Simple Algorithms for Stochastic Score Classification with Small Approximation Ratios

TL;DR

This paper addresses SSC, where $n$ tests with costs $c_j$ and success probabilities $p_j$ yield a score classified by a partition of $\{0,\dots,n\}$ into $B$ intervals, aiming to minimize expected testing cost to identify the interval. It introduces simple non-adaptive strategies built via a weighted RoundRobin interleaving of three sub-algorithms, achieving strong constant-factor guarantees. The main results show a $(3+2\sqrt{2})$-approximation using a weighted three-way RoundRobin (3R) and a $6$-approximation using a two-way RoundRobin (2R); it also establishes a lower bound of $3/2$ on the adaptivity gap for unit-cost SSC, together with an existing upper bound, tightening the adaptivity-gap landscape in this setting. Overall, the work provides elegant, easy-to-implement algorithms with solid theoretical guarantees and clarifies the adaptivity gap in SSC, motivating further refinements and extensions to weighted SSC.

Abstract

We revisit the Stochastic Score Classification (SSC) problem introduced by Gkenosis et al. (ESA 2018): We are given $n$ tests. Each test $j$ can be conducted at cost $c_j$, and it succeeds independently with probability $p_j$. Further, a partition of the (integer) interval $\{0,\dots,n\}$ into $B$ smaller intervals is known. The goal is to conduct tests so as to determine that interval from the partition in which the number of successful tests lies while minimizing the expected cost. Ghuge et al. (IPCO 2022) recently showed that a polynomial-time constant-factor approximation algorithm exists. We show that interweaving the two strategies that order tests increasingly by their $c_j/p_j$ and $c_j/(1-p_j)$ ratios, respectively, -- as already proposed by Gkensosis et al. for a special case -- yields a small approximation ratio. We also show that the approximation ratio can be slightly decreased from $6$ to $3+2\sqrt{2}\approx 5.828$ by adding in a third strategy that simply orders tests increasingly by their costs. The similar analyses for both algorithms are nontrivial but arguably clean. Finally, we complement the implied upper bound of $3+2\sqrt{2}$ on the adaptivity gap with a lower bound of $3/2$. Since the lower-bound instance is a so-called unit-cost $k$-of-$n$ instance, we settle the adaptivity gap in this case.

Figures (4)

• Figure 1: Sample instance with four tests and three intervals ($t_1 = 0, t_2 = 1, t_3 = 3, t_4 = 5$). The tree depicts the unique optimal adaptive evaluation strategy. A path to the left denotes failure, to the right success of the test in the parent node. A circled node specifies the next test to be conducted; a boxed node corresponds to a determined score. Observe that the first test is not the cheapest; for some paths, the cheapest test is performed last.
• Figure 2: Visualization of the end of phase $1$ and the resulting subinstance. The dotted area shows the interval $i$ containing the score according to the condition in \ref{['lemm:RR:Phase2']}. The dark bar displays the successful tests performed so far, the white bar the failed tests. The number $n'$ of tests in the subinstance is obtained by subtracting the number of tests performed so far from $n$.
• Figure 3: Visualization of some of the sets used in the proof of \ref{['lemm:Alg2:Phase1']}. The slanted lines mark a constant $c_j$-$p_j$- or $c_j$-$(1-p_j)$-ratio, the decision criteria for $\mathrm{A}\xspace_{\mathrm{succ}\xspace}$ and $\mathrm{A}\xspace_{\mathrm{fail}\xspace}$. Markers symbolize tests in $c_j$-$p_j$-space: stands for tests in $U\xspace_\mathrm{succ}\xspace\xspace$, for $C\xspace_\mathrm{succ}\xspace\xspace$ and for $P\xspace_\mathrm{fail}\xspace\xspace$. The symbol stands for a test performed by $\mathrm{A}\xspace_{\mathrm{succ}\xspace}$ not in $U\xspace_\mathrm{succ}\xspace\xspace$ and thus contained in $P\xspace_\mathrm{fail}\xspace\xspace$. Combinations of markers stand for tests contained in multiple sets. The set $C\xspace_{\mathrm{succ}\xspace}$ corresponds to $S\xspace_3 \cup S\xspace_5$, $P\xspace_{\mathrm{fail}\xspace}$ to $S\xspace_1 \cup S\xspace_2$, and $U\xspace_{\mathrm{succ}\xspace}$ to $S\xspace_3 \cup S\xspace_4 \cup S\xspace_6$. Here, $n_i = 11$ with $|C\xspace_\mathrm{fail}\xspace\xspace| = 7$ and $|C\xspace_\mathrm{succ}\xspace\xspace| = 4$.
• Figure 4: Visualization of the adaptivity gap instance. The black bar shows $m$ tests with success probability $1$. The white bar shows $m$ tests with success probability $0$. The gray bar corresponds to test $n$ with $p_n = 1/2$. Note that the interval of the overall score depends only on the realization of the gray test. Any algorithm must inspect the gray test and at least $m$ other tests.

Theorems & Definitions (15)

• Lemma 1: Ben-Dov BenDov81
• Theorem 1
• Lemma 2
• proof
• Lemma 3
• proof
• proof : Proof of Theorem \ref{['thm:TwRR']}.
• Theorem 2
• Lemma 4
• proof