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Signed double Roman domination in graphs
Discrete Applied Mathematics, 2019 zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Hossein Abdollahzadeh Ahangar +1 more
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On the signed Roman k-domination in graphs
Quaestiones Mathematicae, 2020 Let k ≥ 1 be an integer and G be a simple and finite graph with vertex set V (G). A signed Roman k-dominating function (SRkDF) on a graph G is a function f : V (G) → {−1, 1, 2} such that (i) every vertex v with f(v) = −1 is adjacent to at least one ...
Lutz Volkmann, S M Sheikholeslami
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[k]-Roman Domination in Digraphs
Symmetry, 2023 Let D=(V(D),A(D)) be a finite, simple digraph and k a positive integer. A function f:V(D)→{0,1,2,…,k+1} is called a [k]-Roman dominating function (for short, [k]-RDF) if f(AN−[v])≥|AN−(v)|+k for any vertex v∈V(D), where AN&
Ruijuan Li, Xinhong Zhang
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Signed Roman domination in digraphs
Journal of Combinatorial Optimization, 2013zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Seyed Mahmoud Sheikholeslami +1 more
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Signed Roman domination in graphs
Journal of Combinatorial Optimization, 2012zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Vladimir Samodivkin +2 more
exaly +4 more sources
Signed Roman kk-Domination in Graphs
Graphs and Combinatorics, 2015zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Michael A. Henning, Lutz Volkmann
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Signed Roman k-domination in Digraphs
Graphs and Combinatorics, 2015zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Lutz Volkmann
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Outer independent signed double Roman domination
Journal of Applied Mathematics and Computing, 2021Suppose $$[3]=\{0,1,2,3\}$$ and $$[3^{-}]=\{-1,1,2,3\}$$ . An outer independent signed double Roman dominating function (OISDRDF) of a graph
Seyed Mahmoud Sheikholeslami +1 more
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Roman domination in signed graphs
2022Summary: Let \(S = (G, \sigma)\) be a signed graph. A function \(f: V \rightarrow \{0,1,2\}\) is a Roman dominating function on \(S\) if (i) for each \(v \in V\), \(f(N[v]) = f(v) + \sum_{u \in N(v)} \sigma (uv) f(u) \geq 1\) and (ii) for each vertex \(v\) with \(f(v) = 0\) there exists a vertex \(u\in N^+(v)\) such that \(f(u) = 2\).
Joseph, James, JOSEPH, MAYAMMA
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