International Journal of Mechanical Engineering and Technology (IJMET) Volume 9, Issue 1, January 2018, pp. 995–1004 Article ID: IJMET_09_01_106 Available online at http://iaeme.com/Home/issue/IJMET?Volume=9&Issue=1 ISSN Print: 0976-6340 and ISSN Online: 0976-6359

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RESULTS ON THE INVERSE DOMINATION NUMBER OF SOME CLASS OF GRAPHS

Jayasree T G Department of Mathematics, Adi Shankara Institute of Engineering and Technology, Kerala, India

Radha R Iyer Department of Mathematics, Amrita School of Engineering, Coimbatore, Amrita Vishwa Vidyapeetham, Amrita University, India

ABSTRACT A set D of vertices in a graph G(V, E) is a dominating set of G, if every vertex of V not in D is adjacent to at least one vertex in D. Let D be a minimum dominating set of G. If V – D contains a dominating set say D’ is called an inverse dominating set with respect to D. The inverse domination number γ’(G) of G is the order of a smallest inverse dominating set of G. In this paper we establish inverse domination number for some families of graphs. Keywords: Dominating sets, Domination number, Inverse Domination number. Cite this Article: Jayasree T G and Radha R Iyer, Results on the Inverse Domination Number of Some Class of Graphs, International Journal of Mechanical Engineering and Technology 9(1), 2018. pp. 995–1004. http://iaeme.com/Home/issue/IJMET?Volume=9&Issue=1

1. INTRODUCTION The graphs considered here are simple, finite, non-trivial, undirected graphs without isolated vertices. The study of domination in graphs was initiated by Ore[13] and Berge. The domination in graphs is one of the vital area in graph theory which has attracted many researchers because of its potentiality to solve and address many real life situations like in the communication and social network and in defence purpose to name a few. In a communication network, let D denote the set of transmitting stations so that every station not belonging to D has a link with at least one station in D. If this set of stations fails or has a fault, then one has to find another disjoint such set of stations. This leads to define the domatic number. The domatic number d(G) of G is the maximum number of disjoint dominating sets in G. This concept was defined by Cockayne and Hedetniemi [11]. In this paper we consider the problem of finding the inverse domination number. We use the following results to prove our later results.

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2. INVERSE DOMINATION NUMBER OF SOME FAMILIES OF GRAPHS In this section, we establish the inverse domination number of some families of graphs. Proposition The cocktail party graph of order 'n', (also called the hyper octahedral graph) is the graph consisting of two rows of paired nodes in which all nodes except the paired ones are connected with an edge. Let Cn be a cocktail party graph. Then γ (Cn) = 2.

2.1. Lemma

Let Cn be a cocktail party graph. Then γ’(Cn) = 2. Proof Let V = {v1, u1, v2, u2, ..., vn\2,un\2} be vertices of Cn. Let V1 = {v1, v2, ... , vn\2 } and V2 = {u1, u2, ..., un\2 } be the two vertex sets of Cn. The necessary and sufficient condition for the existence of at least one inverse dominating set of G is that G contains no isolated vertices. To have a dominating set D', we need to have one vertex from V1 and another from V2. Fig 1 shows that the vertices {v1, u1} form a dominating set of Cn. Let D' = {( vi, ui) / (vi+1, ui+1)  D} is an inverse dominating set of Cn. Thus γ’(Cn) = 2.

Figure 1 Cocktail Party Graph Proposition The Coxeter graph is a 3 regular graph with 28 vertices and 42 edges. Let G is a Coxeter graph. Then γ (G) = 7.

2.2. Lemma If G is a Coxeter graph, γ’ (G) = 9. Proof

Let G be a Coxeter graph with 28 vertices of degree three. Let V = {v1, v2, ..., v7}, U = {u1, u2, ..., u7} and W ={w 1, w 2, ..., w14} be the vertex sets of the graph G. To have a dominating set, we need seven vertices. Fig 2 shows that the vertices {v1, v2, ..., v7 } form a minimum dominating set. Also there exists a set D' such that (V - D) contains another dominating set with    , giving   . Hence γ’ (G) = 9.

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Figure 2 Coxeter graph with 28 vertices. Proposition 0 0 Let Sn be a crown graph with 2n vertices. Then γ (Sn ) = 2.

2.3. Lemma 0 0 Let Sn be a crown graph. Then γ’ (Sn ) = 2. Proof 0 Let Sn be a crown graph of 2n vertices with vertex set {(vi, uj) /          }, of degree (n - 1).

0 Figure 3 Crown graph S4 We need only one vertex from each vertex set to form the dominating set of cardinality 0 two. Fig.3. shows that the vertices (vi, ui) form a minimum dominating set of Sn . But (V - D) 0 contains another dominating set D' such that γ'(D') = 2. Hence in Crown graph Sn , the domination number as well as the inverse domination number is 2.

2.4. Lemma Let G be the Cayley graph. Then γ’(G) = 3. Proof By the definition, the truncated tetrahedron or the Cayley graph is, the graph comprises of 12 vertices each of degree three with number of edges 18. We need three vertices to dominate all the remaining vertices. Suppose V is the set of all vertices and D is the dominating set, then there exists another set D' which still dominates the vertices of V. Thus D' is the inverse dominating set of cardinality again three. Thus γ(G) = γ'(G) = 3.

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Figure 4 Truncated Tetrahedron Proposition The Cubic symmetric graph is a regular graph of order 3. Let G be a cubic symmetric graph. Then γ(G) = 2.

2.5. Lemma Let G be a cubic symmetric graph. Then γ'(G) = 2.

Figure 5 Cubic symmetric graph Proof By the definition of the cubic symmetric graph, it consists of 8 vertices each of degree 3. Only two vertices are enough to dominate all the remaining vertices of G. Suppose V is the set of all vertices and D is the dominating set, then there exists another set D' which still dominates the vertices of G. Thus D' is the inverse dominating set with cardinality 2. Thus γ (G) = γ'(G) = 2. Proposition The Holt graph [5] is a graph on 27 vertices. The Holt graph is also known as Doyle graph. Let Cn be a Doyle graph. Then γ (Cn) = 9.

2.6. Lemma Let G be a Doyle graph. Then γ' (G) = 9. Proof Let G be a Doyle graph with 27 vertices. Let V = {v1, v2... vn}, be the set of vertices of the graph. The vertex set V can be partitioned into three vertex sets V1, V2 and V3, each comprising of nine vertices and the degree of each vertex being four. Fig.6. shows the graph. If V1 is taken as the dominating set whose vertices dominates the other two sets giving γ (G) = 9. Then (V - D) contains another two vertex sets V2 and V3 which dominates the remaining vertices of (V - D). Let us consider V2 as the inverse dominating set. Hence γ'(G) = 9.

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Figure 6 Doyle graph. Proposition The Folkman graph is a bipartite four regular Hamiltonian graph with 20 vertices and 40 edges. It is four vertexes connected and four edge connected perfect graph. Let G be a Folkman graph. Then γ(G) = 6.

2.7. Lemma If G is a Folkman graph, then γ'(G) = 6.

Figure 7 Folkman graph. Proof

Let G be the Folkman graph of 20 vertices with degree 4. Let V = {a0, a1, a2, a 3, a4, a 0, b1, b2, b3, b 4, c0, c1, c2, c3 , c4, d0, d1, d2, d3, d 4} be the vertices of G. We need six vertices to dominate the entire graph. Fig 7 shows that the vertices {a1, a2, b0, b3, c3, d2} form a minimum dominating set of G. Thus γ (G) = 6. There exist one more dominating set D' which has the same number of vertices and this forms an inverse dominating set of G, and hence γ'(G) = 6. Proposition Frucht Graph is a 3-regular graph with 12 vertices and 18 edges. It is planar cubic, Hamiltonian graph. Let G be the Frucht graph. Then γ (G) = 3.

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2.8. Lemma Let G be the Frucht graph. Then γ' (G) = 4.

Figure 8 Frucht graph. Proof Let G be the Frucht graph, which is a cubic graph with 12 vertices and 18 edges. Let V = {v 1, v 2, ... , v7} and U = {u1, u2, ... ,u5} be the two sets of vertices of G. To dominate G, we need three vertices. Fig 8 shows that four vertices are needed in D', to dominate the entire vertex set. Let D be the dominating set and D' be the inverse dominating set. Then there exists at least one D' so that (V - D) contains one inverse dominating set. Hence γ' (G) = 4. Proposition The Levi graph is a graph with 30 nodes and 45 edges. Let G be a Levi graph. Then γ (G) = 10.

2.9. Lemma  γ Let G be a Levi graph, then ' (G) =  . Proof

Figure 9 Levi graph.

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Let G be a Levi graph with 30 vertices and 45 edges. Let V = {v 1, v 2, ..., v10}, U = { u1, u2, ..., u10} and W = {w1, w2, ..., w10} with degree three. To dominate the graph we need 10 vertices. Fig 9 shows γ (G) = γ' (G) = 10.

3. INVERSE DOMINATION NUMBER OF PLATONIC SOLIDS

3.1 Lemma Let G be the tetrahedron. Then γ '(G) = 1.

Figure 10 Tetrahedron Proof Let G be the platonic solid tetrahedron. Then G is a 3 regular graph with 4 vertices and 6 edges. Let D be the dominating set and D' = (V –D) be the inverse dominating set of G. From Fig 10 the dominating set as well as the inverse dominating set, needs only one vertex to dominate G. Hence γ (G) = γ' (G) = 1.

3.2. Lemma Let G be the cube. Then γ' (G) = 2.

Figure 11 Cube Proof Let G be the platonic solid cube. Then G is a 3 regular graph with 8 vertices and 12 edges. Let D be the dominating set and D' = (V -D) be the inverse dominating set of G. From Fig 11 the dominating set as well as the inverse dominating set, needs two vertices to dominate G. Hence γ (G) = γ' (G) = 2.

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3.3. Lemma Let G be the octahedron. Then γ' (G) = 2.

Figure 12 Octahedron Proof Let G be the platonic solid octahedron. Then G is a 4 regular graph with 6 vertices and 12 edges. Let D be the dominating set and D' = (V - D) be the inverse dominating set of G. From Fig 12 the dominating set as well as the inverse dominating set, needs two vertices to dominate G. Hence γ (G) = γ' (G) = 2.

3.4 Lemma Let G be the dodecahedron. Then γ '(G) = 6.

Figure 13 Dodecahedron Proof Let G be the platonic solid dodecahedron. Then G is a 3 regular graph of 20 vertices and 30 edges. Let D be the dominating set and D' = (V - D) be the inverse dominating set of G. From Fig 13 the dominating set D contains six vertices. For D', the inverse dominating set, also needs six vertices to dominate G. Hence γ (G) = γ' (G) = 6.

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3.5. Lemma Let G be a Icosahedral graph. Then γ' (G) = 2.

Figure 14 Icosahedral graph Proof

Let G be an icosahedral graph with 12 vertices each of degree 5. Let V = {v1, v2, ..., v6} and U = { u1, u2, ..., u6 } be the two sets of vertices of G. To dominate G, we need only two vertices. Fig.14. shows that only two vertices are needed to dominate the entire vertex set. Let D be the dominating set and D' be the inverse dominating set. Hence γ (G) = γ' (G) = 2.

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