For Revver and Fevver

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


(click or open in new window for full size)


(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:


point = 1/2


point = 1/4


point = 3/4


point = 1/8


point = 5/8


point = 3/8


point = 7/8


(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]:


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:


x = 1/2, y = 1/3


x = 1/4, y = 2/3


x = 3/4, y = 1/9


x = 1/8, y = 4/9


x = 5/8, y = 7/9


x = 3/8, y = 2/9


x = 7/8, y = 5/9


x = 1/16, y = 8/9


x = 9/16, y = 1/27…


animated square

Read full page: For Revver and Fevver

Pigmental Paradox

From Raymond Smullyan’s Logical Labyrinths (2009):

We now visit another knight/knave island on which, like on the first one, all knights tell the truth and all knaves lie. But now there is another complication! For some reason, the natives refuse to speak to strangers, but they are willing to answer yes/no questions using a secret sign language that works like this:

Each native carries two cards on his person; one is red and the other is black. One of them means yes and the other means no, but you are not told which color means what. If you ask a yes/no question, the native will flash one of the two cards, but unfortunately, you will not know whether the card means yes or no!

Problem 3.1. Abercrombie, who knew the rules of this island, decided to pay it a visit. He met a native and asked him: “Does a red card signify yes?” The native then showed him a red card.

From this, is it possible to deduce what a red card signifies? Is it possible to deduce whether the native was a knight or a knave?

Problem 3.2. Suppose one wishes to find out whether it is a red card or a black card that signifies yes. What simple yes/no question should one ask?

Terminal Breach

It’s said that, if you hear “in terms of” 23 times in 23 hours on the 23rd of the month, the ghost of William Burroughs will appear and offer you a heroin enema.

I don’t know whether this is true.

Elsewhere other-engageable:

William S. Burroughs
Alan Moore, C.B.E.
Michael Moorcock
Will Self
Stewart Home
Cormac McCarthy
Dr Joan Jay Jefferson
Serpent’s Tail
Titans of Trangression

Rep-tilian Rites

A pentomino is one of the shapes created by laying five squares edge-to-edge. There are twelve of them (not counting reflections) and this is the P-pentomino:


But it’s not just a pentomino, it’s also a rep-tile, or a shape that can divided into smaller copies of itself. There are two ways of doing this (I’ve rotated the pentomino 90° to make the images look better):



Once you’ve divided the shape into four copies, you can divide the copies, then the copies of the copies, and the copies of the copies of the copies, and so on for ever:



And if you’ve got a reptile, you can turn it into a fractal. Simply divide the shape, discard one or more copies, and continue:


Pentomino-based fractal stage 1


Pentomino-based fractal stage 2


Pentomino-based fractal stage 3


Stage 4


Stage 5


Stage 6


Stage 7


Stage 8


Stage 9


Stage 10

Here are some more fractals created using the same divide-and-discard process:



Animated version



Animated version













You can also use variants on a standard rep-tile dissection, like rotating the copies or trying different patterns of dissection at different levels to see what new shapes appear:































Rock’n’Roll Suislide

Q. Each face of a convex polyhedron can serve as a base when the solid is placed on a horizontal plane. The center of gravity of a regular polyhedron is at the center, therefore it is stable on any face. Irregular polyhedrons are easily constructed that are unstable on certain faces; that is, when placed on a table with an unstable face as the base, they topple over. Is it possible to make a model of an irregular convex polyhedron that is unstable on every face?

Portrait of Luca Pacioli (1495)

Portrait of Luca Pacioli (1495)

A. No. If a convex polyhedron were unstable on every face, a perpetual motion machine could be built. Each time the solid toppled over onto a new base it would be unstable and would topple over again.

 — From “Ridiculous Questions” in Martin Gardner’s Mathematical Magical Show (1965), chapter 10.

Bestia Bestialissima

Auberon Waugh called himself a “practitioner of the vituperative arts”. Perhaps it was a Catholic thing. And unless you know Latin, you won’t understand. Or you won’t understand as much as you might. I don’t know Latin well, but I can appreciate some of the wonderful vituperation in a book of Latin exorcisms I’ve found scanned at Google Books. The title alone is good: Flagellum Daemonum: Exorcismos Terribiles, Potentissimos et Efficaces, which means (I think) The Flail of Demons: Exorcisms Terrible, Most Potent and Effective. Or is the title Fustis Daemonum: Adiurationes Formidabiles, Potentissimas et Efficaces, meaning The Cudgel of Demons: Adjurations Formidable, Most Potent and Effective?

Vituperation from the Flagellum Daemonum (1644)

Vituperation from the Flagellum Daemonum (1644)

Either way, one of the exorcisms contains a good list of curses directed at the Devil. He’s called Bestia Omnium Bestiarum Bestialissima, meaning “Beast of All Beasts the Most Beastly”. Beside that, there are Dux Hæreticorum and Lupus Rapacissimus, “Duke of Heretics” and “Most Rapacious Wolf”. There’s an odd Sus Macra, Famelica, et Immundissima, which means something like “Scrawny, Famished and Most Filthy Hog”. Lovecraft would have liked Nefandissimus Susurrator, “Most Unspeakable Whisperer”, and Draco Iniquissimus, “Most Iniquitous Dragon”.

Pessimus Dux Tenebrarum is “Most Evil Duke of Darkness” and Janua et Vorago Inferni is “Door and Abyss of Hell”. Seminator Zizaniarum, meaning “Sower of Tares”, refers to Matthew xiii, 25: “But while men slept, his enemy came and sowed tares among the wheat, and went his way.” And those are only a few of the curses poured on the Devil’s head. I’ve turned the full list into plain text. As it says in the book that originally led me to the Flagellum Daemonum, “The following is a specimen of one of these vituperative addresses”:

Audi igitur insensate, false, reprobe, et iniquissime Spiritus. Inimice fidei. Adversarie generis humani. Mortis adductor. Vitæ raptor. Justitiæ declinator. Malorum radix. Fomes vitiorum. Seductor hominum. Proditor gentium. Incitator invidiæ. Origo aravitiæ. Causa discordiæ. Excitator malorum. Dæmonum magister. Miserrima Creature. Tentator Homininum. Deceptor malorum Angelorum. Fallax animarum. Dux Hæreticorum. Pater Mendacii. Fatue Bestialis. Tui creatoris Inimicus. Insipiens ebriose. Inique et iniquorum caput. Prædo infernalis. Serpens iniquissime. Lupe rapacissime. Sus macra, famelica, et immundissima. Bestia eruginosa. Bestia scabiosa. Bestia truculentissima. Bestia crudelis. Bestia cruenta. Bestia omnium Bestiarum Bestialissima. Ejecte de Paradise. De gratiâ Dei. De Cœli fastigio. De loco inerrabili. De Societate et consortia Angelorum. Immundissime Spiritus Initium omnium malorum. Trangressor bonæ vitæ. Veritatis et Justitiæ persecutor. Auctor fornicationum. Seminator zizaniarum. Dissipator pacis. Latro discordiæ. Pessime dux tenebrarum. Mortis inventor. Janua et vorago Inferni. Crudelis devorator animarum omniumque malorum causa. Malignissime Dæmon. Spurcissime Spiritus. Nefandissime susurrator. Nequissima Creatura. Vilissime apostata. Scelestissima latro. Impiissima bestia infernalis. Superbissime et ingratissime Spiritus. Iniquissime refuga. Tyranne, Omni bono vacue. Plene omni dolo et fallaciâ. Hominum exterminator. Derisio totius Angelicæ Naturæ. Maledicte Satana a Deo. Excommunicate a totâ cœlesti curiâ. Blaspheme Dei et omnium Sanctorum. Damnate a Deo atque Damnande. Spiritus Acherontine. Spiritus Tartaree. Fili Perditionis. Fili maledictionis æternæ. Rebellis Dei et totius cœlestis curiæ. Serpens crudelissime. Draco iniquissime. Creatura damnata, reprobata et maledicta a Deo in æternum ob superbiam nequitiam tuam.

The first line, Audi igitur insensate, false, reprobe, et iniquissime Spiritus means something like “Hear, then, Senseless, False, Reprobate and Most Iniquitous Spirit”. Then the Devil is called Inimicus Fidei, “Enemy of the Faith”, Adversarius Generis Humani, “Adversary of the Human Race”, Mortis Adductor, “Dragger to Death”, and Vitæ Raptor, “Snatcher of Life”. Then the vituperation really begins.

Fragic Carpet

Maths is like a jungle: rich, teeming and full of surprises. A waterfall here, a glade of butterflies there, a bank of orchids yonder. There is always something new to see and a different route to try. But sometimes a different route will take you to the same place. I’ve already found two ways to reach this fractal (see Fingering the Frigit and Performativizing the Polygonic):


Fractal Carpet

Now I’ve found a third way. You could call it the rep-tile route. Divide a square into four smaller squares:


Add an extra square over the centre:


Then keep dividing the squares in the same way:


Animated carpet (with coloured blocks)


Animated carpet (with empty blocks)

The colours of the fractal appear when the same pixel is covered repeatedly: first it’s red, then green, yellow, blue, purple, and so on. Because the colours and their order are arbitrary, you can use different colour schemes:


Colour scheme #1


Colour scheme #2


Colour scheme #3

Here are more colour-schemes in an animated gif:


Various colour-schemes

Now try dividing the square into nine and sixteen, with an extra square over the centre:


3×3 square + central square


3×3 square + central square (animated)


4×4 square + central square


4×4 square + central square (animated)

You can also adjust the size of the square added to the 2×2 subdivision:


2×2 square + 1/2-sized central square


2×2 square + 3/4-sized central square

Elsewhere Other-Posted:

Fingering the Frigit
Performativizing the Polygonic

Toxic Turntable #7

Currently listening…

• Slow Exploding Gulls, Salmaris EP (1997)
• Dubioso, Codicil LVI (1968)
• Ubair Yex, Weever (1973)
• Dux Tenebrarum, Quinque Fatuae (2012)
• Arctic Midge, Celsius (1992)
• Ijek Mveodeybda, Terë Conuva (1980)
• Schwarzschrein, Du Bist Dunst (1995)

Previously pre-posted:

Toxic Turntable #1
Toxic Turntable #2
Toxic Turntable #3
Toxic Turntable #4
Toxic Turntable #5
Toxic Turntable #6

Polymorphous Pursuit

Suppose four mice are standing on the corners of a large square. Each mouse begins running at the same speed towards the mouse one place away, reckoning clockwise. The mice will meet at the centre of the square and the path taken by each mouse will be what is known as a pursuit curve:


vertices = 4, mouse-increment = 1


v = 4, mi = 1 (animated)

As I showed in “Persecution Complex”, it’s easy to find variants on the basic pursuit curve. If mi = 2, i.e. each mouse runs towards the mouse two places away, the mice will run in straight lines direct to the centre of the square:


v = 4, mi = 2


v = 4, mi = 2 (animated)

That variant is trivial, but suppose there are eight mice, four starting on the corners of the square and four starting on the midpoints of the sides. Mice starting on the corners will run different pursuit curves to those starting on the midpoints, because the corners are further from the centre than the midpoints are:


v = 4, si = 1, mi = 1


If mi = 3, the pursuit curves look like this:


v = 4, si = 1, mi = 3


v = 4, si = 1, mi = 3 (animated)

Suppose there are twelve mice, four on each corner and two more on each side. If each mouse runs towards the mouse four places away, then the pursuit curves don’t all meet in the centre of the square. Instead, they meet in groups of three at four points equidistant from the centre, like this:



v = 4, si = 2, mi = 4


v = 4, si = 2, mi = 4 (animated)


v = 4, si = 4, mi = 4 (animated)


v = 4, si = 4, mi = 4 (zoom)

Now suppose each mouse become sophisticated and runs toward the combined positions of two other mice, one two places away, the other three places away, like this:


v = 4, si = 1, mi = (2, 3)


v = 4, si = 1, mi = (2, 3) (animated)

These polypursuits, as they could be called, can have complicated central regions:


v = 4, si = 2, mi = (1, 4)


v = 4, si = 2, mi = (1, 4) (animated)


v = 4, si = various, mi = various

And what if you have two teams of mice, running towards one or more mice on the other team? For example, suppose two mice, one from each team, start on each corner of a square. Each mouse on team 1 runs towards the mouse on team 2 that is one place away, while each mouse on team 2 runs towards the mouse on team 1 that is two places away. If the pursuits curves of team 1 are represented in white and the pursuit curves of team 2 in green, the curves look like this:


v = 4 * 2, vmi = 1, vmi = 2


v = 4 * 2, vmi = 1, vmi = 2


v = 4 * 2, vmi = 1, vmi = 2 (animated)

Now suppose the four mice of team 1 start on the corners while the mice of team 2 start at the centre of the square.


v = 4, centre = 4, vmi = 1, cmi = 2 (white team)


v = 4, centre = 4, vmi = 1, cmi = 2 (green team)


v = 4, centre = 4, vmi = 1, cmi = 2 (both teams)


v = 4, centre = 4, vmi = 1, cmi = 2 (animated)

Here are more variants on pursuit curves formed by two teams of mice, one starting on the corners, one at the centre:


v = 4, centre = 4, vmi = (0, 1), cmi = 0


v = 4, centre = 4, vmi = (0, 2), cmi = 0


v = 4, centre = 4, vmi = (0, 3), cmi = 0