We consider the online $k$-taxi problem, a generalization of the $k$-server problem, in which $k$ servers are located in a metric space. A sequence of requests is revealed one by one, where each request is a pair of two points, representing the start and destination of a travel request by a passenger. The goal is to serve all requests while minimizing the distance traveled without carrying a passenger. We show that the classic Double Coverage algorithm has competitive ratio $2^k-1$ on HSTs, matching a recent lower bound for deterministic algorithms. For bounded depth HSTs, the competitive ratio turns out to be much better and we obtain tight bounds. When the depth is $d\ll k$, these bounds are approximately $k^d/d!$. By standard embedding results, we obtain a randomized algorithm for arbitrary $n$-point metrics with (polynomial) competitive ratio $O(k^c\Delta^{1/c}\log_{\Delta} n)$, where $\Delta$ is the aspect ratio and $c\ge 1$ is an arbitrary positive integer constant. The only previous known bound was $O(2^k\log n)$. For general (weighted) tree metrics, we prove the competitive ratio of Double Coverage to be $\Theta(k^d)$ for any fixed depth $d$, but unlike on HSTs it is not bounded by $2^k-1$. We obtain our results by a dual fitting analysis where the dual solution is constructed step-by-step backwards in time. Unlike the forward-time approach typical of online primal-dual analyses, this allows us to combine information from the past and the future when assigning dual variables. We believe this method can be useful also for other problems. Using this technique, we also provide a dual fitting proof of the $k$-competitiveness of Double Coverage for the $k$-server problem on trees.
翻译:我们考虑的是在线的 $k$- taxi 问题, 这是一种通用的 $k$- server 问题, 即 $k$ 服务器位于一个宽度空间中。 逐个逐个显示一系列请求, 每份请求是一对两分, 代表乘客旅行申请的起始点和目的地。 目标是满足所有请求, 同时在不携带乘客的情况下尽可能减少旅行的距离。 我们显示经典的双重覆盖算法在 HST 上具有2 k- 1 美元的竞争比率, 匹配最近较低的确定性算法。 对于受约束的 HST 深度, 竞争比率显示要好得多, 我们得到的比重要好得多。 当深度为$d/ llk, 这些界限是大约两美元, 代表乘客旅行申请的起始点。 标准嵌嵌入结果, 我们得到任意的算法, 与( Polynomialomi) 竞争的 比率为 O( k) $ (k) $ (k $ d) (cel- d) (c) 1/ log_ Delk) (n- lader lader) (n) lax) roder) roder) 分析结果, 。 在前一个已知的 Ris 方法中, 这个比比比比比比值为2 。