Some simple algebraic manipulation tells us that distance between points would then scale as D^(-1/d).
This is a really helpful summary! I’m having trouble following this step here and in the original work—could someone point me in the right direction as to why this follows from needing ~2^d as many points to make the “net” twice as fine?
This is a really helpful summary! I’m having trouble following this step here and in the original work—could someone point me in the right direction as to why this follows from needing ~2^d as many points to make the “net” twice as fine?
Say that at dataset size D0, the distance between points is x0. Now consider a new distance x -- what is the corresponding D we need?
Intuitively, for each factor of 2 that x is smaller than x0 (which we can quantify as log2(x0x)), we need to multiply by another factor of 2d.
So D=D0×(2d)log2(x0x)=D0×(2log2(x0x))d=D0×(x0x)d
(DD0)1d=x0x
x=x0(DD0)−1d
That is, the distance scales as D−1d.
Oh hi! I linked your video in another comment without noticing this one. Great visual explanation!