In conversation with Anne Skeldon

19 April 2026

It’s Monday morning on Surrey campus, and you’ve woken up nice and early for your 10am lecture. It’s the start of a new term and you’re certain that, with your new morning routine, everything will be different this time: you’ll get perfect sleep every night, and your fellow students will all be in awe of how switched on you seem in lectures.

On your way to the lecture, you catch yourself yawning and realise you’re still tired… but, why? You got eight hours of sleep, had a good breakfast, even practised a new meditation routine! OK, you used your phone a bit before bed but you had that night-time setting on, so that’s fine. You even live next to uni and so your commute through the shaded back garden path is short. No wasted energy on moving, right?

As you wonder why on earth one even needs to sleep, you remember that you study at the University of Surrey and, fortunately for you, Anne Skeldon is your lecturer. She might just have the answers to your queries—let’s just hope you don’t sleep through it.

Cause and effect

Anne Skeldon is professor of mathematics at the University of Surrey, whose research in applied maths has, in recent years, focused on sleep, circadian rhythms and data science. That wasn’t, however, how it all started. “I didn’t start doing maths at university; I started doing physics. I actually found maths was a bit boring at school.”

It was only when Anne began studying physics at Oxford that she (somewhat fortunately) realised that mathematics was hard-coded into the subject, and in fact it was through the maths that she was able to understand the physics.

“I reckon you can explain anything with maths. If you think $A$ causes $B$ which causes $C$, then if I change $A$ in some way, then I should be able to predict something about $C$.” Anne now believes that if someone thinks they understand how something works, a mathematician should be able to swoop in, write down some equations and comment on the argument. “If one’s hypothesis is correct, then this is what we should see.” There is a strong proposition that mathematics deserves to be considered a core part of the scientific method. The power of maths to explain the real world is what really captivates Anne: “I love the ability to write down equations and understand the dynamical structure of things, especially if it applies to something; maths can take you anywhere, right?”

Across Anne’s career, she has researched topics such as pattern formation and bifurcation theory, both from a relatively abstract point of view and in terms of how they interact with the real world. The carbon cycle, tumours, and the social politics of recycling are just a few of the areas where her mathematics has found application. Currently, however, she is focused mainly on a particular application area: what makes a good night’s sleep?

Sleep is a perfect storm

Anne hadn’t come to Surrey with an interest in sleep; she had come for its leading reputation in nonlinear dynamics. One day, Derk-Jan Dijk, director of the Surrey Sleep Research Centre, came to give a talk in the mathematics department. They got talking afterwards and ended up co-supervising a masters student. From there, the idea of more modelling problems around sleep “snowballed” into the research area it is today.

Banks of screens on a wall above a desk containing a number of keyboards

Inside the control room of the sleep research centre. Image: Anne Skeldon

The sleep research centre had previously considered more physiological, psychological, cognitive and behavioural patterns within sleep; the mathematical tools which Anne develops provide a new direction by which to unpick sleep’s mysteries. This has built on a rich tradition of using mathematics in sleep and circadian research which dates back to the 70s and 80s.

Though we may not always give it the thought it deserves, sleep is actually a highly complex phenomenon which is deeply interconnected with many aspects of our lives. “There’s a social science aspect too because it affects your relationships; when you go to bed to sleep is also dependent on social, work and study constraints.” In terms of biology, one can study the impact sleep has on health. There are short-term associations like sleep and mood, sleep and performance, sleep and injury risk, but there are also long-term risks associated with cardiovascular disease and certain cancers. Anne’s interest in sleep arises from the implications of these risks—“The question is: what is the biology that sits underneath that? What happens to your brain when you go to sleep?” Anne points out that, rather fundamentally, researchers still don’t even know why we sleep. “Your car doesn’t get tired… so why do we?”

In Anne’s ideal world, nobody would have to struggle to get out of bed when their alarm clock goes off. We agree; in fact, she goes on to suggest that a perfect scenario would have everyone leaping out of bed or not even need an alarm clock. Sounds utopian! But where do we even start?

“You obviously can’t change your physiology, but you can change your environment. What I’ve done has mostly been around when people sleep.” A significant chunk of Anne’s research is based on taking known data and building quantitative models that are able to explain sleep at an individual level, or can be fit to specific individuals. “We aim to design interventions for individuals to help them sleep at the time they want to sleep.”

A woman sleeping in a dark room. She has probes attached to her head.

A participant sleeps soundly while being monitored. Image: Anne Skeldon

Thinking like an oscillator

It’s intuitive that different forms of light will affect how you sleep: how sunny it is outside, for example, or whether you have lamps on in your room, or even the brightness of your phone screen that might be filtered by blue-blocking glasses. “Your eye doesn’t know what the source is. It’s a photon, in the end!” Anne tells us. “But it does know something about the amount of energy and the flux.

“If you were an engineer, and you were designing a solar cell, you’d probably be interested in watts per square metre. But that’s not the unit that biologists use, because your eye has receptors that are responsive to different wavelengths;

The key to understanding sleep patterns is undoubtedly one’s body clock (or biological clock). Our body clocks have a natural period, but, for a given individual, that natural period may not be exactly 24 hours. “Imagine yourself as an oscillator; if you’re trying to live on a 24-hour cycle, which is away from your natural period, your natural rhythm needs to be entrained.” This idea frames each individual as having their own self-sustained internal (circadian) oscillation which is then altered by an input signal from the light-dark cycle.

Cartoon of Anne in front of a whiteboard, with a drawing of the sun next to both a cross and a tick.

Cartoon Anne telling us that light is both the problem and the solution

Whatever helps you sleep at night…

Biologically speaking, an important cue that sends you to sleep is melatonin, a hormone whose concentration in the bloodstream naturally increases during the evening and into the night. Anne tells us light suppresses melatonin—that’s one reason why having bright light in the evening has an acute effect on when you decide to sleep!

“If you want to entrain a rhythm, the input signal needs to be strongly rhythmic with a strong 24-hour component—there has to be significant contrasts between day and night.” Anne says that this entrained oscillator model would work really well if we were still hunter-gatherers. The sun would come up and we’d naturally be outside; the sun would go down and we’d know to go to sleep.

Every individual has their own body clock which, for most of us, runs slightly slow. But for others it can be too fast. The key idea is that depending on the time of day, light can either speed up or slow down your clock. This process can be modelled via velocity response models; for example, in the Winfree model, the rate of change of phase $\theta$ is given by
$$ \frac{\mathrm{d}\theta}{\mathrm{d}t} = \omega + \ell(t)\sin\theta. $$

In the absence of light, phase advances uniformly with angular velocity $\omega$. When light is present, $\ell(t)$ is non-zero, and thus whether light speeds us or slows down the rate of change of phase (in the case of humans, our body clock) depends on the phase (the time of biological day). “If you have a slow internal clock, and you need to speed it up to 24 hours, you need to have lots of light in the morning and not too much light in the evening.” It’s encoded in our biology.

We were also interested to know if these models had actually been tested and matched with data. Returning to her analogy with hunter-gatherers, Anne explained to us that there have actually been experiments to try to recreate those conditions. “There are these neat cave experiments where people hid themselves away from external cues to try and figure out why we actually get up in the morning.” It was a burning question; is it just the sun comes up and we get up, or does there exist some internal rhythm? The experimenters relieved themselves of technology new and old, set down their watches and alarm clocks, removed any thermal cues in order to remove any contrast between night and day, and took themselves away from all natural light and into cave systems.

“And they found that they did still have a rhythm,” Anne concluded, “but it actually wasn’t 24 hours.” There were individual differences amongst the subjects in that most people went to bed slightly later each night and got up slightly later each day, though some people would do the reverse and get up earlier.

Anne’s model as featured on YouTube. Image: Peter Caires

“As a society I think there are a lot of people who struggle on that temptation to go to bed later and wake up later—you often see that on holidays.” It poses an interesting question from a psychological point of view: “What is it that takes us to bed in the end? Do you look at your watch and think, I really ought to go to bed now, it’s three in the morning?”

As the sun set on our conversation with Anne, we were left to reflect on what she had said meant for our own sleep cycles. Of course, sleep is incredibly complex, and many other aspects of our routine can affect our sleeping patterns (caffeine intake, diet and stress to name a few). But perhaps as a society we have come to overestimate our own roles in these matters and forgotten the importance of the natural rhythm influencing all of life on Earth: the solar cycle.

Sweet dreams!

On the cover: A new puzzle every day

Matthew Scroggs

19 April 2026

In January, I paid a visit to MathsWorld, the recently opened maths discovery centre in London, alongside some other members of the Chalkdust team. One of the highlights of the trip was playing the two-player game Genius Square.

In Genius Square, you start with a six-by-six board and roll seven dice. These dice tell you where to place seven cylindrical blocks, for example:

The dice

The board with the cylinders placed

The two players then race to fit the pieces shown above into the remaining space on the board. The pieces that the players have are the five tetrominoes, the two triominoes, a domino, and a single square (or monomino); these are all the shapes you can make by gluing together up to four squares (if rotations and reflections are considered the same shape).

The pieces

There’s some clever design in this game: if, instead of rolling the dice, you were to randomly pick any set of seven spaces to place the cylinders, the puzzle is not guaranteed to have a solution. The locations printed on the dice have been carefully chosen so that any combination that you can roll leads to a solvable puzzle.

A puzzle-a-day

The Genius Square puzzle is similar to another rearrangement puzzle: the puzzle-a-day calendar, created by the Norwegian puzzle makers DragonFjord. In this puzzle, you are given the pieces below and asked to place them on the board to cover everything except today’s date.

The puzzle-a-day board

The puzzle-a-day pieces

For example, on 22 July, you could place the pieces as shown below.

A solution of puzzle-a-day for 22 July

DragonFjord make and sell wooden and plastic version of puzzle-a-day, which you can buy from Maths Gear—who also provide the top prize for the crossnumber (see the Issue 23 crossnumber)—to avoid the cost of shipping directly from Norway.

In puzzle-a-day, it’s possible to arrange the pieces to make every single combination of a number and a month, including days that don’t exist like 31 September and 30 February. While we were considering options for the cover of this issue, we discussed putting something like Genius Square on the cover, and I began to wonder if it would be possible to make a puzzle like puzzle-a-day but where it was only possible to make days that actually appear on the calendar.

A solution of puzzle-a-day for 31 September?!

A new puzzle

After spending a while scribbling on squared paper and getting nowhere, I had an idea: I could put the months in regions that were disconnected from the day numbers. Then, by carefully choosing the shape of the month regions and the arrangement of the dates, I could force the solver to use different combinations of pieces on the day numbers for different months.

Once I’d had this idea, I threw together some Python code that could see which day numbers you could and couldn’t leave uncovered with a set of pieces, and waited for it to find a good set of pieces. It found the board and pieces shown to the right. As in Tetris, I’ve named the pieces after letters that they vaguely resemble.

In January, March, May, July, August, October and December, you have to use a P, an O and the A in the month regions. The remaining pieces can make any day from 1 to 31.

In April, June, September and November, you need to use the C, an O and the A in the month regions. This leaves pieces that can make any day from 1 to 30, but importantly can’t make 31.

In February, you need to use both Os and the C in the month regions. This leaves pieces that can make any day from 1 to 29, but not 30 or 31.

Now all we need to do is find another new arrangement that somehow works differently in leap and non-leap years…

The board for the new puzzle

The pieces for the new puzzle

Cheating at Tetris

Ffreuer Bristow

19 April 2026

Let’s imagine you have a friend who’s very good at Tetris. You’ve challenged them to a game with a twist: you get to pick which pieces they have to play. You agree that if your friend can survive 100,000 blocks then you’ll declare them the winner; but if they get game over before this, you win the challenge. So what do you do?

Tetris is a puzzle game of falling blocks that you can rotate and move as they fall. Forming complete horizontal lines clears all the squares in that line, and the aim is to prevent the blocks from stacking all the way to the top of the playfield, which results in game over.

The playfield is 10 cells wide and 20 cells tall and the game is played with seven different blocks, called one-sided tetrominoes, that are each made up of four squares. They are labelled I, J, L, O, S, T and Z because they look a bit like these letters.

There are many different versions of Tetris but, for the challenge against your friend, you will be playing a very simple version where you can pick any block you like, regardless of what has been picked previously, and the speed of the blocks falling stays the same through the whole game.

How to lose

First let’s look at the worst strategy you could go with: choosing the same block 100,000 times.

It’s easy to see that the I, J, L and O blocks fit very neatly together and can always be arranged to completely clear the playfield. The playfield will be cleared after only ten I blocks, ten J blocks, ten L blocks or five O blocks have been placed.

Ten $J$ blocks.

…or ten $L$ blocks fit together nicely then all disappear.

Ten $I$ blocks also fit together nicely
then all disappear.

For the S, T and Z blocks it’s a little less obvious. All three follow the same idea, so let’s look at how it works for the S block:

You can see that after a number of block placements it loops back to a previous step, but never back to the start. But the stack can always be kept between two cells and four cells tall, so S blocks can be played infinitely without resulting in game over, despite never clearing the playfield.

So, regardless of the block you choose, it can always be played as many times as you like without resulting in game over. Since your friend is the best Tetris player you know, this strategy will pretty certainly result in you losing the challenge.

How to win

Let’s instead consider a combination of S and Z blocks. S blocks fit nicely with other S blocks, and Z blocks fit nicely with other Z blocks; but S and Z blocks don’t fit well together.

Two $S$ blocks or two $Z$ blocks fit together nicely…

…but an $S$ block and a $Z$ block don’t.

Because of this, if you choose an alternating sequence of S and Z blocks, the best way for your friend to avoid game over is to construct columns of only S blocks and columns of only Z blocks. The playfield is 10 cells wide and each column will be two cells wide, so there will be five columns, which is odd, resulting in either three S columns and two Z columns, or two S columns and three Z columns.

Since the S and Z blocks are simply the reflection of each other, consider (without loss of generality) the scenario with three S columns and two Z columns. This will result in the Z columns growing faster than the S columns and eventually both Z columns will reach the top of the playfield. At this point, to avoid game over your friend must place a Z block in an S column, creating a one cell wide and two cell high `hole’.

Eventually, your friend is forced to play a $Z$ block in an $S$ column

Filling these holes causes issues later on in the game since there are a finite number of times they can be filled without creating new holes elsewhere. So new holes will continue to form and your friend can only create so many gaps before they reach game over.

But will this happen within 100,000 blocks?

How many blocks?

A more in-depth explanation can be found in Heidi Burgiel’s paper How to lose at Tetris (1997), but the most important parts are the following facts that were proved:

The maximum number of alternating S and Z blocks that can be placed before a hole must appear is 240 blocks.

The maximum number of times an S or Z block can be used to fill existing holes is 120 times, each time removing two holes.

The maximum number of holes the board can contain without resulting in game over is 50 holes, which is 10 holes per column.

By calculating the total number of holes that can possibly appear on the board before game over, including the ones that get filled in, and then multiplying by the number of block placements required for each hole to appear, we get an upper bound for how many blocks will force game over. The number of holes that can appear is $(120\times2)+50=290$, so game over must occur within $290\times240=69600$ alternating S and Z blocks.

Since this is well within 100,000 blocks, you now have your winning strategy. Choose only alternating S and Z blocks and even if your friend is the Tetris world champion you’ll be sure to beat them!

Infinite Tetris?

What would happen if you were to pick blocks at random instead?

In the finite game that you and your friend are playing, the probability of a sequence of 69,600 alternating S and Z blocks occurring at any point within 100,000 blocks is incredibly small, but not zero.

Despite being very unlikely to happen during your challenge, the game-ending sequence of S and Z blocks is guaranteed to occur at some point in a long enough game of Tetris. Because of this, every game of Tetris must eventually end.

Another way to think about this is in terms of the infinite monkey theorem, which states that a monkey randomly hitting keys on a typewriter for an infinite amount of time will eventually end up typing the entire works of Shakespeare. Instead, imagine that the typewriter only has the letters I, J, L, O, S, T and Z, and the monkey hitting the keys chooses the next block in the Tetris game. Given an infinite amount of time, the monkey will eventually end up typing the sequence of 69,600 alternating Ss and Zs, ending the Tetris game it was choosing blocks for.

Back to reality

In regular Tetris games, the speed at which the blocks fall steadily increases. This causes players to make more and more mistakes and, just like the holes in the S and Z columns, these mistakes will stack up until game over is inevitable. Ultimately, small mistakes due to the time constraints will likely be what causes your friend to lose the challenge, so you probably won’t have to wait 69,600 blocks to beat your friend.

But maybe next time pick someone who isn’t as good at Tetris!

The secret life of words

Barry Pawlik

19 April 2026

Let’s begin gently, with the unassuming question: What is a square? The most precise answer is the favourite one of professional physicists: the universal It depends. On the context, of course.

For many people, the first thing that comes to mind is a geometric figure, like this one: $\square$. A perfect answer is also the number 529 (since we’re in the 23rd issue of Chalkdust, after all), as well as its more recognisable friends: 9 and 25. But not everyone realises that squares can be not only figures, numbers and even logical symbols, but also… words!

The arena of this article is combinatorics on words: a discipline at the intersection of discrete mathematics and theoretical computer science. Combinatorics on words studies patterns and repetitions in sequences of symbols, treating words as purely mathematical objects rather than carriers of meaning. Below we present a brief guide to this world.

We start with a set, preferably nonempty and finite. For the sake of intuition, we will call this set an alphabet, and its elements letters. For a binary alphabet (ie consisting of exactly two letters) we usually take it to be $\{0,1\}$ or $\{a,b\}$ and for $k$-element alphabets we take $\{0,1,\ldots,k-1\}$. We can also consider the whole $26$-letter Latin alphabet $\{a,b,c,\ldots,x,y,z\}$, which is most often used to give fancy examples from natural languages that illustrate the relevant definitions. Finally, by a word we mean a (finite or infinite) sequence of elements of the alphabet. Instead of writing $(w_1,w_2,w_3,\ldots, w_n)$ we will write $w_1w_2w_3\cdots w_n$ so, for example, the word (W,O,R,D) can be written as WORD.
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Coffee problems

Patrick Hedfeld

19 April 2026

Coffee is not only an elixir of life for tired students and overworked mathematicians, but also a fascinating object of scientific consideration. From the optimal brewing temperature to the perfect milk foam geometry—behind every cup of coffee is a world full of formulae, functions and fascinating physical phenomena.

So put the kettle on and join us as we dive into the maths of coffee and discover why a cup of coffee sometimes requires more variables than a partial differential equation.

A question of thermodynamics

The ideal brewing temperature for coffee is 92°C to 96°C. This temperature range maximises the extraction rate of the coffee particles without releasing bitter tannins. This can be described by the Arrhenius equation,
\begin{equation*} k=A\mathrm{e}^{-E_\text{a}/RT}, \end{equation*}
with $k$ the speed constant of the extraction, $A$ a constant factor that is determined by experiment, $E_\text{a}$ the activation energy, $R$ the universal gas constant, and $T$ the temperature in Kelvin.

Studies have shown that at temperatures above 96°C, the extraction of undesirable bitter compounds increases significantly, while at temperatures below 92°C, the extraction of the desired flavouring compounds remains incomplete. So make sure to have your thermometer handy when making a pot.

Voronoi diagrams

A Voronoi diagram divides an area into regions based on a set of points. Each region contains all the points that are closer to a certain point than to any other. These regions are called Voronoi cells. Voronoi diagrams are used in many fields such as geography, biology, robotics and telecommunications. For example, they help in determining catchment areas, modelling cell structures, path planning and optimising the placement of cell towers or even in milk foam over coffee.

The perfect milk foam geometry

Milk foam is a complex colloidal system consisting of gas bubbles in a liquid matrix. The bubbles in the milk foam form a Voronoi diagram-like structure, in which each bubble optimally fills the space allocated to it. The mathematical description of this structure is complex, but we can represent it—very simplified—as a sum over spheres and sum it up as
\begin{equation*} V=\sum_{i=1}^{n}\frac{4}{3}\pi r_{i}^{3}, \end{equation*}
with $V$ the total volume of milk foam and $r_{i}$ the radius of the $i$th bubble.

Research shows that the stability and texture of milk foam depends heavily on the size distribution of the bubbles. An even distribution of small bubbles results in a creamier and more stable foam.

The coffee break: a time optimisation problem

How long should a coffee break last? Mathematically, this is an optimisation problem that aims to maximise the benefits of the break without affecting productivity too much. The optimal duration of breaks, $t$, can be described as
\begin{equation*} t=\frac{T_{W}}{2}\ln\left(\frac{P_{\text{max}}}{P_{\text{min}}}\right), \end{equation*}
where $T_{W}$ is the total working time, $P_{\text{max}}$ is the maximum productivity, and $P_{\text{min}}$ is the minimum productivity, or the absolute minimum of your cognitive function.

The art of taking a break

Studies have shown that short, regular breaks can increase productivity and even creativity. However, taking too long a break can interrupt the flow of work and reduce efficiency. Finding the right time is an optimisation problem with multiple critical points and a strong dependence on coffee quality. For instance if our maximum productivity is twice our minimum, then we should take a three-hour break for every eight hours that we work. This gives plenty of time for working out the mathematics of coffee!

The coffee cup: a problem of geometry

The shape of a coffee cup has a significant influence on the taste experience. Surface area is the one of the main factors for both the release of volatile compounds and temperature retention, though this also depends on the material the cup is made of. A cylindrical mug with a diameter of 7cm and a height of 9cm offers the optimal surface for the aroma development. The volume $V$ of such a mug is given by $V=\pi r^{2} h$, where $r$ is the radius of the mug and $h$ is its height.

Alternatively for a hemispherical cup, as is popular in many high-street coffee shops, the volume is $V=2\pi r^{3}/3$. Both the mug and the cup have an opening with area $2\pi r^2$.

Research suggests that the shape of the cup affects the perception of flavours, as it affects the release of volatile compounds and temperature retention—at least until the drinking process begins.

Coffee dosage: a question of statistics

The optimal amount of coffee per cup is 7–9g per 150ml of water.

This can be described by a normal distribution,
\begin{equation*} P(x)=\frac{1}{\sqrt{2\pi} \sigma}\exp\left[-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right], \end{equation*}
where $\mu= \text{8g}$ is the mean value of coffee per 150ml of water, $\sigma=\text{0.5g}$ is the standard deviation, $x$ is the variable amount of coffee per 150ml of water, and $P(x)$ is the probability density of the amount of coffee being $x$ per 150ml.

Empirical studies have shown that a dosage of 8g provides the best balance between strength and aroma, though that depends on the type of coffee. Deviations from this amount will result in either too weak or too bitter a taste. The barista must have developed ‘the right touch’ or—much easier—must know and use the normal distribution.

Solving a cubic in 1646

Elinor Flavell

19 April 2026

The year is 1646 and, as the nerd you have always been, you are reading a book about how to draw conic sections by the Dutch mathematician Frans van Schooten. However, in the back of the book there is an appendix looking at ways to solve cubic equations. This is strange.

In 2026 if I asked you to solve $x^3 = 13x + 12$, I imagine you could find a solution in just a few steps. But back in 1646 this is hard to do and mathematicians don’t like finding solutions that are negative or, God forbid, complex. In the book, Van Schooten actually describes negative solutions as false and complex solutions as fictitious.

In many ways the cubic equation is the open question of the century. While different approaches have been developed around the world, in 17th century Europe they have been battling with solutions to cubics for over a hundred years—in some cases literally (Google ‘Italian mathematics duel,’ and you won’t be disappointed). Various methods and solutions have been found for special cases, but there isn’t a simple way to find the roots of a general cubic.

So why would cubic equations show up in a book about conic sections?! It turns out that Van Schooten has found a way to represent a certain type of cubic equation using a geometrical drawing. By doing this, he is trying to bridge the gap between an exciting new branch of mathematics called ‘analytic geometry,’ cooked up by Descartes and Fermat, and the deductive geometry you have grown up with in the 1600s.

Van Schoot me a line

In 2026, you would have seen both types of geometry but probably would be far more comfortable using analytic geometry. Analytic geometry is about finding the equations of lines or circles and working out the information you need just using equations. You don’t need images or diagrams to help you. In contrast, deductive geometry relies on the fact that you have a diagram to help find your solution. You will need to construct circles, triangles and other shapes using a compass and a straight edge and then use your knowledge of angles and the properties of shapes to draw conclusions.

You may have come across deductive geometry at about the age of 14 or 15 in school where you were taught about circle theorems, and how to bisect a line or construct an equilateral triangle using a compass. I did not enjoy this topic in school. I struggled to understand why I had to use a compass and couldn’t just use my ruler to measure where the halfway point was on a line.

What my teacher didn’t say (or more probably my teenage self couldn’t comprehend) is that deductive geometry is incredibly useful and has been used by people across the world for thousands of years. It allows you to find an answer by building on what you already know. This is great if you want to find the new area of a field after the Nile floods in ancient Egypt or work out the size of container needed for a certain volume of crops.

Today, analytic geometry is taught and used much more heavily in mathematics classrooms than deductive geometry because it allows us to generalise. You don’t need to carry a compass and straight edge around everywhere you go because you can use equations to find your answer. You don’t need to draw a shape in the complex plane because you can use equations. Mathematicians love this and they often find deductive geometry hard because they just don’t have the practice!

However, I think it’s a shame not knowing what deductive geometry can do. The ideas in Van Schooten’s appendix are important because he is at the start of this change from deductive to analytic geometry. He draws together these two types of mathematics and tries to teach the reader how we can find a solution to a particular type of cubic equation.

So let’s solve a cubic equation like it’s 1646! We want to solve a cubic equation with a particular form: a depressed cubic. This means it looks something like $x^3 = px + q$. Here $x$ is our variable and $p$ and $q$ are positive real numbers. We’ll also throw in the extra condition that the square of half of the last term is less than the cube of one-third of the coefficient of the next to last term.

This is just a rather wordy way of saying that the cubic must satisfy $q^2/4 < p^3/27$. As an example, Van Schooten suggests the cubic equation $x^3 = 13x + 12$, which you can check satisfies the required properties.

Now, this seems pretty specific: no quadratic term, and that strange inequality involving $p$ and $q$. Actually, it’s not too bad: every cubic equation can be linked to a depressed cubic, so it’s only the inequality and the signs of $p$ and $q$ that hold us back.

So looking at our restriction of $q^2/4 < p^3/27$ you may immediately recognise the discriminant for a depressed cubic. You may also not. A depressed cubic (of the form $x^3-px-q = 0$) has as its discriminant the quantity $4p^3-27q^2.$ Like with a quadratic, looking at the sign of the discriminant tells us how many real roots the equation has. In our case, the requirement that $q^2/4 < p^3/27$ means the discriminant, $4p^3-27q^2$, is strictly positive. We should expect that the cubic will have three real roots (and later on we'll see that it does!).

Van Schoot your shot

Now’s the time to dig out that compass you haven’t used since you were 16, find a ruler (or straight edge), a piece of paper, and let’s get stuck in!

Let’s start by drawing a circle. Then pick two points, $F$ and $L$ on the circumference of the circle, and draw a line from $F$ to $L$, and another from $F$ going through the centre, $H$. Label the point opposite $F$ as $K$. Now mark another point, $I$, so that the distance around the circle from $L$ to $I$ is the same as the distance from $K$ to $L$.

Next, Van Schooten tells us to:

Join with straight lines the points KL, LI, and IF and continue the straight line FI outside of the circle to a point S, such that the angle FSL is equal to the angle IFL. From which SL is equal to LF, while SI is equal to FK.

That’s depicted above: the two line segments marked with blue dashes, $SL$ and $LF$, have the same length by construction, and so do the two segments shaded in green, $SI$ and $FK$.

It is less obvious that $SI$ must be the same length as $FK$, so let’s show this. Look at triangles $ILS$ and $KLF$. We know that $IL = LK$ because we have split the arc $IK$ into equal parts, and straight lines formed in this way are the same by proposition 29 in the third book of Euclid (in equal circles, equal circumferences are subtended by equal straight lines).

But also, as the angles $FKL$ and $FIL$ are opposite angles in a cyclic quadrilateral (highlighted in pink, below), they must add to 180 degrees, by proposition 22 in the 3rd book of Euclid.

So angle $LIS$ must also be the same as angle $LKF$, because $FS$ is a straight line! It may seem like a deep cut that Van Schooten is quoting Euclid but you have done the same! Any time you use a circle theorem to support your argument, or argue that a triangle is isosceles because two sides are the same length, then you are using Euclid. You can search for Euclid’s books online and you will see that they are full of these types of geometrical reasoning. Van Schooten just uses references like proposition 29, book 3 because this is how he has been taught.

Now we’ve got two triangles ($LFK$ and $ISL$) with two sides and one angle the same as each other. They must be two copies of the same triangle, which means we know that $FK$ must be the same length as $SI$, and the angles $LFK$, $LSI$, and $LFI$ must be the same as each other.

Next, Van Schooten tells us to add some more to our drawing. We add another point, $G$, to the circle, at the same distance from $I$ as $I$ was from $L$.

Then we continue:

To the same purpose extend FG to the point T until the angle FTI is equal to the angle GFI, in a similar manner TI equals IF, and TG themselves LF.

This is just repeating the logic we used a moment ago, with the triangle $FLS$: this time, it’s $LF$ and $TG$ that have the same length as each other (and also the same length as $SL$).

One more step:

Taking it further, the triangles FHL, FLS and FIT are similar: as before the ratio between HF and FL is the same as the ratio between LF and FS. Likewise for the same reason let the ratio between HF and FL the same as IF and FT.

In other words, the three shaded triangles (below) are similar, and so they all have the same ratios between their side lengths.

So we have all our triangles; it’s time to assign some values to each of the lengths.

Now, if you happen to be an ancient Greek mathematician, you will probably be saying “Elinor, Έλινορ, Elinor: we can’t assign lengths to the shapes we draw! If we do that we lose the abstractness, the essence of the shape!”

The ancient Greeks got really into doing maths in its ‘abstract form’. They wanted to build up all of geometry from only a few axioms. But being so restricted created certain problems for them. (The parallel postulate, squaring the circle and trisecting an angle may ring a bell.)

People had been using deductive geometry to find areas and volumes for thousands of years, and so of course they were using lengths and units. Although Van Schooten grew up learning this Greek mathematics, in this book he is linking deductive and analytic geometry. As such, by assigning lengths to part of the shape we are starting to think in terms of equations rather than diagrams.

Let’s keep following Van Schooten and set $HF = a$, $GF = b$, and $FL = x$.

Once we’re ready to choose them, $a$ and $b$ will be constants based on our equation, and we’ll try to work out the value of our variable $x$. Using our knowledge of the sine rule and similar triangles, we can then find a cubic equation from this diagram.

Firstly, we can conclude that $FS = x^2/a$ by using similar triangles. If the ratio between the long and short sides in triangle $FHL$ (the blue one, below) is $x/a$ and the short side in triangle $FLS$ (the pink one) has length $x$, then its long side has length $x^2/a$.

Next, we can work out the length $FI$. We know $$FS = \frac{x^2}{a},$$ and we worked out earlier that $SI$ was the same as $FK$, which we’ve now decided is equal to $2a$; so $$FI = \frac{x^2}{a}-2a$$:

Using the same ‘similar triangles’ logic, the long side in triangle $FIT$ must have length $$FT = \frac{x}{a} \left(\frac{x^2}{a}-2a\right) = \frac{x^3}{a^2}-2x.$$ Finally(!) we can work out the length of $FG$. Since $GT$ is the same length as $FL$ (in other words, length $x$), we know that, as shown below, $$FG = \frac{x^3}{a^2}-3x.$$

Now, finally, because we assigned $FG =b$, we have our cubic equation! It’s $b =x^3/a^2-3x$, which can be rearranged to $x^3 = 3a^2x + a^2b$. It even looks like the depressed cubics we met earlier, with $p=3a^2$ and $q=a^2b$.

We want to check that it’s of the correct form, that is, $$\frac{1}{4}q^2 < \frac{1}{27}p^3.$$

Substituting in the coefficients from our equation, we can see this as $a^4b^2/4<a^6$, which Van Schooten simplifies (dividing through and taking square roots) to $b<2a$.

This makes perfect sense in the context of our circle! We just need the line segment $FG$ (with length $b$) to be shorter than the circle diameter $FK$ (with length $2a$).

So we have constructed our diagram, assigned some lengths, and found a cubic equation. Great! Let’s now apply it to our particular equation, $x^3 = 13x + 12$.

Matching up $x^3 = 13x + 12$ with $x^3 = 3a^2 x + a^2b$, we need to have $a^2 = 13/3$, or in other words the radius of the circle should be $a = \sqrt{13/3}$.

We also conclude that $13b/3 = 12$, or $b =36/13$, and now all that’s left is to trisect an angle and work out what $x$ is. Easy. Cubic solved. Root found. Time to relax.

But there is one problem.

In 1646, we can’t trisect an angle. Remember earlier when I said that the Greeks struggled with some geometrical questions because they were very strict about what they could do with a ruler and compass? Well, they couldn’t trisect an angle. Mathematicians still hadn’t worked this out in 1646 and by 2026 we know that it is impossible with these strict rules. But all is not lost! Van Schooten embarks on another geometrical quest to find $x$ combining ideas from several different mathematicians. It takes him 17 pages (compared to the three pages of his book we’ve covered so far in this article), so I will spare you the details. Instead, here is another cool thing we can get out of the diagrams we’ve already drawn.

Van rooten

So far, we’ve only found one root of the cubic equation (that’s $x$). But, you might be thinking, surely there are two more? Where do they come into this? Once again, circles have the answers.

At the beginning of the appendix, Van Schooten tells us:

If you have an equilateral triangle MNL, where M, N, and L are all points on a circle and draw a line from L to a point F on the circumference in such a way that the line LF cuts MN at the point O and then draw lines MF, FN. We say that FL is equal to MF plus FN.

In the style of a university textbook, I will leave it to the reader to prove that the light and dark green lines have the same total lengths. One way to do it (the one in the book) involves spotting sets of similar triangles, which have the same ratios between the lengths of pairs of their sides.

For our specific example, $x^3 = 13x + 12$, we want each edge of the triangle to have length $\sqrt{13}$.

This side length of $\sqrt{13}$ might seem very random—but it means that the radius of the circle will be $\sqrt{13/3}$: our value of $a$, from earlier. (That’s another fact you can check: think about isosceles triangles that can be made from the radii and the edges of the original triangle.)

We can set off on another ‘similar triangles’ endeavour to show that $-FM$ and $-FN$ in the picture above are also roots of the equation! So if we can construct this triangle, we won’t just get one root of the cubic: we’ll get all three.

But we also know that the edges $FM$ and $FN$ added together make $FL$; that is, $FL = FM + FN$.

It turns out for a depressed cubic of this form, this will always be the case. This is because of an interesting relationship between the coefficients and roots of a cubic equation. If we take a general cubic of the form \[ ax^3 + bx^2 + cx + d = 0, \] with roots $\alpha, \beta, \gamma$ then $\alpha + \beta+ \gamma = -b/a$. The cubics we are looking at are all of the form $x^3 = px \pm q$. Because is no $x^2$ term, $-b/a = 0$ and so $\alpha + \beta + \gamma = 0$. This lets us conclude that the three roots of a cubic equation in this form will always sum to zero or, in other words, two of the roots sum to the negative of the third.

Van Schoot for the stars

You may think, “that was some cool maths but there is no way I would ever have come up with that approach to solving a cubic equation”. I would agree: at first I found it confusing and at points a little tedious going through every step. But I also haven’t been taught deductive geometry in the same way that Van Schooten was. I have been taught how to approach this type of question using analytic geometry and so of course I will use the method that seems easiest to me. Van Schooten is bringing together two different types of maths. We can see glimpses of the mathematics we are familiar with but there are also lots of different ideas.

This is what I find fascinating about the history of mathematics. You may not have come across this field before or realised that this is something that you can study. But it’s exactly what we’ve been doing here. We have looked into the past and followed the mathematics that somebody was doing in 1646! More generally, the history of maths looks at mathematicians from all over the world, and across time, to try to understand their work. We want to better understand different approaches to mathematics and different mathematical traditions. When you look at a range of methods and ideas, it helps you to better understand the underlying structure.

Next time you have a bit of time to spare, I encourage you to look up your favourite area of mathematics—or your favourite mathematician—to see what they were doing 380 years ago.

The beautiful soaisu partition

Kenichi Takemura

19 April 2026

Take a look at the array of numbers below. Can you tell what it is?

What a throwback

Yep, it’s the good old times table we all know and love.

Turns out, the times table we think we know inside out hides layer upon layer of astonishing structure that almost nobody has noticed. In this article I’m delighted to share just one of them with you. Right away, let’s focus on this blue area. What happens if we multiply every number in it? \[ 5 \cdot 6 \cdot 8 \cdot 14 \cdot 9 \cdot 24 \cdot 8 \cdot 36 \cdot 5 \cdot 50 \cdot 6 \cdot 60 \cdot 14 \cdot 63 \cdot 24 \cdot 64 \cdot 36 \cdot 63 \cdot 50 \cdot 60 \] Doing the calculation gives \[ 173401213127727513600000000, \] which doesn’t look obviously pretty. But if we rewrite it like this: \[ 173401213127727513600000000 = (10!)^4, \] the product is $10!$ raised to the fourth power. That’s far too suggestive to be a coincidence.

For the times tables, they are-a changin’

The key lies in a deeper property of the times table: pick one entry from each row and each column (ten numbers in total) and multiply them. The result is always $(10!)^2$:

\[ 1 \cdot 8 \cdot 9 \cdot 20 \cdot 50 \cdot 54 \cdot 42 \cdot 64 \cdot 63 \cdot 20 = 13168189440000 = (10!)^2. \]

This explains the diamond’s $(10!)^4$. See why? The diamond is not just a random shape; it is actually formed by combining two distinct sets of selections. In each set, we pick exactly one number from every row and every column.

Since the product of any such single selection is always $(10!)^2$, the product of the entire diamond region is simply \[(10!)^2 \times (10!)^2 = (10!)^4.\]

An entry in the $i$th row and $j$th column is simply the product $ij$. Selecting ten numbers such that each row and each column is represented exactly once is mathematically equivalent to choosing a permutation $\sigma$ of the set $\{1, 2, \dots, 10\}$. The product $P$ of these ten selected numbers can be written as \[ P = \prod_{i=1}^{10} (i \, \sigma(i)). \] Since $\sigma$ is a permutation (a reordering), the set $\{\sigma(1), \sigma(2), \dots, \sigma(10)\}$ is identical to the set $\{1, 2, \dots, 10\}$, just in a different order. Therefore, we can rearrange the product as \[ P = \left( \prod_{i=1}^{10} i \right) \left( \prod_{i=1}^{10} \sigma(i) \right) = 10! \times 10! = (10!)^2. \] The rows provide the `components’ $1$ through $10$, and the columns provide another set of $1$ through $10$. No matter how you pick the numbers, you are always multiplying the same set of parts. This is why the result is always $(10!)^2$.

We can pick cells from distinct rows and columns in multiple ways.

Using this fact, we can split the times table into five regions where the product of the numbers in each region is identical. The colouring above corresponds to the table below. Every colour group multiplies to the same value, \[(10!)^4 = 173401213127727513600000000.\]

And even better, the sums of the numbers in each set are also identical—each totals 605.

Isn’t it amazing that the times table can be split equally both by multiplication and addition using the same pattern? And the pattern stays unchanged under multiple $90^{°}$ rotations, left–right flips and up–down flips.

Groovy baby

The term soaisu

The term soaisu is a coinage of my own, derived from the Japanese word soai, which means mutual love or `being mutually affectionate’. I chose this name because these sets of numbers are bound together in a beautiful, harmonious relationship through their power sums—it is truly a mathematical love story! For English speakers, the pronunciation is “soh-eye-su.”

The humble number grid

But the story doesn’t end here. This is only the warm-up… The real thrill comes next. This highly symmetric structure has far more power than we might imagine. Let’s look at something even more familiar: the $10 \times 10$ number grid, where the numbers $1$ to $n^2$ are simply filled in order.

Just the numbers 1 to 100 in order. Looks plain, right? Don’t be fooled. This grid hides breathtaking beauty. To reveal it, we need a new idea: soaisu.

The products of each set in the times table are all equal to $(𝟣𝟢!)^𝟦$ … and the sums of each set are all equal to 605.

Soaisu partitions

Soaisu are special partitions of distinct numbers where the sums of powers from 1 up to $k$ are equal across the sets. Formally, sets $S_1, S_2, \dots, S_m$ (each with $n$ integers) form a strength $k$ $\underbrace{n\text{–}n\text{–}\cdots\text{–}n}_{\text{$m$ times}}$ soaisu if for every $l = 1, 2, \dots, k$,
\[ \sum_{a \in S_1} a^l = \sum_{a \in S_2} a^l = \cdots = \sum_{a \in S_m} a^l. \]

Soaisu extends the classic Prouhet–Tarry–Escott problem (equal power sums for two sets) to three or more sets. I coined the term because I want every maths lover to know about them. The strength $k$ is affectionately written with $k$ hearts, ♥.

Seeing is believing, so here’s an example. Inside a $4 \times 4$ number grid lives a strength 3 8–8 soaisu ♥♥♥:

This particular soaisu has strength 3 since we can see that the sums of the first, second and third powers remain the same for both sets $A$ and $B$ but not for the fourth powers:

Yes—the numbers 1 to 16 split perfectly into two sets with equal power sums up to the cube. Every number lover should know this.

Number grids are soaisu squares

Back to the $10 \times 10$ number grid. Astonishingly, the numbers 1 to 100 can be split into five groups where the sums of the first, second, and third powers are all equal—using the exact same partition pattern from the times table! Let us call this arrangement type I.

Here are the groups:

\begin{align*} A = \{& 1, 10, 15, 16, 24, 27, 33, 38, 42, 49, 52, 59, 63, 68, 74, 77, 85, 86, 91, 100\}, \\ B = \{& 2, 9, 11, 20, 25, 26, 34, 37, 43, 48, 53, 58, 64, 67, 75, 76, 81, 90, 92, 99\}, \\ C = \{& 3, 8, 12, 19, 21, 30, 35, 36, 44, 47, 54, 57, 65, 66, 71, 80, 82, 89, 93, 98\}, \\ D = \{& 4, 7, 13, 18, 22, 29, 31, 40, 45, 46, 55, 56, 61, 70, 72, 79, 83, 88, 94, 97\}, \\ E = \{& 5, 6, 14, 17, 23, 28, 32, 39, 41, 50, 51, 60, 62, 69, 73, 78, 84, 87, 95, 96\}, \end{align*}

and here are the sums of the powers:

When expressed in the language of soaisu, this fact is represented as a 20–20–20–20–20 soaisu ♥♥♥. This notation concisely captures the essence of the structure: five groups, each consisting of 20 elements, are firmly bound together by a `soai power’ of strength 3.

The number grid is coloured in accordingly:

The natural numbers in a grid…

… now with added colour

To think that the numbers 1 to 100 could be partitioned equally into five sets as soaisu ♥♥♥!

When I first discovered this, I could hardly believe my eyes. I was certain that such a miraculous partition must be unique. However, as is often the case in mathematics, my intuition was about to be betrayed.

Now there are two of them!

Incredibly, another version exists: type II, a pattern that appears to be the dual of type I! While the long journey to this discovery is a story for another time, this second miraculous arrangement can be obtained by tiling four copies of the original pattern together.

Four tiles of type I reveal type II.

Take a look at the tiled pattern above. The bold black border marks the type II pattern. Just like type I, this layout is invariant under rotations of $90n^{°}$ (where $n$ is an integer) and reflections.

When we apply this pattern to the $10 \times 10$ number grid, we get the colouring below.

X marks the spot: we can analyse the sets in type II as well.

The groups are

\begin{align*} A = \{&1, 10, 12, 19, 23, 28, 34, 37, 45, 46,\\ & 55, 56, 64, 67, 73, 78, 82, 89, 91, 100\}, \\ B = \{&2, 9, 13, 18, 24, 27, 35, 36, 41, 50,\\ & 51, 60, 65, 66, 74, 77, 83, 88, 92, 99\}, \\ C = \{&3, 8, 14, 17, 25, 26, 31, 40, 42, 49,\\ & 52, 59, 61, 70, 75, 76, 84, 87, 93, 98\}, \\ D = \{&4, 7, 15, 16, 21, 30, 32, 39, 43, 48,\\ & 53, 58, 62, 69, 71, 80, 85, 86, 94, 97\}, \\ E = \{&5, 6, 11, 20, 22, 29, 33, 38, 44, 47,\\ & 54, 57, 63, 68, 72, 79, 81, 90, 95, 96\}. \end{align*}

The sums of the first to third powers remain `pinned’ to the same values, as if by magic. They are:

Who would have thought such elegant symmetry — doubled, no less — lurked behind these numbers? This is pure design crafted by the numbers themselves. One reason we find symmetry beautiful is right here.

Perhaps the robust symmetry embedded within these patterns is the very key to unlocking the secrets of soaisu. Indeed, a group structure known as the dihedral group $D_4$ lies latent here.

A thrilling thought flashed through my mind: `Using this idea, might it be possible to achieve equal partitions of soaisu in grids of sizes other than $10 \times 10$?’ Many of you may have already noticed that the 8–8 soaisu ♥♥♥ pattern also possesses the same kind of symmetry.

I have been travelling through the world of soaisu for over 15 years, and the more I learn, the more profound I find their structure to be. Beyond the dihedral group, cyclic groups and symmetric groups pull the strings behind the scenes in many different ways. Moreover, soaisu find their home not only in number grids like these, but also in magic squares.

Oh dear, I could go on forever. This is becoming a long story, but let’s save that for another time. I hope to see you again soon!

Order, order!

Louie Leventhall

19 April 2026

Democracy isn’t perfect. Don’t worry readers, I won’t get too political! All I’m saying is that, mathematically, democracy isn’t perfect.

What I’m referring to is the difficulty in making group decisions: how to convert differing individual desires and wishes into a group preference. You may have experienced this when deciding with friends what movie to watch, or restaurant to go to, or what music piece is the best. Since unanimity, the agreement of everyone, is very rare, voting can be held to gauge the support of different proposals within the group and from there make a single decision.

Since voting is often synonymous with politics, let us imagine a situation where a committee of 10 people are considering three tax proposals. For the sake of simplicity let us call them $A$, $B$ and $C$. Yes, very imaginative. To make a group decision, the simplest, and perhaps most obvious way, would be using majority vote, where the proposal with at least half of the support of the committee is chosen. However, there are three proposals, so a strict binary choice cannot immediately be obtained. What can be done is pairwise majority vote, where each proposal is pitted against each other, thus ensuring a binary choice in each vote.

To illustrate this better we’ll use the standard preference notation where, for the proposals $A$ and $B$, $A$ is preferred more than $B$ is denoted $A \succ B$. For the group preference a subscript $g$ is given. Suppose the preferences of the 10 committee members are as below:

number of members	preference ordering
3	$A \succ C \succ B$
4	$C \succ B \succ A$
3	$B \succ A \succ C$

Reading this, three members prefer proposal $A$ over $C$, and proposal $C$ over $B$. Four members prefer proposal $C$ over $B$, and proposal $B$ over $A$. And the final three prefer proposal $B$ over $A$, and proposal $A$ over $C$. From this we can deduce that the group prefers $B$ over $A$ by a vote of 7 to 3, denoted $B \succ_g A$. Success! The committee prefers $B$ to $A$.

But hold on, what about proposal $C$? Looking at our table reveals that the group prefers $C$ over $B$ by a vote of 7 to 3 as well. Uh oh. But are they both still preferred over $A$? Consulting the table again reveals that $A$ is in fact preferred to $C$ by a vote of 6 to 4. So $A$ is preferred to $C$ which is preferred to $B$ which is preferred to $A$ which is $\dots$ ah.

This is called a Condorcet cycle, where no single option can win head-to-head against every other option. To visualise this cyclical nature we thankfully have preference notation: $$\cdots \; A \succ_g C \succ_g B \succ_g A \; \cdots$$

Think of a Condorcet cycle like a game of rock paper scissors

While not a common problem within voting, this intransitivity or inconsistency within group preferences can be exploited. Suppose the chair of the committee is aware of this intransitive relation and supports proposal $C$. Could the chair, by majority vote, somehow guarantee $C$ wins but $B$ and $A$ do not?

Using sequential pairwise majority voting, the chair can pair off alternatives to compete in successive runoffs. Under this method, the chair sets an agenda, an ordered list of alternatives, $A$ then $B$ then $C$. $A$ runs off against $B$ and loses, then $B$ runs off against $C$ and loses leaving $C$ the overall winner. By manipulating the order of alternatives or agenda setting, the chair can influence the final result and is therefore an agenda setter.

Arguably, though, this is not a realistic example of agenda setting.

Firstly, the committee members cannot vote on each of the proposals with respect to their views on other issues. Changing the levels of how much the government takes in taxes, will in the long run affect how much it spends and how much it borrows.

Secondly, the members are unable to express the intensity of their feelings towards a proposal. Consider a voter with the preference $A \succ B \succ C$. On the surface this preference suggests that the member prefers $A$ over $B$ just as much as they prefer $B$ over $C$, which might not be exactly what the member feels.

To address these problems, let us imagine a more general situation where voters can consider more than one political issue and express their exact level of support for a proposal. To address the first point, let us use an $n$-dimensional policy space, which houses all the proposals or policies the voters may wish to consider, each dimension representing a political issue.

$n$-dimensional space

Given $n$ is a positive integer, an ordered n-tuple is a sequence of $n$ real numbers $(x_{1}, x_{2},\dots, x_{n})$. The set of all ordered $n$-tuples is a n-dimensional real space, denoted $\mathbb{R}^n$.

Each voter will hold some point within this space to be their ideal policy, the policy they would prefer over all others.

To address the second point, each voter will also hold Euclidean preferences, where a voter’s support of a policy decreases with the distance from an ideal point. So given a choice between two alternatives, they will prefer the alternative closer to their ideal point. What makes this type of preference useful is its simplicity and that being Euclidean, based on distances between policies, preferences can be easily illustrated.

The group preference will be defined by majority vote, where the group prefers one policy over another if a majority prefers one policy over another. Can agenda setting be exerted in this context?

For simplicity, suppose five voters lie in a 2D policy space on the issues of increasing levels of taxation and spending. The status quo point $SQ$ acts as a consensus point that all have initially agreed upon.

Since this is 2D, the voters’ Euclidean preferences can be represented by a circle centred on their ideal points with $SQ$ lying on each curve. Given a choice of $SQ$ or any policy within their circle, a voter prefers the policy within their circle. If a majority of voters prefer a policy, it becomes the new consensus. For instance, any policy in the shaded petal is preferred by a majority coalition over $SQ$. As every policy in a shaded petal can be defeated by a policy in a different petal, the group preference relation is intransitive, just as we saw with the committee example before.

The intransitivity in voter preferences is something an agenda setter can exploit to their advantage. If they wished to shift the consensus from $SQ$ to $D$, since $D$ lies outside the win set, the union of all shaded petals, the group would reject it. This is denoted with the standard group preference notation, $SQ \succ_g D$. Yet, by proposing alternatives sequentially, the consensus can be gradually shifted from $SQ$ to $D$.

The figures on the right show this shift in action. Suppose the agenda setter proposes policy $A$, which lies in the petal of coalition $\{2,4,5\}$, with the aim of getting the consensus policy closer to $D$. In proposing policy $A$, which lies within the win set, $A$ is preferred by the majority $\{ 2,4,5 \}$ over $SQ$ (step 1). With a new consensus, voter preferences and thus the circle shapes change. In particular, as shown in step 2, it changes in such a way that the agenda setter cannot shift the consensus closer to $D$. However, by proposing a series of varying policies, the agenda setter could engulf the proposal $D$ within the win set. Now suppose the agenda setter proposes $B$, which lies in the petal of coalition $\{1,3,4\}$. With a majority holding $B$ over $A$, the win set changes again as shown in step 3 and the agenda setter can propose policy $C$, which lies within the petal of $\{1,2,3\}$. Once more $C$ holds majority support over $B$ so the win set changes such that it encompasses the policy $D$ as shown in step 4. By proposing more policies from the win set, eventually policy $D$ would be accepted by the group!

Preference notation allows us to write this more nicely:$$\cdots\; D \succ_g C \succ_g B \succ_g A \succ_g SQ \succ_g D \;\cdots$$

Of course, this type of manipulation fails to work if the consensus cannot be moved at all. If a policy is unable to be defeated by another, we call it a core point and the set of all core points, the core.

Consider five voters in a 1D policy space, where $SQ$ lies on voter 3’s ideal point.

Here the group would prefer $SQ$ to $A$ since $SQ$ is supported by the majority coalition $\{1,2,3\}$. In fact no other alternative in this dimension can beat $SQ$. The reason being that every potential majority coalition designed to defeat $SQ$ must contain the support of the median voter, but since $SQ$ lies on the median voter, it could never be defeated! So the median voter’s ideal point forms the core.

Beyond the 1D case though, the core is almost always non-existent. To explain this we must consider how the core can be defined through the convex hull of every majority coalition of voters.

Convex hulls

A set of points $S$ is said to be convex if we can draw a line joining any pair of points in $S$ such that the line remains entirely within $S$. The convex hull of $S$ is the smallest possible convex set that contains $S$.

Since every policy outside a winning coalition would be defeated by some point within its convex hull, the common intersection of them must form the core. Yet as the number of dimensions increases, these convex hulls are stretched out across the policy space, making it harder for them to intersect, leaving an empty core.

Under some circumstances, a core can be forced into existence. A Plott configuration is an arrangement of ideal points where pairs of these points are positioned about the median voter’s in order to force a core point. The 1D case is the simplest example of this. To create this configuration in more than one dimension:

Start by placing $n$ voter ideal points in a straight line.
If $n$ is odd, the median voter’s ideal point forms the core, if even, add a fictional median voter.

Split the remaining $n-1$ points into pairs such that the two points lie colinearly on either side of the median voter’s ideal point. The points in the pair need not be equidistant from the median.

Finally, rotate each pair through the dimensions about the median voter’s ideal point.

By overlaying every voter’s preference curve, we find that the win set is empty and can confirm the median voter’s ideal point forms a core point.

Note, however, that any slight change in ideal points can cause the configuration, and thus the core, to collapse.

The importance of the core in group decision making was demonstrated by Richard McKelvey who proved that if the core exists, preferences are transitive and if not, they are intransitive. Through this he established his seminal theorem on agenda setting.

The McKelvey chaos theorem states that given a policy space of at least two dimensions and at least three voters, each with Euclidean preferences, then in the absence of the core, there exists an agenda such that one can get from the first item of the agenda to the last in such a way that every item on the agenda will be defeated by the next item by majority vote.

In other words if the core is empty, any policy is reachable! This is what we saw in the 2D example. An agenda setter was able to gradually shift $SQ$ to $D$ since the core was empty.

This is a very powerful result which shows that even a simple binary choice is not beyond manipulation. However, please do not lose your faith in democratic process! This theorem does rely on a few assumptions for it to work:

The agenda setter holds complete information on everyone’s preferences.
Voters don’t ‘give up’ on assigning preferences and default to being indifferent between policies.
Everyone votes in accordance to their true preferences and without collusion.
Given the first and final items are fixed on the agenda, any ordering is permissible.
No constraint exists on how many items are allowed on the agenda or on how much they may differ from one another.

These conditions are relatively weak and don’t match up exactly with real world decision making.

Yet it provides a useful caution as to the type of manipulation that can occur in group decision making from an agenda setter with a sufficient amount of power. Next time you find yourself in a series of votes, keep an eye on the person leading it. Who knows? Maybe they have an agenda they wish to implement.

Taking the sting out of wicked problems

Isabel Harrison and Quinn Thomazin

19 April 2026

Colony collapse disorder is an increasingly common problem facing honeybees. Many worker bees (small female bees that gather pollen and nectar for the hive) suddenly disappear from an otherwise apparently healthy hive, causing a sudden, drastic decline. Lots of ideas have been suggested by conservationists; some are a mite concerned about parasites, while others drone on about beekeeping practices. But the buzz is that no definitive solution has yet been found to this crisis, which affects honey producers, the bees themselves and the plants that depend on them for pollination.

The dynamics of honeybee populations is an example of what is called a wicked problem: an intractable problem arising in a complex interacting system, with many contributing factors and no clear solution. To address this problem, first you have to factor in that honeybees are not all the same: there are different types of bee in the colony that have different roles and could respond differently to crises. Next comes the inter-dependence of bee and flowering plant populations, since these plants provide food for bees and depend on bees for pollination. Then there are lots of environmental factors to consider, which can come with further complications. For example, including insecticides in a model can be difficult because they affect not only the herbivores that people are trying to eradicate, but also the pollinators. To make matters worse, the ecosystems honeybees live in can be drastically altered by extreme weather and human activities, so you have to consider that as well $\dots{}$ The list of factors seems endless, yet potential intervention measures must be carefully considered from all these angles to prevent unexpected side effects from undermining conservation efforts.
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Significant figures: Emmy Noether

Sonia Balan

19 April 2026

In the early 20th century, mathematics was a world largely reserved for men. Those who sought to break these barriers confronted centuries of systemic inertia, both in institutions and in collective consciousness.

In Göttingen, David Hilbert had just begun popularising his reformist vision: a radical re-examination of the foundations of mathematics as people understood them. But amid this great optimism surrounding mathematics, women were still largely excluded from academic life.

Invited to Göttingen by Hilbert, Emmy Noether was denied a teaching post simply because she was a woman. For four years she lectured under Hilbert’s name rather than her own, as she was not permitted to teach officially. When the faculty objected, Hilbert famously responded: “I do not see that the gender of the candidate is an argument against her admission. We are a university, not a bathing establishment.”
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