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

Back to Drac’ #2

Boring, dull, staid, stiff, everyday, ordinary, unimaginative, unexceptional, crashingly conventional — the only interesting thing about squares is the number of ways you can say how uninteresting they are. Unlike triangles, which vary endlessly and entertainingly, squares are square in every sense of the word.

And they don’t get any better if you tilt them, as here:

Sub-squares from gray square (with corner-numbers)


Nothing interesting can emerge from that set of squares. Or can it? As I showed in Curvous Energy, it can. Suppose that the gray square is dividing into the colored squares like a kind of amoeba. And suppose that the colored squares divide in their turn. So square divides into sub-squares and sub-squares divide into sub-sub-squares. And so on. And all the squares keep the same relative orientation.

What happens if the gray square divides into sub-squares sq2 and sq9? And then sq2 and sq9 each divide into their own sq2 and sq9? And so on. Something very unsquare-like happens:

Square-split stage #1


Stage #2


Square-split #3


Square-split #4


Square-split #5


Square-split #6


Square-split #7


Square-split #8


Square-split #9


Square-split #10


Square-split #11


Square-split #12


Square-split #13


Square-split #14


Square-split #15


Square-split #16


Square-split (animated)


The square-split creates a beautiful fractal known as a dragon-curve:

Dragon-curve


Dragon-curve (red)


And dragon-curves, at various angles and in various sizes, emerge from every other possible pair of sub-squares:

Lots of dragon-curves


And you get other fractals if you manipulate the sub-squares, so that the corners are rotated or reverse-rotated:

Rotation = 1,2 (sub-square #1 unchanged, in sub-square #2 corner 1 becomes corner 2, 2 → 3, 3 → 4, 4 → 1)


rot = 1,2 (animated)


rot = 1,2 (colored)


rot = 1,5 (in sub-square #2 corner 1 stays the same, 4 → 2, 3 stays the same, 2 → 4)


rot = 1,5 (anim)


rot = 4,7 (sub-square #2 flipped and rotated)


rot = 4,7 (anim)


rot = 4,7 (col)


rot = 4,8


rot = 4,8 (anim)


rot = 4,8 (col)


sub-squares = 2,8; rot = 5,6


sub-squares = 2,8; rot = 5,6 (anim)


sub-squares = 2,8; rot = 5,6 (col)


Another kind of dragon-curve — rot = 3,2


rot = 3,2 (anim)


rot = 3,2 (col)


sub-squares = 4,5; rot = 3,9


sub-squares = 4,5; rot = 3,9 (anim)


sub-squares = 4,5; rot = 3,9 (col)


Elsewhere other-accessible…

Curvous Energy — a first look at dragon-curves
Back to Drac’ — a second look at dragon-curves

Think Inc #2

In a pre-previous post called “Think Inc”, I looked at the fractals created by a point first jumping halfway towards the vertex of a square, then using a set of increments to decide which vertex to jump towards next. For example, if the inc-set was [0, 1, 3], the point would jump next towards the same vertex, v[i]+0, or the vertex immediately clockwise, v[i]+1, or the vertex immediately anti-clockwise, v[i]+3. And it would trace all possible routes using that inc-set. Then I added refinements to the process like giving the point extra jumping-targets half-way along each side.

Here are some more variations on the inc-set theme using two and three extra jumping-targets along each side of the square. First of all, try two extra jumping-targets along each side and a set of three increments:

inc = 0, 1, 6


inc = 0, 2, 6


inc = 0, 2, 8


inc = 0, 3, 6


inc = 0, 3, 9


inc = 0, 4, 8


inc = 0, 5, 6


inc = 0, 5, 7


inc = 1, 6, 11


inc = 2, 6, 10


inc = 3, 6, 9


Now try two extra jumping-targets along each side and a set of four increments:

inc = 0, 1, 6, 11


inc = 0, 2, 8, 10


inc = 0, 3, 7, 9


inc = 0, 4, 8, 10


inc = 0, 5, 6, 7


inc = 0, 5, 7, 8


inc = 1, 6, 7, 9


inc = 1, 4, 6, 11


inc = 1, 5, 7, 11


inc = 2, 4, 8, 10


inc = 3, 5, 7, 9


And finally, three extra jumping-targets along each side and a set of three increments:

inc = 0, 3, 13


inc = 0, 4, 8


inc = 0, 4, 12


inc = 0, 5, 11

inc = 0, 6, 9


inc = 0, 7, 9


Previously Pre-Posted

Think Inc — an earlier look at inc-set fractals

Dime Time

Everyone knows the shapes for one and two dimensions, far fewer know the shapes for three and four dimensions, let alone five, six and seven. And what shapes are those? The shapes that answer this question:

• How many equidistant points are possible in 1d, 2d, 3d, 4d…?

In one dimension it’s obvious that the answer is 2. In other words, you can get only two equidistant points, (a,b), on a straight line. Point a must be as far from point b as point b is from point a. You can’t add a third point, c, such that (a,b,c) are equidistant. Not on a straight line in 1d. But suppose you bend the line into a circle, so that you’re working in two dimensions. It’s easy to place three equidistant points, (a,b,c), on a circle.

equidistant points on a circle

Three equidistant points around a circle forming the vertices of an equilateral triangle


And it’s also easy to see that the three points will form the vertices of an equilateral triangle. Now try adding a fourth point, d. If you place it in the center of the triangle, it will be equidistant from (a,b,c), but it will be nearer to (a,b,c) than they are to each other. So you can have only 2 equidistant points in 1d and 3 equidistant points in 2d.

But what are the co-ordinates of the equidistant points in 1d and 2d? Suppose (a,b) in 1d are given the co-ordinates (0) and (1), so that a is 1 unit distant from b. When you move to 2d and add point c, the co-ordinates for (a,b) become (0,0) and (1,0). They’re still 1 unit distant from each other. But what are the co-ordinates for c? Start by placing c exactly midway between a and b, so that it has the co-ordinates (0.5,0) and is 0.5 units distant from both a and b. Now, if you move c in the first dimension, it will become nearer either to a or b: (0.49,0) or (0.51,0) or (0.48,0) or (0.52,0)…

But if you move c in the second dimension, it will always be equidistant from a and b, because (a,b) stay in the first dimension, as it were, and c moves equally away from both into the second dimension. So where in 2d will c be 1 unit distant from both a and b just as a and b are 1 unit distant from each other in 1d? You can see the answer here:

equilateral_triangle heightHeight of an equilateral triangle


The co-ordinates for c are (0.5,√3/2) or (0.5,0.8660254…), because the second co-ordinate satisfies the Pythagorean equation 1^2 = 0.5^2 + (√3/2)^2 = 0.25 + 0.75. That is, to find the second co-ordinate of c for 2d, you find the answer to √(1 – 0.5^2) = √(1-0.25) = √0.75 = √(3/4) = √3/2 = 0.8660254….

But you can’t add a fourth point, d, in 2d such that (a,b,c,d) are equidistant. So let’s move to 3d for the points (a,b,c,d). Begin with point d in the center of the triangle formed by (a,b,c), where it will have the co-ordinates (0.5,√3/6,0) = (0.5,0.28867513…,0) and will be equidistant from (a,b,c). But d will be nearer to (a,b,c) than they are to each other. However, if you move d in the third dimension, it will be moving equally away from (a,b,c). So where in 3d will d be 1 unit from (a,b,c)? By analogy with 2d, the third co-ordinate for d will satisfy the generalized Pythagorean equation √(1 – 0.5^2 – (√3/6)^2). And √6/3 = √(1 – 0.5^2 – (√3/6)^2) = 0.81649658… So point d will have the co-ordinates (0.5,√3/6,√6/3) = (0.5, 0.288675135…, 0.816496581…).

And the four points (a,b,c,d) will be the vertices of a three-dimensional shape called the tetrahedron:

Rotating tetrahedron

Rotating tetrahedron


But you can’t add a fifth point, e, in 3d such that (a,b,c,d,e) are equidistant. So let’s move to 4d, the fourth dimension, for the points (a,b,c,d,e). Begin with point e in the center of the tetrahedron formed by (a,b,c,d), where it will have the co-ordinates (0.5,√3/6,√6/12,0) = (0.5,0.28867513…, 0.2041241…, 0) and will be equidistant from (a,b,c,d). But e will be nearer to (a,b,c,d) than they are to each other. However, if you move e in the fourth dimension, it will be moving equally away from (a,b,c,e). So where in 4d will e be 1 unit from (a,b,c,d)? By analogy with 2d and 3d, the co-ordinate for 4d will satisfy the equation √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2). And √10/4 = √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2) = 0.79056941… So point e will have the co-ordinates (0.5,√3/6,√6/3,√10/4) = (0.5, 0.288675135…, 0.816496581…, 0.79056941…).

And the five points (a,b,c,d,e) will be the vertices of a four-dimensional shape called variously the hyperpyramid, the 5-cell, the pentachoron, the 4-simplex, the pentatope, the pentahedroid and the tetrahedral pyramid. It’s impossible for 3d creatures like human beings (at present) to visualize the hyperpyramid, but we can see its 3d shadow, as it were. And here is the 3d shadow of a rotating hyperpyramid:

Rotating hyperpyramid or 5-cell

Rotating hyperpyramid


N.B. Wikipedia reveals the mathematically beautiful fact that the “simplest set of coordinates [for a hyperpyramid] is: (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (φ,φ,φ,φ), with edge length 2√2, where φ is the golden ratio.”

So that’s the hyperpyramid, with 5 points in 4d. But you can’t add a sixth point, f, in 4d such that (a,b,c,d,e,f) are equidistant. You have to move to 5d. And it should be clear by now that in any dimension nd, the maximum possible number of equidistant points, p, in that dimension will be p = n+1. And here are the co-ordinates for p in dimensions 1 to 10 (the co-ordinates are given in full for 1d to 4d, then for 5d to 10d only the co-ordinates of the additional point are given):

d1: (0), (1)
d2: (0,0), (1,0), (0.5,0.866025404)
d3: (0,0,0), (1,0,0), (0.5,0.866025404,0), (0.5,0.288675135,0.816496581)
d4: 0.5, 0.288675135, 0.204124145, 0.790569415
d5: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.774596669
d6: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.763762616
d7: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.755928946
d8: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.75
d9: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.745355992
d10: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.0745355992, 0.741619849

In each dimension d, the final co-ordinate, cd+1, of the additional point satisfies the generalized Pythagorean equation cd+1 = √(1 – c1^2 – c2^2 – … cd^2).


Readers’ advisory: I am not a mathematician and the discussion above cannot be trusted to be free of errors, whether major or minor.