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

Square Routes Re-Re-Re-Revisited

Discovering something that’s new to you in recreational maths is good. But so is re-discovering it by a different route. I’ve long been passionate about what happens when a point is allowed to jump repeatedly halfway towards the randomly chosen vertices of a square. If the point can choose any vertex any number of times, the interior of the square fills slowly and completely with points, like this:

Point jumping at random halfway towards vertices of a square


However, if the point is banned from jumping towards the same vertex twice or more in a row, an interesting fractal appears:

Fractal #1 — ban on jumping towards vertex vi twice or more


If the point can’t jump towards the vertex one place clockwise of the vertex it’s just jumped towards, this fractal appears:

Fractal #2 — ban on jumping towards vertex vi+1


If the point can’t jump towards the vertex two places clockwise of the vertex it’s just jumped towards, this fractal appears (two places clockwise is also two places anticlockwise, i.e. the banned vertex is diagonally opposite):

Fractal #3 — ban on jumping towards vertex vi+2


Now I’ve discovered a new way to create these fractals. You take a filled square, divide it into smaller squares, then remove some of them in a systematic way. Then you do the same to the smaller squares that remain. For fractal #1, you do this:

Fractal #1, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Fractal #1 (animated)


For fractal #2, you do this:

Fractal #2, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Fractal #2 (animated)


For fractal #3, you do this:

Fractal #3, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Fractal #3 (animated)


If the sub-squares are coloured, it’s easier to understand how, say, fractal #1 is created:

Fractal #1 (coloured), stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Fractal #1 (coloured and animated)


The fractal is actually being created in quarters, with one quarter rotated to form the second, third and fourth quarters:

Fractal #1, quarter









Here’s an animation of the same process for fractal #3:

Fractal #3 (coloured and animated)


So you can create these fractals either with a jumping point or by subdividing a square. But in fact I discovered the subdivided-square route by looking at a variant of the jumping-point route. I wondered what would happen if you took a point inside a square, allowed it to trace all possible routes towards the vertices without marking its position, then imposed the restriction for Fractal #1 on its final jump, namely, that it couldn’t jump towards the vertex it jumped towards on its previous jump. If the point is marked after its final jump, this is what appears (if the routes chosen had been truly random, the image would be similar but messier):

Fractal #1, restriction on final jump


Then I imposed the same restriction on the point’s final two jumps:

Fractal #1, restriction on final 2 jumps


And final three jumps:

Fractal #1, restriction on final 3 jumps


And so on:

Fractal #1, restriction on final 4 jumps


Fractal #1, restriction on final 5 jumps


Fractal #1, restriction on final 6 jumps


Fractal #1, restriction on final 7 jumps


Here are animations of the same process applied to fractals #2 and #3:

Fractal #2, restrictions on final 1, 2, 3… jumps


Fractal #3, restrictions on final 1, 2, 3… jumps


The longer the points are allowed to jump before the final restriction is imposed on their n final jumps, the more densely packed the marked points will be:

Fractal #1, packed points #1


Packed points #2


Packed points #3


Eventually, the individual points will form a solid mass, like this:

Fractal #1, solid mass of points


Fractal #1, packed points (animated)


Previously pre-posted (please peruse):

Square Routes
Square Routes Revisited
Square Routes Re-Revisited
Square Routes Re-Re-Revisited

Fink Frakt

Pre-previously on Overlord-In-Terms-of-Issues-Around-Engagement-with-the-Über-Feral, I’ve looked at various ways of creating fractals by restricting the moves of a point jumping towards the vertices of a polygon. For example, the point can be banned from jumping towards the same vertex twice in a row. This time, I want to look at fractals created not by restriction, but by compulsion. If the point jumps towards vertex v and then tries to jump towards vertex v again, it will be forced to jump towards vertex v+1 instead, and so on.

You could call vv+1 a forced increment or finc. So these are finc fractals. In some cases, restriction and compulsion create the same fractals, but I’ve found some new fractals using compulsion. Consider the fractal created by the rule v[-2]+1, v[-1] → +0,+1, where the subscripts refer to the history of jumps: v[-2] is the jump-before-last, v[-1] is the last jump. If the new vertex, v[0], chosen is the same as v[-2]+1 (e.g., v[0] = 2 = v[-2]+1 = 1+1), then the forced increment is 0, i.e., the point is allowed to choose that jump. However, if v[0] = v[-1], then the forced increment is 1 and the point must jump towards v[-1]+1.

Here is the fractal in question:

v[-2]+1, v[-1] → +0,+1 (black-and-white)


v[-2]+1, v[-1] → +0,+1 (colour)


1,0 → +0,+1 (animated)


1,0 → +1,+0 (bw)


1,0 → +1,+0 (col)


1,0 → +1,+0 (anim)


1,0 → +1,+1 (bw)


1,0 → +1,+1 (col)


1,0 → +1,+1 (animated)


0,1 → +2,+1 (anim)


0,1 → +3,+1


1,0 → +0,+1


1,0 → +1,+0


1,1 → +0,+1


1,1 → +1,+2


1,1 → +1,+3


1,1 → +2,+1


1,2 → +0,+3


1,3 → +0,+1


2,2 → +0,+1


But suppose the history of jumps records not actual jumps, but the jumps the point wanted to make instead. In some cases, the jump made will be the same as the jump originally chosen, but in other cases it won’t. Here are some fractals using this method:

0 → +2


0 → +3


2 → +1


2 → +2