Performativizing the Polygonic #2

Suppose a café offers you free drinks for three days. You can have tea or coffee in any order and any number of times. If you want tea every day of the three, you can have it. So here’s a question: how many ways can you choose from two kinds of drink in three days? One simple way is to number each drink, tea = 1, coffee = 2, then count off the choices like this:


1: 111
2: 112
3: 121
4: 122
5: 211
6: 212
7: 221
8: 222

Choice #1 is 111, which means tea every day. Choice #6 is 212, which means coffee on day 1, tea on day 2 and coffee on day 3. Now look at the counting again and the way the numbers change: 111, 112, 121, 122, 211… It’s really base 2 using 1 and 2 rather than 0 and 1. That’s why there are 8 ways to choose two drinks over three days: 8 = 2^3. Next, note that you use the same number of 1s to count the choices as the number of 2s. There are twelve 1s and twelve 2s, because each number has a mirror: 111 has 222, 112 has 221, 121 has 212, and so on.

Now try the number of ways to choose from three kinds of drink (tea, coffee, orange juice) over two days:


11, 12, 13, 21, 22, 23, 31, 32, 33 (c=9)

There are 9 ways to choose, because 9 = 3^2. And each digit, 1, 2, 3, is used exactly six times when you write the choices. Now try the number of ways to choose from three kinds of drink over three days:


111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333 (c=27)

There are 27 ways and (by coincidence) each digit is used 27 times to write the choices. Now try three drinks over four days:


1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221, 1222, 1223, 1231, 1232, 1233, 1311, 1312, 1313, 1321, 1322, 1323, 1331, 1332, 1333, 2111, 2112, 2113, 2121, 2122, 2123, 2131, 2132, 2133, 2211, 2212, 2213, 2221, 2222, 2223, 2231, 2232, 2233, 2311, 2312, 2313, 2321, 2322, 2323, 2331, 2332, 2333, 3111, 3112, 3113, 3121, 3122, 3123, 3131, 3132, 3133, 3211, 3212, 3213, 3221, 3222, 3223, 3231, 3232, 3233, 3311, 3312, 3313, 3321, 3322, 3323, 3331, 3332, 3333 (c=81)

There are 81 ways to choose and each digit is used 108 times. But the numbers don’t have represent choices of drink in a café. How many ways can a point inside an equilateral triangle jump four times half-way towards the vertices of the triangle? It’s the same as the way to choose from three drinks over four days. And because the point jumps toward each vertex in a symmetrical way the same number of times, you get a nice even pattern, like this:

vertices = 3, jump = 1/2


Every time the point jumps half-way towards a particular vertex, its position is marked in a unique colour. The fractal, also known as a Sierpiński triangle, actually represents all possible choices for an indefinite number of jumps. Here’s the same rule applied to a square. There are four vertices, so the point is tracing all possible ways to choose four vertices for an indefinite number of jumps:

v = 4, jump = 1/2


As you can see, it’s not an obvious fractal. But what if the point jumps two-thirds of the way to its target vertex and an extra target is added at the centre of the square? This attractive fractal appears:

v = 4 + central target, jump = 2/3


If the central target is removed and an extra target is added on each side, this fractal appears:

v = 4 + 4 midpoints, jump = 2/3


That fractal is known as a Sierpiński carpet. Now up to the pentagon. This fractal of endlessly nested contingent pentagons is created by a point jumping 1/φ = 0·6180339887… of the distance towards the five vertices:

v = 5, jump = 1/φ


With a central target in the pentagon, this fractal appears:

v = 5 + central, jump = 1/φ


The central red pattern fits exactly inside the five that surround it:

v = 5 + central, jump = 1/φ (closeup)


v = 5 + c, jump = 1/φ (animated)


For a fractal of endlessly nested contingent hexagons, the jump is 2/3:

v = 6, jump = 2/3


With a central target, you get a filled variation of the hexagonal fractal:

v = 6 + c, jump = 2/3


And for a fractal of endlessly nested contingent octagons, the jump is 1/√2 = 0·7071067811… = √½:

v = 8, jump = 1/√2


Previously pre-posted:

Performativizing the Polygonic

Breeding Bunnies

Front cover of The Golden Ratio by Mario Livio
The Golden Ratio: The Story of Phi, the Extraordinary Number of Nature, Art and Beauty, Mario Livio (Headline Review 2003)

A good short popular guide to perhaps the most interesting, and certainly the most irrational, of all numbers: the golden ratio or phi (φ), which is approximately equal to 1·6180339887498948482… Prominent in mathematics since at least the ancient Greeks and Euclid, phi is found in many places in nature too, from pineapples and sunflowers to the flight of hawks. Livio catalogues its appearances in both maths and nature, looking closely at the Fibonacci sequence and rabbit-breeding, before going on to debunk mistaken claims that phi also appears a lot in art, music and poetry. Dalí certainly used it, but da Vinci, Debussy and Virgil almost certainly didn’t. Nor, almost certainly, did the builders of the Parthenon and pyramids. Finally, he examines what has famously been called (by the physicist Eugene Wiegner) the unreasonable effectiveness of mathematics: why is this human invention so good at describing the behaviour of the Universe? Livio quotes one of the best short answers I’ve seen:

Human logic was forced on us by the physical world and is therefore consistent with it. Mathematics derives from logic. That is why mathematics is consistent with the physical world. (ch. 9, “Is God a mathematician?”, pg. 252)

It’s not hard to recommend a book that quotes everyone from Johannes Kepler and William Blake to Lewis Carroll, Christopher Marlowe and Jef Raskin, “the creator of the Macintosh computer”, whose answer is given above. Recreational mathematicians should also find lots of ideas for further investigation, from fractal strings to the fascinating number patterns governed by Benford’s law. It isn’t just human beings who look after number one: as a leading figure, 1 turns up much more often in data from the real world, and in mathematical constructs like the Fibonacci sequence, than intuition would lead you to expect. If you’d like to learn more about that and about many other aspects of mathematics, hunt down a copy of this book.


Elsewhere other-posted:

Roses Are Golden – φ and floral homicide