ISSN: 2250-3005 - ijcer

International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8Case (i) γ nsdsd (G) = n and χ (G) = n-3. Since γ nsdsd (G) = n, G is a star. Therefore n = 5. Then G K 1,4 . On increasing thedegree no new graph exists.Case (ii) γ nsdsd (G) = n-1 and χ (G) = n-2. Since χ (G) = n-2, G contains a clique K on n-2 vertices. Let S = {v 1 , v 2 } be thevertices other than the clique K n-2 then the possible cases are = K 2 or K 2.Subcase (i) Let = K 2. Since G is connected, either v 1 or v 2 is adjacent to u i for some i in K n-2 , then {v 1 , v 2 , u i, u j } is aγ nsdsd set so that n = 5. Hence K = K 3. If d (v 1 ) = 2 and d (v 2 ) = 1, then G K 3 (P 3 ). On increasing the degree, no graphexists.Subcase (ii) Let = K 2. Since G is connected, v 1 and v 2 is adjacent to u i for some i in K n-2. Then γ nsdsd (G) = 4, so that K= K 3. If d(v 1 ) = d (v 2 ) = 1, then G K 3 (2). If d(v 1 )=1 and d(v 2 ) = 2 then G G 1 . If v 1 is adjacent to u i and v 2 adjacent to u jfor some i ≠ j in K n-2 then γ nsdsd (G) = 4. Hence K = K 3 . If d (v 1 ) = d (v 2 ) = 1, then G K 3 (P 2, P 2, 0). On increasing thedegree, no graph exists.Case (iii) γ nsdsd (G) = n-2 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 .If v is adjacent to u i for some i in K n-1, then γ nsdsd (G) = 3. Hence n = 4. Therefore K= K 4 . Ifd (v) = 1, then G K 4 (P 2 ). If d(v) = 2, then G G 2. On increasing the degree, no new graph exists.Case (iv) γ nsdsd (G) = n-3 and χ (G) = n. Since χ (G) = n, G K n. But for complete Graph K n, γ nsdsd (G) = 2 so that n = 5.Therefore G K 5.Theorem 3.8 For any graph G, γ nsdsd (G) + χ (G) = 2n - 4 if and only if G K 1,5, K 3 (3), C 4 (P 2 ), S(K 1,3 ), K 3 (P 3 ), C 3 (1,1,1),K 3 (2 , 1 , 0) , K 4 (2), K 4 (P 2, P 2, 0,0), K 5 (P 2 ). K 6, or any one of the graphs given in the figure 3.2v 1 v 2v 1v 3v 1u 1u 1v 2u 3v 2u 3v 1u 2 u 3G 1u 2 u 3G 2v 3u 1G 3u 2v 3u 1 u 2 v 2 v 3G 4v 1v 2v 1u 3v 1u 3v 3u 3v 2v 1 v 2u 1 u 2v 2u 2G 5u 1u 1 u 2G 6u 1 u 2G 7v 3u 3u 4vG 8v 1 v 2u 1u 2v 1 v 2u 3 u 4v 1v 2u 3u 4u 5u 1u 3u 2u 3 u 4u 1 u 2 u 1 u 2G 9 G 11G 10u 4G 12Figure 3.2Proof Assume that γ nsdsd (G) + χ (G) = 2n-4. This is possible only if γ nsdsd (G) = n and χ(G) = n-4 (or) γ nsdsd (G) = n-1 andχ(G) = n-3 (or) γ nsdsd (G) = n-2 and χ(G) = n-2 (or) γ nsdsd (G) = n-3 and χ(G) = n-1 (or) γ nsdsd (G) = n-4 and χ(G) = n.Issn 2250-3005(online) December| 2012 Page 44