Some Bounds for the Kirchhoff Index of Graphs

and Applied Analysis 3 components, one is C, and denote the other one by G 1 . Then by Lemma 5, Kf (G 0 ) = Kf (G 1 ) + Kf (C) + |V (C)| Kfx (G1) + 󵄨󵄨󵄨󵄨V (G1) 󵄨󵄨󵄨󵄨 Kfy (C) + 󵄨󵄨󵄨󵄨V (G1) 󵄨󵄨󵄨󵄨 |V (C)| . (10) Since Kf(G 0 ) has the maximum Kirchhoff index and by Lemma 6 and the fact that the Kirchhoff index is equal to the Wiener index for trees, we know thatKf(G 0 ) is maximized if and only if C is a path and y is an end vertex of C; the claim holds. Claim 4 (every vertex of H has degree at most ω in G 0 . Suppose to the contrary that there exists x ∈ H such that the degree of x in G 0 is larger than ω. Then by Claim 1, x is connected to at least two components of G 0 − H. Let C 1 andC 2 be two such components connecting to x by edges xy 1 and xy 2 , respectively. By Claim 3, both C 1 and C 2 are paths with y 1 and y 2 being end vertices of them. Without loss of generality, suppose that the length of C 1 is less than or equal to the length of C 2 . Let z 1 (resp., z 2 ) be the end vertex of C 1 (resp.,C 2 ) different fromy 1 (resp.,y 2 ), and let u be the unique neighbor of z 1 . Now construct a new graph G 1 from G 0 by first deleting the edge z 1 u and then adding a new edge z 1 z 2 . ClearlyG 1 ∈ G ω . We showKf(G 0 ) < Kf(G 1 ) so that Claim 4 is proved by contradiction. It is easily seen that, for any two vertices u, V distinct from z 1 , r uV (G0) = ruV (G1) . (11) Thus to show Kf(G 0 ) < Kf(G 1 ), we need only to show that Kf z 1 (G 0 ) < Kf z 1 (G 1 ). On one hand, it is easily verified that


Introduction
Let  be a connected graph with vertices labeled as 1, 2, . . ., .It is natural to view  as an electrical network by imagining each edge of  to be a (unit) resistor.In this guise, it is reasonable to consider the effective resistance between any two vertices of , and the novel concept of resistance distance [1]   () between any two vertices  and  of  is thus defined as the effective resistance between them.Compared to the (shortest-path) distance   () between  and  in , it is well known that   () ≤   () with equality if and only if  and  are connected by unique path [1].
There are many distance-based molecular structure descriptors, as reviewed in [2], which have played important roles in QSAR and QSPR.Among these structure descriptors, the most famous one is the Wiener index () [3], which is known as the sum of distances between all pairs of vertices.Analogous to the Wiener index, the Kirchhoff index of  [1,4], denoted by (), is defined as the sum of resistance distances between all pairs of vertices; that is, As summarized in [2], much work has been done by many researchers to investigate bounds for the Kirchhoff index.There are not only general bounds that are given in terms of various graph structural parameters like the number of vertices, the number of edges, the matching number, the chromatic number, the maximum degree, and the number of spanning trees [5][6][7][8][9][10][11][12][13][14][15][16][17][18][19][20], but also bounds for some special interesting classes of graphs, such as circulant graphs, unicyclic graphs, and bicyclic graphs [21][22][23][24][25][26][27].Along this line, we consider the relation between the Kirchhoff index and the independence number as well as the clique number, and bounds are obtained for the Kirchhoff index of graphs via the two graph invariants.In addition, lower and upper bounds for the Kirchhoff index of planar graphs and fullerene graphs are investigated.For more information on the Kirchhoff index of graphs, the readers are referred to the most recent papers [28][29][30][31][32][33][34][35][36] and references therein.

A Lower Bound via the Independence Number.
We first introduce some notations.Denote the vertex set and edge set of , respectively, by () and ().A subset  of () is called independent if its vertices are mutually nonadjacent.The independence number () is the largest cardinality among all independent sets of .The clique number () of  is the largest set of mutually adjacent vertices in .The degree of vertex , denoted by   (), is the number of neighbors of .The adjacency matrix () of  is an  ×  matrix with the (, )-entry equal to 1 if vertices  and  are adjacent and 0 otherwise.Let () = diag( 1 (),  2 (), . . .,   ()) be the diagonal matrix of vertex degrees.Then the Laplacian Abstract and Applied Analysis be the eigenvalues of (), called the Laplacian eigenvalues of .Since  is connected,  0 = 0 and   > 0 for  = 1, 2, . . .,  − 1 [37].The spectrum of (), also known as the Laplacian spectrum of , is  () = ( 0 ,  1 , . . .,  −1 ) . ( Zhu et al. [38] and Gutman and Mohar [39] established a classical result for computing the Kirchhoff index via the Laplacian spectrum of .

𝐾𝑓 (𝐺) = 𝑛
The join of  1 and  2 , denoted by  1 +  2 , is the graph obtained from  1 ∪  2 by adding all edges between vertices of  1 and that of  2 , where ∪ denote the disjoint union.The Laplacian eigenvalues of  1 +  2 are characterized in the following result.
The nonincreasing property of the Kirchhoff index, as stated blow, plays an important role in estimating bounds for the Kirchhoff index.
For simplicity, if there is no confusion, we always abbreviate   (),   (),  V (), (), and () to   ,   ,  V , , and , respectively.Throughout the paper, we use   ,   to denote the complete and path graph of order , respectively.We use  to denote the complement of .Then the main result of this subsection is given as follows.Lemma 5 (see [40]).Let  be a graph with a cut edge  =  and  1 and  2 the components of  −  containing  and , respectively.Then where   () denote the sum of resistance distances between  and all the other vertices of .
Lemma 6 (see [41]).Let  be an  vertex tree different from   and   .Then Theorem 7. Let  ∈ G  be different from   .Then Proof.If  = 2, then   is a path.Noticing that the Kirchhoff index is equal to the Wiener index for trees, by Lemmas 3 and 6, the assertion holds.In the following, we assume that  ≥ 3.
Let  0 be the graph with the maximum Kirchhoff index and let  be the subgraph of  0 such that  ≅   .
Claim 1 (every component of  0 −  is connected to  by only one edge).Suppose to the contrary that there exists a component of  0 −  such that it is connected to  with edges  1 ,  2 , . . .,   ( > 1).Let  1 be the graph obtained from  0 by the deletion of  2 ,  3 , . . .,   .Then it is easily seen that  1 ∈ G  and  1 is a proper spanning subgraph of  0 .Thus by Lemma 3, we have ( 1 ) > ( 0 ), contradicting the property that  0 has the maximum Kirchhoff index.
Claim 2 (every component of  0 − is a tree).Suppose to the contrary that there exists a component  of  0 −  such that  is not a tree.Let  be a spanning tree of .Remove from  0 all the edges in  but not in  to obtain  1 .Then  1 ∈ G  and  1 is a proper subgraph of  0 .Again by Lemma 3, we have ( 1 ) > ( 0 ), a contradiction.
By Lemma 5, simple calculation leads to Consequently, the upper bound on the Kirchhoff index is given in terms of the clique number.

Planar Graphs
A planar graph is a graph which can be drawn in the plane without edges crossing.In this section, we investigate bounds for Kirchhoff index of planar graphs.The following lemma is used.
The equality holds if and only if , which implies by Lemma 9 that  is a complete graph.Then the proof is completed by noticing that only  3 and  4 are planar complete graphs for  ≥ 3.
Though the lower bound is not sharp for  ≥ 5, it can be shown that, up to a scale factor, the bound is asymptotically attainable.One example is the star graph   , which has Kirchhoff index ( − 1) 2 .This indicates that the lower bound can be asymptotically attained up to a scale factor of at least 1/6.In fact, the scale factor could be improved to at least √ 3/6.For example, consider the planar graph  =  2 +  −2 .Since the Laplacian eigenvalues of   are [42] by Lemma 2, we readily obtain that the Laplacian eigenvalues of  are 0, , , 2 + 4sin 2  2 ( − 2) ,  = 1, 2, . . .,  − 3.
Therefore, by Theorem 1, we have Now we consider the asymptotic behavior of () as  → ∞: Hence (  ) grows as ( √ 3/6) 2 as  → ∞, and the scale factor is improved to at least √ 3/6.

Fullerene Graphs
A fullerene graph  is a cubic 3-connected planar graph with exactly 12 pentagons and other hexagons.Fullerene graphs are well studied in both mathematical and chemical literatures.To give bounds for fullerene graphs, we need some preparations.
The famous Foster first formula, given by Foster [43], states that where  ∼  means  and  are adjacent.Foster's second formula [44], also given by Foster, perhaps less well known, states that where   is measured across the end vertices of two adjacent edges V and V and the sum is taken over all adjacent edges.Palacios [14] extended Foster's first and second formulae and obtained the so-called Foster third formula, which states that where the sum is taken over all pairs of vertices  and  such that V is a 3-walk.In particular, if  is  regular, then the above equation can be simply written as where   is the number of 3-walks from  to .Now we give lower bounds for resistance distance between any two nonadjacent vertices in  in terms of the distance between them.
Next we introduce a classical result in graph theory-Menger's Theorem.
Theorem 13 (see [48], Menger's Theorem).Let  be an undirected graph, and let  and V be nonadjacent vertices in .Then, the maximum number of pairwise internally disjoint (, V) paths in  equals the minimum number of vertices from () − {, V} whose deletion separates  and V. Now we are ready for our main result.Theorem 14.For an  vertex fullerene graph , one has Proof.We first prove the lower bound.By Foster's first formula, Since  is triangle free, it is obvious that the end vertices of any two adjacent edges in G are at distance 2. Hence by Foster's second formula, Since there are exactly 6 − 60 pairs of vertices at distance 3 and by Lemma 11  ( For the upper bound, we consider any two nonadjacent vertices  and .Since  is 3-connected, by Menger's Theorem,  and  are connected by at least three pairwise internally disjoint paths.Suppose that  1 ,  2 , and  3 are three pairwise internally disjoint paths connecting  and .We consider the graph  * induced by  1 ,  2 , and  3 .By Rayleigh's short-cut principle,   () ≤   ( * ).Suppose that the lengths of  1 ,  2 , and  3 are  1 ,  2 , and  3 , respectively.Then by the series and parallel connection rules of resistors, Since it is obvious that  1 +  2 +  3 ≤  + 1, it follows that Thus   () ≤   ( * ) ≤ ( + 1)/9.By Foster's third formula (30) and noticing that there exist 3-walks which are not 3-path in , we conclude that < ∑ (47)
An Upper Bound via the Clique Number.Let G  be the set of graphs with  vertices and clique number .Let   denote the graph obtained by identifying one end vertex of path  −+1 with any vertex of   .In the following, we have shown that, among all the graphs in G  ,   has the maximum Kirchhoff index.To this end, we need the following two lemmas.
Claim 3 (every component of  0 − is a path, which connects to  via an end vertex).Let  be any component of  0 −.By Claims 1 and 2, we know that  is a tree and  is connected to  by a single edge  =  ( ∈ ).Clearly  −  has two Abstract and Applied Analysis 3 components, one is , and denote the other one by  1 .Then by Lemma 5,  ( 0 ) =  ( 1 ) +  () + | ()|   ( 1 ) ) has the maximum Kirchhoff index and by Lemma 6 and the fact that the Kirchhoff index is equal to the Wiener index for trees, we know that ( 0 ) is maximized if and only if  is a path and  is an end vertex of ; the claim holds.Claim 4 (every vertex of  has degree at most  in  0 .Suppose to the contrary that there exists  ∈  such that the degree of  in  0 is larger than .Then by Claim 1,  is connected to at least two components of  0 − .Let  1 and  2 be two such components connecting to  by edges  1 and  2 , respectively.By Claim 3, both  1 and  2 are paths with  1 and  2 being end vertices of them.Without loss of generality, suppose that the length of  1 is less than or equal to the length of  2 .Let  1 (resp.,  2 ) be the end vertex of  1 (resp.,  2 ) different from  1 (resp.,  2 ), and let  be the unique neighbor of  1 .Now construct a new graph  1 from  0 by first deleting the edge  1  and then adding a new edge  1  2 .
Clearly  1 ∈ G  .We show ( 0 ) < ( 1 ) so that Claim 4 is proved by contradiction.It is easily seen that, for any two vertices , V distinct from  1 , Claim 5 ( −  has only one component).Suppose to the contrary that  −  has at least two components  1 and  2 .Then by Claim 4,  1 and  2 must be connected with  via different vertices of , say  and .Let  1 =  and  2 = V be the two edges connecting  with  1 and  2 , respectively.Then both  1 and  2 are paths with  and V being their end vertices.Suppose that the lengths of  1 and  2 are  1 and  For any two vertices ,  distinct from  1 , it is easily seen that   ( 0 ) =   ( 1 ) .
2, respectively.Without loss of generality, we may assume that  1 ≤  2 .Let  1 (resp.,  2 ) be the other end vertex of  1 (resp.,  2 ) different from  (resp., V), and let  be the unique neighbor of  1 .Construct a new graph  1 from  0 by first deleting the edge  1 and then adding a new edge between  1 and  2 .Now we show ( 0 ) < ( 1 ), which thus gives the desired contradiction.
and only if  ≅   .