نتایج جستجو برای: signed roman edge k dominating function
تعداد نتایج: 1662242 فیلتر نتایج به سال:
In this paper, we present an improved algorithm to decide whether a graph of maximum degree 3 has an edge dominating set of size k or not, which is based on enumerating vertex covers. We first enumerate vertex covers of size at most 2k and then construct an edge dominating set based on each vertex cover to find a satisfied edge dominating set. To enumerate vertex covers, we use a branch-and-red...
Given a graph G = (V,E) together with a nonnegative integer requirement on vertices r : V → Z+, the annotated edge dominating set problem is to find a minimum set M ⊆ E such that, each edge in E −M is adjacent to some edge in M , and M contains at least r(v) edges incident on each vertex v ∈ V . The annotated edge dominating set problem is a natural extension of the classical edge dominating se...
a emph{signed graph} (or, in short, emph{sigraph}) $s=(s^u,sigma)$ consists of an underlying graph $s^u :=g=(v,e)$ and a function $sigma:e(s^u)longrightarrow {+,-}$, called the signature of $s$. a emph{marking} of $s$ is a function $mu:v(s)longrightarrow {+,-}$. the emph{canonical marking} of a signed graph $s$, denoted $mu_sigma$, is given as $$mu_sigma(v) := prod_{vwin e(s)}sigma(vw).$$the li...
Tutte observed that every nowhere-zero k-flow on a plane graph gives rise to a kvertex-coloring of its dual, and vice versa. Thus nowhere-zero integer flow and graph coloring can be viewed as dual concepts. Jaeger further shows that if a graph G has a face-k-colorable 2-cell embedding in some orientable surface, then it has a nowhere-zero k-flow. However, if the surface is nonorientable, then a...
a function $f:v(g)rightarrow {-1,0,1}$ is a {em minusdominating function} if for every vertex $vin v(g)$, $sum_{uinn[v]}f(u)ge 1$. a minus dominating function $f$ of $g$ is calleda {em global minus dominating function} if $f$ is also a minusdominating function of the complement $overline{g}$ of $g$. the{em global minus domination number} $gamma_{g}^-(g)$ of $g$ isdefined as $gamma_{g}^-(g)=min{...
Tutte observed that every nowhere-zero k-flow on a plane graph gives rise to a kvertex-coloring of its dual, and vice versa. Thus nowhere-zero integer flow and graph coloring can be viewed as dual concepts. Jaeger further shows that if a graph G has a face-k-colorable 2-cell embedding in some orientable surface, then it has a nowhere-zero k-flow. However, if the surface is nonorientable, then a...
Let G=(V(G),E(G)) be a connected simple undirected graph with non empty vertex set V(G) and edge set E(G). For a positive integer k, by an edge irregular total k-labeling we mean a function f : V(G)UE(G) --> {1,2,...,k} such that for each two edges ab and cd, it follows that f(a)+f(ab)+f(b) is different from f(c)+f(cd)+f(d), i.e. every two edges have distinct weights. The minimum k for which G ...
A double Roman dominating function on a graph G=(V,E) is f:V?{0,1,2,3} with the properties that if f(u)=0, then vertex u adjacent to at least one assigned 3 or two vertices 2, and f(u)=1, 2 3. The weight of f equals w(f)=?v?Vf(v). domination number ?dR(G) G minimum G. said be ?dR(G)=3?(G), where ?(G) We obtain sharp lower bound generalized Petersen graphs P(3k,k), we construct solutions providi...
Let G = (V,E) be a finite undirected graph. An edge set E ⊆ E is a dominating induced matching (d.i.m.) in G if every edge in E is intersected by exactly one edge of E. The Dominating Induced Matching (DIM ) problem asks for the existence of a d.i.m. in G; this problem is also known as the Efficient Edge Domination problem. The DIM problem is related to parallel resource allocation problems, en...
A Roman domination function on a graph G is a function r : V (G) → {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 a Roman function is the value r(V (G)) = ∑ u∈V (G) r(u). The Roman domination number γR(G) of G is the minimum weight of a Roman domination function on G . "Roman Criticality" has been ...
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