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)


The Hex Fractor

Pre-previously on Overlord-in-terms-of-issues-around-the-Über-Feral, I looked at the fractals created when various restrictions are placed on a point jumping at random half-way towards the vertices of a square. For example, the point can be banned from jumping towards the same vertex twice in a row or towards the vertex to the left of the vertex it has just jumped towards, and so on.

Today I want to look at what happens to a similar point moving inside pentagons and hexagons. If the point can’t jump twice towards the same vertex of a pentagon, this is the fractal that appears:

Ban second jump towards same vertex (v + 0)


Ban second jump towards same vertex (color)


If the point can’t jump towards the vertex immediately to the left of the one it’s just jumped towards, this is the fractal that appears:

Ban jump towards v + 1


Ban jump towards v + 1 (color)


And this is the fractal when the ban is on the vertex two places to the left:

Ban jump towards v + 2


Ban jump towards v + 2 (color)


You can also ban more than one vertex:

Ban jump towards v + 0,1


Ban jump towards v + 1,2


Ban jump towards v + 1,4


Ban jump towards v + 1,4 (color)


Ban jump towards v + 2,3


And here are fractals created in similar ways inside hexagons:

Ban jump towards v + 0,1


Ban jump towards v + 0,3


Ban jump towards v + 0,1,2


Ban jump towards v + 0,1,2 (color)


Ban jump towards v + 0,1,4


Ban jump towards v + 0,1,5


Ban jump towards v + 0,2,4


Ban jump towards v + 0,2,4 (color)


Ban jump towards v + 1,2,3


Ban jump towards v + 1,2,3 (color)


Ban jump towards v + 1,2,4


Ban jump towards v + 1,2,4, (color)


Ban jump towards v + 1,3,5


Ban jump towards v + 1,3,5 (color)


Ban jump towards v + 1,2


Ban jump towards v + 1,2


Ban jump towards v + 1,3


Ban jump towards v + 1,3 (color)


Ban jump towards v + 1,5


Ban jump towards v + 1,5 (color)


Ban jump towards v + 2,3


Ban jump towards v + 2,3 (color)


Ban jump towards v + 2,4


Ban jump towards v + 2,4 (color)


Elsewhere other-accessible:

Square Routes Re-Verticed

Bent for the Pent

A triangle can be tiled with triangles and a square with squares, but a pentagon can’t be tiled with pentagons. At least, not in the same way, using smaller copies of the same shape. The closest you can get is this:

Pentaflake #1


If you further subdivide the pentagon, you create what is known as a pentaflake:

Pentaflake #2


Pentaflake #3


Pentaflake #4


Pentaflake (animated)


Pentaflake (static)


But if you bend the rules and use irregular smaller pentagons, you can tile a pentagon like this, creating what I called a pentatile:

Pentatile stage 1


Further subdivisions create an interesting final pattern:

Pentatile #2


Pentatile #3


Pentatile #4


Pentatile #5


Pentatile #6


Pentatile (animated)


Pentatile (static)


By varying the size of the central pentagon, you can create other patterns:

Pentatile #1 (animated)


Pentatile #2 (animated)

Pentatile #2







Pentatile with no central pentagon


And here are various pentatiles in an animated gif:


And here are some variations on the pentaflake:







Elsewhere other-posted:

Bent for the Rent (1976) — the title of the incendiary intervention above is of course a reference to the “first and last glitter-rock album” by England’s loudest band, Spinal In Terms Of Tap
Phrallic Frolics — more on pentaflakes

Phrallic Frolics

It’s a classic of low literature:

There was a young man of Devizes
Whose balls were of different sizes:
     The one was so small
     ’Twas no use at all;
But t’other won several prizes.

But what if he had been a young man with balls of different colours? This is a core question I want to interrogate issues around in terms of the narrative trajectory of this blog-post. Siriusly. But it’s not the keyliest core question. More corely keyly still, I want to ask what a fractal phallus might look like. Or a phrallus, for short. The narrative trajectory initializes with this fractal, which is known as a pentaflake (so-named from its resemblance to a snowflake):

Pentaflake — a pentagon-based fractal


It’s created by repeatedly replacing pentagons with six smaller pentagons, like this:

Pentaflake stage 0


Pentaflake stage 1


Pentaflake stage 2


Pentaflake stage 3


Pentaflake stage 3


Pentaflake stage 4


Pentaflake (animated)


Pentaflake (static)


This is another version of the pentaflake, missing the central pentagon of the six used in the standard pentaflake:

No-Center Pentaflake stage 0


No-Center Pentaflake stage 1


Stage 2


Stage 3


Stage 4


No-Center Pentaflake (animated)


No-Center Pentaflake (static #1)


No-Center Pentaflake (static #2)


The phrallus, or fractal phallus, begins with an incomplete version of the first stage of the pentaflake (note balls of different colours):

Phrallus stage 1


Phrallus stage 1 (monochrome)


Phrallus stage 2


Phrallus stage 3


Stage 4


Stage 5


Stage 6


Stage 7


Stage 8


And there you have it: a fractal phallus, or phrallus. Here is an animated version:

Phrallus (animated)


Phrallus (static)


But the narrative trajectory is not over. The center of the phrallus can be rotated to yield mutant phralloi. Stage #1 of the mutants looks like this:

Phrallus (mutation #1)


Phrallus (mutation #2)


Phrallus (mutation #3)


Phrallus (mutation #4)


Phrallus (mutation #5)


Mutant phralloi (rotating)


Here are some animations of the mutant phralloi:

Phrallus (mutation #3) (animated)


Phrallus (mutation #5) (animated)


This mutation doesn’t position the pentagons in the usual way:

Phrallus (another upright version) (animated)


The static mutant phralloi look like this:

Phrallus (mutation #2)


Phrallus (mutation #3)


Phrallus (upright #2)


And if the mutant phralloi are combined in a single image, they rotate like this:

Mutant phralloi (rotating)


Coloured mutant rotating phralloi #1


Coloured mutant rotating phralloi #2


The Swing’s the Thing

Order emerges from chaos with a triangle or pentagon, but not with a square. That is, if you take a triangle or a pentagon, chose a point inside it, then move the point repeatedly halfway towards a vertex chosen at random, a fractal will appear:

triangle

Sierpiński triangle from point jumping halfway to randomly chosen vertex


pentagon

Sierpiński pentagon from point jumping halfway to randomly chosen vertex


But it doesn’t work with a square. Instead, the interior of the square slowly fills with random points:

square

Square filling with point jumping halfway to randomly chosen vertex


As I showed in Polymorphous Perverticity, you can create fractals from squares and randomly moving points if you ban the point from choosing the same vertex twice in a row, and so on. But there are other ways. You can take the point, move it towards a vertex at random, then swing it around the center of the square through some angle before you mark its position, like this:

square_sw90

Point moves at random, then swings by 90° around center


square_sw180

Point moves at random, then swings by 180° around center


You can also adjust the distance of the point from the center of the square using a formula like dist = r * rmdist, where dist is the distance, r is the radius of the circle in which the circle is drawn, and rm takes values like 0.1, 0.25, 0.5, 0.75 and so on:

square_dist_rm0_05

Point moves at random, dist = r * 0.05 – dist


square_dist_rm0_1

Point moves at random, dist = r * 0.1 – dist


square_dist_rm0_2

Point moves at random, dist = r * 0.2 – dist


But you can swing the point while applying a vertex-ban, like banning the previously chosen vertex, or the vertex 90° or 180° away. In fact, swinging the points converts one kind of vertex ban into the others.

square_ban0

Point moves at random towards vertex not chosen previously


square_ban0_sw405

Point moves at random, then swings by 45°


square_ban0_sw360

Point moves at random, then swings by 360°


square_ban0_sw697

Point moves at random, then swings by 697.5°


square_ban0_sw720

Point moves at random, then swings by 720°


square_ban0_sw652

Point moves at random, then swings by 652.5°


square_ban0_swing_va_animated

Animated angle swing


You can also reverse the swing at every second move, swing the point around a vertex instead of the center or around a point on the circle that encloses the square. Here are some of the fractals you get applying these techniques.
square_ban0_sw45_rock

Point moves at random, then swings alternately by 45°, -45°


square_ban0_sw90_rock

Point moves at random, then swings alternately by 90°, -90°


square_ban0_sw135_rock

Point moves at random, then swings alternately by 135°, -135°


square_ban0_sw180_rock

Point moves at random, then swings alternately by 180°, -180°


square_ban0_sw225

Point moves at random, then swings alternately by 225°, -225°


square_ban0_sw315

Point moves at random, then swings alternately by 315°, -315°


square_ban0_sw360_rock

Point moves at random, then swings alternately by 360°, -360°


square_swing_vx0_va_animated

Animated alternate swing


square_circle_sw45

Point moves at random, then swings around point on circle by 45°


square_circle_sw67

Point moves at random, then swings around point on circle by 67.5°


square_circle_sw90

Point moves at random, then swings around point on circle by 90°


square_circle_sw112

Point moves at random, then swings around point on circle by 112.5°


square_circle_sw135

Point moves at random, then swings around point on circle by 135°


square_circle_sw180

Point moves at random, then swings around point on circle by 180°


square_circle_sw_animated

Animated circle swing


For Revver and Fevver

This shape reminds me of the feathers on an exotic bird:

feathers

(click or open in new window for full size)


feathers_anim

(animated version)


The shape is created by reversing the digits of a number, so you could say it involves revvers and fevvers. I discovered it when I was looking at the Halton sequence. It’s a sequence of fractions created according to a simple but interesting rule. The rule works like this: take n in base b, reverse it, and divide reverse(n) by the first power of b that is greater thann.

For example, suppose n = 6 and b = 2. In base 2, 6 = 110 and reverse(110) = 011 = 11 = 3. The first power of 2 that is greater than 6 is 2^3 or 8. Therefore, halton(6) in base 2 equals 3/8. Here is the same procedure applied to n = 1..20:

1: halton(1) = 1/10[2] → 1/2
2: halton(10) = 01/100[2] → 1/4
3: halton(11) = 11/100[2] → 3/4
4: halton(100) = 001/1000[2] → 1/8
5: halton(101) = 101/1000[2] → 5/8
6: halton(110) = 011/1000 → 3/8
7: halton(111) = 111/1000 → 7/8
8: halton(1000) = 0001/10000 → 1/16
9: halton(1001) = 1001/10000 → 9/16
10: halton(1010) = 0101/10000 → 5/16
11: halton(1011) = 1101/10000 → 13/16
12: halton(1100) = 0011/10000 → 3/16
13: halton(1101) = 1011/10000 → 11/16
14: halton(1110) = 0111/10000 → 7/16
15: halton(1111) = 1111/10000 → 15/16
16: halton(10000) = 00001/100000 → 1/32
17: halton(10001) = 10001/100000 → 17/32
18: halton(10010) = 01001/100000 → 9/32
19: halton(10011) = 11001/100000 → 25/32
20: halton(10100) = 00101/100000 → 5/32…

Note that the sequence always produces reduced fractions, i.e. fractions in their lowest possible terms. Once 1/2 has appeared, there is no 2/4, 4/8, 8/16…; once 3/4 has appeared, there is no 6/8, 12/16, 24/32…; and so on. If the fractions are represented as points in the interval [0,1], they look like this:

line1_1_2

point = 1/2


line2_1_4

point = 1/4


line3_3_4

point = 3/4


line4_1_8

point = 1/8


line5_5_8

point = 5/8


line6_3_8

point = 3/8


line7_7_8

point = 7/8


line_b2_anim

(animated line for base = 2, n = 1..63)


It’s apparent that Halton points in base 2 will evenly fill the interval [0,1]. Now compare a Halton sequence in base 3:

1: halton(1) = 1/10[3] → 1/3
2: halton(2) = 2/10[3] → 2/3
3: halton(10) = 01/100[3] → 1/9
4: halton(11) = 11/100[3] → 4/9
5: halton(12) = 21/100[3] → 7/9
6: halton(20) = 02/100 → 2/9
7: halton(21) = 12/100 → 5/9
8: halton(22) = 22/100 → 8/9
9: halton(100) = 001/1000 → 1/27
10: halton(101) = 101/1000 → 10/27
11: halton(102) = 201/1000 → 19/27
12: halton(110) = 011/1000 → 4/27
13: halton(111) = 111/1000 → 13/27
14: halton(112) = 211/1000 → 22/27
15: halton(120) = 021/1000 → 7/27
16: halton(121) = 121/1000 → 16/27
17: halton(122) = 221/1000 → 25/27
18: halton(200) = 002/1000 → 2/27
19: halton(201) = 102/1000 → 11/27
20: halton(202) = 202/1000 → 20/27
21: halton(210) = 012/1000 → 5/27
22: halton(211) = 112/1000 → 14/27
23: halton(212) = 212/1000 → 23/27
24: halton(220) = 022/1000 → 8/27
25: halton(221) = 122/1000 → 17/27
26: halton(222) = 222/1000 → 26/27
27: halton(1000) = 0001/10000 → 1/81
28: halton(1001) = 1001/10000 → 28/81
29: halton(1002) = 2001/10000 → 55/81
30: halton(1010) = 0101/10000 → 10/81

And here is an animated gif representing the Halton sequence in base 3 as points in the interval [0,1]:

line_b3_anim


Halton points in base 3 also evenly fill the interval [0,1]. What happens if you apply the Halton sequence to a two-dimensional square rather a one-dimensional line? Suppose the bottom left-hand corner of the square has the co-ordinates (0,0) and the top right-hand corner has the co-ordinates (1,1). Find points (x,y) inside the square, with x supplied by the Halton sequence in base 2 and y supplied by the Halton sequence in base 3. The square will gradually fill like this:

square1

x = 1/2, y = 1/3


square2

x = 1/4, y = 2/3


square3

x = 3/4, y = 1/9


square4

x = 1/8, y = 4/9


square5

x = 5/8, y = 7/9


square6

x = 3/8, y = 2/9


square7

x = 7/8, y = 5/9


square8

x = 1/16, y = 8/9


square9

x = 9/16, y = 1/27…


square_anim

animated square


Read full page: For Revver and Fevver