Tri-Way to L

The name is more complicated than the shape: L-triomino. The shape is simply three squares forming an L. And it’s a rep-tile — it can be divided into four smaller copies of itself.

l-triomino

An L-triomino — three squares forming an L


l-triomino_anim

L-triomino as rep-tile


That means it can also be turned into a fractal, as I’ve shown in Rep-Tiles Revisited and Get Your Prox Off #2. First you divide an L-triomino into four sub-copies, then discard one sub-copy, then repeat. Here are the standard L-triomino fractals produced by this technique:

l-triomino_123_134

Fractal from L-triomino — divide and discard


l-triomino_234


l-triomino_124


l-triomino_124_upright


l-triomino_124_upright_static1

(Static image)


l-triomino_124_upright_static2

(Static image)


But those fractals don’t exhaust the possibilities of this very simple shape. The standard L-triomino doesn’t have true chirality. That is, it doesn’t come in left- and right-handed forms related by mirror-reflection. But if you number its corners for the purposes of sub-division, you can treat it as though it comes in two distinct orientations. And when the orientations are different in the different sub-copies, new fractals appear. You can also delay the stage at which you discard the first sub-copy. For example, you can divide the L-triomino into four sub-copies, then divide each sub-copy into four more sub-copies, and only then begin discarding.

Here are the new fractals that appear when you apply these techniques:

l-triomino_124_exp

Delay before discarding


l-triomino_124_exp_static

(Static image)


l-triomino_124_tst2_static1

(Static image)


l-triomino_124_tst2_static2

(Static image)


l-triomino_124_tst1


l-triomino_124_tst1_static1

(Static image)


l-triomino_124_tst1_static2

(Static image)


l-triomino_134_adj1

Adjust orientation


l-triomino_134_adj2


l-triomino_134_adj3


l-triomino_134_adj3_tst3

(Static image)


l-triomino_134_adj4


l-triomino_134_exp_static

(Static image)


l-triomino_234_exp

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Performativizing Papyrocentricity #51

Papyrocentric Performativity Presents:

Bits of the Best – The Shorter Strachey, Lytton Strachey, ed. Michael Holroyd and Paul Levy (Oxford University Press 1980)

Shaman On U!Copendium: An Expedition into the Rock’n’Roll Underworld, Julian Cope (Faber and Faber 2012)

Scorpions and Sea-LordsPhilip’s Guide to Seashells, A.P.H. Oliver, illustrated by James Nicholls (various)

Spike-U-LikeThe Cactus Handbook, Erik Haustein, translated by Pamela Marwood (Cathay Books 1988)

GlasguitargangDog Eat Dog: A Story of Survival, Struggle and Triumph by the Man Who Put AC/DC on the World Stage, Michael Browning (Allen & Unwin 2014)


Or Read a Review at Random: RaRaR

He Say, He Sigh, He Sow #40

Muerto, no faltarán manos piadosas que me tiren por la baranda; mi sepultura será el aire insondable; mi cuerpo se hundirá largamente y se corromperá y disolverá en el viento engendrado por la caída, que es infinita. — «La biblioteca de Babel» (1941), Jorge Luis Borges (1899–1986).

When I die, there shall be no lack of pious hands to cast me over the railing; my grave shall be the fathomless air; my body shall fall for ever and rot and dissolve in the wind generated by the fall, which is everlasting. — “The Library of Babel”, Jorge Luis Borges.

Oneiric Ocean

20000-leagues-under-the-sea


I like this illustration of a scene in Jules Vernes’s Twenty Thousand Leagues Under the Sea (1870) even more because it has at least one mistake in it. At least, I think it’s a mistake: the jellyfish on the upper left are two Portuguese men-o’-war (really colonial hydrozoans, not jellyfish). They have gas-filled float-bladders, so in reality you see them only on the surface, not hanging in midwater like that. The mistake makes the scene like a dream. The absence of colour is good too: it fixes the illustration firmly in the past and the colours you imagine are more vivid. The artist is imagining, dreaming, conjuring a vision of an oneiric ocean.

Can You Dij It? #2

It’s very simple, but I’m fascinated by it. I’m talking about something I call the digit-line, or the stream of digits you get when you split numbers in a particular base into individual digits. For example, here are the numbers one to ten in bases 2 and 3:

Base = 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010…
Base = 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101…


If you turn them into digit-lines, they look like this:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0… (A030190 in the Online Encyclopedia of Integer Sequences)
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 2, 1, 0, 0, 1, 0, 1… (A003137 in the OEIS)


At the tenth digit of the two digit-lines, both digits equal zero for the first time:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0


When the binary and ternary digits are represented together, the digit-lines look like this:

(1,1), (1,2), (0,1), (1,0), (1,1), (1,1), (0,1), (0,2), (1,2), (0,0)


But in base 4, the tenth digit of the digit-line is 1. So when do all the digits of the digit-line first equal zero for bases 2 to 4? Here the early integers in those bases:

Base 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101…

Base 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002…

Base 4: 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 200…


And here are the digits of the digit-line in bases 2 to 4 represented together:

(1,1,1), (1,2,2), (0,1,3), (1,0,1), (1,1,0), (1,1,1), (0,1,1), (0,2,1), (1,2,2), (0,0,1), (1,2,3), (1,1,2), (1,2,0), (0,2,2), (1,1,1), (1,0,2), (1,0,2), (1,1,2), (0,0,3), (0,1,3), (0,1,0), (1,0,3), (0,2,1), (0,1,3), (1,1,2), (1,0,3), (0,1,3), (1,1,1), (0,1,0), (1,1,0), (0,1,1), (1,2,0), (1,1,1), (1,2,1), (1,0,0), (0,1,2), (0,2,1), (1,1,0), (1,1,3), (0,2,1), (1,2,1), (1,2,0), (1,0,1), (1,0,1), (0,2,1), (1,0,1), (1,1,1), (1,2,2), (1,0,1), (1,2,1), (0,2,3), (0,1,1), (0,0,2), (0,2,0), (1,1,1), (0,1,2), (0,2,1), (0,1,1), (1,2,2), (1,2,2), (0,2,1), (0,0,2), (1,2,3), (0,2,1), (1,1,3), (0,2,0), (0,2,1), (1,2,3), (1,1,1), (1,0,1), (0,0,3), (1,0,2), (0,1,1), (0,0,3), (1,0,3), (0,1,2), (1,1,0), (0,0,0)

At the 78th digit, all three digits equal zero. But the 78th digit of the digit-line in base 5 is 1. So when are the digits first equal to zero in bases 2 to 5? It’s not difficult to find out, but the difficulty of the search increases fast as the bases get bigger. Here are the results up to base 13 (note that bases 11 and 12 both have zeroes at digit 103721663):

dig=0 in bases 2 to 3 at the 10th digit of the digit-line
dig=0 in bases 2 to 4 at the 78th digit of the digit-line
dig=0 in bases 2 to 5 at the 182nd digit of the digit-line
dig=0 in bases 2 to 6 at the 302nd digit of the digit-line
dig=0 in bases 2 to 7 at the 12149th digit of the digit-line
dig=0 in bases 2 to 8 at the 45243rd digit of the digit-line
dig=0 in bases 2 to 9 at the 255261st digit of the digit-line
dig=0 in bases 2 to 10 at the 8850623rd digit of the digit-line
dig=0 in bases 2 to 12 at the 103721663rd digit of the digit-line
dig=0 in bases 2 to 13 at the 807778264th digit of the digit-line


I assume that, for any base b > 2, you can find some point in the digit-line at which d = 0 for all bases 2 to b. Indeed, I assume that this happens infinitely often. But I don’t know any short-cut for finding the first digit at which this occurs.


Previously pre-posted:

Can You Dij It? #1