the resulting “cells” that approximates the norm $\| \cdot \|$ on the unit torus. We call this the $s$-graded model.

Fix an integer $s$, and define $$m :=  \lfloor s/r \rfloor,$$ so that $$\frac{s}{r} - 1 \leq m \leq \frac{s}{r}.$$ Let $T = \{1,2, \dots m\}^d$. Pick $I = (i_1, i_2, \dots, i_d) \in T$, and define $$A_I = \left[\frac{i_1-1}{m},\frac{i_1}{m}\right] \times \dots \times \left[\frac{i_d-1}{m},\frac{i_d}{m}\right].$$ The $A_I$’s partition the unit torus into $m^d$ cubes (ignoring sets of measure $0$), each of volume $1/m^d$, and therefore, $X_I = |\chi(A_I)|$ is a Poisson random variable of mean $$\mathcal{D} := \frac{n}{m^d}.$$ As a function of $n$, $m^d$ grows as $n^{2-p}$, up to a possible multiplicative factor that varies more slowly than any polynomial or rational function in $n$ (recall that $s$ is fixed); this implies that $\mathcal{D}$ grows as $n^{p-1}$. Note that $s$ is exactly the minimal number of cells that must be traversed in any one axial direction to ensure that we exceed the distance $r$. This parameter takes the place of $r$ in the $s$-graded model.

We now define a metric on $T$, induced by the norm on torus: $$d(I,J) = \inf_{x \in A_I^o,\,  y \in A_J^o}  \left\lceil m \|  x - 