On $k$-rainbow domination in middle graphs

Let $G$ be a finite simple graph with vertex set $V(G)$ and edge set $E(G)$. A function $f : V(G) \rightarrow \mathcal{P}(\{1, 2, \dotsc, k\})$ is a \textit{$k$-rainbow dominating function} on $G$ if for each vertex $v \in V(G)$ for which $f(v)= \emptyset$, it holds that $\bigcup_{u \in N(v)}f(u) = \{1, 2, \dotsc, k\}$. The weight of a $k$-rainbow dominating function is the value $\sum_{v \in V(G)}|f(v)|$. The \textit{$k$-rainbow domination number} $\gamma_{rk}(G)$ is the minimum weight of a $k$-rainbow dominating function on $G$. In this paper, we initiate the study of $k$-rainbow domination numbers in middle graphs. We define the concept of a middle $k$-rainbow dominating function, obtain some bounds related to it and determine the middle $3$-rainbow domination number of some classes of graphs. We also provide upper and lower bounds for the middle $3$-rainbow domination number of trees in terms of the matching number. In addition, we determine the $3$-rainbow domatic number for the middle graph of paths and cycles.


Introduction
Let G = (V, E) be a connected undirected graph with the vertex set V = V (G) and edge set E = E(G). The order of G is defined as the cardinality of V . The open neighborhood of v ∈ V (G) is the set N (v) = {u ∈ V (G) | uv ∈ E(G)} and the closed neighborhood of v ∈ V (G) is the set N [v] := N (v) ∪ {v}. The degree of v ∈ V (G) is defined as the cardinality of N (v), denoted by deg G (v). When no confusion arises, we may delete the subscript G in deg G (v). The maximum degree and minimum degree of G are denoted by ∆(G) and δ(G), respectively We write P n , C n and K n for a path, a cycle and a complete graph, respectively.
In [8], Hamada and Yoshimura defined the middle graph of a graph. The middle graph M (G) of a graph G is the graph obtained by subdividing each edge of G exactly once and joining all these newly introduced vertices of adjacent edges of G. The precise definition of M (G) is as follows. The vertex set V (M (G)) is V (G) ∪ E(G). Two vertices v, w ∈ V (M (G)) are adjacent in M (G) if (i) v, w ∈ E(G) and v, w are adjacent in G or (ii) v ∈ V (G), w ∈ E(G) and v, w are incident in G.
In the graph domination, a set of vertices is selected as guards such that each vertex not selected has a guard as a neighbor. As a generalization   [3]. In the k-rainbow domination, k-different types of guards are required in the neighborhood of a non-selected vertex. Let [k] be the set of positive integers at most k. A function f : V (G) → P([k]) is a k-rainbow dominating function on G if for each vertex v ∈ V (G) for which f (v) = ∅, it holds that u∈N (v) f (u) = [k]. The weight of a k-rainbow dominating function is the value v∈V (G) |f (v)|. The k-rainbow domination number γ rk (G) is the minimum weight of a k-rainbow dominating function on G. In [4], Chang et al. proved that the k-rainbow domination is NP-complete. So, it is worthwhile to determine the k-rainbow domination numbers of some classes of graphs. Indeed, there are many papers on the 2-rainbow domination. The latest survey of the 2-rainbow domination is introduced in [2]. For k ≥ 3, it is more difficult to determine the k-rainbow domination number of a graph. The following are a few results on the 3-rainbow domination number. In [10], Shao et al. determined the 3-rainbow domination numbers of paths, cycles and generalized Petersen graphs P (n, 1). In [12], Wang et al. determined the 3-rainbow domination number of P 3 P n . In [7], Gao et al. determined the 3-rainbow domination numbers of C 3 C m and C 4 C m . In [5], Cynthia et al. determined the 3-rainbow domination number of circulant graph G(n; ±{1, 2, 3}). In [6], Furuya et al. proved that for every connected graph G of order n ≥ 8 with δ(G) ≥ 2, γ r3 (G) ≤ 5n 6 . To study the k-rainbow domination number in the class of middle graphs, we define the following concept.
The middle k-rainbow domination number γ ⋆ rk (G) of G is the minimum weight of a middle k-rainbow dominating function of G. A γ ⋆ rk (G)-function is a MkRDF on G with weight γ ⋆ rk (G). We remark that γ ⋆ rk (G) = γ rk (M (G)) for any graph G. In [9], it was considered only 2-rainbow domination numbers of the middle graphs. In this paper, we initiate the study of the middle k-rainbow domination in graphs. In particular, we determine the exact value of middle 3-rainbow domination numbers of some classes of graphs. A matching in a graph G is a set of pairwise nonadjacent edges. The maximum number of edges in a matching of a graph G is called the matching number of G and denoted by α ′ (G). We provide upper and lower bounds for the middle 3-rainbow domination number of trees in terms of the matching . The maximum number of functions in a k-rainbow dominating family on G is the k-rainbow domatic number of G, denoted by d rk (G). It is known that k-rainbow domatic number is well-defined and d rk (G) ≥ k for every graph G (See [11]). We determine the 3-rainbow domatic number for the middle graph of paths and cycles.
The rest of this section, we present some necessary terminology and notation. For terminology and notation on graph theory not given here, the reader is referred to [1]. Let T be a (rooted) tree. A leaf of T is a vertex of degree one. A pendant edge is an edge incident with a leaf. A support vertex is a vertex adjacent to a leaf. For a vertex v, C(v) denote the set of the children of v. D[v] denote the set of the descendants and v. The subtree induced by D[v] is denoted by T v . We write K 1,n−1 for the star of order n ≥ 3. The double star DS p,q , where p, q ≥ 1, is the graph obtained by joining the centers of two stars K 1,p and K 1,q . A healthy spider S t,t is the graph from a star K 1,t by subdividing each edges of K 1,t . A wounded spider S t,r is the graph from a star K 1,t by subdividing r edges of K 1,t , where r ≤ t − 1. Note that a star K 1,t is a wounded spider S t,0 . For a graph G and its subset S, G − S denotes the subgraph of G induced by V (G) \ V (S). A diametral path of G is a path with the length which equals the diameter of

General bounds of the middle k-rainbow domination number
In this section, we obtain general bounds of the middle k-rainbow domination number. First, we begin by giving a simple lower bound on the middle k-rainbow domination number. [3], then x ∈ E(G) for otherwise x can not dominate the other vertices. Thus, G = P 2 . Now assume that there is no element with weight 3. If f (v) = ∅ for some v ∈ V (G), then there exist at least two edges e 1 , e 2 incident to v such that f (e 1 ) and f (e 2 ) are not empty. But, end vertices of e 1 , e 2 except for v are not dominated, a contradiction. Thus, every vertex in V (G) has non-zero weight so that there are at most three vertices in the graph G. One can easily check that G = K 3 or P 2 .
We proceed by induction on the order n of T . Obviously, the statement is true for a path P 3 .
Let T be a tree of order n ≥ 4. Suppose that every tree T ′ of order Assume that T is a double star DS p,q with p ≥ q ≥ 1. Then p + q + 3 = n + 1 = γ ⋆ r3 (T ) ≤ 3n 2 . Now we assume that T is neither a star or a double star. Then it is easy to see that T has diameter at least four. Among all of diametrical paths in T , we choose x 0 x 1 . . . x d so that it maximizes the degree of x d−1 . Root T at x 0 . We divide our consideration into three cases.
Applying the induction hy- where y ∈ C(x), and g(x) = f (x) otherwise. Then clearly g is a M3RDF of T and so Applying the induction hypothesis 3. The middle 3-rainbow domination number of paths, cycles and complete graphs In this section, we determine the middle 3-rainbow domination number of paths, cycles and complete graphs.

Now assume that
Clearly, g is a M3KDF of P n−3 with weight at most ω(f ) − 4. By the induction hypothesis, we have for 0 ≤ i ≤ n−4 3 and h(x) = ∅ otherwise. It is easy to see that h is a M3RDF of P n with weight 4n−1 3 . Thus, we have γ ⋆ r3 (P n ) = 4n−1 3 . Case 2. n ≡ 2 (mod 3). By the same argument as in Case 1, we can show that γ ⋆ r3 (P n ) ≥ 4n+1 3 .
It is easy to see that g is a M3RDF of P n with weight 4n+1 3 . Thus, we have γ ⋆ r3 (P n ) = 4n+1 3 . Case 3. n ≡ 0 (mod 3). By the same argument as in Case 1, we can show that γ ⋆ r3 (P n ) ≥ 4n 3 .
It is easy to see that h is a M3RDF of P n with weight 4n 3 . Thus, we have γ ⋆ r3 (P n ) = 4n 3 . Proposition 3.2. For n ≥ 3, where the subscript k of v k is read by modulo n. Let f be a γ ⋆ r3 (P n )-function such that the size of N := {v i | f (v i ) = ∅} is as small as possible. We divide our consideration into three cases.
otherwise. It is easy to see that g is a M3RDF of P n−2 with weight at most γ ⋆ r3 (C n ) − 3. By Proposition 3.1, we have Assume that [3]. Thus, it follows from n ≥ 4 that where t = |N |. By Theorem 2.4, γ ⋆ r3 (C n ) ≤ γ ⋆ r3 (P n ) + 1. Thus, we have γ ⋆ r3 (C n ) = 4n+2 3 . Case 2. n ≡ 2 (mod 3). If there exists some k such that |f otherwise. It is easy to see that g is a M3RDF of P n−2 with weight at most γ ⋆ r3 (C n ) − 3. By Proposition 3.1, we have

Now suppose that
Clearly, g is a M3RDF of P n−1 with weight at most γ ⋆ r3 (C n ) − 2. By Proposition 3.1, we have Assume that |f (v k )| ≤ 1 for each k ∈ [n]. By the same argument as Case [3] and h(x) = ∅ otherwise. It is easy to see that h is a M3RDF of C n with weight 4n+1 3 . Thus, we have γ ⋆ r3 (C n ) = 4n+1 3 . Case 3. n ≡ 0 (mod 3). If there exists some k ∈ [n] such that |f (v k )| ≥ 1, then define g : 3 and h(x) = ∅ otherwise. It is easy to see that h is a M3RDF of C n with weight 4n 3 . Thus, we have γ ⋆ r3 (C n ) = 4n 3 .
if n is even; if n is odd.
for 1 ≤ i ≤ n−1 2 and g(x) = ∅ otherwise. It is easy to see that g is a M3RDF of K n with weight 3n−1 2 . If n is even, then define h : It is easy to see that h is a M3RDF of K n with weight 3n 2 . Now we claim that γ ⋆ r3 (K n ) ≥ 3n 2 if n is even and γ ⋆ r3 (K n ) ≥ 3n−1 2 if n is odd. One can easily check that γ ⋆ r3 (K 2 ) = 3, γ ⋆ r3 (K 3 ) = 4, and γ ⋆ r3 (K 4 ) = 6. We proceed by induction on n. Assume that n ≥ 5.
Case 1. n is odd.
Assume that |f (v i )| ≤ 1 for each v i ∈ V (K n ). Let N := {v i ∈ V (K n ) | f (v i ) = ∅}. For a fixed v i ∈ N , suppose that there exists no v j ∈ V (K n ) such that f (v i v j ) = ∅. Without loss of generality, we may assume that f (v i ) = {1}. To dominate elements in f (x) should contain 2 and 3 for each v k ∈ V (K n ) \ {v i }. Thus, we have s 2 , s 3 ≥ n 2 , where s j := |{x ∈ Then to dominate such an element x, y∈N M (x) f (y) should contain 1. Thus, we have s 1 ≥ n 2 so that ω(f ) ≥ |f (v i )| + s 2 + s 3 + s 1 = 1 + 3n 2 . Assume that for each v i ∈ N there exists a vertex v j ∈ V (K n ) such that f (v i v j ) = ∅. Let t be the size of set N . For v j ∈ V (K n ) \ N , x∈N M [v j ] f (x) should contain [3]. For v i ∈ N , x∈N M (v i ) f (x) contains at least one element.