On the difference between the (revised) Szeged index and the Wiener index of cacti

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1 On the difference between the revised) Szeged index and the Wiener index of cacti Sandi Klavžar a,b,c Shuchao Li d, Huihui Zhang d a Faculty of Mathematics and Physics, University of Ljubljana, Slovenia sandi.klavzar@fmf.uni-lj.si b Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia c Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia d Faculty of Mathematics and Statistics, Central China Normal University, Wuhan 30079, P.R. China lscmath@mail.ccnu.edu.cn S.C. Li), zhanghhmath@163.com H. Zhang) Abstract A connected graph is said to be a cactus if each of its blocks is either a cycle or an edge. Let C n be the set of all n-vertex cacti with circumference at least, and let C n,k be the set of all n-vertex cacti containing exactly k 1 cycles where n 3k +1. In this paper, lower bounds on the difference between the revised) Szeged index and Wiener index of graphs in C n resp. C n,k ) are proved. The minimum and the second minimum values on the difference between the Szeged index and Wiener index of graphs among C n are determined. The bound on the minimum value is strengthened in the bipartite case. A lower bound on the difference between the revised Szeged index and Wiener index of graphs among C n,k is also established. Along the way the corresponding extremal graphs are identified. Key words: Wiener index; Szeged index; revised Szeged index; extremal problem; isometric cycle AMS Subj. Class: 05C35, 05C1 1 Introduction All graphs considered in this paper are finite, undirected and simple. If G is a graph, then its vertex set and edge set will be denoted V G and E G, respectively. The distance, d G u,v), between S.K. acknowledge the financial support from the Slovenian Research Agency research core funding No. P1-097). S.L. acknowledge the financial support from the National Natural Science Foundation of China Grant Nos , , ), the Program for New Century Excellent Talents in University Grant No. NCET ) and the Special Fund for Basic Scientific Research of Central Colleges Grant No. CCNU15A005)). H.Z. acknowledge the financial support from the excellent doctoral dissertation cultivation grant from Central China Normal University Grant No. 016YBZZ0). Corresponding author 1

2 vertices u and v of G is the length of a shortest u,v-path in G. The celebrated Wiener index or transmission) WG) of G is the sum of distances between all pairs of vertices of G, that is, WG) {u,v} V G d G u,v). 1) This distance-based graph invariant was in chemistry introduced back in 197 [3] and in mathematics about 30 years later [9]. Nowadays, the Wiener index is a well-established and much studied graph invariant; see the reviews [6, 7], a collection of papers dedicated to a half century of investigations of the Wiener index [1], and recent papers [, 6, 33, 36]. Given an edge e uv of a graph G, set N u e) {w V G : d G u,w) < d G v,w)}, N v e) {w V G : d G v,w) < d G u,w)}, N 0 e) {w V G : d G u,w) d G v,w)}. Clearly, N u e) N v e) N 0 e) is a partition of V G with respect to e, where N 0 e) if G is bipartite. For convenience, denote by n u e), n v e) and n 0 e) the number of vertices of N u e), N v e) and N 0 e), respectively. Thus, one has n u e)+n v e)+n 0 e) V G. From [13, 3] we know that for a tree T, WT) euv E T n u e)n v e). Inspired by this fact, Gutman [10] introduced the Szeged index SzG) of a graph G as SzG) euv E G n u e)n v e). Furthermore, in order to involve also those vertices that are at equal distance from the endpoints of an edge, Randić [31] proposed the revised Szeged index Sz G) of a graph G as follows: Sz G) n u e)+ n ) 0e) n v e)+ n ) 0e). euv E G Since SzT) WT) holds for any tree T, a lot of research has been done on the difference between the Szeged index and the Wiener index on general graphs. If G is a graph, then SzG) WG) 0 holds, a result conjectured in [10] and proved in [3]. Moreover, SzG) WG) holds if and only if every block of G is a complete graph [], see [16] for an alternative proof. Nadjafi-Arani et al. [] investigated the structure of graphs G with SzG) WG) k, where k is a positive integer. In particular, in [9] they characterized the graphs for which the difference is and 5. The difference between SzG) and WG) in networks was investigated in [19]. Based on the computer program AutoGraphiX, Hansen [1] presented nine conjectures on relations between the revised) Szeged index and the Wiener index. Chen, Li and Liu [3, ] proved three of these conjectures, while Li and Zhang [7] confirmed three additional above conjectures; these results deal with quotients between the revised) Szeged index and the Wiener index of unicyclic graphs. Motivated by these conjectures, further relationship between the Wiener index and the revised) Szeged index was established in [37]. For additional results on

3 relations between the revised) Szeged index and Wiener index see [, 0, 1], and for more information about the revised) Szeged index in general we refer to [1, 3, 15, 5, 30, 3, 35]. In this paper we continue the above direction of research by considering the difference between the Szeged index and the Wiener index on cacti. Since this difference is 0 if the circumference of a cactus is 3, we may and will restrict our attention to cacti with circumference at least. In the next section we give necessary definitions and state the main results of the paper. The first of them determines the minimum value on the difference between the Szeged index and the Wiener index of cacti, the second result strengthens this result in the bipartite case, while the third result determines the second minimum value on the difference between the Szeged index and the Wiener index of cacti. These three theorems are then proved in Section. The last main theorem that establishes a sharp lower bound on the difference between the revised Szeged index and the Wiener index is proved in Section 5. In Section 3 we recall some known results and prove new results that are needed for the proofs of the main results, while in the concluding section a couple of consequences are listed and a couple of problems are posed. Main results A cactus is a connected) graph in which any two cycles have at most one common vertex, that is, a graph whose every block is either an edge or a cycle. A cycle in a cactus is called an endblock cycle if all but one vertex of this cycle have degree. The circumference of a graph is the length of its longest cycle. As already mentioned, since SzG) WG) 0 if the circumference of G is 3, hence we set: C n {G : G is a cactus of order n and circumference at least }. In addition, for integers 3 r n, let C r n be the subset of n-vertex cacti defined as follows. If r n, then set C n n {C n }. Otherwise, C r n consists of the n-vertex cacti each of which is obtained from the cycle C r, and either a cactus G rooted at a vertex of the C r or cacti G and G rooted at two adjacent vertices of the C r, where G, G, and G are cacti whose blocks are only K or C 3. Using this notation our first main result reads as follows. Theorem.1. If G C n, then with equality if and only if G C 5 n. SzG) WG) n 5 In the bipartite case this result can be strengthen as follows. For a graph G let lg) denote the sum of the lengths of the cycles of G. Then: Theorem.. If G C n is bipartite, then SzG) WG) lg)n ), with equality if and only if each block of G is either a K or an end-block C. We note that a result closely related to Theorem. has been proved in [3, ]. Namely, if G is a connected bipartite graph of order at least, then SzG) WG) n. Moreover, the 3

4 bound is best possible when the graph is composed of C and a tree on n 3 vertices sharing a single vertex. In the case of cacti this result coincides with Theorem. for lg). In our next result we establish a sharp lower bound on the difference between the Szeged index and the Wiener index of graphs from the family C n \Cn 5. In other words, the next theorem gives the second minimum value on the difference between Sz and W in the class of cacti. For the equality case we define H to be the set of graphs isomorphic to some graph from the two families of graphs that are schematically presented in Fig. 1. Theorem.3. If G C n \C 5 n, then with equality if and only if G H. SzG) WG) n 10 H 1 H H 3 H H 5 H 6 Figure 1: Two families of graphs that constitute H; here each H i, 1 i 6, is either a trivial graph or each block of H i is K or C 3. To formulate our last main result that deals with the difference between the revised Szeged index and the Wiener index in cacti, we define C n,k {G : G is a cactus of order n containing at least k cycles}. Then our result reads as follows: Theorem.. Let G C n,k, where k 1 and n 3k +1. i) If n 9, then Sz G) WG) kn +n 6) with equality if and only if each block of G is either a K or an end-block C 3. ii) If n 10, then Sz G) WG) kn ) with equality if and only if each block of G is either a K or an end-block C. It is interesting to observe that the extremal graphs of Theorem. and of Theorem. ii) are the same.

5 3 Preliminary results In this section, we give some preliminary results which will be used in the subsequent sections. Simić et al. [3] rewrote the Szeged index as: SzG) euv E G {x,y} V G µ x,y e), ) where µ x,y e) is the contribution of the vertex pair x and y to n u e)n v e), that is, Setting x N u e) and y N v e), 1, if or µ x,y e) x N v e) and y N u e), 0, otherwise. πx,y) it then follows from Equalities 1) and ) that SzG) WG) e E G µ x,y e) d G x,y), {x,y} V G πx,y). 3) Recall that a subgraph H of a graph G is called isometric if the distance between any two vertices of H is independent of whether it is computed in the subgraph H or in G. Lemma 3.1 [37]). Let C r be an isometric cycle of a connected graph G, and let x,y V Cr. i) If r is even, then πx,y) e E Cr µ x,y e) d Cr x,y) d Cr x,y). ii) If r is odd, then πx,y) e E Cr µ x,y e) d Cr x,y) d Cr x,y) 1. Lemma 3.. Let G 1,G be vertex-disjoint, connected graphs of order at least. Let G be the graph obtained from G 1,G by identifying one vertex of G 1 with one vertex of G, denote the new vertex by u 0. Then for all x V G1 \{u 0 } and y V G \{u 0 }, one has πx,y) πx,u 0 )+πu 0,y). Proof. Consider a vertex pair {x, y} with x V G1 \{u 0 },y V G \{u 0 }. Note that u 0 is a cut vertex of G, we have d G x,y) d G x,u 0 ) + d G u 0,y). Next, we show that for any edge e uv E G1, µ x,y e) 1 if and only if µ x,u0 e) 1. ) First, suppose that µ x,y e) 1. Then we may assume without loss of generality that x N u e) and y N v e). Let P k be a shortest path connecting y and u, and P l be a shortest path joining y and v. As y N v e), one has l < k. Since u 0 is a cut vertex of G, we have u 0 V Pk V Pl. Thus, P k and P l can be written as P k yp a u 0 P b u and P l yp a u 0 P c v, where P a resp. P b, P c ) is a shortest path joining y and u 0 resp. u 0 and u, u 0 and v). Therefore, k a+b,l a+c. As l < k, we have c < b, i.e., d G u 0,v) < d G u 0,u). This implies that u 0 N v e). Note that x N u e), hence µ x,u0 e) 1. Similarly one shows that if µ x,u0 e) 1, then µ x,y e) 1. 5

6 In view of ) and the definition of µ x,y e), it is clear that e E G1 µ x,y e) e E G1 µ x,u0 e). By a similar discussion as above, e E G µ x,y e) e E G µ u0,ye) holds. Thus, one has as desired. πx,y) µ x,y e) d G x,y) e E G µ x,y e)+ µ x,y e) d G x,u 0 ) d G u 0,y) e E G1 e E G µ x,u0 e) d G x,u 0 ) + µ u0,ye) d G u 0,y) e E G1 e E G πx,u 0 )+πu 0,y), Lemma 3.3. Let G be an n-vertex cactus containing an even cycle C r. Then SzG) WG) nr r 3 with equality if and only if G C r or G is composed from C r and a graph G on n r + 1 vertices sharing a single vertex, where each block of G is a K or a C 3. Proof. For convenience, let C r v 1 v...v r v 1. Clearly, C r is an isometric cycle. Let G i be the component of G E Cr containing the vertex v i, 1 i r. Thus, V Gi 1 for all 1 i r. For each edge e uv E Gi, 1 i r, and every vertex pair {x,y} V Cr, it is straightforward to check that N u e), if v i N u e); x,y N v e), if v i N v e); N 0 e), if v i N 0 e). This implies that µ x,y e) 0. Therefore, for every vertex pair {x,y} V Cr, we have This gives πx,y) e E G µ x,y e) d G x,y) x,y V Cr πx,y) e E Cr µ x,y e) d Cr x,y) d Cr x,y). 5) x,y V Cr d Cr x,y) r3. 6) If V Gi, then, for every vertex pair {x,y} with x V Cr,y V Gi \V Cr, one has d G y,v i ) min z VCr d G y,z). Note that v i is a cut vertex of G. By Lemma 3., πx,y) πx,v i )+πv i,y). 6

7 Thus, x V Cr r πx,y) πx,v i )+πv i,y)) y V G \V Cr i1 x V Cr y V Gi \V Cr r πx,v i ) since πv i,y) 0) 7) i1 x V Cr y V Gi \V Cr n r) d Cr x,v i ) by 5)) x V Cr n r)r, ) where the equality in 7) holds if and only if πv i,y) 0 for all y V Gi \V Cr,1 i r. Note that πx,y) 0 for every vertex pair {x,y} V G \V Cr. Together with 3), 6) and ), we obtain that SzG) WG) πx,y)+ πx,y)+ πx,y) x,y V Cr x V Cr,y V G \V Cr x,y V G \V Cr r3 + n r)r +0 9) nr r 3, where the equality in 9) holds if and only if πv i,y) 0 for all y V Gi \V Cr, 1 i r, and πx,y) 0 for every vertex pair {x,y} V G \V Cr. Now, we show that if SzG) WG) nr r 3, then G has exactly one cycle whose length is at least. Supposeon the contrary that G contains a cycle C k, k, different from C r. Because G is a cactus, V Cr V Ck 1. This implies that V Ck \V Cr 3. Thus, there exist two vertices u,v V Ck \V Cr suchthatd Ck u,v).bylemma3.1, wehaveπu,v) d Ck u,v) 1 1,which contracts the assumption that πx,y) 0 for every vertex pair {x,y} V G \V Cr. Therefore, G contains only one cycle whose length is at least. If there are two components, say G a,g b, of G E Cr such that V Ga, V Gb, then consider x V Ga \{v a } and y V Gb \{v b }. Note that v a and v b are cut vertices of G. By Lemma 3., we have πx,y) πx,v a ) + πv a,v b ) + πv b,y) πv a,v b ) 1, which is also a contradiction to the fact that πx,y) 0 for every vertex pair {x,y} V G \V Cr. Therefore, there is at most one nontrivial component G i. In this case, by direct calculation, we have SzG) WG) nr r 3. Hence, x,y V G πx,y) nr r 3 if and only if G is composed of a cycle C r on r vertices and a graph G on n r +1 vertices sharing a single vertex, where each block of G is either a K or a C 3. Lemma 3.. Let G be an n-vertex cactus containing an odd cycle C r v 1 v...v r v 1 of length at least 5. Then SzG) WG) with equality if and only if G C r n. r 1)r 3)n r) 7

8 Proof. Clearly, C r is an isometric cycle. Let G i be the component of G E Cr containing the vertex v i,1 i r. Then V Gi 1 for any 1 i r. For any edge e uv E Gi, 1 i r, and every vertex pair {x,y} V Cr, by a discussion similar to the one from the beginning of the proof of Lemma 3.3, we obtain µ x,y e) 0. Thus, for every vertex pair {x,y} V Cr, we have πx,y) e E G µ x,y e) d G x,y) e E Cr µ x,y e) d Cr x,y) d Cr x,y) 1. 10) So, πx,y) x,y V Cr x,y V Cr d Cr x,y) 1) rr 1)r 3). 11) For every vertex pair {x,y} with x V Cr, y V Gi \V Cr, 1 i r, we have d G y,v i ) min z VCr d G y,z). Since v i is a cut vertex of G, Lemma 3. implies that πx,y) πx,v i ) + πv i,y). Then x V Cr r πx,y) πx,v i )+πv i,y)) y V Gi \V Cr i1 x V Cr y V Gi \V Cr r πx,v i ) since πv i,y) 0) 1) i1 x V Cr y V G \V Cr n r) d Cr x,v i ) 1) by 10)) x V Cr n r)r 1)r 3), 13) where the equality in 1) holds if and only if πv i,y) 0 for all y V Gi \V Cr, 1 i r. Note that πx,y) 0 for every vertex pair {x,y} V G \V Cr. Combining this fact with 3), 11) and 13), it follows that SzG) WG) πx,y)+ πx,y)+ πx,y) x,y V Cr x V Cr,y V G \V Cr x,y V G \V Cr rr 1)r 3) n r)r 1)r 3) + r 1)r 3)n r), where the equality in 1) holds if and only if πv i,y) 0 for all y V Gi, 1 i r and πx,y) 0 for every vertex pair {x,y} V G \V Cr. We next show that if SzG) WG) r 1)r 3)n r), then C r is the only cycle of G with length at least. Otherwise, there exists another cycle C k of length k. Since G is a cactus, V Cr V Ck 1, which in turn implies that V Ck \V Cr 3. Thus, there are two vertices u,v V Ck \V Cr such that d Ck u,v). By Lemma 3.1, we have πu,v) d Ck u,v) 1 1, this is a contradiction to πx,y) 0 for every vertex pair {x,y} V G \V Cr. Therefore, C r is the only cycle of G with length at least. 1)

9 Suppose that there exist two nontrivial components G i,g j of G E Cr such that i j 1, r 1. Then, take two vertices x,y with x V Gi \{v i },y V Gj \{v j }. By 10), we have πv i,v j ) 1. Note that v i,v j are cut vertices of G. By Lemma 3., we have πx,y) πx,v i )+πv i,v j )+πv j,y) πv i,v j ) 1, a contradiction to the fact πx,y) 0 for every vertex pair {x,y} V G \V Cr. Therefore, there are just two nontrivial G i,g i+1 or there is only one nontrivial G i. For each of the above cases, by direct calculation, we have SzG) WG) r 1)r 3)n r). Hence, SzG) WG) r 1)r 3)n r) holds if and only if G Cn r, as desired. Proofs of Theorems Proof of Theorem.1 Recall the statement of the theorem to be proved in this subsection: If G C n, then SzG) WG) n 5 with equality if and only if G C 5 n. Let C r be a longest cycle of G. Then r since the circumference of G is at least. If r is even, then by Lemma 3.3, SzG) WG) nr r 3 because n r. If r is odd, then by Lemma 3., > n 5 r 1)r 3)n r) SzG) WG) 15) n 5. since r 5) 16) Based on Lemma 3., the equality in 15) holds if and only if G Cn r ; whereas the equality in 16) holds if and only if r 5. Hence, SzG) WG) n 5 holds if and only if G Cn, 5 as claimed.. Proof of Theorem. Recall the statement of Theorem.: If G C n is bipartite, then SzG) WG) lg)n ) with equality if and only if each block of G is either a K or an end-block C, where lg) is the sum of the lengths of the cycles of G. We firstnote that a bipartite cactus is a partial cube, that is, isometrically embeddable into a hypercube. One way to see it is by applying Djoković s characterization of partial cubes from [5] asserting that G is a partial cube if and only if G is bipartite and for any edge e uv the subgraphs of G induced by N u e) and by N v e) are convex. Let F be the partition of E G that consists of the singletons corresponding to the K -blocks of G, while each cycle C k contributes k pairs of opposite edges to F. The partition F thus contains lg)/ sets of cardinality, the other sets are singletons. Then, applying the main theorem of [17], WG) F Fn u e)n v e), 9

10 where, for a given F F, the edge e uv is an arbitrary, fixed edge from F. Similarly, applying the main theorem from [11], SzG) F n u e)n v e), F F where again the edge e uv is an arbitrary but fixed edge from F. These two results are instances of the so called standard cut method, see the recent survey [] for more information on the method.) Therefore, SzG) WG) F F F 1)n u e)n v e) F 1)n u e)n v e) F F, F n u e)n v e) F F, F F F, F n ) lg) n ) lg)n ). The above inequality turns into equality if and only if every cycle C of G is a -cycle and {n u e),n v e)} {,n } holds for any edge e of C. That is, the equality holds if and only if every cycle of G is an end-block C..3 Proof of Theorem.3 We next prove Theorem.3 which asserts the following: If G C n \Cn 5, then SzG) WG) n 10 with equality if and only if G H. If G contains an even cycle C r, then by Lemma 3.3, we have SzG) WG) nr r 3 > n 10, the last inequality follows since r. Hence, we may assume in what follows that the lengths of all the cycles in G are odd. Let C r be one of the longest odd cycles of G. Then clearly r 5. If r 7, then by Lemma 3., we have SzG) WG) r 1)r 3)n r) > n 10, as desired. Thus, it suffices to consider the remaining case r 5. Note that G C 5 n. Hence, there exist at least two cycles of length 5, say C, and C in G. For convenience, let C v 1 v...v 5 v 1 and C u 1 u...u 5 u 1. Since G is a cactus, we have V C V C 1. Thus, we proceed by considering the following two possible cases. Case 1. V C V C 0. In view of 11), we have x,y V C πx,y) x,y V C πx,y) 5. 17) As G is a cactus, we may without loss of generality assume that u 1,v 1 are the vertices in G such that d G v i,u 1 ) min x VC d G v i,x) and d G u j,v 1 ) min x VC d G u j,x) for all 1 i,j 5. Based on 10), we have x V C πx,v 1 ) and y V πy,u C 1). Since u 1 and v 1 are cut vertices of G, using Lemma 3. we infer that πx,y) πx,v 1 )+πv 1,u 1 )+πu 1,y)) x V C,y V C x V C,y V C 5 x V C πx,v 1 )+5 y V C πu 1,y) since πv 1,u 1 ) 0 ) 1) 0, 19) 10

11 where the equality in 1) holds if and only if πv 1,u 1 ) 0. Consider every vertex pair {x,y} with x V C, y V G \V C V C ). Assume that v i0 is the vertex of C such that d G y,v i0 ) min z VC d G y,z). In view of Lemma 3., we have x V C πx,y) x V C πx,v i0 )+πv i0,y)) x V C πx,v i0 ). Thus, πx,y) n 10). 0) x V C y V G \V C V C ) y V G \V C V C ) The equality in 0) holds if and only if x V C πx,y) for all y V G \V C V C ). Similarly, we can also obtain that x V C y V G \V C V C ) πx,y) n 10) with equality if and only if x V πx,y) for all y V C G\V C V C ). Note that πx,y) 0 for every vertex pair {x,y} V G \V C V C ). Combined with 17), 19) and 0), it follows that SzG) WG) πx,y)+ πx,y)+ πx,y)+ πx, y) x,y V C x,y V C x V C,y V C x V C y V G\V C V C ) + πx,y)+ πx, y) x V C y V G\V C V C ) x,y V G\V C V C ) n 10)+n 10) 1) n 10, where the equality in 1) holds if and only if πv 1,u 1 ) 0, x V C πx,y), and x V C πx,y) for all y V G \V C V C ), as well as πx,y) 0 for every vertex pair {x,y} V G \V C V C ). Next, weshowthat ifszg) WG) n 10,then Gcontainsjust twocyclesoflength5. Otherwise, G contains a third cycle C of length 5. Note that V C V C 1 and V C V C 1. Then there exist two vertices x,y V C \V C V C ) such that d C x,y). Since C is an isometric cycle, by Lemma 3.1ii) we have πx,y) d C x,y) 1 1. This is a contradiction to the fact πx,y) 0 for every vertex pair {x,y} V G \V C V C ). Let G i resp. G i ) be the component of G E C resp. G E C ) that contains the vertex v i resp. u i ), 1 i r. Then V G1 and V Gi 1 for i 5. Suppose that there exist components G i and G j with V Gi and V Gj, where v i and v j are not adjacent. Select arbitrary vertices x V Gi \{v i } and y V Gj \{v j }. Because v i and v j are cut vertices of G, applying Lemma 3. we get πx,y) πx,v i )+πv i,v j )+πv j,y) πv i,v j ) 1, which is a contradiction to the fact πx,y) 0 for every vertex pair {x,y} V G \V C V C ). Combined with V G1, we obtain V G V G3 V G 1 or V G3 V G V G5 1. Similarly, we can also show that V G V G 3 V G 1 or V G 3 V G V G 5 1. In each of the above subcases, by direct calculation we have SzG) WG) n 10. Hence, SzG) WG) n 10 if and only if G H, where H is depicted in Fig. 1. Case. V C V C 1. By a similar discussion as in the proof of Case 1, we can show that SzG) WG) n 10 with equality if and only if G H; see Fig. 1 again. This completes the proof of Theorem.3. 5 Proof of Theorem. In this section we prove Theorem.. Since n u e)+n v e)+n 0 e) n for e uv E G, it is a routine to check that Sz G) WG) SzG) WG)+ ) n n 0e). ) 11 e E G

12 In order to prove the theorem, we first demonstrate a couple of claims. Claim 1. If G C n,k is such that Sz G) WG) is as small as possible, then each cycle of G is an end-block. Proof. Suppose on the contrary that G contains a cycle C r v 1 v...v r v 1 which is not an end-block. Let G i be the component of G E Cr containing the vertex v i, 1 i r. As C r is not an end-block, G E Cr contains two nontrivial components, say G a and G b. We construct a new graph G as follows: G G r {v i x : x N Gi v i )}+ i r {v 1 x : x N Gi v i )}. Then G is in C n,k. By direct calculation based on 3) and Lemma 3.), one has i SzG) WG) r r πx,y)+ i1 x,y V Gi j1 + πx, y) 1 i<j r x V Gi \{v i}, y V Gj \{v j} y V Gj \{v j} i j πv i,y)+ x,y V Cr πx,y) r r πx,y)+ πv i,y)+ πx,y) i1 x,y V Gi j1 y V Gj \{v j} i j x,y V Cr + πx,v i )+πv i,v j )+πv j,y)). 3) 1 i<j r x V Gi \{v i}, y V Gj \{v j} Similarly, we have SzG ) WG ) r r r πx,y)+ πv i,y)+ πx,y) i1 x,y V Gi j1 y V Gj \{v j} i x,y V Cr + πx,v 1 )+πv 1,y)). ) 1 i<j r x V Gi \{v i}, y V Gj \{v j} For convenience, set and : 1 : SzG) WG)) SzG ) WG )) e E G n n 0 e) ) n n 0 e) ). e E G As V Ga, V Gb, we have V Ga \{v a } 1, V Gb \{v b } 1. In view of 3) and ), we have 1 πv i,v j ) 1 i<j r x V Ga \{v a}, y V Gb \{v b } x V Gi \{v i}, y V Gj \{v j} πv a,v b ) πv a,v b ). 5) In what follows, we consider two possible cases according to the parity of r. 1

13 Case 1. r is even. In this case, on the one hand, it is routine to check that 0. On the other hand, by Lemma 3.1i), we have πv a,v b ) d Cr v a,v b ) 1. In view of 5), we have 1 πv a,v b ) 1. Combined with ), we have Sz G) WG)) Sz G ) WG )) 1 + > 0, a contradiction to the choice of G. Case. r is odd. Since C r is an isometric cycle, by Lemma 3.1ii) we have πv a,v b ) d G v a,v b ) 1 0. In view of 5), we have 1 0. Next, we show that > 0. For the graph G, it is straightforward to check that e E G n n 0 e) ) Similarly, for the graph G, we have e E G n n 0 e) ) r i1 e E Gi r i1 e E Gi r i1 e E Gi Then, together with 6) and 7), it follows that r 1+n r+1) n n 0 e) ) + n n 0 e) ) e E Cr r n n 0e) )+ n i1 V G i. 6) n n 0 e) )+ n r 1+n r +1). 7) r i1 V G i. Suppose that there exist two components G p,g q satisfying V Gp, V Gq for 1 p,q r. Let r 1+n r+1) i p,q VG i V Gp + V Gq 1) 1. Then, V Gp + V Gq 1) +1 V Gp V Gq V Gp 1) V Gq 1) > 0. The last inequality follows because V Gp, V Gq. Thus, attains its minimum 0 if and only if V G1 n r+1, V Gi 1 for i r up to isomorphism. Bearing in mind that V Ga, V Gb, one can easily obtain that > 0. Inviewof), wehavesz G) WG)) Sz G ) WG )) 1 + > 0,i.e., Sz G ) WG ) < Sz G) WG), which is a contradiction to the choice of G. This completes the proof of Claim 1. Let each cycle of a graph G 1 C n,k be an end-block cycle. Let C p,c q be two disjoint cycles of G 1 containing cut vertices u 0,v 0, respectively. Construct a new graph G as follows: G G 1 {v 0 x : x N Cq v 0 )}+{u 0 x : x N Cq v 0 )}. In other words, if v 0 and v 0 are the neighbors of v 0 on C q, then G is obtained from G 1 by removing the edges v 0 v 0 and v 0 v 0 and adding the edges u 0 v 0 and u 0 v 0. Then G is in C n,k and each cycle of G is also an end-block. Here, we show that this graph transformation keeps the value of Sz G) WG) unchanged. Claim. Let G 1 and G be the graphs as defined above. Then Sz G 1 ) WG 1 ) Sz G ) WG ). Proof. First, we show that πx,y) 0 for every pair of cut vertices x,y V G1. For such cut vertices x,y V G1, there is a shortest path P t connecting x and y. It is routine to check that for any edge e E Pt, 13

14 µ x,y e) 1. Recall that each cycle of G 1 is an end-block. Assume that x 0 is a cut vertex of the cycle C r in G 1. Then, for an edge e uv E Cr, we have N u e), if x 0 N u e); x,y N v e), if x 0 N v e); N 0 e), if x 0 N 0 e). This implies that µ x,y e) 0. For the remaining cut edge e uv E G1 \E Pt, it is routine to check that x,y N u e)orx,y N v e),whichimpliesthatµ x,y e) 0.Thus,πx,y) e E G1 µ x,y e) d G1 x,y) e E Pt µ x,y e) t+1 0. Bearing in mind that u 0,v 0 are cut vertices, we have πu 0,v 0 ) 0. For convenience, denote by V 1 V G1 \V Cp V Cq ). In view of 3), one can obtain that SzG 1 ) WG 1 ) πx,y)+ πx,y)+ πx,y)+ πx,y) x,y V Cp x,y V Cq x,y V 1 x V Cp,y V 1 + πx,y)+ πx,y) x V Cp,y V Cq x V Cq,y V 1 πx,y)+ πx,y)+ πx,y)+ πx,y) x,y V Cp x,y V Cq x,y V 1 x V Cp,y V 1 + πx,u 0 )+πv 0,y))+ πx,v 0 )+πv 0,y)) x V Cp,y V Cq x V Cq,y V 1 and SzG ) WG ) πx,y)+ πx,y)+ πx,y)+ πx,y) x,y V Cp x,y V Cq x,y V 1 x V Cp,y V 1 + πx,u 0 )+πu 0,y))+ πx,u 0 )+πu 0,y)). x V Cp,y V Cq x V Cq,y V 1 Thus, based on 5) or 10), we have SzG 1 ) WG 1 )) SzG ) WG )) q y V 1 πv 0,y) πu 0,y)). ) Now, we provethat πv 0,y) πu 0,y) for all y V 1. If y V Cl for some cycle C l in G 1 except C p,c q, and denote the unique cut vertex of C l by w. By Lemma 3., we have πv 0,y) πv 0,w)+πw,y) πw,y) and πu 0,y) πu 0,w) + πw,y) πw,y). Thus, πv 0,y) πu 0,y). Otherwise, y isn t contained in any cycle. Then y is a cut vertex of G 1 or d G1 y) 1. In this case, we have πv 0,y) πu 0,y) 0. Therefore, by ), we know that SzG 1 ) WG 1 ) SzG ) WG ). Note that ) n n 0e) ) n n 0e). e E G e E G1 In view of ), we have Sz G 1 ) WG 1 ) Sz G ) WG ), as desired. Now all is ready for the proof of Theorem.. Choose a graph G in C n,k such that Sz G) WG) is as small as possible. By Claim 1, each cycle of G is an end-block cycle. By a repeated application of the construction of Claim, we may assume that all the cycles of G have a common vertex. Denote the common vertex by u 0. For convenience, let C r1,c r,...,c rp be the even cycles and C t1,c t,...,c tq the odd cycles of G. Then p+q k. In what follows, we first determine the lower bound on SzG) WG). 1

15 For every vertex pair {x,y} V Cri, 1 i p, by 6) we have πx,y) d Cri x,y) r3 i. 9) x,y V Cri x,y V Cri Consider every vertex pair {x, y} with x V Cri \{u 0 }, y V Crj \{u 0 }, i j. Since u 0 is a cut vertex of G, by 5) and Lemma 3. we have πx,y) πx,u 0 )+πu 0,y) d Cri x,u 0 )+d Crj u 0,y). Then x V Cri \{u 0} y V Crj \{u 0} πx, y) x V Cri \{u 0} y V Crj \{u 0} r i r j 1) d Cri x,u 0 )+d Crj u 0,y)) + r j r i 1). 30) For convenience, denote by n 0 : r 1 +r + +r p. Then, in view of 30), we have ri πx, y) r j 1) + r j r ) i 1) 1 i<j p x V Cri \{u 0} y V Crj \{u 0} 1 i<j p 1 i p ri n 0 r i p+1). 31) Consider the remaining vertex pairs {x, y} with x p i1 V C ri and y V G \ p i1 V C ri ). Then from 5) and Lemma 3. we get πx,y) πx,u 0 )+πu 0,y) πx,u 0 ) d G x,u 0 ). Consequently, πx,y) d G x,u 0 ) since πu 0,y) 0) x p i1 VCr y V p i G\ i1 VCr ) x p i i1 VCr y V p i G\ i1 VCr ) i n n 0 +p 1) d G x,u 0 ) x p i1 VCr i n n 0 +p 1) p i1 r i. 3) Note that πx,y) 0 for every vertex pair {x,y} V G \ p i1 V C ri ). Together with 9), 31) and 3), it follows that SzG) WG) p πx,y)+ i1 x,y V Cri 1 i<j p + πx,y)+ x p i1 VCr y V p i G\ i1 VCr ) i p ri 3 p + ri n 0 r i p+1) p + i1 p i1 i1 r 3 i +nr i x V Cri \{u 0} y V Crj \{u 0} i1 πx, y) x,y V G\ p i1 VCr i ) πx,y) n n 0 +p 1)r i 33) pn ), since r i ) 3) where the equality in 33) holds if and only if πu 0,y) 0 for any vertex y V G \ p i1 V C ri ) and πx,y) 0 for every vertex pair {x,y} V G \ p i1 V C ri ); whereas the equality in 3) holds if and only if r 1 r r p. 15

16 Next, we consider the value of ) n0e) e E G n n 0 e). Bear in mind that, for an edge e, if it is not contained in any odd cycle, then n 0 e) 0. So we consider it is in some odd cycle, say C ti. One has n 0 e) n t i +1 if u 0 N 0 e) and n 0 e) 1 otherwise. Thus, we have e E G ) n n 0e) q i1 e E Cti q n i1 where the equality in 35) holds if and only if t 1 t t q 3. Together with ), 3) and 35), it follows that Sz G) WG) SzG) WG)+ ) n n 0e) t i 1+n t i +1) qn +n 6), since t i 3) 35) e E G ) ) n n 0e) pn )+ qn +n 6). 36) Based on 3) and 35) we obtain that the equality in 36) holds if and only if r 1 r r p and t 1 t t q 3. Now, we give the proofs of i) and ii), respectively. i) If n 9, then by direct calculation, we have n > n +n 6. In view of 36), one has Sz G) WG) kn +n 6) with equality if and only if t 1 t t q 3 and q k, i.e., G is a graph satisfying each block of G being a K or a C 3 and each cycle of G being an end-block. Thus, i) holds. ii) If n 10, then it is routine to check that n < n +n 6. In view of 36), one has Sz G) WG) kn ) with equality if and only if r 1 r r p and p k, i.e., G is a graph satisfying each block of G being a K or an end-block C. Hence, ii) holds. 6 Concluding remarks In this paper we have established lower bounds on the difference between the revised) Szeged index and Wiener index of graphs in C n resp. C n,k ) are established. To conclude the paper we state two corollaries and two problems. The following result follows from Theorem.1 and can also be deduced from [, Theorems 3.1 and 3.]. Corollary 6.1. Let G be an n-vertex unicyclic graph with circumference at least. Then SzG) WG) n 5 with equality if and only if G is a graph which is composed from C 5 and either a tree on n vertices rooted at a vertex of C 5, or two trees rooted at two adjacent vertices of C 5. Similarly, the following result follows from Theorem. and can alternatively be obtained from [, Theorems 3. and.3]. Corollary 6.. Let G be a unicyclic graph on n vertices. 16

17 i) If n 9, then Sz G) WG) n +n 6 with equality if and only if G is composed from C 3 and a tree T on n vertices sharing a single vertex. ii) If n 10, then Sz G) WG) n with equality if and only if G is composed from C and a tree T on n 3 vertices sharing a single vertex. In Subsection. we have demonstrated that the cut method provides an efficient method to bound the difference between SzG) and WG) when G is a bipartite cactus. Extensions of the cut method to general graphs are known, see [1, ], hence the following problem appears natural in this context. Problem 6.3. Can the standard cut method for the Szeged index be extended to general graphs or to arbitrary cacti? We have identified the graphs G from C n,k such that Sz G) WG) attains its minimum value. Hence we pose: Problem 6.. Determine the second minimum value on the difference between the revised Szeged index and the Wiener index among the graphs from C n,k. References [1] M. Aouchiche, P. Hansen, On a conjecture about the Szeged index, European J. Combin ) [] M. Bonamy, M. Knor, B. Lužar, A. Pinlou, R. Škrekovski, On the difference between the Szeged and Wiener index, arxiv: [math.co]. [3] L.L. Chen, X.L. Li, M.M. Liu, I. Gutman, On a relation between the Szeged and the Wiener indices of bipartite graphs, Trans. Comb. 1 01) 3-9. [] L.L. Chen, X.L. Li, M.M. Liu, The revised) Szeged index and the Wiener index of a nonbipartite graph, European J. Combin ) [5] D. Djoković, Distance preserving subgraphs of hypercubes, J. Combin. Theory Ser. B ) [6] A.A. Dobrymin, R. Entriger, I. Gutman, Wiener index of trees: theory and applications, Acta Appl. Math ) [7] A.A. Dobrymin, I. Gutman, S. Klavžar, P. Žigert, Wiener index of hexagonal systems, Acta Appl. Math. 7 00) 7 9. [] A. Dobrynin, I. Gutman, Solving a problem connected with distances in graphs, Graph Theory Notes N. Y. 1995) 1 3. [9] R.C. Entringer, D.E. Jackson, D.A. Snyder, Distance in graphs, Czech. Math. J ) [10] I. Gutman, A formula for the Wiener number of trees and its extension to graphs containing cycles, Graph Theory Notes N. Y ) [11] I. Gutman, S. Klavžar, An algorithm for the calculation of the Szeged index of benzenoid hydrocarbons, J. Chem. Inf. Comput. Sci )

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