A Roman dominating function (RDF) on a graph \(G = (V;E)\) is a function \(f : V \to \{0, 1, 2\}\) satisfying the condition that every vertex \(u\) for which \(f(u) = 0\) is adjacent to at least one vertex \(v\) for which \(f(v) = 2\). The weight of an RDF is the value \(f(V(G)) = \sum _{u \in V (G)} f(u)\). An RDF \(f\) in a graph \(G\) is independent if no two vertices assigned positive values are adjacent. The Roman domination number \(\gamma _R (G)\) (respectively, the independent Roman domination number \(i_{R}(G)\)) is the minimum weight of an RDF (respectively, independent RDF) on \(G\). We say that \(\gamma _R (G)\) strongly equals \(i_R (G)\), denoted by \(\gamma _R (G) \equiv i_R (G)\), if every RDF on \(G\) of minimum weight is independent. In this note we characterize all unicyclic graphs \(G\) with \(\gamma _R (G) \equiv i_R (G)\)

Mustapha Chellali

Nader Jafari Rad

A Roman dominating function (RDF) on a graph G= (V, E) is a function &fnof; : V &rarr; {0, 1, 2} satisfying the condition that every vertex u for which &fnof;(u) = 0 is adjacent to at least one vertex v for which &fnof;(v)=2. The weight of an RDF is the value [formula]. An RDF &fnof; in a graph G is independent if no two vertices assigned positive values are adjacent. The Roman domination number &Upsilon;R (G) (respectively, the independent Roman domination number &Upsilon;R(G) is the minimum weight of an RDF (respectively, independent RDF) on G. We say that &Upsilon;R(G) strongly equals iR(G), denoted by &Upsilon;R(G) &equiv; iR(G), if every RDF on G of minimum weight is independent. In this note we characterize all unicyclic graphs G with &Upsilon;R(G) &equiv; iR(G)

Chellali, M.

Rad, N. J.

Biblioteka Nauki - repozytorium artykuÅÃ³w

A note on the independent Roman domination in unicyclic graphs

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Opuscula Mathematica

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Opuscula Mathematica  Vol. 32  No. 4  2012A NOTE ON THE INDEPENDENT ROMANDOMINATION IN UNICYCLIC GRAPHSMustapha Chellali and Nader Jafari RadAbstract. A Roman dominating function (RDF) on a graph G = (V;E) is a functionf : V  ! f0;1;2g satisfying the condition that every vertex u for which f(u) = 0 isadjacent to at least one vertex v for which f(v) = 2. The weight of an RDF is the valuef(V (G)) =Pu2V (G) f(u): An RDF f in a graph G is independent if no two vertices as-signed positive values are adjacent. The Roman domination number R(G) (respectively,the independent Roman domination number iR(G)) is the minimum weight of an RDF(respectively, independent RDF) on G. We say that R(G) strongly equals iR(G); denotedby R(G)  iR(G); if every RDF on G of minimum weight is independent. In this note wecharacterize all unicyclic graphs G with R(G)  iR(G):Keywords: Roman domination, independent Roman domination, strong equality.Mathematics Subject Classiﬁcation: 05C69.1. INTRODUCTIONWe consider ﬁnite, undirected, and simple graphs G with vertex set V = V (G) andedge set E = E(G). The open neighborhood of a vertex v 2 V is N(v) = NG(v) =fu 2 V j uv 2 Eg and the degree of v, denoted by dG(v), is the cardinality of itsopen neighborhood. A vertex of degree one is called a leaf, and its neighbor is calleda support vertex. If v is a support vertex; then v is called strong if v is adjacent to atleast two leaves.For a graph G, let f : V (G) ! f0;1;2g be a function, and let (V0;V1;V2) be theordered partition of V = V (G) induced by f, where Vi = fv 2 V (G) : f(v) = ig fori = 0;1;2. There is a 1 1 correspondence between the functions f : V (G) ! f0;1;2gand the ordered partitions (V0;V1;V2) of V (G). So we will write f = (V0;V1;V2).A function f : V (G) ! f0;1;2g is a Roman dominating function (RDF) on Gif every vertex u of G for which f(u) = 0 is adjacent to at least one vertex v of Gfor which f(v) = 2. The weight of an RDF is the value f(V (G)) =Pu2V (G) f(u):An RDF f in a graph G is independent if no two vertices assigned positive values715http://dx.doi.org/10.7494/OpMath.2012.32.4.715716 Mustapha Chellali and Nader Jafari Radare adjacent. The Roman domination number R(G) (respectively, the independentRoman domination number iR(G)) is the minimum weight of an RDF (respectively,independent RDF) on G. A function f = (V0;V1;V2) is called a R(G)-function orR-function for G if it is a Roman dominating function on G and f(V (G)) = R(G).An iR(G)-function or iR-function for G is deﬁned similarly. Let f be a R(G)-function,and f(x) = 0 for some vertex x. Then we say that x is a private neighbor of a vertex ywith f(y) = 2 if f is not an RDF for G xy. Roman domination has been introduced byCockayne et al. [3] and has been studied for example in [7]. The study of independentRoman domination has been initiated in [1].We say that R(G) and iR(G) are strongly equal for G; denoted by R(G)  iR(G);if every R(G)-function is an iR(G)-function. In [2] a constructive characterization ofall trees T with R(T)  iR(T) is provided. Note that strong equality between twoparameters was considered ﬁrst by Haynes and Slater [6]. Later Haynes, Henning andSlater gave in [4] and [5] constructive characterizations of trees with strong equalitybetween some domination parameters.In this note we characterize all unicyclic graphs G with R(G)  iR(G):2. MAIN RESULTWe ﬁrst describe the procedure given in [2] to built trees T with R(T)  iR(T).Let T be the family of trees T that can be obtained from k (k  1) disjoint stars ofcenters x1;x2;:::;xk; where each star has order at least three, attached by edges fromtheir center vertices either to a single vertex or to the same leaf of a path P2: Such avertex is called a special vertex of T: Let F be the collection of trees T that can beobtained from a sequence T1, T2, :::, Tk (k  1) of trees, where T1 is a star K1;t witht  2; T = Tk, and, if k  2, Ti+1 can be obtained recursively from Ti by one of thefollowing operations:— Operation O1 : Assume y is a leaf of Ti with fi(y) = 0 and whose support vertexz is either strong or satisﬁes R(Ti   z) > R(Ti): Then Ti+1 is obtained from Tiby adding a new vertex x and adding the edge xy:— Operation O2 : Assume y is a vertex of Ti: Then Ti+1 is obtained from Ti byadding a tree T 2 T of special vertex x and adding the edge xy with the conditionthat if x is a support vertex; then y satisﬁes R(Ti   y)  R(Ti):— Operation O3 : Assume y is a vertex of Ti assigned 0 or 1 for everyR(Ti)-function. Then Ti+1 is obtained from Ti by adding a path P3 = u-v-wand adding the edge wy:Theorem 2.1 (Chellali and Jafari Rad [2]). Let T be a tree. Then R(T)  iR(T) ifand only if T = K1 or T 2 F.Let H be the class of all graphs G such that G is obtained from a tree T 2 F byjoining two non-adjacent vertices v1;v2 such that:(1) For every R(T)-function f, 0 2 ff(v1);f(v2)g,A note on the independent Roman domination in unicyclic graphs 717(2) For 1  i 6= j  2, there is no non-independent RDF f for T   vi with weightR(T) such that f(vj) = 2.Now we are ready to state our main result.Theorem 2.2. Let G be a unicyclic graph. Then R(G)  iR(G) if and only if G 2 H.Proof. Let G be a unicyclic graph, where C is its unique cycle. Assume that R(G) iR(G) and let f = (V0;V1;V2) be a R(G)-function. By assumption f is independent.Let x 2 V (C) \ V0, and let N(x) \ V (C) = fy;zg. Clearly x cannot be a privateneighbor for both y and z: Hence we assume that x is not a private neighbor of y andlet T = G   xy. Then f is an IRDF for T, and so R(T)  iR(T)  R(G) = iR(G).If R(T) < iR(G), and f1 is a R(T)-function, then f1 is an RDF for G with weightless than R(G), a contradiction. Thus R(T) = iR(T) = iR(G) = R(G). Next weshow that any R(T)-function is independent. Assume to the contrary that f is aR(T)-function and f is not independent. Since f is an RDF for G and R(G) =R(T), we obtain that f is a R(G)-function, contradicting the fact that R(G) iR(G). Thus f is independent and consequently, R(T)  iR(T). We deduce thatT 2 F.Next we prove (1). Suppose that there is a R(T)-function f such that 0 62ff(x);f(y)g. If ff(x);f(y)g = f2;1g and f(x) = 1, then g deﬁned on G by g(x) = 0and g(u) = f(u) if u 6= x is an RDF for G with weight less than R(G), a contradiction.Thus ff(x);f(y)g 6= f2;1g but then f would be a non-independent R(G)-function,a contradiction since R(G)  iR(G).Finally, let us prove (2). Assume that there is a non-independent RDF f for T  xwith weight R(T) such that f(y) = 2. Then f is a R(G)-function which is notindependent, a contradiction.Conversely, assume that G 2 H. Let G be obtained from a tree T 2 F by joiningtwo vertices x and y such that (1) and (2) hold. First notice that R(G)  R(T).Assume to the contrary that R(G) < R(T), and let f = (V0;V1;V2) be aR(G)-function. If ff(x);f(y)g 6= f0;2g, then f is an RDF for T with weight lessthan R(T); a contradiction. Thus ff(x);f(y)g = f0;2g. Suppose that f(y) = 0.Then N(y) \ V2 = fxg. Now g deﬁned on T by g(y) = 1 and g(u) = f(u) if u 6= y, isan RDF for T. Then w(g) = R(T) for otherwise g is an RDF for T with weight lessthan R(T) which is impossible. Hence g is a R(T)-function and 0 62 fg(x);g(y)g,contradicting (1). Therefore R(G) = R(T). Now let h be an iR(T)-function. Notethat h is a R(T)-function since R(T)  iR(T). If h is not an IRDF for G, then0 62 fh(x);h(y)g, and h is a R(T)-function that does not satisfy (1), a contradiction.Thus h is an IRDF for G, and so iR(G)  R(T) = R(G)  iR(G), implying thatiR(G) = R(G) = R(T) = iR(T): So h is an iR(G)-function. We next show thateach R(G)-function is independent. Assume to the contrary that f = (V0;V1;V2)is a R(G)-function and f is not independent. If 0 62 ff(x);f(y)g, then f is aR(T)-function which is not independent, contradicting the fact that T 2 F . Thus0 2 ff(x);f(y)g; and we may assume that f(y) = 0. Furthermore, N(y) \ V2 = fxg.Then fjT y is an IRDF for T   y with weight R(T) and f(x) = 2, a contradictionwith (2). We deduce that R(G)  iR(G).718 Mustapha Chellali and Nader Jafari RadAcknowledgementsThis research was supported by “Programmes Nationaux de Recherche: Code8/u09/510”.This research of the second author was supported by Shahrood University of Tech-nology.REFERENCES[1] M. Adabi, E. Ebrahimi Targhi, N. Jafari Rad, M. Saied Moradi, Properties of independentRoman domination in graphs, Australasian Journal of Combinatorics 52 (2012), 11–18.[2] M. Chellali, N. Jafari Rad, Strong equality between the Roman domination and indepen-dent Roman domination numbers in trees, Discussiones Mathematicae Graph Theory, toappear.[3] E.J. Cockayne, P.M. Dreyer Jr., S.M. Hedetniemi, S.T. Hedetniemi, On Roman domina-tion in graphs, Discrete Mathematics 278 (2004), 11–22.[4] T.W. Haynes, M.A. Henning, P.J. Slater, Strong equality of domination parameters intrees, Discrete Mathematics 260 (2003), 77–87.[5] T.W. Haynes, M.A. Henning P.J. Slater, Strong equality of upper domination and inde-pendence in trees, Utilitas Math. 59 (2001), 111–124.[6] T.W. Haynes, P.J. Slater, Paired-domination in graphs, Networks 32 (1998), 199–206.[7] M.A. Henning, A characterization of Roman trees, Discussiones Mathematicae GraphTheory 22 (2002), 325–334.Mustapha Chellalim_chellali@yahoo.comUniversity of BlidaLAMDA-RO Laboratory, Department of MathematicsB.P. 270, Blida, AlgeriaNader Jafari Radn.jafarirad@shahroodut.ac.irShahrood University of TechnologyDepartment of MathematicsShahrood, IranReceived: January 10, 2011.Revised: May 2, 2012.Accepted: May 7, 2012.

A note on the independent roman domination in unicyclic graphs

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A note on the independent roman domination in unicyclic graphs

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