## 1. Introduction

All graphs in this paper are undirected, finite and simple. A graph is said to be *subcubic* if has maximum degree at most , and *cubic* if is -regular. A subset of edges of a graph is a *matching* if no two edges of are adjacent in . A *maximal matching* is a matching that cannot be extended to a larger matching. For a graph , let denote the minimum size of a maximal matching of . The parameter is sometimes also called the *edge domination number*.

The problem of finding a minimum maximal matching goes back a long way [10] and is known to be NP-hard, even when restricted to cubic bipartite or cubic planar graphs [20, 13, 9]. Approximating the problem within a factor smaller than is NP-hard as well [8]. On the other hand, it is well known that every matching of provides a -approximation for . There also exist algorithms that approach within a factor strictly smaller than , see e.g. [6, 7, 12, 19], yet most of this work seems to have focused on the general case or on very dense graphs. In this paper, we instead focus on sparse graphs. Concretely, we study the following recent conjecture of Baste, Fürst, Henning, Mohr and Rautenbach.

###### Conjecture 1 ([2]).

Let be a connected -regular graph of order for some , then

For every , the authors of [2] provided evidence for Conjecture 1 in the form of a weaker upper bound which for cubic graphs (i.e. ) specialises to . Furthermore, for cubic graphs they confirmed Conjecture 1 under the additional conditions that the graph is bipartite and does not contain a certain -vertex tree as an induced subgraph. In this paper, we unconditionally prove Conjecture 1 for cubic graphs:

###### Theorem 2.

Let be a connected cubic graph of order . Then

In doing so, we derive the following slightly more general result, which also applies to non-regular graphs.

###### Theorem 3.

Let be a connected subcubic graph on vertices and edges. Then Moreover, if is not cubic then

An effective and widely used approach to prove theorems on subcubic graphs is to take a hypothetical minimum counterexample and then analyse small cut sets as well as the local structure around vertices of small degree, followed by an analysis of the local structure in a highly connected cubic graph, eventually leading to a contradiction (see e.g. [4, 3, 14] for some recent examples). This paper is no exception. However, one way in which perhaps our approach stands out is how we carefully almost avoid the need to analyse any cubic graph (at the cost of a slightly worse result than is perhaps possible; see the discussion of Conjecture 7). Relatedly, to make the ‘induction’ work, we have devised the following technical theorem, which has Theorems 2 and 3 as direct corollaries.

###### Theorem 4.

Let be a connected subcubic graph with vertices, edges and vertices of degree . Let be the indicator variable being if is cubic, and equal to otherwise. Let be the indicator variable being if is isomorphic to , and equal to otherwise. Then

where

Moreover, if is not isomorphic to and has a degree- vertex with neighbour , then has a maximal matching of size at most which avoids the edge .

Tightness

Theorem 3 is best possible insofar that the four-cycle and the complete bipartite graph attain the bound. Likewise, Theorem 2 is best possible due to . A small computer search did not reveal any other graph that attains these bounds. In particular there exists no other among the cubic graphs on at most vertices.

Let denote the graph that is obtained from by deleting an edge. In [2], it was conjectured that for cubic connected graphs , it holds that if and only if

has a spanning subgraph that is the union of an odd number of copies of

. (In particular this would have implied that infinitely many graphs attain the bounds of Theorems 2 and 3.) However, this turns out to be false, as we will now detail. Let be vertex-disjoint copies of . Writing for the degree- vertices of , we let be the graph obtained by adding the edges for all (where should be understood to mean ); see Figure 1. Then is a cubic connected graph on vertices, which however has^{1}

^{1}1The equality follows from an induction applied to ; we omit the formal proof. For the upper bound, see the pattern emerging in Figure 1. To quickly see that , observe that for every maximal matching of , every three consecutive copies together with the edges must contain at least edges of . a minimum maximal matching of size , so that . For every , this is smaller than . This raises the question whether asymptotically, as the number of vertices increases, the bound of Theorem 2 is improvable.

###### Question 5.

What is the infimum over all constants for which there exists an such that every connected cubic graph on vertices satisfies

The graphs and Theorem 2 certify that . We suspect that and that moreover the following is true.

###### Conjecture 6.

Let be a connected cubic graph of order . Then

In a different direction, we believe the following minor strengthening of Theorem 3 is plausible.

###### Conjecture 7.

Let be a connected subcubic graph with vertices and edges. Then unless is isomorphic to the complete graph or the complete bipartite graph .

Despite the small gap with Theorem 3, we would not be surprised if considerable more work is required to prove Conjecture 7. This is because in the current paper we were able to almost avoid the analysis of cubic graphs. In particular, once we have deduced that our minimum counterexample to Theorem 4 is cubic, we are done in only a few extra lines, due to the extra room in the induction hypothesis arising from the additive factor . This ‘shortcut’ approach turned out to be fruitful but does not seem possible for an attack on Conjecture 7.

Efficient approximation algorithm

The proof of Theorem 4 is of an inductive nature. As such, the proof can be viewed as a polynomial-time algorithm that constructs for every connected subcubic graph a maximal matching of of size at most . Indeed, we repeatedly identify a set of at most nine vertices and apply induction to the graph obtained from by deleting and possibly adding one or two edges. In determining the appropriate set and the edges that are to be added back, the only external procedures that we require are (i) find a bridge (if present) and determine the number of vertices and edges in each component of and (ii) determine which pairs of neighbours of are connected in . For both of these subroutines, there exist time algorithms. The number of iterations required is and therefore in total the required time is at most .

Until now, we have solely focused on upper bounds on . On the other hand, it is not hard to see that for every cubic graph on vertices (see e.g. Lemma 3.(i) in [2]). Since , it follows that our proof constitutes a approximation algorithm for minimum maximal matching in connected cubic graphs.

Some remarks on maximal independent sets and line graphs

Before going to the proof of Theorem 4, we would like to point out that the problem of determining can be viewed as a special case of a more general problem. These remarks can be freely skipped.

First some definitions. An *independent set* of a graph is a set of vertices that are pairwise non-adjacent. A *maximal independent set* is an independent set that cannot be extended to a larger one. Correspondingly, the *independent domination number* of a graph is the size of the smallest maximal independent set of . The well-studied *domination number* of is the size of a smallest subset of such that every vertex of is either in or adjacent to it. Finally, the *line graph* of a graph is the graph on vertex set that has an edge between if and only if and are adjacent in .

In this terminology, the edge domination number of can be rewritten as . Furthermore, for claw-free graphs (and hence for line graphs) it is known [1] that the domination number and independent domination number are equal, so that in fact .

For every cubic graph , its line graph is -regular and has vertices. Therefore an equivalent statement of Theorem 2 is that for every (-regular) line graph of a connected cubic graph, one has . One may wonder to what extent the condition of being a line graph is essential here. Often problems on claw-free graphs actually reduce to the case of line graphs (e.g. see [5] for a somewhat similar problem where this was the case), so it is natural to expect that holds for every -regular connected claw-free graph , with equality if is the line graph of . Assuming Conjecture 6, even the stronger bound can be expected. A proof of such a generalization would be nice.

To put this into perspective, let us now consider a connected graph on vertices that is not necessarily claw-free. Then if has minimum degree at least three [18], if and is cubic [15], and if is -regular [17]. The first of these two bounds are attained, while for the third bound it is not clear from the literature to what extent it is best possible. On the other hand, the parameter can attain much higher values than , as demonstrated by the balanced complete bipartite graphs. There also exists a -regular graph on vertices with and it is conjectured [11] that this is best possible among all -regular graphs other than . For cubic graphs more is known in this respect. For instance, if is an -vertex connected cubic graph other than then , as proved in [16], and it is conjectured [11] that up to two exceptions, which would be sharp due to infinitely many graphs. For various other results regarding and , we refer the reader to the survey [11] and citing papers.

## 2. A technical lemma that helps to avoid cubic graphs

The following technical lemma will be used a few times in the proof of Theorem 4. It allows us to only apply induction to non-cubic graphs, which is of considerable benefit due to the factor in our definition of .

###### Lemma 8.

Let be a bridgeless connected subcubic graph. Let . Let denote the set of neighbours of that are not in . Let be another graph on of which each component is either a singleton or contains at least three vertices. Suppose furthermore that there is at least one edge in which is not an edge of . Then there exists such that and the graph has no cubic component (and is subcubic).

###### Proof.

For each , let denote the component of that contains . Let be such that . If possible, we choose such that additionally .

Consider the reduced graph and let . Since has a neighbour in in the graph , it has degree less than three in and so the component of in is not cubic.

It remains to show that the component of containing the edge is not cubic. Note that is the union of and and the edge . We may assume that in , the vertex has three neighbours of which exactly one is in (otherwise would have degree less than three in , implying that is not cubic); let be the neighbour of that is in .

If then the facts that is connected and bridgeless (in particular is not a bridge) imply that for some . On the other hand, if , then by the choice of and we have that for all vertices that belong to the component of in the graph . By assumption, the component of in has at least three vertices, so again we obtain for some . In both cases and so is not cubic. ∎

## 3. Proof of the main theorem

In this section, we prove Theorem 4. Let be a hypothetical counterexample to Theorem 4 that minimises . We proceed by a series of lemmas that describe properties of which ultimately lead to a contradiction.

In various cases, we will delete a set of vertices from and possibly add some edges, and then apply induction to the resulting graph . When the graph is clear from the context, we let and denote the number of vertices, edges respectively degree- vertices of . Furthermore, will denote the number of cubic components of and will denote the number of components of that are isomorphic to . We then write , , and and for the respective differences. Since, by induction, we will arrive at a contradiction (as desired) if we can show that . It will therefore always be sufficient to show that

(1) |

where

and

Remark
*Since is connected, whenever only vertices are deleted (and no edge added) to construct from , then cannot be a cubic graph.
Hence in most of our applications, will be .*

Remark
* At some parts in the proof of Theorem 4, we deduce that must be a specific graph on at most nine vertices.
In those cases it is easily seen that is not a counterexample to Theorem 4, so we have decided to not explicitly specify a matching certifying this. We remark that we have also performed a small computer search to double-check that indeed no graph on nine vertices or less is a counterexample to Theorem 4.*

The bulk of the proof is focused on the structure around vertices of degree smaller than three, starting with the following lemma.

###### Lemma 9.

has minimum degree at least .

###### Proof.

If there is a vertex of degree , then since is connected, must be a singleton, which is not a counterexample to Theorem 4. Next, suppose that has a vertex of degree . Let be the unique neighbour of . Since the complete graph on two vertices is not a counterexample, must have a neighbour which is distinct from . Then consider the graph . Let be a minimum maximal matching of . Then is a maximal matching of of order , so . We have that and . Furthermore, since is connected, every component of that is isomorphic to contains a vertex that has degree larger than in , and hence we have (unless equals the graph induced by , which clearly cannot yield a counterexample).

Thus

so must be a maximal matching of of size at most . Because avoids , we obtain the desired contradiction. ∎

As does not have any degree- vertices, we know from now on that . Moreover, in all our forthcoming constructions of a reduced graph , we will make sure that does not have any cubic components, so that in fact we will always have

(2) |

This simplifies our computations ever so slightly.

###### Lemma 10.

has no bridge

###### Proof.

Suppose has a bridge . For , let be the component of that contains . Furthermore, let be the graph induced by . Let and . Observe that and .

For each we apply induction to and . Because is a a degree- vertex of , there is a maximal matching of that avoids and has size at most .

On the other hand, by induction has a maximal matching of size at most . So is a maximal matching of of size at most . Considering the minimum of and , we obtain that has a maximal matching of size at most

(3) |

If at least one of and does *not* equal mod , then due to the rounding in expression (3) there is a maximal matching of size at most

Thus mod . Next, for each , we consider the graph . Note that has one more vertex and (depending on the degree of ) either one or two more edges than . So while either equals or . It follows that is either mod or mod . Hence has a maximal matching of size at most . We conclude that has a maximal matching of size at most , a contradiction. ∎

###### Lemma 11.

There are no two adjacent degree- vertices in .

###### Proof.

Suppose for a contradiction that there are two adjacent degree- vertices and . Let be the other neighbour of and let be the other neighbour of . First, if then consider . Then a maximal matching of can be extended to a maximal matching of , so . Therefore

So we may assume that . Next, suppose that . Since the graph induced by is not a counterexample, without loss of generality has a neighbour distinct from . Define . Then a maximal matching of can be extended to a maximal matching of , so . So

Therefore . Next, consider the graph . (Equivalently: is obtained from by contracting the edges and .) Let be a maximal matching of . If , then is a maximal matching of . On the other hand, if , then is a maximal matching of . In both cases the new matching of has size , so we have . Furthermore, and . If is not cubic, then so it follows that

a contradiction. Thus, must be cubic. Therefore from now on we may assume that all vertices of other than and have degree three. In particular, has neighbours and has neighbours such that is disjoint from .

Next, suppose that and have a common neighbour, without loss of generality it is . Then a maximal matching of can be extended to the maximal matching of , which implies that . Thus again We conclude that must be pairwise distinct. Our next goal is to show that these four vertices form an independent set.

First, assume that or is an edge, without loss of generality the former. Then a maximal matching of yields a maximal matching of , so that . Thus ; contradiction.

Second, assume that there is an edge joining and ; without loss of generality is such an edge. Observe that then at least one neighbour of is not equal to or , nor adjacent to (because otherwise would have more than three distinct neighbours). We define the reduced graph . Let be a maximal matching of . If , then is a maximal matching of , and otherwise is a maximal matching of . In both cases the new matching contains two more edges than , so . Furthermore, and (here we use that is an edge in but not in ). Crucially, does not contain any cubic component, so that . Indeed, the only component of that could a priori be cubic is the component belonging to the added edge . But due to the edge (which we assumed to exist), we know that also contains the vertex , which has degree less than in . We obtain ; contradiction. This concludes the proof that is independent.

In summary, we have deduced so far that induces a tree in , and all vertices of other than and have degree three. Let and denote the two neighbours of that are distinct from .

Now let us define the vertex set and the set which consists of pairs of neighbours of . If each pair in is present as an edge in , then it follows that is a bridge, contradicting Lemma 10. Thus we can apply Lemma 8 to conclude that there exists such that and the reduced graph has no cubic component. This means that with respect to . Furthermore, by exactly the same analysis as before (by symmetry of , we may assume that for some neighbour of ), we again obtain and . Therefore ; contradiction.

∎

###### Lemma 12.

Every degree- vertex of has at most one degree- neighbour.

###### Proof.

We suppose for a contradiction that a degree- vertex of has two degree- neighbours and .

If , then every maximal matching of can be extended to a maximal matching of , so . So ; contradiction. Therefore . So has neighbours and has neighbours such that is disjoint from .

Case : and have a common neighbour that is distinct from .

Without loss of generality this common neighbour is . Consider the graph with a maximal matching . Then is a maximal matching of , so . Moreover, and . If , then it follows that . Thus must equal , which implies that are distinct vertices, that and have degree three and that . By symmetry, must have degree three as well, and . Thus, are distinct degree- vertices that either form an independent set or induce the graph with the edge .

In particular, has a neighbour which is distinct from . For , let and be the two neighbours of that are distinct from .

Suppose for a contradiction that is a neighbour of both and . If additionally , then the structure of is fully determined; it is the graph induced by the seven vertices , which clearly is not a counterexample. Thus and so has a neighbour that is distinct from and . In that case, a maximal matching of can be extended to the maximal matching of , so that , and . Hence ; contradiction. We have thus derived that is not a neighbour of both and .

By symmetry, we may assume that is not a neighbour of . We then consider the graph . Then . Note that to obtain from , we not only delete a vertex, but we also add an edge, so . Consider a maximal matching of . If then is a maximal matching of , and otherwise is a maximal matching of . In both cases . To arrive at a contradiction, we still need to check that has no cubic components (so that ). For this, it suffices to demonstrate that the component of containing the added edge is not cubic. If , then it is immediate that is in and has degree less than three in . Thus we may assume that . Because has no bridge (in particular is not a bridge), there exists a path in that joins with while avoiding . (Here we allow to be a single vertex.) Note that is also a path in and hence or . Since and have degree less than three in , it follows that is not cubic.

Case : is the only common neighbour of and .

From now on, we know that are pairwise distinct. Our next task is to show that they form an independent set.

Case : or .

By symmetry, it suffices to consider the case that . Suppose for a contradiction that .

If additionally there is an edge between and (say is an edge), then a maximal matching of can be extended to the maximal matching of . Then ; contradiction. Thus there is no edge between and .

By applying induction to the graph , it is easily seen that each of needs to have degree three. For , let denote the neighbour of that is distinct from .

Let and note that is the set of neighbours of . (Possibly has less than four distinct elements since it could be that .) If is complete to , then must be the graph induced by , which is a graph on at most nine vertices; contradiction. Thus there must exist such that . Then by Lemma 8 applied to and , there exists an edge such that the graph has no cubic component. By symmetry, we may assume that . We then apply induction to . By the choice of we have . Moreover and

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