Category Archives: Accessible

Ideally accessible to those without specific mathematical knowledge

About time

Sometimes I start newcomers to MathsJam on timing puzzles. Here are five classics in increasing order of difficulty and implausibility, each with a distinct flavour. The first is distinctly steaky…

1. Some friends are coming over for a steak dinner. You want to cook three steaks as quickly as possible, but your grill pan only holds two at a time. Each steak must be cooked for five minutes on each side. What is the fastest you can have all three ready? Show that you can’t do any better.

Alas, when this precise situation once arose with my family at home (who says puzzles are never useful?), I was unable to convince my father not to cook the steaks in the obvious order. But at least that meant I got my steak sooner.

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More on Gale–Shapley Nobel prize

I’ve long been a fan of the Gale–Shapley matching algorithm, and related problems, so was happy to see that a Nobel Prize was awarded for it. Having seen Peter Rowlett’s article that laid down the following gauntlet:

“I see ‘Nobel week’ as an opportunity for mathematicians to go in search of the mathematics behind each prize, rather than to retreat and complain about the lack of a prize specifically for mathematics”,

I was surprised that none of the mathsy types in my tiny corner of internet seemed to have noticed that a mathematician won a Nobel prize essentially for mathematics. After growing slightly impatient, I realised I only had myself to blame for not acting earlier, so I sketched a quick news story contribution over at the Aperiodical (it’s short and so reproduced here in full):
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Post on Aperiodical

I’ve written an article for the Aperiodical entitled “Ask a mathematician: where should we live?”.

Dear Mathematician,

My partner and I are trying to buy a house. We both work in different places, and neither of us enjoys commuting. How could we decide where to live?

Fictionally yours,

Norman Mettrick

Norman,

Thank you for your intriguing and entirely imaginary letter. The short and not terribly useful answer would be…

Want to know? Read the rest of it there.

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Jenga blocks and Gowers’ hyperplane space

This post is an attempt to communicate some of the feel of Banach space theory to those who aren’t familiar with it. I once tried to explain my research to a six year old using Jenga blocks, but fortunately only got as far as the triangle inequality. Near the end of my Phd, at my supervisor’s suggestion, I started to explore the complicated Banach space that is Timothy Gowers’ solution to Banach’s hyperplane problem. These experiences inspired the following explanation of one relatively simple observation (that I included as an example in my thesis) through the delightful medium of building blocks.

Our object of study are towers of good old-fashioned building blocks. Each block has a number written on its side, so each tower built from these blocks gives a sequences of numbers (x_1, x_2, x_3, \ldots). These don’t have to be positive natural numbers, but you won’t lose much by pretending, in this post, that they are. There are innumerably many different brands of towers, but we’ll concentrate on one particular brand: the ‘Gowers Towers’. Let’s say the number written on each block represents how heavy the block is, and is inversely proportional to the length of the block. So we’d represent the sequence (0, 1, 5, 0, 3, 5, 5, 1, 10) with the Gowers Tower pictured.

It’s worth mentioning that the Gowers Towers include every individual tower of finite height that you can build with your unlimited set of Gowers branded building blocks (and lots of infinite height, but you don’t really need to worry about those here).

Let’s pretend we’ve got a measure of the instability of a tower (the norm of the sequence), and whenever we increase the instability beyond a certain threshold, K, the tower collapses.

Blocks with higher numbers are heavier, as well as narrower and perhaps inherently more unstable. How the blocks of different weights at different heights affect the stability of the Towers of Gowers is extremely complicated. However, the towers do have some nice, intuitive properties.

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Getting into Norms: Part II

In Part I of Getting into Norms, I talked about three different ways of measuring distance (I also considered  the accuracy of a series of guesses to be a ‘distance’). All three of these were norms, but there are many ways of measuring distances that aren’t norms.

So to study norms, mathematicians must define them really rigourously, using something known as axioms. These are the basic assumptions and definitions of mathematics. Once we’ve made these assumptions we can prove what has to follow from them.

We can think of norms as a measure of distance from the origin. If you think about it in this way, the following seem quite obvious, and appeal well to our instincts. A norm satisfies the following three axioms.

  1. Distances are always positive!
  2. If the distance from your location to the origin is zero, then you must be at the origin. Or alternatively, if two points are separate then the distance between them isn’t zero. Conversely, the distance from any point to itself is zero.
  3. Taking a detour is always longer than travelling in a straight line. This is the triangle inequality: the sum of the length of any two sides of a triangle is longer than the length of the third.
  4. Now we come to axiom four. This one is tough to describe in words. Here goes. If you walk a pace forwards and then take another in the same direction, then you will have walked twice the distance of the original pace. Also it doesn’t matter whether you take a pace forwards or backwards: they will give you the same distance.

When mathematicians want to be precise, we use symbols. The distance between points x and y is written as \|x-y\|. The distance from x to the origin is \|x-\underline{0}\|=\|x\|. We say that \| \cdot \| is a norm if whenever we pick vectors x and y, and a number \lambda, then the following axioms hold:

  1. \|x\| \geq 0.
  2. If \|x\| = 0 then x = \underline{0}. And visa-versa.
  3. \|x+y\| \leq \|x\|+\|y\|.
  4. \|\lambda \cdot x\| = |\lambda| \|x\|.

These four conditions should match with our verbal descriptions above. You may recognise them from this blog’s exquisitely hand-drawn logo.

They were pretty trivial intuitions, once we thought of \|x\| as being the distance of a point x from the origin (the origin above \underline{0} is underlined to distinguish it from the normal 0, though we don’t choose a different notation because the origin behaves a lot like the number zero). Continue reading

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Dice and Dissection: a puzzle

This is a puzzle, first appearing in Martin Gardner’s column in 1978, with a new way of thinking about the solution. Before the puzzle, though, a cultural diversion.

Rolling it in

In the modern classic board game The Settlers of Catan, it’s very important to know, when you roll a pair of dice, the frequency with which each number occurs. Resources are given players only if they have a settlement adjacent to those tiles whose number is rolled. If you build your settlements next to a tile labelled 2 or 12, it will, on average, only be productive once every 36 rolls. Tiles labelled 6 or 8 will produce resources five times in 36 rolls. It’s so fundamental to the gameplay that the relative frequencies are visualised as dots on the pieces: sixes and eights are so important, they are marked in red (rolling a seven does something different).

Puzzle

The well-known puzzle is:

By relabelling the faces of two dice, can you design a new, unusual pair of six-sided dice that achieves rolls with the same frequencies as a pair of normal dice? All the faces must have a positive number of spots.

If I didn’t require you to use a positive number of spots on each face: then dice labelled {0, 1, 2, 3, 4, 5} and {2, 3, 4, 5, 6, 7} would work as a pair. If you allow negative numbers, there’s infinitely many solutions!

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Percentages for sceptics: part III

I wanted to do some self-criticism of my previous two posts in this series:

  1. You can calculate the minimum of responses from a single percentage by hand (no need for computer programmes or look-up tables).
  2. I’ve made a very rough model to estimate how many people the program typically returns when fed six percentages (as I did several times here).
In between, I’ve collected some links to demonstrate how great continued fractions can be.

Handy calculations

There are many ways of writing real numbers (fractions and irrationals) apart from in decimal notation. You can represent them in binary, for instance \pi = 11.001001000011\ldots, or in other bases. These have their uses: there is a formula to calculate the nth binary digit of \pi without calculating all the preceding digits.

For our purposes we will use continued fractions. People write \pi as [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, ...]: this notation means that 3 is a good first approximation to \pi, the well-known 3+\frac{1}{7}=\frac{22}{7} is the closest you can be with any fraction \frac{p}{q} with q \leq 7. Then 3+\frac{1}{ 7 + \frac{1}{15}}=\frac{333}{106} is the best with q\leq 106, and the fourth term 

3+\frac{1}{7+\frac{1}{15+\frac{1}{1}}}=\frac{355}{113}

is a very good approximation, as the next number in the square brackets, 292, is very large (I’ll motivate this observation at the end of the section). The golden ratio is sometimes called the ‘most irrational’ number because it has a continued fraction expansion with all ones, so the sequence converges slowly.

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