In this paper‎, ‎we present the upper bounds for the product version of reciprocal degree distance of the tensor product‎, ‎join and strong product of two graphs in terms of other graph invariants including the Harary index and Zagreb indices‎

K‎. ‎Pattabiraman

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CCOCommun. Comb. Optim.c© 2018 Azarbaijan Shahid Madani University. All rights reserved.Communications in Combinatorics and OptimizationVol. 3 No. 1, 2018 pp.25-35DOI: 10.22049/CCO.2017.26018.1067Product version of reciprocal degree distance ofcomposite graphsK. Pattabiraman11Department of Mathematics, Annamalai University, Annamalainagar 608 002, Indiapramank@gmail.comReceived: 7 September 2017; Accepted: 21 October 2017Published Online: 25 October 2017Communicated by Ivan GutmanAbstract: In this paper, we present the upper bounds for the product version ofreciprocal degree distance of the tensor product, join and strong product of two graphsin terms of other graph invariants including the Harary index and Zagreb indices.Keywords: Degree distance, reciprocal degree distance, composite graphAMS Subject classification: 05C12, 05C761. IntroductionAll the graphs considered in this paper are simple and connected. For vertices u, v ∈V (G), the distance between u and v in G, denoted by dG(u, v), is the length of ashortest (u, v)-path in G and let dG(v) be the degree of a vertex v ∈ V (G). Fortwo simple graphs G and H their tensor product, denoted by G × H, has vertexset V (G) × V (H) in which (g1, h1) and (g2, h2) are adjacent whenever g1g2 is anedge in G and h1h2 is an edge in H. Note that if G and H are connected graphs,then G × H is connected only if at least one of the graph is non-bipartite. Thestrong product of graphs G and H, denoted by G  H, is the graph with vertex setV (G) × V (H) = {(u, v) : u ∈ V (G), v ∈ V (H)} and (u, x)(v, y) is an edge whenever(i) u = v and xy ∈ E(H), or (ii) uv ∈ E(G) and x = y, or (iii) uv ∈ E(G) andxy ∈ E(H). The join G + H of graphs G and H is obtained from the disjoint unionof the graphs G and H, where each vertex of G is adjacent to each vertex of H.A topological index of a graph is a real number related to the graph; it does not depend26 Product version of reciprocal degree distance of composite graphson labeling or pictorial representation of a graph. In theoretical chemistry, molecularstructure descriptors (also called topological indices) are used for modeling physic-ochemical, pharmacologic, toxicologic, biological and other properties of chemicalcompounds. There exist several types of such indices, especially those based on ver-tex and edge distances. One of the most intensively studied topological indices is theWiener index; for other related topological indices see [2].Let G be a connected graph. The Wiener index of G is defined as W (G) =12∑u, v ∈V (G)dG(u, v) where the summation goes over all pairs of distinct vertices ofG. Similarly, the Harary index of G is defined as H(G) = 12∑u, v ∈V (G)1dG(u,v). Gut-man et al. [7, 8] were introduced the product version of Wiener index as followsW ∗(G) =∏{u,v}⊆V (G)dG(u, v). Dobrynin and Kochetova [4] and Gutman [6] inde-pendently proposed a vertex-degree-weighted version of Wiener index called degreedistance or Schultz molecular topological index, which is defined for a connected graphG as DD(G) = 12∑u,v∈V (G)(dG(u) + dG(v))dG(u, v), where dG(u) is the degree of thevertex u in G. Note that the degree distance is a degree-weight version of the Wienerindex. Hua and Zhang [10] introduced a new graph invariant named reciprocal degreedistance, which can be seen as a degree-weight version of Harary index, that is,HA(G) =12∑u,v∈V (G), u 6=v(dG(u) + dG(v))dG(u, v).Hua and Zhang [10] have obtained lower and upper bounds for the reciprocal degreedistance of graph in terms of other graph invariants including the degree distance,Harary index, the first Zagreb index, the first Zagreb coindex, pendent vertices, in-dependence number, chromatic number and vertex and edge-connectivity. In thissequence, the product version of reciprocal degree distance is defined asH∗A(G) =∏{u,v}⊆V (G), u 6=vdG(u) + dG(v)dG(u, v).The first Zagreb index and second Zagerb index are defined asM1(G) =∑u∈V (G)dG(u)2 and M2(G) =∑uv∈E(G)dG(u)dG(v).Similarly, the first Zagreb coindex and second Zagerb coindex are defined asM1(G) =∑uv/∈E(G)(dG(u) + dG(v)) and M2(G) =∑uv/∈E(G)dG(u)dG(v).K. Pattabiraman 27The Zagreb indices are found to have applications in QSPR and QSAR studies aswell, see [3]. Various topological indices on tensor product, strong product have beenstudied by several authors [1, 5, 9, 11–15].In this paper, we present upper bounds for the product version of reciprocal degreedistance of the tensor product, join and strong product of two graphs in terms ofother graph invariants including the Harary index and Zagreb indices.2. Tensor productIn this section, we compute the product version of the reciprocal degree distance ofG×Kr.The proof of the following lemma follows easily from the properties and structure ofG×Kr. The lemma is used in the proof of the main theorem of this section.Lemma 1. Let G be a connected graph on n ≥ 2 vertices. For any pair of verticesxij , xkp ∈ V (G×Kr), r ≥ 3, i, k ∈ {1, 2, . . . , n} j, p ∈ {1, 2, . . . , r}. Then(i) If uiuk ∈ E(G), thendG×Kr (xij , xkp) =1 if j 6= p,2 if j = p and uiuk is on a triangle of G,3 if j = p and uiuk is not on a triangle of G.(ii) If uiuk /∈ E(G), then dG×Kr (xij , xkp) = dG(ui, uk).(iii) dG×Kr (xij , xip) = 2.Proof. Let V (G) = {u1, u2, . . . , un} and V (Kr) = {v1, v2, . . . , vr}. Let xij denotethe vertex (ui, vj) of G×Kr. We only prove the case when uiuk /∈ E(G), i 6= k andj = p. The proofs for other cases are similar.We may assume j = 1. Let P = uius1us2 . . . uspuk be the shortest path oflength p + 1 between ui and uk in G. From P we have a (xi1, xk1)-path P1 =xi1xs12 . . . xsp−12xsp3xk1 if the length of P is odd, and P1 = xi1xs12 . . . xsp−12xsp2xk1if the length of P is even.Obviously, the length of P1 is p+ 1, and thus dG×Kr (xi1, xk1) ≤ p+ 1 ≤ dG(ui, uk). Ifthere were a (xi1, xk1)-path in G×Kr that is shorter than p+1 then it is easy to find a(ui, uk)-path in G that is also shorter than p+1 in contrast to dG(ui, uk) = p+1.Remark 1. (Arithmetic Geometric Inequality) Let a1, a2, . . . , an be non-negative num-bers. Then n√a1a2 . . . an ≤ a1+a2+...+ann .Theorem 1. Let G be a connected graph with n ≥ 2 vertices and m edges. ThenH∗A(G×Kr) ≤(r − 1)5nrmnrn3nr[HA(G)(HA(G)−M1(G)2− t)]nr,28 Product version of reciprocal degree distance of composite graphswhere t =∑uiuk∈E2dG(ui)+dG(uk)6and r ≥ 3.Proof. Set V (G) = {u1, u2, . . . , un} and V (Kr) = {v1, v2, . . . , vr}. Let xij denotethe vertex (ui, vj) of G×Kr. The degree of the vertex xij in G×Kr is dG(ui)dKr (vj),that is dG×Kr (xij) = (r − 1)dG(ui). By the definition of H∗A, we haveH∗A(G×Kr) =∏xij , xkp ∈V (G×Kr)dG×Kr (xij) + dG×Kr (xkp)dG×Kr (xij , xkp)=n−1∏i=0r−1∏j, p=0j 6= pdG×Kr (xij) + dG×Kr (xip)dG×Kr (xij , xip)×n−1∏i, k=0i 6= kr−1∏j =0dG×Kr (xij) + dG×Kr (xkj)dG×Kr (xij , xkj)×n−1∏i, k=0i 6= kr−1∏j, p=0j 6= pdG×Kr (xij) + dG×Kr (xkp)dG×Kr (xij , xkp). (1)We shall calculate the sums of (1) are separately.First we computen−1∏i=0r−1∏j, p=0j 6= pdG×Kr (xij)+dG×Kr (xip)dG×Kr (xij ,xip).n−1∏i=0r−1∏j, p=0j 6= pdG×Kr (xij) + dG×Kr (xip)dG×Kr (xij , xip)=n−1∏i=0r−1∏j, p=0j 6= p2(r − 1)dG(ui)2(by Lemma 1)≤[ 12 n−1∑i=0r−1∑j, p=0j 6= p(r − 1)dG(ui)nr]nr(by Rem. 1)=[2r(r − 1)2m2nr]nr=[ (r − 1)2mn]nr. (2)Next we computer−1∏j =0n−1∏i, k=0i 6= kdG×Kr (xij)+dG×Kr (xkj)dG×Kr (xij ,xkj). By Remark 1, we haver−1∏j =0n−1∏i, k=0i 6= kdG×Kr (xij) + dG×Kr (xkj)dG×Kr (xij , xkj)≤[ 12 r−1∑j =0 n−1∑i, k=0i 6= kdG×Kr (xij)+dG×Kr (xkj)dG×Kr (xij ,xkj)nr]nr=[ 12 r−1∑j =0Snr]nr. (3)K. Pattabiraman 29Now we compute S. For that, let E1 = {uv ∈ E(G) |uv is on a C3 in G} and E2 =E(G)− E1.S =( n−1∑i, k=0i 6= kuiuk /∈E(G)+n−1∑i, k=0i 6= kuiuk∈E1+n−1∑i, k=0i 6= kuiuk∈E2)(dG×Kr (xij) + dG×Kr (xkj)dG×Kr (xij , xkj))=(n−1∑i, k=0i 6= kuiuk /∈E(G)(r − 1)(dG(ui) + dG(uk))dG(ui, uk)+n−1∑i, k=0i 6= kuiuk∈E1(r − 1)(dG(ui) + dG(uk))2+n−1∑i, k=0i 6= kuiuk∈E2(r − 1)(dG(ui) + dG(uk))3)(by Lemma 1)= (r − 1){(n−1∑i, k=0i 6= kuiuk /∈E(G)dG(ui) + dG(uk)dG(ui, uk)+n−1∑i, k=0i 6= kuiuk∈E1dG(ui) + dG(uk)dG(ui, uk)+n−1∑i, k=0i 6= kuiuk∈E2dG(ui) + dG(uk)dG(ui, uk))−n−1∑i, k=0i 6= kuiuk∈E1dG(ui) + dG(uk)2− 2n−1∑i, k=0i 6= kuiuk∈E2dG(ui) + dG(uk)3}= (r − 1){2HA(G)−n−1∑i, k=0i 6= kuiuk∈E(G)dG(ui) + dG(uk)2−n−1∑i, k=0i 6= kuiuk∈E2dG(ui) + dG(uk)6}= (r − 1)2{2HA(G)−M1(G)−∑uiuk∈E2dG(ui) + dG(uk)3}. (4)Now summing (4) over j = 0, 1, . . . , r − 1, we get,r−1∑j =0S = r(r − 1)(2HA(G)−M1(G)−∑uiuk∈E2dG(ui) + dG(uk)3). (5)Hencer−1∏j =0n−1∏i, k =0i 6= kdG×Kr (xij) + dG×Kr (xkj)dG×Kr (xij, xkj)≤[ r(r−1)2 (2HA(G) −M1(G) − ∑uiuk∈E2dG(ui)+dG(uk)3)nr]nr=[ (r − 1)(HA(G) − M1(G)2 − ∑uiuk∈E2dG(ui)+dG(uk)6)n]nr. (6)Next we computen−1∏i, k=0i 6= kr−1∏j, p=0j 6= pdG×Kr (xij)+dG×Kr (xkp)dG×Kr (xij ,xkp). By Lemma 1 and Remark 1, we30 Product version of reciprocal degree distance of composite graphshaven−1∏i, k=0i 6= kr−1∏j, p=0,j 6= pdG×Kr (xij) + dG×Kr (xkp))dG×Kr (xij , xkp≤[ 12 n−1∑i, k=0i 6= kr−1∑j, p=0,j 6= p(r−1)(dG(ui)+dG(uk)dG(ui,uk)nr]nr,=[ r(r − 1)2HA(G)nr]nr=[ (r − 1)2HA(G)n]nr. (7)Using (1) and the sums A1,A2 and A3 in (2),(5) and (7), respectively, we have,H∗A(G×Kr) ≤(r − 1)5nrmnrn3nr[HA(G)(HA(G)−M1(G)2− t)]nrwhere t =∑uiuk∈E2dG(ui)+dG(uk)6 .Using Theorem 1, we have the following corollaries.Corollary 1. Let G be a connected graph on n ≥ 2 vertices with m edges. If each edgeof G is on a C3, thenH∗A(G×Kr) =≤(r − 1)5nrmnrn3nr[HA(G)(HA(G)−M1(G)2)]nrwhere r ≥ 3.For a triangle-free graph,∑uiuk∈E2dG(ui)dG(uk) = M2(G).Corollary 2. If G is a connected triangle free graph on n ≥ 2 vertices and m edges, thenH∗A(G×Kr) ≤(r − 1)5nrmnrn3nr[HA(G)(HA(G)−2M1(G)3)]nrwhere r ≥ 3.By direct calculations we obtain expressions for the values of the Harary indices of Knand Cn. H(Kn) =n(n−1)2 and H(Cn) = n( n2∑i=11i)− 1 when n is even, and n( n−12∑i=11i)otherwise. Similarly, HA(Kn) = n(n− 1)2 and HA(Cn) = 4H(Cn).Using Corollaries 1 and 2, we obtain the H∗A of the graphs Kn ×Kr and Cn ×Kr.Example 1. (i) H∗A(Kn ×Kr) ≤ (r−1)5nr(n−1)3nr2nr.(ii)H∗A(Cn ×Kr) ≤(r−1)15r(48)3r36r, if n = 3,(16)nr(r−1)5nrn2nr[(H(Cn))2 − 2nH(Cn)3]nr, if n > 3.K. Pattabiraman 313. Join of graphsIn this section, we compute the product version of reciprocal degree distance of joinof two graphs.Theorem 2. Let G1 be a graph of order n and size p and let G2 be a graph of order mand size q. Then H∗A(G1+G2) ≤[(M1(G1)+2mp)(M1(G2)+2nq)(M1(G1)+m(n(n−1)−2p))(M1(G2) + n(m(m− 1)− 2q))(2mp+ 2nq +mn(m+ n))]nm.Proof. Let V (G1) = {u1, u2, . . . , un} and V (G2) = {v1, v2, . . . , vm}. By definition ofthe join of two graphs, one can see that, dG1+G2 (x) ={dG1 (x) + |V (G2)| , if x ∈ V (G1)dG2 (x) + |V (G1)| , if x ∈ V (G2)anddG1+G2 (u, v) =0, if u = v1, if uv ∈ E(G1) or uv ∈ E(G2) or (u ∈ V (G1) and v ∈ V (G2))2, otherwise.Therefore,H∗A(G1 +G2) =∏{u,v}⊆V (G1+G2)dG1+G2 (u) + dG1+G2 (v)dG1+G2 (u, v)=∏uv∈E(G1)((dG1 (u) +m) + (dG1 (v) +m))×∏uv/∈E(G1)(dG1 (u) +m) + (dG1 (v) +m)2×∏uv∈E(G2)((dG2 (u) + n) + (dG2 (v) + n))×∏uv/∈E(G2)(dG2 (u) + n) + (dG2 (v) + n)2×∏u∈V (G1), v∈V (G2)((dG1 (u) +m) + (dG2 (v) + n))≤[ ∑uv∈E(G1)(dG1 (u) + dG1 (v) + 2m)nm]nm[ ∑uv/∈E(G1) dG1 (u)+dG1 (v)+2m2nm]nm[ ∑uv∈E(G2)(dG2 (u) + dG2 (v) + 2n)nm]nm[ ∑uv/∈E(G2) dG2 (u)+dG2 (v)+2n2nm]nm[ ∑u∈V (G1), v∈V (G2)(dG1 (u) + dG2 (v) + (m+ n))nr]nm(by Remark 1)=[M1(G1) + 2mp]nm[M1(G2) + 2nq]nm[M1(G1) +m(n(n− 1)− 2p)]nm[M1(G2) + n(m(m− 1)− 2q)]nm[2mp+ 2nq +mn(m+ n)]nm.This completes the proof.32 Product version of reciprocal degree distance of composite graphsOne can observe that M1(Cn) = 4n, n ≥ 3, M1(P1) = 0, M1(Pn) = 4n−6, n > 1 andM1(Kn) = n(n−1)2. Similarly, M1(Kn) = M2(Kn) = 0. Moreover M2(Pn) = 4(n−2)and M2(Cn) = 4n. Using Theorem 2, we have the following corollaries.Corollary 3. Let G be graph on n vertices and p edges. ThenH∗A(G+Km) ≤[(M1(G1)+2mp)(M1(G1) +m(n(n− 1)− 2p))(m(m− 1)(m+ n− 1))(2mp+ nm(2m+ n− 1))]nm.Let Kn,m be the bipartite graph with two partitions having n and m vertices. Notethat Kn,m = Kn + Km.Corollary 4. H∗A(Kn,m) = H∗A(Kn +Km) ≤ (nm)3nm[(n− 1)(m− 1)(m+ n)]nm.4. Strong productIn this section, we obtain the product version of reciprocal degree distance of GKr.Theorem 3. Let G be a connected graph with n vertices and m edges. ThenH∗A(G Kr) = H∗A(G Kr) ≤(r − 1)2nrn3nr[(2rm+ n(r − 1)]nr[rHA(G) + 2(r − 1)H(G))]2nr.Proof. Let V (G) = {u1, u2, . . . , un} and V (Kr) = {v1, v2, . . . , vr}. Let xij denote thevertex (ui, vj) of GKr. The degree of the vertex xij in GKr is dG(ui)+dKr (vj)+dG(ui)dKr (vj), that is dGKr (xij) = rdG(ui) + (r− 1). One can observe that for anypair of vertices xij , xkp ∈ V (G  Kr), dGKr (xij , xip) = 1 and dGKr (xij , xkp) =dG(ui, uk).H∗A(G Kr) =∏xij , xkp ∈V (GKr)dGKr (xij) + dGKr (xkp)dGKr (xij , xkp)=n−1∏i=0r−1∏j, p=0j 6= pdGKr (xij) + dGKr (xip)dGKr (xij , xip)×n−1∏i, k=0i 6= kr−1∏j =0dGKr (xij) + dGKr (xkj)dGKr (xij , xkj)×n−1∏i, k=0i 6= kr−1∏j, p=0j 6= pdGKr (xij) + dGKr (xkp)dGKr (xij , xkp). (8)K. Pattabiraman 33We shall obtain the products of (8) separately. First we computen−1∏i=0r−1∏j, p=0j 6= pdGKr (xij)+dGKr (xip)dGKr (xij ,xip). By Remark 1, we haven−1∏i=0r−1∏j, p=0j 6= pdGKr (xij) + dGKr (xip)dGKr (xij , xip)=n−1∏i=0r−1∏j, p=0j 6= p(2dG(ui) + 2(r − 1) + 2(r − 1)dG(ui))≤[ 12 n−1∑i=0r−1∑j, p=0j 6= p(2dG(ui) + 2(r − 1) + 2(r − 1)dG(ui))nr]nr=[2r2(r − 1)m+ nr(r − 1)2nr]nr=[2r(r − 1)m+ n(r − 1)2n]nr. (9)Next we computer−1∏j =0n−1∏i, k=0i 6= kdGKr (xij)+dGKr (xkj)dGKr (xij ,xkj). We haver−1∏j =0n−1∏i, k=0i 6= kdGKr (xij) + dGKr (xkj)dGKr (xij , xkj)=r−1∏j =0n−1∏i, k=0i 6= k(dG(ui) + (r − 1)dG(ui) + dG(uk) + (r − 1)dG(uk) + 2(r − 1))dG(ui, uk)≤[ r2 r−1∑j =0∑n−1i, k=0i 6= k dG(ui)+dG(uk)dG(ui,uk) + 12r−1∑j =0n−1∑i, k=0i 6= k2(r−1)dG(ui,uk)nr]nr(by Remark 1)=[ r2HA(G) + 2r(r − 1)H(G)nr]nr=[ rHA(G) + 2(r − 1)H(G)n]nr. (10)Finally we determinen−1∏i, k=0i 6= kr−1∏j, p=0,j 6= pdGKr (xij)+dGKr (xkp)dGKr (xij ,xkp).n−1∏i, k=0i 6= kr−1∏j, p=0,j 6= pdGKr (xij) + dGKr (xkp)dGKr (xij , xkp)≤[ r2(r−1)2 n−1∑i, k=0i 6= kdG(ui)+dG(uk)dG(ui,uk)+ r(r − 1)2n−1∑i, k=0i 6= k1dG(ui,uk)nr]nr(by Remark 1)=[ r2(r − 1)HA(G) + 2r(r − 1)2H(G)nr]nr=[ r(r − 1)HA(G) + 2(r − 1)2H(G)n]nr. (11)34 Product version of reciprocal degree distance of composite graphsUsing (9), (10) and (11) in (8), we haveH∗A(GKr) =(r − 1)2nrn3nr[(2rm + n(r − 1)]nr[rHA(G) + 2(r − 1)H(G))]2nr.Using Theorem 3, we obtain the following corollary.Corollary 5. H∗A(Cn Kr) ≤ (3r − 1)nr[2(r−1)(3r−1)H(Cn)n]2nr.As an application we present formula for product version of reciprocal degree distanceof closed fence graph, Cn K2.Example 2. By Corollary 5, we haveH∗M (Cn K2) ≤52n[10n(nn2∑i=11i− 1)]4n, if n is even52n[10n−12∑i=11i]4n, if n is odd.References[1] Y. Alizadeh, A. Iranmanesh, and T. Došlić, Additively weighted Harary index ofsome composite graphs, Discrete Math. 313 (2013), no. 1, 26–34.[2] K.C. Das, B. Zhou, and N. Trinajstić, Bounds on Harary index, J. Math. Chem.46 (2009), no. 4, 1377–1393.[3] J. Devillers and A.T. Balaban, Topological indices and related descriptors inQSAR and QSPAR, Gordon and Breach, Amsterdam, The Netherlands, 2000.[4] A.A. Dobrynin and A.A. Kochetova, Degree distance of a graph: A degree analogof the Wiener index, J. Chem. Inf. Comput. Sci. 34 (1994), no. 5, 1082–1086.[5] T. Došlić, Vertex-weighted Wiener polynomials for composite graphs, Ars Math.Contemp. 1 (2008), no. 1.[6] I. 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Product version of reciprocal degree distance of composite graphs

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Product version of reciprocal degree distance of composite graphs

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