k-Rainbow Domination Number of P 3 P n

Let k be a positive integer, and set [k] := {1, 2, . . . , k}. For a graph G, a k-rainbow dominating function (or kRDF) of G is a mapping f : V(G) → 2[k] in such a way that, for any vertex v ∈ V(G) with the empty set under f , the condition ⋃ u∈NG(v) f (u) = [k] always holds, where NG(v) is the open neighborhood of v. The weight of kRDF f of G is the summation of values of all vertices under f . The k-rainbow domination number of G, denoted by γrk(G), is the minimum weight of a kRDF of G. In this paper, we obtain the k-rainbow domination number of grid P3 Pn for k ∈ {2, 3, 4}.


Introduction
For a graph G, we denote by V(G) and E(G) the vertex set and the edge set of G, respectively.For a vertex v ∈ V(G), the open neighborhood of v, denoted by N G (v), is the set {u ∈ V(G) : uv ∈ E(G)} and the closed neighborhood of v, denoted by N G [v], is the set N G (v) ∪ {v}.The degree of a vertex v ∈ V(G), denoted by d G (v), is defined by d G (v) = |N G (v)|.We let δ(G) and ∆(G) denote the minimum degree and maximum degree of a graph G, respectively.
For graphs F and G, we let F G denote the Cartesian product of F and G. Vizing [16] conjectured that for arbitrary graphs F and G, γ(F G) ≥ γ(F)γ(G).This conjecture is still open, and the domination number or its related invariants of F G are extensively studied with the motivation from Vizing's conjecture.
Concerning the k-rainbow domination number of F G, one problem naturally arises: Given two graphs F and G under some conditions, determine γ rk (F G) for all k.In [3], the authors determined γ rk (P 2 P n ) for k = 3, 4, 5.
In this paper, we examine grid graphs P 3 P n , and determine the value γ rk (P P n ) for k ∈ {2, 3, 4} and all n, where P m is the path of order m.
To provide a complete answer, we need the following fact that can easily be proved as an exercise.
Theorem 1.For n ≥ 8, γ r2 (P Proof.First, we present constructions of a 2RDF of P 3 P n of the desired weight.
To show that these are also lower bounds, we prove there is a γ r2 (P is as small as possible.We claim that |S| = 0. Suppose, to the contrary, that |S| ≥ 1 and let s be the smallest positive integer for which ω( f s ) = 0.Then, ω( f s−1 ) + ω( f s+1 ) ≥ 6.Then, we consider the following cases.Case 1. s = 1 (the case s = n − 1 is similar).
Then, we have f (v 1 ) = f (u 1 ) = f (w 1 ) = {1, 2} and the function g defined by g(u 0 ) = {1}, otherwise, is a 2RDF of P 3 P n of weight at most ω( f ), which contradicts the choice of f .
. We may assume that {1} ⊆ f (w s−2 ).Hence, the function g defined by g ) = ∅ and g(x) = f (x) otherwise, is an 2RDF of P 3 P n of weight ω( f ), which is contradicting the choice of f .Therefore, |S| = 0.
We can see that for every 0
To provide a complete answer, we need the following fact.
To provide a complete answer, we need the following fact.
Proof.First, we show that γ r4 (P 3 P n ) ≤ 2n.To do this, we present constructions of a 4RDF of P 3 P n of the desired weight.