## Lattices and shortest vector

In my book about Lattice basis reduction I found the following:

Gaussian Heuristic: for $r$ big enough, there are approximately $v_m r^m /vol(L)$ lattice points in $L$ of norm $\le r$, where $v_m$ denotes the volume of the m-dimensional unit ball.

Corollary: for a lattice $L$ of rank $m$ the shortest nonzero lattice vector has norm at most approximately $vol(L)^{1/m}$.

Then they give this example:

Given $\alpha, \beta, \gamma \in \mathbb{R}$, and $X$ > $0$ with $X$ large. Small values for the linear form $f(x,y,z)=\alpha x + \beta y + \gamma z$ with $x,y,z \in \mathbb{Z}$ corresponds to lattices

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ C\alpha & C\beta & C\gamma \end{pmatrix}$

How large should $C$ be taken to compute $x,y,z \in \mathbb{Z}$ with $max(\mid x \mid,\mid y \mid,\mid z \mid) \le X$ and $\mid \alpha x + \beta y + \gamma z \mid$ minimal?

The Gaussian heuristic with $C=X^3$ expects solutions of size approximately $X$, with $\mid \alpha x + \beta y + \gamma z \mid$ approximately $X^{-2}$.

How do they come up with $C=X^3$ ?

Using the above Corallary I have that the shortest nonzero lattice vector has norm at most approximately $vol(L)^{1/m} = \mid determinant(L) \mid ^{1/3}=(C\gamma)^{1/3}$

Then since $max(\mid x \mid,\mid y \mid,\mid z \mid) \le X$ we have that the norm of $(\alpha x, \beta y, \gamma z) \le \sqrt(\alpha^2 + \beta^2 + \gamma^2) X$.

So $(C\gamma)^{1/3} \le \sqrt(\alpha^2 + \beta^2 + \gamma^2) X$and thus $C$ is approximately $X^3$?

Am I doing this right? I hope someone can explain where the $C$ is coming from.

Thank you!