On the Inverse Sum Indeg Index

Transcription

1 On the Inverse Sum Indeg Index Jelena Sedlar a, Dragan Stevanović b,c,, Alexander Vasilyev c a University of Split, Faculty of Civil Engineering, Matice Hrvatske 15, HR Split, Croatia b Mathematical Institute, Serbian Academy of Science and Arts, Knez Mihajlova 36, RS Belgrade, Serbia c University of Primorska, Institute Andrej Marušič, Muzejski trg 2, SI-6000 Koper, Slovenia Abstract Discrete Adriatic indices have been defined by Damir Vukičević and Marija Gašperov in Croat. Chem. Acta 83 (2010), , as a way of generalizing well-known molecular descriptors defined as the sums of individual bond contributions, such as the second Zagreb index or the Randić index. Among the 18 discrete Adriatic indices analyzed by Vukičević and Gašperov on the benchmark datasets of the International Academy of Mathematical Chemistry, 20 indices were selected as significant predictors of physicochemical properties. Here we study graph-theoretical properties of the Inverse Sum Indeg index, which was selected as a significant predictor of total surface area of octane isomers, and determine extremal values of this index across several graph classes, including connected graphs, chemical graphs, trees and chemical trees. Keywords: Molecular descriptors; Bond-additive descriptors; Discrete Adriatic indices. 1. Introduction Molecular descriptors, results of functions mapping molecule s chemical information into a number [1], have found applications in modeling many physicochemical properties in QSAR and QSPR studies [2, 3]. A particularly common Corresponding author addresses: jelena.sedlar@gradst.hr (Jelena Sedlar), dragan.stevanovic@upr.si (Dragan Stevanović), alexander.vasilyev@upr.si (Alexander Vasilyev) Preprint submitted to Discrete Applied Mathematics November 6, 201

2 type of molecular descriptors are those that are defined as functions of the structure of the underlying molecular graph, such as the Wiener index [], the Zagreb indices [5], the Randić index [6] or the Balaban J-index [7]. Damir Vukičević and Marija Gašperov [8] observed that many of these descriptors are defined simply as the sum of individual bond contributions (actually, all previously mentioned descriptors except for the Wiener index). In order to study whether there are other possibly significant descriptors of this form, they have introduced a class of discrete Adriatic indices in [8], generally defined as Adr(G) = uv E(G) f(g(p u ), g(p v )), where G is the molecular graph, E(G) is the set of its bonds, p u is either the degree d u of a vertex u V (G) or the sum D u of distances from u to all other vertices in V (G), while f and g are suitably chosen functions. Among the 18 discrete Adriatic indices studied in [8], whose predictive properties were evaluated against the benchmark datasets of the International Academy of Mathematical Chemistry [9], 20 indices were selected as significant predictors of physicochemical properties. Graph-theoretical properties of one of these indices, the max-min rodeg index, were studied previously in [10]. In order to study graph-theoretical properties of other discrete Adriatic indices, we have implemented all of them in MathChem, an open source Python package for calculating topological indices [11, 12], and obtained a number of conjectures about their extremal values in several graph classes. We study here the properties of the inverse sum indeg index, the descriptor that was selected in [8] as a significant predictor of total surface area of octane isomers and for which the extremal graphs obtained with the help of MathChem have a particularly simple and elegant structure. The inverse sum indeg index is defined as ISI(G) = uv E(G) 1 1 d u + 1 = d v uv E(G) d u d v d u + d v. (1) After establishing basic properties of the inverse sum indeg index in Section 2, in the following sections we determine extremal values and extremal graphs of 2

3 20 the inverse sum indeg index in several classes of graphs (with given number of vertices): the class of all connected graphs, the class of all trees, the class of all chemical graphs, the class of all chemical trees, the class of all graphs with given maximum degree, the class of all graphs with given minimum degree, the class of all graphs with given number of pendant vertices, and the class of all trees with given number of pendant vertices. We give some complete and some partial solutions for these problems. 2. Preliminaries 25 We assume in the sequel that the graph G does not contain isolated vertices, although we allow the possibility that G may have several connected components. The basic property of the inverse sum indeg index is that it is monotone with respect to addition of edges. Lemma 1. Let u and v be two nonadjacent vertices of graph G, and let G + uv be the graph obtained from G by adding edge uv to it. Then ISI(G) < ISI(G + uv). Proof. Note first that for x, y > 0 (x + 1)y (x + 1) + y xy x + y = y 2 (x + y)(x + y + 1) > Now, let u 1,..., u k be the neighbors of u in G for k = d u, and let v 1,..., v l be the neighbors of v for l = d v. Then ISI(G + uv) ISI(G) = (d u + 1)(d v + 1) d u + d v + 2 k [ (du + 1)d ui + d ] u d ui (d i=1 u + 1) + d ui d u + d ui l [ (dv + 1)d vj + d ] v d vj (d v + 1) + d vj d v + d vj j=1 > 0, 3

4 where the first summand above is the (positive) contribution of edge uv in G + uv. The following corollary is an immediate consequence of previous lemma. 35 Corollary 2. If a connected graph G itself is not a tree, then ISI(T ) < ISI(G) holds for any spanning tree T of G. The next result establishes useful bound on the inverse sum indeg index in terms of the numbers of vertices and edges of a graph. Theorem 3. If G is a graph with n vertices and m edges, then ISI(G) 2m n, with equality if and only if G is a union of cycles. Proof. Since the degree of any vertex u V (G) is positive, we have n = 1 = 1 d u = ( ) = d u + d v. d u d u d v d u d v Then u V (G) u V (G) ISI(G) + n = uv E(G) uv E(G) uv E(G) ( du d v + d ) u + d v. (2) d u + d v d u d v Since x + 1 x 2 for any positive x, with equality if and only if x = 1, we have from (2), by substituting x with dudv d u+d v ISI(G) + n for each edge uv E(G), that uv E(G) 2 = 2m. 0 Equality holds if and only if d u d v = d u + d v for each edge uv E(G) if and only if d u = 2 for each u V (G), i.e., if and only if G is a union of cycles. The next theorem gives a relation between the inverse sum indeg index and the first Zagreb index M 1 (G) = u V (G) d2 u. Theorem. For any graph G holds ISI(G) M 1(G), with equality if and only if G is a union of regular graphs.

5 Proof. Let uv E(G) be an arbitrary edge in G (so that d u, d v 0). From 0 (d u d v ) 2 we get, after adding d u d v to both sides, that d u d v (d u + d v ) 2, so that after division with (d u + d v ), it follows that Now we have ISI(G) = uv E(G) d u d v d u + d v d u d v d u + d v d u + d v. uv E(G) d u + d v = u V (G) d 2 u = M 1(G). 5 Equality holds if and only if d u = d v for each edge uv E(G), i.e., if and only if each component of G is a regular graph. 3. Minimum values of the inverse sum indeg index 50 Since the minimum value of the inverse sum indeg index is, by Corollary 2, necessarily obtained by some tree, we put focus first on the sets of trees. Let S n be the star on n vertices, a tree consisting of a central vertex adjacent to n 1 leaves. Theorem 5. If T is a tree with n vertices, then ISI(T ) n + 1 n 2, with equality if and only if T is isomorphic to S n. 55 Proof. Since ISI(S n ) = (n 1) (n 1) 1 (n 1)+1 = n + 1 n 2, this theorem states that ISI(S n ) ISI(T ) for any tree T, with equality if and only if T = S n. The only tree with one, two or three vertices is exactly the star, so that the statement follows trivially for n 3. 5

6 60 65 The proof for larger trees is by induction on n. The basis of induction for n 3 is proved above, so that we may make an inductive hypothesis that, for some n, ISI(S n 1 ) ISI(T ) holds for each (n 1)-vertex tree T with equality if and only if T = S n 1. Now, suppose that T = Sn is an arbitrary tree with n vertices. Let u be a pendant vertex of T, with v as its only neighbor. Since T has at least four vertices, v cannot be a leaf, so that d v 2. Further, since T is not isomophic to a star, we have that there exists at least one neighbor w of v in T with d w 2. Let z 1,..., z dv 2 be the remaining neighbors of v in T, other than u and w. Let T = T u. Since the edges of T, which are not incident to v, contribute equally to both ISI(T ) and ISI(T ), we have that where ISI(T ) ISI(T ) = d v 1 d v f(x) = d v 2 i=1 The first derivative of f(x) is equal to f (x) = ( dw d v d w(d v 1) d w + d v d w + d v 1 ) ( dzi d v d zi + d v d z i (d v 1) d zi + (d v 1) = d d v 1 d v f(d v 2 w) + f(d zi ), i=1 xd v x + d v x(d v 1) x + d v 1. d 2 v (x + d v ) 2 (d v 1) 2 (x + d v 1) 2 = x2 (2d v 1) + 2xd v (d v 1) (x + d v ) 2 (x + d v 1) 2. Since for x > 0 we have that x 2 (2d v 1) + 2xd v (d v 1) > 0, it follows that f (x) > 0 and the function f is strictly increasing for x > 0. Together with d w 2 and d zi 1, this implies that ISI(T ) ISI(T ) d v 1 d v f(2) + (d v 2)f(1) Since d v 2, we have that ISI(T ) ISI(T ) 1 > 1 = 1 + 2(d v 2) d v (d v + 1)(d v + 2). (3) 1 n(n 1) = ISI(S n) ISI(S n 1 ), ) 6

7 from where it follows that ISI(T ) ISI(S n ) > ISI(T ) ISI(S n 1 ) 0, 70 due to the inductive hypothesis. This concludes the proof by induction and also shows that ISI(T ) = ISI(S n ) holds if and only if T = S n. Let us recall that a tree (a graph) is called chemical if the degree of each of its vertices is at most four. Theorem 6. If T is a chemical tree with n vertices, then n + 1 n 2 if n 5, ISI(T ) n 5 3 if n 6, with equality if and only if T is isomorphic to the star S n if n 5 and to the path P n if n 6. Proof. For n 5, the star S n is a chemical tree and the result follows from Theorem 5. We will prove the statement for larger values of n by induction. The basis of induction for n 5 is proved above. Suppose that the statement of the theorem is proved for all chemical trees with less than n vertices for some n 6, and let T be a chemical tree with n vertices. Suppose first that T contains a vertex u of degree four. Then at least one neighbor of u has degree at least two. Denote the neighbors of u by u 1, u 2, u 3, u and suppose, without loss of generality, that d u 2. Form trees T and T from T by splitting the vertex u into two new, nonadjacent vertices u and u, such that u is adjacent to u 1 and u 2 in T, while u is adjacent to u 3 and u in T. The edges of T that are not incident to u contribute equally to ISI(T ) and ISI(T ) + ISI(T ), so that ISI(T ) = ISI(T ) + ISI(T ) + 7 i=1 ( dui d ui + 2 d ) u i. d ui + 2

8 The function p(x) = x x+ 2x x+2 is strictly increasing for x > 0, as p (x) = 12x 2 +32x (x+2) 2 (x+) 2 > 0, so that from d u1, d u2, d u3 1 and d u 2 it follows that ISI(T ) ISI(T ) + ISI(T ) + 3p(1) + p(2) = ISI(T ) + ISI(T ) () Let T have n vertices and T have n vertices. Then 3 n, n < n, so that the inductive hypothesis holds for both T and T. If n 6, then ISI(T ) n 5 3 by the inductive hypothesis. If n 5, then T cannot be isomorphic to either of the stars S and S 5, as it contains vertex u of degree two, so that T is either P 3, P, P 5 or a unique tree with degree sequence 3, 2, 1, 1, 1, and ISI(T ) n 5 3 holds for each of these cases as well. Inequality ISI(T ) n 5 3 follows analogously. Taking into account that n + n = n + 1, as the vertex u was split into two new vertices, () implies that ( ISI(T ) n 5 ) + 3 ( n 5 3 ) = n 8 5. Hence, if T contains a vertex of degree four, then ISI(T ) n 8 5 > n 5 3. Suppose next that T contains a vertex u of degree three. Then at least one neighbor of u has degree at least two. Denote the neighbors of u by u 1, u 2, u 3 and suppose, without loss of generality, that d u3 2. Form trees T and T from T by splitting the vertex u into two new, nonadjacent vertices u and u, such that u is adjacent to u 1 and u 2 in T, while u is adjacent to u 3 in T. The edges of T that are not incident to u contribute equally to ISI(T ) and ISI(T ) + ISI(T ), so that ISI(T ) = ISI(T ) + ISI(T ) + The functions q(x) = 3x x+3 2x x+2 2 ( 3dui d i=1 ui + 3 2d ) u i d ui + 2 ( 3du3 + d u3 + 3 d ) u 3. d u3 + 1 and r(x) = 3x x+3 x x+1 for x > 0, as q (x) = 5x2 +12x (x+3) 2 (x+2) 2 > 0 and r (x) = 8x2 +12x (x+3) 2 (x+1) 2 from d u1, d u2 1 and d u3 2 we have that (5) are strictly increasing > 0, so that ISI(T ) ISI(T ) + ISI(T ) + 2q(1) + r(2) = ISI(T ) + ISI(T ) (6) 8

9 Let T have n vertices and T have n vertices. Then 3 n, n < n, so that 85 the inductive hypothesis holds for both T and T. Inequality ISI(T ) n 5 3 follows as in the earlier case d u =, as T contains a vertex u of degree two. If n 6, then ISI(T ) n 5 3 by the inductive hypothesis, so that (6) implies that ISI(T ) ( n 5 ) ( + n 5 ) = n (7) For n 5, if T is one of P 3, P, P 5 and a unique tree with degree sequence 3, 2, 1, 1, 1, then ISI(T ) n 5 3 and inequality (7) holds as well. However, if T is one of S and S 5 (in which case n = d u3 + 1 and T = S du3 +1), then (5) implies that ISI(T ) ISI(T ) + ISI(T ) + 2q(1) + r(d u3 ) ( n 5 ) ( + n + 1 ) 3 n r(d u 3 ) = n d2 u 3 + d u3 + 3 (d u3 + 1)(d u3 + 3) The function s(x) = 2x2 +x+3 (x+1)(x+3) is strictly increasing for x > , as s (x) = 7x 2 +6x 9 (x+1) 2 (x+3) 2 > 0, so that from d u3 2 we have that ISI(T ) n s(2) = n Hence, if T contains a vertex of degree three, then ISI(T ) n 9 30 > n 5 3. Finally, if T does not contain vertices of degrees three or four, then T is isomorphic to the path P n and ISI(T ) = n 5 3, finishing the proof of the inductive step. Let B n,k be a tree obtained from the star S k+1 by identifying one of its leaves with a leaf of the path P n k. The tree B n,k has n vertices, among which there are k pendant vertices, and it is usually named the broom in the literature. Theorem 7. If T is a tree with n vertices and maximum vertex degree 2, then n + 1 n 2, if = n 1, ISI(T ) ISI(B n, ) = (8) n 3 2 ( +1)( +2), if < n 1, 9

10 with equality if and only if T is isomorphic to the broom B n, Proof. We prove the inequality (8) by induction on n, n + 1. If = 2, then the path P n is the only tree with maximum vertex degree two. By definition P n = B n,2 and the inequality (8) holds for = 2 and all n + 1. Suppose, therefore, that 3. The basis of the induction are the cases n = + 1 and n = + 2. A unique tree with + 1 vertices and maximum vertex degree is the star S +1 and since S +1 = B +1, by definition, the inequality (8) holds for n = + 1. Similarly, a unique tree with + 2 vertices and maximum vertex degree is the broom B +2,, so that the inequality (8) holds for n = + 2 as well. Let us, therefore, make the inductive hypothesis that the inequality (8) holds for all trees with n 1 vertices with maximum vertex degree, for some n > + 2 (with equality for and only for the broom B n 1, ). Let T be a tree with n vertices and maximum vertex degree. Since n > + 2, T cannot be a star and, therefore, there exists a pendant vertex u in T such that T = T u also has maximum vertex degree. Then the inductive hypothesis holds for T and since d v 2 for the unique neighbor v of u, we have from (3) ISI(T ) ISI(T ) (d v 2) d v (d v + 1)(d v + 2) (n 1) (9) ( + 1)( + 2) Equality in this chain of inequalities is attained only if T satisfies equality in (8) and d v = 2. The inductive hypothesis implies that T is then isomorphic to the broom B n 1,, while d v = 2 (which holds for v as a vertex of T ) implies that v is one of the leaves of B n 1,. Closer inspection of the argument used to derive (3) further reveals that equality may hold in (9) only if the other neighbor of v in T has degree two, i.e., if T itself is the broom B n,. With small modifications of the proof of Theorem 7, we get the following 10

11 Theorem 8. If T is a tree with n vertices, among which there are p 2 pendant vertices, then n + 1 n 2, if p = n 1, ISI(T ) ISI(B n,p ) = (10) n 3 2p (p+1)(p+2), if p < n Proof. We prove inequality (10) by double induction, first on the value of p 2 (the outer induction), and then on the value of n p + 1 (the inner induction). The basis of the outer induction is p = 2. As the path P n is the only tree with exactly two leaves and P n = B n,2 by definition, inequality (10) holds for p = 2 and all n p + 1. Let us, therefore, make the outer inductive hypothesis that inequality (10) holds for all trees with p 1 pendant vertices, for some p 3. We will prove that inequality (10) then holds for all trees with p pendant vertices by the inner induction on n p + 1. The basis of the inner induction are the cases n = p + 1 and n = p + 2. Unique tree with p + 1 vertices, of which p are pendant vertices, is the star S p+1 and since S p+1 = B p+1,p by definition, inequality (10) holds for n = p + 1. Similarly, unique tree with p + 2 vertices, of which p are pendant vertices, is the broom B p+2,p, so that inequality (10) holds for n = p + 2 as well. Let us, therefore, make the inner inductive hypothesis that inequality (10) holds for all trees with n 1 vertices, of which p are pendant vertices, for some n > p + 2. Let T be a tree with n vertices, of which p are pendant vertices. Let u be an arbitrary pendant vertex of T, with v as its only neighbor. Here 2 d v p: firstly, 2 d v because T has more than two vertices and, secondly, d v p because each of d v subtrees of T obtained by removing v contains at least one pendant vertex of T. Now, let T = T u. Since n > p + 2, T cannot be a star, so that from (3): ISI(T ) ISI(T ) (d v 2) d v (d v + 1)(d v + 2). Tree T has n 1 vertices, of which either p or p 1 are pendant vertices, 11

12 10 depending on whether d v = 2 or d v > 2, respectively. If d v = 2, then T has p pendant vertices and, by the inductive hypothesis of the inner induction, ISI(T ) (n 1) 3 ISI(T ) n 3 2p (p+1)(p+2), so that 2p (p + 1)(p + 2). If 3 d v p, then T has p 1 pendant vertices and, by the inductive hypothesis of the outer induction, ISI(T ) (n 1) 3 2(p 1) p(p+1), so that ISI(T ) n 3 2(p 1) p(p + 1) + 2(d v 2) d v (d v + 1)(d v + 2). (11) Let us consider the function f(x) = 2(x 2) x(x+1)(x+2). It has the first derivative f (x) = 2(2x3 3x 2 12x ) x 2 (x+1) 2 (x+2). The roots of 2x 3 3x 2 12x are, approximately, , and , so that f (x) < 0 and f(x) is strictly decreasing for x > Since also f(3) = 1 30 = f(), from 3 d v p we conclude that f(d v ) f(p) (with equality either if d v = p or d v = 3, p = ). Therefore, from (11) and f(d v ) f(p) we have ISI(T ) n 3 2(p 1) p(p + 1) + 2(p 2) p(p + 1)(p + 2) = n 3 2p (p + 1)(p + 2) A careful reader will notice that the previous theorem does not fully characterize the case of equality in ISI(T ) ISI(B n,p ). The reason for this is that there exist trees that are not brooms and still have the same value of the inverse sum indeg index as the corresponding broom. They are obtained in the previous proof whenever ISI(T ) = ISI(B n 1,p 1 ) and d v = p 3 or d v = 3, p =. One such example is shown in Fig. 1, while the remaining such graphs could be characterized with little extra effort. Previous results about trees may be directly extended to graphs, due to Corollary 2. Specifically, Theorems 5-8 give rise to the following Corollary 9. If G is a connected graph with n vertices, then ISI(G) n + 1 n 2, with equality if and only if G is isomorphic to S n. 12

13 Figure 1: Two trees with the same value of the inverse sum indeg index, exactly one of which is a broom. Corollary 10. If G is a connected chemical graph with n vertices, then n + 1 n 2 if n 5, ISI(G) n 5 3 if n 6, with equality if and only if G is isomorphic to the star S n if n 5 and to the path P n if n 6. Corollary 11. If G is a connected graph with n vertices and maximum vertex degree, then ISI(G) n 3 2 ( + 1)( + 2). Corollary 12. If G is a connected graph with n vertices, among which there are p 2 pendant vertices, then ISI(G) n 3 2p (p + 1)(p + 2). 155 The following theorem describes minimum value of the inverse sum indeg index for graphs with given minimum vertex degree. Theorem 13. Let G be a graph with n vertices and minimum vertex degree δ. Then n + 1 n 2, if δ = 1, ISI(G) nδ 2, if δ 2 and nδ is even, nδ 2 + δ 2 + 2δ 8(2δ+1), if δ 2 and nδ is odd. Proof. If δ = 1, the statement follows from Corollary 9. If δ 2, then the contribution of each edge uv E(G) to ISI(G) is at least δ 2 : from d u, d v δ it follows that (d u δ 2 )(d v δ 2 ) δ2, which is, after 13

14 rearranging the terms and division by d u + d v, equivalent with Therefore d u d v d u + d v δ 2. ISI(G) mδ 2 nδ2, as 2m nδ holds in a graph in which every vertex has degree at least δ. Note, however, that equality 2m = nδ may hold only if nδ is even. If nδ is odd, then at least one vertex of G has degree at least δ + 1, so that there are at least δ + 1 edges whose contribution is at least δ(δ+1) 2δ+1, and the total number of edges is at least nδ+1 2. Hence, in such case ( ) δ(δ + 1) nδ + 1 δ ISI(G) (δ + 1) 2δ (δ + 1) 2 2 = nδ2 + δ 2 + 2δ 8(2δ + 1).. Maximum values of the inverse sum indeg index Our first result on the maximum value of the inverse sum indeg index is a direct corollary of Lemma 1 stating that the value of the inverse sum indeg index increases with addition of new edges. Namely, as each n-vertex graph G can be made into a complete graph by adding edges between all pairs of nonadjacent vertices, we obtain that ISI(G) ISI(K n ) = n(n 1)2 160 holds for each graph with n vertices. Another useful auxiliary result is the following Lemma 1. Let G be a graph with m edges and the maximum degree. Then ISI(G) m 2, with equality if and only if G is -regular graph. 1

15 Proof. Since d u for each u V (G), we have that 1 1 d u + 1 d v Then, directly from (1), we have that ISI(G) m 2. = 2. The equality holds if and only if d u = for each u V (G). Previous lemma directly implies the following Corollary 15. If G is a graph with n vertices and maximum vertex degree, then ISI(G) n 2, if n is even, (n 1) 2 + ( 1)2 2(2 1), if n is odd. Proof. Since 2m is equal to the sum of the vertex degrees in a graph, the fact that the maximum vertex degree is implies both that m n 2 ISI contribution of each edge is at most 2. Then and that the ISI(G) m 2 Equality above is attained only if m = n 2 possible only if n is even. n 2. and G is -regular graph, which is If n is odd, then m n 1 2 and at least one vertex of G has degree at most 1, while the remaining vertices have degrees at most. Hence, at most (n 1) 2 edges have the ISI contribution equal to 2, while the remaining m (n 1) 2 edges have the ISI contribution at most ( 1) 2 1. In total, ISI(G) (n 1) 2 (n 1) 2 ( + m ) (n 1) ( 1) ( 1) When applied to chemical graphs, which have maximum vertex degree at most four, previous corollary further implies 15

16 Corollary 16. If G is a chemical graph with n vertices, then n(n 1) 2, if n, ISI(G) n if n Proof. If n 5, the upper bound follows directly from Corollary 15 and the fact that. If n, then no vertex of G can have degree equal to four, so that the maximum value of the inverse sum indeg index is obtained by the complete graph K n, for which ISI(K n ) = n(n 1)2. Let KD n,δ be the graph obtained from the complete graph K n 1 by adding to it a new vertex, adjacent to exactly δ vertices of K n 1. Another corollary of Lemma 1 is the following Corollary 17. If G is a graph with n vertices and minimum vertex degree δ, then ISI(G) ISI(KD n,δ ) with equality if and only if G is isomorphic to KD n,δ Proof. Let u be the vertex of G having degree δ. As the part of G, induced by vertices different from u, can be made into the complete graph K n 1 by adding edges between all pairs of nonadjacent vertices, we immediately obtain that ISI(G) ISI(KD n,δ ), with equality if and only if no edges were added to G, i.e., if and only if G = KD n,δ. In order to characterize the graph with the maximum inverse sum indeg index among graphs with given number of pendant vertices, we need to define an auxiliary type of graphs. For k 1 and the sequence of nonnegative integers q 1,..., q k, the graph H q1,...,q k is obtained from the complete graph K k on the vertex set {1,..., k} by attaching q i new pendant vertices to vertex i for each i = 1,..., k. Further, for given k 1 and p 0 let KP k,p = H q1,...,q k where q 1,..., q k are chosen such that k i=1 q i = p and q i { p k, p k } for each i = 1,..., k. 16

17 Theorem 18. If G is a graph with n vertices, among which there are p 0 pendant vertices, then ISI(G) ISI(KP n p,p ), with equality if and only if G is isomorphic to KP n p,p. Proof. From Lemma 1 it is apparent that the inverse sum indeg index gets increased if one adds edges between all pairs of nonadjacent vertices in the part of G induced by nonpendant vertices. Therefore, the graph with the maximum value of the inverse sum indeg index is necessarily of type H q1,...,q k for k = n p and some nonnegative integers q 1,..., q k such that k i=1 q i = p. Let H = H p1,...,p k graphs H q1,...,q k have the maximum inverse sum indeg index among all such that k i=1 q i = p. Note that the degree of the vertex i, 1 i k, belonging to the complete part K k of H is equal to d i = k 1 + p i. Suppose that p i p j 2 for some j i and, without loss of generality, suppose that p i +2 p j. Let H = H p1,...,p i+1,...,p j 1,...,p k be the graph obtained by reattaching one of the pendant vertices from j to i. Since the edges of H, which are not adjacent to either i or j, contribute equally to both ISI(H) and ISI(H ), we have that ( ISI(H di + 1 ) ISI(H) = d i + 2 d ) ( j (di + 1)(d j 1) + d ) id j d j + 1 d i + d j d i + d j ( dj 1 +(p j 1) d ) ( j di p i d j d j + 1 d i + 2 d ) i d i ( (dj 1)d k d ) jd k + ( (di + 1)d k d ) id k. d j 1 + d k d j + d k d i d k d i + d k k i,j k i,j This difference may be written more compactly using the function as f(x, y) = ISI(H ) ISI(H) = xy (x 1)y x + y x 1 + y d i + 1 d j (d i + 2)(d j + 1) + d j d i 1 d i + d j (p j 1)f(d j, 1) + p i f(d i + 1, 1) f(d j, d k ) + f(d i + 1, d k ). k i,j k i,j 17

18 For constant y 1, the function f(x, y) is strictly decreasing for x > 0 since f x(x, y) = y 2 (y 1)2 (x + y) 2 (x + y 1) 2 = x2 (2y 1) + 2xy(y 1) (x + y) 2 (x + y 1) 2 > 0. Therefore, from the initial assumption that p i + 2 p j (which translates to d i + 2 d j as d j d i = p j p i ) we have that f(d i + 1, d k ) > f(d j, d k ) for each k i, j and f(d i + 1, 1) > f(d j, 1). Hence ISI(H ) ISI(H) > d i + 1 d j (d i + 2)(d j + 1) + d j d i 1 (p j p i 1)f(d j, 1) d i + d j ( ) 1 1 = (d j d i 1) d i + d j (d i + 2)(d j + 1) 1 d j (d j + 1) = (d j d i 1) (d2 j d2 i ) + d i(d 2 j d j d i 2). (d i + d j )(d i + 2)d j (d j + 1) Since d j d i + 2, we have that d j (d j 1) d i 2 (d i + 2)d i, so that the difference ISI(H ) ISI(H) is strictly positive, which is a contradiction to the maximality of H. This contradiction shows that for each i j it has to hold p i p j 1, which is, due to the condition k i=1 p i = p, equivalent to p i { p k, p k } for each i = 1,..., k. Hence, the graph H with the maximum inverse indeg index is isomorphic to KP k,p = KP n p,p. 215 At the end, we have to leave tasks of determining maximum inverse sum indeg index in the classes of trees and chemical trees as open problems. Experiments with MathChem with trees and chemical trees with up to 20 vertices suggest the emergence of a common structure see extremal trees in Fig. 2 and extremal chemical trees in Fig. 3. However, in order to formulate sound conjectures on their structure, one would need to be able to process sets of (chemical) trees with much larger numbers of vertices, which is not easily attainable with computers at our disposal. 18

19 18 vertices 19 vertices 20 vertices Figure 2: Trees on 18, 19 and 20 vertices having maximum inverse sum indeg index. 18 vertices 19 vertices 20 vertices Figure 3: Chemical trees on 18, 19 and 20 vertices having maximum inverse sum indeg index. Acknowledgements J.S. was supported by EUROCORES Programme EUROGIGA (project GReGAS) of the European Science Foundation and by Croatian Ministry of Science, Education and Sports. D.S. was supported by Research Programme P and Research Project J1-021 of the Slovenian Research Agency and Research Grant No. ON17033 of the Ministry of Education and Science of Serbia. A.V. was supported by the Young Researchers programme of the Slovenian Research Agency. The authors would like to thank Damir Vukičević for valuable discussions. [1] R. Todeschini, V. Consonni, Handbook of Molecular Descriptors, Wiley- VCH, Weinheim,

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