Counting graph orientations with no directed triangles

05/27/2020 ∙ by Pedro Araújo, et al. ∙ UFRJ IMPA Universidade Federal do ABC 0

Alon and Yuster proved that the number of orientations of any n-vertex graph in which every K_3 is transitively oriented is at most 2^⌊ n^2/4⌋ for n ≥ 10^4 and conjectured that the precise lower bound on n should be n ≥ 8. We confirm their conjecture and, additionally, characterize the extremal families by showing that the balanced complete bipartite graph with n vertices is the only n-vertex graph for which there are exactly 2^⌊ n^2/4⌋ such orientations.

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1. Introduction

Given a graph and an oriented graph , we say that is an -free orientation of if contains no copy of . We denote by the family of -free orientations of and we write . In 1974, Erdős [Er74] posed the problem of determining the maximum number of -free orientations of , for every -vertex graph . Formally, we define

Since every orientation of an -free graph does not contain any orientation of , it is fairly straightforward to see that , where is the maximum number of edges in an -free graph on vertices. For a tournament on vertices, Alon and Yuster [AlYu06] proved that  for with a very large , as they use the Regularity Lemma [Sz75]. For tournaments with three vertices, they avoid using the regularity lemma to prove that for , where is slightly less than . Furthermore, for the strongly connected triangle, denoted by , using a computer program they verified that and for . In view of this, Alon and Yuster posed the following conjecture.

Conjecture 1 (Alon and Yuster [AlYu06]).

For , we have

Using a simple computer program, we checked that is the only -vertex graph that maximizes . This fact together with the verification made by Alon and Yuster for graphs with at most seven vertices implies the following proposition.

Proposition 1.1.

and among all graphs with vertices, if and only if . Furthermore, for .

In this paper we prove the following result that confirms Conjecture 1 and states that the balanced complete bipartite graph is the only -vertex graph for which there are exactly orientations with no copy of .

Theorem 1.2.

For , we have Furthermore, among all graphs  with vertices, if and only if .

Overview of the paper. Our proof is divided into two parts. Proposition 3.3 deals with graphs with at most vertices, and its proof is given in the appendix (Section 5); and Theorem 1.2 deals with general graphs (Section 3). The proofs of these results are somehow similar and consist of an analysis of the size of a maximum clique of the given graph. In each step, we partition the vertices of a graph  into a few parts and, using the results presented in Section 2, explore the orientations of the edges between these parts that lead to -free orientations of . Our proof is then reduced to solving a few equations which, in the case of the proof of Proposition 3.3, can be checked by straightforward computer programs. In Section 4 we present some open problems. The reader is referred to [Bo78, Di10] for standard terminology on graphs.

2. Extensions of -free orientations

In this section we provide several bounds on the number of ways one can extend a -free orientation of a subgraph of a graph to a -free orientation of .

Given subgraphs and of , we write for the subgraph of with vertex set and edge set . Let and be orientations, respectively, of and with the property that any edge of gets the same orientation in and . We denote by the orientation of following the orientations and .

Let be a graph and . For simplicity, we say that an orientation of the subgraph of induced by the set of edges is an orientation of . The next definition is a central concept of this paper.

Definition (Compatible orientations).

Given a graph , disjoint sets , and orientations of and of , we say that and are compatible if is -free.

Given a graph and disjoint sets , denote by the set of edges of between and and by the spanning subgraph of induced by . It is useful to have an upper bound on number of -free orientations of that are compatible with a fixed orientation of . This quantity is precisely the maximum number of ways one can extend a -free orientation of to a -free orientation of .

Definition.

Given a graph and disjoint sets , , let . We define as follows:

For simplicity, when , we write instead of . In the rest of this section we give upper bounds for for specific graphs and subgraphs and . If induces a complete graph with vertices, then we remark that any -free orientation of is a transitive orientation, which thus induces a unique ordering of the vertices of , called the transitive ordering of , such that every edge () is oriented from to in .

Given a graph , a vertex and a clique , we denote by the number of neighbors of in . Consider a -free orientation of and note that if we have a transitive ordering of , then there are exactly ways to extend this ordering to a transitive ordering of , as it depends only on the position in which we place in with respect to its neighbors in  (there are such positions). We summarize this discussion in the following proposition.

Proposition 2.1.

Given a graph , and . If is a complete graph, then .

In the next two results, we give an upper bound for when induces a complete graph and is an edge. We denote by the neighborhood of in and denotes the number of common neighbors of and in .

Lemma 2.2.

Let be an integer and let be a graph. If induce disjoint cliques with and such that , then

Proof.

Let and be arbitrary -free orientations of and respectively. Suppose without loss of generality that assigns the orientation of from to  and consider the transitive ordering of

. We estimate in how many ways one can include

and in the ordering while keeping it transitive. Since and are cliques, by Proposition 2.1 we have and , which gives at most positions to put the vertices and in the transitive ordering of . Note that there are ways to place in the transitive ordering of such that appears after and they have a common neighbor between them. But each such ordering induces a . This finishes the proof. ∎

The following corollary bounds the number of extensions when is a maximum clique of and is an edge.

Corollary 2.3.

Let be an integer and let be a -free graph. If are disjoint cliques with and , then

Proof.

Let and , and put . If , then by applying Proposition 2.1 twice, with and , we have

Therefore, we assume that . Note that since is -free, we have . Applying Lemma 2.2 and using the fact that, for , we have , we obtain

(1)

One can check that the right-hand side of (1) is a polynomial on of degree with negative leading coefficient and it is a growing function in the interval . Since , we have

Given a graph , an edge , and an orientation of , we say that the orientation of is forced if there is only one orientation of compatible with . In the next two lemmas we provide bounds for the number of -free orientations of -free graphs.

Lemma 2.4.

Let be a -free graph and let , be disjoint cliques of size . Then .

Proof.

First, note that if , then the trivial bound implies . Also, since is -free, we have . Thus, we may assume that , i.e., is a . Let and so that is not an edge and consider an arbitrary orientation of the edges and .

If the oriented edges are and (or, by symmetry, and ), then for the two possible orientations of , the orientation of one of the two remaining edges in is forced. Thus, since there is only one edge left to orient in , which can be done in two ways, we have .

It remains to consider the case where the oriented edges are and (or, by symmetry, and ). If is oriented from to , then the orientation of the two remaining edges in are forced, which gives us one -free orientation. On the other hand, if is oriented from to , then one can orient the both remaining edges in in two ways, which in total gives that . ∎

Lemma 2.5.

Let be a -free graph and let and with . If induces a copy of , then .

Proof.

Consider an arbitrary orientation of the edges of . We may assume that , as otherwise we have . Since is -free and induces a copy of , the vertex must have exactly three neighbors in , which span an induced path . By symmetry, we assume that is oriented from to . If we orient from to , then the orientation of is forced, which leaves two possible orientations for the edge . On the other hand, if we orient from to , we just apply Proposition 2.1 to conclude that . Combining the possible orientations, we obtain . ∎

We now provide an upper bound for (see Lemma 2.7 below) in a specific configuration of a -free graph , and subsets of vertices and , which is proved using the following proposition.

Proposition 2.6.

Let be a path , and let . Given an orientation of , there are at most eight orientations of compatible with . Moreover, if the edges and are oriented, respectively, towards and , then there are at most  such orientations.

Proof.

By Proposition 2.1, there are three orientations of (resp. ) compatible with , and hence there are at most nine orientations of compatible with . In these orientations, each direction of and appears at least once. If is oriented towards (resp. towards ), then the orientations in which and are oriented, respectively, towards and (resp. and ) are not compatible with . Therefore, there are at most eight orientations of compatible with . Now, suppose that and are oriented towards and . If we orient towards (resp. ), then must be oriented towards (resp. must be oriented towards ), and there are three orientations of (resp. four orientations of ) from which we can complete a compatible orientation of . Therefore, there are at most seven orientations of . ∎

Lemma 2.7.

Let be a -free graph and let , be disjoint cliques of size . Then .

Proof.

Let and . Since is -free, cannot be adjacent to every vertex of , for . This implies that , for . Analogously, we have , for . Thus the set of edges in joining and induces a set of paths and cycles. Since is -free, does contain a cycle of length . If , then , as desired. If , then some vertex, say , is incident to two edges of , say and , which implies that , and hence , as desired. If , then induces a path of length , say . In this case, note that and are disjoint cliques of size , and hence, by Lemma 2.4, we have . Since each edge in either joins to , or is adjacent to , we have , as desired.

Thus, we may assume , and hence induces the cycle . By symmetry, we may assume that , are both oriented towards , and is oriented towards . Suppose is oriented towards . If we orient towards , then must be oriented towards , and, since and are oriented towards and , by Proposition 2.6, there are compatible orientations of the edges in the path (see Figure 0(a)). If we orient towards , then must be oriented towards , and by Proposition 2.6, there are compatible orientations of the edges in the path (see Figure 0(b)). Thus, there are compatible orientations of , as desired. Thus, we may assume that is oriented towards , and hence must be oriented towards . If we orient towards , then must be oriented towards , and by Proposition 2.6, there are compatible orientations of the edges in the path (see Figure 0(c)). If we orient towards , then must be oriented towards , and, since and are oriented towards , regardless of the orientation of , by Proposition 2.6, there are compatible orientations of the edges in the path (see Figure 0(d)) Thus, there are compatible orientations of , as desired. ∎

(a)

(b)

(c)

(d)
Figure 1. Compatible orientations between two cliques of size in a -free graph.

3. Proof of the main theorem

In this section we prove our main result, Theorem 1.2. In order to bound the number of -free orientations of a graph , we decompose it into disjoint cliques of different sizes and we use the results of Section 2 to bound the number of extensions of -free orientations between those cliques. Before moving to the proof of the main theorem though, we need bounds on the number of -free orientations of some small graphs. The first one concerns the complete tripartite graph .

Proposition 3.1.

For any positive integer , we have

Proof.

Let be a complete tripartite graph with vertex partition . For there are orientations of the edges incident to with exactly out-neighbors of in , and in-neighbors of in , sets which we denote by and respectively. For each of those orientations, the edges between and and between and are forced in any -free orientation. Since any of the other edges can be oriented in two ways, we sum over and to get

In the rest of the paper, we count the number of -free orientations of a graph by decomposing its vertex set and we often use the following inequality without explicit reference. For a partition of the vertices of a graph into sets and we have, from the definition of , that

When is a clique, we define and use the bound

In the proof of Theorem 1.2, we first show that for every graph containing a . For -free graphs we may use Lemma 2.7 to bound the number of extensions between two triangles. But when considering graphs with no two disjoint triangles, we need the following result.

Lemma 3.2.

Let be a -free graph with vertices that contains a triangle , a matching such that and are not incident to the vertices of , and that does not contains two vertex-disjoint triangles. Then, .

Proof.

Let be as in the statement. Recall that and . Moreover, for , by Corollary 2.3. Also, since is triangle-free, and hence . Throughout the proof, we use each of these bounds unless the structure of allows us to obtain a better bound.

First note that if there is at most one edge between and , then . In this case we use the bound

to obtain , which allows us to restrict to graphs such that .

We count the number of orientations by considering different values of . In particular, since is -free, we have that for . First note that if for , then . Therefore, if there are at most six edges between and , then either there are at most three edges between each and , which implies ; or, without loss of generality, there are at most two edges between and , which implies . In both cases we have that and consequently that .

Thus, we assume that . Then, without loss of generality, we have and, by Turán’s Theorem, is isomorphic to . If , the aforementioned bounds and Lemma 3.1 yields

Finally, if for both , then the graphs are isomorphic to with being the vertex of degree in . Since does not contain two disjoint triangles, then and since , we have in fact . Finally, Lemma 3.1 yields . ∎

In the remainder of this section we prove Theorem 1.2, which follows by induction on the number of vertices. Unfortunately, we need a slightly stronger base of induction than the one given by Proposition 1.1, which is the content of the next proposition. We present its proof in the Appendix (Section 5). Recall that the clique number of a graph , denoted by , is the size of a clique in with a maximum number of vertices.

Proposition 3.3.

Let be an -vertex graph. If , then . Furthermore, if and only if .

We are now ready to prove Theorem 1.2, which is rewritten as follows:

Theorem (Theorem 1.2).

Let be an -vertex graph. If , then Furthermore, if and only if .

Proof.

Let . The proof follows by induction on . By Proposition 1.1, the statement holds for . If , then the result follows from Mantel’s Theorem (see [Ma1907]) for , and from Proposition 3.3 for , as . Thus, assume and suppose that the statement holds for any graph with less than vertices (but at least 8 vertices).

Let be a clique of of size . If , then the result follows from Proposition 3.3, so we may assume that . Thus, we can apply the induction hypothesis for any subgraph of with at least vertices.

If , then we have . By Proposition 1.1, we have and, by Proposition 2.1, for each vertex we have . Therefore, applying the induction hypothesis to we have

(2)

where we used that . From now on we assume that and consequently that . Due to the different structure of the graphs with small clique numbers, we divide the rest of the proof according to the value of .

Case . Let . Since is -free, every vertex of is adjacent to at most vertices of . Then, by Proposition 2.1, we have for every . Therefore, the following holds for and .

(3)

Case . Let . By the induction hypothesis, for any , we have . If contains a vertex with degree smaller than , then

(4)

Thus, we may assume that . Since , we have and since is -free, each vertex in contains at most neighbors in . Hence, we have