A Graph Theoretic Analysis of Leverage Centrality

0.18 MB·

by  Roger Vargas,

A Graph Theoretic Analysis of Leverage Centrality
7 Roger Vargas, Jr.∗ Abigail Waldron† Anika Sharma‡
1
0 Rigoberto Flo´rez§ Darren A. Narayan¶
2

n January 18, 2017
a
J

7
1 Abstract

] In 2010, Joyce et. al defined the leverage centrality of vertices in a graph as a means to analyze
O
functional connections within the human brain. In this metric a degree of a vertex is compared to
C the degrees of all it neighbors. We investigate this property from a mathematical perspective. We first
. outlinesomeofthebasicpropertiesandthencomputeleveragecentralitiesofverticesindifferentfamilies
h
ofgraphs. Inparticular,weshowthereisasurprisingconnectionbetweenthenumberofdistinctleverage
t
a centralities in theCartesian product of paths and thetriangle numbers.
m
[
1 Introduction

1
v
In a social network people influence each other and those with lots of friends often have more leverage (or
0
influence) than those with fewer friends. However the true influence of a person not only depends on the
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7 numberoffriendsthatthey have,butalsoonthe number offriends thattheirfriends have. Apersonthatis
4 wellconnected canpass informationto many friends, but if their friends are also receivinginformationfrom
0 others, their influence on others is lessened. The extreme cases of influence occurs with a person who has a
.
1 large number of friends, and for each of the friends, their only source of information is the original person.
0 In this situation, the original person has the highest possible influence and all of the others have the lowest
7 possible influence.
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The level of influence can be quantified by a property defined by Joyce et al. [6] known as leverage
:
v centrality. Werecallthatthe degreeofavertexv isthenumberofedgesincidenttov andisdenoteddeg(v).
i We next give a formal definition of leverage centrality [6].
X
r
a Definition 1 (leverage centrality) Leverage centrality is a measure of the relationship between the degree of
a given node v and the degree of each of its neighbors v , averaged over all neighbors N , and is defined as
i v
shown below:
1 deg(v)−deg(v )
l(v)= i .
deg(v) deg(v)+deg(v )
i
viX∈Nv
ThispropertywasusedbyJoyceetal. [6]intheanalysisoffunctionalmagneticresonanceimaging(fMRI)
data[6]andhasalsobeenappliedtoreal-worldnetworksincludingairlineconnections,electricalpowergrids,
andcoauthorshipcollaborations[8]. Howeverdespite thesestudies leveragecentralityhasyettobe explored
from a mathematical standpoint. The formula gives a measure of the relationship between a vertex and its
∗Mathematics andStatistics,WilliamsCollege,Williamstown,MA01267USA
†Mathematics Department,PresbyterianCollege,Clinton,SC29325
‡DepartmentofComputer ScienceandDepartmentofMathematics, UniversityofBuffalo,Buffalo,NY14260
§DepartmentofMathematics andComputer Science,TheCitadel,Charleston,SC29409
¶SchoolofMathematical Sciences,Rochester InstituteofTechnology, Rochester,NY14623-5604
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neighbors. A positive leverage centrality means that this vertex has influence over its neighbors, where as a
negative leverage centrality indicates that a vertex is being influenced by its neighbors.
We begin with an elementary result involving the bounds of leverage centrality (Li et al. [8]).
Lemma 2 Let G be a graph with n vertices. For any vertex v, |l(v)| ≤ 1− 2. Furthermore, these bounds
n
are tight in the cases of stars and complete graphs.
We note that the bounds are also tight for regular graphs.
ThereexistgraphsGwheretheleveragecentralityofallverticesisequalandwheretheleveragecentrality
of vertices is distinct. It is clear that if G is a regular graph than l(v) = 0 for every v ∈ G. We give an
example below of a graph that has distinct leverage centralities.
0 1 5−4 = 1
5 5+4 45
1 1 (cid:16)5−6 (cid:17)=− 1
5 5+6 55
2 0 (cid:16) (cid:17)
3 1 3 6−5 + 6−3 = 10
6 6+5 6+3 99
4 1(cid:16) (cid:16)4−6 (cid:17)+2(cid:16)4−5(cid:17)(cid:17)+ 4−3 =− 22
4 4+6 4+5 4+3 315
5 12(cid:16)(cid:16)3−6(cid:17)=−(cid:16)2 (cid:17) (cid:16) (cid:17)(cid:17)
3 3+6 9
6 1 (cid:16)2 6−(cid:17)5 + 6−4 + 6−3 = 59
6 6+5 6+4 6+3 495
7 1(cid:16)2(cid:16)5−6(cid:17)+(cid:16)5−4(cid:17) =(cid:16)− 7(cid:17)(cid:17)
5 5+6 5+4 495
8 1−(cid:16)6 =(cid:16) −5(cid:17) (cid:16) (cid:17)(cid:17)
1+6 7
9 1 3 6−5 + 6−1 = 38
6 6+5 6+1 231
(cid:16) (cid:16) (cid:17) (cid:16) (cid:17)(cid:17)
Figure 1. A graph with distinct leverage centralities
Intuitively one would think that the sum of the leverage centralities over a graph would be zero. This is
in fact the case when a graph is regular. However,for non-regular graphs the sum of leverage centralities is
negative. Thisarisessinceeachedgebetweentwoverticesofdifferentdegreescontributesanegativeamount
to the sum of the leverage centralities. Let G be the graph K with a pendant edge (see Figure 2).
3
Figure 2. Calculating leverage centrality
2
Then l(v ) = 1 2−3 + 1(2−2); l(v ) = 1 3−1 + 1(3−2)+ 1(3−2); l(v ) = 1 1−3 ; and l(v ) =
1 2 2+3 2 2+1 2 3 3+1 3 3+2 3 3+2 3 1 1+3 4
1 2−3 + 1(2−2). (cid:16)We c(cid:17)an regroup the sum to(cid:16)be (cid:17) l(v ) = 1 2−3 + 1(2−2)+ 1(cid:16) 3−(cid:17)1 + 1(3−2)+
2 2+3 2 2+2 i 2 2+3 2 2+1 3 3+1 3 3+2
(cid:16) (cid:17) vXi∈G (cid:16) (cid:17) (cid:16) (cid:17)
1(3−2)+ 1 1−3 + 1 2−3 + 1(2−2)
3 3+2 1 1+3 2 2+3 2 2+2
= 1 2−(cid:16)3 +(cid:17)1(3−(cid:16)2) +(cid:17) 1 3−1 + 1 1−3 + 1(3−2)+ 1 2−3 + 1(2−2)+ 1(2−2)
2 2+3 3 3+2 3 3+1 1 1+3 3 3+2 2 2+3 2 2+1 2 2+1
Sin(cid:16)ce(cid:16)the fi(cid:17)rst three pa(cid:17)rts(cid:16)are(cid:16)nega(cid:17)tive an(cid:16)d th(cid:17)e(cid:17)last(cid:16)part is zero,(cid:16)the s(cid:17)um(cid:17) m(cid:16)ust be negative. (cid:17)
Proposition 3 For any graph G, l(v)≤0.
v∈G
X
Proof. If G is a regular graph, then l(v)=0 for all v, and hence l(v)=0. If G is not regular,there
v∈G
X
mustexistanedgeewithendverticesuandvwhered(u)>d(v). Wenotethatthecontributionofeachedge
uv to the sum of the leveragecentralities is 1 d(u)−d(v) − 1 d(u)−d(v) <0. Hence for a non-regular
d(v) d(u)+d(v) d(u) d(u)+d(v)
graph, the sum of the leverage centralities is (cid:16)l(v)= (cid:17) 1 (cid:16) d(u)−d(v)(cid:17) − 1 d(u)−d(v) <0.
d(v) d(u)+d(v) d(u) d(u)+d(v)
vX∈G (uX,v)∈G (cid:16) (cid:17) (cid:16) (cid:17)
2 Vertices with positive / negative leverage centrality
A vertex of lowest degree cannot have a positive leverage centrality and a vertex of highest degree cannot
have a negative leverage centrality. However it is possible to have all the vertices in a graph except for one
to have negative leverage centrality, or all but one have positive leverage centrality. The star graph K
1,n−1
has n−1 vertices with negative leverage centrality. We show in the next theorem there exist graphs where
there are n−1 vertices with positive leverage centrality.
Theorem 4 The maximum number of vertices with positive leverage centrality is n−1.
Proof. Since the sum of leverage centralities over all vertices in a graph is less than or equal to zero, it
is impossible for a graph to have n vertices with positive leveragecentrality. Let G be a graph with vertices
v ,...,v ,wheren≥11,andedges{v v |1≤i<j ≤n−4}∪{v v |1≤i≤n−4andn−3≤j ≤n−1}∪{v v
1 n i j i j i n
| n−3≤i≤n−1}. We note that deg(v )=n−2 for 1≤i≤n−4, deg(v )=n−3, for n−3≤i≤n−1,
i i
and deg(v ) = 3. Then l(v ) > 0 for all 1 ≤ i ≤ n−4 since these vertices have the largest degree in G.
n i
Then for n−3≤i≤n−1, l(v )= 1 (n−4)(n−3)−(n−2) + (n−3)−3 = 1 (n−10). Here l(v )>0
i n−3 (n−3)+(n−2) (n−3)+3 n(2n−5) i
whenever n≥11. Hence we have n−1 v(cid:16)ertices with positive leverage c(cid:17)entrality.
We present a second example. Let G be a graph with n ≥ 12 vertices v ,v ,...,v and edges: {v v |
1 2 n i j
1 ≤ i < j ≤ n−5} ∪ {v v | 1 ≤ i ≤ n−5 and n−4 ≤ j ≤ n−1} ∪ {v v } ∪ {v v } ∪ {v v |
i j n−4 n−2 n−3 n−1 i n
n−3 ≤ i ≤ n−1}. It is clear that l(v ) > 0 for all 1 ≤ i ≤ n−5 since these vertices have the maximum
i
degree. Then for n−4≤i≤n−1, l(v )= 1 (n−5)(n−3)−(n−2) + (n−3)−4 = n2−15n+40 which is
i n−3 (n−3)+(n−2) (n−3)+4 2n3−9n2+4n+15
positive when n>11.531. (cid:16) (cid:17)
2.1 Leverage Centrality vs. Degree Centrality
Degree centrality weights a vertex based on its degree. A vertex with higher (lower) degree is deemed
more (less) central. This property has been well-studied (for early works see Czepiel [1], Faucheaux and
Moscovici[2],Freeman[3],Garrison,[4],HannemanandNewman[5],KajitaniandMaruyama[7],Mackenzie
[9], Nieminen [10], [11], Pitts [12], Rogers [13], and Shaw [14]). For some families of graphs the leverage
centrality and degree centralities of vertices are closely related. For example, in scale-free networks where
thedistributionofdegreesfollowsthe powerlaw,verticeswithlargedegreewillbeadjacenttomanyvertices
with much lower degrees. Hence the leverage centrality of these vertices will also be high.
3
However,forotherfamilies ofgraphsleveragecentralityanddegreecentralityarenotcloselyrelated. We
showinthefollowingexampleitispossibletoconstructinfinitefamiliesofgraphswherethevertexoflargest
degree does not have the highest leverage centrality. We do this by connecting nearly complete graphs as
shown in Figure 3.
Figure 3. A family of connected nearly complete graphs
For all n≥5, we have deg(u)>deg(v), however l(u)<l(v).
LetubeavertexinK thathasaneighborvertexontheK graph. Then,deg(u)=nandasn→∞,
n+1 n
it follows that deg(u) → ∞. Let v be the vertex that is the base of the claw graph found on the right side
of the graph shown in Figure 2. Thus, the degree of v will always equal 4 and therefore, for all n ≥ 5,
deg(u)>deg(v).
Since we know the degree of the neighbors of u, we can calculate the leverage centrality of u as shown:
1 n−(n−1) n−n 1
l(u)= +(n−1) = .
n n+(n−1) n+n 2n2−n
(cid:18) (cid:18) (cid:19)(cid:19)
Thus, if we take the limit of the leverage centrality of u as n→∞ we get:
1
lim =0.
n→∞2n2−n
We can also calculate the leverage centrality of v:
1 4−2 4−1 8
l(v)= +(3) = .
4 4+2 4+1 15
(cid:18) (cid:18) (cid:19)(cid:19)
Since the leverage centrality of u converges to 0 as n → ∞, and the leverage centrality of v is equal to 8 ,
15
then l(v)>l(u) ∀n≥5.
2.2 Leverage Centrality Zero
We note that bounds given in Lemma 2 are tight for regular graphs, where the leverage centrality of all
vertices is zero. In fact, it is straightforward to show that l(v) = 0 for every vertex v if and only if G is a
regular graph. It is also clear that for a vertex v with degree k that if all of the neighbors of v have degree
k, then l(v) = 0. However, it is possible for a vertex to have a leverage centrality of zero without all of its
neighbors having the same degree as the original vertex. We investigate this property below.
Example 5 Let G be a graph containing a vertex v of degree k where k−1 of v”s neighbors have degree
k =2 and the remaining neighbor has degree 1. Then l(v)= 1 k−1 +(k−1) k−(k+2) =0.
k k+1 k+(k+2)
(cid:16) (cid:16) (cid:17)(cid:17)
Wealsogiveanexampleofagraphwithavertexvwhoseneighborsallhavedistinctdegreesandl(v)=0.
Example 6 Let G be a graph containing a vertex v of degree 3 and the neighbors of v have degrees 1, 2, and
17. The leverage centrality of v is l(v)= 1 3−1 + 3−2 + 3−17 =0.
3 3+1 3+2 3+17
(cid:16) (cid:17)
4
Itwouldbe aninteresting problemindeedto determine necessaryandsufficient conditions fora vertex v
to have leverage centrality zero, particularly when the neighbors of v all have distinct degrees. A computer
search gives several examples for vertices with small degree.
d(v) degrees of the neighbors of v
7 1,2,3,7,11,33,77
7 1,2,3,9,11,33,41
d(v) degrees of the neighbors of v 7 1,2,3,11,13,23,33
3 1,2,17 7 1,2,5,7,11,21,49
3 1,3,9 7 1,2,5,7,11,28,33
4 1,2,5,41 7 1,2,5,8,13,17,38
5 1,2,4,13,37 7 1,2,5,9,11,13,73
5 1,2,5,10,37 7 1,2,5,11,14,17,21
5 1,3,5,7,35 7 1,3,4,5,8,37,81
6 1,2,3,6,36,66 7 1,3,5,7,8,21,49
7 1,3,5,7,8,28,33
7 1,3,5,8,9,13,73
7 1,3,5,8,14,17,21
3 Complete Multipartite Graphs
We use K to denote the complete multipartite graph with parts of sizes t ,t ,..,t and each vertex
t1,t2,...,tr 1 2 r
in a part is adjacent to every vertex in each of the other parts. As noted in [8] for vertices in the star graph
K the leveragecentrality meets the two extremes. The vertex in a part by itself has leveragecentrality
1,n−1
1 (n−1)(n−1)−1 =1− 2 and all other vertices have a leverage centrality of 1 1−(n−1) =−1+ 2.
n−1 (n−1)+1 n 1 1+(n−1) n
W(cid:16)e can extend th(cid:17)e same idea to the general case of complete multipartite gra(cid:16)phs. We(cid:17)will use G =
K to denote a complete multipartite graph with r parts n ,n ,...,n where each part n has order
t1,t2,...,tr 1 2 r i
t for all 1≤i≤r.
i
Theorem 7 Let G=K where t is the order of part n . Then
t1,t2,...,tr i i
1 t −t
l(v )= t k i
i k
j6=itj k6=i j6=itj + j6=ktj!
X
P  P P 
Proof. Let v be a vertex in part n with degree t . Due to the nature of a complete multipartite
i i j6=i j
graph, it follows that v will have t neighbors in part n , t neighbors in part n , t neighbors in part n ,
i 1 1 2 2 i i
P
andthe patterncontinuesfor all1≤i≤r groups. Notethateveryvertexv ∈n willhavedegree t .
k k j6=k k
Thus the leverage centrality of v can be calculated as follows:
i
P
l(v )= 1 t j6=itj − j6=1tj +t j6=itj − j6=2tj +···+t j6=itj − j6=rtj
i t 1 t + t 2 t + t r t + t
j6=i j Pj6=i j Pj6=1 j! Pj6=i j Pj6=2 j! Pj6=i j Pj6=r j!!
P P P P P P P
= 1 t j6=itj − j6=ktj
k
j6=itj k6=i Pj6=itj +Pj6=ktj!
X
P  P P 
1 t −t
= t k i
k
j6=itj k6=i j6=itj + j6=ktj!
X
This completes the proof. P  P P 
5
4 Cartesian Product of Graphs
Definition 8 Given a graph F with vertex set V(F) and edge set E(F), and a graph H with vertex set
V(H) and edge set E(H) we let G define the Cartesian Product of F and H to be the graph G = F ×H
which is defined as follows: V(G) = {(u,v)|u ∈ V(F) and v ∈ V(H)} and E(G) = {(u ,v ),(u ,v ) where
1 1 2 2
u =u and (v ,v )∈E(H) or v =v and (u ,u )∈E(F)}. We use ×G to denote the Cartesian product
1 2 1 2 1 2 1 2 i
m
of m copies of a graph G .
i
We next present an elementary result from graph theory.
Lemma 9 If G=F ×H, then the degree of a vertex (u,v) in G is the sum of the degrees of vertices u and
v, where u∈V(F) and v ∈V(H).
Theorem 10 Let G be a graph and let G be a regular graph where each vertex has degree r. Let u∈V(G )
r r
and let v and v be vertices in G with degrees k and k respectively. For each vertex (u,v ) ∈ V (G ×G)
i j i j i r
we have
l(u,v )= 1 ki−kj .
i r+ki 2r+ki+kj
j6=i
X
Proof. Consider a vertex (u,v ) ∈ V (G ×G). We note that deg((u,v )) = deg(u)+deg(v ) = r+k .
i r i i i
Then
l(u,v )= 1 (r+ki)−(r+kj) = 1 ki−kj .
i r+ki 2r+ki+kj r+ki 2r+ki+kj
j6=i j6=i
X X
Corollary 11 Let (u,v ) be a vertex in K ×G where u∈V (K ) and v ∈V(G). Then for all (v ,v )∈
i m m i i j
E(G)
l(u,v )= 1 (m−1)+deg(vi)−((m−1)+deg(vj))
i (m−1)+deg(vi) (m−1)+deg(vi)+((m−1)+deg(vj))
j
X
= 1 deg(vi)−deg(vj) .
(m−1)+deg(vi) (2m−2)+deg(vi)+deg(vj)
j
X
Proof. By Lemma 9 we have that deg((u,v )) = m−1+deg(v ) and for all neighbors v of vertex v
i i j i
we have that deg((u,v ))=m−1+deg(v ). The result then follows.
j j
4.1 Cartesian Products of Pn
In this section we will consider the lattice, ×P . As the calculation of the degrees of vertices in a lattice
n
m
is straightforward we will present results only involving the degrees without proof. We continue with some
definitions.
Definition 12 Any vertex of ×P can be defined by an m-tuple:
n
m
v =(v ,v ,··· ,v ) such that v ∈{1,...,n} ∀i∈{1,...,m}.
1 2 m i
Definition 13 We define a corner vertex of ×P to be
n
m
v =(v ,v ,...,v ) such that v ∈{1,n} ∀i∈{1,...,m}.
c 1 2 m i
A non-corner vertex is a vertex v =(v ,v ,...,v ) of ×P such that at least one v ∈{2,...,n−1}.
1 2 m n i
m
6
An inner corner vertex of ×P is defined as follows.
n
m
v =(v ,v ,...,v ) such that v ∈{2,n−1} ∀i∈{1,...,m}.
ic 1 2 m i
It follows by definition that all vertices that are inner corner vertices are also non-corner vertices.
We note that
m
1 if v ∈{1,n}
i
deg(v)= x such that x =
i i
i=1 (2 if vi ∈{2,...,n−1}
X
We also observe that neighbor v′ of vertex v = (v ,v ,...,v ) is defined as v′ = (v′,v′,...,v′ ) such that
1 2 m 1 2 m
′ ′ ′ ′
v =v for m−1 elements of (v ,v ,...,v ) and
i i 1 2 m
′
v −v =1
i i
(cid:12) (cid:12)
fortheremainingelementofthem-tuple(v′,v′,.(cid:12)..,v′ )(cid:12). Notice,therearetwospecialcasesforthisremaining
1 2 (cid:12) m (cid:12)
′ ′
element. If the remaining element v =1, then v =2 and if the remaining element v =n, then v =n−1.
i i i i
4.1.1 General Lemmas
We begin with a basic result involving the degrees of vertices and its neighbors in a lattice.
Lemma 14 Let G be a lattice ×P . Any vertex adjacent to a vertex with degree k must have degree k−1,
n
m
k, or k+1.
4.1.2 Extreme Leverage Centralities
We next identify vertices with the minimum and maximum leverage centralities. We will show that the
vertices with the minimum leverage centrality are the corners and the vertices with the maximum leverage
centrality are the inner corners. Furthermore, we will show that for any vertex v in the lattice G = ×P ,
n
m
− 1 ≤l(v)≤ 1 .
2m+1 8m−2
Minimum Leverage Centrality Wefirstcharacterizetheverticeswiththeminimumleveragecentrality.
We begin by stating two elementary lemmas involving degrees of vertices in a lattice.
Lemma 15 Any corner vertex v in G = ×P will have a degree of m. Furthermore, each neighbor of v
c n c
m
will have degree of m+1.
Lemma 16 Let G be the lattice ×P . A vertex v that is non-corner vertex of G must have at least one
n
m
neighbor u such that:
deg(u)≤deg(v).
Theorem 17 Let v be a corner vertex of G= ×P . Then,
c n
m
1
l(v )=− .
c
2m+1
Proof. By Lemma 15 we have that for G = ×P , deg(v ) = m and that for a neighbor u of v ,
n c c
m
deg(u)=m+1. We can compute the leverage centrality of v with Definition 1.
c
m m
1 m−(m+1) 1 −1 1 −1 1
l(v )= = = ·m =− .
c
m m+(m+1) m 2m+1 m 2m+1 2m+1
i=1 i=1 (cid:18) (cid:19)
X X
7
Theorem 18 (Minimum Leverage Centrality) Let u be any vertex in G = ×P that is not a corner vertex
n
m
and let v be a corner vertex in G. Then, l(v )<l(u).
c c
Proof. Let v be a non-corner vertex in G with degree k. We know from Lemma 16 that at least one
adjacent node has degree at most k. We know from Lemma 14 that the remaining adjacent nodes can have
degree at most k+1.
Letv haveoneadjacentnodewithdegreek andk−1adjacentnodeswithdegreek+1. Wenowcalculate
the leverage centrality of v.
1 k−(k+1) k−k 1−k
l(v)= ·(k−1)+ = .
k k+(k+1) k+k k(2k+1)
(cid:18) (cid:19) (cid:18) (cid:19)
From Theorem 17, we have that for a corner vertex v of degree k, the leverage centrality is:
c
1
l(v )= − .
c
2k+1
(cid:18) (cid:19)
Given that the degree of any adjacent node must be greater than 0, we know that 0 ≤ k−1 < 1. It follows
k
that − 1 < 1−k and hence l(v )<l(v).
2k+1 k(2k+1) c
If(cid:16)the neig(cid:17)hbors of any non-cornervertex u differ from that of v, then it follows from our constructionof
v and Lemma 14 that for any corresponding neighbors u from u and v from v, that deg(u )≤deg(v ) and
i i i i
hence, l(v)≤l(u). So we have that l(v )<l(v)≤l(u).
c
This implies that l(v )<l(u) which completes the proof.
c
Maximum Leverage Centrality We next characterize the vertices with the largest leverage centrality,
beginning with two elementary results involving degrees of vertices in a lattice.
Lemma 19 Let v be an inner vertex of G = ×P . Then v has 2m neighbors, such that m neighbors
ic n ic
m
have degree 2m and the remaining m neighbors have degree 2m−1.
Lemma 20 Let v be a vertex in G = ×P . Then, deg(v)≤2m.
n
m
Theorem 21 (Maximum leverage centrality) Let u be a vertex in G = ×P that is not an inner corner
n
m
vertex of G, and let v be an inner corner vertex in G. Then, l(u)<l(v ). Furthermore, l(v )= 1 .
ic ic ic 8m−2
Proof. Let v be an inner corner vertex of G. We have that
ic
v =(v ,v ,...,v ) such that v ∈{2,n−1} ∀i∈{1,...,m}.
ic 1 2 m i
By Lemma 19, we know that deg(v )=2m. We are also giventhat m neighbors of v have degree 2m and
ic ic
that m neighbors of v have degree 2m−1. The leverage centrality of v is
ic ic
1 2m−2m 2m−2m 2m−(2m−1) 2m−(2m−1)
l(vic)=  +···+ + +···+ .
2m 2m+2m 2m+2m 2m+(2m−1) 2m+(2m−1)
   
 mterms   mterms 
   
By rearranging terms w|e get: {z } | {z }
1 2m−2m 2m−(2m−1) 2m−2m 2m−(2m−1)
 + +···+ + . (1)
2m 2m+2m 2m+(2m−1) 2m+2m 2m+(2m−1)
(cid:18) (cid:19) (cid:18) (cid:19)
 
 mterms 
 
| {z }
8
By distributing 1 we get that each term of the sum for l(v ) can be expressed as:
2m ic
1 2m−2m 2m−(2m−1)
+
2m 2m+2m 2m+(2m−1)
(cid:20) (cid:21)
and since there are m terms in the sum, we can express l(v ) as:
ic
1 2m−2m 2m−(2m−1)
l(v )=m· + . (2)
ic
2m 2m+2m 2m+(2m−1)
(cid:20) (cid:21)
We simplify this to get:
1 2m−2m 2m−(2m−1) 1 2m−2m 2m−(2m−1) 1
l(v )=m· + = + = ,
ic
2m 2m+2m 2m+(2m−1) 2 2m+2m 2m+(2m−1) 8m−2
(cid:20) (cid:21) (cid:20) (cid:21)
which proves the second part of the theorem.
Let u be a vertex in G that is not an inner corner vertex of G. We have that ∃u∗ ∈u=(u ,u ,...,u )
i 1 2 m
suchthat u∗ ∈{1,3,...,n−2,n}. Without lossofgenerality,we canassumethat u =v whenu 6=u∗ and
i i i i i
thus u and v differ only in one element, u∗ ∈u and v∗ ∈v where u∗ 6=v∗.
ic i i ic i i
We see that two cases arise in calculating the leverage centrality of u.
(i) Let u∗ ∈{1,n} and v∗ ∈{2,n−1}
i i
By Lemma 19, we have that deg(u)=2m−1. In calculating the leverage centrality of u, we see that
l(u) and l(v ) can differ only in one term of Equation 1 such that:
ic
1 2m−2m 2m−(2m−1) 1 (2m−1)−2m
l(u)=(m−1)· + +
2m 2m+2m 2m+(2m−1) 2m−1 (2m−1)+2m
(cid:20) (cid:21) (cid:20) (cid:21)
1 2m−2m 2m−(2m−1) 1 2m−2m 2m−(2m−1)
l(v )=(m−1)· + + +
ic 2m 2m+2m 2m+(2m−1) 2m 2m+2m 2m+(2m−1)
(cid:20) (cid:21) (cid:20) (cid:21)
1 2m−2m 2m−(2m−1)
Let q =(m−1)· + .
2m 2m+2m 2m+(2m−1)
(cid:20) (cid:21)
1 (2m−1)−2m 1 1
Then l(u)=q+ =q−
2m−1 (2m−1)+2m 2m−1 4m−1
(cid:20) (cid:21) (cid:20) (cid:21)
1 2m−2m 2m−(2m−1) 2−m
and l(v )=q+ + = .
ic
2m 2m+2m 2m+(2m−1) 2m(1−4m)
(cid:20) (cid:21)
For the differing terms for the expressions for leverage centrality of u and v we see that
ic
1 1 1 1
− < .
2m−1 4m−1 2m 4m−1
(cid:20) (cid:21) (cid:20) (cid:21)
and it follows that
l(u)<l(v ).
ic
(ii) If u∗ ∈{3,n−2} and v∗ ∈{2,n−1}
i i
By Lemma 19, we know that deg(u)=2m. Incalculating the leveragecentrality ofu, we see thatl(u)
and l(v ) can differ only in one term of Equation 1 such that:
ic
1 2m−2m 2m−(2m−1) 1 2m−2m
l(u)=(m−1)· + +
2m 2m+2m 2m+(2m−1) 2m 2m+2m
(cid:20) (cid:21) (cid:20) (cid:21)
1 2m−2m 2m−(2m−1) 1 2m−2m 2m−(2m−1)
l(v )=(m−1)· + + +
ic
2m 2m+2m 2m+(2m−1) 2m 2m+2m 2m+(2m−1)
(cid:20) (cid:21) (cid:20) (cid:21)
1 2m−2m 2m−(2m−1) m−1
Let q =(m−1)· + =
2m 2m+2m 2m+(2m−1) 2m(4m−1)
(cid:20) (cid:21)
9
From the proof of Case (i), we already have l(v )
ic
1 2m−2m
l(u)=q+ =q, and
2m 2m+2m
(cid:20) (cid:21)
1 1
l(v )=q+ .
ic
2m 4m−1
(cid:20) (cid:21)
For the differing terms for the expressions for leverage centrality of u and v we see that
ic
1 1
0<
2m 4m−1
(cid:20) (cid:21)
and it follows that
l(u)<l(v ).
ic
In both Cases (i) and (ii), we find that l(u) < l(v ) which proves that first part of the theorem and
ic
completes the proof.
4.1.3 Convergence of Leverage Centrality as m→∞
We next consider the leverage centrality of different vertices as the number of dimensions is increased.
Theorem 22 As the number of paths in the Cartesian product increases (m→∞), the leverage centralities
of all of the vertices of G= ×P converge to 0.
n
m
Proof. Let G= ×P . From Theorem 21, we know that for any m, the maximum leverage centrality of
n
m
any vertex v of G is:
1
max(l(v))=
8m−2
From Theorem 18, we know that for any m, the minimum leverage centrality of any vertex v of G is:
1
min(l(v))=− .
2m+1
Therefore, for any vertex v in G the leverage centrality is bounded as follows:
1 1
− ≤l(v)≤ .
2m+1 8m−2
We see that
1 1
lim − = lim =0.
m→∞ 2m+1 m→∞ 8m−2
(cid:18) (cid:19) (cid:18) (cid:19)
It follows that
lim l(v)=0.
m→∞
which completes the proof.
5 Leverage Centralities in Lattices and Triangle Numbers
In this section we investigate the number of distinct leverage centralities for lattices and show there is a
surprising connection to the triangle numbers m+2 where m ≥ 1. We can label the vertices of ×P with
2 n
m
using m-tuples where v =(v1,v2,··· ,vm) such(cid:0)that(cid:1)vi ∈{1,...,n} ∀i∈{1,...,m}. For simplicity we will
denote v by (r,s,t).
r,s,t
10