Approximating Klee's Measure Problem and a Lower Bound for Union Volume Estimation

TL;DR

This work resolves a longstanding question on the optimality of the classic three-query approach to union volume estimation and Klee's measure problem by proving a tight $\Omega\left(\frac{n}{\varepsilon^2}\right)$ lower bound. It then delivers an algorithmic improvement that exploits geometric structure by classifying boxes into shape-based classes, sampling from class unions, and using orthogonal range searching to join results, achieving a running time of $O\left(\left(n+\frac{1}{\varepsilon^2}\right) \cdot \log^{O(d)} n\right)$ for constant dimension $d$. The paper also provides a near-linear practical algorithm with provable guarantees, and a robust discrete-to-continuous reduction showing the lower bound extends from discrete grids to continuous space, reinforcing the tightness of the known upper bounds. Overall, these results deepen our understanding of the Klee's measure problem and unify lower-bound methodology with geometry-driven algorithmic improvements, with implications for DNF counting, network reliability, and related geometric problems.

Abstract

Union volume estimation is a classical algorithmic problem. Given a family of objects $O_1,\ldots,O_n \subseteq \mathbb{R}^d$, we want to approximate the volume of their union. In the special case where all objects are boxes (also known as hyperrectangles) this is known as Klee's measure problem. The state-of-the-art algorithm [Karp, Luby, Madras '89] for union volume estimation and Klee's measure problem in constant dimension $d$ computes a $(1+\varepsilon)$-approximation with constant success probability by using a total of $O(n/\varepsilon^2)$ queries of the form (i) ask for the volume of $O_i$, (ii) sample a point uniformly at random from $O_i$, and (iii) query whether a given point is contained in $O_i$. We show that if one can only interact with the objects via the aforementioned three queries, the query complexity of [Karp, Luby, Madras '89] is indeed optimal, i.e., $Ω(n/\varepsilon^2)$ queries are necessary. Our lower bound already holds for estimating the union of equiponderous axis-aligned polygons in $\mathbb{R}^2$, and even if the algorithm is allowed to inspect the coordinates of the points sampled from the polygons, and still holds when a containment query can ask containment of an arbitrary (not sampled) point. Guided by the insights of the lower bound, we provide a more efficient approximation algorithm for Klee's measure problem improving the $O(n/\varepsilon^2)$ time to $O((n+\frac{1}{\varepsilon^2}) \cdot \log^{O(d)}n)$. We achieve this improvement by exploiting the geometry of Klee's measure problem in various ways: (1) Since we have access to the boxes' coordinates, we can split the boxes into classes of boxes of similar shape. (2) Within each class, we show how to sample from the union of all boxes, by using orthogonal range searching. And (3) we exploit that boxes of different classes have small intersection, for most pairs of classes.

Figures (4)

• Figure 1: When the side lengths of two boxes differ a lot in at least one of their dimensions (in our examples, the $y$-axis), their intersection is small compared to their union.
• Figure 2: We sample points in the grid cells $\mathcal{G}$ that are intersected by a box $\mathcal{O}_i$ from a fixed class. We then use orthogonal range searching to determine whether a sampled point is in a box from the class and should be kept ($\bullet$), or is not and should be discarded ($\times$).
• Figure 3: The vector $x=\left(+1,-1,+1,+1,-1,-1 \right)$ represented as the set $\{ (j,x_j) : j \in [6] \}$, where each point is drawn as a rectangle.
• Figure 4: The vector $y$ or $x=\left(+1,-1,+1,+1,-1,-1 \right)$ gives rise to $n$ polygons; one of these polygons is illustrated in dark blue. The light blue area indicates the union of all these $n$ polygons.

Theorems & Definitions (27)

• Theorem 1
• Theorem 2
• Theorem 2
• Definition 1: $p$-sample
• Lemma 1
• proof
• Lemma 2: adapted from KarpL85
• proof
• Definition 2
• Lemma 3