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Strengthening the Murty-Simon conjecture on diameter 2 critical graphs

A graph is diameter-2-critical if its diameter is 2 but the removal of any edge increases the diameter. A well-studied conjecture, known as the Murty-Simon conjecture, states that any diameter-2-critical graph of order n has at most n^2/4 edges, with equality if and only if G is a balanced complete bipartite graph. Many partial results about this conjecture have been obtained, in particular it is known to hold for all sufficiently large graphs, for all triangle-free graphs, and for all graphs with a dominating edge. In this paper, we discuss ways in which this conjecture can be strengthened. Extending previous conjectures in this direction, we conjecture that, when we exclude the class of complete bipartite graphs and one particular graph, the maximum number of edges of a diameter-2-critical graph is at most ((n -- 1)^2/4) + 1. The family of extremal examples is conjectured to consist of certain twin-expansions of the 5-cycle (with the exception of a set of thirteen special small graphs). Our main result is a step towards our conjecture: we show that the Murty-Simon bound is not tight for non-bipartite diameter-2-critical graphs that have a dominating edge, as they have at most (n^2/4) -- 2 edges. Along the way, we give a shorter proof of the Murty-Simon conjecture for this class of graphs, and stronger bounds for more specific cases. We also characterize diameter-2-critical graphs of order n with maximum degree n -- 2: they form an interesting family of graphs with a dominating edge and 2n -- 4 edges.

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1 Introduction

A graph is called diameter--critical if its diameter is , and the deletion of any edge increases the diameter. Diameter--critical graphs and their extremal properties have been studied since at least the 1960s, see for example the few selected references [4, 5, 7, 22, 25, 26, 27].

The case of diameter--critical graphs (D2C graphs for short), being the simplest nontrivial case, has been the focus of much work. Many famous and beautiful graphs are D2C. Such examples are: complete bipartite graphs; the -cycle (more generally, the set of Moore graphs of diameter , the others being the Petersen graph, the Hoffman-Singleton graph, and a hypothetical graph of order  [21]); the Wagner graph; the Chvátal graph; the Grötzsch graph; the Clebsch graph… When the order is at least , D2C graphs coincide with minimally triangle-saturated graphs, see [23].

In the 1960s and 1970s, Ore, Murty, Pleśnik and Simon independently stated the following conjecture, generally known under the name of Murty-Simon conjecture (see [5, 25, 26]). A complete bipartite graph is balanced if the sizes of the two parts differ by at most .

Conjecture 1.

Any D2C graph of order has at most edges, with equality if and only if is a balanced complete bipartite graph.

Many partial results towards Conjecture 1 have been obtained. The best general upper bound is due to Fan [8], who showed that any D2C graph of order has less than edges. In the same paper, Fan also proved the conjecture for graphs of order at most  and . On the other hand, using the Regularity Lemma, Füredi [9] proved Conjecture 1 for graphs with order , where is a gigantic number: roughly a tower of ’s of height .

It is not difficult to observe that any bipartite graph of diameter  must be a complete bipartite graph. Thus, Conjecture 1 restricted to bipartite graphs is well-understood; when studying D2C graphs, we can consider only non-bipartite graphs.

Hanson and Wang observed in [11] that a non-bipartite graph is D2C if and only if its complement is -total domination critical. This launched a fruitful research path that led to many results around Conjecture 1: see the papers [1, 12, 13, 14, 16, 17, 18, 19] and the survey [15].

Conjecture 1 holds for triangle-free graphs: this is exactly Mantel’s theorem [24], which states that a triangle-free graph of order has at most edges, with equality if and only if is a balanced complete bipartite graph. It is not difficult to observe that a graph is both D2C and triangle-free if and only if it is maximal triangle-free [3] (and equivalently, triangle-free with diameter ). Maximal triangle-free graphs are widely studied, see for example [2, 6, 10, 23].

It is known that no triangle-free non-bipartite D2C graph has a dominating edge [1] (a dominating edge is a pair of adjacent vertices that have no common non-neighbour). Thus, another case of interest is the set of D2C graphs with a dominating edge.666Equivalently, the set of D2C graphs with total domination number . A non-bipartite D2C graph has a dominating edge if and only if its complement has diameter  [11] (note that the complement of a non-bipartite D2C graph has diameter either or  [19]). Conjecture 1 was proved for D2C graphs with a dominating edge in the series of papers [11, 13, 16, 28].

It has been observed that the bound of Conjecture 1 might be strengthened. Such a claim was made by Füredi in the article [9] containing his proof of the conjecture for very large graphs. In the conclusion of [9] (Theorem 7.1), he claimed, without proof, that his method can be extended to show the following: any sufficiently large non-bipartite D2C graph of order has at most edges, with equality if and only if the graph is obtained from an even-order balanced complete bipartite graph by subdividing an edge once. Such a graph is also obtained from a -cycle by replacing two adjacent vertices by two same-size independent sets of twins (vertices with the same open neighbourhood). Nevertheless, one might perhaps need to consider Füredi’s claim with caution, as it appears to be partly false. Indeed, it was observed in [1] that one can also obtain a D2C graph with the same number of edges by replacing three vertices , , of a -cycle by three independent sets , , of twins, under the following conditions: (1) , and are consecutive on the -cycle; (2) , where is the number of vertices of the obtained graph. Let us call , the family of these “expanded -cycles” that satisfy (1) and (2): the construction is depicted in Figure 1.

Figure 1: The infinite family of expanded -cycles: we have .

The authors of [1], unaware of Füredi’s claim, focused their attention on the case of non-bipartite D2C graphs without a dominating edge (recall that this class includes all non-bipartite D2C triangle-free graphs).777Note that the terminology used in [1] is the equivalent one of total domination criticality for the complement of the graphs. They conjectured that, for this class, the corrected version of Füredi’s claim is true:

Conjecture 2 (Balbuena, Hansberg, Haynes, Henning [1]).

Let be a non-bipartite D2C graph without a dominating edge. Then, has at most edges. For sufficiently large , equality holds if and only if belongs to .

Conjecture 2 was proved in [1] for triangle-free graphs without the restriction on the order (the first part of the statement was also proved independently for triangle-free graphs in [3]).

One non-bipartite D2C graph that does not satisfy the bound is known: it is the graph of Figure 2 (which has six vertices, eight edges, and a dominating edge). This fact was observed in [3, 12]; the authors of [3] asked whether this graph is the only exception to the bound.

Figure 2: The graph , a D2C graph with a dominating edge (in bold).

We have performed a computer search on all graphs of order up to . This search has found exactly thirteen non-bipartite D2C graphs that are not members of but nevertheless reach the bound (see Figure 3).888Note that the complete list of D2C graphs of order at most was given in [23]; Graph (m) is studied in [27] as a planar D2C graph with every edge in a triangle. These graphs have order , or : there are no examples of order and . Only three of these graphs have no dominating edge.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)
Figure 3: The thirteen non-bipartite D2C graphs of order with edges that are not in the family . Bold edges are dominating. Only the graphs (d), (e) and (f) have no dominating edge.

Closer scrutiny and the aforemetioned computer search suggest that could well be the only small non-bipartite D2C exception to the bound of Conjecture 2 (this was also suggested in [3]). Moreover, it seems likely that only the thirteen graphs not in from Figure 3 reach this bound. This leads us to propose a stronger version of Conjecture 2, as follows.

Conjecture 3.

Let be a non-bipartite D2C graph of order . If is not , then has at most edges, with equality if and only if belongs to or is one of the thirteen graphs from Figure 3.

In this paper, we focus on the case of non-bipartite D2C graphs with a dominating edge. Note that the graphs in do not have a dominating edge. As a piece of evidence towards Conjecture 3, our main theorem shows that the bound of Conjecture 1 is not tight for this case:

Theorem 4.

Let be a non-bipartite D2C graph with vertices having a dominating edge. If is not , then has at most edges.

Along the way, we will also give some stronger bounds for special cases. Our proof technique is an extension of the technique introduced by Hanson and Wang in [11], where the bound of Conjecture 1 was proved for graphs with a dominating edge. The full statement of Conjecture 1 was later proved for this case in the papers [13, 16, 28] using a different method, but the proof is quite involved. As a side result, our new approach also gives a simpler and shorter proof of Conjecture 1 for graphs with a dominating edge.

We also study non-bipartite D2C graphs with maximum degree (note that the only D2C graphs with maximum degree are stars). It turns out that these graphs can be described precisely, and they all have a dominating edge.

We prove Theorem 4 (and a stronger version for special cases) in Section 2 and obtain, in passing, a shorter proof of Conjecture 1 for graphs with a dominating edge. We then characterize D2C graphs with maximum degree in Section 3. Finally, we conclude in Section 4.

2 Conjecture 1 and beyond for graphs with a dominating edge

In this section, we further develop a proof technique of Hanson and Wang [11], who showed the bound of Conjecture 1 for D2C graphs having a dominating edge. Their result was extended (using a different technique) by Haynes et al. and Wang (see the series of papers [13, 16, 28]) to prove the full conjecture for this class. Nevertheless, it turns out that when excluding bipartite graphs, the Murty-Simon bound is not tight for this class of graphs. To show this, we use the original idea of Hanson and Wang [11] and extend it by a finer analysis. To demonstrate the potential strength of this technique, we give, along the way, a shorter proof of Conjecture 1 for graphs with a dominating edge.

In order to have a smoother presentaion, we split this section into several parts. We start with establishing the general setting of the proof technique in Section 2.1. Then, we prove some general lemmas in Sections 2.2 and 2.3. We use some of these lemmas to give our new proof of Conjecture 1 for graphs with a dominating edge in Section 2.4. Then, in Section 2.5, we prove a stronger bound than the one of Theorem 4 for some special cases. Finally, in Section 2.6, we conlclude the proof of Theorem 4.

2.1 Preliminaries: notations and setting for the proofs

Let us first fix our notation. Given a vertex from a graph , we denote by and the open and closed neighbourhoods of , respectively. An edge between vertices and is denoted , while a non-edge between and is denoted . An oriented graph is a graph where edges have been given an orientation; oriented edges are called arcs. If is oriented towards , we denote the arc from to by . In an oriented graph, we denote by , , and the out-neighbourhood, closed out-neighbourhood, in-neighbourhood, and closed in-neighbourhood of vertex . In an oriented graph, a directed cycle is a cycle such that all arcs are oriented in the same cyclic direction. We say that a source is a vertex with and while a sink is a vertex with and (we consider that an isolated vertex is neither a source nor a sink). A triangle on an oriented graph is transitive if it induces a subgraph with a source and a sink.

We start with the following definition, which is fundamental to our study.

Definition 5.

Let be a D2C graph. An edge is critical for a pair of vertices if the only path of length 1 or 2 from to uses the edge .

The following observation is easy but important.

Observation 6.

An edge in a D2C graph is critical for a pair with or with .

We are ready to describe the setting for the proofs of this section. Let be a D2C graph with vertices and edges, and let be a dominating edge of . We split the other vertices of into four sets (see Figure 4 for an illustration):

Figure 4: The structure of a D2C graph with the dominating edge (the only edges that are depicted are those incident with or ). Lemma 7 allows us to represent all vertices in as adjacent to .

We have the following fact.

Lemma 7.

Either , or (or both).

Proof.

Let , and assume by contradiction that there is . We have since is critical for the pairs and . Thus, and have a common neighbour . However, since is a dominating edge, . Suppose without loss of generality that , then there are two paths of length  between and : one going through and one going through . Thus, the edge is not critical for the pair , a contradiction. ∎

Because of Lemma 7, in the whole section, without loss of generality, we will always assume that . We next prove the following lemma.

Lemma 8.

The following properties hold.

  • There is no edge between and .

  • If , then .

  • If , then every vertex in (resp. ) has a neighbour in (resp. ).

  • If , then every vertex in (resp. ) that has at least one neighbour in (resp. ) has a non-neighbour in (resp. ).

Proof.

(a) Let and assume by contradiction that there is such that . Then there are two paths of length 2 between and : one going through and one going through . Thus, the edge is not critical for the pair , a contradiction.

(b) If , then the edge can only be critical for the pair . This implies that and have no common neighbour, that is, .

(c) Assume by contradiction that there is a vertex (without loss of generality) such that . Then since . This implies that the edge is critical for the pair , and thus , a contradiction.

(d) Assume by contradiction that there is a vertex (without loss of generality) such that . Then, the edge is not critical. Indeed, it cannot be critical for the pair since has a neighbour in . It cannot be critical for a pair with : since we have by (b), and by (c), has a neighbour in . So, there is a path of length 2 from to going through . Finally, it cannot be critical for a pair with since every neighbour of is either in (thus, a neighbour of ) or in (and a neighbour of ). Observation 6 ensures that we considered all the cases, and reached a contradiction which proves the claim. ∎

Following the proof of Hanson and Wang [11], we will next partition the vertices of into two parts and , and prove that every edge within or within can be assigned injectively to a non-edge between and . This will prove that has at most as many edges as the complete bipartite graph with parts and .

We define the partition as follows:

Lemma 9.

For every edge (resp. ), there exists (resp. ) such that is critical for either the pair or the pair .

Proof.

Assume without loss of generality that . This implies that both and are neighbours of . Then, the edge cannot be critical for the pair . Without loss of generality, we assume it is critical for , where . However, if , then and are neighbours of , and then the edge is not critical for this pair. Hence, . ∎

We use Lemma 9 to define a function assigning the edges of and to non-edges between and , as follows. For every edge (resp. ), we select one vertex (resp. ) such that is critical for the pair , where (such vertex exists by Lemma 9). We let . Note that is well-defined, since is critical for the pair and thus . This construction is depicted in Figure 5.

Figure 5: The construction of the function . The edge is critical for the pair , thus .
Lemma 10.

The function is injective.

Proof.

Assume by contradiction that is not injective. Without loss of generality, let be the non-edge between and such that there are two edges and in verifying .

By definition of , both and are critical for the pair . This implies that and form the unique path of length  from to . Thus, one of or is in , a contradiction. ∎

We saw in Lemma 10 that is injective. Moreover, we will show later that, if is not bipartite, then is not surjective. We call any non-edge in that has no preimage by , an -free non-edge. We also let be the number of -free non-edges.

Lemma 11.

We have .

Proof.

By the injectivity of (Lemma 10) and the definition of , there are exactly more non-edges between and than edges inside and inside . Thus, has exactly edges.

Without loss of generality, we assume that and we pose . By the above paragraph, we have . Since , this implies that . In particular, we now have:

Because and are integers, we have . Moreover, since , we obtain that . ∎

Lemma 11 implies that has at most edges: this is the result of Hanson and Wang [11]. It will require some more effort to prove the whole Conjecture 1: our aim will be to show that has at least a certain size. First, we will prove some general lemmas.

2.2 Preliminaries for the case

We now prove a useful lemma about , which is illustrated in Figure 6.

Lemma 12.

Let be a vertex in , and let be the set of vertices such that the non-edge is not -free. Then, for each vertex , there is a vertex in such that . Denote by the set of vertices of such that for some vertex of . Then, the following holds.

  • We have (that is, is injective).

  • The only edges in are those of the form .

  • For any two vertices of , if one of the edges or exists, then one of the non-edges and is -free. If both edges and exist, then both non-edges and are -free.

  • We have .

Proof.

Let . By Lemma 8(a), has no neighbour neither in nor in . Thus, and have a common neighbour, , in , and . We let . Now, if for some pair of distinct vertices of , we had , then one of the non-edges and would be -free (since both can only be assigned to by ), a contradicton. Thus, and (a) is true.

Moreover, there is no edge for two distinct vertices in , since otherwise and would have two common neighbours ( and ), contradicting the fact that . Thus, (b) holds.

Finally, assume that there is an edge in (the proof is the same for the edge ). Then, both non-edges and can only be assigned to the edge , so one of them is -free. If we have both edges and , then both endpoints of and have two common neighbours, so both are -free, and (c) is true.

To prove (d), we let be the number of -free non-edges incident with a vertex of . Let be some vertex of . By the previous parts of the lemma, we have:

which completes the proof of (d). ∎

Figure 6: Illustration of Lemma 12. We have and . Each edge in and induces an -free non-edge between the two sets (-free non-edges are depicted in bold).

2.3 Preliminaries for the case : the -orientation and related lemmas

In this section, we gather some lemmas about the structure of and when . They will be useful to our proofs but we feel that they could perhaps be used again. So we assume from here on in this in this section that is a D2C graph with . Observe that, by Lemma 8, .

We will use to define an orientation, called -orientation, of the edges induced by and by , as follows. Let be an edge within or within with . Then, we orient towards and we denote the resulting arc by . This construction is shown in Figure 7. Since is injective (Lemma 10), each edge of and receives exactly one orientation. From now on, all arcs considered are those of this -orientation. We denote by , , and the out-neighbourhood, closed out-neighbourhood, in-neighbourhood, and closed in-neighbourhood of vertex with respect to the -orientation, while and continue to denote the neighbourhood and closed neighbourhood of in .

Figure 7: The -orientation is constructed from the function : we orient all edges within and , and an edge is oriented from to if .

We will now study the properties of the -orientation. The first important lemma is the following (see Figure 8 for an illustration).

Lemma 13.

Let (resp. ) be two vertices such that is an arc of the -orientation. If neither nor is incident with an -free non-edge, then there exists a vertex (resp. ) such that (resp. ).

Proof.

Assume without loss of generality that . First, if there exists a vertex of that is not adjacent to , then and have as a common neighbour and thus the non-edge could only be assigned by to , contradicting the -orientation of . Thus, we have . Now, by the -orientation of , there exists a vertex of with . Assume now, for a contradiction, that there exists another vertex in that is adjacent to but not to . Then, and have as a common neighbour, so the non-edge can only be assigned by to . This is a contradiction and proves the claim. ∎

Figure 8: Illustration of Lemma 13: if and are two vertices in (resp. ) such that is an arc of the -orientation and neither nor is incident with an -free non-edge, then the neighbourhood of in (resp. ) is exactly the neighbourhood of in (resp. ) plus one vertex.

The next lemma states that directed cycles in yield many -free non-edges.

Lemma 14.

Let be a cycle of that is directed with respect to the -orientation. Then, there are at least -free non-edges incident with the vertices of .

Proof.

Let be a cycle of that is directed with respect to the -orientation (with the arc for each in ). Without loss of generality, is in . In this proof, we consider the addition modulo .

By definition of the -orientation, for all , there exists a vertex such that . Note that we may have for some .

Let , and let be the first predecessor of in the cyclic order of such that . This clearly happens at some point since . Note that we have , since otherwise and would have two common neighbours: and , a contradiction since implies that is the unique common neighbour of and .

We now prove that is -free. Assume by contradiction that it is not -free. Then, there exists an edge such that . However, since and have a common neighbour which is , we necessarily have . But by definition of , we already have the vertex such that , and thus . This is a contradiction, which implies that is -free. This is illustrated in Figure 9.

Finally, we prove that any two -free non-edges found with this method are distinct. Assume by contradiction that there are two vertices , in with which lead to the same -free non-edge (with ). Then, this means that is the first predecessor of such that . In particular, since , this implies that , which contradicts the fact that .

Thus, there are at least -free non-edges incident with the vertices of . ∎

Figure 9: Illustration of the proof of Lemma 14: is the first predecessor of in the cycle such that . The non-edge is then -free.

We now show that transitive triangles also induce -free non-edges.

Lemma 15.

Let (resp. ) be three pairwise adjacent vertices such that , and are oriented edges. Then there is an -free non-edge incident with .

Proof.

Without loss of generality, assume that . Assume by contradiction that there is no -free non-edge incident with . By Lemma 8(c), has a neighbour in , and by definition of the -orientation, we can assume that . Similarly, there is a vertex of adjacent to with (clearly ). Then neither nor is adjacent to since is the unique common neighbour of and . Now similarly, and cannot be adjacent, the only common neighbour of and is . But then the non-edge is assigned to by , which contradicts the injectivity of . This is illustrated in Figure 10. ∎

Figure 10: Illustration of the proof of Lemma 15: there is an -free non-edge incident with the ancestor in a transitive triangle.

The next two lemmas state that each source and each sink of has an -free non-edge in its closed neighbourhood. Recall that we do not consider isolated vertices as sources or sinks.

Lemma 16.

Let be a sink of the -orientation. Then, there is at least one -free non-edge incident with the vertices of .

Proof.

Let (without loss of generality) be a sink of the -orientation, and let be the in-neighbours of (recall that they all belong to ).

By Lemma 8(c), has a neighbour . Let be an in-neighbour of . If is not adjacent to , then the non-edge is -free, since otherwise we should have and the arc , a contradiction since is a sink. Thus, from now on we may assume that all in-neighbours of are adjacent to , for otherwise the statement of the claim holds.

Like every edge, the edge is critical. It cannot be critical for the pair , since (by the previous paragraph) these two vertices have all vertices in as common neighbours. It also cannot be critical for a pair consisting of and a neighbour of : if then is an ; and if then . Hence, must be critical for a pair with a neighbour of . Since , we necessarily have , and for every . However, among all the non-edges and the non-edge , all but one are -free: their only possible preimage by the function if . This is depicted in Figure 11. Since at least one exists, there is at least one -free non-edge incident with , and the claim follows. ∎

Figure 11: If is a sink, then its closed in-neighbourhood is incident with -free non-edges: among all the bolded non-edges, only one can have a preimage by the function , this preimage being .
Lemma 17.

Let be a source of the -orientation. Then, there is at least one -free non-edge incident with the vertices of .

Proof.

Let (without loss of generality) be a source of the -orientation, and let be the out-neighbours of (recall that they all belong to ). Assume by contradiction that no -free non-edge is incident with .

By Lemma 8(d), has a non-neighbour . Lemma 13 ensures that no is adjacent to . The non-edge has a preimage by the function , and this preimage is necessarily an arc with , since the are the only neighbours of in and none of those are adjacent to .

Let . If , then , and since no non-edge incident with a is -free we necessarily have , a contradiction to the fact that . Thus, is a non-edge incident with , thus it has a preimage by the function . This preimage is necessarily the edge .

Furthermore, by Lemma 8(c), has a neighbour . Lemma 13 ensures that . Now, the edge is critical. It cannot be critical for the pair since those vertices share as a common neighbour. It cannot be critical for a pair with since by definition of critical edges we would have , and one of the two non-edges ,