通道图上的加速路由调度

Francesco Betti Sorbelli, Federico Coró, Sajal K. Das, A. Navarra, M. C. Pinotti
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引用次数: 3

摘要

本文研究了定向通道图单列问题(OASP),该问题是实体/机器人沿特定通道图移动的路径规划问题的一个变体,该通道图由一组行组成,在这些行的一个端点仅通过一列连接。这种受约束的通道图可以建模,例如,一个葡萄园或仓库,其中每个顶点被分配一个奖励,当机器人完成任务访问它时获得奖励。由于机器人的能量有限,它必须在返回仓库充电之前访问一个顶点子集,同时最大化获得的总奖励。已知由m行长度为n的受限通道图的OASP在$\mathcal{O}\left( {{m^2}{n^2}} \right)$时间内多项式可解,这对于大维度的图来说可能是禁止的。为了设计更省时的解决方案,我们提出了四种算法,以贪婪的方式迭代构建解决方案。这些解决方案最多花费$\mathcal{O}(mn(m + n))$时间,从而将最优解决方案提高了n倍。实验表明,这些算法收集了80多个% of the optimum reward. For two of them, we also guarantee an approximation ratio of $\frac{1}{2}\left( {1 - \frac{1}{e}} \right)$ on the reward function by exploiting the submodularity property, where e is the base of the natural logarithm.
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Speeding-up Routing Schedules on Aisle-Graphs
In this paper, we study the Orienteering Aisle-graphs Single-column Problem (OASP), which is a variant of the route planning problem for an entity/robot moving along a specific aisle-graph consisting of a set of rows connected via just one column at one endpoint of the rows. Such constrained aisle-graph may model, for instance, a vineyard or warehouse, where each vertex is assigned with a reward that a robot gains when visiting it for accomplishing a task. As the robot is energy limited, it must visit a subset of vertices before going back to the depot for recharging, while maximizing the total reward gained. It is known that the OASP for constrained aisle-graphs composed by m rows of length n is polynomially solvable in $\mathcal{O}\left( {{m^2}{n^2}} \right)$ time, which can be prohibitive for graphs of large dimensions. With the goal of designing more time efficient solutions, we propose four algorithms that iteratively build the solution in a greedy manner. These solutions take at most $\mathcal{O}(mn(m + n))$ time, thus improving the optimal solution by a factor of n. Experimentally, we show that these algorithms collect more than 80% of the optimum reward. For two of them, we also guarantee an approximation ratio of $\frac{1}{2}\left( {1 - \frac{1}{e}} \right)$ on the reward function by exploiting the submodularity property, where e is the base of the natural logarithm.
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