Recovering Assignments with One-Sided Noise

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 $1 - \delta$. While the naive counting argument may suggest a query complexity of $\log_2 \binom{n}{n/2}=\Theta(n)$, we show that the query complexity is $(1+o(1))c(p) \frac{n}{2} \left( \log_2 n + \log_2(1/\delta)\right)$, 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 $1 - \delta$. The counting baseline is $\Theta(n)$, yet we prove a query complexity of $(1+o(1))\,c(p,k)\, \frac{n}{2}\left( \log_2 n + \log_2(1/\delta)\right)$, 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 $\exp(\Theta(k))$ 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.