Vertex and edge PI indices of Cartesian product graphs

Vertex and edge PI indices of Cartesian product graphs

Discrete Applied Mathematics 156 (2008) 1780 – 1789 www.elsevier.com/locate/dam Vertex and edge PI indices of Cartesian product graphs M.H. Khalifeha...

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Discrete Applied Mathematics 156 (2008) 1780 – 1789 www.elsevier.com/locate/dam

Vertex and edge PI indices of Cartesian product graphs M.H. Khalifeha , H. Yousefi-Azaria , A.R. Ashrafib,∗ a School of Mathematics, Statistics and Computer Science, University of Tehran, Tehran, Islamic Republic of Iran b Department of Mathematics, Faculty of Science, University of Kashan, Kashan 87317-51167, Islamic Republic of Iran

Received 24 March 2007; received in revised form 22 June 2007; accepted 26 August 2007 Available online 24 October 2007

Abstract The Padmakar–Ivan (PI) index of a graph G is the sum over all edges uv of G of the number of edges which are not equidistant from u and v. In this paper, the notion of vertex PI index of a graph is introduced. We apply this notion to compute an exact expression for the PI index of Cartesian product of graphs. This extends a result by Klavzar [On the PI index: PI-partitions and Cartesian product graphs, MATCH Commun. Math. Comput. Chem. 57 (2007) 573–586] for bipartite graphs. Some important properties of vertex PI index are also investigated. © 2007 Elsevier B.V. All rights reserved. MSC: 05C12; 05A15; 05A20; 05C05 Keywords: Edge PI index; Vertex PI index; Cartesian product graph

1. Introduction Let G be a connected graph with vertex and edge sets V (G) and E(G), respectively. As usual, the distance between the vertices u and v of G is denoted by d(u, v) and it is defined as the number of edges in a minimal path connecting the vertices u and v. A topological index is a real number related to a graph. It must be a structural invariant, i.e., it preserves by every graph automorphisms. There are several topological indices have been defined and many of them have found applications as means to model chemical, pharmaceutical and other properties of molecules. The Wiener index W is the first topological index to be used in chemistry. Usage of topological indices in chemistry began in 1947 when chemist Harold Wiener developed the most widely known topological descriptor, the Wiener index, and used it to determine physical properties of types of alkanes known as paraffins [19]. In a graph theoretical language, the Wiener index is equal to the count of all shortest distances in a graph. We encourage the reader to consult the special issues of MATCH Communication in Mathematics and in Computer Chemistry [10], Discrete Applied Mathematics [11,7,8], for information on results on the Wiener index, the chemical meaning of the index and its history. Let G be a graph and e = uv an edge of G. neu (e|G) denotes the number of edges lying closer to the vertex u than the vertex v, and nev (e|G) is the number of edges  lying closer to the vertex v than the vertex u. The Padmakar–Ivan (PI) index of a graph G is defined as PI(G) = e∈E(G) [neu (e|G) + nev (e|G)], see for details [1–5,13–15]. In this ∗ Corresponding author.

E-mail address: ashrafi@kashanu.ac.ir (A.R. Ashrafi). 0166-218X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.dam.2007.08.041

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definition, edges equidistant from both ends of the edge e = uv are not counted. We call this index the edge PI index and denote by PIe (G). We also define the vertex PI index of G, PIv (G), as the sum of [meu (e|G) + mev (e|G)] over all edges of G, where meu (e|G) is the number of vertices lying closer to the vertex u than the vertex v and mev (e|G) is the number of vertices lying closer to the vertex v than the vertex u. The Cartesian product G × H of graphs G and H has the vertex set V (G × H ) = V (G) × V (H ) and (a, x)(b, y) is an edge of G × H ifa = b and xy ∈ E(H ), or ab ∈ E(G) and x = y. If G1 , G2 , . . . , Gn are graphs then we denote G1 × · · · × Gn by ni=1 Gi . The Wiener index of Cartesian product graphs was studied in [9,20]. In [16], Klavzar, Rajapakse and Gutman computed the Szeged index of Cartesian product graphs. Here we continue this progress to compute the PI index of Cartesian product graphs. The main result of this paper is as follows: n

Theorem 3. Let G1 , G2 , . . . , Gn be connected graphs. Then PIv ( and,  PIe

n  i=1

 Gi

=

n  i=1

⎛ ⎝

n

j =1,j =i

⎞ |V (Gj )|2 ⎠ PIe (Gi ) +

n  i=1

i=1 Gi )

PIv (Gi )

=

n 

n

n

i=1 (

j =1,j =i

|V (Gj )||E(Gj )|

j =1,j =i

|V (Gj )|2 )PIv (Gi ), n

|V (Gk )|2 .

k=1,k=i,j

Corollary. If G is a connected graph then PIe (Gn ) = |V (G)|2(n−1) (PIe (G) + n(n − 1)(|E(G)|/|V (G)|)PIv (G)) and PIv (Gn ) = n|V (G)|2(n−1) PIv (G). In [17], Klavzar presented a formula for calculating the PI index of the product of two bipartite graphs. In [21], Yousefi, Manoochehrian and Ashrafi independently from Klavzar, computed an exact formula for the product of n bipartite graphs. In what follows, we obtain Klavzar’s result by our main theorem. We first prove the following simple lemma. Lemma 1. Let G be a graph. Then PIv (G) |E(G)||V (G)| with equality if and only if G is bipartite.   Proof. Clearly, PIv (G) = e∈E(G) [meu (e|G) + mev (e|G)]  e∈E(G) |V (G)| = |V (G)||E(G)|, which completes the first part of our lemma. If G is bipartite then G does not have cycles of odd length and so meu (e|G)+mev (e|G)=|V (G). Thus PIv (G) = |V (G)||E(G)|. Conversely, suppose that PIv (G) = |V (G)||E(G)| and G is not bipartite. Then G has a cycle of odd length. We assume that G has girth k and choose T and e = xy such that T is a cycle of length k and e is one of the edges of T. Then mex + mey < |V (G)|, a contradiction. This implies that G is bipartite.  The previous lemma shows that for a tree T with exactly n vertices, PIv (T ) = n(n − 1). Corollary (Klavzar [17]). If G is a bipartite connected graph then PIe (Gn ) = n|V (G)|2(n−1) PIe (G) + n(n − 1)| E(G)|2 |V (G)|2(n−1) . Proof. The proof is straightforward and follows from our main theorem and Lemma 1.



Throughout this paper, we only consider connected graphs. Our notation is standard and taken mainly from [6,12,18]. Kn denotes a complete graph on n vertices. Suppose G is a graph, e = xy, f = uv ∈ E(G) and w ∈ V (G). Define d(w, e)=min{d(w, x), d(w, y)}. We say that e is parallel to f and write e||f , if d(x, f )=d(y, f ). Note that parallelism is reflexive but it is neither symmetric nor transitive. Since bipartite graphs do not have cycles of odd length, parallelism in bipartite graphs is symmetric. 2. Examples In this section we calculate the vertex and edge PI indices of some well-known graphs. Suppose G is a graph. Define NG (e) = |E| − (neu (e|G) + nev (e|G)), where e is an arbitrary edge of the graph G. Clearly, PIe (G) = |E|2 −  e∈E(G) NG (e). We use this simple equation freely throughout the paper.

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M.H. Khalifeh et al. / Discrete Applied Mathematics 156 (2008) 1780 – 1789 x1

x2

xi

xn-1

xn

y1

y2

yi

yn-1

yn

Fig. 1. The ladder graph with 2n vertices.

Example 1. Let Cn be a cycle graph with n vertices. Then  PIv (Cn ) =

n2 , n(n − 1),

2|n, 2n.

On the other hand, by [4, Lemma 2],  PIe (Cn ) =

n(n − 2), n(n − 1),

2|n, 2n.

Example 2. Consider the ladder graph Ln , Fig. 1. Clearly, Ln = Pn × P2 , where Pn is a path with n vertices. So, by our main theorem PIv (Ln ) = 6n2 − 4n and by [4, Example 1], PIe (Ln ) = 8(n − 1)2 . Consider the graph G whose vertices are the N -tuples b1 b2 · · · bN with bi ∈ {0, 1, . . . , ni − 1}, ni 2, and let two vertices be adjacent if the corresponding tuples differ in precisely one place. Such a graph is called a Hamming graph.  It is well-known fact that a graph G is a Hamming graph if and only if it can be written in the form G = N K i=1 ni . In the following example, the vertex and edge PI indices of a Hamming graph is computed. Lemma 2. Let G be a Hamming graph with above parameter. Then   N N  1 2 (a) , PIv (G) = ni N − ni i=1 i=1 ⎛ ⎞ N N n N    n 1 i (b) PIe (G) = n2i ⎝(N + 1) ni + (N + 3) ni − − N 2 − 4N ⎠ . 2 nj i=1

i=1

i=1

i,j =1

Proof. (a) It is easy to see that PIv (Km ) = m(m − 1). Since Hamming graph is a product of complete graphs, by Theorem 3,  PIv (G) = PIv

N 

 Kni

i=1

=

N 

PIv (Kni )

j =1,j =i

i=1

= =

N i=1 N i=1

N

n2i

N  i=1

 n2i

ni − 1 ni

N  1 N− ni i=1

n2j

 .

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This completes part (a). Since PIe (Km ) = m(m − 1)(m − 2), by Theorem 3 and part (a), we have

PIe (G) =

N 

ni (ni − 1)(ni − 2)

j =1,j =i

i=1

=

N i=1

=

N

N

n2j +

N 

 n2i

i=1

nj

j =1,j =i

i=1

⎡ ⎛ N N   (n − 1)(n − 2) i i ⎝ ni − 1 + n2i ⎣ ni ni i=1

i=1

N 

ni (ni − 1)

N  j =1,j =i

n 

N

j

2

k=1,k=i,j

n2k

⎞⎤ nj − 1 ⎠⎦ 2

 N  N N N   (ni − 1)(ni − 2) 1  (ni − 1)2 1  ni − 1 − + (ni − 1) ni 2 ni 2 ni i=1

i=1

i=1

i=1

N   N    N N   1 2  (ni − 1)(ni − 3) 1 = ni + N− ni − N 2 ni ni i=1

i=1

i=1

i=1

N  N   N N N N N N     1   1 1 1 2  4+3 +N ni − N 2 − ni + N = ni ni − ni ni ni 2 i=1

i=1

i=1

i=1

i=1

i=1

i=1

i=1

⎡ ⎤ N N N N    1 ni 1 2⎣ ni (N + 1) ni + (N + 3) − − N 2 − 4N ⎦ . = 2 ni nj i=1

This completes the lemma.

i=1

i=1

i,j =1



Corollary. Let Qn denote the hypercube of dimension n then PIv (Qn ) = n22n−1 and PIe (Qn ) = n(n − 1)22(n−1) . 3. Main results In this section, we prove the main result of this paper. In the following lemma, some well-known properties of Cartesian product graphs are introduced. We encourage the reader to consult the book of Imrich and Klavzar [12], for more details. Lemma 3. Let G and H be graphs. Then we have: (a) (b) (c) (d)

|V (G × H )| = |V (G)| · |V (H )| and |E(G × H )| = |E(G)| · |V (H )| + |V (G)| · |E(H )|. G × H is connected if and only if G and H are connected. If (a, x) and (b, y) are vertices of G × H then dG×H ((a, x), (b, y)) = dG (a, b) + dH (x, y). The Cartesian product is commutative and associative.

In what follows, we introduce the main properties of vertex PI index. We begin with an equality between two indices. Lemma 4. Suppose G is a connected graph with exactly m edges and n vertices, m 3. Then 2|E(G)| PIe (G)| 2 E(G)|(|E(G)| − 1), 2|E(G)| v (G) |V (G)||E(G)|. Moreover, 3|E(G)| − |E(G)| PIv (G) − PIe (G)|E(G)| PI n 2 (|V (G)| − 2) and PIe (G) i=1 deg(vi ) − 2|E(G)|. Proof. Suppose uv = e ∈ E(G). If deg(u), deg(v) > 1 then  N (e) |E(G)| − 2. If u or v is an end vertex then N(e)=1|E(G)|−2. This implies that PIe (G)=|E(G)|2 − e∈E(G) N (e) |E(G)|2 −|E(G)|(|E(G)|−2)=2|E(G)|.

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On the other hand, PIv (G) = 

PIe (G) =



e∈E(G) [meu (e|G) + mev (e|G)] 2|E(G)|.

We also note that

[neu (e|G) + nev (e|G)]

e∈E(G)



 =

(deg(u) + deg(v) − 2)

uv∈E(G) n 

deg(vi )2 − 2|E(G)|.



i=1

Let G be a graph and e = uv ∈ E(G).Define M(e) = |{w ∈ V (G) | d(u, w) = d(v, w)}|. Then  M(e) = |V (G)| − meu (e|G) − mev (e|G) and so PIv (G) = e∈E(G) [meu (e|G) + mev (e|G)] = |V (G)||E(G)| − e∈E(G) M(e). We use this relation freely throughout this section. Theorem 1. Suppose G and H are arbitrary graphs. Then PIe (G × H ) = |V (G)||E(G)|PIv (H ) + |V (H )||E(H )|PIv (G) + PIe (G)|V (H )|2 + PIe (H )|V (G)|2 . Proof. Suppose V (G) = {u1 , . . . , us }, V (H ) = {v1 , . . . , vt } and define Ak = {(uk , vm )(uk , vn ) | vm vn ∈ E(H )}, Bk = {(ui , vk )(uj , vk ) | ui uj ∈ E(G)}, Ck = {(uk , vi ) | vi ∈ V (H )}, Dk = {(ui , vk ) | ui ∈ V (G)}, A=

s  i=1

Ak ,

B=

t 

Bk

i=1

    Then it is clear that ( sk=1 Ak )∪( tk=1 Bk )=E(G×H ) and sk=1 Ck = tk=1 Dk =V (G×H ). On the other hand, by [12, Corollary 1.35] dG×H ((ui , vm ), (uj , vn ))=dG (ui , uj )+dH (vm , vn ) and so dG×H ((ui , vm ), (ui , vn ))=dH (vm , vn ). Claim 1. Suppose (um , vi )(um , vj ), (um , vi )(um , vj ) ∈ Am . Then we have (um , vi )(um , vj )(um , vi )(um , vj ) if and only if vi vj vi vj . To prove, it is enough to look at the following equalities: dH (vj , vi vj ) = min{dH (vj , vi ), dH (vj , vj )} = min{dG×H ((um , vj ), (um , vi )), dG×H ((um , vj ), (um , vj ))} = dG×H ((um , vj ), (um , vi )(um , vj )),

dH (vi , vi vj ) = min{dH (vi , vi ), dH (vi , vj )} = min{dG×H ((um , vi ), (um , vi )), dG×H ((um , vi ), (um , vj ))} = dG×H ((um , vi ), (um , vi )(um , vj )).

M.H. Khalifeh et al. / Discrete Applied Mathematics 156 (2008) 1780 – 1789

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Claim 2. Suppose ((um , vi )(um , vj ) ∈ Am , (uk , vi )(uk , vj )) ∈ Ak . Then (um , vi ) (um , vj ) (uk , vi )(uk , vj ) if and only if vi vj vi vj . Consider the following equalities: dH (vj , vi vj ) + dG (um , uk ) = min{dH (vj , vi ), dH (vj , vj )} + dG (um , uk )

= min{dG×H ((um , vj ), (uk , vi )), dG×H ((um , vj ), (uk , vj ))}

= dG×H ((um , vj ), (uk , vi )(uk , vj )),

dH (vi , vi vj ) + dG (um , uk ) = min{dH (vi , vi ), dH (vi , vj )} + dG (um , uk )

= min{dG×H ((um , vi ), (uk , vi )), dG×H ((um , vi ), (uk , vj ))} = dG×H ((um , vi ), (uk , vi )(uk , vj )).

Then, dH (vj , vi vj ) = dH (vi , vi vj ) if and only if (um , vi )(um , vj )||(uk , vi )(uk , vj ). Claim 3. Suppose (um , vi )(um , vj ) ∈ Am . Then |{e ∈ A | (um , vi )(um , vj )e}| = s · NH (vi vj ). By Claim 2, we have |{e ∈ Ak | (um , vi )(um , vj )e}| = |{e ∈ E(G) | vi vj e}| = NH (vi vj ). The proof is now follows from the fact that the sets Ak ’s, 1 k s, are disjoint. Claim 4. Suppose (um , vi )(um , vj ) ∈ Am and (up , vk )(uq , vk ) ∈ Bk . Then (um , vi )(um , vj )(up , vk )(uq , vk ) if and only if dH (vi , vk ) = dH (vj , vk ). Consider the following equalities: dG (um , up uq ) + dH (vj , vk ) = min{dG (um , up ), dG (um , uq )} + dH (vj , vk ) = min{dG×H ((um , vj ), (up , vk )), dG×H ((um , vj ), (uq , vk ))} = dG×H ((um , vj ), (up , vk )(uq , vk )), dG (um , up uq ) + dH (vi , vk ) = min{dG (um , up ), dG (um , uq )} + dH (ui , vk ) = min{dG×H ((um , vi ), (up , vk )), dG×H ((um , vi ), (uq , vk ))} = dG×H ((um , vi ), (up , vk )(uq , vk )). Then, dH (vj , vk ) = dH (vi , vk ) if and only if (um , vi )(um , vj )(up , vk )(uq , vk ). Claim 5. Suppose (um , vi )(um , vj ) ∈ Am . Then |{(up , vr )(uq , vr ) ∈ B | (um , vi )(um , vj )| |(up , vr )(uq , vr )}| = |E(G)|MH (vi vj ). Choose an element up uq ∈ E(G). Since the sets Bi are disjoint, by Claim 3, MH (vi , vj ) = |{vk ∈ V (H ) | d(vi , vk ) = d(vj , vk )}| = |{(up , vk )(uq , vk ) | (um , vi )(um , vj )| |(up , vk )(uq , vk ); 1 k t}|. We now vary up uq on E(G). We have |E(G)|MH (vi vj ) = |E(G)|.|{(up , vk )(uq , vk ) | (um , vi )(um , vj )(up , vk )(uq , vk ); 1 k t}| = |{(up , vk )(uq , vk ) | (um , vi )(um , vj )||(up , vk )(uq , vk ); 1 k t; up uq ∈ E(G)} = |{e ∈ B | (um , vi )(um , vj )||e}|.

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Claim 6. Suppose (um , vi )(um , vj ) ∈ E(G × H ). Then NG×H ((um , vi )(um , vj )) = |E(G)|MH (vi vj ) + |V (G)| 2 (v v ). Moreover, N e∈A NG×H (e) = |E(G)| |E(H )| |V (G)| |V (H )| − |E(G)| |V (G)|PIv (H ) + |V (G)| H i j e∈E(H ) NH (e). Since A ∩ B = ∅, NG×H ((um , vi )(um , vj )) = |{e ∈ A | (um , vi )(um , vj )||e}| + |{e ∈ B | (um , vi )(um , vj )| |e}| = |E(G)|MH (vi vj ) + |V (G)|NH (vi vj ). Using this relation, we have 

NG×H (e) =

e∈A

s  

NG×H ((um , vi )(um , vj ))

m=1 e∈Am

=

s 



(|E(G)|MH (vi vj ) + |V (G)|NH (vi vj ))

m=1 vi vj ∈E(H )

=

s 





⎣|E(G)|

MH (vi vj ) + s

vi vj ∈E(H )

m=1

⎤ NH (vi vj )⎦

vi vj ∈E(H )



= |E(G)| |V (G)|





MH (vi vj ) + s 2

vi vj ∈E(H )

NH (vi vj )

vi vj ∈E(H )

= |E(G)| |E(H )| |V (G)| |V (H )| − |E(G)| |V (G)|PIv (H ) + |V (G)|2



NH (e).

e∈E(H )

 one can see that e∈A NH ×G ((vm , ui )(vm , uj )) = Since G × H H × G, using a similar argument as above,  e∈B NG×H ((ui , vm )(uj , vm )). Therefore for computing e∈B NG×H (e) it is enough to interchange G and H in the second part of Claim 6. We now compute PIe (G × H ).  PIe (G × H ) = |E(G × H )|2 − NG×H (e) e∈E(G×H )

= |E(G × H )|2 −



NG×H (e) −

e∈A



NG×H (e)

e∈B

= |V (G)|2 |E(G)|2 + |V (H )|2 |E(H )|2 + |E(G)||V (G)|PIv (H ) + |V (G)|2





NH (e) + |V (H )|2

e∈E(H )

NG (e) + |E(H )||V (H )|PIv (G)

e∈E(G)

⎛ = |V (H )|2 ⎝|E(G)|2 −



⎞ NG (e)⎠ + |E(G)| |V (G)|PIv (H )

e∈E(G)

⎛ + |V (G)|

2⎝

|E(H )| − 2



⎞ NH (e)⎠ + |E(H )| |V (H )|PIv (G)

e∈E(H )

= PIe (G)|V (H )|2 + PIe (H )|V (G)|2 + |V (G)| |E(G)|PIv (H ) + |V (H )| |E(H )|PIv (G). This completes the proof.



Theorem 2. Suppose G and H are arbitrary graphs. Then PIv (G × H ) = PIv (G)|V (H )|2 + PIv (H )|V (G)|2 .

M.H. Khalifeh et al. / Discrete Applied Mathematics 156 (2008) 1780 – 1789

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Proof. Consider the notation of Theorem 1. Suppose (um , vi )(um , vj ) ∈ Am . We first prove that dG×H ((um , vi ), (ur , vk )) = dG×H ((um , vj ), (ur , vk )) if and only if dH (vi , vk ) = dH (vj , vk ). To see this, we have dG×H ((um , vi ), (ur , vk )) = dG (um , ur ) + dH (vi , vk ) = dG (um , ur ) + dH (vj , vk ) = dG×H ((um , vj ), (ur , vk )). Since vi vj ∈ E(H ) and Ck ={uk }×V (H ), |{(uk , vr ) ∈ Ck |dG×H r ))=dG×H ((um , vj ), (uk , vr ))}|= ((um , vi ), (uk , v |{v ∈ V (H )|dH (v, vi ) = dH (v, vj )}| = MH (vi vj ). Suppose C = si=1 Ci and D = ti=1 Di . Since Ci ’s and also Dj ’s are disjoint, |{v ∈ C | dG×H ((um , vi ), v) = dG×H ((um , vj ), v)}| = |V (G)|MH (vi vj ). Thus, 

MG×H (e) =

e∈A

=

s  

MG×H ((um , vi )(um , vj ))

m=1 e∈Am s 



|V (G)|

MH (vi , vj )

vi vj ∈E(H )

m=1



= |V (G)|2

MH (e).

e∈E(H )

Again since G × H H × G,



e∈B MG×H

(e) = |V (H )|2

PIv (G × H ) = |E(G × H )||V (G × H )| −





e∈E(G) MG (e).

MG×H (e)

e∈E(G×H )

= |E(G × H )||V (G × H )| −



Therefore,

MG×H (e) −

e∈A



MG×H (e)

e∈B

= |V (H )|2 |V (G)||E(G)| + |V (G)|2 |V (H )||E(H )|   − |V (H )|2 MG (e) − |V (G)|2 MH (e) e∈G

e∈H

= PIv (G)|V (H )|2 + PIv (H )|V (G)|2 .   G1 , G2 , . . . , Gn be n graphs and G = ni=1 Gi . Then by Lemma 3, |V (G)| = ni=1 |V (Gi )| and |E(G)| = Suppose n n j =1 |E(Gj )| i=1,i=j |V (Gi )|. We now are ready to state our main result. We have: Proof of the Theorem 3. In Theorem 1, we proved the case of n = 2 for PIv . We continue our argument by induction. Suppose the result is valid for n graphs. Then we have  n+1    n   Gi = PIv Gn+1 × Gi PIv i=1

i=1



= |V (Gn+1 )| PIv 2

= |V (Gn+1 )|

2

n 

n 

=

i=1

PIv (Gi )

Gi

i=1

PIv (Gi )

i=1 n+1 



n+1

  n 2      + V Gi  PIv (Gn+1 )   n

j =1,j =i

i=1

|V (Gj )| + PIv (Gn+1 ) 2

n

|V (Gi )|2

i=1

|V (Gj )|2 ,

j =1,j =i

as desired. To prove the second part of the theorem, we apply again an inductive argument. In Theorem 1, we proved the case of n = 2. Suppose the result is valid for n graphs. Then by our assumption, the first part of this theorem and

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Theorem 1, we have  n+1    n   PIe Gi = PIe Gn+1 × Gi i=1

i=1

  n 2      Gi  PIe (Gn+1 ) = |V (Gn+1 )| PIe Gi + V   i=1  i=1  n  Gi + |V (Gn+1 )| · |E(Gn+1 )| · PIv i=1   n    n            Gi  · E Gi  · PIv (Gn+1 ) + V     i=1 ⎛ i=1 n n  |V (Gj )|2 PIe (Gi ) = |V (Gn+1 )|2 ⎝ 

2

n 



j =1,j =i

i=1

+

n 

n 

PIv (Gi )



n

|E(Gj )| · |V (Gj )|

|V (Gk )|2 ⎠

j =1,j =i

k=1,k=i,j   n   n             2 + PIe (Gn+1 ) |V (Gi )| + V Gi  · E Gi  · PIv (Gn+1 )     i=1 i=1  ni=1   Gi + |V (Gn+1 )| · |E(Gn+1 )| · PIv i=1

=

n

n

|V (Gi )| · PIv (Gn+1 ) ·

n 

n+1 

n 

PIv (Gi )

i=1

+

i=1

PIe (Gi )

j =1,j =i n+1

n

|E(Gj )|

j =1

i=1

+

n 

|V (Gi )|

i=1,i=j n+1

|E(Gj )| · |V (Gj )|

|V (Gk )|2

k=1,k=i,j



|V (Gj )| + |V (Gn+1 )| · |E(Gn+1 )| · PIv 2

j =1,j =i

     n   V (Gi )|2  = PIv (Gn+1 ) · |E(Gj )| |V (Gj )    j =1 i=1,i=j  i=1

n 

 Gi

i=1

n 

+

n 

PIv (Gi )

i=1

+

n+1 

PIe (Gi )

=

j =1,j =i n+1

|E(Gj )| · |V (Gj )|

PIv (Gi )

2

n 

PIv (Gi )

n+1  j =1,j =i

i=1



n 

 Gi

i=1 n+1

|E(Gj )| · |V (Gj )| n 

k=1,k=i,j n

PIv (Gi )

i=1

This completes the proof.



|V (Gj )| + |V (Gn+1 )| · |E(Gn+1 )| · PIv

+ |V (Gn+1 )| · |E(Gn+1 )| · =

|V (Gk )|2

k=1,k=i,j

j =1,j =i

i=1

n+1 

n+1

j =1,j =i

i=1 n+1 

n 

|E(Gj )| · |V (Gj )|

j =1,j =i n+1 k=1,k=i,j

|V (Gk )|2 |V (Gj )|2 +

n+1 

PIe (Gi )

i=1

|V (Gk )|2 +

n+1  i=1

PIe (Gi )

n+1 j =1,j =i n+1 j =1,j =i

|V (Gj )|2 |V (Gj )|2 .

M.H. Khalifeh et al. / Discrete Applied Mathematics 156 (2008) 1780 – 1789

1789

Acknowledgment We are greatly indebted to the referees, whose valuable criticisms and suggestions led us to correct the paper. The research of the third author was in part supported by the Center of Excellence of Algebraic Methods and Applications of Isfahan University of Technology. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21]

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