Title: IFS and Minkowski Dimension Post by: Crucifixio on February 11, 2009, 09:04:11 PM I'm writing program to generate fractals using IFS and caluclating Minkowski dimension. I've got question how to calculate Minkowski dimension from Image (2d fractals) using box counting with some Epsilon (size of the box) and how to calculate exact dimension from IFS to find error between count-box dimension and exact dimension from IFS ?
Title: Re: IFS and Minkowski Dimension Post by: David Makin on February 11, 2009, 10:21:24 PM I haven't looked into using box counting to get the fractal dimension but I did search a lot on the internet a while ago for documents on finding the "correct" fractal dimension of a given affine IFS from the transformations and I found:
"A Multifractal Analysis of IFSP Invariant Measures With Application to Fractal Image Generation" by J.M. Gutierrez and A. Iglesias and M.A. Rodrıguez I found it on the web as "fractals96.pdf". I highly recommend it as a read (if your maths is up to it) but basically it concludes that for a given IFS: log(p0)/log(s0) = log(p1)/log(s1) = log(p2)/log(s2) =.... = Constant where the pn are the (correct) probabilities for the scales sn and Constant is basically the fractal dimension. Normally it's used to get the correct probabilities for a given set of transforms. The only problem with this is how you decide the scale value to use for transforms that have non-uniform scales. I basically used the determinant values but other methods may yield better results in some situations (from a rendering point of view) e.g. min or max scale along any one axis. If you want the document but can't find it then just drop me a line at makinmagic@tiscali.co.uk Title: Re: IFS and Minkowski Dimension Post by: David Makin on March 29, 2009, 12:42:13 AM <snip> ...basically it concludes that for a given IFS: log(p0)/log(s0) = log(p1)/log(s1) = log(p2)/log(s2) =.... = Constant where the pn are the (correct) probabilities for the scales sn and Constant is basically the fractal dimension. Normally it's used to get the correct probabilities for a given set of transforms. <snip> Something interesting just struck me about this - the standard "Measure Theory" version of fractal dimension is dimension=log(new length)/log(old length) so in fact here the probablity is analagous to "new length" and/if the scale is analagous to "old length". Title: Re: IFS and Minkowski Dimension Post by: paxinum on July 08, 2010, 05:31:03 PM You mean Hausdorff dimension? Minkowski developed the 4-dimensional 3-space + time used by Einstein in relativity theory... |