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Seymour's Second Neighborhood Conjecture (SNC) asserts that every oriented graph has a vertex whose first out-neighborhood is at most as large as its second out-neighborhood. In this paper, we prove that if $G$ is a graph containing no…

Combinatorics · Mathematics 2020-10-22 Darine Al Mniny , Salman Ghazal

Seymour's Second Neighborhood Conjecture asserts that every digraph (without digons) has a vertex whose first out-neighborhood is at most as large as its second out-neighborhood. It is proved for tournaments, tournaments missing a matching…

Combinatorics · Mathematics 2015-09-08 Salman Ghazal

A vertex in a directed graph is said to have a large second neighborhood if it has at least as many second out-neighbors as out-neighbors. The Second Neighborhood Conjecture, first stated by Seymour, asserts that there is a vertex having a…

Discrete Mathematics · Computer Science 2021-10-25 Suresh Dara , Mathew C. Francis , Dalu Jacob , N. Narayanan

Seymour's Second-Neighborhood Conjecture states that every directed graph whose underlying graph is simple has at least one vertex $v$ such that the number of vertices of out-distance $2$ from $v$ is at least as large as the number of…

Combinatorics · Mathematics 2019-07-31 Farid Bouya , Bogdan Oporowski

Seymour conjectured that every oriented simple graph contains a vertex whose second neighborhood is at least as large as its first. In this note, we put forward a conjecture that we prove is actually equivalent: every oriented simple graph…

Combinatorics · Mathematics 2019-04-15 Tyler Seacrest

Seymour's second neighborhood conjecture states that every simple digraph (without digons) has a vertex whose first out-neighborhood is at most as large as its second out-neighborhood. Such a vertex is said to have the second neighborhood…

Combinatorics · Mathematics 2016-02-09 Salman Ghazal

Seymour's Second Neighborhood Conjecture asserts that every digraph (without digons) has a vertex whose first out-neighborhood is at most as large as its second out-neighborhood. We prove its weighted version for tournaments missing a…

Combinatorics · Mathematics 2015-05-11 Salman Ghazal

A longstanding conjecture of Seymour states that in every oriented graph there is a vertex whose second outneighbourhood is at least as large as its outneighbourhood. In this short note we show that, for any fixed $p\in[0,1/2)$, a.a.s.…

Combinatorics · Mathematics 2024-08-12 Alberto Espuny Díaz , António Girão , Bertille Granet , Gal Kronenberg

Seymour's second neighbourhood conjecture asserts that every oriented graph has a vertex whose second out-neighbourhood is at least as large as its out-neighbourhood. In this paper, we prove that the conjecture holds for quasi-transitive…

Discrete Mathematics · Computer Science 2017-11-21 Gregory Gutin , Ruijuan Li

Seymour's Second Neighborhood Conjecture (SSNC) asserts that every oriented finite simple graph (without digons) has a vertex whose second out-neighborhood is at least as large as its first out-neighborhood. Such a vertex is said to have…

Combinatorics · Mathematics 2024-06-07 Moussa Daamouch , Salman Ghazal , Darine Al-Mniny

Seymour's Second Neighborhood Conjecture (SNC) states that every oriented graph contains a vertex whose second neighborhood is as large as its first neighborhood. We investigate the SNC for orientations of both binomial and pseudo random…

Combinatorics · Mathematics 2022-11-15 Fábio Botler , Phablo F. S. Moura , Tássio Naia

Seymour conjectured that every oriented simple graph contains a vertex whose second neighborhood is at least as large as its first. Seymour's conjecture has been verified in several special cases, most notably for tournaments by Fisher. One…

Combinatorics · Mathematics 2012-12-11 Tyler Seacrest

The Second Neighborhood Conjecture of Seymour asserts that every oriented graph contains a vertex~$v$ satisfying $|\Npp(v)|\ge|\Np(v)|$. We introduce \emph{Pisa graphs} -- strongly connected oriented graphs~$D$ with $\Delta(D)=\max_{v\in…

Combinatorics · Mathematics 2026-05-25 Stanisław M. S. Halkiewicz

Let D be a simple digraph without loops or digons. For any v in V(D) let N_1(v) be the set of all nodes at out-distance 1 from v and let N_2(v) be the set of all nodes at out-distance 2. We provide sufficient conditions under which there…

Combinatorics · Mathematics 2008-08-19 James N. Brantner , Greg Brockman , Bill Kay , Emma E. Snively

Seymour Second Neighborhood Conjecture (SSNC) asserts that every finite oriented graph has a vertex whose second out-neighborhood is at least as large as its first out-neighborhood. Such a vertex is called a Seymour vertex. A digraph $D =…

Combinatorics · Mathematics 2024-05-29 Dania Mezher , Moussa Daamouch

The Second Neighborhood Conjecture states that every simple digraph has a vertex whose second out-neighborhood is at least as large as its first out-neighborhood, i.e. a vertex with the Second Neighborhood Property. A cycle intersection…

Combinatorics · Mathematics 2019-11-28 Michael Cary

Seymour's celebrated second neighborhood conjecture, now more than thirty years old, states that in every oriented digraph, there is a vertex $u$ such that the size of its second out-neighborhood $N^{++}(u)$ is at least as large as that of…

Combinatorics · Mathematics 2024-12-31 Hao Huang , Fei Peng

We prove Seymour's Second Neighborhood Conjecture when the missing graph is disjoint stars under some conditions. Weaker conditions are required when n=2 or 3. In some cases, we exhibit two vertices with the desired property.

Combinatorics · Mathematics 2016-10-13 Salman Ghazal

For a vertex $x$ of a digraph, $d^+(x)$ ($d^-(x)$, resp.) is the number of vertices at distance 1 from (to, resp.) $x$ and $d^{++}(x)$ is the number of vertices at distance 2 from $x$. In 1995, Seymour conjectured that for any oriented…

Combinatorics · Mathematics 2023-06-07 Jiangdong Ai , Stefanie Gerke , Gregory Gutin , Shujing Wang , Anders Yeo , Yacong Zhou

This paper gives an approximate result related to Seymour's Second Neighborhood conjecture, that is, for any $m$-free digraph $G$, there exists a vertex $v\in V(G)$ and a real number $\lambda_m$ such that $d^{++}(v)\geq \lambda_m d^+(v)$,…

Combinatorics · Mathematics 2017-01-03 Hao Liang , Jun-Ming Xu
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