Twi-Phi

Here’s a pentagon:

Stage #1


And here’s the pentagon with smaller pentagons on its vertices:

Stage #2


And here’s more of the same:

Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Animated fractal


At infinity, the smaller pentagons have reached out like arms to exactly fill the gaps between themselves without overlapping. But how much smaller is each set of smaller pentagons than its mother-pentagon when the gaps are exactly filled? Well, if the radius of the mother-pentagon is r, then the radius of each daughter-pentagon is r * 1/(φ^2) = r * 0·38196601125…

But what happens if the radius relationship of mother to daughter is r * 1/φ = r * 0·61803398874 = r * (φ-1)? Then you get this fractal:

Stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Stage #8


Stage #9


Animated 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.

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.

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


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 #3

Pre-previously in my passionate portrayal of polygonic performativity, I showed how a single point jumping randomly (or quasi-randomly) towards the vertices of a polygon can create elaborate fractals. For example, if the point jumps 1/φth (= 0.6180339887…) of the way towards the vertices of a pentagon, it creates this fractal:

Point jumping 1/φth of the way to a randomly (or quasi-randomly) chosen vertex of a pentagon


But as you might expect, there are different routes to the same fractal. Suppose you take a pentagon and select a single vertex. Now, measure the distance to each vertex, v(1,i=1..5), of the original pentagon (including the selected vertex) and reduce it by 1/φ to find the position of a new vertex, v(2,i=1..5). If you do this for each vertex of the original pentagon, then to each vertex of the new pentagons, and so on, in the end you create the same fractal as the jumping point does:

Shrink pentagons by 1/φ, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Shrink by 1/φ (animated) (click for larger if blurred)


And here is the route to a centre-filled variant of the fractal:

Central pentagon, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Central pentagon (animated) (click for larger if blurred)


Using this shrink-the-polygon method, you can reach the same fractals by a third route. This time, use vertex v(1,i) of the original polygon as the centre of the new polygon with its vertices v(2,i=1..5). Creation of the fractal looks like this:

Pentagons over vertices, shrink by 1/φ, stage #1 (no pentagons over vertices)


Stage #2


Stage #3


Stage #4


Stage #4


Stage #5


Stage #7


Pentagons over vertices (animated) (click for larger if blurred)


And here is a third way of creating the centre-filled pentagonal fractal:

Pentagons over vertices and central pentagon, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Stage #6


Stage #7


Pentagons over vertices with central pentagon (animated) (click for larger if blurred)


And here is a fractal created when there are three pentagons to a side and the pentagons are shrunk by 1/φ^2 = 0.3819660112…:

Pentagon at vertex + pentagon at mid-point of side, shrink by 1/φ^2


Final stage


Pentagon at vertex + pentagon at mid-point of side (animated) (click for larger if blurred)


Pentagon at vertex + pentagon at mid-point of side + central pentagon, shrink by 1/φ^2 and c. 0.5, stage #1


Stage #2


Stage #3


Stage #4


Stage #5


Pentagon at vertex + mid-point + center (animated) (click for larger if blurred)


Previously pre-posted:

Performativizing the Polygonic #2
Performativizing the Polygonic #1

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

Horn Again

Pre-previously on Overlord-in-terms-of-Core-Issues-around-Maximal-Engagement-with-Key-Notions-of-the-Über-Feral, I interrogated issues around this shape, the horned triangle:

unicorn_reptile_static

Horned Triangle (more details)


Now I want to look at the tricorn (from Latin tri-, “three”, + -corn, “horn”). It’s like a horned triangle, but has three horns instead of one:

Tricorn, or three-horned triangle


These are the stages that make up the tricorn:

Tricorn (stages)


Tricorn (animated)


And there’s no need to stop at triangles. Here is a four-horned square, or quadricorn:

Quadricorn


Quadricorn (animated)


Quadricorn (coloured)


And a five-horned pentagon, or quinticorn:

Quinticorn, or five-horned pentagon


Quinticorn (anim)


Quinticorn (col)


And below are some variants on the shapes above. First, the reversed tricorn:

Reversed Tricorn


Reversed Tricorn (anim)


Reversed Tricorn (col)


The nested tricorn:

Nested Tricorn (anim)


Nested Tricorn (col)


Nested Tricorn (red-green)


Nested Tricorn (variant col)


The nested quadricorn:

Nested Quadricorn (anim)


Nested Quadricorn


Nested Quadricorn (col #1)


Nested Quadricorn (col #2)


Finally (and ferally), the pentagonal octopus or pentapus:

Pentapus (anim)


Pentapus


Pentapus #2


Pentapus #3


Pentapus #4


Pentapus #5


Pentapus #6


Pentapus (col anim)


Elsewhere other-engageable:

The Art Grows Onda — the horned triangle and Katsushika Hokusai’s painting The Great Wave off Kanagawa (c. 1830)

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)