Minkowski Dimension
In the study of fractals, Minkowski dimension (a.k.a. box-counting dimension) is a notion of dimension for fractals, measuring how complexity of detail changes with the scale at which one views the fractal. Very roughly, the larger a fractal's Minkowski dimension, the more self-similar the fractal is, in the sense that the fractal contains many constricted copies of itself.
Motivation and Definition
Consider a square $I^2$ of side length 1. For each $k\in \mathbb{N}$, the square may be covered by smaller squares of side length $s = \frac1k$. Each of these squares is a scaled copy of $I^2$ $($where the smaller squares are scaled down by a factor of $k),$ and it takes $k^2 = \frac1{s^2}$ smaller squares to cover all of $I^2$. Similarly, if one were to cover the cube $I^3$ of side length 1 with cubes of side length $s = \frac1k$, it would take $k^3 = \frac1{s^3}$ to do so. This suggests that the exponent of this "box-counting function," which takes in side length and returns how many such boxes one would need to cover the whole figure, can be thought of as a sort of dimensional measurement. For instance, the dimension of the cube $I^3$ (to which any reasonable notion of dimension should assign the number 3) can be retrieved as $\frac{\log\left(\frac1{s^3}\right)}{\log\left(\frac1s\right)} = \frac{-3 \log(s)}{-\log(s)} = 3.$ One can formalize this observation as follows: Let $X\subset \mathbb{R}^n$ be the object whose fractal dimension is to be measured. For each $\epsilon >0$, let $f(\epsilon)$ denote the minimum number of non-overlapping $n$-dimensional boxes of side length $\epsilon$ necessary to cover $X$. By $n$-dimensional box of side length $\epsilon$, we mean a translated copy of the $n$-fold product of intervals $[0,\epsilon]^n \subset \mathbb{R}^n$.
The upper Minkowski dimension of $X$ is $\overline{\dim}_{\text{box}}(X) := \limsup_{\epsilon \to 0} \frac{ \log\big(f(\epsilon)\big)}{\log\left(\frac1{\epsilon}\right)}$ and the lower Minkowski dimension of $X$ is $\underline{\dim}_{\text{box}}(X) := \liminf_{\epsilon \to 0} \frac{ \log\big(f(\epsilon)\big)}{\log\left(\frac1{\epsilon}\right)}.$ If $\overline{\dim}_{\text{box}}(X) = \underline{\dim}_{\text{box}}(X)$, then the common value is denoted $\dim_{\text{box}}(X)$ and called the Minkowski dimension of $X$.
Examples/Computations
The following computations will be rough and some steps will remain unjustified; to give careful proofs would require more details than are instructive.
Let $X = \{p_1, \ldots, p_d\} \subset \mathbb{R}^n$ be a finite set of points. Compute the Minkowski dimension of $X$.
We may cover $X$ by simply placing a box on each point, so $f(\epsilon) = d$ for all $\epsilon > 0$. Thus, the Minkowski dimension of $X$ is $\lim_{\epsilon \to 0} \frac{\log(d)}{\log\left(\frac1\epsilon\right)} = 0.$ This is consistent with our intuition that points (and finite sets of them) should have zero dimension. $_\square$
Let $X = \left\{0, 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \right\}$. Compute the upper Minkowski dimension of $X$.
Write $X = A \cup B$, where $A =\left \{1, \frac{1}{2}, \ldots, \frac{1}{n} \right\}$ and $B = \left \{\frac{1}{n+1}, \frac{1}{n+2}, \ldots\right\}$. For $\epsilon < \frac{1}{n} - \frac{1}{n+1} < \frac{1}{n^2}$, we can cover $A$ with $n$ intervals of length $\epsilon$, one centered at each point in $A$. And, we can cover $B$ with approximately $n$ intervals of length $\epsilon$, since all elements of $B$ are contained in $\left[0,\frac1n\right]$. Thus, $f(\epsilon) \approx 2n \approx \frac2{\sqrt{\epsilon}}$. It follows that $\overline{\dim}_{\text{box}}(X) = \limsup_{\epsilon \to 0} \frac{\log\left(\frac2{\sqrt{\epsilon}}\right)}{\log\left(\frac1\epsilon\right)} = \frac{1}{2}.\ _\square$
Let $X \subset \mathbb{R}$ be the middle-thirds Cantor set $\mathcal{C}$. Compute the upper Minkowski dimension of $X$.
Note that $\mathcal{C}$ is the union of two copies of itself scaled down by a factor of $\frac13$. Thus, if $\mathcal{C}$ can be covered by $f(\epsilon)$ intervals of length $\epsilon$, then $\mathcal{C}$ can be covered by $2f(\epsilon)$ intervals of length $\frac{\epsilon}3$, and hence $f\left(\frac{\epsilon}3\right) \approx 2f(\epsilon)$.
Certainly, $f(1) = 1$, so $f\left(\frac1{3^n}\right) \approx 2^n$. Accordingly, we have $\overline{\dim}_{\text{box}}(X) = \lim_{n\to\infty} \frac{\log\Big(f\left(\frac1{3^n}\right)\Big)}{\log(3^n)} = \frac{\log(2)}{\log(3)}.\ _\square$
References
- Prokofiev, . Estimating the box-counting dimension of the coast of Great Britain. Retrieved September 7, 2016, from https://en.wikipedia.org/wiki/Minkowski%E2%80%93Bouligand_dimension#/media/File:Great_Britain_Box.svg