Here's the abstract:

We analyze linkage strategies for a set I of webpages for which the webmaster wants to maximize the sum of Google's PageRank scores. The webmaster can only choose the hyperlinks starting from the webpages of I and has no control on the hyperlinks from other webpages. We provide an optimal linkage strategy under some reasonable assumptions.

And the conclusion:

Conclusions

In this paper we provide the general shape of an optimal link structure for a website in order to maximize its PageRank. This structure with a forward chain and every possible backward links may be not intuitive. At our knowledge, it has never been mentioned, while topologies like a clique, a ring or a star are considered in the literature on collusion and alliance between pages [3, 8]. Moreover, this optimal structure gives new insight into the affirmation of Bianchini et al. [5] that, in order to maximize the PageRank of a website, hyperlinks to the rest of the webgraph “should be in pages with a small PageRank and that have many internal hyperlinks”. More precisely, we have seen that the leaking pages must be choosen with respect to the mean number of visits before zapping they give to the website, rather than their PageRank.

I have no clue whatsoever as to what that means. I've posted the full text below:

arXiv:0711.2867v1 [cs.IR] 19 Nov 2007

Maximizing PageRank via outlinks

Cristobald de Kerchove Laure Ninove Paul Van Dooren

CESAME, Universit´e catholique de Louvain,

Avenue Georges Lemaˆıtre 4–6, B-1348 Louvain-la-Neuve, Belgium

{c.dekerchove, laure.ninove, paul.vandooren}@uclouvain.be

Abstract

We analyze linkage strategies for a set I of webpages for which the

webmaster wants to maximize the sum of Google’s PageRank scores.

The webmaster can only choose the hyperlinks starting from the webpages

of I and has no control on the hyperlinks from other webpages.

We provide an optimal linkage strategy under some reasonable assumptions.

Keywords: PageRank, Google matrix, Markov chain, Perron vector,

Optimal linkage strategy

AMS classification: 15A18, 15A48, 15A51, 60J15, 68U35

1 Introduction

PageRank, a measure of webpages’ relevance introduced by Brin and Page, is

at the heart of the well known search engine Google [6, 15]. Google classifies

the webpages according to the pertinence scores given by PageRank, which

are computed from the graph structure of the Web. A page with a high

PageRank will appear among the first items in the list of pages corresponding

to a particular query.

If we look at the popularity of Google, it is not surprising that some

webmasters want to increase the PageRank of their webpages in order to

get more visits from websurfers to their website. Since PageRank is based

on the link structure of the Web, it is therefore useful to understand how

addition or deletion of hyperlinks influence it.

Mathematical analysis of PageRank’s sensitivity with respect to perturbations

of the matrix describing the webgraph is a topical subject of interest

(see for instance [2, 5, 11, 12, 13, 14] and the references therein). Normwise

and componentwise conditioning bounds [11] as well as the derivative [12, 13]

are used to understand the sensitivity of the PageRank vector. It appears

that the PageRank vector is relatively insensitive to small changes in the

graph structure, at least when these changes concern webpages with a low

1

Preliminary version – February 2, 2008

PageRank score [5, 12]. One could think therefore that trying to modify

its PageRank via changes in the link structure of the Web is a waste of

time. However, what is important for webmasters is not the values of the

PageRank vector but the ranking that ensues from it. Lempel and Morel [14]

showed that PageRank is not rank-stable, i.e. small modifications in the link

structure of the webgraph may cause dramatic changes in the ranking of the

webpages. Therefore, the question of how the PageRank of a particular page

or set of pages could be increased–even slightly–by adding or removing links

to the webgraph remains of interest.

As it is well known [1, 9], if a hyperlink from a page i to a page j is

added, without no other modification in the Web, then the PageRank of j

will increase. But in general, you do not have control on the inlinks of your

webpage unless you pay another webmaster to add a hyperlink from his/her

page to your or you make an alliance with him/her by trading a link for a

link [3, 8]. But it is natural to ask how you could modify your PageRank by

yourself. This leads to analyze how the choice of the outlinks of a page can

influence its own PageRank. Sydow [17] showed via numerical simulations

that adding well chosen outlinks to a webpage may increase significantly its

PageRank ranking. Avrachenkov and Litvak [2] analyzed theoretically the

possible effect of new outlinks on the PageRank of a page and its neighbors.

Supposing that a webpage has control only on its outlinks, they gave the

optimal linkage strategy for this single page. Bianchini et al. [5] as well as

Avrachenkov and Litvak in [1] consider the impact of links between web

communities (websites or sets of related webpages), respectively on the sum

of the PageRanks and on the individual PageRank scores of the pages of

some community. They give general rules in order to have a PageRank as

high as possible but they do not provide an optimal link structure for a

website.

Our aim in this paper is to find a generalization of Avrachenkov–Litvak’s

optimal linkage strategy [2] to the case of a website with several pages. We

consider a given set of pages and suppose we have only control on the outlinks

of these pages. We are interested in the problem of maximizing the sum of

the PageRanks of these pages.

Suppose G = (N, E) be the webgraph, with a set of nodes N = {1, . . . , n}

and a set of links E ⊆ N × N. For a subset of nodes I ⊆ N, we define

EI = {(i, j) ∈ E : i, j ∈ I} the set of internal links,

Eout(I) = {(i, j) ∈ E : i ∈ I, j /∈ I} the set of external outlinks,

Ein(I) = {(i, j) ∈ E : i /∈ I, j ∈ I} the set of external inlinks,

EI = {(i, j) ∈ E : i, j /∈ I} the set of external links.

If we do not impose any condition on EI and Eout(I), the problem of

maximizing the sum of the PageRanks of pages of I is quite trivial and does

2

Preliminary version – February 2, 2008

not have much interest (see the discussion in Section 4). Therefore, when

characterizing optimal link structures, we will make the following accessibility

assumption: every page of the website must have an access to the rest

of the Web.

Our first main result concerns the optimal outlink structure for a given

website. In the case where the subgraph corresponding to the website is

strongly connected, Theorem 10 can be particularized as follows.

Theorem. Let EI, Ein(I) and EI be given. Suppose that the subgraph (I, EI)

is strongly connected and EI 6= ∅. Then every optimal outlink structure

Eout(I) is to have only one outlink to a particular page outside of I.

We are also interested in the optimal internal link structure for a website.

In the case where there is a unique leaking node in the website, that is only

one node linking to the rest of the web, Theorem 11 can be particularized

as follows.

Theorem. Let Eout(I), Ein(I) and EI be given. Suppose that there is only one

leaking node in I. Then every optimal internal link structure EI is composed

of together with every possible backward link.

Putting together Theorems 10 and 11, we get in Theorem 12 the optimal

link structure for a website. This optimal structure is illustrated in Figure 1.

Theorem. Let Ein(I) and EI be given. Then, for every optimal link structure,

EI is composed of a forward chain of links together with every possible

backward link, and Eout(I) consists of a unique outlink, starting from the last

node of the chain.

I

Figure 1: Every optimal linkage strategy for a set I of five pages must

have this structure.

This paper is organized as follows. In the following preliminary section,

we recall some graph concepts as well as the definition of the PageRank, and

3

Preliminary version – February 2, 2008

we introduce some notations. In Section 3, we develop tools for analysing the

PageRank of a set of pages I. Then we come to the main part of this paper:

in Section 4 we provide the optimal linkage strategy for a set of nodes. In

Section 5, we give some extensions and variants of the main theorems. We

end this paper with some concluding remarks.

2 Graphs and PageRank

Let G = (N, E) be a directed graph representing the Web. The webpages

are represented by the set of nodes N = {1, . . . , n} and the hyperlinks are

represented by the set of directed links E ⊆ N × N. That means that

(i, j) ∈ E if and only if there exists a hyperlink linking page i to page j.

Let us first briefly recall some usual concepts about directed graphs (see

for instance [4]). A link (i, j) is said to be an outlink for node i and an

inlink for node j. If (i, j) ∈ E, node i is called a parent of node j. By

j ← i,

we mean that j belongs to the set of children of i, that is j ∈ {k ∈ N : (i, k) ∈

E}. The outdegree di of a node i is its number of children, that is

di = |{j ∈ N : (i, j) ∈ E}|.

A path from i0 to is is a sequence of nodes hi0, i1, . . . , isi such that (ik, ik+1) ∈

E for every k = 0, 1, . . . , s − 1. A node i has an access to a node j if there

exists a path from i to j. In this paper, we will also say that a node i has an

access to a set J if i has an access to at least one node j ∈ J . The graph G

is strongly connected if every node of N has an access to every other node

of N. A set of nodes F ⊆ N is a final class of the graph G = (N, E) if the

subgraph (F, EF) is strongly connected and moreover Eout(F) = ∅ (i.e. nodes

of F do not have an access to N \ F).

Let us now briefly introduce the PageRank score (see [5, 6, 12, 13, 15]

for background). Without loss of generality (please refer to the book of

Langville and Meyer [13] or the survey of Bianchini et al. [5] for details),

we can make the assumption that each node has at least one outlink, i.e.

di 6= 0 for every i ∈ N. Therefore the n×n stochastic matrix P = [Pij ]i,j2N

given by

Pij =

(

di

−1 if (i, j) ∈ E,

0 otherwise,

is well defined and is a scaling of the adjacency matrix of G. Let also

0 < c < 1 be a damping factor and z be a positive stochastic personalization

vector, i.e. zi > 0 for all i = 1, . . . , n and zT 1 = 1, where 1 denotes the

vector of all ones. The Google matrix is then defined as

G = cP + (1 − c)1zT .

4

Preliminary version – February 2, 2008

Since z > 0 and c < 1, this stochastic matrix is positive, i.e. Gij > 0 for all

i, j. The PageRank vector is then defined as the unique invariant measure

of the matrix G, that is the unique left Perron vector of G,

T = TG,

T 1 = 1.

(1)

The PageRank of a node i is the ith entry i = T ei of the PageRank

vector.

The PageRank vector is usually interpreted as the stationary distribution

of the following Markov chain (see for instance [13]): a random surfer moves

on the webgraph, using hyperlinks between pages with a probability c and

zapping to some new page according to the personalization vector with a

probability (1−c). The Google matrix G is the probability transition matrix

of this random walk. In this stochastic interpretation, the PageRank of a

node is equal to the inverse of its mean return time, that is −1

i is the mean

number of steps a random surfer starting in node i will take for coming back

to i (see [7, 10]).

3 PageRank of a website

We are interested in characterizing the PageRank of a set I. We define this

as the sum

T eI =

X

i2I

i,

where eI denotes the vector with a 1 in the entries of I and 0 elsewhere.

Note that the PageRank of a set corresponds to the notion of energy of a

community in [5].

Let I ⊆ N be a subset of the nodes of the graph. The PageRank of I can

be expressed as T eI = (1−c)zT (I−cP)−1eI from PageRank equations (1).

Let us then define the vector

v = (I − cP)−1eI. (2)

With this, we have the following expression for the PageRank of the set I:

T eI = (1 − c)zT v. (3)

The vector v will play a crucial role throughout this paper. In this

section, we will first present a probabilistic interpretation for this vector

and prove some of its properties. We will then show how it can be used in

order to analyze the influence of some page i ∈ I on the PageRank of the

set I. We will end this section by briefly introducing the concept of basic

absorbing graph, which will be useful in order to analyze optimal linkage

strategies under some assumptions.

5

Preliminary version – February 2, 2008

3.1 Mean number of visits before zapping

Let us first see how the entries of the vector v = (I − cP)−1eI can be

interpreted. Let us consider a random surfer on the webgraph G that,

as described in Section 2, follows the hyperlinks of the webgraph with a

probability c. But, instead of zapping to some page of G with a probability

(1 − c), he stops his walk with probability (1 − c) at each step of

time. This is equivalent to consider a random walk on the extended graph

Ge = (N ∪ {n + 1}, E ∪ {(i, n + 1) : i ∈ N}) with a transition probability

matrix

Pe =

cP (1 − c)1

0 1

.

At each step of time, with probability 1−c, the random surfer can disappear

from the original graph, that is he can reach the absorbing node n + 1.

The nonnegative matrix (I −cP)−1 is commonly called the fundamental

matrix of the absorbing Markov chain defined by Pe (see for instance [10,

16]). In the extended graph Ge, the entry [(I − cP)−1]ij is the expected

number of visits to node j before reaching the absorbing node n + 1 when

starting from node i. From the point of view of the standard random surfer

described in Section 2, the entry [(I − cP)−1]ij is the expected number of

visits to node j before zapping for the first time when starting from node i.

Therefore, the vector v defined in equation (2) has the following probabilistic

interpretation. The entry vi is the expected number of visits to the

set I before zapping for the first time when the random surfer starts his

walk in node i.

Now, let us first prove some simple properties about this vector.

Lemma 1. Let v ∈ Rn

0 be defined by v = cPv + eI. Then,

(a) maxi/2I vi ≤ c maxi2I vi,

(b) vi ≤ 1 + c vi for all i ∈ N; with equality if and only if the node i does

not have an access to I,

(c) vi ≥ minj i vj for all i ∈ I; with equality if and only if the node i

does not have an access to I;

Proof. (a) Since c < 1, for all i /∈ I,

max

i/2I

vi = max

i/2I

c

X

j i

vj

di

≤ cmax

j

vj .

Since c < 1, it then follows that maxj vj = maxi2I vi.

(b) The inequality vi ≤ 1

1−c follows directly from

max

i

vi ≤ max

i

1 + c

X

j i

vj

di

≤ 1 + cmax

j

vj .

6

Preliminary version – February 2, 2008

From (a) it then also follows that vi ≤ c

1−c for all i /∈ I. Now, let

i ∈ N such that vi = 1

1−c . Then i ∈ I. Moreover,

1 + c vi = vi = 1 + c

X

j i

vj

di

,

that is vj = 1

1−c for every j ← i. Hence node j must also belong to I.

By induction, every node k such that i has an access to k must belong

to I.

(c) Let i ∈ I. Then, by (b)

1 + c vi ≥ vi = 1 + c

X

j i

vj

di

≥ 1 + cmin

j i

vj ,

so vi ≥ minj i vj for all i ∈ I. If vi = minj i vj then also 1+c vi = vi

and hence, by (b), the node i does not have an access to I.

Let us denote the set of nodes of I which on average give the most visits

to I before zapping by

V = argmax

j2I

vj .

Then the following lemma is quite intuitive. It says that, among the nodes

of I, those which provide the higher mean number of visits to I are parents

of I, i.e. parents of some node of I.

Lemma 2 (Parents of I). If Ein(I) 6= ∅, then

V ⊆ {j ∈ I : there exists ℓ ∈ I such that (j, ℓ) ∈ Ein(I)}.

If Ein(I) = ∅, then vj = 0 for every j ∈ I.

Proof. Suppose first that Ein(I) 6= ∅. Let k ∈ V with v = (I − cP)−1eI. If

we supposed that there does not exist ℓ ∈ I such that (k, ℓ) ∈ Ein(I), then

we would have, since vk > 0,

vk = c

X

j k

vj

dk

≤ cmax

j /2I

vj = cvk < vk,

which is a contradiction. Now, if Ein(I) = ∅, then there is no access to I

from I, so clearly vj = 0 for every j ∈ I.

Lemma 2 shows that the nodes j ∈ I which provide the higher value

of vj must belong to the set of parents of I. The converse is not true, as

we will see in the following example: some parents of I can provide a lower

mean number of visits to I that other nodes which are not parents of I. In

other word, Lemma 2 gives a necessary but not sufficient condition in order

to maximize the entry vj for some j ∈ I.

7

Preliminary version – February 2, 2008

1

2

3

4 5

6

7

8

9

10

11

I

Figure 2: The node 6 /∈ V and yet it is a parent of I = {1} (see Example

1).

Example 1. Let us see on an example that having (j, i) ∈ Ein(I) for some

i ∈ I is not sufficient to have j ∈ V. Consider the graph in Figure 2. Let

I = {1} and take a damping factor c = 0.85. For v = (I −cP)−1e1, we have

v2 = v3 = v4 = 4.359 > v5 = 3.521 > v6 = 3.492 > v7 > · · · > v11,

so V = {2, 3, 4}. As ensured by Lemma 2, every node of the set V is a parent

of node 1. But here, V does not contain all parents of node 1. Indeed, the

node 6 /∈ V while it is a parent of 1 and is moreover its parent with the

lowest outdegree. Moreover, we see in this example that node 5, which is a

not a parent of node 1 but a parent of node 6, gives a higher value of the

expected number of visits to I before zapping, than node 6, parent of 1.

Let us try to get some intuition about that. When starting from node 6,

a random surfer has probability one half to reach node 1 in only one step.

But he has also a probability one half to move to node 11 and to be send

far away from node 1. On the other side, when starting from node 5, the

random surfer can not reach node 1 in only one step. But with probability

3/4 he will reach one of the nodes 2, 3 or 4 in one step. And from these

nodes, the websurfer stays very near to node 1 and can not be sent far away

from it.

In the next lemma, we show that from some node i ∈ I which has an

access to I, there always exists what we call a decreasing path to I. That is,

we can find a path such that the mean number of visits to I is higher when

starting from some node of the path than when starting from the successor

of this node in the path.

Lemma 3 (Decreasing paths to I). For every i0 ∈ I which has an access

to I, there exists a path hi0, i1, . . . , isi with i1, . . . , is−1 ∈ I and is ∈ I such

8

Preliminary version – February 2, 2008

that

vi0 > vi1 > ... > vis .

Proof. Let us simply construct a decreasing path recursively by

ik+1 ∈ argmin

j ik

vj ,

as long as ik ∈ I. If ik has an access to I, then vik+1 < vik < 1

1−c by

Lemma 1(b) and (c), so the node ik+1 has also an access to I. By assumption,

i0 has an access to I. Moreover, the set I has a finite number of elements,

so there must exist an s such that is ∈ I.

3.2 Influence of the outlinks of a node

We will now see how a modification of the outlinks of some node i ∈ N can

change the PageRank of a subset of nodes I ⊆ N. So we will compare two

graphs on N defined by their set of links, E and e E respectively.

Every item corresponding to the graph defined by the set of links e E will

be written with a tilde symbol. So e P denotes its scaled adjacency matrix,

e the corresponding PageRank vector, e di = |{j : (i, j) ∈ e E}| the outdegree

of some node i in this graph, ev = (I − c e P)−1eI and eV = argmaxj2I evj .

Finally, by jf←i we mean j ∈ {k : (i, k) ∈ e E}.

So, let us consider two graphs defined respectively by their set of links E

and e E. Suppose that they differ only in the links starting from some given

node i, that is {j : (k, j) ∈ E} = {j : (k, j) ∈ e E} for all k 6= i. Then their

scaled adjacency matrices P and e P are linked by a rank one correction. Let

us then define the vector

=

X

jf i

ej

e di

−

X

j i

ej

di

,

which gives the correction to apply to the line i of the matrix P in order to

get e P.

Now let us first express the difference between the PageRank of I for two

configurations differing only in the links starting from some node i. Note

that in the following lemma the personalization vector z does not appear

explicitly in the expression of e.

Lemma 4. Let two graphs defined respectively by E and e E and let i ∈ N

such that for all k 6= i, {j : (k, j) ∈ E} = {j : (k, j) ∈ e E}. Then

eT eI = T eI + ci

T v

1 − c T (I − cP)−1ei

.

9

Preliminary version – February 2, 2008

Proof. Clearly, the scaled adjacency matrices are linked by e P = P + ei T .

Since c < 1, the matrix (I − cP)−1 exists and the PageRank vectors can be

expressed as

T = (1 − c)zT (I − cP)−1,

eT = (1 − c)zT (I − c (P + eiT ))−1.

Applying the Sherman–Morrison formula to ((I − cP) − ceiT )−1, we get

eT = (1 − c)zT (I − cP)−1 + (1 − c)zT (I − cP)−1ei

cT (I − cP)−1

1 − cT (I − cP)−1ei

,

and the result follows immediately.

Let us now give an equivalent condition in order to increase the PageRank

of I by changing outlinks of some node i. The PageRank of I increases

essentially when the new set of links favors nodes giving a higher mean

number of visits to I before zapping.

Theorem 5 (PageRank and mean number of visits before zapping). Let

two graphs defined respectively by E and e E and let i ∈ N such that for all

k 6= i, {j : (k, j) ∈ E} = {j : (k, j) ∈ e E}. Then

eT eI > T eI if and only if T v > 0

and eT eI = T eI if and only if T v = 0.

Proof. Let us first show that T (I − cP)−1ei ≤ 1 is always verified. Let

u = (I −cP)−1ei. Then u−cPu = ei and, by Lemma 1(a), uj ≤ ui for all

j. So

Tu =

X

jf i

uj

e di

−

X

j i

uj

di

≤ ui −

X

j i

uj

di

≤ ui − c

X

j i

uj

di

= 1.

Now, since c < 1 and > 0, the conclusion follows by Lemma 4.

The following Proposition 6 shows how to add a new link (i, j) starting

from a given node i in order to increase the PageRank of the set I. The

PageRank of I increases as soon as a node i ∈ I adds a link to a node j

with a larger or equal expected number of visits to I before zapping.

Proposition 6 (Adding a link). Let i ∈ I and let j ∈ N be such that

(i, j) /∈ E and vi ≤ vj . Let e E = E ∪ {(i, j)}. Then

eT eI ≥ T eI

with equality if and only if the node i does not have an access to I.

10

Preliminary version – February 2, 2008

Proof. Let i ∈ I and let j ∈ N be such that (i, j) /∈ E and vi ≤ vj . Then

1 + c

X

k i

vk

di

= vi ≤ 1 + cvi ≤ 1 + cvj ,

with equality if and only if i does not have an access to I by Lemma 1(b).

Let eE = E ∪ {(i, j)}. Then

T v =

1

di + 1

vj −

X

k i

vk

di

≥ 0,

with equality if and only if i does not have an access to I. The conclusion

follows from Theorem 5.

Now let us see how to remove a link (i, j) starting from a given node i in

order to increase the PageRank of the set I. If a node i ∈ N removes a link

to its worst child from the point of view of the expected number of visits to

I before zapping, then the PageRank of I increases.

Proposition 7 (Removing a link). Let i ∈ N and let j ∈ argmink i

vk.

Let eE = E \ {(i, j)}. Then

eT eI ≥ T eI

with equality if and only if vk = vj for every k such that (i, k) ∈ E.

Proof. Let i ∈ N and let j ∈ argmink i

vk. Let e E = E \ {(i, j)}. Then

T v =

X

k i

vk − vj

di(di − 1)

≥ 0,

with equality if and only if vk = vj for all k ← i. The conclusion follows by

Theorem 5.

In order to increase the PageRank of I with a new link (i, j), Proposition

6 only requires that vj ≤ vi. On the other side, Proposition 7 requires

that vj = mink i vk in order to increase the PageRank of I by deleting link

(i, j). One could wonder whether or not this condition could be weakened

to vj < vi, so as to have symmetric conditions for the addition or deletion

of links. In fact, this can not be done as shown in the following example.

Example 2. Let us see by an example that the condition j ∈ argmink i

vk

in Proposition 7 can not be weakened to vj < vi. Consider the graph in

Figure 3 and take a damping factor c = 0.85. Let I = {1, 2, 3}. We have

v1 = 2.63 > v2 = 2.303 > v3 = 1.533.

As ensured by Proposition 7, if we remove the link (1, 3), the PageRank of

I increases (e.g. from 0.199 to 0.22 with a uniform personalization vector

z = 1

n1), since 3 ∈ argmink 1

vk. But, if we remove instead the link (1, 2),

the PageRank of I decreases (from 0.199 to 0.179 with z uniform) even if

v2 < v1.

11

Preliminary version – February 2, 2008

1

2

3 4

5

6

7

I

Figure 3: For I = {1, 2, 3}, removing link (1, 2) gives eT eI < T eI,

even if v1 > v2 (see Example 2).

Remark. Let us note that, if the node i does not have an access to the set I,

then for every deletion of a link starting from i, the PageRank of I will not

be modified. Indeed, in this case T v = 0 since by Lemma 1(b), vj = 1

1−c

for every j ← i.

3.3 Basic absorbing graph

Now, let us introduce briefly the notion of basic absorbing graph (see Chapter

III about absorbing Markov chains in Kemeny and Snell’s book [10]).

For a given graph (N, E) and a specified subset of nodes I ⊆ N, the basic

absorbing graph is the graph (N, E0) defined by E0

out(I) = ∅, E0

I = {(i, i) : i ∈

I}, E0

in(I) = Ein(I) and E0

I = EI. In other words, the basic absorbing graph

(N, E0) is a graph constructed from (N, E), keeping the same sets of external

inlinks and external links Ein(I), EI, removing the external outlinks Eout(I)

and changing the internal link structure EI in order to have only self-links

for nodes of I.

Like in the previous subsection, every item corresponding to the basic

absorbing graph will have a zero symbol. For instance, we will write 0

for the PageRank vector corresponding to the basic absorbing graph and

V0 = argmaxj2I[(I − cP0)−1eI]j .

Proposition 8 (PageRank for a basic absorbing graph). Let a graph defined

by a set of links E and let I ⊆ N. Then

T eI ≤ T0

eI,

with equality if and only if Eout(I) = ∅.

Proof. Up to a permutation of the indices, equation (2) can be written as

I − cPI −cPout(I)

−cPin(I) I − cPI

vI

v

I

=

1

0

,

12

Preliminary version – February 2, 2008

so we get

v =

vI

c(I − cPI)−1Pin(I)

vI

. (4)

By Lemma 1(b) and since (I − cPI)−1 is a nonnegative matrix (see for

instance the chapter on M-matrices in Berman and Plemmons’s book [4]),

we then have

v ≤

1

1−c 1

c

1−c (I − cPI)−1Pin(I)1

= v0,

with equality if and only if no node of I has an access to I, that is Eout(I) = ∅.

The conclusion now follows from equation (3) and z > 0.

Let us finally prove a nice property of the set V when I = {i} is a

singleton: it is independent of the outlinks of i. In particular, it can be

found from the basic absorbing graph.

Lemma 9. Let a graph defined by a set of links E and let I = {i} Then there

exists an α 6= 0 such that (I −cP)−1ei = α(I −cP0)−1ei. As a consequence,

V = V0.

Proof. Let I = {i}. Since vI = vi is a scalar, it follows from equation (4)

that the direction of the vector v does not depend on EI and Eout(I) but

only on Ein(I) and EI.

4 Optimal linkage strategy for a website

In this section, we consider a set of nodes I. For this set, we want to choose

the sets of internal links EI ⊆ I × I and external outlinks Eout(I) ⊆ I × I

in order to maximize the PageRank score of I, that is T eI.

Let us first discuss about the constraints on E we will consider. If we do

not impose any condition on E, the problem of maximizing T eI is quite

trivial. As shown by Proposition 8, you should take in this case Eout(I) =

∅ and EI an arbitrary subset of I × I such that each node has at least

one outlink. You just try to lure the random walker to your pages, not

allowing him to leave I except by zapping according to the preference vector.

Therefore, it seems sensible to impose that Eout(I) must be nonempty.

Now, let us show that, in order to avoid trivial solutions to our maximization

problem, it is not enough to assume that Eout(I) must be nonempty.

Indeed, with this single constraint, in order to lose as few as possible visits

from the random walker, you should take a unique leaking node k ∈ I (i.e.

Eout(I) = {(k, ℓ)} for some ℓ ∈ I) and isolate it from the rest of the set I

(i.e. {i ∈ I : (i, k) ∈ EI} = ∅).

Moreover, it seems reasonable to imagine that Google penalizes (or at

least tries to penalize) such behavior in the context of spam alliances [8].

13

Preliminary version – February 2, 2008

All this discussion leads us to make the following assumption.

Assumption A (Accessibility). Every node of I has an access to at least

one node of I.

Let us now explain the basic ideas we will use in order to determine an

optimal linkage strategy for a set of webpages I. We determine some forbidden

patterns for an optimal linkage strategy and deduce the only possible

structure an optimal strategy can have. In other words, we assume that

we have a configuration which gives an optimal PageRank T eI. Then we

prove that if some particular pattern appeared in this optimal structure,

then we could construct another graph for which the PageRank eT eI is

strictly higher than T eI.

We will firstly determine the shape of an optimal external outlink structure

Eout(I), when the internal link structure EI is given, in Theorem 10.

Then, given the external outlink structure Eout(I) we will determine the possible

optimal internal link structure EI in Theorem 11. Finally, we will put

both results together in Theorem 12 in order to get the general shape of an

optimal linkage strategy for a set I when Ein(I) and EI are given.

Proofs of this section will be illustrated by several figures for which we

take the following drawing convention.

Convention. When nodes are drawn from left to right on the same horizontal

line, they are arranged by decreasing value of vj . Links are represented

by continuous arrows and paths by dashed arrows.

The first result of this section concerns the optimal outlink structure

Eout(I) for the set I, while its internal structure EI is given. An example of

optimal outlink structure is given after the theorem.

Theorem 10 (Optimal outlink structure). Let EI, Ein(I) and EI be given.

Let F1, . . . ,Fr be the final classes of the subgraph (I, EI). Let Eout(I) such

that the PageRank T eI is maximal under Assumption A. Then Eout(I) has

the following structure:

Eout(I) = Eout(F1) ∪ · · · ∪ Eout(Fr),

where for every s = 1, . . . , r,

Eout(Fs) ⊆ {(i, j) : i ∈ argmin

k2Fs

vk and j ∈ V}.

Moreover for every s = 1, . . . , r, if EFs 6= ∅, then |Eout(Fs)| = 1.

14

Preliminary version – February 2, 2008

Proof. Let EI, Ein(I) and EI be given. Suppose Eout(I) is such that T eI is

maximal under Assumption A.

We will determine the possible leaking nodes of I by analyzing three

different cases.

Firstly, let us consider some node i ∈ I such that i does not have children

in I, i.e. {k ∈ I : (i, k) ∈ EI} = ∅. Then clearly we have {i} = Fs for some

s = 1, . . . , r, with i ∈ argmink2Fs

vk and EFs = ∅. From Assumption A, we

have Eout(Fs) 6= ∅, and from Theorem 5 and the optimality assumption, we

have Eout(Fs) ⊆ {(i, j) : j ∈ V} (see Figure 4).

i ℓ j

I

Figure 4: If vj < vℓ, then eT eI > T eI with e Eout(I) = Eout(I)∪{(i, ℓ)}\

{(i, j)}.

Secondly, let us consider some i ∈ I such that i has children in I, i.e.

{k ∈ I : (i, k) ∈ EI} 6= ∅ and

vi ≤ min

k i

k2I

vk.

Let j ∈ argmink i

vk. Then j ∈ I and vj < vi by Lemma 1(c). Suppose

by contradiction that the node i would keep an access to I if we took

e Eout(I) = Eout(I) \ {(i, j)} instead of Eout(I). Then, by Proposition 7, considering

e Eout(I) instead of Eout(I) would increase strictly the PageRank of I

while Assumption A remains satisfied (see Figure 5). This would contradict

i j

I

Figure 5: If vj = mink i vk and i has another access to I, then eT eI >

T eI with eEout(I) = Eout(I) \ {(i, j)}.

the optimality assumption for Eout(I). From this, we conclude that

• the node i belongs to final class Fs of the subgraph (I, EI) with EFs 6= ∅

for some s = 1, . . . , r;

• there does not exist another ℓ ∈ I, ℓ 6= j such that (i, ℓ) ∈ Eout(I);

• there does not exist another k in the same final class Fs, k 6= i such

that such that (k, ℓ) ∈ Eout(I) for some ℓ ∈ I.

Again, by Theorem 5 and the optimality assumption, we have j ∈ V (see

Figure 4).

15

Preliminary version – February 2, 2008

Let us now notice that

max

k2I

vk < min

k2I

vk. (5)

Indeed, with i ∈ argmink2I

vk, we are in one of the two cases analyzed above

for which we have seen that vi > vj = argmaxk2I

vk.

Finally, consider a node i ∈ I that does not belong to any of the final

classes of the subgraph (I, EI). Suppose by contradiction that there exists

j ∈ I such that (i, j) ∈ Eout(I). Let ℓ ∈ argmink i

vk. Then it follows

from inequality (5) that ℓ ∈ I. But the same argument as above shows

that the link (i, ℓ) ∈ Eout(I) must be removed since Eout(I) is supposed to

be optimal (see Figure 5 again). So, there does not exist j ∈ I such that

(i, j) ∈ Eout(I) for a node i ∈ I which does not belong to any of the final

classes F1, . . . ,Fr.

Example 3. Let us consider the graph given in Figure 6. The internal link

structure EI, as well as Ein(I) and EI are given. The subgraph (I, EI) has two

final classes F1 and F2. With c = 0.85 and z the uniform probability vector,

this configuration has six optimal outlink structures (one of these solutions

is represented by bold arrows in Figure 6). Each one can be written as

Eout(I) = Eout(F1) ∪ Eout(F2), with Eout(F1) = {(4, 6)} or Eout(F1) = {(4, 7)}

and ∅ 6= Eout(F2) ⊆ {(5, 6), (5, 7)}. Indeed, since EF1 6= ∅, as stated by

Theorem 10, the final class F1 has exactly one external outlink in every

optimal outlink structure. On the other hand, the final class F2 may have

several external outlinks, since it is composed of a unique node and moreover

this node does not have a self-link. Note that V = {6, 7} in each of these six

optimal configurations, but this set V can not be determined a priori since

it depends on the chosen outlink structure.

Now, let us determine the optimal internal link structure EI for the set

I, while its outlink structure Eout(I) is given. Examples of optimal internal

structure are given after the proof of the theorem.

Theorem 11 (Optimal internal link structure). Let Eout(I), Ein(I) and EI

be given. Let L = {i ∈ I : (i, j) ∈ Eout(I) for some j ∈ I} be the set of

leaking nodes of I and let nL = |L| be the number of leaking nodes. Let

EI such that the PageRank T eI is maximal under Assumption A. Then

there exists a permutation of the indices such that I = {1, 2, . . . , nI}, L =

{nI − nL + 1, . . . , nI},

v1 > · · · > vnI−nL > vnI−nL+1 ≥ · · · ≥ vnI ,

and EI has the following structure:

EL

I ⊆ EI ⊆ EU

I ,

16

Preliminary version – February 2, 2008

1

2

3

4

5

6

7

8

I

F1

F2

Figure 6: Bold arrows represent one of the six optimal outlink structures

for this configuration with two final classes (see Example 3).

where

EL

I = {(i, j) ∈ I × I : j ≤ i} ∪ {(i, j) ∈ (I \ L) × I : j = i + 1},

EU

I = EL

I ∪ {(i, j) ∈ L × L: i < j}.

Proof. Let Eout(I), Ein(I) and EI be given. Suppose EI is such that T eI is

maximal under Assumption A.

Firstly, by Proposition 6 and since every node of I has an access to I,

every node i ∈ I links to every node j ∈ I such that vj ≥ vi (see Figure 7),

that is

{(i, j) ∈ EI : vi ≤ vj} = {(i, j) ∈ I × I : vi ≤ vj}. (6)

i

I

Figure 7: Every i ∈ I must link to every j ∈ I with vj ≥ vi.

Secondly, let (k, i) ∈ EI such that k 6= i and k ∈ I \L. Let us prove that,

if the node i has an access to I by a path hi, i1, . . . , isi such that ij 6= k for

all j = 1, . . . , s and is ∈ I, then vi < vk (see Figure 8). Indeed, if we had

vk ≤ vi then, by Lemma 1(c), there would exists ℓ ∈ I such that (k, ℓ) ∈ EI

and vℓ = minj k vj < vi ≤ vk. But, with e EI = EI \ {(k, ℓ)}, we would

have eT eI > T eI by Proposition 7 while Assumption A remains satisfied

since the node k would keep access to I via the node i (see Figure 9). That

17

Preliminary version – February 2, 2008

i k j

I

Figure 8: The node i can not have an access to I without crossing k

since in this case we should then have vi < vk.

i k ℓ

I

Figure 9: If vℓ = minj k vj, then eT eI > T eI with e Eout(I) = Eout(I) \

{(k, ℓ)}.

contradicts the optimality assumption. This leads us to the conclusion that

vk > vi for every k ∈ I\L and i ∈ L. Moreover vi 6= vk for every i, k ∈ I\L,

i 6= k. Indeed, if we had vi = vk, then (k, i) ∈ EI by (6) while by Lemma 3,

the node i would have an access to I by a path independant from k. So we

should have vi < vk.

We conclude from this that we can relabel the nodes of N such that

I = {1, 2, . . . nI}, L = {nI − nL + 1, . . . , nI} and

v1 > v2 > · · · > vnI−nL > vnI−nL+1 ≥ · · · ≥ vnI . (7)

It follows also that, for i ∈ I \ L and j > i, (i, j) ∈ EI if and only if j =

i + 1. Indeed, suppose first i < nI − nL. Then, we cannot have (i, j) ∈ EI

with j > i+1 since in this case we would contradict the ordering of the nodes

given by equation (7) (see Figure 8 again with k = i+1 and remember that

by Lemma 3, node j has an access to I by a decreasing path). Moreover,

node i must link to some node j > i in order to satisfy Assumption A, so

(i, i+1) must belong to EI. Now, consider the case i = nI −nL. Suppose we

had (i, j) ∈ EI with j > i+1. Let us first note that there can not exist two or

more different links (i, ℓ) with ℓ ∈ L since in this case we could remove one

of these links and increase strictly the PageRank of the set I. If vj = vi+1,

we could relabel the nodes by permuting these two indices. If vj < vi+1,

then with e EI = EI ∪ {(i, i + 1)} \ {(i, j)}, we would have eT eI > T eI

by Theorem 5 while Assumption A remains satisfied since the i would keep

access to I via node i + 1. That contradicts the optimality assumption. So

we have proved that

{(i, j) ∈ EI : i < j and i ∈ I \ L} = {(i, i + 1) : i ∈ I \ L}. (8)

Thirdly, it is obvious that

{(i, j) ∈ EI : i < j and i ∈ L} ⊆ {(i, j) ∈ L × L: i < j}. (9)

18

Preliminary version – February 2, 2008

The announced structure for a set EI giving a maximal PageRank score

T eI under Assumption A now follows directly from equations (6), (8)

and (9).

Example 4. Let us consider the graphs given in Figure 10. For both cases,

the external outlink structure Eout(I) with two leaking nodes, as well as Ein(I)

and EI are given. With c = 0.85 and z the uniform probability vector, the

optimal internal link structure for configuration (a) is given by EI = EL

I ,

while in configuration (b) we have EI = EU

I (bold arrows), with EL

I and EU

I

defined in Theorem 11.

I

L

(a)

I

L

(b)

Figure 10: Bold arrows represent optimal internal link structures. In (a)

we have EI = EL

I , while EI = EU

I in (b).

Finally, combining the optimal outlink structure and the optimal internal

link structure described in Theorems 10 and 11, we find the optimal linkage

strategy for a set of webpages. Let us note that, since we have here control

on both EI and Eout(I), there are no more cases of several final classes or

several leaking nodes to consider. For an example of optimal link structure,

see Figure 1.

Theorem 12 (Optimal link structure). Let Ein(I) and EI be given. Let EI

and Eout(I) such that T eI is maximal under Assumption A. Then there

exists a permutation of the indices such that I = {1, 2, . . . , nI},

v1 > · · · > vnI > vnI+1 ≥ · · · ≥ vn,

19

Preliminary version – February 2, 2008

and EI and Eout(I) have the following structure:

EI = {(i, j) ∈ I × I : j ≤ i or j = i + 1},

Eout(I) = {(nI, nI + 1)}.

Proof. Let Ein(I) and EI be given and suppose EI and Eout(I) are such that

T eI is maximal under Assumption A. Let us relabel the nodes of N such

that I = {1, 2, . . . , nI} and v1 ≥ · · · ≥ vnI > vnI+1 = maxj2I

vj . By

Theorem 11, (i, j) ∈ EI for every nodes i, j ∈ I such that j ≤ i. In particular,

every node of I has an access to node 1. Therefore, there is a unique final

class F1 ⊆ I in the subgraph (I, EI). So, by Theorem 10, Eout(I) = {(k, ℓ)}

for some k ∈ F1 and ℓ ∈ I. Without loss of generality, we can suppose that

ℓ = nI + 1. By Theorem 11 again, the leaking node k = nI and therefore

(i, i + 1) ∈ EI for every node i ∈ {1, . . . , nI − 1}.

Let us note that having a structure like described in Theorem 12 is a

necessary but not sufficient condition in order to have a maximal PageRank.

Example 5. Let us show by an example that the graph structure given in

Theorem 12 is not sufficient to have a maximal PageRank. Consider for instance

the graphs in Figure 11. Let c = 0.85 and a uniform personalization

vector z = 1

n1. Both graphs have the link structure required Theorem 12 in

order to have a maximal PageRank, with v

(a) =

6.484 6.42 6.224 5.457

T

and v

(b) =

6.432 6.494 6.247 5.52

T

. But the configuration (a) is

not optimal since in this case, the PageRank T

(a)

eI = 0.922 is strictly

less than the PageRank T

(b)

eI = 0.926 obtained by the configuration (b).

Let us nevertheless note that, with a non uniform personalization vector

z =

0.7 0.1 0.1 0.1

T

, the link structure (a) would be optimal.

1 2 3 4

I

(a)

2 1 3 4

I

(b)

Figure 11: For I = {1, 2, 3}, c = 0.85 and z uniform, the link structure

in (a) is not optimal and yet it satisfies the necessary conditions of

Theorem 12 (see Example 5).

20

Preliminary version – February 2, 2008

5 Extensions and variants

Let us now present some extensions and variants of the results of the previous

section. We will first emphasize the role of parents of I. Secondly, we will

briefly talk about Avrachenkov–Litvak’s optimal link structure for the case

where I is a singleton. Then we will give variants of Theorem 12 when

self-links are forbidden or when a minimal number of external outlinks is

required. Finally, we will make some comments of the influence of external

inlinks on the PageRank of I.

5.1 Linking to parents

If some node of I has at least one parent in I then the optimal linkage strategy

for I is to have an internal link structure like described in Theorem 12

together with a single link to one of the parents of I.

Corollary 13 (Necessity of linking to parents). Let Ein(I)6= ∅ and EI be

given. Let EI and Eout(I) such that T eI is maximal under Assumption A.

Then Eout(I) = {(i, j)}, for some i ∈ I and j ∈ I such that (j, k) ∈ Ein(I)

for some k ∈ I.

Proof. This is a direct consequence of Lemma 2 and Theorem 12.

Let us nevertheless remember that not every parent of nodes of I will

give an optimal link structure, as we have already discussed in Example 1

and we develop now.

Example 6. Let us continue Example 1. We consider the graph in Figure 2 as

basic absorbing graph for I = {1}, that is Ein(I) and EI are given. We take

c = 0.85 as damping factor and a uniform personalization vector z = 1

n1.

We have seen in Example 1 than V0 = {2, 3, 4}. Let us consider the value of

the PageRank 1 for different sets EI and Eout(I):

Eout(I)

∅ {(1, 2)} {(1, 5)} {(1, 6)} {(1, 2), (1, 3)}

EI = ∅ 0.1739 0.1402 0.1392 0.1739

EI = {(1, 1)} 0.5150 0.2600 0.2204 0.2192 0.2231

As expected from Corollary 15, the optimal linkage strategy for I = {1} is

to have a self-link and a link to one of the nodes 2, 3 or 4. We note also that

a link to node 6, which is a parent of node 1 provides a lower PageRank that

a link to node 5, which is not parent of 1. Finally, if we suppose self-links

are forbidden (see below), then the optimal linkage strategy is to link to one

or more of the nodes 2, 3, 4.

In the case where no node of I has a parent in I, then every structure

like described in Theorem 12 will give an optimal link structure.

21

Preliminary version – February 2, 2008

Proposition 14 (No external parent). Let Ein(I) and EI be given. Suppose

that Ein(I) = ∅. Then the PageRank T eI is maximal under Assumption A

if and only if

EI = {(i, j) ∈ I × I : j ≤ i or j = i + 1},

Eout(I) = {(nI, nI + 1)}.

for some permutation of the indices such that I = {1, 2, . . . , nI}.

Proof. This follows directly from T eI = (1 − c)zT v and the fact that, if

Ein(I) = ∅,

v = (I − cP)−1eI =

(I − cPI)−11

0

,

up to a permutation of the indices.

5.2 Optimal linkage strategy for a singleton

The optimal outlink structure for a single webpage has already been given

by Avrachenkov and Litvak in [2]. Their result becomes a particular case of

Theorem 12. Note that in the case of a single node, the possible choices for

Eout(I) can be found a priori by considering the basic absorbing graph, since

V = V0.

Corollary 15 (Optimal link structure for a single node). Let I = {i} and

let Ein(I) and EI be given. Then the PageRank i is maximal under Assumption

A if and only if EI = {(i, i)} and Eout(I) = {(i, j)} for some j ∈ V0.

Proof. This follows directly from Lemma 9 and Theorem 12.

5.3 Optimal linkage strategy under additional assumptions

Let us consider the problem of maximizing the PageRank T eI when selflinks

are forbidden. Indeed, it seems to be often supposed that Google’s

PageRank algorithm does not take self-links into account. In this case,

Theorem 12 can be adapted readily for the case where |I| ≥ 2. When I is

a singleton, we must have EI = ∅, so Eout(I) can contain several links, as

stated in Theorem 10.

Corollary 16 (Optimal link structure with no self-links). Suppose |I| ≥ 2.

Let Ein(I) and EI be given. Let EI and Eout(I) such that T eI is maximal

under Assumption A and assumption that there does not exist i ∈ I such

that {(i, i)} ∈ EI. Then there exists a permutation of the indices such that

I = {1, 2, . . . , nI}, v1 > · · · > vnI > vnI+1 ≥ · · · ≥ vn, and EI and Eout(I)

have the following structure:

EI = {(i, j) ∈ I × I : j < i or j = i + 1},

Eout(I) = {(nI, nI + 1)}.

22

Preliminary version – February 2, 2008

Corollary 17 (Optimal link structure for a single node with no self-link).

Suppose I = {i}. Let Ein(I) and EI be given. Suppose EI = ∅. Then the

PageRank i is maximal under Assumption A if and only if ∅ 6= Eout(I) ⊆ V0.

Let us now consider the problem of maximizing the PageRank T eI

when several external outlinks are required. Then the proof of Theorem 10

can be adapted readily in order to have the following variant of Theorem 12.

Corollary 18 (Optimal link structure with several external outlinks). Let

Ein(I) and EI be given. Let EI and Eout(I) such that T eI is maximal under

Assumption A and assumption that |Eout(I)| ≥ r. Then there exists a

permutation of the indices such that I = {1, 2, . . . , nI}, v1 > · · · > vnI >

vnI+1 ≥ · · · ≥ vn, and EI and Eout(I) have the following structure:

EI = {(i, j) ∈ I × I : j < i or j = i + 1},

Eout(I) = {(nI, jk) : jk ∈ V for k = 1, . . . , r}.

5.4 External inlinks

Finally, let us make some comments about the addition of external inlinks to

the set I. It is well known that adding an inlink to a particular page always

increases the PageRank of this page [1, 9]. This can be viewed as a direct

consequence of Theorem 5 and Lemma 1. The case of a set of several pages

I is not so simple. We prove in the following theorem that, if the set I has

a link structure as described in Theorem 12 then adding an inlink to a page

of I from a page j ∈ I which is not a parent of some node of I will increase

the PageRank of I. But in general, adding an inlink to some page of I from

I may decrease the PageRank of the set I, as shown in Examples 7 and 8.

Theorem 19 (External inlinks). Let I ⊆ N and a graph defined by a set

of links E. If

min

i2I

vi > max

j /2I

vj ,

then, for every j ∈ I which is not a parent of I, and for every i ∈ I, the

graph defined by e E = E ∪ {(j, i)} gives eT eI > T eI.

Proof. This follows directly from Theorem 5.

Example 7. Let us show by an example that a new external inlink is not

always profitable for a set I in order to improve its PageRank, even if I has

an optimal linkage strategy. Consider for instance the graph in Figure 12.

With c = 0.85 and z uniform, we have T eI = 0.8481. But if we consider

the graph defined by e Ein(I) = Ein(I)∪{(3, 2)}, then we have eT eI = 0.8321 <

T eI.

23

Preliminary version – February 2, 2008

1 2 3

I

Figure 12: For I = {1, 2}, adding the external inlink (3, 2) gives eT eI <

T eI (see Example 7).

Example 8. A new external inlink does not not always increase the PageRank

of a set I in even if this new inlink comes from a page which is not already a

parent of some node of I. Consider for instance the graph in Figure 13. With

c = 0.85 and z uniform, we have T eI = 0.6. But if we consider the graph

defined by eEin(I) = Ein(I) ∪ {(4, 3)}, then we have eT eI = 0.5897 < T eI.

1 2 3

5 4

I

Figure 13: For I = {1, 2, 3}, adding the external inlink (4, 3) gives

eT eI < T eI (see Example 8).

6 Conclusions

In this paper we provide the general shape of an optimal link structure

for a website in order to maximize its PageRank. This structure with a

forward chain and every possible backward links may be not intuitive. At

our knowledge, it has never been mentioned, while topologies like a clique,

a ring or a star are considered in the literature on collusion and alliance

between pages [3, 8]. Moreover, this optimal structure gives new insight

into the affirmation of Bianchini et al. [5] that, in order to maximize the

PageRank of a website, hyperlinks to the rest of the webgraph “should be

in pages with a small PageRank and that have many internal hyperlinks”.

More precisely, we have seen that the leaking pages must be choosen with

respect to the mean number of visits before zapping they give to the website,

rather than their PageRank.

Let us now present some possible directions for future work.

We have noticed in Example 5 that the first node of I in the forward

chain of an optimal link structure is not necessarily a child of some node of

I. In the example we gave, the personalization vector was not uniform. We

wonder if this could occur with a uniform personalization vector and make

the following conjecture.

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Preliminary version – February 2, 2008

Conjecture. Let Ein(I)6= ∅ and EI be given. Let EI and Eout(I) such that

T eI is maximal under Assumption A. If z = 1

n1, then there exists j ∈ I

such that (j, i) ∈ Ein(I), where i ∈ argmaxk

vk.

If this conjecture was true we could also ask if the node j ∈ I such that

(j, i) ∈ Ein(I) where i ∈ argmaxk

vk belongs to V.

Another question concerns the optimal linkage strategy in order to maximize

an arbitrary linear combination of the PageRanks of the nodes of I.

In particular, we could want to maximize the PageRank T eS of a target

subset S ⊆ I by choosing EI and Eout(I) as usual. A general shape for

an optimal link structure seems difficult to find, as shown in the following

example.

Example 9. Consider the graphs in Figure 14. In both cases, let c = 0.85

and z = 1

n1. Let I = {1, 2, 3} and let S = {1, 2} be the target set. In the

configuration (a), the optimal sets of links EI and Eout(I) for maximizing

T eS has the link structure described in Theorem 12. But in (a), the

optimal EI and Eout(I) do not have this structure. Let us note nevertheless

that, by Theorem 12, the subsets ES and Eout(S) must have the link structure

described in Theorem 12.

1 2 3 4 5

6

7

8

S

I

(a)

1 2 3 4

S

I

(b)

Figure 14: In (a) and (b), bold arrows represent optimal link structures

for I = {1, 2, 3} with respect to a target set S = {1, 2} (see Example 9).

Acknowledgements

This paper presents research supported by a grant “Actions de recherche

concert´ees – Large Graphs and Networks” of the “Communaut´e Fran¸caise

de Belgique” and by the Belgian Network DYSCO (Dynamical Systems,

Control, and Optimization), funded by the Interuniversity Attraction Poles

25

Preliminary version – February 2, 2008

Programme, initiated by the Belgian State, Science Policy Office. The second

author was supported by a research fellow grant of the “Fonds de la

Recherche Scientifique – FNRS” (Belgium). The scientific responsibility

rests with the authors.

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