路径上不可分割流的常因子逼近算法

P. Bonsma, J. Schulz, Andreas Wiese
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引用次数: 78

摘要

本文给出了一种求解路径上不可分流问题的常因子近似算法。这改进了之前最著名的近似因子O(log n)。对于任何e>0,我们算法的近似比为7+e。在路径上的不可分割流问题中,我们有P和n个可容路径任务,每个任务都有需求、利润、起始点和结束点。目标是计算任务的最大利润集,使得对于P的每条边e,使用e的选定任务的总需求不超过e的容量。这是一个很好的研究问题,在各种设置中自然发生,因此它已经在其他名称下进行了研究,例如资源分配,带宽分配,资源约束调度,时间背包和间隔打包。该问题的多项式时间常数因子近似算法以前只在无瓶颈假设下已知(其中最大任务需求必须不大于最小边缘容量)。我们介绍了一些新的算法技术,它们可能是独立的兴趣:一个框架,它将问题简化为具有有限容量范围的实例,一个新的几何启发的动态规划,它将最大权无关矩形集问题的特殊情况解决为最优性。此外,我们证明了即使所有边缘容量相等并且所有需求为1、2或3,问题也是强np困难的。
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A Constant Factor Approximation Algorithm for Unsplittable Flow on Paths
In this paper, we present a constant-factor approximation algorithm for the unsplittable flow problem on a path. This improves on the previous best known approximation factor of O(log n). The approximation ratio of our algorithm is 7+e for any e>0. In the unsplittable flow problem on a path, we are given a capacitated path P and n tasks, each task having a demand, a profit, and start and end vertices. The goal is to compute a maximum profit set of tasks, such that for each edge e of P, the total demand of selected tasks that use e does not exceed the capacity of e. This is a well-studied problem that occurs naturally in various settings, and therefore it has been studied under alternative names, such as resource allocation, bandwidth allocation, resource constrained scheduling, temporal knapsack and interval packing. Polynomial time constant factor approximation algorithms for the problem were previously known only under the no-bottleneck assumption (in which the maximum task demand must be no greater than the minimum edge capacity). We introduce several novel algorithmic techniques, which might be of independent interest: a framework which reduces the problem to instances with a bounded range of capacities, and a new geometrically inspired dynamic program which solves a special case of the maximum weight independent set of rectangles problem to optimality. In addition, we show that the problem is strongly NP-hard even if all edge capacities are equal and all demands are either 1, 2, or 3.
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