Dan Q

Tag: maths

Build Colors from Colors with CSS Relative Color Syntax

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The feature here is that you can take a color you already have and manipulate its components. Which things you can change vary by the color space you choose, so for an RGB color you can change the red, green, blue, and alpha channels, for an HSL color you can change hue, saturation, lightness, and alpha, and for my beloved OKLCH you can change lightness, chroma, hue, and yes, opacity.

The syntax if you wanted to use this and not change anything about the color is: 

oklch(from var(--color) l c h / 1)

But of course you can change each component, either swapping them entirely as with this which sets the lightness to 20%: 

oklch(from var(--color) 20% c h / 1)

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This is really something. I was aware that new colour functions were coming to CSS but kinda dropped the ball and didn’t notice that oklch(...) is, for the most part, usable in any modern browser. That’s a huge deal!

The OKLCH colour model makes more sense than RGB, covers a wider spectrum than HSL, and – on screens that support it – describes a (much) larger spectrum, providing access to a wider array of colours (with sensible fallbacks where they’re not supported). But more than that, the oklch(...) function provides good colour adaptation.

If you’ve ever used e.g. Sass’s darken(...) function and been disappointed when it seems to have a bigger impact on some colours than others… that’s because simple mathematical colour models don’t accurately reflect the complexities of human vision: some colours just look brighter, to us, thanks quirks of biochemistry, psychology, and evolution!

This colour vision curve feels to me a little like how pianos aren’t always tuned to equal-temper – i.e. how the maths of harmonics says that should be – but are instead tuned so that the lowest notes are tuned slightly flat and the highest notes slightly sharp to compensate for inharmonicity resulting from the varying stiffness of the strings. This means that their taut length alone doesn’t dictate what note humans think they hear: my understanding is that at these extremes, the difference in the way the wave propagates within the string results in an inharmonic overtone that makes these notes sound out-of-tune with the rest of the instrument unless compensated for with careful off-tuning! Humans experience something other than what the simple maths predicts, and so we compensate for it! (The quirk isn’t unique to the piano, but it’s most-obvious in plucked or struck strings, rather than in bowed strings, and for instruments with a wide range, of which a piano is of course both!)

OKLCH is like that. And with it as a model (and a quick calc(...) function), you can tell your CSS “make this colour 20% lighter” and get something that, for most humans, will actually look “20% lighter”, regardless of the initial hue. That’s cool.

I spent way too long playing with this colour picker while I understood this concept. And now I want to use it everywhere!

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Hero of Alexandria

A little under two millennia ago1 there lived in the Egyptian city of Alexandria a Greek mathematician and inventor named Hero2, and he was a total badass who invented things that you probably thought came way later, and come up with mathematical tricks that we still use to this day3.

Inventions

Illustration of an aeolipile, showing a heated sphere or cylinder of water spraying steam in two different directions, causing it to rotate.
If you know of Hero’s inventions because of his aeolipile4,known as “Hero’s engine”, I’ve got bad news: it was probably actually invented by his predecessor Vitruvius. But Hero did come up with a way to use the technique to make a pneumatic temple door that automatically opened when you lit the fires alongside it.
Hero is variably credited with inventing, in some cases way earlier than you’re expecting:
  • automatic doors (powered either by pressure plates or by lit fires),
  • vending machines, which used the weight of a dispensed coin to open a valve and dispense holy water,
  • windmills (by which I mean wind-powered stationary machines capable of performing useful work),
  • the force pump – this is the kind of mechanism found in traditional freestanding village water pumps – for use in a fire engine,
  • float-valve and water-pressure based equilibrium pumps, like those found in many toilet cisterns, and
  • a programmable robot: this one’s a personal favourite of mine because it’s particularly unexpected – Hero’s cart was a three-wheeled contraption whose wheels were turned by a falling weight pulling on a rope, but the rope could be knotted and looped back over itself (here’s a modern reimplementation using Lego) to form a programmed path for the cart
Illustration showing mechanism of action of a fire-powered automatic door: heated air expands to displace water into a vessel whose now-increased weight pulls ropes wrapped around two wheels which pull open a pair of temple doors.
It’s just headcanon, but I choose to believe that the reason Hero needed to invent the fire extinguisher might have involved the number of “attempting to make fire do work” inventions that he came up with.

Mathematics

If you know of Hero because of his mathematical work, it’s probably thanks to his pre-trigonometric work on calculating the area of a triangle based only on the lengths of its sides.

But I’ve always been more-impressed by the iterative5 mechanism he come up with by which to derive square roots. Here’s how it works:

  1. Let n be the number for which you want to determine the square root.
  2. Let g1 be a guess as to the square root. You can pick any number; it can be 1.
  3. Derive a better guess g2 using g2 = ( g1 + n / g1 ) / 2.
  4. Repeat until gN ≈ gN-1, for a level of precision acceptable to you. The algorithm will be accurate to within S significant figures if the derivation of each guess is rounded to S + 1 significant figures.

That’s a bit of a dry way to tell you about it, though. Wouldn’t it be better if I showed you?

Put any number from 1 to 999 into the box below and see a series of gradually-improving guesses as to its square root6.

Interactive Widget

(There should be an interactive widget here. Maybe you’ve got Javascript disabled, or maybe you’re reading this post in your RSS reader?)

Maths is just one of the reasons Hero is my hero. And now perhaps he can be your hero too.

Footnotes

1 We’re not certain when he was born or died, but he wrote about witnessing a solar eclipse that we know to have occurred in 62 CE, which narrows it down a lot.

2 Or Heron. It’s not entirely certain how his name was pronounced, but I think “Hero” sounds cooler so I’m going with that.

3 Why am I blogging about this? Well: it turns out that every time I speak on some eccentric subject, like my favourite magic trick, I come off stage with like three other ideas for presentations, which leads to an exponential growth about “things I’d like to talk about”. Indeed, my OGN talk on the history of Oxford’s telephone area code was one of three options I offered to the crowd to vote on at the end of my previous OGN talk! In any case, I’ve decided that the only way I can get all of this superfluity of ideas out of my head might be to blog about them, instead; so here’s such a post!

4 If the diagram’s not clear, here’s the essence of the aeolipile: it’s a basic steam reaction-engine, in which steam forces its way out of a container in two different directions, causing the container to spin on its axis like a catherine wheel.

5 You can also conceive of it as a recursive algorithm if that’s your poison, for example if you’re one of those functional purists who always seem somehow happier about their lives than I am with mine. What’s that about, anyway? I tried to teach myself functional programming in the hope of reaching their Zen-like level of peace and contentment, but while I got reasonably good at the paradigm, I didn’t find enlightenment. Nowadays I’m of the opinion that it’s not that functional programming leads to self-actualisation so much as people capable of finding a level of joy in simplicity are drawn to functional programming. Or something. Anyway: what was I talking about? Oh, yeah: Hero of Alexandria’s derivation of square roots.

6 Why yes, of course I open-sourced this code.

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D20 with Advantage

Dungeons & Dragons players spend a lot of time rolling 20-sided polyhedral dice, known as D20s.

In general, they’re looking to roll as high as possible to successfully stab a wyvern, jump a chasm, pick a lock, charm a Duke1, or whatever.

A 'full set' of white polyhedral dice commonly-used by roleplayers - a D4, D6, D8, two D10s, a D12, and a D20 - sit half-submerged in a red liquid.
Submerging your dice set in the blood of a halfling is a sure-fire way to get luckier rolls.

Roll with advantage

Sometimes, a player gets to roll with advantage. In this case, the player rolls two dice, and takes the higher roll. This really boosts their chances of not-getting a low roll. Do you know by how much?

I dreamed about this very question last night. And then, still in my dream, I came up with the answer2. I woke up thinking about it3 and checked my working.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
2 2 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
3 3 3 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
4 4 4 4 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
5 5 5 5 5 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
6 6 6 6 6 6 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
7 7 7 7 7 7 7 7 8 9 10 11 12 13 14 15 16 17 18 19 20
8 8 8 8 8 8 8 8 8 9 10 11 12 13 14 15 16 17 18 19 20
9 9 9 9 9 9 9 9 9 9 10 11 12 13 14 15 16 17 18 19 20
10 10 10 10 10 10 10 10 10 10 10 11 12 13 14 15 16 17 18 19 20
11 11 11 11 11 11 11 11 11 11 11 11 12 13 14 15 16 17 18 19 20
12 12 12 12 12 12 12 12 12 12 12 12 12 13 14 15 16 17 18 19 20
13 13 13 13 13 13 13 13 13 13 13 13 13 13 14 15 16 17 18 19 20
14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 16 17 18 19 20
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 16 17 18 19 20
16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 17 18 19 20
17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 18 19 20
18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 19 20
19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 20
20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20
Table illustrating the different permutations of two D20 rolls and the “advantage” result (i.e. the higher of the two).

The chance of getting a “natural 1” result on a D20 is 1 in 20… but when you roll with advantage, that goes down to 1 in 400: a huge improvement! The chance of rolling a 10 or 11 (2 in 20 chance of one or the other) remains the same. And the chance of a “crit” –  20 – goes up from 1 in 20 when rolling a single D20 to 39 in 400 – almost 10% – when rolling with advantage.

You can see that in the table above: the headers along the top and left are the natural rolls, the intersections are the resulting values – the higher of the two.

The nice thing about the table above (which again: was how I visualised the question in my dream!) is it really helps to visualise why these numbers are what they are. The general formula for calculating the chance of a given number when rolling D20 with advantage is ( n2 – (n-1)2 ) / 400. That is, the square of the number you’re looking for, minus the square of the number one less than that, over 400 (the total number of permutations)4.

Why roll two dice when one massive one will do?

Knowing the probability matrix, it’s theoretically possible to construct a “D20 with Advantage” die5. Such a tool would have 400 sides (one 1, three 2s, five 3s… and thirty-nine 20s). Rolling-with-advantage would be a single roll.

'400-sided die' shown on Numberwang.
I don’t think anybody’s ever built a real 400-sided die, but Numberwang! claimed to have one.

This is probably a totally academic exercise. The only conceivable reason I can think of would be if you were implementing a computer system on which generating random numbers was computationally-expensive, but memory was cheap: under this circumstance, you could pre-generate a 400-item array of possible results and randomly select from it.

But if anybody’s got a 3D printer capable of making a large tetrahectogon (yes, that’s what you call a 400-sided polygon – you learned something today!), I’d love to see an “Advantage D20” in the flesh. Or if you’d just like to implement a 3D model for Dice Box that’d be fine too!

Footnotes

1 Or throw a fireball, recall an anecdote, navigate a rainforest, survive a poisoning, sneak past a troll, swim through a magical swamp, hold on to a speeding aurochs, disarm a tripwire, fire a crossbow, mix a potion, appeal to one among a pantheon of gods, beat the inn’s landlord at an arm-wrestling match, seduce a duergar guard, persuade a talking squirrel to spy on some bandits, hold open a heavy door, determine the nature of a curse, follow a trail of blood, find a long-lost tome, win a drinking competition, pickpocket a sleeping ogre, bury a magic sword so deep that nobody will ever find it, pilot a spacefaring rowboat, interpret a forgotten language, notice an imminent ambush, telepathically commune with a distant friend, accurately copy-out an ancient manuscript, perform a religious ritual, find the secret button under the wizard’s desk, survive the blistering cold, entertain a gang of street urchins, push through a force field, resist mind control, and then compose a ballad celebrating your adventure.

2 I don’t know what it says about me as a human being that sometimes I dream in mathematics, but it perhaps shouldn’t be surprising given I’m nerdy enough to have previously recorded instances of dreaming in (a) Perl, and (b) Nethack (terminal mode).

3 When I woke up I also found that I had One Jump from Disney’s Aladdin stuck in my head, but I’m not sure that’s relevant to the discussion of probability; however, it might still be a reasonable indicator of my mental state in general.

4 An alternative formula which is easier to read but harder to explain would be ( 2(n – 1) + 1 ) / 400.

5 Or a “D20 with Disadvantage”: the table’s basically the inverse of the advantage one – i.e. 1 in 400 chance of a 20 through to 39 in 400 chance of a 1.

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Sisyphus: The Board Game (Digital Edition)

I’m off work sick today: it’s just a cold, but it’s had a damn good go at wrecking my lungs and I feel pretty lousy. You know how when you’ve got too much of a brain-fog to trust yourself with production systems but you still want to write code (or is that just me?), so this morning I threw together a really, really stupid project which you can play online here.

Screenshot showing Sisyphus carrying a rock up a long numbered gameboard; he's on square 993 out of 1000, but (according to the rules printed below the board) he needs to land on 1000 exactly and never roll a double-1 or else he returns to the start.
It’s a board game. Well, the digital edition of one. Also, it’s not very good.

It’s inspired by a toot by Mason”Tailsteak” Williams (whom I’ve mentioned before once or twice). At first I thought I’d try to calculate the odds of winning at his proposed game, or how many times one might expect to play before winning, but I haven’t the brainpower for that in my snot-addled brain. So instead I threw together a terrible, terrible digital implementation.

Go play it if, like me, you’ve got nothing smarter that your brain can be doing today.

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Mathematics of Mid-Journey Refuelling

I love my electric car, but sometimes – like when I need to transport five people and a week’s worth of their luggage 250 miles and need to get there before the kids’ bedtime! – I still use our big ol’ diesel-burning beast. And it was while preparing for such a journey that I recently got to thinking about the mathematics of refuelling.

Car display showing "Please refuel. Range: 40ml"
I don’t know why you’d measure range in millilitres in the first place, but I’m hearing that I ought to fill up the car before we go.

It’s rarely worth travelling out-of-your-way to get the best fuel prices. But when you’re on a long road trip anyway and you’re likely to pass dozens of filling stations as a matter of course, you might as well think at least a little about pulling over at the cheapest.

You could use one of the many online services to help with this, of course… but assuming you didn’t do this and you’re already on the road, is there a better strategy than just trusting your gut and saying “that’s good value!” when you see a good price?

It turns out this is an application for the Secretary Problem (and probably a little more sensible than the last time I talked about it!).

A woman's hand reaches for one of four fuel pump nozzles. Photo by Gustavo Fring, used under the Pexels License.
If you can’t decide which nozzle is best, mix up a cocktail of them all. #TerribleProTip

Here’s how you do it:

  1. Estimate your outstanding range R: how much further can you go? Your car might be able to help you with this. Let’s say we’ve got 82 miles in the tank.
  2. Estimate the average distance between filling stations on your route, D. You can do this as-you-go by counting them over a fixed distance and continue from step #4 as you do so, and it’ll only really mess you up if there are very few. Maybe we’re on a big trunk road and there’s a filling station about every 5 miles.
  3. Divide R by D to get F: the number of filling stations you expect to pass before you completely run out of fuel. Round down, obviously, unless you’re happy to push your vehicle to the “next” one when it breaks down. In our example above, that gives us 16 filling stations we’ll probably see before we’re stranded.
  4. Divide F by e to get T (use e = 2.72 if you’re having to do this in your head). Round down again, for the same reason as before. This gives us T=5.
  5. Drive past the next T filling stations and remember the lowest price you see. Don’t stop for fuel at any of these.
  6. Keep driving, and stop at the first filling station where the fuel is the same price or cheaper than the cheapest you’ve seen so far.
Dan sitting in the driving seat of a car, doing maths on a portable whiteboard.
Obviously you should take care doing maths on the road. Don’t drink and derive!

This is a modified variant of the Secretary Problem because it’s possible for two filling stations to have the same price, and that’s reflected in the algorithm above by the allowance for stopping for fuel at the same price as the best you saw during your sampling phase. It’s probably preferable to purchase sub-optimally than to run completely dry, right?

Of course, you’re still never guaranteed a good solution with this approach, but it maximises your odds. Your own risk-assessment might rank “not breaking down” over pure mathematical efficiency, and that’s on you.

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Here’s What a Googol-to-One Gear Ratio Looks Like

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To celebrate being alive for a billion seconds, Daniel Bruin built a machine with 100 gears with a 10-to-1 gear ratio…meaning that the overall gear ratio is a googol-to-one. (A googol is 1 with 100 zeros.)

To turn the last gear in this train one full revolution, you’d need to turn the first gear 10,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000,​000 times.

By my estimation, that’s enough gearing to allow you to winch the entire solar system, by hand, with ease. Assuming you can find a tow hitch on it somewhere.

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Who finished second?

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Three athletes (and only three athletes) participate in a series of track and field events. Points are awarded for 1st, 2nd, and 3rd place in each event (the same points for each event, i.e. 1st always gets “x” points, 2nd always gets “y” points, 3rd always gets “z” points), with x > y > z > 0, and all point values being integers.
The athletes are named: Adam, Bob, and Charlie.

  • Adam finished first overall with 22 points.
  • Bob won the Javelin event and finished with 9 points overall.
  • Charlie also finished with 9 points overall.

Question: Who finished second in the 100-meter dash (and why)?

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I enjoyed this puzzle so much that I shared it with (and discussed it at length with) my smartypants puzzle-sharing group. Now it’s your turn. The answer, plus a full explanation, can be found on the other side of the link, but I’d recommend that you try to solve it yourself first. If it seems impossible at first glance, start by breaking it down into what you can know, and what you can almost know, and work from there. Good luck!

And if anybody feels like hiring Nick to come and speak anywhere near where I am, that’d be awesome of you.

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This equation will change how you see the world

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Normally I find Veritasium’s videos to be… less mindblowing than their titles would aim to have me believe. But I found this one pretty inspiring; the first Feigenbaum constant is a proper headtrip. And I feel like I’ve got new insights into the Mandelbrot set too.

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Elf Chalkboard Puzzle

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I write the integers 1-9999 (inclusive) on a huge chalkboard. Each number is written once.

During the night the board is visited by a series of naughty math elves (it’s a thing!)

Each elf approaches the board, selects two numbers at random, erases them, and replaces them with a new number that is the absolute difference of the two numbers erased.

This vandalism continues all night until there is just one number remaining.

I return to the board the next morning and find the single number of the board. The question is: Is this remaining number odd or even?

Elf Chalkboard…

A fun, lightweight maths puzzle for your amusement. I was able to find the right answer pretty quickly by spotting the pattern; it took me longer to find the words to adequately explain the pattern.

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Going Critical

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If you’ve spent any time thinking about complex systems, you surely understand the importance of networks.
Networks rule our world. From the chemical reaction pathways inside a cell, to the web of relationships in an ecosystem, to the trade and political networks that shape the course of history.
Or consider this very post you’re reading. You probably found it on a social network, downloaded it from a computer network, and are currently deciphering it with your neural network.
But as much as I’ve thought about networks over the years, I didn’t appreciate (until very recently) the importance of simple diffusion.
This is our topic for today: the way things move and spread, somewhat chaotically, across a network. Some examples to whet the appetite:
  • Infectious diseases jumping from host to host within a population
  • Memes spreading across a follower graph on social media
  • A wildfire breaking out across a landscape
  • Ideas and practices diffusing through a culture
  • Neutrons cascading through a hunk of enriched uranium
A quick note about form.
Unlike all my previous work, this essay is interactive. There will be sliders to pull, buttons to push, and things that dance around on the screen. I’m pretty excited about this, and I hope you are too.
So let’s get to it. Our first order of business is to develop a visual vocabulary for diffusion across networks.

A simple model

I’m sure you all know the basics of a network, i.e., nodes + edges.
To study diffusion, the only thing we need to add is labeling certain nodes as active. Or, as the epidemiologists like to say, infected:
This activation or infection is what will be diffusing across the network. It spreads from node to node according to rules we’ll develop below.
Now, real-world networks are typically far bigger than this simple 7-node network. They’re also far messier. But in order to simplify — we’re building a toy model here — we’re going to look at grid or lattice networks throughout this post.
(What a grid lacks in realism, it makes up for in being easy to draw ;)
Except where otherwise specified, the nodes in our grid will have 4 neighbors, like so:
And we should imagine that these grids extend out infinitely in all directions. In other words, we’re not interested in behavior that happens only at the edges of the network, or as a result of small populations.
Given that grid networks are so regular, we can simplify by drawing them as pixel grids. These two images represent the same network, for example:
Alright, let’s get interactive.
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Fabulous (interactive! – click through for the full thing to see for yourself) exploration of network interactions with applications for understanding epidemics, memes, science, fashion, and much more. Plus Kevin’s made the whole thing CC0 so everybody can share and make use of his work. Treat as a longread with some opportunities to play as you go along.

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Follow-up: I found two identical packs of Skittles, among 468 packs with a total of 27,740 Skittles

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Two identical packs of Skittles

Applied mathematics at its… best? After predicting statistically that it would take 400-500 packets of Skittles before you’d expect to find the same permutation of colours, an experiment finds empirical backing for this answer at pack number 464.

Somebody get the Ig Nobel Prize folks on the line.

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Going beyond the Golden Ratio

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I show that for the same reason that the golden ratio, ϕ=1.6180334.., can be considered the most irrational number, that 1 + √2 can be considered the 2nd most irrational number, and indeed why (9 + √221)/10 can be considered the 3rd most irrational number.

Let us imagine a game between two kids, Emily and Sam – both strong and determined in their own way who spend their entire lives trying to outwit each other, instead of doing their homework. (A real life Generative Adversial Network…)

Emily, proudly reminds us that she simultaneously bears the same first name as Emily Davison, the most famous of British suffragettes; Emily Balch, Nobel Peace Prize laureate; Emilie du Chatelet, who wrote the first French translation and commentary of Isaac Newton’s “Principia”; Emily Roebling, Chief Engineer of New York’s iconic Brooklyn Bridge; Emily Bronte author of Wuthering Heights; Emily Wilson, the first female editor of ‘New Scientist‘ publication; and also Emmy Noether, who revolutionized the field of theoretical physics.

On the other side we have Sam (and all his minion friends, who are aptly called Sam-002, Sam-003, Sam-004) who is part human / part robot and plays Minecraft or watches Youtube, 24/7.

They agree to play a game where Emily thinks of a number, and then Sam (with the possible help of his minions) has 60 seconds to find any fractions that are equal to Emily’s number.

And so the game begins…

Emily says “8.5″.

Sam & friends quickly reply with “85/10″,… “34/4″,…  “17/2″,… “425/50″,…

They soon realize that all these answers are equally valid because they are all equivalent fractions. Being competitive they want to pick a single winner, so they all agree that the best answer is the one with the lowest denominator. And so, 17/2 is deemed the best answer.

This time, Emily tries to make it harder by picking ‘0.123456‘. After only a slight pause, Sam  slyly says “123456/1000000“.

Emily’s annoyed with this answer. She knows that although the best answer would be the irreducible fraction 1929/15625, Sam’s answer is still valid answer, and furthermore he will always be able to instantly answer like this if she picks any number with a terminating decimal expansion.

So this time, Emily picks “π”.

…

Delightful exploration of the idea that while all irrational numbers are irrational, some can be considered more-irrational than others if you consider the complexity of the convergent series of fractions necessary to refine the representation of it. Some of this feels to me like the intersection between meta-mathematics and magic, but it’s well-written enough that I was able to follow along all the way to the end and I think that you should give it a go, too.

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The most unexpected answer to a counting puzzle

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Summary: if an idealised weight slides into another, bouncing it off a wall then back into itself, how many times will the two collide? If the two weights are the same then the answer is 3: the first collision imparts all of the force of the first into the second, the second collision is the second bouncing off the wall, and the third imparts the force from the second back into the first. If the second weight weighs ten times as much as the first, the answer turns out to be 31. One hundred times as much, and there are 314 bounces. One thousand times, and there are 3,141. Ten thousand times, and there are 31,415… spot the pattern? The number of bounces are the digits of pi.

Why? This is mindblowing. And this video doesn’t answer the question (completely): it only poses it. But I’ll be looking forward to the next episode’s explanation…

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How Few Votes Can Win a UK General Election?

Here’s a thought: what’s the minimum number of votes your party would need to attract in order to be able to secure a majority of seats in the House of Commons and form a government? Let’s try to work it out.

The 2017 general election reportedly enjoyed a 68.8% turnout. If we assume for simplicity’s sake that each constituency had the same turnout and that votes for candidates other than yours are equally-divided amongst your opposition, that means that the number of votes you need to attract in a given constituency is:

68.8% × the size of its electorate ÷ the number of candidates (rounded up)

For example, if there was a constituency of 1,000 people, 688 (68.8%) would have voted. If there were 3 candidates in that constituency you’d need 688 ÷ 3 = 229⅓, which rounds up to 230 (because you need the plurality of the ballots) to vote for your candidate in order to secure the seat. If there are only 2, you need 345 of them.

Breakdown of a 1,000-person electorate in which 688 vote, of which 345 vote for your candidate.
It would later turn out that Barry and Linda Johnson of 14 West Street had both indented to vote for the other candidate but got confused and voted for your candidate instead. In response, 89% of the nation blame the pair of them for throwing the election.

The minimum number of votes you’d need would therefore be this number for each of the smallest 326 constituencies (326 is the fewest number of seats you can hold in the 650-seat House of Commons and guarantee a strict majority; in reality, a minority government can sometimes form a government but let’s not get into that right now). Constituencies vary significantly in size, from only 21,769 registered voters in Na h-Eileanan an Iar (the Western Isles of Scotland, an SNP/Labour marginal) to 110,697 in the Isle of Wight (which flip-flops between the Conservatives and the Liberals), but each is awarded exactly one seat, so if we’re talking about the minimum number of votes you need we can take the smallest 326.

Map showing the 326 UK constituencies with the smallest electorate size
Win these constituencies and no others and you control the Commons, even though they’ve tiny populations. In other news, I think this is how we end up with a SNP/Plaid coalition government.

By my calculation, with a voter turnout of 68.8% and assuming two parties field candidates, one can win a general election with only 7,375,016 votes; that’s 15.76% of the electorate (or 11.23% of the total population). That’s right: you could win a general election with the support of a little over 1 in 10 of the population, so long as it’s the right 1 in 10.

Part of my spreadsheet showing how I calculated this
I used a spreadsheet and everything; that’s how you know you can trust me. And you can download it, below, and try it for yourself.

I’ll leave you to decide how you feel about that. In the meantime, here’s my working (and you can tweak the turnout and number-of-parties fields to see how that affects things). My data comes from the following Wikipedia/Wikidata sources: [1], [2], [3], [4], [5] mostly because the Office of National Statistics’ search engine is terrible.

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