Seymour’s Second Neighborhood Conjecture for <i>m</i>-Free Oriented Graphs

Ma, Huawen

doi:https://doi.org/10.1155/2022/1463016

Mathematical Problems in Engineering

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Research Article | Open Access

Volume 2022 | Article ID 1463016 | https://doi.org/10.1155/2022/1463016

Seymour’s Second Neighborhood Conjecture for m-Free Oriented Graphs

Huawen Ma¹

Academic Editor: Ganesh Ghorai

Received27 May 2022

Revised16 Jul 2022

Accepted18 Jul 2022

Published11 Aug 2022

Abstract

Let be an oriented graph with minimum out-degree . For , let and be the out-degree and second out-degree of in , respectively. For a directed graph , we say that a vertex is a Seymour vertex if . Seymour in 1990 conjectured that each oriented graph has a Seymour vertex. A directed graph is called -free if there are no directed cycles with length at most in . A directed graph is called -transitive if, for any directed -path of length , there exists . In this paper, we show that (1) each -free oriented graph has a Seymour vertex and (2) each vertex with minimum out-degree in -free and -transitive oriented graph is a Seymour vertex. The latter result improves a theorem of Daamouch (2021).

1. Introduction

Let be a directed graph (digraph) with vertex set and edge set . For any , we write meaning that and meaning that . For a vertex , let and be the out-neighborhood and second out-neighborhood of , respectively. Let and be out-degree and second out-degree of , respectively. Let be the minimum out-degree of . For any , let be the subdigraph of induced by . Let and . Denote by and a directed -path of length and a directed -cycle, respectively. We may omit the subscript if it is clear from the context. For a vertex , if , then we say is a . Seymour in 1990 proposed the following conjecture.

Conjecture 1 (Seymour’s second neighborhood conjecture). Each oriented graph has a Seymour vertex.
This conjecture has not been well solved since it was proposed. This problem is widely used in the field of computer network, especially, in fuzzy information and complex network. On the application of this problem, we refer readers to [1–3]. One direction on this conjecture is to consider some specific classes of digraphs. Fisher [4] verified the SSNC for tournaments using a certain probability distribution on the vertices. Havet and Thomassé [5] gave a short proof of SSNC for tournaments using a tool called median orders. Kaneko and Locke [6] proved SSNC for oriented graphs with minimum out-degree smaller than 7. Fidler and Yuster [7] considered the case of tournaments missing a matching. Ghazal [8–10] showed that SSNC holds for tournaments missing -generalized stars and other class of oriented graphs. A digraph is -transitive (resp., -quasi-transitive, resp., -antitransitive) if for any , the existence of a directed path of length from to implies that (resp., or , resp., ). García-Vásquez and Hernández-Cruz [11] proved that SSNC holds for 4-transitive oriented graphs. Gutin and Li [12] proved SSNC for 2-quasi-transitive oriented graphs. Moussa Daamoch [13–15] proved SSNC for 5-antitransitive oriented graphs, -transitive oriented graphs , and 3-quasi-transitive oriented graphs.
The other direction on SSNC is to bound such that for each oriented graph . Then, Conjecture 1 is . Chen, Shen, and Yuster [16] showed that is the unique real root in (0, 1) of . A digraph is -free if contains no directed cycles with length at most . Liang and Xu [17] showed that if is an -free oriented graph and , then .
Recently, Moussa Daamoch [14] proved that if is an -free digraph, and , then has a Seymour vertex. Naturally, we may ask the following problem: if is -free, whether there exists a Seymour vertex in or not. In this paper, we first show that SSNC holds for -free oriented graphs.

Theorem 1. Let be an oriented graph with the minimum out-degree . If is -free, then has a Seymour vertex.

A related problem to SSNC is to find more than one Seymour vertex. As we know, a transitive tournament has exactly one Seymour vertex. This problem was proved for some classes of oriented graphs, for example, tournaments, quasitransitive oriented graphs, bipartite tournaments, and -transitive oriented graphs (see [5,11,12,14,18–21]). Recently, in [14], Daamouch gave the following two results.

Theorem 2 (see [4]). Let be an -free and -transitive oriented graph. If , then has a Seymour vertex.

Theorem 3 (see [4]). Let be a -transitive oriented graph. If , then every vertex with minimum out-degree in is a Seymour vertex.

Combining with the above two theorems, we may ask the following natural problem: for an -free and -transitive oriented graph , whether each vertex with minimum out-degree in is a Seymour vertex or not. We definitely answered this problem in this paper.

Theorem 4. Let be an -free and -transitive oriented graph. Then, every vertex with a minimum out-degree in is a Seymour vertex.

In the remainder of this paper, the proof of Theorem 1 is given in Section 2, and the proof of Theorem 4 is postponed in Section 3. In Sections 4 and 5, the advantages, limitations, and innovations of this article are summarized. Finally, we sum up a conclusion and propose some open problems.

2. Proof of Theorem 1

We first introduce a Lemma due to M. Daamouch [13] before proving Theorem 1, which plays a key role in the proofs of Theorems 1 and 4.

Lemma 1 (see Daamouch [13]). Let be an oriented graph with the minimum out-degree and let . If there exists such that , then is a Seymour vertex.

Proof. .
Let be an oriented graphs with minimum out-degree . By the theorem of Kaneko and Locke [6], we may assume that . Let such that and . If there is such that , then is a Seymour vertex by Lemma 1. So we may assume that for each , which implies that contains at least 1 directed cycle. Note that contains no directed cycles with length at most . Then, must contain a directed cycle with length or . We split it into the following two cases.

Case 1. contains a directed cycle of length .
Without loss of generality, assume that is a directed cycle of length in . Recall that is -free. Then, (1) for each , , , and . (2) If , then . Hence, there exists such that . Without loss of generality, we may suppose that . If , then , is a Seymour vertex. So we suppose that . Then, . We claim that , . For otherwise, , is a Seymour vertex.
If there exists such that , then is a Seymour vertex by Lemma 1. Now, we may suppose that for each which implies that contains a directed cycle , with length or . Note that which implies that . (For otherwise, it yields a directed cycle in of length smaller than .) Thus, , and the length of is . Note that and . Then, . Hence, there exists such that and . That is to say, and . Then, . Thus, , and is a Seymour vertex.

Case 2. contains a directed cycle of length and contains no directed cycle of length .
Without loss of generality, we may assume that is a directed cycle in of length . Note that . We may suppose that . Then, . For otherwise, contains a directed cycle of length which implies that . If , then , and is a Seymour vertex. Hence, we may suppose that . And similarly, . Then, . We claim that . For otherwise, , and is a Seymour vertex.
By Lemma 1, we may suppose that for each , which implies that contains a directed cycle , with length . Clearly, . Recall that . Then, . If there exists such that , then , is a Seymour vertex. Hence, we may assume that for each , . We claim that . For otherwise, , is a Seymour vertex. Thus, there exists such that . That is to say, . Note that . Hence, . Thus, is a Seymour vertex, and we are done.

3. Proof of Theorem 4

We now give the proof of Theorem 4. We spilt this proof into two cases according to whether larger than and give the following two lemmas.

Lemma 2. Let be an -free and -transitive oriented graph. If , then every vertex with minimum out-degree in is a Seymour vertex.

Proof. If , then by Theorem 3, we are done. So we may suppose that . It is sufficient to consider . Let be a vertex with minimum out-degree. Suppose to contrary that is not a Seymour vertex. Set . Clearly, is -free. Then, by Lemma 1, is a directed cycle of length which implies that and for .
We first claim that for any , . Since otherwise, . Thus, is a Seymour vertex, a contradiction.
Clearly, there exists a vertex, say , such that . (In fact, for any , .) Also, there exists a vertex, say , such that . We can apply this process until , and thus, there exists . We claim that . For otherwise, there exists a vertex, say such that . Thus, is a directed path of length which implies that , a contradiction. Note that and . We next split it into the following two cases according to whether directs to .

Case 3. .
Denote . In , directs to and may directs to . Recall that which implies that there exists a vertex, say , such that . Thus, is a directed path of length . Recall that is -transitive. Hence, we have , a contradiction.

Case 4. .
Recall that and which implies that . Since is -free, does not direct to any vertex except and in . Then, there exists a vertex, say , such that . Thus, is a directed path of length which implies that , a contradiction. □

Lemma 3. Let be an -free and -transitive oriented graph. If , then every vertex with minimum out-degree in is a Seymour vertex.

Proof. As the same proof of Lemma 2, we may suppose that . Let be a vertex with minimum out-degree. Suppose to contrary that is not a Seymour vertex. Set . Since is -free, is also -free. By Lemma 1, we have for each , , which implies that contains directed cycle of length at least . That is to say, there is a directed path of length in . Clearly, there exists such that . Note that and for each since is -free. But , hence, there exists such that . We can apply this process until ; thus, there exists . We claim that there exists such that . For otherwise, . Note that and , which implies that . Recall that . That is to say, there is a vertex, say such that . Clearly, there exists such that . Since is -free and , . Thus, is a directed path of length . Since is -transitive, . This contradicts with .
Consider the directed path of length . Since is -transitive, . Note that and . We now split it into the following two cases according to whether directs to .

Case 5. .
Recall that . Note that is a directed path of length which implies that . Since is -free, . Recall that and . Hence, . Recall that . That is to say, there is a vertex, say such that . Clearly, there exists such that . Thus, is a directed path of length . Since is -transitive, . This contradicts with .

Case 6. .
Recall that and . Then, we have , and there is a vertex, say , such that . Clearly, for since is -free. Note that is a directed path of length which implies that . Thus, we have which implies that there exists a vertex, say , such that . Since is -free, for . Thus, is a directed path of length which implies that . This contradicts with .

4. Comparison and Analysis

Seymour’s second neighborhood conjecture has always been a classical problem. Much research was done in this field, but it still remains open. It was proven only for some very specific classes of directed graphs. It can be seen that this is not an easy problem. Daamoch [14] showed that if is an -free digraph, and , then has a Seymour vertex. We in this paper improve this result in oriented graphs and show that if is an -free oriented graph, then has a Seymour vertex. In addition, the condition of transitivity is widely used as a limiting condition in this problem. Daamoch [14] showed that if is an -free and -transitive oriented graph, then has a Seymour vertex. We in this paper also improve this result and show that each vertex with minimum out-degree in such is a Seymour vertex. And we believe that the transitivity conditions can be reduced. In this paper, we combine the condition of transitivity with the condition of cycles-free well, and some interesting results are given.

5. Advantages and Limitations

Now, we analyze some advantages and limitations of this paper. Firstly, we have indeed improved some important results of predecessors although the improvement is not too heavy. Second, in the proofs of theorems, we used some innovative methods, which provide some new ideas for future proofs. However, our method is only applicable to digraphs without short cycles. For digraphs with short cycles, new ideas are still needed. Third, this thought can be applied to other problems of graph theory, such as graph partitioning and coloring. Given an oriented graph , we can partition into and by placing each vertex into if there exists a directed path of length between and a fixed vertex , and placing into otherwise. Also, the thought can be applied to solving the indexes of graphs [22–28].

6. Conclusion and Future Research

Based on Theorems 1 and 4, we propose the following natural problems.

Problem 1. What is the largest positive integer , such that each -free-oriented graph has a Seymour vertex?
Obviously, Theorem 1 shows that . When , we believe that the conclusion is also true, but more details need to be considered in the proof. For Theorem 4, we propose the following problem.

Problem 2. Let be an -free and -transitive oriented graph. Then, whether each vertex with minimum out-degree in is a Seymour vertex or not?
Clearly, here, we reinforce the condition of transitivity. Then, a lot of structures will be added to the oriented graphs, and we also need to consider more structures of oriented graphs.

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares no conflicts of interest.

Acknowledgments

This work was supported by the Scientific Research Program Funded by Shaanxi Provincial Education Department (Program No. 21JK0994) and Natural Science Basic Research Program of Shaanxi (Program Nos. 2022JQ-026 and YDBK 2019-71).

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Copyright © 2022 Huawen Ma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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