excerpt of a Burning Ship (quasi-)Julia set rendered in the same way

trying to figure out a heuristic for limiting iteration count to prevent the cells becoming too small and numerous, nothing working yet

Playing around with low-iteration count, binary decomposition, edge detection renderings of the Burning Ship. Maybe something to turn into a colouring book?

claude boosted

OK, let's try to make this happen (again).

Enter THE POLL'S DILEMMA 2: Pareto Edition!

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(a,b)=(0.25,-0.75) results in a dense pattern

trouble is to show that it remains bounded...


plot "data.txt" u 1:2:0 w dots lc palette


Some progress: (a,b)=(-0.75,-0.25) gives a non-repeating but apparently bounded collection of iterates.

Need to improve my gnuplot skills to colour the points in a comprehensible gradient from first point to last

I got checking working for the , based on Xavier Buff's method for the Mandelbrot set presented by Arnaud Cheritat: math.univ-toulouse.fr/~cherita

Replace der by the Jacobian $L$ w.r.t. $(x_1, y_1)$. Replace squared_modulus(der) with $|\det{L}|$. Arbitrarily use the pixel spacing for eps.

Should be straightforward to generalize the idea for other formulas.

The red zone of unknown is troubling - I wonder what is really going on in there.

x/(√(a+x)+√(a)) seems to be a worthwhile rewrite, haven't spotted a fatal flaw. considering |x|<<|a| in general, intending to use it for perturbation rendering of fractals in the future, perhaps

mandelbulb autostereogram (cross-eyed) using a slice from images.nasa.gov/details-GSFC_2 made seamless in gimp

. image
o stereo image pair
O autostereogram image
( ) autostereogram image pair

got centred pattern fill working

print at 180dpi (A4) and view cross-eyed from about 7 inches

texture is based on rdex.mathr.co.uk/kiblix/textur

party like it's 1979


// gcc -std=c99 -Wall -Wextra -pedantic -O3 -o autostereo autostereo.c -lm && ./autostereo > autostereo.ppm
<math.h>
<stdio.h>
<stdlib.h>
<time.h>
typedef int N;typedef double R;typedef unsigned char B;N main(){srand(time(0));R e=3,a=8,r=100,Z=2,D=18,b=18;N w=1920,h=1080;printf("P6\n%d %d\n255\n",w,h);for(N j=0;j<h;++j){R y=(h*0.5-(j+0.5))/r;B p[w];for(N i=0;i<w;++i){R x=(i+0.5-w*0.5)/r;R S=Z*Z-x*x-y*y;R d=S>0?D-sqrt(S):b;N o=floor(fabs(d-a)*e/d*r+rand()/(R)RAND_MAX);if(i-o<0||o==0)for(N c=0; c<3; ++c)p[i][c]=rand();else for(N c=0;c<3;++c)p[i][c]=p[i-o][c];}fwrite(&p,w*3,1,stdout);}return 0;}


(I have a less obfuscated version but it's too long to toot)

claude boosted

let's make this happen

I think i figured out a method to read off external angles of islands from binary decomposition images of the Mandelbrot set.

First, draw a line from a tip of a filament in the period doubling cascade following the partition until you can go no further. Then do the same for the two next-largest filaments either side.

From the top of each smaller filament, collect cells until you reach the top of the longer filament in between. The number of cells you collect is the period of the island, and the two bit-strings are the possibly-rotated external angle pair.

To work out the rotation offset needed, note that the last digit of a pair that lands together must differ. This allows many to be rejected. Note also that if the first digits differ, the island must be on the real axis and so those that are not complements of each other can be rejected. Finally, if they are complement, the island must be in the wake of the Feigenbaum point with an external angle .01 10 1001 10010110 ... (the other is its complement)., which allows more cases to be rejected.

Still need to prove that it works in all cases. Probably not cheaper than tracing external rays (it is still O(period^2), not counting image rendering time, but more parallelism is possible).

claude boosted

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pdfimages -all input.pdf tmp
ls tmp-* > tmp.list
tesseract tmp.list output pdf 