Containment and Evasion in Stochastic Point Data

Abstract

Given two disjoint and finite point sets $A$ and $B$ in $\mathrm{I\! R}^d$, we say that $B$ is contained in $A$ if all the points of $B$ lie within the convex hull of $A$, and that $B$ evades $A$ if no point of $B$ lies inside the convex hull of $A$. We investigate the containment and evasion problems of this type when the set $A$ is stochastic, meaning each of its points $a_i$ is present with an independent probability $\pi (a_i)$. Our model is motivated by situations in which there is uncertainty about the set $A$, for instance, due to randomized strategy of an adversarial agent or scheduling of monitoring sensors. Our main results include the following: (1) we can compute the exact probability of containment or evasion in two dimensions in worst-case $O(n^4 + m^2)$ time and $O(n^2 + m^2)$ space, where $n = | A |$ and $m = | B |$, and (2) we prove that these problems are #P-hard in 3 or higher dimensions.

This research was partially supported by NSF grants CCF-1161495 and CCF-1525817.

A The Containment Problem Is #P-hard for \(d \ge 3\)

Theorem 4

Given a stochastic point set $A$, and another point set $B$, in $\mathrm{I\! R}^d$ for $d \ge 3$, the problem of computing the probability that $A$ contains $B$ is #P-hard.

Proof

We reduce from the #P-hard problem of counting vertex covers in planar graphs. In any graph, a subset of vertices is a vertex cover iff its complement is an independent set. As such the problem of counting vertex covers is equivalent to counting independent sets, and moreover, the problem is hard on any class of graphs on which the problem of counting independent sets is hard. In particular, it is hard on planar graphs. We use the transformation used in the proof of Theorem 3. More specifically, given a planar graph $G = (V, E)$ on $V = [n]$, it constructs a graph $G = (V, E')$ where $E \subseteq E'$ and a set $A$ of n points $\{a_1, \ldots , a_n \}$, which are the vertices of a convex body, such that every edge (i, j) in $E'$ corresponds to an edge $(a_i, a_j)$ of the polytope. We assign the probability 1/2 to each of these stochastic points. Let $P= \mathsf {CH}(A)$ denote the polytope. We now construct a new polytope $P' \subseteq P$, where each edge e of $P$ will have a corresponding (2 dimensional) facet $f_e$ in $P'$ “close” to e, i.e., lying inside $P$ and in the close vicinity of e. To this end, we consider a supporting plane $H_e$ such that $H_e \cap P= e$. We now move all such planes $H_e$, for each edge e slightly, shifting it parallel to itself, inside the polytope by a distance that is smaller than 1/2 of the minimum nonzero distance of any vertex of $P$ to any of the planes $H_e$. Let $H'_e$ denote the shifted plane. As a result, we get a new polytope, $\bigcap H'_e$ which lies inside $P$ and where now for each edge e of $P$, the portion of $H'_e$ lying inside $P'$ is a facet. This is the facet $f_e$ corresponding to e. Clearly, $P'$ as defined, can be computed in polynomial time given $P$.

Consider an edge $(i,j) \in E$, and the corresponding edge $(a_i, a_j)$ of $P$. We now add 3 points to $A$ and one point to $B$, for each such edge, as follows. Choose 3 points $b'_{ij}, b''_{ij}, b'''_{ij}$ in general position on the facet $f_e$ of $P'$.

We add them to $A$ and assign them the probability 1 each. The simplices $b'_{ij} b''_{ij} b'''_{ij} a_i$, and $b'_{ij} b''_{ij} b'''_{ij} a_j$ share a base, and are on the same side of it, and therefore they share a point that is interior to both of them. We choose such a point $b_{ij}$ and include it in $B$. See Figure on the right for an example of this construction in 2 dimensions.

It is clear that if $A' \subseteq A$ excludes both $a_i, a_j$, then $H'_e$ is a hyperplane that separates $b_{ij}$ from $\mathsf {CH}(A')$. Conversely, if at least one of $a_i, a_j$ are included, then $b_{ij}$ must be in $\mathsf {CH}(A')$. Therefore, if $\mathsf {CH}(A')$ contains all such points $b_{ij}$, the corresponding vertices forms a vertex cover in G. It may be observed that since all the new points added in $A$ have probability 1 they do not affect our counting argument. $\square$