Prescriptive version of counting hyperplane arrangements In Hyperplane arrangement theory, Zaslavsky's Theorem necessarily bounds 
the number of bounded and unbounded regions in the complement of a real hyperplane arrangement. While this counting theorem is great, is there some thing more prescriptive which tells us which elements of the intersection lattice of the halfspaces resulting from the original arrangement are empty specifically? 
Given some affine arrangement $A$ and the exact normal and offset of the planes in $A$, and a corresponding halfspace intersection lattice $L$, if I know some subset of $L$ is non-empty then is there an algorithm or some algebraic formulation, which prescribes another subset of $L$ which is empty?
As an example, if I have some affine arrangement in $\mathbb{R}^1$ of two hyperplanes (points in this case)  $A,B$, then I know that either $A_l \cap B_r = \emptyset$ or $A_r \cap B_l = \emptyset$ where $A_r$ denotes the halfspace to the right of $A$ and $A_l$ notes that to the left.
 A: The question is confused.  The intersection poset of an affine arrangement is a semilattice, not necessarily a lattice.  It is defined as the set of non-empty intersections of subsets of A, so nothing in the semilattice is empty.
The example suggests that the question is about the intersections of half-spaces.  (I assume $R_r$ means $B_r$.)  This is separate from the intersection semilattice.
I am not sure, but I think that from the normal vectors and the intersection semilattice it is not possible to say which half-space intersections are empty.  It is possible for central arrangements (all hyperplanes contain 0) since the normal vectors determine everything, but I think for affine arrangements there is not enough information.  That is only an opinion.  
In the central case, the intersection is empty if and only if the normal vectors positively span the origin (if I remember correctly).  However, that seems neither necessary nor sufficient for affine hyperplanes.
This is a good question.  I'm sorry I can't answer it.
