Think Inc

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


Boole(b)an #3

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

Boole(b)an #1
Boole(b)an #2

Root Routes

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

This Charming Dis-Arming

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

Boole(b)an #2

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

Boole(b)an

Suppose you allow a point to jump at random half-way towards one of the four vertices of a square. But not entirely at random — you ban the point from jumping towards the same vertex twice (or more) in a row. You get this pattern:

ban on v(i) + 0


It’s a fractal, that is, a shape that contains smaller and smaller copies of itself. Next you ban the point from jumping towards the vertex one place clockwise of the vertex it last jumped towards (i.e., it can jump towards, say, vertex 2 as many times as it likes, but it can’t jump towards vertex 2+1 = 3, and so on). You get this fractal:

ban on v(i) + 1


Now ban it from jumping towards the vertex two places clockwise of the vertex it last jumped towards (i.e., it can’t jump towards the diagonally opposite vertex). You get this fractal:

ban on v(i) + 2


And if you ban the point from jumping towards the vertex three places clockwise of the last vertex, you get a mirror-image of the v(i)+1 fractal (see above):

ban on v(i) + 3


The fractals above have a memory one vertex into the past: the previous vertex. Let’s try some fractals with a memory two vertices into the past: the previous vertex and the pre-previous vertex (and even the pre-pre-previous vertex).

But this time, let’s suppose that sometimes the point can’t jump if the previous or pre-previous isn’t equal to v(i) + n. So sometimes the jump is banned when the test is true, sometimes when the test is false — you might call it a boolean ban or boole(b)an. Using boole(b)ans, you can get this set of fractals:
















With these fractals, the boolean test sends the point back to the center of the square:











Posteriously post-posted

Boole(b)an #2 — a later look at ban-fractals

Controlled Chaos

The chaos game is a simple mathematical technique for creating fractals. Suppose a point jumps over and over again 1/2 of the distance towards a randomly chosen vertex of a triangle. This shape appears, the so-called Sierpiński triangle:

Sierpiński triangle created by the chaos game


But the jumps don’t have to be random: you can use an array to find every possible combination of jumps and so create a more even image. I call this controlled chaos. However, if you try the chaos game (controlled or otherwise) with a square, no fractal appears unless you restrict the vertex chosen in some way. For example, if the point can’t jump towards the same vertex twice or more in a row, this fractal appears:

Ban on jumping towards previously chosen vertex, i.e. v + 0


And if the point can’t jump towards the vertex one place clockwise of the previously chosen vertex, this fractal appears:

Ban on v + 1


If the point can’t jump towards the vertex two places clockwise of the previously chosen vertex, this fractal appears:

Ban on v + 2


If the point can’t jump towards the vertex three places clockwise, or one place anticlockwise, of the previously chosen vertex, this fractal appears (compare v + 1 above):

Ban on v + 3


You can also ban vertices based on how close the point is to them at any given moment. Suppose that the point can’t jump towards the nearest vertex, which means that it must choose to jump towards either the 2nd-nearest, 3rd-nearest or 4th-nearest vertex. A fractal we’ve already seen appears:

Must jump towards vertex at distance 2, 3 or 4


In effect, not jumping towards the nearest vertex means not jumping towards a vertex twice or more in a row. Another familiar fractal appears if the point can’t jump towards the most distant vertex:

d = 1,2,3


But new fractals also appear when the jumps are determined by distance:

d = 1,2,4


d = 1,3,4


And you can add more targets for the jumping point midway between the vertices of the square:

d = 1,2,8


d = 1,4,6


d = 1,6,8


d = 1,7,8


d = 2,3,6


d = 2,3,8


d = 2,4,8


d = 2,5,6


And what if you choose the next vertex by incrementing the previously chosen vertex? Suppose the initial vertex is 1 and the possible increments are 1, 2 and 2. This new fractal appears:

increment = 1,2,2 (for example, 1 + 1 = 2, 2 + 2 = 4, 4 + 2 = 6, and 6 is adjusted thus: 6 – 4 = 2)


And with this set of increments, it’s déjà vu all over again:

i = 2,2,3


And again:

i = 2,3,2


With more possible increments, familiar fractals appear in unfamiliar ways:

i = 1,3,2,3


i = 1,3,3,2


i = 1,4,3,3


i = 2,1,2,2


i = 2,1,3,4


i = 2,2,3,4


i = 3,1,1,2


Now try increments with midpoints on the sides:

v = 4 + midpoints, i = 1,2,4


As we saw above, this incremental fractal can also be created from a square with four vertices and no midpoints:

i = 1,3,3; initial vertex = 1


But the fractal changes when the initial vertex is set to 2, i.e. to one of the midpoints:

i = 1,3,3; initial vertex = 2


And here are more inc-fractals with midpoints:

i = 1,4,2 (cf. inc-fractal 1,2,4 above)


i = 1,4,8


i = 2,6,3


i = 3,2,6

<hr

i = 4,7,8


i = 1,2,3,5


i = 1,4,5,4


i = 6,2,4,1


i = 7,6,2,2


i = 7,8,2,4


i = 7,8,4,2


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

For a good example of how more can be less, try the chaos game. You trace a point jumping repeatedly 1/n of the way towards a randomly chosen vertex of a regular polygon. When the polygon is a triangle and 1/n = 1/2, this is what happens:

Chaos triangle #1


Chaos triangle #2


Chaos triangle #3


Chaos triangle #4


Chaos triangle #5


Chaos triangle #6


Chaos triangle #7


As you can see, this simple chaos game creates a fractal known as the Sierpiński triangle (or Sierpiński sieve). Now try more and discover that it’s less. When you play the chaos game with a square, this is what happens:

Chaos square #1


Chaos square #2


Chaos square #3


Chaos square #4


Chaos square #5


Chaos square #6


Chaos square #7


As you can see, more is less: the interior of the square simply fills with points and no attractive fractal appears. And because that was more is less, let’s see how less is more. What happens if you restrict the way in which the point inside the square can jump? Suppose it can’t jump twice towards the same vertex (i.e., the vertex v+0 is banned). This fractal appears:

Ban on choosing vertex [v+0]


And if the point can’t jump towards the vertex one place anti-clockwise of the currently chosen vertex, this fractal appears:

Ban on vertex [v+1] (or [v-1], depending on how you number the vertices)


And if the point can’t jump towards two places clockwise or anti-clockwise of the currently chosen vertex, this fractal appears:

Ban on vertex [v+2], i.e. the diagonally opposite vertex


At least, that is one possible route to those three particular fractals. You see another route, start with this simple fractal, where dividing and discarding parts of a square creates a Sierpiński triangle:

Square to Sierpiński triangle #1


Square to Sierpiński triangle #2


Square to Sierpiński triangle #3


Square to Sierpiński triangle #4


[…]


Square to Sierpiński triangle #10


Square to Sierpiński triangle (animated)


By taking four of these square-to-Sierpiński-triangle fractals and rotating them in the right way, you can re-create the three chaos-game fractals shown above. Here’s the [v+0]-ban fractal:

[v+0]-ban fractal #1


[v+0]-ban #2


[v+0]-ban #3


[v+0]-ban #4


[v+0]-ban #5


[v+0]-ban #6


[v+0]-ban #7


[v+0]-ban #8


[v+0]-ban #9


[v+0]-ban (animated)


And here’s the [v+1]-ban fractal:

[v+1]-ban fractal #1


[v+1]-ban #2


[v+1]-ban #3


[v+1]-ban #4


[v+1]-ban #5


[v+1]-ban #6


[v+1]-ban #7


[v+1]-ban #8


[v+1]-ban #9


[v+1]-ban (animated)


And here’s the [v+2]-ban fractal:

[v+2]-ban fractal #1


[v+2]-ban #2


[v+2]-ban #3


[v+2]-ban #4


[v+2]-ban #5


[v+2]-ban #6


[v+2]-ban #7


[v+2]-ban #8


[v+2]-ban #9


[v+2]-ban (animated)

And taking a different route means that you can find more fractals — as I will demonstrate.


Previously pre-posted (please peruse):

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

Dilating the Delta

A circle with a radius of one unit has an area of exactly π units = 3.141592… units. An equilateral triangle inscribed in the unit circle has an area of 1.2990381… units, or 41.34% of the area of the unit circle.

In other words, triangles are cramped! And so it’s often difficult to see what’s going on in a triangle. Here’s one example, a fractal that starts by finding the centre of the equilateral triangle:

Triangular fractal stage #1


Next, use that central point to create three more triangles:

Triangular fractal stage #2


And then use the centres of each new triangle to create three more triangles (for a total of nine triangles):

Triangular fractal stage #3


And so on, trebling the number of triangles at each stage:

Triangular fractal stage #4


Triangular fractal stage #5


As you can see, the triangles quickly become very crowded. So do the central points when you stop drawing the triangles:

Triangular fractal stage #6


Triangular fractal stage #7


Triangular fractal stage #8


Triangular fractal stage #9


Triangular fractal stage #10


Triangular fractal stage #11


Triangular fractal stage #12


Triangular fractal stage #13


Triangular fractal (animated)


The cramping inside a triangle is why I decided to dilate the delta like this:

Triangular fractal

Circular fractal from triangular fractal


Triangular fractal to circular fractal (animated)


Formation of the circular fractal (animated)


And how do you dilate the delta, or convert an equilateral triangle into a circle? You use elementary trigonometry to expand the perimeter of the triangle so that it lies on the perimeter of the unit circle. The vertices of the triangle don’t move, because they already lie on the perimeter of the circle, but every other point, p, on the perimeter of the triangles moves outward by a fixed amount, m, depending on the angle it makes with the center of the triangle.

Once you have m, you can move outward every point, p(1..i), that lies between p on the perimeter and the centre of the triangle. At least, that’s the theory between the dilation of the delta. In practice, all you need is a point, (x,y), inside the triangle. From that, you can find the angle, θ, and distance, d, from the centre, calculate m, and move (x,y) to d * m from the centre.

You can apply this technique to any fractal created in an equilateral triangle. For example, here’s the famous Sierpiński triangle in its standard form as a delta, then as a dilated delta or circle:

Sierpiński triangle

Sierpiński triangle to circular Sierpiński fractal


Sierpiński triangle to circle (animated)


But why stop at triangles? You can use the same elementary trigonometry to convert any regular polygon into a circle. A square inscribed in a unit circle has an area of 2 units, or 63.66% of the area of the unit circle, so it too is cramped by comparison with the circle. Here’s a square fractal that I’ve often posted before:

Square fractal, jump = 1/2, ban on jumping towards any vertex twice in a row


It’s created by banning a randomly jumping point from moving twice in a row 1/2 of the distance towards the same vertex of the square. When you dilate the fractal, it looks like this:

square_fractal_circ_i0

Circular fractal from square fractal, j = 1/2, ban on jumping towards vertex v(i) twice in a row


Circular fractal from square (animated)


And here’s a related fractal where the randomly jumping point can’t jump towards the vertex directly clockwise from the vertex it’s previously jumped towards (so it can jump towards the same vertex twice or more):

Square fractal, j = 1/2, ban on vertex v(i+1)


When the fractal is dilated, it looks like this:

Circular fractal from square, i = 1


Circular fractal from square (animated)


In this square fractal, the randomly jumping point can’t jump towards the vertex directly opposite the vertex it’s previously jumped towards:

Square fractal, ban on vertex v(i+2)


And here is the dilated version:

Circular fractal from square, i = 2

Circular fractal from square (animated)


And there are a lot more fractals where those came from. Infinitely many, in fact.

Square Routes Re-Re-Re-Re-Revisited

Pre-previously in my post-passionate portrayal of polygonic performativity, I’ve usually looked at what happens when a moving point is banned from jumping twice-in-a-row (and so on) towards the same vertex of a square or other polygon. But what happens when the point isn’t banned but compelled to do something different? For example, if the point usually jumps 1/2 of the distance towards the vertex for the second (third, fourth…) time, you could make it jump 2/3 of the way, like this:

usual jump = 1/2, forced jump = 2/3


And here are the fractals created when the vertex currently chosen is one or two places clockwise from the vertex chosen before:

usual jump = 1/2, forced jump = 2/3, vertex-inc = +1


j1 = 1/2, j2 = 2/3, vi = +2


Or you can make the point jump towards a different vertex to the one chosen, without recording the different vertex in the history of jumps:

v1 = +0, v2 = +1, j = 1/2


v1 = +0, v2 = +1, vi = +2


v1 = 0, v2 = +2


v1 = 0, v2 = +2, vi = +1


Or you can make the point jump towards the center of the square:

v1 = 0, v2 = center, j = 1/2


v1 = 0, v2 = center, vertex-inc = +1


v1 = 0, v2 = center, vertex-inc = +2


And so on:

v1 = +1, v2 = +1, vi = +1


v1 = +1, v2 = +1, vi = +2


v1 = +0, v2 = +1, reverse test


v1 = +0, v2 = +1, vi = +1, reverse test


v1 = +0, v2 = +1, vi = +2, reverse test


v1 = +0, v2 = +2, reverse test


v1 = +0, v2 = +2, vi = +1, reverse test


v1 = +2, v2 = +2, vi = +1, reverse test


j1 = 1/2, j2 = 2/3, vi = +0,+0 (record previous two jumps in history)


j1 = 1/2, j2 = 2/3, vi = +0,+2


j1 = 1/2, j2 = 2/3, vi = +2,+2


j1 = 1/2, j2 = 2/3, vi = +0,+0,+0 (previous three jumps)


Previously pre-posted (please peruse):

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