Smallest hexagonal lattice with no more than one point per unit Cartesian cell, as a function of rotation angle? This answer to How to randomly but evenly distribute nodes on a plane introduces Bridson's Algorithm for Poisson-disc sampling (original paper: Fast Poisson Disk Sampling in Arbitrary Dimensions (also here) by Robert Bridson, University of British Columbia) Also see this and animation and explanation here.
I'm seeding the initial field with some regions with points on periodic lattices which will be both square and hexagonal, and of arbitrary rotation.
Bridson's method is characterized by a minimum distance $r$ that new points can be generated
For the algorithm to go fast, one chooses a square grid of side $r/\sqrt{2}$ which is small enough that there will always be either zero or one point in each cell. This way when a new point is introduced:

*

*it must be introduced in a currently empty cell

*only eight adjacent cells must be checked to evaluate the minimum distance to all existing points.

When I seed the space with preexisting array regions, I need to make sure that they never put more than one point in any cell.
Question: What are the constraints on lattice spacing $a$ and rotation angle $\theta$ for a hexagonal lattice with unit vectors:
$$\mathbf{a_1} =  a\cos \left(\theta \right) \ \mathbf{\hat{x}} + a\sin \left(\theta \right) \ \mathbf{\hat{y}}$$
$$\mathbf{a_2} =  a\cos \left(\theta +\frac{\pi}{3} \right) \ \mathbf{\hat{x}} + a\sin \left(\theta +\frac{\pi}{3} \right) \ \mathbf{\hat{y}}$$
such that there will never be more than one point in a unit square cartesian grid with any origin? (In other words, no matter how you translate the origin of the hexagonal lattice with respect to the cartesian grid.)

Rough example of application, for background only. Left: real space, right: reciprocal space (Fourier transform).

 A: There are three regular tilings of the Euclidean plane, and they correspond to regular triangular lattice, regular square lattice, and regular hexagonal lattice.
In order for a lattice to be able to describe any point configuration with at most one point per cell, with pairwise distances greater than $r_\min$, the radius of the lattice cell circumcircle must be at most $2 r_\min$.  In other words, the maximum distance between lattice cell vertices, or "cell diameter", must be $r_\min$ or less.  Otherwise, a single lattice cell could contain two points (near vertices of the cell) with their distance greater than $r_\min$.
The properties of the three lattices with the same "cell diameter" (cell circumcircle diameter $d = r_\min$) are
$$\begin{array}{l|ll|ll}
\text{Lattice} & \text{Side length} & ~ & \text{Cell area} & ~ \\
\hline
\text{Triangular} & a_△ = r_\min & \approx 1.0000 \, r_\min & A_△ = \frac{\sqrt{3}}{4} r_\min^2 & \approx 0.433 \, r_\min^2 \\
\text{Square} & a_□ = \frac{r_\min}{\sqrt{2}} & \approx 0.7071 \, r_\min & A_□ = \frac{1}{2} r_\min^2 & \approx 0.500 \, r_\min^2 \\
\text{Hexagonal} & a_⬡ = \frac{r_\min}{2} & \approx 0.5000 \, r_\min & A_⬡ = \frac{3 \sqrt{3}}{8} r_\min^2 & \approx 0.650 \, r_\min^2 \\
\end{array}$$
Maximum density of points is reached with a hexagonal lattice with $a = 2/\sqrt{3} \; r_\min$, where each cell has area $A_\min = \sqrt{3}/2 \, r_\min^2 \approx 0.866 \, r_\min^2$.  From this we can estimate that for triangular lattices, at most half the cells will be populated with points; for square lattices, $1/\sqrt{3} \approx 0.5/0.866 = 58\%$ of cells will be populated; and for hexagonal lattices, $3/4 = 75\%$ of cells will be populated.
The following illustration shows these lattices, with shaded areas showing the region where existing points can limit where/whether the center white cell can contain a new point or not.

Without subdividing the cell further, one needs to examine up to 24 triangular cells, 20 square cells, or 18 hexagonal cells.
If a larger side length $a$ is used for a lattice, some valid point configurations cannot be described using only one point per lattice cell.  (That is, with larger $a$, not all valid point sets can be generated when limiting to at most a single point per cell.)
The problem at hand is to find $a_x$ for one lattice, given a larger than minimum other lattice $a_y$ and the relative rotation $\theta$ between the two.  Because of the above, I believe this is the wrong way to try to fix the underlying problem.
What I would suggest, is to use the desired lattices (with arbitrary spacings, as long as they are not less than the ones stated above wrt. the minimum pairwise distance $r_\min$), but additionally verify and add each point to a minimal square lattice (which is computationally the simplest to implement, and likely most efficient on current hardware).
That is, an arbitrary lattices can be used to generate the points, but only when the minimal square lattice also allows each new point.  While the minimal square lattice has smaller cells than the lattices used to generate points, it allows all possible point configurations (with pairwise distances greater than $r_\min$), and is therefore well suited for verification (that a point to be added does not break the minimum pairwise distance rule).
A: The two furthest points in a square are a pair of opposite vertices, i.e. the end points of a diagonal. A square with diagonal $r$ has side $r/\sqrt{2}$.
This ensures that two points with distance of at least $r$ will be in two different gridcells. Note that if you have a cell with a point, you do need to check more than just the 8 immediately neighbouring cells to be sure that there is no other point within a distance of $r$. In fact you need to check 20 cells - a 5x5 square without its corner cells and without the centre cell itself.
The two furthest points in a regular hexagon are a pair of opposite vertices, i.e. the end points of a diagonal through the centre. A regular hexagon with diameter $r$ has side $r/2$.
This side length is enough to ensure that two points of distance $r$ or more will land in different cells, but as with the squares checking the immediate neighbouring cells is not sufficient to ensure that minimum distance. You'd need to check the 6 immediate neighbours, and also the next layer of 12 neighbours.
A: I don't think one needs to go to tiling theory to solve this. It's just a question of the lattice constant being longer than the length of the chord of a unit square, with some fancy modulo 12 footwork, 12 being the least common multiple of the 4-fold and 6-fold symmetries.
$$a_{0°} = \frac{2}{\sqrt{3}} = \frac{1}{\cos30°} \approx 1.1547$$
$$a_{15°} = \frac{4}{\sqrt{6}+\sqrt{2}} = \frac{1}{\cos15°} \approx 1.0353$$
In general:
$$a(\theta) = \sec\left(30- |(\theta+15) \bmod 30 -15| \right)$$

