Geography is just physics slowed down, with a couple of trees stuck in it.

Terry Pratchett


Way back when I was doing my undergrad degree, I was lucky enough that most of it made sense, usually straight away. A lot of math is just learning what the weird notation means. It’s like learning a new language – you need a bunch of vocabulary.

Not everything was easy though, and one result that sent me into a spin was Green’s Theorem. I’m sure it isn’t the most surprising theorem for everyone but it left me pretty dumbstruck. I’ll give you the math in a moment, but intuitively what it says is that we can learn about the interior of a region just by examining its boundary. That’s magic. Can we really see inside without looking inside?

Of course it isn’t magic, and we can’t see whatever we like. But we can calculate the area of a region looking only at the boundary. I thought it would be fun to try this out on some non-trivial data, and what better than a map. But the map needs to be relevant to the theme of the blog. Enter the Terry Pratchett’s Discworld.

Terry Pratchett was one of the most prolific and successful fantasy authors ever. You could argue that Tolkein was more influential or that Rowling made more money, but Pratchett wrote more and IMHO did it better than anyone else in the genre.

Most importantly, from my point of view, Terry understood how important stories are. He even said it in his own writing:

“All right,” said Susan. “I’m not stupid. You’re saying humans need… fantasies to make life bearable.”


“Tooth fairies? Hogfathers? Little—”


“So we can believe the big ones?”


“They’re not the same at all!”


“Yes, but people have got to believe that, or what’s the point—”


Terry Pratchett, Hogfather (Susan is talking to Death)

Terry Pratchett started writing fantasy satire set on a world that really was flat, carried on the back of backs of four elephants which in turn stand on the back of the Great A’Tuin, a giant turtle swimming through space. But the novels grew into much more than just satire.

They still have a strong sense of humour, but the novels (47 of them – see here for the ‘correct’ reading order) became the canonical description of a whole world – the Discworld.

Pratchett also wrote several other novels and series, some jointly with other authors such as Neil Gaiman.

Quite a few of Terry’s works have been adapted into graphic novel form or produced as movies and TV series, the most recent being the aforementioned collaboration with Gaiman – Good Omens. The Discworld novels are so popular that they spawned a side industry of books about the world, its science and its geography.

It’s the latter that I am focusing on today. In particular the beautiful map from The Compleat Discworld Atlas and the earlier Discworld Mapp1, devised by Terry Pratchett and Stephen Briggs, and illustrated in glorious thaumicolour by Stephen Player. The featured image in this blog is my humble reproduction of the map, whose only virtue lies in that it is an SVG file, and thus possible for me to analyse (I am not trying to violate anyone’s copyright here, just get the data in a form that I and others can work with).

The end result (if you can’t bear to read your way through, then follow this link) is that I can calculate not just the area of the whole Discworld, but more interestingly the land area of its continents and islands.

Green’s Theorem 

Green’s theorem is named after George Green (1793-1841) (not my colleague Ed who is as good a mathematician as you are ever likely to meet, but was born too late to get credit for this little gem). George’s version of the theorem wasn’t quite like the modern version, and in fact there are quite a few alt versions and generalisations of the theorem that we might also look at if we had time. Here is the modern version (the theorem has a few technical assumptions2; I won’t delve into here. I’m just giving you the key equation.)

\[\displaystyle \oint_C ( L d x + M d y ) = \iint_{D} \left({\frac{\partial M}{\partial x}}-{\frac{\partial L}{\partial y}}\right)\,dx\,dy \]

Unless you are a serious mathematician or engineer this probably looks pretty strange. As I said before, some of the confusion is just understanding what the notation means. Here

  • C is a curve that surrounds a region D

  • The symbols like \( \int \) are called integrals. But the symbol itself is an elongated S, standing for sum. An integral is just a special sort of sum. We use two types of integrals here.

    • The first integral, \( \oint_C \) , means sum along the path of the curve C.

    • The second integral, \( \iint_D \) , means sum over the region D.

  • The \(dx\) and \(dy\) tell you how to do the sum.

  • The \(L\) and \(M\) and their partial derivatives (slopes) \(\frac{\partial M}{\partial x}\) and \(\frac{\partial L}{\partial y}\) are the things being summed.

It’s a complicated calculus formula – we don’t even teach it until the second year of a degree in mathematics. I don’t really expect you to remember it or understand my fairly basic explanation.

The really exciting thing about it is that the left-hand side only needs us to perform a sum around the edge of the region, and from that we can calculate the right-hand side, which is a sum over the whole interior of the region. So we can learn about the interior, just by looking at the exterior.

There’s an episode of The Big Bang Theory where they get quite excited about this idea as a model of the Universe. Because it’s a cool idea. I have no idea how Green discovered it – apparently he hadn’t even had a very good mathematical education in his early life.

The result has been useful in both classical and quantum mechanics, in electromagnetism and in analysing superconductivity. In a practical setting, it has been used by engineers to calculate areas of complex shapes, which is what we are doing here.


So, now we come to calculating areas. Green’s Theorem works for almost any reasonable functions \(L\) and \(M\) so we need to choose the right functions. To get there we need to understand that area is the simplest region integral: it looks like

\[\displaystyle \iint_{D} 1 \,dx\,dy \]

Hence we can get away with pretty simple functions namely:

  • \(L = -y/2\)
  • \(M = x/2\)

So the left-hand side becomes:

\[\displaystyle \frac{1}{2} \oint_C ( -y d x + x d y ) \]

Now we just need a way to evaluate the integral. If you’ve done a course in advanced calculus, you probably learnt 12 different tricks to do this for theoretical problems where the region is nice and simple, e.g., a circle or a square, But we need to do it on a large irregular shape (a continent). There are some tricks we can pull out of the bag that sometimes aren’t taught.


The first trick is to realise that “adding up” along a curve is an analogue task, and so we can build a mechanism to do it. Before the days of Computer Aided Design (CAD) draftsmen would all have a device called a planimeter. It’s a physical device that you trace across around a shape in a blueprint, and it mechanically calculates the area of the region.

I have one (somewhere) but I’m not using it these days. It’s much easier to calculate integrals with a computer. But our computers are digital. What that means is that all of my data will come in little discrete chunks (like your digits, your fingers). That means the data is an approximation (a pretty good one) but it also means we can replace the integrals above with more straightforward summations as we will do below.


The discrete, digital version of Green’s theorem is called the Shoelace Theorem (the name is a mnemonic that is meant to help you remember the result, but my memory is so bad it doesn’t help).

The theorem says that for any polygon (the discrete version of our curve C), with vertices \(\displaystyle (x_i, y_i) \) for \(\displaystyle i=1,\ldots,n \) the area is given by

\[\displaystyle A = \frac{1}{2} \sum_{i=1}^{n} x_i y_{i+1} - x_{i+1} y_{i} \]

Here the \(\Sigma\) is a Greek sigma, or S, i.e., just another symbol used to mean sum. There are some nice examples on Wikipedia so I don’t feel the need to go through one here. There are also some variant formulas but this is the one I like, and have implemented as follows (in Julia):

function area_contour_integral( points::Array{Float64,2} )
    n = size(points,1) - 1 # assumes closed curve
    x = points[:,1]
    y = points[:,2]
    area = 0.5*abs( sum( x[1:n].*y[2:n+1] .- x[2:n+1].*y[1:n] ) )

The array points contains the (x,y) coordinates of the polygon describing (for instance) a continent. It has to be closed, i.e., the last point repeats the first, so n is one smaller than the size of array.

So we go from a (for me) highly non-intuitive calculus formula to one of the easiest pieces of code you could write.

One final note: the most cluey of you will have noticed the abs function hidden in there. That’s because the sum above assumes I go anti-clockwise around the polygon. If you go clockwise, you get a negative area. I don’t know what order I will be given vertices on an arbitrary map, so I use the abs function to ensure I get a positive area.

A Mapp and a Julia 

The map I am using is my transcription of the original. It isn’t compleat and there are most likely some mistakes, and certainly some noise in the data. But it’s in a vector graphics format (SVG) which makes it possible for me to read it into [Julia](

It’s not quite trivial to read SVG files. Superficially SVG is an XML variant, so we can use standard packages to get them into Julia (I use EzXML.jl and XMLDict.jl at present). The net effect is a very easy bit of Julia code to read in an SVG file:

using EzXML 
using XMLDict

filename = "discworld.svg"
s = read(filename, String) # read the file into a string
xml = xml_dict(s) # parse the string, and put the results in a dictionary          

The code gets the data into a generic Dictionary (see this post for more detail), but you still need to interpret the information you have just slurped in. This is not the place for a long discussion of SVG formats, and I won’t share my code just yet (it is not very robust or complete). However, I will share the data in the form of a set of CSV files (in addition to the map itself). These don’t have all the information in the map, but include (x,y) coordinates of the main two continents so you can potentially reproduce most of the results. Here is the first few lines – click on them to get the files.



The task at hand is calculating land area. There are a few assumptions built into the calculation3,4,5, but more importantly, what are we looking at? What land area will we calculate? We need to have a quick look at what we know about the Discworld’s land masses.

The Discworld has four continents, but as in our world, several of these are defined by social divisions, not clear geographic boundaries. For instance, in our world Europe, Africa and Asia form a super-continent: Afro-Eurasia. In the Discworld there is a similar super-continent composed of Klatch, the unnamed continent (containing Anhk-Morpork, the Hub and so on), and the Counterweight Continent). As the boundary of this super-continent is well defined in the map (see data above), but its subcontinents are not, we shall look at the super-continent’s land area.

There is an additional smaller continent called Fourecks or “XXXX”. Pratchett thought my own country was fairly amusing (I don’t doubt he was correct).

There are also many islands visible on the map. We will consider these as a whole, but sunken lands (e.g. Ku and Leshp are excluded).

Finally, there are three large lakes on the super-continent. As we are interested in land area, we will remove the area of these lakes from the total for the super-continent.

The results 

First up, all units are in miles (or square miles) as this was the unit used in the map and in many of Pratchett’s works. We cannot confirm that a Discworld mile is exactly the same as a modern imperial mile4, but the manner in which it is used in speech suggests they are at least approximately the same.

All areas assume that the disc is 10,000 miles in diameter. This may be an approximation, however, if it is inaccurate, all areas scale as the square of the ratio of change, and so proportional coverage will be the same.

The disc itself is around 79 million square miles (obviously this is not “land” area). Compare this to our Earth’s 196.9 million mi², and we can see that it is about 40% of the area of the Earth (or Roundworld), but of course the usable land area depends on how large a proportion is sea. Only 29.2\% of our Earth is covered by land (57 million mi²). The Discworld has about 30 million mi² of land, closer to half that of Earth.

That’s important! It has geographic constraints similar to ours in terms of time taken to get places. And the disc has enough space for as many stories as our world (almost). It has room to breathe. Pratchett took advantage of that scope to create many new places, and satirise many more (Fourecks is his satirical version of my own Australia).

On our world Afro-Eurasia has a land area of 33 million square miles, which is surprisingly close to that of the Discworld super-continent, given the large differences in shape (and underlying geometry). And Fourecks’ land area is comparable to Australia (it’s about half the size).

A much larger proportion of the Discworld’s total land area lies in the single supercontinent than on Earth. Here we have two super-continents (Afro-Eurasia and America) and two smaller continents (Australia and Antarctica), whereas almost all of the Discworld’s land area lies in the single land mass.

Does that matter from a story telling point of view? I guess it shows a bias in Pratchett’s worldview towards the old world. Although many of the topics covered in his books are those I consider “American” in the sense that the US is the place where those ideas really germinated, Pratchett didn’t create an American analogue in his world. Perhaps that was unconscious, or perhaps it was deliberate. It is after all a rather well-trod path for a British writer to take, to satirise the US, and maybe one of the hardest working writers in fantasy thought it a little lazy.


The main point of this blog: the Discworld is big. It’s not so far from our own world in land area. Very few created worlds are so big and actually fleshed out in any detail (though there are world builders out there who do that for a hobby). Terry Pratchett created something magnificent.

How is this important for data science? Well, too often data-scientists crunch numbers directly. A common approach to the above problem might have been to create a really big array of (x,y) points (pixels) and to test if each is inside the continent. Then add up the total. That’s nice in one respect – it parallelises, and so we can use the Hadoop et al. of the world to calculate the resulting big sums. But I don’t need a supercomputer. I can barely measure the amount of time it takes to do the trivial calculation above (Julia is very fast).

So I guess the point is that if you know the right tricks, sometimes you don’t need big data, or big computing, or big anything.

One final disclaimer, I have read a lot of the Discworld books, but not all. As long as there is still at least one I haven’t read, Terry Pratchett isn’t really dead. We love him for his charm and his humour and his insight. He understood the world as it is, but wrote about it as it should be. I don’t want to live in a universe where he isn’t there writing books, so I’ll try to fool myself for a little longer.


Just a quiet thanks for the people who have been helping me edit these blogs, notably my wife and Giang Nguyen.


  1. Mapp, compleat, etc. are correct (thank you for noticing though). Pratchett often liked to use archaic or whimsical spellings.
  2. Green's Theorem assumptions: the curve C must be be a positively oriented (counter-clockwise), piecewise smooth, simple closed curve in a plane, and let D is the region bounded by C. The functions L and M are functions of (x, y) defined on the open region containing D and must have continuous partial derivatives there.
  3. The key assumptions throughout this discussion are that the disc is (1) circular (it isn't exactly, but the approximation seems good), and (2) that it is flat (it isn't exactly, but Cori Celesti, the most magnificent peak of the world is 10 miles high. That sounds very, very tall, but on the scale of the Discworld it is almost invisible).
  4. A "mile" has varied in meaning across time and space. Historically the term derives from the Roman mile - mille passus - which means a thousand steps. In 29 BCE Agrippa standardised the meaning of a step as 5 feet. There have been many variants though, and the mile is still not completely standard. The US survey mile is subtly different from the British mile, and the nautical mile is different again. If you want to measure distances (outside of a story) use SI units.
  5. I am aware that some aspects of the geography of the Discworld and the text describing it are at odds. A good discussion is here.