On the Balaban Index of Trees

In this paper, we give a new proof that among all trees with n vertices, the star Sn and the path Pn have the maximal and the minimal Balaban index, respectively. This corrects some errors of proofs in [H. Dong, X. Guo, Character of graphs with extremal Balaban index, MATCH Commun. Math. Comput. Chem. 63 (2010) 799–812] and [L. Sun, Bounds on the Balaban index of trees, MATCH Commun. Math. Comput. Chem. 63 (2010) 813–818]. We also characterize the trees with the second minimal and maximal Balaban index, respectively. For a simple and connected graph G with vertex–set V (G) and edge–set E(G), dG(u, v) denotes the distance between vertices u and v in G, and DG(u) = ∑ v∈V (G) dG(u, v) is the distance sum of vertex u in G, i. e., the row sum of distance matrix of G corresponding to u. The Balaban index of G is defined as J(G) = m μ+ 1 ∑ uv∈E(G) 1 √ DG(u)DG(v) where m is the number of edges and μ is the cyclomatic number of G, respectively. The Balaban index was proposed by A. T. Balaban [1,2], which also called the average distance–sum connectivity or J index. It appears to be a very useful molecular descriptor with attractive properties. Recently, it was showed in [3] that Theorem 12 ([3]). If T is a tree with n ≥ 2 vertices, then J(Pn) ≤ J(T ) ≤ J(Sn) with left (or right) equality if and only if T = Pn (or T = Sn), where Pn and Sn are the path and the star on n vertices. The proof of theorem above is dependent on Lemma 6 in [3]. But, the following example 1 shows that Lemma 6 is incorrect. †This work was supported by Scientific Research Fund of Hunan Provincial Education Department (09A057), Hunan Provincial Natural Science Foundation of China (09JJ6009) and the Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province. MATCH Communications in Mathematical and in Computer Chemistry MATCH Commun. Math. Comput. Chem. 66 (2011) 253-260 ISSN 0340 6253 Example 1. (see Figure 1) Let P = p1p2p3 be a path in a graph G0, where G0 = P is the path itself, and Gj(1 ≤ j ≤ 3) are the component of G0 − E(P ) containing pj. H1 = u1u2u3, H2 = v1v2v3, H ′ 1 = u1u2 and H ′ 2 = v1v2v3u3 are also paths. We have |V (H1)| + |V (H2)| = |V (H ′ 1)| + |V (H ′ 2)| and |V (H ′ 1)| < |V (H1)| ≤ |V (H2)| < |V (H ′ 2)|. Construct the graph G by identifying p1 to u2 in H1 and p3 to v1 in H2, and the graph G ′ by identifying p3 to u2 in H ′ 1 and p3 to v1 in H ′ 2. Clearly, |V (Gi)| ≤ |V (Gl+1−i)| since Gi is an isolated vertex, l = 3 and 1 ≤ i ≤ l 2 − 1. However, we have JG(P ) > JG′(P ), where JG(P ) = ∑ uv∈E(P ) 1 √ DG(u)DG(v) = 1 √ 12×11 + 1 √ 11×12 and JG′(P ) = ∑ uv∈E(P ) 1 √ DG′ (u)DG′ (v) = 1 √ 16×13 + 1 √ 13×12 . Also, it was showed in [4] that Theorem 2.2 ([4]). If T is a tree with n ≥ 2 vertices, then J(T ) ≥ J(Pn) with equality if and only if T = Pn. But the following example 2 shows that its proof in [4, lines 1-3 on page 818] is incorrect. Example 2. (see Figure 2) Let P = v1v2v3v4 be a longest path in T , and v1 the root of T . Then the other vertices of T can be divided into levels by the distance on v1. Let Li be the set of vertices in level i (1 ≤ i ≤ 3), where L1 = {v2}, L2 = {v3, x} and L3 = {v4}. However, for x ∈ L2, Dx = DT (x) = 8, and D3 = 6 since Di = (n−i+1)(n−i) 2 + (i−1)i 2 and -254-