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AIJRSTEM19-

Abstract

In 1962 ,Ore used the name " dominating set" and "domination number". In 1977 , Cockayne and Hedetniemi made an interesting and extensive survey of the results known at that time about dominating sets in graphs. The survey paper of cockayne and Hedetniemi has generated a lot of interest in the study of domination in graphs. Domination has a wide range of applications in Radio station, modeling social networks, coding theory, nuclear power plants problems. A non-empty subset D of vertices in a graph G=(V,E) is a "Dominating set", if every vertex in V-D is adjacent to some vertex in D. The domination numberγ(G) of G is the minimum cardinality of a minimal dominating set. A total dominating set Dt of G is a dominating set such that the induced subgraph <D> has no isolated vertices. The total domination number γt(G) of G is the minimum cardinality of a minimal total dominating set of G. Let D be the minimum dominating set of G. If V-D contains a dominating set say ′,then ′ is called an inverse dominating set with respect to D. The invese domination number ′( )of G is the order of the smallest inverse dominating set of G. "Kulli and Sigarkandi" introduced the concept of inverse domination in Graphs. Let G = (V,E) be a simple graph. Let Dt be the minimum total dominating set of G. If V-Dt contains a total dominating set ′ of G, then ′ is called an inverse total dominating set with respect to D. The inverse total domination number ′( ) of G is the minimum number of vertices in an inverse total dominating set of G. The concept of total domination is first introduced by Cockayne, Dawes and Hedetniemi in 1980. V.R. Kulli and RadhaRajamaniIyer introduced the concept of inverse total domination in graphs. In this paper, we found the total domination number and inverse total domination number of some special classes of graphs.

III. Results on Total Domination and Inverse Total Domination in Some Special Classes of Graphs Theorem 3.1

If G is a Paley graph, then γt(G)=4 = ′( ) Proof:

Figure1: Paley Graph

Step(i)

Let G=(V,E) be a Paley graph with 13 vertices of degree 6.Let V={v0,v1,…….v12} be the vertex set of the graph G. To have a dominating set D, it is found that, the vertices {v0 ,v3 ,v7 } forms a minimum dominating set. Thus, │D│=3 .Hence ( )=3. Step(ii) Now, the total dominating set Dt is a dominating set such that the induced subgraph {v0 ,v1 ,v6,v7} has no isolated vertices. Therefore │Dt│=4.Hence γt(G) =4.

Step(iii)

Also, there exists a set Dt′such that (V-Dt ) contains a total dominating set with │Dt′│=│Dt│,giving │Dt′│=4. Hence ′( )=4. Observation 3.1. 1 In a Paley graph, (i) γ(G)<γt(G) and (ii) γt(G)= ′( ) Theorem 3.2 If G is a Shrikhande graph with 16 vertices, then γ(G)=4= γt(G)= ′( ) Proof:

Also, there exists a set ′={v4,v5,v6} in (V-Dt) which contains a total dominating set with │Dt′│=│Dt│, giving │Dt′│=3,Hence ′(G)=3.Therefore, γ(G)= γt(G)= ′(G).

Figure 2: Shrikhande graph

(i) Consider the Shrikhande graph G with 16 vertices and G is a 6-regular graph. Let {v0,v1,v2,v3,v4,v5,v6,v7}be the outer vertices and {v8,v9,v10,v11,v12,v13,v14,v15}be the corresponding inner vertices. In the above figure, the vertices v11 and v2 dominate the vertices {v13,v14,v8,v9,v4,v0,v3,v1} and similarly the vertices v15 and v6 dominate the vertices{v10,v8,v7,v5}. Therefore, the minimum dominating set is {v11,v2,v15,v6}.Hence γ(G)=4. (ii) Here, the total dominating set is a dominating set such that the induced subgraph {v11,v2,v15,v6} has no isolated vertices. Thus, the minimum total dominating set is {v11,v15,v2,v6}.Therefore, the minimum cardinality of the total dominating set is 4.Hence γt(G)=4. (iii) There exists a non-empty subset ′ such that (V-Dt) contains a total dominating set with │Dt′│=4.Hence γt′(G)=4. Thus, In a Shrikhande graph,(i) γ(G)=4 and (ii) γ(G)= γt(G)= ′ ) Theorem 3.3 If G is Clebsch graph, then γt(G)=5= ′ ( ) Proof:

From the below figure,V={v1,v2,………………………………v16} D={v11,v7,v8,v9}

By the definition of Clebsch graph, the graph comprises of 16 vertices and G is a 5-regular graph. Here, there are 4 vertices which dominate all the remaining vertices. Suppose V={v1,v2,………v16} is the set of all vertices and D={ v11,v7,v8,v9} is the manimum dominating set , then there exists the induced subgraph <Dt>={ v7,v16,v8,v9,v10} which dominate the remaining vetices of V and <Dt> has no isolated vertices. Thus Dt is the total dominating set of minimum cardinality 5.Hence γt(G) =5. Also, there exists another set D ′ ={v11,v6,v4,v12,v2} in (V-Dt ) which dominates the remaining vertices of V. Thus Dt′ is the inverse total dominating set of minimum cardinality 5.Hence ′( )=5. Step(i) Let G be the Andrasfai graph with 17 vertices and G is a 6-regular graph. Let V={v1,v2,………….v17} be the vertex set of the graph G. In the above figure, the vertex set {v1,v2,v3} forms a minimum dominating set D. Thus │D│=3. Hence γ(G) =3.

Step(ii)

Now, the total dominating set Dt be the minimum dominating set such that the induced subgraph {v1,v2,v3} has no isolated vertices. Therefore,│Dt│=3,Hence γt(G) =3.

Theorem 3.5

If G is Chvatal graph, then γ(G)=4= γt(G). Proof:

If G be the Soifer graph, then γt(G)=3= γt′(G) Proof: In a Moser graph, (i) γ(G)=2 (ii) γt′(G)=4 (iii) γ(G)<γt(G)<γt′(G) Theorem 3.8 If G be the Franklin graph, then γt(G)=6= γt′(G) Proof:

Figure 8:Franklin graph

Let G be the Franklin graph with 12 vertices and G is a Cubic graph. Let V={v1,v2,………v6}and U={v7,v8,……..v12} be the two subset of vertices of G.Here, the set of 4 vertices {v1,v4,v7,v8} dominates the entire vertex set without isolated vertices. Here,the set Dt={v1,v2,v3,v4,v5,v6}is the total dominating set and Dt′={v7,v8,v9,v10,v11,v12} is the inverse total dominating set. Hence γt(G)=6= γt′(G). Note 3.8.1

In a Franklin graph, (i)γ(G)=4 (ii)γ(G)<γt(G)= γt′(G)

Theorem 3.9

If G be the Golomb graph, then γt(G)=3 and γt′(G)=4.

Proof:

By the definition of Golomb graph , the graph G comprises of 10 vertices.The set D={v1,v4,v5} to dominates all the remaining vertices of V. Suppose V={v1,v2,………..,v10}is the set all vertices and D={v1,v4,v5} is the minimum dominating set of G. Then, γ(G)=3.This dominating set is the minimum total dominating set such that the induced subgraph <Dt>={v1,v5,v4} is adjacent to every vertex of G and it has no isolated vertices. Hence γt(G)=3.Also,there exists an inverse total dominating set Dt′={v2,v7,v9,v10} which dominates the vertices of V with minimum cardinality 4.Thus γt′ (G)=4. In the above figure, it is found that the set{v1,v10,v6} forms the minimum dominating set. Thus │D│=3, Hence γ(G)=3.

Case (ii)

In Poussin graph, the total dominating set Dt is a dominating set such that the induced subgraph <Dt>={v3,v6,v10,v14} has no isolated vertices. Therefore,│Dt│=4. Hence γt(G)=4.

Case (iii)

Also, there exists a set Dt′ such that (V-Dt) contains a total dominating set. In the above figure, the another minimum total dominating set Dt′={v1,v7,v13,v9 } with minimum cardinality │Dt′│=4.Hence γt′(G)=4. Note 3.11.1

In a Poussin graph, (i) γt(G)= γt′(G) Theorem 3.12 Let G be the Mobius-Kantor graph with 16 vertices, then γt(G)= γt′(G)=8. (i.e.)V1={v1,v2,…………….v8} V2={v9,v10,………….v16} each one comprising of 8 vertices and the degree of each vertex being three. In the above figure, V1 is taken as the minimum total dominating set Dt={v1,v2,………v8} whose vertices dominates the remaining vertices, giving γt(G)=8. Also, there exists a set Dt′={v9,v10,………v16} such that (V-Dt) contains a total dominating set V2 which dominates the remaining vertices of (V-Dt). Thus, the vertex subset V2 is the inverse total dominating set with minimum cardinality 8. Hence γt′(G)=8. Consider the Icosahedral graph G with 12 vertices of degree 5.Let {u1,u2,…………u6} be the inner vertices and {v7,v8 ,……...v12} be the corresponding outer vertices. Now, the vertex v7 is adjacent to the vertices {u1,u2,u6,v9,v11} and the vertex v10 is adjacent to the vertices {u3,u4,u5,v8,v12}.

Therefore, these two vertices {v7,v10} dominate the entire vertex set. But it is not a total dominating set, because {v7,v10} are not adjacent in D.

Consider another subset Dt={u1,u4,v7,v10}which is a total dominating set. Therefore, Every vertex in V is adjacent to atleast one vertex in D. Thus, Dt={u1,u4,v7,v10} is the minimum total dominating set with cardinality 4.Hence γt(G)=4 Note 3.13.1 In a Icosahedral graph, (i) γ(G) =3 (ii) γt(G) = γt′(G)=4 (iii) γ(G) <γt(G)= γt′(G) Theorem 3.14 If G be the Octahedron graph, then γ(G)= γt(G)= γt′(G)=2 Proof: Figure 14: Octahedron graph Let G be the Platonic Solid Octahedron. Then G is a 4-regular graph with 6 vertices and 12 edges. Let D={1,4} be the minimum dominating set and Dt={1,4} be the minimum total dominating set and also Dt'={2,5} be the minimum inverse total dominating set of G. In the above figure, the domination number as well as total domination number and the inverse total dominating number, are the same. Hence γ(G)=2; γt(G)=2; γt′(G)=2.

Figure 14

Theorem 3.15

If G be the Desargues graph with 20 vertices, then γt(G)=9 and γt′(G) doesnot exists.