Discrete Applied Mathematics 166 (2014) 115–122

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Discrete Applied Mathematics

journal homepage: www.elsevier.com/locate/dam

Rainbow domination numbers on graphs with given radiusShinya Fujita a, Michitaka Furuya b,∗

a International College of Arts and Sciences, Yokohama City University, 22-2 Seto, Kanazawa-ku, Yokohama, Kanagawa 236-0027, Japanb Department of Mathematical Information Science, Tokyo University of Science, 1-3 Kagurazaka, Sinjuku-ku, Tokyo 162-8601, Japan

a r t i c l e i n f o

Article history:Received 20 February 2011Received in revised form 5 October 2013Accepted 11 October 2013Available online 30 October 2013

Keywords:Rainbow domination numberk-rainbow dominating functionRadius

a b s t r a c t

For k ≥ 1, r ≥ 1 and n ≥ 1, let t∗(k, r; n) be the minimum value satisfying thatγrk(G) ≤ t∗(k, r; n) · n for any connected graph G of order nwith radius r; if no such graphexists, we set t∗(k, r; n) = ∞. For k ≥ 1 and r ≥ 1, let t∗(k, r) = lim supn→∞ t∗(k, r; n). Inthis paper, we investigate the behavior of the function t∗(k, r) and determine some exactvalues of t∗(k, r) when k or r is small.

© 2013 Elsevier B.V. All rights reserved.

1. Introduction

All graphs considered here are simple, finite, and undirected graphs. A k-rainbow dominating function f of a graph G, asintroduced in [1], is a function f : V (G) → 2[k] such that, for each vertex v ∈ V (G), either f (v) = ∅ or

u∈N[v]

f (u) = [k].Naturally, an isolated vertex v must have f (v) = ∅. The k-rainbow domination number γrk(G) of G is the minimum weightw(f ) =

v∈V (G) |f (v)| of a k-rainbow dominating function f . A k-rainbow dominating function f of G is γrk(G)-function

if w(f ) = γrk(G). Please see [3] for standard graph theory notation not defined here. In general, determining γrk(G) isNP-complete (see [2]) so it is important to find good bounds on γrk.

Let G be a graph, and let u, v ∈ V (G). Define distG(u, v) to be the length of the shortest path from u to v in G.For i ≥ 0, define the ith neighborhood of v in G to be Ni(v) = {x ∈ V (G) | distG(v, x) = i}. In particular, defineN0(v) = {v}. A vertex is central in G if its greatest distance from any other vertex is as small as possible. This distanceis called the radius of G (denoted by rad(G)). A rooted tree T with root x is called a complete l-ary tree with height h ifV (T ) =

0≤i≤h Ni(x) and every vertex y ∈ V (T ) − Nh(x) has exactly l children. For a graph G, a γrk(G)-function f is called

flat if |{v ∈ V (G) | f (v) = ∅}| ≥ |{v ∈ V (G) | f ′(v) = ∅}| for every γrk(G)-function f ′.For k ≥ 1 and n ≥ 1, let t(k; n) be the minimum value satisfying that γrk(G) ≤ t(k; n) · n for any connected graphs G of

order n. For k ≥ 1, let

t(k) = lim supn→∞

t(k; n).

This value is studied in [4–6]. In order to present the value in a concise manner, Fujita et al. [4] provided the following table.

k 1 2 3 ≥ 4t(k) 1/2 3/4 8/9 1

∗ Corresponding author. Tel.: +81 3 3260 4271; fax: +81 3 3260 4271.E-mail addresses: shinya.fujita.ph.d@gmail.com (S. Fujita), michitaka.furuya@gmail.com (M. Furuya).

0166-218X/$ – see front matter© 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.dam.2013.10.020

116 S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122

Note that we can easily show t(k) ≤ 1. To see this, it suffices to show that γrk(G) ≤ |V (G)| for every graph G. We definea function f : V (G) → 2[k] as f (u) = {1} for every u ∈ V (G). We see that f is a k-rainbow dominating function of G becausethere exists no vertex assigned ∅ by f . This implies that γrk(G) ≤ w(f ) = |V (G)|. Hence, as is shown in the above table, wecannot hope for non-trivial upper bounds on rainbow domination numbers in general graphs for any k ≥ 4. But what willhappen if we consider this problem on some restricted graph classes? Motivated by this question, we focus on a family ofgraphs with given radius. We introduce the following function: for k ≥ 1, r ≥ 1 and n ≥ 1, let t∗(k, r; n) be the minimumvalue satisfying that γrk(G) ≤ t∗(k, r; n) · n for any connected graph G of order n with radius r; if no such graph exists, weset t∗(k, r; n) = ∞. For k ≥ 1 and r ≥ 1, let

t∗(k, r) = lim supn→∞

t∗(k, r; n).

Our purpose of this paper is to investigate the behavior of t∗(k, r).First we consider the case r = 1. Let k ≥ 1 be an integer. Let G be a connected graph of order n with radius 1, and x be a

central vertex of G. Note that N[x] = V (G). We define a function f : V (G) → 2[k] as

f (u) =

[k] (u = x)∅ (Otherwise).

We see that f is a k-rainbow dominating function of G and w(f ) = k(= knn). This implies that t∗(k, 1; n) ≤

kn . Thus

0 ≤ t∗(k, 1) = lim supn→∞ t∗(k, 1; n) ≤ lim supn→∞kn = 0, and hence t∗(k, 1) = 0. Thus we will hereafter consider

t∗(k, r) for r ≥ 2.In Section 2, wewill prepare for several useful lemmas in our argument. In Section 3, wewill show that t∗(k, r) < 1 holds

in general. In Section 4, we give some lower bounds on t∗(k, r) and also determine the exact value of t∗(k, r) for k = 1, 2, 3.Finally, in Section 5, we determine the exact value of t∗(k, 2).

2. Preliminaries

Lemma 2.1. Let k ≥ 1 be an integer. Let G be a graph and x be a vertex of G. Let f be a flat γrk(G)-function. If |f (x)| ≥ 2, then|{y ∈ N(x) | f (y) = ∅}| ≥ |f (x)|.

Proof. We define a function f ′: V (G) → 2[k] as

f ′(u) =

{1} (u ∈ {x} ∪ {y ∈ N(x) | f (y) = ∅})f (u) (Otherwise).

We see that f ′ is a k-rainbow dominating function of G because f ′(u) = ∅ implies u ∈ N[x], meaning

u′∈N[u] f′(u′) ⊇

u′∈N[u] f (u′) = [k] by the definition of f . We have w(f ′) = w(f ) − (|f (x)| − 1) + |{y ∈ N(x) | f (y) = ∅}|. Since

w(f ) ≤ w(f ′) and |f (x)| ≥ 2,

w(f ) ≤ w(f ′)

= w(f ) − (|f (x)| − 1) + |{y ∈ N(x) | f (y) = ∅}|

≤ w(f ) − 1 + |{y ∈ N(x) | f (y) = ∅}|,

and hence 1 ≤ |{y ∈ N(x) | f (y) = ∅}|. This implies |{v ∈ V (G) | f ′(v) = ∅}| > |{v ∈ V (G) | f (v) = ∅}|. Since f is flat, f ′ isnot a γrk(G)-function, and so w(f ) + 1 ≤ w(f ′). Thus w(f ) + 1 ≤ w(f ′) = w(f ) − (|f (x)| − 1) + |{y ∈ N(x) | f (y) = ∅}|,and hence |f (x)| ≤ |{y ∈ N(x) | f (y) = ∅}|, as desired. �

As an immediate corollary, we obtain the following:

Corollary 2.2. Let k ≥ 1 be an integer. Let G be a graph and x be a leaf of G. Write N(x) = {x′}. Let f be a flat γrk(G)-function.

Then the following hold.(i) |f (x)| ≤ 1.(ii) If f (x) = ∅, then d(x′) ≥ |{y ∈ N(x′) | f (y) = ∅}| ≥ |f (x′)| = k.

Lemma 2.3. Let k ≥ 1 be an integer and G be a graph of order at most k. Then γrk(G) = |V (G)|.

Proof. Let f be a γrk(G)-function. We may assume that there exists a vertex v ∈ V (G) such that f (v) = ∅. Thenu∈N[v]

f (u) = [k], and hence γrk(G) = w(f ) ≥ k ≥ |V (G)|. Since γrk(G) ≤ |V (G)|, this implies that γrk(G) = |V (G)|. �

It is easy to verify the following lemma.

Lemma 2.4. Let k ≥ 1 be an integer. Let G be a graph, and let G1, . . . ,Gm be all of the components of G. Then γrk(G) =1≤i≤m γrk(Gi).

Lemma 2.5 (Fujita et al. [4]). For every k ≥ 4 and every n ≥ 1, γrk(Pn) = n.

S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122 117

3. An upper bound of t∗(k, r)

Theorem 3.1. Let k ≥ 3 and r ≥ 2 be two integers. Then there exists a number 0 < ϵ < 1 and an integer n0 such that

γrk(G) ≤ ϵn

for every connected graph G of order n ≥ n0 and rad(G) = r.

In particular, this means that γrk(G) < n.

Proof. Let l = ⌊r/2⌋. We can choose a number ϵ such that(k − 1)2l+1

− k + 1k(k − 2)

− l

(1 − ϵ)kk − 2

+k − 2ϵk − 1

< ϵ < 1.

Let n0 be an integer such that(k − 1)2l+1

− k + 1k(k − 2)

− l

(1 − ϵ)kk − 2

+k − 2ϵk − 1

n0 +

((k − 1)2l+1− k + 1)

(k − 2)≤ ϵn0.

Let G be a graph of order n ≥ n0 and rad(G) = r . Suppose that γrk(G) > ϵn. Let x be a central vertex of G. For each0 ≤ i ≤ 2l + 1, let mi = |Ni(x)|. Note that m2l+1 = 0 if r is even. For each 1 ≤ i ≤ 2l − 1, we define a functionfi : V (G) → 2[k] as

fi(u) =

[k] (u ∈ Ni−1(x))∅ (u ∈ Ni(x)){1} (Otherwise).

We see that fi is a k-rainbow dominating function of G because fi(u) = ∅ implies that there exists a vertex u′∈ Ni−1(x) such

that uu′∈ E(G), meaning that

v∈N[u] fi(v) = [k].

Fix an integer i with 1 ≤ i ≤ 2l − 1. Then w(fi) =

j∈{i−1,i} mj + kmi−1 so, by assumption, ϵn < γrk(G) ≤j∈{i−1,i} mj + kmi−1. Recall that n =

0≤j≤r mj. Thus, mi < (1 − ϵ)n + (k − 1)mi−1 and hence,

mi +1 − ϵ

k − 2n < (k − 1)

mi−1 +

1 − ϵ

k − 2n

.

This together with the fact thatmi−1 ≥ m0 = 1 leads to

mi < (k − 1)i1 +

1 − ϵ

k − 2n

−1 − ϵ

k − 2n. (3.1)

For each i = 1, 2, we define a function gi : V (G) → 2[k] as

g1(u) =

[k]

u ∈

0≤j≤l

N2j(x)

∅ (Otherwise)

and

g2(u) =

[k]

u ∈

1≤j≤l

N2j−1(x)

{1} (u ∈ N2l+1(x))∅ (Otherwise).

By the definition of gi, if gi(u) = ∅ for some u ∈ V (G), then there exists a vertex u′∈ V (G) such that gi(u′) = [k] and

uu′∈ E(G). This implies that gi is a k-rainbow dominating function of G. By the definition of g1 and g2, we have

ϵn < γrk(G) ≤ w(g1) =

0≤j≤l

km2j (3.2)

and

ϵn < γrk(G) ≤ w(g2) =

1≤j≤l

km2j−1 + m2l+1. (3.3)

Recall again that n =

0≤j≤r mj. By Inequalities (3.2) and (3.3), we get 2ϵn <

0≤j≤2l kmj + m2l+1 and hence

m2l+1 <k − 2ϵk − 1

n. (3.4)

118 S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122

By Inequalities (3.1), (3.3) and (3.4), we have

ϵn <1≤j≤l

k

(k − 1)2j−11 +

1 − ϵ

k − 2n

−1 − ϵ

k − 2n

+k − 2ϵk − 1

n

= k1 +

1 − ϵ

k − 2n

1≤j≤l

(k − 1)((k − 1)2)j−1−

k(1 − ϵ)lk − 2

n +k − 2ϵk − 1

n

= k1 +

1 − ϵ

k − 2n

·(k − 1)(((k − 1)2)l − 1)

(k − 1)2 − 1−

k(1 − ϵ)lk − 2

n +k − 2ϵk − 1

n

=((k − 1)2l+1

− k + 1)(k − 2)

+

(k − 1)2l+1

− k + 1k(k − 2)

− l

(1 − ϵ)kk − 2

+k − 2ϵk − 1

n,

which contradicts the choice of n0. �

Recall that t∗(k, 1) < 1 for every k ≥ 1. Since t(k) < 1 for k ∈ {1, 2}, we get the following corollary.

Corollary 3.2. Let k and r be two positive integers. Then t∗(k, r) < 1.

4. Some lower bounds on t∗(k, r)

In this section, to see lower bounds on t∗(k, r), we give some constructions of graphs. Let G and H be two graphs and letv be a vertex of H . Let {Hu | u ∈ V (G)} be a family consisting of |V (G)| copies of H . For each u ∈ V (G), let vu be a vertex ofHu corresponding to v. For a fixed vertex v ∈ V (H), let G(H; v) be the graph obtained from G and

u∈V (G) Hu by identifying

two vertices u and vu for each u ∈ V (G). Note that |V (G(H; v))| = |V (G)| |V (H)|.First we give a lower bound of t∗(k, r) which is monotone increasing on k and r . Let k ≥ 4 and h ≥ 0 be two integers. Let

Tk,h be a complete ⌊k−22 ⌋-ary tree with height h. Note that if k = 4, 5, then Tk,h ≃ Ph+1. For every k ≥ 6, note that

|V (Tk,h)| =

0≤i≤h

k − 22

i

=

k−22

h+1− 1 k−2

2

− 1

.

We fix the root vk,h of Tk,h as follows: if k = 4, 5, we let vk,h be one of the vertices of degree at most one; if k ≥ 6, we let vk,hbe the unique central vertex.

Proposition 4.1. Let k ≥ 4 and h ≥ 1 be two integers. Then γrk(Tk,h) = |V (Tk,h)|.

Proof. By Lemma 2.5, we may assume that k ≥ 6. We proceed by induction on h. If h ≤ 1, then |V (Tk,h)| < k and henceγrk(Tk,h) = |V (Tk,h)| by Lemma 2.3. Thus we may assume h ≥ 2. Let f be a flat γrk(Tk,h)-function. Since γrk(Tk,h) ≤ |V (Tk,h)|,it suffices to showw(f ) ≥ |V (Tk,h)|. Note that every vertex of Nh(vk,h) is a leaf of Tk,h. Since∆(Tk,h) < k, |f (x)| = 1 for everyx ∈ Nh(vk,h) by Corollary 2.2(i)(ii). Suppose that f (x) = ∅ for some x ∈ Nh−1(vk,h). Write N(x) ∩ Nh−2(vk,h) = {y}. Since|

u∈N(x)∩Nh(vk,h)f (u)| ≤ ⌊

k−22 ⌋, |f (y)| ≥ k − ⌊

k−22 ⌋. By Lemma 2.1, this implies ⌊

k−22 ⌋ + 1 ≥ d(y) ≥ k − ⌊

k−22 ⌋, which is a

contradiction. Thus f (x) = ∅ for every x ∈ Nh−1(vk,h). This implies that the function f restricted to V (Tk,h) − Nh(vk,h) is ak-rainbow dominating function of Tk,h − Nh(vk,h)(≃ Tk,h−1), and hence

u∈V (Tk,h)−Nh(vk,h)

|f (u)| ≥ |V (Tk,h−1)| by induction.Therefore w(f ) =

u∈V (Tk,h)−Nh(vk,h)

|f (u)| +

u∈Nh(vk,h)|f (u)| ≥ |V (Tk,h−1)| + |Nh(vk,h)| = |V (Tk,h)|. �

Let k ≥ 4, r ≥ 2 andm ≥ 2 be three integers. LetK be a graphwhich is isomorphic toKm and letGk,r,m = K(Tk,r−1; vk,r−1).Then we see that rad(Gk,r,m) = r .

Proposition 4.2. Let k ≥ 4, r ≥ 2 and m ≥ 2 be three integers. Then γrk(Gk,r,m) >|V (Tk,r−1)|−1|V (Tk,r−1)|

|V (Gk,r,m)|.

Proof. Recall that Gk,r,m is obtained from K and

u∈V (K)(Tk,r−1)u where (Tk,r−1)u (u ∈ V (K)) are copies of Tk,r−1. Let f be aγrk(Gk,r,m)-function.

Claim 4.1. Let u ∈ V (K).(i) If f (u) = ∅, then

v∈V ((Tk,r−1)u)

|f (v)| ≥ |V ((Tk,r−1)u)|.(ii) If f (u) = ∅, then

v∈V ((Tk,r−1)u)

|f (v)| ≥ ⌊k−22 ⌋|V (Tk,r−2)| = |V ((Tk,r−1)u)| − 1.

Proof. If f (u) = ∅, then the function f restricted to V ((Tk,r−1)u) is a k-rainbow dominating function of (Tk,r−1)u(≃ Tk,r−1),and hence

v∈V ((Tk,r−1)u)

|f (v)| ≥ |V ((Tk,r−1)u)| by Proposition 4.1. If f (u) = ∅, then the function f restricted toV ((Tk,r−1)u) − {u} is a k-rainbow dominating function of (Tk,r−1)u − u(≃ ⌊

k−22 ⌋Tk,r−2), and hence

v∈V ((Tk,r−1)u)

|f (v)| ≥

⌊k−22 ⌋|V (Tk,r−2)| = |V ((Tk,r−1)u)| − 1 by Proposition 4.1. �

S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122 119

Suppose that f (u) = ∅ for every u ∈ V (K). Then, for each u ∈ V (K), the function f restricted to V ((Tk,r−1)u) is ak-rainbow dominating function of (Tk,r−1)u, and hence w(f ) ≥

u∈V (K) γrk((Tk,r−1)u) =

u∈V (K) |V ((Tk,r−1)u)| =

|V (Gk,r,m)| by Proposition 4.1, as desired. Thus we may assume that there exists a vertex u0 ∈ V (K) such that f (u0) = ∅.Then

v∈V ((Tk,r−1)u0 ) |f (v)| ≥ |V ((Tk,r−1)u0)| by Claim 4.1(i). Again by Claim 4.1,

v∈V ((Tk,r−1)u)

|f (v)| ≥ |V ((Tk,r−1)u)| − 1for every u ∈ V (K) − {u0}. Hence

w(f ) =

v∈V ((Tk,r−1)u0 )

|f (v)| +

u∈V (K)−{u0}

v∈V ((Tk,r−1)u)

|f (v)|

≥ |V ((Tk,r−1)u0)| +

u∈V (K)−{u0}

(|V ((Tk,r−1)u)| − 1)

>

u∈V (K)

(|V ((Tk,r−1)u)| − 1)

=|V (Tk,r−1)| − 1

|V (Tk,r−1)||V (Gk,r,m)|,

as desired. �

The above proposition leads to the following corollary.

Corollary 4.3. Let r ≥ 2 be an integer.

(i) For k ∈ {4, 5}, t∗(k, r) ≥r−1r .

(ii) For k ≥ 6,

t∗(k, r) ≥

k−22

r− k−2

2

k−22

r− 1

.

We improve the bound of Corollary 4.3 for small r . For each k ≥ 1, let T (1)k be a tree which is isomorphic to K1,k with

partite sets {v′

k} and Xk = {x1, . . . , xk}. Next, for each k ≥ 2, let T (2)k be a tree obtained from T (1)

k by adding a new vertex setYk = {yi,j | 1 ≤ i ≤ k−1, 1 ≤ j ≤ k− i} and an edge set {xiyi,j | 1 ≤ i ≤ k−1, 1 ≤ j ≤ k− i}. Finally, for each k ≥ 3, let T (3)

k

be a tree obtained from T (2)k by adding a new vertex set Zk = {zi,j,l | 1 ≤ i ≤ k − 1, 1 ≤ j ≤ k − i, 1 ≤ l ≤ k − j − 1} and an

edge set {yi,jzi,j,l | 1 ≤ i ≤ k−1, 1 ≤ j ≤ k−i, 1 ≤ l ≤ k−j−1} (see Fig. 1). By the definition, every vertex of {xk, y1,k−1}∪Zkis a leaf of T (3)

k . Note that |V (T (1)k )| = k+1, |V (T (2)

k )| = |V (T (1)k )|+

1≤i≤k−1(k− i) = (k+1)+k(k−1)/2 = k2/2+k/2+1

and

|V (T (3)k )| = |V (T (2)

k )| +

1≤i≤k−1

1≤j≤k−i

(k − j − 1)

=12k2 +

12k + 1 +

(k − 1)

1≤i≤k−1

(k − i) −

1≤i≤k−1

12(k − i)(k − i + 1)

=12k3 −

32k2 + 2k + 1 +

32

1≤i≤k−1

i −12

1≤i≤k−1

i2

=12k3 −

32k2 + 2k + 1 +

32

·12(k − 1)k −

12

·16(k − 1)k(2k − 1)

=12k3 −

32k2 + 2k + 1 +

34k2 −

34k −

16k3 +

14k2 −

112

k

=13k3 −

12k2 +

76k + 1.

Lemma 4.4. Let k ≥ 1 and 1 ≤ m ≤ min{k, 3} be two integers. Then γrk(T(m)k ) = γrk(T

(m)k − v′

k) = |V (T (m)k )| − 1.

Proof. First we prove γrk(T(m)k − v′

k) = |V (T (m)k )| − 1. Suppose m ∈ {1, 2}. Then we can check that each component

T ′ of T (m)k − v′

k has order at most k, and by Lemma 2.4, γrk(T ′) = |V (T ′)|. This, together with Lemma 2.4, impliesγrk(T

(m)k − v′

k) = |V (T (m)k )| − 1. Thus, we may assume m = 3 (and so k ≥ 3). Let f be a flat γrk(T

(3)k − v′

k)-function. Sinceγrk(T

(3)k −v′

k) ≤ |V (T (3)k −v′

k)| = |V (T (3)k )|−1, it suffices to showw(f ) ≥ |V (T (3)

k )|−1. Since∆(T (3)k −v′

k) = k−1, |f (x)| = 1for every leaf x of T (3)

k − v′

k by Corollary 2.2(i)(ii), and hence, |f (z)| = 1 for every z ∈ Zk. Suppose that f (yi,j) = ∅ for some1 ≤ i ≤ k−1, and let j0 = max{j | f (yi,j) = ∅}. Recall dT (3)

k −v′k(yi,j0) = k− j0. Since |

1≤l≤k−j0−1 f (zi,j0,l)| ≤ k− j0−1, we get

120 S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122

Fig. 1. A graph T (3)5 .

|f (xi)| ≥ j0 + 1(≥ 2). By Lemma 2.1, this implies that f (yi,j) = ∅ for some j ≥ j0 + 1, which contradicts the definition of j0.Thus, f (y) = ∅ for every y ∈ Yk. Recall that f (z) = ∅ for every z ∈ Zk. Hence, the function f restricted to V (T (3)

k )− ({v′

k}∪Zk)is a k-rainbow dominating function of T (3)

k − ({v′

k} ∪ Zk)(≃ T (2)k − v′

k), and hence,

u∈V (T (3)k )−({v′

k}∪Zk)|f (u)| ≥ |V (T (2)

k )| − 1.

Therefore, w(f ) =

u∈V (T (3)k )−({v′

k}∪Zk)|f (u)| +

u∈Zk

|f (u)| ≥ |V (T (2)k )| − 1 + |Zk| = |V (T (3)

k )| − 1.

Next, we prove that γrk(T(m)k ) = |V (T (m)

k )| − 1. We define a function f ′: V (T (m)

k ) → 2[k] as

f ′(u) =

∅ (u = v′

k){i} (u = xi, 1 ≤ i ≤ k){1} (Otherwise).

Since we can easily check that f ′ is a k-rainbow dominating function of T (m)k and w(f ′) = |V (T (m)

k )| − 1, γrk(T(m)k ) ≤

w(f ′) = |V (T (m)k )| − 1. Hence, it suffices to prove γrk(T

(m)k ) ≥ |V (T (m)

k )| − 1. Let f ′′ be a flat γrk(T(m)k )-function. We show

w(f ′′) ≥ |V (T (m)k )| − 1. If f ′′(v′

k) = ∅ or f ′′(x) = ∅ for every x ∈ Xk, then the function f ′′ restricted to V (T (m)k ) − {v′

k} is ak-rainbow dominating function of T (m)

k − v′

k, and so w(f ′′) ≥ |V (T (m)k )| − 1 because γrk(T

(m)k − v′

k) = |V (T (m)k )| − 1. Thus,

we may assume that f ′′(v′

k) = ∅ and f ′′(x) = ∅ for some x ∈ Xk. If m = 1, then this implies that f ′′(v′

k) = [k], and hencew(f ′′) ≥ k = |V (T (1)

k )| − 1. Thus, we may assume m ∈ {2, 3}. Since f ′′(v′

k) = ∅, there exists no vertex x ∈ V (T (m)k ) − {v′

k}

such that |{y ∈ N(x) | f ′′(y) = ∅}| ≥ k. This implies that |f ′′(x)| = 1 for every leaf x ∈ V (T (m)k ) − {xk} by Corollary 2.2.

Supposem = 2 and let p = |f ′′(v′

k)|. For each p+ 1 ≤ i ≤ k, since |

u∈N(xi)f ′′(u)| ≤ |N(xi)∩ Yk|+ |f ′′(v′

k)| = (k− i)+ p ≤

k − (p + 1) + p = k − 1, f ′′(xi) = ∅. Hence, w(f ) ≥ |f ′′(v′

k)| +

p+1≤i≤k |f ′′(xi)| +

y∈Yk|f ′′(y)| ≥ p + (k − p) + |Yk| =

|V (T (m)k )| − 1. Thus, the desired result holds if m ≤ 2 and hence, we may assume m = 3. Suppose that f ′′(yi,j) = ∅ for

some 1 ≤ i ≤ k − 1, and let j0 = max{j | f ′′(yi,j) = ∅}. Recall that dT (3)k −v′

k(yi,j0) = k − j0. Since |

1≤l≤k−j0−1 f

′′(zi,j0,l)| ≤

k − j0 − 1, |f ′′(xi)| ≥ j0 + 1(≥ 2). By Lemma 2.1, this implies that f ′′(yi,j) = ∅ for some j ≥ j0 + 1, which contradicts thedefinition of j0. Thus, f ′′(y) = ∅ for every y ∈ Yk. Recall that f ′′(z) = ∅ for every z ∈ Zk. Hence, the function f ′′ restrictedto V (T (3)

k ) − Zk is a k-rainbow dominating function of T (3)k − Zk(≃ T (2)

k ) so

u∈V (T (3)k )−Zk

|f ′′(u)| ≥ |V (T (2)k )| − 1. Therefore,

w(f ′′) =

u∈V (T (3)k )−Zk

|f ′′(u)| +

u∈Zk|f ′′(u)| ≥ |V (T (2)

k )| − 1 + |Zk| = |V (T (3)k )| − 1. �

S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122 121

Lemma 4.5. Let k ≥ 1 be an integer. Let G and H be two graphs and let v be a vertex of H. If γrk(H) = γrk(H − v), thenγrk(G(H; v)) = |V (G)|γrk(H).

Proof. For each u ∈ V (G), let fu be a γrk(Hu)-function. Let f : V (G(H; v)) → 2[k] be a function satisfying that f (w) = fu(w)for every u ∈ V (G) and every w ∈ V (Hu). Then f is a k-rainbow dominating function of G(H; v) with weight |V (G)|γrk(H).Thus, it suffices to show γrk(G(H; v)) ≥ |V (G)|γrk(H). Let f ′ be a γrk(G(H; v))-function and let u ∈ V (G). If f ′(u) = ∅,then the function f ′ restricted to V (Hu) is a k-rainbow dominating function of Hu and hence,

w∈V (Hu)

|f ′(w)| ≥ γrk(H); iff ′(u) = ∅, then the function f ′ restricted to V (Hu)−{u} is a k-rainbow dominating function of Hu −u so

w∈V (Hu)

|f ′(w)| ≥

γrk(Hu − u) + |f ′(u)| = γrk(H). Since u is arbitrarily chosen, this leads to

w(f ′) =

u∈V (G)

w∈V (Hu)

|f (w)| ≥

u∈V (G)

γrk(H) = |V (G)|γrk(H). �

Theorem 4.6. (i) For k ≥ 1 and r ≥ 2, t∗(k, r) ≥k

k+1 .

(ii) For k ≥ 2 and r ≥ 3, t∗(k, r) ≥k2+k

k2+k+2.

(iii) For k ≥ 3 and r ≥ 4, t∗(k, r) ≥2k3−3k2+7k

2k3−3k2+7k+6.

Proof. Let k ≥ 1, 1 ≤ m ≤ min{k, 3} and r ≥ m + 1 be three integers. Let G be a connected graph with rad(G) = r − m.Then we can easily see that rad(G(T (m)

k ; v′

k)) = r . Note that γrk(G(T (m)k ; v′

k)) = |V (G)|γrk(T(m)k ) = (|V (T (m)

k )| − 1)|V (G)| by

Lemmas 4.4 and 4.5. Hence γrk(G(T (m)k ; v′

k)) =|V (T (m)

k )|−1

|V (T (m)k )|

|V (G(T (m)k ; v′

k))|. This yields the desired results. �

Since t(1) = 1/2, t(2) = 3/4 and t(3) = 8/9, this theorem leads to the following corollary.

Corollary 4.7. (i) For r ≥ 2, t∗(1, r) =12 .

(ii) For r ≥ 3, t∗(2, r) =34 .

(iii) For r ≥ 4, t∗(3, r) =89 .

5. Exact value of t∗(k, 2)

Theorem 5.1. Let k ≥ 2 be an integer and ϵ > 0 be a number. Then there exists an integer n0 such that γrk(G) < ( kk+1 + ϵ)n

for every connected graph G of order n ≥ n0 and rad(G) = 2.

Proof. Let n0 be an integer such that k < (k+1)ϵk n0. Let G be a connected graph of order n ≥ n0 with rad(G) = 2, and x be a

central vertex ofG. Suppose that γrk(G) ≥ ( kk+1 +ϵ)n. Let X1 = N(x)∩N(N2(x)) and X2 = N(x)−X1. Letm1 = |X1|,m′

1 = |X2|

and m2 = |N2(x)|. For each i = 1, 2, we define a function fi : V (G) → 2[k] as

f1(u) =

[k] (u = x)∅ (u ∈ N(x)){1} (Otherwise)

and

f2(u) =

[k] (u ∈ {x} ∪ X1)∅ (Otherwise).

Note that if fi(u) = ∅, then there exists a vertex u′∈ V (G) such that fi(u′) = [k] and uu′

∈ E(G). This implies that fi is ak-rainbow dominating function of G. By the definition of f1, we have

kk + 1

+ ϵ

n ≤ γrk(G) ≤ w(f1) = k + m2,

and hence,

m2 ≥

k

k + 1+ ϵ

n − k. (5.1)

By the definition of f2, we havek

k + 1+ ϵ

n ≤ γrk(G) ≤ w(f2) = k(1 + m1),

122 S. Fujita, M. Furuya / Discrete Applied Mathematics 166 (2014) 115–122

and hence

1 + m1 ≥

1

k + 1+

ϵ

k

n. (5.2)

Recall again that n = 1 + m1 + m′

1 + m2. By Inequalities (5.1) and (5.2),

n − m′

1 = 1 + m1 + m2 ≥

1 +

k + 1k

ϵ

n − k.

This implies that k ≥ m′

1 +(k+1)ϵ

k n ≥(k+1)ϵ

k n0, which contradicts the choice of n0. �

This, together with Theorem 4.6(i) and Corollary 4.7, leads to the following corollary.

Corollary 5.2. For k ≥ 1, t∗(k, 2) =k

k+1 .

Acknowledgments

We would like to thank Colton Magnant for his assistance in correction of this paper.The first author was supported by the Japan Society for the Promotion of Science Grant-in-Aid for Young Scientists (B)

(20740095).

References

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