Sheikholeslami, Seyed Mahmoud

Volkmann, Lutz

English

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The signed Roman domatic number of agraphSeyed Mahmoud Sheikholeslamia, Lutz VolkmannbaResearch Group of Processing and CommunicationandDepartment of MathematicsAzarbaijan Shahid Madani UniversityTabriz, I.R. Irans.m.sheikholeslami@azaruniv.edubLehrstuhl II für MathematikRWTH Aachen University52056 Aachen, Germanyvolkm@math2.rwth-aachen.deSubmitted March 30, 2012 — Accepted November 28, 2012AbstractA signed Roman dominating function (SRDF) on a graph G is a functionf : V (G) → {−1, 1, 2} such that ∑u∈N [v] f(u) ≥ 1 for every v ∈ V (G), andevery vertex u ∈ V (G) for which f(u) = −1 is adjacent to at least onevertex w for which f(w) = 2. A set {f1, f2, . . . , fd} of distinct signed Romandominating functions on G with the property that∑di=1 fi(v) ≤ 1 for eachv ∈ V (G), is called a signed Roman dominating family (of functions) on G.The maximum number of functions in a signed Roman dominating family onG is the signed Roman domatic number of G, denoted by dsR(G). In thispaper we initiate the study of signed Roman domatic number in graphs andwe present some sharp bounds for dsR(G). In addition, we determine thesigned Roman domatic number of some graphs.Keywords: signed Roman dominating function, signed Roman dominationnumber, signed Roman domatic numberMSC: 05C69Annales Mathematicae et Informaticae40 (2012) pp. 105–112http://ami.ektf.hu1051. IntroductionIn this paper, G is a simple graph with vertex set V = V (G) and edge set E = E(G).The order |V | of G is denoted by n = n(G). For every vertex v ∈ V , the openneighborhood N(v) is the set {u ∈ V (G) | uv ∈ E(G)} and the closed neighborhoodof v is the set N [v] = N(v) ∪ {v}. The degree of a vertex v ∈ V is d(v) = |N(v)|.The minimum and maximum degree of a graph G are denoted by δ = δ(G) and∆ = ∆(G), respectively. A graph G is k-regular if d(v) = k for each vertex v of G.The open neighborhood of a set S ⊆ V is the set N(S) = ∪v∈SN(v), and the closedneighborhood of S is the set N [S] = N(S)∪S. A tree is an acyclic connected graph.The complement of a graph G is denoted by G. A cactus graph is a connectedgraph in which any two cycles have at most one vertex in common. We write Knfor the complete graph of order n and Cn for a cycle of length n.A Roman dominating function (RDF) on a graph G = (V,E) is defined in[6, 8] as a function f : V → {0, 1, 2} satisfying the condition that every vertex vfor which f(v) = 0 is adjacent to at least one vertex u for which f(u) = 2. Theweight of an RDF f is the value ω(f) =∑v∈V f(v). The Roman dominationnumber of a graph G, denoted by γR(G), equals the minimum weight of an RDFon G. The Roman domination number has been studied by several authors (see forexample [2, 3, 4]). A set {f1, f2, . . . , fd} of distinct Roman dominating functionson G with the property that∑di=1 fi(v) ≤ 2 for each v ∈ V (G), is called a Romandominating family (of functions) on G. The maximum number of functions in aRoman dominating family (RD family) on G is the Roman domatic number of G,denoted by dR(G). The Roman domatic number was introduced by Sheikholeslamiand Volkmann [7] and has been studied by several authors (see for example [5]).A signed Roman dominating function (SRDF) on a graph G = (V,E) is definedin [1] as a function f : V → {−1, 1, 2} such that ∑u∈N [v] f(u) ≥ 1 for each v ∈V (G), and such that every vertex u ∈ V (G) for which f(u) = −1 is adjacent toat least one vertex w for which f(w) = 2. The weight of an SRDF f is the valueω(f) =∑v∈V f(v). The signed Roman domination number of a graph G, denotedby γsR(G), equals the minimum weight of an SRDF on G. A γsR(G)-function isa signed Roman dominating function of G with weight γsR(G). A signed Romandominating function f : V → {−1, 1, 2} can be represented by the ordered partition(V−1, V1, V2) (or (Vf−1, Vf1 , Vf2 ) to refer f) of V , where Vi = {v ∈ V | f(v) = i}. Inthis representation, its weight is ω(f) = |V1|+ 2|V2| − |V−1|.A set {f1, f2, . . . , fd} of distinct signed Roman dominating functions on G withthe property that∑di=1 fi(v) ≤ 1 for each v ∈ V (G), is called a signed Romandominating family (of functions) on G. The maximum number of functions in asigned Roman dominating family (SRD family) on G is the signed Roman domaticnumber of G, denoted by dsR(G). The signed Roman domatic number is well-defined anddsR(G) ≥ 1 (1.1)for all graphs G since the set consisting of the SRDF with constant value 1 formsan SRD family on G. If G1, G2, . . . , Gk are the connected components of G, then106 S. M. Sheikholeslami, L. Volkmannobviously dsR(G) = min{dsR(Gi) | 1 ≤ i ≤ k}. Hence, we only consider connectedgraphs.Our purpose in this paper is to initiate the study of signed Roman domaticnumber in graphs. We first study basic properties and bounds for the signedRoman domatic number of a graph. In addition, we determine the signed Romandomatic number of some classes of graphs.We make use of the following results in this paper.Proposition A ([1]). If Kn is the complete graph of order n ≥ 1, then γsR(Kn) =1, unless n = 3 in which case γsR(Kn) = 2.Proposition B ([1]).1. For n ≥ 3, γsR(Cn) = d 2n3 e,2. For n ≥ 2, γsR(Pn) = b 2n3 c.Proposition C ([1]). Let G be a graph of order n ≥ 1. Then γsR(G) = n if andonly if G = Kn.Proposition D ([1]). If G is a δ-regular graph of order n with δ ≥ 1, thenγsR(G) ≥ dn/(δ + 1)e.2. Properties of the signed Roman domatic numberIn this section we present basic properties of dsR(G) and sharp bounds on thesigned Roman domatic number of a graph.Theorem 2.1. For every graph G,dsR(G) ≤ δ(G) + 1.Moreover, if dsR(G) = δ(G) + 1, then for each SRD family {f1, f2, . . . , fd} on Gwith d = dsR(G) and each vertex v of minimum degree,∑u∈N [v] fi(u) = 1 for eachfunction fi and∑di=1 fi(u) = 1 for all u ∈ N [v].Proof. If dsR(G) = 1, the result is immediate. Let now dsR(G) ≥ 2 and let{f1, f2, . . . , fd} be an SRD family on G such that d = dsR(G). Assume that vis a vertex of minimum degree δ(G). We haved ≤d∑i=1∑u∈N [v]fi(u) =∑u∈N [v]d∑i=1fi(u) ≤∑u∈N [v]1 = δ(G) + 1.Thus dsR(G) ≤ δ(G) + 1.If dsR(G) = δ + 1, then the two inequalities occurring in the proof becomeequalities. Hence for the SRD family {f1, f2, . . . , fd} on G and for each vertex vof minimum degree,∑u∈N [v] fi(u) = 1 for each function fi and∑di=1 fi(u) = 1 forall u ∈ N [v].The signed Roman domatic number of a graph 107The next results are immediate consequences of Proposition C and Theorem 2.1.Corollary 2.2. For n ≥ 1, dsR(Kn) = 1.Corollary 2.3. For any tree T of n ≥ 3, dsR(T ) ≤ 2. The bound is sharp for adouble star obtained from two vertex disjoint stars K1,3 by connecting their centers.Problem 2.4. Characterize all trees T for which dsR(T ) = 2.Corollary 2.5. For n ≥ 2, dsR(K1,n) = 1.Proof. It follows from Theorem 2.1 that dsR(K1,n) ≤ 2. Suppose to the contrarythat dsR(K1,n) = 2 and assume that {f1, f2} is an SRD family on K1,n. LetV (K1,n) = {v, u1, . . . , un} and E(K1,n) = {vui | 1 ≤ i ≤ n}. Theorem 2.1 impliesthat f1(v) + f2(v) = 1. Since fj(x) ∈ {−1, 1, 2} for each j and each vertex x, wededuce that f1(v) = −1 and f2(v) = 2 or f1(v) = 2 and f2(v) = −1. Assume,without loss of generality, that f1(v) = −1 and f2(v) = 2. By Theorem 2.1, wemust have f2(ui) +f2(v) = 1 for each 1 ≤ i ≤ n and therefore f2(ui) = −1 for each1 ≤ i ≤ n. Since n ≥ 2, we obtain the contradiction 1 ≤∑x∈N [v] f2(x) = 2−n ≤ 0.Thus dsR(K1,n) = 1.Theorem 2.6. If G is a graph of order n, thenγsR(G) · dsR(G) ≤ n.Moreover, if γsR(G) · dsR(G) = n, then for each SRD family {f1, f2, . . . , fd} on Gwith d = dsR(G), each function fi is a γsR(G)-function and∑di=1 fi(v) = 1 for allv ∈ V .Proof. Let {f1, f2, . . . , fd} be an SRD family on G such that d = dsR(G) and letv ∈ V . Thend · γsR(G) =d∑i=1γsR(G) ≤d∑i=1∑v∈Vfi(v) =∑v∈Vd∑i=1fi(v) ≤∑v∈V1 = n.If γsR(G) · dsR(G) = n, then the two inequalities occurring in the proof be-come equalities. Hence for the SRD family {f1, f2, . . . , fd} on G and for each i,∑v∈V fi(v) = γsR(G). Thus each function fi is a γsR(G)-function, and∑di=1 fi(v)= 1 for all v ∈ V .The next two results are immediate consequences of Propositions B, C andTheorem 2.6.Corollary 2.7. For n ≥ 3, dsR(Cn) = 1.Corollary 2.8. Let G be a graph of order n ≥ 1. Then γsR(G) = n and dsR(G) = 1if and only if G = Kn.Corollary 2.9. For n ≥ 1, dsR(Pn) = 1, unless n = 2 in which case dsR(Pn) = 2.108 S. M. Sheikholeslami, L. VolkmannProof. If follows from Proposition B and Theorem 2.6 that dsR(Pn) = 1, unlessn = 2 or n = 4. Let Pn := v1v2 . . . vn. First let n = 2. Define the functionsfi : {v1, v2} → {−1, 1, 2} for i = 1, 2 by f1(v1) = 2, f1(v2) = −1, f2(v1) = −1 andf2(v2) = 2. Obviously f1 and f2 are signed Roman dominating functions of P2and {f1, f2} is a signed Roman dominating family on P2. Hence dsR(P2) ≥ 2.Therefore dsR(P2) = 2 by Theorem 2.1.Now let n = 4. It follows from Theorem 2.1 that dsR(P4) ≤ 2. Suppose tothe contrary that dsR(P4) = 2 and let {f1, f2} be a signed Roman dominatingfamily on P4. By Theorem 2.1, we must have fi(v1) + fi(v2) = 1 for i = 1, 2 andf1(v2) + f2(v2) = 1. By Theorem 2.1, f1(v1) + f2(v1) = 1. Similarly, we havef1(v4) + f2(v4) = 1. Thus f1(vi) + f2(vi) = 1 for 1 ≤ i ≤ 4. Since f1(vi), f2(vi) ∈{−1, 1, 2} and f1(vi) + f2(vi) = 1, we deduce that f1(vi) = −1, f2(vi) = 2 orf1(vi) = 2, f2(vi) = −1 for 1 ≤ i ≤ 4. Assume, without loss of generality, thatf1(v1) = 2 and f2(v1) = −1. Since fi(v1) + fi(v2) = 1 for i = 1, 2, we must havef1(v2) = −1 and f2(v2) = 2. If f1(v3) = −1, then we have∑u∈N [v2] f1(u) ≤ 0which is a contradiction. Thus, f1(v3) = 2 and hence f2(v3) = −1 which impliesthat∑u∈N [v2] f2(u) ≤ 0 which is a contradiction again. Therefore dsR(P4) = 1and the proof is complete.Theorem 2.10. If Kn is the complete graph of order n ≥ 1, then dsR(Kn) = n,unless n = 3 in which case dsR(Kn) = 1.Proof. If n = 3, the the result follows from Proposition A and Theorem 2.6. Nowlet n 6= 3 and let V (Kn) = {v0, v1, . . . , vn−1} be the vertex set of Kn. Considertwo cases.Case 1. Assume that n is even. Define the functions f1, f2, . . . , fn as follows.f1(vn−1) = 2, f1(vi) = −1 if 0 ≤ i ≤ n−22 and f1(vi) = 1 if n2 ≤ i ≤ n− 2, and for2 ≤ j ≤ q and 0 ≤ i ≤ n− 1,fj(vi) = fj−1(vi+j−1),where the sum is taken modulo n. It is easy to see that fj is a signed Romandominating function of Kn of weight 1 and for each 1 ≤ j ≤ n and {f1, f2, . . . , fn}is a signed Roman dominating family on Kn. Hence dsR(Kn) ≥ n. ThereforedsR(Kn) = n by Proposition A and Theorem 2.6.Case 2. Assume that n is odd. Define the functions f1, f2, . . . , fn as follows.f1(vn−1) = f(vn−2) = 2, f1(vi) = −1 if 0 ≤ i ≤ n−12 and f1(vi) = 1 if n+12 ≤ i ≤n− 3, and for 2 ≤ j ≤ q and 0 ≤ i ≤ n− 1,fj(vi) = fj−1(vi+j−1),where the sum is taken modulo n. It is easy to see that fj is a signed Romandominating function of Kn of weight 1, for each 1 ≤ j ≤ n and {f1, f2, . . . , fn}is a signed Roman dominating family on Kn. Hence dsR(Kn) ≥ n. ThereforedsR(Kn) = n by Proposition A and Theorem 2.6.The signed Roman domatic number of a graph 109For some regular graphs we will improve the upper bound given in Theorem 2.1.Theorem 2.11. Let G be a δ-regular graph of order n such that δ ≥ 1. If n 6≡0 (mod (δ + 1)), then dsR(G) ≤ δ.Proof. Since n 6≡ 0 (mod (δ + 1)), we deduce that n = p(δ + 1) + r with integersp ≥ 1 and 1 ≤ r ≤ δ. Let {f1, f2, . . . , fd} be an SRD family on G such thatd = dsR(G). It follows thatd∑i=1ω(fi) =d∑i=1∑v∈Vfi(v) =∑v∈Vd∑i=1fi(v) ≤∑v∈V1 = n.Proposition D implies ω(fi) ≥ γsR(G) ≥ p + 1 for each i ∈ {1, 2, . . . , d}. If wesuppose to the contrary that d ≥ δ + 1, then the above inequality chain leads tothe contradictionn ≥d∑i=1ω(fi) ≥ d(p+ 1) ≥ (δ + 1)(p+ 1) = p(δ + 1) + δ + 1 > n.Thus d ≤ δ, and the proof is complete.Theorem 2.10 demonstrates that Theorem 2.11 is not valid in general whenn ≡ 0 (mod (δ + 1)).Theorem 2.12. If G is a graph of order n ≥ 1, thenγsR(G) + dsR(G) ≤ n+ 1 (2.1)with equality if and only if G ' Kn or G ' Kn (n 6= 3).Proof. It follows from Theorem 2.6 thatγsR(G) + dsR(G) ≤ ndsR(G)+ dsR(G). (2.2)According to Theorem 2.1, we have 1 ≤ dsR(G) ≤ n. Using these bounds, and thefact that the function g(x) = x+ n/x is decreasing for 1 ≤ x ≤ √n and increasingfor√n ≤ x ≤ n, the last inequality leads to the desired bound immediately.If G ' Kn (n 6= 3) then it follows from Proposition A and Theorem 2.10 thatγsR(G) + dsR(G) = n + 1. If G ' Kn, then it follows from Proposition C andCorollary 2.2 that γsR(G) + dsR(G) = n+ 1.Conversely, let equality hold in (2.1). It follows from (2.2) thatn+ 1 = γsR(G) + dsR(G) ≤ ndsR(G)+ dsR(G) ≤ n+ 1,which implies that γsR(G) = ndsR(G) and dsR(G) = 1 or dsR(G) = n. If dsR(G) = n,then δ(G) = n − 1 by Theorem 2.1 and hence G is a complete graph Kn. Sincealso γsR(G) = 1, we deduce that n 6= 3 and hence G ' Kn (n 6= 3). If dsR(G) = 1,then γsR(G) = n, and it follows from Proposition C that G ' Kn. This completesthe proof.110 S. M. Sheikholeslami, L. VolkmannAs an application of Theorems 2.1 and 2.11, we will prove the following Nordhaus-Gaddum type result.Theorem 2.13. For every graph G of order n,dsR(G) + dsR(G) ≤ n+ 1. (2.3)Furthermore, dsR(G) + dsR(G) = n + 1 if and only if n 6= 3 and G ' Kn orG ' Kn.Proof. It follows from Theorem 2.1 thatdsR(G) + dsR(G) ≤ (δ(G) + 1) + (δ(G) + 1)= (δ(G) + 1) + (n−∆(G)− 1 + 1) ≤ n+ 1.If G is not regular, then ∆(G)−δ(G) ≥ 1, and hence the above inequality chainimplies the better bound dsR(G) + dsR(G) ≤ n.If n 6= 3 and G ' Kn or G ' Kn, then Corollary 2.2 and Theorem 2.10 lead todsR(G) + dsR(G) = n+ 1.Conversely, assume that dsR(G) + dsR(G) = n + 1. Then G is δ-regular andthus G is (n− δ − 1)-regular. If δ = 0 or δ = n− 1, then G ' Kn or G ' Kn, andwe obtain the desired result.Next assume that 1 ≤ δ ≤ n− 2 and 1 ≤ δ(G) = n− δ− 1 ≤ n− 2. We assume,without loss of generality, that δ ≤ (n−1)/2. If n 6≡ 0 (mod (δ+1)), then it followsfrom Theorems 2.1 and 2.11 thatdsR(G) + dsR(G) ≤ δ(G) + (δ(G) + 1)= δ(G) + (n− δ(G)− 1 + 1) = n,a contradiction. Next assume that n ≡ 0 (mod (δ + 1)). Then n = p(δ + 1) withan integer p ≥ 2. If n 6≡ 0 (mod (n − δ)), then it follows from Theorems 2.1 and2.11 thatdsR(G) + dsR(G) ≤ (δ(G) + 1) + δ(G)= δ(G) + 1 + (n− δ(G)− 1) = n,a contradiction. Therefore assume that n ≡ 0 (mod (n − δ)). Then n = q(n − δ)with an integer q ≥ 2. Since δ ≤ (n− 1)/2, this leads to the contradictionn = q(n− δ) ≥(n− n− 12)=q(n+ 1)2≥ n+ 1,and the proof is complete.The next result is a generalization of Corollary 2.3.Theorem 2.14. If G is a connected cactus graph, then dsR(G) ≤ 2.The signed Roman domatic number of a graph 111Proof. Let d = dsR(G). If δ(G) ≤ 1, then Theorem 2.1 implies the desired boundd ≤ 2 immediately.It remains the case that δ(G) = 2. If G is a cycle, then the result follows fromCorollary 2.7. Otherwise, the cactus graph G contains a cycle v1v2 . . . vtv1 as anend block with exactly one cut vertex, say v1. Applying Theorem 2.1, we see thatd ≤ 3. Suppose to the contrary that d = 3. Let {f1, f2, f3} be a signed Romandominating family on G.Claim. If fi(vj) = 2 for 1 ≤ i ≤ 3 and 2 ≤ j ≤ t, then d ≤ 2.Proof of claim. Assume, without loss of generality, that f1(v2) = 2. Becauseof f1(v2) + f2(v2) + f3(v2) ≤ 1, we deduce that f2(v2) = f3(v2) = −1. Sincefi is a signed Roman dominating function, we see that fi(v1) = 2 or fi(v3) = 2for 2 ≤ i ≤ 3. Assume, without loss of generality, that f2(v1) = 2. It followsas above that f1(v1) = f3(v1) = −1. Hence we obtain the contradiction 1 ≤∑x∈N [v2] f3(x) = −2 + f3(v3) ≤ 0, and the claim is proved.Thus we assume that fi(vj) ≤ 1 for 1 ≤ i ≤ 3 and 2 ≤ j ≤ t. If t ≥ 4, then weconclude that fi(v3) = 1 for 1 ≤ i ≤ 3, a contradiction to f1(v3)+f2(v3)+f3(v3) ≤1. Finally, assume that t = 3. If fi(v1) ≤ 1 for 1 ≤ i ≤ 3, then fi(v2) = 1 for1 ≤ i ≤ 3, a contradiction. Now assume, without loss of generality, that f1(v1) = 2.This implies that f2(v1) = f3(v1) = −1 and therefore f2(v2) = f3(v2) = f2(v3) =f3(v3) = 1. This leads to f1(v2) = f1(v3) = −1. Thus we obtain the contradiction1 ≤∑x∈N [v2] f1(x) = f1(v1) + f1(v2) + f1(v3) = 0, and the proof is complete.References[1] Ahangar, H. A., Henning, M. A., Zhao, Y., Lo¨wenstein, C., Samodivkin,V., Signed Roman domination in graphs, J. Comb. Optim., (to appear).[2] Chambers, E. W., Kinnersley, B., Prince, N., West, D. B., Extremal prob-lems for Roman domination, SIAM J. Discrete Math., 23 (2009) 1575–1586.[3] Cockayne, E. J., Dreyer Jr., P. M., Hedetniemi, S. M., Hedetniemi, S. T.,On Roman domination in graphs, Discrete Math., 278 (2004) 11–22.[4] Favaron, O., Karami, H., Sheikholeslami, S. M., On the Roman dominationnumber in graphs, Discrete Math., 309 (2009) 3447–3451.[5] Kazemi, A. P., Sheikholeslami, S. M., Volkmann, L., Roman (k, k)-domaticnumber of a graph, Ann. Math. Inform., 38 (2011) 45–57.[6] ReVelle, C. S., Rosing, K. E., Defendens imperium romanum: a classical prob-lem in military strategy, Amer. Math. Monthly, 107 (2000) 585–594.[7] Sheikholeslami, S. M., Volkmann, L., Roman domatic number of a graph, Appl.Math. Letters, 23 (2010) 1295–1300.[8] Stewart, I., Defend the Roman Empire, Sci. Amer., 281 (1999) 136–139.[9] West, D.B, Introduction to Graph Theory, Prentice-Hall, Inc, 2000.112 S. M. Sheikholeslami, L. Volkmann

The signed Roman domatic number of a graph

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The signed Roman domatic number of a graph

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