Journal of Mathematical Chemistry Vol. 38, No. 3, October 005 ( 005) DOI: 0.007/s090-005-584-7 On the Randić index Huiqing Liu School of Mathematics and Computer Science, Hubei University, Wuhan 43006, China E-mail: liuhuiqing@eyou.com Mei Lu Department of Mathematical Sciences, Tsinghua University, Beijing 00084, China Feng Tian Institute of Systems Science, Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences, Beijing 00080, China Received 6 March 003; April 005 The Randić index of an organic molecule whose molecular graph is G is defined as the sum of (d(u)d(v)) / over all pairs of adjacent vertices of G, where d(u) is the degree of the vertex u in G. In Discrete Mathematics 57, 9 38 by Delorme et al. gave a best-possible lower bound on the Randić index of a triangle-free graph G with given minimum degree δ(g). In the paper, we first point out a mistake in the proof of their result (Theorem of [00]), and then we will show that the result holds when δ(g). KEY WORDS: connectivity index, Randić index, triangle-free graph, minimum degree AMS subject classification: 05C8. Introduction A single number which characterizes the graph of a molecular is called a graphtheoretical invariant or topological index. The structure property relationships quantity the connection between the structure and properties of molecules. The connectivity index is one of the most popular molecular-graph-based structure-descriptors (see [3]), and is defined in [] as C(λ) = C(λ; G) = (d(u)d(v)) λ, Corresponding author 345 059-979/05/000-0345/0 005 Springer Science+Business Media, Inc.
346 H. Liu et al. / On the Randić index where d(u) denotes the degree of the vertex u of the molecular graph G, where the summation goes over all pairs of adjacent vertices of G and where λ is a pertinently chosen exponent. The respective structure-descriptor was introduced a quarter of century ago by Randić, who chose λ =(/), and now is referred to as the Randić index or molecular connectivity index or simply connectivity index. The Randić index has been closely correlated with many chemical properties [9]. It is viewed as a measure of branching of the carbon-atom skeleton, and hence the ordering of isomeric alkanes with respect to decreasing C(/)-values basically represents their ordering according to the increasing extent of branching. This index was found to parallel closely the boiling point, Kovats constants, and a calculated surface. However, other choice of λ were also considered (see [,7,8,]) and the exponent λ was treated (see [5,4,]) an adjustable parameter, chosen so as to optimize the correlation between C(λ) and some selected class of organic compounds. Comparing with other topological indexes reported by Amidon and Anik (see [0]), the Randić index appears to predict the boiling points of alkanes more closely, and only it takes into account the bonding or adjacency degree among carbons in alkanes. More data and additional references on C(λ) can be found in [6,7]. In order to discuss the Randić index of the molecular graph, we first introduced some terminologies and notations of graphs. Let G = (V, E) be a graph. For a vertex x of G, we denote the neighborhood and the degree of x by N(x) and d(x), respectively. The minimum degree of G is denoted by δ(g). We will use G x or G xy to denote the graph that arises from G by deleting the vertex x V (G) or the edge xy E(G). Similarly, G + xy is a graph that arises from G by adding an edge xy / E(G), where x,y V (G). Let G be a graph and uv E(G). The Randić weight or simply weight of the edge uv is R(uv) = / d(u)d(v). Then, the Randić index of a graph G, R(G) = C(/; G), is the sum of the weights of its edges. In [], Bollobás and Erdös show that R(G) n ifg is a connected graph of order n. In Delorme et al. [3], prove a lower bound on R(G) for δ(g). Theorem [3]. Let G = (V, E) be a graph of order n with δ(g). Then R(G) (n ) + n n with equality if and only if G = K,n. (Where K,n is a graph that arises from a complete bipartite graph K,n by joining the vertices in the part with vertices by a new edge). Note that K,n contains triangles and (n ) > (n )+ n n for n 4, and hence in [3], Delorme et al. gave a best-possible lower bound on R(G) in terms of δ when G is triangle-free as follows.
H. Liu et al. / On the Randić index 347 (Theorem of [3]). Let G = (V, E) be a triangle-free graph of order n with δ(g) δ. Then R(G) δ(n δ) with equality if and only if G = K δ,nδ. Although we find a mistake in the proof of Theorem of [3], we still believe that Theorem of [3] is correct. In section 3, we will show (Theorem of [3]) which supports Theorem. Theorem. Let G = (V, E) be a triangle-free graph of order n with δ(g). Then R(G) (n ).. Examples Let G be a triangle-free graph of order n with δ(g) δ and v 0 a vertex of minimum degree of G. In the proof of Theorem, the authors claimed that R(G v o ) δ(n δ ). ( ) We note that ( ) is not always true for all triangle-free graphs. The following graphs are the counterexamples. In order to depict construction of the counterexamples, we first define one kind of graph operation as follows. For a given graph H with V(H) = {v,...,v s }, we define the graph G(H, m)(m ) as follows. Take m disjoint copies H,...,H m of H, with the vertex vj i in H i corresponding to the vertex v j in H( j s, i m). Let G(H, m) be the graph obtained from H I H m by joining vj k and vk j+ (k k, j s(vs+ k = vk ) for k, k =,,...,m. Example. Regular graph. Let H = (V, E) be a graph with V ={v,v,v 3,v 4 } and E(H) ={v v,v 3 v 4 }. Denote G = G(H, m). Obviously, G is triangle-free. In the graph G, wehavethatn = 4m, δ(g) = m. Let v 0 be any vertex of G. It is easy to see that m (m )(m ) R(G v 0 )= + = + (m )(m ) (m )(m ) (m )(m ) and δ(n δ ) = m(m ).
348 H. Liu et al. / On the Randić index Since (m)(m ) (m )(m ) ( m ) ( m ) m m = > 0, m + m we have R(G v 0 )< δ(n δ ). Example. Non-regular graph. Let P 4 = v v v 3 v 4, a path of order 4, and let G = G(P 4,m). Obviously, G is triangle-free. In the graph G, wehavethatn = 4m, δ(g) = m. Let v 0 be any vertex of G with d(v 0 ) = δ(g) = m. Then and R(G v 0 ) = m(m + ) m(m ) + m(m ) + + m(m ) δ(n δ ) = m(m ). (m + )(m ) (m )(m ) (m )(m ) (m )(m ) It is checked (in Appendix A) that R(G v 0 )< δ(n δ ). Remark. Note that there is NO vertex with minimum degree such that ( ) holds in these graphs. 3. Proof of Theorem In order to prove Theorem, we first need some lemmas. Lemma []. Let x x be an edge of maximal weight in a graph G. Then R(G x x ) < R(G). Lemma. Let d,d,d be positive integers and f(d,d ) = ( d ) d + ( d ) d ( + ) ( ) d d ( )( ). d d If 3 d,d, d, then f(d,d ) ( d d ).
H. Liu et al. / On the Randić index 349 Proof. From the proof of Lemma in [3], we have f(d,d ) f(d,d) = ( ) ( d d + )( ). d d d d Obviously, d d > 0. On the other hand, by d 3, we have d d 3 > 0, hence f(d,d ) ( d d ). The following result holds from the proof of Theorem of [3]. Lemma 3. Let G = (V, E) be a triangle-free graph of order n with δ(g) δ. Let V δ ={v : d(v) = δ}. If there exists a vertex v V δ such that N(v) V δ =, then Now, we will prove our theorem. R(G) δ(n δ). Proof. We assume that G is a counterexample of minimal order for which R(G) is minimal. Since G is a triangle-free graph and δ(g),n 4. If δ(g) >, then, by Lemma, we have a triangle-free graph G of minimum degree at least and with R(G )<R(G)by deleting the maximal weight edge, a contradiction with the choice of G. Hence δ(g) =. Denote T ={v V (G) : d(v) = }. If there exists a vertex v T such that N(v) T =, then R(G) (n ) by Lemma 3, a contradiction with the choice of G. Thus we can assume that N(v) T for any v T. Choose a vertex u T such that N(u) T is as small as possible. Let N(u) ={u,u } with u l T and d(u ) = d. Claim. (N(u ) N(u )) {u}. Proof. Suppose that (N(u ) N(u ))\{u} =. Then n 5 and G = Gu+u u is no counterexample, R(G ) (n 3). Thus R(G) = R(G ) + d + d (n 3) + > (n ), which is a contradiction.
350 H. Liu et al. / On the Randić index By Claim and u T, we can assume that (N(u ) N(u )) ={u, v}. Denote d = d(v). Since G is a triangle-free graph, d + d n. Let S,S be the sums of the weights of the edges incident with v and u except for the edges u v, u v and uu,vu, respectively. Then S i d i di for i =,. Claim. v T. Proof. Suppose v/ T. Then by the choice of u, u / T. Thus d,d 3,n 6 and G = G u u is no counterexample, R(G ) (n 4). InG,if we denote S,S the sums of the weights of the edges incident with v and u except for the edge u v, respectively. Then S i = S i di /(d i ) for i =,. Since d + d n and d i 3,d i n 3fori =,. Then R(G) = R(G ) + + + + + S + S d d d d (d )(d ) S d d S d d (n 4) + + + + d d d d ( ) ( ) (d )(d ) d d d d d d d d = (n 4) + + ( d ) d + ( d ) d ( + )( ) ( )( ). d d d d Using the same notation as in Lemma, we have R(G) (n 4) + + f(d,d ). Since 3 d,d n 3, we have f(d,d ) ( n 3 n 4) by Lemma. Thus R(G) (n 4) + + ( n 3 n 4) = (n ) + + ( n 3 n ) Since + ( n 3 n ) + ( 6 3 6 ) >0 we have R(G) (n ), which is a contradiction.
H. Liu et al. / On the Randić index 35 Claim 3. u / T. Proof. Suppose u T.Ifn = 4, then R(G) = = (4 ), a contradiction with the choice of G. Therefore n 4. By G being a triangle-free graph and δ(g) =, we have n 8. So G = G u u u v is no counterexample. Thus which is a contradiction. R(G) = R(G ) + (n 6) + (n ), By Claims and 3, we have v T and d 3, Thus d n byg being triangle-free. Now, we will complete our proof by considering the following two cases. Case. d(u ) = d 4. In the case, n 6 and G = G u u v is no counterexample, and then R(G ) (n 5). Thus ( ) R(G) = R(G ) + + + d d + + S d d d d (n 5) + + n n 4 (n 5) + + = + n 5 n n 4 (n 5) +. For n 6, + n 5 n n 4 + 6 5 6 6 4 = 0, and hence R(G) (n ), which is a contradiction. Case. d(u ) = d = 3. Let N(u ) \{u, v} ={x}, d(x) = d and y N(x) \{u }.Ifd =, then (N(y) N(u ))\{x} = by d(u) = d(v) = and d = 3. Thus we can derive a contradiction by the same argument as the proof of Claim. Hence we can assume that d 3 and then n 8. Let S be the sum of the weights of the edges incident with x different from u x. Then S d d. Since G = G u u u v is no counterexample, R(G ) (n 6), wehave
35 H. Liu et al. / On the Randić index ( R(G) = R(G ) + + + + S 6 3d (n 6) + + 6 + d d d d ( ) d 3 (n 6) + + 6 d ( 3 ) (n ) + (n 6) + (n ) + + 6 3 ( 3 ). Since (n 6) ( ) (n ) + + 6 3 3 3 3 + 6 + 3 > 0, we have R(G) > (n ), which is a contradiction. Hence the proof of our Theorem is completed. ) Appendix A Let G be the graph of Example and v 0 be any vertex of G with d(v 0 ) = δ(g). In the following, we will show that δ(n δ ) R(G v0 )>0. We have δ(n δ ) R(G v0 ) 3m(m ) (m + )(m ) = m(m ) m ( = (m ) 3m m + m(m ) m Since m > 0, it just needs to check that 3m m + m(m ) m Noting that 3m m(m ) m(m ) m(m ) m m(m ) m m(m ) m > m + (m )(m ) m + (m )(m ) (m )(m ) ) m. (m )(m ) m (m )(m ) > 0, m m(m ). 3m m > 3m m(m ) (m )(m ) m m m = m m > 0
H. Liu et al. / On the Randić index 353 and m + m + m > 0, m(m ) we will check that ( 3m m(m ) ) m ( m + > (m )(m ) m + m m(m ) ), 9m (m ) + (m ) (m )(m ) 6m(m ) (m ) m(m ) (m + ) > (m ) + m 4(m ) + m(m + ) (m ) m(m ), 5m 4 (m + ) (m ) (m ) m 4(m ) > m(4m 5) (m ) m(m ), 3m 3 7m + 45m 4 > 4(m )(4m 5) m(m ). Since m, 3m 3 7m + 45m 4 = (m )(m(3m 40) + 5) + > 0 and (m )(4m 5) m(m ) >0. Hence we just need to check that Note that (3m 3 7m + 45m 4) > ( 4(m )(4m 5) m(m )), 04m 6 4608m 5 + 8064m 4 6736m 3 + 60m 360m + 6 > 6(m m)(64m 4 4m 3 + 76m 40m + 5), 64m 4 80m 3 39m + 40m + 6 > 0. 64m 4 80m 3 39m + 40m + 6 = m (8m 3)(8m + 3) + 40m + 6 > 0 by m, and hence δ(n δ ) R(G v 0 )>0 holds.
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