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Author Topic: Coloring algorithm to allow for multi-colored Mandelbrot edge?  (Read 11813 times)
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twinbee
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« on: March 12, 2008, 07:36:06 PM »

Despite appearances to the contrary, the whole of the stalk (taken from the Mandelbrot) is 'coated' in the same bright white that the tip is at the top right...



In fact, the whole of the Mandelbrot is like that - zoom in far enough, and the edge is always the same colour. Is there a coloring algorithm so that different edges of the Mandelbrot have different colours? In the stalk example, I want the lower parts of the stalk to be more green.
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cKleinhuis
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« Reply #1 on: March 12, 2008, 09:46:19 PM »

hmm, this is a very good question on how to do it generally, because this is a coloring made up from iteration depth, the easyiest way would be to simnply apply a gradient from the upper right to the lower left ... ,)

applied gradually to iteration depth, e.g.

gradientcolor*(iteration/iterationmax)*iterationcolor

 afro
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divide and conquer - iterate and rule - chaos is No random!
Duncan C
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« Reply #2 on: March 13, 2008, 01:49:40 PM »

Despite appearances to the contrary, the whole of the stalk (taken from the Mandelbrot) is 'coated' in the same bright white that the tip is at the top right...

<Quoted Image Removed>

In fact, the whole of the Mandelbrot is like that - zoom in far enough, and the edge is always the same colour. Is there a coloring algorithm so that different edges of the Mandelbrot have different colours? In the stalk example, I want the lower parts of the stalk to be more green.

Twinbee,

You need some way to characterize the different areas so you can color them differently. Off the top of my head, I'm thinking that you could use the average iteration value for a circular area around a point in question in order to color it. Areas deep in a "groove" are surrounded by high iteration pixels, even if you exclude neighbors that are Mandelbrot points. Areas that are out on the periphery of the set should be surrounded by pixels with lower average iteration values.

Unless you're writing your own fractal-rendering application I'm not sure how you'd accomplish this though.

I am writing my own fractal app, so I might give this a try. It occurs to me that averaging iteration values could produce smooth color transitions without the performance hit of using a continuous function (like distance estimates, continuous potential, fractional iterations values, etc) for coloring.*

Hmm....


Duncan


*My app uses a boundary following algorithm to skip iterating the interior of contiguous areas with the same iteration value. Some continuous functions like distance estimates are computationally expensive on their own, and others that aren't so bad still force me to iterate every pixel, so I lose the speedup of the boundary following optimization when I calculate.
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Regards,

Duncan C
twinbee
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« Reply #3 on: March 26, 2008, 02:20:05 AM »

Hi Duncan,

Averaging iteration values eh? That sounds interesting if you'd like to elaborate.

At first, I thought this idea would be useful to make a hypothetical 3D Mandelbrot fractal more colourful. Otherwise, it would look pretty boring if the mandelbrot cave inside was all one colour - you'd have to 'chop' the object to see the colours then.

But then it occurred to me that even for say, the mandelbrot mountain, you'd get some distinctly different types of images using coloring in this way.

My idea to solve the problem was to 'chop away' slightly at the outline, and then obviously, the tip of that image would remain white, while the rest would gradually turn green. However, you'd need to keep adjusting the chopping degree level as you zoom into the fractal. Also some detail is lost.

The other idea is to look at the iteration level perpendicular to that point in the Mandelbrot curve (at a given distance), and 'check' what colour/iteration it is. I think this what you thought of, and if so, I bet it would be horribly expensive computationally. Also, there's the problem that it might go inside the curve, and 'back out' again.
« Last Edit: March 26, 2008, 02:39:25 AM by twinbee » Logged
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