Plus Blog

December 5, 2013

Triangles do it, squares do it, even hexagons do it — but pentagons don't. They just won't fit together to tile your bathroom wall. That's the reason why you'll find it very difficult to find pentagonal tiles in any hardware shop.

It's actually really easy to see why pentagons won't tile the plane. We are talking regular pentagons here, shapes that have five sides of equal length and angles between them. In a regular pentagon the internal angle between two sides is 108°. In a regular tiling, adjacent tiles share whole edges, rather than just parts of edges, so corners of tiles meet corners of other tiles. To fit a number of tiles around a corner point, their internal angles must add up to 360°, since that's a full turn. If you try to fit three pentagons, you only get 3 x 108° = 324°, so there is a gap. If you fit four pentagons, you get 4 x 108° = 432°, so two of them overlap. The same isn't true for equilateral triangles, squares, or regular hexagons. Here the internal angles are 60°, 90° and 120° respectively, so you can fit six triangles, four squares and three hexagons around a corner point. (You can try and work out for yourself if any other regular polygons can give you a regular tiling.)

Figure 2: Three pentagons arranged around a point leave a gap, and four overlap.

Three pentagons arranged around a point leave a gap, and four overlap. Image: Craig Kaplan.

If pentagons don't work can we perhaps use other shapes with five-fold symmetry to tile the plane? Find out more in The trouble with five.

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December 4, 2013

Remember maps? In the old days, before smart phones and satnavs, we used to use them to get around. They posed quite a few problems, including folding them and understanding them. To help with the latter, map makers often use different colours to colour adjacent regions, be they countries, counties, or parts of a city. And pretty much since the beginning of map making, they have known that they will never need more than four colours to colour a whole map. It's something that's so blindingly obvious once you start playing with a few maps, it should be easy to prove mathematically: when colouring a map drawn on a flat piece of paper, four colours will always suffice.


The US in four colours. Image: Wikipedia.

Mathematicians first set themselves the task back in 1852, and it proved a massive headache. It took 24 years for the first proof to be announced, only to be found faulty another eleven years later. Then it took another 86 years, until 1976, until the mathematicians Kenneth Appel and Wolfgang Haken came up with a second proof. They first supposed that there are maps that need five colours, from which you can then choose one that has the smallest number of countries. They then showed that such a minimal map must contain one of 1,936 possible configurations; and they also proved that every one of these possible configurations can be reduced into a smaller configuration which also needs five colours. This is a contradiction because they assumed that they had already started with the smallest five-colour map. When you find a contradiction you can deduce that your initial assumption, that there are maps that need five colours, is false.

This sounds good, but there was a hitch. The part of their proof which showed that these 1,936 configurations could be reduced was done by a computer, which simply ploughed through every configuration and checked it. No human being could in their lifetime ever actually read the entire proof to check that it was correct. This caused an outcry: if nobody can check it, how can we ever know it doesn't contain a mistake? Can mathematics ever be done by computers?

The proof has since been improved and verified independently (by computers), so most people are quite happy to consider the result proven. The question of whether computer proofs can ever be considered sound, however, is still a hot one.

To find out more, read The philosophy of proof and Welcome to the maths lab.

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December 3, 2013

Here's a little game to play. Take the interval from 0 to 1, including the end points. Now delete the middle third of this interval, from 1/3 to 2/3, but without deleting the two end points 1/3 and 2/3. Then delete the middle thirds of the two intervals that are left, [0,1/3] and [2/3,1], again without deleting the end points. Keep going, deleting middle thirds ad infinitum — well, you can't really do this in practice, but you can imagine it. What will you be left with in the end?

Cantor set

Seven steps towards C.

Let's call the set of left-over points C. Clearly, C contains the end points of middle third intervals because these don't get removed: 0, 1, 1/3, 2/3, 1,9, 2/9, 7/9, 8/9, and so on. But other points are left too. For example, 1/4 does not get removed at the first step because it's less than 1/3, it does not get removed at the second step either because it's bigger than 2/9, and so on.

Does C contain any intervals at all? It's pretty easy to convince yourself that it doesn't: any line segment, no matter how tiny, will have been broken up by having its middle third removed. C is what's called totally disconnected because all its connected components are just points.

But these points are not isolated. If you draw a little circle around any point x in C, then no matter how small the circle, it will contain other points of C. For example, take the point 0. Arbitrarily close to it there is a point of the form 1/3k for some k, which is in C because it's the end point of a middle third interval. The fact that every point has other points arbitrarily close to it makes C into a perfect set.

All this means that C inhabits a strange world between dimensions. It's more than a collection of isolated points, which has dimension 0, and less than a line, which has dimension 1. If you calculate its Hausdorff dimension, which is an extension of our usual concept of dimension, the answer is ln(2)/ln(3)=0.63, which is between 0 and 1 as expected.

Menger sponge

The Menger sponge. Image: Niabot.

The set C is also self-similar. For example, if you take the part of C contained in some tiny interval [0,1/3k] and blow it up, what you get looks exactly the same as C itself. Being self-similar and having a dimension that's not a whole number makes C into a fractal.

C has a name too: it's called the middle-thirds Cantor set, after Georg Cantor, who wrote about it in 1883, although it had previously been discovered by Henry John Stephen Smith. Cantor was one of the first people to give a rigorous description of a fractal, though the term "fractal" was not coined until almost 100 years later by Benoit Mandelbrot.

Cantor sets often turn up in mathematics when things are being broken up according to the same recipe ad infinitum. For example, you can create a similar construct in two dimensions by dividing a square up into nine smaller squares and repeatedly removing the middle one. The result is called the Sierpinski carpet, which has a Hausdorff dimension of ln(8)/ln(3)=1.892. The three-dimensional analogue is the Menger sponge.

You can read more about Cantor sets and fractals in

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December 2, 2013
Rocket tortoise!

Will he rocket to the lead? (Most excellent tortoise suit by Katie Bradley.)

Achilles and a tortoise are competing in a 100m sprint. Confident in his victory, Archilles lets the tortoise start 10m ahead of him. The race starts, Achilles zooms off and the tortoise starts bumbling along. When Achilles has reached the point A from where the tortoise started, it has crawled along by a small distance to point B. In a flash Achilles reaches B, but the tortoise is already at point C. When he reaches C, the tortoise is at D. When he's at D, the tortoise is at E. And so on. He's never going to catch up with the tortoise, so he has no chance of winning the race.

Something's wrong here, but what? Let's assume that Achilles is ten times faster than the tortoise and that both are moving at constant speed. In the times it takes Achilles to travel the first 10m to point A, the tortoise, being ten times slower, has only moved by 1m to point B. By the time Achilles has travelled 1m to point B, the tortoise has crawled along by 0.1m to point C. And so on. After n such steps the tortoise has travelled

  \[  1+1/10+1/100+1/1000+ \ldots +1/10^{(n-1)} \mbox{metres}.  \]    
And this is where the flaw of the argument lies. The tortoise will never cover the 90m it has to run using steps like these, no matter how many of them it takes. In fact, the distance covered in this way will never exceed $10/9=1.111\ldots $m. This is because the geometric progression
  \[  1+1/10+1/100+1/1000+\ldots  \]    
converges to $10/9$. Since the tortoise is travelling at constant speed, it covers this distance in a finite time, and it's precisely when it's done that that Achilles overtakes it.

This problem is known as one of Zeno's paradoxes after the ancient Greek philosopher Zeno, who used paradoxes like this one to argue that motion is just an illusion.

Find out more about Zeno's paradoxes and infinite series on Plus.

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December 1, 2013

Here's a strange fact: if you look up some numbers, for example the numbers in your tax return, population sizes of Chinese provinces, or the length of the world's rivers, then most likely around 30% of these numbers start with the digit 1, around 18% start with the digit 2, 12.5% start with a 3, and so on, all the way to 9 (which only heads up around 5% of the numbers) - the larger the digit, the fewer numbers in your list start with it. This fact, known as Benford's law, applies to so many different kinds of data sets that it's often used to detect fraud. But why does it work?

Well, if the processes that give rise to your list of numbers do produce a universal distribution of first digits, then this distribution should apply no matter what units you use. It should work no matter if you do your tax return in pounds or in euros, or measure your rivers in metres or miles - it's universal after all. This means that the distribution of first digits remains the same when you multiply your numbers by whatever constant you need to change between units. And it turns out that the only distribution with this property of scale invariance is precisely the Benford distribution.

As an example, imagine that your first digits are distributed equally (roughly the same proportion of numbers begin with the digit 1, 2, 3...) – so NOT according to the Benford distribution. Is this distribution scale invariant? Let's see what happens when we multiply by 2. All numbers starting with 5, 6, 7, 8, and 9, when multiplied by 2, give a number starting with 1. By contrast, the only way to end up with a number beginning with, say, 3, is to start out with a number starting with 1. In other words, the resulting distribution of first digits, after multiplying by 2, is skewed towards 1. It's not uniform, so your original distribution is not scale invariant. It's not too hard to show that in order to be scale invariant, the first digits have to be distributed in the way stipulated by the Benford distribution. It's worth noting though that Benford's law only applies to data sets that are neither too random, nor too constrained: alas, it doesn't work for lottery numbers.

To find out more about Benford's law read our article Looking out for number one. This news story explores an application of Benford's law to uncover potentially fraudulent elections.

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November 29, 2013
christmas mystery

Oooo.... We're getting excited! It's nearly time for our Plus Advent Calendar! The first door won't open until Sunday 1 December, but here's a sneek peek!

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