Intrinsic dimension calculated via 2_NN method vs Hausdroff dimension

27 Views Asked by At

I followed the following video: https://www.youtube.com/watch?v=zZ_NmaMeblU&ab_channel=YoavFreund to understand the intrinsic dimension. He took a shape that looks like a blob ($2$ dimensions) and a line rope-like figure ($1$ dimension) in a square $2d$ box of size $[0,1] \times [0,1]$. The partition of the square in a grid with $n_1^2$ and sides of length $\epsilon_1$, let's denote the tuple by $(n_1, \epsilon_1)$. Let $N_1$ denote the number of squares that intersect the blob, with scale $(n_1 , \epsilon_1)$, let $N_2$ denote the square the intersect the blob with scale $$n_2 > n_1 \, \epsilon_1 > \epsilon_2 $$ then intrinsic dimension is given by $$ d = -\frac{\log(N_2) - \log(N_1)}{\log(\epsilon_2) - \log(\epsilon_1 )} $$

So, I haven't found a proof, but I am guessing as the scale get finer the ratio converges to some number. With this definition the limitation is that we can't go arbitrarily fine as the data set is always finite and hence it will converge to zero.

I saw the method for calculating intrinsic dimension here:

https://www.nature.com/articles/s41598-017-11873-y

but they never define intrinsic dimension? Also, it is not clear for me how the ratio of two nearest nbd would satisfy pareto distribution? So, what is the relation between intrinsic dimension and Hausdroff dimension? It seems only the closest defined definition matches the computation.

Also, I cannot relate the way of computation of intrinsic dimension in the video and by tw neighborhood method.