## martes, 7 de enero de 2014

### Blender Cycles OSL: Mandelbrot Fractal

Hi. Today I'm going to show you an OSL code to get a Mandelbrot Fractal, like this:

This fractal uses the following complex numbers formula:

Z(n+1) = Z(n) ^ 2  + C

I have parameterized the exponent of the formula and the bailout value, that is, the value to stop the iterations. So the code is this:

```shader mandelbrot_fractal_zn_zoom(
float Scale = 1.0,
float Zoom = 3.5,
vector Vector = (vector)P,
float CenterX = 0.0,
float CenterY = 0.0,
int MaxIterations = 64,
int Exponent = 2,
float Bailout = 2.0,
output float Fac = 0.0)
{
point p = (point)Vector*Scale/(2*Zoom);
int i = 0;
int j = 0;
int prevIteration = MaxIterations;
float factor = 0.0;
float foo = 0.0;

// Starting point
p += CenterX;
p += CenterY;

// Initialize
point pa = point(0.0);
point pb = point(0.0);
point pc = point(0.0);

// Mandelbrot algorithm
do
{
// z^j
for(j = 1; j < Exponent; j++) {
pc = pa * pb - pa * pb;
pc = pa * pb + pa * pb;

// for next product
pb = pc;
pb = pc;
}

// z^exponent + c
pa = pc + p;
pb = pa;
pb = pa;

i++;

if ((prevIteration == MaxIterations) && (sqrt((pa*pa) + (pa*pa)) > Bailout)) {
prevIteration = i + 1;
}
} while (i < prevIteration);

// Factor to colorize
factor = float(i);
foo = sqrt((pa*pa) + (pa*pa));
factor = factor - (log(log(foo))/log(Bailout));
factor = factor / float(MaxIterations);

Fac = factor;
}```

And this is the node setup to get the image before:

As it is using a exponential formula, if you change the exponent you can get another fractals (the power node and the color ramp could be slightly different):
• Exponent 3

• Exponent 4
• Exponent 6
• Exponent 7
• Exponent 8
• Exponent 9
• Exponent 11

You can see other fractal renders using this code here.

Bye

#### 2 comentarios:

Spirou4D dijo...

Hi Friend,

You are interested about Fractal and me too: a crop-circle draw a radio telescope in a field near a terrian radio-telescope, do you know what I mean?
Your fractal "Exponent 3" is his plan. Do you know how calculate the perimetre of this line, may be please?

Bye
Spirou4D

falckonnahabedian dijo...

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