Disproof of a conjecture on the edge mostar index

Fazal Hayat1, Shou-Jun Xu1, Bo Zhou2
1School of Mathematics and Statistics, Gansu Center for Applied Mathematics, Lanzhou University, Lanzhou 730000, China
2School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China

Abstract

For a connected graph G, the edge Mostar index Moe(G) is defined as Moe(G)=e=uvE(G)|mu(e|G)mv(e|G)|, where mu(e|G) and mv(e|G) are respectively, the number of edges of G lying closer to vertex u than to vertex v and the number of edges of G lying closer to vertex v than to vertex u. We determine a sharp upper bound for the edge Mostar index on bicyclic graphs and identify the graphs that achieve the bound, which disproves a conjecture proposed by Liu et al. [Iranian J. Math. Chem. 11(2) (2020) 95–106].

Keywords: Mostar index, Edge Mostar index, Bicyclic graph, Distance-balanced graph

1. Introduction

Let G be a graph with vertex set V(G) and edge set E(G). The order and size of G are the cardinality of V(G) and E(G), respectively. The distance between u and v in G, denoted by dG(u,v), is the shortest path connecting u and v in G. For a vertex x and edge e=uv of a graph G, the distance between x and e, denoted by dG(x,e) , is defined as dG(x,e)=min{dG(x,u),dG(x,v)}.

A molecular graph is a simple graph such that its vertices correspond to the atoms and the edges to the bonds of a molecule. A topological graph index, also called a molecular descriptor, is a mathematical formula that can be applied to any graph which models some molecular structure. From this index, it is possible to analyse mathematical values and further investigate some physicochemical properties of a molecule. Therefore, it is an efficient method in avoiding expensive and time-consuming laboratory experiments.

Došlić et al. [7] introduced the Mostar index, which is a measure of peripherality in graphs. It can also be considered as a bond-additive structural invariant as a quantitative refinement of the distance non-balancedness. For a graph G, the Mostar index is defined as Mo(G)=e=uvE(G)|nu(e|G)nv(e|G)|. where nu(e|G) is the number of vertices of G closer to u than to v and nv(e|G) is the number of vertices closer to v than to u.

Since its introduction in 2018, the Mostar index has already incited a lot of research, mostly concerning trees [1,6,5,4,8,12,14], unicyclic graphs [18], bicyclic graphs [20], tricyclic graphs [11], cacti [13] and chemical graphs [3,16,21,22].

Arockiaraj et al. [2], introduced the edge Mostar index as a quantitative refinement of the distance non-balancedness, also measure the peripherality of every edge and consider the contributions of all edges into a global measure of peripherality for a given chemical graph. The edge Mostar index of G is defined as Moe(G)=e=uvE(G)ψG(uv), where ψG(uv)=|mu(e|G)mv(e|G)|, mu(e|G) and mv(e|G) are respectively, the number of edges of G lying closer to vertex u than to vertex v and the number of edges of G lying closer to vertex v than to vertex u. We use ψ(uv)=|mu(e)mv(e)| for short, if there is no ambiguity.

The problem of determining which graphs uniquely maximize (resp. minimize) the edge Mostar index in various classes of graphs has received much attention. For example, Imran et al. [17] studied the edge Mostar index of chemical structures and nanostructures using graph operations. Havare [10] computed the edge Mostar index for several classes of cycle-containing graphs. Hayat et al. [15] obtained a sharp upper bound for the edge Mostar index on tricyclic graphs and identified the graphs that attain the bound. Liu et al. [19] determined the extremal values of the edge Mostar index among trees, unicyclic graphs, cacti and posed two conjectures for the extremal edge Mostar index among bicyclic graphs. Ghalavand et al. [9] gave proof to a conjecture in [19], and obtained the graph that minimizes the edge Mostar index among bicyclic graphs.

Motivated directly by [19] and [9], we determine the unique graph that maximizes the edge Mostar index over all bicyclic graphs with fixed size, which disproves the following conjecture.

Conjecture 1.1.

[19] If the size m of bicyclic graphs is large enough, then the bicyclic graphs Bm (see Figure 1) and Bm5 (see Figure 2) have the maximum edge Mostar index.

We disprove Conjecture 1.1, by presenting the following result.

Theorem 1.2.

Let G be a bicyclic graph of size m5. Then Moe(G){4,if m=5, and equality holds iff GBm3,Bm4;m23m6,if 6m8, and equality holdsiff GBm1,Bm3;48,if m=9, and equality holds iff GBm,Bm1,Bm2,Bm3,Bm4;m2m24,if m10, and equality holds iff GBm. (Where Bm,Bm1,Bm2 and Bm3,Bm4 are depicted in Figure 1 and Figure 2, respectively).

Based on the above Theorem, if the size m of bicyclic graphs is large, then Bm is the unique graph that maximizes the edge Mostar index. Therefore, Conjecture 1.1 is disproved.

In Section 2, we provide some definitions and preliminary results. The proof of Theorem 1.2 is presented in Section 3.

2. Preliminaries

In this section, we present some basic notations and elementary results, which will be useful in the proof of our main result.

For vV(G), let NG(v) be the set of vertices that are adjacent to v in G. The degree of vV(G), denoted by dG(v), is the cardinality of NG(v). A vertex with degree one is called a pendent vertex and an edge incident to a pendent vertex is called a pendent edge. A graph G with n vertices is a bicyclic graph if |E(G)|=n+1. As usual by Sn, Pn, and Cn we denote the star, path, and cycle graph on n vertices, respectively.

The join of two graphs G1 and G2 is denoted by G1G2, is obtained by identifying one vertex from G1 and G2. Let u be the identified vertex in both G1 and G2. If G1 contains a cycle and u belongs to some cycle, while G2 is a tree, then we call G2 a pendent tree in G1G2 associated with u.

For each eE(G1), every path from e to some edges in G2 pass through u. Therefore, the contribution of G2 to eE(G1)ψ(e) totally depends on the size of G2. In other words, changing the structure of G2 does not affect the value of eE(G1)ψ(e).

Lemma 2.1.

[19] Let G be a graph of size m with a cut edge e=uv. Then ψ(e)m1 with equality if and only if e=uv is a pendent edge.

By Lemma 2.1, if e is a pendent edge, then ψ(e) is maximum. Therefore, it is easy to verify the following result.

Lemma 2.2.

Let H be a graph of size m. Then Moe(H1H)Moe(H1Sm), where the common vertex of H1 and Sm is the center of Sm, while the common vertex of H1 and H can be chosen arbitrarily.

Let Sm,r=SmrCr, where the common vertex of Smr and Cr is the center of Smr.

Lemma 2.3.

[19] Let G be a unicyclic graph of size m3. Then Moe(G){m22m3,if 3m8, and equality holdsiff GSm,3;60,if m=9, and equality holds iff GSm,3,Sm,4;m2m12,if m10, and equality holds iff GSm,4.

Lemma 2.4.

Let G1 be a connected graph of size m1 and G2 be a unicyclic graph of size m2. Then Moe(G1G2){Moe(G1Sm2,3)for m1+m28;Moe(G1Sm2,3)=Moe(G1Sm2,4)for m1+m2=9;Moe(G1Sm2,4)for m1+m210; where the common vertex of G1 and Sm2,3 (resp. G1 and Sm2,4) is the center of Sm2,3 (resp. Sm2,4), while the common vertex of G1 and G2 can be chosen arbitrarily.

Proof. If m1+m28, then by Lemma 2.3, we have Moe(G1G2)=e=uvE(G1G2)ψG1G2(uv)=e=uvE(G1)ψG1G2(uv)+e=uvE(G2)ψG1G2(uv)=e=uvE(G1)ψG1G2(uv)+Moe(Sm1G2)m1(m1)e=uvE(G1)ψG1Sm2,3(uv)+Moe(Sm1Sm2,3)m1(m1)=e=uvE(G1)ψG1Sm2,3(uv)+e=uvE(Sm2,3)ψG1Sm2,3(uv)=Moe(G1Sm2,3).

Similarly, if m1+m2=9, then Moe(G1G2)Moe(G1Sm2,3)=Moe(G1Sm2,4); if m1+m210, then Moe(G1G2)Moe(G1Sm2,4)◻

3. Proof of theorem 1.2

The Θ-graph is a graph that connects two fixed vertices with three internally disjoint paths. If the lengths of these paths are l1,l2, and l3 where l1l2l3, the graph is denoted as Θ(l1,l2,l3).

Let Gm1 represent the set of bicyclic graphs of size m that contain exactly two edge-disjoint cycles. Let Gm2 denote the set of bicyclic graphs of size m that contain exactly three cycles. Moreover, if GGm2, then G contains a Θ-graph as a subgraph, which is called as the brace of G.

For the proof of Theorem 1.2, we first develop several lemmas. In the first step, we establish a sharp upper bound for Moe(G) on the set Gm1.

Lemma 3.1.

Let GGm1 . Then Moe(G){m23m6,if 6m8, and equality holdsiff GBm2;48,if m=9, and equality holds iff GBm,Bm1,Bm2;m2m24,if m10, and equality holds iff GBm.

Proof. Let GGm1. Then there are two unicyclic graphs G1 and G2 with size m1 and m2, respectively such that G=G1G2. Then, in view of Lemma 2.4. If 6m8, we get Moe(G)=Moe(G1G2)Moe(G1Sm2,3)Moe(Sm1,3Sm2,3)=Moe(Bm2); if m=9, then Moe(G)=Moe(G1G2)Moe(G1Sm2,3)=Moe(G1Sm2,4)Moe(Sm1,3Sm2,3)=Moe(Sm1,3Sm2,4)=Moe(Bm2)=Moe(Sm1,4Sm2,4)=Moe(Bm)=Moe(Bm1)=Moe(Bm2).

If m10, we have Moe(G)=Moe(G1G2)Moe(G1Sm2,4)Moe(Sm1,4Sm2,4)=Moe(Bm).

By simple calculation, we get Moe(Bm)=m2m24, Moe(Bm1)=m22m15, and Moe(Bm2)=m23m6. If m=9, then Moe(Bm)=Moe(Bm1)=Moe(Bm2)=48. This completes the proof. ◻

In the following, we determine a sharp upper bound for Moe(G) on the set Gm2.

Lemma 3.2.

Let GGm2 with brace Θ(1,2,2). Then Moe(G){4,if m=5, and equality holds iff GBm3,Bm4;m23m6,if 6m8, and equality holdsiff GBm3;48,if m=9, and equality holds iff GBm3,Bm4;m22m15,if m10, and equality holds iff GBm4.

Proof. Let GGm2 with brace Θ(1,2,2). Suppose that v1,v2,v3,v4 be the vertices in Θ(1,2,2) of G with dG(v1)=dG(v2)=3, and dG(v3)=dG(v4)=2. Let ai be the number of pendent edges of vi (i=1,2,3,4). Suppose that a1a2 and a3a4. Let G1 be the graph obtained from G by shifting a2 pendent edges from v2 to v1. We have Moe(G1)Moe(G)=(a1+a2+22)(a1+2a22)+(a1+a2+a4+2a31)(a1+a4+2a31)+(a1+a2+a3+2a41)(a1+a3+2a41)+(a4+2a31)(a2+a4+2a31)+(a3+2a41)(a2+a3+2a41)=2a20.

Hence, Moe(G1)Moe(G) and equality holds if and only if a2=0, i.e., GG1.

Let G2 be the graph obtained from G1 by shifting a4 pendent edges from v4 to v3. We have Moe(G2)Moe(G1)=(a1+22)(a1+22)+(a3+a4+1a12)(a3+1a1a42)+(a1+a3+a4+21)(a1+a3+2a41)+(a3+a4+12)(a3+1a42)+(a3+a4+21)(a3+2a41)=8a40.

Hence, Moe(G2)Moe(G1) and equality holds if and only if a4=0, i.e., G2G1.

Let G3 be the graph obtained from G2 by shifting a1 pendent edges from v1 to v3. We have Moe(G3)Moe(G2)=(a1+22)+(a1+a3+12)(a3+1a12)+(a1+a3+21)(a1+a3+21)+(a1+a3+12)(a3+12)+(a1+a3+21)(a3+21)=3a10.

Hence, Moe(G3)Moe(G2) and equality holds if and only if a1=0, i.e., G3G2. Clearly, G2Bm3 with a3=0, and G3Bm4. Also, by simple calculation, we have Moe(Bm3)=m23m6, Moe(Bm4)=m22m15◻

Lemma 3.3.

Let GGm2 with brace Θ(2,2,2). If m6, then Moe(G)m2m28 with equality if and only if GBm5 (see Figure 2).

Proof. Let GGm2 with brace Θ(2,2,2). Let v1,v2,v3,v4,v5 be vertices in Θ(2,2,2) of G with dG(v1)=dG(v2)=3 and dG(v3)=dG(v4)=dG(v5)=2. Let ai be the number of pendent edges of vi (i=1,2,3,4,5). Suppose that a1a2 and a3a4a5. Let G1 be the graph obtained from G by shifting a2 pendent edges from v2 to v1. We have Moe(G1)Moe(G)=(a1+a2+a4+a5+2a31)(a1+a4+a5+2a2a31)+(a1+a2+a3+a5+2a41)(a1+a3+a5+2a2a41)+(a1+a2+a3+a4+2a51)(a1+a3+a4+2a2a51)+(a1+a2+a3+1a4a52)(a1+a3+1a2a4a52)+(a1+a2+a4+1a3a52)(a1+a4+1a2a3a52)+(a1+a2+a5+1a3a42)(a1+a5+1a2a3a42)=12a20.

Hence, Moe(G1)Moe(G) and equality holds if and only if a2=0, i.e., GG1.

Let G2 be the graph obtained from G1 by shifting a5 pendent edges from v5 to v3. We get Moe(G2)Moe(G1)=(a3+a5+1a1a42)(a3+1a1a4a52)+(a1+a3+a5+2a41)(a1+a3+a5+2a41)+(a1+a3+a4+a5+21)(a1+a3+a4+2a51)+(a1+a3+a5+1a42)(a1+a3+1a4a52)+(a3+a5+2a1a41)(a3+a5+2a1a41)+(a3+a4+a5+2a11)(a3+a4+2a1a51)=8a50.

Hence, Moe(G2)Moe(G1) and equality holds if and only if a5=0, i.e., G2G1.

Let G3 be the graph obtained from G2 by shifting a4 pendent edges from v4 to v3. By symmetry, Moe(G3)Moe(G2), and equality holds if and only if G3G2. Let G4 be the graph obtained from G3 by shifting a1 pendent edges from v1 to v3. We get Moe(G4)Moe(G3)=(a1+a3+12)(a3+1a12)+(a1+a3+21)(a1+a3+21)+(a1+a3+21)(a1+a3+21)+(a1+a3+12)(a1+a3+12)+(a1+a3+21)(a3+2a11)+(a1+a3+21)(a3+2a11)=6a10.

Hence, Moe(G4)Moe(G3) and equality holds if and only if a1=0, i.e., G4G3. Note that G4Bm5. Clearly, Moe(Bm5)=m2m28◻

Lemma 3.4.

Let GGm2 with brace Θ(1,2,3). If m6, then Moe(G)m22m16 with equality if and only if GBm6 (see Figure 2).

Proof. Let GGm2 with brace Θ(1,2,3). Let v1,v2,v3,v4,v5 be vertices in Θ(1,2,3) of G with dG(v1)=dG(v2)=3 and dG(v3)=dG(v4)=dG(v5)=2. Let ai be the number of pendent edges of vi (i=1,2,3,4,5). Suppose that a1+a4a2+a5 . Let G1 be the graph obtained from G by shifting a2 (resp. a5) pendent edges from v2 (resp. v5) to v1 (resp. v4). We deduce that Moe(G1)Moe(G)=(a1+a2+a4+a5+22)(a1+a4+2a2a52)+(a1+a2+a4+a5+3a31)(a1+a4+3a31)+(3a31)(a2+a5+3a31)+(a1+a2+a3+3a4a51)(a1+a2+a3+3a4a51)+(a1+a2+a3+3a4a51)(a1+a2+a3+3a4a51)+(a1+a2+a4+a5+22)(a1+a4+2a2a52)=4(a2+a5)0.

Hence, Moe(G1)Moe(G) and equality holds if and only if a2=a5=0, i.e., GG1.

Let G2 be the graph obtained from G1 by shifting a4 pendent edges from v4 to v1. We obtain Moe(G2)Moe(G1)=(a1+a4+22)(a1+a4+22)+(a1+a4+3a31)(a1+a4+3a31)+(3a31)(3a31)+(a1+a3+a4+31)(a1+a3+3a41)+(a1+a3+a4+31)(a1+a3+3a41)+(a1+a4+22)(a1+a4+22)=4a4+a1a30.

Hence, Moe(G2)Moe(G1) and equality holds if and only if a4=0,a1=a3, i.e., G2G1.

Let G3 be the graph obtained from G2 by shifting a3 pendent edges from v3 to v1. We have Moe(G3)Moe(G2)=(a1+a3+22)(a1+22)+(a1+a3+31)(a1+3a31)+(a1+a3+31)(a1+a3+31)+(a1+a3+22)(a1+22)+(31)(3a31)+(a1+a3+31)(a1+a3+31)=5a30.

Hence, Moe(G3)Moe(G2) and equality holds if and only if a3=0, i.e., G3G2. Note that G3Bm6. Clearly, Moe(Bm6)=m22m16◻

Lemma 3.5.

Let GGm2 with brace Θ(a,b,c). Then Moe(G){4,if m=5, and equality holds iff GBm3,Bm4;m23m6,if 6m8, and equality holdsiff GBm3;48,if m=9, and equality holds iff GBm3,Bm4;m22m15,if 10m12, and equalityholds iff GBm4;128,if m=13, and equality holds iff GBm4,Bm5;m2m28,if m14, and equality holds iff GBm5.

Proof. Suppose that x and y are the vertices with degree 3 in Θ(a,b,c) of G. Let Pa+1,Pb+1,Pc+1 be the three disjoint paths connecting x and y. We now proceed by considering the following three possible cases.

Case 1. cba3.

Subcase 1.1. c=b=a3.

We choose six edges such that each one is incident to x or y in Θ(a,b,c). Let e be one of these six edges. Then ψ(e)m7. This property holds for the remaining five edges as well. Therefore, Moe(G)6(m7)+(m6)(m1)<m2m28.

Subcase 1.2. cb4,a=3.

Let e1 be one of the four edges incident to x or y in the paths Pb+1 and Pc+1, and e2 be one of the two edges incident to x or y in the path Pa+1. Then ψ(e1)m8, and ψ(e2)m9. Hence, Moe(G)4(m8)+2(m9)+(m6)(m1)<m2m28.

Case 2. cba=2.

Subcase 2.1. c=b=a=2.

The Subcase follows from Lemma 3.3.

Subcase 2.2. c3,b=a=2.

Let e1 be one of the four edges in the paths Pa+1 and Pb+1, and e2 be one of the two edges incident to x or y in the path Pc+1. Then ψ(e1)m7, and ψ(e2)m6. Hence, Moe(G)4(m7)+2(m6)+(m6)(m1)<m2m28.

Subcase 2.3. cb3,a=2.

Let e1 be one of the four edges incident to x or y in the paths Pb+1 and Pc+1, and e2 is an edge in the path Pa+1. Then ψ(e1)m6, and ψ(e2)m9. Hence, Moe(G)4(m6)+2(m9)+(m6)(m1)<m2m28.

Case 3. cb2a=1.

Subcase 3.1. c=b=3,a=1.

By Lemma 3.2 and simple calculation, we have if m14, then Moe(Bm5)>Moe(Bm4); if m=13, then Moe(Bm3)=Moe(Bm4)>Moe(Bm5); if 10m12, then Moe(Bm4)>Moe(Bm5); if m=9, then Moe(Bm3)=Moe(Bm4)>Moe(Bm5); if 6m8, Moe(Bm3)>Moe(Bm4)>Moe(Bm5); if m=5, then Moe(Bm3)=Moe(Bm4).

Subcase 3.2. c=3,b=2,a=1.

This Subcase follows from Lemma 3.4.

Subcase 3.3. c4,b=2,a=1.

Let e1=xy, e2 be one of the four edges in the paths Pb+1 and Pc+1 that are incident to x or y, and e3 be one of the two edges in the middle of the path Pc+1. Then ψ(e1)=0, ψ(e2)m5, and ψ(e3)m6. Thus, Moe(G)4(m5)+2(m6)+(m7)(m1)<m2m28.

Subcase 3.4. c4,b=3,a=1.

Let e1=xy, e2 be one of the two edges in the path Pb+1 that are incident to x or y, e3 be one of the two edges in the path Pc+1 that are incident to x or y, and e4 be one of the two edges in the middle of the path Pc+1. Then ψ(e1)=0, ψ(e2)m4, ψ(e2)m5, and ψ(e4)m6. Hence, Moe(G)2(m4)+2(m5)+2(m6)+(m7)(m1)<m2m28. ◻

The proof of Theorem 1.2 directly follows from Lemmas 3.1 and 3.5.

Acknowledgements

The authors express their gratitude to the anonymous referee for their insightful comments and helpful suggestions. This work was supported by the National Natural Science Foundation of China (Grant Nos. 12071194, 11571155 and 12071158).

Declarations

The authors declare no conflict of interest.

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