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

Back to Drac’

draconic, adj. /drəˈkɒnɪk/ pertaining to, or of the nature of, a dragon. [Latin draco, -ōnem, < Greek δράκων dragon] — The Oxford English Dictionary

In Curvous Energy, I looked at the strange, beautiful and complex fractal known as the dragon curve and showed how it can be created from a staid and sedentary square:

A dragon curve

Here are the stages whereby the dragon curve is created from a square. Note how each square at one stage generates a pair of further squares at the next stage:

Dragon curve from squares #1

Dragon curve from squares #2

Dragon curve from squares #3

Dragon curve from squares #4

Dragon curve from squares #5

Dragon curve from squares #6

Dragon curve from squares #7

Dragon curve from squares #8

Dragon curve from squares #9

Dragon curve from squares #10

Dragon curve from squares #11

Dragon curve from squares #12

Dragon curve from squares #13

Dragon curve from squares #14

Dragon curve from squares (animated)

The construction is very easy and there’s no tricky trigonometry, because you can use the vertices and sides of each old square to generate the vertices of the two new squares. But what happens if you use lines rather than squares to generate the dragon curve? You’ll discover that less is more:

Dragon curve from lines #1

Dragon curve from lines #2

Dragon curve from lines #3

Dragon curve from lines #4

Dragon curve from lines #5

Each line at one stage generates a pair of further lines at the next stage, but there’s no simple way to use the original line to generate the new ones. You have to use trigonometry and set the new lines at 45° to the old one. You also have to shrink the new lines by a fixed amount, 1/√2 = 0·70710678118654752… Here are further stages:

Dragon curve from lines #6

Dragon curve from lines #7

Dragon curve from lines #8

Dragon curve from lines #9

Dragon curve from lines #10

Dragon curve from lines #11

Dragon curve from lines #12

Dragon curve from lines #13

Dragon curve from lines #14

Dragon curve from lines (animated)

But once you have a program that can adjust the new lines, you can experiment with new angles. Here’s a dragon curve in which one new line is at an angle of 10°, while the other remains at 45° (after which the full shape is rotated by 180° because it looks better that way):

Dragon curve 10° and 45°

Dragon curve 10° and 45° (animated)

Dragon curve 10° and 45° (coloured)

Here are more examples of dragon curves generated with one line at 45° and the other line at a different angle:

Dragon curve 65°

Dragon curve 65° (anim)

Dragon curve 65° (col)

Dragon curve 80°

Dragon curve 80° (anim)

Dragon curve 80° (col)

Dragon curve 135°

Dragon curve 135° (anim)

Dragon curve 250°

Dragon curve 250° (anim)

Dragon curve 250° (col)

Dragon curve 260°

Dragon curve 260° (anim)

Dragon curve 260° (col)

Dragon curve 340°

Dragon curve 340° (anim)

Dragon curve 340° (col)

Dragon curve 240° and 20°

Dragon curve 240° and 20° (anim)

Dragon curve 240° and 20° (col)

Dragon curve various angles (anim)

Previously pre-posted:

Curvous Energy — a first look at dragon curves

Rep-Tile Reflections


Various rep-tiles

Various rep-tiles — click for larger image



The rep-tile known as the fish

Credo in Piscem…
