Duncan Adamson , Nathan Flaherty , Igor Potapov , Paul G. Spirakis
{"title":"Collision-free Robot Scheduling","authors":"Duncan Adamson , Nathan Flaherty , Igor Potapov , Paul G. Spirakis","doi":"10.1016/j.ic.2025.105294","DOIUrl":null,"url":null,"abstract":"<div><div>In this paper, we investigate the problem of designing <em>schedules</em> for completing a set of tasks at fixed locations with multiple robots in a laboratory. We represent the laboratory as a graph with tasks placed on fixed vertices and robots represented as agents, with the constraint that no two robots may occupy the same vertex at any given timestep. Each schedule is partitioned into a set of timesteps, corresponding to a walk through the graph (allowing for a robot to wait at a vertex to complete a task), with each timestep taking time equal to the time for a robot to move from one vertex to another and each task taking some given number of timesteps during the completion of which a robot must stay at the vertex containing the task. The goal is to determine a set of schedules, with one schedule for each robot, minimising the number of timesteps taken by the schedule taking the greatest number of timesteps within the set of schedules. We show that this problem is NP-complete for both star graphs (for <span><math><mi>k</mi><mo>≥</mo><mn>2</mn></math></span> robots), and planar graphs (for any number of robots). Finally, we provide positive results for path, cycle, and tadpole graphs, showing that we can find an optimal set of schedules for <em>k</em> robots completing <em>m</em> tasks of equal duration of a path of length <em>n</em> in <span><math><mi>O</mi><mo>(</mo><mi>k</mi><mi>m</mi><mi>n</mi><mo>)</mo></math></span>, <span><math><mi>O</mi><mo>(</mo><mi>k</mi><mi>m</mi><msup><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>)</mo></math></span> time, and <span><math><mi>O</mi><mo>(</mo><msup><mrow><mi>k</mi></mrow><mrow><mn>3</mn></mrow></msup><msup><mrow><mi>m</mi></mrow><mrow><mn>4</mn></mrow></msup><mi>n</mi><mo>)</mo></math></span> time respectively.</div></div>","PeriodicalId":54985,"journal":{"name":"Information and Computation","volume":"304 ","pages":"Article 105294"},"PeriodicalIF":0.8000,"publicationDate":"2025-03-06","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Information and Computation","FirstCategoryId":"94","ListUrlMain":"https://www.sciencedirect.com/science/article/pii/S0890540125000306","RegionNum":4,"RegionCategory":"计算机科学","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"COMPUTER SCIENCE, THEORY & METHODS","Score":null,"Total":0}
引用次数: 0
Abstract
In this paper, we investigate the problem of designing schedules for completing a set of tasks at fixed locations with multiple robots in a laboratory. We represent the laboratory as a graph with tasks placed on fixed vertices and robots represented as agents, with the constraint that no two robots may occupy the same vertex at any given timestep. Each schedule is partitioned into a set of timesteps, corresponding to a walk through the graph (allowing for a robot to wait at a vertex to complete a task), with each timestep taking time equal to the time for a robot to move from one vertex to another and each task taking some given number of timesteps during the completion of which a robot must stay at the vertex containing the task. The goal is to determine a set of schedules, with one schedule for each robot, minimising the number of timesteps taken by the schedule taking the greatest number of timesteps within the set of schedules. We show that this problem is NP-complete for both star graphs (for robots), and planar graphs (for any number of robots). Finally, we provide positive results for path, cycle, and tadpole graphs, showing that we can find an optimal set of schedules for k robots completing m tasks of equal duration of a path of length n in , time, and time respectively.
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