I just finished reading China Mieville's novel Kraken. It was really cool, it did take me a while to get into Mieville's voice, but once I got into the swing of it I really enjoyed it. Here is the blurb in case you are interested:

"In the Darwin Centre at London’s Natural History Museum, Billy Harrow, a cephalopod specialist, is conducting a tour whose climax is meant to be the Centre’s prize specimen of a rare Architeuthis dux--better known as the Giant Squid. But Billy’s tour takes an unexpected turn when the squid suddenly and impossibly vanishes into thin air."

One of the lines in the book struck me as surprisingly familiar. Here is the quote from the book:

"She flicked through a pad by her bed, where she made notes of various summonings. A spaceape, all writhing tentacles, to stimulate her audio nerve directly? Too much attitude."

After thinking about this for a moment, it clicked that this is a reference to a Burial song called Spaceape (from his self titled album). The line goes:

"Living spaceapes, creatures, covered, smothered in writhing tentacles
Stimulating the audio nerve directly"

I couldn't find any reference online to the inclusion of Burial lyrics in Mieville's novels. Okay I thought, that's a cool a easter egg, but it got my thinking, are there any other song lyrics buried in Mieville's books? And if so, is there any way we can scan them automatically?

Collecting Lyrics

The first step was to get a database of song lyrics which we can use to scan the novels for, Unfortunately there is no easy place to find a database of song lyrics, so I was forced to scrape them from a lyrics site.

I used the following free chrome extension web scraper which is very easy to use, and in my experience very reliable:
webscraper.io/

After about 10 minutes of setting it up, and about an hour of leaving it to run. I had managed to scrape most of Burial's lyrics in to csv files.

I also scraped lyrics by Kode9 and Spaceape so I could see if they were referenced anywhere. It's hard to know which artist I should look for, but both of these have been mentioned by Mieville in interviews.

The web scraping add-in has an easy to use GUI. Here is a screenshot of what it looks like to set it up:
​

Analysing the text

This is now the hardest part of the problem. We have electronic copies of China Mieville's novels in .txt format, and we have a collection of lyrics in .txt format which we would like to compare them against, how can we programmatically analyse whether Mieville references other Burial lyrics in one of his books?

If we simply attempt to match entire strings, then we have the issue that we might miss out on some references due to small changes in word ordering or punctuation. For example, in the example above using Burial's Spaceape, the wording is slightly different and the tenses of some of the words have been changed, therefore looking for an exact match between lyrics and text will probably not work. If on the other hand we don't match complete strings, but just try to match words, then we will be overwhelmed by small words like 'the' and 'a' which will be used multiple times in both Burial's song lyrics, and in China Mieville's novels.

There are two main approaches I came up with.to solve this problem. My first thought was to match individual words, generating a huge list of matches, and then to count the number of uses of each word in Mieville's novels, and then sorting by the words that match but which are also the most uncommon. For example I would imagine that Spaceape is only ever used once in all of Mieville's novels, giving us information about how unusual this word is. Combined with the fact that this word is also used in a Burial lyric, gives us enough information to assume that there is a high probability of a match, at this point we could investigate manually. 

I ultimately didn't go down this road. Instead, I had the idea to try to adapt plagiarism detection software to this problem. When you think about it, the two problems are actually quite similar. Plagiarism detection is about trying to automatically check two documents for similar phrases, without relying on complete matches.

Open Source Plagiarism Detection

I found the following free-to-use program created by Lou Bloomfield of the University of Virginia which is perfect for what I was trying to do.
plagiarism.bloomfieldmedia.com/wordpress/

It compares two sets of files and then creates a side by side hyperlinked comparison, which can be viewed in chrome, highlighting the parts of the documents where a possible match has been detected. There are various settings you can tweak to specify how close of a match you are interested in. 

I have included a screenshot below of the section where the Spaceape line is detected. There were about 500 matches detected when I ran this, but it only took about a minute to scroll through and check up on the ones that looked significant.
​

Results

Ultimately, this analysis felt like a bit of a failure. These were the only lyrics I could find in all of his novels and while  there is always the chance that I need to expand the number of artists I'm looking at, or refine my detection methods I imagine this is all there is. I still thought the process was quite cool so I thought I'd write up what I had done anyway.

If you have any thoughts, let me know by leaving a comment.

When you are playing a drinking game, do you ever catch yourself calculating the probabilities of various events happening? If you are like most people the answer to that is probably "..... um..... no....... why would I ever do that...."

Okay, maybe that's not the kind of thing you think about when playing a drinking game. But I bet you think it would be interesting to know what the odds are anyway? No? Really? Still not that interested? .... Okay... well I find it interesting, so I'm going to write about it anyway. 

Eyes up/Eyes down

I've only played this game a couple of times but it's pretty fun, it requires no equipment, and has quite a lot of drinking. All players sit in a circle, one player calls out "Eye's down" at which point everyone looks down, the same player then says "Eye's up" at which point everyone looks up at a random person in the circle. If the person you are looking at is also looking at you, then you both drink. Pretty simple.

What is the probability that you will drink? Let's denote the event that you have matched someone with $M$. Then:

$$P(M) = \frac{1}{n-1} $$
Where $n$ is the number of players.

​Of course people in real life tend not to pick a random person to look at, and even if they attempt to pick a random person, people have been shown to be terrible at picking randomly. But for the purposes of this analysis, unless we make this assumption, the only alternative would be to play the game hundreds of times, record how often matches are made, and then make an empirical claim. As fun as it sounds to play this game hundreds of times in a row, it would be better for your health to just assume a uniform spread of guesses.

The fact that you have a $ \frac{1}{n-1} $ chance of getting a match means that the more people are playing the game, the less likely each person is to drink. Does this apply to the group as a whole though? What is the probability that someone in the group drinks? If we had a hundred people playing, then each individual would hardly ever drink. In fact you would expect them to drink once every $99$ goes. But would you expect it to also be unlikely that anyone at all drinks?  I spent about an hour trying to calculate this and didn't get very far.

Calculating through conditional probabilities doesn't seem to help, and I couldn't come up with a decent approach to count the permutations of all the possible ways of selecting pairings for elements in a set, such that there are no 2-cycles. In the end I gave up and just wrote a program to calculate the answer stochastically. Monte Carlo analysis really is such a versatile tool for problems like these. Anytime you can formulate the problem easily, but struggle to come up with an analytical solution, Monte Carlo analysis will probably work.

Here is the table I generated of the probability of someone in the group drinking for a group of size n:

There is a definite trend here, if we plot the values on the chart it looks pretty clear that out solution converges to $0.6$. No idea why though! I might come back to this problem another time, but if anyone has any thoughts then please let me know.
​

Odds

So this is not technically a drinking game, but is played quite often by people who are out drinking. The rules for this game are pretty simple, anytime you would like to dare someone to do something you say "odds of you doing this". The dare might be anything, from 'eating a spoon full of chilli powder, downing your drink, or even getting a tattoo (that last one is a real one I heard about from a friend who playing the game while travelling in Eastern Europe). The person you have challenged then gives you a number. They might say $20$, then each of you count down from $3$ and say an integer from $1$ to $20$ (or whatever number they selected). If you both said the same number, then they have to do the dare. Therefore the higher number you pick, the less likely it is that you will have to do the dare. 

What are the odds of you having to do whatever it is you are betting on, based on you selecting the number $n$? This is quite a straightforward problem, the answer is just $\frac{1} {n} $. I was playing this game last week with someone who thought the answer might be $\frac{1} {n^2} $. This would give a very different likelihood of having to do the dare. For example if you selected $20$, then you would have a $5 \%$ chance of having to do the dare, but according to my friend's calculation he thought you would have $0.25 \%$ chance of doing the dare.

Here is a table showing the correct probabilities for various values of $n$.

Dirty Pint Flip

I wouldn't recommend playing this game to be honest. I played it in uni once, but it's a bit grim. All the players sit in a circle with a central pint glass in the middle, the players take it in turn to pour any amount of their own drink into the central pint glass as they would like, they then have to flip a coin and guess whether it will be heads or tails. If they get it right, then the game moves on to the person to their left, if they get it wrong, then they have to down the dirty pint. When I played it some people were drinking wine, some people were drinking beer... it was not a good combination.

What is the probability that you will have to drink? This is quite straight forward, it is simply the probability that you will guess the coin flip correctly. Which is just $\frac{1}{2} $.

What about the question of how much you will drink on average? That is, on average if you have to drink, how many people will have poured their drink into the glass before you drink? We can model this as a geometric distribution and calculate the probabilities of each possible number of people. Let's denote the number of people who have poured in before you, given you are about to drink as $N$, then:

The expected value of the number of drinks is then the sumproduct of these two columns . This gives us a value of $2$. Therefore, if you have to drink, the average number of drinks that will have been poured into the pint is $2$.

Two Dice Nominate

This is another game I've only played a couple of times. And I'm not sure if it has a better name than the one I've given it, if you know of one then let me know. In this game, you sit in a circle with a range of drinks on the table, you take it in turns to nominate a person, a drink, and a number between $2$ and $12$. You then roll two dice and add them together. If you have rolled the correct number, the person you nominated drinks the drink you selected, if however you roll a double, then you drink that drink. If it is both a double and you get it correct, then you both have to drink.

The reason that this game does not have much depth to it is that you can always just pick the most likely number to come up when rolling two dice. Here is a table showing the probability of each value coming up:

Therefore you might as well pick $7$ every single time! Is there anything else we can say about this game? We know that if we roll  a double, then we have to drink the drink. What is the probability of you rolling a double in this game? The probability of this is always just $\frac{1}{6} $. Interesting this is the same probability as rolling the dice such that the sum is $7$. Therefore, if you do pick $7$, there is an equal chance of you and the person you are nominating drinking the drink.
​

Recently I've been reading The Mathematics of Poker (2009, Bill Chen and Jerrod Ankenman) and I came across an interesting idea that I thought I'd write about. For me, understanding how to analyse this situation really gets to the heart of how I think about poker. I'd love to spend more time playing and studying poker but it's such a time-sink and I don't really have the oodles of free time it would require, but every now and again I'll still open up a poker book and read about something, this is one of those interesting topics I was reading about hopefully you find it as interesting as I do.

Calling a shove pre-flop in heads up

The scenario being analysed in the book is a relatively common situation, particularly in online poker where people are more inclined to shove than in real life games. The question is:

How should we analyse whether to call when we have a moderately strong hand against an opponent who has gone all-in pre flop.

Let's set up an example so that we have something concrete to talk about. Say we have pocket Kings pre flop, and our opponent goes all-in, how should we decide whether to call? Obviously without knowing our opponent's hand there is no 100% correct answer. There is  however one very useful way of analysing the situation.

Equity against a range

We need to ask ourselves - what cards would are opponent go all-in with, and how does my current hand fare against that range? i.e. we need to calculate our equity against our opponent's range. We are adding another layer of stochastic uncertainty on the event, instead of trying to guess what hand our opponent has (which is almost impossible) we are trying to guess what kind of hands he might go all-in with (which is very much possible). We then take this extra level of uncertainty and calculate the correct mathematical way to proceed.

On the one extreme, let's suppose that based on our read of how our opponent is playing, we might think that they would only go all-in with very strong hands, in this case just pocket Aces. We would then be a 4:1 underdog if we call with Ks, and we should definitely fold. (In order to calculate this we could use any standard poker calculator like the following)

www.cardplayer.com/poker-tools/odds-calculator/texas-holdem

One the other hand, suppose we know for a fact that our opponent has not looked at their cards at all but has still decided to go all in. In this case we should definitely call. The only cards we will be behind are pocket Aces, which make up a small fraction of the possible hands that our opponent could shove with, and we will be ahead or equal against all other possible hands. Therefore we would have a positive EV when calling.

What if our read on our opponent's range is somewhere in between though?

What we need to do is calculate our equity against each individual hand in our opponent's range, and then calculate the probability of our opponent having a given hand from that range. That is to say, in order to calculate the conditional expectation, we need to calculate the unconditional expectations against each hand and then multiply by the conditional probability of our opponent having that hand, given our belief about our opponent's range. 

Numerical Example

Let's go back to our numerical example, and suppose that we have pocket Kings, and we put our opponent on either Pocket Aces, Kings, Queens, or Jacks. All of these hands are equally likely, so there is a 25% chance of our opponent having each hand. We can look up our equity against each hand (after you've been playing for a while, you naturally start to memorise hand equities anyway)

Our probability of winning is then:

$$P(A) * P(beating A) + P(K)*P(beating K) + P(Q)*P(beating Q) + P(J) * P(beating J)$$
Putting in our values:

$$ 0.25*0.2 + 0.25*0.5 + 0.25*0.8 + 0.25*0.8 = 0.575.$$

​We therefore see we have a positive expectation against this range, and should call.

No one actually thinks like this in real games?

It is a general misconception that professional poker is a game where players are trying to guess exactly what hand their opponent has, are constantly trying to bluff each other, or trying to pick up on subtle tells or signs that their opponent is or isn't bluffing.

The more mundane truth is that poker is ultimately a game of imperfect information, where the better player is the one who can correctly interpret the betting information their opponent is giving them, and can then quickly and accurately make the kind of judgements described above during a game.

Obviously poker players are not carrying out these calculations in their head to multiple decimal places in real time, what they will do though is review their hands after a game, calculate exactly what they should have done, and try to build up an intuition as to what the correct answer is, so that in the middle of a game they can quickly make decisions.

Software to Analyse this situation

Is there an easy software based method way of calculating our equity against a range? After I did a quick google there are a few programs that offer this type of analysis. For example:

combonator.com/
www.power-equilab.com/

More interestingly though, I also found the following open source software, that can be adapted to carry out this type of analysis:

github.com/zekyll/OMPEval

At some point, I might try to use this code to set up a page on this website to let people analyse this situation.

Over the Christmas break, I finally got around to playing some Computer Games, which I unfortunately haven't done in ages. One really fun game (if you're like me and love difficult puzzles) is a game called Myst.

It was originally released in 1993 as a point and click puzzle adventure game, but it has subsequently been remade, using the Unity engine, as a full 3d interactive game. I really liked the fact that there is no time limit, no danger of dying, just a cool story, interesting puzzles and relaxing music.

One puzzle that I thought was cool involved aligning gear wheels. If you haven't played the game and don't want spoilers, then stop reading now.

The Gear Puzzle

​We are given the contraption in the image below, and are told that we need to end up with the numbers 2,2,1 (reading from top to bottom). We have three levers we can pull, the left lever moves the top two gears left once, the right lever moves the bottom two gears right once. The lever on the back wall resets the machine. The starting values of the gears are 3,3,3. So how many times do we need to pull each lever in order to end up at 2,2,1?

After playing around with it for a while I began to think that it might not even be possible to get to 2,2,1 using the two levers. Then I thought if that's true, can I prove it as well?

Proof that the puzzle is impossible

Instead of thinking of the value as $3$, let's call it $0$ instead. We can then think of the starting state as a vector:

To see that we are not able to do this with just the basic operations from the starting point we are given, note that the the sum of the first and third elements of our starting vector is equal to the second vector mod $3$, that is, $0 + 0 \equiv 0$ $(mod$ $3)$. For the vector we trying to create, the first and third is not congruent to the second element mod $3$. Instead we have $ 2 + 1 \equiv 0$ $(mod$ $3) $ rather than $2$.

The two operations we are applying to our starting vector preserve this congruence property, as it applies to both the vectors we are adding on as well. Therefore, we can conclude that this property is an invariant of the system, and because our goal does not satisfy this condition, we will never be able to reach:

by just using simple pulls of the lever. We need to realise this in order to force us to look for an alternative way to solve the puzzle. Otherwise we would probably just keep yanking on the levers in the hope that we will eventually get there. Once we realise this and start looking around, we are able to find another way to manipulate the machine, allowing us to make the numbers we need.

Designing a new number system

I got interested in alternative number systems when I read Malcolm Gladwell's book Outliers back in uni. You might be surprised that Gladwell would write about this topic, but he actually uses it to attempt to explain why Asian students tend to do so well at maths in the US. I was flicking through an old notebook the other day and I can across my attempt at designing such a system. I thought it might be interesting to write up my system. 

To set the scene, here is the relevant extract from the book:

"Take a look at the following list of numbers: 4,8,5,3,9,7,6. Read them out loud to yourself. Now look away, and spend twenty seconds memorizing that sequence before saying them out loud again.If you speak English, you have about a 50 percent chance of remembering that sequence perfectly If you’re Chinese, though, you’re almost certain to get it right every time. Why is that? Because as human beings we store digits in a memory loop that runs for about two seconds. We most easily memorize whatever we can say or read within that two second span. And Chinese speakers get that list of numbers—4,8,5,3,9,7,6—right every time because—unlike English speakers—their language allows them to fit all those seven numbers into two seconds.

That example comes from Stanislas Dehaene’s book “The Number Sense,” and as Dehaene explains:
Chinese number words are remarkably brief. Most of them can be uttered in less than one-quarter of a second (for instance, 4 is ‘si’ and 7 ‘qi’) Their English equivalents—”four,” “seven”—are longer: pronouncing them takes about one-third of a second. The memory gap between English and Chinese apparently is entirely due to this difference in length. In languages as diverse as Welsh, Arabic, Chinese, English and Hebrew, there is a reproducible correlation between the time required to pronounce numbers in a given language and the memory span of its speakers. In this domain, the prize for efficacy goes to the Cantonese dialect of Chinese, whose brevity grants residents of Hong Kong a rocketing memory span of about 10 digits.

It turns out that there is also a big difference in how number-naming systems in Western and Asian languages are constructed. In English, we say fourteen, sixteen, seventeen, eighteen and nineteen, so one would think that we would also say one-teen, two-teen, and three-teen. But we don’t. We make up a different form: eleven, twelve, thirteen, and fifteen. Similarly, we have forty, and sixty, which sound like what they are. But we also say fifty and thirty and twenty, which sort of sound what they are but not really. And, for that matter, for numbers above twenty, we put the “decade” first and the unit number second: twenty-one, twenty-two. For the teens, though, we do it the other way around. We put the decade second and the unit number first: fourteen, seventeen, eighteen. The number system in English is highly irregular. Not so in China, Japan and Korea. They have a logical counting system. Eleven is ten one. Twelve is ten two. Twenty-four is two ten four, and so on.
That difference means that Asian children learn to count much faster. Four year old Chinese children can count, on average, up to forty. American children, at that age, can only count to fifteen, and don’t reach forty until they’re five: by the age of five, in other words, American children are already a year behind their Asian counterparts in the most fundamental of math skills.

The regularity of their number systems also means that Asian children can perform basic functions—like addition—far more easily. Ask an English seven-year-old to add thirty-seven plus twenty two, in her head, and she has to convert the words to numbers (37 + 22). Only then can she do the math: 2 plus 7 is nine and 30 and 20 is 50, which makes 59. Ask an Asian child to add three-tens-seven and two tens-two, and then the necessary equation is right there, embedded in the sentence. No number translation is necessary: It’s five-tens nine.

“The Asian system is transparent,” says Karen Fuson, a Northwestern University psychologist, who has done much of the research on Asian-Western differences. “I think that it makes the whole attitude toward math different. Instead of being a rote learning thing, there’s a pattern I can figure out. There is an expectation that I can do this. There is an expectation that it’s sensible. For fractions, we say three fifths. The Chinese is literally, ‘out of five parts, take three.’ That’s telling you conceptually what a fraction is. It’s differentiating the denominator and the numerator.”

The much-storied disenchantment with mathematics among western children starts in the third and fourth grade, and Fuson argues that perhaps a part of that disenchantment is due to the fact that math doesn’t seem to make sense; its linguistic structure is clumsy; its basic rules seem arbitrary and complicated.

Asian children, by contrast, don’t face nearly that same sense of bafflement. They can hold more numbers in their head, and do calculations faster, and the way fractions are expressed in their language corresponds exactly to the way a fraction actually is—and maybe that makes them a little more likely to enjoy math, and maybe because they enjoy math a little more they try a little harder and take more math classes and are more willing to do their homework, and on and on, in a kind of virtuous circle.

When it comes to math, in other words, Asians have built-in advantage. . ."

Here's a link to Gladwell's website which contains the extract.
gladwell.com/outliers/rice-paddies-and-math-tests/

Base-10 System

Gladwell mainly talks about the words that the Chinese use for numbers and the structure inherent in them, but there is actually another more interesting way we can embed structure in our number system.

Our current number system is base-10, which means that we have different symbols for all the numbers up to 10 (0,1,2,3,4,...,9), and then when we get to the number 10, we use the first number again but we move it over one place, and then put a zero in it's place. When we are teaching a child how to write numbers, this is exactly how we explain it to them. For example to write two hundred and seven, we would tell them that they need to put a 2 in the hundreds columns a 0 in the tens column and a 7 in the units column - 207.

The fact that we use a base-10 number system is actually just a historical quirk. The only reason humans started to use it is just that we have 10 fingers, there's no particular mathematical benefit to a base-10 system. We could actually base our number system on any integer. The most commonly used alternative is the binary number system used extensively in computing. The binary number system is a base-2 number system where we only have 2 symbols - 0 and 1. The trick is that instead of reusing the first symbol when we get to ten, we do it when we get to two. So the sequence of numbers up to 10 in binary is:

In fact we don't even need to restrict ourselves to integers as the base of our number system. For example, we could even use base $ \pi $! Under this system if we want to write $ \pi $, then we would just write 1, to write $ \pi^2 $ it would just be 10. Writing one in base $ \pi $ would be quite difficult though, and we would need to either write $ 1 / {\pi} $ or invent a new character.



Base 12 number system

So what would be the ideal number on which to base our number system?

I'm going to make the argument that a base 12 number system would be the best option. 12 has a large number of small factors and this makes it ideal as a base.

• 12 has almost twice as many prime factors as 10. When working in base 10, it is relatively easy to multiply a number by either 2, 5, or 10, this is because these are all factors of 10. When we are working in base 12, it will be easy to multiply by 2, 3, 4, 6, and 12. Division by these numbers is also much easier. 
• It is the smallest number divisible by 1,2,3, and 4.
• We seem to intuitively favour the number 12 in many common circumstances. It has it's own name - a dozen, we also used it to split inches into feet. It's actually the largest integer that has still has it's own name. Once we get to 13, it's just a combination of three and teen, and so on for larger numbers, but eleven and twelve have their own names.
• We have 12 knuckles on the four fingers of each hand (ignoring thumbs) making counting on our hands easy still.

Returning back to Gladwell's idea, another change we could make to the number system to help make it more structured is to change the actual symbols we use for the integers. Here was my attempt from 2011 for alternative symbols we could use.

My approach was to embed as much structure into the numbers as possible, so the symbol for three is actually just a combination of the symbols for one and two. This applies for all the numbers other than one, two, four, and eight.

I wonder if it would be possible to come up with some sort of rule about how the symbols rotate to further reduce the total number of symbols and also to add some additional structure to the symbols.

Let me know if you can think of any additional improvements that could be made to our number system.
​

Fact 1 - If Airline Companies were ran like Rocket Companies, a trip from London to New York would cost about $300,000.

Prior to SpaceX, launching a a Satellite into orbit was expensive. Really, really expensive. The most commonly used launch vehicle was the Ariane 5, manufactured by the French Company Arianespace, and costing somewhere in the region of USD75 million per launch. 

SpaceX have stated that their eventual hope is target a cost of USD6 million per launch. How do they hope to achieve a more than 10 times cost reduction?

Better Economies of Scale?
Improved Manufacturing Processes?
Cheap Labour?
3d Printing?

Actually none of the above. The answer is actually kind of obvious once you hear it though. They want to make rockets that can be used more than once!

As a though experiment, let's apply the concept of reusability to a familiar mode of transport. If we take the most common airliner in the world - the Boeing 737 - then we can compare the cost of a flight with full reusability against the cost of the flight without reusability.

A new Boeing 737 costs in the region of USD75 million (depending on the exact model) whereas a return ticket to New York on a Boeing 737 only costs about USD600 per seat.  Multiplying the number of seats (~250) by the cost of a one way ticket (~USD300) we see that the total cost of the flight is somewhere around the USD75,000. So each flight costs about 0.1% of the total cost of the airliner.

Suppose though, that like a rocket, a 737 could only be flown once. Then all of a sudden, instead of costing less than USD75,000 per flight, the flight would include the full cost of the 737, meaning each ticket would cost around USD300,000!

By creating rockets that are reusable, SpaceX hopes to get similar cost savings in the arena of launch vehicles.

SpaceX is actually quite close to achieving full reusability, here is a cool video of a Falcon 9 rocket making a vertical landing on a drone ship in April 2016.

Fact 2 - It is easier to land the rocket on a Drone Ships than the original launch site

In the video above, the Falcon 9 lands on a floating platform in the ocean. Given landing on a ship in the ocean seems to add another layer of complexity to an already difficult task why did SpaceX decide to do this? A few ideas spring to mind, you might think that this is because the rocket is not sufficiently accurate and the drone ship has to be able to move around to get into the right place, or possibly that the ocean is a safer environment to land on given the lack of nearby people or property to damage. Or perhaps SpaceX are doing this as a form of marketing given the fact that it looks pretty cool.

However, the actual reason is surprisingly prosaic, and also, once you think about it, kind of obvious again. It'all has to do with how you get a satellite into orbit in the first place.
​
When a rocket attempts to get a satellite into orbit it does not just go vertically up in the air, in order for a satellite to be in a stable orbit, it actually needs to be moving perpendicular to the earth's surface at a very high speed (which can be calculated based on the height of the orbit) in order to not fall back to earth.

I wrote a model in Python of a two stage rocket launching a satellite into orbit around earth. I then ran the model multiple times keeping the overall thrust of the rocket exactly the same every time, but varying the initial direction of acceleration. Using this model we can see the large effect that varying the path the rocket takes has on the ease in which a satellite can be put into orbit.

In the first video, the launch rocket is sent almost vertically up in the air, and the satellite (the blue part) makes very little progress before it crashes back down to earth.

In the second attempt, we angle the rocket towards the horizon more, in this case, by 55 degrees. And we can see that whilst the satellite does not make it into orbit, it does come closer than the first attempt.

In the final attempt, we angle the rocket at a 45 degree angle to the horizon. Now we see that the satellite does in fact make it into orbit. Note that the only thing we have changed in all these attempts is the angle of launch, the acceleration of the rocket has not been changed at all.

The point to note from all this is that in order to get a satellite into orbit, the rocket that was used to put it into orbit also needs to have a lot of horizontal velocity.

There are quite a few simplifications in our model. For example, we have not included air resistance which will have an impact as in practice there will be a benefit in gaining as much vertical height as possible initially so as to reduce air resistance. The model also does not take account of the fact that the mass of the rocket will reduce as it gets higher due to the fuel expended. However neither of these simplifications effects the the general principle of needing horizontal acceleration.

To tie it back to our original point, from watching the videos you might be able to tell the problem that SpaceX found themselves dealing with when using this approach. The first stage of the rocket ends up miles away from the launch site!

It turns out that for SpaceX, given the location of their launch site, the landing site for these rockets tends to be in the ocean, and this is why that they land the rockets on barges in the ocean.

Fact 3 - ​The floating ships are named after ships in Iain M. Banks novels

SpaceX have two landing barges (also known as Autonomous Spaceport Drone Ships) that they use to land the rockets on. The barges are called 'Just Read The Instructions' and 'Of Course I Still Love You'.

These may seem like very unusual names for SpaceX to pick, unless you spot that they are in fact names of spaceships in Iain M. Banks' Culture series, a series of sci-fi books set in a Utopian future with an interstellar human race.

Python Code

In case anyone is interested, I have pasted the Python code for the two stage rocket model below. My code was partially based on the following model, which simulates the n-body problem in Python. 
fiftyexamples.readthedocs.io/en/latest/gravity.html

import turtle
import math

turtle.color("White")
turtle.sety(100)
turtle.color("Black")
turtle.circle(-100)
turtle.seth(90)
turtle.color("Red")

Vx = 0.3
Vy = 0.3

SecondStage = False

for sec in range(3500):
    CurrentX = turtle.xcor()
    CurrentY = turtle.ycor()
    d = math.sqrt(CurrentX ** 2 + CurrentY ** 2)
    f = 17 / (d ** 2)
    theta = math.atan2(CurrentY, CurrentX)
    fx = math.cos(theta) * f
    fy = math.sin(theta) * f

    Vx += -fx
    Vy += -fy
   
    if turtle.pencolor() == "Red":
        turtle.goto(turtle.xcor() + Vx, turtle.ycor() + Vy)

    if d < 100:
        turtle.color("white")

    if sec == 250:
        SecondStage = True
        New = turtle.Turtle()
        New.color("white")
        NewCurrentX = CurrentX
        NewCurrentY = CurrentY
        New.goto(NewCurrentX,NewCurrentY)
        New.color("blue")
        VNx = Vx + 0.05
        VNy = Vy - 0.2

    if SecondStage == True:
        NewCurrentX = New.xcor()
        NewCurrentY = New.ycor()
        d = math.sqrt(NewCurrentX ** 2 + NewCurrentY ** 2)
        f = 17 / (d ** 2)
        theta = math.atan2(NewCurrentY, NewCurrentX)
        fx = math.cos(theta) * f
        fy = math.sin(theta) * f
        VNx += -fx
        VNy += -fy
        
        New.goto(New.xcor() + VNx, New.ycor() + VNy)

        if d < 100:
            New.color("white")

turtle.exitonclick()