粗心优惠券收集问题
The Careless Coupon Collector's Problem
AI总结：
研究了粗心优惠券收集问题，分析了在不同丢失概率下完成收集所需轮次，并提出了计算预期完成时间的算法。
摘要：
我们首次研究了粗心优惠券收集问题（CCCP），这是经典优惠券收集问题的一种新变种，我们设想它可作为信息系统模型，如网络爬虫、动态缓存和容错网络。在CCCP中，收集者通过在每个离散轮次中随机获得一种优惠券类型来收集n种不同的优惠券类型，然而收集者是粗心的：在每个轮次结束时，每种已收集的优惠券类型以概率p独立丢失。我们分析了完成收集所需的轮次数作为n和p函数。特别是，我们证明了它从当p=o(ln n/n²)时的Θ(n ln n)过渡到当p=ω(1/n)时的Θ((np/(1-p))ⁿ)的多个不同阶段。有趣的是，当p=c/n时，过程仍处于亚稳态阶段，其中已收集的优惠券类型的分数集中在1/(1+c)周围，概率为1-o(1)，持续时间窗口长度为e^{Θ(n)}。最后，我们给出一个在O(n²)时间内计算CCCP预期完成时间的算法。
英文摘要：
We initiate the study of the Careless Coupon Collector's Problem (CCCP), a novel variation of the classical coupon collector, that we envision as a model for information systems such as web crawlers, dynamic caches, and fault-resilient networks. In CCCP, a collector attempts to gather $n$ distinct coupon types by obtaining one coupon type uniformly at random in each discrete round, however the collector is \textit{careless}: at the end of each round, each collected coupon type is independently lost with probability $p$. We analyze the number of rounds required to complete the collection as a function of $n$ and $p$. In particular, we show that it transitions from $Θ(n \ln n)$ when $p = o\big(\frac{\ln n}{n^2}\big)$ up to $Θ\big((\frac{np}{1-p})^n\big)$ when $p=ω\big(\frac{1}{n}\big)$ in multiple distinct phases. Interestingly, when $p=\frac{c}{n}$, the process remains in a metastable phase, where the fraction of collected coupon types is concentrated around $\frac{1}{1+c}$ with probability $1-o(1)$, for a time window of length $e^{Θ(n)}$. Finally, we give an algorithm that computes the expected completion time of CCCP in $O(n^2)$ time.
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。
