Delta Skelta

“When I get to the bottom I go back to the top of the slide,
Where I stop and I turn and I go for a ride
Till I get to the bottom and I see you again.” — The Beatles, “Helter Skelter” (1968)


First stage of fractal #1











Animated fractal #1


First stage of fractal #2













Animated fractal #2

See-Saw Jaw

From Sierpiński triangle to T-square to Mandibles (and back again) (animated)
(Open in new window if distorted)


Elsewhere other-accessible…

Mandibular Metamorphosis — explaining the animation above
Agnathous Analysis — more on the Sierpiński triangle and T-square fractal

Middlemath

Suppose you start at the middle of a triangle, then map all possible ways you can jump eight times half-way towards one or another of the vertices of the triangle. At the end of the eight jumps, you mark your final position with a dot. You could jump eight times towards the same vertex, or once towards vertex 1, once towards vertex 2, and once again towards vertex 1. And so on. If you do this, the record of your jumps looks something like this:


The shape is a fractal called the Sierpiński triangle. But if you try the same thing with a square — map all possible jumping-routes you can follow towards one or another of the four vertices — you simply fill the interior of the square. There’s no interesting fractal:


So you need a plan with a ban. Try mapping all possible routes where you can’t jump towards the same vertex twice in a row. And you get this:

Ban on jumping towards same vertex twice in a row, v(t) ≠ v(t-1)


If you call the current vertex v(t) and the previous vertex v(t-1), the ban says that v(t) ≠ v(t-1). Now suppose you can’t jump towards the vertex one place clockwise of the previous vertex. Now the ban is v(t)-1 ≠ v(t-1) or v(t) ≠ v(t-1)+1 and this fractal appears:

v(t) ≠ v(t-1)+1


And here’s a ban on jumping towards the vertex two places clockwise (or counterclockwise) of the vertex you’ve just jumped towards:

v(t) ≠ v(t-1)+2


And finally the ban on jumping towards the vertex three places clockwise (or one place counterclockwise) of the vertex you’ve just jumped towards:

v(t) ≠ v(t-1)+3 (a mirror-image of v(t) ≠ v(t-1)+1, as above)


Now suppose you introduce a new point to jump towards at the middle of the square. You can create more fractals, but you have to adjust the kind of ban you use. The central point can’t be included in the ban or the fractal will be asymmetrical. So you continue taking account of the vertices, but if the previous jump was towards the middle, you ignore that jump. At least, that’s what I intended, but I wonder whether my program works right. Anyway, here are some of the fractals that it produces:

v(t) ≠ v(t-1) with central point (wcp)


v(t) ≠ v(t-1)+1, wcp


v(t) ≠ v(t-1)+2, wcp


And here are some bans taking account of both the previous vertex and the pre-previous vertex:

v(t) ≠ v(t-1) & v(t) ≠ v(t-2), wcp


v(t) ≠ v(t-1) & v(t-2)+1, wcp


v(t) ≠ v(t-1)+2 & v(t-2), wcp


v(t) ≠ v(t-1) & v(t-2)+1, wcp


v(t) ≠ v(t-1)+1 & v(t-2)+1, wcp


v(t) ≠ v(t-1)+2 & v(t-2)+1, wcp


v(t) ≠ v(t-1)+3 & v(t-2)+1, wcp


v(t) ≠ v(t-1) & v(t-2)+2, wcp


v(t) ≠ v(t-1)+1 & v(t-2)+2, wcp


v(t) ≠ v(t-1)+2 & v(t-2)+2, wcp


Now look at pentagons. They behave more like triangles than squares when you map all possible jumping-routes towards one or another of the five vertices. That is, a fractal appears:

All possible jumping-routes towards the vertices of a pentagon


But the pentagonal-jump fractals get more interesting when you introduce jump-bans:

v(t) ≠ v(t-1)


v(t) ≠ v(t-1)+1


v(t) ≠ v(t-1)+2


v(t) ≠ v(t-1) & v(t-2)


v(t) ≠ v(t-1)+2 & v(t-2)


v(t) ≠ v(t-1)+1 & v(t-2)+1


v(t) ≠ v(t-1)+3 & v(t-2)+1


v(t) ≠ v(t-1)+1 & v(t-2)+2


v(t) ≠ v(t-1)+2 & v(t-2)+2


v(t) ≠ v(t-1)+3 & v(t-2)+2


Finally, here are some pentagonal-jump fractals using a central point:








Post-Performative Post-Scriptum

I’m not sure if I’ve got the order of some bans right above. For example, should v(t) ≠ v(t-1)+1 & v(t-2)+2 really be v(t) ≠ v(t-1)+2 & v(t-2)+1? I don’t know and I’m not going to check. But the idea of jumping-point bans is there and that’s all you need if you want to experiment with these fractal methods for yourself.

Six Mix Trix

Here’s an equilateral triangle divided into six smaller triangles:

Equilateral triangle divided into six irregular triangles (Stage #1)


Now keep on dividing:

Stage #2


Stage #3


Stage #4


Stage #5


Equilateral triangle dividing into six irregular triangles (animated)


But what happens if you divide the triangle, then discard some of the sub-triangles, then repeat? You get a self-similar shape called a fractal:

Divide-and-discard stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Triangle fractal (animated)


Here’s another example:

Divide-and-discard stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Triangle fractal (animated)


You can also delay the divide-and-discard to create a more symmetrical fractal, like this:

Delayed divide-and-discard stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Triangle fractal (animated)


What next? You can use trigonometry to turn the cramped triangle into a circle:

Triangular fractal

Circular fractal
(Open in new window for full image)


Triangle-to-circle (animated)


Here’s another example:

Triangular fractal

Circular fractal


Triangle-to-circle (animated)


And below are some more circular fractals converted from triangular fractals. Some of them look like distorted skulls or transdimensional Lovecraftian monsters:

(Open in new window for full image)


















Previous Pre-Posted

Circus Trix — an earlier look at sextally-divided-equilateral-triangle fractals

Square’s Flair

If you want to turn banality into beauty, start here with three staid and static squares:

Stage #1


Now replace each red and yellow square with two new red and yellow squares orientated in the same way to the original square:

Stage #2


And repeat:

Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Stage #9


Stage #10


Stage #11


Stage #12


Stage #13


Stage #14


Stage #15


Stage #16


Stage #17


Stage #18


And you arrive in the end at a fractal called a dragon curve:

Dragon curve


Dragon curve (animated)


Elsewhere other-engageable

Curvous Energy — an introduction to dragon curves
All Posts — about dragon curves

Spiral Artefact

What’s the next number in this sequence of integers?


5, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55... (A227793 at the OEIS)

It shouldn’t be hard to work out that it’s 64 — the sum-of-digits of n is divisible by 5, i.e., digsum(n) mod 5 = 0. Now try summing the numbers in that sequence:


5 + 14 = 19
19 + 19 = 38
38 + 23 = 61
61 + 28 = 89
89 + 32 = 121
121 + 37 = 158
158 + 41 = 199
199 + 46 = 245
[...]

Here are the cumulative sums as another sequence:


5, 19, 38, 61, 89, 121, 158, 199, 245, 295, 350, 414, 483, 556, 634, 716, 803, 894, 990, 1094, 1203, 1316, 1434, 1556, 1683, 1814, 1950, 2090, 2235, 2389, 2548, 2711, 2879, 3051, 3228, 3409, 3595, 3785, 3980, 4183, 4391, 4603, 4820, 5041, 5267, 5497, 5732, 5976, 6225...

And there’s that cumulative-sum sequence represented as a spiral:

Spiral for cumulative sum of n where digsum(n) mod 5 = 0


You can see how the spiral is created by following 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E… from the center:


ZYXWVU
GFEDCT
H432BS
I501AR
J6789Q
KLMNOP

What about other values for the cumulative sums of digsum(n) mod m = 0? Here’s m = 2,3,4,5,6,7:

Spiral for cumulative sum of n where digsum(n) mod 2 = 0
s1 = 2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22…
s2 = 2, 6, 12, 20, 31, 44, 59, 76, 95, 115… (cumulative sum of s1)


sum of digsum(n) mod 3 = 0
s1 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33…
s2 = 3, 9, 18, 30, 45, 63, 84, 108, 135, 165…


sum of digsum(n) mod 4 = 0
s1 = 4, 8, 13, 17, 22, 26, 31, 35, 39, 40, 44…
s2 = 4, 12, 25, 42, 64, 90, 121, 156, 195, 235…


sum of digsum(n) mod 5 = 0
s1 = 5, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55…
s2 = 5, 19, 38, 61, 89, 121, 158, 199, 245, 295…


sum of digsum(n) mod 6 = 0
s1 = 6, 15, 24, 33, 39, 42, 48, 51, 57, 60, 66…
s2 = 6, 21, 45, 78, 117, 159, 207, 258, 315, 375…


sum of digsum(n) mod 7 = 0
s1 = 7, 16, 25, 34, 43, 52, 59, 61, 68, 70, 77…
s2 = 7, 23, 48, 82, 125, 177, 236, 297, 365, 435…


The spiral for m = 2 is strange, but the spirals are similar after that. Until m = 8, when something strange happens again:

sum of digsum(n) mod 8 = 0
s1 = 8, 17, 26, 35, 44, 53, 62, 71, 79, 80, 88…
s2 = 8, 25, 51, 86, 130, 183, 245, 316, 395, 475…


Then the spirals return to normal for m = 9, 10:

sum of digsum(n) mod 9 = 0
s1 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99…
s2 = 9, 27, 54, 90, 135, 189, 252, 324, 405, 495…


sum of digsum(n) mod 10 = 0
s1 = 19, 28, 37, 46, 55, 64, 73, 82, 91, 109, 118…
s2 = 19, 47, 84, 130, 185, 249, 322, 404, 495, 604…


Here’s an animated gif of m = 8 at higher and higher resolution:

sum of digsum(n) mod 8 = 0 (animated gif)


You might think this strange behavior is dependant on the base in which the dig-sum is calculated. It isn’t. Here’s an animated gif for other bases in which the mod-8 spiral behaves strangely:

sum of digsum(n) mod 8 = 0 in base b = 5, 6, 7, 9, 11, 12, 13 (animated gif)


But the mod-8 spiral stops behaving strangely when the spiral is like this, as a diamond:


   W
  XIV
 YJ8HU
ZK927GT
LA3016FS
 MB45ER
  NCDQ
   OP

Now the mod-8 spiral looks like this:

sum of digsum(n) mod 8 = 0 (diamond spiral)


But the mod-4 and mod-9 spirals look like this:

sum of digsum(n) mod 4 = 0 (diamond spiral)


sum of digsum(n) mod 9 = 0 (diamond spiral)


You can also construct the spirals as a triangle, like this:


     U
    VCT
   WD2CS
  XE301AR
 YF456789Q
ZGHIJKLMNOP

Here’s the beginning of the mod-5 triangular spiral:

sum of digsum(n) mod 5 = 0 (triangular spiral) (open in new window for full size)


And the beginning of the mod-8 triangular spiral:

sum of digsum(n) mod 8 = 0 (triangular spiral) (open in new window for full size)


The mod-8 spiral is behaving strangely again. So the strangeness is partly an artefact of the way the spirals are constructed.


Post-Performative Post-Scriptum

“Spiral Artefact”, the title of this incendiary intervention, is of course a tip-of-the-hat to core Black-Sabbath track “Spiral Architect”, off core Black-Sabbath album Sabbath Bloody Sabbath, issued in core Black-Sabbath success-period of 1973.

Bent Pent

This is a beautiful and interesting shape, reminiscent of a piece of jewellery:

Pentagons in a ring


I came across it in this tricky little word-puzzle:

Word puzzle using pentagon-ring


Here’s a printable version of the puzzle:

Printable puzzle


Let’s try placing some other regular polygons with s sides around regular polygons with s*2 sides:

Hexagonal ring of triangles


Octagonal ring of squares


Decagonal ring of pentagons


Dodecagonal ring of hexagons


Only regular pentagons fit perfectly, edge-to-edge, around a regular decagon. But all these polygonal-rings can be used to create interesting and beautiful fractals, as I hope to show in a future post.

Performativizing Polyhedra

Τα Στοιχεία του Ευκλείδου, ια΄

κεʹ. Κύβος ἐστὶ σχῆμα στερεὸν ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενον.
κϛʹ. ᾿Οκτάεδρόν ἐστὶ σχῆμα στερεὸν ὑπὸ ὀκτὼ τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.
κζʹ. Εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.
κηʹ. Δωδεκάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ δώδεκα πενταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον.

Euclid’s Elements, Book 11

25. A cube is a solid figure contained by six equal squares.
26. An octahedron is a solid figure contained by eight equal and equilateral triangles.
27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles.
28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.