The hardest maze in the world...

[twinbee]

...is the Mandelbrot set! Just think; all of of the tiny paths are connected, and only the slightest hints are given at any junction as to what the correct path to choose is to (say...) find the main core from one of the tiny offshoots.

But it got me thinking. For any given iteration depth, what is the maximum number of straight connected lines (to form a path) one would take to connect a point to any other point in the set? (One is not allowed to cross outside the set.) Is it possible to formalize this somehow? Is the growth of lines exponential or double exponential with the iteration level maybe?

An example for iteration 10 would be seven connected lines. (going from the offshoot near the very bottom of the Mandelbrot to the symmetrical offshoot near the very top).

[lkmitch]

If I remember correctly, the boundary of the Mandelbrot set can be formed from a really warped circle.  In that case, maybe it's not a maze at all, just an infinitely-long direct path?

I don't understand your example; why would 7 be the maximum number of lines at iteration 10?  In fact, don't you mean minimum number of lines?  Wouldn't the maximum number always be infinity?

[twinbee]

Interesting. Yes you could trail the edge or 'wall' of the path of the Mandelbrot, and reach the core eventually, but then the same goes for any maze. Example, following the inside of the wall from this super-easy maze would always reach the exit.

However, it take many times longer to complete any maze this way. That's why it's best to look for clues as to the right path.

Here's an example from the Mandelbrot:


You could 'trail the inside', but it would takes ages. Best to follow your instincts and check out the most likely path to the 'exit' (main mandelbrot bulb, not seen here obviously), which is probably the big path at the bottom.

On a related note, I wonder what the perimeter of the Mandelbrot set is given an iteration level.

I don't understand your example; why would 7 be the maximum number of lines at iteration 10?  In fact, don't you mean minimum number of lines?  Wouldn't the maximum number always be infinity?

Nope, I mean the maximum. The idea is to reach any part of the Mandelbrot from any other part. Some paths only require two straight lines, some only one. At iteration 10, the 'hardest' path that you can choose which maximizes the number of connected straight lines is 7, which is from the bottom to the top.

Examples:
At iteration 1 or iteration 2: Max = 1 line (since it's a circle, any part can be reached from a single line max to any other part)

At iteration 3: Max = 2 lines (slightly warped circle, so to get from the top to the left, you need to connect two straight lines. Most other paths would just take one line, but the top (or bottom) to the left is the most difficult 'route'). See below. The red line is not allowed. However, you can connect the two points with the 2 connected green lines.

[gandreas]

Interesting. Yes you could trail the edge or 'wall' of the path of the Mandelbrot, and reach the core eventually, but then the same goes for any maze.

That's not exactly true - the Mandelbrot set has infinite circumference (and so does every branch of it), so while that strategy would work in theory, in practice, not so much.

For that matter, it's not even clear that you could actually find the exact edge to trail...

[twinbee]

That's not exactly true - the Mandelbrot set has infinite circumference (and so does every branch of it), so while that strategy would work in theory, in practice, not so much.

I meant at a finite iteration depth.

[gandreas]

That's not exactly true - the Mandelbrot set has infinite circumference (and so does every branch of it), so while that strategy would work in theory, in practice, not so much.

I meant at a finite iteration depth.

It becomes a much more interesting problem if you have to work at infinite iteration, since, even if you find a point that is on the edge (i.e., a points such that (x,y) is inside and (x+epsilon*cos(theta),y+epsilon*sin(theta)), I'm not convinced there is even a way to find the next point (assuming that theta is the direction your hand is in, and your "facing" theta + pi / 4).  Since the edge is rough (i.e., there is no normal for any given point on the edge) a whole lot of things stop working.

Regardless, it's an interesting problem, and I'm thinking that the "best" solution should work for either finite or infinite iterations (though the latter is far more difficult due to the inability to calculate meaningful values for limits of various values that might be useful in the finite case).

[twinbee]

Hmm, I wonder if in the same way the series 0.5+0.25+0.125+0.0625... converges to 1, and therefore = 1, that infinite iterations of the Mandelbrot would effectively produce pure circles (and one cardioid) without any of the tiny branches at all. Therefore the parameter is basically the circumference of all the 'circles' and cardioid.

In any case, I'd still love to know the perimeter of the Mandelbrot set given x iterations. Had a look on Google yesterday, and there was zero info on it.

[lkmitch]

Hmm, I wonder if in the same way the series 0.5+0.25+0.125+0.0625... converges to 1, and therefore = 1, that infinite iterations of the Mandelbrot would effectively produce pure circles (and one cardioid) without any of the tiny branches at all. Therefore the parameter is basically the circumference of all the 'circles' and cardioid.

In any case, I'd still love to know the perimeter of the Mandelbrot set given x iterations. Had a look on Google yesterday, and there was zero info on it.

The problem with the circumference "just" being the sum of the circumferences of the disks and cardioids is that there are infinitely many dendrites--infinitely thin stalks like the spike out to (-2, 0) that have length but no interior.

In concept, you can find the perimeter for finite iterations by using a variation of a boundary-tracing routine.  Start at a known boundary point (say, the tip of the spike) for a given number of iterations and trace the boundary.  Keep a tally of the sum of the segment lengths.  Repeat with smaller segment sizes until your sum convergences.  Since you'd be using finite iterations, the sum should converge nicely once your segments became smaller than the smallest crook or bend.

Kerry