Recovering Assignments with One-Sided Noise
2026-07-27 • Data Structures and Algorithms
Data Structures and AlgorithmsComputational Complexity
AI summaryⓘ
The authors analyze how many questions you need to ask to figure out a hidden pattern in a set of binary values when the answers can be noisy or uncertain. For the simple case with single variables (1-CNF), they provide a precise formula for the number of queries needed, which grows slightly faster than just the size of the problem. They then extend this to more complex group queries (k-CNF) where groups of variables form clauses, showing that adaptive questioning can be much more efficient than nonadaptive approaches. Their findings also apply to related problems like noisy sorting of zero-one strings and involve studying cases with negations.
query complexityplanted assignment1-CNFk-CNFone-sided noiseadaptive algorithmsnonadaptive algorithmsconstraint satisfaction problemsnoisy sortingprobabilistic queries
Authors
Cassandra Marcussen, Elchanan Mossel, Colin Sandon
Abstract
We study the query complexity of recovering a planted assignment from a random constraint-satisfaction instance with one-sided noise. We consider the following 1-CNF recovery problem: an unknown binary string with $n/2$ ones and $n/2$ zeros is queried at individual variables. A query to a $1$-variable returns "$1$" with probability $p$ and "$0$" otherwise, while a $0$-variable always returns "$0$" (each query is a fresh noisy draw). The goal is to recover the binary string with probability at least $1 - δ$. While the naive counting argument may suggest a query complexity of $\log_2 \binom{n}{n/2}=Θ(n)$, we show that the query complexity is $(1+o(1))c(p) \frac{n}{2} \left( \log_2 n + \log_2(1/δ)\right)$, where $c(p) = \tfrac{1}{-\log_2(1-p)}$. We then study planted $k$-CNF satisfaction with one-sided noise. Each $k$-set containing a $1$-variable is included as a clause independently with probability $p$, and an algorithm may ask whether any given $k$-set is a clause. Unlike the $1$-CNF case, a clause-existence query is one-shot: each $k$-set either is or is not a clause, so repeating yields no new information. The model is one-sided because an observed clause certifies that at least one queried variable is assigned 1, whereas its absence does not certify all are assigned 0. The goal is to recover the planted assignment with probability at least $1 - δ$. The counting baseline is $Θ(n)$, yet we prove a query complexity of $(1+o(1))\,c(p,k)\, \frac{n}{2}\left( \log_2 n + \log_2(1/δ)\right)$, where $c(p,k) = \tfrac{1}{k(-\log_2(1-p))}$. These bounds are for adaptive algorithms. We also prove bounds for nonadaptive algorithms, showing that for fixed $p$, adaptivity gives a factor $\exp(Θ(k))$ improvement. Our results also imply lower bounds for noisy sorting of $\{0,1\}$-valued strings, and we study a variant of the model with negations.