The Degree Distance of Certain Particular Graphs 1 Introduction

Applied Mathematical Sciences, Vol. 6, 2012, no. 18, 857 - 867

The Degree Distance of Certain Particular Graphs G. Al Hagri, M. El Marraki and M. Essalih Department of Computer Sciences, Faculty of Sciences University of Mohamed V, P.O. Box 1014, Rabat, Morocco chima [email protected], [email protected] Abstract If G is a connected graph with vertex set V (G), then the degree dis tance of G is defined as DD(G) = {u,v}∈V (G) (deg(u) + deg(v))d(u, v) Where deg(u) is the degree of vertex u, and d(u, v) denotes the distance between u and v. Where deg(u) is the degree of vertex u, and d(u, v) denotes the distance between u and v. The degree distance is a frequently and successfully used structure descriptor in studies, based on molecular topology, of quantitative relations between structure and property and between structure and activity. Its numerous physicochemical applications range from the prediction of the boiling points to the calculation velocity of ultrasound in organic materials. There are a number of chemical application.

Mathematics Subject Classification: 05C12, 05C05 Keywords: Graph, distance, degree, wiener index

1

Introduction

A graph G is a triple consisting of a vertex set V (G), an edge set E(G), and a relation that associates with each edge two vertices called its end points. A connected graph is a graph such that for each pair u, v (two distinct vertices), there is a path of vertices connecting these two points. The distance between two vertices vi and vj in a graph G is the length of the shortest path (number of edge ) connecting these two vertices, and we note d(vi , vj ). The degree or the valency deg(v) of a vertex v is the number of edges incident to it. The Wiener index W (G) a connected graph G is the sum of all distances between pairs of vertices of G [2, 1]: W (G) = {u,v}⊆V (G) d(u, v). The wiener index of a vertex v in G is defined as: w(v, G) = u∈V (G) d(u, v). The degree distance of G is defined as: DD(G) = {u,v}∈V (G) (deg(u) + deg(v)) d(u, v). Where deg(u)

The degree distance of certain particular graphs

858

is the degree of vertex u and d(u, v) denotes the distance between u and v. As well as the degree distance of u is defined as dd(u, G) = v∈V (G) d(u, v) deg v. The relations between the degree distance and the Wiener index were studied in [9, 5, 7]

2

Main Results

In this section, we will give the formulas for the degree distance of some particular graph, The interested readers for more information on topological indices of graph operations can be referred to the papers [4, 6, 2, 1] and their references. Lemma 1 The degree distance of T is defined as DD(T ) = 4 W (T )−n(n−1). Which holds whenever T is a tree of order n (see [8]) Lemma 2 The degree distance of G is defined as DD(G) = u∈V (G) w(u, G) deg u.(see [3]) Let Gm1 · Gm2 be the Connected simple graph

Figure 1: Graphs Gm1 . Gm2

Let Gm1 · Gm2 be the graphs composed by graph Gm1 , Gm2 that possess respectively m1 , m2 vertices connected by a vertex s (s is an articulation vertex) (see Figure 1). Lemma 3 Suppose Gm1 and Gm2 are connected simple graphs. Then the following are hold: (a) V (Gm1 · Gm2 ) = V (Gm1 ) ∪ V (Gm2 ) (b) |V (Gm1 · Gm2 )| = |V (Gm1 )| + |V (Gm2 )| − 1 (c)If u, v ∈ V (G) then d(u, v) ≤ d(u, s) + d(s, v) with s ∈ V (Gm1 ) ∩V (Gm2 ) (d) If s is an articulation vertex, then for u ∈ V (Gm1 ) and v ∈ V (Gm2 ), we find that d(u, v) = d(u, s) + d(s, v) w(s, Gm1 · Gm2 ) = w(s, Gm1 ) + w(s, Gm2 ).

859


Lemma 4 If u ∈ V (Gm1 )\{s}, then W (u, Gm1 · Gm2 ) = w(u, Gm1 ) + (m2 − 1) d(u, s) + w(s, Gm2 ).

(1)

If u ∈ V (Gm2 )\{s}, then W (u, Gm1 · Gm2 ) = w(u, Gm2 ) + (m1 − 1) d(u, s) + w(s, Gm1 ).

(2)

Proof: W (u, Gm1 · Gm2 ) =

v∈V (Gm1 ·Gm2 )

=

d(u, v) =

d(u, v) +

v∈V (Gm1 )

= w(u, Gm1 ) +

d(u, v)

v∈V (Gm1 )∪V (Gm2 )

d(u, v) − d(u, s)

v∈V (Gm2 )

(d(u, s) + d(s, v)) − d(u, s)

v∈V (Gm2 )

= w(u, Gm1 ) + d(u, s)

1+

v∈V (Gm2 )

d(s, v) − d(u, s)

v∈V (Gm2 )

= w(u, Gm1 ) + (m2 − 1) d(u, s) + w(s, Gm2 ). (1) We use the same technique for (2). Lemma 5 If u ∈ V (Gmi )\{s}, then W (u, Gm1 .Gm2 ...Gmn ) = 1) d(u, s) + w(s, Gm,n ) − w(s, Gmi ) + w(u, Gmi )

Figure 2: Graphs (Gm1 .Gm2 . ...Gmn )

Proof: W (u, Gm1 .Gm2 . ... .Gmn ) =

u∈V (Gm1 )∪V (Gm2 )∪..cupV (Gmn )

d(u, v)

n

j=1 j=i

(mj − n +

860


=

u∈V (Gm1 )

(d(u, s) + d(s, v)) + ... +

u∈V (Gm1 )

+w(u, Gmi ) +

+.. +

d(u, v)

u∈V (Gmi )

d(u, v) − (n − 1) d(u, s)

u∈V (Gmn )

=

d(u, v) + ... +

u∈V (Gm2 )

+

d(u, v) +

(d(u, s) + d(s, v))

u∈V (Gmi−1 )

(d(u, s) + d(s, v))

u∈V (Gmi+1 )

(d(u, s) + d(s, v)) − (n − 1) d(u, s)

u∈V (Gmn )

= m1 d(u, s) + w(s, Gm1 ) + m2 d(u, s) + w(s, Gm2 ) + ... + mi−1 d(u, s) +w(s, Gmi−1 ) + w(u, Gmi ) + mi+1 d(u, s) + w(s, Gmi+1 ) + ... + +mn d(u, s) + w(s, Gmn ) − (n − 1)d(u, s) = (m1 + m2 + ... + mi−1 + mi+1 + ... + mn − n + 1) d(u, s) n + w(s, GN ) − w(s, Gmi ) + w(u, Gmi ) i=1

=

n

(mj − n + 1) d(u, s) + w(s, Gm,n ) − w(s, Gmi ) + w(u, Gmi ).

j=1 j=i

Lemma 6 The degree distance of Gm1 · Gm2 is: DD(Gm·G 1 m2 ) = DD(Gm1 ) + DD(Gm2 ) + (n2 − 1) dd(s, Gm1 ) +(2|E(Gm1 )|w(s, Gm2 ) − w(s, Gm2 ) degGm1 (s)) +(n1 − 1) dd(s, Gm2 ) + (2|E(Gm2 )w(s, Gm1 ) − w(s, Gm1 ) degGm2 (s)) − w(s, Gm1 ) deg (s) − w(s, Gm2 ) deg (s).

Proof:

DD(Gm1 ·Gm2 ) =

w(u, Gm1 · Gm2 ) deg (u)

u∈V (Gm1 )∪V (Gm2 )

=

w(u, Gm1 · Gm2 ) deg (u) +

u∈V (Gm1 )

w(u, Gm1 · Gm2 ) deg (u)

u∈V (Gm2 )

−w(s, Gm1 · Gm2 ) deg (s) (we use lemma (4)) = (w(u, Gm1 ) + (n2 − 1)d(u, s) + w(s, Gm2 )) deg (u) u∈V (Gm1 )\{s}

861


+

(w(u, Gm2 ) + (n1 − 1) d(u, s) + w(s, Gm1 )) deg (u)

u∈V (Gm2 )\{s}

−w(s, Gm1 · Gm2 ) deg (s) = DD(Gm1 ) + DD(Gm2 ) + (n2 − 1)

deg(u) + (n1 − 1)

u∈V (Gm1 )\{s}

+w(s, G1)

d(u, s) deg (u)

u∈V (Gm1 )

+w(s, Gm2 )

d(u, s) deg (u)

u∈V (Gm2 )

deg (u) − w(s, Gm1 · G2 ) deg (s)

u∈V (Gm2 )\{s}

The hence result. We generalize the previous result. Let Gm,n be the graph composed by n graphs Gm1 .Gm2 · · · Gmn that possess respectively 1 , m2 , ..., mn vertices connected by a vertex s (see Figure 2) such m n that N = i=1 mi − n + 1 Lemma 7 The degree = Gm1 .Gm2 .....Gmn is DD(Gm,n ) = n distance of graph Gm,n n n n DD(G )− w(s, G ) deg (s)+ ( j=1 mj −n+1) dd(s, Gmi )+ mi mi Gmi i=1 i=1 i=1 j=i n w(s, Gm,n ) degGN (s) + i=1 (w(s, Gm,n ) − w(s, Gmi ))(2|EGmi | − degGmi (s)). Proof:

DD(Gm1 .Gm2 .....Gmn ) =

w(u, Gm1 .Gm2 .....Gmn ) deg u

u∈V (Gm1 .Gm2 .....G mn )

=


u∈V (Gm1 )∪..cupV (Gmn )

=


u∈V (Gm1 )\{s}

+... +

w(u, Gmn .Gm2 .....Gmn ) deg u

u∈V (Gmn )\{s}

+w(u, Gm1 .Gm2 .....Gmn ) deg s + w(s, Gm,n ) degGm,n (s) n n [( mj − n + 1)) d(u, s) = i=1 u∈V (Gmi )\{s}

j=1 j=i

+w(s, Gm,n ) − w(s, Gmi ) + w(u, Gmi )] deg u +w(s, Gm,n ) degGm,n (s) n n [( mj − n + 1) dd(s, Gmi ) = i=1

j=1 j=i


862

+(w(s, Gm,n) − w(s, Gmi ))(2|EGmi | − degGmi (s))

+DD(Gmi ) − w(s, Gmi ) degGmi (s)] + w(s, Gm,n ) degGm,n (s) The hence result. Corollary 8 If the graphs Gm have the some number of vertices m(mi = m) for i = 1, ..., n we have ⎧ ⎨ dd(s, Gmi ) = dd(s, Gmj ), (1) Gmi = Gmj = Gmn , for i, j∈ {1, 2, ..., n} ; ⎩ N=nm-n+1, The degree distance of star graph Gm,n = Gm · Gm · · · Gm is: DD(Gm,n ) = n DD(Gm ) + n(n − 1)(m − 1) dd(s, Gm) + n(w(s, Gm,n) −w(s, Gm ))(2|EGm | − degGm (s)) +n w(s, Gm )(degGm,n s − degGm (s)). With |EGm | the number of edges. Proof: We use lemma 7 and (1) in Corollary we obtains the formula.

3

Application

3.1 Case of a Tent Tm

Figure 3: Tent-graph

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Lemma 9 Let Tm be the Tent, m is the vertices (m ≥ 5) (see Figure 3 (a)) we have : m−1 2 m−1 ) + , dd(s, Tm ) = 3( 2 2 m2 − 1 w(s, Tm ) = , 4 (m − 1) (m − 2)(m + 3) + 3(m + 1) . W (Tm ) = 12 Lemma 10 The degree distance of Tm is DD(Tm ) =

2m3 +3m2 −14m+9 ,m 4

≥ 5.

Let Gm,n be the tent formed by n stars Tm connected by a vertex s. Gm,n is Tm . Tm ...Tm (see Figure 3 (b)). Theorem 11 The degree distance of tent-graph Gm,n = Tm .Tm ...Tm is DD(Gm,n ) =

2

2 3 2 m−1 m −1 +n(n−1) 3(m− + n 2m +3m4−14m+9 +n(n−1)(m−1) 3 m−1 2 2 4 1) . Proof: We use the corollary 1. 3.2 Case of a cycle Cm

Figure 4: Cycle-graph

Lemma 12 Let Cm be the cycle (see Figure 4 (a)) We have: m2 if m is odd, m ≥ 2, 2 dd(s, Cm ) = m2 −1 if m is even, m ≥ 3, 2 m2 if m is odd, m ≥ 2, 4 w(s, Cm) = m2 −1 if m is even, m ≥ 3, 4


W (Cm ) =

m3 if 8 m(m2 −1) 8

864

m is odd, m ≥ 2, if m is even, m ≥ 3,

Lemma 13 The degree distance of Cm is DD(Cm ) = 4W (Cm ) Let Gm,n be the graphs composed by a cycle Cm as a star.GN = Cm .Cm ...Cm (n times) (see Figure 4 (b)). Theorem 14 The degree distance of cycle-graph Gm,n = Cm .Cm ...Cm is: nm2 (2mn − n − m + 1) if m is odd, m ≥ 2, 2 DD(Gm,n ) = n(m2 −1) (2mn − n − m + 1) if m is even, m ≥ 3, 2 Proof: We use the corollary. 3.3 Case of a Benzenoid graph Lk In this section we will apply Corollary 8 to a particular, but chemically very important class of partial cubes-benzenoid graphs.The term benzenoid graph is used for graphs constructed in the following manner.Consider the hexagonal lattice k. Let Z be a circuit on this lattice.Then benzenoid graphs are formed by the vertices and edges of k lying on some circuit Z or in its interior. The vertices and edges belonging to Z form the perimeter of the respective benzenoid graph, while the vertices (if any) not belonging to the perimeter are said to be the internal vertices.

Figure 5: The Benzenoid graph Lk -graph

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Lemma 15 Let Lk be The Benzenoid graph (see Figure 5(a)) we have (k is hexagons): dd(s, Lk ) = 10 k 2 + 7k + 1, k ≥ 2 w(s, Lk ) = 4 k 2 + 4k + 1, k ≥ 2

1 W (Lk ) = 16k 3 + 132k 2 + 362k + 327 , 3

k≥2

Lemma 16 The degree distance of Lk (see Figure 5(b))is DD(Lk ) =

72k 2 + 47k + 3 , k ≥ 2

2 3

40k 3 +

Let Gk,n be the graphs composed by The Benzenoid graph Lk as a star.Gk,n = Lk .Lk ...Lk (n times) (see Figure 5(c)). Theorem 17 The degree distance of The Benzenoid graph-graph Gk,n = 2 3 2 Lk .Lk ...Lk is DD(Gk,n) = n 3 (40k +72k +47k +3) +n(n−1)(k −1)(10 k 2 + 7k + 1) + 2n(n − 1)(4k 2 + 4k + 1)(5k + 1). k ≥ 2

Proof: We use the corollary 8. 3.4 Case of a Linear phenylenes P Hk Consider the linear phenylene P Hk , consisting of k − 1 four-membered (cyclobutadiene) and k six-membered (benzene) rings, in which each cyclobutadiene unit is adjacent to two benzene rings, whereas benzene rings are not adjacent to each other.

Figure 6: The linear phenylene P Hk -graph

866


Lemma 18 Let P Hk be The linear phenylene we have: The number of vertices equals mk = 6k + 2, with k is hexagons (see Figure 6(a)) dd(s, P Hk ) = 24k 2 − 36k + 13, w(s, P Hk ) = 9k 2 − 12k + 4,

k≥2 k≥2

W (P Hk ) = 2(8k 3 − 8k 2 − k − 1),

k≥2

Lemma 19 The degree distance of P Hk is DD(P Hk ) = 12k 2 (8k+1),

k≥2

Let Gk,n be the graphs composed by a linear phenylene P Hk as a star.Gk,n = P Hk .P Hk ...P Hk (n times) (see Figure 6(b)). Theorem 20 The degree distance of linear phenylene -graph Gk,n = P Hk .P Hk

...P Hk (see Figure 6(c)) is DD(Gk,n) = n 12k 2 (8k+1) +n(n−1)(k−1)(24k 2 − 36k + 13) + 2n(n − 1)(9k 2 − 12k + 4)(8k − 7)

k≥2

Proof: We use the corollary.

4

Conclusion

In this article we give the formulas for the degree distance of certain particular graphs. As well as some applications for the calculation of the degree distances frequently used by chemists We have mentioned some such as Linear benzenoid chains graph There are numerous physical and chemical applications that make use of calculating the degree distances.

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