While we’re on the subject of sorting things online, we might as well talk about Google: the 93-billion dollar company whose main export is taking all the things ever and putting them in the right order. If there’s one thing Google knows best, it’s sorting stuff.

I was curious how it all works, and it turned out really interesting, plus I got to learn a bit about Markov chains. It all starts with an algorithm called PageRank¹. Accodring to Wikipedia:

Pagerank uses a model of a random surfer who gets bored after several clicks and switches to a random page. It can be understood as a Markov chain in which the states are pages, and the transitions are the links between pages. When calculating PageRank, pages with no outbound links are assumed to link out to all other pages in the collection (the random surfer chooses another page at random).

The PageRank values are the entries of the dominant eigenvector of the modified adjacency matrix.

In this post I’ll try to break that down and provide some of the background necessary to understand Google PageRank. My main reference for making sense of it all was the book Probability, Markov Chains, Queues, And Simulation by William J. Stweart, my linear algebra textbook, and Google.

Graphs as Matrices

A graph is a collection of nodes joined by edges. If the edges are arrows that flow in one direction, we call that a directed graph. A graph whose edges have each been assigned a “weight” (usually some real number) is a weighted graph.

A graph of n nodes can be represented in the form of an n x n adjacency matrix, $M = [m_{ij}]$% such that $m_{ij}$ is equal to the weight of the edge going from node $j$ to node $i$:

[0, 1, 0, 0]
[1, 0, 2, 0]
[2, 1, 0, 1]
[0, 0, 4, 0]

Stochastic Matrices

The term “stochastic” is used to describe systems whose state can only be described in probabilistic terms (i.e: the likelihood of some event happening at any given time).

Scenario:

Consider two competing websites. Every month, the first website loses 30% of its audience to the second website, while the second website loses 60% of its audience to the first.

If the two websites start out with 50% of the global audience each, how many users will each website have after a month? After a year?

This scenario can be represented as the following system:

P = [0.7, 0.6],   x_0 = [0.5, 0.5]
    [0.3, 0.4]

This is a Markov chain with transition matrix $P$ and a state vector $\mathbf{ x^{(0)} }$.

The transition matrix is called a stochastic matrix; it represents the likelihood that some individual in a system will transition from one state to another. The columns on a stochastic matrix are always non-negative numbers that add up to 1 (i.e: the probability of at least one of the events occurring is always 1 – the likelihood of a user either staying on the same website, or leaving, is always 100%. He must choose one of the two).

The state after the first month is

$$ \begin{gather} \mathbf{x^{ (1) }} = P \mathbf{ x^{ (0) } } \cr = [(0.7 + 0.6)\times0.5, (0.3 + 0.4)\times0.5] \cr = [0.65, 0.35] \cr \end{gather} $$

So, after the first month, the second website will have only 35% of the global audience.

To get the state of the system after two months, we simply apply the transition matrix again, and so on. That is, the current state of a Markov chain depends only on its previous state. Thus, the state vector at month $k$ can be defined recursively:

$$ \mathbf{ x^{(k)} } = P\mathbf{ x^{ (k - 1) } } $$

From which, through substitution, we can derive the following equation:

$$\mathbf{ x^{(k)} } = P^k \mathbf{ x^{(0)} }$$

Using this information, we can figure out the state of the system after a year, and then again after two years (using the (Sage)[http://www.sagemath.org/], a mathematical library for python):

P = Matrix([[0.70, 0.60], [0.30, 0.40]])
x = vector([0.5,0.5])

P^12 * x
# -> (0.666666666666500, 0.333333333333500)

P^24 * x
# -> (0.666666666666666, 0.333333333333333)

So it seems like the state vector is “settling” around those values. It would appear that, as $n \to \infty$, $P^n\mathbf{x^{(0)}}$ is converging to some $\mathbf{x}$ such that $P\mathbf{x} = \mathbf{x}$. As we’ll see below, this is indeed the case.

We’ll call this $\mathbf{x} $ the steady state vector.

Eigenvectors!

Recall from linear algebra that an eigenvector of a matrix $A$ is a vector $\mathbf{x}$ such that:

$$A\mathbf{x} = \lambda \mathbf{x}$$

for some scalar $\lambda$ (the eigenvalue). A leading eigenvalue is an eigenvalue $\lambda_{1}$ such that its absolute value is greater than any other eigenvalue for the given matrix.

One method of finding the leading eigenvector of a matrix is through a power iteration sequence, defined recursively like so:

$$ \mathbf{ x_k } = \cfrac{A\mathbf{x_{k-1}}} {| A\mathbf{x_{ k-1 }} |} $$

Again, by noting that we can substitute $A\mathbf{x_{k-1}} = A(A\mathbf{x_{k-2}}) = A^2\mathbf{x_{k-2}}$, and so on, it follows that:

$$ \mathbf{x_k} = \cfrac{A^k \mathbf{x_0}} {| A^k \mathbf{x_0} |} $$

This sequence converges to the leading eigenvector of $A$.

Thus we see that the steady state vector is just an eigenvector with the special case $\lambda = 1$.

Stochastic Matrices that Don’t Play Nice

Before we can finally get to Google PageRank, we need to make a few more observations.

First, it should be noted that power iteration has its limitations: not all stochastic matrices converge. Take as an example:

P = Matrix([ [0, 1, 0], [1, 0, 0], [0, 0, 1]])
x = vector([0.2, 0.3, 0.5])

P * x
# -> (0.3, 0.2, 0.5)

P^2 * x
# -> (0.2, 0.3, 0.5)

P^3 * x
# -> (0.3, 0.2, 0.5)

The state vectors of this matrix will oscillate in such a way forever. This matrix can be thought of as the transformation matrix for reflection about a line in the x,y axis… this system will never converge (indeed, it has no leading eigenvalue: $|\lambda_1| = |\lambda_2| = |\lambda_3| = 1$).

Another way of looking at $P$ is by drawing its graph:

Using our example of competing websites, this matrix describes a system such that, every month, all of the first website’s users leave and join the seconds website, only to abandon the second website again a month later and return to the first, and so on, forever.

It would be absurd to hope for this system to converge to a steady state.

States 1 and 2 are examples of recurrent states. These are states that, once reached, there is a probability of 1 (absolute certainty) that the Markov chain will return to them infinitely many times.

A transient state is such that the probability is $> 0$ that they will never be reached again. (If the probability is 0, we call such a state ephemeral – in terms of Google PageRank, this would be a page that no other page links to):

There are two conditions a transition matrix must meet if we want to ensure that it converges to a steady state:

It must be irreducible: an irreducible transition matrix is a matrix whose graph has no closed subsets. (A closed subset is such that no state within it can reach a state outside of it. 1, 2 and 3 above are closed from 4 and 5.)

It must be primitive: A primitive matrix $P$ is such that, for some positive integer $n$, $P^n$ is such that $p_{ij} > 0$ for all $p_{ij} \in P$ (that is: all of its entries are positive numbers).

More generally, it must be positive recurrent and aperiodic.

Positive recurrence means that it takes, on average, a finite number of steps to return to any given state. Periodicity means the number of steps it takes to return to a particular state is always divisible by some natural number $n$ (its period).

Since we’re dealing with finite Markov chains, irreducibility implies positive recurrence, and primitiveness ensures aperiodicity.

Google PageRank

We are now finally ready to understand how the PageRank algorithm works. Recall from Wikipedia:

The formula uses a model of a random surfer who gets bored after several clicks and switches to a random page. The PageRank value of a page reflects the chance that the random surfer will land on that page by clicking on a link. It can be understood as a Markov chain in which the states are pages, and the transitions, which are all equally probable, are the links between pages.

So, for example, if we wanted to represent our graph above, we would start with the following adjacency matrix:

[0, 0, 0.5, 0,   0],
[0, 0, 0.5, 0.5, 0],
[1, 1, 0,   0,   0],
[0, 0, 0,   0,   0],
[0, 0, 0,   0.5, 0]

For the algorithm to work, we must transform this original matrix in such a way that we end up with an irreducible, primitive matrix. First,

If a page has no links to other pages, it becomes a sink and therefore terminates the random surfing process. If the random surfer arrives at a sink page, it picks another URL at random and continues surfing again.

When calculating PageRank, pages with no outbound links are assumed to link out to all other pages in the collection.

    [0, 0, 0.5, 0,   0.2],
    [0, 0, 0.5, 0.5, 0.2],
S = [1, 1, 0,   0,   0.2],
    [0, 0, 0,   0,   0.2],
    [0, 0, 0,   0.5, 0.2]

We are now ready to produce $G$, the Google Matrix, which is both irreducible and primitive. Its steady state vector gives us the final PageRank score for each page.

The Google Matrix

The [Google Matric](http://en.wikipedia.org/wiki/Google_matrix] for an $n \times n$ matrix $S$ is derived from the equation

$$ G = \alpha S + (1 - \alpha) \frac{1}{n} E$$

Where $E = \mathbf{ e }\mathbf{ e }^T$ is an $n \times n$ matrix whose entries are all 1, and $0 \le \alpha \le 1$ is referred to as the damping factor.

If $\alpha = 1$, then $G = S$. Meanwhile, if $\alpha = 0$ all of the entries in $G$ are the same (hence, the original structure of the network is “dampened” by $\alpha$, until we lose it altogether).

So the matrix $(1 - \alpha) \frac{1}{n} E$ is a matrix that represents a “flat” network in which all pages link to all pages, and the user is equally likely to click any given link (with likelihood $\frac{1-\alpha}{n}$), while $S$ is dampened by a factor of $\alpha$.

Google uses a damping factor of 0.85. For more on this, I found this paper.

tl;dr: the second eigenvalue of a Google matrix is $|\lambda_2| = \alpha \le |\lambda_1| = 1 $, and the rate of convergence of the power iteration is given by $\frac{|\lambda_2|}{|\lambda_1|} = \alpha$. So higher values of $\alpha$ imply better accuracy but worse performance.

With some moving stuff around, we can see that

$$ \left(\alpha s_{ 1j } + \frac{1-\alpha}{ n }\right) + \left(\alpha s_{ 2j } + \frac{1-\alpha}{ n }\right) + … + \left(\alpha s_{ nj } + \frac{1-\alpha}{ n }\right) = 1 $$

For all $j$ up to $n$, which means that $G$ is indeed stochastic, irreducible, and primitive. Cool.

In conclusion,

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1. Actually, it all started with the HITS algorithm, which PageRank is based off of. More details here.