Results 21 to 30 of about 990 (254)

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

Bounds on the Locating Roman Domination Number in Trees

open access: yesDiscussiones Mathematicae Graph Theory, 2018
A Roman dominating function (or just RDF) on a graph G = (V, E) is a function f : V → {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. The weight of an RDF f is the value f(
Jafari Rad Nader, Rahbani Hadi
doaj   +1 more source

Algorithmic Aspects of the Independent 2-Rainbow Domination Number and Independent Roman {2}-Domination Number

open access: yesDiscussiones Mathematicae Graph Theory, 2022
A 2-rainbow dominating function (2RDF) of a graph G is a function g from the vertex set V (G) to the family of all subsets of {1, 2} such that for each vertex v with g(v) =∅ we have ∪u∈N(v) g(u) = {1, 2}.
Poureidi Abolfazl, Rad Nader Jafari
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

On the total Roman domination stability in graphs

open access: yesAKCE International Journal of Graphs and Combinatorics, 2021
A total Roman dominating function on a graph G is a function satisfying the conditions: (i) every vertex u with f(u) = 0 is adjacent to at least one vertex v of G for which f(v) = 2; (ii) the subgraph induced by the vertices assigned non-zero values has ...
Ghazale Asemian   +3 more
doaj   +1 more source

Roman {2}-Bondage Number of a Graph

open access: yesDiscussiones Mathematicae Graph Theory, 2020
For a given graph G=(V, E), a Roman {2}-dominating function f : V (G) → {0, 1, 2} has the property that for every vertex u with f(u) = 0, either u is adjacent to a vertex assigned 2 under f, or is adjacent to at least two vertices assigned 1 under f. The
Moradi Ahmad   +2 more
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Bounds on the Double Italian Domination Number of a Graph

open access: yesDiscussiones Mathematicae Graph Theory, 2022
For a graph G, a Roman {3}-dominating function is a function f : V → {0, 1, 2, 3} having the property that for every vertex u ∈ V, if f(u) ∈ {0, 1}, then f(N[u]) ≥ 3.
Azvin Farzaneh, Rad Nader Jafari
doaj   +1 more source

Weak signed Roman domination in graphs [PDF]

open access: yesCommunications in Combinatorics and Optimization, 2020
A weak signed Roman dominating function (WSRDF) of a graph $G$ with vertex set $V(G)$ is defined as a function $f:V(G)\rightarrow\{-1,1,2\}$ having the property that $\sum_{x\in N[v]}f(x)\ge 1$ for each $v\in V(G)$, where $N[v]$ is the closed ...
Lutz Volkmann
doaj   +1 more source

Total double Roman domination in graphs [PDF]

open access: yesCommunications in Combinatorics and Optimization, 2020
Let $G$ be a simple graph with vertex set $V$. A double Roman dominating function (DRDF) on $G$ is a function $f:V\rightarrow\{0,1,2,3\}$ satisfying that if $f(v)=0$, then the vertex $v$ must be adjacent to at least two vertices assigned $2$ or one ...
Guoliang Hao   +2 more
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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