On points arranged in a parallelogram lattice, like on the image in this Wikipedia article, how to calculate the maximal distance any point on the plane may have to its closest point from the lattice. Or alternately, the maximal radius of a disk that can be placed on the plane such that it does not contain any point of the lattice.
As input I have the side length and both diagonals of one possible parallelogram that fits the lattice.
Edit: I meant the lattice not grid, i.e. only the sparse set of intersection points of a parallelogram grid.
Two vectors ${\bf a}$, ${\bf b}\in{\mathbb R}^2$ representing the sides of the given parallelogram determine the lattice $$\Lambda:={\mathbb Z}{\bf a}+{\mathbb Z}{\bf b}:=\bigl\{j {\bf a}+k {\bf b}\bigm| j, \ k\in{\mathbb Z}\bigr\}\ ,\tag{1}$$ but the representation $(1)$ of $\Lambda$ is not uniquely determined: The lattice ${\mathbb Z}^2$, for example, is generated by the pair $(1,0)$, $(0,1)$ as well as by the pair ${\bf a}:=(3,16)$, ${\bf b}:=(5,27)$.
In order to solve the problem at hand we have to determine a certain "standard presentation" of $\Lambda$: Find the shortest vector $${\bf p}=p\> {\bf u}, \quad |u|=1, \quad p>0,$$ occurring in $\Lambda$ (this is a standard problem in computational geometry). Then $\Lambda$ contains the one-dimensional lattice $\Lambda':={\mathbb Z}{\bf p}$, and is the union of translated copies of $\Lambda'$. Denote the distance between two successive such copies by $h>0$, and let ${\bf v}$ be a unit vector orthogonal to ${\bf u}$. There is a unique vector ${\bf q}\in\Lambda$ having a representation of the form $${\bf q}=c\>{\bf u}+h\>{\bf v}, \qquad 0\leq c<p\ ,$$ and $\Lambda$ can then be presented in the form $$\Lambda:={\mathbb Z}{\bf p}+{\mathbb Z}{\bf q}\ .$$ Let $\rho$ be the circumradius of the triangle with vertices ${\bf 0}$, ${\bf p}$, ${\bf q}$. Then any point ${\bf x}\in{\mathbb R}^2$ has a distance $\leq\rho$ from $\Lambda$, and there are points for which this bound is realized.