The Power of Powder

• Racine carrée de 2, c’est 1,414 et des poussières… Et quelles poussières ! Des grains de sable qui empêchent d’écrire racine de 2 comme une fraction. Autrement dit, cette racine n’est pas dans Q. — Rationnel mon Q: 65 exercices de styles, Ludmilla Duchêne et Agnès Leblanc (2010)

• The square root of 2 is 1·414 and dust… And what dust! Grains of sand that stop you writing the root of 2 as a fraction. Put another way, this root isn’t in Q [the set of rational numbers].

Root Routes

Suppose a point traces all possible routes jumping half-way towards the three vertices of an equilateral triangle. A special kind of shape appears — a fractal called the Sierpiński triangle that contains copies of itself at smaller and smaller scales:

Sierpiński triangle, jump = 1/2


And what if the point jumps 2/3rds of the way towards the vertices as it traces all possible routes? You get this dull fractal:

Triangle, jump = 2/3


But if you add targets midway along each side of the triangle, you get this fractal with the 2/3rds jump:

Triangle, jump = 2/3, side-targets


Now try the 1/2-jump triangle with a target at the center of the triangle:

Triangle, jump = 1/2, central target


And the 2/3-jump triangle with side-targets and a central target:

Triangle, jump = 2/3, side-targets, central target


But why stop at simple jumps like 1/2 and 2/3? Let’s take the distance to the target, td, and use the function 1-(sqrt(td/7r)), where sqrt() is the square-root and 7r is 7 times the radius of the circumscribing circle:

Triangle, jump = 1-(sqrt(td/7r))


Here’s the same jump with a central target:

Triangle, jump = 1-(sqrt(td/7r)), central target


Now let’s try squares with various kinds of jump. A square with a 1/2-jump fills evenly with points:

Square, jump = 1/2 (animated)


The 2/3-jump does better with a central target:

Square, jump = 2/3, central target


Or with side-targets:

Square, jump = 2/3, side-targets


Now try some more complicated jumps:

Square, jump = 1-sqrt(td/7r)


Square, jump = 1-sqrt(td/15r), side-targets


And what if you ban the point from jumping twice or more towards the same target? You get this fractal:

Square, jump = 1-sqrt(td/6r), ban = prev+0


Now try a ban on jumping towards the target two places clockwise of the previous target:

Square, jump = 1-sqrt(td/6r), ban = prev+2


And the two-place ban with a central target:

Square, jump = 1-sqrt(td/6r), ban = prev+2, central target


And so on:

Square, jump = 1-sqrt(td/6.93r), ban = prev+2, central target


Square, jump = 1-sqrt(td/7r), ban = prev+2, central target


These fractals take account of the previous jump and the pre-previous jump:

Square, jump = 1-sqrt(td/4r), ban = prev+2,2, central target


Square, jump = 1-sqrt(td/5r), ban = prev+2,2, central target


Square, jump = 1-sqrt(td/6r), ban = prev+2,2, central target


Elsewhere other-accessible

Boole(b)an #2 — fractals created in similar ways

Squooh You

Here’s an interesting little puzzle:

Winnie-the-Pooh and Piglet set out to visit one another. They leave their houses at the same time and walk along the same road. But Piglet is absorbed in counting the birds overhead, and Winnie-the-Pooh is composing a new “hum,” so they pass one another without noticing. One minute after the meeting, Winnie-the-Pooh is at Piglet’s house, and 4 minutes after the meeting Piglet is at Winnie-the-Pooh’s. How long has each of them walked? — “A puzzle by S. Sefibekov” viâ Futility Closet

If you’re good at maths, you should see the answer in an intuitive instant. I’m not good at maths, so it took me much longer, because I didn’t understand what was going on. But I can explain the answer like this. Pooh is obviously walking faster than Piglet. Therefore, Pooh and Piglet can’t have met after one minute, because that would mean Pooh takes one minute to walk the distance walked by Piglet in one minute.

So let’s suppose Pooh and Piglet met after two minutes. If Pooh takes one minute to walk the distance walked by Piglet in two minutes, then Pooh is walking twice as fast as Piglet. Does that work? Yes, because Piglet walks Pooh’s distance in four minutes, which is twice as long as Pooh took. Therefore Piglet is walking twice as slowly as Pooh. It’s symmetrical and we can conclude that they did indeed meet after two minutes. Pooh then walks another minute, for three minutes in total, and Piglet walks another four minutes, for six minutes in total.

But guessing is not a good way to find the answer to the puzzle. Let’s try to reason it through properly. Suppose that Pooh and Piglet meet after one unit of time, during which Piglet has walked one unit of distance and Pooh has walked x units of distance, where x > 1. In other words, Pooh is walking x times faster than Piglet. The distances they walk before meeting are therefore in the ratio:

1 : x

Next, note that Pooh will cover the distance Piglet has already walked in 1 unit / x = 1 minute, while Piglet covers the distance Pooh has already walked in x / 1 = 4 minutes. The times they take are therefore in the ratio:

1 / x : x → 1 : x^2 → 1 : 4

And if 1 : x^2 is 1 : 4 (the ratio of the minutes they walk after meeting), then 1 : x (the ratio of the distances they walk before meeting) = 1 : √(x^2) = 1 : √4 = 1 : 2. Pooh is therefore walking 2x faster than Piglet and Piglet is walking 2x slower than Pooh. If Pooh covers Piglet’s distance in 1 minute, Piglet must have taken 2 minutes to walk that distance. And if Piglet covers Pooh’s distance in 4 minutes, Pooh must have taken 2 minutes to walk that distance.

Therefore, when they meet, each of them has been walking for 2 minutes. Pooh therefore walks 2 + 1 = 3 minutes in total and Piglet walks 2 + 4 = 6 minutes in total.

The result can be generalized for all relative speeds. Suppose that Pooh and Piglet meet after m1 minutes and that Pooh then takes m2 minutes to walk the distance Piglet walked in m1 minutes, while Piglet takes m3 minutes to walk the distance Pooh walked in m1 minutes. The time they walk before they meet, m1 minutes, is therefore supplied by this simple equation:

m1 = √(m3 / m2)

And you can call √(m3 / m2), the square root of m3 / m2, the squooh function:

m1 = √(m3 / m2) = squooh(m2,m3)

Now suppose the distance between Pooh’s and Piglet’s houses houses is 12 units of distance and that Piglet always walks at 1 unit a minute. If Pooh walks at the same speed as Piglet, i.e. 1 unit a minute, then:

After they meet, Pooh walks 6 more min = m2, Piglet walks 6 more min = m3.

How long do they walk before they meet?

m1 = m3 / m2 = 1, √1 * 6 = 6

They meet after 6 min.

Now suppose that after they meet, Pooh walks 2 more min, Piglet walks 8 more min.

Therefore, m3 / m2 = 4, √4 * 2 = 2 * 2 = 4 = m1

Therefore they walk for 4 min before they meet and Pooh walks 2x faster than Piglet.

After they meet, Pooh walks 1 more min, Piglet walks 9 more min (m3 / m2 = 9, √9 * 1 = 3)

Therefore they walk for 3 min before they meet and Pooh walks 3x faster than Piglet.

After they meet, Pooh walks 0·6 more min, Piglet walks 9·6 more min (m3 / m2 = 16, √16 * 0·6 = 4 * 0·6 = 2·4)

Therefore they walk for 2·4 min before they meet and Pooh walks 4x faster than Piglet:

After they meet, Pooh walks 0·4 more min, Piglet walks 10 more min (m3 / m2 = 25, √25 * 0·4 = 5 * 0·4 = 2)

Therefore they walk for 2 min before they meet and Pooh walks 5x faster than Piglet.

And so on.