AI 中文总结
本文提出一个黑盒提升定理，将针对互不相同输入的奇偶检验树转化为任意输入上的多项式决策树，仅增加O(log n)深度，从而在无并列承诺下实现最大值查找和top-k选择，保持随机化误差不变。
AI 中文摘要
在无限制多项式决策树模型中，仅对多项式符号检验的次数进行计数。奇偶检验询问成对差值的乘积的符号。此类检验支撑了最大值查找和top-$k$选择的低深度随机化算法，但其通常的分析假设输入互不相同，因为并列会使乘积为零。我们给出了一个黑盒提升定理，消除了这一假设。在$O(\log n)$次多项式检验确定非零成对差值的数量之后，每个后续的奇偶检验都由一个多项式检验模拟，并与固定的字典序并列打破顺序保持一致。该模拟器是所有成对差值的掩蔽一次幂和二次幂的初等对称多项式。因此，在互不相同输入上的深度为$D$的奇偶检验树，在任意输入上变为深度为$D+O(\log n)$的多项式决策树，且对于顺序选择问题，随机化逐点误差不增加。我们实现了深度为$O(\log n[\log n+\log(1/\delta)])$、误差为$\delta$的最大值查找，以及深度为$O(\log^2 n+k\log n)$、误差为逆多项式误差的top-$k$选择，两者均无需对并列做任何承诺。
英文摘要
In the unrestricted polynomial decision-tree model, only the number of polynomial sign tests is charged. A parity test asks for the sign of a product of pairwise differences. Such tests underlie low-depth randomized algorithms for maximum finding and top-$k$ selection, but their usual analysis assumes distinct inputs because a tie makes the product vanish. We give a black-box lifting theorem that removes this assumption. After $O(\log n)$ polynomial tests determine the number of nonzero pairwise differences, every subsequent parity test is simulated by one polynomial test, consistently with a fixed lexicographic tie-breaking order. The simulator is an elementary symmetric polynomial in masked first and second powers of all pairwise differences. Thus a depth-$D$ parity-test tree on distinct inputs becomes a polynomial decision tree of depth $D+O(\log n)$ on arbitrary inputs, with no increase in randomized pointwise error for order-selection problems. We obtain maximum finding in depth $O(\log n[\log n+\log(1/δ)])$ with error $δ$, and top-$k$ selection in depth $O(\log^2 n+k\log n)$ with inverse-polynomial error, both without any promise on ties.
