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

Seems simple to do by trial and error. Assume you can define HSV1 and HSV2 (hue, saturation, value) of the endpoint colors you wish to use (black and white; blue and yellow; dark red and light green; etc.), and assume you have an algorithm to assign a value P between 0.0 and 1.0 to each of your pixels. Then that pixel's color becomes

(H2 - H1) * P + H1 = HP
(S2 - S1) * P + S1 = SP
(V2 - V1) * P + V1 = VP

With that done, just observe the results and see how you like them. If the algorithm to assign P is continuous, then the gradient should be smooth as well.

Answer 2

Use the smooth coloring algorithm to calculate all of the values within the viewport, then map your palette from the lowest to highest value. Thus, as you zoom in and the higher values are no longer visible, the palette will scale down as well. With the same constants for n and B you will end up with a range of 0.0 to 1.0 for a fully zoomed out set, but at deeper zooms the dynamic range will shrink, to say 0.0 to 0.1 at 200% zoom, 0.0 to 0.0001 at 20000% zoom, etc.

Answer 3

My eventual solution was to create a nice looking (and fairly large) palette and store it as a constant array in the source, then interpolate between indexes in it using the smooth coloring algorithm. The palette wraps (and is designed to be continuous), but this doesn't appear to matter much.

Answer 4

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 wolud be

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

This only works for mandelbrot, if you want to compute a smoot 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 passed to 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.