Non-parametric color mapping techniques

[tomnzed]

Thanks a lot for the post, lots of useful info

[SeryZone]

thank you very much!!! I experiment with my Iterations Equalization and I use logarithmic, because histogram take too much time for delphi programms...

[SeryZone]

I've written a short comparison of the rank-ordering and histogram methods (and some other typical ones) of mapping counts to colors

http://www.hpdz.net/TechInfo_Colorizing.htm.

I made some test images and graphs showing the effect of each method on some different images. I call them "non-parametric" because they are not based on any analytic model of how the counts are distributed (e.g. linear, log); a similar name is used for statistical tests that do not assume how the data is distributed.

This is a follow-up to a conversation between Duncan C and me in a different thread. Continuing to go into more detail on this issue in that thread seemed to off-topic, so I am starting a new one here.

Hello! how to realize rank-order equalization??? Please, even logarithmic do not give smooth results! Please, show me C code or Delphi - I can understand all!

[johandebock]

Hello! how to realize rank-order equalization??? Please, even logarithmic do not give smooth results! Please, show me C code or Delphi - I can understand all!

Check my code here:
https://github.com/johandebock/BuddhaBrot-MT/blob/master/BuddhaBrot-MT.cpp
coloring method = cm = 0 is my implementation of rank-order mapping
coloring method = cm = 1 is my implementation of histogram mapping

Easiest way is to search for hits on the normalisation value cm0n/cm1n in the code.
1st step is building up the histogram from the counts.
2nd step is using it to map the counts to colors in the color table.

[FractalStefan]

One way to do it with logarithms would be with the function 1 + 1/(log(1-x)-1). This will still map [0,1] to [0,1] monotonically increasing, also with infinite derivative at x=1.

There must be a miskake in this formula, since log(1-x) gives an error for x=1.

The function

x = log(x*4ⁿ+1) / log(4ⁿ+1)

with n = [0...10] (set by a slider control) lets the user determine the degree of the gradient.

Example images:

Original (linear gradient)


n = 0.00


n = 1.00


n = 1.50


n = 2.00


n = 2.50


Using a gamma function instead of a logarithmic one, the gradient becomes a bit softer:

x = x^1/Gamma

(Gamma = 0.1...10)

Example images:

Original (linear gradient)


Gamma = 0.50


Gamma = 1.00 (this is equal to a linear gradient)


Gamma = 1.50


Gamma = 2.00


Gamma = 2.50


Sometimes a simple 1/4 sinus curve yields good results:

x = sin(x * Pi / 2)

Example image:

Original (linear gradient)


1/4-Sinus


This function automatically adapts the degree of the gradient to the number of visible iterations/colors:

x = (log(iterations) - log(minIterations)) / (log(maxIterations) - log(g_minIterations))

Example image:

Original (linear gradient)


Auto-Logarithm


The example images were created by this JavaScript application.

Stefan

[FractalStefan]

I just wrote a more detailed article about some non-linear color mapping functions including an comparison chart with example images:

Color Mapping – Mapping an iteration count to a color in the Mandelbrot set