General problem statement
Find the 2D regions $A$ and $B$ on the cartesian plane, where the relationship $R(a, b)$ is satisfied for all points $a \in A$ and $b \in B$.
I believe a more formal way to state this is (correct me if I'm wrong): $$ A = \{a \in \mathbb{R}^2 \mid R(a, b) \forall b \in B\} \\ B = \{b \in \mathbb{R}^2 \mid R(a, b) \forall a \in A\} $$
Obviously any pair of valid regions $A$ and $B$ aren't necessarily unique.
A trivial solution could be fixing $A$ to be a single point, and then finding $B$ becomes a simple inequality.
Instead, I'm looking for a way to generate and parameterise all possible valid pairs of regions $A$ and $B$ (and count the degrees of freedom).
Furthermore, if the relationship is symmetric $R(a, b) = R(b, a)$, I would like to find "special" pairs of $A$ and $B$ that are rotationally or reflectionally symmetric around a point such as the origin.
Is there a general method to do this?
Specific examples
All pairs of points are at most $d$ distance apart
In this case, $R(a, b) = |a - b| \le d$.
Some possible solutions include:
- Trivial $A$ is the origin, and $B$ is a disc of radius $d$: $A = \{(0, 0)\}; B = \{(x, y) \mid x^2 + y^2 \le d^2\}$
- Symmetric $A$ and $B$ are the same disc of radius $\frac{d}{2}$: $A = B = \left\{(x, y) \mid x^2 + y^2 \le \left(\frac{d}{2}\right)^2\right\}$
All pairs of points are at least $d$ distance apart
In this case, $R(a, b) = |a - b| \ge d$.
Some possible solutions include:
- Trivial $A$ is the origin, and $B$ is the plane with a disc of radius $d$ punched out: $A = \{(0, 0)\}; B = \{(x, y) \mid x^2 + y^2 \ge d^2\}$
- Symmetric $A$ and $B$ are opposite half-planes $d$ distance apart: $A = \left\{(x, y) \mid x \le \frac{-d}{2}\right\}; B = \left\{(x, y) \mid x \ge \frac{d}{2}\right\}$
- $A$ is a line, and $B$ is the two half-planes $d$ distance from the line: $A = \{(x, y) \mid x = 0\}; B = \{(x, y) \mid |x| \ge d\}$
(The above solutions can also be freely translated and rotated anywhere on the plane to produce a new solution)