google and the page algortimo

7/28/2019 Google and the Page Algortimo

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Definition of PageRankComputation of PageRank

The effect of new linksChoosing PageRank damping factor

Reducing the value of c

Mathematical Analysis of Google PageRank

Konstantin Avrachenkov

INRIA Sophia Antipolis, France

Konstantin Avrachenkov Mathematical Analysis of Google PageRank
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Ranking Answers to User Query

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Ranking Answers to User Query

How a search engine should sort the retrieved answers?

Possible solutions: (a) use the frequency of the searched terms inthe Web page, (b) analyse the log files,... These solutions mightbe not objective.

An original idea of Google is based on two observations:

1 The more pages point to a Web page, the more important thepage is.

2 If more important Web pages point to the page, the page iseven more important.

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Web Graph

Consider the Web as a directed graph:


D fi i i f P R k
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Random surfer PageRank Definition

Consider a random surfer who, with probability c (=0.85) follows arandomly chosen outgoing link, otherwise, with probability 1 c

jumps to a completely random page.

Then, PageRank i of page i is the long run fraction of time thata random surfer spends on page i.

The dynamics of random surfer can be described using Markovchains.

Markov chains definition


D fi iti f P R k
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Formal PageRank Definition

Let n be the total number of pages on the Web (n 8 109).

Define the hyperlink matrix P = {pij}ni,j=1 as follows:

pij = 1/di, if j is one of the di outgoing links of i,

pij = 1/n, if di = 0 (dangling node),

pij = 0, otherwise.

The transitions of easily bored surfer corresponds to the

following perturbed Google matrix

P = cP + (1 c)(1/n)E,

where E is an n n matrix consisting of ones, c = 0.85.


Definition of PageRank


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Formal PageRank Definition

Then, the PageRank vector is a solution of

P = , 1 = 1,

or, equivalently, in the component form

i = j i

c

dj

j +1 c

n

, i

i = 1.




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Example

P =

0 0.5 0.5 0 00.2 0.2 0.2 0.2 0.20.2 0.2 0.2 0.2 0.2

0.5 0 0 0 0.50 0 0 1 0

=

0.1982 0.1731 0.1731 0.2573 0.1982


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Power Method

Even though this is a well kept secret, it seems that Google stilluses the simple power iteration method for PageRank computation

(k+1) = c(k)P + (1 c) 1n

1T, (0) = 1n

1T.

It can be easily estimated that using the constant c = 0.85 Googleachieves the tolerance level (measured by the residual(k+1) (k)) of 103 105 for only 50-100 iterations.

But even this small number of iterations takes Google about aweek to update the PageRank...


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Monte Carlo Method

Run random surfer process (Xt)t0, m times from each page,

terminating at each step with probability 1 c. Evaluate j asj =[fraction of time spent in j].

It turns out the one iteration of Monte Carlo method (m = 1) issufficient to estimate well the PageRank of important pages.


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Monte Carlo Method


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gComputation of PageRank



Advantages of MC Method in respect to PI Method

Monte Carlo method has natural parallel implementation;

Monte Carlo method provides good estimation of thePageRank for important pages already after one iteration;

Monte Carlo method allows one to perform continuous update

of the PageRank as the structure of the Web changes.


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Computation of PageRankThe effect of new links

Choosing PageRank damping factorReducing the value of c

Decomposition based on SCC

It is known that the Web Graph consistes of many disjoint StronglyConnected Components (SCCs). This fact implies that thehyperlink matrix has the following form

P =

P1 0...

. . ....

0 PN

,

where the elements of diagonal blocks PI, I = 1,..., N, correspondto links inside the I-th SCC. Denote by nI the size of the I-th SCC.


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Decomposition based on SCC

For each block I, define the Google matrixPI = cPI + (1 c)(1/nI)E,and let vector I be the PageRank of SCC I such that

IPI = , I1 = 1.

Then the following theorem holds.

Theorem

The PageRank is given by

= ((n1/n)1, (n2/n)2, . . . , (nN/n)N). (1)


Definition of PageRankC i f P R k


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To what extend a page can control its PageRank?

Let us give a rough estimation by how much a page can control itsPageRank by modifying its outgoing links.

Define a discrete-time absorbing Markov chain {Xt, t = 0, 1, . . .}with the state space {0, 1 . . . , n}, where transitions between thestates 1, . . . , n are conducted by the matrix cP, and the state 0 isabsorbing.

Let Nj be the number of visits to state j = 1, . . . , n beforeabsorption. Then, denote zij := E(Nj|X0 = i).


Definition of PageRankC t ti f P R k
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Let qji be the probability to reach the state i before absorption ifthe initial state is j.

We have the following decomposition result:

Theorem

The PageRank of page i = 1, . . . , n is given by

i = 1 cn

zii1 +

nj=1j=i

qji , i = 1, . . . , n. (2)

Proof


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The decomposition formula (2) represents the PageRank of page ias a product of three multipliers where only the term zii dependson the outgoing links of page i.

Hence, by changing the outgoing links, a page can control itsPageRank up to a multiple factor

zii = 1/(1 qii) [1, 1/(1 c2)],

where qii [0, c2

] is a probability to return back to i starting fromi before absorption.

Note that the upper bound 1/(1 c2) (approximately 3.6 forc = .85) is hard or rather not possible to achieve...


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We note that even a threefold increase of the PageRank might not

be considered as a significant improvement, since Google measuresthe PageRank on a logarithmic scale.

Next we show how a Web page should use its scarce resources toincrease its PageRank.


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Optimal Linking Strategy

Let us show that there exists in fact an optimal linking strategy.

Consider a page i = 1, . . . , n and assume that i has links to thepages i1, . . . , ik where il = i for all l = 1, . . . , k.

Then for the mean return time, we have

ii = 1 +c

k

kl=1

ili +1

n(1 c)

nj=1j=i

ji, (3)

where ij is the mean first passage time from page i to page j andc is the Google constant.

Since i = 1/ii, the objective now is to choose k and i1, . . . , iksuch that ii becomes as small as possible.




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Optimal Linking Strategy

From (3) one can see that i is a linear function of jis.Moreover, outgoing links from i do not affect jis.

Thus, the best what one can do is to link only to one Web page j

such thatji = min

j{ji}.

Note that (surprisingly) the PageRank of j

plays no role here.

Still, as was already mentioned, we need to admit that a Web pageowner has very limited control of his/her PageRank.


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p gThe effect of new links


The Bowtie structure of the Web graph

A. Broder et al. 2000 and R. Kumar et al. 2000 have observedthat the Web Graph has a Bowtie structure.


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p gThe effect of new links


Maximizing the mass of SCC

One can choose the damping factor c to maximize the totalPageRank mass of SCC:

However, the factor c becomes to close to one. Is it good?


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More detailed structure of the Web graph


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By renumbering the nodes, the transition matrix P can be thentransformed to the following form

P =

Q 0R T

, (4)

where

the block T corresponds to the Extended SCC,

the block Q corresponds to the part of the OUT component

without dangling nodes and their predecessors,and the block R corresponds to the transitions from ESCC to thenodes in block Q.



Th ff f li k
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As was observed by Moler 2003, the PageRank vector can beexpressed by the following formula

=1 c

n1T[I cP]1. (5)

If we substitute the expression (4) for the transition matrix P into(5), we obtain the following formula for the part of the PageRankvector corresponding to the nodes in ESCC:

T =

1 c

n1T

[I

cT]

1

= (1 c

)uT[

I

cT]

1

, (6)

where = nT/n and nT is the number of nodes in ESCC, andwhere uT is the uniform distribution over all ESCC nodes.



Th ff t f li k
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First, we note that since matrix T is substochastic, the inverse[I T]1 exists and consequently T 0 as c 1.

Clearly, it is not good to take the value of c too close to one.



The effect of new links


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It follows that the value of c should not be chosen in the criticalregion where the PageRank mass of the ESCC component israpidly decreasing.

Luckily, the shape of the function ||T(c)||1 is such that itdecreases drastically only when c is really close to one, whichleaves a lot of freedom for choosing c.

In particular, the famous Google constant c = 0.85 is small enoughto ensure a reasonably large PageRank mass of ESCC.



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However, as we have observed in numerical experiments, evenmoderately large values of c result in an unfairly large PageRankmass of the Pure OUT component.

Now, our goal is to find the values of c that lead to a fairdistribution of the PageRank mass between the Pure OUT and theESCC components.



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Let v be some probability vector over ESCC. We would like to

choose c = c that satisfies the condition

||T(c)|| = ||vT||, (7)

that is, starting from v, the probability mass preserved in ESCC

after one step should be equal to the PageRank of ESCC.

Reasonable choices of v:

1 T, the quasi-stationary distribution of T,

2 the uniform vector u,

3 the normalized PageRank vector T(c)/||T(c)||.

All three criteria indicate that c = 1/2 seems to be quite a goodchoice.



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Experiments with the log files

c PR rank w/o link PR rank with link rank by no. of clicks

Node A

0.5 1648 2307 25880.85 731 2101 25880.95 226 2116 2588

Node B

0.5 1648 4009 36490.85 731 3279 3649

0.95 226 3563 3649

Table: Comparison between PR and click based rankings.



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Recommendations for Web Page Design

Now we can suggest the following recommendations for Web PageDesign:

1 The more pages a Web site has, the better.

2 Link all pages inside a Web site to the main page. This waythe main page will have a significant weight.

3 Give hyperlinks to the Departement and Institution Web sites.

4 Do not make inappropriate links.

And, of course, one should not forget that content still matters forGoogle. There is really no substitute for good original content...



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Thank you!



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Markov Chain Definition

Discrete-time discrete-state Markov chain is a stochastic process{Xn}

n=0 on the set of states S = {1, 2,..., |S|} such that

P{Xn+1 = j} =iS

P{Xn+1 = j|Xn = i}P{xn = i}.

We denote pij := P{Xn+1 = j|Xn = i} and call {pij}|S|i,j=1 the

matrix of transition probabilities.return



The effect of new linksC f
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Proof: It follows from (??) that

i =1 c

n

1T[I cP]1ei =1 c

n

n

j=1

zji. (8)

Next, we note that for any i,j = 1, . . . , n; i= j, we have

zji = qjizii,

and consequently, substituting the last equation in (8) we obtain(2). Q.E.D.

return

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google and the page algortimo

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