This blog is about Simon, a young gifted mathematician and programmer, who had to move from Amsterdam to Antwerp to be able to study at the level that fits his talent, i.e. homeschool. Visit https://simontiger.com
At the beginning of each year, I make a very long, ridiculous video (which I like to call a “magnum opus”). Last year it was the 2048 cookies project. This year, it got so long that I had to split it into parts. In this video (parts 1 to 4), I set up to prove that there are 13 Archimedean solids. Link to my text file with valid and invalid corners: https://repl.it/@simontiger/ArchimedeanSolids#corners.txt
Simon had been working on this proof for many days. At first he didn’t know where to start, he only knew he wanted to prove that there can be only 13 Archimedean solids by building them with Geomag. He then came up with the idea to write a Python code that would search for all the possible corners that solids could have. It was a project in its own right to come up with a good algorithm for such a code! And after his code delivered a little over 120 possible corners, Simon started building solids with those corners coming up with logical arguments to eliminate invalid ones along the way. For example, he discovered this general rule to build an Archimedean solid:
Geomag has been ruling the house for a while here, allowing Simon to both explore shapes and electromagnetism in a tactile way and just try out some architectural concepts:
The above shape looks like two pentagonal bipyramids combined but they’re not! That’s because the height of a pentagonal bipyramid is not the same as its length, it’s actually slightly longer.
Reading up on why triangles make stable constructions and why truss is better than beam in our currently favorite bedtime book, Math with Bad Drawings by Ben Orlin.
And this is probably the closest we’re getting to New York this year, Manhattan Bridge and Statue of Liberty visualized below (our trip booked for May last year got cancelled) 😏
Simon’s latest obsession is assembling gears so that they speed up exponentially. He has been inspired by this creator from Finland called Brick Experiment Channel, especially this presentation visualizing Googol (showing how a gear train using 186 Lego gears can slow the last gear by over a Googol times). It’s mind blowing that a physical and relatively simple visualization of this giant number is even possible. Simon has been fantasizing that if one would power a similar gear train the other way around (powering the slowest gear), one would theoretically be able to speed the fastest gear to the speed of light. Realizing, of course, that the more plausible scenario would be the gear train breaking down 😃
Simon blew the dust off of his old LegoWeDo set to use its parts to build gear trains powered by programmable motors, but that’s way too few parts. So now he’s busy with scavenging the internet for more LEGO Technic parts like gears and motors and power units.
Our hour of snow this winter. Had been anticipating this for months. I really miss snow, considering my Russian childhood/youth and all, but the kids treat it more like an interesting rare phenomenon, en passant. Half an hour later, Simon was already saying he’d had enough and could we please go home already
Simon’s got a new collab going and possibly what is a beginning of a real friendship with a peer from Bangalore, India. They have already finished several wonderful projects together that have been merged on The Coding Train GitHub:
Yesterday we visited the forest lake where Steven’s grandparents’ and, since recently, his dear aunt’s ashes are spread. To me, it feels like such a peaceful and dignified way to honor the dead, so much more so than visiting a cemetery and consoling in an almost regenerative way.
I was looking up the lyrics of The Twelve Days of Christmas song (you know, the one that goes “On the first day of Christmas, my true love sent to me…”) and Simon said he actually wanted to know how many presents the lucky singer got from her true love. Because she got one partridge in a pear tree on the first day, then two turtle doves and another partridge in a pear tree on the second day, then three French hens, another pair of turtle doves and yet another partridge in a pear tree on the third day, and then on the fourth day, she got four calling birds, and – you guessed it – three more French hens, two more doves and – why did he keep doing that to her? – another freaking partridge! In another pear tree! In the end, she must have had 12 partridges in 12 trees, 22 doves, 30 French hens and 36 calling birds. That’s a lot of bird poop! And the true love was only just getting started, that was all just a warm-up. Luckily, he stepped over from poultry to gold on day five, so the girl at least got some durable assets added in there: 40 golden rings (which is also strange as she couldn’t possibly wear the rings on her toes). But what did that bastard do next? He started throwing birds at her again! 42 geese a-laying (so more geese coming on top of our calculation!) and 42 swans a-swimming (I guess they came with an inflatable pool of some sort). By this time it was already one loud stinking mess, but it’s what he did next that was absolutely unacceptable: he started propping her home with hoards of strangers! He gave her 40 maids a-milking (did they come with a cow each?), 36 ladies dancing (was that a form of human trafficking?) and topped those with 30 lords a-leaping (I don’t even dare to picture those, especially since they didn’t get enough ladies). And just when she couldn’t take it anymore and probably tried begging him for a quantum of solace and quiet, he shove 22 pipers piping in there and 12 drummers! At that point they probably didn’t even hear each other anymore. It’s actually quite a nightmare of a story, if you think about it. Possibly even an early attempt of critique on consumerism.
But funnies aside, Simon came up with a beautiful geometrical visualization of how to calculate the total number of gifts that quickly yielded a formula (the tetrahedral numbers formula!), akin to the famous Gauss trick (Simon did the writing from here):
With the Gauss trick, when you have a sum like 1 + 2 + 3 + 4 + 5 + 6, you first write it down backwards, like 6 + 5 + 4 + 3 + 2 + 1. Then you add the two sums together, and then you get 7 + 7 + 7 + 7 + 7 + 7. That’s just 6 * 7! Though we do have to divide through by 2, because, remember, this was not the original sum. It was the original sum PLUS the original sum backwards. Anyway, this gives 6*7/2, which is 28. If you do this again, but in general, you get a formula of n(n+1)/2.
Now, how can we extend this to the 12 days of Christmas? It’s actually a bit more complicated than you think. For the 12 days of Christmas, it’s not just a sum of the numbers 1 through 12. No, on the first day you get just 1 present, a partridge. On the second day, you get 2 turtle doves and 1 partridge, which is 1 + 2. On the third day, you get 3 French hens, 2 turtle doves, and 1 partridge in a pear tree. That’s 1 + 2 + 3. And so on. So the actual sum we want to calculate is 1 plus 1 + 2 plus 1 + 2 + 3 plus 1 + 2 + 3 + 4 plus and so on.
Now we’re ready to apply the modified version of the Gauss trick. We can arrange the sum in a triangle:
Now, remember how we wrote the sum backwards with the normal Gauss trick? Well, here, we can rotate the triangle in 2 different ways:
3 triangles. We can add them together:
We get all 6s! How many 6s are there? Well, it’s a triangle of 6s, where the side length is the number that we started with, which in this case is 4. Applying the normal Gauss trick to that we get 10. So there are 10 6s! We can calculate that 10*6 is 60. This is not the answer, though. Remember in the normal Gauss trick when we had to divide by 2, because we added 2 copies of the sum? Well, here, we added 3 copies of the triangle, so we have to divide through by 3. This gives 10*6/3, which is 20.
We can do this in general, with n instead of 4. In that case, we get n(n+1)/2 (n+2)’s. So we get n(n+1)(n+2)/2. After dividing by 3, we get n(n+1)(n+2)/6. So that’s the general formula!
So, drumroll (that’s those 12 drummers still going!) and let’s plug the song’s numbers in: 12(12+1)(12+2)/6 equals… 364!
That’s like getting a present every day of the year. Except for one lucky day. Because, let’s face it, you would return most of those presents, wouldn’t you?
\* I hope no one notices the photoshops that we did on some of the images.
Happy Winter Solstice! Last night, we made these solstice cookies, one for every hour. Symbolizing 7 hours 52 minutes of light and 16 hours 08 minutes of dark at where we are. We also made Saturn to mark the rare Saturn-Jupiter conjunction observable these few nights, but it had already been eaten by the time we took this picture.
Neva invented this game for us and we have had a late night candle light “party” playing it (just the three of us, Dad already gone to bed). We laughed so hard our belly muscles ached. It’s so simple but so much fun and a guaranteed cosy together time.
To play the game, give every player two identical sheets of paper and have them write a word or a phrase that fills the gap in this sentence: Write as many _____ as possible. This same sentence is pictured in Dutch below:
So, for example, Simon wrote down “ways to know down a snow wall” and Neva wrote down “ways to warm up your water”. All the topics written down should have something to do with winter.
The sheets of paper are then collected and shuffled. Every round, players pull one sheet from the pile, at random, and that’s what they’ve got to do: write down as many (whatever this particular sheet says) as possible! Just one player pulls the sheet and reads it aloud and then all the players quickly write down as many things as possible that fit that particular category. They can be whimsical and unrealistic, too. Like I wrote down “use a fire spitting dragon” for warming up water and Neva had “walk around with a radiator”. For younger kids, no writing down is necessary, as long as they remember a few of their options after the others are done writing. Whoever has more options, wins, but it’s more about coming up with the weirdest scenarios than points, as you’ll soon see!
And, of course, it doesn’t have to be winter themed. It can be anything, but do keep it themed. As Richard Feynman famously reminded us, constraints unleash creativity.
Last night, using simple logic, Simon proved to me why a two-candidate plurality voting system is better than many others.
There’s a theorem called Arrow’s Theorem which says that any ranked voting system with 3 or more candidates cannot have all of the following three properties: * Non-dictatorship: A dictatorship is a system where one person decides the entire election, even if everyone else is against them. This should obviously not be the case. * Unanimity: If literally everybody ranks candidate A over candidate B, the results should also rank candidate A over candidate B. * Independence of Irrelevant Alternatives: If some people just change the ranking of C, that shouldn’t affect A or B in the results. The pictures are of me trying to prove this theorem. Each of these properties are great for a voting system to have. Removing any one of these properties will lead to strategic voting (intentionally lying in the hope that that helps your favorite candidate).
Note that plurality voting is also a ranked voting system; it just only takes advantage of voters’ 1st-place candidates, not the full rankings.
So what to do if you have more than 2 candidates?
Score voting (based on giving every candidate an individual rating) is the optimal system in that case. There is also approval voting (which is actually just a simpler version of score voting), but people tend to mistake it for plurality voting, which, as I said already, is bad.
In The Netherlands (elections coming up in March!), we have sort of a rescaled version of plurality voting (as Simon has defined it), when the seats in the parliament are distributed among many parties according to their ranking across the country. So the Dutch system doesn’t withstand the criticism of Arrow’s Theorem!
I just want to be clear here. It doesn’t matter how you try to improve the voting system (for example with rounds, or like in the US, the Electoral College), as long as it’s a ranked system with more than two candidates, it’ll fall victim to Arrow’s Theorem.