The PI Index of the Deficient Sunflowers Attached with Lollipops

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 6, 011, no. 41, The PI Index of the Deficient Sunflowers Attached with Lollipops Min Huang Institute of Mathematics Physics and Information Sciences Zhejiang Normal University Jinhua 31004, China Lili He Institute of Mathematics Physics and Information Sciences Zhejiang Normal University Jinhua 31004, China Abstract The Padmaar-Ivan (PI) index is a Wiener-Szeged-lie topological index which reflects certain structural features of organic molecules, in this paper we study the PI index of the deficient sunflowers attached with lollipops. Mathematics Subject Classification: 05C1, 05C90 Keywords: PI index, deficient sunflowers, lollipops 1 Introduction The topological index is introduced to reflect certain structural features of organic molecules in chemical graph theory. The topological indices based on the distance of vertexes in a graph have played an important role in describing molecular graph and establish the relationship between the structure of molecular and its features in chemistry theory. At the same time it was extensively used to predict the physical and chemical properties of compounds and their biological activity. The wiener index proposed by American chemist Wiener in 1947 was the first topological index proposed [1-]. It was used to investigate the relationship between the boiling point and its molecular structures of Alanes. Then P.V.Khadiar proposed another topological index: the

2 00 Min Huang and Lili He Padmaar-Ivan (PI) index [3] which has many superior properties to wiener index [4]. The PI index of a graph G is defined as follows: PI = PI(G) = [n eu (e G)+n ev (e G)], where for edge e = uv n eu (e G) is the number of edges of G lying closer to u than v, n ev (e G) is the number of edges of G lying closer to v than u and the summation goes over all edges of G. The edges which are equidistant from u and v are not considered for the calculation of PI index [5]. Preliminary Notes We called a vertex with degree a -vertex. For further details, please see [6]. Definition.1. [7]Suppose every edge of a cycle C ( 3)is adjacent with a triangle, every triangle except the cycle C has a -vetex, we called this graph a sunflower and is denoted by SF. All these triangles except C are called the petals of this sunflower, every petal has and only has one -vertex. Definition.. When we delete λ petals from SF, we obtain a new graph which is called deficient sunflower, denote it DSF,λ, where λ is the number of edges in C which are not adjacent to a triangle except C, we called λ the deficient value of DSF,λ. Obviously, 0 λ. Forv i v i+1 E(C ), suppose v i v i+1 is adjacent with a triangle v i v i+1 u i v i except C, we say there is a petal at edge v i v i+1. Definition.3. For the deficient sunflower DSF,λ, let all the vertex of C uniformly distributed on a cycle, draw a diameter through the midpoint of edge v i v i+1, when is odd, this diameter must pass through a vertex v j of C, we called this vertex v j the corresponding vertex of edge v i v i+1. Suppose v j is a -vertex, we say the corresponding vertex of edge v i v i+1 is a -vertex. Similarly we called the edge v i v i+1 the corresponding edge of vertex v j. When is even, this diameter must intersect with another edge v j v j+1 of C,wesayv j v j+1 the corresponding edge of edge v i v i+1. Definition.4. [8]Let C = v 1 v v v 1, P n = w 1 w w n. The lollipop L(n, ) is obtained by connecting v 1 and w 1 with an edge v 1 w 1, where n>. Definition.5. Attaching a lollipop graph L(n i, i ) to each vertex v i of cycle C in DSF,λ and mae the vertex v i coincide with the vertex w ni i in L(n i, i ), attaching a lollipop graphl(n i, i ) to each -vetex u i of a petal and mae the vertex u i coincide with the vertex w ni i, we get a deficient sunflower attached with lollipops and is denoted by LDSF,λ.

3 PI index of deficient sunflowers 003 We use (H) to express the contributions of edges in subgraph H of G to the PI index of graph G and use Δ i to express the triangle v i v i+1 u i v i which is adjacent with the edge v i v i+1 of C. If there is a deficit at the edge of v i v i+1, Δ i degenerate into an edge, then the contribution of Δ i to the PI index of graph G equals to the contribution of edge v i v i+1 to the PI index of graph G. 3 Main Results These are the main results of the paper. Lemma 3.1. [9]Let T n be a tree with n vertices, n, we have PI(T n )= (n 1)(n ). Lemma 3.. [9]Let C n be a cycle with n vertices, we have { n(n 1), n is odd; PI(C n )= n(n ), n is even. Theorem 3.3. Let G be a graph with n vertices u 1, u,..., u n. We define G as follows: attaching the end vertex w ni i of the lollipops graph L(n i, i ) to the vertex u i, where,,..., n. We have PI(G )= n (L(n i, i )) + PI(G)+N E(G) (m 1 n 1 + m n + + m n n n ) where N = n 1 + n + + n n,{ m i = {xy E(G) d(x, u i )=d(y, u i )}, t = (ni 1)(t 1) + ( E(G) + N, (L(n i, i )) = i 1), i is odd; n i (t 1) i, i is even. Proof. Obviously, when i is odd (L(n i, i )) = (n i 1)(t 1) + ( i 1). When i is even (L(n i, i )) = i (t ) + (n i i )(t 1) = n i (t 1) i. And because (G) =PI(G)+N E(G) (m 1 n 1 + m n + + m n n n ). The theorem follows.

4 004 Min Huang and Lili He Theorem 3.4. When is even, we have PI(DSF,λ )=6 7λ +λ λ. When is odd, let the corresponding vertex of edge v i v i+1 be v j, we have PI(DSF,λ )=6 7λ +λ + λ d(v j ), where d(v j ) is the degree of vertex v j in DSF,λ. Proof. Let x = E(DSF,λ ) =3 λ Case 1. Let be even Subcase 1.1. Δ i is a degenerate triangle. The corresponding edge v i+ v i+1+ v i and v i+1. Other edges except v i v i+1 and v i+,,,; no petal at v i v i+1 of edge v i v i+1 would be equidistant from are not equidistant from v i v i+1+ and v i+1. Thus, the contribution of edge v i v i+1 to PI(DSF,λ )isx. Subcase 1.. Δ i is a non-degenerate triangle. Let Δ i be v i v i+1 u i v i, the corresponding edge of edge v i v i+1 be v j v j+1. Similar to Subcase 1.1, the contribution of v i v i+1 to PI(DSF,λ )isx. edge v i v i+1, we have d(u i,v i )=d(v i+1,v i ), For d(u i,v i 1 )=d(v i+1,v i 1 ), d(u i,v j+1 )=d(v i+1,v j+1 ), d(u i,v j ) >d(v i+1,v j ), d(u i,v j 1 ) >d(v i+1,v j 1 ), d(u i,v i+ ) >d(v i+1,v i+ ). Hence, the edges whose two ends are in set {v i,u i 1,v i 1,,v j+1,u j } are equidistant from v i+1 and u i. The remaining edges are not equidistant from

5 PI index of deficient sunflowers 005 v i+1 and u i. For edge u i v i, we have similar results. Hence, the contributions of edges in Δ i to PI(DSF,λ ) are x. The first half part of the theorem follows. Case. Let be odd Subcase.1. Δ i is a degenerate triangle. Let the corresponding vertex of edge v i v i+1 be v j. Clearly, d(v i,v j )=d(v i+1, v j ). If there existed {u j 1,u j } V(DSF,λ ), edges v j u j 1 and v j u j would be equidistant from v i and v i+1 respectively. Similarly, the edges except v j u j 1, v j u j and v i v i+1 are not equidistant from v i and v i+1. Hence, the contribution of edge in a degenerate triangle to PI(DSF,λ ) isx d(v j ) + 1. There are all λ degenerate triangles. Subcase.. Δ i is a non-degenerate triangle. Let Δ i be v i v i+1 u i v i, the corresponding vertex of edge v i v i+1 be v j. When d(v i ) = 4, similar to Subcase.1 we have the contribution of edge v i v i+1 to PI(DSF,λ )isx d(v j )+1. For edge u i v i+1, similar to Subcase 1. we now the edges whose two ends are in set {v i,u i 1,v i 1, v j+1,u j,v j } except v j u j are equidistant from v i+1 and u i, the remaining edges except u i v i+1 are not equidistant from v i+1 and u i. For edge u i v i, we have similar results. Hence, the sum of the contributions of edges u i v i and u i v i+1 to PI(DSF,λ ) equals x + d(v j ) 1. Thus, the contributions of edges in Δ i to PI(DSF,λ ) are x. When d(v i )= 3 and d(v i ) =, we can discuss similarly. Hence, PI(DSF,λ )=x( λ)+ (x d(v j )+1),,,; no petal at v i v i+1 Substitute 3 λ to x. The theorem follows. For the deficient sunflower attached with lollipops LDSF,λ defined in definition.5, in the following let N =(n 1 +n + +n )+(n 1 +n + +n ), x = E(DSF,λ ) =3 λ, t = x + N. Theorem 3.5. Let S = { i i is odd} { i i is odd}, P = i is even i + i, Q = i + i, PI(DSF,λ ) is provided by Theorem i is even i is odd i is odd 3.. When is odd, let the corresponding edge of vertex v i be v j v j+1 and T =( λ)n n i.,,,; no petal at v j v j+1 When is even, let the corresponding edge of edge v i v i+1 be v j v j+1 and T =( λ)n n i.,,,; no petal at v j v j+1

6 006 Min Huang and Lili He We have PI(LDSF,λ )=Q ts P +(t 1)N + PI(DSF,λ )+T. Proof. Let M = Q S, M = P (t ), M =(t 1)(N S P ). Claim 1. PI LDSF,λ [L(n i, i )] + PI LDSF,λ [L(n i, i )] = M + M + M In fact, we can classify all the edges of lollipops L(n i, i ) and L(n i, i ) in LDSF,λ into three classes according to their contributions to LDSF,λ. If i (or i ) is odd, the contribution of the corresponding edge of vertex w 1 in C i (or C i )to PI(LDSF,λ )is i 1(or i 1). All these edges are in the first class.the sum of the contributions of all these edges to PI(LDSF,λ ) are i is odd ( i 1) + i is odd ( i 1) = M. If i (or i ) is even, the contribution of every edge in C i (or C i )topi(ldsf,λ ) is t. All these edges are in the second class. The sum of the contributions of all these edges to PI(LDSF,λ ) are (t )( i is even i + i is even i )=M. The other edges except the edges in class one and class two are in the third class. Each of them has t 1 contribution to PI(LDSF,λ ). The sum of the contributions of all these edges to PI(LDSF,λ ) are (t 1)[N S ( i is even i + i is even i )] = M. Thus Claim 1 follows. Claim. Let m i and m i be defined in Theorem 3.1. When is odd, let the corresponding edge of vertex v i be v j v j+1, we have m i = λ; m i = λ if there is a petal at v j v j+1, m i = λ + 1 if there is no petal at v j v j+1. When is even, let the corresponding edge of edge v i v i+1 be v j v j+1, we have m i = λ ; m i =0ifn i =0,m i = λ if n i > 0 and there is a petal at v j v j+1, m i = λ +1ifn i > 0 and there is no petal at v j v j+1. In fact, we can prove Claim as follows: Case.1. is odd.

7 PI index of deficient sunflowers 007 For the vertex u i V (DSF,λ ), there existed one and only one edge in each petal whose two ends are equidistant from u i.thusm i = λ. For the vertex v i V (DSF,λ ), clearly, d(v j,v i )=d(v j+1,v i ). Other petals except the one at v j v j+1 have and only have one edge whose two ends are equidistant from v i. Case.. is even. We can get the conclusion similar to Case.1. Thus, Claim follows. By Theorem 3.1 the theorem follows. References [1] H Wiener. Structural determination of paraffin boiling points, J. Am. Chem. Soc,69 (1949), [] H Wiener. Co-relation: a heat of isomerization and differences in heats of vaporization of isomers, among the paraffin hydrocarbons, J. Am. Chem. Soc,69 (1947), [3] P. V. Khadiar. on a novel structural descriptor PI, Nat. Acad. Sci. Lett,3 (000), [4] M. Jaiswal, P. V. Khadiar. QSAR study on tadpole narcosis using PI index: a case of hetergenous set of compounds, Biorg. Med. Chem Soc,1 (004), [5] P. E. John, P. V. Khadiar, J.Singh. A method of computing the PI index of benzenoid hydrocarbons using orthogonal cuts, J. Math. Chem,4 (006), [6] J. A. Bondy, U. S. R. Murty. Graph Theory with Applications, Macmillan Press Ltd, London, [7] Oleg V. Borodin, Anna O. Ivanova. Planar graphs decomposable into a forest and a matching, Discrete Mathematics,309 (009), [8] Junfeng Li, Fangli Xia. The degree distance of a class of unicyclic graph, Journal of Hunan University of Technologies, (010), 1-3. [9] P. V. Khadiar, S. Karmarar, V. K. Agrawal. A novel PI index and its applications to QSPR/QSAR studies, J. Chem. Inf. Comput. Sci,41 (001), Received: May, 011