Suppose that a group test operation is available for checking order relations in a set, can this speed up problems like finding the minimum/maximum element, determining the rank of element, and computing order statistics? We consider a one-sided group inequality test to be available, where queries are of the form $u \le_Q V$ or $V \le_Q u$, and the answer is `yes' if and only if there is some $v \in V$ such that $u \le v$ or $v \le u$, respectively. We restrict attention to total orders and focus on query-complexity; for min or max finding, we give a Las Vegas algorithm that makes $\mathcal{O}(\log^2 n)$ expected queries. We observe that rank determination can be solved with existing ``defect-counting'' algorithms, but also give a simple Monte Carlo approximation algorithm with expected query complexity $\tilde{\mathcal{O}}(\frac{1}{δ^2} \log \frac{1}ε)$, where $1-ε$ is the probability that the algorithm succeeds and we allow a relative error of $1 \pm δ$ for $δ> 0$ in the estimated rank. We then give a Monte Carlo algorithm for approximate selection that has expected query complexity $\tilde{\mathcal{O}}(\frac{1}{δ^4}\log \frac{1}{εδ^2} )$; it has probability at least $\frac{1}{2}$ to output an element $x$, and if so, $x$ has the desired approximate rank with probability $1-ε$. Keywords: Order statistics, Group inequality testing, Randomized algorithms

翻译：假设存在一种群组测试操作可用于检查集合中的顺序关系，这是否能加速诸如寻找最小/最大元素、确定元素秩以及计算顺序统计量等问题？我们考虑一种单边群组不等式测试的可用性，其中查询形式为$u \le_Q V$或$V \le_Q u$，当且仅当存在某个$v \in V$使得$u \le v$（或$v \le u$）时，答案才为“是”。我们将注意力限制在全序关系上，并聚焦于查询复杂度；对于最小或最大元素查找，我们提出了一种拉斯维加斯算法，其期望查询次数为$\mathcal{O}(\log^2 n)$。我们观察到秩确定问题可通过现有的“缺陷计数”算法解决，但也提出了一种简单的蒙特卡洛近似算法，其期望查询复杂度为$\tilde{\mathcal{O}}(\frac{1}{δ^2} \log \frac{1}ε)$，其中$1-ε$为算法成功的概率，且允许估计秩的相对误差为$1 \pm δ$（$δ> 0$）。随后我们提出了一种近似选择问题的蒙特卡洛算法，其期望查询复杂度为$\tilde{\mathcal{O}}(\frac{1}{δ^4}\log \frac{1}{εδ^2} )$；该算法以至少$\frac{1}{2}$的概率输出元素$x$，且若成功输出，则$x$以$1-ε$的概率具有所需的近似秩。关键词：顺序统计量，群组不等式测试，随机化算法