Global efficiency of graphs

Accepted Manuscript Global efficiency of graphs Bryan Ek, Caitlin VerSchneider, Darren A. Narayan PII: DOI: Reference:

S0972-8600(15)00002-X http://dx.doi.org/10.1016/j.akcej.2015.06.001 AKCEJ 1

To appear in:

AKCE International Journal of Graphs and Combinatorics

Received date: 11 February 2013 Revised date: 7 November 2013 Accepted date: 21 November 2013 Please cite this article as: B. Ek, C. VerSchneider, D.A. Narayan, Global efficiency of graphs, AKCE International Journal of Graphs and Combinatorics (2015), http://dx.doi.org/10.1016/j.akcej.2015.06.001 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

Global e¢ ciency of graphs Bryan Ek, Caitlin VerSchneidery, and Darren A. Narayan

Abstract The distance d(i; j) between any two vertices i and j in a graph is the number of edges in a shortest path between i and j. If there is no path connecting i and j, then d(i; j) = 1. In 2001, Latora and Marchiori introduced the measure of e¢ ciency between vertices in a graph [8]. The e¢ ciency between two vertices i 1 for all i 6= j. The global e¢ ciency of a graph is the average e¢ ciency over and j is de…ned to be 2i;j = d(i;j) all i 6= j. The concept of global e¢ ciency has been applied to optimization of transportation systems and brain connectivity. In this paper we determine the global e¢ ciency for complete multipartite graphs Km;n , star and subdivided star graphs, and the Cartesian Products Kn Pnm , Kn Cnm , Km Kn , and Pm Pn .

1

Introduction

The distance d(i; j) between any two vertices i and j in a graph is the number of edges in a shortest path between i and j. If there is no path connecting i and j, then d(i; j) = 1. In 2001, Latora and Marchiori introduced the measure of e¢ ciency between vertices in a graph [8]. The (unweighted) e¢ ciency between two 1 vertices i and j is de…ned to be 2i;j = d(i;j) for all i 6= j. The global e¢ ciency of a graph G (with n vertices) is X 1 2 (vi ; vj ), which is simply the average of the e¢ ciencies over all pairs of vertices. denoted Eglob (G) = n(n 1) i6=j

Global e¢ ciency has emerged in a plethora of real world applications including optimization of transportation systems [8, 9, 3, 4], as well as brain connectivity [2], [5] and [11]. Recently, Ek, Verschneider, and Narayan applied the concept of global e¢ ciency to the metro system in Atlanta [3]. The concept of reciprocal distance has been studied previously. In 1993, Plavši´c, Nikoli´c, Trinajsti´c, and Mihali´c introduced the Harary index of a simple X graph [10]. For a simple graph G with vertices v1 ; v2 ; :::; vn 1 the Harary index is denoted H(G) and equals d(vi ;vi ) . We note the close relationship between global 1 i
e¢ ciency and the Harary index, Eglob (G) = n(n2 1) H(G). There also have been other studies involving the Harary index and reciprocal distances [7], [12] and [13]. In this paper, we determine the global e¢ ciency for several families of graphs, including powers of graphs and Cartesian Products. Recall the kth power of a graph G, which is denoted Gk , where V Gk = V (G) and (u; v) 2 E(Gk ) if and only if the distance between u and v in G is less than or equal to k. We determine the global e¢ ciency for complete multipartite graphs, star and subdivided star graphs, and the Cartesian Products Kn Pnm , Kn Cnm , Km Kn , and Pm Pn . As a consequence, we determine new results involving the Harary index for these families of graphs.

1.1

Complete Multipartite Graphs

We will use Kt1 ;t2 ;:::;tr to denote the complete multipartite graph with r parts with ti vertices where 1 i r. We note that the distance between any pair of vertices in di¤erent classes is 1 and the distance between any pair of vertices in the same class is 2. This leads to our next theorem. School of Mathematical Sciences, Rochester Institute of Technology, Rochester, NY 14623 Department, Nazareth College, Rochester, NY 14618

y Mathematics

1

Theorem 1 Let G = Kt1 ;t2 ;:::;tr where n =

r X

ti . Then

i=1

Eglob (G) =

1 n(n

1)

r X

ti

(ti

i=1

1) 2

+ (n

ti )

(1)

Proof. Let vi be a vertex in a part with ti vertices. Then the shortest path from vi to any vertices in the same part is 2 and vi is adjacent to all vertices in other parts. Hence the e¢ ciencies of all pairs containing vi is r h i X ti (ti 2 1) + (n ti ) . Averaging over all vertices gives the desired result. i=1

1.2

Star graphs and subdivided star graphs

A star graph is a complete bipartite graph where one part is a single vertex, and the other part has at least one vertex. We will use S(k; l) to denote the star graph with k spokes each of length l. We next recall the well known operation of an edge subdivision. De…nition 2 An edge subdivision is an operation that is applied to an edge (u; v) where a new vertex w is inserted, and the edge (u; v) is replaced by edges (u; w) and (w; v). A subdivision H of a graph G is a graph that can be obtained by performing a sequence of edge subdivisions. De…nition 3 Let Sd;l be the subdivision of the star K1;d where each of the d edges is replaced by a path with l vertices. The graph S4;3 is shown below in Figure 1.

Figure 1. The star graph S4;3 The e¢ ciency matrix for S4;3 is given below.

2

Figure 2. The e¢ ciency matrix for S4;3 . We note that the patches of four identical entries across the top row begin a downward diagonal with an equivalent entry. On each side of the diagonals, there are patches of identical entries. The sum of these diagonals PL D(L+1 i) 4(2) 4(1) equals 4(3) , which we divide into two sums. The …rst i=1 1 + 2 + 3 . In the general case we have i sum includes the largest number of patches; the second sum begins with the smallest e¢ ciency patch and 1 includes the patches up to but not including the largest section. In our example the …rst sum equals 4(3) 2 2 4(3) 2 4(3) 3 4(3) 1 4(3) 2 + 2 3 + 2 4 . The second sum equals 2 6 + 2 5 . Now we proceed with the general case. We will use SD;L to denote a subdivided star graph with D arms of length L. Let vd;l denote the lth vertex along the dth arm. The center vertex is denoted v0 . The distance between non-center vertices is then d(vd;l ; vd0 ;l0 ) =

l + l0 if d 6= d0 jl l0 j if d = d0

(2)

The sum of e¢ ciencies can be broken up into two di¤erent sections: e¢ ciencies along the same arm and e¢ ciencies between arms. First, we consider the e¢ ciencies along the same arm. Note each arm is isomorphic to every other arm so the e¢ ciencies along one arm are identical to all other arms. A single arm, including v0 , is isomorphic to PL+1 so we have this sum to be " # L L L X k X X X L+1 k 1 1 =D = D (L + 1) L = D [(L + 1) HL L] (3) D i k k i=1 k=1

k=1

k=1

where HL is the Lth harmonic number. Second, we consider the e¢ ciencies along di¤erent arms. There are D 2 di¤erent pairs of arms. Since we are considering unweighted star graphs, each pair is isomorphic. Using

3

equation 2, the sum between two arms is: L X L X j=1 i=1

L+1 2L Xk 1 X 1 2L = + i+j k

=

k=2

k=L+2

L X

L X1

k + k+1

k=1

=

L X

k + k+1

k=1

=

L X

L X

[(k + L + 1) k+L+1

1]

L X 2L

[k + L] k+L+1

k=1

1 k+1

2L + 1 k+L+1

k=1

=

k=1

1)

L k k + k+1 k+L+1

k=1

=

2L

(k k

L X

L(2k + 1) (k + L + 1)(k + 1)

k=1

The last few expressions were given as options. The total from di¤erent arms is then L

D(D 1) X 2L + 1 2 k+L+1 k=1

1 k+1

(4)

Finally we can combine the two sums: equations 3 and 4, and normalize with n = DL + 1 to obtain the global e¢ ciency given in equation 5. Theorem 4 2 Eglob (SD;L ) = DL(DL + 1) =

2 L(DL + 1)

D

L X L+1

k

k=1

L X

L+1 k

k=1

k

k

L

D(D 1) X 2L + 1 + 2 k+L+1

1 k+1

k=1

+

D

1 2

2L + 1 k+L+1

!

1 k+1

(5)

The sum of e¢ ciencies for a star graph with D spokes, each of L length is: X

ij

=2

i;j

L X D(L + 1 i=1

= 2D

i

L X (L + 1 i i=1

i)

+

L X D(D

2

i=1

i)

+

1)

(D

1)

i i + i + 1 2L + 1

i i + i + 1 2L + 1

2

i

L X1

D(D 1) L + i 2 L +1 i=1 ! (D 1) L 2 L+1

! (6)

The unweighted global e¢ ciency, with n = D (L) + 1, is: Eglob ((D; L)) 2D = n(n 1) 2 = L(DL + 1)

L X (L + 1 i i=1

L X (L + 1 i i=1

i)

+ i)

(D

1)

i i + i + 1 2L + 1

2 +

(D

1) 2

4

i i + i + 1 2L + 1

(D i

1) 2

(D i

L L+1 1)

2

!

L L+1

(7) !

(8)

We assume that our graph is embedded in the plane. Then we consider the Euclidean distances between vertices where the distance between two adjacent vertices is de…ned to be 1. Furthermore, we consider all spokes to be linear and evenly spaced around the center vertex, v0 . Creating a weighted e¢ ciency can e¤ectively approximate real life networks such as a subway system [3]. This is found by taking the unweighted global e¢ ciency and dividing by the ideal global e¢ ciency. A partial completion is shown in Figure 3.

Figure 3. ‘Ideal’connections in a star graph The matrix in Figure 4 gives the e¢ ciencies of a star graph where each pair of vertices is connected with an edge weighted by the p p Euclidean distance between them. For example, the Euclidean distance between v8 and v11 is 22 + 32 = 13. Hence 8;11 = p113 .

5

Figure 4. The Eulcidean e¢ ciency matrix of the star graph S4;3 Notice that the diagonal terms equal to 1 are identical to the matrix of the subdivided star graph. As a PL D(L+1 i) result, the term for this sum is . In this case, we are considering the patches to be squares i=1 i indexed by i and j. The sum of the entries in the block i = 1, j = 2 equals 8 p15 + 4 13 = p45 + 34 + p45 . A path from v1 to v8 includes an angle of 2 . Identical angles within a block are shaded alike. The terms can be expressed using the law of cosines, Sum = p 2 1 + 22

4

4 +p 2 1 2 cos( 2 ) 12 + 22

2 1 2 cos( )

+q

4 12

D i +j 2 2ij cos( 2D )

The generic terms in a given block can be written as p 2

+

22

where

2 1 2 cos( 32 ) varies from 1 to D

1. We

then sum over all of the blocks and add in the diagonal terms to yield Equation 9. The sum of Euclidean (ideal) e¢ ciencies for a star graph with D spokes, each of L length is given by: X i;j

ij

=2

L X D(L + 1

i)

i

i=1

+

L D L X X1 X i=1 j=1

=1

q

D i2 + j 2

(9)

2ij cos( 2D )

We de…ne the global e¢ ciency ratio ERglob to be the ratio of the unweighted global e¢ ciency to the Euclidean (ideal) e¢ ciency. The global e¢ ciency ratio of the star graph is found by dividing Equation. 6 by Equation 9. Theorem 5 ERglob ((D; L)) =

PL h 2(L+1 i=1

i

PL

i=1

i)

+ (D

2(L+1 i) i

1)

+

PL

i i+1

j=1

+

PD

i 2L+1 i

1 =1

p2

i

i +j 2

(D

1)

L L+1

(10)

1 2ij cos( 2D )

We note that as D increases, the global e¢ ciency ratio decreases as a path between two vertices on di¤erent spokes must pass through the center vertex. However as L increases the graph bears a closer resemblance to a path. Hence the weighted global e¢ ciency will increase with a limit of 1.

1.3

Cartesian Products

We next investigate the e¢ ciency of powers of a path Pn . Su, Xiong, and Gutman obtained the Harary index k k of Pm , from which Eglob (Pm ) can easily be obtained. However, we include a computation of Eglob (Pnm ), as it is useful for obtaining the global e¢ ciency for the families Kn Pnm and Kn Cnm . For the global e¢ ciency of path power graphs: Eglob (Pnm ), each element of the e¢ ciency matrix is given by 2ij = l

1 ji jj m

m

(11)

where i is the row and j is the column of the entry. This value corresponds to the e¢ ciency between vertices m i and j. The distance between the vertices in Pn lis simply m ji jj. In Pn , each step can be up to m vertices ji jj away. Hence the distance between vertices equals . Taking the inverse gives the formula in equation 11. m Hence the matrix is:

6

v1

v1 0

v2 1

v3 1

v2

1

0

1

v3 .. .

1

1 .. .

0

1 d nm4 e

1 d nm5 e

vn

2

vn

1

1 d nm3 e 1 d nm2 e

1 d nm2 e

1 d nm3 e

n X j=2

For the second vertex we have, 1

j2

jj n X j=3

21;j = n

i=1

1 i m

+

n X2 i=1

1 i m

vn 1 d nm1 e

1 d nm5 e

1 d nm4 e

1 d nm3 e

1 d nm3 e

.. .

1 d nm2 e

0

1

1

1

0

1

1

1

0

1) other vertices to compute the e¢ ciency with. The sum

n X

l

j=2

1 j1 jj m

2 since 3

22;j =

n X

+

+

j=3

Summing the terms for all vertices gives: n X1

1

1 d nm2 e

.

1 d nm4 e

Consider the …rst vertex of Pnm . There are (n of e¢ ciencies from the …rst vertex as:

vn

2

1 d nm3 e 1 d nm4 e

..

1 d nm3 e

1 d nm1 e

vn

vn

l

m=

1 j2 jj m

j

m=

2 X 1

i m

i=1

+

n X1 i=1

1 i m

:

n, which yields: n X2 i=1

1 i m

1 X 1 i=1

i m

=

k n X1 X k=1 i=1

1 i m

Finally, we divide this term to get the following result. We note as the matrix is symmetric, we can sum over the upper half of the matrix (ordered pairs) and then multiply our result by 2. Theorem 6 Let 1

m

n. Then Eglob (Pnm ) =

2 n(n

1)

n k X1 X k=1 i=1

1 i m

(12)

De…nition 7 The Cartesian product of two graphs G and H is a graph G H, with the vertex set V (G) V (H), where vertices f(i1 ; i2 ) ; (j1 ; j2 )g are adjacent if fi1 ; j1 g 2 E(G) and i2 = j2 , or fi2 ; j2 g forms an edge in H and i1 = j1 . In the …gure below, we show the graph of the Cartesian product of K4 and P42 .

7

Figure 5. The Cartesian product of K4 and P42 The e¢ ciency matrix is given in Figure 6 below.

P42

Figure 6. The e¢ ciency matrix for K4 We next extend to the general case. Theorem 8 Eglob (Kr

Pnm )

2 = nr(nr

1)

"n 1 k XX k=1 i=1

8

1 i m

r (r + 1+

1) i m

!

nr(r 1) + 2

#

(13)

Proof. Notice that the matrix is very similar to that of a path power. Each i now corresponds to a block in the k th row from the bottom. Each block has r terms on the main diagonal of a block and these correspond to the pairs of vertices in the r separate path Kr powers. All other terms correspond to the distance between vertices in di¤erent path powers and then within the copies of the complete subgraphs. For this, we have r vertices in the initial class to choose from and r 1 vertices in the Kr . Since each class is complete, it will only take one additional step to reach the …nal vertex, and so the e¢ ciency is only slightly smaller. There are also ‘triangles’ of 1s next to the main diagonal; these correspond to movements within a single Kr . The number of 1s is then n times the number of edges in Kr which equals r(r2 1) . Averaging the e¢ ciencies over all pairs yields Equation 13. We next investigate the global e¢ ciency of a Cartesian product of a complete graph and a cycle. The graph of the Cartesian product of K3 and C62 is shown in Figure 7.

Figure 7. The Cartesian product of K3 and C62 The e¢ ciency matrix is found below.

9

Figure 8. The e¢ ciency matrix of K3

C6 .

For a Cartesian product between a complete graph Kr and a cycle power Cnm , we must divide the global e¢ ciency into two cases where n is either odd or even. Theorem 9 If n is odd then Cnm ) =

Eglob (Kr

1 rn

1

If n is even then Eglob (Kr

Cnm ) =

1 rn

1

2

42

n=2 X i=1

1 i m

2

n

42

1

2 X

i=1

+

1 i m

r 1 1 + mi

+

!

r 1 1 + mi

+r

1

!

+r

1 n 2m

3

15 .

+

(14)

r n 2m

!3

1 5. +1

(15)

Proof. Each i corresponds to a single line of each block. Also, each i has one entry that falls on the main diagonal of an r r block that corresponds to pairs of vertices within a cycle power. All other terms correspond to pairs of vertices that are in di¤erent cycle powers but in di¤erent copies of Kr . There are also 1s next to the main diagonal; these correspond to movements within a single complete graph. The number of 1s is then r 1: the number of vertices that are available for the …nal position. Averaging the e¢ ciencies over all pairs of

10

vertices yields Equation 14 and 15.

Eglob (Kr

1 Cnm ) = nr(nr =

1 nr

1

2

1) 2

42

n

n

nr 42

1

2 X

1

r 1 + 1 + mi

i m

i=1

1

2 X

i=1

1

i m

r 1 + 1 + mi

!

+r

!

3

15

+r

3

15

For the even case, note that the term corresponding to the e¢ ciency of moving across the diameter is counted twice, so it must be subtracted to obtain Equation 15. Theorem 10 Eglob (Km We obtain the global e¢ ciency for Km Eglob (Km

Kn ) =

nm + m + n 3 2(nm 1)

Kn using Eglob Km

Pnn

1

(16)

and Eglob Km

Pnn 1 ) 1 2 0 n k X1 X 2 m m(m 1) A + nm(m 4 @l m+l m = i i nm(nm 1) 2 k=1 i=1 n 1 n 1 +1 "n 1 k # XX 1 m 1 2 n(m 1) = + + n(nm 1) 1 1 + 1 2 k=1 i=1 "n 1 k # XX 1 (m + 1) + n(m 1) = n(nm 1) k=1 i=1 "n 1 # X 1 = (k(m + 1)) + n(m 1) n(nm 1)

Kn ) = Eglob (Km

bnc Cn 2 .

3 1) 5

k=1

1 (n 1)(n 1 + 1) (m + 1) + n(m n(nm 1) 2 1 n 1 (m + 1) + (m 1) = nm 1 2 1 nm m n 1 = + +m 1 nm 1 2 2 2 2 nm + m + n 3 = 2(nm 1) =

We note that the ceiling functions were dropped since 1

11

i

n

1 implies 0 <

1 n 1

1)

i n 1

1 which makes

the ceiling terms always equal to 1. For the Cartesian product of a complete graph and an odd cycle, we have: n

Eglob (Km

1

Kn ) = Eglob Km Cn 2 2 n 10 2 X 1 42 @l = nm 1 i=1

1 i (n 1)=2

m+l

m i (n 1)=2

2 n 1 2 X 1 1 m 1 42 = + +m nm 1 1 1+1 i=1 2 n 1 3 2 X 1 m + 1 42 = + m 15 nm 1 2 i=1 1 n 1 (m + 1) + m nm 1 2 nm + m + n 3 = 2(nm 1)

=

1 m

3

+1

1

A+m

3

15

15

1

For the Cartesian product of a complete graph and an even cycle, we have: n

Eglob (Km

Kn ) = Eglob Km Cn2 1 0 13 2 n 0 2 X 1 42 @ l 1 m + l m m 1 A + m 1 @ l 1 m + l m m1 A5 = i i n n nm 1 i=1 n=2 n=2 + 1 2(n=2) 2(n=2) + 1 2 n 3 2 X 1 1 m 1 1 m 1 42 5 = + +m 1 + nm 1 1 1+1 1 1+1 i=1 2 n 3 2 X m + 1 1 1 42 +m 1 (m + 1)5 = nm 1 2 2 i=1 1 n (m + 1) + m 1 nm 1 2 1 n 1 = (m + 1) + m nm 1 2 nm + m + n 3 = 2(nm 1) =

Thus Eglob (Km

1.3.1

1 (m + 1) 2

1

Kn ) is given by Equation 16.

Grid Graphs: Pm

Pn

Consider the grid graph Pm Pn which is embedded in the plane. The vertex in the upper left corner is labeled v1;1 and vi;j is used to label vertex that is obtained by starting at vertex v1;1 and travelling i 1 positions to the right and then j 1 units downward.

12

Figure 9. A grid graph Now consider the graph P3

P6 .

Figure 10. The graph P3

P6

The initial block of 9 vertices from v1;1 to v3;3 creates the graph P3 P3 . Adding sets of 3 additional vertices, v1;4 to v3;4 up to v1;6 to v3;6 we obtain the entire graph of P3 P6 . This can be seen in Figure 8. The e¢ ciency matrix below is divided into sections of P3 Pn where n 6.

13

Figure 11. The e¢ ciency matrix for P3

P6

Our …rst goal is to sum the e¢ ciencies of Pm Pn . We shall consider the copies of Pm to be ‘vertical’and the Pn copies to be ‘horizontal’. To sum theP e¢ ciencies we begin by considering the n copies of Pm sum P. mThe m 1 1 of e¢ ciencies between a single Pm is simply k=1 mk k . So our total for vertical connections is n k=1 mk k . Pn 1 Similarly, our total for horizontal connections is m k=1 n k k . Next we determine the remaining e¢ ciencies. Consider two copies of Pm . There are n i pairs of Pm that are separated by a horizontal distance of i n 1. There are 2(m j) pairs of points in separate Pm that are separated by a vertical distance of j m 1. Thus the sum of e¢ ciencies of the cross terms is Pn 1 Pm 1 2(n i)(m j) . Since the total number of vertices is nm, our global e¢ ciency is: i=1 j=1 i+j Eglob (Pm

Pn ) =

2 mn(mn

2 = mn(mn =

2 mn(mn

1)

2

1)

2

1)

2

4n

m X1 k=1

m X1

4

m

k=1

nm k

4m + n

k k

+m

n X1

n

k=1

n(m

k k

1) +

n X1 k=1

2nm +

m X1 k=1

+

n X1 m X1 i=1 j=1

mn k

m(n

2(n

i)(m i+j

1) + 2

j) 5

n X1 m X1

(n

i=1 j=1

n 1 n X1 m X1 (n nm X nm + +2 k k i=1 j=1 k=1

3

i)(m i+j

i)(m i+j 3 j) 5

3 j) 5

Using a weight corresponding to the Euclidean distance, we can obtain the global e¢ ciency ratio which compares the e¢ ciency using distances along the lines of the grid versus the ideal Euclidean distance.

14

Theorem 11 The global e¢ ciency ratio is given by: h Pm 1 2 2nm + k=1 mn(mn 1) m + n ERglob (Pm Pn ) = Pm 1 2 2nm + k=1 mn(mn 1) m + n

nm k

+

nm k

+

Pn

1 nm k=1 k

Pn

1 nm k=1 k

+2 +2

Pn

1 i=1

Pn

1 i=1

Pm

1 (n i)(m j) j=1 i+j

Pm

1 (n i)(m j) p2 2 j=1 i +j

We can use the close relationship between global e¢ ciency and the Harary index, Eglob (G) = to obtain new results. Corollary 12 Let H(G) be the Harary index of a graph G. Then we have: Pn 1 Pk 2 1 2 + r(r 1) + nr(r2 1) (i) H (Kn Pnm ) = n2 k=1 i=1 nr(nr 1) d mi e 1+d mi e P n2 1 2 1 1 (ii) H (Kn Cnm ) = n2 + r 1 +r 1 i=1 nr 1 2 d mi e 1+d mi e h i 3 (iii) H (Km Kn ) = n2m nm+m+n h 2(nm 1)h Pn 1 nm Pn 1 Pm 1 Pm 1 2 (iv) H (Pm Pn ) = n2m mn(mn 2nm + k=1 nm k=1 k + 2 i=1 j=1 1) m + n k +

i

(17)

2 n(n 1) H(G),

(n i)(m j) i+j

ii

Acknowledgments The authors are grateful for an anonymous referee, for their valuable comments and alerting us that the global e¢ ciency measure is the Harary index divided by n2 . Research was supported by an National Science Foundation Research Experience for Undergraduates Site Award #1062128.

References [1] K. C. Das, K. Xu, I.N. Cangul, A. S Cevik, and A. Graovac, On the Harary index of graph operations, Journal of Inequalities and Applications (2013), 339. [2] B. Ek, C. VerSchneider, J. Lind and D. A. Narayan, Real world graph e¢ ciency, to appear in the Bulletin of the Institute of Combinatorics and Its Applications [3] B. Ek, C. VerSchneider, and D. A. Narayan, E¢ ciency of star-like networks and the Atlanta subway network, Physica A, 392 (2013) 5481-5489. [4] B. Ek, C. VerSchneider, and D. A. Narayan, Local E¢ ciency and Clustering Coe¢ cients of Graphs, submitted. [5] C. Honey, R. Kötter, M. Breakspear, and O. Sporns. Network structure of cerebral cortex shapes functional connectivity on multiple time scales, PNAS, Vol. 104, No. 24, (2007), 10240-10245. [6] A. Ili´c, G. Yu, and L. Feng, The Harary index of trees, arXiv:1104.0920v3 [math.CO]. [7] O. Ivanciuc, T.S. Balaban, and A.T. Balaban, J. Math. Chem, 1993, 12, 309-318. [8] V. Latora and M. Marchiori, E¢ cient Behavior of Small-World Networks, Physical Review Letters E, Vol. 87, No. 19, (2001). [9] V. Latora and M. Marchiori, Is the Boston subway a small-world network?, Physica A 314 (2002) 109 – 113:

15

[10] D. Plavši´c, S. Nikoli´c, N. Trinajsti´c, and Z. Mihali´c, On the Harary index for the characterization for chemical graphs, J. Math. Chem., 12 (1993), 235-250. [11] O. Sporns, Networks of the Brain, MIT Press, 2010. [12] G. Su, L. Xiong, and I. Gutman, Harary index of the k-th power of a graph, Appl. Anal. Discrete Math. 7 (2013), 94–105. [13] H. Wang and L.Kang, More on the Harary index of cacti. J. Appl. Math. Comput. 43 (2013), no. 1-2, 369–386.

16