volume, denoted by $V(\Lambda)$, is said to be the *volume of* $\Lambda$. The *first minimum* $\lambda_1(\Lambda)$ of $\Lambda$ is the shortest Euclidean norm of non-zero vectors in $\Lambda$. In general, the $i$-th minima of $\Lambda$ are defined as $$\lambda_i(\Lambda) = \min\left\{r : \dim\left( \mbox{span}\left\{\mathcal{B}_r \cap \Lambda\right\}\right) = i\right\}.$$ The *packing density* of a rank $m$ lattice $\Lambda$ is defined as $$\Delta(\Lambda) = \frac{\mbox{vol }\mathcal{B}_{\lambda_1/2}}{V(\Lambda)}.$$ We say that a point in ${\mathbf{x}}\in \Lambda$ is *primitive* if the intersection between $\Lambda$ and the open line segment $\left\{\alpha {\mathbf{x}}: \alpha \in (0,1)\right\}$ is the empty set. The set of all primitive points in $\Lambda$ is denoted by $\Lambda^\prime$.

Theorem \[thm:2\] implies the Minkowski-Hlawka lower bound in the following fashion. From the average result, it follows that it must exist at least one $\Lambda \in \mathbb{L}$ such that $\mathcal{N}_{\Lambda}(r) \leq (\zeta(m)V)^{-1} \text{vol } B_r^m$. Now if we force the right-hand side to be equal to $2(1-\varepsilon)$, for some small $\varepsilon >0$, then, since a lattice has at least two minimum vectors, we must have $\mathcal{N}_{\Lambda}(r) = 0$. Therefore $\Lambda$ can pack balls of radius $r/2$; rearanging the terms gives us, up to $\varepsilon$, the Minkowski-Hlawka bound. If $\Lambda$ is a lattice with guaranteed number 