# CLUSTERfoo!

## Eigengoogle. How the Google PageRank Algorithm Works

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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.

And it all started with an algorithm called PageRank1. According 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.

### 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]$M = [m_{ij}]$ such that m_ij$m_{ij}$ is equal to the weight of the edge going from node j$j$ to node i$i$:

[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$P$ and a state vector x_0$\mathbf{ x^{(0)} }$.

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

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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:

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

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Using this information, we can figure out the state of the system after a year, and then again after two years (using the Sage mathematical library for python):

1P = Matrix([[0.70, 0.60],
2            [0.30, 0.40]])
3x = vector([0.5,0.5])
4P^12*x
5# -> (0.666666666666500, 0.333333333333500)
6P^24*x
7# -> (0.666666666666666, 0.333333333333333)


So it seems like the state vector is “settling” around those values. It would appear that, as EQ$n \to \infty$, EQ$P^n\mathbf{ x^{ (0) } }$ is converging to some $\mathbf{ x }$ such that EQ$P\mathbf{ x } = \mathbf{ x }$. 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:

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for some scalar EQ$\lambda$ (the eigenvalue). A leading eigenvalue is an eigenvalue EQ$\lambda_{ 1 }$ 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:

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

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

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

### 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:

 1P = Matrix([ [0, 1, 0],
2             [1, 0, 0],
3             [0, 0, 1]])
4
5x = vector([0.2, 0.3, 0.5])
6
7P * x
8# -> (0.3, 0.2, 0.5)
9P^2 * x
10# -> (0.2, 0.3, 0.5)
11P^3 * x
12# -> (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: EQ$|\lambda_1| = |\lambda_2| = |\lambda_3| = 1$).

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 EQ$> 0$ 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 EQ$n$, EQ$P^n$ is such that EQ$p_{ ij } > 0$ for all EQ$p_{ ij } \in P$ (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.

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 Matrix for an $n \times n$ matrix $S$ is derived from the equation

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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. If you would like to research further, this paper is a good place to start and is quite readable.

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 elementary algebra we can see that

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

In conclusion,

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

## Sorting Posts by User Engagement Level (With Elasticsearch Implementation)

At Functional Imperative we’re building the new CanLII Connects website (a social portal for Canada’s largest database of legal cases), and this week I was given the task of figuring out a sensible way of sorting posts.

Figuring out how to sort user-generated content is a common problem that many social websites face.

Here’s Reddit’s scoring equation for ‘Best’ (source and explanation):

Not all scoring equations are that hairy, here are a few more.

Interestingly enough, Reddit’s ‘Hot’ scoring function (explained in link above):

is quite flawed.

Sidenote: One observation not mentioned in that first article is that, while all other equations use some form of time_now - time_posted to calculate how old a post is, the clever guys at Reddit use time_posted - some_old_date.

The advantage of this is that the post’s score need only be calculated once, whereas the value of scores calculated with time_now will change on every request.

Anyway, while all those scoring functions work pretty well, they didn’t quite fit the requirements for CanLII Connects.

In this post, I’ll walk through the decision process of creating a scoring function. Hopefully this will be useful if you encounter a similar feature to implement.

### Requirements:

CanLII Connects links to a database of legal cases, and users can post opinions on those cases:

1. A user can post.
2. A user can upvote a post.
3. A user can comment on a post.
4. A user can comment on a comment on a post.

So what’s a sensible way of sorting posts?

Right away, we’re dealing with a different problem than Reddit or HN: while it makes sense to slowly degrade the score of a post on those sites over time, the same does not make sense for CanLII. Old cases might be cited at any time, no matter how old they are, so what matters is not how old a discussion is, but rather how actively engaged users are within a given discussion.

### Initial Score

Ok, so first we have to give each post an initial score. I like Reddit’s approach of taking the base-10 log of its upvotes. This makes sense because, the more popular a post already is, the more likely people are to see it, and therefore upvote it, which gives it an unfair advantage.

In our case, we’re not only trying to measure how much people “like” a post, but rather how engaged they are with it. It makes sense that, while an upvote is a fine indicator of “engagedness”, a user actually bothering to comment on a post is even more of an indicator. I’ll count that as equivalent to two upvotes, and a user commenting on a comment will count as three upvotes (the 2 is so we don’t take the log of 1 or 0):

log_10(2 + u + 2c + 3cc)

### Frequency

Next, we need the post’s position to degrade as it becomes less active. It makes sense to divide the intial score by some factor of time:

log_10(score)/t_ave

Now we need a reasonable value for t_ave$\bar{ t }$. A good start is the average time, in seconds, between the three most recent user interactions with a post.

We define a user interaction to be: a user creates a post, a user comments on a post, or a user upvotes a post.

Also, we want the most recent interactions to weigh more than older interactions. So let’s say each t weighs twice as much as the previous:

t_ave = sum(0.5^i-1 * (t_i - t_i-1)) / sum(o.5^i-1) for i = 1..3 -- ENABLE JAVASCRIPT TO VIEW RENDERED EQUATION

Where

t_0$t_0$ = UNIX timestamp, at now, in seconds.

t_n$t_n$ = UNIX timestamp of nth interaction.

### One Final Detail

There is one last property we want this function to have, which is the following: if interactions are very frequent right now (within a timeframe of, say, 10 days), then clearly the post is “hot”, and its score should be boosted. But as time passes, it really doesn’t matter as much how much distance there is between interactions. If a post has already gone a full year without anyone commenting on it, does it really make that much difference if it goes another month without a comment?

To accomplish the first property, all we do is divide t_ave$\bar{ t }$ by the number of seconds in 10 days: 60*60*24*10.

To accomplish the second property, what we are looking for is some sort of always-increasing, concave function (positive derivative, negative second derivative). The first thing that comes to mind is the square-root function, which is good enough.

### Result

And thus we have our final scoring function:

log(u + 2c + 3cc)/(sqrt(t_ave/60*60*24*10)); t_ave defined above -- ENABLE JAVASCRIPT FOR PRETTY EQUATIONS

If we plot this equation for x = number of points and y = time, we can see the shape of this function and check for different values if they make sense:

As expected, there is a steep 10-day “boost” period, followed by an increasingly slower decline in the value as more and more time passes.

The function is also heavily biased toward very new posts, which will always come out on top, giving them a chance. This might be a bad idea if posting becomes frequent, but user interaction is low (many summaries a day, few votes or comments), and might have to be changed.

There are many ways to tweak this equation (changing the boost period, for example) to make it more or less biased towards either time or user interaction.

### Bonus Round: Implementing in ElasticSearch

Implementing a custom scoring function in Elasticsearch, though easy once it’s all set up, was rather frustrating because of the poor documentation.

For our implementation, we’re using the tire gem (a wrapper around the Elasticsearch API). This is where we call the custom scoring script:

1query do
2  #custom_score script: "parseInt(doc['upvote_count'].value)", lang: "javascript" do
3  custom_score script: script, lang: 'javascript' do
4    string query.join(" OR ")
5  end
6end


Where script is simply a variable holding the contents of a javascript file as a string. Note the option lang: 'javascript'. This lets us use javascript as our language of choice, as opposed to mvel, the most poorly documented scripting language on the face of the earth. To enable this option, we’ll also require the elasticsearch-lang-javascript plugin.

Here is our script:

Sidenote: Notice the logger function. This enables us to implement a sort of “console.log” which we can read using the following shell command tail -f /var/log/elasticsearch/elasticsearch.log.

 1// Logger function:
2var logger = org.elasticsearch.common.logging.Loggers.getLogger("rails_logger");
3// Example usage:
4logger.info("========= NEW CALC ===========");
5
6var points_log = parseFloat(doc.points_log.value);
7var now = Math.round(new Date().getTime() / 1000);
8
9/**
10* NOTE: doc.ts.values is not actually an array,
11* here I create an array out of it:
12**/
13var ts = [];
14for (var i = 0; i < doc.ts.values.length; i++) ts[i] = doc.ts.values[i];
15ts.push(now);
17ts.reverse();
18
19/**
20* Boost period.
21**/
22var ten_days = 60*60*24*10;
23
24/**
25* The scoring function
26**/
27function score() {
28  /**
29  * Weighed average numerator
30  **/
31  var times_num = (function() {
32    var val = 0;
33    for (var i = 1; i < ts.length; i++) {
34      var exp = i - 1;
35      val += Math.pow(0.5, exp) *
36             (parseFloat(ts[i]) -
37             parseFloat(ts[i - 1]));
38    }
39    return val;
40  })();
41
42  /**
43  * Weighed average denominator
44  **/
45  var times_denom = (function() {
46    var val = 0;
47    for (var i = 1; i < ts.length; i++) {
48      var exp = i - 1;
49      val += Math.pow(0.5, exp);
50    }
51    return val;
52  })();
53
54  var t_ave = (times_num/times_denom);
55
56  return points_log/Math.sqrt(t_ave/ten_days);
57};
58
59score();


## Amdahl's Law

The mathematical equations in this post require Javascript and will not render if you are on RSS or email.

As multicore computing becomes the norm (even my phone is dual core!), it’s important to understand the benefits and also the limitations of concurrency. Amdahl’s Law addresses the latter.

Let’s imagine a simple program. It prints “Hello World” $100$ times, then quits.

Our first version of the program is written as a single sequential task: it prints one “Hello World”, then another, then another, $100$ times, then quits. This program takes some unit of time, $t$ to execute.

Now say we have a dual-core machine at hand. (My phone, perhaps).

Cool, now we can spawn two tasks that print “Hello World” $50$ times each. And, because our magical imaginary computer experiences no overhead, it takes us exactly $\frac{ t }{ 2 }$ units of time to run our second program.

So we keep adding more and more processors, until we have $100$ concurrent threads printing one “Hello World” each, and our program runs $100$ times faster.

At this point we stop: “Ah, the trend is clear: more processors equals more speed! No point in continuing this experiment.”

A naive (wrong) first guess: Given $n$ processors executing a program, the maximum boost in speed is $n$. (That is, we can get our program to run $n$ times faster).

Cool! This means that, given enough processors, we could make any program run almost instantly. Right?

Of course this is not the case! Enough daydreaming. Let’s figure out a more realistic estimate.

Let $P$ be the proportion of our program that can run in parallel. Then it follows that $1 - P$ is the proportion that cannot be broken up into independent tasks.

For example, since our program can be broken up into $100$ independent tasks, then $1 - P = \frac{ 1 }{ 100 }$.

It follows that the maximum boost in speed (denoted $S(n)$) that we can expect out of assigning concurrent tasks to $n$ parallel processors can be represented by the following equation:

This is, in fact, Amdahl’s equation.

Uh-oh… do you see it? As we add more and more processors to our computer, and $n \to \infty$, we are left with $S = \frac{ 1 }{ 1 - p }$.

What we have here is a clear case of diminishing returns.

How bad is it? Let’s add one million cores to our imaginary computer, and measure its performance at $P = 99\%$:

Well, for our imaginary software, $99\%$ of which can be parallelized, we can expect a maximum boost of $S = 100$.

What about a program with $P = 90\%$?

There’s that same plateau again. But this time we’re only seeing a maximum performance boost of $S = 10$.

By $P = 50\%$, we’re down to a program that can only be boosted to run twice as fast no matter how much parallel processing your machine is capable of!

Final Note: In fact, Amdahl’s Law is not exclusive to concurrency, but applies to any speed-boosting strategy that only affects some portion of a program.

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