Advances in Domination Theory I, ed VR Kulli Vishwa International Publications (2012) 63-70 THE SEMIENTIRE DOMINATING GRAPH VRKulli Department of Mathematics Gulbarga University, Gulbarga - 585 106, India e-mail: vrkulli@gmailcom ABSTRACT Let G=(V, E) be a graph Let S be the collection of all minimal dominating sets of G The semientire dominating graph Ed(G) of a graph G is the graph with the vertex set V S and with two vertices u, v V S adjacent if u, v D where D is a minimal dominating set or u V and v = D is a minimal dominating set of G containing u In this paper, we initiate a study of this new class of intersection graphs in the field of domination theory and present some properties of semientire dominating graphs Keywords : common minimal dominating graph, dominating graph, semientire dominating graph Mathematics Subject Classification: 05C 1 INTRODUCTION The graphs considered here are finite, undirected without loops or multiple edges Any undefined term here may be found in [1] Let G = (V, E) be graph A set D V is a dominating set of G if every vertex in V D is adjacent to some vertex in D A dominating set D of G is a minimal dominating set if for every vertex v D, D {v} is not a dominating set of G The domination number γ(d) of a graph G is the minimum cardinality of a minimal dominating set of G Let A be a finite set Let F = A 1, A 2,, A n be a partition of A Then the intersection graph Ω(F) of F is the graph whose vertices are the
64 Advances in Domination Theory I subsets in F and in which two vertices A i and A j are adjacent if and only if A i A j φ The common minimal dominating graph CD(G) of a graph G is the graph having the same vertex set as G with two vertices adjacent in CD(G) if and only if there exists a minimal dominating set in G containing them This concept was introduced by Kulli and Janakiram [2] The dominating graph D(G) of a graph G is the graph with vertex set V S where S is the set of all minimal dominating sets of G and with two vertices u, v in V S adjacent if u V and v=d is a minimal dominating set of G containing u This concept was introduced by Kulli et al [3] The purpose of this paper is to introduce another new class of intersection graphs The semientire dominating graph Ed(G) of a graph G is the graph with the vertex set V S where S is the set of all minimal dominating sets of G and with two vertices u, v in V S adjacent if u, v D where D is a minimal dominating set or u V and v = D is a minimal dominating set of G containing u In Figure 1, a graph G and its semientire dominating graph Ed(G) are shown {1, 3} 1 2 3 4 G 1 3 {1, 4} {2, 3} 4 2 {2, 4} VRKulli, The Semientire Dominating Graph 65 Remark 1 For any graph G, CD(G) and D(G) are edge disjoint subgraphs of Ed(G) Remark 2 For any graph G, CD(G) is an induced subgraph of Ed(G) The following results are useful in the proof of our results Theorem A [3] For any graph G with p 2 vertices, D(G) is connected if and only if (G) < p 1 Theorem B A nontrivial graph is bipartite if and only if all its cycles are even 2 MAIN RESULTS We characterize graphs whose semientire dominating graphs Ed(G) are connected Theorem 1 For any graph G with at least two vertices, Ed(G) is connected if and only if (G) < p 1 Proof: Suppose (G) < p 1 By Theorem A, D(G) is connected By Remark 1, D(G) Ed(G) Also V(Ed(G))=V(D(G)) Since every two vertices of D(G) are connected, every two vertices of Ed(G) are also connected Thus Ed(G) is connected Conversely, suppose Ed(G) is connected Assume (G) = p 1 Let u be a vertex of degree p 1 Then {u} is a minimal dominating set of G Thus u and v={u} are adjacent in Ed(G) Since G has at least two vertices, Ed(G) has at least two components, a contradiction Thus (G)<p 1 We now characterize graphs whose semientire dominating graphs are complete Figure 1 Ed(G) Theorem 2 The semientire dominating graph Ed(G) of a graph G is complete if and only if G is totally disconnected
66 Advances in Domination Theory I Proof: Suppose G is totally disconnected Then G has exactly one minimal dominating set D containing all vertices of G Let u be the corresponding vertex of D in Ed(G) Thus V(Ed(G))=V(G) {u} Since D contains all vertices of G, every two vertices of Ed(G) are adjacent in Ed(G) Thus Ed(G) is complete Conversely, suppose Ed(G) is complete Assume G is not totally disconnected Then there exist minimal dominating sets D 1 and D 2 in G Thus D 1 and D 2 are nonadjacent vertices in Ed(G), a contradiction Hence G is totally disconnected Theorem 3 Ed(G)=K p if and only if G = K p 1 p 2 Theorem 4 If G is not a complete graph with p 2 vertices, then Ed(G) contains a triangle Proof: Suppose G K p, p 2 Then G has at least one minimal dominating set D containing two or more vertices Let v 1, v n D, n 2 Then v 1, v n and u=d are mutually adjacent vertices in Ed(G) Thus Ed(G) contains a triangle Theorem 5 Ed(G) = pk 2 if and only if G=K p Proof: Suppose G=K p Then each vertex v i of K p forms a minimal dominating set {v i } Thus v i and {v i } are adjacent vertices in Ed(G) Since each minimal dominating set {v i } contains only one vertex, no two vertices of V(G) are adjacent in Ed(G) and no two corresponding vertices of minimal dominating sets are adjacent in Ed(G) Thus Ed(G)=pK 2 Conversely, suppose Ed(G)=pK 2 We now prove that G is K p Assume G K p By Theorem 4, Ed(G) contains a triangle, which is a contradiction Thus G=K p VRKulli, The Semientire Dominating Graph 67 The following result gives the existence of the semientire minimal dominating graph of a graph Theorem 6 For any graph G, Ed(G) is either connected or has at most one component that is not K 2 Proof: We consider the following three cases Case 1 If (G) < p 1, then by Theorem 1, Ed(G) is connected Case 2 If δ(g) = (G) = p 1, then G=K p By Theorem 5, it follows that Ed(G)=pK 2 Case 3 Suppose δ(g) < (G) = p 1 Let u 1, u 2,,u n be the vertices of degree p 1 in G Let H = G {u 1, u 2,,u n } Then clearly (H)<V(H) 1 Thus by Theorem 1, Ed(G) is not connected Since Ed(G) = Ω(V(Ed(G))) ({u 1 }+u 1 ) ({u 2 }+u 2 ) ({u n }+u n ), exactly one component of Ed(G) is not K 2 and each of the remaining components is K 2 We now prove the following result Theorem 7 Ed(G)=K 2 K p if and only if G=K 1, p 1 Proof: Suppose G=K 1, p 1 Let v be the vertex of degree p 1 in G Then {v} and the set {v 1,, v p 1 } are the only two disjoint minimal dominating sets in G and they are nonadjacent vertices in Ed(G) The vertices v and u={v} are adjacent vertices in Ed(G); and the vertices v 1,, v p 1 and w={v 1,, v p 1 } are mutually adjacent vertices in Ed(G) Thus Ed(G) = K 2 K p Conversely, suppose Ed(G) = K 2 K p Then Ed(G) is disconnected By Theorem 1, (G) = p 1 Then G has a vertex v of degree p 1 Thus {v} is a minimal dominating set in G Since Ed(G)
68 Advances in Domination Theory I has p+2 vertices, G has exactly one more minimal dominating set containing all remaining vertices v 1,, v p 1 which are mutually nonadjacent vertices Thus G=K 1, p 1 From Theorem 7, the following result follows Theorem 8 If G is a connected graph with p vertices and deg u i =p 1, 1 i n and deg u i =n, n+1 i p, then Ed(G) = nk 2 K p n+1 The following result is a characterization of graphs whose semientire dominating graphs are bipartite Theorem 9The semientire dominating graph Ed(G) of a graph G is bipartite if and only if G is complete Proof: Suppose Ed(G) is bipartite If Ed(G) = K 2, then clearly G=K 1 and is complete Suppose G has p 2 vertices We now prove that G is complete Assume G is not complete By Theorem 4, Ed(G) contains a triangle Thus by Theorem B, Ed(G) is not bipartite, which is a contradiction Hence G is complete Conversely, suppose G is complete and G=K p Then by Theorem 4, Ed(G) = pk 2 Thus Ed(G) is bipartite By Theorem 9, the following result follows Theorem 10 The semientire dominating graph Ed(G) of a graph G is complete bipartite if and only if G=K 1 The domatic number d(g) of G is the maximum order of a partition of the vertex set of G into disjoint dominating sets VRKulli, The Semientire Dominating Graph 69 Theorem 11 For any graph G, d(ed(g))=2 if and only if G is complete Proof: Suppose d(ed(g))=2 We now prove that G is complete Assume G K p Then by Theorem 4, Ed(G) contains a triangle Then d(ed(g)) 3, which is a contradiction Thus G is complete Conversely, suppose G=K p Then by Theorem 5, Ed(G)=pK 2 Thus d(pk 2 )=2 and hence d(ed(g))=2 Theorem 12 For any graph G, d(ed(g))=p if and only if G = K p 1 Proof: Suppose d(ed(g))=p Then Ed(G)=K p By Theorem 3, G is K p 1 Conversely, suppose G = K p 1 Then by Theorem 3, Ed(G)=K p Thus d(k p )=p and hence d(ed(g))=p one We characterize the graphs G for which γ(ed(g)) is equal to Theorem 13 For any graph G, γ(ed(g))=1 if and only if G is totally disconnected Proof: Suppose γ(ed(g))=1 Suppose there exist two adjacent vertices u, v in G Then G has two minimal dominating sets D 1 and D 2 such that u D 1 and v D 2 Thus u and v are not adjacent vertices in Ed(G) and also D 1, D 2 are not adjacent vertices in Ed(G) Hence γ(ed(g))>1, a contradiction Thus every two vertices in G are not adjacent This implies that G is totally disconnected Conversely, suppose G is totally disconnected Then by Theorem 2, Ed(G) is complete Hence γ(ed(g))=1
70 Advances in Domination Theory I Theorem 14 For any graph G, D(G) Ed(G) Furthermore, D(G)=Ed(G) if and only if every minimal dominating set of G contains exactly one vertex Proof: By Remark 1, D(G) Ed(G) Suppose D(G)=Ed(G) It implies that two vertices of G are not adjacent in D(G) Therefore two vertices of G are not in the same minimal dominating set Thus every minimal dominating set of G contains exactly one vertex Conversely, suppose every minimal dominating set of G contains exactly one vertex Then two vertices of G are not adjacent in Ed(G) Thus Ed(G) D(G) and since D(G) Ed(G), we see that D(G)=Ed(G) REFERENCES [1] V R Kulli, Theory of Domination in Graphs, Vishwa International Publications, Gulbarga, India (2010) [2] VRKulli and B Janakiram, The common minimal dominating graph, Indian J Pure Appl Math 27(1996) 193-196 [3] VRKulli, B Janakiram and K M Niranjan, The dominating graph, Graph Theory Notes of New York, New York Academy of Sciences, XLVI(2004) 5-8