再论自适应采样
The Adaptive Sampling Revisited
摘要：
估计大量$N$数据的不同键的数目$n$的问题在计算机科学中是众所周知的。一个经典的算法是自适应采样（AS）。$n$可以通过$R.2^D$估计，其中$R$是最终存储桶（缓存）大小，$D$是过程结束时的最终深度。关于AS可以提出几个新的有趣的问题（其中一些是由P.Flajolet提出的，并由J.Lumbroso推广）。$W=\log (R2^D/n)$的分布是已知的，我们重新推导这个分布在一个更简单的方法。我们提供了关于$D$和$W$矩的新结果。我们还分析了最终的缓存大小$R$分布。我们考虑彩色键：假设在$n$不同的键中，$n_C$确实有彩色$C$。我们展示了如何估计$p=\frac{n_C}{n}$。我们还研究了由一些分布函数给出的多重性的色密钥。我们要估计这个分布的均值和方差.最后，我们考虑的情况下，既不知道颜色也不知道多重性。在这里，我们要估计相关参数。一个附录专门讨论散列函数提供的比特的概率不同于$1/2$的情况。
英文摘要：
The problem of estimating the number $n$ of distinct keys of a large collection of $N$ data is well known in computer science. A classical algorithm is the adaptive sampling (AS). $n$ can be estimated by $R.2^D$, where $R$ is the final bucket (cache) size and $D$ is the final depth at the end of the process. Several new interesting questions can be asked about AS (some of them were suggested by P.Flajolet and popularized by J.Lumbroso). The distribution of $W=\log (R2^D/n)$ is known, we rederive this distribution in a simpler way. We provide new results on the moments of $D$ and $W$. We also analyze the final cache size $R$ distribution. We consider colored keys: assume that among the $n$ distinct keys, $n_C$ do have color $C$. We show how to estimate $p=\frac{n_C}{n}$. We also study colored keys with some multiplicity given by some distribution function. We want to estimate mean an variance of this distribution. Finally, we consider the case where neither colors nor multiplicities are known. There we want to estimate the related parameters. An appendix is devoted to the case where the hashing function provides bits with probability different from $1/2$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
