Worst-Case Adaptive Submodular Cover

TL;DR

This paper introduces a new approach to the worst-case adaptive submodular cover problem, providing approximation algorithms for minimizing worst-case costs and maximizing worst-case utility under budget constraints.

Contribution

It develops tight approximation policies for the worst-case adaptive submodular cover and a dual maximum-coverage problem, extending prior work to worst-case scenarios.

Findings

Achieves a $( ext{log}(Q/\eta)+1)$-approximation for the cover problem.

Provides a $(1-1/e)/2$-approximation for the maximum-coverage problem.

Generalizes previous models like active learning and stochastic submodular set cover.

Abstract

In this paper, we study the adaptive submodular cover problem under the worst-case setting. This problem generalizes many previously studied problems, namely, the pool-based active learning and the stochastic submodular set cover. The input of our problem is a set of items (e.g., medical tests) and each item has a random state (e.g., the outcome of a medical test), whose realization is initially unknown. One must select an item at a fixed cost in order to observe its realization. There is an utility function which maps a subset of items and their states to a non-negative real number. We aim to sequentially select a group of items to achieve a ``target value'' while minimizing the maximum cost across realizations (a.k.a. worst-case cost). To facilitate our study, we assume that the utility function is \emph{worst-case submodular}, a property that is commonly found in many machine learning applications. With this assumption, we develop a tight $(\log (Q/\eta)+1)$-approximation policy, where $Q$ is the ``target value'' and $\eta$ is the smallest difference between $Q$ and any achievable utility value $\hat{Q}<Q$. We also study a worst-case maximum-coverage problem, a dual problem of the minimum-cost-cover problem, whose goal is to select a group of items to maximize its worst-case utility subject to a budget constraint. To solve this problem, we develop a $(1-1/e)/2$-approximation solution.