# Mostar Index of Cycle-Related Structures

Authors: Fatima Asif; Agha Kashif; Sohail Zafar; Michael Onyango Ojiema
Journal: Journal of Chemistry
 (2022)

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## Abstract

A topological index is a numerical quantity associated with the molecular structure of a chemical compound. This number remains fixed with respect to the symmetry of a molecular graph. Diverse research studies have shown that the topological indices of symmetrical graphs are interrelated with several physiochemical properties such as boiling point, density, and heat of formation. Peripherality is also an important tool to study topological aspects of molecular graphs. Recently, a bond-additive topological index called the Mostar index that measures the peripherality of a graph is investigated which attained wide attention of researchers. In this article, we compute the Mostar index of cycle-related structures such as the Jahangir graph and the cycle graph with chord.


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## Body

## 1. Introduction and Preliminaries

Graph theory is being extensively used in mathematical chemistry for the numerical formulation of chemical compounds by representing atoms as vertices and bonds as edges. The topological index (TI) of a molecular graph is a numerical quantity associated with the molecular structure of a chemical compound [1, 2]. These quantities are well correlated with physicochemical properties and are used as a tool to predict quantitative structure-activity relationships and quantitative structure-property relationships (QSAR/QSPR) [3, 4]. QSAR and QSPR techniques have been widely used to study the structural properties of a molecule and its biological activity [5, 6]. The TIs are majorly classified into two types, namely, degree-based and distance-based or bond-additive topological indices. The degree-based TIs focus on the role of incident bonds towards the molecular structure [7, 8] whereas the distance-based TIs emphasize on the contribution of distances between atoms towards the structure of a compound [9]. The introduction of the first distance-based TI by Wiener in [10] pointed its significance towards the physicochemical properties of the compound. The physicochemical properties such as boiling point and melting point were shown in correlation with the Wiener index [11]. Afterwards, distance-based TIs such as the Hosaya, Shultz, and Szeged indexes were introduced [12, 13]. The bond-additive TIs have several applications; for example, they are useful in pharmaceutical sciences, in the prediction of physical properties of a molecule, and in complex network theory [14, 15]. To study further topological properties of various chemical compounds, we refer to [16, 17].In this article, we will consider a simple and finite graphG with a vertex set VG and an edge set EG. The degree of any vertex x, denoted by dx, is the number of its incident vertices. The distance between any two vertices x, y∈VG, denoted by dx,y, is the length of a shortest path connecting them. For any edge xy∈EG, the collection of all vertices that are nearer to x than y will be called the resolving neighborhood of x,y, denoted by ηx,y, and its cardinality is denoted by ηxy.Recently, to measure the peripherality of graphs, Doslic et al. proposed a new bond-additive topological index called the Mostar index [18]. It is defined as(1)MoG=∑xy∈EGnx−ny.The Mostar index measures peripheral atoms and bonds to determine the physical and chemical properties of a molecular graph. Each edge (bond) is peripheral if there are an unequal number of vertices (atoms) in each neighborhood of its end vertices. In chemistry, peripherality is used to measure the nonbalancedness among the bonds of a chemical graph. Recently, many researchers took the initiative to investigate the chemical properties and mathematical perspectives of the Mostar index. Arockiaraj et al. defined the edge Mostar index and computed the Mostar index and edge Mostar index of carbon nanostructures [19]. Hayat and Zhou determined the maximum Mostar index of all n− vertex cacti and upper bound for the Mostar index of n− vertex cacti that contains k− cycles [20]. Huang et al. computed the Mostar index to find the extremal hexagonal chains [21]. The conjecture about the bicyclic graphs characterized by Doslic et al. has been proved by Tepeh [22].The Mostar index of a graphG can be defined as(2)MoG=∑xy∈EGMoxy,where(3)Moxy=ηxy−ηyx.It is an NP-hard problem to derive general formulas for the Mostar index; therefore, Doslic et al. computed the Mostar index for different classes of graphs. Further, topological properties of some interesting symmetrical structures are computed in the articles [23, 24]. This motivates us to compute the Mostar index of cycle-related graphs such as the Jahangir graph and the cycle graph with chord.


### 1.1. Main Results

The main results computed in this article are as follows.Theorem 1.
Consider the graphsJn,m and Cn,t; then,(1)
Forn,m≥5,(4)MoJn,m=m2n2+2m2n−mn2+m2−6mn−5m,ifn=odd,m2n2+2m2n−mn2+m2−5mn−m,ifn=even.(2)
Forn≥9 and t≥3,(5)MoCn,t=2t−2n−t,ifn=odd,2n−tt−2,ifn=even,t=odd,2n−tt−2−2,ifn=even,t=even.


## 1.1. Main Results

The main results computed in this article are as follows.Theorem 1.
Consider the graphsJn,m and Cn,t; then,(1)
Forn,m≥5,(4)MoJn,m=m2n2+2m2n−mn2+m2−6mn−5m,ifn=odd,m2n2+2m2n−mn2+m2−5mn−m,ifn=even.(2)
Forn≥9 and t≥3,(5)MoCn,t=2t−2n−t,ifn=odd,2n−tt−2,ifn=even,t=odd,2n−tt−2−2,ifn=even,t=even.


## 2. The Mostar Index of the Jahangir Graph

In this section, the Mostar index of the family of the Jahangir graphJn,m for n,m≥5 is computed. The Jahangir graph Jn,m consists of a cycle Cnm with one additional vertex which is adjacent to m vertices of Cnm at distance n to each other on Cnm as shown in Figure 1. The vertex set of Jn,m is VJn,m=u,w1,w2,…,wm∪v1,v2,…,vnm. The edge set of Jn,m is EJn,m=uwk|1≤k≤m∪w1v1,w1vnm,wkvk−1n,wkvk−1n+1|2≤k≤m∪vni+1vni+2,…,vni+n−1vni+n|0≤i≤m−1. The edge set of Jn,m is partitioned as follows.Figure 1
The Jahangir graphJ3,m.Forn=even,(6)E1Jn,m=uwk|1≤k≤m,E2Jn,m=wkvk−1n,wkvk−1n+1|2≤k≤m∪w1v1,w1vnm,E3Jn,m=vni+1vni+2,…,v2i+1n−4/2v2i+1n−2/2,v2i+1n+4/2v2i+1n+6/2−,…,vni+n−1vni+n|0≤i≤m−1,E4Jn,m=v2i+1n−2/2v2i+1n/2,v2i+1n+2/2v2i+1n+4/2|0≤i≤m−1,E5Jn,m=v2i+1n/2v2i+1n+2/2|0≤i≤m−1.Forn=odd,(7)E1′Jn,m=uwk|1≤k≤m,E2′Jn,m=wkvk−1n,wkvk−1n+1|2≤k≤m∪w1v1,w1vnm,E3′Jn,m=vni+1vni+2,…,v2i+1n−3/2v2i+1n−1/2,v2i+1n+3/2v2i+1n+5/2−,…,vni+n−1vni+n|0≤i≤m−1,E4′Jn,m=v2i+1n−1/2v2i+1n+1/2,v2i+1n+1/2v2i+1n+3/2|0≤i≤m−1.In the following lemmas, the Mostar index corresponding to equations (6) and (7) of Jn,m is computed.Lemma 1.
Forxy∈EJn,m where n is even and n,m≥5, then(8)Moxy=mn−2n+m−3,ifxy∈E1Jn,m,mn+m−n−1,ifxy∈E2Jn,m,mn−n+m−2,ifxy∈E3Jn,m,2mn+2m−3n−42,ifxy∈E4Jn,m,0,ifxy∈E5Jn,m.Proof.
The symmetry of graphJn,m allows us to discuss the Mostar index of only the following types of edges:Case (i) (forxy∈E1Jm,n): let x=u and y=w1, the resolving neighborhoods of u,w1 and w1,u are u,w2,…,wm,vn+4/2,…,vmn−n+2/2 and w1,v1,…,vn/2,vmn−n−2/2−,…,vmn, respectively. Hence, ηuw1=mn−n+m−2 and ηw1u=n+1. Therefore, in view of equation (3), Mouw1=mn−2n+m−3.Case (ii) (forxy∈E2Jm,n): let x=v1 and y=w1, by means of Figure 1, ηv1,w1=v1,…,vn+2/2 and ηw1,v1=u,w1,…,wm,vn+5/2,…,vmn. Hence, ηv1w1=n+2/2 and ηw1v1=2mn−n+2m−1/2. Therefore, using equation (3), Mov1w1=mn+m−n−1.Case (iii) (forxy∈E3Jm,n): let x=vi and y=vi+1 for 1≤i≤n−3/2, the resolving neighborhoods of vi,vi+1 and vi,vi+1 are u,w1,…,wm,v1,…,vi,vn+2i+6/2,…,vmn and vi+1,…,vn+2i+2/2, respectively. Hence, ηvivi+1=2mn−n+2m−2/2 and ηvi+1vi=n+2/2. Therefore, using equation (3), Movivi+1=mn−n+m−2.Case (iv) (forxy∈E4Jm,n): let x=vn−2/2 and y=vn/2, by means of Figure 1, we obtain ηvn−2/2,vn/2=u,w1,w3,…,wm,v1,…,vn−2/2,v3n+2/2,…,vmn and ηvn/2,vn−2/2=vn/2,…,vn. Therefore, ηvn−2/2vn/2=mn−n+m−1 and ηvn/2vn−2/2=n+2/2. Hence, using equation (3), Movn−2/2vn/2=2mn+2m−3n−4/2.Case (v) (forxy∈E5Jm,n): let x=vn/2 and y=vn+2/2, the resolving neighborhoods of vn/2,vn+2/2 and vn+2/2,vn/2 are w1,v1,…,vn/2,vn2m−1/2…,vmn and w2,vn+2/2,…,v3n+2/2, respectively. Hence, ηvn/2vn+2/2=n+2 and ηvn+2/2vn/2=n+2. Therefore, in view of equation (3), Movn+2/2vn/2=0.Lemma 2.
Forxy∈EJn,m where n is odd and n,m≥5, then(9)Moxy=mn−2n+m−3,ifxy∈E1′Jn,m,mn−n+m−2,ifxy∈E2′Jn,m,mn−n+m−2,ifxy∈E3′Jn,m,mn−2n+m−3,ifxy∈E4′Jn,m.Proof.
The symmetry of the Jahangir graphJn,m allows us to compute the Mostar index of the following type of edges:Case (i) (forxy∈E1Jm,n): let x=u and y=w1, the resolving neighborhoods of u,w1 and w1,u are u,w2,…,wm,vn+3/2,…vmn−n+1/2 and w1,v1,…,vn+1/2,vmn−n−1/2−,…,vmn, respectively. Hence, ηuw1=mn−n+m−1 and ηw1u=n+2. Therefore, applying equation (3), Mouw1=mn−2n+m−3.Case (ii) (forxy∈E2Jm,n): let x=v1 and y=w1, by means of Figure 1, ηv1,w1=v1,…,vn+3/2 and ηw1,v1=u,w1,…,wm,vn+5/2,…,vmn. Therefore, ηv1w1=n+3/2 and ηw1v1=2mn−n+2m−1/2. Hence, using equation (3), Mov1w1=mn+m−n−2.Case (iii) (forxy∈E3Jm,n): let x=vi and y=vi+1 for 1≤i≤n−4/2, the resolving neighborhoods of vi,vi+1 and vi+1,vi are u,w1,…,wm,v1,…,vi,vn+2i+5/2…,vmn and vi+1,…,vn+2i+1/2, respectively. Hence, ηvivi+1=2mn−n+2m−1/2 and ηvi+1vi=n+3/2. Therefore, using equation (3), Movivi+1=mn−n+m−2.Case (iv) (forxy∈E4Jm,n): let x=vn−1/2 and y=vn+1/2, the resolving neighborhoods of vn−1/2,vn+1/2 and ηvn+1/2,vn−1/2 are u,w1,w3,…,wm,v1,…,vn−1/2,v3n+1/2,…,vmn and w1,vn+1/2,…,v3n+1/2, respectively. Therefore, ηvn−1/2vn+1/2=mn−n+m−1 and ηvn+1/2vn−1/2=n+2. Hence, in view of equation (3), Movn−1/2vn+1/2=mn−2n+m−3.In the following theorem, the Mostar index of the Jahangir graphJn,m for n,m≥5 is computed.Theorem 2.
Forxy∈EJn,m, then(10)MoJn,m=m2n2+2m2n−mn2+m2−5mn−m,ifn=even,m2n2+2m2n−mn2+m2−6mn−5m,ifn=odd.Proof.
Case (i)n=even: in view of Lemma 1 and equation (2), we have(11)MoJn,m=∑xy∈E1Jn,mnxy−nyx+∑xy∈E2Jn,mnxy−nyx+∑xy∈E3Jn,mnxy−nyx+∑xy∈E4Jn,mnxy−nyx+∑xy∈E5Jn,mnxy−nyx=mmn−2n+m−3+2mmn+m−n−1−+mn−4mn−n+m−2+m2mn+2m−3n−4+0,=m2n2+2m2n−mn2+m2−5mn−m.Case (ii)n=odd: it is easy to see from Lemma 2 and equation (2) that(12)MoJn,m=∑xy∈E1′Jn,mnxy−nyx+∑xyE2′Jn,mnxy−nyx+∑xy∈E3′Jn,mnxy−nyx+∑xy∈E4′Jn,mnxy−nyx.=mmn−2n+m−3+2mmn+m−n−2+mn−3mn−n+m−2+2mmn−2n+m−3,=m2n2+2m2n−mn2+m2−6mn−5m.Here, we present a numerical example in support of the above theorem.Example 1.
Consider the Jahangir graphJ5,5; then, the sets E1′J5,5=uw1,uw2,…,uw5, E2′J5,5=w1v25,w2v5,…,w5v21, E3′J5,5=v1v2,v4v5,…,v24v25, and E4′J5,5=v2v3,v3v4,…,v23v24 form the edge partition of J5,5. Now for respective edges xy, st, uv, and pq in these partitions, Lemma 1 implies that ηxy=24, ηyx=7, and Moxy=17; ηst=27, ηts=4, and Most=23; ηuv=27, ηvu=4, and Mouv=23; ηpq=24, ηqp=7, and Mopq=17. Therefore, using Theorem 2, MoJ5,5=5×17+2×5×23+5×5−4×23+2×5×17=600.


## 3. The Mostar Index of Cycle with a Chord Graph

In this section, the Mostar index of a cycle with a chord graphCn,t is computed. The graph Cn,t is obtained from a cycle Cn by joining its two vertices at a distance t−1 as shown in Figure 2. The vertex set and edge set of Cn,t are VCn,t=v1,v2,…,vn and ECn,t=vivi+1|1≤i≤n∪v1vt|3≤t≤n+3/2, respectively, where subscripts are taken as mod n.Figure 2
The cycle with a chord graphCn,7.The edge set ofCn,t is partitioned as follows.Forn=odd and t=even,(13)E11Cn,t=vt/2vt+2/2,E21Cn,t=vivi+1|1≤i≤vt−2/2,vt+2/2≤i≤t−1,E31Cn,t=vn+t−1/2vn+t+1/2,vn+t+1/2vn+t+3/2,E41Cn,t=vn+t−3/2vn+t−1/2,vn+t+3/2vn+t+5/2,E51Cn,t=vivi+1|t≤i≤n+t−52,n+t+52≤i≤n,E61Cn,t=v1vt.Forn=odd, t=odd:(14)E12Cn,t=vt/2vt+2/2,E22Cn,t=vivi+1|1≤i≤t−32,t+32≤i≤t−1,E32Cn,t=vn+t/2vn+t+2/2,E42Cn,t=vn+t−2/2vn+t/2,vn+t+2/2vn+t+4/2,E52Cn,t=vivi+1|t≤i≤n+t−42,n+t+42≤i≤n,E62Cn,t=v1vt.Forn=even and t=even,(15)E13Cn,t=vt/2vt+2/2,E23Cn,t=vivi+1|1≤i≤t−22,t+22≤i≤t−1,E33Cn,t=vn+t/2vn+t+2/2,E43Cn,t=vn+t−2/2vn+t/2,vn+t/2vn+t+2/2,E53Cn,t=vivi+1|t≤i≤n+t−42,n+t+22≤i≤n,E63Cn,t=v1vt.Forn=even and t=odd,(16)E14Cn,t=vt−1/2vt+1/2,vt+1/2vt+3/2,E24Cn,t=vivi+1|1≤i≤t−32,t+32≤i≤t−1,E34Cn,t=vn+t−1/2vn+t+1/2,vn+t+1/2vn+t+3/2,E44Cn,t=vn+t−3/2vn+t−1/2,vn+t+3/2vn+t+5/2,E54Cn,t=vivi+1|t≤i≤n+t−42,n+t+22≤i≤n,E64Cn,t=v1vt.\In the following lemmas, the Mostar index corresponding to equations (13)–(16) of Cn,t for n≥9 and t≥3 is computed.Lemma 3.
Letxy∈ECn,t where n is odd and t is even and 4≤t<n+3/2, then(17)Moxy=0,ifxy∈E11Cn,t,n−t,ifxy∈E21Cn,t,t−22,ifxy∈E31Cn,t,t−2,ifxy∈E41Cn,t,t−2,ifxy∈E51Cn,t,0,ifxy∈E61Cn,t.Proof.
Case (i) (forxy∈E11Cn,t): let x=vt/2 and y=vt+2/2, the resolving neighborhoods of vt/2,vt+2/2 and vt+2/2,vt/2 are v1,…,vt/2,vn+t+3/2,…,vn and vt+2/2,…,vn+t−1/2, respectively. Hence, ηvt/2vt+2/2=ηvt+2/2vt/2=n−1/2. Therefore, in view of equation (3), Movt/2vt+2/2=0.Case (ii) (forxy∈E21Cn,t): let x=vi and y=vi+1 for 1≤i≤t−2/2, by means of Figure 2, ηvi,vi+1=v1,…,vi,vt+2i+2/2,…,vn and ηvi+1,vi=vi+1,…,vt+2i/2. Hence, ηvivi+1=2n−t/2 and ηvi+1vi=t/2. Therefore, equation (3) implies Movivi+1=n−t.Case (iii) (forxy∈E31Cn,t): let x=vn+t−1/2 and y=vn+t+1/2, then ηvn+t−1/2,vn+t+1/2=vt+2/2,…,vn+t−1/2 and ηvn+t+1/2,vn+t−1/2=vn+t−1/2,…,vn. Hence, ηvn+t−1/2vn+t+1/2=n−1/2 and ηvn+t+1/2vn+t−1/2=n−t+1/2. Therefore, applying equation (3), Movn+t−1/2vn+t+1/2=t−2/2.Case (iv) (forxy∈E41Cn,t): let x=vn+t−3/2 and y=vn+t−1/2, the resolving neighborhoods of vn+t−3/2,vn+t−1/2 and vn+t−1/2,vn+t−3/2 are v1,…,vn+t−3/2 and vn+t−1/2,…,vn−1, respectively. Therefore, ηvn+t−3/2vn+t−1/2=n+t−3/2 and ηvn+t−1/2vn+t−3/2=n+t−1/2. Hence, using equation (3), Movn+t−3/2vn+t−1/2=t−2.Case (v) (forxy∈E51Cn,t): let x=vi and y=vi+1 for t≤i≤n+t−5/2, the resolving neighborhoods of vi,vi+1 and vi+1,vi are v1,…,vi,vn+t+2i+4/2,…,vn and vi+1,…,vn−t+2i+2/2, respectively. Hence, ηvivi+1=n+t−2/2 and ηvi+1vi=n−t+2/2. Therefore, in view of equation (3), Movivi+1=t−2.Case (vi) (forxy∈E61Cn,t): let x=v1 and y=vt, then we have ηv1,vt=v1,…,vt/2,vn+t+3/2,…,vn and ηvt,v1=vt+2/2,…,vn+t−1/2. Hence, ηv1vt=ηvtv1=n−1/2. Therefore, using equation (3), Mov1vt=0.Lemma 4.
Letxy∈ECn,t where n and t are odd and 3≤t≤n+1/2, then(18)Moxy=n−t2,ifxy∈E12Cn,t,n−t,ifxy∈E22Cn,t,0,ifxy∈E32Cn,t,t−2,ifxy∈E42Cn,t,t−2,ifxy∈E52Cn,t,0,ifxy∈E62Cn,t.Proof.
Case (i) (forxy∈E12Cn,t): let x=vt−1/2 and y=vt−1/2, the resolving neighborhoods of vt−1/2,vt+1/2 and vt+1/2,vt−1/2 are v1,…,vt−1/2,vn+t+2/2,…,vn and vt−1/2,…,vt−1, respectively. Hence, ηvt−1/2vt+1/2=n−1/2 and ηvt+1/2vt−1/2=t−1/2. Therefore, using equation (3), Movt−1/2vt+1/2=n−t/2.Case (ii) (forxy∈E22Cn,t): let x=vi and y=vi+1 for 1≤i≤t−3/2, by means of Figure 2, ηvi,vi+1=v1,…,vi,vt+2i+3/2,…,vn and ηvi+1,vi=vi+1,…,vt+2i−1/2. Hence, ηvivi+1=2n−t−1/2 and ηvi+1vi=t−1/2. Therefore, in view of equation (3), Movivi+1=n−t.Case (iii) (forxy∈E32Cn,t): let x=n+t/2 and y=vn+t+2/2, then ηvn+t/2,vn+t+2/2=v1,…,vt−1/2,vn+t+2/2,…,vn and ηvn+t+2/2,vn+t/2=vt+3/2,…,vn+t/2. Hence, ηvn+t/2vn+t+2/2=n−1/2 and ηvn+t+2/2vn+t/2=n−1/2. Hence, equation (3) implies Movn+t/2vn+t+2/2=0.Case (iv) (forxy∈E42Cn,t): let x=vn+t−2/2 and y=vn+t/2, the resolving neighborhoods of vn+t−2/2,vn+t/2 and vn+t/2,vn+t−2/2 are v1,…,vn+t−2/2 and vn+t/2,…,vn, respectively. Therefore, ηvn+t−2/2vn+t/2=n+t−2/2 and ηvn+t/2vn+t−2/2=n−t+2/2. Henceforth, applying equation (3), Movn+t−2/2vn+t/2=t−2.Case (v) (forxy∈E52Cn,t): let x=vi and y=vi+1 where t≤i≤n+t−4/2, the resolving neighborhoods of vivi+1 and vi+1vi are v1,…,vi,vn−t+2i+4/2,…,vn and vi+1,…,vn−t+2i+2/2, respectively. Hence, ηvivi+1=n+t−2/2 and ηvi+1vi=n−t+2/2. Therefore, using equation (3), Movivi+1=t−2.Case (vi) (forxy∈E62Cn,t): let x=v1, y=vt, the resolving neighborhoods of v1,vt and vt,v1 are v1,…,vt−1/2,vn+t+2/2,…,vn and vt+3/2,…,vn+t/2 respectively. Hence, ηv1vt=n−1/2 and ηvtv1=n−1/2. Therefore, equation (3) implies Mov1vt=0.Lemma 5.
Letxy∈ECn,t where n and t are even and 4≤t≤n+2/2, then(19)Moxy=0,ifxy∈E13Cn,t,n−t,ifxy∈E23Cn,t,0,ifxy∈E33Cn,t,t−2,ifxy∈E43Cn,t,t−2,ifxy∈E53Cn,t,0,ifxy∈E63Cn,t.Proof.
Case (i) (forxy∈E13Cn,t): let x=vt/2 and y=vt+2/2, the resolving neighborhoods of vt/2,vt+1/2 and vt+1/2,vt/2 are v1,…,vt/2,vn+t+2/2,…,vn and vt+2/2,…,vn+t/2, respectively. Hence, ηvt/2vt+1/2=n/2 and ηvt+1/2vt/2=n/2. Therefore, using equation (3), Movt/2vt+2/2=0.Case (ii) (forxy∈E23Cn,t): let x=vi and y=vi+1 for 1≤i≤t−2/2, the resolving neighborhoods of vi,vi+1 and vi+1,vi are v1,…,vi,vt+2i+2/2,…,vn and vi+1,…,vt+2i/2, respectively. Therefore, ηvivi+1=2n−t/2 and ηvi+1,vi=t/2. Hence, equation (3) implies Movivi+1=n−t.Case (iii) (forxy∈E33Cn,t): let x=vn+t/2 and y=vn+t+2/2, by means of Figure 2, ηvn+t/2,vn+t+2/2=vt+2/2,…,vn+t/2 and ηvn+t+2/2,vn+t/2=v1,…,vt/2,vn+t+2/2,…,vn. Therefore, ηvn+t/2vn+t+2/2=n/2 and ηvn+t+2/2vn+t/2=n/2. Hence, applying equation (3), Movn+t/2vn+t+2/2=0.Case (iv) (forxy∈E43Cn,t): let x=vn+t−2/2 and y=vn+t/2, then ηvn+t−2/2,vn+t/2=v1,…,vn+t−2/2 and ηvn+t/2,vn+t−2/2=vn+t/2,…,vn. Hence, ηvn+t−2/2vn+t/2=n+t−2/2 and ηvn+t/2vn+t−2/2=n−t+2/2. Therefore, in view of equation (3), Movn+t−2/2vn+t/2=t−2.Case (v) (forxy∈E53Cn,t): let x=vi and y=vi+1 for t≤i≤n+t−4/2, the resolving neighborhoods of vi,vi+1 and vi+1,vi are v1,…,vi,vn−t+2i+4/2,…,vn and vi+1,…,vn−t+2i+2/2, respectively. Hence, ηvivi+1=n+t−2/2 and ηvi+1vi=n−t+2/2. Therefore, Movivi+1=t−2.Case (vi) (forxy∈E63Cn,t): let x=v1 and y=vt, then resolving neighborhoods are ηv1,vt=v1,…,vt/2,vn+t+2/2,…,vn and ηvt,v1=vt+2/2,…,vn+t/2. Hence, ηv1vt=n/2 and ηvtv1=n/2. Therefore, using equation (3), Mov1vt=0.Lemma 6.
Letxy∈ECn,t where n is even and t is odd and 3≤t≤n+2/2, then(20)Moxy=n−t+12,ifxy∈E14Cn,t,n−t,ifxy∈E24Cn,t,t−12,ifxy∈E34Cn,t,t−2,ifxy∈E44Cn,t,n−t−3,ifxy∈E54Cn,t,0,ifxy∈E64Cn,t.Proof.
Case (i) (forxy∈E14Cn,t): let x=vt−1/2 and y=vt+1/2, the resolving neighborhoods of vt−1/2,vt+1/2 and vt+1/2,vt−1/2 are v1,…,vt−1/2,vn+t+1/2…,vn and vt+1/2,…,vt−1, respectively. Hence, ηvt−1/2vt+1/2=n/2 and ηvt+1/2vt−1/2=t−1/2. Therefore, using equation (3), Movt−1/2vt+1/2=2n−t+1/2=n−t+1/2.Case (ii) (forxy∈E24Cn,t): let x=vi and y=vi+1 for 1≤i≤t−3/2, by means of Figure 2, ηvi,vi+1=v1,…,vi,vt+2i+3/2,…,vn and ηvi+1,vi=vi+1,…,vt+2i−1/2. Hence, ηvivi+1=2n−t−1/2 and ηvi+1vi=t−1/2. Therefore, in view of equation (3), Movivi+1=n−t.Case (iii) (forxy∈E34Cn,t): let x=vn+t−1/2 and y=vn+t−1/2, the resolving neighborhoods of vn+t−1/2,vn+t+1/2 and vn+t−1/2,vn+t+1/2 are vt+1/2,…,vn+t−1/2 and vn+t+1/2,…,vn, respectively. Therefore, ηvn+t−1/2vn+t+1/2=n/2 and ηvn+t+1/2vn+t−1/2=n−t+1/2. Hence, equation (3) implies Movn+t−1/2vn+t+1/2=2t−1/2=t−1/2.Case (iv) (forxy∈E44Cn,t): let x=vn+t−3/2 and y=vn+t−1/2, we have ηvn+t−3/2,vn+t−1/2=v1,…,vn+t−3/2 and ηvn+t−1/2,vn+t−3/2=vn+t−1/2,…,vn−1. Hence, ηvn+t−3/2vn+t−1/2=n+t−3/2 and ηvn+t−1/2vn+t−3/2=n−t+1/2. Therefore, applying equation (3), Movn+t−3/2vn+t−1/2=t−2.Case (v) (forxy∈E54Cn,t): let x=vi and y=vi+1 where t≤i≤n+t−5/2, the resolving neighborhoods of vi,vi+1 and vi+1,vi are v1,…,vi,vn−t+2i+5/2,…,vn and vi+1,…,vn−t+2i+1/2, respectively. Hence, ηvivi+1=n+t−3/2 and ηvi+1vi=n+t−1/2. Therefore, using equation (3), Movivi+1=t−2.Case (vi) (Forxy∈E64Cn,t): Let x=v1 and y=vt, then resolving neighborhoods are ηv1,vt=v1,…,vt/2,vn+t+3/2,…,vn and ηvt,v1=vt+3/2,…,vn+t−1/2. Hence, ηv1vt=n−2/2 and ηvtv1=n−2/2. Therefore, in view of equation (3), Mov1vt=0.In the following theorem, the Mostar index of cycle with a chord graphCn,t is computed.Theorem 3.
Forxy∈ECn,t where n≥9 and t≥3, then(21)MoCn,t=2t−2n−t,ifn=odd,2n−tt−2,ifn=even,t=odd,2n−tt−2−2,ifn=even,t=even,Proof.
Case (i)n=odd,t=odd: in view of Lemma 3 and equation (2), we have(22)MoCn,t=∑xy∈E11Cn,tnxy−nyx+∑xyE21Cn,tnxy−nyx+∑xy∈E31Cn,tnxy−nyx+∑xy∈E41Cn,tnxy−nyx+∑xy∈E51Cn,tnxy−nyx+∑xy∈E61Cn,tnxy−nyx,=1×0+t−2×n−t+2×t−22+2×t−2+n−t−3×t−2+1×0,=2n−tt−2.Case (ii)n=odd,t=even: in view of Lemma 4 and equation (2), we have(23)MoCn,t=∑xy∈E12Cn,tnxy−nyx+∑xyE22Cn,tnxy−nyx+∑xy∈E32Cn,tnxy−nyx+∑xy∈E42Cn,tnxy−nyx+∑xy∈E52Cn,tnxy−nyx+∑xy∈E62Cn,tnxy−nyx,=2×n−t2+t−3×n−t+1×0+2×t−2+n−t−2×t−2+1×0,=2n−tt−2.Case (iii)n=even,t=odd: in view of Lemma 5 and equation (2), we have(24)MoCn,t=∑xy∈E13Cn,tnxy−nyx+∑xyE23Cn,tnxy−nyx+∑xy∈E33Cn,tnxy−nyx+∑xy∈E43Cn,tnxy−nyx+∑xy∈E53Cn,tnxy−nyx+∑xy∈E63Cn,tnxy−nyx,=1×0+t−2×n−t+2×t−22+2×t−2+n−t−3×t−2+1×0,=2n−tt−2.Case (iv)n=even,t=even: by using Lemma 6 and equation (2),(25)MoCn,t=∑xy∈E14Cn,tnxy−nyx+∑xyE24Cn,tnxy−nyx+∑xy∈E34Cn,tnxy−nyx+∑xy∈E44Cn,tnxy−nyx+∑xy∈E54Cn,tnxy−nyx+∑xy∈E64Cn,tnxy−nyx,=2×n−t+12+t−3×n−t+2×t−12+2×t−2+n−t−3×t−2+1×0,=2n−tt−2−2.Now, we present a numerical example in support of the above theorem.Examples 2.
Consider a cycle with a chord graphC14,7; then, the sets E14C14,7=v3v4,v4v5, E24C14,7=v1v2,v2v3,v5v6,v6v7, E34C14,7=v10v11,v11v12, E44C14,7=v9v10,v12v13, E54C14,7=v7v8,v8v9,v13v14,v14v1, and E64C14,7=v1v7 form the edge partition of C14,7. Now for respective edges xy, st, uv, pq, ef, and jk in these partitions, Lemma 6 implies that ηxy=7, ηyx=3, and Moxy=4; ηst=10, ηts=3, and Most=7; ηuv=7, ηvu=4, and Mouv=3; ηpq=9, ηqp=4, and Mopq=5; ηef=9, ηfe=4, and Moef=5; ηjk=6, ηkj= 6 and Mojk=0. Therefore, using Theorem 3, MoC14,7=1×0+4×5+2×5+2×3+4×7+2×4=70.


## 4. Conclusion

In this article, by considering cycle-related graphs, we give explicit expressions of the Mostar index of the Jahangir graph and the cycle with a chord graph. The paper is concluded by the following open problems:Open Problem 1: compute the Mostar index of certain classes of cycle-related structures such as convex polytopesQn, Rn, and AnOpen Problem 2: compute the Mostar index of some rotationally symmetric graphs andSn



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