نتایج جستجو برای: signed total roman k domination number
تعداد نتایج: 2157122 فیلتر نتایج به سال:
The Fibonacci cube Γn is the subgraph of the n-dimensional cube Qn induced by the vertices that contain no two consecutive 1s. Using integer linear programming, exact values are obtained for γt(Γn), n ≤ 12. Consequently, γt(Γn) ≤ 2Fn−10 + 21Fn−8 holds for n ≥ 11, where Fn are the Fibonacci numbers. It is proved that if n ≥ 9, then γt(Γn) ≥ d(Fn+2 − 11)/(n− 3)e−1. Using integer linear programmin...
For a positive integer k, a total {k}-dominating function of a graph G without isolated vertices is a function f from the vertex set V (G) to the set {0, 1, 2, . . . , k} such that for any vertex v ∈ V (G), the condition ∑ u∈N(v) f(u) ≥ k is fulfilled, where N(v) is the open neighborhood of v. The weight of a total {k}-dominating function f is the value ω(f) = ∑ v∈V f(v). The total {k}-dominati...
A double Roman dominating function of a graph $G$ is $f:V(G)\rightarrow \{0,1,2,3\}$ having the property that for each vertex $v$ with $f(v)=0$, there exists $u\in N(v)$ $f(u)=3$, or are $u,w\in $f(u)=f(w)=2$, and if $f(v)=1$, then adjacent to assigned at least $2$ under $f$. The domination number $\gamma_{dR}(G)$ minimum weight $f(V(G))=\sum_{v\in V(G)}f(v)$ among all functions $G$. An outer i...
A function f : V (G) → {0, 1, 2} is a Roman dominating function if for every vertex with f(v) = 0, there exists a vertex w ∈ N(v) with f(w) = 2. We introduce two fractional Roman domination parameters, γR ◦ f and γRf , from relaxations of two equivalent integer programming formulations of Roman domination (the former using open neighborhoods and the later using closed neighborhoods in the Roman...
a set $s$ of vertices of a graph $g=(v,e)$ without isolated vertex is a {em total dominating set} if every vertex of $v(g)$ is adjacent to some vertex in $s$. the {em total domatic number} of a graph $g$ is the maximum number of total dominating sets into which the vertex set of $g$ can be partitioned. we show that the total domatic number of a random $r$-regular graph is almost...
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