What's the (closest) analogue of Sierpinski triangle to disk?
What's the analogue of Sierpinski triangle to disk?
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2It probably depends on your definition of "closest", "analogue", and, quite possibly, "of". :) – 2012-07-27
3 Answers
You could replace each disk with seven disks of $1/3$ the radius packed inside it, like this:
After four iterations, it looks like this:
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2@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(j*Pi/3),sin(j*Pi/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); – 2012-07-27
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.
I've always though of this having fairly strong links:
I'm not sure what it's called though.
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0@achillehui thanks! – 2013-10-27