Strong Triple Connected Domination Number of a Graph - ijcer

International Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue. 11) For any path of order p ≥ 3, stc(P p ) =2) For any cycle of order p ≥ 3, stc(C p ) =3) For any complete bipartite graph of order p ≥ 4, stc(K m,n ) =(where m, n ≥ 2 and m + n = p ).4) For any star of order p ≥ 3 , stc(K 1, p-1 ) = 3.5) For any complete graph of order p ≥ 3, stc(K p ) = 3.6) For any wheel of order p ≥ 4, stc(W p ) = 3.7) For any helm graph of order p ≥ 7, stc(H n ) = (where 2n – 1 = p).8) For any bistar of order p ≥ 4, stc(B(m, n)) = 3 (where m, n ≥ 1 and m + n + 2= p).9) For the friendship graph, , stc(F n ) = 3.Observation 2.12 For any connected graph G with p vertices, stc(G) = p if and only if G P 3 or C 3 . .Theorem 2.13 For any connected graph G with p > 3, we have 3 stc(G) ≤ p 1 and the bounds are sharp.Proof The lower bound follows from Definition 2.1 and the upper bound follows from Observation 2.12. The lower boundis attained for C 5 and the upper bound is attained for K 1,3 .Theorem 2.14 For a connected graph G with 5 vertices, stc(G) = p – 2 if and only if G is isomorphic to P 5 , C 5 , W 5 , K 5 ,K 2,3 , F 2 , K 5 – {e}, K 4 (P 2 ), C 4 (P 2 ), C 3 (P 3 ), C 3 (2P 2 ), C 3 (P 2 , P 2 , 0), P 4 (0, P 2 , 0, 0) or any one of the graphs shown in Figure 2.4.v 2v 2v 2v 2H 1 :v 1v 1v 3 H H 3 : v 12 :v 3 v 3 Hv 1 4 :v 3v 4 v 5v 4 v 5v 4 v 5v 4 v 5v 2v 2v 2H 5 :v 1v 3H 6 : v 1v 3vH 1 7 :v 3v 4 v 5v 4 v 5v 4 v 5Figure 2.4 : Graphs with stc = p – 2.Proof Suppose G is isomorphic to P 5 , C 5 , W 5 , K 5 , K 2,3 , F 2 , K 5 – {e}, K 4 (P 2 ), C 4 (P 2 ), C 3 (P 3 ), C 3 (2P 2 ), C 3 (P 2 , P 2 , 0), P 4 (0, P 2 ,0, 0) or any one of the graphs H 1 to H 7 given in Figure 2.4., then clearly stc(G) = p – 2. Conversely, let G be a connectedgraph with 5 vertices and stc(G) = 3. Let S = {v 1 , v 2 , v 3 } be a stc -set, then clearly = P 3 or C 3 . Let V – S = V(G) –V(S) = {v 4 , v 5 }, then = K 2 or 2.Case (i) = P 3 = v 1 v 2 v 3 .Subcase (i) = K 2 = v 4 v 5 .Since G is connected, there exists a vertex say v 1 (or v 3 ) in P 3 which is adjacent to v 4 (or v 5 ) in K 2 . Then S = {v 1 , v 2 , v 4 }forms a stc -set of G so that stc(G) = p – 2. If v 4 is adjacent to v 1 , if d(v 1 ) = d(V 2 ) = 2, d(v 3 ) = 1, then G P 5 . Since G isconnected, there exists a vertex say v 2 in P 3 is adjacent to v 4 (or v 5 ) in K 2 . Then S = {v 2 , v 4 , v 5 } forms a stc -set of G so thatstc(G) = p – 2. If d(v 1 ) = d(v 3 ) = 1, d(v 2 ) = 3, then G P 4 (0, P 2 , 0, 0). Now by increasing the degrees of the vertices, bythe above arguments, we have G C 5 , W 5 , K 5 , K 2,3 , K 5 – {e}, K 4 (P 2 ), C 4 (P 2 ), C 3 (P 3 ), C 3 (2P 2 ), C 3 (P 2 , P 2 , 0) and H 1 to H 5 andH 7 in Figure 2.4. In all the other cases, no new graph exists.Subcase (ii) = 2.||Issn 2250-3005(online)|| ||January|| 2013 Page 245