ISSN: 2250-3005 - ijcer

International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8Notations 3.1P k (m 1, m 2 ) where k 2, m 1 ,m 2 1 be the graph obtained by identifying centers of the stars K 1,.m1 and K 1,.m2 at theends of P K respectively. The graph C 3 (m 1 , m 2 , 0) is obtained from C 3 by identifying the centers of stars K 1,.m1 and K 1,.m2 atany two vertices of C 3. The graph K n (m 1 , m 2 , m 3 , m 4 , m 5,…….. m n ) denote the graph obtained from K n by pasting m 1 edges toany one vertex u i of K n , m 2 edges to any vertex u j of K n , for i j, m 3 edges to any vertex u k for i j k, m 4 edges to u li j k l,……….m n edges to all the distinct vertices of K n . C n (P k ) is the graph obtained from C n by attaching the endvertex of P k to any one vertices of C n . K n (P k ) is the graph obtained from K n by attaching the end vertex of P k to any onevertices of K n .Theorem 3.2 For any graph G, γ nsdsd (G) ≤ n.Theorem 3.3 For any connected graph G, χ (G) ≤ ∆ (G) +1.Theorem 3.4 For any graph, γ nsdsd (G) + χ (G) ≤ 2n and equality holds if and only if G K 2.Proof By theorem 3.2 and 3.3, it follows that γ nsdsd (G) + χ (G) n + ∆ + 1 n + n – 1 +1 2n. Now we assume thatγ nsdsd (G) + χ (G) = 2n. This is possible only if γ nsdsd (G) = n and χ (G) = n. Since χ (G) = n, G is complete. But for completegraph, γ nsdsd (G) = 2. Hence G K 2 . Converse is obvious.Theorem 3.5 For any graph G, γ nsdsd (G) + χ (G) = 2n-1 if and only if G P 3, K 3.Proof If G is either P 3 or K 3 , then clearly γ nsdsd (G) + χ (G) = 2n-1. Conversely, assume that γ nsdsd (G) + χ(G) = 2n-1. Thisis possible only if γ nsdsd (G) = n and χ(G) = n-1 (or) γ nsdsd (G) = n-1 and χ (G) = n.Case (i) γ nsdsd (G) = n and χ(G) = n-1. Since γ nsdsd (G) = n, G is a star. Therefore n=3. Hence G P 3. On increasing thedegree we get a contradiction.Case (ii) γ nsdsd (G) = n-1 and χ (G) = n. Since χ (G) = n, G is complete. But for K n , γ nsdsd (G) = 2, so that n = 3. HenceG K 3 .Theorem 3.6 For any graph G, γ nsdsd (G) + χ (G) = 2n-2 if and only if G K 1,3 , K 3 (P 2 ), K 4.Proof If G is any one of the following graphs K 1,3, K 3 (P 2 ), K 4, then clearly γ nsdsd (G) + χ (G) = 2n-2. Conversely, assumethat γ nsdsd (G) + χ (G) = 2n-2. This is possible only if γ nsdsd (G) = n and χ(G) = n-2 (or) γ nsdsd (G) = n-1and χ(G) = n-1 (or)γ nsdsd (G) = n-2 and χ(G) = nCase (i) γ nsdsd (G) = n and χ(G) = n-2. Since γ nsdsd (G) = n, G is a star. Therefore n = 4. Hence G K 1,3. On increasing thedegree we get a contradiction.Case (ii) γ nsdsd (G) = n-1 and χ (G) = n-1. Since χ(G) = n-1, G contains a clique K on n-1 vertices. Let S = {v} be thevertex other than the clique K n-1 .Then v is adjacent to u i for some i in K n-1 . Then { v 1 ,u i, u j } is a γ nsdsd set. Hence n = 4.Therefore K= K 3 . If d(v 1 ) = 1 then G K 3 ( P 2 ). On increasing the degree of v 1 , no graph exists.Case (iii) γ nsdsd (G) = n-2 and χ(G) = n. Since χ(G) = n, G K n . But for K n , γ nsdsd (G) = 2. Therefore n = 3. Hence G K 4 .Theorem 3.7 For any graph G, γ nsdsd (G) + χ (G) = 2n-3 if and only if G K 1,4 , K 3 (P 3 ), K 3 (2), K 3 (P 2, P 2, 0), K 5, or anyone of the graphs in the figure 3.1v 1u 3v 2u 1 u 2vu 1Gu 21u 3Figure 3.1G 2u 4Proof If G is any one of the above graphs then clearly γ nsdsd (G) + χ (G) = 2n-3. Conversely, assume that γ nsdsd (G) + χ (G)= 2n-3. This is possible only if γ nsdsd (G) = n, χ (G) = n-3 (or) γ nsdsd (G) = n-1, χ (G) = n-2 (or) γ nsdsd (G) = n-2, χ(G) = n-1(or) γ nsdsd (G) = n-3, χ(G) = n.Issn 2250-3005(online) December| 2012 Page 43