Explicit formula for IFS fractal dimesnion

Question

Is there an explicit formula for finding the box counting dimension of an arbitrary IFS fractal, such as the IFS fern or any other random IFS fractal? If not, is there at least a summation, or recurrence relation that could find the fractal dimension? An example of how this would work would also be appreciated. If there is really no formula, is it just because we don't know it or because it cannot be expressed?

Answer 1

No - there is no such formula as discussed in this MathOverflow post.

The most important work in this area has been done by Ken Falconer. His first paper on the subject "The Hausdorff dimension of self-affine fractals" Math. Proc. Cambr. Phil. Soc. 103 (1988) 339-350 contains an algorithm that yields an upper bound on the box-counting dimension which almost always yields the Hausdorff dimension as well. Using these ideas, I estimate the dimension of the fern to be about 1.84. I'm not certain, though, as the technique can fail in special cases. Later papers by Falconer and others explore situations where the technique is guaranteed to work.

Answer 2

For graph-directed IFS of similarities satisfying an open set condition with every cycle contracting, it's possible to compute the Hausdorff dimension:

Hausdorff Dimension in Graph Directed Constructions
R Daniel Mauldin, S C Williams
Transactions of AMS, vol309 no2, October 1988, 811-829

A regular IFS of can be considered as a graph-directed IFS with one node, so every similarity must be contracting.

However, similarities are not very general (you can have IFS with all sorts of transformation functions), box-dimension is sometimes different from Hausdorff dimension, and I don't know if any algorithm exists for verifying that the IFS passes the open set condition.

I have a Javascript implementation of the algorithm described in the paper (view page source), with some more information in two blog posts.