Properties of independent Roman domination in graphs

Transcription

1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 5 (01), Pages Properties of independent Roman domination in graphs M. Adabi E. Ebrahimi Targhi N. Jafari Rad M. Saied Moradi Department of Mathematics Shahrood University of Technology Shahrood Iran Abstract A Roman dominating function on a graph G is a function f : V (G) {0, 1, } satisfying the condition that every vertex u for which f(u) =0 is adjacent to at least one vertex v for which f(v) =. The weight of a Roman dominating function is the value f(v (G)) = u V (G) f(u). The Roman domination number of G, γ R (G), is the minimum weight of a Roman dominating function on G. In this paper, we study independent Roman domination in graphs and obtain some properties, bounds and characterizations for the independent Roman domination number of a graph. 1 Introduction Let G =(V (G),E(G)) be a simple graph of order n. We denote the open neighborhood of a vertex v of G by N G (v), or just N(v), and its closed neighborhood by N[v]. For a vertex set S V (G), N(S) = v S N(v) andn[s] = v S N[v]. The degree deg(x) of a vertex x denotes the number of neighbors of x in G. A set of vertices S in G is a dominating set, ifn[s] =V (G). The domination number, γ(g) of G, is the minimum cardinality of a dominating set of G. For a graph G and a subset of vertices S we denote by G[S] the subgraph of G induced by S. A subset S of vertices is independent if G[S] has no edge. A set S V (G) isanindependent dominating set if S is independent and dominating set. The minimum cardinality of such a set is the independent domination number i(g). For notation and graph theory terminology in general we follow []. The research is supported by Shahrood University of Technology. Corresponding author; n.jafarirad@shahroodut.ac.ir

2 1 M. ADABI, E.E. TARGHI, N.J. RAD AND M.S. MORADI With K n we denote the complete graph on n vertices and with C n the cycle of length n. The cartesian product of two graphs G 1 and G is the graph G 1 G with vertex set V (G 1 ) V (G ), such that two vertices (u 1,u )and(v 1,v ) are adjacent if and only if either u 1 = v 1 and u v E(G ), or u = v and u 1 v 1 E(G 1 ). An r-partite graph G is a graph whose vertex set V (G) can be partitioned into r sets of pairwise nonadjacent vertices. For positive integers p 1,p,...,p r,thecomplete r- partite graph K p1,p,...,p r is the r-partite graph with partition V (G) =V 1 V V r such that V i = p i for 1 i r and such that every two vertices belonging to different partite sets are adjacent to each other. We recall that a leaf in a graph is a vertex of degree one, and a support vertex is one that is adjacent to a leaf. Let L(G) bethesetofallleavesinagraphg. For a graph G, let f : V (G) {0, 1, } be a function, and let (V 0 ; V 1 ; V )be the ordered partition of V (G) induced by f, where V i = {v V (G) :f(v) =i} and V i = n i for i =0, 1,. There is a 1 1 correspondence between the functions f : V (G) {0, 1, } and the ordered partitions (V 0 ; V 1 ; V )ofv(g). So we will write f =(V 0 ; V 1 ; V ). A function f : V (G) {0, 1, } is a Roman dominating function, orjustrdf, if every vertex u for which f(u) = 0 is adjacent to at least one vertex v for which f(v) =. The weight of an RDF is the value f(v (G)) = u V f(u). The Roman domination number of a graph G, denoted by γ R (G), is the minimum weight of an RDF on G. A function f =(V 0 ; V 1 ; V ) is called a γ R -function if it is an RDF and f(v (G)) = γ R (G),[1,5,4]. Cockayne et al. in [1] introduced the concept of independent Roman domination in graphs. An RDF f =(V 0 ; V 1 ; V ) is called an independent Roman dominating function if the set V 1 V is an independent set. The independent Roman domination number, i R (G), is the minimum weight of an independent RDF on G. McRae [3] has shown that the decision problem corresponding to independent Roman dominating functions is NP-complete, even when restricted to bipartite graphs. In this paper we study the independent Roman domination number in a graph G. We give some properties, bounds and characterizations for this new parameter. We make use of the following: Proposition A [1]: Let f =(V 0 ; V 1 ; V ) be any γ R -function. Then (a) G[V 1 ], the subgraph induced by V 1, has maximum degree 1; (b) no edge of G joins V 1 and V. We call an RDF f =(V 0 ; V 1 ; V ) in a graph G an i R -function if it is an independent RDF and f(v (G)) = i R (G). Let v S V (G). A vertex u is called a private neighbor of v with respect to S if u N[v] N[S v]. General results and bounds It is obvious that any independent RDF on G is also an RDF. So for any graph G, i R (G) γ R (G). Our first aim is to give a characterization of all graphs with equal

3 INDEPENDENT ROMAN DOMINATION 13 Roman domination number and independent Roman domination number. Lemma 1 Let f =(V 0 ; V 1 ; V ) be an RDF for a graph G. IfV is independent, then there is an independent RDF g for G such that w(g) w(f). Proof. Let G be a graph and let f =(V 0 ; V 1 ; V )beanrdfforg such that V is independent. Let f 1 =(V0 1; V 1 1; V 1) where V 0 1 = N(V ), V1 1 = V 1 and V 1 = V. If V 1 is independent then f 1 is an independent RDF for G. So suppose that V 1 is not independent. Let x 1 V 1 be a vertex such that deg G[V1](x 1 ) 1, and let f = (V0 ; V1 ; V ), where V0 = V 0 N G[V1](x 1 ), V1 = V 1 \ N G[V1][x 1 ]andv = V {x 1 }. It is obvious that V is independent. If V1 is not independent, we continue the above process. So there is an integer k such that f k =(V0 k; V 1 k; V k)andv 1 k V k is independent. Since for i =1,,..., k 1, w(f i+1 ) w(f i ), we deduce that f k is an independent RDF for G with w(f k ) w(f). Corollary For a graph G, i R (G) =γ R (G) if and only if there is a γ R -function f =(V 0 ; V 1 ; V ) such that V is independent. Applying Corollary, the following is easily verified. Proposition 3 (1) i R (K n )=γ R (K n ). () i R (P n )=i R (C n )=γ R (P n )=γ R (C n ). (3) i R (P P n )=γ R (P P n ). Theorem 4 i R (K m,n )=min{m, n} +1. Proof. Let X and Y be the partite sets of K m,n. First notice that if x X, then (Y ; X \{x}; {x}) is an independent RDF for G, andsoi R (K m,n ) min{m, n} +1. Now let f =(V 0 ; V 1 ; V ) be an independent RDF for K m,n. It is obvious that either V 1 V X or V 1 V Y. Further, V. We deduce that the weight of f is at least min{m, n} + 1. Thus the result follows. In the following we prove that for any graph G with maximum degree at most three, i R (G) =γ R (G). Theorem 5 For any graph G with Δ(G) 3, γ R (G) =i R (G). Proof. Let G be a graph with Δ(G) 3. Suppose to the contrary that γ R (G) < i R (G). Let f = (V 0 ; V 1 ; V )beaγ R -function for G. By Corollary, V is not independent. If G[V ] has a component containing a path P 3 on three vertices, then we label the central vertex of P 3 by 0, and its private neighbor with respect to V by 1, and obtain an RDF for G with weight less than γ R (G). This contradiction leads that any component of G[V ]isak.letg 1,G,..., G r be the components of G[V ] with V (G i )={v i,u i } for 1 i r. It follows that g =(V 0 {u 1,..., u r }; V 1 ( r i=1n(u i ) \{v i }); V \ r i=1{u i }) is an RDF for G with weight γ R (G). By Lemma 1, there is a i R (G)-function g 1 such that w(g 1 ) w(g). We deduce that γ R (G) =i R (G), a contradiction.

4 14 M. ADABI, E.E. TARGHI, N.J. RAD AND M.S. MORADI Corollary 6 If γ R (G) <i R (G), thenδ(g) 4. Next we give some upper bounds. Theorem 7 For any graph G of order n, i R (G) n. Further, the equality holds if and only if G = mk K l for some integers m, l with n =m + l. Proof. Let G be a graph of order n. It is obvious that ( ; V (G); ) isanrdffor G. So Lemma 1 shows that i R (G) n. For the next part, first it is obvious that if G = mk K l for some integers m, l with n = m + l, theni R (G) = n. Suppose that G is a graph of order n and i R (G) =n. For n =1, the statement is obviously true. Thus assume that n>. We show that G has maximum degree at most one. Suppose to the contrary, that v is a vertex of degree at least. Let G 1 be the graph obtained from G by removing G[N[v]]. By the first part of the current theorem, i R (G 1 ) V (G 1 ). Let f =(V 0 ; V 1 ; V )beani R -function for G 1. Then g =(V 0 N(v); V 1 ; V {v}) is an independent RDF for G with weight less than n, a contradiction. Thus G has maximum degree at most one, and so the result follows. Theorem 8 For any graph G with Δ(G) 3, i R (G) γ R (G)+ γ R(G) (Δ(G) 3). Proof. Let G be a graph with Δ(G) 3. If i R (G) =γ R (G), then the inequality follows since γ R (G). Suppose that i R (G) >γ R (G). Let f =(V 0 ; V 1 ; V )bea γ R -function for G. By Corollary, V is not independent. Let x 1 y 1 E(G[V ]), and let Y 1 be the set of all private neighbors of y 1 with respect to V. Then f 1 = (V (1) 0 ; V (1) 1 ; V (1) )isanrdfforg, where V (1) 0 =(V 0 \ Y 1 ) {y 1 }, V (1) 1 = V 1 Y 1 and V (1) = V \{y 1 }.IfV (1) is not independent, then we let x y E(G[V (1) ]), and let Y be the set of all private neighbors of y with respect to V (1). Then f =(V () 0 =(V (1) 0 \ Y ) {y }; V () 1 = V (1) 1 Y ; V () = V (1) \{y }) is an RDF for G. Continuing this process produces an RDF f m =(V (m) 0 ; V (m) 1 ; V (m) ) for some integer m such that V (m) is independent. By Lemma 1, there is an i R - function g such that i R (G) =w(g) w(f m ). It follows that i R (G) γ R (G) + m(δ(g) 3). But m V 1 γ R(G) 1= γ R(G). We deduce that i R (G) γ R (G)+ γ R(G) (Δ(G) 3). We remark that for a graph G with Δ(G) = 3, Theorem 8 implies that i R (G) = γ R (G). This fact is also stated in Theorem 5. In the following we characterize all graphs G with Δ(G) > 3 achieving equality in Theorem 8. Theorem 9 For a graph G with Δ(G) > 3, i R (G) =γ R (G)+ γ R(G) (Δ(G) 3) if and only if either γ R (G) =i R (G) =,orγ R (G) =4, i R (G) =1+Δ(G).

5 INDEPENDENT ROMAN DOMINATION 15 Proof. Let G be a graph with Δ(G) > 3. First notice that if either γ R (G) =i R (G) =, or γ R (G) =4,i R (G) =1+Δ(G), then i R (G) =γ R (G)+ γ R(G) (Δ(G) 3). Suppose that i R (G) =γ R (G)+ γ R(G) (Δ(G) 3). The result is obvious for γ R (G) =. So suppose that γ R (G) 3. Since Δ(G) > 3, i R (G) γ R (G), and so by Corollary, for any γ R -function f =(V 0 ; V 1 ; V )ong, V is not independent. Let f =(V 0 ; V 1 ; V ) be a γ R -function on G, and let A be the a maximum independent subset of V.We observe that V \ A. We proceed with Fact 1. Fact 1. N[V ] \ N[A] (Δ(G) 1) V \ A. To see this notice that N[V ] \ N[A] =N[V \ A] \ N[A]. Also N[V ] \ N[A] = N[u] \ N[A]. Now u V \A N[V ] \ N[A] = u V \A u V \A N[u] \ N[A] N[u] N[A]. But for any u V \ A by maximality of A we find that u is adjacent to a vertex v 1 A. Thus N[u] N[A]. We deduce that N[u] \ N[A] = N[u] N[A] N[u] Δ(G) 1. This completes a proof for Fact 1. Let g = (N(A); V (G) \ N[A]; A). Then g is an RDF for G. It follows that w(g) = A + V (G)\N[A]. Since V 1 V (G)\N[A], we observe that V (G)\N[A] = V 1 (N[V ] N[A]). Now w(g) = A + V (G) \ N[A] = A + V 1 (N[V ] N[A]) = A + V 1 + N[V ] N[A]. Applying Fact 1, we obtain that w(g) A + V 1 +(Δ(G) 1) V \ A = A + V 1 +(Δ(G) 1)( V A ) = A V + V + V 1 +(Δ(G) 1)( V A ) = γ R (G)+( V A )(Δ(G) 3) = γ R (G)+( γ R(G) V 1 A )(Δ(G) 3) = γ R (G)+( γ R(G) V 1 A )(Δ(G) 3). Since Δ(G) > 3 we find that γ R (G). It is a routine matter to see that V 1 + A. This leads that w(g) γ R (G)+( γ R(G) )(Δ(G) 3). By Lemma 1,

6 16 M. ADABI, E.E. TARGHI, N.J. RAD AND M.S. MORADI there is an i R -function h for G such that i R (G) =w(h) w(g) γ R (G)+( γ R(G) )(Δ(G) 3) = i R (G). We conclude that γ R (G)+( γ R(G) and )(Δ(G) 3) = γ R (G)+( γ R(G) V 1 A )(Δ(G) 3) (1) N[V ] \ N[A] =(Δ(G) 1) V \ A () The relation (1) implies that V 1 + A =, and therefore V 1 =0and A =1. Then G[V ] is a complete graph. Let G[V ]=K t where t. Let u V \ A. It follows that N[u] \ N[A] Δ(G)+1 t, andso u V \A N[u] \ N[A] u V \A N[u] \ N[A] (Δ(G)+1 t) V \ A (Δ(G) 1) V \ A. But relation () implies that (Δ(G) +1 t) V \ A =(Δ(G) 1) V \ A, and therefore t =. Since V 1 = 0, we deduce that γ R (G) =4andsoi R (G) =1+Δ(G). 3 Relations with independent domination It is obvious that for any graph G, i(g) i R (G) i(g). In this section we give a characterization of all graphs G which i R (G) =i(g)+k for 0 k i(g). Theorem 10 For a graph G of order n, i R (G) =i(g) if and only if G = K n Proof. It is obvious that i R (K n ) = i(k n ). Let G be a graph of order n and i R (G) =i(g). Let f =(V 0 ; V 1 ; V )beani R -function for G. It follows that i(g) V 1 + V V 1 + V = i R (G). We deduce that V =, andsoi(g) =n. Thus,G = K n. The next theorem gives a characterization for all graphs G with i R (G) =i(g)+1. Theorem 11 For a graph G of order n, i R (G) =i(g)+1 if and only if G has a vertex of degree n i(g).

7 INDEPENDENT ROMAN DOMINATION 17 Proof. Let G has a vertex v with deg(v) =n i(g). First let G = G[V (G) \ N[v]]. By Theorem 7, i R (G ) V (G ). Let(V 0; V 1; V ) beani R -function for G. It follows that (N(v) V 0 ; V 1 ; {v} {V } ) is an independent RDF G. We conclude that i R (G) i(g)+1. For the converse suppose that i R (G) =i(g)+1. Letg =(V 0 ; V 1 ; V )beani R - function. Since V 1 V is an independent dominating set, we deduce that V 1. If V =, thenv 1 is not independent, which is a contradiction. So V =1. Let V = {u}. Then any vertex of V 0 is adjacent to u. Sodeg(u) = V 0 = n V 1 V = n i(g). Now we are ready to give the main result of this section. Theorem 1 For a graph G of order n and an integer k with k i(g), i R (G) =i(g)+k if and only if the following hold: (i) For any integer s with 1 s k 1, there is no independent set U t of cardinality t such that 1 t s and v U t N[v] = n i(g) s +t, (ii) There is an independent set W l of cardinality l for some integer l with 1 l k such that v W l N[v] = n i(g) k +l. Proof. Let G be a graph of order n and k be an integer. ( ) Leti R (G) =i(g)+k. First we prove (i). Suppose to the contrary that there exist two integers s 0 and t 0 with 1 t 0 s 0 k 1andG has an independent set U t0 of cardinality t 0 such that v U t N[v] = n i(g) s +t. Then f = (N(U t0 ); V (G)\N[U t0 ]; U t0 )isanrdfforg such that U t0 is independent. By Lemma 1, there is a i R (G) function g such that w(g) w(f) =i(g)+s 0 i(g)+k 1. This is a contradiction, since i R (G) =i(g)+k. Thus (i) is true. Now we prove (ii). From V + V 1 = i(g)+k we have that V 1 + V i(g). By Theorem 10, V 0. Then V = l (1 l k), and so V 1 = i(g)+k l. Let W l = V. There exists an independent set W l of cardinality l such that v W l N[v] = n V 1 = n (i(g)+k l) =n i(g) k +l. ( =) Suppose to the contrary that i R (G) =i(g) +m where m k 1. If m = 1 then Theorem 11 implies that G has a vertex of degree n i(g), and putting s = t = 1 in (i) produce a contradiction. So suppose that m. By part (ii) of ( ) G has a set W l of cardinality l with 1 l k 1 such that N[v] = n i(g) m +l. v W l By (i) for any integer s with 1 s k 1, G does not have a set U t of cardinality t (1 t s) such that v U t N[v] = n i(g) s +t, a contradiction. On the other hand if i R (G) =i(g)+m where m k +1, then bypart(i)of( ) forany integer s with 1 s m 1, G does not have a set U t of cardinality t (1 t s) such that v U t N[v] = n i(g) s +t.

8 18 M. ADABI, E.E. TARGHI, N.J. RAD AND M.S. MORADI This contradicts (ii). We deduce that i R (G) =i(g)+k. Acknowledgments We would like to thank the referees for their careful review of our manuscript and some helpful suggestions. References [1] E. J. Cockayne, P. M. Dreyer Jr., S. M. Hedetniemi and S. T. Hedetniemi, On Roman domination in graphs, Discrete Math. 78 (004), 11. [] T. W. Haynes, S. T. Hedetniemi and P. J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, NewYork, [3] A. A. McRae, Private communication, March 000. [4] C. S. ReVelle and K. E. Rosing, Defendens imperium romanum: a classical problem in military strategy, Amer Math. Monthly 107 (000), [5] I. Stewart, Defend the Roman Empire!, Sci. Amer. 81 (6) (1999), (Received 9 Dec 009; revised 7 Sep 011)