Strong connectivity and directed triangles in oriented graphs. Partial results on a particular case of the Caccetta-Häggkvist conjecture

preprint2011arXivOpen access

Abstract

A particular case of Caccetta-Häggkvist conjecture, says that a digraph of order nn with minimum out-degree at least 1/3n1/3n contains a directed cycle of length at most 3. Recently, Kral, Hladky and Norine proved that a digraph of order nn with minimum out-degree at least 0.3465n0.3465n contains a directed cycle of length at most 3 (which currently is the best result). A weaker particular case says that a digraph of order nn with minimum semi-degree at least 1/3n1/3n contains a directed triangle. In a recent paper, by using the result of Kral et al, the author proved that for β≥0.343545β\geq 0.343545, any digraph DD of order nn with minimum semi-degree at least βnβn contains a directed cycle of length at most 3 (which currently is the best result). This means that for a given integer d≥1d\geq 1, every digraph with minimum semi-degree dd and of order mdmd with m≤2.91082m\leq 2.91082, contains a directed cycle of length at most 3. In particular, every oriented graph with minimum semi-degree dd and of order mdmd with m≤2.91082m\leq 2.91082, contains a directed triangle. In this paper, by using the result of Kral et al, we prove that every oriented graph with minimum semi-degree dd, of order mdmd with 2.91082<m≤32.91082< m\leq 3 and of strong connectivity at most 0.679d0.679d, contains a directed triangle. This will be implied by a more general and more precise result, valid not only for 2.91082<m≤32.91082< m\leq 3 but also for larger values of mm. As application, we improve two existing results. The first result (Authors Broersma and Li), concerns the number of the directed cycles of length 4 of a triangle free oriented graph of order nn and of minimum semi-degree at least n3\frac{n}{3}. The second result (Authors Kelly, Kühn and Osthus), concerns the diameter of a triangle free oriented graph of order nn and of minimum semi-degree at least n5\frac{n}{5}.

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