We revisit the knapsack-secretary problem (Babaioff et al.; APPROX 2007), a generalization of the classic secretary problem in which items have different sizes and multiple items may be selected if their total size does not exceed the capacity $B$ of a knapsack. Previous works show competitive ratios of $1/(10e)$ (Babaioff et al.), $1/8.06$ (Kesselheim et al.; STOC 2014), and $1/6.65$ (Albers, Khan, and Ladewig; APPROX 2019) for the general problem but no definitive answers for the achievable competitive ratio; the best known impossibility remains $1/e$ as inherited from the classic secretary problem. In an effort to make more qualitative progress, we take an orthogonal approach and give definitive answers for special cases. Our main result is on the $1$-$2$-knapsack secretary problem, the special case in which $B=2$ and all items have sizes $1$ or $2$, arguably the simplest meaningful generalization of the secretary problem towards the knapsack secretary problem. Our algorithm is simple: It $\textit{boosts}$ the value of size-$1$ items by a factor $\alpha>1$ and then uses the size-oblivious approach by Albers, Khan, and Ladewig. We show by a nontrivial analysis that this algorithm achieves a competitive ratio of $1/e$ if and only if $1.40\lesssim\alpha\leq e/(e-1)\approx 1.58$. Towards understanding the general case, we then consider the case when sizes are $1$ and $B$, and $B$ is large. While it remains unclear if $1/e$ can be achieved in that case, we show that algorithms based only on the relative ranks of the item values can achieve precisely a competitive ratio of $1/(e+1)$. To show the impossibility, we use a non-trivial generalization of the factor-revealing linear program for the secretary problem (Buchbinder, Jain, and Singh; IPCO 2010).
翻译:我们重新审视了Knappsack保密问题(Babaioff等人;APROX 2007),这是对传统秘书问题的概括化,其中项目规模不同,如果总规模不超过一个 knapscack 的容量,则可能选择多个项目;以前的工作显示1美元/(10美元)(Babaioff等人)的竞争性比率(Kesselheim等人);1/8.06美元(Kesselheim等人;STOC 2014)和1/6.65美元(Abers、Khan和Ladewig;APROX 2019),涉及一般问题,但对于可实现的竞争比率没有确定答案;最已知的不可能仍然是1美元/e美元,这是从经典秘书问题中继承下来的。为了取得更高质量的进展,我们采取了一种或多的方法,我们只能用1美元-knassack秘书的问题,如果我们用1美元/美元,那么我们用1美元/40美元,那么所有项目就达到1美元或2美元这个特殊的情况。 可以说,最简单的秘书问题就是向Knabkh schedeal a licess a cros deal licult laus a proal macial ex ex ex ex max a pal laus a pal lax a pal a procuilate.