of that join and to . Then every edge with and every edge with does not belong to . Let be the set of size in formed by the edges of . Hence is a feedback vertex set of that has size and that contains all vertices of the form and except the vertices and . We need to show that induces a connected subgraph of . This follows from the fact that every vertex is adjacent to and , and at least one of them belongs to . Moreover, the set of vertices is a clique in . Finally, every vertex with is adjacent to in .
Now suppose that has a connected feedback vertex set of size at most . Then has at least vertices and must be a disjoint union of paths (as is a line graph and thus clawfree). Every induced path in corresponds to a path in with one more vertex. Moreover, has vertices in total. Hence must induce a single path on vertices. The corresponding path in has vertices and is thus a Hamiltonian path of .∎
We now show that the restriction of Connected Feedback Vertex Set to graphs of arbitrarily large girth is NPhard. We note that it is known that Feedback Vertex Set is NPcomplete for graphs of girth at least for every constant ; this follows immediately from the wellknown observation (see for example [3, 22]) that if a graph is obtained from a graph by subdividing an edge, then has a feedback vertex set of size at most if and only if does. In order to prove the result for Connected Feedback Vertex Set, we reduce from Connected Vertex Cover restricted to graphs of girth at least using essentially the same construction and arguments as in the proof of Theorem 2.4. In this case, we can simply say that for an instance of Connected Vertex Cover, we construct an instance of Connected Feedback Vertex Set by adding a path of length between each pair of adjacent vertices in (so that each edge of corresponds to a cycle of length in ).
Theorem 2.6
For every constant , Connected Feedback Vertex Set is NPcomplete for graphs of girth at least .
Corollary 2
Let be a graph. Then Connected Feedback Vertex Set is NPcomplete for free graphs if contains a cycle or a claw.
3 PolynomialTime Results
In this section we prove our polynomialtime results (see also Table 1). The following lemma provides a useful technique that we will be able to apply several times. We say that a set of vertices in a graph is connected if it induces a connected subgraph.
Lemma 4
Let be a property of a set of vertices in a graph that if it holds for a set also holds for all supersets. Given a graph , let be the minimum size of a set of vertices in with property . Let be a graph class and let be a constant. Suppose that for every graph , all minimal sets with property can be enumerated in polynomial time, and the minimum size of a connected set of vertices with property is at most . Then a minimum size connected set of vertices of with property can be found in polynomial time.
Proof
The algorithm for finding a minimum size connected set with property is simple. Enumerate the minimal sets of with property , and, for each such set consider all the sets that can be obtained by adding or fewer additional vertices and check whether each is connected. Return the smallest connected set found (if any). It is clear that the algorithm runs in polynomial time. We must prove correctness.
Let be a minimum size connected set of vertices in with property . We will show that the algorithm finds and so returns it (or another connected set with property of the same size). We know that . From remove as many vertices as possible until a minimal set with property is obtained. Then is one of the sets that will be enumerated and as , we have that and so will be found. ∎
For an algorithm that enumerates all solutions to a problem, the delay is the maximum of the time taken before finding the first solution and that required between any pair of consecutive outputs. We need two wellknown results. The first one is due to Balas and Yu (who gave a sharper bound, which we do not need for our purposes) and the second one is due to Tsukiyama, Ide, Ariyoshi, and Shirakawa.
Theorem 3.1 ([1])
For every constant , the number of maximal independent sets of an free graph on vertices is at most .
Theorem 3.2 ([30])
For every constant , it is possible to enumerate all maximal independent sets of a graph on vertices and edges with a delay of .
It is wellknown that Vertex Cover is polynomialtime solvable for free graphs; in fact this follows immediately from Theorems 3.1 and 3.2. Our proof of the same result for Connected Vertex Cover can be found in Section 3.1, for Feedback Vertex Set and Connected Feedback Vertex Set in Section 3.2, and for Odd Cycle Transversal and Connected Odd Cycle Transversal in Section 3.3.
3.1 Connected Vertex Cover
We consider the Connected Vertex Cover problem. We have the following result on the price of connectivity for vertex cover restricted to free graphs by Lemma 10 of a previous paper of four of the current authors. The constant is determined in the proof of Lemma 10 of [2] (the statement of the lemma only mentions the existence of a constant without further specification).
Theorem 3.3 ([18])
Let and let be a connected free graph. Let be the size of a minimum vertex cover of . Then the size of a minimum connected vertex cover of is at most .
We can use Theorem 3.3 to determine the complexity of Connected Vertex Cover for free graphs.
Theorem 3.4
For every constant , Connected Vertex Cover can be solved in polynomial time for free graphs.
Proof
We know by Theorems 3.1 and 3.2 that all maximal independent sets, and thus all minimal vertex covers, of free graphs can be enumerated in polynomial time. Thus, as the property of being a vertex cover holds for supersets, the theorem follows from Theorem 3.3 and Lemma 4 (note that we may restrict ourselves to connected input graphs). ∎
3.2 Feedback Vertex Set and Connected Feedback Vertex Set
To prove our results for Feedback Vertex Set and Connected Feedback Vertex Set we need the following result of Schwikowski and Speckenmeyer [28].
Theorem 3.5 ([28])
It is possible to enumerate all minimal feedback vertex sets of a graph on vertices and edges with a delay of .
We also need the following lemma.
Lemma 5
For every constant there is a constant such that the number of minimal feedback vertex sets of an free graph on vertices is .
Proof
Let be a minimal feedback vertex set of . Let , which, by definition, is a forest. Let contain each vertex of degree at least 2 in plus one vertex (arbitrarily chosen) from each component of isomorphic to (if there are any). We say that is a skeleton of . Let and notice that is an independent set and each of its vertices has at most one neighbour in . We say that is the leave of .
Claim
.
To prove the claim, first consider the following subset of : all isolated vertices of , one arbitrarily chosen vertex from each , and every vertex of degree 1 from each other tree. Pair each vertex in with one of its neighbours in (it must have at least one by the definition of ). These pairs induce a collection of pairwise nonadjacent ’s, and, as is free, . Notice that from a vertex in , it is not possible to find a path containing more than vertices within since a longer path contains as an induced subgraph. We choose a path of length at most from each vertex to a fixed vertex of the component of containing . In this way the union of all these paths cover all the vertices of , and the claim follows.
Let be the set of vertices in that have at most one neighbour in . Notice that .
Claim
is a maximal independent set in .
To prove the claim, suppose instead that there is a vertex that has no neighbour in . Then is a forest as has at most one neighbour in , and so is a feedback vertex set of , contradicting the minimality of and proving the claim.
We can find minimal feedback vertex sets of as follows. First we choose a skeleton on at most vertices, and then, from amongst those vertices that have at most one neighbour in the chosen skeleton, we choose a maximal independent set. Then the set of vertices in neither the forest nor the independent set is tested to see whether it forms a minimal feedback vertex set. If we consider all possible choices of forest and independent set, then each minimal feedback set will be found as at some point a corresponding skeleton and its leave will be chosen. By our first claim, there are choices for the forest, and, by Theorem 3.1, there are choices for the independent set. The theorem is proved. ∎
We also need the following result on the price of connectivity for feedback vertex set due to Belmonte, van ’t Hof, Kamiński, and Paulusma. Notice that this result holds even for free graphs. The constant is determined in the proof of Lemma 9 of [2].
Theorem 3.6 ([2])
Let and let be a connected free graph. Let be the size of a minimum feedback vertex set of . Then the size of a minimum connected feedback vertex set of is at most .
We are now ready to prove our result on the complexities of Feedback Vertex Set and Connected Feedback Vertex Set restricted to free graphs.
Theorem 3.7
For every constant , Feedback Vertex Set and Connected Feedback Vertex Set can be solved in polynomial time for free graphs.
Proof
We will show that there are polynomialtime algorithms that find the minimum size sets. The theorem, which concerns decision problems, follows.
Let be an free graph. We may assume without loss of generality that is connected. By the combination of Lemma 5 with Theorem 3.5, we can in fact enumerate all minimal feedback vertex sets of in polynomial time. Thus the result for Feedback Vertex Set follows.
For Connected Feedback Vertex Set, we can apply Lemma 4. We need only note that the property of being a feedback vertex set is inherited by supersets and that, by Theorem 3.6, within the class of graphs, the minimum size of a connected feedback vertex set differs from the size of a minimum size set by at most a constant. ∎
3.3 Odd Cycle Transversal and Connected Odd Cycle Transversal
We now consider the Odd Cycle Transversal problem and its connected variant. Again we can use a result on the price of connectivity, which is also covered by Lemma 10 of [18].
Theorem 3.8 ([18])
Let and let be a connected free graph. Let be the size of a minimum odd cycle transversal of . Then the size of a minimum connected odd cycle transversal of is at most .
We can now prove our final result.
Theorem 3.9
For every constant , Odd Cycle Transversal and Connected Odd Cycle Transversal can be solved in polynomial time for free graphs.
Proof
Let be a minimal odd cycle transversal of an free graph . Let . Then is a bipartite graph. We choose a bipartition of such that has maximum size (so every vertex in has at least one neighbour in ). Then is a maximal independent set of , as otherwise there exists a vertex not adjacent to any vertex of , and thus is an odd cycle transversal of , contradicting the minimality of . Moreover, by a similar argument, is a maximal independent set of .
We describe a procedure to find all minimal odd cycle transversals of , and, as the minimum size transversals of will be amongst them this provides an algorithm for Odd Cycle Transversal. We enumerate all maximal independent sets of , and for each maximal independent set , we enumerate all maximal independent sets of . For each such set , we note that is an odd cycle transversal of . By the arguments above, we will find every minimal odd cycle transversal in this way, and, by Theorems 3.1 and 3.2, this takes polynomial time.
Remark 1. Just as for minimal vertex covers, we can compute all minimal feedback vertex sets and all minimal odd cycle transversals of an free graph in polynomial time for every constant .
4 Conclusions
We first showed that just like the Connected Vertex Cover problem, the Connected Feedback Vertex Set and Connected Odd Cycle Transversal problems are NPcomplete for free graphs if contains a cycle or a claw. Hence, we could restrict ourselves to graphs that are linear forests. We then proved that all three problems can be solved in polynomial time for free graphs.
Our main goal was to show an application of the price of connectivity for cycle transversals (Theorems 3.3, 3.6, and 3.8). In fact these three structural results on the price of connectivity hold even for free graphs. Hence, a natural future research direction would be to try to extend our algorithmic results to free graphs. However, then we can no longer bound the number of minimal transversals as before, so the situation is not clear.
It would also be interesting to research whether it is possible to compute, just as for the three original problems (see Remark 1), all minimal connected transversals in polynomial time for the three variants studied, even when we restrict the input to free graphs. With respect to this question, we recall an open problem of Golovach, Heggernes, and Kratsch [16]. They asked if there exists a graph class with a polynomial number of minimal vertex covers but an exponential number of minimal connected vertex covers.
Acknowledgement
We are grateful to Andrea Munaro and an anonymous referee for helpful remarks.
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