Photo of developing ferns by the German nature photographer Karl Blossfeldt (1866-1932)
(open in new window for full image)
Post-Performative Post-Scriptum
“Farnsicht” is a pun on German Farn, meaning “fern”, and Fernsicht, meaning “view” or “visibility” (literally fern, “far”, + Sicht, “visibility”).
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
In a prev-previous post, I looked at this interesting fractal image on the front cover of a Ray Bradbury book:
It seems obvious that the image is created from photographs: only the body of the centaur is drawn by hand. And here’s my attempt at extending the fractality of the image:
• Mythical Mathical — Man-Horse! — the pre-previous post about the fractal centaur
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
That’s a striking cover — and more than that. The blog where I found the cover says this: “This very odd cover clearly features a heavily rouged glam rock centaur with a rather natty feather-cut hairstyle flexing his biceps, his forearms transmogrifying into miniature bicep flexing glam rock figures. I think I’m slowly losing the plot here.” Losing the plot? No, losing the mathematicality in the mythical. The artist has started to make the centaur into a fractal. Or rather, the artist has started to make more explicit what is already there in the human body. As I wrote in “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.
This is a T-square fractal:
T-square fractal
Or you could say it’s a T-square fractal with the scaffolding taken away, because there’s nothing to show how it was made. And how is a T-square fractal made? There are many ways. One of the simplest is to set a point jumping 1/2 of the way towards one or another of the four vertices of a square. If the point is banned from jumping towards the vertex two places clockwise (or counter-clockwise) of the vertex, v[i=1..4], it’s just jumped towards, you get a T-square fractal by recording each spot where the point lands.
You also get a T-square if the point is banned from jumping towards the vertex most distant from the vertex, v[i], it’s just jumped towards. The most distant vertex will always be the diagonally opposite vertex, or the vertex, v[i+2], two places clockwise of v[i]. So those two bans are functionally equivalent.
But what if you don’t talk about bans at all? You can also create a T-square fractal by giving the point three choices of increment, [0,1,3], after it jumps towards v[i]. That is, it can jump towards v[i+0], v[i+1] or v[i+3] (where 3+2 = 5 → 5-4 = 1; 3+3 = 6 → 2; 4+1 = 5 → 1; 4+2 = 6 → 2; 4+3 = 7 → 3). Vertex v[i+0] is the same vertex, v[i+1] is the vertex one place clockwise of v[i], and v[i+3] is the vertex two places clockwise of v[i].
So this method is functionally equivalent to the other two bans. But it’s easier to calculate, because you can take the current vertex, v[i], and immediately calculate-and-use the next vertex, without having to check whether the next vertex is forbidden. In other words, if you want speed, you just have to Think Inc!
Speed becomes important when you add a new jumping-target to each side of the square. Now the point has 8 possible targets to jump towards. If you impose several bans on the next jump, e.g the point can’t jump towards v[i+2], v[i+3], v[i+5], v[i+6] and v[i+7], you will have to check for five forbidden targets. But using the increment-set [0,1,4] you don’t have to check for anything. You just inc-and-go:
inc = 0, 1, 4
Here are more fractals created with the speedy inc-and-go method:
inc = 0, 2, 3
inc = 0, 2, 5
inc = 0, 3, 4
inc = 0, 3, 5
inc = 1, 4, 7
inc = 2, 4, 7
inc = 0, 1, 4, 7
inc = 0, 3, 4, 5
inc = 0, 3, 4, 7
inc = 0, 4, 5, 7
inc = 1, 2, 6, 7
With more incs, there are more possible paths for the jumping point and the fractals become more “solid”:
inc = 0, 1, 2, 4, 5
inc = 0, 1, 2, 6, 7
inc = 0, 1, 3, 5, 7
Now try applying inc-and-go to a pentagon:
inc = 0, 1, 2
(open in new window if blurred)
inc = 0, 2, 3
And add a jumping-target to each side of the pentagon:
inc = 0, 2, 5
inc = 0, 3, 6
inc = 0, 3, 7
inc = 1, 5, 9
inc = 2, 5, 8
inc = 5, 6, 9
And add two jumping-targets to each side of the pentagon:
inc = 0, 1, 7
inc = 0, 2, 12
inc = 0, 3, 11
inc = 0, 3, 12
inc = 0, 4, 11
inc = 0, 5, 9
inc = 0, 5, 10
inc = 2, 7, 13
inc = 2, 11, 13
inc = 3, 11, 13
After the pentagon comes the hexagon:
inc = 0, 1, 2
inc = 0, 1, 5
inc = 0, 3, 4
inc = 0, 3, 5
inc = 1, 3, 5
inc = 2, 3, 4
Add a jumping-target to each side of the hexagon:
inc = 0, 2, 5
inc = 0, 2, 9
inc = 0, 6, 11
inc = 0, 3, 6
inc = 0, 3, 8
inc = 0, 3, 9
inc = 0, 4, 7
inc = 0, 4, 8
inc = 0, 5, 6
inc = 0, 5, 8
inc = 1, 5, 9
inc = 1, 6, 10
inc = 1, 6, 11
inc = 2, 6, 8
inc = 2, 6, 10
inc = 3, 5, 7
inc = 3, 6, 9
inc = 6, 7, 11
In the posts “Boole(b)an #1″ and “Boole(b)an #2” I looked at fractals created by certain kinds of ban on a point jumping (quasi-)randomly towards the four vertices, v=1..4, of a square. For example, suppose the program has a vertex-history of 2, that is, it remembers two jumps into the past, the previous jump and the pre-previous jump. There are sixteen possible combinations of pre-previous and previous jumps: [1,1], [1,2], 1,3] … [4,2], [4,3], [4,4].
Let’s suppose the program bans 4 of those 16 combinations by reference to the current possible jump. For example, it might ban [0,0]; [0,1]; [0,3]; [2,0]. To see what that means, let’s suppose the program has to decide at some point whether or not to jump towards v=3. It will check whether the combination of pre-previous and previous jumps was [3+0,3+0] = [3,3] or [3+0,3+1] = [3,4] or [3+0,3+3] = [3,2] or [3+2,3+0] = [1,3] (when the sum > 4, v = sum-4). If the previous combination is one of those pairs, it bans the jump towards v=3 and chooses another vertex; otherwise, it jumps towards v=3 and updates the vertex-history. This is the fractal that appears when those bans are used:
ban = [0,0]; [0,1]; [0,3]; [2,0]
And here are more fractals using a vertex-history of 2 and a ban on 4 of 16 possible combinations of pre-previous and previous jump:
ban = [0,0]; [0,1]; [0,3]; [2,2]
ban = [0,0]; [0,2]; [1,0]; [3,0]
ban = [0,0]; [0,2]; [1,1]; [3,3]
ban = [0,0]; [0,2]; [1,3]; [3,1]
ban = [0,0]; [1,0]; [2,2]; [3,0]
ban = [0,0]; [1,1]; [1,2]; [3,2]
Continue reading “Boole(b)an #3”…
Elsewhere other-engageable
Suppose a point traces all possible routes jumping half-way towards the three vertices of an equilateral triangle. A special kind of shape appears — a fractal called the Sierpiński triangle that contains copies of itself at smaller and smaller scales:
Sierpiński triangle, jump = 1/2
And what if the point jumps 2/3rds of the way towards the vertices as it traces all possible routes? You get this dull fractal:
Triangle, jump = 2/3
But if you add targets midway along each side of the triangle, you get this fractal with the 2/3rds jump:
Triangle, jump = 2/3, side-targets
Now try the 1/2-jump triangle with a target at the center of the triangle:
Triangle, jump = 1/2, central target
And the 2/3-jump triangle with side-targets and a central target:
Triangle, jump = 2/3, side-targets, central target
But why stop at simple jumps like 1/2 and 2/3? Let’s take the distance to the target, td, and use the function 1-(sqrt(td/7r)), where sqrt() is the square-root and 7r is 7 times the radius of the circumscribing circle:
Triangle, jump = 1-(sqrt(td/7r))
Here’s the same jump with a central target:
Triangle, jump = 1-(sqrt(td/7r)), central target
Now let’s try squares with various kinds of jump. A square with a 1/2-jump fills evenly with points:
Square, jump = 1/2 (animated)
The 2/3-jump does better with a central target:
Square, jump = 2/3, central target
Or with side-targets:
Square, jump = 2/3, side-targets
Now try some more complicated jumps:
Square, jump = 1-sqrt(td/7r)
Square, jump = 1-sqrt(td/15r), side-targets
And what if you ban the point from jumping twice or more towards the same target? You get this fractal:
Square, jump = 1-sqrt(td/6r), ban = prev+0
Now try a ban on jumping towards the target two places clockwise of the previous target:
Square, jump = 1-sqrt(td/6r), ban = prev+2
And the two-place ban with a central target:
Square, jump = 1-sqrt(td/6r), ban = prev+2, central target
And so on:
Square, jump = 1-sqrt(td/6.93r), ban = prev+2, central target
Square, jump = 1-sqrt(td/7r), ban = prev+2, central target
These fractals take account of the previous jump and the pre-previous jump:
Square, jump = 1-sqrt(td/4r), ban = prev+2,2, central target
Square, jump = 1-sqrt(td/5r), ban = prev+2,2, central target
Square, jump = 1-sqrt(td/6r), ban = prev+2,2, central target
Elsewhere other-accessible
• Boole(b)an #2 — fractals created in similar ways
One of the charms of living in an old town or city is finding new routes to familiar places. It’s also one of the charms of maths. Suppose a three-armed star sprouts three half-sized arms from the end of each of its three arms. And then sprouts three quarter-sized arms from the end of each of its nine new arms. And so on. This is what happens:
Three-armed star
3-Star sprouts more arms
Sprouting 3-Star #3
Sprouting 3-Star #4
Sprouting 3-Star #5
Sprouting 3-Star #6
Sprouting 3-Star #7
Sprouting 3-Star #8
Sprouting 3-Star #9
Sprouting 3-Star #10
Sprouting 3-Star #11 — the Sierpiński triangle
Sprouting 3-star (animated)
The final stage is a famous fractal called the Sierpiński triangle — the sprouting 3-star is a new route to a familiar place. But what happens when you trying sprouting a four-armed star in the same way? This does:
Four-armed star #1
Sprouting 4-Star #2
Sprouting 4-Star #3
Sprouting 4-Star #4
Sprouting 4-Star #5
Sprouting 4-Star #6
Sprouting 4-Star #7
Sprouting 4-Star #8
Sprouting 4-Star #9
Sprouting 4-Star #10
Sprouting 4-star (animated)
There’s no obvious fractal with a sprouting 4-star. Not unless you dis-arm the 4-star in some way. For example, you can ban any new arm sprouting in the same direction as the previous arm:
Dis-armed 4-star (+0) #1
Dis-armed 4-Star (+0) #2
Dis-armed 4-Star (+0) #3
Dis-armed 4-Star (+0) #4
Dis-armed 4-Star (+0) #5
Dis-armed 4-Star (+0) #6
Dis-armed 4-Star (+0) #7
Dis-armed 4-Star (+0) #8
Dis-armed 4-Star (+0) #9
Dis-armed 4-Star (+0) #10
Dis-armed 4-star (+0) (animated)
Once again, it’s a new route to a familiar place (for keyly committed core components of the Overlord-of-the-Über-Feral community, anyway). Now try banning an arm sprouting one place clockwise of the previous arm:
Dis-armed 4-Star (+1) #1
Dis-armed 4-Star (+1) #2
Dis-armed 4-Star (+1) #3
Dis-armed 4-Star (+1) #4
Dis-armed 4-Star (+1) #5
Dis-armed 4-Star (+1) #6
Dis-armed 4-Star (+1) #7
Dis-armed 4-Star (+1) #8
Dis-armed 4-Star (+1) #9
Dis-armed 4-Star (+1) #10
Dis-armed 4-Star (+1) (animated)
Again it’s a new route to a familiar place. Now trying banning an arm sprouting two places clockwise (or anti-clockwise) of the previous arm:
Dis-armed 4-Star (+2) #1
Dis-armed 4-Star (+2) #2
Dis-armed 4-Star (+2) #3
Dis-armed 4-Star (+2) #4
Dis-armed 4-Star (+2) #5
Dis-armed 4-Star (+2) #6
Dis-armed 4-Star (+2) #7
Dis-armed 4-Star (+2) #8
Dis-armed 4-Star (+2) #9
Dis-armed 4-Star (+2) #10
Dis-armed 4-Star (+2) (animated)
Once again it’s a new route to a familiar place. And what happens if you ban an arm sprouting three places clockwise (or one place anti-clockwise) of the previous arm? You get a mirror image of the Dis-armed 4-Star (+1):
Dis-armed 4-Star (+3)
Here’s the Dis-armed 4-Star (+1) for comparison:
Dis-armed 4-Star (+1)
Elsewhere other-accessible
• Boole(b)an #2 — other routes to the fractals seen above
In “Boole(b)an”, I looked at some of the things that happen when you impose bans of different kinds on a point jumping half-way towards a randomly chosen vertex of a square. If the point can’t jump towards the same vertex twice (or more) in a row, you get the fractal below (or rather, you get a messier version of the fractal below, because I’ve used an algorithm that finds all possible routes to create the fractals in this post):
ban = v(i) + 0
If the point can’t jump towards the vertex one place clockwise of the point it has just jumped towards, you get this fractal:
ban = v(i) + 1
If the point can’t jump towards the vertex two places clockwise (or anti-clockwise) of the point it has just jumped towards, you get this fractal:
ban = v(i) + 2
Finally, you get a mirror-image of the one-place-clockwise fractal when the ban is on jumping towards the vertex three places clockwise (or one place anti-clockwise) of the previous vertex:
ban = v(i) + 3
Now let’s introduce the concept of “vertex-history”. The four fractals above use a vertex-history of 1, vh = 1, because they look one step into the past, at the previously chosen vertex. Because there are four vertices, there are four possible previous vertices. But when vh = 2, you’re taking account of both the previous vertex, v(i), and what you might call the pre-previous vertex, v(i-1). There are sixteen possible combinations of previous vertex and pre-previous vertex (16 = 4 x 4).
Now, suppose the jump-ban is imposed when one of two conditions is met: the vertex is 1) one place clockwise of the previous vertex and the same as the pre-previous vertex; 2) three places clockwise of the previous chosen vertex and the same as the pre-previous vertex. So the boolean test is (condition(1) AND condition(2)) OR (condition(3) AND condition(4)). When you apply the test, you get this fractal:
ban = v(i,i-1) + [0,1] or v(i,i-1) + [0,3]
The fractal looks more complex, but I think it’s a blend of some combination of the four classic fractals shown at the beginning of this post. Here are more multiple-ban fractals using vh = 2 and bani = 2:
ban = [0,1] or [1,1]
ban = [0,2] or [2,0]
ban = [0,2] or [2,2]
ban = [1,0] or [3,0]
ban = [1,1] or [3,3]
ban = [1,2] or [2,2]
ban = [1,2] or [3,2]
ban = [1,3] or [2,0]
ban = [1,3] or [3,1]
ban = [2,0] or [2,2]
ban = [2,1] or [2,3]
For the fractals below, vh = 2 and bani = 3 (i.e., bans are imposed when one of three possible conditions is met). Again, I think the fractals are blends of some combination of the four classic ban-fractals shown at the beginning of this post:
ban = [0,0] or [1,2] or [3,2]
ban = [0,0] or [1,3] or [3,1]
ban = [0,0] or [2,1] or [2,3]
ban = [0,1] or [0,2] or [0,3]
ban = [0,1] or [0,3] or [1,1]
ban = [0,1] or [0,3] or [2,0]
ban = [0,1] or [0,3] or [2,2]
ban = [0,1] or [1,1] or [1,2]
ban = [0,1] or [1,1] or [3,0]
ban = [0,1] or [1,2] or [3,2]
ban = [0,2] or [1,0] or [3,0]
ban = [0,2] or [1,1] or [3,3]
ban = [0,2] or [1,2] or [2,2]
ban = [0,2] or [1,2] or [3,1]
ban = [0,2] or [1,2] or [3,2]
ban = [0,2] or [1,3] or [2,0]
ban = [0,2] or [1,3] or [3,1]
ban = [0,2] or [2,0] or [2,2]
ban = [0,2] or [2,1] or [2,3]
ban = [0,2] or [2,2] or [3,2]
ban = [0,3] or [1,0] or [2,0]
ban = [1,0] or [1,2] or [3,0]
ban = [1,0] or [2,2] or [3,0]
ban = [1,1] or [2,0] or [3,3]
ban = [1,1] or [2,1] or [3,1]
ban = [1,1] or [2,2] or [3,3]
ban = [1,1] or [2,3] or [3,3]
ban = [1,2] or [2,0] or [3,1]
ban = [1,2] or [2,0] or [3,2]
ban = [1,2] or [2,1] or [2,3]
ban = [1,2] or [2,3] or [3,2]
ban = [1,2] or [3,2] or [3,3]
ban = [1,3] or [2,0] or [2,2]
ban = [1,3] or [2,0] or [3,0]
ban = [1,3] or [2,0] or [3,1]
ban = [1,3] or [2,2] or [3,1]
ban = [2,0] or [3,1] or [3,2]
ban = [2,1] or [2,3] or [3,2]
Previously pre-posted
Boole(b)an — an early look at ban-fractals
