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What's the (closest) analogue of Sierpinski triangle to disk?

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Since a disk is topologically homeomorphic to a triangle, you can just apply a homeomorphism from the disk to the triangle, apply the process of Sierpinski to the triangle, and then take the inverse of the homeomorphism. –  Thomas Andrews Jul 25 '12 at 20:14
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The three (so far) answers have each elected to maintain different parts of the Sierpinski triangle in making the analogy. Douglas Hofstadter has written much on the various ways of making analogies like this, most directly in his Metamagical Themas book. –  Ross Millikan Jul 25 '12 at 20:46
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It probably depends on your definition of "closest", "analogue", and, quite possibly, "of". :) –  Rahul Jul 27 '12 at 16:06

3 Answers 3

up vote 12 down vote accepted

You could replace each disk with seven disks of $1/3$ the radius packed inside it, like this:

enter image description here

After four iterations, it looks like this:

enter image description here

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Very nice. If I'm not mistaken, it looks like each of the holes tends to a Koch snowflake. –  Ilmari Karonen Jul 25 '12 at 21:00
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+1 for pretty pictures. –  Rick Decker Jul 26 '12 at 0:15
    
@RickDecker: Agreed. Robert, how did you make these? –  akkkk Jul 27 '12 at 14:19
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Here's some code for doing this disk fractal in Mathematica: With[{n = 4}, Nest[(# /. Disk[c_?VectorQ, r_] :> Append[Map[Function[p, Disk[p + c, r/3]], 2 r Map[Composition[Through, {Cos, Sin}], Pi Range[0, 5]/3]/3], Disk[c, r/3]]) &, Graphics[Disk[{0, 0}, 1]], n]] –  J. M. Jul 27 '12 at 15:43
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@Auke: I used Maple. > with(plots): with(plottools): P:= [disk([0,0],1,colour=red,numpoints=12)]: F:= p -> (homothety(p,1/3), seq(homothety(p,1/3,evalf([cos(jPi/3),sin(jPi/3)])),j=0..5)): for i from 1 to 4 do P:= map(F,P) end do: display(P,circle([0,0],1),axes=none,scaling=constrained); –  Robert Israel Jul 27 '12 at 17:03

The difficulty is that a triangle naturally decomposes into four similar triangles, while a disk doesn't decompose into disks of any size. You have to give up something. A sector of a disk looks a lot like a triangle, so you could divide your disk into sectors (any number that you like) and carry out the Sierpinski construction in each sector, bending the bases of the triangles around the circle at each radius. As the Sierpinski triangle is usually done for an equilateral triangle, I would do six sectors of $\frac \pi 3$ each and call it the closest analogue.

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I've always though of this having fairly strong links:

enter image description here

I'm not sure what it's called though.

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Apollonian gasket –  achille hui Oct 23 '13 at 14:16
    
@achillehui thanks! –  will Oct 27 '13 at 19:36

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