Active-Learning a Convex Body in Low Dimensions

Abstract

Consider a set $P\subseteq \mathbb {R}^d$ of n points, and a convex body $C$ provided via a separation oracle. The task at hand is to decide for each point of $P$ if it is in $C$ using the fewest number of oracle queries. We show that one can solve this problem in two and three dimensions using queries, where is the size of the largest subset of points of $P$ in convex position. In 2D, we provide an algorithm that efficiently generates these adaptive queries. Furthermore, we show that in two dimensions one can solve this problem using oracle queries, where is a lower bound on the minimum number of queries that any algorithm for this specific instance requires. Finally, we consider other variations on the problem, such as using the fewest number of queries to decide if $C$ contains all points of $P$. As an application of the above, we show that the discrete geometric median of a point set P in $\mathbb {R}^2$ can be computed in expected time.

Expected Separation Price for Random Points

We first extend the notion of separation price (see Sect. 3.1) to higher dimensions. For a closed convex d-dimensional polytope F, we let $f_k(F)$ denote the number of k-dimensional faces of F.

Definition 6

(Separation price in higher dimensions) Let $P$ be a set of points and $C$ be a convex body in $\mathbb {R}^d$. The inner fence $F_{\mathrm {in}}$ is a closed convex d-dimensional polytope with the minimum number of vertices, such that $F_{\mathrm {in}}\subseteq C$ and $C\cap P= F_{\mathrm {in}}\cap P$. Similarly, the outer fence $F_{\mathrm {out}}$ is a closed convex d-dimensional polytope with the minimum number of facets, such that $C\subseteq F_{\mathrm {out}}$ and $C\cap P= F_{\mathrm {out}}\cap P$. The separation price is defined as .

By extending the argument of Lemma 13 to use Definition 6, one can prove the following.

Lemma 27

Given a point set $P$ and a convex body $C$ in $\mathbb {R}^d$, any algorithm that classifies the points of $P$ in relation to $C$, must perform at least separation oracle queries.

Informally, for any fixed convex body $C$ and a set of n points $P$ chosen uniformly at random from the unit cube, the separation price is sublinear (approaching linear as the dimension increases).

Lemma 28

Let $P$ be a set of n points chosen uniformly at random from the unit cube $[0,1]^d$, and let $C$ be a convex body in $\mathbb {R}^d$, with $\textsf {vol}(C) \ge c$ for some constant $c \le 1$. Then , where O hides constants that depend on d and $C$.

Proof

It is known that for convex bodies $C$, the expected number of vertices of the convex hull of $P\cap C$ is $O(n^{1 - 2/(d+1)})$. Indeed, since $\textsf {vol}(C) \ge c$, the expected number of points of $P$ that fall inside $C$ is $m = \varTheta (n)$ (and these bounds hold with high probability by applying any Chernoff-like bound). It is known that for m points chosen uniformly at random from $C$, the expected size of the convex hull of points inside $C$ is $O(m^{1 - 2/(d+1)}) = O(n^{1 - 2/(d+1)})$ [28]. This readily implies that .

To bound , we apply a result of Dudley [9] which states the following. Given a convex body $C$ and a parameter $\varepsilon > 0$, there exists a convex body D, which is a polytope formed by the intersection of $O(\varepsilon ^{-(d-1)/2})$ halfspaces, such that $C\subseteq D \subseteq (1+\varepsilon )C$, where .

We claim that the number of points of $P$ that fall inside $D \setminus C$, plus the number of halfspaces defining D, is an upper bound on the size of the outer fence. Indeed, for each point $p$ that falls in inside $D \setminus C$, let $q$ be its nearest neighbor in $C$ (naturally $q$ lies on $\partial C$). Let be the hyperplane that is perpendicular to the segment $pq$ and passing through the midpoint of pq. Next, let be the halfspace bounded by such that . If H is the collection of $O(\varepsilon ^{-(d-1)/2})$ halfspaces defining D, then it is easy to see that the polytope defined by

separates the boundary of $C$ from $P\setminus C$ (i.e., it is an outer fence). See the figure below.

We now bound the size of this inner fence. Since $\textsf {vol}(D) - \textsf {vol}(C) \le \textsf {vol}((1+\varepsilon )C) - \textsf {vol}(C) \le O(\varepsilon )$, we have that . Combining both inequalities,

Choose $\varepsilon = 1/n^{2/(d+1)}$ to balance both terms, so that . $\square$

The next Lemma shows that the bound of Lemma 28 is tight in the worst case.

Lemma 29

Let $P$ be a set of n points chosen uniformly at random from the hypercube $[-2,2]^d$, and let $C$ be a unit radius ball centered at the origin. Then , where $\varOmega$ hides constants depending on d.

Proof

For a parameter $\delta$ to be chosen, let $Q \subseteq \partial C$ be a maximal set of points such that:

• (i) for any $p\in \partial C$, there is a point $q\in Q$ such that $\Vert p- q\Vert \le \delta$, and

• (ii) for any two points $p, q\in Q$, $\Vert p- q\Vert \ge \delta$.

Note that $\left| {Q} \right| = \varOmega (1/\delta ^{d-1})$. For each $p\in Q$, we let $\gamma _{p}$ be the spherical cap that is “centered” at $p$ (in the sense that the center of the base of $\gamma _{p}$, $p$, and the origin are collinear) and has base radius $2\delta$. Let . By construction, the caps of $\varGamma$ cover the surface of $C$.

By setting $\delta = 1/n^{1/(d+1)}$, we claim that for each cap $\gamma \in \varGamma$, in expectation $\varOmega (1)$ points of $P$ fall inside $\gamma$. This implies that there must be a vertex of the inner fence inside $\gamma$, and this holds for all caps in $\varGamma$. As such, the size of the inner fence is at least $\left| {Q} \right| = \varOmega (1/\delta ^{d-1}) = \varOmega (n^{1 - 2/(d+1)})$.

To prove the claim, for all $\gamma \in \varGamma$, we show that $\textsf {vol}(\gamma ) = \varOmega (1/n)$, and hence . By construction, the cap has a polar angle of $\theta = \varOmega (\delta )$. Indeed, we have that $\theta \ge \sin (\theta ) = 2\delta$ for $\theta \in [0,\pi /2]$ (which holds when n is sufficiently large). Let t denote the distance from the origin to the center of the base of $\gamma$. Then the height h of the spherical cap is $h = 1 - t = 1 - \cos (\theta ) \ge \theta ^2/6 = \varOmega (\delta ^2)$ (using the inequality $\cos (x) \le 1 - x^2/6$). Since the volume of the base of $\gamma$ is $\varOmega (\delta ^{d-1})$, we have that $\textsf {vol}(\gamma ) = \varOmega (h \delta ^{d-1}) = \varOmega (\delta ^{d+1}) = \varOmega (1/n)$, as required. $\square$