Get Your Prox Off #3

I’ve looked at lot at the fractals created when you randomly (or quasi-randomly) choose a vertex of a square, then jump half of the distance towards it. You can ban jumps towards the same vertex twice in a row, or jumps towards the vertex clockwise or anticlockwise from the vertex you’ve just chosen, and so on.

But you don’t have to choose vertices directly: you can also choose them by distance or proximity (see “Get Your Prox Off” for an earlier look at fractals-by-distance). For example, this fractal appears when you can jump half-way towards the nearest vertex, the second-nearest vertex, and the third-nearest vertex (i.e., you can’t jump towards the fourth-nearest or most distant vertex):

vertices = 4, distance = (1,2,3), jump = 1/2


It’s actually the same fractal as you get when you choose vertices directly and ban jumps towards the vertex diagonally opposite from the one you’ve just chosen. But this fractal-by-distance isn’t easy to match with a fractal-by-vertex:

v = 4, d = (1,2,4), j = 1/2


Nor is this one:

v = 4, d = (1,3,4)


This one, however, is the same as the fractal-by-vertex created by banning a jump towards the same vertex twice in a row:

v = 4, d = (2,3,4)


The point can jump towards second-nearest, third-nearest and fourth-nearest vertices, but not towards the nearest. And the nearest vertex will be the one chosen previously.

Now let’s try squares with an additional point-for-jumping-towards on each side (the points are numbered 1 to 8, with points 1, 3, 5, 7 being the true vertices):

v = 4 + s1 point on each side, d = (1,2,3)


v = 4 + s1, d = (1,2,5)


v = 4 + s1, d = (1,2,7)


v = 4 + s1, d = (1,3,8)


v = 4 + s1, d = (1,4,6)


v = 4 + s1, d = (1,7,8)


v = 4 + s1, d = (2,3,8)


v = 4 + s1, d = (2,4,8)


And here are squares where the jump is 2/3, not 1/2, and you can choose only the nearest or third-nearest jump-point:

v = 4, d = (1,3), j = 2/3


v = 4 + s1, d = (1,3), j = 2/3


Now here are some pentagonal fractals-by-distance:

v = 5, d = (1,2,5), j = 1/2


v = 5 + s1, d = (1,2,7)


v = 5 + s1, d = (1,2,8)


v = 5 + s1, d = (1,2,9)


v = 5 + s1, d = (1,9,10)


v = 5 + s1, d = (1,10), j = 2/3


v = 5 + s1, d = (various), j = 2/3 (animated)


And now some hexagonal fractals-by-distance:

v = 6, d = (1,2,4), j = 1/2


v = 6, d = (1,3,5)


v = 6, d = (1,3,6)


v = 6, d = (1,2,3,4)


v = 6 + central point, d = (1,2,3,4)


v = 6, d = (1,2,3,6)


v = 6, d = (1,2,4,6)


v = 6, d = (1,3,4,5)


v = 6, d = (1,3,4,6)


v = 6, d = (1,4,5,6)


Elsewhere other-accessible:

Get Your Prox Off — an earlier look at fractals-by-distance
Get Your Prox Off # 2 — and another

Fractal + Star = Fractar

Here’s a three-armed star made with three lines radiating at intervals of 120°:

Triangular fractal stage #1


At the end of each of the three lines, add three more lines at half the length:

Triangular fractal #2


And continue like this:

Triangular fractal #3


Triangular fractal #4


Triangular fractal #5


Triangular fractal #6


Triangular fractal #7


Triangular fractal #8


Triangular fractal #9


Triangular fractal #10


Triangular fractal (animated)


Because this fractal is created from a series of star, you could call it a fractar. Here’s a black-and-white version:

Triangular fractar (black-and-white)


Triangular fractar (black-and-white) (animated)
(Open in a new window for larger version if the image seems distorted)


A four-armed star doesn’t yield an easily recognizable fractal in a similar way, so let’s try a five-armed star:

Pentagonal fractar stage #1


Pentagonal fractar #2


Pentagonal fractar #3


Pentagonal fractar #4


Pentagonal fractar #5


Pentagonal fractar #6


Pentagonal fractar #7


Pentagonal fractar (animated)


Pentagonal fractar (black-and-white)


Pentagonal fractar (bw) (animated)


And here’s a six-armed star:

Hexagonal fractar stage #1


Hexagonal fractar #2


Hexagonal fractar #3


Hexagonal fractar #4


Hexagonal fractar #5


Hexagonal fractar #6


Hexagonal fractar (animated)


Hexagonal fractar (black-and-white)


Hexagonal fractar (bw) (animated)


And here’s what happens to the triangular fractar when the new lines are rotated by 60°:

Triangular fractar (60° rotation) #1


Triangular fractar (60°) #2


Triangular fractar (60°) #3


Triangular fractar (60°) #4


Triangular fractar (60°) #5


Triangular fractar (60°) #6


Triangular fractar (60°) #7


Triangular fractar (60°) #8


Triangular fractar (60°) #9


Triangular fractar (60°) (animated)


Triangular fractar (60°) (black-and-white)


Triangular fractar (60°) (bw) (animated)


Triangular fractar (60°) (no lines) (black-and-white)


A four-armed star yields a recognizable fractal when the rotation is 45°:

Square fractar (45°) #1


Square fractar (45°) #2


Square fractar (45°) #3


Square fractar (45°) #4


Square fractar (45°) #5


Square fractar (45°) #6


Square fractar (45°) #7


Square fractar (45°) #8


Square fractar (45°) (animated)


Square fractar (45°) (black-and-white)


Square fractar (45°) (bw) (animated)


Without the lines, the final fractar looks like the plan of a castle:

Square fractar (45°) (bw) (no lines)


And here’s a five-armed star with new lines rotated at 36°:

Pentagonal fractar (36°) #1


Pentagonal fractar (36°) #2


Pentagonal fractar (36°) #3


Pentagonal fractar (36°) #4


Pentagonal fractar (36°) #5


Pentagonal fractar (36°) #6


Pentagonal fractar (36°) #7


Pentagonal fractar (36°) (animated)


Again, the final fractar without lines looks like the plan of a castle:

Pentagonal fractar (36°) (no lines) (black-and-white)


Finally, here’s a six-armed star with new lines rotated at 30°:

Hexagonal fractar (30°) #1


Hexagonal fractar (30°) #2


Hexagonal fractar (30°) #3


Hexagonal fractar (30°) #4


Hexagonal fractar (30°) #5


Hexagonal fractar (30°) #6


Hexagonal fractar (30°) (animated)


And the hexagonal castle plan:

Hexagonal fractar (30°) (black-and-white) (no lines)


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

Jumping Jehosophracts!

As I’ve shown pre-previously on Overlord-in-terms-of-issues-around-the-Über-Feral, you can create interesting fractals by placing restrictions on a point jumping inside a fractal towards a randomly chosen vertex. For example, the point can be banned from jumping towards the same vertex twice in a row, and so on.

But you can use other restrictions. For example, suppose that the point can jump only once or twice towards any vertex, that is, (j = 1,2). It can then jump towards the same vertex again, but not the same number of times as it previously jumped. So if it jumps once, it has to jump twice next time; and vice versa. If you use this rule on a pentagon, this fractal appears:

v = 5, j = 1,2 (black-and-white)


v = 5, j = 1,2 (colour)


If the point can also jump towards the centre of the pentagon, this fractal appears:

v = 5, j = 1,2 (with centre)


And if the point can also jump towards the midpoints of the sides:

v = 5, j = 1,2 (with midpoints)


v = 5, j = 1,2 (with midpoints and centre)


And here the point can jump 1, 2 or 3 times, but not once in a row, twice in a row or thrice in a row:

v = 5, j = 1,2,3


v = 5, j = 1,2,3 (with centre)


Here the point remembers its previous two moves, rather than just its previous move:

v = 5, j = 1,2,3, hist = 2 (black-and-white)


v = 5, j = 1,2,3, hist = 2


v = 5, j = 1,2,3, hist = 2 (with center)


v = 5, j = 1,2,3, hist = 2 (with midpoints)


v = 5, j = 1,2,3, hist = 2 (with midpoints and centre)


And here are hexagons using the same rules:

v = 6, j = 1,2 (black-and-white)


v = 6, j = 1,2


v = 6, j = 1,2 (with centre)


And octagons:

v = 8, j = 1,2


v = 8, j = 1,2 (with centre)


v = 8, j = 1,2,3, hist = 2


v = 8, j = 1,2,3, hist = 2


v = 8, j = 1,2,3,4 hist = 3


v = 8, j = 1,2,3,4 hist = 3 (with center)


The Hex Fractor

Pre-previously on Overlord-in-terms-of-issues-around-the-Über-Feral, I looked at the fractals created when various restrictions are placed on a point jumping at random half-way towards the vertices of a square. For example, the point can be banned from jumping towards the same vertex twice in a row or towards the vertex to the left of the vertex it has just jumped towards, and so on.

Today I want to look at what happens to a similar point moving inside pentagons and hexagons. If the point can’t jump twice towards the same vertex of a pentagon, this is the fractal that appears:

Ban second jump towards same vertex (v + 0)


Ban second jump towards same vertex (color)


If the point can’t jump towards the vertex immediately to the left of the one it’s just jumped towards, this is the fractal that appears:

Ban jump towards v + 1


Ban jump towards v + 1 (color)


And this is the fractal when the ban is on the vertex two places to the left:

Ban jump towards v + 2


Ban jump towards v + 2 (color)


You can also ban more than one vertex:

Ban jump towards v + 0,1


Ban jump towards v + 1,2


Ban jump towards v + 1,4


Ban jump towards v + 1,4 (color)


Ban jump towards v + 2,3


And here are fractals created in similar ways inside hexagons:

Ban jump towards v + 0,1


Ban jump towards v + 0,3


Ban jump towards v + 0,1,2


Ban jump towards v + 0,1,2 (color)


Ban jump towards v + 0,1,4


Ban jump towards v + 0,1,5


Ban jump towards v + 0,2,4


Ban jump towards v + 0,2,4 (color)


Ban jump towards v + 1,2,3


Ban jump towards v + 1,2,3 (color)


Ban jump towards v + 1,2,4


Ban jump towards v + 1,2,4, (color)


Ban jump towards v + 1,3,5


Ban jump towards v + 1,3,5 (color)


Ban jump towards v + 1,2


Ban jump towards v + 1,2


Ban jump towards v + 1,3


Ban jump towards v + 1,3 (color)


Ban jump towards v + 1,5


Ban jump towards v + 1,5 (color)


Ban jump towards v + 2,3


Ban jump towards v + 2,3 (color)


Ban jump towards v + 2,4


Ban jump towards v + 2,4 (color)


Elsewhere other-accessible:

Square Routes Re-Verticed

Holey Trimmetry

Symmetry arising from symmetry isn’t surprising. But what about symmetry arising from asymmetry? You can find both among the rep-tiles, which are geometrical shapes that can be completely replaced by smaller copies of themselves. A square is a symmetrical rep-tile. It can be replaced by nine smaller copies of itself:

Rep-9 Square

If you trim the copies so that only five are left, you have a symmetrical seed for a symmetrical fractal:

Fractal cross stage #1


Fractal cross #2


Fractal cross #3


Fractal cross #4


Fractal cross #5


Fractal cross #6


Fractal cross (animated)


Fractal cross (static)


If you trim the copies so that six are left, you have another symmetrical seed for a symmetrical fractal:

Fractal Hex-Ring #1


Fractal Hex-Ring #2


Fractal Hex-Ring #3


Fractal Hex-Ring #4


Fractal Hex-Ring #5


Fractal Hex-Ring #6


Fractal Hex-Ring (animated)


Fractal Hex-Ring (static)


Now here’s an asymmetrical rep-tile, a nonomino or shape created from nine squares joined edge-to-edge:

Nonomino


It can be divided into twelve smaller copies of itself, like this:

Rep-12 Nonomino (discovered by Erich Friedman)


If you trim the copies so that only five are left, you have an asymmetrical seed for a familiar symmetrical fractal:

Fractal cross stage #1


Fractal cross #2


Fractal cross #3


Fractal cross #4


Fractal cross #5


Fractal cross #6


Fractal cross (animated)


Fractal cross (static)


If you trim the copies so that six are left, you have an asymmetrical seed for another familiar symmetrical fractal:

Fractal Hex-Ring #1


Fractal Hex-Ring #2


Fractal Hex-Ring #3


Fractal Hex-Ring #4


Fractal Hex-Ring #5


Fractal Hex-Ring (animated)


Fractal Hex-Ring (static)


Elsewhere other-available:

Square Routes Re-Re-Visited

Squaring and Paring

Squares are often thought to be the most boring of all shapes. Yet every square holds a stunning secret – something that in legend prompted a mathematical cult to murder a traitor. If each side of a square is one unit long, how long is the square’s diagonal, that is, the line from one corner to the opposite corner?

By Pythagoras’ theorem, the answer is this:

• x^2 = 1^2 + 1^2
• x^2 = 2
• x = √2

But what is √2? Pythagoras and his followers thought that all numbers could be represented as either whole numbers or ratios of whole numbers. To their dismay, so it’s said, they discovered that they were wrong. √2 is an irrational number – it can’t be represented as a ratio. In modern notation, it’s an infinitely decimal that never repeats:

• √2 = 1·414213562373095048801688724209698…

A modern story, unattested in ancient records, says that the irrationality of √2 was a closely guarded secret in the Pythagorean cult. When Hippasus of Metapontum betrayed the secret, he was drowned at sea by enraged fellow cultists. Apocryphal or not, the story shows that squares aren’t so boring after all.

Nor are they boring when they’re caught in the fract. Divide one square into nine smaller copies of itself:


Discard three of the copies like this:

Stage 1
Retain squares 1, 2, 4, 6, 8, 9 (reading left-to-right, bottom-to-top)


Then do the same to each of the sub-squares:

Stage 1


And repeat:

Stage 3


Stage 4


Stage 5


Stage 6


The result is a fractal of endlessly subdividing contingent hexagons:

Animated vesion


Retain squares 1, 2, 4, 6, 8, 9 (reading left-to-right, bottom-to-top)


Here are a few more of the fractals you can create by squaring and paring:

Retain squares 1, 3, 5, 7, 9 (reading left-to-right, bottom-to-top)


Retain squares 2, 4, 5, 6, 8


Retain squares 1, 2, 4, 5, 6, 8, 9


Retain squares 1, 4, 6, 7, 10, 11, 13, 16


Retain squares 1, 3, 6, 7, 8, 9, 10, 11, 14, 16


Retain squares 2, 3, 5, 6, 8, 9, 11, 12, 14, 15


Retain squares 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25


Retain squares 1, 3, 7, 8, 11, 12, 14, 15, 18, 19, 23, 25


Retain squares 1, 5, 7, 8, 9, 12, 14, 17, 18, 19, 21, 25


Retain squares 2, 3, 4, 6, 7, 9, 10, 11, 15, 16, 17, 19, 20, 22, 23, 24


Retain squares 1, 2, 5, 6, 7, 9, 13, 17, 19, 20, 21, 24, 25


Previously pre-posted (please peruse):

M.i.P. Trip

Performativizing the Polygonic

Maths is a mountain: you can start climbing in different places and reach the same destination. There are many ways of proving the irrationality of √2 or the infinitude of the primes, for example. But you can also arrive at the same destination by accident. I’ve found that when I use different methods of creating fractals. The same fractals appear, because apparently different algorithms are actually the same underneath.

But different methods can create unique fractals too. I’ve found some new ones by using what might be called point-to-point recursion. For example, there are ten ways to select three vertices from the five vertices of a pentagon: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5). Find the midpoint of the first three-point set, (1, 2, 3). Then select two vertices to go with this midpoint, creating a new three-point set, and find the midpoint again. And so on. The process looks like this, with the midpoints shown for all the three-point sets found at each stage:

v5_p3_stage1

vertices = 5, choose sets of 3 points, find mid-point of each

v5_p3_stage2

v5_p3_stage3


At stage 5, the fractal looks like this:

v5_p3_static

v = 5, p = 3


Note that when pixels are used again, the colour changes. That’s another interesting thing about maths: limits can sometimes produce deeper results. If these fractals were drawn at very high resolution, pixels would only be used once and the colour would never change. As it is, low resolution means that pixels are used again and again. But some are used more than others, which is why interesting colour effects appear.

If the formation of the fractal is animated, it looks like this (with close-ups of even deeper stages):
v5_p3


Here are some more examples:

v4c_p2_static

v = 4 + central point, p = 2 (cf. Fingering the Frigit)

v4c_p2

v = 4c, p = 2 (animated)


v4_p3_static

v = 4, p = 3

v4_p3


v5_p4_static

v = 5, p = 4

v5_p4


v5c_p3_static

v = 5 + central point, p = 3

v5c_p3


v5c_p4

v = 5c, p = 4


v5c_p5

v = 5c, p = 5


v6_1_p6

v = 6 + 1 point between each pair of vertices, p = 6


v6_p2

v = 6, p = 2


v6_p3_static

v = 6, p = 3

v6_p3


v6_p4

v = 6, p = 4


v6c_p2_static

v = 6c, p = 2 (cf. Fingering the Frigit)

v6c_p2


v6c_p3_static

v = 6c, p = 3

v6c_p3


v6c_p4

v = 6c, p = 4


v7_p3

v = 7, p = 3


v7_p4_static

v = 7, p = 4

v7_p4


v7_p5_static

v = ,7 p = 5

v7_p5


v7_p4

v = 7c, p = 4


v3_1_p2

v = 3+1, p = 2


v3_1_p3

v = 3+1, p = 3


v3_1_p4

v = 3+1, p = 4


v3_2_p5

v = 3+2, p = 5


v3c_1_p2

v = 3c+1, p = 2


v3c_1_p4

v = 3c+1, p = 4


v3c_p2

v = 3c, p = 2


v3c_p3

v = 3c, p = 3


v4_1_p3

v = 4+1, p = 3


v4_1_p4

v = 4+1, p = 4


v4_1_p5

v = 4+1, p = 6


v4_1_p6

v = 4+1, p = 2


v4c_1_p4

v = 4c+1, p = 4


v4c_p3_static

v = 4c, p = 3

v4c_p3


v5_1_p4_va

v = 5+1, p = 4 (and more)


v5_p2

v = 5, p = 2


Polymorphous Perverticity

As I’ve explained before on Overlord of the Über-Feral, the planet’s premier purveyor of polygonic performativity (probably (possibly (perspectivistically))), it works with triangles and pentagons, but not with squares. And what is “it”? A simple procedure in which you create a polygon, choose a point inside it, then repeatedly move half-way towards a vertex chosen at random, marking each new position as you go.

pol3_4_5

When the polygon has three vertices, you get a Sierpiński triangle. When it has five, you get what might be called a  Sierpiński pentagon. When it has four, you get nothing. Or rather: you get everything, because the whole interior of the square gradually fills with points. But, as I’ve also explained before, there’s a simple way to change this. You can adapt the procedure so that a vertex can’t be chosen twice in a row, and so on.

When the rule is “No vertex twice in a row”, you get this fractal (colours change as a pixel is selected again):

pol4_0

But you can also use what might be a vertex increment, or vi, whereby you disallow vertices that are next to the previously chosen vertex, or two positions away, and so on. When the rule is “No vertex twice in a row”, the disallowed vertex is (v + 0), that is, vi = 0. If vi = 2 and the rule is disallow(v + 2), this fractal appears (when vi = 1, there’s no fractal):

pol4_2

v = 4, vi = 2

pol4_2_anim


You can extend these rules to apply not just to the previously chosen vertex, but also to the vertex chosen before that. Here are some fractals produced by the rule disallow(v[1] + vi[1], v[2] + vi[2]), where v[1] is the vertex previously chosen and v[2] is the vertex chosen before that:

pol4_1_2

v = 4, vi[1] = 1, vi[2] = 2

pol4_1_2_anim


pol4_2_0

v = 4, vi[1] = 2, vi[2] = 0

pol4_2_0_anim

pol4_2_0_white


pol4_2_1

v = 4, vi[1] = 2, vi[2] = 1

pol4_2_1_anim


pol4_2_2

v = 4, vi[1] = 2, vi[2] = 2

pol4_2_2_anim


And here are some fractals produced by the rule disallow(v[1] + vi[1], v[2] + vi[2], v[3] + vi[3]):

pol4_1_1_0

v = 4, vi[1] = 1, vi[2] = 1, vi[3] = 0

pol4_1_1_0_anim


pol4_1_1_2

v = 4, vi[1] = 1, vi[2] = 1, vi[3] = 2

pol4_1_1_2_anim


Applying these rules to pentagons rather than squares doesn’t produce such a dramatic difference, because the original procedure – choose any vertex at random, taking no account of previous choices – produces a fractal when v = 5, as noted above, but not when v = 4. Nevertheless, here are some fractals for v > 4:

pol5_0

v = 5, vi = 0


pol5_1

v = 5, vi = 1

pol5_1_anim


pol5_2

v = 5, vi = 2

pol5_2_anim


pol5_0_0

v = 5, vi[1] = 0, vi[2] = 0


pol5_1_0

v = 5, vi[1] = 1, vi[2] = 0


pol5_2_0

v = 5, vi[1] = 2, vi[2] = 0

pol5_2_0_anim


pol5_1_1

v = 5, vi[1] = 1, vi[2] = 1

pol5_1_1_anim


pol5_1_1_1

v = 5, vi[1] = 1, vi[2] = 1, vi[3] = 1


pol5_va2

v = 5, vi = various


pol6_1

v = 6, vi = 1

pol6_1_anim

Fingering the Frigit

Fingers are fractal. Where a tree has a trunk, branches and twigs, a human being has a torso, arms and fingers. And human beings move in fractal ways. We use our legs to move large distances, then reach out with our arms over smaller distances, then move our fingers over smaller distances still. We’re fractal beings, inside and out, brains and blood-vessels, fingers and toes.

But fingers are fractal are in another way. A digit – digitus in Latin – is literally a finger, because we once counted on our fingers. And digits behave like fractals. If you look at numbers, you’ll see that they contain patterns that echo each other and, in a sense, recur on smaller and smaller scales. The simplest pattern in base 10 is (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). It occurs again and again at almost very point of a number, like a ten-hour clock that starts at zero-hour:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9…
10, 11, 12, 13, 14, 15, 16, 17, 18, 19…
200… 210… 220… 230… 240… 250… 260… 270… 280… 290…

These fractal patterns become visible if you turn numbers into images. Suppose you set up a square with four fixed points on its corners and a fixed point at its centre. Let the five points correspond to the digits (1, 2, 3, 4, 5) of numbers in base 6 (not using 0, to simplify matters):

1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55, 61, 62, 63, 64, 65… 2431, 2432, 2433, 2434, 2435, 2441, 2442, 2443, 2444, 2445, 2451, 2452…

Move between the five points of the square by stepping through the individual digits of the numbers in the sequence. For example, if the number is 2451, the first set of successive digits is (2, 4), so you move to a point half-way between point 2 and point 4. Next come the successive digits (4, 5), so you move to a point half-way between point 4 and point 5. Then come (5, 1), so you move to a point half-way between point 5 and point 1.

When you’ve exhausted the digits (or frigits) of a number, mark the final point you moved to (changing the colour of the pixel if the point has been occupied before). If you follow this procedure using a five-point square, you will create a fractal something like this:
fractal4_1single

fractal4_1
A pentagon without a central point using numbers in a zero-less base 7 looks like this:
fractal5_0single

fractal5_0
A pentagon with a central point looks like this:
fractal5_1single

fractal5_1
Hexagons using a zero-less base 8 look like this:
fractal6_1single

fractal6_1


fractal6_0single

fractal6_0
But the images above are just the beginning. If you use a fixed base while varying the polygon and so on, you can create images like these (here is the program I used):
fractal4


fractal5


fractal6789