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Algorithms for Covering Barrier Points by Mobile Sensors with Line Constraint 移动传感器覆盖障碍点的线性约束算法
Pub Date : 2024-02-27 DOI: 10.1142/s0218195924500018
Princy Jain, Haitao Wang
We study the problem of covering barrier points by mobile sensors. Each sensor is represented by a point in the plane with the same covering range [Formula: see text] so that any point within distance [Formula: see text] from the sensor can be covered by the sensor. Given a set [Formula: see text] of [Formula: see text] points (called “barrier points”) and a set [Formula: see text] of [Formula: see text] points (representing the “sensors”) in the plane, the problem is to move the sensors so that each barrier point is covered by at least one sensor and the maximum movement of all sensors is minimized. The problem is NP-hard. In this paper, we consider two line-constrained variations of the problem and present efficient algorithms that improve the previous work. In the first problem, all sensors are given on a line [Formula: see text] and are required to move on [Formula: see text] only while the barrier points can be anywhere in the plane. We propose an [Formula: see text] time algorithm for the problem. We also consider the weighted case where each sensor has a weight; we give an [Formula: see text] time algorithm for this case. In the second problem, all barrier points are on [Formula: see text] while all sensors are in the plane but are required to move onto [Formula: see text] to cover all barrier points. We also solve the weighted case in [Formula: see text] time.
我们研究的是移动传感器覆盖障碍点的问题。每个传感器由平面上的一个点表示,该点的覆盖范围[公式:见正文]相同,因此传感器可以覆盖距离[公式:见正文]内的任何一点。给定平面上一组[公式:见正文][公式:见正文]点(称为 "障碍点")和一组[公式:见正文][公式:见正文]点(代表 "传感器"),问题是移动传感器,使每个障碍点至少被一个传感器覆盖,并使所有传感器的最大移动量最小。该问题具有 NP 难度。在本文中,我们考虑了该问题的两个线性约束变体,并提出了改进前人工作的高效算法。在第一个问题中,所有传感器都给定在一条直线上[计算公式:见正文],并且只要求在[计算公式:见正文]上移动,而障碍点可以是平面上的任何位置。我们为这个问题提出了一种[公式:见正文]时间算法。我们还考虑了加权情况,即每个传感器都有一个权重;我们给出了这种情况下的[公式:见正文]时间算法。在第二个问题中,所有障碍点都在[公式:见正文]上,而所有传感器都在平面上,但需要移动到[公式:见正文]上以覆盖所有障碍点。我们也可以在[公式:见正文]时间内解决加权情况。
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引用次数: 0
Maximum Bipartite Subgraphs of Geometric Intersection Graphs 几何交点图的最大双方子图
Pub Date : 2023-12-16 DOI: 10.1142/s021819592350005x
Satyabrata Jana, Anil Maheshwari, Saeed Mehrabi, Sasanka Roy
We study the Maximum Bipartite Subgraph ([Formula: see text]) problem, which is defined as follows. Given a set [Formula: see text] of [Formula: see text] geometric objects in the plane, we want to compute a maximum-size subset [Formula: see text] such that the intersection graph of the objects in [Formula: see text] is bipartite. We first give an [Formula: see text]-time algorithm that computes an almost optimal solution for the problem on circular-arc graphs. We show that the [Formula: see text] problem is [Formula: see text]-hard on geometric graphs for which the maximum independent set is [Formula: see text]-hard (hence, it is [Formula: see text]-hard even on unit squares and unit disks). On the other hand, we give a [Formula: see text] for the problem on unit squares and unit disks. Moreover, we show fast approximation algorithms with small-constant factors for the problem on unit squares, unit disks, and unit-height axis parallel rectangles. Additionally, we prove that the Maximum Triangle-free Subgraph ([Formula: see text]) problem is NP-hard for axis-parallel rectangles. Here the objective is the same as that of the [Formula: see text] except the intersection graph induced by the set [Formula: see text] needs to be triangle-free only (instead of being bipartite).
我们研究的是最大双方子图([公式:见正文])问题,其定义如下。给定一组[公式:见正文]平面内的[公式:见正文]几何物体,我们想计算一个最大尺寸的子集[公式:见正文],使得[公式:见正文]中物体的交集图是双分部的。我们首先给出了一种[公式:见正文]时间算法,它能在圆弧图上计算出该问题的几乎最优解。我们证明,在最大独立集是[公式:见正文]的几何图形上,[公式:见正文]问题是[公式:见正文]困难的(因此,即使在单位正方形和单位圆盘上,它也是[公式:见正文]困难的)。另一方面,我们给出了单位正方形和单位磁盘上问题的[公式:见正文]。此外,我们还展示了单位正方形、单位磁盘和单位高度轴平行矩形上问题的小常数快速近似算法。此外,我们还证明了最大无三角形子图([公式:见正文])问题对于轴平行矩形来说是 NP 难的。这里的目标与[公式:见正文]的目标相同,只是由集合[公式:见正文]诱导的交集图只需要是无三角形的(而不是双方形的)。
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引用次数: 0
Tiling the Plane with a Set of Ten Polyominoes 用一组10个多边形平铺平面
Pub Date : 2023-04-04 DOI: 10.1142/s0218195923500012
Chao Yang
There exists a linear algorithm to decide whether a polyomino tessellates the plane by translation only. On the other hand, the problem of deciding whether a set of [Formula: see text] or more polyominoes can tile the plane by translation is undecidable. We narrow the gap between decidable and undecidable by showing that it remains undecidable for a set of [Formula: see text] polyominoes, which partially solves a conjecture posed by Ollinger.
存在一种线性算法来确定多边形是否只通过平移来镶嵌平面。另一方面,决定一组[公式:见文本]或更多多边形是否可以通过平移平铺平面的问题是无法确定的。我们缩小了可判定和不可判定之间的差距,证明了一组[公式:见文本]多项式仍然是不可判定的,这部分解决了Ollinger提出的一个猜想。
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引用次数: 0
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International Journal of Computational Geometry & Applications
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