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.

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.

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.

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.

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.

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…

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.

Lucy`s Secret Number puzzle

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Lucy’s Secret Number puzzle (datagenetics.com)

You are at a party and overhear a conversation between Lucy and her friend.

In the conversation, Lucy mentions she has a secret number that is less than 100.

She also confesses the following information:

“The number is uniquely describable by the answers to the following four questions:”

Q1) Is the number divisible by two?
Q2) Is the number divisible by three?
Q3) Is the number divisible by five?
Q4) Is the number divisible by seven?

I loved this puzzle. I first solved it a brute-force way, with Excel. Then I found increasingly elegant and logical solutions. Then I shared it with some friends: I love it! Go read the whole thing.

Voronoi Diagrams

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Imagine I’m in a desert, and there are two wells where I can obtain water. If I want to go to the nearest well, which well do I visit? Clearly, it depends one where I am standing. It’s possible to draw a line dividing the desert. To the ‘left’ of the line, it’s nearer to go to the well on the well on the ‘left’, to the ‘right’ of the line, it’s closer to go to the well on the ‘right’…

Calculating Pi (when you’re ill)

So, I’ve not been well lately. And because a few days lying on my back with insufficient mental stimulation is a quick route to insanity for me, I’ve been trying to spend my most-conscious moment doing things that keep my brain ticking over. And that’s how I ended up calculating pi.

Dan, shortly before inpatient admission but already recovering from the worst parts of his hospital visit, last week.
When I say I’ve been unwell, that might be an understatement. But we’ll get to that another time.

Pi (or π) is, of course, the ratio of the circumference of a circle to its diameter, for every circle. You’ll probably have learned it in school as 3.14, 3.142, or 3.14159, unless you were one of those creepy kids who tried to memorise a lot more digits. Over the years, we’ve been able to calculate it to increasing precision, and although there’s no practical or theoretical reason that we need to know it beyond the 32 digits worked out by Ludolph van Ceulen in the 16th Century, it’s still a fascinating topic that attracts research and debate.

Graph illustrating the calculation of digits of pi over the millenia. Note the logarithmic scale on the left and the staggered scale on the bottom axis.
Our calculation of pi has rocketed since the development of the digital computer.

Most of the computer-based systems we use today are hard to explain, but there’s a really fun computer-based experimental method that can be used to estimate the value of pi that I’m going to share with you. As I’ve been stuck in bed (and often asleep) for the last few days, I’ve not been able to do much productive work, but I have found myself able to implement an example of how to calculate pi. Recovery like a nerd, am I right?

A "pi pie", from a Pi Day celebration.
Pi goes on forever. Pie, sadly, comes to an end.

Remember in school, when you’ll have learned that the formula to describe a circle (of radius 1) on a cartesian coordinate system is x2 + y2 = 1? Well you can work this backwards, too: if you have a point on a grid, (x,y), then you can tell whether it’s inside or outside that circle. If x2 + y2 < 1, it’s inside, and if x2 + y2 > 1, it’s outside. Meanwhile, the difference between the area of a circle and the area of a square that exactly contains it is π/4.

A circle of radius 1 at the intersection of the axes of a cartesian coordinate system.
Think back to your school days. Ever draw a circle like this? Do the words “cartesian coordinates” ring any bells?

Take those two facts together and you can develop an experimental way to determine pi, called a Monte Carlo method. Take a circle of radius 1 inside a square that exactly contains it. Then randomly choose points within the square. Statistically speaking, these random points have a π/4 chance of occuring within the circle (rather than outside it). So if we take the number of points that lie within the circle, divide that by the total number of points, and then multiply by 4, we should get something that approaches the value of pi. You could even do it by hand!

Output of Dan's demonstration of the Monte Carlo method as used to approximate the value of pi.
I wrote some software to do exactly that. Here’s what it looks like – the red points are inside the circle, and the black points are outside.

The software illustration I’ve written is raw Javascript, HTML, and SVG, and should work in any modern web browser (though it can get a little slow once it’s drawn a few thousand points!). Give it a go, here! When you go to that page, your browser will start drawing dots at random points, colouring them red if the sum of the squares of their coordinates is less than 1, which is the radius of the circle (and the width of the square that encompasses it). As it goes along, it uses the formula I described above to approximate the value of pi. You’ll probably get as far as 3.14 before you get bored, but there’s no reason that this method couldn’t be used to go as far as you like: it’s not the best tool for the job, but it’s super-easy to understand and explain.

Oh, and it’s all completely open-source, so you’re welcome to take it and do with it what you wish. Turn off the graphical output to make it run faster, and see if you can get an accurate approximation to 5 digits of pi! Or slow it down so you can see how the appearance of each and every point affects the calculation. Or adapt it into a teaching tool and show your maths students one way that pi can be derived experimentally. It’s all yours: have fun.

And I’ll update you on my health at some other point.