# Narratives and Moral Membership

1. You subscribe to narratives that are closest fit to your limited scope of experience. (Hence, shifts in narrative are predated by events, not deliberation.)

2. You are mostly exposed to others whose experiential reality matches yours. (Either by choice or proximity.)

3. Homogeneity reinforces narratives.

4. Language is contextual. Therefore narrative shapes meaning. (You are not reading what I’m writing; you are reading what you think I’m writing.) (To understand a narrative you must adopt it.)

5. Narrative shapes experience.

6. GOTO 1.

Do your beliefs closely match (or merely appear to) those of your in-group? Do you think this is a coincidence? — You give yourself too much credit for being a moral being.

Now consider a difficult exercise: make an attempt to detach your thinking from narratives. Now try to evaluate the world only in terms of causal relationships between actions. So that you say “This is a description of this event, or of this person’s action, and only what I know about it, and this is the result I observe (intended or unintended), and that is all I know; and anything else is unknown and pure conjecture”. Observe actions and their effects only, and subdue the instinct to form a story around it.

If the observed act was beneficial, does embellishing and forming a story around it, and saying “This is the sort of person that does such an act, and this is the sort of person that doesn’t”, make it more so? Or is it enough to study the act, its benefits, and act according to it. If the act was harmful, does it serve any purpose to distort its causes and effects through your preferred narrative lens, and dress it up in idioms so to signal to others your membership to this or that moral group, or is it enough to study the causes and effects dispassionately, and act according to your own reason.

Note then how moral narratives serve only the purpose of gatekeeping social memberships. Otherwise it’s enough to do actions that are beneficial, and refrain from actions that are harmful, without inventing stories and adorning yourself with vapid mannerisms that signal your adherence to this or that social group and their narrative.

# The Enemy of Your Enemy

1. If the enemy of your enemy is your friend, then by the transitive property, if any of your enemies are enemies, you are your enemy.

# A Pair of Dirty Shoes

A second pair stacked on top. Dirty socks. Dirty shirts. Piles of them.
That chair I don’t even like. Too many cups. A clean bedroom, the bed made. Clean dishes, one of those pluggable scented candles. His records. My records. Too much space.

# Books

I find there’s something about learning from a physical object (with volume, texture, scent) that makes things “click” better in my head; and I doubt I’m an oddity in this respect — the root of this effect may be more biological than preferential: For example, when recalling some piece of information, often I can remember where — within some textbook I read half a decade ago — the subject is explained. And I do mean, a literal, physical “where”: a page, (a thing,) which at all times can be found tucked somewhere within a physical stack of bound paper, which I can touch and feel; I might even remember where I was sitting (or standing, or lying), as I held that particular book, open at that particular page, when I studied the subject, and what it felt like to the touch, and how the lighting in the room fell on the paper; maybe I jotted some note on the margin, maybe I spilled tea on the pages before, giving that “where” its own unique irregular texture and color — all of which my brain will forever associate with that information. Digital can’t do any of that.

In this sense, there is nothing left to improve about the physical book.

On the other hand, if once I open this book I find myself confused, (maybe because the book was written by a guy named named Spivak, who does not like to explain what happens in between those little “=” symbols,) I can’t press on anything and expect it to expand magically and reveal more detail. It’s at this point that I will turn to my laptop, and search for more information, maybe even ask a stranger half a world away, or watch an interactive video. My copy of “Calculus on Manifolds” can’t do any of that.

So even though ebooks do offer some perks, in many ways the printed book remains superior. I think it’s a safe bet that books aren’t going anywhere. Radio did not replace print, TV did not replace radio. (Interactive did not replace passive; print did not replace the spoken word.) Each of these media is better suited for different tasks.

(I also still find that the best way to solve a problem (even a programming problem), is with pen and paper. Physically jotting down an idea beats pressing buttons while staring at an obnoxious, headache-inducing glowing screen any time (which I try to avoid as much as possible, even as a software developer). To me flipping pages still beats searching and clicking — what a dreadful user experience is a digital textbook!)

# Sorting By Relative Popularity

Hey, looks like I’m sorting user content again! last time, I sorted user posts by “interestingness”; this time around I’ll be sorting players from a set of sports teams. Once again we’ll look at why sorting things based on popularity alone is a bad idea, we’ll get a primer on standard deviation, and finally a bit of scripting to put it all together.

To drive up user engagement at theScore, we introduced an onboarding screen that’s shown when you open the app for the first time.

First, you get a list of sports teams that are popular in your area, and the option to subscribe to some of them.

Now based on the teams you choose, it would be nice to also recommend some players for you to follow. So how would one go about choosing which players to recommend out of all those teams?

Let’s assume we have three six-player teams, where each player has the following number of subscribers:

big_team_1:

1. 100 000
2. 110 000
3. 90 000
4. 80 500
5. 140 000
6. 140 500

big_team_2:

1. 120 000
2. 250 000
3. 180 000
4. 135 000
5. 157 000
6. 202 000

small_team:

1. 3 000
2. 100
3. 234
4. 301
5. 250
6. 400


Now let’s consider some properties we want from our algorithm:

• A compute-once, static value: we don’t want to run our algorithm on every user. We want to give each player in our database a static recommendation_score; a numeric value that is cheap to index.
• Simple: the algorithm should be simple and use elementary methods. Recommending players is a small part of the app; there should be little code maintenance involved.

• Variety: The purpose of the onboarding process is to get you engaged, so we want to recommend a variety of players from different sports and leagues.

### The Naive Approach

NOTE: I will be using the Julia programming language for all my examples. You can find the complete implementation at the bottom of this post.

The first obvious solution is to simply sort players by their subscription_count. The more popular the player, the more recommendable he is. Here is our naive sorting function:

function naive_sort(teams)
Base.sort(
[[team.players for team in teams]...],
by=x -> x.subscribers,
rev=true
)[1:5]
end


Which yields:

5-element Array{PlayerRecommender.Player,1}:
Big Team 2 Player | subscribers: 250000
Big Team 2 Player | subscribers: 202000
Big Team 2 Player | subscribers: 180000
Big Team 2 Player | subscribers: 157000
Big Team 1 Player | subscribers: 140500


The problem with this approach is that we only get results from the most popular teams. So if you’re a fan of both a very popular NFL team and another team that is not as popular (your local basketball team, perhaps), even the least popular player from the NFL team will be recommended to you, whereas the most popular player from your favorite local basketball team will not show up in your list at all!

### A Better Approach

What are we really looking for?

Well, I think the players we want to recommend are the not necessarily the ones who are most famous, but rather, the ones who are most popular compared to the rest of their team, regardless of how popular that teams is (we already know you’re a fan of the team or you wouldn’t have subscribed to it in the first place).

In other words, we want an algorithm that answers the following question:

Which players deviate the most in popularity from the rest of their team?

Luckily, there’s a mathematical tool for figuring out just this: standard deviation.

#### Standard Deviation tl;dr

Standard deviation is a pretty straight-forward concept: Take a set of values, and figure out the average value for that set (also known as the arithmetic mean). The standard deviation simply tells us by how much the value of the typical element in our set deviates from the average.

The mathematical representation of this calculation is:

$s_N = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}$

Where $N$ is the population size, and $\overline{x}$ is the arithmetic mean, which itself is represented by:

$\overline{x}_N=\frac{1}{N}\sum_{i=1}^{N} a_i$

For example, the following two sets have the same average value, but clearly the values are spread out differently:

close_to_average = [11,8,9,12]

average(close_to_average) == 10

standard_deviation(close_to_average) == 1.59


Thus we arrive at our simple scoring function. All we need to do is find players whose deviation from the average is substantially higher than that of their teammates:

function recommend(player, team)
player_dev = player.subscribers - team.average
if player_dev == 0
player.recommendation_score = 0
else
player.recommendation_score = player_dev / team.std_dev
end
end


Using this scoring function on our original teams, we get the following results:

5-element Array{PlayerRecommender.Player,1}:
Small Team Player | subscribers: 3000 | score: 2.2276261544470644
Big Team 2 Player | subscribers: 250000 | score: 1.749555170961297
Big Team 1 Player | subscribers: 140500 | score: 1.3134034577576315
Big Team 1 Player | subscribers: 140000 | score: 1.291753950212176
Big Team 2 Player | subscribers: 202000 | score: 0.6445729577225832


Much better. This time around, our top player actually has the least number of subscribers, but this makes sense, because even though he belongs to a team that is not very popular, his subscription count is tenfold that of his teammates; clearly someone to keep an eye on! (perhaps a rising star in a college league? Certainly wouldn’t want our recommendation script to ignore that one.)

### Limitations

This algorithm has many limitations.

• New players won’t be very well represented (they will by nature have low subscription counts).
• All-star teams might result in nobody being particularly recommendable. Though this one might be less of a problem: thanks to our good ol’ friend the bell curve, even among rare anomalies, there are rare anomalies.

While there are ways to address these limitations and improve the accuracy of the algorithm (for example, taking into account the rate of change in subscription_count), one has to remember the purpose of this feature: to drive up user engagement during onboarding. Is the added complexity of such changes worth the minimal improvement in the recommendations?

Point is, it’s Friday night and I should go out for a beer now. I’m also looking forward to testing out the enormous Chinese fermentation jug I bought yesterday. It looks something like this, but a LOT bigger:

And it was only \$30. What a bargain.

Here is the code used in these examples (working as of Julia 0.4.1). Our actual code at theScore is in Ruby.

# The Diminishing Returns of Focusing on the Right Thing

There was a recent thread on Hacker News about the search for the “cure for aging”:

The thought has recently occurred to me that we should all be working on the “aging problem”, until it’s solved, and then spending all our extra time on other pursuits. — b_emery

I’m sure you’ve heard (or expressed) a similar sentiment about other pressing Big Problems.

After all, why should we spend millions in tax dollars funding “useless”, or at the very least, non-vital research, when there are real problems to be solved? Why do we pay pure mathematicians to dream up infinite-dimensional spheres and Hilbert spaces, when that money could be better spent training engineers to build next-generation sustainable transportation?

This line of thinking may be noble in principle (“Let’s prioritize research that focuses on finding solutions to our most urgent real-world problems, instead of wasting resources on useless/unprofitable endeavours”), but it tends to lead to stagnation, not innovation.

Focusing on big problems is important, but there are diminishing returns to focusing on the right thing, possibly even negative returns. That is, focusing too much on solutions makes us less likely to find them.

While, improvements and advancements come from focused research, paradigm-shifting discoveries tend to come from unexpected places; often based on information that is initially discarded as not useful, even by experts.

By taking away funding from “useless” pursuits (like pure maths, arts, theoretical research) and focussing solely on the bottom line (“real-world problems”, what is profitable, what is immediately useful), we make great advancements less likely, not more.

This is why we can’t simply pour infinite money into “Finding the cure for x” and expect results; we can’t pour infinite money on engineers, while defunding basic research, and expect innovation.

Reminds me of the Louie bit where David Lynch instructs Louie to “Be funny: 3… 2… 1… Go”. — Well, that’s not how funny work. Achieving that level of mastery of his craft is 90% non-funny related activities (life experiences, personal growth, self-reflection etc.), and 10% actually focusing on “being funny” in itself.

What if the key to unlocking “disease x” comes from unrelated research in metabolism, what if the techniques needed to sort through the data come from an applied mathematician investigating economic trends? What if the technology required to model the problem in order to even ask the right question in the first place is developed by computer scientists modelling data for a social network? There’s just no way to know but to explore.

# Stones

I was at the lake last Summer. I picked a mossy old bench to sit, when a group of adolescents gathered to throw stones into the water. For an hour they busied around, making loud gestures and arguing over whose stone made the bigger splash. They would try every angle, and then bicker about which one spattered out the widest, or which one sent the water flying the highest, and how could they agree on what they’d seen if in a moment it was gone? Finally they approached me hoping I would settle their dispute, but I’d not seen the water splash– they scoffed, frustrated, and went on their way, and I was free to continue watching the ripples that now adorned the surface of the lake.

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. 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

In this post I’ll try to break that down and provide some of the background

## 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

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

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 mathematical library for python):

P = Matrix([[0.70, 0.60],
[0.30, 0.40]])
x = vector([0.5,0.5])
P^12*x
# -&amp;gt; (0.666666666666500, 0.333333333333500)
P^24*x
# -&amp;gt; (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
# -&amp;gt; (0.3, 0.2, 0.5)
P^2 * x
# -&amp;gt; (0.2, 0.3, 0.5)
P^3 * x
# -&amp;gt; (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.

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 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,

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