2 β 0 -excellent graphs

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1 β 0 -excellent graphs A. P. Pushpalatha, G. Jothilakshmi Thiagarajar College of Engineering Madurai India gjlmat@tce.edu, appmat@tce.edu S. Suganthi, V. Swaminathan Ramanujan Research Centre Saraswathi Narayanan College Madurai India sulnesri@yahoo.com Abstract: Claude Berge [1] in 1980, introduced B graphs. These are graphs in which every vertex in the graph is contained in a maximum independent set of the graph. Fircke et al [] in 00 made a beginning of the study of graphs which are excellent with respect to various graph parameters. For example, a graph is domination excellent if every vertex is contained in a minimum dominating set. The B-graph of Berge was called β 0 excellent graph. β 0 excellent trees were characterized []. A graph is just β 0 excellent if every vertex belongs to exactly one maximum independent set of the graph.this paper is devoted to the study of β 0 excellent graphs and just β 0 excellent graphs. Key Words: β 0 -excellent and just β 0 excellent, Harary graphs, Generalized Petersen graph 1 Introduction Let µ be a parameter and let G = (V, E) be simple graph. A vertex v V (G) is said to be µ-good if v belongs to a µ-minimum (µ-maximum) set of G according as µ is a super hereditary (hereditary) parameter. v is said to be µ-bad if it is not µ-good. A graph G is said to be µ-excellent if every vertex of G is µ-good. G is µ-commendable if number of µ-good vertices in G is strictly greater than the number µ-bad vertices of G and there should be at least one µ-bad vertex in G. G is said to be µ-fair if number of µ-good vertices in G is equal to the number of µ-bad vertices in G and G is said to be µ-poor if number of µ-bad vertices in G is strictly greater than the number of µ-good vertices in G. γ-excellent trees and total domination excellent trees have been studied in [], [8]. β 0 -excellent trees was also dealt with in some of the theorems in []. Continuing the study on γ-excellent graphs, N.Sridharan and Yamuna [4, 5, 6], made an extensive work in this area. They have defined excellent, very excellent, just total excellent, rigid very excellent graphs with respect to the domination parameter and made a substantial contribution in this area. This paper starts with the definition of β 0 - excellent graphs. In the first section, general results on β 0 - excellent graphs are proved.the second section is devoted to β 0 -excellence in Cartesian Product of graphs. The third section deals with β 0 -excellence of Harary graphs. The fourth section is devoted to the study of just β 0 -excellent graphs. Definition 1.1. Double star is a graph obtained by taking two stars and joining the vertices of maximum degrees with an edge. If the stars are K 1,r and K 1,s, then the double star is denoted by D r,s. Definition 1.. A fan F n is defined as the graph join P n 1 + K 1,where n and P n 1 is the path graph on n 1 vertices. β 0 -excellent graphs Definition.1. Let G = (V, E) be a simple graph. Let u V (G). u is said to be β 0 -good if u is contained in a β 0 -set of G. Definition.. u is said to be β 0 -bad if there exists no β 0 -set of G containing u. Definition.. A graph G is said to be β 0 -excellent if every vertex of G is β 0 -good. Example The β 0 -sets of G are {1,, 6, 8}, {5, 6, 7, 8}, {, 4, 5, 7}. Hence all the vertices are β 0 -good. Hence G is β 0 - Theorem.5. (1) K n is β 0 - ()The central vertex of K 1,n is β 0 -bad and every other vertex is β 0 -good. () C n is β 0 - ISSN: Issue, Volume 10, February 011

2 (4) P n is β 0 -excellent if and only if n is even. (5) In a Double star D r,s, all the pendent vertices are β 0 -good but the two supporting vertices are β 0 - bad. Hence D r,s is not a β 0 -excellent graph. (6) K m,n is β 0 -excellent if and only if m = n. (7) In W n, the central vertex is β 0 -bad, while other vertices are β 0 -good. (8) K n is a β 0 -excellent graph. (9) F n, n is not β 0 - Remark.6. Suppose G has a unique β 0 -set. Then G is β 0 -excellent if and only if G = K n. Remark.7. If G has a full degree vertex and if G K n, then G is not β 0 - Theorem.8. For any graph G, G K 1 is β 0 - Definition.9. A graph is said to be β 0 -fair(β 0 -poor) graph if the number of β 0 -good vertices is greater than(less than) the number of β 0 -bad vertices. Example.10. Let G be the graph obtained from K 1, by subdividing all pendent edges exactly once. Then G is β 0 -fair. Example.11. In G = K 4 {e}, exactly two vertices are β 0 -good and remaining vertices are β 0 -bad. If n 5, then G = K n {e} is β 0 -poor, since the number of β 0 -bad vertices is greater than number of β 0 -good vertices. Theorem.1. Every non β 0 - excellent graph can be embedded in a β 0 -excellent graph. If G is a non β 0 -excellent graph, then G K 1 is a β 0 -excellent graph in which G is embedded. Remark.1. Suppose G = K n+1. Then β 0 (G K 1 ) β 0 (G) = n, which means the difference between the independence number of the graph, in which the given graph is embedded and the given graph is large. Definition.14. A graph G is said to be vertex transitive if given any two vertices u, v(u v) of G, there is an automorphism φ of G such that φ(u) = v. If G is vertex transitive, then it is regular. Theorem.15. Any vertex transitive graph is β 0 - Proof. Let G be a vertex transitive graph. Let S be a β 0 -set of G. Let u V (G). Suppose u / S. Select any vertex v S. As G is vertex transitive, there exists an automorphism φ of G which maps v to u. Let S = {φ(w) : w S}. Since S is a β 0 -set and φ is an automorphism, S is a β 0 -set. Since v S, φ(v) = u S. Therefore G is β 0 - Theorem.16. Let G be a non β 0 -excellent graph. Then there exists a graph H in which the following conditions are true. (i) H is β 0 - (ii) G is an induced subgraph of H. (iii) β 0 (H) = β 0 (G). Proof. Let G be a non -β 0 -excellent graph. Let B = {b 1, b,..., b k } be the set of all β 0 -bad vertices of G. Let V 1, V,..., V k be a set of independent sets of maximum cardinalities containing b 1, b,..., b k respectively. Let V i = t i, 1 i k. Then t i < β 0 (G), for } all i, 1 i k. Let W i = {w i1, w i,..., w iβ0 ti, 1 i k. Add each element of W i, 1 i k as a vertex to the vertex set of G. Let the new sets of vertices W 1, W,...,, W k be made a complete k- partite graph. Join each vertex of W i with every vertex of V V i, 1 i k. Let H be the resulting graph. Then V i W i is an independent set of H of cardinality β 0. Any β 0 -set of G continues to be an independent set of H of cardinality β 0. There is no other independent set of H of cardinality greater than β 0. Therefore β 0 (H) = β 0 (G). Each new vertex added to G and each b i is contained in a maximum independent set of H. Therefore H is a β 0 -excellent graph. Clearly, G is an induced subgraph of H and β 0 (H) = β 0 (G). Theorem.17. Let G, H be β 0 -excellent graphs with V (G) V (H) = φ. Then (i) G H is β 0 - (ii) G + H is β 0 -excellent if and only if β 0 (G) = β 0 (H). Proof. (i) Any β 0 -set of G H is of the form S 1 S, where S 1 is a β 0 -set of G and S is a β 0 -set of H. Hence G H is β 0 - (ii) Let β 0 (G) < β 0 (H). Then any β 0 -set of G + H is a β 0 -set of H and conversely. If β 0 (G) = β 0 (H), then any β 0 -set of G and any β 0 -set of H are β 0 -sets of G + H and conversely. Therefore G + H is β 0 - excellent if and only if β 0 (G) = β 0 (H). Definition.18. Let G 1 = (V 1, E 1 ) and G = (V, E ) be any two graphs Then their Cartesian Product G 1 G is defined to be the graph whose vertex set is V 1 V and edge set is {((u 1, v 1 ), (u, v )) : either u 1 = u and v 1 v E or v 1 = v and u 1 u E 1 }. Theorem.19. Let H be a graph. (i) Let n χ(h). Then β 0 (K n H) = V (H) and K n H is β 0 - (ii) Let n < χ(h). Then β 0 (K n H) = t, where t is the maximum cardinality of an union of n-disjoint independent sets in H. ISSN: Issue, Volume 10, February 011

3 Proof. { (i) Let } n χ(h). Let = V1, V,..., V χ(h) be a chromatic partition of H. Let V (K n ) = {u 1, u,..., u n }. Then S { = {(u 1, v) : v V 1 } {(u }, v) : v V }... (uχ(g), v) : v χ(g) V χ(g) is an independent set of K n H. Therefore β 0 (K n H) V (H). But β 0 (K n H) β 0 (K n ) V (H) = V (H). Therefore β 0 (K n H) = V (H). Any set of χ-vertices of K n will produce a β 0 -set of K n H. Hence K n H is β 0 - (ii) Let n < χ(h). Let S 1, S,..., S n be disjoint independent sets in H such that S i is n maximum. Let t = n i=1 i=1 S i. Then T = {(u 1, v) : v S 1 } {(u, v) : v S }... {(u n, v) : v S n } is an independent set of K n H. Therefore β 0 (K n H) n S i = T = t. Let S be a maximum i=1 independent set of K n H. Let X i = S ({u i } V (H)), 1 i n. Let Y i = {v V (H) : (u i, v) X i, 1 i n). Then Y i s are independent and disjoint in H. S = n X i = i=1 n Y i n S i = T. Therefore t = T i=1 i=1 β 0 (K n H) = T = t. Illustration.0. Let H be K 5,,5,. Then K H is not β 0 - (Here χ(h) = 4 > ). Theorem.1. K n H is β 0 -excellent if and only if every vertex of H belongs to the union of disjoint independent sets of H of maximum cardinality. Proof. Suppose every vertex of H belongs to the union of disjoint independent sets of H of maximum cardinality. Let V (K n ) = {u 1, u,..., u n }. (u i, v) V (K n H), 1 i n. Then v V (H). Then there exist disjoint independent sets S 1, S,..., S n of n H such that S i = t is maximum and v S j, for i=1 some j, 1 j n. Then T = {(u 1, v) : v S 1 } {(u, v) : v S }... {(u i, v) : v S j } {(u j, v) : v S i }... {(u n, v) : v S n } is a maximum independent set of K n H containing (u i, v). Therefore K n H is β 0 - Conversely, Suppose every vertex of H belongs to the union of disjoint independent sets of H of maximum cardinality. Then there exists a vertex v H such that v does not belong to any union of n disjoint independent sets of H of maximum cardinality. Since any maximum independent set of K n H is obtained from n disjoint independent sets of H, with the union having maximum cardinality, (u i, v), 1 i n will not belong to any maximum independent set of K n H. Therefore K n H is not β 0 - Theorem.. Let H be a graph. Then K n H is β 0 -excellent if and only if H is β 0 - Proof. Suppose H is β 0 - Then β 0 (K n H) = n.β 0 (H). Any β 0 -set of H gives rise to a β 0 -set of K n H. Therefore K n H is β 0 - Suppose H is not β 0 - Let u V (H) be such that u is not contained in any β 0 -set of H. Suppose S is a β 0 -set of K n H containing (v, u), for some v V (K n ). Therefore S = n.β 0 (H). Also S is of the form V (G) T, where T is a β 0 -set of H. Therefore u T, a contradiction. Theorem.. Let G K n and let G be a β 0 - excellent graph. Let H = P n. Then G H is β 0 -excellent if (i) or (ii)is satisfied. G H is not β 0 - excellent if (iii) is satisfied. (i) For any β 0 -set S of G, there exists a β 0 -set of G in V S (ii) Let the cardinality of the union of any two disjoint non-maximum independent set of G S + β 0 (< V S >), for any β 0 -set S of G. For every β 0 -set S of G, V S does not contain β 0 -set of G and for any β 0 -set S of G, the maximum number of independent elements in V S is the same. (iii) If any two β 0 -sets of G are not disjoint and there exists a β 0 -set S of G such that the maximum number of independent elements in V S is greater than the maximum number of independent elements in the complement of any other β 0 -set, then G H is not β 0 - Proof. (i) Let G have two disjoint β 0 -sets. Then β 0 (G P n ) = nβ 0 (G). For: Let S 1, S be two disjoint β 0 -sets of G. Let V (P n ) = {v 1, v,..., v n }. {(x i, v i ) : x i S 1 } {(y i, v i+1 ) : y i S } is an independent set in G P n. Thus {(x i, v 1 ) : x i S 1 } {(y i, v ) : y i S } {(x i, v ) : x i S 1 } {(y i, v 4 ) : y i S }... {(x i, v n 1 ) : x i S 1 } {(y i, v n ) : y i S } is an independent set of G P n. Therefore β 0 (G P n ) nβ 0 (G). But β 0 (G P n ) β 0 (G) V (P n ) = nβ 0 (G). Hence β 0 (G P n ) = nβ 0 (G). Let (x, y) V (G P n ). Then there exists a β 0 -set S 1 of G containing x. Also by hypothesis, V S 1 contains a β 0 - n set of G,say S. (S 1 {v t 1 }) n (S {v t }) t=1 t=1 and n (S {v t 1 }) n (S 1 {v t }) are β 0 -sets t=1 t=1 ISSN: Issue, Volume 10, February 011

4 of G P n. Hence there exists a β 0 - set of G P n containing (x, y). Therefore G P n is β 0 - (ii) It can be easily proved that β 0 (G P n ) = (β 0 (G) + k)n, where k is the maximum number of independent elements in the complement of any β 0 - set of G. In this case, G P n is β 0 -[For: Let (x, y) be any element of V (G P n ). Then there exists a β 0 -set S 1 of G containing x. Also V S 1 contains an independent set of cardinality k. Let S be a maximum independent set in V S 1. n (S 1 {v t 1 }) n (S {v t }) and n (S t=1 t=1 t=1 {v t 1 }) n (S 1 {v t }) are the β 0 -elements of t=1 G P n. Hence there exists a β 0 -set of G P n containing (x, y). Therefore G P n is β 0 - (iii) Suppose there exists a β 0 -set S 1 of G such that the maximum number of independent elements say k in V S 1 is greater than the maximum number of independent elements in the complement of any other β 0 -set of G. β 0 (G P n ) = (β 0 (G) + k)n. Let u V (G) S 1. Then there exists a β 0 -set S of G containing u. The maximum number of independent elements in V (G) S is less than k. Therefore (u, v), where v V (P n ) is not contained in any β 0 -set of G P n. Hence G P n is not β 0 - Remark.4. There exist graphs in which the maximum number of independent elements in the complement of any β 0 -set of G is greater than the maximum number of independent elements in the complement of any other β 0 -set of G. Example The β 0 -sets of G are S 1 = {1,,, 4, 5, 8, 9, 10}, S = {, 4, 5, 6, 7, 10, 11, 1}, S = {8, 9, 10, 11, 1, 1, 14, 15}. Then V S 1 = {6, 7, 11, 1, 1, 14, 15}, V S = {1,, 8, 9, 1, 14, 15}, V S = {1,,, 4, 5, 6, 7}. The set {11, 1, 1, 14, 15} is a β 0 -set in V S 1 ; {8, 9, 1, 14, 15} is a β 0 -set in V S and {,, 4, 5, 6, 7} is a β 0 -set in V S. Hence S satisfies the property described in the remark.4 Example The β 0 -sets of G are S 1 = {1,, 4, 6, 7}, S = {1,, 8, 6, 7}. Clearly G is not β 0 - The maximum number of independent sets in V S 1 and in V S is one. The sets S = {1,, 4} and S 4 = {8, 6, 7} are not β 0 -sets. The maximum number of independent sets in V (S S 4 ) is one. That is, there exist two disjoint independent sets of cardinality each and the maximum number of independent elements in complement of their union is one. S + S 4 + β 0 (V (S S 4 )) = 7 > S 1 + β 0 (V S 1 ) = S + β 0 (V S ) = 6. Example The β 0 -set of G is S = {1,,, 6, 7, 8}. The 4-element disjoint independent sets are {1,,, 5}, {4, 6, 7, 8}. β 0 (G P n ) = 8n. It can be shown that there is a β 0 -set of G and a set of maximum number of elements in the complement, such that independent set generated contains 7n elements. Though G is not β 0 excellent, G P n is β 0 excellent here. For: Example Let V (P n ) = {u 1, u,..., u n }. Let A = {1,,, 5} and B = {4, 6, 7, 8}. Two maximum independent sets of G P n are {(u i, 1), (u i, ), (u i, ), (u i, 5)} i 1(mod),1 i n and j 0(mod ), j n i 1(mod),1 i n 6 7 {(u j, 4), (u j, 6), (u j, 7), (u j, 8)} {(u i, 4), (u i, 6), (u i, 7), (u i, 8)} ISSN: Issue, Volume 10, February 011

5 j 0(mod ), j n {(u j, 1), (u j, ), (u j, ), (u j, 5)}. Hence G P n is β 0 -excellent, but G is not β 0 - Remark.9. Suppose G is a graph in which V (G) = A B, where A, B are independent and disjoint subsets are V (G). Then G P n is β 0 - (or) equivalently if G is bipartite graph, then G P n is β 0 - Hence T P n is β 0 -excellent, for any tree T and C n P n is β 0 - Theorem.0. Suppose G is of even order in which V (G) = A B, A B = φ, A, B are independent and A = B. Then G P n+1 is β 0 - Proof. The following are β 0 -sets of G P n+1. P and i 1(mod ), 1 i n+1 i 1(mod ), 1 i n+1 j 0(mod ), 1 j n R j 0(mod ), 1 j n where P = {(v, u i ) : v A)}, Q = {(v, u j ) : v B)}, R = {(v, u i ) : v B)} and S = {(v, u j ) : v A)}. Hence G P n+1 is β 0 - Corollary.1. If G is bipartite graph with equicardinal bipartition, then G P n+1 is β 0 - Corollary.. (1) C n P n+1 is β 0 - () If T is a tree with equi-cardinal bipartition, then T P n+1 is β 0 - Example The graph G is of even order in which V (G) = A B, A B = φ, A, B are independent and A = B, where A = {1,, }, B = {4, 5, 6}. Here G is not β 0 -excellent (since β 0 (G) = 4 and {1,,, 4} is the unique β 0 -set of G). Example.4. D r,r is a graph of even order in which V (G) = A B, A B = φ, A, B are independent and A = B, where A, B are respectively the set of pendents adjacent to each of the two centers. D r,r is not β 0 -excellent, since the two centers are β 0 -bad vertices. Q S, Remark.5. Consider the path P t with each vertex of P t as centers, add r-pendent vertices. Let the resulting graph be denoted by M r,r,...,t times. Then M r,r,...,r P n+1 is β 0 -excellent, but M r,r,...,r is not β 0 - Illustration M (,,,) 1: A = {1,,, 5, 10, 11, 1, 1}, B = {4, 6, 7, 8, 9, 14, 15, 16} are two disjoint independent sets and A B = V (G). M,,, is not β 0 - But M,,, P n+1 is β 0 - Theorem.7. G P n is β 0 -excellent if and only if there exists an independent partition π = {V 1, V,..., V k } of G such that max 1 i,j k,i j { V i V j } is attained for pairs (i, j) with V i V j {i, j} = {1,,..., k}. Proof. Any maximum independent subset of G P n is of the form X 1 X X n where X i = A {u i } if i is odd and X i = B {u i } if i is even, A, B being disjoint independent sets of G such that A B has maximum cardinality. Suppose G has an independent partition satisfying the hypothesis. Then clearly, G P n is β 0 - Conversely, suppose G P n is β 0 - Then every vertex {u, v}, u V (G) and v V (P n ) belongs to a β 0 -set of G P n. The structure of β 0 - sets of G P n imply that there exist disjoint independent sets V 1, V,..., V k in G whose union is V (G), satisfying the condition in the theorem. Corollary.8. Let χ(g). Then G P n is β 0 -excellent if there exists a chromatic partition π = { } V 1, V,..., V χ(g) of G such that max 1 i,j χ,i j { V i V j } is attained for pairs (i, j) with V i V j {i, j} = {1,,..., χ(g)}. Remark.9. The converse of the above corollary is not true. Consider the graph G A chromatic partition of G is given by {{, 4, 5}, {, 6}, {1}}. ISSN: Issue, Volume 10, February 011

6 Corollary.40. If G is a complete r-partite(r )graph with equi-cardinal partite sets, then G P n is β 0 - Remark.41. Let G = K + K + K. Then G is not β 0 -excellent, but G P n is β 0 - Corollary.4. Q n is β 0 -excellent,since Q n = Q n 1 P. Moreover Q n P n is also β 0 - Remark.4. Let G = K 4 + K + K. Then G and G P n are not β 0 - Theorem.44. There exists a regular graph which is not β For this graph G, the β 0 -set is {, 5, 6, 8, 11, 1, 16} consisting of 7 vertices. The remaining vertices are contained in the independent sets {1,, 6, 8, 1, 14}, {1,, 7, 9, 1, 14}, {1,, 7, 9, 1, 15} of cardinality 6 each. Thus this graph is -regular but not β 0 - Theorem.45. Let G be a bipartite graph with bipartition V 1, V. Then G C m is β 0 - Proof. Case(i): Let m = n. Let V (C n ) = {u 1, u,..., u n }. The maximum independent sets of G C n are P Q j 0(mod), 1 j n j 0(mod), j n and R S, j 0(mod), 1 j n j 0(mod), j n where P = {(v i, u j ) : v i V 1 )}, Q = {(v i, u j ) : v i V )}, R = {(v i, u j ) : v i V )} and S = {(v i, u j ) : v i V 1 )}. Hence G C m is β 0 - Case(ii): Let m = n + 1. Let V (C n+1 ) = {u 1, u,..., u n+1 }. β 0 (G C n+1 ) = n V (G). The following are β 0 -sets of G C n+1. P 1 P ; j 1(mod ), 1 j n j 0(mod ), j n and j 1(mod ), 1 j n j 1(mod ), j n+1 j 1(mod ), j n+1 Q 1 S 1 R 1 Q ; j 0(mod ), j n R j 0(mod ), j n j 0(mod ), j n S, wherep 1 = {(v i, u j ) : v i V 1 )}, P = {(v i, u j ) : v i V )}, Q 1 = {(v i, u j ) : v i V )}, Q = {(v i, u j ) : v i V 1 )}, R 1 = {(v i, u j ) : v i V 1 )}, R = {(v i, u j ) : v i V )}, S 1 = {(v i, u j ) : v i V )} and S = {(v i, u j ) : v i V 1 )}. Hence G C m is β 0 - Theorem.46. Let G be a β 0 -excellent graph. For any β 0 -set S of G, let V S contain a β 0 -set of G. Then G C n is β 0 - Proof. The proof follows from the fact that for any β 0 -set S of G and a β 0 -set S 1 of G in V S, P Q, where i 1(mod ), 1 i n j 0(mod ), 1 j n P = {(v, u i ) : v S)}, Q = {(v, u j ) : v S 1 )} and R S, i 1(mod ), 1 i n j 0(mod ), 1 j n where R = {(v, u i ) : v S 1 )}, S = {(v, u j ) : v i S)} are β 0 -sets of G C n. Theorem.47. (i) C n C k+1 is β 0 - (ii)c n C m is β 0 - (iii) C k+1 C n+1, n k is β 0 - Corollary.48. The following graphs are β 0 - (i) P n P k+1. Result follows from the fact that G P k+1 is β 0 -excellent if G is of even order and V (G) = A B, A B = φ, A = B and A, B are independent. (ii) P n C k+1. (G C m is β 0 -excellent if G is bipartite with partition V 1, V.) (iii) P n+1 C k+1. (since G C m is β 0 -excellent if G is bipartite with partition V 1, V.) (iv) P n C k. (since G C m is β 0 -excellent if G is bipartite with partition V 1, V.) (v) P n+1 C k. (since G C m is β 0 -excellent if G is bipartite with partition V 1, V.) Definition.49. Mycielski Graphs Let G = (V, E) be a simple graph. The Mycielskian of G is the graph µ(g) with vertex set equal to the disjoint union V V {u} where V = {x : x V } and the edge set E {xy, x y : xy E} {y u : y V }. The ISSN: Issue, Volume 10, February 011

7 vertex x is called the twin of the vertex and the vertex u is called the root of µ(g). Theorem.50. Let G K be a graph. Then µ(g) is not β 0 - Proof. Let V (G) = {u 1, u,..., u n }. Let V (µ(g)) = {u 1, u,..., u n, u 1,..., u n, v}. Then E(µ(G)) = n i=1 {u i u j : u j N G (u i ), 1 j n} {u iv : 1 i n}. It has been proved that β 0 (µ(g)) = max {β 0 (G), V (G) }. V (G) Suppose β 0 (G) <. Then {u 1, u,..., u n} is the only β 0 -set of µ(g). V (G) Suppose β 0 (G) > }. Let {u i1, u i,..., u iβ0 be a β 0 -set of } G.Then {u i1, u i,..., u iβ0, u, u,..., u is i1 i iβ0 a β 0 -set of µ(g). Clearly v is not in any β 0 -set of µ(g). V (G) Suppose β 0 (G) =. Suppose β 0 (G) = 1 and V (G) =. Then G = K in which case µ(g) = C 5 which is β 0 - Suppose β} 0 (G) > 1. Then for any β 0 -set {u i1, u i,..., u iβ0 of G, } {u i1, u i,..., u iβ0, u, u,..., u is a β i1 i iβ0 0 -set of µ(g). Also {u 1,..., u n} is a β 0 -set of µ(g). v does not belong to any of these β 0 -sets. Therefore µ(g) is not β 0 -excellent, when G K. Definition.51. Let G be a graph. G is said to be β 1 -excellent if every edge of G belongs to a β 1 -set of G. Remark.5. G is β 1 -excellent if and only if L(G) is β 0 - β 0 -excellence of Harary graphs Definition.1. Harary graphs H n,m with n vertices and m < n is defined as follows: Case(i): n is even and m = r. Then H n,r has n vertices 0, 1,,, n 1 and i, j are joined if i r j i + r, where the addition is taken under modulo n. Case(ii):m is odd and n is even. Let m = r + 1. Then H n, is constructed by first drawing H n,r and then adding edges joining vertex i to the vertex i + n, for 0 i n. Case(iii): m and n are odd. Let m = r + 1. Then H n, is constructed by drawing H n,r and then adding edges joining vertex 0 to the vertices n 1 and n 1. and vertex i to i + n+1 n 1, for 1 i Theorem.. Let n > r. β 0 (H r,n ) = { if divides n-r + 1 if does not divide n-r Proof. Let V (H r,n ) = {0, 1,,..., n 1}. Case(i): Let r + 1 divides n r. Consider S = {i, r + i + 1, r + i +,..., tr + t + i}, where t = 1. tr + t + i = (n r) r 1 + i = n r 1 + i. Suppose 1+i = i s (or) i s+n, according as i s 0 (or) otherwise. Then n r 1 = s (or)n s. That is r + 1 = s + n(or)s. Since s r, r + 1 s. Therefore r + 1 = s + n. But s + n > r + 1. Since s 1, n > r, a contradiction. Therefore S is an independent set in H r,n. Therefore β 0 (H (r,n) ) t + 1. Suppose S 1 is an independent set of H r,n of cardinality t + l, l. Let S 1 = {a 1, a,... a t+l }. Let a 1 < a <..., < a t+l. t + l = 1 + l + 1 (since l ). Let a 1 = i. Then a > i + r, a > i + r,..., a t+l > i + (t + l 1)r. That is a t+l > ( ) i + r. Let 1 s r. a t+l ( is adjacent ) to a 1 if and only if i s or i s + n > i + r. That is if and only ) r, a contradiction since right hand ( if s < side is negative and s is positive. (or) i s + n > i + ( ) r. This implies n s > ( ) r. n r = q(r + 1) q(r + 1) + r s > qr. qr + q + (r s) > qr. Since s r, r s 0, one has qr + q + (r s) > qr (since q 1), which is true. a t+l is adjacent to a 1. Therefore S 1 is not independent. So β 0 (H r,n ) t + 1. Therefore β 0 (H r,n ) = t + 1. Case(ii): Let r + 1 do not divide n r. Consider S = {i, r i, r + + i,..., tr + t + i}, where t =. Let n r = q(r + 1) + α, α > 0, α < r + 1. Therefore t = q. tr+t+i = qr+q+i = q()+i = α+i. Let 1 s r. If tr + t + i = i s(or)i s + n, according as i s 0(or) otherwise, then n α r + i = i s(or)i s + n. n α r + i = i s (or)i s + n. n α r = s (or) α r = s. That implies r + α n = s (or) s = r + α i.e, s < 0 or s > r(since α + r < n), a contradiction. [ r + α < n, because ISSN: Issue, Volume 10, February 011

8 n = r +q(r +1)+α. If q = 0, then n = r +α, where α < r + 1. That is n r, a contradiction. So q 1. Therefore n > r +α.] Thus S is an independent set in H r,n. Therefore β 0 (H r,n ) t+1. Suppose S 1 is an independent set of H r,n of cardinality t+l, l. Let S 1 = {a 1, a,..., a t+l }. Let a 1 < a <... < a t+l. t + l = q + l = + l > + 1. Let a 1 = i. Then a > i + r, a > i + ( ) r,..., a t+l > i + (t + l 1)r > i + r. a t+l is adjacent to a( 1 if and ) only if i s(or)i s + n is greater than i + r. That is if and only if s ( ) (or) s + n > r. ( ) But s > r is not possible, since the right hand side is positive and ( left ) hand side is negative. Therefore s + n > r. That is n s > ( q + α ) r. That leads to q(r + 1) + α + r s > qr + rα rα which means q(r + 1) + r s > qr + α = qr α. That is q(r + 1) + r s qr [ since α < 1 ], which is true, since q(r + 1) + r s = qr + q + r s qr, as r s 0. Therefore a t+l is adjacent to a 1. Therefore S 1 is not an independent set. Therefore β 0 (H r,n ) t + 1. Therefore β 0 (H r,n ) = t + 1. Theorem.. Consider H,n, where n is even. Then (i) If (r + 1) does not divide n, then {, if divides n-r β 0 (H,n ) =. + 1, otherwise (ii)if (r + 1) divides n,then β 0 (H,n ) =. Proof. We observe the following (i) Suppose (r + 1) divides n. Then r + 1 does not divide n r. Let r + 1 divide n r. Let n = q(r + 1) and = q 1 (). Therefore q() = r+q 1 (). That is (q q 1 )(r + 1) = r, a contradiction. Hence (i). (ii) Suppose (r + 1) does not divide n. Then r + 1 divides n r if and only if n = q(r + 1) + r, for some positive integer q. Let n = q(r + 1) + α, where 0 < α < (r + 1). Then n r = q(r + 1) + α r. Suppose divides. Then α r is divisible by. Let α r = k(). If k < 0, α = r+k(r+ 1) implies that α < 0, a contradiction. Hence k 0. Thus α = k(r + 1) + r. Since α < (r + 1), k = 1, one has α =. Therefore n = q()+. That means that n is odd, a contradiction. Therefore k = 0. That is, α = r. Therefore n = q(r + 1) + r. Conversely, if n = q(r + 1) + r, then clearly n r is divisible by r + 1. Case(i): Subcase (i): Suppose (r + 1) does not divide n and r + 1 divides n r. Then by observation (ii), n = q(r + 1) + r, for some positive integer q. For any integer i, i + n is of the form lr + l + i if and only if l(r + 1) = n. That is if and only if () divides n, a contradiction. Let S = {i, r i, r + + i,... tr + t + i}, where t = 1. That is t = q 1. tr + t + i = (n r) r 1 + i = n r 1 + i. Suppose n r 1 + i = i s (or) i s + n according as i s 0 (or) otherwise. Then n r 1 = s(or) n s. That is r + 1 = n + s (or) s. Since s r, r + 1 s. Therefore r + 1 = n + s. But s + n > r + 1, since s 1 and n > r, a contradiction. Therefore S is an independent set in H,n. Therefore β 0 (H,n ) t+1 =. Since H r,n is a spanning subgraph of H,n, we get that β 0 (H,n ) β 0 (H r,n ) =. Therefore β 0 (H,n ) =. Subcase (ii): (r + 1) does not divide n and r + 1 does not divide n r. By observation (ii), n = q(r + 1) + α, where 0 < α < (r + 1) and α r. Proceeding as in case (ii) of theorem., we get that S = {i, r i, r + + i,... tr + t + i}, where t = is an independent set of H,n. Therefore β 0 (H,n ) + 1. But β 0 (H,n ) β 0 (H (r,n) ) = + 1. Therefore β 0 (H,n ) = + 1. Case (ii): (r + 1) divides n. By observation (i), r + 1 does not divide n r. n Let () = l. Then i is adjacent to i + n gives that i is adjacent to lr + l + i. Let S be the set of all elements i, r i, r + + i,..., (l 1)(r + 1) + i, l(r + 1) i, (l + 1)(r + 1) i,..., t(r + 1) + l + i, where t = 1. Let n r = q(r + 1) + α, where 0 < α < r + 1. Therefore t = q 1. tr + t i = t(r + 1) i = (q 1)(r + 1) i = q(r + 1) + i r = n r α + i r = n r α + i. ISSN: Issue, Volume 10, February 011

9 Let 1 s r. If t(r + 1) i = i s (or) i s + n (according as i s 0 (or) otherwise), then n r α + i = i s(or) i s + n. That is n r α = s (or) n s. That is n r α = s (or) n r α = n s. If n r α = n s, then s = r + α, a contradiction, since s r. If n r α = s, then s = r + α n = r + α q(r + 1) α r. That is s = r q(r + 1) < 0, a contradiction. Therefore S is an independent set in H,n. Therefore β 0 (H,n ) Let S 1 = {i, r i, r + + i,..., (l 1)(r + 1) + i, l(r + 1) i, (l + 1)(r + 1) i,..., t(r + 1) + l + i}, where t =. Let. n r = q(r + 1) + α, 0 < α < r + 1. Let 1 s r. Then t = q. If t(r + 1) i = i s (or) i s + n (according as i s 0 (or) otherwise.). Then q(r + 1) i = i s (or)i s + n. That is q(r + 1) + 1 = s(or) n s. If q(r + 1) + 1 = s, then a contradiction, since L.H.S is positive. If q(r + 1) + 1 = n s, then n r α + 1 = n s. That is s = r + α 1. But s r. Therefore α 1. But α > 0. Therefore α = 1. Therefore, = q()+1 = t()+1. In this case, t(r +1)+1+i is adjacent with i. Therefore S 1 is not independent. Therefore β 0 (H,n ) Therefore β 0 (H,n ) =.. Observation.4. (i) (r + 1) can not divide both n + 1, n 1. This is because in such a case (r + 1) divides, a contradiction, since (r + 1) 4. (ii) If (r + 1) divides n 1, then r + 1 does not divide n r. This is because if divides, then = a(). n = a()+r. Let n 1 = ()l. Then ()l+1 = a()+r. ()l = a()+r 1, a contradiction. (iii) Suppose (r + 1) divides n 1. Let r + 1 do not divide n r. Then t(r + 1) + 1 < n r, where t =. Since r + 1 does not divide n r, t(r + 1) + 1 n r.suppose t(r + 1) + 1 = n r. Let n 1 = q(). Therefore t()+r = n 1 = q(). Thus r + 1 divides r, a contradiction. (iv) Suppose (r + 1) divides n + 1. Then r + 1 divides n r. Let n + 1 = q(r + 1). Therefore n r = q(r + 1) r 1 = (r + 1)(q 1). Therefore r + 1 divides n r. Theorem.5. Consider H,n, where n is odd. (i) (r +1) does not divide n 1 as well as n+1. Then β 0 (H,n ) = {, if divides n-r + 1, otherwise (ii) (r + 1) divides n 1 but not n + 1. β 0 (H,n ) = + 1. (iii) (r + 1) divides n + 1 but not n 1. Then β 0 (H,n ) =. Proof. Case (i): (r + 1) does not divide n 1 as well as n + 1. Let 0 i n 1. Let S = {i, r i, (r + 1) + i,..., t(r + 1) + i}. l(r + 1) + i = i + n+1, (i 0). This implies r + 1 divides n+1, a contradiction. l(r + 1) + 0 = 0 + n 1. This implies r + 1 divides n 1, a contradiction. l(r + 1) + i is adjacent to m()+i, if l()+i+ ( ) n+1 = m()+i. This implies (m l)() = n+1. This implies divides n+1, a contradiction. Subcase(i): Let r + 1 divide n r. Let t = 1. t(r + 1) + i = n r r 1 + i = n r 1 + i. Suppose t(r + 1) + i = i s(or) i s + n, (1 s n) according as i s 0 (or) otherwise. Then 1+i = i s (or) i s+n. That is 1 = s (or)n s. Since n >, n () is positive and s is negative. Therefore n r 1 = s is not possible. n r 1 = n 1 gives s = r + 1, 1 s n, a contradiction. Therefore S = t + 1 =. Therefore β 0 (H,n ). H r,n is a spanning subgraph of H,n. β 0 (H,n ) β 0 (H r,n ) β 0 (H,n ) =. Subcase(ii): Let r + 1 do not divide n r. Let t =. Therefore Proceeding as in case (ii) of theorem., we get that β 0 (H,n ) H r,n is a spanning subgraph of H,n. Therefore β 0 (H,n ) β 0 (H r,n ) = + 1. Therefore β 0 (H,n ) = + 1. Case (ii): (r + 1) divides n 1 but not n + 1. By observation(ii), r +1 does not divide and t(r +1)+1 < n r, where t =. Let 0 i n 1. ISSN: Issue, Volume 10, February 011

10 Let i > 0 and let S = {i, r i, (r + 1) + i,..., t(r + 1) + i}, where t =. l(r + 1) + i = i + n+1. This implies r + 1 divides n+1, a contradiction. l(r + 1) + i is adjacent to m(r + 1) + i, if l(r + 1) + i + ( ) n+1 = m()+i. This implies (m l)() = n+1. This implies divides n+1, a contradiction. Proceeding as in Case(ii), we get that S is an independent set of cardinality + 1. Thus β 0 (H,n ) + 1. H r,n is a spanning subgraph of H,n. Therefore β 0 (H,n ) β 0 (H r,n ) = 1. Therefore β 0 (H,n ) = Case(iii): (r + 1) divides n + 1 but not n 1. Then r + 1 divides n r. 0 is adjacent to n 1 and n+1. Let l 1 = n+1 (). 0 is adjacent to l 1() and l 1 () 1. Let S 0 be the set of all elements 0, r + 1,..., (l 1 1)(r + 1), l 1 (r + 1) + 1,..., t(r + 1) + 1. If a(r + 1) = b(r + 1) + n+1, where a, b (l 1 1). (a b)(r + 1) = n+1 l 1, a contradiction.. This implies a b = n+1 () = If a(r + 1) + n+1 = b(r + 1) + 1,where a l 1 1, b l 1, then (b a)(r + 1) = n 1. That is r + 1 divides n 1, a contradiction. If a(r + 1) + 1 = b(r + 1) n+1, where a, b l 1 and a > b, then a b = n+1 () = l 1. Therefore a = b + l 1 l 1 + l 1 = l 1 = n+1 (). Therefore a(r + 1) n + 1.That is t(r + 1) a(r + 1) n + 1. Therefore t n+1 (), a contradiction, since t = 1. Therefore S 0 is an independent set of cardinality t + 1 =. Let i 0. i is adjacent to n+1 + i. Let l 1 = n+1 (). Therefore i is adjacent to l 1(r + 1) + i. Let S i = {i, i + r + 1, i + (r + 1),..., (l 1 1)(r + 1) + i, l 1 (r + 1) i,..., t(r + 1) i}. If a(r + 1) + i = b(r + 1) + i + n+1 where a, b l 1 1, then a b)(r +1) = n+1 n+1. a b = () = l 1, a contradiction. If a(r + 1) + i + n+1 = b(r + 1) i, where a l 1 1, b l 1, then (b a)(r + 1) = n 1, a contradiction. If a(r +1)+1+i = b(r +1)+1+ n+1 +i where a, b l 1 and a > b,then (a b)(r + 1) = n+1. This implies a b = n+1 () = l 1. Therefore a = l 1 + b l 1 + l 1 = l 1 = n+1 and a(r + 1) n + 1. That is t(r + 1) a(r + 1) n + 1, which implies t n+1, a contradiction, since t = 1. So S i is independent set of cardinality t + 1 =. Therefore β 0(H,n ). H r,n is a spanning subgraph of H,n. Therefore β 0 (H,n ) β 0 (H r,n ) =. Therefore β 0 (H,n ) =. Theorem.6. Consider H,n, where n is odd. Let (r + 1) divide (n 1) but not n + 1. If t(r + 1) + n r, where t =, then H,n is not β 0 - Proof. (r + 1) divides n 1 but not n + 1. Let l 1 = n 1 (. 0 is adjacent to l 1(r + 1). Also 0 is adjacent to l 1 ()+1,since l 1 ()+1 = n 1 S 1 be the set of all elements 0, r + 1,..., (l 1 1)(r + 1), l 1 (r +1) =, (l 1 +1)(r +1)+,..., t(r +1)+. If t(r + 1) + < n r, then S 1 is an independent set of cardinality +1 = n+1. Let + 1. Suppose t(r + 1) + =. Then S 1 is not independent. Let S be the set of all elements 0, r + 1,..., (l 1 1)(r + 1), l 1 (r + 1) +, (l 1 + 1)(r + 1) +,..., (t 1)(r + 1) + be an independent set. Therefore S = t = < β 0. Let S 1 be the set of all elements 0, (r + 1), )r + 1),..., (l 1 1)(r + 1), l 1 (r + 1),..., t(r + 1). If t(r + 1) + < n r, then t(r + 1) > r. Therefore t(r + 1) is not adjacent to0. Therefore S 1 is an independent set of cardinality + 1 = β 0. Suppose t(r + 1) + = n r. Then S 1 is not independent. Also S be the set of all elements 0, (r +1), (r +1),..., (l 1 1)(r +1), l 1 (r + 1),..., (t 1)() is an independent set of cardinality set. < β 0. Therefore 0 does not belong to any β 0 - Illustration.7. Consider H 5,7. r =, n = 7, (r + 1) = 6 does not divide n + 1 = 8, but 6 divides n 1 = 7 1 = 6. t = = 1, β 0 = + 1 = 5 =. S 0 = {0} is a maximal independent set containing 0 and there is no β 0 -set containing 0. S 1 = {1, 4}, S = {, 5}, S = {, 6} are the β 0 -sets of H 5,7. ISSN: Issue, Volume 10, February 011

11 Remark.8. H r,n is β 0 - H,n is not β 0 -excellent if and only if n is odd and () divides n 1. 4 JUST β 0 - EXCELLENT GRAPHS N. Sridharan and M. Yamuna [10] initiated the study of just excellence in graphs with respect to the domination parameter. A graph G is just γ excellent if every vertex is contained in a unique minimum dominating set. In this section, just β 0 - excellent graphs are defined and studied. 4.1 Introduction Partition of V (G) into independent sets is the same as proper coloring of the graph. A chromatic partition is a partition of the vertex set into minimum number of independent sets. Such a partition may not contain a maximum independent set. For example, a double star contains a unique chromatic partition of cardinality two in which both the independent sets are not maximum. The question that naturally arises is that Does there exist a graph in which the vertex set can be partitioned into maximum independent sets?. This leads to the concept of just β 0 - excellent graphs. It is shown in this chapter that a graph of order n is just β 0 - excellent if and only if β 0 (G) divides n, G has exactly n β 0 distict β 0 sets and the maximum cardinality of a partition of V (G) into independent sets is n β 0. This section is devoted to the definition and properties of just β 0 - excellent graphs, just β 0 excellence in product graphs, just β 0 excellence in Generalized Petersen graphs and just β 0 excellence in Harary graphs. 4. Definitions and Properties of just β 0 - excellent graphs Definition 4.1. A graph G is said to be just β 0 - excellent graph if for each u V, there exists a unique β 0 -set of G containing u. Examples of just β 0 -excellent graphs (1) C n () K n () K n,n (4) P m P n, if mn 0(mod). Examples of not just β 0 -excellent graphs (1) C n+1 () K 1,n ()P n (4) The subdivision graph of K 1,n (5) Petersen graph (6) W n, n 5 (7) D r,s (8) G K 1, for any connected graph G. (9) F n = P n 1 + K 1. Properties of just β 0 -excellent graphs 1. Every just β 0 -excellent graph is a β 0 -excellent graph.. If G is justβ 0 -excellent and G K n,then there is no vertex u such that < V N[u] > contains at least two maximum independent sets. Proof. Since G is just β 0 -excellent, given u V (G), there exists a unique β 0 -set S of G containing u. Suppose V N[u] contains at least two maximum independent sets. G K n. Therefore β 0 (G) and β 0 (< V N[u] >) 1. S {u} is an independent set of < V N[u] > and hence β 0 (< V N[u] >) β 0 (G) 1. If β 0 (< V N[u] >) = β 0 (G), then any β 0 -set of < V N[u] > together with u is an independent set of G of cardinality β 0 (G)+1, a contradiction. Let T 1, T be two maximum independent sets of V N[u]. Then T 1 {u} and T {u} are maximum independent sets of G, a contradiction.. Let G be just β 0 - Then there exists a unique partition of V (G) into β 0 -sets of G. Proof. Let u V (G). Let S 1 be the unique β 0 -set of G containing u. If V S 1 = φ, then there is nothing to prove. Otherwise consider a vertex v V S 1. v is contained in a unique β 0 -set say S of G. S 1 S = φ, since G is just β 0 - If V (S 1 S ) = φ, the process stops. Otherwise there exists w V (S 1 S ). There exists a unique β 0 -set say S of G containing w. Clearly S i S j = φ, i j, 1 i, j. Proceeding like this, we get a partition of V (G) into β 0 -sets of G. 4. β 0 (G) is a factor of n. Proof. From the previous property, n = mβ 0 (G), where m is the cardinality of the partition of V (G) into β 0 -sets. 5. Let G be a just β 0 -excellent graph. Let V (G) = n. Then n = χ(g)β 0 (G). n From property 4, n = dβ 0 (G). Also β 0 (G) χ(g) and hence d χ(g). Clearly χ(g) d. Hence χ(g) = d. 6. In a just β 0 -excellent graph G, V (G) = β 0 (G).χ(G). The converse is not true. Consider P 6. β 0 (P 6 ) =, χ(p 6 ) =. V (P 6 ) = 6 = β 0 (P 6 ).χ(p 6 ). But P 6 is not a just β 0 -excellent graph. 7. δ(g) n β 0 (G) 1. Proof. Let Π = {S 1, S,..., S m } be a β 0 -set partition of V (G). Let u S i. Then u is adjacent to at least one vertex in each S j, j i. Therefore deg(u) m 1. Therefore δ(g) m 1 = ISSN: Issue, Volume 10, February 011

12 n β 0 (G) n β 0 (G) = 1 if and only if G = K n. 9. If G has two or more disjoint β 0 -sets, then G has no isolates. Proof. Suppose G has two or more disjoint β 0 -sets. Let S 1, S,..., S t be the disjoint β 0 -sets. Then t. Suppose G has an isolate, say u. Let u S 1. Then S {u} is an independent set of cardinality β 0 (G)+1, a contradiction. (or) [ Equivalently, any isolate vertex is contained in every β 0 -set and hence if there are isolates, there can not be two or more disjoint β 0 -sets.] Thus, if G is just β 0 -excellent and G has an isolate, then G = K n and conversely. 10. Let G be a just β 0 -excellent graph. If G K and G K n, then δ(g). Proof. Since G K n and since G is a just β 0 - excellent graph, δ(g) 1. Suppose u is a pendent vertex of G. Let N(u) = {v}. Since G is just β 0 -excellent, there exists a β 0 - set of D containing v. Therefore v D and u / D. Suppose β 0 (G) = 1. Then G is a complete graph. Since G K and δ(g) 1, G = K n, n. Therefore δ(g). Therefore u is not a pendent vertex, a contradiction. Suppose β 0 (G). Then D. Therefore there exist w D, w v. Let D 1 = (D {v}) {u}. Then D 1 ia a β 0 -set of G and w is contained in two β 0 -sets of G namely D and D 1, a contradiction. Therefore δ(g). Remark 4.. Any even cycle G is a just β 0 -excellent graph with δ(g) =. Any tree is not a just β 0 - excellent graph. 11. A graph G has exactly two disjoint β 0 -sets whose union is V (G) say V 1, V if and only if for every non empty proper subset A of V 1 or V, N(A) > A. Proof. Suppose G has exactly two disjoint β 0 -sets whose union is V (G) say V 1, V. Let A V 1. Suppose N(A) A. Let C = V N(A). If C = φ, then N(A) = V. Thus A N(A) = V = β 0 (G). But A V 1, a contradiction. Thus C φ. A C is an independent set of G and A C = A + C = A +β 0 (G) N(A) β 0 (G), a contradiction, since G has exactly two disjoint β 0 - sets whose union is V (G). Therefore, N(A) > A. Conversely, let there be two disjoint β 0 -sets whose union is V (G) say V 1, V and for any proper subset A of V 1 or V, N(A) > A. Let W be a β 0 -set of G. W V 1, and W V. Let W V 1 = W 1, W V = W. Then W 1 φ, W φ. N(W 1 ) > W 1. N(W 1 ) W = φ. (For: if x N(W 1 ) W, then x N(W 1 ) and x W. That is x is adjacent to every vertex in W 1 and x W. But W 1 W = W is an independent set, a contradiction.) W 1 + W = β 0 (G). Therefore N(W 1 ) + W > β 0 (G). That is V > β 0 (G), a contradiction. Hence the theorem. Corollary 4.. A graph G has exactly two disjoint β 0 - sets whose union is V (G) if G is of even order and contains a spanning cycle u 1, u,..., u n such that whenever u i, u j are adjacent, then i, j are of opposite parity. Proof. Suppose G is of even order and contains a spanning cycle u 1, u,..., u n such that whenever u i, u j are adjacent, then i, j are of opposite parity. Then {u 1, u,..., u n 1 }, {u, u 4,..., u n } are the only β 0 -sets of G whose union is V (G). The converse is not true. (i) Consider G u u u u There are exactly two disjoint β 0 -sets {1,, 5, 7, 9, 11, u }, {, 4, 6, 8, 10, 1, u} whose union is V (G). G has no spanning cycle. For: consider S = {4, 8, 5, 11}. ω(g S) = 5 and the five components are {u}, {u }, {9, 10}, {6, 7}, {1,,, 1}. (If G has spanning cycle, then for any S V (G), ω(g S) S.) (ii) Consider C n, (n 6). Let V (C n ) = {u 1, u,..., u n }. Add two more vertices u, u. Join u with u n 1, u n and u with u n 4, u n 6. Let G be the resulting graph. Let S = {u n 1, u n, u n 4, u n 6 }. The components of G S are {u}, {u }, {u n }, {u n 5 }, {u n, 1,..., u n 7 }. Therefore ω(g S) > S. Therefore G can not contain a spanning cycle. But G has exactly two β 0 -sets namely {1,, 5,..., (n 1), u }, {, 4,..., n, u}. 1. Let G have two disjoint β 0 -sets V 1, V whose union is V (G). Then (a) G has no isolates. (b) N(V 1 ) = V and N(V ) = V 1. ISSN: Issue, Volume 10, February 011

13 (c)if G K, then δ(g). Proof. (a) Suppose G has an isolate say u. Let u V 1. Then V {u} is an independent set of cardinality β 0 (G) + 1, a contradiction. Therefore G has no isolates. (b) Suppose N(V 1 ) V. Let v V N(V 1 ). Then v is an isolate of G, a contradiction. Therefore N(V 1 ) = V. Similarly, N(V ) = V 1. (c) Let u V 1. (Similar proof holds if u V ). If V 1 = 1, then G = K, a contradiction. Therefore V 1 > 1. Let A = {u}. Since N(A) > A, N(A). Therefore deg(u). Therefore δ(g). 1. Let G have exactly two disjoint β 0 -sets V 1 (G) and V (G) whose union is V (G). Then G is connected. Proof. Suppose G is disconnected. Let G 1 be a component of G and G =< V (G) V (G 1 ) >. Let V 1 V (G 1 ) = A, V V (G 1 ) = D, V 1 V (G ) = C and V V (G ) = B. Since φ A V 1. Then N(A) > A, ( using property 11). N(A) D. Therefore V N(A) B (since B D = V ). Therefore V N(A) B. V = N(A) + V N(A) and hence β 0 (G) > A + B. Similarly, V 1 = β 0 > C + D. Therefore A + B + C + D < β 0 (G). But V 1 = A + C. V = B + D. V 1 + V = A + B + C + D. Then β 0 (G) = A + B + C + D, a contradiction. Therefore G is connected. 14. Every just β 0 -excellent graph G K n is connected. Proof. Suppose G is not connected. Since G K n, one of the connected components of G, say G 1, has at least two vertices. Claim: G 1 is a just β 0 -excellent graph. Let u V (G 1 ). Then there exists a unique β 0 -set say S of G containing u. Let S 1 = S V (G 1 ). Then S 1 is an independent set of G 1 containing u. Suppose S 1 is not a β 0 -set of G 1. Then S 1 < β 0 (G 1 ). Let S = S V (G G 1 ). Then S = S 1 S and S 1 and S are disjoint. Therefore β 0 (G) = S = S 1 + S < β 0 (G 1 ) + β 0 (< G G 1 >). But β 0 (G) = β 0 (G 1 )+β 0 (< G G 1 >), a contradiction. Therefore S 1 is a β 0 -set of G. G 1 is β 0 -excellent graph. Let u V (G 1 ). Suppose A and B are β 0 -sets of G 1 containing u. Let C be any β 0 -set of < G G 1 >. Then A C, B C are β 0 -sets of G containing u, a contradiction, since G is just β 0 - Therefore G 1 is just β 0 - Since G 1 is connected and of order, there are at least two β 0 -sets in G 1. Let A 1, B 1 be two β 0 -sets of G 1. Let C be a β 0 -set of < G G 1 >. Then C A 1, C B 1 are two β 0 -sets of G containing C which is non empty, a contradiction. Therefore G is connected. 15. Let G be a just β 0 -excellent graph. Let u V (G). Let S be the unique β 0 -set of G containing u. Then < pn[u, S] > is complete and pn[u, S] χ(g). Proof. Let x, y pn(u, S). Then u is adjacent to x, y. Also x, y are not adjacent to any vertex of S {u}. If x, y are not adjacent, then (S {u}) {x, y} is an independent set of G of cardinality β 0 (G) + 1, a contradiction. Therefore N[u] is complete. Since G is just β 0 -excellent, there exist at least N[u] β 0 -sets in G. Therefore pn[u, S] number of β 0 -sets of G = χ(g). 16. There are graphs for which pn[u, S] = χ(g). Consider K 4,4,4. Let V (K 4,4,4 ) be the set of all elements u 1, u,u, u 4, v 1, v, v, v 4, w 1, w, w, w 4 where {u 1, u, u, u 4 }, {v 1, v, v, v 4 }, {w 1, w, w, w 4 } are β 0 -sets. Remove the edges v 1 u, v 1 u, v 1 u 4, w 1 u, w 1 u, w 1 u 4. Let G be the resulting graph. G is just β 0 -excellent having the three β 0 - sets, {u 1, u, u, u 4 }, {v 1, v, v, v 4 }, and {w 1, w, w, w 4 }. Let S = {u 1, u, u, u 4 }. pn[u, S] = {u 1, v 1, w 1 }. Then pn[u, S] = = χ(g). 17. Let G be a bipartite just β 0 -excellent graph and G K. Let u V (G). Let S be the unique β 0 -set of G containing u. Then pn[u, S] = {u}. Proof. Since G is bipartite, χ(g) =. Since G is just β 0 -excellent, number of β 0 -sets of G = χ(g) =. If for any u V (G), pn[u, S] {u}, then there exists u pn[u, S], v u. Also, if V 1, V is the bipartition and if u V 1, then (V 1 {u}) {v} is a β 0 -set, contradicting the fact that there are exactly two β 0 - sets. Therefore pn[u, S] = {u}. Remark 4.4. pn[u, S] = 1 < = χ(g). Example: 1 G 1 : 1 4 G 1 is γ-excellent but not β 0 - Example : 5 ISSN: Issue, Volume 10, February 011

14 G : G is neither γ nor β 0 - Example: 1 G : G is β 0 -excellent but not γ- Example:4 G 5 : G 5 is just β 0 -excellent but not γ-just Example:5 G 6 : G 6 is neither just γ-excellent nor β 0 - Example:6 C 9 is just γ-excellent but not just β 0 - C 9 is β 0 - Example:7 K n is both just γ-excellent and just β 0 - Remark 4.5. Q n is just β 0 -excellent (β 0 (Q n ) = n 1, each vertex is n-regular and χ(q n ) = ). Theorem 4.6. A graph G is just β 0 -excellent if and only if (i) β 0 (G) divides n. n (ii) G has exactly β 0 (G) distinct β 0-sets. (iii) The maximum cardinality of a partition of V (G) into independent sets is n β 0 (G). 7 Proof. Let G be a just β 0 - Let S 1, S,..., S m be the collection of distinct β 0 -sets of G. Since G is just β 0 -excellent, these sets are pairwise disjoint and their union is V (G). Therefore (i),(ii) and (iii) follows. Conversely, let G be a graph satisfying the conditions (i), (ii) and (iii). Let n = mβ 0 (G). By condition (iii), there exist independent sets V 1, V,..., V m such that they are pairwise disjoint and V 1 V... V m = V. Therefore n = m i=1 V i mβ 0 (G). Since n = mβ 0 (G), each V i is a maximum independent sets of G. Therefore V = V 1 V... V m and V i s are pairwise disjoint β 0 -sets. Therefore G is β 0 - n Since G has exactly β 0 (G) (= m) distinct β 0-sets, V 1, V,..., V m are the only β 0 -sets of G. Therefore G is just β 0 - Observation 4.7. Let G be a just β 0 -excellent graph. Then (G) (χ(g) 1)β 0 (G). Proof. Let u V (G). Let deg(u) > (χ(g) 1)β 0 (G). u is not adjacent to at least β 0 (G) 1 vertices. deg G (u) + deg G (u) = n 1. Therefore n 1 > (χ(g) 1)β 0 (G) + β 0 (G) 1 = χ(g)β 0 (G) 1 = n 1, a contradiction. Therefore deg G (u) (χ(g) 1)β 0 (G). Therefore (G) (χ(g) 1)β 0 (G). Remark 4.8. The upper bound is reached in G = K n1,n,...,n r, where n 1 = n,... = n r = n.( χ(g) = r, β 0 (G) = n, deg(u) = (r 1)n) = (χ(g) 1)β 0 (G). Theorem 4.9. Let G, H be just β 0 -excellent graphs and G K n, H K n. Then (i) G H is not just β 0 - (ii) G + H is just β 0 -excellent if and only if β 0 (G) = β 0 (H). Proof. (i) Since G K n, H K n, G has at least two β 0 -sets and H has at least two β 0 -sets. Let u V (G). Then there exists a unique β 0 -set S in G containing u. Let T 1, T be two β 0 - sets of H. Then S T 1, S T are two β 0 -sets of G H containing u. Therefore G H is not just β 0 - (ii) Suppose G + H is just β 0 - Then G + H is β 0 - Therefore β 0 (G) = β 0 (H). Conversely, let β 0 (G) = β 0 (H). Any β 0 -set of G+H is either a β 0 -set of G or a β 0 -set of H. Since G, H are just β 0 -excellent, we get that G + H is just β 0 - Theorem Let G be a just β 0 -excellent and let G K n and G be not a bipartite graph. Then β 0 (G) n, where n = V (G). ISSN: Issue, Volume 10, February 011