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Convex Roman Dominating Function in Graphs

open access: yesEuropean Journal of Pure and Applied Mathematics, 2023
Let $G$ be a connected graph. A function $f:V(G)\rightarrow \{0,1,2\}$ is a \textit{convex Roman dominating function} (or CvRDF) if every vertex $u$ for which $f(u)=0$ is adjacent to at least one vertex $v$ for which $f(v)=2$ and $V_1 \cup V_2$ is convex.
Rona Jane Fortosa, Sergio Canoy
openaire   +1 more source

Total Protection of Lexicographic Product Graphs

open access: yesDiscussiones Mathematicae Graph Theory, 2022
Given a graph G with vertex set V (G), a function f : V (G) → {0, 1, 2} is said to be a total dominating function if Σu∈N(v) f(u) > 0 for every v ∈ V (G), where N(v) denotes the open neighbourhood of v. Let Vi = {x ∈ V (G) : f(x) = i}. A total dominating
Martínez Abel Cabrera   +1 more
doaj   +1 more source

A note on k-Roman graphs [PDF]

open access: yesOpuscula Mathematica, 2013
Let \(G=\left(V,E\right)\) be a graph and let \(k\) be a positive integer. A subset \(D\) of \(V\left( G\right) \) is a \(k\)-dominating set of \(G\) if every vertex in \(V\left( G\right) \backslash D\) has at least \(k\) neighbours in \(D\).
Ahmed Bouchou   +2 more
doaj   +1 more source

Calculating Modern Roman Domination of Fan Graph and Double Fan Graph [PDF]

open access: yesJournal of Applied Sciences and Nanotechnology, 2022
This paper is concerned with the concept of modern Roman domination in graphs. A Modern Roman dominating function on a graph is labeling such that every vertex with label 0 is adjacent to two vertices; one of them of label 2 and the other of label 3 and ...
Saba Salah, Ahmed Omran, Manal Al-Harere
doaj   +1 more source

A characterization of trees with equal Roman $\{2\}$-domination and Roman domination numbers [PDF]

open access: yesCommunications in Combinatorics and Optimization, 2019
Given a graph $G=(V,E)$ and a vertex $v \in V$, by $N(v)$ we represent the open neighbourhood of $v$. Let $f:V\rightarrow \{0,1,2\}$ be a function on $G$. The weight of $f$ is $\omega(f)=\sum_{v\in V}f(v)$ and let $V_i=\{v\in V \colon f(v)=i\}$, for $i=0,
Abel Cabrera Martinez, Ismael G. Yero
doaj   +1 more source

Bounds on the restrained Roman domination number of a graph

open access: yesCommunications in Combinatorics and Optimization, 2016
A {\em Roman dominating function} on a graph $G$ is a function‎ ‎$f:V(G)\rightarrow \{0,1,2\}$ satisfying the condition that every‎ ‎vertex $u$ for which $f(u) = 0$ is adjacent to at least one vertex‎ ‎$v$ for which $f(v) =2$‎.
H‎. ‎Abdollahzadeh Ahangar   +1 more
doaj   +1 more source

The Roman domination and domatic numbers of a digraph [PDF]

open access: yesCommunications in Combinatorics and Optimization, 2019
Let $D$ be a simple digraph with vertex set $V$. A Roman dominating function (RDF) on a digraph $D$ is a function $f: V\rightarrow \{0,1,2\}$ satisfying the condition that every vertex $v$ with $f(v)=0$ has an in-neighbor $u$ with $f(u)=2$. The weight
Z.Xie1, G. Hao, Sh. Wei
doaj   +1 more source

A note on the Roman domatic number of a digraph [PDF]

open access: yesCommunications in Combinatorics and Optimization, 2020
A {\em Roman dominating function} on a digraph $D$ with vertex set $V(D)$ is a labeling $f\colon V(D)\to \{0, 1, 2\}$ such that every vertex with label $0$ has an in-neighbor with label $2$. A set $\{f_1,f_2,\ldots,f_d\}$ of Roman dominating functions
Lutz Volkmann, D. Meierling
doaj   +1 more source

On trees with equal Roman domination and outer-independent Roman domination number [PDF]

open access: yesCommunications in Combinatorics and Optimization, 2019
A Roman dominating function (RDF) on a graph $G$ is a function $f : V (G) \to \{0, 1, 2\}$ satisfying the condition that every vertex $u$ for which $f(u) = 0$ is adjacent to at least one vertex $v$ for which $f(v) = 2$.
S. Nazari-Moghaddam, S.M. Sheikholeslami
doaj   +1 more source

Double Roman reinforcement number in graphs

open access: yesAKCE International Journal of Graphs and Combinatorics, 2021
For a graph a double Roman dominating function is a function having the property that if f(v) = 0, then vertex v must have at least two neighbors assigned 2 under f or one neighbor w with f(w) = 3, and if f(v) = 1, then vertex v must have at least one ...
J. Amjadi, H. Sadeghi
doaj   +1 more source

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