Strong Triple Connected Domination Number of a Graph - ijcer

International Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue. 1Since G is connected, there exists a vertex say v 1 (or v 3 ) in P 3 is adjacent to v 4 and v 5 in 2. Then S = {v 1 , v 2 , v 3 } forms astc -set of G so that stc(G) = p – 2. Since G is connected, there exists a vertex say v 2 in P 3 which is adjacent to v 4 and v 5 in2. Then S = {v 1 , v 2 , v 3 } forms a stc -set of G so that stc(G) = p – 2. Since G is connected, there exists a vertex say v 1 in P 3which is adjacent to v 4 in 2 and v 2 in P 3 is adjacent to v 5 in 2. Then S = {v 1 , v 2 , v 3 } forms a stc -set of G so thatstc(G) = p – 2. Since G is connected, there exists a vertex say v 1 in P 3 which is adjacent to v 4 in 2 and v 3 in P 3 which isadjacent to v 5 in 2. Then S = {v 1 , v 2 , v 3 } forms a stc -set of G so that stc(G) = p – 2. In all the above cases, no new graphexists.Case (ii) = C 3 = v 1 v 2 v 3 v 1 .Subcase (i) = K 2 = v 4 v 5 .Since G is connected, there exists a vertex say v 1 (or v 2 , v 3 ) in C 3 is adjacent to v 4 (or v 5 ) in K 2 . Then S = {v 1 , v 2 , v 4 } forms astc -set of G so that stc(G) = p – 2. If d(v 1 ) = 4, d(v 2 ) = d(v 3 ) = 2, then G F 2 . In all the other cases, no new graph exists.Subcase (ii) = 2.Since G is connected, there exists a vertex say v 1 (or v 2 , v 3 ) in C 3 is adjacent to v 4 and v 5 in 2. Then S = {v 1 , v 2 , v 3 } forms astc -set of G so that stc(G) = p – 2. If d(v 1 ) = d(v 2 ) = 4, d(v 3 ) = 2, then G H 6 . In all the other cases, no new graph exists.Since G is connected, there exists a vertex say v 1 (or v 2 , v 3 ) in C 3 is adjacent to v 4 in 2 and v 2 (or v 3 ) in C 3 is adjacent to v 5in 2. Then S = {v 1 , v 2 , v 3 } forms a stc -set of G so that stc(G) = p – 2. In this case, no new graph exists.Theorem 2.15 Let G be a connected graph with p ≥ 3 vertices and has exactly one full vertex. Then stc = 3 .For, let v be the full vertex in G. Then S = {v, v i , v j } is a minimum strong triple connected dominating set of G, where v iand v j are in N(v). Hence stc(G) = 3.Theorem 2.16 For any connected graph G with p ≥ 3 vertices and exactly one vertex has (G) = p – 2, stc(G) = 3.Proof Let G be a connected graph with p ≥ 3 vertices and exactly one vertex has maximum degree (G) = p – 2. Let v bethe vertex of maximum degree (G) = p – 2. Let v 1 , v 2 , ….. and v p-2 be the vertices which are adjacent to v, and let v p-1 bethe vertex which is not adjacent to v. Since G is connected, v p-1 is adjacent to a vertex v i for some i. Then S = {v, v i , v p-1 }forms a minimum strong triple connected dominating set of G. Hence stc(G) = 3.The Nordhaus – Gaddum type result is given below:Theorem 2.18 Let G be a graph such that G and have no isolates of order p > 3. Then stc(G) + stc( ) ≤ 2(p – 1) andstc(G). stc( ) ≤ (p – 1) 2 and the bound is sharp.Proof The bounds directly follows from Theorem 2.13. For the path P 4 , the bounds are sharp.3 Relation With Other Graph Theoretical ParametersTheorem 3.1 For any connected graph G with p > 3 vertices, stc(G) + κ(G) ≤ 2p – 2 and the bound is sharp if and only ifG K 4 .Proof Let G be a connected graph with p > 3 vertices. We know that κ(G) ≤ p – 1 and by Theorem 2.13, stc(G) ≤ p – 1.Hence stc(G) + κ(G) ≤ 2p – 2. Suppose G is isomorphic to K 4 . Then clearly stc(G) + κ(G) = 2p – 2. Conversely, Letstc(G) + κ(G) = 2p – 2. This is possible only if stc(G) = p – 1 and κ(G) = p – 1. But κ(G) = p – 1, and so G K p forwhich stc(G) = 3 = p – 1 so that p = 4. Hence G K 4 .Theorem 3.2 For any connected graph G with p > 3 vertices, stc(G) + (G) ≤ 2p – 1 and the bound is sharp if and only ifG K 4 .Proof Let G be a connected graph with p > 3 vertices. We know that (G) ≤ p and by Theorem 2.13, stc(G) ≤ p – 1. Hencestc(G) + (G) ≤ 2p – 1. Suppose G is isomorphic to K 4 . Then clearly stc(G) + (G) = 2p – 1. Conversely, letstc(G) + (G) = 2p – 1. This is possible only if stc(G) = p – 1 and (G) = p. Since (G) = p, G is isomorphic to K p forwhich stc(G) = 3 = p – 1 so that p = 4. Hence G K 4 .||Issn 2250-3005(online)|| ||January|| 2013 Page 246