Is all grid based noise inevitably anisotropic? I'm interested in how this applies to higher numbers of dimensions too, but for this question I will focus solely on 2D grids.



I know that Perlin noise is not isotropic (direction invariant), and that the underlying square grid shows up enough to be able to identify its orientation. Simplex noise is an improvement on this but its underlying equilateral triangle grid is still not completely obscured.

My intuition is that any attempt to make noise of a particular frequency on a grid will result in a lower frequency in directions not aligned to the grid. So while attempts can be made to disguise this, the noise cannot in principle be isotropic unless it is generated without reference to a grid, allowing the average frequency to be the same in all directions.

For example, with a square grid without noise, with square side length $n$, the frequency of vertices horizontally or vertically is $\frac1n$, whereas the frequency of vertices at 45 degrees (through opposite corners of the squares) is $\frac1{\sqrt{2}n}$. 



Is there a random distribution that could be applied to offset the vertex positions that would result in the frequency becoming identical in all directions? My suspicion is that there is no such distribution, but I don't have a way of proving either way.

In short, is there a way of making perfect grid based noise of a given frequency, or should I be focused on other approaches (non-grid based noise or ways of disguising artifacts)?
As usual with numerical methods and samplings, it also depends of your quality threshold of what you consider &quot;isotropic&quot;. And of what you would consider as a being or not a &quot;grid-based noise algorithm&quot;.
For instance Gabor Noise reproduces a target spectrum, for instance blue noise, which in Fourier domain is a simple isotropic ring.
Now if you consider that this ring is not analytic but rasterized, as such it is not perfectly symmetrical. Also if the ring radius (i.e., frequency) get too close to the window size (i.e., maximum frequency), it will be truncated (and thus no longer symmetrical). Up to you to accept or not these as anisotropic ;-)


&quot;This is not a circle&quot; - Magritte . . . . . . . . . . . . . . . .   &quot;This is not a circle&quot; - Nyquist
You might or might not accept a rasterized ring in Fourier space to be &quot;isotropic&quot;. Still, in the extreme cases where the ring get thinner than the resolution, or larger than the window, isotropy is objectively lost.