Let $S \subset \mathbf{R}^3$ be a set of points in 3D and let $O=(x_0,y_0,z_0)$ be the origin/point of reference.
We consider a convex polytope $P$ good / interior if:
$P$ is wholly contained within the interior of the convex hull of $S$: $P \subsetneq \text{convexhull}(S)$.
$O$ is contained within the interior of $P$, namely: $O \in P$.
- none of the points in $S$ are in $P$ namely $S \cap P = \emptyset$.
Out of all good convex polytopes, we want to find either the maximum (supremum) of their volumes, i.e. $\sup \text{Vol}(P)$ where $P$ ranges over good polytopes.
Here is a 2D modest illustration (the green polytope might not be really the optimal):
The black dots are the set $S$, the origin $O$ is in the middle. The red lines go through the points in $S$ which define the boundary of $S$ (not a good polytope) and the green lines go through a good polytope.
Is there an efficient algorithm to compute this best convex and non-convex polytopes for a given set $S$ of points and origin $O$?
I would like to formulate this problem as an optimization problem or to come up with some algorithm using triangulation, $\alpha$-shapes or $\beta$-skeleton geometric transformation + convex hull, linear programming, maybe working in the dual space? etc.
One idea which I have is "pumping down" the convex hull of Delaunay Triangulation:
DT $\leftarrow$ delaunayTriangulation($S \cup O$)
CH $\leftarrow$ convexHull(DT)
while Not empty(DT) and $O \notin CH$:
$\quad$ i. previousDT $\leftarrow$ DT
$\quad$ ii. update DT $\leftarrow$ DT $\setminus$ CH
return previousDT
EDIT: I think what I am looking for in a one-liner of math symbols is:
$$\text{argmax}_{P} \text\{\text{Vol}(P) \, | \, O \in P \setminus \partial P \, ; \, S\cap P = \emptyset \, ; \, P \subsetneq \text{convexhull}(S) \} $$
Where $\partial P$ is the boundary of a polytope $P$.
good
polyhedron to go on and ask for best convex and non-convex. Removing the convex hull seems a reasonable first step. A pity that convert to polar and add, in increasing distance from the origin, every point that doesn't make the polyhedron non-convex (one of the points formerly added interior) doesn't work: you might want to construct an example showing that. (I'm down to 3 (hull) + 1 (shared) + 2 (suboptimal) + 3 points.) $\endgroup$not sure I get your point
the point of my 1st comment is that I can't see from your sketch that this problem isn't trivial. $\endgroup$