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I would like to construct a fractal (traditional, self-affine, and fat fractal) with a given embedding and fractal dimension, but I don't know how to do it programmatically. The shape of the fractal is negligible.

Mark McClure
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Alíz
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    Motivation? I assume you just want a graphic showing the transition between all the dimensions continuously? If so, This has sliders for Wierstrass fractals. – Zach466920 Mar 28 '16 at 17:43
  • Motivation is for example to draw or give the iteration rule of an arbitrary fractal with D = 3/2 fractal dimension and d=3 embedding dimension. – Alíz Mar 29 '16 at 09:43
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    The graph of $x(t)=\sum_{n=1}^{\infty} 2^{-n \cdot (2-d)} \cdot \cos(2^n \cdot t)$ with $y(t)$ and $z(t)$ being differentiable and finite and $d=3/2$ is an example. – Zach466920 Mar 29 '16 at 14:35

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Wikipedia describes a generalized Cantor set with parameter $0 < \gamma < 1$ giving fractal dimension

$$0 < D = \frac{-\log(2)}{\log\left(\displaystyle\frac{1-\gamma}{2}\right)} < 1$$

Built by removing at the $m$th iteration the central interval of length $\gamma\,l_{m-1}$ from each remaining segment (of length $l_{m-1}=(1-\gamma)^{m-1}/2^{m-1}$). At $\gamma=\frac{1}{3}$ one obtains the usual Cantor set. Varying $\gamma$ between $0$ and $1$ yields any fractal dimension $0 < D < 1$

Generalized Cantor set with $\gamma = 0.2$

The Cartesian product of Cantor set fractals satisfies (Falconer "Fractal Geometry: Mathematical Foundations and Applications" chapter 7):

$$\dim(F \times G) = \dim(F) + \dim(G)$$

In an $n$-dimensional space one can fit a product of $n$ one-dimensional Cantor sets (each with $0 < D_i < 1$) to get any dimension $0<D<n$

The product of the middle-thirds Cantor set with itself, aka "Cantor dust": Cantor dust

Claude
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  • Thank you, the key will be that if a fractal is a direct product of two fractals then its dimension is the sum of the dimensions of the two fractals. – Alíz Mar 29 '16 at 09:50