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I was wondering if it was possible to sort of enumerate all points on the border of the Mandelbrot set. Similar to the idea of external angles, which map (at least the rational points) of an infinitely large surrounding circle to points on the border.

But external angles are related to the idea of (electrical) field lines, which leads to a somewhat uneven distribution. Simply speaking, approximately all field lines starting infinitely far away will land on antennas and spikes of the Mandelbrot border, and approximately zero field lines will land deep in spirals.

Ever since I first read about external angles, their uneven distribution has irked me. If there existed a more evenly distributed mapping, we could do the following: generate a random number in [0 .. 1], evenly distributed (possibly with very high precision), and map it to the Mandelbrot border. Zoom in, and be surprised!

This algorithm does not work all that well with external angles, because those tend to cover only the same boring few regions over and over.


I guess maybe one could use the level sets to construct some sort of "equidistant" mapping. Basically the borders of all level sets are smooth curves, and their arc length can be determined (albeit not necessarily in closed form). So an equidistant mapping from [0 .. 1] to the corresponding fraction of total arc length can be defined for each and every level set.

The question is: would the series of mappings at all converge as the number of iterations approaches infinity? Would the resulting mapping indeed be somewhat equidistant (in the sense that it could enable a random selection of targets to cover all of the border with equal probability)?

I don't really know where to start looking for an answer.

I don't think external angles are very uneven... it's just the rational ones land on "special points". It would be interesting to see which structures appear at an external angle of the Golden Ratio (one of the "most irrational" algebraics), e, 1/e, pi, 1/pi, etc....

Even so, it would be hard to talk about a mapping from [0...1] to the arc length along the Mset, because the boundary is so fractal that it has 2 dimensions. It's possible to create a mapping from a 1-dimensional space to a full 2-dimensional space, such as the even/odd digit proof goes. External angles also do this. So you might end up wanting a mapping from [0...1] to an external angle, which you then look for on the set...

I very much like the idea of generating a random number and "being surprised"! I would love to see some little applet that does this...  :D

Strictly speaking, no, it's not possible.  The boundary contains an uncountable infinity of points, so it's impossible to enumerate them all, even with infinite time and memory.

One simple alternative is to tick off key points of the boundary of some components.  For example, the main cardioid and the large disk centered at (-1,0) both have simple equations parameterized by an angle (around (0,0) for the cardioid and around (-1,0) for the disk).  Then, one could create lists of {angle, x, y, interesting note} for each piece.

For other components, the equations are substantially messier, but I think one can find points numerically based on the period of the component and the derivative of the composition of the iterates.  But, I haven't tried it.

Yes, I realize my terminology was wrong for the cardinalities involved.

To clear things up, what I am dreaming of is some way to more evenly cover the Mandelbrot sets border with any kind of sample points. For the purpose of breaking my own habit of always ending up zooming into the same old areas. :)

The external angles don't work for that as well as I would want. So I came up with this suggestion about the level set curves. But I don't know how to formally reason about those curves. I could try numerical experiments, but even then I'd need to know something abstract about the level set curves. Say, bounds of their curvature, or some other measure of what the smallest detail is of any given level set.

For the level set thing, you could try following the path given by directional estimation... I posted about that in another thread. It's like distance estimation, but instead to estimates the direction to the nearest point on the set. By following these lines - which would, admittedly, be a computationally intensive process - you're traveling directly at the Mandelbrot. Just by choosing an initial point anywhere on the circle (radius=2). The one problem I can think of is how to get "into" some parts of the set - the points where two protruding stalks produce a saddle node, if the path you take is viewed as a vector field. Because the distance estimation does not have a continuous derivative, it is not very possible to make useful decision making about navigating between two spikes. Using the derivative of the iteration count in conjunction, though, could help guide the path an potentially lead to useful navigating. First one would need to work out what the equations for directional estimation are, because as far as I know, they don't exist anywhere yet.  :embarrass:

I like the idea of mapping the unit interval to the border of the M-set, but I'm fairly sure an explicit formula is impossible. I bet you could do it with iteration, though--for example, starting with a circle of radius 2 centered at the origin and doing some sort of reverse-iteration process. Though reverse-iteration doesn't really work as well for the Mandelbrot as for Julias...