On the relationship between PageRank and
automorphisms of a graph
Modjtaba Ghorbania;1, Matthias Dehmerb;c, Abdullah Lota, Najaf Amraeia,
Abbe Mowshowitzd, Frank Emmert-Streibe;f
a Department of Mathematics, Faculty of Science, Shahid Rajaee
Teacher Training University, Tehran, 16785-136, I. R. Iran
bSwiss Distance University of Applied Sciences, Department of Computer
Science, Brig, Switzerland
cDepartment of Biomedical Computer Science and Mechatronics,
UMIT, Hall in Tyrol, Austria
dDepartment of Computer Science, The City College of
New York (CUNY), New York, NY, USA
ePredictive Society and Data Analytics Lab, Faculty of Information
Technology and Communication Sciences, Tampere University, Finland
f Institute of Biosciences and Medical Technology, Tampere 33520, Finland
Abstract
PageRank is an algorithm used in Internet search to score the importance of
web pages. The aim of this paper is demonstrate some new results concerning
the relationship between the concept of PageRank and automorphisms of a
graph. In particular, we show that if vertices u and v are similar in a graph G
(i.e., there is an automorphism mapping u to v), then u and v have the same
PageRank score. More generally, we prove that if the PageRanks of all vertices
in G are distinct, then the automorphism group of G consists of the identity
alone. Finally, the PageRank entropy measure of several kinds of real-world
networks and all trees of orders 10-13 and 22 is investigated.
Keywords: PageRank, eigenvalue, graph automorphism, web page.
2010 AMS: 05C70, 05C07, 05C35, 92E10.
1Author to whom correspondence should be addressed(E-mail: mghorbani@sru.ac.ir)
1. Introduction1
The eigenvalues and eigenvectors of the adjacency matrix of a graph oer2
necessary conditions for a graph to possess certain properties. In particular,3
they have been found useful in studies of graphs associated with web searches.4
The world wide web can be modeled as a directed graph in a natural way by5
interpreting web pages as vertices and links between web pages as directed edges6
in the graph.This model provides a basis for ranking web pages by means of the7
PageRank (PR) algorithm. The algorithm was developed by Brin and Page in8
1998 [2].9
The PageRank (PR) algorithm provides a mechanism for scoring the im-10
portance of web pages. PR has applications in such diverse elds such as neu-11
roscience [32], bioinformatics [16, 30], sports [3, 25], trac modeling [5, 29],12
chemistry [28] and social network analysis [12, 23], as well as others [22, 26].13
Also, PR has been used extensively for improving the quality of search engines14
such as google and so forth, see [5]15
The research reported here is especially relevant for chemical database ap-16
plications. Searching for compounds with special properties can be aided by17
making use of page rank, and the automorphism group is useful for computing18
page rank. For other possible applications of the results in this paper (see [4]).19
In this paper, we establish connections between the PageRank concept and20
automorphisms of a graph. The motivation to do so is to get deeper insights into21
graph-theoretical properties of graphs (here symmetry) in conjunction with PR.22
First, we dene the PageRank (PR) vector and show how it can be computed.23
In Section 3, we establish new results concerning the concept of PageRank and24
automorphisms of a graph. In section 4, the PageRank entropy measure is25
dened In other words, analyzing the reported data shows that the PR-entropy26
measure is not highly correlated with the size of automorphism group and hence27
it can be regarded as a new measure to study the algebraic properties of the28
automorphism group.29
Finally, in Section 5, we dene the notion of a Co-PageRank graph and30
2
oer a conjecture concerning PageRank scores of vertices in non-Co-PageRank31
graphs. The notation used in this paper mainly follows [24].32
2. PageRank Vector33
The following discussion makes use of the model of the web as a directed34
graph. Let n be the number of all web pages, and suppose Pnn is the Markov35
transitions matrix associated with the web graph dened as follows:36
pij =
8<: 1di if page i and page j are linked0 otherwise ;
where di is the degree of vertex i. In other words, pij is the probability of navi-37
gating from vertex i to vertex j. For a dangling vertex (one with outdegree 0),38
a zero row appears in the matrix P which violates the condition of a transition39
matrix. To overcome this violation and obtain a transition matrix, we dene40
P + luT where u is the probability distribution vector, u = [1=n; 1=n; :::; 1=n]T ,41
and l is an n-dimensional vector as follows:42
li =
8<: 1 if i is a dangling node0 otherwise :
A PR vector [24], is an n-dimensional vector satisfying the following:43
8<: T = T eGT j = 1 ; (1)
where eG = (P + luT )+(1)jvT , j = [1; 1; : : : ; 1] and 2 (0; 1) (typically44
= 0:85). In the present paper, we focus on graphs without dangling vertices.45
Hence, the vector can be derived from the following equation:46
T = TP + (1 )vT ; (2)
or equivalently,47
(I PT ) = (1 )v; (3)
3
in which v = [1=n; 1=n; :::; 1=n]T .48
The PageRank (PR) score of vertex i is the ith entry of the vector [6]. An49
example will help to x ideas.50
The Google matrix eG of a directed network is a stochastic square matrix
with non-negative matrix elements and the sum of elements in each column
being equal to unity. By above notation, the elements of the Google matrix are
dened as eGij = Pij + (1 ) 1
n
:
Proposition 2.1. [24] If f1; 2; : : : ; ng are all eigenvalues of transitions ma-51
trix P , then f1; 2; : : : ; ng are all eigenvalues of eG.52
Let G be a graph with adjacency eigenvalues 1; 2; ; n. The graph
energy of G is dened as
E(G) =
nX
i=1
jij;
see [18, 19, 20, 21]. Following Gutman denition, if f1; 2; : : : ; ng are all eigen-
values of transitions matrix P , then the transition energy can be dened as
E(P ) =
nX
i=1
jij:
Corollary 2.1. Suppose G is a graph with transitions matrix P . Then
E( eG) = (E(P ) 1) + 1:
Proof. By Proposition 2.1, the proof is straightforward.53
54
Example 2.1. The following is the adjacency matrix of the graph G1 in Figure55
1.56
A =
26666666664
0 0 0 1 1
0 0 1 0 0
0 1 0 1 0
1 0 1 0 1
1 0 0 1 0
37777777775
;
4
The transition matrix of this graph is57
P =
26666666664
0 0 0 1=2 1=2
0 0 1 0 0
0 1=2 0 1=2 0
1=3 0 1=3 0 1=3
1=2 0 0 1=2 0
37777777775
:
With = 0:85 Eq. 3 gives the following linear system58
8>>>>>>>>><>>>>>>>>>:
1 0:853 4 0:852 5 = 0:03
2 0:852 3 = 0:03
0:852 + 3 0:853 4 = 0:03
0:852 1 ( 0:852 )3 + 4 ( 0:852 )5 = 0:03
( 0:852 )1 ( 0:853 )4 + 5 = 0:03
: (4)
Solving Eq. 4 we obtain the PR vector of G1:59
PR = [0:1918; 0:1204; 0:2126; 0:2834; 0:1918]:
2.1. PageRank score of a vertex60
The concept of PageRank score at a vertex is needed to determine the rela-61
tionship between PageRank and automorphisms of a graph.62
Denition 2.1. Let A = (aij) be an nn matrix. Then the 1-Norm of matrix63
A is dened as [27]64
jjAjj1 =Maxj
nX
i=1
jaij j:
Denition 2.2. The spectral radius (A) of an square matrix A is the largest65
absolute value of eigenvalues of A, see [27].66
Theorem 2.1. [1] Let A be an arbitrary square matrix. Then67
(A) jjAjj1:
5
Theorem 2.2. [1] (Geometric series) Let A be an square matrix. If (A) < 1,68
then (I A)1 exists, and it can be expressed as a convergent series,69
(I A)1 = I +A+A2 + +Am + =
1X
k=0
Ak: (5)
Lemma 2.1. Let G be a graph of order n and be the PR vector of G. The70
PR of vertex vi can be determined from the following equation:71
i =
(1 )
n
1X
k=0
k
nX
t=1
P kti; (6)
where 2 (0; 1) and P is the transition matrix.72
Proof. Since
nP
i=1
PTij = 1, we see that jjPT jj1 = 1. Consequently, we have73
jjPT jj1 = jjPT jj1 = < 1. Theorem 2.1 implies that (PT ) < 1, and74
Theorem 2.2 implies that the inverse matrix (I PT )1 exists and thus75
(I PT )1 =
1X
k=0
(PT )
k
: (7)
From Eq. 3 and Eq. 7 we conclude that,76
= (1 )(I PT )1v = (1 )(
1X
k=0
(PT )
k
)v
= (1 )(I + PT + 2PT 2 + :::)v:
Since i is the ith row of the matrix (1 )(I PT )1v, it is clear that77
i is the ith row of column matrix
(1)
n (I PT )1e. This means that78
i =
(1 )
n
nX
t=1
(I PT )1it
=
(1 )
n
nX
t=1
1X
k=0
(PT )kit: (8)
Hence79
i =
(1 )
n
1X
k=0
k
nX
t=1
P kti: (9)
80
6
According to the denition of matrix P , P kti is the transition probability from81
vertex t to vertex i in k steps.82
Theorem 2.3. Let G be a graph and i; j 2 V (G). If
nP
t=1
P kti =
nP
t=1
P ktj, (for all83
k 2 N ), then i = j.84
Proof. Suppose
nP
t=1
P kti =
nP
t=1
P ktj and 2 (0; 1). Then85
nX
t=1
kP kti =
nX
t=1
kP ktj (for all k 2 N);
and consequently86
1X
k=0
k
nX
t=1
P kti =
1X
k=0
k
nX
t=1
P ktj :
Eq. 9 implies that87
i =
(1 )
n
1X
k=0
k
nX
t=1
P kti =
(1 )
n
1X
k=0
k
nX
t=1
P ktj = j :
88
In light of Theorem 2.3, consider the tree T1 shown in Figure 2.89
Example 2.2. The sums of the entries in each column of matrices P; P 2; P 3,90
respectively, of graph T1, are shown in the end of each column. Consider also the91
vertices 1; 2 or 3; 4 of T1 and their corresponding columns in matrices P; P
2; P 392
as follows:93
A =
26666666666664
0 0 1 0 0 0
0 0 0 1 0 0
1 0 0 0 1 0
0 1 0 0 1 0
0 0 1 1 0 1
0 0 0 0 1 0
37777777777775
; P =
26666666666666664
0 0 1 0 0 0
0 0 0 1 0 0
1/2 0 0 0 1/2 0
0 1/2 0 0 1/2 0
0 0 1/3 1/3 0 1/3
0 0 0 0 1 0
1=2 1=2 4=3 4=3 2 1=3
37777777777777775
;94
7
P 2 =
26666666666666664
1/2 0 0 0 1/2 0
0 1/2 0 0 1/2 0
0 0 2/3 1/6 0 1/6
0 0 1/6 2/3 0 1/6
1/6
1/6 0 0 2/3 0
0 0 1/3 1/3 0 1/3
2=3 2=3 7=6 7=6 5=3 2=3
37777777777777775
;95
P 3 =
26666666666666664
0 0 2/3 1/6 0 1/6
0 0 1/6 2/3 0 1/6
1/3 1/12 0 0 7/12 0
1/12 1/3 0 0 7/12 0
0 0 7/18 7/18 0 2/9
1/6 1/6 0 0 5/9 0
7=12 7=12 22=18 22=18 72=36 5=9
37777777777777775
:96
The sums of columns 1; 2 or 3; 4 of P k (for all k ) are the same, and thus97
the PR scores of corresponding vertices are the same. This means that98
6X
t=1
P kt1 =
6X
t=1
P kt2;
and thus 1 = 2. A similar argument shows that 3 = 4. Hence, the PR99
vector of this tree is100
T = [0:1090; 0:1090; 0:1975; 0:1975; 0:2821; 0:1049]:
3. PageRank Vector and Graph Automorphisms101
An identity graph or asymmetric graph is a graph whose automorphism102
group consists of the identity element alone. An example of such a graph is T2103
shown in Figure 4. Note that all entries of the PR vector of this graph are104
distinct. The aim of this section is to prove that if the PageRank scores of all105
vertices are distinct, then the graph must be asymmetric.106
8
Lemma 3.1. Every vertex vi in a regular graph G of order n has PR score107
i =
1
n .108
Proof. Let G be a regular graph of degree r. Then for every vertex vi 2 V (G),109
we have110
nX
t=1
Pti = r:
1
r
= 1;
and111
nX
t=1
P 2ti = 1:
1
r
:r = 1:
Hence, for each k 2 N,112
nX
t=1
P kti = 1: (10)
Using Eq. 10 and Lemma 2.1 implies that113
i =
1
n
1X
k=0
k
nX
t=1
P kti
=
1
n
1X
k=0
k =
1
n
(
1
1 )
=
1
n
This completes the proof.114
Let T be a tree on n vertices, and denote the degree of a vertex v by dv. A115
non-pendant vertex v of T is adjacent to dv > 1 vertices in T .116
Theorem 3.1. Let i; j be two vertices in a graph G. If there exists an auto-117
morphism 2 Aut(G) such that (i) = j, then i = j.118
Proof. Suppose Ni denotes the set of neighbors of vertex i, namely Ni =119
ft 2 V jti 2 Eg, see Figure 5. For every vertex i1 in Ni there is a vertex120
9
j1 2 Nj such that (i1) = j1. Since i1 and j1 are similar, di1 = dj1 , and121
Pi1i =
1
di1
= 1dj1
= Pj1j . Hence,122
nX
t=1
Pti =
nX
t=1
Ptj :
Continuing the method illustrated in Figure 6, for given vertex i2 2 Ni1 ,123
there exists a vertex j2 2 Nj1 such that (i2) = j2, since maps the edge Bi to124
Bj . This implies that di2 = dj2 and thus Pi2i1 =
1
di2
= 1dj2
= Pj2j1 . Therefore,125
(P 2)i2i = (P
2)j2j and thus,126
nX
t=1
P 2ti =
nX
t=1
P 2tj : (11)
In general, we have,127
nX
t=1
P kti =
nX
t=1
P ktj (k 2 N): (12)
From Eq. 12 and Theorem 2.3 it follows that i = j , and the assertion is128
proved.129
Theorem 3.1 says that if an automorphism maps a vertex x to vertex y,130
they must have the same PR score. However, the converse does not hold. A131
counterexample is the Frucht graph shown in Figure 7. The Frucht graph is132
regular of degree 3 with 12 vertices and 18 edges and is asymmetric, see [13].133
Since it is a regular graph, Lemma 3.1 shows the PR-vector is [1=12; : : : ; 1=12],134
while the automorphism group of this graph consists of the identity element135
alone.136
In what follows, we prove that a graph whose vertices have distinct PageRank137
scores is asymmetric. First, consider the following example.138
Example 3.1. The following is the adjacency matrix of the tree T2 shown in139
Figure 4:140
10
A =
26666666666666664
0 1 0 0 0 0 0
1 0 1 0 1 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 1 0 0 0 1 0
0 0 0 0 1 0 1
0 0 0 0 0 1 0
37777777777777775
;
In the matrix P associated with A, the sums of 4th and 7th columns are equal,141
but in P 2 and P 3 these column sums are not equal.142
P =
26666666666666666664
0 1 0 0 0 0 0
1/3 0 1/3 0 1/3 0 0
0 1/2 0 1/2 0 0 0
0 0 1 0 0 0 0
0 1/2 0 0 0 1/2 0
0 0 0 0 1/2 0 1/2
0 0 0 0 0 1 0
1=3 2 4=3 1=2 5=6 3=2 1=2
37777777777777777775
;143
P 2 =
26666666666666666664
1=3 0 1=3 0 1=3 0 0
0 2=3 0 1=6 0 1=6 0
1=6 0 2=3 0 1=6 0 0
0 1=2 0 1=2 0 0 0
1=6 0 1=6 0 5=12 0 1=4
0 1=4 0 0 0 3=4 0
0 0 0 0 1=2 0 1=2
2=3 7=12 7=6 4=6 17=12 11=12 3=4
37777777777777777775
;144
11
P 3 =
26666666666666666664
0 2=3 0 1=6 0 1=6 0
2=9 0 5=6 0 11=36 0 1=12
0 7=12 0 1=3 0 1=12 0
1=6 0 2=3 0 1=6 0 0
0 11=24 0 1=12 0 11=24 0
1=12 0 1=12 0 11=24 0 3=8
0 1=4 0 0 0 3=4 0
17=36 47=24 19=12 7=12 67=72 35=24 11=24
37777777777777777775
.145
On the other hand, we have,146
7X
t=1
Pt4 =
7X
t=1
Pt7;
while147
7X
t=1
P 2t4 6=
7X
t=1
P 2t7 and
7X
t=1
P 3t4 6=
7X
t=1
P 3t7:
The graph T2 has no vertices for which corresponding column sums are the148
same. This means that their PR scores are not equal and the entries of the PR149
vector are all distinct. Finally, the PR vector of this tree is150
T = [0:0878; 0:2343; 0:1660; 0:0920; 0:1592; 0:1680; 0:0928]:
On the other hand, the automorphism group of T2 consists of the identity element151
alone.152
Corollary 3.1. Let G be a graph. If the PR scores of all the vertices are153
distinct, then G is asymmetric.154
Proof. For two arbitrary vertices u; v 2 V (G), if u 6= v, then by Theorem 3.1,155
there is no an automorphism that maps u to v and the assertion follows.156
Corollary 3.2. Let T be a tree in which no two pendant vertices have the same157
PR scores. Then the automorphism group of T consists of the identity element158
alone.159
12
Proof. For the non-identity automorphism of Aut(T ), there are at least two160
pendant vertices i; j such that (i) = j and thus i = j . But the pendant161
vertices have dierent PR scores from which the result follows.162
Denition 3.1. Let G be a graph with automorphism group Aut(G), and denote163
the orbit of a vertex u 2 V (G) by uAut(G) or [u]. Note that uAut(G) is the set164
f(u) : 2 Aut(G)g.165
A graph G is called vertex-transitive, if it has exactly one orbit. In other166
words, for any two vertices u; v 2 V (G), there is an automorphism 2 Aut(G)167
such that (u) = v.168
The PR complexity, PRC(G), is the number of dierent values of PR vector.169
Theorem 3.2. Let V1; V2; V3; : : : ; Vk be all the orbits of Aut(G). Then for two170
vertices x; y 2 Vi(1 6 i 6 k), x = y. In particular, if G is vertex-transitive,171
then PRC(G) = 1.172
Proof. If two vertices are in the same orbit, there is an automorphism mapping173
one to the other. The assertion follows from Theorem 3.1.174
Corollary 3.3. Let #O be the number of distinct orbits of a graph G. Then
PRC(G) #O:
An illustration of this corollary is given by the tree T1 shown in Figure 2.175
This graph has four orbits f1; 2g, f3; 4g, f5g and f6g. By Theorem 3.2, 1 = 2176
and 3 = 4. This means that the PR vector has at most four distinct entries.177
Example 3.2. Suppose t denotes the number of orbits of graph G. It should be178
noted here that there are graphs with k < t: For example consider the graph K179
in Figure 3. This graphs has three orbits while k = 2, the vertices in an orbit180
are colored by the same colors.181
This example shows that determining graphs with k = t is a hard task. We182
Solve this problem for graphs with exactly two orbits.183
13
Lemma 3.2. The connected graph G is regular if and only if = j, where184
2 R.185
Proof. If G is regular, then by Lemma 3.1, = 1n j. Conversely, if = j for a
scaler 2 R, then all entries of are the same. Since for two vertices vi and
vj , we have
i j = (j
dj
i
di
);
necessarily di = dj and thus the graph is regular.186
Theorem 3.3. Let G be a graph with two distinct orbits. Then either G is a187
regular graph or k = 2.188
Proof. Since G has two orbits, it follows that k 2. If k 6= 2, then by Lemma189
3.2, G is regular. This completes the proof.190
Corollary 3.4. Let G be an edge-transitive graph. Then either G is a regular191
graph or k = 2.192
Example 3.3. Consider the complete graph Km;n(m 6= n). It is a well-known193
fact that Km;n has two orbitse. Since, m 6= n, by Theorem 3.3, we obtain k = 2.194
In addition, the matrix P associated to the adjacency matrix of G is195
P =
0@ 0nn 1m jnm
1
n jmn 0mm
1A :
Hence,
Spec(P ) = f1; 0; 0; :::; 0; 1g
and thus for the Google matrix, we have
Spec( eG) = f1; 0; 0; :::; 0;g:
Example 3.4. Let Sn denotes to the star graph with n vertices. The bistar196
graph Bm;n is a graph obtained from union of Sn+1 and Sm+1 by joining their197
central vertices. For the star graph, we obtain198
P (Sn+1) =
0@ 011 1n j1n
jn1 0nn
1A :
14
This yields that PR = [1; 2; : : : ; 2; 2], where 1 = (
1
n+1 + ) 11+ and199
2 =
n+
n(n+1)(1+) . Also, for the bistar graph, it yields200
P (Bm;n) =
0@ A B
BT C
1A ;
where C = 0m+n,201
A =
0@ 0 1n+1
1
m+1 0
1A ; and B =
0@ 1n+1 j1n 01m
01n 1m+1 j1m
1A :
Lemma 3.3. Let G be a graph and i; j be two distinct vertices having the same202
neighbors. Then i = j.203
Proof. Two following cases hold:204
a) Suppose vertices i and j are adjacent. According to the denition of PR205
score, we have,206
i =
X
k2Nifjg
k
dk
+
j
dj
+
1
n
;
and207
j =
X
k2Njfig
k
dk
+
i
di
+
1
n
:
Thus208
i j = (j
dj
i
di
);
and therefore209
i(1 +
di
) = j(1 +
dj
):
Since jNij = jNj j; we have di = dj which implies i = j :210
b) Now suppose i and j are not adjacent. Then i =
P
k2Ni
k
dk
+ (1)n and211
j =
P
k2Nj
k
dk
+ (1)n . Since Ni = Nj , we conclude i j = 0 and thus212
i = j :213
15
Lemma 3.4. Let i; j be two adjacent vertices of a graph G. If i < j, then214
Nj * Ni215
Proof. Suppose to the contrary that Nj Ni. Hence, we obtain216
j =
X
k2Nj
k
dk
+
1
n
X
k2Ni
k
dk
+
1
n
= i;
a contradiction.217
Lemma 3.5. Let G be a graph. If i is a pendant vertex adjacent to vertex j,218
then i < j.219
Proof. Clearly dj 2 and thus 1dj 12 . This implies220
(i 1
dj
j) (i 1
2
j): (13)
From the denition of PR and Eq. 13, we have221
j = (
i
1
) +
X
k 2 Nj
k 6= i
k
dk
+
1
n
; i = (
j
dj
) +
1
n
:
Hence222
16
j i = (i 1
dj
j) +
X
k 2 Nj
k 6= i
k
dk
+ (
1
n
1
n
)
(i 1
2
j) +
X
k 2 Nj
k 6= i
k
dk
= (i j) + 1
2
j +
X
k 2 Nj
k 6= i
k
dk
> (i j) + 1
2
j ;
and thus223
(j i) >
1
2j
1 +
> 0: (14)
224
4. Graph Entropy Measure225
The general Shannon entropy [5] is dened by I(p) =
nX
i=1
pi log(pi) for
nite probability vector p and the symbol log is the logarithm on the basis 2.
Let =
Pn
j=1 j and pi = i=; (i = 1; 2; : : : ; n): Generally, the entropy of an
n-tuple (1;2; : : : ;n) of real numbers is given by
I(1;2; : : : ;n) = log
nX
i=1
i
!
nX
i=1
iPn
j=1 j
log i: (15)
There are many dierent ways to associate an n-tuple (1;2; : : : ;n) to226
a graph G (see [1, 6, 7, 8, 9, 10, 11, 14, 22, 24, 32]). A graph entropy measure227
due to PageRank vector [15] is dened as228
I(G) = log
nX
i=1
i
!
nX
i=1
iPn
j=1 j
log i: (16)
17
This phrase reduces the complexity of the graph G into a single quantity: I(G)229
bits of information. This means that the PR-entropy I, forms a simple and230
graceful discriminant statistic for determining the topology of a graph. This231
metric is the subject of the present section. The entropy function maximizes the232
freedom in choosing the pij 's. The theory tell us that the entropy function gives233
the best unbiased probability assignment to the variables given the restriction.234
Example 4.1. Consider the Karate graph K [31] as depicted in Figure 10. It235
has 34 vertices and 78 edges and the PageRank vector is as follows:236
= [ 0:097; 0:053; 0:057; 0:036; 0:022; 0:029; 0:029; 0:024; 0:029; 0:014; 0:022;
0:009; 0:015; 0:029; 0:014; 0:014; 0:017; 0:014; 0:014; 0:019; 0:014; 0:015;
0:014; 0:031; 0:021; 0:021; 0:015; 0:026; 0:019; 0:026; 0:025; 0:037; 0:072;
0:101]:
The interpretation of 1 = 0:097 is that 9.7 percent of the time the random237
surfer visits page 1. Therefore, the pages in this tiny web can be ranked by238
their importance. Hence, page 34 is the most important page and page 12 by239
12 = 0:009 is the least important page, according to the PageRank denition of240
importance. Also its PR-entropy is I(K) = 4:78.241
Example 4.2. Consider the graph G as depicted in Figure 11. It presents a242
typical arrangement of symmetric subgraphs found in many real world networks.243
It has 33 vertices and 37 edges. The PageRank vector is as follows:244
= [ 0:04; 0:031; 0:018; 0:031; 0:018; 0:064; 0:031; 0:031; 0:031; 0:04; 0:031;
0:016; 0:018; 0:035; 0:027; 0:075; 0:017; 0:017; 0:017; 0:017; 0:045; 0:046;
0:037; 0:015; 0:037; 0:015; 0:04; 0:046; 0:046; 0:017; 0:017; 0:017; 0:017]:
The PR-entropy for graph G is I(G) = 4:89.245
In continuing, ve classes of trees of orders 10-13, and 22, were choosen246
and the results indicated a weak correlation between jAut(G)j and I(G). These247
values are given in Figures 12,13, Figure 14, 15, and Figure 16. In other words,248
18
analyzing the reported data shows that the PR-entropy measure is not highly249
correlated with the size of automorphism group and hence it can be regarded as250
a new measure to study the algebraic properties of the automorphism group.251
It is clear that if in the Shannon entropy denition, all pi's are equal, then252
I achieves the maximum value which is log(n). By Lemma 3.2, if G is regular,253
then I = log(n). Graphs with minimum value of PR-entropy are more dicult254
to characterize. We conjecture that for a given number n, the star graph Sn has255
the minimum PR-entropy. To do this, three classes of graphs, namely all graphs256
of orders 5-6 and all trees of order 12 were choosen and the results conrm our257
following conjecture.258
Conjecture 4.1. Among all connected graphs on n vertices, the star graph Sn259
has the minimum value of PR-entropy.260
In [14], it is proved that if T is a tree with two orbits and n 3 vertices,261
then T is isomorphic with either the star graph Sn or bistar graph Bm;m. By262
Example 3.4, we conclude the following result.263
Theorem 4.1. Let T be a tree with two orbits and n 3 vertices. Then one of264
the following cases hold:265
i) T = Sn and I(T ) 0:55 log n+ 0:91.266
ii) T = Bm;m and I(T ) 0:6 log n+ 0:93.267
Many networks can be modeled as a star graph. For example, an inwardly268
directed star graph may be used to represent retweet activity on Twitter and269
an outwardly directed star graph can be used to represent a hub authority. One270
may see that the star graph is a special case of G+ fug in which G is a vertex-271
transitive graph. Here, we explain how one can the PR-vector of G + fug by272
having the PR-vector of G.273
Lemma 4.1. Let G be an r-regular graph on n vertices. Then the PageRank274
vector of graph G+ fug is = [1; : : : ; n; n+1], where n+1 = ( 1n+1 + r+1 )275
( r+1+r+1 ) and 1 = : : : = n =
1n+1
n .276
19
Proof. Suppose G is a regular graph with P (G) associated to its adjacency277
matrix. For an arbitrary vertex u, the matrix eP = P (G+ fug) can be regarded278
as follows:279
eP =
0@ 1r+1A 1r+1 jn1
1
n j1n 011
1A ;
where A is the adjacency matrix of G. By replacing eP with P in Eq. 3 the280
result follows.281
5. Co-PageRank Graphs282
There exist non-isomorphic graphs with the same PR vectors; these graphs283
are said to be Co-PageRank (or Co-PR). For example, the two graphs G and H284
shown in Figure 8 have the same PR-vector, namely,285
[0:185065; 0:185065; 0:129870; 0:185065; 0:185065; 0:129870]:
but they are not isomorphic. In general, suppose = 1; ; n and =286
1; ; n the PR vectors of two graphs G and H, respectively, where 1 287
2 n and 1 2 n. If = , then G and H are Co-PR;288
if , on the other hand, and dier in at least one entry, then G and H289
are non-Co-PR. Two graphs G and H are completely non-Co-PR if for each i290
(1 i n) i 6= i. For example, the two graphs L and K shown in Figure 9,291
are non-Co-PR, with292
PR(L) = [0:143736; 0:209536; 0:143736; 0:209536; 0:146727; 0:146727];
PR(K) = [0:161121; 0:237500; 0:177757; 0:100546; 0:161121; 0:161954]:
We end this paper with the following conjecture.293
Conjecture 5.1. Suppose G and H are two non-Co-PR graphs. Then for each294
vertex u 2 V (G) and each vertex v 2 V (H), u 6= v. More generally G and H295
are completely non-Co-PR.296
20
Conclusion297
In this paper, we have investigated the relationship between the concept of298
PageRank and automorpisms of a graph. In particular, we proved that if the299
pendant vertices of a tree T have distinct PRs, then T is asymmetric. Results300
regarding symmetry relations for trees as well as graphs can be useful to design301
new graph measures. Moreover, we established conditions for which two distinct302
vertices of a graph have the same PageRank. The main result in this paper is303
that two vertices in the same orbit have the same PR score. As future work, we304
hope to determine the structure of automorphism groups of well-known graphs305
in terms of PR vectors.306
Acknowledgements307
The authors would like to thank Prof. Paolo Boldi for his valuable comments308
and suggestions to improve the quality of the paper.309
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Figure 1: Graph G1 in Example 2.1.
Figure 2: The tree T1 in Example 2.2.
25
Figure 3: The graph K with three orbits and k = 2.
Figure 4: The tree T2 in Example 3.1.
Figure 5: The neighbors of two adjacent vertices i; j.
26
Figure 6: The neighbors of neighbors of vertices i; j.
Figure 7: The Frucht graph.
27
Figure 8: Two Co-PR graphs.
Figure 9: Two non-Co-PR graphs.
28
Figure 10: Zachary's Karate graph K.
Figure 11: The graph G.
Figure 12: All trees of order 10. The correlation between jAut(T )j and I(T ) is -0.60.
29
Figure 13: All trees of order 11. The correlation between jAut(T )j and I(T ) is -0.50.
Figure 14: All trees of order 12. The correlation between jAut(T )j and I(T ) is -0.34.
Figure 15: All trees of order 13. The correlation between jAut(T )j and I(T ) is -0.46.
Figure 16: All trees of order 22. The correlation between jAut(T )j and I(T ) is -0.29.
30
Figure 17: The value of I(T ) for a star graph with at most 872 vertices.
Figure 18: The value of I(T ) for a bistar graph with at most 467 vertices.
31