Game of Throwns

In “Scaffscapes”, I looked at these three fractals and described how they were in a sense the same fractal, even though they looked very different:

Fractal #1


Fractal #2


Fractal #3


But even if they are all the same in some mathematical sense, their different appearances matter in an aesthetic sense. Fractal #1 is unattractive and seems uninteresting:

Fractal #1, unattractive, uninteresting and unnamed


Fractal #3 is attractive and interesting. That’s part of why mathematicians have given it a name, the T-square fractal:

Fractal #3 — the T-square fractal


But fractal #2, although it’s attractive and interesting, doesn’t have a name. It reminds me of a ninja throwing-star or shuriken, so I’ve decided to call it the throwing-star fractal or ninja-star fractal:

Fractal #2, the throwing-star fractal


A ninja throwing-star or shuriken


This is one way to construct a throwing-star fractal:

Throwing-star fractal, stage 1


Throwing-star fractal, #2


Throwing-star fractal, #3


Throwing-star fractal, #4


Throwing-star fractal, #5


Throwing-star fractal, #6


Throwing-star fractal, #7


Throwing-star fractal, #8


Throwing-star fractal, #9


Throwing-star fractal, #10


Throwing-star fractal, #11


Throwing-star fractal (animated)


But there’s another way to construct a throwing-star fractal. You use what’s called the chaos game. To understand the commonest form of the chaos game, imagine a ninja inside an equilateral triangle throwing a shuriken again and again halfway towards a randomly chosen vertex of the triangle. If you mark each point where the shuriken lands, you eventually get a fractal called the Sierpiński triangle:

Chaos game with triangle stage 1


Chaos triangle #2


Chaos triangle #3


Chaos triangle #4


Chaos triangle #5


Chaos triangle #6


Chaos triangle #7


Chaos triangle (animated)


When you try the chaos game with a square, with the ninja throwing the shuriken again and again halfway towards a randomly chosen vertex, you don’t get a fractal. The interior of the square just fills more or less evenly with points:

Chaos game with square, stage 1


Chaos square #2


Chaos square #3


Chaos square #4


Chaos square #5


Chaos square #6


Chaos square (anim)


But suppose you restrict the ninja’s throws in some way. If he can’t throw twice or more in a row towards the same vertex, you get a familiar fractal:

Chaos game with square, ban on throwing towards same vertex, stage 1


Chaos square, ban = v+0, #2


Chaos square, ban = v+0, #3


Chaos square, ban = v+0, #4


Chaos square, ban = v+0, #5


Chaos square, ban = v+0, #6


Chaos square, ban = v+0 (anim)


But what if the ninja can’t throw the shuriken towards the vertex one place anti-clockwise of the vertex he’s just thrown it towards? Then you get another familiar fractal — the throwing-star fractal:

Chaos square, ban = v+1, stage 1


Chaos square, ban = v+1, #2


Chaos square, ban = v+1, #3


Chaos square, ban = v+1, #4


Chaos square, ban = v+1, #5


Game of Throwns — throwing-star fractal from chaos game (static)


Game of Throwns — throwing-star fractal from chaos game (anim)


And what if the ninja can’t throw towards the vertex two places anti-clockwise (or two places clockwise) of the vertex he’s just thrown the shuriken towards? Then you get a third familiar fractal — the T-square fractal:

Chaos square, ban = v+2, stage 1


Chaos square, ban = v+2, #2


Chaos square, ban = v+2, #3


Chaos square, ban = v+2, #4


Chaos square, ban = v+2, #5


T-square fractal from chaos game (static)


T-square fractal from chaos game (anim)


Finally, what if the ninja can’t throw towards the vertex three places anti-clockwise, or one place clockwise, of the vertex he’s just thrown the shuriken towards? If you can guess what happens, your mathematical intuition is much better than mine.


Post-Performative Post-Scriptum

I am not now and never have been a fan of George R.R. Martin. He may be a good author but I’ve always suspected otherwise, so I’ve never read any of his books or seen any of the TV adaptations.

Hedonic Hierairchy

Pilot-thrilling… 10 flyers’ favourites

1. Vickers VC10 — A pilot’s airliner that also beguiled its passengers
2. North American F-86 Sabre — The classic second-generation jet fighter
3. Hawker Hart — An elegant two-seat day bomber of the 1930s that delighted its pilots
4. de Havilland Hornet — The ultimate twin-piston fighter, evolved from the classic Mosquito
5. Sopwith Pup — A First World War fighter beloved by all who flew it
6. Armstrong Whitworth Siskin — Agile and manouverable radial-engined inter-war fighter
7. Stampe SV-4 — A classic sporting biplane renowned for its aerobatic qualities
8. Hawker Hurricane — A distant descendant of the Pup with similarly wide pilot appeal
9. North American P-51 Mustang — One of the greatest Second World War single-seat fighters, an ace-maker
10. Avro Lancaster — A pilot’s favourite with exceptional handling qualities for a multi-engine type

• From Aeroplane magazine, December 2003, viâ a VC10 fan-site

Scaffscapes

A fractal is a shape that contains copies of itself on smaller and smaller scales. You can find fractals everywhere in nature. Part of a fern looks like the fern as a whole:

Fern as fractal (source)


Part of a tree looks like the tree as a whole:

Tree as fractal (source)


Part of a landscape looks like the landscape as a whole:

Landscape as fractal (source)


You can also create fractals for yourself. Here are three that I’ve constructed:

Fractal #1


Fractal #2


Fractal #3 — the T-square fractal


The three fractals look very different and, in one sense, that’s exactly what they are. But in another sense, they’re the same fractal. Each can morph into the other two:

Fractal #1 → fractal #2 → fractal #3 (animated)


Here are two more fractals taken en route from fractal #2 to fractal #3, as it were:

Fractal #4


Fractal #5


To understand how the fractals belong together, you have to see what might be called the scaffolding. The construction of fractal #3 is the easiest to understand. First you put up the scaffolding, then you take it away and leave the final fractal:

Fractal #3, scaffolding stage 1


Fractal #3, stage 2


Fractal #3, stage 3


Fractal #3, stage 4


Fractal #3, stage 5


Fractal #3, stage 6


Fractal #3, stage 7


Fractal #3, stage 8


Fractal #3, stage 9


Fractal #3, stage 10


Fractal #3 (scaffolding removed)


Construction of fractal #3 (animated)


Now here’s the construction of fractal #1:

Fractal #1, stage 1


Fractal #1, stage 2


Fractal #1, stage 3

Construction of fractal #1 (animated)


Fractal #1 (static)


And the constructions of fractals #2, #4 and #5:

Fractal #2, stage 1


Fractal #2, stage 2


Fractal #2, stage 3

Fractal #2 (animated)


Fractal #2 (static)


Fractal #4, stage 1


Fractal #4, stage 2


Fractal #4, stage 3

Fractal #4 (animated)


Fractal #4 (static)


Fractal #5, stage 1


Fractal #5, stage 2


Fractal #5, stage 3

Fractal #5 (animated)


Fractal #5


Root Pursuit

Roots are hard, powers are easy. For example, the square root of 2, or √2, is the mysterious and never-ending number that is equal to 2 when multiplied by itself:

• √2 = 1·414213562373095048801688724209698078569671875376948073...

It’s hard to calculate √2. But the powers of 2, or 2^p, are the straightforward numbers that you get by multiplying 2 repeatedly by itself. It’s easy to calculate 2^p:

• 2 = 2^1
• 4 = 2^2
• 8 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
[...]

But there is a way to find √2 by finding 2^p, as I discovered after I asked a simple question about 2^p and 3^p. What are the longest runs of matching digits at the beginning of each power?

131072 = 2^17
129140163 = 3^17
1255420347077336152767157884641... = 2^193
1214512980685298442335534165687... = 3^193
2175541218577478036232553294038... = 2^619
2177993962169082260270654106078... = 3^619
7524389324549354450012295667238... = 2^2016
7524012611682575322123383229826... = 3^2016

There’s no obvious pattern. Then I asked the same question about 2^p and 5^p. And an interesting pattern appeared:

32 = 2^5
3125 = 5^5
316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185

The digits 31622 rang a bell. Isn’t that the start of √10? Yes, it is:

• √10 = 3·1622776601683793319988935444327185337195551393252168268575...

I wrote a fast machine-code program to find even longer runs of matching initial digits. Sure enough, the pattern continued:

• 316227... = 2^2728361
• 316227... = 5^2728361
• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 3162277660... = 2^961700165
• 3162277660... = 5^961700165

But why are powers of 2 and 5 generating the digits of √10? If you’re good at math, that’s a trivial question about a trivial discovery. Here’s the answer: We use base ten and 10 = 2 * 5, 10^2 = 100 = 2^2 * 5^2 = 4 * 25, 10^3 = 1000 = 2^3 * 5^3 = 8 * 125, and so on. When the initial digits of 2^p and 5^p match, those matching digits must come from the digits of √10. Otherwise the product of 2^p * 5^p would be too large or too small. Here are the records for matching initial digits multiplied by themselves:

32 = 2^5
3125 = 5^5
• 3^2 = 9

316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
• 31^2 = 961

3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
• 3162^2 = 9998244

3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185
• 31622^2 = 999950884

• 316227... = 2^2728361
• 316227... = 5^2728361
• 316227^2 = 99999515529

• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 3162277^2 = 9999995824729

• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 31622776^2 = 999999961946176

• 3162277660... = 2^961700165
• 3162277660... = 5^961700165
• 3162277660^2 = 9999999998935075600

The square of each matching run falls short of 10^p. And so when the digits of 2^p and 5^p stop matching, one power must fall below √10, as it were, and one must rise above:

3 162266908803418110961625404267... = 2^127185
3·162277660168379331998893544432... = √10
3 162288411569894029343799063611... = 5^127185

In this way, 2^p * 5^p = 10^p. And that’s why matching initial digits of 2^p and 5^p generate the digits of √10. The same thing, mutatis mutandis, happens in base 6 with 2^p and 3^p, because 6 = 2 * 3:

• 2.24103122055214532500432040411... = √6 (in base 6)

24 = 2^4
213 = 3^4
225522024 = 2^34 in base 6 = 2^22 in base 10
22225525003213 = 3^34 (3^22)
2241525132535231233233555114533... = 2^1303 (2^327)
2240133444421105112410441102423... = 3^1303 (3^327)
2241055222343212030022044325420... = 2^153251 (2^15007)
2241003215453455515322105001310... = 3^153251 (3^15007)
2241032233315203525544525150530... = 2^233204 (2^20164)
2241030204225410320250422435321... = 3^233204 (3^20164)
2241031334114245140003252435303... = 2^2110415 (2^102539)
2241031103430053425141014505442... = 3^2110415 (3^102539)

And in base 30, where 30 = 2 * 3 * 5, you can find the digits of √30 in three different ways, because 30 = 2 * 15 = 3 * 10 = 5 * 6:

• 5·E9F2LE6BBPBF0F52B7385PE6E5CLN... = √30 (in base 30)

55AA4 = 2^M in base 30 = 2^22 in base 10
5NO6CQN69C3Q0E1Q7F = F^M = 15^22
5E63NMOAO4JPQD6996F3HPLIMLIRL6F... = 2^K6 (2^606)
5ECQDMIOCIAIR0DGJ4O4H8EN10AQ2GR... = F^K6 (15^606)
5E9DTE7BO41HIQDDO0NB1MFNEE4QJRF... = 2^B14 (2^9934)
5E9G5SL7KBNKFLKSG89J9J9NT17KHHO... = F^B14 (15^9934)
[...]
5R4C9 = 3^E in base 30 = 3^14 in base 10
52CE6A3L3A = A^E = 10^14
5E6SOQE5II5A8IRCH9HFBGO7835KL8A = 3^3N (3^113)
5EC1BLQHNJLTGD00SLBEDQ73AH465E3... = A^3N (10^113)
5E9FI455MQI4KOJM0HSBP3GG6OL9T8P... = 3^EJH (3^13187)
5E9EH8N8D9TR1AH48MT7OR3MHAGFNFQ... = A^EJH (10^13187)
[...]
5OCNCNRAP = 5^I in base 30 = 5^18 in base 10
54NO22GI76 = 6^I (6^18)
5EG4RAMD1IGGHQ8QS2QR0S0EH09DK16... = 5^1M7 (5^1567)
5E2PG4Q2G63DOBIJ54E4O035Q9TEJGH... = 6^1M7 (6^1567)
5E96DB9T6TBIM1FCCK8A8J7IDRCTM71... = 5^F9G (5^13786)
5E9NM222PN9Q9TEFTJ94261NRBB8FCH... = 6^F9G (6^13786)
[...]

So that’s √10, √6 and √30. But I said at the beginning that you can find √2 by finding 2^p. How do you do that? By offsetting the powers, as it were. With 2^p and 5^p, you can find the digits of √10. With 2^(p+1) and 5^p, you can find the digits of √2 and √20, because 2^(p+1) * 5^p = 2 * 2^p * 5^p = 2 * 10^p:

•  √2 = 1·414213562373095048801688724209698078569671875376948073...
• √20 = 4·472135954999579392818347337462552470881236719223051448...

16 = 2^4
125 = 5^3
140737488355328 = 2^47
142108547152020037174224853515625 = 5^46
1413... = 2^243
1414... = 5^242
14141... = 2^6651
14142... = 5^6650
141421... = 2^35389
141420... = 5^35388
4472136... = 2^162574
4472135... = 5^162573
141421359... = 2^3216082
141421352... = 5^3216081
447213595... = 2^172530387
447213595... = 5^172530386
[...]

A Vulcar Display of Power


Post Performative Post-Scriptum

The title of this incendiary intervention is a paronomasia on Pantera’s Vulgar Display of Power (1992). I don’t like Pantera, but that’s a good title.

Performativizing Papyrocentricity #72

Papyrocentric Performativity Presents…

Homing in the Gloaming – Homing: On Pigeons, Dwellings and Why We Return, Jon Day (John Murray 2019)

Niceberg – The Storyteller: Tales of Life and Music, Dave Grohl (Simon & Schuster 2021)

Nasty Lastly – Nasty Endings 1, compiled by Dennis Pepper (Oxford University Press 2001)

Daysed and ConfusedHawkwind: Days of the Underground: Radical Escapism in the Age of Paranoia, Joe Banks (Strange Attractor 2020)

World-Wide Wipe-Out – Empty World, John Christopher (1977)

Chuck Off – Post Office, Charles Bukowski (1971)

#AllDayDong – Dong, Peter Sotos and Sam Salatta (TransVisceral Books 2022)

Meet the Maverick Munch-Bunch… – Naked Krunch: The Sinister, Sordid and Strangely Scrumptious Story of SavSnaq, Dr David M. Mitchell (Savoy Books 2022)


Or Read a Review at Random: RaRaR

Les Mains de Montesquiou

Comte Robert de Montesquiou-Fézensac, Henri Lucien Doucet (1879)


Passionately Pre-Posted…

Portrait of a Peacock — Cornelia Otis Skinner’s biographical sketch of Montesquiou
Le Paon dans les Pyrénées — review of Julian Barnes’ The Man in the Red Coat (2019)

Elsewhere Other-Accessible…

The hands of Robert de Montesquiou at Strange Flowers

Toxic Turntable #25

Currently listening…

• Lupine Katie, Dark Calico (1976)
• Ô Grecq, Mouse Makes Cop (2005)
• Julia Sedburgh, Vacant Reef (1975)
• Fox in Quake, Drifting on Sonic Seas (2002)
• الاصداف, ميوريكس ﮪاوستلم (1995)
• Pidita Kick, I Leap for Hawaii (1998)
• Yuperfen Lagekim, Jeptic (2013)
• Quiggy and the Grin, Soda Family (1983)
• Oxton Bvulpsi, Dealt a Devil’s Hand (2017)
• მარმარილი მანქანა, აბრეშუმის ჭია (1997)
• Dogmatic Cajuns, Little Bit (1996)
• Pamela Kelpstan, Episodic Outtakes (1970)
• Okeäp V, Räucherfisch (1994)
• Bells in the Battery, Hue You (1990)
• Showt Nowt, Our Wild Folkloric Dreams (2012)
• ¡XtrmDrm!, LIVE in ¡Thunderland! (2004)
• Jumjin, Mqoag i Hio (1979)
• Ѫй Ѯяоѱ Ѩѽ, Koxiio (1997)
• Edgar Roslin Quintet, Zeppelin Zoo (2017)
• Uh Bpuerw, Sqiaamn (1983)
• Stellae Vulpis, Umbrian Folk Revival (2006)
• Milirad im Untergang, Ab’xu Mim Myaceleg (1973)


Previously pre-posted:

Toxic Turntable #1#2#3#4#5#6#7#8#9#10#11#12#13#14#15#16#17#18#19#20#21#22#23#24