I got #interior checking working for the #BurningShip , based on Xavier Buff's method for the Mandelbrot set presented by Arnaud Cheritat: https://www.math.univ-toulouse.fr/~cheritat/wiki-draw/index.php/Mandelbrot_set#The_idea

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.

How many Taylor series terms are needed to accurately approximate √(a+x)−√(a)? #math #maths

blog post with formulas and not much else

https://mathr.co.uk/blog/2019-05-13_autostereograms.html

got centred pattern fill working

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

texture is based on https://rdex.mathr.co.uk/kiblix/texture/1324 #reactiondiffusion

party like it's 1979

#C99 #autostereogram #sphere

https://en.wikipedia.org/wiki/Autostereogram

```

// gcc -std=c99 -Wall -Wextra -pedantic -O3 -o autostereo autostereo.c -lm && ./autostereo > autostereo.ppm

#include <math.h>

#include <stdio.h>

#include <stdlib.h>

#include <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][3];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[0][0],w*3,1,stdout);}return 0;}

```

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

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).

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How to #OCR a #PDF with #tesseract :

```

pdfimages -all input.pdf tmp

ls tmp-* > tmp.list

tesseract tmp.list output pdf

```

adds selectable #text to #scanned #images #cheatsheet #commandline

making art with maths and algorithms

Joined May 2018