Adaptivity Gaps for Stochastic Probing with Subadditive Functions

TL;DR

The paper resolves major open questions on adaptivity gaps in stochastic probing with general monotone norms, proving an $O(\log^2 n)$ bound that refines to $O\left(\dfrac{\log r\log n}{\log\log n}\right)$, where $r$ bounds the maximum probing length, and establishing a tight $\Theta\left(\dfrac{\log n}{\log\log n}\right)$ gap for Bernoulli probing with binary-XOS objectives. By reducing general randomness to Bernoulli settings and further to a structured XOS norm, the authors design a greedy labeling algorithm that balances adaptive and non-adaptive rewards; they also derive an $O(\log^3 n)$ bound for Bernoulli probing with general subadditive objectives and show a constant adaptivity gap for monotone symmetric norms. The results settle conjectures about polylog bounds for general monotone norms, demonstrate tightness in the Bernoulli-binary-XOS case, and significantly improve understanding of how adaptivity impacts performance in stochastic probing across a broad class of objectives. Collectively, these insights have substantial implications for designing near-optimal non-adaptive strategies in uncertainty-aware optimization problems under prefix-closed constraints.

Abstract

In this paper, we study the stochastic probing problem under a general monotone norm objective. Given a ground set $U = [n]$, each element $i \in U$ has an independent nonnegative random variable $X_i$ with known distribution. Probing an element reveals its value, and the sequence of probed elements must satisfy a prefix-closed feasibility constraint $\mathcal{F}$. A monotone norm $f: \mathbb{R}_{\geq 0}^n \to \mathbb{R}_{\geq 0}$ determines the reward $f(X_P)$, where $P$ is the set of probed elements and $X_P$ is the vector with $X_i$ for $i \in P$ and 0 otherwise. The goal is to design a probing strategy maximizing the expected reward $\mathbb{E}[f(X_P)]$. We focus on the adaptivity gap: the ratio between the expected rewards of optimal adaptive and optimal non-adaptive strategies. We resolve an open question posed in [GNS17, KMS24], showing that for general monotone norms, the adaptivity gap is $O(\log^2 n)$. A refined analysis yields an improved bound of $O(\log r \log n / \log\log n)$, where $r$ is the maximum size of a feasible probing sequence. As a by-product, we derive an asymptotically tight adaptivity gap $Θ( \log n/\log\log n)$ for Bernoulli probing with binary-XOS objectives, matching the known lower bound. Additionally, we show an $O(\log^3 n)$ upper bound for Bernoulli probing with general subadditive objectives. For monotone symmetric norms, we prove the adaptivity gap is $O(1)$, improving the previous $O(\log n)$ bound from [PRS23].

Figures (7)

• Figure 1: Roadmap for \ref{['main:conjecture']}.
• Figure 2: In the decision tree, red edges indicate Yes arcs, and blue edges indicate No arcs. We suppose nodes labeled with the same letters (but with different number of primes) represent the same element: $\mathsf{elt}(v_1)=\mathsf{elt}(v_1')=\mathsf{elt}(v_1")=\mathsf{elt}(v_1"')$, $\mathsf{elt}(v_2)=\mathsf{elt}(v_2')=\mathsf{elt}(v_2")$, etc. $v_4$ is the root of the whole tree. Suppose that the specific outcome $R_0$ includes all green nodes in $R_0$, but excludes all white nodes. The algorithm with input $(\ell,R_0)$ first visits $\ell$ and next discovers that $v_1\in A_\ell$ satisfies both conditions. Thus, $B\gets \{\mathsf{elt}(v_1)\}$. After moving up one step without updating, the algorithm detects $v_2\in R_0\cap A_{\ell_1}$ has a descendant $v_1'$ such that $v_1,v_1'$ represent the same element. Then $B\gets \{\mathsf{elt}(v_1),\mathsf{elt}(v_2)\}$. One should note that we do not require $v_2 \in A_\ell$, so we are considering all nodes in $P_\ell$ from the leaf to the root, not just the nodes in $A_\ell$. After two steps, $v_3\in R_0\cap A_{\ell_2}$ is detected, and $B\gets \{\mathsf{elt}(v_1), \mathsf{elt}(v_2), \mathsf{elt}(v_3)\}$. The same thing happens for $v_4\in R_0\cap A_{\ell_3}$, where $B\gets \{\mathsf{elt}(v_1), \mathsf{elt}(v_2), \mathsf{elt}(v_3),\mathsf{elt}(v_4)\}$, and the algorithm terminates with output $B(\ell,R_0 ):=B = \{\mathsf{elt}(v_1), \mathsf{elt}(v_2), \mathsf{elt}(v_3),\mathsf{elt}(v_4)\}$.
• Figure 3: The decision tree is essentially the same as \ref{['fig:binary_tree_part']}, where nodes of the same letters stand for the same nodes. The notation $\triangle$ means an abbreviated subtree. Suppose $B_0=\{\mathsf{elt}(v_1),\mathsf{elt}(v_2),\mathsf{elt}(v_3),\mathsf{elt}(v_4)\}$ is the fixed sequence. By the execution $B(\ell,R_0) = B_0$ shown in \ref{['fig:binary_tree_part']}, we have $\ell \in S(B_0)$. Now, nodes $u_1,u_2,u_3$ are marked black (they are in $A_\ell$ and represent elements $\notin B_0$). Recall that the nodes $v_j$ for $1\leq j\le 4$ trigger updates for $B$ in the execution $B(\ell,R_0)=B_0$. Therefore, we can deduce that among $\mathsf{elt}(v_j)$'s ($1\leq j\le 4$), only $\mathsf{elt}(v_3), \mathsf{elt}(v_4)$ are represented by nodes above $u_1$, because different nodes in $P_\ell$ represent distinct elements. Similarly, only $\mathsf{elt}(v_4)$ is represented by nodes above $u_2,u_3$. Now, we focus on an input $(\ell^*, R)$ that satisfies $B(\ell^*,R)=B_0$, where $\ell^*$ is an arbitrary descendant leaf of $u_1$. In such execution, some nodes that represent$\mathsf{elt}(v_j)$ for $1\leq j\le 4$ must update $B$. Since only $\mathsf{elt}(v_3), \mathsf{elt}(v_4)$ are represented by nodes above $u_1$, when the iterator approaches $u$ in the execution $B(\ell^*,R)$, the current $B$ must be $\{\mathsf{elt}(v_1),\mathsf{elt}(v_2)\}$, confirming \ref{['ob:overview']}. Now, $u_1$ satisfies condition (2) by taking $\ell' = \ell_1$ and hence is an impossible node. Similarly, $u_2,u_3$ are impossible nodes by taking $\ell' = \ell_2$. It is straightforward to see $u_1\notin R$, since otherwise $u_1$ would trigger update for $B$. Similarly, for any descendant leaf $\ell_0$ of $u_1$ (where $\ell_0$ can be $\ell_1$ or any leaf in the left subtree of $v_1^*$), if $B(\ell_0,R) = B_0$ then $u_1\notin R$. Similarly, for any descendant leaf $\ell_0$ (including all leaves drawn in the figure, and possibly leaves in the subtree $\mathcal{T}_1$) of impossible nodes $u_2$ or $u_3$, if $B(\ell_0,R) = B_0$, then $u_2,u_3\notin R$.
• Figure 4: A binary decision tree of height $3$ is transformed to a layered tree of depth $3$. Note that $A$ is an active ancestor of $B$ and $C$, but not for $D$. The active ancestor set $A_C=\{\textnormal{root},A\}$.
• Figure 5: In the decision tree, red edges indicate Yes arcs, and blue edges indicate No arcs. Nodes with the same depth are connected using dashed lines, and their depth is labeled on them. Nodes higher than $v_4$ and in the $\mathsf{Yes}$ subtree of $v_4$ are not drawn. Let us assume that $K$ is large and $|A_{\ell}'| = 4K$ and the depths of the $K$-th, $2K$-th, $3K$-th, $4K$-th nodes of $A_{\ell}'$ (counted from $\ell$) are $4K+10,3K+10,2K+10,K+10$ for all leaves $\ell = \ell_0,\ell',\ell",\ell"'$. Moreover, we assume $v_2 \in A_{\ell_0}'$, $v_2' \in A_{\ell'}'$, $v_2" \in A_{\ell"}'$, $v_2"' \in A_{\ell"'}'$, $v_3\in A_{\ell'}',A_{\ell"}'$, but $v_3\notin A_{\ell_0}',A_{\ell"'}'$. Regarding elements, $\mathsf{elt}(v_2)=\mathsf{elt}(v_2')=\mathsf{elt}(v_2") = \mathsf{elt}(v_2"')$. Consider the label $B_0=(4;4K;6K,6K-1,6K-3,6K-4;0,\mathsf{elt}(v_2),\mathsf{elt}(v_3),\mathsf{elt}(v_4);4K+10,3K+10,2K+10,K+10)$. $\bullet$ All leaves $\ell_0,\ell',\ell",\ell"'$ are in $L_1(B_0)$, and all $v_2,v_2',v_2",v_2"'$ are in $L_2(B_0)$. $v_3$ is in $L_3(B_0)$. Now consider $\mathcal{A}_{v_3}(B_0)$. Note that the satisfying $\ell$ in the definition of $\mathcal{A}_{v_3}(B_0)$ can only be $\ell'$ or $\ell"$ (it needs $v_3\in A_\ell'$), and hence $\mathcal{A}_{v_3}(B_0)$ is defined to be the nodes of depth less than $6K-3$ in $A_{\ell'}'$ (as $\ell'$ is on the 'left' of $\ell"$). $\bullet$ Suppose that the specific outcome $R_0$ includes all green nodes, but does not include white nodes and the nodes above $v_4$. The algorithm with input $(\ell_0,R_0)$ updates $B$ at $\ell_0,v_2,v_3,v_4$, and hence $B(\ell_0,R_0) = B_0$ (note that $v_3\in L_3(B_0)$ can trigger an update, even though $v_3 \notin A_{\ell_0}'$). $\bullet$ Note that $u$ is a descendant of $v_3=v_3(\ell_0,B_0)$, and $u\in \mathcal{A}_{v_2'}(B_0)$ for $v_2'\in L_2(B_0)$. These facts imply the existence of the label $B_1 =(3;4K;6K,6K-1,6K-2;0,\mathsf{elt}(v_2),\mathsf{elt}(u);4K+10,3K+10,2K+10,K+10)$ that $\mathsf{Pref}_2(B_1) = \mathsf{Pref}_2(B_0)$ by \ref{['reduction:unique']} (c). This means that $u\in \mathsf{Imp}(\ell_0, B_0)$. Intuitively $u\notin R_0$, since otherwise, it can update $B$ at $u$ before $v_3$.

Theorems & Definitions (83)

• Definition 1.1: Stochastic Probing
• Definition 1.2: Bernoulli Stochastic Probing
• Theorem 1.1: See also \ref{['thm:6.8']}
• Theorem 1.2: See also \ref{['thm:6.6']}
• Theorem 1.3: See also \ref{['thm:tight']}
• Theorem 1.4: See also \ref{['thm:6.7']}
• Theorem 1.5: See also \ref{['thm:geneXOS']}
• Lemma 1.6: See also \ref{['lm:reduce']}
• Lemma 1.7: See also \ref{['liu:induction']}
• Definition 2.1: (Monotone) Norm Objective