Independent Rainbow Domination of Graphs

Bulletin of the Malaysian Mathematical Sciences Society, Apr 2017

Given a positive integer t and a graph F, the goal is to assign a subset of the color set \(\{1,2,\ldots ,t\}\) to every vertex of F such that every vertex with the empty set assigned has all t colors in its neighborhood. Such an assignment is called the t-rainbow dominating function (\(t\mathrm{RDF}\)) of the graph F. A \(t\mathrm{RDF}\) is independent (\(It\mathrm{RDF}\)) if vertices assigned with non-empty sets are pairwise non-adjacent. The weight of a \(t\mathrm{RDF}\) g of a graph F is the value \(w(g) =\sum _{v \in V(F)}|g(v)|\). The independent t-rainbow domination number \(i_{rt}(F)\) is the minimum weight over all \(It\mathrm{RDF}\)s of F. In this article, it is proved that the independent t-rainbow domination problem is NP-complete even if the input graph is restricted to a bipartite graph or a planar graph, and the results of the study provide some bounds for the independent t-rainbow domination number of any graph for a positive integer t. Moreover, the exact values and bounds of the independent t-rainbow domination numbers of some Petersen graphs and torus graphs are given.

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Independent Rainbow Domination of Graphs

Independent Rainbow Domination of Graphs Zehui Shao 0 1 2 3 4 5 Zepeng Li 0 1 2 3 4 5 Aljoša Peperko 0 1 2 3 4 5 Jiafu Wan 0 1 2 3 4 5 Janez Žerovnik 0 1 2 3 4 5 B Janez Žerovnik 0 1 2 3 4 5 0 School of Electronic Engineering and Computer Science, Peking University , Beijing 100871 , China 1 School of Information Science and Engineering, Chengdu University , Chengdu 610106 , China 2 School of Computer Science and Educational Software, Guangzhou University , Guangzhou 510006 , China 3 School of Mechanical and Automotive Engineering, South China University of Technology , Guangzhou , China 4 Institute of Mathematics , Physics and Mechanics, Jadranska 19, 1000 Ljubljana , Slovenia 5 FME, University of Ljubljana , Aškercˇeva 6, 1000 Ljubljana , Slovenia Given a positive integer t and a graph F , the goal is to assign a subset of the color set {1, 2, . . . , t } to every vertex of F such that every vertex with the empty set assigned has all t colors in its neighborhood. Such an assignment is called the t - rainbow dominating function (t RDF) of the graph F . A t RDF is independent (I t RDF) if vertices assigned with non-empty sets are pairwise non-adjacent. The weight of a t RDF g of a graph F is the value w(g) = v∈V (F) |g(v)|. The independent t -rainbow domination number ir t (F ) is the minimum weight over all I t RDFs of F . In this article, it is proved that the independent t -rainbow domination problem is NP-complete even - Received: 22 August 2016 / Revised: 19 March 2017 © Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2017 if the input graph is restricted to a bipartite graph or a planar graph, and the results of the study provide some bounds for the independent t -rainbow domination number of any graph for a positive integer t . Moreover, the exact values and bounds of the independent t -rainbow domination numbers of some Petersen graphs and torus graphs are given. Mathematics Subject Classification 05C69 · 05C15 · 05C76 · 05C85 1 Introduction As a combinatorial optimization issue, ordinary domination consists of determining the minimum number of places in which to keep a resource such that every place either is adjacent to the place in which a resource exists or has a resource. In practical applications, some additional constraints or desires must be taken into account. For example [ 12 ], if we are given a large computer network which consists of some clients and servers with t distinct resources s1, s2, . . . , st , we need to seek the minimum number of servers each one possessing a non-empty subset of these resources in order that any client can be connected directly to a subset of servers that together have each resource si (1 ≤ i ≤ t ). On the assumption that all resources have an identical cost, the goal is to seek the minimum value of the number of copies of such t resources. This application naturally can be modeled by the concept of t -rainbow domination. In addition, if a constraint prevents any pair of servers from occupying adjacent locations, then we have the independent t -rainbow domination problem. For a graph F , S ⊆ V (F ) and w ∈ V (F ), let NS(w) denote the open neighborhood of w in S, i.e., {u|uw ∈ E (F ), u ∈ S}, and let NS[w] denote the closed neighborhood of w, i.e., NS[w] = {w} ∪ NS(w). If S = V (F ) and no confusion can occur, NS(w) and NS[w] will be denoted shortly by N (w) and N [w], respectively. If S ⊆ V (F ), then the definition N (S ) = ∪x∈S N (x ) is applied. The degree of a vertex w is the total number of edges incident to w, and in this paper, (F ) denotes the maximum degree of a vertex in the graph F . Inspired by several facility location problems, Brešar, Henning and Rall [ 1–3 ] initiated the study of the k-rainbow domination problem, and such a problem is proved to be NP-complete even if the input graph is a chordal graph or a bipartite graph (see Chang [ 4 ]). This problem has attracted considerable attention (see [ 19,21,25 ]), and many other types of domination are widely applied to real-world scenarios, see, for example, [ 5,7,9,10,13,14,22,24,26 ]. An independent set S of a graph F is a subset of V (F ) for which vertices are pairwise non-adjacent. The independence number of F , denoted as α(F ), is the maximum size of an independent set in F . Given a positive integer t and a graph F , the goal is to assign a subset of the color set {1, 2, . . . , t } to every vertex of F such that every vertex with the empty set assigned has all t colors in its neighborhood. Such an assignment is called the t -rainbow dominating function (t RDF) of the graph F . A t RDF is independent (I t RDF) if vertices assigned non-empty sets are pairwise non-adjacent. The weight of a t RDF g of a graph F is the value w(g) = v∈V (F) |g(v)|. If H is a vertexinduced subgraph of V (F ), the weight restricted to H is wH (g) = v∈H |g(v)|. The independent t -rainbow domination number irt (F ) is the minimum weight of an I t RDFs in F . The upper t -rainbow domination number of F , denoted by Irt (F ), is the maximum weight of a minimal t -rainbow dominating function. (In other words, there is a dominating function f of weight Irt (F ) on F such that no proper restriction of f is dominating.) We use i (F ) to denote the independent domination number, i.e., the size of a smallest independent dominating set, of a graph F . The rest of the paper is organized as follows: In Sect. 2, the independent k-rainbow domination problem is proved to be NP-complete. Section 3 provides some general bounds for the independent t -rainbow domination number. In Sect. 4, the independent 2-rainbow domination problem is studied on generalized Petersen graphs and tori. Closed expressions for infinite families are provided on the basis of constructions for which we conjecture that they are optimal. In Sect. 5, the independent 3-rainbow domination of trees is considered and a tight upper bound for ir3(T ) of a tree T is given. Conclusions are summarized in the last section. 2 Complexity of Independent Rainbow Domination It is well known that determining whether F has an independent dominating set with at most k vertices (where k is a positive integer) is NP-complete for a graph F even when F is a bipartite graph or a planar graph, or belong to some other classes of graphs; see [ 8 ]. In this section, we will prove that it is NP-complete to determine whether F has an independent t -rainbow dominating function of weight at most k for a positive integer k and a given graph F even if F is bipartite or planar. Formally, the problem can be stated as PROBLEM: INDEPENDENT t -RAINBOW DOMINATION INSTANCE: Graph G = (V , E ). A natural number B. QUESTION: Decide whether the independent t -rainbow domination number irt (G) is at most B. Theorem 1 Given a positive integer t , the independent t -rainbow domination problem is NP-complete for general graphs. Proof Theorem 1 can be proved via reducing to the independent t -rainbow domination problem from the independent domination problem. With a graph F on n vertices, the graph F with the vertex set V (F ) = V (F ) ∪ {v2, v3, . . . , vt : v ∈ V (F )} ∪ {vx , vy : v ∈ V (F )} and edge set E (F ) = E (F ) ∪ {vvi , vi vx : v ∈ V (F ), 2 ≤ i ≤ t } ∪ {vx vy : v ∈ V (F )} is considered (see Fig. 1). We claim that F has an independent dominating set ID with |ID| ≤ s if and only if F has an ItRDF f of weight w( f ) ≤ s + nt . Assume F contains an independent dominating set ID of cardinality at most s. Consider the following function f from V (F ) to P({1, 2, . . . , t }) by setting {1}, if u ∈ ID or u = vy for some v ∈ V (F )\ID, ∅, if u ∈ V (G)\ID u = vi for some v ∈ ID and 2 ≤ i ≤ t or u = vy for some v ∈ ID or ⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪ {1, 2,{i.}.,. , t }, iiuff =uu==vxvvfixofrfoosrrossmoomemvee v∈v ∈∈V V(IFD( )F.\)I\DI,D and 2 ≤ i ≤ t, If f (u) = ∅, the following four cases are considered: Case 1 u ∈ V (F )\ID. From the definition of f , we obtain f (ui ) = i for 2 ≤ i ≤ t . Moreover, u has a neighbor v ∈ ID. Hence, f (v) = {1}, and therefore, f (NF (u)) = {1, 2, . . . , t }. Case 2 u = vi for some v ∈ ID and 2 ≤ i ≤ t . From the definition of f , we obtain f (vx ) = {1, 2, . . . , t }. Since vx is a neighbor of vi , we have f (NF (u)) = {1, 2, . . . , t }. Case 3 u = vy for some v ∈ ID . From the definition of f , we obtain f (vx ) = {1, 2, . . . , t }. Since vx is a neighbor of vy , we have f (NF (u)) = {1, 2, . . . , t }. Case 4 u = vx for some v ∈ V (F )\ID. From the definition of f , we obtain f (vi ) = {i } for any 2 ≤ i ≤ t and f (vy ) = {1}. Since vi (2 ≤ i ≤ t ) and vy are the neighborhood of vx , we have f (NF (u)) = {1, 2, . . . , t }. Since any two vertices u1, u2 ∈ V (F ) (with f (ui ) = ∅ for i = 1, 2) are not adjacent in F , it follows that f is an ItRDF of F . Also the weight of f is |ID| + nt ≤ s + nt . Suppose F has an ItRDF f of weight at most s + nt . Since vx vy ∈ E (F ) for some v ∈ V (F ), together with the definition of independent rainbow dominating function, we have f (vx ) = ∅ or f (vy ) = ∅. Therefore, | f (vy )| + | f (vx )| + t i=2 | f (vi )| ≥ t . Let ID = {v ∈ V (F ): f (v) = ∅}. Then we obtain v∈V (F ) = t i=2 | f (vi )|) ≤ s + nt . Therefore, |ID| ≤ s, |ID| + v∈V (F)(| f (vy )| + | f (vx )| + as needed. Noting that F is bipartite or planar if F is a bipartite or planar graph, together with the fact that the independent domination problem is NP-complete for bipartite or planar graphs, we have Corollary 1 The independent t -rainbow domination problem is NP-complete for bipartite graphs or planar graphs for any positive integer t . 3 Bounds for the Independent t-Rainbow Domination Number Theorem 2 For any graph F with maximum degree Δ, we have |V (F)|t Δ+t irt (F ) ≤ t α(F ) ≤ Irt (F ). ≤ γrt (F ) ≤ Proof The result |V (F)t|t ≤ γrt (F ) is proved in [ 19 ]. Since any I t RDF of a graph F is a t RDF of F , wΔe+have γrt (F ) ≤ irt (F ). Let I be a maximal independent set of F with |I | = α(F ). Let f be a function from V (F ) to P({1, 2, . . . , t }) defined by f (v) = ∅ if v ∈/ I and f (v) = {1, 2, . . . , t } if v ∈ I . Then f is an I t RDF of F with weight w( f ) = t I | | = t α(F ). Therefore, irt (F ) ≤ t α(F ). By the definition of the upper t -rainbow domination number, we have t α(F ) = w( f ) ≤ Irt (F ). Similarly to [19, Theorem 1 and Corollary 1], we have the following two results. The proof of Theorem 3 is left to the reader. Theorem 3 For positive integers t ≥ t and a connected graph F , we have i) irt (F ) ≤ irt (F ) + (t − t ) irt (F) ; t ii) irt (F ) ≤ t irt (tF) . Theorem 4 Let F be a non-trivial graph. Then max{i (F ), t } ≤ irt (F ) ≤ ti (F ), and these bounds are sharp. Proof On the assumption that f is an ItRDF of F , and D = {u ∈ V (F ) : f (u) = ∅}. Then D is an independent dominating set of F . So i (F ) ≤ |D| ≤ irt (F ). Note that F is a non-trivial graph, so F has at least one vertex v for which f (v) = ∅. Hence, t ≤ irt (F ) and the inequality max{i (F ), t } ≤ irt (F ) holds. Moreover, for any independent domination set D of F , we can obtain an ItRDF f of F with f (u) = {1, 2, . . . , t } if u ∈ D and f (u) = ∅ otherwise. Hence, the inequality irt (F ) ≤ ti (F ) holds. Furthermore, it can be seen that max{i (F ), t } = irt (F ) for any complete bipartite graph F and irt (F ) = ti (F ) for any star. Therefore, these bounds are sharp. 4 The Independent 2-Rainbow Domination Numbers of Some Classes of Graphs Let n > 0, b ≥ 0, 0 ≤ k < n be integers, we denote by Zkn,b = {x |x ∈ Z, x ≥ b, x ≡ k (mod n)} 4.1 Generalized Petersen Graphs For positive integers n ≥ 3 and k ∈ {1, 2, . . . , n − 1}, the generalized Petersen graph P(n, k) is defined to be a graph with the vertex set {hi1, hi2 | i ∈ {0, 1, . . . , n − 1}} and the edge set {hi1hi2, hi1hi1+k , hi2hi2+1 | i ∈ {0, 1, . . . , n − 1}}, in which the subscripts are computed modulo n (see [ 6,23 ]). An ItRDF f is given by a pattern of two lines, where the values of the upper line are values of { f (h20), f (h21), . . . , f (h2n−1)}, and the values of the bottom line are values of { f (h10), f (h11), . . . , f (h1n−1)}, i.e., hi2 lies exactly above hi1 for each i . Moreover, the sets ∅, {1}, {2} and {1, 2} are encoded with 0, 1, 2 and 3, respectively. Let f be an I2RDF of P(n, k), we define the weight sequence S f = s1s2 . . . sn of P(n, k) under f the following way: si = | f (hi1)|+| f (hi2)| for each i . For example, let h20 h12 h22 h23 h24 0 0 3 0 1 be an I2RDF of P(5, 1), then S f = 20211. = Proof First the upper bounds are proved by giving constructions of I2RDF of P(n, 1) with the weight n if n ≡ 0 (mod2) and n + 1 if n ≡ 1 (mod2) as follows. 1. n ∈ Z40,4: 0202...0202 1010...1010 2. n ∈ Z41,9: 0202...0202 00301 1010...1010 30020 3. n ∈ Z42,6: 4. n ∈ Z43,7: 0202...0202 02 1010...1010 10 0030...0030 003 3000...3000 300 0 0 3 0 1 The pattern 3 0 0 1 0 yields an I2RDF of P(5, 1) of the weight 6. Thus, ir2( P(5, 1)) ≤ 6 and so all the upper bounds are established. Note that P(n, 1) is the ladder graph, which is the Cartesian product P2 Pn. Now, we will show the lower bounds. Let f be an I2RDF of P(n, 1) with minimum weight. For i = 0, 1, 2, 3, 4, let Li denote a set of columns of weight i and let |Li | = ni . Since f is independent, we have n3 = n4 = 0. We have w( f ) = n1 + 2n2, and n = n0 + n1 + n2. Claim 1 If f (hi1) = f (hi2) = ∅, then | f (hi1−1)| + | f (hi2−1)| = 2 and | f (hi1+1)| + | f (hi2+1)| = 2. From Claim 1, we have n2 ≥ n0. Hence, w( f ) − n = n2 − n0 ≥ 0. Therefore, the theorem holds for n ≡ 0 (mod2). We now consider n ≡ 1 (mod 2). Let S be a segment of a weight sequence of P(n, 1). We define three types of subsequences (segments) in the following way: Type 1: S = a1, a2, . . . , a2 +1 with ai = 2 for odd i , and ai = 0 for even i . Type 2: S = a1, a2, . . . , a with ai = 1 for each i . Type 3: S = a1, a2, · · · , a with ai = 2 for each i . From Claim 1, it can be seen that the weight sequence S f of P(n, 1) under f can be decomposed into disjoint segments of Type 1, Type 2 and Type 3. Let pi be the number of maximal segments of Type i for i ∈ {1, 2, 3}. Then w( f ) − n = n2 − n0 = p1 + p3. Since n is odd and f is independent, then there exists an i such that ai = 0. So we have p1 > 0 and w( f ) > n, and so the proof is completed. Theorem 6 Let n ≥ 7. 4n 5 + 1, n ≡ 9 mod 10 Proof First the upper bounds are proved by giving constructions of I2RDF of P(n, 2) of the desired weight as follows. 1. n ∈ Z100,10: 2. n ∈ Z11,021: 3. n ∈ Z120,22: 2010010200...2010010200 0002200011...0002200011 0200201001...0200201001 02000301001 0011000220...0011000220 00330000320 2010010200...2010010200 201001030200 0002200011...0002200011 000230000031 0010200201...0010200201 0030030200201 2200011000...2200011000 2300200011000 4. n ∈ Z13,023: 5. n ∈ Z140,14: 6. n ∈ Z15,015: 7. n ∈ Z16,016: 8. n ∈ Z17,017: 9. n ∈ Z18,018: 10. n ∈ Z19,019: 03 00 01 30 20 01 00 02 10 30 20 00 . Then the pattern Pn yields an I2RDF of P(n, 2) with the desired weight for each n ∈ {7, 8, 9, 11, 12, 13}, and so all the upper bounds are established. Theorem 7 Let n ≥ 7. ⎧ 78n , n ≡ 0, 2, 4, 14 mod 16 ⎪⎪⎪ ir2(P(n, 3)) ≤ ⎨⎪ 7n 8 + 1, n ≡ 5, 7, 8, 10, 12, 13, 15 mod 16 ⎪⎪⎪⎪⎩ 78n + 2, n ≡ 1, 3, 6, 9, 11 mod 16 Proof First the upper bounds are proved by giving constructions of I2RDF of P(n, 3) of the desired weight. 3002020030010100...3002020030010100 0010101000202020...0010101000202020 1003002020030010...1003002020030010 10101030010 0200010101000202...0200010101000202 02020000303 0200300101003002...0200300101003002 010201010202 1010002020200010...1010002020200010 101020202010 an I2RDF of P(n, 3) with the desired weight for each n ∈ {7, 8, 9, 11, 12, 13, 14, 15, 17, 19, 21, 23}, and this completes the proof. Remark We believe that the bounds given in Theorems 6 and 7 are the exact values. A method for the proofs may follow the path algebra approach [ 11,15–17,27 ]. As the corresponding matrices are rather large already for P(n, 2) and P(n, 3), we do not go in more detail here and leave it for future work. In [ 20 ], it was shown that Theorem 8 Let m > 3 and k, relatively prime to m with k ≡ 1 mod m. Then P(m, k) is isomorphic to P(m, ). By Theorem 8, we have P(2k + 1, k + 1) ∼= P(2k + 1, 2) and P(3k + 2, k + 1) ∼= P(3k + 2, 3) for any k ≥ 2. Therefore, we have Corollary 2 Let k ≥ 2. 8k+4 5 8k+4 5 4.2 Torus 21k+14 8 The Cartesian product of graphs F1 and F2 is defined to be the graph F1 F2 with vertex set F1 × F2 = {x y|x ∈ V (F1) and y ∈ V (F2)} such that (x1, y1)(x2, y2) ∈ E(F1 F2) if x1x2 ∈ E(F1) and y1 = y2, or y1 y2 ∈ E(F2) and x1 = x2 (see [ 15 ]). The torus is the graph Tm,n = Cm Cn, which can be viewed as a kind of grid graph with n columns and m rows. Theorem 9 Let n ≥ 3. ⎧ n, n ≡ (0 mod 6) ⎪⎪⎪⎪ ir2(T3,n)) = ⎨ n + 1, n ≡ (3 mod 6) ⎪⎪ ⎪⎪⎩ n + 2, n ≡ 1, 2, 4, 5 (mod6) Proof First the upper bounds are proved by giving constructions of I2RDF of T3,n of the desired weight. 1. n ∈ Z60,6: 2. n ∈ Z61,13: 100200...100200 10100300 001002...001002 00020002 020010...020010 03001010 3. n ∈ Z62,14: an I2RDF of T3,n with the desired weight for each n ∈ {3,4,5,7, 8, 9}, and all the upper bounds are established. The lower bounds can be determined by the dynamic algorithm used in [ 18 ], and the implementation is left to the reader. Theorem 10 If m and n are positive integers, then ir2(T3m,6n) = 6mn. Proof By Theorem 2, we have ir2(T3m,6n) ≥ 6mn. Let P = ⎣ 02 00 02 10 00 01 ⎦ , Q = [P, . . . , P], and R = ⎢⎣ Q... ⎥ , ⎦ n times ⎡ Q ⎤ where Q is repeated m times in R. Then R yields an I2RDF of T3m,6n with weight 6mn, which completes the proof. 5 The Independent 3-Rainbow Domination Numbers of Trees The following result for paths and stars is easily proved: Proposition 1 If n ≥ 3, then + 1, + 2, We first define the spider graph as follows. A spider S = S(a1, . . . , ar ) is a tree formed by joining r ≥ 1 vertex-disjoint paths of orders a1, . . . , ar as pendent paths to a single vertex b, which is called the anchor of S, e.g., Fig. 2 is the spider S(2, 2, 2). We denote by H (k) = S(2, 2, . . . , 2) with a total of k 2s. Lemma 1 Let integers t, k ≥ 1 and T belong to one of the following trees with n vertices: (i) a vertex u connected to every anchor of t spiders H (k). (ii) a vertex u connected to another vertex and each anchor of t spiders H (k). (iii) a vertex u connected to a pendent P2 = v1v2 and each anchor of t spiders H (k). Then T has an I3RDF g with wT (g) ≤ n such that g(u) = ∅ . Proof i) Assume T is a tree which is a vertex z connected to each anchor yi (1 ≤ i ≤ s) of s spiders H (k). Define a function g : V (T ) → P({3, 2, 1}) as follows. If k ≥ 2, let g(z) = ∅, g(y1) = {1, 2}. The vertices of P2 attached to y1 is assigned with ∅ and {3}, respectively. For i ≥ 2, g(yi ) = {2, 3}. The vertices of P2 attached to yi are assigned with ∅ and {1}, respectively. Then it is easy to see that wT (g) ≤ n. If k = 1, let g(z) = ∅, g(y1) = {3, 2, 1}. The vertices of P2 attached to y1 are assigned with ∅ and {3}, respectively. Then wT (g) ≤ n. ii) Consider a function g : V (T ) → P({3, 2, 1}) by letting the same color to the vertices as case i except v and let g(v) = 1. Then we obtain wT (g) ≤ n. iii) Consider a function g : V (T ) → P({3, 2, 1}) by letting the same color to the vertices as case i except v1, v2. Assume v2 be the leaf, and let g(v1) = {3, 2, 1} and g(v2) = ∅. Then we obtain wT (g) ≤ n. Remark Figures 3, 4 and 5 show examples of I3RDFs of graphs of such three cases, where we use 0 to denote the empty set ∅. Theorem 11 If T is an n-vertex tree with at least three vertices, then ir3(T ) ≤ n. Proof Assume that there exists an n-vertex tree T with minimum n such that ir3(T ) > n, i.e., ir3(T R ) ≤ n for each n -vertex tree T R if n < n. Then we have Claim 2 T has no vertex which is adjacent to at least two leaves. Proof of Claim 1. Assume that u ∈ V (T ) and u1 and u2 are two leaves adjacent to u. Let T R = T − u1 and f be an optimal I3RDF of T R. Then we have w( f ) = ir3(T R) ≤ n −1. If f (u) = ∅, then f (NT R (u)) = {1, 2, 3}. Extend f to g by putting g(u1) = {1} and g(x) = f (x) for x ∈ V (T )\{u1}. Then g is an I3RDF of T with w(g) ≤ n, a contradiction. If f (u) = ∅, then f (u2) = ∅ and so f (u) = {1, 2, 3}. Extend f to g by putting g(u1) = ∅ and g(x) = f (x) for x ∈ V (T )\{u1}. Then g is an I3RDF of T with w(g) ≤ n − 1, a contradiction. Claim 3 T has no vertex u which is adjacent to both a leaf and a pendent P2. Proof of Claim 2. Suppose to the contrary that u ∈ V (T ) and there exist vertices u1, u2, u3 ∈ V (T ) such that {uu1, uu2, u2u3} ∈ E(T ), u1, u3 are leaves and dT (u2) = 2. Let T R = T − u1 and f be an optimal I3RDF of T R. Then we have w( f ) = ir3(T R ) ≤ n − 1. If f (u) = ∅, then f (NT R (u)) = {1, 2, 3}. Extend f to g by putting g(u1) = {1} and g(x ) = f (x ) for x ∈ V (T )\{u1}. Then g is an I3RDF of T with w(g) ≤ n, a contradiction. If f (u) = ∅, then f (u1) = f (u2) = ∅ and f ({u, u3}) = {1, 2, 3}. We may w.l.o.g assume that f (u) = {1, 2}. Extend f to g by putting g(u1) = ∅, g(u) = {1, 2, 3} and g(x ) = f (x ) for x ∈ V (T )\{u1, u}. Then we have w(g) ≤ n, a contradiction. Claim 4 T has no vertex u which is adjacent to a pendent P3. Proof of Claim 3. Suppose to the contrary that u ∈ V (T ) and u is adjacent to a pendent P3 = u1u2u3, say uu1, u1u2, u2u3 ∈ E (T ). Let T R f be an optimal I3RDF of T R . Then we have w( f ) = ir3=(T TR )−≤{un1,−u23,. uE3x}teanndd f to g by putting g(u1) = g(u3) = ∅, g(2) = {1, 2, 3} and g(x ) = f (x ) for x ∈ V (T )\{u1, u2, u3}. Then g is an I3RDF of T with w(g) ≤ n, a contradiction. Let P = x1, x2, . . . , xm be a longest path in T . By Claims 2, 3 and 4 , we have that x1 is a leaf and d(x2) = 2, x3 is adjacent to at least two pendent P2s and x4 is adjacent to some anchors of pendent spiders H (k). Let T1 and T2 be two trees obtained by deleting the edge x4x5 from T and T1 be the tree rooted at x4. Let f be an optimal I3RDF of T2. We now consider the following possible cases: Case 1 There exists no leaf or pendent P2 adjacent to x4 in T1 (see, e.g., Fig. 6). Let f be an I3RDF of T1 described in case i of Lemma 1. Then we obtain an I3RDF of T by putting g(x ) = f (x ) if x ∈ V (T1) and g(x ) = f (x ) if x ∈ V (T2). By Lemma 1, g(x4) = ∅ and so g is independent. Therefore, wT (g) ≤ wT1 (g)+wT2 (g) ≤ n, a contradiction. Case 2 There exists exactly one leaf z adjacent to x4 in T1 (see, e.g., Fig. 7). Let f be an I3RDF of T1 described in case ii of Lemma 1. Then we obtain an I3RDF of T by putting g(x ) = f (x ) if x ∈ V (T1) and g(x ) = f (x ) if x ∈ V (T2). By Lemma 1, g(x4) = ∅ and so g is independent. Therefore, wT (g) ≤ wT1 (g)+wT2 (g) ≤ n, a contradiction. Case 3 There exists exactly one pendent P2 = w1w2 adjacent to x4 (see, e.g., Fig. 8). Let f be an I3RDF of T1 described in case iii of Lemma 1. Then we obtain an I3RDF of T by putting g(x ) = f (x ) if x ∈ V (T1) and g(x ) = f (x ) if x ∈ V (T2). By Lemma 1, g(x4) = ∅ and so g is independent. Therefore, wT (g) ≤ wT1 (g)+wT2 (g) ≤ n, a contradiction. Case 4 There exist at least two pendent P2s adjacent to x4 (see, e.g., Fig. 9). Let u1u2 be a P2 adjacent to x3, w1w2 be a P2 adjacent to x4 and T R = T − {u1, u2, w1, w2}. Let f1 be an optimal I3RDF of T R . Then it is impossible that f1(x3) = f1(x4) = ∅ (otherwise, recolor x3 and the pendent P2s adjacent to x3 as described in Lemma 1 to obtain an I3RDF with smaller weight). Assume that f1(x3) = ∅ and f1(x4) = ∅. If | f1(x4)| = 1, we can adjust the coloring of T1 as described in Lemma 1 such that | f1(x4)| = 2. Let u2 and w2 be leaves. We can assume w.l.o.g. that f1(x4) = {2, 3}. Extending f1 to g by putting g(u1) = {1, 2, 3}, g(u2) = g(w1) = ∅, g(w2) = {1} and g(x ) = f1(x ) for x ∈ V (T )\{u1, u2, w1, w2}. Therefore, wT (g) ≤ wT R (g) + 4 ≤ n, a contradiction. The case when f1(x3) = ∅ and f1(x4) = ∅ is similar. Case 5 d(x4) = 2. In this case, we have that x3 is adjacent to at least two pendent P2s by Claim 4. We define a function f : V (T1) → P({1, 2, 3}) as follows. f (x4) = ∅, f (x3) = {1, 2, 3}. For each pendent P2 = v1v2 (v2 is a leaf) adjacent to x3, let f (v2) = {1} and f (v1) = ∅. Then we obtain an I3RDF of T by putting g(x ) = f (x ) if x ∈ V (T1) and g(x ) = f (x ) if x ∈ V (T2). Therefore, wT (g) ≤ wT1 (g) + wT2 (g) ≤ n, a contradiction. Theorem 12 If n ≥ 6, there exists an n-vertex tree T for which ir3(T ) = n. Proof Let P4 be a path with V ( P4) = {v1, x , v3, v4} and E ( P4) = {v1x , x v3, v3v4}, and P3 be a path with V ( P3) = {x , v2, v3} and E ( P3) = {x v2, v2v3}. Let Lk be the set of graphs each of which consists of the disjoint union of k copies of P4 or P3 plus a path through the vertex x of these copies. Figure 10 shows a graph in L5. It can be seen that each tree T in Lk satisfies that ir 3(T ) = n. By adding some P3s or P4s to a path, we can obtain such a tree of any order n for n ≥ 6. 6 Conclusion In this paper, we investigate the complexity of the independent t -rainbow domination problem. More precisely, we prove that the independent t -rainbow domination problem is NP-complete if the input is a bipartite graph or planar graph. 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Zehui Shao, Zepeng Li, Aljoša Peperko, Jiafu Wan, Janez Žerovnik. Independent Rainbow Domination of Graphs, Bulletin of the Malaysian Mathematical Sciences Society, 2017, 1-19, DOI: 10.1007/s40840-017-0488-6