Smooth spectrum for Mandelbrot Set rendering

Question

I'm currently writing a program to generate really enormous (65536x65536 pixels and above) Mandelbrot images, and I'd like to devise a spectrum and coloring scheme that does them justice. The wikipedia featured mandelbrot image seems like an excellent example, especially how the palette remains varied at all zoom levels of the sequence. I'm not sure if it's rotating the palette or doing some other trick to achieve this, though.

I'm familiar with the smooth coloring algorithm for the mandelbrot set, so I can avoid banding, but I still need a way to assign colors to output values from this algorithm.

The images I'm generating are pyramidal (eg, a series of images, each of which has half the dimensions of the previous one), so I can use a rotating palette of some sort, as long as the change in the palette between subsequent zoom levels isn't too obvious.

Answer 1

This is the smooth color algorithm:

Lets say you start with the complex number z0 and iterate n times until it escapes. Let the end point be zn.

A smooth value would be

nsmooth := n + 1 - Math.log(Math.log(zn.abs()))/Math.log(2)

This only works for mandelbrot, if you want to compute a smooth function for julia sets, then use

Complex z = new Complex(x,y);
double smoothcolor = Math.exp(-z.abs());

for(i=0;i<max_iter && z.abs() < 30;i++) {
    z = f(z);
    smoothcolor += Math.exp(-z.abs());
}

Then smoothcolor is in the interval (0,max_iter).

Divide smoothcolor with max_iter to get a value between 0 and 1.

To get a smooth color from the value:

This can be called, for example (in Java):

Color.HSBtoRGB(0.95f + 10 * smoothcolor ,0.6f,1.0f);

since the first value in HSB color parameters is used to define the color from the color circle.