## 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.

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.

## 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.

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!).

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.

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.

× × ×

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

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.

## Who finished second?

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)?

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.

## This equation will change how you see the world

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.

## Elf Chalkboard Puzzle

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?

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

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
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

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

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

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.

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.

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.

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

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.