Optimal Coverage in Wireless Sensor Networks [1st ed.] 9783030528225, 9783030528249

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Springer Optimization and Its Applications 162

Weili Wu · Zhao Zhang Wonjun Lee · Ding-Zhu Du

Optimal Coverage in Wireless Sensor Networks

Springer Optimization and Its Applications Volume 162 Series Editors Panos M. Pardalos , University of Florida My T. Thai , University of Florida Honorary Editor Ding-Zhu Du, University of Texas at Dallas Advisory Editors Roman V. Belavkin, Middlesex University John R. Birge, University of Chicago Sergiy Butenko, Texas A&M University Franco Giannessi, University of Pisa Vipin Kumar, University of Minnesota Anna Nagurney, University of Massachusetts Amherst Jun Pei, Hefei University of Technology Oleg Prokopyev, University of Pittsburgh Steffen Rebennack, Karlsruhe Institute of Technology Mauricio Resende, Amazon Tamás Terlaky, Lehigh University Van Vu, Yale University Guoliang Xue, Arizona State University Yinyu Ye, Stanford University

Aims and Scope Optimization has continued to expand in all directions at an astonishing rate. New algorithmic and theoretical techniques are continually developing and the diffusion into other disciplines is proceeding at a rapid pace, with a spot light on machine learning, artificial intelligence, and quantum computing. Our knowledge of all aspects of the field has grown even more profound. At the same time, one of the most striking trends in optimization is the constantly increasing emphasis on the interdisciplinary nature of the field. Optimization has been a basic tool in areas not limited to applied mathematics, engineering, medicine, economics, computer science, operations research, and other sciences. The series Springer Optimization and Its Applications (SOIA) aims to publish state-of-the-art expository works (monographs, contributed volumes, textbooks, handbooks) that focus on theory, methods, and applications of optimization. Topics covered include, but are not limited to, nonlinear optimization, combinatorial optimization, continuous optimization, stochastic optimization, Bayesian optimization, optimal control, discrete optimization, multi-objective optimization, and more. New to the series portfolio include Works at the intersection of optimization and machine learning, artificial intelligence, and quantum computing. Volumes from this series are indexed by Web of Science, zbMATH, Mathematical Reviews, and SCOPUS.

More information about this series at http://www.springer.com/series/7393

Weili Wu • Zhao Zhang • Wonjun Lee Ding-Zhu Du

Optimal Coverage in Wireless Sensor Networks

Weili Wu Department of Computer Science University of Texas at Dallas Richardson, TX, USA

Zhao Zhang Department of Computer Science Zhejiang Normal University Jinhua, Zhejiang, China

Wonjun Lee School of Cybersecurity Korea University Seoul, Korea (Republic of)

Ding-Zhu Du Department of Computer Science University of Texas at Dallas Richardson, TX, USA

ISSN 1931-6828 ISSN 1931-6836 (electronic) Springer Optimization and Its Applications ISBN 978-3-030-52822-5 ISBN 978-3-030-52824-9 (eBook) https://doi.org/10.1007/978-3-030-52824-9 © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Since the fabric of the world is the most perfect and was established by the wisest Creator, nothing happens in this world in which some reason of maximum or minimum would not come to light. —Euler The only difference between suicide and martyrdom is press coverage. —Chuck Palahniuk God used beautiful mathematics in creating the world. —Paul Dirac

Preface

The motivation for writing this book came from our previous research works on sensor coverage. They received a large number of citations and this number grows rapidly each year. This means that this is a hot research subject currently. In fact, the coverage and connectivity are the two fundamental subjects in the study of wireless sensor networks. As the technology of wireless sensor network has developed quite fast, these two subjects get more and more attention from researchers. Especially, many works are made on optimization issues, which raise many optimization problems and generate a lot of optimization techniques. This is a reference book which presents the state of the art in research on theoretical aspect of the optimal sensor coverage problems. The book can be divided into three parts. The first part consists of Chapters 1–8 which mainly contain classical optimal sensor coverage problems, such as the minimum sensor cover problem, the minimum connected cover problem, the maximum lifetime coverage problem, etc. The second part consists of Chapters 9–12 which give different types of sensor coverage. The third part contains Chapters 13–17; different types of sensors are studied in this part. Most contents in this book have served as teaching materials for advanced topics in a graduate course in computer science in University of Texas at Dallas, as well as many graduate summer schools in China. Therefore, we appreciate very much our colleagues and graduate students for their course organization and comments on teaching materials, including Hejiao Huang, Zhenhua Duan, Cong Tian, Tiende Guo, Suixianng Gao, Xiaodong Hu, Deying Li, Yuexuan Wang, Jianzhong Li, Hong Gao, Yingshu Li, Zhipeng Cai, Peng-Jun Wan, Hongwei Du, Donghyun Kim, Manki Min, Xiaofeng Gao, Feng Zou, Yaochun Huang, Zaixin Lu, Xiuzhen Cheng, Mihaela Cardei, Ionut E. Cardei, My Thai, Maggie Xiaoyan Cheng, Chih-Hao Huang, Feng Wang, Ling Ding, James Willson, Xianyue Li, Jiaofei Zhong, Yan Shi, Lidong Wu, Shan Shan, Zhongnan Zhang, Heejun Roh, Kai Xing, Lidan Fan, Wen Xu, Yuanjun Bi, Yuqing Zhu, Huan Ma, Nassim Sohaee, Wei Wang, and Wei Zhang, who have made direct or indirect contributions in the process of writing this book. Especially, we wish to thank Xiaohui Huang for drawing all the figures in this book. vii

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Moreover, we are grateful for the support of Professors Andy Yao, Francis Yao, Xiaohua Jia, and Jiannong Cao. Actually, a lot of writing was done during some authors visited at Institute for Interdisciplinary Information Sciences, Tsinghua University, City University of Hong Kong, and Hong Kong Polytechnic University. Finally, we would like to acknowledge the support in part by NSF of USA under grants 1747818 and 1907472, by NSF of China under grants 11771013, 11531011, and 61751303, by NRF of Korea (MSIT) under grants 2019R1A2C2088812 and 2017M3C4A7083676, and by ZJNSF of China under grants LD19A010001 and LY19A010018. Richardson, TX, USA

Weili Wu

Jinhua, China

Zhao Zhang

Seoul, Republic of Korea

Wonjun Lee

Richardson, TX, USA January 2020

Ding-Zhu Du

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Sensor and Sensor Network. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Sensing Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Communication Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 4

2

Fundamental Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Energy Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Deployment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 7 14 16 17 19

3

Sensor Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Planar Expansion Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 PTAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23 23 24 26 29

4

Connected Sensor Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 NP-Hardness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 O(r)-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Network Steiner Tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 LP Relaxation for NST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 The Algorithm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Bridge Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 Performance Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Metric Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Randomized O(log2 n log m)-Approximation . . . . . . . . . . . . . . . . . . . . . . 4.7 O((Rs /Rc )2 )-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 33 34 36 37 37 40 40 43 50 54 61

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5

Lifetime of Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Garg–Könemann Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Lifetime of Connected Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Weighted Connected Sensor Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 A Framework of Calnescu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

67 67 70 71 76 82 85

6

Weighted Sensor Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6.2 Partition: 28-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 6.3 Double Partition: 6-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 6.4 Jigsaw Puzzle: 4-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 6.5 Magic Transformation: 3.63-Approximation . . . . . . . . . . . . . . . . . . . . . . . 107 6.6 PTAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

7

k-Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Reduction to Weighted Sensor k-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Parity Strip Multi-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 (4 + ε)-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 (3 + ε)-Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

117 117 118 119 125 129

8

Heterogeneous Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Minimum Weight Set Multi-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Minimum Weight Disk Multi-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Shallow Cell Complexity and ε-Net . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 ε-Net and Minimum Cardinality Disk Cover . . . . . . . . . . . . . . . 8.3.3 Quasi-Uniform Sampling and Minimum Weight Disk Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.4 Minimum Weight Disk Multi-Cover . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Minimum Power Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Composite Event Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

135 135 137 140 141 143

9

Grid-Based Deployment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Coverage with Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Coverage with Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

153 153 154 155

10

Barrier Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Disjoint Barrier Covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Secure Schedule with Homogeneous Sensors . . . . . . . . . . . . . . . . . . . . . . 10.4 Secure Schedule with Heterogeneous Sensors . . . . . . . . . . . . . . . . . . . . . . 10.5 Quality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Weak Barrier Cover and Local Barrier Cover. . . . . . . . . . . . . . . . . . . . . . .

159 159 163 165 171 173 179

144 147 148 150

Contents

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Sweep-Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Min-Sensor Sweep-Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Min-Sweep-Period Sweep-Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183 183 184 185 187 190

12

Partial Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Geometric Maximum Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Set k-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Minimum Weight Partial Set Multi-Cover . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Minimum Partial Sensor Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 p-Percent Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

193 193 194 196 198 200 202

13

Probabilistic Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Random Deployment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Probabilistic Sensing Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

203 203 204 206

14

Mobile Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 An Observation on Cascade Healing Paths . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Maximum Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Greedy Is Exponentially Bad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Online Algorithm with Tight Competitive Ratio . . . . . . . . . . . . . . . . . . . 14.6 Consider Longest Moving Distance, Again . . . . . . . . . . . . . . . . . . . . . . . . . 14.7 Remark. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

209 209 211 213 215 216 224 225

15

Camera Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Sensing Direction Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Group Set Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Directional Targets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Target Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

227 227 230 231 240 242

16

Energy-Harvesting Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Mule Scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Sleep/Wakeup Scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Hybrid Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Energy-Data Dual Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

245 245 246 251 252 253

17

Underwater Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 3D Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Underwater Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

257 257 257 259

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Contents

Crowdsensing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Motivation and Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Time Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Reward Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4 Continuous Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

261 261 262 264 265

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

Chapter 1

Introduction

Parking is a nightmare for me. . . I still have sensors on my car that help me park. Jordana Brewster

1.1 Sensor and Sensor Network Do you know the movie Twister directed by Jan de Bont? The movie grossed $494.5 million worldwide, and became the second-highest-grossing film of 1996. The part of story described in Wikipedia is as follows. In June 1969 Oklahoma, young Jo Thornton and her family are awakened by an approaching F5 tornado. The family seeks refuge in their storm cellar, but the tornado rips the cellar door off and pulls Jo’s father to his death while Jo and her mother look on. The next morning, they awake to find their farmhouse destroyed. In the present day, 27 years after the death of Jo’s father, the National Severe Storms Laboratory predicts a record outbreak of tornadoes in Oklahoma over a 24-hour period. Now an adult, Jo, a meteorologist and storm chaser, is reunited with her estranged husband, Bill Harding, a former weather researcher and fellow storm chaser, who has since moved on to become a popular television weather reporter. He has a brand new Dodge Ram pickup truck and is planning to marry reproductive therapist Melissa Reeves. However, his plans are delayed until Jo signs her long overdue divorce papers. Arriving to Jo’s camp in order to get her to complete the forms, he finds that she and her team have built four tornado research devices called DOROTHY based on Bill’s ground breaking design. Each unit contains hundreds of sensors that, if picked up by a tornado, will provide data that could lead to revolutionary breakthroughs in meteorological research.

At the end of the movie, sensors are successfully deployed into tornado. They are flying in the sky. They collect data and send data to computer for investigating the tornado. Often, sensors are small electronic devices. Each sensor consists of several components, including sensing unit, communication unit, storage unit, processing unit, embedded software, and power unit (Fig. 1.1). These components enable sensor to have three functions, sensing, communication, and computation. The sensing function is used for collecting information from environment, the communication function is used for data delivery between sensors © Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_1

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1 Introduction

Fig. 1.1 Components of a sensor

Fig. 1.2 Wireless sensor network

or from sensor to sink. The computation function is usually quite weak, which can be used for performing only simple computation task. The communication function connects sensors together with other device, such as sink, into a network. This network is called the wireless sensor network (Fig. 1.2). The wireless sensor network has a lot of applications [75]. When sensors are deployed into a battlefield, they can monitor activity of armies and help general make decision. When sensors are deployed underwater, they can be used to detect submarines and missiles launched from submarines (Fig. 1.3). When sensors are deployed into concrete structure of a bridge, they can measure the stability of the bridge during earthquake. When smart phones are used as sensors for crowdsensing, they can report traffic information and provide advice to drivers (Fig. 1.4). When sensors are deployed into field with pollution danger, they can monitor the changes of the field condition and report the information about environment. When sensors are installed into a human body, they can monitor biological system and report health status of the human (Fig. 1.5).

1.2 Sensing Model

3

Fig. 1.3 Underwater sensor network

Fig. 1.4 Traffic control

Nowadays, we meet sensors everyday and everywhere, in house and public service, in school and research labs, in industry and technology development. Therefore, the study of fundamental issues and problems on sensors and wireless sensor networks has great broader impact in social and economic developments.

1.2 Sensing Model There are two quite different sensing models, deterministic model and probabilistic model. The deterministic model is also called the classic 0/1 model or the binary model. In this model, each sensor has a sensing area. A target can be detected by a sensor if and only if the target moves into the sensing area of the sensor.

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1 Introduction

Fig. 1.5 Biological system

The sensing area can be a disk with center at the sensor’s location. Such a sensor is said to have an undirected sensing power. The sensing area can also be a sector. Such a sensor is said to have a directed sensing power. In probabilistic model, there is a probability associated with the event that a target is detected by a sensor. This probability is a function with respect to the distance between the target and the sensor. For example, Pr(s, p) =

C d k (s, p),

where s is a sensor, p is a target, d(s, p) is the Euclidean distance between s and p, k and C are constants related to characteristic of the sensor and environment condition, respectively. In this function, the distance d(s, p) gets larger, the detection probability gets smaller. In this book, our study is mainly under the binary model. However, in Chapter 13, we will pay more attention on the probabilistic model.

1.3 Communication Model The sensing and the communication are the similar signal processing in nature. In sensing, the signal comes from target and is received by the sensor. Meanwhile, in communication, the signal comes from the sensor and is received by other devices. Therefore, the communication also has two models, the binary model and the SINR (signal-to-interference-and-noise ratio) model.

1.3 Communication Model

5

In binary model, each sensor has a communication area. A device is able to receive the signal from the sensor if and only if the device is located at the communication area. The communication area can be a disk with center at the sensor’s location. Such a sensor is said to have an undirected communication power. The communication area can also be a sector. Such a sensor is said to have a directed communication power. A sensor is called a directional sensor if it has either directed sensing power or directed communication power, or both. Unless special mentioning (e.g., in Chapter 15), sensors studied in this book are not directional. In SINR model, the power of signal received by a receiver is a function with respect to the distance between the receiver and the sensor. The reader may find out more from [517]. We would not discuss further because throughout this book, our study is under the binary communication model.

Chapter 2

Fundamental Issues

Most practical questions can be reduced to problems of largest and smallest magnitudes . . . and it is only by solving these problems that we can satisfy the requirements of practice which always seeks the best, the most convenient. ˘ P. L. Ceby˘ sev

2.1 Coverage Nowadays, sensors exist everywhere. They are used for monitoring battlefield, controlling traffic, watching environment, managing manufacture process, detecting disasters, examining human’s bodies, and collecting data from hostile area, etc. In many applications, the coverage is a fundamental requirement [365, 408, 574] and hence becomes an important issue in study of sensor systems, especially wireless sensor networks. For example, when a request comes to ask for information on enemy’s activities in a certain area of battlefield, a set of sensors are required to activate for covering (sensing) the target area. Given a target area or a set of target points, the coverage problem usually aims at finding a set of sensors for covering given target area or all given target points. A set of sensors is called a sensor cover if it can realize the coverage, i.e., covers a given target area or all given target points. In the literature, a related problem is often called a coverage problem or a sensor cover problem, e.g., the minimum connected sensor cover problem and the maximum lifetime coverage problem. The coverage of target area can often be converted into the coverage of a set of target points. For a sensor o, its sensing area is a disk, denoted by disks (o). Its boundary is called the sensing circle, denoted by circles (o). Please note that disks (o) and circles (o) are simplified notations of diskRs (o) and circleRs (o), respectively, where Rs is the sensing radius. In general, diskr (o) (circler (o)) is the disk with center o and radius r. Therefore, at subscript of disk (circle), a symbol other that s and c would represent the radius. For example, disk0.5 (s) is a disk with center s and radius size 0.5 unit.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_2

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Fig. 2.1 In each small area, choose an interior point as a target point

The area coverage problem can be formulated as follows: Given a set D of n disks with possibly different radius, and a target area , find a set P of target points such that a subset D of D covers target area  if and only if D covers all target points in P. There are several ways in the literature for choosing the set of target points, P. For any area A, denote by ∂A the boundary of A. Note that circles ∂D for all D ∈ D divide the area  into many small areas. The first way is to choose an interior point (Fig. 2.1) from each small area. Let a1 , . . . , ama be all chosen points. The following theorem indicates that they can be target points to replace the target area . Theorem 2.1.1 (Wu et al. [567]) Suppose ∂ has O(n2 ) intersection points with ∂D for all D ∈ D. Then ma = O(n2 ). Moreover, any disk subset D ⊆ D covers area  if and only if every ai is an interior point of D for some D ∈ D . Proof By Euler’s formula, (ma + 1) + (v  + v  ) = e + 2, where v  = |{x ∈  | x is an intersection of ∂D and ∂D  for some D, D  ∈ D}|, v  = |{x | x is an intersection of ∂ and ∂D for some D ∈ D}|, and e is the number of segments on the boundaries of small areas in ; they result from ∂ and ∂D for D ∈ D, cutting by intersection points. Note that v  + v  = O(n2 ). To show ma = O(n2 ), it suffices to prove e = O(n2 ). First, note that since v  = O(n2 ), the number of segments on ∂ is O(n2 ). Now, we show that the number of segments on ∂D for D ∈ D is also O(n2 ). To do so, we first remove ∂ and consider only ∂D for D ∈ D. We claim that the number of pieces resulting from cutting all ∂D for D ∈ D with intersection points is at most 2n(n − 1). We show this claim by induction on n. For n = 1, it is trivially true. Next, consider n disks with assumption that the claim for n − 1 disks is true. A new circle ∂D can intersect n − 1 circles with at most 2(n − 1) intersection points. Those intersection points would cut ∂D into at most 2(n − 1) segments. Each of those intersection points may also break some segments, on other n − 1 circles, into more

2.1 Coverage

9

Fig. 2.2 The second way is to choose all intersection points of disk boundaries lying in the interior of area

segments. However, each intersection point can break at most one such segment into two. Therefore, at most 4(n − 1) segments would be increased. The total number of segments on n circles is at most 2(n − 1)(n − 2) + 4(n − 1) = 2n(n − 1). Although those segments can be further cut by ∂, the increasing number of segments is at most O(n2 ). Therefore, e = O(n2 ). This completes the proof of the first half of this theorem. For the second half of this theorem, it suffices to note that no boundary cuts any small area. Thus, each ai is covered by a disk if and only if the small area with representative ai is covered by the disk. Thus, D covers area  if and only if every small area is covered by a disk in D if and only if all representatives a1 , a2 , . . . , ama are covered by the disk subset D . Moreover, each ai is covered by a disk if and only if it is covered by the interior of the disk.   The second way is to choose intersection points of circles ∂D for D ∈ D. Let b1 , . . . , bmb be all such intersection points lying in the interior of  (Fig. 2.2). Since two circles have at most two intersection points, we have mb = O(n2 ). The following theorem indicates that those intersection points can be used for target points to replace the target area. Theorem 2.1.2 (Wu et al. [567]) Suppose mb ≥ 1 and for every disk D, every maximal connected piece of ∂D lying in the interior of  contains at least one bi . Then a disk subset D covers area  if and only if every bi lies in the interior of the union ∪D∈D D. Proof Note that all b1 , .., bmb lie in the interior of . If D covers , then clearly, every bi is in the interior of the union ∪D∈D D. Thus, the necessity is true. Next, we show the sufficiency. For contradiction, suppose the sufficiency is not true. Then there exists a point p in , not covered by the union ∪D∈D D. However, since mb ≥ 1 and b1 is covered by the union ∪D∈D D, there must exist a boundary piece of this union separating p and b1 , and lying in the interior of . If this piece comes from boundaries of at least two disks (Fig. 2.3a), then it must contain an intersection point bi which is not an interior point of the union ∪D∈D D, contradicting the theorem condition.

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2 Fundamental Issues

Fig. 2.3 A piece of the boundary of the union of disks in D separates p and b1

Fig. 2.4 A counterexample

If this piece of the boundary of ∪D∈D D comes from one disk (Fig. 2.3b), then it must be a maximal connected piece lying in the interior of  and hence contains some point bi which is not an interior point of the union ∪D∈D D, again contradicting the theorem condition.   When a large amount of sensors deployed into an area , the assumption in Theorem 2.1.2 holds easily. However, the assumption may not be held when the number of sensors is small. In such a case, we can consider to use the first method. It is necessary for Theorem 2.1.2 to have the condition that every maximal connected piece of disk boundary lying in the interior of  contains at least one bi . A counterexample in Fig. 2.4 shows that if this condition does not hold, then D may not cover whole area although all bi are covered in the interior of the union of all disks. This condition also has some relationship with the condition in Theorem 2.1.1 that the number of intersection points of ∂ and boundaries of disks is O(n2 ). Indeed, when the number of intersection points of ∂ and boundaries of disks is more than O(n2 ). There may exist more than O(n2 ) maximal connected pieces of boundaries of disks, lying in the interior of  (Fig. 2.5), which implies the existence of at least one such maximal connected piece not containing any intersection point ai because ma = O(n2 ). Hall [243] gave an interesting sufficient condition for convex area  and a set D of disks with the same radius to satisfy the condition in Theorem 2.1.2. Theorem 2.1.3 (Hall [243]) Consider a convex area  and a set D of disks with the same radius. Suppose convex area  contains one or more disk in D. If for each D ∈ D, either ∂D intersects at least one another ∂D  lying the interior of  for some D  ∈ D, or D ∩  = ∅. Then convex area  and disk set D satisfy the

2.1 Coverage

11

Fig. 2.5 A circle may introduce m  n2 maximal pieces in a regular polygon with m edges

condition that for every disk, every maximal connected piece of its boundary lying in the interior of  contains at least one bi . Proof First, we note that since  contains at least one disk in D, each ∂D can have at most one maximal connected piece lying in the interior of , which must contain an intersection point bi since it intersects at least one disk lying in the interior of .   When the number n of sensors is large, it may not be easy to compute points a1 , a2 , . . . , ama because determination of all nonempty small areas may require to study 2n systems of quadratic inequalities and for each system, determine whether there exists or not a solution of the system. However, computing b1 , b2 , .., bmb may be relatively easy since we need only to solve n(n − 1)/2 systems each of two quadratic equations. Therefore, in practice, we often use b1 , b2 , . . . , bmb instead of a1 , a2 , . . . , ama . A valuable tip is how to deal with the case that there exists a maximal piece of sensing circle of a sensor not containing any ai . The following theorem suggests that we may arbitrarily choose an interior point on the maximal piece. Theorem 2.1.4 Consider an area  and a disk set D. Suppose c1 , c2 , . . . , cmc are interior points of an area , satisfying the following two conditions: (a) {b1 , b2 , . . . , bmb } ⊆ {c1 , c2 , . . . , cmc }. (b) Every maximal piece of disk boundary lying in the interior of area  contains some point ci . Then, a disk subset D covers  if and only if all c1 , c2 , . . . , cmc lie in the interior of the union ∪D∈D D. Proof Similar to the proof of Theorem 2.1.2.

 

Zhang and Hou [615] gave the following result. Theorem 2.1.5 (Zhang and Hou [615]) Consider a convex area  and a homogeneous disk set D. Let bmb +1 , . . . , bmb be all intersection points of ∂ and circle ∂D for D ∈ D. Assume mb ≥ 1. If every point of b1 , b2 , . . . , bmb is covered by the interior of a disk in disk subset D , then  is covered by D .

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2 Fundamental Issues

Fig. 2.6 Graph G() with source nodes circled

Proof Suppose  contains a point p not covered by D . Then between p and b1 , there is a boundary of ∪D∈D D. This boundary cannot contain anyone of b1 , b2 , . . . , bmb since everyone of them is covered by the interior of a disk in D . Thus, this boundary must be the boundary of a disk D in D . However, this is impossible because D must contain b1 and the circle ∂D  passing through b1 must intersect ∂D since D and D  have the same size of radius.   In each of above methods, the number of selected target points is O(n2 ). Is it possible to select less number of target points to replace the target area? This is an interesting open problem. Du and Wu [175] proposed an improvement for the first method as follows. First, construct a directed graph on points a1 , a2 , . . . , ama selected in the first method described as above. Two points ai and aj are said to be adjacent if two small areas represented by ai and aj have a boundary in common. This boundary must be a piece of ∂D for some disk D ∈ D. If ai lies outside of D and aj lies inside of D, then we add an edge from ai to aj . The directed graph constructed in this way is denoted by G() (Fig. 2.6). Theorem 2.1.6 (Du [175]) G() contains no cycle. Proof Consider an edge (ai , aj ). By construction of edge (ai , aj ), we know that there exists a disk D ∈ D such that ai is not in D and aj is in D. Note that every edge crossing ∂D has a direction from outside to inside of D. Therefore, no path exists from aj to ai . This means that no cycle exists in G().   A point ai is called a source node if every edge incident to ai is going out from ai . Let As be the set of all source nodes in G (Fig. 2.6). Corollary 2.1.7 As = ∅. Proof Clearly, ma ≥ 1. If a1 is not a source node, then there exists an edge (ai2 , a1 ). If ai2 is not a source node, then there exists an edge (ai3 , ai2 ). This process ends at finding either a source node or a cycle, contradicting to Theorem 2.1.6.   Theorem 2.1.8 (Du [175]) A disk subset D of D covers target area  if and only if D covers all points in As .

2.1 Coverage

13

Fig. 2.7 Nonseparable source nodes

Fig. 2.8 |As | = (n2 )

Proof The “only if” part is trivial. We show the “if” part as follows. For contradiction, suppose D does not cover . We take disks in D − D one by one and add to D until D becomes a maximal subset not covering , that is, D does not cover  and however, for any D ∈ D − D , D ∪ {D} covers . In this case, the uncovered area of  must contain a source node outside all disks in D , contradicting to the assumption that D covers all points in S.   How large is |As |? In many examples, |As | is significantly smaller than ma . However, Guoli Ding and Wenan Zang (private communication) indicated that |As | = (n2 ). As shown in Fig. 2.7, all n disks are divided into four groups and each contains n/4 disks. They form n2 /16 source nodes. Motivated from this fact, we have a conjecture as follows. Conjecture 2.1.9 For any area , there exists a set of n disks, D such that if a finite set of points, A has the property that any disk subset D ⊆ D covers  if and only if D covers A, then |A| = (n2 ). Sometime the boundary of ∂ may introduce too many source nodes (Fig. 2.8). Du and Wu [175] also proposed a method to give an improvement in such a case as follows. Two source nodes ai and aj in As are separable if there is a disk D in D such that ai is not in D and aj is in D, or vice versa. For any two source nodes in As , if they are not separable, then we delete one from As . Suppose A∗s is obtained from As through this operation until no source node can be deleted any more, that is, every two source nodes in A∗s are separable (Fig. 2.8).

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Theorem 2.1.10 (Du and Wu [175]) A disk subset D of D covers target area  if and only if D covers all points in A∗s . Proof The proof is similar to the proof of Theorem 2.1.8. However, we should note that at end of the proof. The uncovered part of  may contain several source nodes which are not separable, one of which should belongs to A∗s .  

2.2 Connectivity The connectivity is another important issue in the study of wireless sensor networks. In fact, after collect information, sensors need to deliver information to certain station, which requires the existence of a communication network connecting sensors. Actually, each sensor has a communication area, which is also a disk. The radius of communication disk may be different from the radius of sensing disk. A sensor can send information to another one if the latter lies inside the communication disk of the former. For a set of sensors, S, the communication network for S is the directed graph with node set S and edge set {(s, s  ) | s  lies in the communication disk of s}. Since a set of sensors owns a communication network, we may call a set of sensors as a sensor network or wireless sensor network to emphasize that the communication is wireless. A sensor network is homogeneous if all sensors have the same communication radius Rc and sensing radius Rs . For a homogeneous sensor network, the communication network can be looked as an undirected graph. Please recall. In general, we denote by diskr (o) a disk with center o and radius r, and by circler (o) a circle with center o and radius r. Moreover, for simplicity of notation, denote by diskc (o) (instead of diskRc (o)) the communication disk of sensor o, and by circlec (o) (instead of circleRc (o)) the boundary of the communication disk of sensor o. Similarly, denote by disks (o) the sensing disk of sensor o and by circles (o) the boundary of the sensing disk of sensor o. Therefore, for r in diskr (o) and circler (o), if r is not s or c, then r denotes a radius. Although in general the communication network of a sensor network is a directed graph, an undirected communication model is also considered in the literature [291, 454]. In such a model, a communication link only in one direction would be ignored. Surprisingly, the coverage and the connectivity sometimes have a close relationship. In the following, we would like to present two examples. Theorem 2.2.1 (Zhang and Hou [615]) Consider a homogeneous sensor network S with Rc ≥ 2Rs . If S is a sensor cover for a connected area  such that every sensor s in S has a sensing point in , then S is a connected network.

2.2 Connectivity

15

Fig. 2.9 A curve C lies in area 

Proof For any two sensors s and s  , choose two points p and p in  such that p is in the sensing area of s and p is in the sensing area of s  . Since  is a connected area, we can draw a curve C connecting p and p , lying inside of  (Fig. 2.9). Suppose C is covered by sensing areas of a sequence of sensors s1 = s, s2 , . . . , sk = s  . Without loss of generality, we may assume that for any two distinct sensors si and sj , disks (si ) ∩ C ⊆ disks (sj ) ∩ C and disks (si ) ∩ C ⊇ disks (sj ) ∩ C. We may also assume that along curve C from s to s  , a point would meet sensing areas of sensors in ordering s1 , s2 , . . . , sk . This means that disks (si ) ∩ disks (si+1 ) = ∅. Since Rc ≥ 2Rs , we would have si ∈ diskc (si+1 ) and si+1 ∈ diskc (si ), that is, edge (si , si+1 ) exists in communication network. Hence, s and s  is connected by path (s = s1 , s2 , . . . , sk = s  ). Therefore, the communication network of S is connected.   A sensor cover S for a target area  is called a sensor k-cover if every point in  is covered by at least k sensors in S. When an area has a sensor k-cover, we say that the area receives k-coverage. A graph G = (V , E) is k-connected if removal k − 1 nodes from G cannot destroy the connectivity, i.e., for any subset V  of at most k − 1 nodes, G[V − V  ], the subgraph induced by V − V  , is still connected. Theorem 2.2.1 can be generalized as follows. Theorem 2.2.2 (Zhou, Das and Gupta [649]) Consider a homogeneous sensor network S with Rc ≥ 2Rs . If S is a sensor k-cover for a connected area  such that every sensor s in S has a sensing point in , then S is a k-connected network. Proof Since S is a sensor k-cover for , every point in  can still be covered after delete at most k − 1 sensors from S. By Theorem 2.2.1, the sensor network is still connected after delete at most k − 1 sensors.   In Theorems 2.2.1 and 2.2.2, the condition that every sensor s in S has a sensing point in  is quite natural since a sensor which does not cover any point in  can be deleted from a sensor cover or k-cover. Theorem 2.2.3 (Xing et al. [575]) Consider a homogeneous sensor network S with Rc ≥ 2Rs . If S is a sensor k-cover for a convex area , then for every two sensors u and v with their sensing areas lying in the interior of , u and v are 2k-connected in the communication network of S, i.e., there are at least 2k nodedisjoint paths between u and v.

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Fig. 2.10 circles (u) and circles (v) lie in the interior of 

Proof Let (w, x) be the diameter of circles (u) perpendicular to line (u, v) and (y, z) the diameter of circles (v) perpendicular to line (u, v) (Fig. 2.10). Since circles (u) and circles (v) lie in the interior of , the rectangle wxyz lies in the interior of . For contradiction, suppose u and v are not 2k-connected. Then there exists a subset M of 2k − 1 sensors in S such that removal M would break all connections between u and v. This implies that segment wz contains a point p not covered by the sensing disk of any sensor not in M and segment xy also contains a point q not covered by the sensing disk of any sensor not in M. Note that the area A covered by sensing disks of sensors in S \ M is a closed area and hence the area A¯ not covered by any sensing disk of sensor in S \ M is an open area. Moreover, p is an interior point of . Therefore, p has an open neighborhood lying in the interior of ¯ From this neighborhood, choose a point p not in rectangle wxyz.  and also in A.  Then p and q have distance larger than 2Rs where Rs is the sensing radius of every sensor in S. This means that each sensor in M can cover at most one of p and q. Thus, one of p and q cannot be sensed by k sensors in M, a contradiction.  

2.3 Energy Efficiency The energy efficiency is an important issue in study of wireless sensor networks due to the following: (a) Each senor is usually equipped with batteries and hence has a limited energy supply. (b) In many applications, sensors are deployed into a hostile or dangerous region so that changing batteries is a mission impossible. A lot of optimization problems stem from consideration of energy saving [496]. When a target area is requested to be monitored for a long time, one may maximize the lifetime of coverage by making a proper schedule of active and sleeping modes of sensors. However, when a duty can be finished within the lifetime of every sensor, the minimization on total energy consumption may be performed. If all active sensor have the same energy consumption rate, that is, consume the same amount of energy

2.4 Deployment

17

during the same time period, then the minimization of total energy is equivalent to the minimization of the number of active sensors. The following are examples of optimization problems with objective functions coming from above consideration. Problem 2.3.1 (Minimum Sensor Cover) Given a set of sensors and a set of target points in the Euclidean plane, find a sensor cover with minimum cardinality. Problem 2.3.2 (Minimum Connected Sensor Cover) Given a set of sensors and a set of target points in the Euclidean plane, find a connected sensor cover with minimum cardinality. Problem 2.3.3 (Maximum Lifetime Coverage) Given a set of sensors with unit lifetime and a set of target points in the Euclidean plane, find the maximum lifetime of coverage, where the lifetime of coverage is a time period in which every target point is covered by at least one active sensor and every sensor is scheduled to have at most unit total time in active state. Problem 2.3.4 (Minimum Weight Sensor Cover) 2.3.4 Given a set A of target points and a set S of sensors with positive weight S → R + in the Euclidean plane, find a sensor cover with minimum total weight. In above problems, we consider only target points because by Theorems 2.1.1 and 2.1.2, target area may be transformed into an equivalent set of target points in many cases. However, Theorems 2.2.1 and 2.2.2 indicate that sometime, such a transformation may loss the information on connectivity. When this happens, that is, a result holds only for connected target area, we will mention it specially. All optimization problems in above are classic in the literature and will be studied in this book.

2.4 Deployment There are two different ways for deploying sensors. One is to deploy sensors randomly in a certain field. The other one is to put sensors in designated locations [45, 52, 168, 351, 519, 530]. These two ways may produce different problems, different algorithms, and different analysis methods. Therefore, some research results are related to certain deployment. For example, when sensors are randomly deployed into area , the probability for three sensors to have their sensing circles intersecting at a point is close to zero since the measure for such sensor positions is zero. We may take this advantage to give a simpler way to implement Theorem 2.1.2. In fact, Theorem 2.1.2 requires that each bi is covered by the interior of the union of sensing disks of considered sensors, which means that such a covering for bi may be realized by not only one sensing disks. The following result indicates that this situation can be changed under certain condition.

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2 Fundamental Issues

Theorem 2.4.1 (Wu et al. [567]) Suppose mb ≥ 1 and for every sensor s, every maximal connected piece of its sensing circle, lying in the interior of , contains at least one bi . If any three sensing circles of sensors in S  do not intersect at a point, then  is covered by sensor subset S  if and only if every bi for 1 ≤ i ≤ mb is covered by the interior of sensing area of a sensor in S  . Proof Suppose bi is the intersection point of boundaries of disks of sensors o and o . Note that the union of sensing disks of o and o cannot cover a neighbor area of bi completely, i.e., bi cannot be an interior point of the union. This means that bi must be covered by the sensing area of the third sensor o . Since the sensing circle of o cannot pass bi , bi must be covered by the interior of sensing disk of o . Therefore, bi is covered by the interior of sensing disks of sensors in S  if and only if bi is covered by the interior of sensing disk of some sensor in S  . Now, this theorem follows immediately from this fact and Theorem 2.1.2.   Therefore, we have Corollary 2.4.2 Suppose mb ≥ 1 and for every sensor s, every maximal connected piece of its sensing circle, lying in the interior of , contains at least one bi . Then for random deployment of sensors, with probability one, the target area  is covered by a subset of sensors, S  if and only if every bi for 1 ≤ i ≤ mb is covered by the interior of sensing area of a sensor in S  . Let us look at another example. Suppose sensor locations can be designed at any points in the Euclidean plane to monitor a set of given target points. Then we may obtain a minimization problem different from previous ones on coverage as follows. Problem 2.4.3 (Minimum Disk Cover) Given a set of target points in the Euclidean plane and a set of disks, find a way to place the minimum number of disks in order to cover all target points. Solution of this problem consists of a set of locations for disk centers, i.e., sensors. How do we deploy those sensors to those designated locations? One way is to use mobile sensors. First, deploy mobile sensors near designated locations, and then move them to those locations by solving the following problem. Problem 2.4.4 Given a set of m mobile sensors, S and a set of n locations, L in a region, find a one-to-one mapping f : S → L to minimize maxs∈S d(s, f (s)) where d(·, ·) is the Euclidean distance. The purpose for scheduling moving in this way is to maximize the minimum remaining energy of each sensor so that the lifetime of coverage of sensor system gets maximized. Problem 2.4.4 is not hard to be solved. Theorem 2.4.5 Problem 2.4.4 is polynomial-time solvable. Proof First, sort all mn distances d(s, l) for s ∈ S, l ∈ L into nondecreasing ordering, d1 ≤ d2 ≤ · · · ≤ dmn . Define a0 = −1. Then carry out the following algorithm to find an optimal solution.

2.5 Approximation

Initially, set a ← 0 and b ← dmn ; while b − a > 1 do c ← (a + b)/2; if Gdc contains a matching covering S then b ← c else a ← c; output db .

19

 

Often, expensive sensors, such as camera sensors, are not deployed randomly. Actually, different sensors may promote different optimization problems. They form special topics in this book in later chapters.

2.5 Approximation Most of optimization problems studied in this book are NP-hard. Unlike Problem 2.4.4, NP-hard problems may not have a polynomial-time solution. To find efficient solutions, one usually designs approximation algorithms[177] or heuristics [102, 234, 410, 416, 481, 605]. In this book, we emphasize on design and analysis of approximation algorithms. By approximation, we mean that there exists theoretical analysis for establishing the performance. An algorithm is said to be an ρ-approximation for a minimization problem if it produces a solution with the objective function value within a factor of ρ from the objective function value of the minimum solution for every input. An algorithm is said to be an ρ-approximation for a maximization problem if it produces a solution with the objective function value exceed a factor of ρ from the objective function value of the minimum solution for every input. Here, ρ is called a performance ratio of the algorithm. For any minimization problem, we usually have 1 < ρ and for any maximization problem, we often have 0 < ρ < 1. Following is an example. Theorem 2.5.1 For any ε > 0, the minimum disk cover problem has a (1 + ε)2 approximation with running time nO(1/ε ) where n is the number of target points. Proof First, We use a square Q to cover all input points and divide the square Q into a grid of m × m squares, called cells (Fig. 2.11). Then for each cell e, we find the minimum number of disks for covering all target points in cell e, denoted by D(e) the set of those disks. We claim that computing 4m2 ) time where cell e contains n target points. In fact, a D(e) needs at most O(n e √ e √ unit disk can cover a 22 × 22 square. Since a cell can be partitioned into at most 2m2 such squares, at most 2m2 unit disks are required to cover given points in a cell. If there is a point with a distance at least one from other point, then we have to spend a disk to cover it. For disk covering such a point, we count as a unique choice. For disk covering other points, We may move each of them to a canonical position

20

2 Fundamental Issues

Fig. 2.11 Grid

that there are two target points lie on the boundary of the disk. For ne target points,   2 there are at most 2 n2e locations (Fig. 2.12). Hence, there are at most (ne (ne −1))2m possible ways to cover all ne target points in cell e. This means that the exhausted2 to find a minimum solution for the cell. search takes time at most n4m e  2 2 Let D = ∪e D(e). Then computing D would take time e n4m ≤ n4m . e Next, we estimate the size of D. Consider an optimal solution D ∗ for the minimum disk cover problem. Let us make some modification on D ∗ copies for each disk lying on the boundary of cells. For a disk intersecting two cells, make one additional copy and for a disk intersecting with four cells, make three more copies. For each cell e, denote by D ∗ (e) the subset of disks in D ∗ , intersecting with cell e. Then |D(e)| ≤ |D ∗ (e)| and hence |D| ≤



|D ∗ (e)| ≤ |D ∗ |(1 + 3d/|D ∗ |),

e

where d is the number of disks hitting some cutlines of the partition. Now, without of loss of generality, assume Q = {(x, y) | 0 ≤ x ≤ q, 0 ≤ ¯ = {(x, y) | −m ≤ x ≤ y ≤ q}. Let p = q/m + 1. Consider the square Q ¯ into (p + 1) × (p + 1) grid so that each cell is mp, −m ≤ y ≤ mp}. Partition Q ¯ is denoted by P (0, 0) (Fig. 2.13). In general, a m × m square. This partition of Q the partition P (a, b) is obtained from P (0, 0) by shafting the left-bottom corner of S from (−m, −m) to (−m + a, −m + b). Note that for every a, 0 ≤ a < m, the interior of square P (a, a) contains square Q. For simplicity of speaking, by cell we mean an m × m square excluding the top and the right boundaries so that no two cells are overlapping each other and no unit disk intersects with three parallel strips where each strip consists of p +1 cells lying along a vertical or horizontal line (Fig. 2.14). Now, for each cell e of P (a, a), denote by Da the disk set D as above, constructed in time   2 2 2 neO(m ) ≤ ( ne )O(m ) = nO(m ) . e

e

Let Ha (Va ) be the set of all disks in D ∗ each overlapping with two horizontal (vertical) strips in P (a, a). Note that every unit disk can overlap with at most four

2.5 Approximation

21

Fig. 2.12 There are at most two possible positions for a unit disk with two given points on its boundary

Fig. 2.13 Square Q and Partition P (0, 0)

Fig. 2.14 Strips

cells in P (a, a). If a unit disk in D ∗ overlaps with more than two cells in P (a, a), then it must belong to both Ha and Va since no unit disk is able to overlap with three parallel strips. It follows that  |D ∗ (e)| ≤ |D ∗ | + |Ha | + 2|Va |. e over cells in P (a,a) Now, note that all Ha for a = 0, 1, . . . , m − 1 are disjoint. Thus, m−1 

|Ha | ≤ |D ∗ |.

a=0

Similarly, m−1  a=0

|Va | ≤ |D ∗ |.

22

2 Fundamental Issues

Therefore, m−1 

|Aa | ≤

a=0

m−1 

(|D ∗ | + |Ha | + 2|Va |) ≤ (m + 3)|D ∗ |.

a=0

Hence m−1 m−1  1  3 ε |Aa | ≤ (|D ∗ | + |Ha | + 2|Va |) ≤ (1 + )|D ∗ | ≤ (1 + )|D ∗ |. m m 2 a=0

a=0

This implies that at least a half number of a = 0, 1, . . . , m − 1 satisfy that |Da | ≤ (1 + ε)|D ∗ |.

 

A minimization problem is said to have a PTAS (polynomial-time approximation scheme) if for any ε > 0, it has a polynomial-time (1+ε)-approximation. Hence, we may simply say that the minimum disk cover problem has a PTAS. A maximization problem is said to have a PTAS if for any ε > 0, it has a polynomial-time (1 − ε)approximation. Almost every geometric optimization problem has a PTAS since there exist quite powerful geometric tools to construct PTAS [177]. However, it does not mean that the PTAS can be obtained easily for the geometric optimization problem. Actually, every optimization problem on sensor cover has geometric structure. However, there exist many open problems about their approximation algorithms. Following are two well-known examples. Open Problem 1 Is there a polynomial-time constant-approximation for the minimum connected sensor cover problem? Open Problem 2 Is there a PTAS for the maximum lifetime connected coverage problem? Open Problem 3 Is there a PTAS for the maximum lifetime k-coverage (k ≥ 2)? Chapters 3 and 4 are written with Open Problem 1 in mind and Chapters 5–8 are written with Open Problems 2 and 3 in mind. Chapters 9–18 are contributed to new research subjects on sensor coverage.

Chapter 3

Sensor Cover

Don’t Judge a Girl by Her Cover Ally Carter

3.1 Motivation and Overview In this chapter, we study the minimum sensor cover problem (Problem 2.3.1). This is an NP-hard problem [156, 404]. There exist a lot of interesting issues about its approximation solution. First, consider an instance consisting of a set of target points, A, and a set of sensors, S, lying in the Euclidean plane. For each s ∈ S, let A(s) denote the subset of target points lying in the sensing area of sensor s. Then the minimum sensor cover problem can be seen as a special case of the following set cover problem. Problem 3.1.1 (Minimum Set Cover) Given a collection C of subsets of a finite set X with property that ∪A∈C A = X, find a subcollection C  ⊆ C with minimum cardinality and with property that ∪A∈C  A = X. A subcollection C  is called a set cover if ∪A∈C  A = X. Therefore, we may simply say that this problem is to find a minimum set cover from a given set cover. The minimum set cover problem is a well-known NP-hard combinatorial optimization problem. Here, we list some important results about its approximation solutions as follows. Theorem 3.1.2 (Chvátal [155], Johnson [278], Lovász [378]) The minimum set cover problem has a polynomial-time H (n)-approximation where n = |X| and  H (n) = ni=1 1i < 1 + ln n is called a harmonic function. Theorem 3.1.3 (Dinur and Steurer [170]) For any 0 < ρ < 1, there is no polynomial-time (ρ ln n)-approximation for the minimum set cover problem unless NP = P . The low bound stated in Theorem 3.1.3 was obtained through a sequence of efforts [170, 192, 438, 503]. © Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_3

23

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3 Sensor Cover

Clearly, the approximation algorithm for the minimum set cover also works for the minimum sensor cover problem. Therefore, the minimum sensor cover has a polynomial-time (1 + ln n)-approximation. By Theorem 3.1.3, the polynomialtime (1 + ln n)-approximation is most likely the best possible for the minimum set cover problem. However, the minimum sensor cover problem has a geometric background and is also called a geometric set cover problem. Hence, it has better approximations. Actually, in study of the minimum set cover, if X is a geometric universe and each subset in C is a geometric area such as a square or a disk, then the problem is called the minimum geometric set cover problem. More precisely, the minimum sensor cover problem is a minimum geometric set cover problem with disks. For unit disks (or say for homogeneous wireless sensor networks), it has been known for many year that there are several polynomial-time O(1)-approximations for the minimum sensor cover problem [4, 81, 90, 99, 415]. A big progress was made by Wan et al. [516]. Based on a result of Mustafa and Ray [413], Wan et al. showed that the minimum sensor cover problem has PTAS (PolynomialTime Approximation Scheme). More importantly, they showed a planar expansion theorem which has many applications. However, the PTAS often has a high running time and hence has a limitation in practice. Therefore, it is still worth studying O(1)-approximations with fast running time. The following problem is still open. Open Problem 4 Is there an approximation algorithm with performance ratio at most two and running time O(n2 )? It is worth mentioning that in the construction for faster O(1)-approximation for the minimum sensor cover problem, the ε-net (see Chapter 7 for definition) is often involved. All developments are closely related to new results on the ε-net [16, 42, 85, 187, 248, 316, 319, 405, 424, 427] and its connection to the linear programming approach [110, 157, 306]. There exist some variations of the sensor cover problem in the literature. For examples, the sensor location is variable or the radius of sensing disk is variable or the objective function is different. They are motivated from different applications. We may study them in later chapters.

3.2 Complexity In this section, we show the NP-hardness of the minimum sensor cover problem. First, let us introduce a result of Valiant [505]. Lemma 3.2.1 (Valiant [505]) A planar graph G with maximum degree 4 can be embedded in the Euclidean plane within O(|V |) area such that all vertices are located at integer points and all edges are consisting of vertical or horizontal line segments passing through integer points.

3.2 Complexity

25

Fig. 3.1 Graphs G and G

Now, we present the proof of NP-hardness. Theorem 3.2.2 The minimum sensor cover problem is NP-hard. Proof From [215], we can find that the following problem is NP-complete. Problem 3.2.3 Given a planar graph G with maximum degree 3 and an integer k > 0, determine whether G contains a dominating set of size at most k. (A subset of vertices is called a dominating set if every vertex is either in it or adjacent to a vertex in it.) We construct a polynomial-time many-one reduction from this problem to the decision version of the minimum sensor cover problem as follows. For each instance of Problem 3.2.3, which consists of a planar graph G with maximum degree 3 and an integer k > 0, we first embedded G into the Euclidean plane by Lemma 3.2.1. As described in Lemma 3.2.1, every edge can be represented by line segments parallel to the x- or y-axis and each line segment has integer length. Moreover, it is easy to make that no two parallel lines are closer than two units apart and each edge (u, v) is represented by line segments with total length 1 + 3kuv for some integer kuv . (As shown in Fig. 3.1, it needs a little trick to realize such a goal. This may be a good exercise for the reader.) A unit disk graph G is induced by this drawing, containing exactly those integer points lying on a line in the drawing. It is easy to show that G has a dominating  set  of size at most k if and only if G has a dominating set of size at most k + (u,v)∈E(G) kuv . Finally, note that if at every vertex of G , there is a sensor located there and there is a target point located there. Then a subset of sensors is a sensor cover if and only if their locations form a dominating set of G . Therefore, above reduction actually induces a reduction from Problem 3.2.3 to the decision version of the minimum sensor cover problem.  

26

3 Sensor Cover

3.3 Planar Expansion Theorem The minimum sensor cover problem has a PTAS. However, this PTAS cannot be designed by the partition technique as showed in the proof of Theorem 2.5.1. We will employ a new technique, the local search method. To explore it, we show the following theorem. Theorem 3.3.1 (Planar Expansion Theorem, Mustafa and Ray [413]) There exist two universal positive constants K ≥ 3 and c such that for any k ≥ K, the following holds: Let G = (R, B, E) be a planar bipartite graph on red and blue vertex sets R and B with |R| ≥ 2. If for every B  ⊆ B with |B  | ≤ k, |NG (B  )| ≥ |B  |, then √ |B| ≤ (1 + c/ k)|R|,

(3.1)

where NG (B  ) = {u ∈ R | (u, v) ∈ E for some v ∈ B}. To prove this Planar Expansion Theorem, we need first to show the following results. Theorem 3.3.2 (Chrobak and Eppstein [154]) Every planar graph can be oriented such that every node has an in-degree at most three. Proof We show, by induction, a result a little stronger than the theorem as follows. Every planar graph can be oriented such that every node has an in-degree at most three and each exterior node has an in-degree at most two. The theorem holds trivially for every graph with at most three nodes. Consider a graph G with n ≥ 4 nodes. Then G must contain an exterior node v with at most two adjacent exterior nodes. Let G be the graph obtained from G by deleting v. By induction hypothesis, G has the required orientation. Now, for each edge (v, x), if x is an exterior node of G, then assign direction from x to v to this edge and if x is not an exterior node of G, then assign direction from v to x to the edge. Note that the existence of edge (v, x) implies that x is an exterior node of G . (See Fig. 3.2.) Thus, we obtain a required orientation for G.   Theorem 3.3.3 (Koutis and Miller [308]) Any planar graph with n nodes can be partitioned into groups of size√at most k such that the number of edges between different groups is at most γ n/ k where γ is a fixed constant. The proof of this theorem is a little longer. The reader may read the original reference if interested. Now, we start the proof of Planar Expansion Theorem. Proof of Planar Expansion Theorem First, one may assume |B| > |R| since, otherwise, one has |B| ≤ |R| and hence (3.1) holds. One may also assume that G is a maximal planar bipartite graph since, otherwise, one can add edges which

3.3 Planar Expansion Theorem

27

Fig. 3.2 Planar Graphs G and G

Fig. 3.3 Graphs G (black solid and dashed edges), G (black solid edges), Gr (red edges) and T = G ∪ Gr

do not effect the condition and the conclusion in Planar Expansion Theorem. As a maximal planar graph, G must be connected and has no leaf. Let B  be the set of all blue nodes with degree at least three. Let G = (R, B  , E  ) be the graph obtained from G by deleting all blue nodes with degree two and edges incident to them. Then every face of G is a quadrangle with two red nodes and two blue nodes appearing alternatively. In each face of G , connecting the two red nodes by an edge lying inside of the face would result a triangulation T . Let Gr = (R, Er ) be the subgraph of T induced by all red nodes.(See Fig. 3.3.) Then each face of Gr contains exactly one blue node. For each blue node b ∈ B  , let f (b) denote the face of Gr where b lies. Then a red node is adjacent to b if and only it is on the boundary of f (b). Next, for each b ∈ B \ B  , one would also like to assign b to a face of Gr and this assignment is required to satisfy the following conditions:

28

3 Sensor Cover

(a) All red neighbors of b are in f (b ). (b) Each face of Gr receives at most seven assignments (including the one in B  ). Note that in G, each b ∈ B \ B  is adjacent to two red nodes r1 and r2 which are two endpoints of an edge between two faces of Gr . b can be assigned to either one to satisfy condition (a). Moreover, for each pair of such red nodes r1 and r2 , B \ B  contains at most two blue nodes b1 and b2 adjacent to both r1 and r2 since, otherwise, the existence of the third one b3 would imply |NG ({b1 , b2 , b3 })| = |{r1 , r2 }| < |{b1 , b2 , b3 }|, contradicting assumption of the theorem. Next, we determine where b1 and b2 are assigned. Define graph Gb = (B  , Eb ) by defining (b1 , b2 ) ∈ Eb if and only if f (b1 ) and f (b2 ) are adjacent in Gr . Clearly, Gb is the planar dual graph of Gr . By Theorem 3.3.2, Gb can be oriented such that each node has in-degree at most three. Now, when b ∈ B \ B  is required to be assigned f (b1 ) or f (b2 ), choose f (b1 ) if edge (b1 , b2 ) is oriented from b2 to b1 , and choose f (b2 ), otherwise. This assignment meets conditions (a) and (b). By Theorem 3.3.3, graph Gb can be partitioned into several parts B1 , B2 , . . . , Bh , each of size at√most k/7 such that the total number of edges between different parts is at most (γ / k/7)|B  |. Let Bi be the set of blue nodes assigned to some face f (b) for b ∈ Bi . Then by condition (b), |Bi | ≤ 7|Bi | ≤ k. Thus, |Bi | ≤ |NG (Bi )|. All nodes in NG (Bi ) lie in area Qi = ∪b∈Bi f (b). They can be divided into two parts Ri and Ri where Ri is the set of red nodes lying in the interior of area Qi and Ri is the set of red nodes lying on the boundary of Qi . Clearly h 

|Ri | ≤ |R|.

i=1

 Next, we want to establish an upper bound for hi=1 |Ri |. Note that each f (b) is a simply connected area with a cycle as its boundary. The boundary of Qi is obtained from boundaries of f (b)’s for b ∈ Bi through an operation ⊕. For any two cycles C and C  , C ⊕ C  = (C ∪ C  ) \ (C ∩ C  ) is an edge-disjoint union of cycles (Fig. 3.4). Therefore, the boundary of Qi is an edge-disjoint union of cycles. Let Ei be the number of edges on the boundary of Qi . Then |Ri | ≤ |Ei |. By the duality relationship of Gr and Gb (Fig. 3.5), it is easy to see that |Ei | is also the number of edges between part Bi and other parts. Therefore, h  i=1

 √ √ |Ri | ≤ 2 · (γ / k/7)|B  | ≤ (2γ 7/ k)|B|.

3.4 PTAS

29

Fig. 3.4 C ⊕ C  is an edge-disjoint union of cycles

Fig. 3.5 The duality of Gr and Gb

It follows that |B| =

h  i=1

|Bi | ≤

h 

|NG (Bi )| ≤

i=1

√ h  2γ 7 (|Ri | + |Ri |) ≤ |R| + √ · |B|. k i=1

Hence, |B| ≤

1 1−

√ 2γ√ 7 k

√ 4γ 7 · |R| ≤ (1 + √ )|R|, k

√ 2γ√ 7 k

≤ 1/2, i.e., k ≥ 102γ 2 . This means that Planar Expansion Theorem √   holds when choose K ≥ max(2, 102γ 2 ) and c = 4γ 7. when

3.4 PTAS In this section, we show that the minimum sensor cover problem has a PTAS. Theorem 3.4.1 (Mustafa and Ray [413]) The minimum sensor cover problem has a PTAS.

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3 Sensor Cover

To do so, we first show a corollary of Plana Expansion Theorem. Consider an instance of the minimum sensor cover problem, consisting of a set of target points, A, and a set of sensors, S, lying in the Euclidean plane. S is a sensor cover for A. A sensor cover S  ⊆ S for A is said to be k-tight if for any subset U ⊆ S  with |U | ≤ k, there is no subset V ⊂ S with |V| < |U | such that (S  \ U ) ∪ V is still a sensor cover. Corollary 3.4.2 Let K and c be two universal constants in Planar Expansion Theorem. Suppose k ≥ K. If S  is a k-tight sensor cover, then √ |S  | ≤ (1 + c/ k) · opt, where opt is the size of the minimum sensor cover. Proof If there is a sensor s covering all target points, then for k ≥ 2, every k-tight sensor cover S  must contain only one sensor and clearly the inequality holds. Next, assume that no sensor covers all target points so that opt ≥ 2. Let S ∗ be a minimum sensor cover. Denote X = S  \ S ∗ , Y = S  ∩ S ∗ , and Z = S ∗ \ S  . Let A be the set of target points not covered by any sensor in Y . Construct a bipartite graph G = (X, Z, E) by connecting x ∈ X and z ∈ Z with an edge (x, z) if and only if there exists a target point a ∈ A covered by both sensors x and z. Then for every subset U ⊆ X with |U | ≤ k, we must have |NG (U )| ≥ |U | since, otherwise, one has |NG (U )| < |U | and (Y ∪ (X \ U ) ∪ NG (U ) is still a sensor cover, which contradicts assumption that S  is a k-tight sensor cover. Can we apply Planar Expansion Theorem to get the following? √ |X| ≤ (1 + c/ k) · |Z|. If yes, then this inequality implies √ |S  | ≤ (1 + c/ k) · S ∗ . However, we cannot because G may not be planar and hence Planar Expansion Theorem cannot apply to G. To use Planar Expansion Theorem, we need to find a planar subgraph H of G such that for each target point a ∈ A, H contains an edge whose two endpoints (two sensors) both cover a. With this property, for any U ⊆ X, Y ∪ ((X \ U ) ∪ NH (U )) is also a sensor cover. Hence, for U ⊆ X with |U | ≤ k, |NH (U )| < |U |. Then, by Planar Expansion Theorem, √ |X| ≤ (1 + c/ k) · |Z|. Next, we show the existence of subgraph H . The subgraph H is constructed as follows. Draw a variation of Voronoi diagram for X ∪ Z by defining V oro(o) = {u | u, o − Rs (o) = min (u, v − Rs (v))}, v∈X∪Z

3.4 PTAS

31

where Rs (o) is the sensing radius of sensor o ∈ X ∪ Z. It is not hard to verify that V oro(o) is a closed convex region. We may still call V oro(o) the Voronoi cell of o. Connect x ∈ X and z ∈ Z with an edge (x, z) if and only if Voronoi cell of x and Voronoi cell of z are adjacent. Clearly, H is a planar subgraph of G. The remaining is to show that H satisfies the required condition that for any target point a ∈ A, H contains an edge with two endpoints both covering a. For a target point a ∈ A, a must be covered by a sensor x in X and a sensor z in Z. In the Euclidean plane where they lie, let [x, a] and [a, z] be two straight line segments connecting x and a, and a and z, respectively. Note that for any sensor w ∈ X ∪ Z, if its Voronoi cell V oro(w) intersects [x, a] or [a, z], then w covers a. In fact, suppose without loss of generality that V oro(w) ∩ [x, a] = ∅. Let u be a point in V oro(w) ∩ [x, a]. Then a, w − Rs (w) ≤ a, u + u, w − Rs (w) ≤ a, u + u, x − Rs (x) = a, x − Rs (x). Since x covers a, one has a, x − Rs (x) ≤ 0. Hence a, w − Rs (w) ≤ 0. This means that w covers a. Now, going from x to z along [x, a] and [a, z], one would find a sequence of sensors in X ∪ Z, (x = s1 , s2 , . . . , sk = z), such that (a) every si for i = 1, 2, . . . , k covers a and (b) V oro(si ) and V oro(si+1 ) are adjacent for i = 1, 2, k − 1. In this sequence, there exists i such that si ∈ X and si+1 ∈ Z. Thus, (si , si+1 ) is an edge in H , meeting the requirement for target point a.   The following algorithm with local search finds a k-tight sensor cover. input a set of target points, √ A, and a set of sensors, S. Choose k ≥ K such that c/ k ≤ ε. S  ← S; while S  is not k-tight do find a subset U ⊆ S  with |U | ≤ k and V ⊆ S with |V| < |U | such that (S  \ U ) ∪ V is a sensor cover, and set S  ← (S  \ U ) ∪ V; output S  . Let n = |S|. Since the “while” loop runs at most n time and each iteration takes at most nk · nk−1 time, the running time of this algorithm is n2k . By√ Corollary 3.4.2, this algorithm gives an approximation with performance ratio 1+c/ k for k ≥ K and hence gives a PTAS for the minimum sensor cover problem.

Chapter 4

Connected Sensor Cover

I realize love isn’t about sex. It’s about connection. Chuck Palahniuk

4.1 Motivation and Overview The connected sensor cover was first studied by Cardei et al. [97]. The minimum connected sensor cover problem (Problem 1.3.2) was first proposed by Gupta, Das, and Gu [240]. They presented a greedy algorithm with performance ratio O(r ln n) where n is the number of sensors and r is the link radius of the sensor network, i.e., for any two sensors s and s  with a sensing point in common, there exists a path between s and s  with hop distance at most r in communication network. Zhang and Hou [615] studied the minimum connected sensor cover problem in homogeneous wireless sensor networks with property that Rc ≥ 2Rs . They showed that in this case, the coverage of a connected target area implies the connectivity. This result was generalized by Zhou, Das, and Gupta [649] to the m-connectivity that if every point in a connected target area is covered by at least m sensors, then those sensors induce an m-connected sensor network. Xing et al. [575] presented a coverage configuration protocol which can give different degree of coverage requested by applications. Bai et al. [46] studied a sensor deployment problem regarding the coverage and connectivity. Alam and Haas [13] studied this problem in threedimensional sensor networks. Funke et al. [200] improved approximation algorithms for the minimum connected sensor cover problem by allowing sensors to vary their sensing radius. With variable sensing radius and communication radius, Zhou, Das, and Gupta [650] designed a polynomial-time approximation with performance ratio O(log n). Chosh and Das [219] designed a greedy approximation using maximal independent set and Voronoi diagram. They determined the size of connected sensor cover produced by their algorithm. However, no comparison with optimal solution, that is, no analysis on approximation performance ratio is given.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_4

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34

4 Connected Sensor Cover

For homogeneous wireless sensor networks, Wu et al. [562, 567] improved the approximation performance ratio O(r ln n) of Gupta, Das, and Gu [240] by presenting two approximation algorithms with performance ratios O(r) and O(log3 n), respectively. They also made the following conjecture. Conjecture 4.1.1 (Wu et al. [567]) There exists a polynomial-time constantapproximation for the minimum connected sensor cover problem. Li et al. [267] made efforts on this conjecture and showed that there exists a polynomial-time approximation for the minimum connected sensor cover problem with performance ratio O((Rs /Rc )2 ). This result is one more evident for the truth of Conjecture 4.1.1. There also exist some efforts [30, 31, 108, 334, 335, 446, 447] on study of variations of connected sensor covers. Especially, the connected dominating set is closely related to the connected sensor cover. Actually, if we consider all sensors themselves as target points, then the connected sensor cover would become the connected dominating set. The connected dominating set is a quite important research topic in wireless sensor networks [83, 84, 86, 178, 180]. However, it has been discussed quite well in [174]. Therefore, we would not include it here. Another important variation is the coverage with consideration on data reporting latency [34, 148, 149]. This consideration may generate a lot of interesting work later. A survey in earlier stage [220] still contains a lot of useful information.

4.2 NP-Hardness In this section, we show the NP-hardness of the minimum connected sensor cover problem. Theorem 4.2.1 The minimum connected sensor cover problem is NP-hard. Proof The following NP-complete problem can be found in [215, 218]. Problem 4.2.2 (Planar-4-CVC) Given a planar graph G = (V , E) with node degree at most 4 and a positive integer k ≤ |V |, determine whether there is a connected vertex cover of size k, i.e., a subset V  ⊆ V with |V  | = k such that for each edge {u, v} ∈ E at least one of u and v belongs to V  and the subgraph induced by V  is connected. We are going to construct a polynomial-time many-one reduction from this problem to the following decision version of the minimum connected sensor cover problem. Problem 4.2.3 (Connected Sensor Cover) Given a set of target points, A, a set of homogeneous sensors, S on the Euclidean plane, and a integer k > 0, is there a sensor cover of size at most k?

4.2 NP-Hardness

35

Consider an instance of the planar-4-CVC problem, consisting of a graph G = (V , E) with vertex degree at most 4 and a positive integer k. First, we construct a new graph as follows. Note that we can embed G into the plane so that all edges consist of horizontal and vertical segments of lengths being an integer at least 4, so that every two edges meet at an angle of 90◦ or 180◦ . Add new vertices on the interior of each edge in G to divide the edge into a path of many edges, each of length exactly one. Denote by W the set of all such new vertices. (See Fig. 4.1. New vertices are light circled points.) At each vertex of G , put a sensor with sensing radius one and communication radius one. Now, for each sensor w at a location in W , put a target w as shown in Fig. 4.2 such that w can be covered by only sensor w. Then every connected sensor cover must contain W . Now, it is easy to see the following facts: (1) W is a sensor cover. (2) C is a connected vertex cover of G if and only if C ∪ W is a connected sensor cover. Therefore, G has a connected vertex cover of size at most k if and only if there exists a connected sensor cover of size at most |W | + k.  

Fig. 4.1 (a) A planar graph G. (b) The constructed new graph G . The dark circled points are vertices of G and the light circled points are new vertices

Fig. 4.2 Add wi ’s

36

4 Connected Sensor Cover

4.3 O(r)-Approximation In this section, we present a polynomial-time O(r)-approximation for the connected sensor cover problem given by Wu et al. [567]. This approximation is constructed by using an approximation algorithm for the following network Steiner tree problem. Problem 4.3.1 (Network Steiner Tree) Given a graph G = (V , E) with nonnegative edge weight c : E → , and a node subset P ⊆ V , find a minimum total edge-weight tree interconnecting all nodes in P . The network Steiner tree is a classic combinatorial optimization problem. Currently, the best known polynomial-time approximation algorithm has the performance ratio 1.39 [87] which will be introduced in next section. The O(r)approximation designed by Wu et al. [567] for the minimum connected sensor cover problem is as follows. Step 1.

Step 2.

Let A be the target set and S the sensor set in an input of the minimum connected sensor cover problem. Compute a (1+ε)-approximation S  for the minimum sensor cover problem on input target set A and input sensor set S. Consider the communication network G on S. Assign each edge with weight one. Compute a 1.39-approximation T for the network Steiner tree problem on S  as terminal set. Output the vertex set ST of T .

Theorem 4.3.2 (Wu et al. [567]) ST is a polynomial-time O(r)-approximation for the minimum connected sensor cover problem where r is the link radius of the sensor network, that is, for any two sensors s and s  with sensing areas intersecting, there exists a path between s and s  with hop distance at most r. Proof Note that the cost of T is |ST | − 1. Suppose Optcsc is an optimal solution for the minimum connected sensor cover problem. Let S  ⊆ S be a (1 + ε)approximation for the minimum sensor cover problem on input target set A and sensor set S. For each sensor s ∈ S  , find a sensor s  ∈ Optcsc such that diskRs (s) ∩ diskRs (s  ) = ∅ and connect s and s  by a shortest path. Then we obtain a tree with total cost at most |Optcsc | − 1 + |S  |r ≤ |S  |(r + 1) − 1 ≤ (4 + ε)|Optcsc |(r + 1) − 1. Therefore, |ST | − 1 ≤ 1.39 · ((1 + ε)|Optcsc |(r + 1) − 1). Hence |ST | ≤ 1.39(r + 1)(1 + ε) · |Optcsc |.

 

4.4 Network Steiner Tree

37

For homogeneous wireless networks with property, if Rc ≥ 2Rs , then r = 1. Therefore, the minimum connected sensor cover problem has a polynomial-time constant-approximation. From the proof of Theorem 4.3.2, this constant could be at most 1.39(2 + 2ε). Corollary 4.3.3 When Rc ≥ 2Rs , the minimum connected sensor cover problem has a polynomial-time (2.78 + ε)-approximation for any ε > 0.

4.4 Network Steiner Tree One of the classical Steiner tree problems is the network Steiner tree problem (Problem 3.3.1 or 3.4.1). There are many publications in the literature on network Steiner trees [268]. In this section, we would like to introduce the 1.39approximation algorithm designed by Byrka et al. [87]. This is currently the best known approximation for the network Steiner tree problem. Problem 4.4.1 (Network Steiner Tree (NST)) Given a graph G = (V , E) with edge cost c : E → R and a node subset R ⊆ V , find a minimum cost tree interconnecting all nodes in R. Nodes in R are called terminals and the other nodes are called Steiner nodes. Before currently best known ratio 1.39 [87], the previously best known ratio 1.55 was achieved using the concept of k-restricted Steiner tree. By splitting off at nonleaf terminals, a Steiner tree can be decomposed into subtrees in which terminals only appear as leaves. Such subtrees are called full components. The size of a full component is the number of terminals in it. For a fixed integer k, a Steiner tree is a k-restricted Steiner tree if every full component has size at most k. The kSteiner ratio ρk is defined to be the supremum of the ratio between the optimal value of a k-restricted Steiner tree and the optimal value of NST. The exact value of ρk was determined in [77]. As a consequence, ρk ≤ 1 + 1/log2 k, and thus a minimum cost Steiner tree can be approximated within arbitrary accuracy using k-restricted Steiner tree when k is large enough. By presenting a (1 + ln 3/2)approximation algorithm for the minimum cost k-restricted Steiner tree problem, Robins and Zelikovsky [443, 444] obtained their 1.55-approximation for NST. This idea is also used in the 1.39-approximation algorithm for NST, but in a different way: it is used to approximate the optimal solution to the LP relaxation.

4.4.1 LP Relaxation for NST There are many LP models for NST, one of which is the following BCR model. For a digraph on node set V and a node subset U ⊆ V , denote by δ + (U ) the set of arcs with tails in U and heads in V \ U .

38

4 Connected Sensor Cover

Definition 4.4.2 (Bidirected Cut Relation (BCR)) Let D(G) be the digraph obtained from G by replacing each edge with two opposite arcs. Fix an arbitrary root r ∈ R. The following is an LP relaxation for NST. min



ce xe

e∈E

s.t.



xe ≥ 1, ∀U ⊆ V \ {r} with U ∩ R = ∅,

e∈δ + (U )

xe ≥ 0, ∀e ∈ E The following theorem shows that BCR has no integrality gap for the minimum spanning tree problem. Theorem 4.4.3 (Edmonds 1967) For R = V , the polyhedron of BCR is integral. The 1.39-approximation for NST is based on another model called DCR. The readers may refer to Fig. 4.3 for an illustration of the concepts defined in the following. Definition 4.4.4 (Directed Cut Relation (DCR)) For a subset of terminals R  ⊆ R, a directed component C with respect to R  is obtained from a minimum cost Steiner tree on terminal set R  by directing every edge towards an arbitrarily selected root r  ∈ R  . The root r  is called the sink of C and all those nodes in R  \ {r  } are called sources of C. Denote by C the set of all directed components. A directed component C ∈ C is said to cross a node set U if the sink of C is outside of U and U contains at least one source of C. Denote δC+ (U ) = {C ∈ C : C crosses U }. Choose an arbitrary terminal node r ∈ R as the root. The following is an LP relaxation for NST.  min c(C)xC C∈C



s.t.

xC ≥ 1, ∀U ⊆ R \ {r}, U = ∅

C∈δC+ (U )

xC ≥ 0, ∀C ∈ C, Fig. 4.3 Illustration of directed component. Square nodes are terminals in R  . The other nodes are Steiner nodes. The blackened square node is an arbitrarily chosen root. All edges are directed towards this root

4.4 Network Steiner Tree

where c(C) =



39

e∈E(C) ce .

To see that DCR is indeed an LP relaxation for NST, let C ∗ be the directed component obtained from a minimum cost Steiner tree on R by directing every edge towards r. Then C ∗ crosses every node set U with ∅ = U ⊆ R \ {r}. Hence setting xC ∗ = 1 and xC = 0 for all C ∈ C with C = C ∗ is a feasible solution to DCR. Unfortunately, the set C consists of exponentially many directed components and the number of constraints in DCR is also exponential, which makes it difficult to be solved optimally in polynomial-time. However, DCR can be solved nearly optimally by considering k-DCR defined as follows. Denote by Ck the set of directed components of size at most k, where the size of a directed component is the number of terminals contained in it. Replacing C by Ck in DCR, call the resulting LP formulation as k-DCR. In the following, denote by n the number of nodes in the graph, and denote by RC = V (C) ∩ R for a subgraph C and the terminal set R. Lemma 4.4.5 For any fixed integer k, an optimal solution to k-DCR can be found in polynomial-time. Furthermore, optDCRk ≤ ρk optDCR , where optDCRk and optDCR are the optimal values of k-DCR and DCR, respectively, and ρk is the k-Steiner ratio. Proof A polynomial-time algorithm exists for k-DCR by the following three observations. 

(i) An optimal Steiner tree on terminal set R  can be computed in time O(3|R | n +  2|R | n2 + n3 ) [172], which is polynomial for those R  with |R  | ≤ k. (ii) |Ck | = O(knk ). This is because for any terminal set R  , there are exactly |R  | directed components with respect to R  , each node of R  serving as a root once. (iii) Separation oracle exists by implementing minimum cut algorithm. To prove the second half of the lemma, consider an optimal solution x∗ to DCR. For each directed component C ∈ C, view C as an undirected Steiner tree and use the method in [77] to find a k-restricted Steiner tree C  on terminal set RC such that c(C  ) ≤ ρk c(C). Suppose the full components of C  are C1 , . . . , Ct . We may assume that every full component Ci is a minimum cost Steiner tree on terminal set RCi (recall that a minimum cost Steiner tree on a given terminal set of at most k nodes can be found in polynomial-time). Transforming C  into a directed tree by orienting every edge of C  towards the sink of C, then full components C1 , . . . , Ct are transformed into directed components. For each Ci , let xC  = xC∗ . Notice that i

C crosses U implies that at least one of C1 , . . . , Ct crosses U . Hence  the resulting    c(Ci )xC  = x is a feasible solution to DCRk whose objective value is C∈C i      c(Ci ) xC∗ = C∈C c(C  )xC∗ ≤ ρk C∈C c(C)xC∗ = ρk · optDCR .   C∈C

As a corollary of Lemma 4.4.5, for any ε > 0, the optimal solution to DCR can be approximated within factor (1 + ε) in polynomial-time.

40

4 Connected Sensor Cover

4.4.2 The Algorithm The 1.39-approximation algorithm for NST is the first one using a new technique called iterative random rounding. A series of DCR’s are iteratively solved, serving as a basis for rounding. In each iteration, the algorithm samples a directed component with a probability which is determined by the nearly optimal fractional solution to DCR. Contract the sampled directed component to a super terminal node. Construct a DCR on the contracted instance, and iterates. When arriving at an instance with only one super terminal node, the algorithm expands those contracted directed components to obtain a Steiner tree (which can be viewed as piecing up those sampled direction components and then ignoring directions on the edges). Iterative Rounding Algorithm for NST t ← 1. R  ← R. while |R  | > 1 t to DCRt . Compute a (1 + ε/2)-approximation x t t t Sample C ∈ C with probability xC t / C∈C t xCt . Contract C t to a super node v(C t ). R  ← (R  \ RC t ) ∪ {v(C t )}. t ← t + 1. end-while output Sapx ← Expansion({C 1 , . . . , C t }).

4.4.3 Bridge Lemma An important tool to the analysis of the algorithm is the Bridge Lemma, which bridges the cost of a terminal spanning tree and the solution to DCR. A terminal spanning tree is a Steiner tree without Steiner nodes. It should be remembered that when talking about NST, we are always assuming that the graph is complete and the edge cost function is metric, because the computation on a general case is equivalent to the computation on a metric case [268]. Under such an assumption, a terminal spanning tree always exists.   In the following, R2 is used to denote the set of edges with both ends in R  . These edges form a complete graph on node set R  , denoted as KR  . For a Steiner tree S and a terminal subset R  ⊆ RS , call a subset B ⊆ E(S) a bridge set with     respect to S and R  if S + R2 − B is connected and S + R2 − B − e is disconnected for any edge e ∈ E(S) \ B. For example, the blackened edges in Fig. 4.4a form a bridge set with respect to the blackened terminal nodes. The concept of bridge set can be comprehended in the following way (see Fig. 4.4b): contracting nodes in R  creates some cycles, a bridge set is a set of edges the deletion of which breaks all cycles, resulting in a tree spanning those nodes of S in the contracted graph. Clearly, any bridge set with respect to R  contains exactly |R  | − 1 edges. Let BS (R  ) be the

4.4 Network Steiner Tree

41

Fig. 4.4 An illustration of bridge set. In (a), blackened square nodes form a terminal subset R  , and the blackened edges form a bridge set B with respect to R  (in fact, they form BrS (R  )). The tree in (b) is obtained from (a) by contracting nodes in R  and then deleting edges in B. (c) illustrates the proof of Lemma 4.4.6. The dashed lines form T 

set of all bridge sets with respect to S and R  , BrS (R  ) be a maximum cost bridge set in BS (R  ), and brS (R  ) = c(BrS (R  )). For a Steiner tree S on terminal set R and two nodes u, v ∈ R, let PS (u, v) be the unique path on S connecting u and v, and let wS (u, v) = max{ce : e∈ PS (u, v)}. The wS defined in this way is called the bridge weight function on R2 with respect to S. Lemma 4.4.6 Let S be a Steiner tree on terminal set R. For any R  ⊆ R, there is a spanning tree T  of KR  such that wS (T  ) = brS (R  ). Proof Notice that S − BrS (R  ) has exactly |R  | connected components, each component contains exactly one terminal node in R  , and each edge e ∈ BrS (R  ) connects two connected components of S − BrS (R  ) (see Fig. 4.4c). For each edge e ∈ BrS (R  ), suppose ue , ve are the two terminal nodes contained in the two connected components of S − BrS (R  ) which are connected by e (for example, for the edge e of cost 5 in Fig. 4.4a, ue , ve are the first and the third blackened terminal nodes in Fig. 4.4c). Notice that edge ue ve plays the same role as e in connecting different connected components of S − BrS (R  ). Also notice that wS (ue ve ) = ce . In fact, if this is not true, then there is an edge e ∈ PS (u, v) with ce > ce . But then (BrS (R  ) \ e) ∪ {e } will be a bridge set with larger cost, contradicting the definition of BrS (R  ).   Now, let  E(T  ) = {ue ve : e ∈ BrS (R tree of KR  with  )}. Then T is a spanning   )). The lemma is wS (T ) = w (u v ) = c = c(Br (R   S e e e S e∈BrS (R ) e∈BrS (R ) proved.   Which bridge set is BrS (R  )? It depends on the cost function c. So, we shall use BrSc (R  ) and brSc (R  ) to emphasize that it is the cost function c which is under consideration. And when there is no ambiguous, the superscript c is omitted.

42

4 Connected Sensor Cover

Lemma 4.4.7 (Bridge Lemma) Let T be a terminal spanning tree and x = (xC )C∈C be a feasible solution to DCR. Then c(T ) ≤



xC · brTc (RC ).

C∈C

Proof To prove the lemma, we first construct a vector z = (ze )e∈KR satisfying constraint (4.1) of BCR on graph KR . For each directed component C ∈ C, let TC be the spanning tree of KRC with wT (TC ) = brT (RC ),

(4.1)

− → whose existence is guaranteed by Lemma 4.4.6. Let T C be the arborescence  obtained from TC by directing every edge towards the sink of C. For each arc e ∈ KR (view every edge of KR as two opposite arcs), define ze =



xC .

− → C : e∈ T C

Then for any node subset U ⊆ V \ {r} with U ∩ R = ∅,  e∈δ + (U )

ze ≥





− → |δ + (U ) ∩ T C |xC ≥

C∈δC+ (U )

xC ≥ 1,

C∈δC+ (U )

− → where the second inequality is true because any C ∈ δC+ (U ) has δ + (U ) ∩ T C = ∅, and the third inequality follows from the feasibility of x to DCR. Hence (ze )e∈KR is a feasible solution to BCR on graph KR .

(4.2)

Let F be a minimum weight spanning tree of graph KR with respect to weight function wT . By Theorem 4.4.3, F is an optimal solution to BCR on instance (KR , wT ). Combining this with (4.2), wT (F ) ≤



ze wT (e).

(4.3)

e∈KR

Notice that both T and F are spanning trees of graph T ∪ F ⊆ KR . We claim that T is a minimum spanning tree of T ∪ F with respect to weight function wT . To see this claim, first notice that wT (e) = ce for any edge e ∈ E(T ). Furthermore, any edge uv ∈ E(F ) \ E(T ) has wT (uv) = max{ce : e ∈ PT (u, v)} and thus has the largest cost on the fundamental cycle of T + e. The claim follows from the classic characterization of a minimum spanning tree [76]. As a consequence, c(T ) = wT (T ) ≤ wT (F ).

(4.4)

4.4 Network Steiner Tree

43

Combining (4.1), (4.3), (4.4), we have ⎛  ⎜  c(T ) ≤ ⎝ e∈KR

=



⎞

− → C : e∈ T C

⎛ ⎞   ⎟ ⎜ ⎟ xC ⎠ wT (e) = wT (e)⎠ xC ⎝ C∈C



wT (TC )xC =

C∈C

− → e∈ T C

brT (RC )xC .

C∈C

 

The lemma is proved.

4.4.4 Performance Ratio The Outline of the Analysis  Observe that there exists a constant M such that C∈C t xCt ≤ M holds for any iteration (in fact, M can be |R| by Chakrabarty et al. [105]).  Introduce a dummy directed component {r} with cost zero, and add constraint C∈C t xCt = M to the LP formulation of DCRt . Then in the t-th iteration, a directed component C ∈ C t is sampled with probability xCt /M. Denote by opt t the optimal value of DCRt . Then the output Sapx of Iterative Rounding Algorithm has expected cost E[c(Sapx )] =



E[c(C )] = t

t≥1



 E

t≥1

  xt 1 + ε/2  C c(C) ≤ E[opt t ] M M t

C∈C

t≥1

(4.5) (recall that xt is a (1 + ε/2)-approximation to DCRt ). Suppose S ∗ is an optimal Steiner tree. Let S1 = E(S ∗ ). In Subsection 4.4.4, a series of edge sets S1 ⊇ S2 ⊇ · · · are constructed such that each St corresponds to a feasible integral solution to DCRt . So, opt t ≤ c(St ).

(4.6)

For each edge e ∈ E(S ∗ ), let D(e) = max{t : e ∈ S t }. Observe that 

c(St ) =

t≥1



D(e)ce .

e∈S ∗

Combining this with (4.5) and (4.6), E[c(Sapx )] ≤

1 + ε/2  E[D(e)]ce . M ∗ e∈S

(4.7)

44

4 Connected Sensor Cover

In Subsubsection “Upper Bound for E[D(e)],” it is proved that the series of sets S1 ⊇ S2 ⊇ · · · can be constructed in such a way that E[D(e)] ≤ M ln 4. Then E[c(Sapx )] ≤ (1 + ε/2) ln 4



ce ≤ (ln 4 + ε)c(S ∗ ) < 1.39c(S ∗ ).

e∈S ∗

The desired ratio 1.39 is proved. In the following two subsections, we focus on the construction of S1 ⊇ S2 ⊇ · · · and the proof of (4.7). Construction of S1 ⊇ S2 ⊇ · · · Let S ∗ be an optimal Steiner tree. Notice that any Steiner tree can be changed into a rooted binary tree with the same cost in which a node is terminal if and only if it is a leaf [282]. So, in the following, assume that S ∗ is such a Steiner tree. For each Steiner node in S ∗ , choose uniformly at random one of the two  be the set of chosen edges. Define set W = {uv ∈ edges to its children. Let B  = 1}. These definitions, as well as the following observations KR : |PS ∗ (u, v) ∩ B| and properties can be illustrated by Fig. 4.5. Observation 4.4.8 Any Steiner node of S ∗ has a unique path connecting it to a  terminal node through edges not in B. For a Steiner node s in S ∗ , let ts be the unique terminal node which is connected  For a terminal node u, let tu = u. to s using edges not in B. Observation 4.4.9 For any two terminals u, v ∈ R, uv is an edge in W if and only  if u = ts and v = ts  for some edge ss  ∈ B. Fig. 4.5 The blackened  The dashed edges are in B. edges are in W

4.4 Network Steiner Tree

45

Fig. 4.6 An illustration for the proof of Lemma 4.4.10

For example, for the edge u2 u7 ∈ W in Fig. 4.5, u2 = ts , u7 = tr , and rs is an  As a consequence of Observation 4.4.9, edge set W has the following edge in B. property. Lemma 4.4.10 Set W form a terminal spanning tree. Proof For any two terminal nodes u, v ∈ R, suppose si1 , si2 , . . . , siq are the nodes  in the order that they are met on PS ∗ (u, v) which are incident to some edges in B, when walking along PS ∗ (u, v) from u to v (see Fig. 4.6 for an illustration). Then tsi1 = u, tsiq = v, and by observation 4.4.9, tsi1 tsi2 . . . tsiq consists of a path in W connecting u and v. So, W is connected. Furthermore, by the assumption on S ∗ , it  = |R| − 1. A can be seen that S ∗ has exactly |R| − 1 Steiner nodes and thus |B| classic exercise in graph theory says that a graph on n nodes is a tree if and only if it is connected and has n − 1 edges [76]. Hence W form a spanning tree on terminal set R.   For each edge e ∈ S ∗ , define W (e) = {uv ∈ W : e ∈ PS ∗ (u, v)}. For example, in Fig. 4.5, W (e1 ) = {u1 u2 }, W (e2 ) = {u1 u2 , u2 u4 , u2 u7 }, W (e3 ) = {u2 u4 , u2 u7 }, etc. A more illuminating point of view is that for an edge uv ∈ W , the set of edges e with uv ∈ W (e) is exactly PS ∗ (u, v). The sequence of sets S1 ⊇ S2 ⊇ · · · is constructed by the following algorithm. Notice that the algorithm depends on some distribution whose existence is guaranteed by Lemma 4.4.12. But one does not know what the distribution looks like. However, this does not matter since this algorithm is only used for analysis. The existence of such a construction suffices. Construction of S1 ⊇ S2 ⊇ · · · S1 ← S ∗ for t = 1, 2, . . . Let C t be the sampled directed component in the t-th iteration of Iterative Rounding Algorithm.

46

4 Connected Sensor Cover

Choose a bridge set B t ∈ BW (RC t ) according to the distribution in Lemma 4.4.12. Mark all edges in B t . Let St+1 be the set of edges e for which not all edges in W (e) are marked. end-for The following lemma shows that St interconnects R t−1 in Gt−1 , where Gt−1 is the contracted graph before the t-th iteration, and R t−1 is the terminal set in Gt−1 . Hence St corresponds to an integral feasible solution to DCRt .   t−1 s Lemma 4.4.11 St ∪ interconnects R. s=1 C   t−1 s t−1 s Proof Let B = s=1 B . By the explanation for bridge sets, (W \ B) ∪ C s=1 interconnects R. So, to prove the lemma, it suffices to show that nodes connected by W \ B are also connected by St . For this purpose, it suffices to show that nodes adjacent in W \ B are connected by St . In fact, for any edge uv ∈ W \ B, it is an unmarked edge in every W (e) with e ∈ PS ∗ (u, v). By the construction of St in the above algorithm, path PS ∗ (u, v) is contained in St . The lemma is proved.   Upper Bound for E[D(e)] The estimation of E[D(e)] depends on the distribution by which the sequence S1 ⊇ S2 ⊇ · · · is constructed. Lemma 4.4.12 For each directed component C ∈ C, there is a distribution wC : BW (RC ) → [0, 1] such that if B is chosen from BW (RC ) according to this distribution (that is, with probability wC (B)), then each edge e ∈ W is marked with probability at least 1/M. Proof To prove this lemma, it suffices to show that there exists a family of distributions {wC }C∈C such that for any edge e ∈ W , 

xC · wC (B) ≥ 1.

(C,B) : B∈BW (RC ),e∈B

In fact, if this is true, then P r[e is marked] =

 (C,B) : B∈BW (RC ),e∈B

xC 1 · wC (B) ≥ . M M

Suppose such a family of distributions does not exist, then the following linear system has no solution:

4.4 Network Steiner Tree

47

 

wC (B) = 1, ∀C ∈ C,

B∈BW (RC )

xC · wC (B) ≥ 1, ∀e ∈ W,

(C,B) : B∈BW (RC ),e∈B

wC (B) ≥ 0, ∀C ∈ C, ∀B ∈ BW (RC ), where the first and the third constraint indicate that wC is a distribution. Notice that constraint (4.8) can be changed to 

wC (B) ≤ 1, ∀C ∈ C.

(4.8)

B∈BW (RC )

 In fact, if (4.8) to (4.8) has a solution {wC (B)}, then {wC (B)/ B∈BW (RC ) wC (B)} will be a solution to (4.8) to (4.8). Hence, (4.8) to (4.8) having no solution implies that (4.8) to (4.8) do not have solution too. Recall that Farkas’ Lemma says that for any matrix A of dimension m × n and vector b of length m, exactly one of the following two situations occurs: (a) there exists a vector x ≥ 0 such that Ax ≤ b, (b) there exists a vector z ≥ 0 such that zT A ≥ 0 and zT b < 0. So, (4.8) to (4.8) having no solution implies that there is a vector (y, z) ≥ 0 with  yC ≥ ze xC , ∀C ∈ C, ∀B ∈ BW (RC ), and  e∈B  yC < ze . C∈C

(4.9)

e∈W

View z as a cost function on the edges of W . Then (4.9) become yC ≥ xC · z(B), ∀C ∈ C, ∀B ∈ BW (RC ), and  C∈C yC < z(W ). z (RC ) is the bridge set in BW (RC ) with the maximum cost with Recall that BrW z z respect to weight function z, and brW (RC ) = z(BrW (RC )). Since (4.10) holds for any B ∈ BW (RC ), we have z (RC ). yC ≥ xC · brW

Combining this with inequality (4.10),  C∈C

z xC · brW (RC ) ≤

 C∈C

yC < z(W ),

48

4 Connected Sensor Cover

Fig. 4.7 An illustration of how to count |W (e)|. Curves represent paths from Steiner nodes to  (recall Observation 4.4.8) terminals through edges not in B

 

contradicting the Bridge Lemma.

Recall that it is assumed that S ∗ is a rooted binary tree. The level of an edge e is the distance of e from the root, the two edges incident with the root having level 1. Lemma 4.4.13 For any edge e ∈ S ∗ at level le ⎧ ⎨ 1/2q , P r[|W (e)| = q] = 2/2q , ⎩ 0,

1 ≤ q ≤ le − 1, q = le , otherwise.

Proof Let P = v0 v1 . . . vle be the path from edge e to the root, where e = v0 v1  Let (see Fig. 4.7a). Suppose vq−1 vq is the first edge on P which belongs to B.  ui be the other child of vi not on P . Then vi ui ∈ B for i = 1, . . . , q − 1. By Observation 4.4.9, we have W (e) = {tv1 tu1 , . . . , tv1 tuq−1 , tv1 tvq } and thus |W (e)| = q. Notice that the probability that vq−1 vq is the first edge on P which  is 1/2q (with probability 1/2 edge vi−1 vi does not belong to B  for belongs to B  i = 1, . . . , q − 1, and with probability 1/2 edge vq−1 vq belongs to B). If no edge  which happens with probability 1/2le , then similarly to the on P belongs to B, above, W (e) = {tv1 tu1 , . . . , tv1 tule } (see Fig. 4.7b), and thus |W (e)| = 1/2le . Notice that there are two events leading to |W (e)| = le , one is that vle −1 vle is the first edge  the other is that P has no edge belonging to B.  The lemma on P belonging to B, follows.    ⊆ W , let x(W  ) be the number of Lemma 4.4.14 For any subset of edges W   )] ≤ H|W iterations needed for all edges in W to be marked. Then E[x(W  | · M, 1 1 where H|W = 1 + + . . . + is the Harmonic number. | | 2 |W

4.4 Network Steiner Tree

49

 )] : |W  | = i}. We prove Ei ≤ Hi · M by Proof Denote by Ei = max{E[x(W induction on i. The basic case for i = 1 follows from Lemma 4.4.12. For the  | = t and Ei ≤ Hi · M holds for any i < t. For induction step, suppose |W i = 1, . . . , t, denote by qi the probability that exactly i edges are marked in the first iteration and denote by pi the probability that at least i edges are marked in the first iteration. So, pi = ij =1 qj and E[the number of edges marked in the first iteration] =

t 

iqi =

i=0

t 

i(pi − pi+1 ) =

i=0

t 

pi

i=1

(by noticing that pt+1 = 0). Combining this with Lemma 4.4.12, t 

pi ≥

i=1

t . M

(4.10)

Then, Et ≤ 1 +

t 

qi Et−i

i=0

≤ 1+

t 

qi Ht−i M + q0 Et

i=1

≤ 1+M

t  (pi − pi+1 )Ht−i + q0 Et i=1

= 1+M

t 

pi (Ht−i − Ht−i+1 ) + p1 MHt + q0 Et

i=1

≤ 1−

t M pi + p1 MHt + q0 Et t i=1

≤ p1 MHt + (1 − p1 )Et . In the first inequality, 1 represents the first iteration, and if i edges are marked in the first iteration, then the expected number of remaining iterations is at most Et−i . The second inequality follows from induction. Notice that the summation excludes the case i = 0 because induction hypothesis cannot be applied on Et . Inequality (4.10) is used in deriving the fifth inequality. Then, Et ≤ Ht M. This finishes the induction step.   Lemma 4.4.15 For any edge e ∈ W, E[D(e)] ≤ ln 4 · M.

50

4 Connected Sensor Cover

Proof By Lemmas 4.4.13 and 4.4.14, E[D(e)] =



P r[|W (e)| = q] · E[D(e) | |W (e)| = q]

q≥1

≤

le 

P r[|W (e)| = q] · Hq M

q=1

⎛

=M⎝

l e −1  q=1

≤M

1 2

q

le   q  1 q=1

2

 Hq +

2 2le

le

⎞ Hle ⎠

Hq .

It can be calculated that the sum of the right-hand side is at most ln 4.

 

4.5 Metric Approximation Before giving another approximation algorithm for the minimum connected sensor cover problem. Let us first introduce a technique for design of approximation algorithm. The metric approximation is an important technique in design and analysis of approximation algorithms. There are two tools, spanner and tree metric. In this section, we introduce the tree metric which is used for design of randomized approximation. Especially, in the work of Garg, Konjevod, and Ravi [216] for group Steiner tree, it plays an important role. First, let us introduce two concepts about metric spaces. Consider two metric spaces (X, dx ) and (Y, dy ). Let f : X → Y be a one-to-one mapping. (X, dx ) is said to be dominated by (Y, dy ) via f if for any s, t ∈ X with x = y, dx (s, t) ≤ dy (f (s), f (t)). (X, dx ) is said to be embedded with distortion α into (Y, dy ) if for any s, t ∈ X, dy (f (s), f (t)) ≤ α · dx (s, t). Theorem 4.5.1 (Fakcharoenphol et al. [188]) Any metric space with n points can be embedded into randomly chosen dominating tree metric space with expected distortion O(log n). To show this theorem, let us describe what type of trees would be used in the theorem. A tree is called a 2-HWS (2-Hierarchically Well-Separated) tree if it satisfies the following conditions: 1. It is rooted and all leaves lie in the same level. 2. Each node v has the same distance to all of its sons.

4.5 Metric Approximation

51

Fig. 4.8 A 2-HWS tree

3. Let p(v) and c(v) be the father of v and a son of v. Then d(p(v), v) = 2 · d(v, c(v)) if c(v) is not a leaf, and d(p(v), v) = d(v, c(v)) = 2 if c(v) is a leaf. 4. Every leaf has the distance 2 to its father. An example of 2-HWS tree is shown in Fig. 4.8. Let (X, d) be a metric space with |X| = n. We are going to map X into the leaf set of a 2-HWS tree T . Next, we describe the construction of T , from which we can also see the mapping from X to leaves of T . Let the unit distance be the minimum distance between two points in metric space (X, d) and the maximum distance between two point. Denote by δ the integer satisfying 2δ−1 < ≤ 2δ . The algorithm will construct a randomized 2HWS tree with root X such that each level consists of a partition of X and leaves are singletons. The constructure contains two parts. The first part is randomly to select a permutation of all points in X and a constant β ∈ [0, 1]. The second part is to partition X step by step. In each step, the algorithm partition all nodes, which are subsets of X, at current level to smaller ones. In each level, all nodes form a partition of X, denoted by Di (Fig. 4.9). The detail is as follows. Randomized Selection: Choose a random permutation π of points in X. Choose β uniformly at random in [1,2]. Partition: Dδ+1 ← {X}; for i = δ to 0 do βi ← 2i−1 β; for k = 1 to n do for every part S in Di+1 do create a subset of S consisting of all unassigned points within distance βi from π(k); (This is a part of partition Di ) i ← i − 1.

52

4 Connected Sensor Cover

Fig. 4.9 All nodes at each level form a partition of X

Let T be the tree constructed as above and d T (·, ·) the distance induced by tree T . We first show that the distance d is dominated by d T , that is, for any u, v ∈ X, d(u, v) ≤ d T (u, v). To do so, assume βh < d(u, v) ≤ βh+1 for some integer h. Note that if u and v lie in the same part of Di , then there exists a center w such that d(u, w) ≤ βi and d(v, w) ≤ βi . This implies that d(u, v) ≤ d(u, w) + d(v, w) ≤ 2βi = βi+1 . Therefore, u and v lie in different parts of partition Dh−1 . Thus, d T (u, v) ≥ 2(1 + 1 + 2 + · · · + 2h−1 ) ≥ 2h β = βh+1 ≥ d(u, v). Next, we show the randomized distortion, that is, there exists a positive constant α such that for any u, v ∈ X, E(d T (u, v)) ≤ α · d(u, v). (u, v) for u, v ∈ X is cut at level i if u and v lie in different parts of partition Di and in the same part of partition Di+1 . If (u, v) is cut at level i, then d T (u, v) ≤ 2(1 + 1 + 2 + · · · + 2i ) = 2i+2 . Therefore, E[d (u, v) ≤ T

δ  i=0

P r[(u, v) is cut at level i] · 2i+2 .

4.5 Metric Approximation

53

Let h be an integer such that 2h < d(u, v) ≤ 2h+1 . Then for any β ∈ [0, 1], βh < d(u, v) ≤ βh+2 . Therefore, Dh−1 separates u and v. Thus, for i ≤ h − 2, P r[(u, v) is cut at level i] = 0. Hence, E[d (u, v) ≤ T

δ 

P r[(u, v) is cut at level i] · 2i+2 .

(4.11)

i=h−1

Let x1 , x2 , . . . , xn be all elements of X in the nondecreasing ordering of the distance from xs to {u, v}, d(xs , {u, v}) = min(d(xs , u), d(xs , v)). Let ti be the largest s such that d(xs , {u, v}) ≤ βi . (u, v) is cut at level i if and only if there exists s ≤ ti such that (u, v) is cut by xs at level i, that is, the following two facts hold. (1) xs is at the first position among x1 , x2 , . . . , xti in permutation π and (2) βi lies in interval [d(xs , {u, v}), max(d(xs , u), d(xs , v))). Since π is randomly chosen, we have P r[(1) occurs] ≤ 1/ti ]. Since βi has uniform distribution in [2i−1 , 2i ] and max(d(xs , u), d(xs , v)) − d(xs , {u, v}) ≤ d(u, v), we have P r[(2) occurs] ≤ d(u, v)/2i−1 . Moreover, for i ≥ h + 3 and s ≤ ki−1 , we have max(d(xs , u), d(xs , v)) ≤ d(xs , {u.v}) + d(u, v) ≤ βi−1 + 2h+1 ≤ βi , that is, P r[(2) occurs] = 0. Therefore, for i ≥ h + 3, P r[Di separates u and v] ≤

ti  s=ti−1 +1

d(u, v) 1 d(u, v) ti − ti−1 1 · = i−1 · . i 2 ti ti 2

(4.12)

For i = h − 1, h, h + 1, h + 2, we have P r[Di separates u and v] ≤

ti  d(u, v) s=1

2i−1

·

1 d(u, v) = i−1 . ti 2

Substituting (4.12) and (4.13) into (4.11), we obtain E[d T (u, v)] ≤

δ−1  i=h+3

8d(u, v) ·

h+2  ti − ti−1 + 8d(u, v) ti i=h−1

(4.13)

54

4 Connected Sensor Cover

≤ 8d(u, v) · (4 +

n  1 i=1

i

)

≤ 8d(u, v) · (5 + ln n) = d(u, v) · O(log n). This completes the proof of Theorem 4.5.1.

4.6 Randomized O(log2 n log m)-Approximation Suppose that in the minimum connected sensor cover problem, input consists of a set of n sensors, S and a set of m target points, A on the Euclidean plane. Then it can be easily reduced to the group Steiner tree problem defined as follows. Problem 4.6.1 (Group Steiner Tree) Consider a graph G = (V , E) with positive edge length w : E → R+ . Given m subsets of vertices, g1 , . . . , gm , find the shortest tree on a vertex subset which hits every gi for i = 1, . . . , m, where by hitting, we mean that the intersection of two sets is not empty. To see the reduction, set V = S, w(s, s  ) = 1 for every edge (s, s  ) in the communication network G of all sensors, and for each target point ai , set a group gi consisting of all sensors each covering ai . Since every edge has length one, the total edge length of a tree on a vertex subset S  is equal to |S  | − 1, the objective function value of the minimum connected sensor cover problem is minus one. Note that for positive numbers a > b > 1, we have inequality a/b ≤ (a − 1)/(b − 1). It follows that the following lemma holds. Lemma 4.6.2 If the group Steiner tree problem has a polynomial-time ρapproximation, then so does the minimum connected sensor cover problem. Remark Actually, the group Steiner tree problem is equivalent to the connected set cover problem [570, 629], which is a generalization of the connected sensor cover problem. Garg, Konjevod, and Ravi [216] showed that for any ε > 0, there exists a polynomial-time algorithm which can produce an approximation solution for the group Steiner tree problem such that with probability 1 − ε, the solution is O(log2 n log log n log m)-approximation where n is the number of vertices and m is the number of groups in input. One of the computation steps in this algorithm has been improved by Fakcharoenphol et al. [188] from running time O(log n log log n) to O(log n). Therefore, we now have the following. Theorem 4.6.3 (Garg, Konjevod, and Ravi [216]) For any 0 < ε < 1, there exists a polynomial-time approximation algorithm for the group Steiner tree problem, which with probability 1 − ε, produces a O(log2 n log m)-approximation.

4.6 Randomized O(log2 n log m)-Approximation

55

This implies the following. Theorem 4.6.4 For any 0 < ε < 1, there exists a polynomial-time approximation for the minimum connected sensor cover problem such that with probability 1 − varepsilon, this algorithm produces a O(log2 n log m)-approximation solution. In the following, we prove Theorem 4.6.3. To do so, it is sufficient to show the following. Lemma 4.6.5 (Garg, Konjevod, and Ravi [216]) With underline tree background, there exists a polynomial-time approximation algorithm for the group Steiner tree problem, which with probability at least 3/4, produces a O(log n log m)approximation solution. In fact, Theorem 4.6.3 can result from Lemma 4.6.5 as follows. Proof of Theorem 4.6.3 Consider a graph G = (V , E) with positive edge length w : E → R+ and m subsets of vertices, g1 , . . . , gm . For any two nodes u and v in V , denote by d(u, v) the length of the shortest path between u and v. Then (V , d) is a metric space. By Theorem 4.5.1, a tree T together with distance d T on T can be randomly chosen such that there exists a one-one onto mapping f from V to the leaf set of T satisfying that for any u, v ∈ V , d(u, v) ≤ d T (u, v) and E[d T (f (u), f (v))] ≤ O(log n) · d(u, v). By Lemma 4.6.5, with probability at least 3/4, a O(log n log m)-approximation solution S can be computed in polynomial-time for the group Steiner tree problem on input groups f (g1 ), f (g2 ), . . . , f (gm ) in tree T . S may have some nodes not in f (V ). Consider S as a Steiner tree on its nodes in f (V ). We can find a spanning tree S  on those nodes of S in f (V ) such that d T (S  ) ≤ 2 · d T (S). Since d(f −1 (S  )) ≤ d T (S  ), we have d(f −1 (S  )) ≤ 2 · d T (S).

(4.14)

Let Opt be the minimum group Steiner tree for g1 , g2 , . . . gm in graph G. Then E[d T (f (Opt)] ≤ O(log n) · d(Opt). By Markov’s inequality, P r[d T (f (Opt)) ≥ 4E[d T (f (Opt))]] ≤ 1/4. This means that with probability at least 3/4, we can find a tree T such that d T (f (Opt)) ≤ 4E[d T (f (Opt))]] ≤ O(log n) · d(Opt).

(4.15)

Since S is an O(log n log m)-approximation solution and f (Opt) is a feasible solution for the group Steiner tree problem on groups f (g1 ), f (g2 ), . . . , f (gm ), we have

56

4 Connected Sensor Cover

d T (S) ≤ O(log n log m) · d T (f (Opt)).

(4.16)

By combining (4.14), (4.15), and (4.16), we obtain that with probability at least 9/16, we can find a polynomial-time approximation solution f −1 (S  ) such that d(f −1 (S  ) ≤ O(log2 n log m) · d(Opt).

(4.17)

Now, consider a fixed ε ∈ (0, 1). Let k be an integer such that (7/16)k ≤ ε. Repeat above computation k times which produce k solutions. Choose the shortest one among them. Note that P r[(4.17) holds once] ≥ 1 − (1 − 9/16)k ≥ 1 − ε. Therefore, with probability at least 1 − ε, the shortest one satisfies (4.17). This completes the proof of Theorem 4.6.3.   Next, we want to give a proof of Lemma 4.6.5. To do so, we first introduce an equivalent formulation of the group Steiner tree problem. Problem 4.6.6 (Group Steiner Tree) Consider a graph G = (V , E) with positive edge length w : E → R+ . Given a node r and m subsets of nodes, g1 , . . . , gm , find the shortest tree with root r, interconnecting g1 , g2 , . . . , gm , i.e., the tree contains at least one node from each gi for i = 1, . . . , m. Lemma 4.6.7 Problems 4.6.1 and 4.6.6 are equivalent, in the sense that Problem 4.6.1 has a polynomial-time ρ-approximation if and only if Problem 4.6.6 has a polynomial-time ρ-approximation. Proof Suppose Problem 4.6.1 has a polynomial-time ρ-approximation. Consider r as a group consisting of only one node r. Then we know that Problem 4.6.6 has a polynomial-time ρ-approximation. Conversely, suppose Problem 4.6.6 has a polynomial-time ρ-approximation. For each r ∈ g1 , computer a polynomial-time ρ-approximation solution Tr for Problem 4.6.6 on groups g2 , . . . , gm . Choose the shortest one from Tr for r over g1 . Then this shortest one should be a polynomial-time ρ-approximation for Problem 4.6.1 on groups g1 , g2 , . . . , gm .   Moreover, we may assume that all groups g1 , g2 , . . . , gm are disjoint. In fact, if there is a node v appearing in more than one groups, say k groups, then we may introduce k new nodes v1 , v2 , . . . , vk to replace v in the k groups and isolate v to form a new group, meanwhile add edges (v1 , v), (v2 , v), . . . , (vk , v) with zero weight. In the following, we will show Lemma 4.6.5 based on definition Problem 4.6.6 for the group Steiner tree problem with property that all groups are disjoint. Consider a tree T = (V , E) with root r and edge weight w : E → R+ . For any S ⊆ V , let ∂S denote the set of edges each having exactly one endpoint in S. Then the following is a LP relaxation of the group Steiner tree problem:

4.6 Randomized O(log2 n log m)-Approximation

min



57

(4.18)

we xe

e∈E

subject to



xe ≥ 1

e∈∂S

∀S ⊂ V : (r ∈ S) ∧ (∃i)[S ∩ gi = ∅] 0 ≤ xe ≤ 1, ∀e ∈ E. First, we note the following. Lemma 4.6.8 The linear programming (4.18) is polynomial-time solvable although it has an exponential number of constraints. Proof By the max flow-min cut theorem, the first group of constraints means that if each edge e has capacity xe , then the flow from the root r to all nodes in each group g has total value at least one. This means that the linear programming (4.18) is equivalent to the following: min



(4.19)

xuv

(u,v)∈E

subject to yvp =

 u

p

xuv −



p xvw ∀v ∈ V

w

p yr

= −1 ∀p = 1, 2, . . . , m,  yvp ≥ 1, ∀p = 1, 2, . . . , m,

v∈gp p

yu = 0 ∀u ∈ gp , u = r, p

xuv ≤ xuv ∀u, v ∈ V ∀p = 1, 2, . . . , m, p

p

0 ≤ xuv ≤ 1, 0 ≤ yuv ≤ 1 ∀u, v ∈ V ∀p = 1, 2, . . . , m. Clearly, (4.19) is polynomial-time solvable.

 

Let xˆ be the optimal solution of the linear programming (4.18). It is easy to see that all edges e with xˆe > 0 form a tree T  with root r. Moreover, we may assume that (a) all group nodes are leaves, and (b) T  is binary, since, otherwise, those properties can be made by adding some edges with zero weight. Now, we carry out a randomized rounding as follows. Initially T a contains all nodes of T  . If an edge e is incident to r in T  , then put e into T a with probability xˆe . If an edge of T  is not incident to r, then put e into T a with probability xˆe /xˆf where f is the edge adjacent to e and closer to r. Delete all connected components of T a not containing r. Let us still use T a to denote the resulting tree.

58

4 Connected Sensor Cover

Lemma 4.6.9 E[w(T a )] ≤ optLP where optLP the objective function value of optimal solution of the linear programming (4.18). Proof Note that an edge e ∈ T  is contained in T a if and only if all edges in the path from r to this edge, (e1 , e2 , . . . , ep ) and including this edge e are appearing in T a . Hence, this event occurs with probability xˆe1 · (xˆe2 /xˆe1 ) · · · (xˆe /xˆep ) = xˆe . Therefore,  E[wT a ] = we xˆe = optLP .   e∈T  We will also show the following lemma. Lemma 4.6.10 For any group g, P r[T a connect r to g] ≥ c/ log n where c is a positive constant. Before give the proof, let us show that our required result can be derived from Lemmas 4.6.9 and 4.6.10. Theorem 4.6.11 There exists a polynomial-time randomized algorithm which produces a group Steiner tree T such that E[w(T )] ≤ O(log n log m) · optgst , where optgst is the objective function value of the optimal solution for the group Steiner tree problem. Proof Let L = log n log m/c . With randomized rounding L times, we would i obtain tree Ta1 , Ta2 , . . . , TaL . Let T u = ∪L i=1 Ta . If T does not connect r to some u group g, then add to T a shortest path from r to g. Denote by T the final result. Clearly, the length of this shortest path is at most optgst . By Lemma 4.6.10, P r[T u does not connect r to g] ≤ (1 − 1/ log n)L ≤ e−L/ log n ≤ 1/m. By Lemma 4.6.9, E(Tai )] ≤ optLP ≤ optgst . Thus, E[w(T )] ≤

L 

E[w(Tai )] +

i=1

≤ (L + 1) · optgst .

m 

P r[T u does not connect r to gp ] · optgst

p=1

 

4.6 Randomized O(log2 n log m)-Approximation

59

Now, Lemma 4.6.5 is a corollary of Theorem 4.6.11 by employing Markov’s inequality. Theorem 4.6.12 (Markov’s Inequality) Let X be a nonnegative random variable and suppose that E[X] exists. Then for t > 0, P r[X > t] ≤

E[X] . t

Proof of Lemma 4.6.5 Let L be defined in the proof of Theorem 4.6.11, P r[w(T ) > 4(L + 1)optgst ] ≤

E[w(T )] 1 ≤ . 4(L + 1)optgst 4

This completes the proof of Lemma 4.6.5.

 

Next, we come back to the proof of Lemma 4.6.10. To do so, we consider Ta again and show a monotone property of Ta . Lemma 4.6.13 For any group g and any edge e, P r[Ta connects r to g] is monotone nonincreasing as xˆe decreases. Proof Suppose e has two children e1 and e2 as shown in Fig. 4.10. Their endpoints are roots of subtree T1 and T2 , respectively. For i = 1, 2, Denote pi = P r[in Ti , the root connects a node in g]. Then the probability that e connects a node in g ∩ (T1 ∪ T2 ) is as follows: 1 − (1 − xˆe + xˆe ((1 − = xˆe1 p1 + xˆe2 p2 −

xˆe1 xˆe xˆe xˆe ) + 1 · (1 − p1 ))((1 − 2 ) + 2 · (1 − p1 ))) xˆe xˆe xˆe xˆe

xˆe1 xˆe2 p1 p2 . xˆe

This is clearly monotone nonincreasing as xˆe decreases. Since the probability that the root r connects other nodes in g does not effected by xˆe , this lemma is proved.   Fig. 4.10 Subtrees T1 and T2

60

4 Connected Sensor Cover

Based on the monotone property stated in Lemma 4.6.13, we modify Ta as follows. 1. Delete all edges not on any path from the root r to a node in considered group g. This is equivalent to decrease xˆe to 0 for those edge e. 2. Reduce each xˆe as small as possible preserving that the flow from r to g has value exactly one. 3. Round each xˆe down to the nearest power of 2. This would reduce the network flow at most half, i.e., the remaining flow is at least 1/2. 4. Delete all edges with new x-value at most 1/(4|g|) and clear up again those edges not on any path from the root r to a node in g. After this deletion, the flow would get reduction at most |g| · (1/(4|g|)) = 1/4. Hence, the remaining flow has at least 1/4. 5. Denote by p(e) the parent edge of edge e. If p(e) has only one child e, then shrink e into one point. Actually, in this case, by step 2, we already have xˆp(e) = xˆe . Hence, e would be selected with probability one. Lemma 4.6.14 After above five steps, the height of tree Ta is at most log2 (4|g|). Proof Note that for each edge e in Ta , xˆe = xˆp(e) /2 and xˆe ≥ 1/(4|g|). Therefore, the height of Ta is at most log2 (4|g|).   Now, we are going to show Lemma 4.6.10 by employing Janson’s inequality as follows. Theorem 4.6.15 (Janson’s Inequality [17]) Let  be a universe set, and R a subset of , in which each element e is randomly, independently selected from  with probability pe . For a subset Ai of , denote by Bi the  event that Ai ⊆ R. Denote by i ∼ j if Bi and Bj are not independent. Define = i ∼ j P r[Bi ∩Bj ] and μ = i P r[Bi ]. If δ ≥ max( , μ(1 − )), then P r[∩i B¯ i ] ≤ exp(

−μ2 (1 − ) ), 2δ

where P r[Bi ] ≤ for all i. Consider a group g. Let us set  = E(T  ) and pe = xe /xp(e) where p(e) is the parent edge of e and for e incident to the root r, consider xp(e) = 1). Let each Ai be an edge set of path from r to a leaf i in g. Thus, ∩i B¯ i is the event that g would not be reached from r in the random selection. Jonson’s inequality gives an upper bound for the probability that this event occurs, which implies a lower bound for the probability that group g can be reached. Next, we estimate μ and . First, note thatP r[Bi ] = xˆei where ei is the edge incident to leaf i. This implies that 1/4 ≤ μ = i P r[Bi ] ≤ 1.  To estimate , we fix a node u in g and estimate u = w∈g,w =u P r[Bw ∩ Bu ]. Suppose the edge e is the lowest common edge on the paths from r to u and w, respectively (Fig. 4.11). Let W be the set of nodes in g, each of which has e as the lowest common edge on the path from r to it and on the path from r to u.

4.7 O((Rs /Rc )2 )-Approximation

61

Fig. 4.11 e is the lowest common edge on the paths from r to u and w, respectively

Then, P r[Bw ∩ Bu ] =

xˆeu xˆew . xˆe

Note that for fixed e, as w is over all nodes in W , the sum of xˆw cannot exceed xˆe . Therefore, u ≤ xˆu · h, where h is the height of Ta . Hence, =

 1 u ≤ h = xˆu h ≤ log(4|g|). 2 u u∈g

Taking = 1/2 in Janson’s Inequality, we have P r[∩B¯ i ] ≤ exp{− ≤ 1−

1 } 16 log(4|g|)

1 . 32 log(4|g|)

The second inequality follows from the fact that e−x ≤ 1 − x/2 for small positive x. This completes the proof of Lemma 4.6.10.

4.7 O((Rs /Rc )2 )-Approximation Recall that when the connected sensor cover problem is transformed into the group Steiner tree problem, each group is a subset of all sensors covering a certain target point. Hence, sensors in each group are contained in a disk with radius Rs and the target point as the center, where Rs is the sensing radius of each sensor. In this section, this geometric information would be emerged in construction of approximation solution for the group Steiner tree problem.

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4 Connected Sensor Cover

Note that in the formulation of the group Steiner tree, the root is a sensor covering a target point. Since the root is always connected into a group Steiner tree, we may exclude the target point covered by the root from our consideration. For simplicity of speaking, we talk “all target points” which means “all target points other than the one covered by the root.” Denote C = Rs /Rc . Without loss of generality, assume Rc = 1. Hence, Rs = C. Denote by g(p) the group induced by target point p. Then g(p) is the set of all sensors covering target point p; they are located within a disk D(p) with center p and radius C. With a grid, D(p) can√be partitioned into at most 8C 2 cells each of which is a square with edge length 2/2. Since every two sensors in a cell have distance at most 1, there is an edge between them. p p Let (xuv , xuv , yv ) be an optimal solution of the linear programming (4.19). Note that for each target point p, the total flow of commodity p from r to g(p) is at least one. Thus, for each target point p, we may find a cell c such that for sensors in cell c, the total flow from r to cell c is at least 1/(8C 2 ), i.e., 

p

ys ≥ 1/(8C 2 ).

s∈c

Denote by cel(p) the subset of all sensors in this cell c. Consider S  = ∪p cel(p) where p is over all target points. Now, we construct a connected sensor cover in two steps. 1. Using algorithm in Chapter 2, find a 2-approximation solution H for the minimum sensor cover problem with input sensors in S  and all target points. Assume H is a minimal solution, that is, for each s ∈ H , there is a target point p covered only by s. 2. In the communication network of sensors, consider each edge with weight one. Connect H ∪{r} with a 2-approximation T for the Steiner minimum tree problem. Let V (T ) be the set of nodes in T . Then V (T ) ∪ H is a connected sensor cover. We are going to prove |V (T ) ∪ H | ≤ O(C 2 ) · optcsc , where optcsc is the objective function value of the optimal solution for the minimum connected sensor cover problem, that is, the following holds. Theorem 4.7.1 (Huang, Li, Shi [267]) There exists a polynomial-time O((Rs /Rc )2 )approximation solution for the minimum connected sensor cover problem. To this end, we show the following two lemmas. Lemma 4.7.2 |H | ≤ O(C 2 ) · optLP where optLP is the objective function value of the optimal solution for the linear programming (4.19).

4.7 O((Rs /Rc )2 )-Approximation

63

Proof Huang et al. [267] showed that there exists a sensor cover S  ⊆ S  such that |S  | ≤ O(C 2 ) · optLP and hence, |H | ≤ 2|S  | ≤ O(C 2 ) · optLP . We sketch the proof of Huang et al. [267] as follows. First, construct an instance of the hitting set problem, (U , F), where U = ∪p cel(p) and F = {cel(p) | p is a target point}. The problem is to find a minimum subset of U to hit every cel(p). This hitting set problem has the following LP relaxation.  min zu (4.20) u∈U



subject to

zu ≥ 1 ∀cel(p) ∈ F,

u∈cel(p)

0 ≤ zu ≤ 1 ∀u ∈ U . Let optLP −H S be the optimal value of LP (4.20). We claim that optLP −H S ≤ O(C 2 ) · optLP . p

p

To show this, consider an optimal solution of LP (4.19), (xuv , xuv , yv ). We can construct a feasible solution of LP (4.20) by setting p

zu = min(1, 8C 2 · max yu ). p

This is feasible because 



zu ≥ min(1, 8C 2 ·

u∈cel(p)

p

yu ) ≥ 1.

u∈cel(p)

Now, the claim can be proved by optLP −H S ≤



zu

u∈U

=



p

min(1, 8C 2 · max yu )

u∈U

≤ 8C 2 ·

p

 u∈U

≤ 8C 2 ·



u∈U

≤ 8C 2 ·

p

max yu p



max p

p

xwu

(w,u)∈E

 

u∈U (w,u)∈E

p

max xwu p

64

4 Connected Sensor Cover

≤ 8C 2 ·

 

xwu

u∈U (w,u)∈E



≤ 8C 2 ·

xuv

(u,v)∈E

= 8C 2 · optLP . It is showed in [81] that the optimal solution for LP (4.20) can be rounded into an integer solution with objective function value O(optLP −H S ). The lemma is proved by combining this fact and the claim.   Lemma 4.7.3 |V (T )| ≤ O(C 2 ) · optLP . Proof For each target point p, there exists a sensor sp ∈ H ∩ cel(p). If for different target points p and p , sp = sp , then we delete one of p and p and consider only one. Since H is minimal, H = {sp | p is a considered target point}. Following is a LP relaxation for finding the Steiner minimum tree interconnecting r and sp for all considered target points p. min



(4.21)

xuv

(u,v)∈E

subject to yvp =



p

xuv −

u



p xvw ∀v ∈ V

w

p yr = −1 p ysp ≥ 1,

∀p = 1, 2, . . . , m, ∀p = 1, 2, . . . , m,

p yu = 0 ∀u ∈ gp , u = r, p xuv ≤ xuv ∀u, v ∈ V ∀p = 1, 2, . . . , m, p p 0 ≤ xuv ≤ 1, 0 ≤ yuv ≤ 1 ∀u, v ∈ V ∀p p

= 1, 2, . . . , m.

p

Let (xu,v , xuv , yv ) be an optimal solution of LP (4.19). We use it to construct a feasible solution of LP (4.21) as follows. 1. For each commodity p, decompose flow from r to g(p) into path-flows and then delete all path-flows not coming to cel(p). p p 2. Move all commodity p to node sp . This means that we modify xuv and yv by the following p

xvsp ← xvsp + yvp ∀v ∈ cel(p)  p ysp ← yvp v∈cel(p)

yvp ← 0 for v = sp .

4.7 O((Rs /Rc )2 )-Approximation

65

Fig. 4.12 Initially, store commodity p all at sp

p

3. Scale every commodity p by 1/ysp , i.e., make every commodity to have flow value one. p

p

Let (xˆuv , xˆuv , yˆv ) be the feasible solution of LP (4.21) obtained as above. Let p p (xuv , xuv , yv ) denote still the optimal solution of LP (4.19). Then this feasible solution has objective function value (Fig. 4.12): 

xˆuv =

(u,v)∈E



p

max xˆuv

(u,v)∈E

≤



p

max 

(u,v)∈E

≤ 8C 2

p

p



(u,v)∈E

xuv p v∈cel(p) yv

+

 v∈cel(p)

p



yv

p v∈cel(p) yv

p

max xuv + |H | p

≤ O(C 2 ) · optLP . Using the primal–dual method, a Steiner tree can be obtained with total length at most twice of optLP −ST , the optimal value of LP relaxation of the Steiner minimum tree, that is, the optimal value of LP (4.21). Therefore, the Steiner minimum tree for H has length upper bounded by 2 · O(C 2 )optLP and hence, |V (T )| ≤ 4 · O(C 2 )optLP + 1 = O(C 2 ) · optLP . By Lemmas 4.7.2 and 4.7.3, we have |H ∪ V (T )| ≤ (O(C 2 ) + O(C 2 )) · optLP ≤ O(C 2 ) · optcsc .

 

Chapter 5

Lifetime of Coverage

When life gives you a hundred reasons to cry, show life that you have a thousand reasons to smile. Unknown

5.1 Motivation and Overview When a very large number of sensors are randomly deployed into a region by an aircraft in order to monitor a certain set of targets (target points), there are often a lot of redundant sensors. A better usage of those redundant sensors is to make a sleep/wakeup scheduling for sensors, which would increase the lifetime of coverage. The lifetime of coverage is the length of a time period at every moment of which every target is monitored by an active sensor. This is the motivation to study the maximization lifetime coverage problem (Problem 1.3.3). Especially, when a system is expected to work for a time period longer than the lifetime of each sensor, the maximization of the lifetime of coverage has more impact. In study of the lifetime of coverage, there are two milestones. The first one is about mathematical formulation. At earlier stage, to increase the lifetime of coverage, one tried to find as many as possible disjoint sensor covers, i.e., the following formulation is studied. Problem 5.1.1 (Maximum Sensor Cover Partition) Given m targets a1 , . . . , am and n sensors s1 , . . . , sn in the Euclidean plane, find the maximum number of disjoint sensor covers. This problem is NP-hard and various heuristics or approximation algorithms were designed [89, 93, 292–294, 333, 408, 464, 474]. When the target point set and the sensor set are identical, a sensor cover is also called a dominating set. It has been known that for the maximum dominating set partition problem (similar to the maximum sensor cover partition problem) in general graphs, there is no polynomialtime (1−ε) ln n)-approximation for any ε > 0 unless N P ⊆ DT I ME(nO(log log n) ) [192] and there exist polynomial-time O(log n)-approximations [66, 192]. However, © Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_5

67

68

5 Lifetime of Coverage

Fig. 5.1 Three sensors s1 , s2 , s3 and three targets a1 , a2 , a3

those results cannot extended to the maximum sensor cover partition problem in Euclidean plane. Actually, not many theoretical results exist in the literature for the maximum sensor cover partition problem since a new formulation came out in 2005, which made ones lost interesting on this formulation. In the maximum sensor cover partition problem, the sensor is activated only once, i.e., once the sensor is activated, it keeps active until died. Cardei et al. [95] found that it is possible to increase the lifetime of coverage if each sensor is allowed to make changes between active and inactive states. For example, consider three homogeneous sensors s1 , s2 , s3 with unit lifetime and three targets a1 , a2 , a3 . Suppose s1 can cover a1 and a2 , s2 can cover a2 and a3 , and s3 can cover a3 and a1 (Fig. 5.1). If three sensors are partitioned into disjoint sensor covers, then only one sensor cover can be obtained. Therefore after sensors in the sensor cover are died, the coverage cannot be realized by remaining sensor, i.e., the lifetime of coverage is one unit. However, the lifetime of coverage can be one and a half unit time if three sensors work in the following way: In the first period of a half unit time, sensors s1 and s2 are active while sensor s3 sleeps; in the second half unit time period, sensors s2 and s3 are active while sensor s1 sleeps; in the third half unit time period, sensors s3 and s1 are active while sensor s2 has no lifetime left. Would the lifetime of a sensor be changed when it changes its state between active and sleeping more times? The answer is yes. However, the change is toward a better direction. It is an interesting fact discovered in [242] that putting a battery alternatively in active and sleep states in a proper way may double its lifetime since the battery could be recovered in certain level in sleep state. Based on above considerations, Cardei et al. [95] proposed a new formulation for maximization of coverage lifetime, i.e., the maximum lifetime coverage problem (Problem 2.3.3). They also showed the NP-hardness of the set-relaxation of the maximum lifetime coverage problem and proposed a heuristic by using an integer programming. This formulation is widely accepted in the literature. The first approximation solution with theoretically guaranteed performance was given by Berman et al. [66, 67]. They showed that there exists a polynomialtime approximation for the maximum lifetime coverage problem with performance ratio O(log n) where n is the number of sensors. This work introduced a result established by Garg and Könemann [217], which is the second milestone in study of the maximum lifetime coverage problem. With a theorem of Garg and Könemann

5.1 Motivation and Overview

69

[217], the maximum lifetime coverage problem can be reduced to the minimum weight sensor cover problem (Problem 2.3.4). Actually, by this theorem, if the minimum weight sensor cover problem (Problem 2.3.4) has a polynomial-time ρ-approximation, then the maximum lifetime coverage problem has a polynomial-time (1 + ε)ρ-approximation for any ε > 0. Note that the minimum weight sensor cover problem is a special case of the minimum weight set cover problem which has a well-known greedy approximation with performance ratio 1 + ln n. Hence, the maximum lifetime coverage problem has a polynomialtime (1 + ε)(1 + ln n)-approximation for any ε > 0. The application of Garg–Könemann method is the second milestone in study of lifetime of coverage. After the work of Berman et al. [66, 67], the minimum weight sensor cover problem gets ones’ attention. The first polynomial-time constantapproximation for the minimum weight sensor cover problem in unit disk graphs was obtained by Ding et al. [169]. Actually, they extended an approximation algorithm designed for the minimum weight dominating set problem to the minimum weight sensor cover problem. This finding is quite interesting because it could not do a similar extension from the PTAS for the minimum dominating set problem to the minimum sensor cover problem. Actually, it was known for a long time the existence of PTAS for the minimum dominating set problem in unit disk graphs before a PTAS was designed successfully for the minimum weight sensor cover problem for homogeneous wireless sensor networks. The work of Ding et al. speeds up the progress on the study of approximation algorithms for the minimum weight sensor cover problem. Finally, with a quite long analysis, Li and Jin [326] designed a PTAS for the minimum weight sensor cover problem. Theorem 5.1.2 (Li and Jin [326]) There exists a PTAS for the minimum weight sensor cover problem in homogeneous wireless sensor networks. As a corollary, there exists a PTAS for the maximum lifetime coverage problem. Du et al. [179] extended this approach to study the coverage with connectivity requirement, i.e., the maximum lifetime connected coverage problem. In their model, a sensor has two active phases: (1) only do data transmission (communication); (2) collect data (sensing) and transmit data (communication). They established a result with Grag–Könemman method. With this result and a result of Zuo et al. [654], they constructed a polynomial-time constant-approximation in homogeneous wireless sensor networks. The lifetime of coverage and connected coverage has also been studied in [92, 379, 381, 385, 391, 442, 590]. Finally, we would like to remark that it is in a work of Calnescu et al. [91] in which Garg–Könemann method is first employed in study of wireless sensor networks when they studied wireless sensor networks with adjustable energy consumption at each node.

70

5 Lifetime of Coverage

5.2 Complexity If remove the geometric structure from the maximum lifetime coverage problem, then we may obtain a set-relaxation version as follows. Problem 5.2.1 (Set-Relaxation of Maximum Lifetime Coverage) Given a collection C of subsets of a finite set R, find a family of set covers S1 , . . . ., Sp ; with time weights t1 , . . . , tp , respectively, and maximum t1 + · · · + tp such that for each subset s in C, s appears in S1 , . . . , S, with total weight at most 1, where 1 is the life time of each sensor. Cardei et al. [95] showed that this problem is NP-hard. Theorem 5.2.2 (Cardei et al. [95]) The set-relaxation of maximum lifetime coverage problem is NP-hard. Proof Consider the decision version of the set-relaxation of maximum lifetime coverage problem as follows: Given a collection C of subsets of a finite set R and a number k, is there a family of set covers S1 , . . . , Sp ; with time weights tl , . . . , tp , such that tl + · · · + tp ≥ k and for each subset s in C, s appears in S1 , . . . , Sp , with total weight at most 1? We will construct a polynomial-time many-one reduction from the 3SAT problem to this decision version. The 3SAT is a well-known NP-complete problem as follows. Problem 5.2.3 (3SAT) Given a 3CNF F , determine whether F has a satisfiable assignment. Consider an instance of the 3SAT problem, a 3CNF F , i.e., F is a Boolean formula in conjunctive normal form such that each clause contains exactly three literals. Suppose F contains m clauses c1 , . . . , cm and n variables x1 , . . . , xn . Let R = {u, c1 , . . . , cm , x1 , . . . , xn } where u is an element other than clauses and variables. Construct a collection C consisting of the following subsets of R: Pi = {xi } ∪ {cj | cj contains xi }, Qi = {xi } ∪ {cj | cj contains x¯i }, for i = 1, . . . , n; U = {u, c1 , . . . , cm }, V = {u, x1 , . . . , xn }. Choose k = 2. First, suppose F has a satisfiable assignment τ : {x1 , . . . , xn } → {0, 1}. Define S1 = {V } ∪ {Pi | τ (xi ) = 1} ∪ {Qi | τ (xi ) = 0} S2 = {V } ∪ {Qi | τ (xi ) = 1} ∪ {Pi | τ (xi ) = 0}.

5.3 Garg–Könemann Method

71

Then S1 and S2 are two disjoint set covers. Assign each set cover with weight 1, i.e., t1 = t2 = 1. Then we obtain t1 + t2 = k = 2. Conversely, suppose there exists a family of set covers S1 , . . . , Sp ;, with time weights tl , . . . , tp , such that tl + · · · + tp ≥ 2 and for each subset s in C, s appears in S1 , . . . , Sp , with total weight at most 1. Since u appears in only two subsets U and V . Every set cover Si for i = 1, . . . , p contains exactly one of subsets U and V and have to satisfy the following properties: 1. If a set cover contains U , then it must contain either Pi or Qi for any i = 1, . . . , n. 2. If a set cover contains V , then for any j = 1, . . . , m, it must contain either a Pi for xi in cj or a Qi for x¯i in cj . From the first property, we see that if a set cover contains V , then for any i = 1, . . . , n, it contains at most one of Pi and Qi . Fixed a set cover Sh containing V . Define an assignment τ as follows.  τ (xi ) =

1 if Sh contains Pi ; 0 otherwise.

Then by the second property, τ is a satisfiable assignment.

 

So far, we are unable to find, in the literature, a proof for the NP-hardness of the maximum lifetime coverage problem with geometric structure. In other words, we still have an open problem. Open Problem 5 Is the maximum lifetime coverage problem NP-hard?

5.3 Garg–Könemann Method In this section, we use Garg–Könemann Method to show the following theorem: Theorem 5.3.1 (Berman et al. [66]) If the minimum weight sensor cover problem has a polynomial-time ρ-approximation, then the maximum lifetime coverage problem has a polynomial-time (1 + ε)ρ-approximation for any ε > 0. Let S be the set of all sensors. Assume all sensors are uniform, that is, they have the same communication radius Rc , the same sensing radius Rs , and the same life time, say unit time. For any sensor cover p and a sensor s, define  as,p =

1 if s ∈ p, 0 otherwise.

Suppose C is the collection of all sensor covers. Then the maximum lifetime coverage problems can be formulated as the following linear programming:

72

5 Lifetime of Coverage

max



(5.1)

xp

p∈C

subject to



as,p xp ≤ 1 for s ∈ S

p∈C

xp ≥ 0

for p ∈ C.

Its dual is as follows. min



(5.2)

ys

s∈S

subject to



as,p ys ≥ 1 for p ∈ C,

s∈S

ys ≥ 0

for s ∈ S.

To prove Theorem 5.3.1, consider the following Primal–Dual Algorithm. Primal–Dual Algorithm A Initially, set xp = 0 for all p ∈ C and ys = δ for all s ∈ S where δ is a positive constant which will be chosen later. In each iteration, carry out the following steps until (ys , s ∈ S) becomes dual-feasible, that is, all constraints in dual LP (5.2) are satisfied: 1. Compute a ρ-approximation solution p∗ for the minimum weight sensor cover problem as follows. min



as,p ys

s∈S

subject to p ∈ C. 2. Update xp and ys as follows: xp does not change for p = p∗ , and xp∗ ← xp∗ + 1. ys does not change for s ∈ p∗ , and ys ← ys (1 + θ ) for s ∈ p∗ where θ is a constant chosen later. The following lemmas give two important properties at the end of above algorithm.

5.3 Garg–Könemann Method

73

Lemma 5.3.2 At the end of Primal–Dual Algorithm A, (xp , p ∈ C) may not be a primal-feasible solution. However, (xp /τ, p ∈ C) is a primal-feasible solution ln(δ/ρ) where τ = 1 − ln(1+θ) . Proof Note that when ys gets updated, the following facts must hold: (a) (ys , s ∈ S) is not dual-feasible. (b) s ∈ p∗ .

 It follows immediately from (a) that s∈S as,p∗ ys < ρ, which together with (b) yields that ys < ρ before ys receives any value change. After ys is updated, we have ys < (1 + θ )ρ. Therefore, at the end of Primal–Dual Algorithm, ys < (1 + θ )ρ. Now, consider a constraint in the primal linear programming, 

as,p xp ≤ 1,

p∈C

which  may not be satisfied after xp is updated. If updating xp increases the value of p∈C as,p xp by adding as,p∗ , then  the value of ys is increased by multiplying a factor 1 + θ . Suppose the value of p∈C as,p xp is increased  times, i.e., it receives an increase of total value . Correspondingly, ys changes  times. Since initially ys = δ, we have (1 + θ ) ≤

(1 + θ )ρ δ

 ln(δ/ρ) = τ . Moreover, initially, p∈C as,p xp = 0. Thus, at the end that is,  ≤ 1 − ln(1+θ)  of Primal–Dual Algorithm, the value of p∈C as,p xp is  ≤ τ . Therefore, 

as,p xp /τ ≤ 1.

p∈C

  Lemma 5.3.3 At the end of Primal–Dual Algorithm A,  p∈C

xp /τ ≥

− ln(|S|δ) · optlc τ θρ

where optlc is the objective function value of optimal solution for the maximum ln(δ/ρ) lifetime coverage problem and τ = 1 − ln(1+θ) .

74

5 Lifetime of Coverage

Proof Denote by xp (0) the initial value of xp and by ys (0) the initial value of ys . Denote by xp (i) and ys (i), respectively, the values of xp and ys after the ith ∗ iteration.Denote by p∗ (i) the values  of p in the ith iteration. Furthermore, denote X(i) = p∈C xp (i) and Y (i) = s∈S ys (i). Then, for i ≥ 1, we have Y (i) =



ys (i − 1) + θ

s∈S



as,p∗ (i) ys (i − 1)

s∈S

≤ Y (i − 1) + θ (X(i) − X(i − 1))ρ min p∈C



as,p ys (k − 1).

s∈S

Thus, Y (i) ≤ Y (0) + θρ

i   ((X(k) − X(k − 1)) min as,p ys (k − 1). p∈C

k=1

s∈S

By the duality theory of linear programming, optlc is also the objective function value of optimal solution for the dual linear programming (5.2). Therefore, optlc = min ys

 s∈S ys 

minp∈C

s∈S

as,p ys

,

where the minimization is subject to ys ≥ 0 for s ∈ S. Hence, min p∈C



as,p ys (k − 1) ≤

s∈S

Y (k − 1) . optlc

Therefore, Y (i) ≤ |S|δ +

i θρ  (X(k) − X(k − 1))Y (k − 1). optlc k=1

Define w(0) = |S|δ and w(i) = |S|δ +

i θρ  (X(k) − X(k − 1))w(k − 1). optlc k=1

It is easy to prove by induction on i that Y (i) ≤ w(i). Moreover,

5.3 Garg–Könemann Method

75

w(i) = (1 + θρ

≤ e optlc ≤e =e

θρ (X(i) − X(i − 1)))w(i − 1) optlc (X(i)−X(i−1))

θρ optlc

X(i)

θρ optlc

X(i)

w(i − 1)

w(0) |S|δ.

Suppose the Primal–Dual Algorithm stops at the mth iteration. Note that optlc ≥ 1. Then Y (m) ≥ 1. Hence θρ

1 ≤ Y (m) ≤ w(m) ≤ |S|δe optlc

X(m)

.

Therefore, τ θρ optlc ≤ . X(m)/τ − ln(|S|δ)   Now, we are ready to prove Theorem 5.3.1. Proof of Theorem 5.3.1 Set δ = (ρ(1 + θ )|S|)1/θ (ρ(1 + θ ))−1 , that is, 1 τ ln(1 + θ ) = . − ln(δ|S|) 1−θ Note that (1 + θ )1+1/θ > e, i.e., ln(1 + θ ) >

θ 1+θ .

Therefore,

τ θρ θρ 1+θ = ≤ρ· . − ln(|S|δ) (1 − θ ) ln(1 + θ ) 1−θ Select θ such that 1+θ < 1 + ε. 1−θ Then 

optlc ≤ (1 + ε)ρ. p∈C xp /τ

We next show that the algorithm runs in polynomial-time. Note that every iteration can be carried out in polynomial-time since p∗ is a polynomial-time ρapproximation for the weighted sensor cover problem. Therefore, it suffices to show

76

5 Lifetime of Coverage

that the number of iterations has a polynomial upper bound. Note that at each iteration, at least one of ys has its value increased. In the proof of Lemma 5.3.2, it showed that at the end of the algorithm, each ys has its value increased by multiplying at most (1 + θ )/δ. Therefore, the number of iterations is at most |S|(1 +

− ln δ ) = O(|S| log |S|), ln(1 + θ )

where − ln(δ) = O(ln |S|) and θ is fixed as ε is fixed.

 

5.4 Lifetime of Connected Coverage The lifetime of connected coverage has been an important research issue existing in the literature for a long time [47, 50, 97, 281]. Zhang and Li [616] proposed a quite general model. In this model, each sensor has two modes, active mode and sleep mode, and the active mode has two phases, the full-active phase and the semiactive phase. A full-active sensor is able to sense, i.e., collect data, and to transmit, to receive, and to relay the data packets. A semi-active sensor is unable to sense, but it is able to transmit, to receive, and to relay data packets. In different phases, the sensor consumes different quantities of energy. Usually, a sensor in the full-active phase needs more energy than in the semi-active phase. Consider a set S of homogeneous sensors randomly deployed into hostile environment, such as battlefield and inaccessible area with chemical or nuclear pollution, so that recharging batteries of sensors is a mission impossible. Let Rc and Rs denote the communication radius and the sensing radius, respectively. Suppose every sensor contains batteries with totally unit amount of energy. A full-active sensor would consume u amount in unit time period and a semi-active sensor would consume v amount in unit time period. Also, assume u ≥ v. Du et al. [179] studied the maximum lifetime connected coverage problem in homogeneous wireless sensor networks with two active phases. Problem 5.4.1 (Maximum Lifetime Connected Coverage) Consider a set of targets, A and a set of sensors, S. Each sensor has two active phases as described as above. Find a sleep/wakeup schedule for sensors to maximize the lifetime of connected coverage, where the lifetime of connected coverage is a time period such that in every moment, the following conditions are satisfied: (A1) Every target is monitored by a full-active sensor. (A2) All (full-/semi-) active sensors induce a connected subgraph. Let p1 be a set of full-active sensors and p2 a set of semi-active sensors. Then the p = (p1 , p2 ) is called an active sensor set pair. For any active sensor set pair p, define

5.4 Lifetime of Connected Coverage

77

as,p

⎧ ⎨ u if s ∈ p1 , = v if s ∈ p2 , ⎩ 0 otherwise.

Suppose C is the collection of all active sensor set pairs satisfying conditions (A1) and (A2). Then the maximum lifetime connected coverage problem can be formulated as the following linear programming: max



(5.3)

xp

p∈C

subject to



as,p xp ≤ 1 for s ∈ S

p∈C

xp ≥ 0

for p ∈ C.

Its dual is as follows. min



(5.4)

ys

s∈S

subject to



as,p ys ≥ 1 for p ∈ C,

s∈S

ys ≥ 0

for s ∈ S.

Motivated from the work of Garg and Könemann [217], Du et al. [179] studied the relationship of the maximum lifetime connected coverage problem and the following problem. Problem 5.4.2 (Minimum Weight Connected Sensor Cover) Consider a set of targets and a set of homogeneous sensors with two active phases, lying in the Euclidean plane. Suppose sensors may have different weights. The problem is to find the minimum weight connected sensor cover. Du et al. [179] also established the following result. Theorem 5.4.3 (Du et al. [179]) If the minimum weight connected sensor cover problem has a polynomial-time ρ-approximation, then the maximum lifetime connected coverage problem has a polynomial-time ρ(1 + ε)-approximation for any ε > 0. To show this result, let us study the following Primal–Dual Algorithm for the maximum lifetime connected coverage problem. Primal–Dual Algorithm B Initially, set xp = 0 for all p ∈ C and ys = δ for all s ∈ S where δ is a positive constant which will be chosen later. In each iteration, carry out the following steps until (ys , s ∈ S) becomes dual-feasible, that is, all constrains in dual linear programming (5.4) are satisfied:

78

5 Lifetime of Coverage

1. Compute a ρ-approximation solution p∗ for the following minimum weight connected sensor cover problem: min



as,p ys

s∈S

subject to p ∈ C. 2. Let s ∗ = arcmaxs∈S as,p∗ . 3. Update xp and ys as follows: xp does not change for p = p∗ , and xp∗ ← xp∗ +

1 . as ∗ ,p∗

ys does not change for s ∈ p1∗ ∪ p2∗ , and ys ← ys (1 + θ

as,p∗ ) as ∗ ,p∗

for s ∈ p1∗ ∪ p2∗ where θ is a constant which will be chosen later. The following lemmas give two important properties at the end of above algorithm. Lemma 5.4.4 At the end of Primal–Dual Algorithm B, (xp /τ, p ∈ C) is a primalfeasible solution where τ =

(v/u) ln (1+θ)ρ vδ ln(1+θv/u)

.

Proof Note that ys gets updated only if the following facts hold: (a) (ys , s ∈ S) is not dual-feasible. (b) s ∈ p1∗ ∪ p2∗ .  It can be seen from (a) that s∈S as,p∗ ys < ρ, which together with (b) yields that ys < ρ/v before ys changes its value. After ys is updated, we have ys < (1 + θ

as,p∗ )ρ/v ≤ (1 + θ )ρ/v. as ∗ ,p∗

Therefore, at the end of Primal–Dual Algorithm B, ys < (1 + θ )ρ/v. Now, consider a constraint in the primal linear programming (5.3), 

as,p xp ≤ 1,

p∈C

which may not be satisfied after the value of xp is updated. If updating xp increases  as,p∗ the value of p∈C as,p xp by adding a ∗ ∗ , then the value of ys is increased a

∗

s ,p

by multiplying a factor 1 + θ a s,p . Note that the value of ∗ ∗ s ,p

as,p∗ as ∗ ,p∗

has only two

5.4 Lifetime of Connected Coverage

79 a

∗

possibilities, v/u and 1. Suppose a s,p takes value v/u for k times and 1 for  s ∗ ,p∗  times. Then the value of p∈C as,p xp receives an increase in k(v/u) +  and (1 + θ v/u)k (1 + θ ) δ ≤

(1 + θ )ρ v

 since initially ys = δ. Moreover, initially, p∈C as,p xp = 0. Thus, at the end of  Primal–Dual Algorithm B, the value of p∈C as,p xp is k(v/u) + . The maximum value of k(v/u)+ can be obtained from the optimal solution of the following linear programming with respect to k and : max k(v/u) +  subject to k ln(1 + θ v/u) +  ln(1 + θ ) ≤ ln

(1 + θ )ρ vδ

k ≥ 0,  ≥ 0. It is well-known from theory of the linear programming that the maximum value of objective function can be achieved at an extreme point. For above one, the feasible domain has three extreme points (0, 0), (0,

ln (1+θ)ρ ln (1+θ)ρ vδ vδ ), ( , 0). ln(1 + θ ) ln(1 + θ v/u)

Their objective function values are ln (1+θ)ρ ln (1+θ)ρ v vδ vδ , · , ln(1 + θ ) u ln(1 + θ v/u)

0, respectively. Since

is strictly monotone decreasing for z ≤ 1, we have

z ln(1+θz)

0
e, which implies ln(1 + θ v/u) >

vθ u+vθ .

Thus,

τ θρ 1 + θ v/u θρ ≤ρ· . = −1 (1 − θ ) ln(1 + θ v/u) 1−θ ln(v|S|δ) Next, choose θ such that 1 + θ v/u < 1 + ε. 1−θ

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5 Lifetime of Coverage

Then we have 

opt ≤ (1 + ε)ρ. p∈C xp /τ

To show that Primal–Dual Algorithm B runs in polynomial-time, let p∗ be a polynomial-time ρ-approximation solution for the minimum weight connected sensor cover problem. Note that every iteration runs in polynomial-time. Therefore, it suffices to show that the number of iterations is upper bounded by a polynomial. Note that at each iteration, at least one of ys has its value increased. In the proof of Lemma 5.4.4, it is already proved that at the end of the Algorithm B, each ys has its value increased by multiplying at most (1 + θ )ρ/(vδ). Therefore, the number of iterations is at most |S|

ln((1 + θ )ρ) − ln(vδ) = O(|S| log |S|), ln(1 + θ v/u)

where − ln(vδ) = O(ln |S|) and θ is fixed as ε is fixed.

 

5.5 Weighted Connected Sensor Cover First, consider the node-weighted Steiner tree problem as follows. Problem 5.5.1 (Node-Weighted Steiner Tree) Given a connected graph G = (V , E) with nonnegative node weight w : V → R, and a node subset P ⊆ V , find a tree T interconnecting nodes in P to minimize the total weight of Steiner nodes where nodes in P are called terminals and nodes of T not in P are called Steiner nodes. In general graphs, the node-weighted Steiner tree problem has a polynomialtime (1.35 log n + O(1))-approximation [232] and no polynomial-time ρ ln napproximation for 0 < ρ < 1 unless NP ⊆ DT I ME(nO(log log n) ) [303]. However, in unit disk graphs, there is a better result given by Zuo et al. [654]. This result is obtained based on the following lemma. Lemma 5.5.2 In a unit disk graph, for any terminal set P , there exists an optimal solution T for the node-weighted Steiner tree problem such that every node has degree at most five. Proof Among all optimal trees for the node-weighted Steiner tree problem, we consider the one with the shortest total Euclidean edge length, called the shortest optimal tree. First, note that the shortest optimal tree must have the following properties: (a1) (a2)

No two edges cross each other. Two edges meet at a node with an angle of at least 60◦ .

5.5 Weighted Connected Sensor Cover

83

(a3) If two edges meet with an angle of exactly 60◦ , then they have the same length. Indeed, if anyone of the above three condition does not hold, then we can easily find another optimal tree with shorter length. Now, consider a shortest optimal tree T . By (a2), every node has degree at most six. Suppose T has a node u with degree exactly six, i.e., u has six neighbors v1 , v2 , . . . , v6 . By (a2), v1 uv2 = v2 uv3 = · · · = v6 uv1 = 60◦ . By (a3), |uv1 | = |uv2 | = · · · = |uv6 |. Moreover, v2 must have degree at most four since replacing (u, v1 ) and (u, v3 ) by (v1 , v2 ) and (v2 , v3 ), the result should still be a shortest optimal tree. Now, replace (u, v1 ) by (v1 , v2 ) and do similar replacement at all nodes with degree six. Then one would obtain a shortest optimal tree with node degree at most five.   Theorem 5.5.3 (Zou et al. [654]) There exists a polynomial-time 3.475approximation for the node-weighted Steiner tree problem in unit disk graphs. Proof Assign each edge (u, v) with the following weight: w(u, v) =

1 ((1 − χP (u))c(u) + (1 − χP (v))c(v)), 2

where  χP (u) =

1, if u ∈ P , 0, otherwise.

Following algorithm gives a polynomial-time 3.475-approximation for the nodeweighted Steiner tree problem. 3.475-Approximation input: unit disk graph G = (V , E) with node weight c : V → R + and a terminal set P ⊆ V . compute 1.39-approximation T for the network Steiner tree problem on terminal set P in graph G with edge weight w(u, v) = 12 ((1 − χP (u))c(u) + (1 − χP (v))c(v)) for (u, v) ∈ E; output T . ∗ To analyze this algorithm, let Tnode be the optimal solution for the node-weighted Steiner tree problem in unit disk graph G, with the property that every node has degree at most five. Then ∗ ∗ w(Tnode ) ≤ 2.5c(Tnode ). ∗ be an optimal solution for the network Steiner tree problem on terminal Let Tedge set P (see Section 3.3). Then ∗ ∗ ) ≤ w(Tnode ) w(Tedge

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5 Lifetime of Coverage

∗ and from [87], one can compute a 1.39-approximation T for Tedge in polynomialtime. Therefore ∗ ∗ ) ≤ 3.475 · c(Tnode ). w(T ) ≤ 1.39 · w(Tedge

Moreover, c(T ) ≤ w(T ) since each Steiner node has degree at least two in T . Therefore, ∗ ). c(T ) ≤ 3.475 · c(Tnode

 

Now, we come back to the minimum weight connected sensor cover problem. In Chapter 3, we know that it is an open problem whether there exists or not a polynomial-time constant-approximation for the minimum connected sensor cover problem. Of course, the problem is also open for the weighted version. Open Problem 6 Is there a polynomial-time constant-approximation for the minimum weight connected sensor cover problem in homogeneous wireless sensor networks lying the Euclidean plane? Here, we would like to consider a special case that Rc ≥ 2Rs . Corollary 5.5.4 If Rc ≥ 2Rs , then the minimum weight connected sensor cover problem in homogeneous wireless sensor networks has a polynomial-time (4.475 + ε)-approximation for any ε > 0. Proof Let Optwcsc be the optimal solution for the minimum weight connected sensor cover problem. By Theorem 5.1.2, it is able to compute a polynomial-time (1+0.5ε)-approximation solution A for the minimum weight sensor cover problem. Since Rc ≥ 2Rs , Optwcsc ∪ A induces a connected subgraph, that is, Optwcsc is a feasible solution for the node-weighted Steiner tree problem with input terminal set A. Let B be a (3.475 + 0.5ε)-approximation solution for the node-weighted Steiner tree problem with input terminal set A. Then |B| ≤ (3.475 + 0.5ε) · optwcsc where optwcsc = |Optwcsc . Therefore |A ∪ B| ≤ |A| + |B| ≤ (4.475 + ε) · optwcsc .

 

Combining Corollary 5.5.4 and Theorem 5.4.3, we would have Corollary 5.5.5 In homogeneous wireless sensor networks in the Euclidean plane with Rc ≥ 2Rs , the maximum lifetime connected coverage problem has a polynomial-time (4.475 + ε)-approximation for any ε > 0.

5.6 A Framework of Calnescu

85

5.6 A Framework of Calnescu Calnescu et al. [91] gave a framework in study of the lifetime maximization problem with Garg–Könemann method. Consider a graph G = (V , E). A function c : V × V → R + is called a power requirement if edge (u, v) is established if and only if both the power at u and the power at v are not less than c(u, v). With a power requirement c, G = (V , E, c) is called a power requirement graph. Calnescu et al. [91] consider the following general form of the minimum power problem and the maximum lifetime problem. Problem 5.6.1 (Power Assignment) Given a power requirement graph G = (V , E, c) and a connectivityconstraint Q, find a power assignment p : V → R + to minimize the total power u∈V p(u) under condition that supported subgraph H satisfies the connectivity constraint Q where H consists of those edges (u, v) with p(u) ≥ c(u, v) and p(v) ≥ c(u, v). Problem 5.6.2 (Network Lifetime) Given a power requirement graph G = (V , E, c), a battery supply b : V → R + , and a connectivity constraint Q, find afeasible power schedule P T = {(p1 ,  t1 ), (p2 , t2 ), . . . , (pm , tm )} to maximize m m t , where schedule P T is feasible if i i=1 i=1 pi (u) ≤ b(u) for every u ∈ V . Suppose S is the set of all subgraphs satisfying constraint Q. For each H ∈ S, let pH : V → R + be the power assignment required by establishing subgraph H . Then the maximum network lifetime problem can be formulated into a linear programming as follows. max



tH

(5.5)

H ∈S

subj ectto



pH (u)tH ≤ b(u)

H ∈S

tH ≥ 0, ∀H ∈ S. Now, the sum of coefficients for each column is exactly the total power for establishing H , which is called the length of the column. Computing the minimum column length is exactly the minimum power assignment problem. Consider the following weighted version of the minimum power assignment problem. Problem 5.6.3 (Weighted Power Assignment) Given a power requirement graph + G = (V , E, c), a connectivity constraint Q, and a node weight  y : V → R , find a + power assignment p : V → R to minimize the total power u∈V p(u)y(u) under condition that supported subgraph H satisfies the connectivity constraint Q. Suppose the minimum weight power assignment problem has a polynomial-time f -approximation. Then there exists a polynomial-time (1 + ε)f -approximation for the maximum network lifetime problem for any ε > 0. Indeed, the linear

86

5 Lifetime of Coverage

programming formulation (5.5) of the maximum network lifetime problem falls in a general form of linear programming studied by Garg and Könemann [217] as follows: max{cT x | Ax ≤ b, x ≥ 0}, where A is an m × n positive matrix and b and c are m-dimensional and ndimensional positive vectors, respectively. n can be exponentially large than m. However, the following Garg and Könemann Algorithm [217] assumes that the linear programming is actually implicitly given by vector b. Garg–Könemann Algorithm input: A vector b ∈ R m , ε > 0, an algorithm F  y(i) for computing f -approximation of minj m i=1 A(i, j ) c(j ) . Initially, δ = (1 + ε)((1 + ε)m)−1/ε . δ for i = 1, . . . , m, y(i) ← b(i) , D ← mδ, j = 0; while D < 1 do find a column Aq using f -approximation F choose index p to minimize Ab(i) q (i) j ←j +1 x j ← Ab(p) q (p) Aj ← Aq for i = 1, . . . , m, y(i) ← y(i)(1 + ε · Ab(p) / b(i) ) q (p) Aq (i) D ← bT y xj k output {(Aj , 1+ε )}j =1 . log1+ε

δ

This algorithm would give a (1 + ε)f -approximation for the maximum network lifetime problem. It is also proved in [217] that k ≤ m. Theorem 5.6.4 Consider a connectivity constraint Q and a power requirement graph G = (V , E, c). For any polynomial-time f -approximation algorithm F for the minimum weight power assignment, there exists a polynomial-time (1 + ε)f approximation algorithm for the corresponding maximum network lifetime problem. Garg–Könemann Algorithm successfully reduces the lifetime maximization problem to the minimum weighted power assignment problem. This gives a quite useful technique to study approximation of the lifetime maximization problem. We may find a lot of applications [66, 67, 80, 169] or possible applications [94, 96, 97] in the literature.

Chapter 6

Weighted Sensor Cover

I think the weight really got the best of him today. Jody Petty

6.1 Motivation and Overview It was open for many years whether the minimum weight dominating set problem has a polynomial-time constant-approximation or not in unit disk graphs. Problem 6.1.1 (Minimum Weight Dominating Set) Given a unit disk graph G = (V , E) with nonnegative vertex weight w : V → R+ , find a dominating set with minimum total weight. Ambühl et al. [19] made the break through. They designed a polynomial-time 72approximation with partition technique. Huang et al. [206, 265] discovered a new technique on partition, called double partition. With double partition, they gave a big improvement, that is, a polynomial-time (6 + ε)-approximation. Later, with some modification on double partition, Dai and Yu [163] designed a polynomial-time (5 + ε)-approximation, Zou et al. [655] and independently, [185] gave polynomialtime (4 + ε)-approximation, Willson et al. [552] obtained a polynomial-time 3.63-approximation. Ding et al. [169, 560] observed that all above approximations for the minimum weight dominating set problem can be extended to the minimum weight sensor cover problem, so that they found polynomial-time constant-approximation for the minimum weight sensor cover problem and the maximum lifetime coverage problem. Finally, Li and Jin [326] ended the story by giving a PTAS for the minimum weight sensor cover problem and the maximum lifetime coverage problem. In this chapter, we will discuss above works for dealing with coverage instead of dominating set. For simplicity, we assume in this chapter that the studied sensors are homogeneous with Rs = 1, and all targets and all sensors lie in the Euclidean plane. Actually, in 3-dimensional Euclidean space, corresponding problem is still open.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_6

87

88

6 Weighted Sensor Cover

Open Problem 7 Does the minimum weight sensor cover problem have a polynomial-time constant-approximation for all target points and all homogeneous sensors lying in the 3-dimensional Euclidean space? In this chapter, to simplify our statement, we will employ a sentence format “choose something from a pool of polynomial size to reach a goal.” For example, choose x ∈ X to minimize f (x). If X has a polynomial size and f (x) can be computed in polynomial-time, then this sentence can be easily implemented in polynomial-time. Please note that we will use the following general strategy a lot in this chapter. • In a pool P of polynomial size, we show that there exists someone x satisfying an inequality f (x) ≤ a. • Compute x ∗ = argminx∈P . Then f (x ∗ ) ≤ a. In this strategy, we do not care how to compute the x in the first step.

6.2 Partition: 28-Approximation The partition is a classical technique to design approximation algorithms [177]. In Section 2.4, this technique has been used to design a PTAS for the minimum sensor cover problem. One already saw that the approximation performance ratio and the running time have a tradeoff relationship. Actually, the running time of (1 + ε)2 approximation is nO(1/ε ) . As the ε goes to 0, the performance ratio goes to 1 and the running time increases rapidly. Meanwhile, the size of cells, O(1/ε) in the partition also increases quickly. The design of PTAS is based on the fact that the minimum sensor cover problem is polynomial-time solvable within any constant-size cell. When use the partition to design approximation for the minimum weight sensor cover problem, one meets a trouble that for a small constant-size cell, no polynomial-time algorithm for the optimal solution has been√found so far. The best known result is that within a square of edge size at most 2/2, there exists a polynomial-time 2-approximation. Let us first use this result to design a polynomialtime 28-approximation for the minimum weight sensor cover problem and then establish the 2-approximation on the small square. Initially, put all target points into a square. Then, the square is partitioned into √ √ 2 2 × cells by a grid. To have all cells nonoverlapping, assume that each cell has 2 2 open boundary on the right and on the top, and close on the left and on the bottom. To avoid any complication, we also assume that no target point lies on the boundary of a cell. The minimum weight sensor cover problem on a cell can be formally described as follows. Problem 6.2.1 √ (Minimum Weight Sensor Cover on a Cell) Consider a cell e with edge length 2/2. Given a set A of target points lying in e and a sensor cover S of

6.2 Partition: 28-Approximation

89

Fig. 6.1 For two cells at ends of a diagonal, at most one has its interior intersecting a disk of radius one

sensors with nonnegative weight c : S → R+ , find a sensor cover with minimum total weight. The following result will be proved later. Lemma 6.2.2 There is a polynomial-time 2-approximation for the problem of √ minimum weight sensor cover on a cell with side length 2/2. Using this lemma, we design a 28-approximation for the minimum weight sensor cover problem. Consider Fig. 6.1. All cells intersecting a vertical straight line form a vertical strip and all cells intersecting a horizontal straight line form a horizontal strip. The sensing area of a sensor s is a disk with center s and radius 1, denoted by disk(s), which intersects at most four horizontal strips and at most four vertical strips. Hence, it intersects at most 16 cells. Furthermore, consider two cells at two ends of a diagonal. Only one of them intersects disk(s) (Fig. 6.1). Therefore, disk(s) intersects at most 14 cells. Theorem 6.2.3 There is a polynomial-time 28-approximation for the minimum weight sensor cover problem. Proof Let A(e) be a 2-approximation for the minimum sensor cover for target points lying in cell e. Let Opt be a minimum sensor cover for all target points. Let Opt (e) be the set of sensors, in Opt, which covers at least one target point in cell e. Then each sensor s ∈ Opt appears in Opt (e) for at most 14 cells e. Note that c(A(e)) ≤ 2 · c(Opt (e)). Therefore, we have c(∪e∈B A(e)) ≤

 e∈B

where B is the set of all cells.

c(A(e)) ≤



2 · c(Opt (e)) ≤ 28c(Opt),

e∈B

 

90

6 Weighted Sensor Cover

Fig. 6.2 Sensor Cover with target points lying in a strip

The proof of Lemma 6.2.2 is based on the minimum weight sensor cover problem on a (vertical or horizontal) strip and a partition of target point in the cell e into two parts such that for each part there exists a polynomial-time algorithm for computing the minimum weight sensor cover. We first study the problem on a strip. Problem 6.2.4 (Minimum Weight Sensor Cover on a Strip) Consider a set A of m target points lying in a horizontal strip and a set S of n sensors lying outside of the strip with weight c :→ R+ (Fig. 6.2). Assume that S is a sensor cover for A. Find a minimum weight sensor cover for all target points in A. Let S + = {s ∈ S | s lies above the strip} ∪ {+∞} and S − = {s ∈ S | s lies below the strip} ∪ {−∞}. For simplicity, we will treat +∞ and −∞ as two sensors. Assume c(+∞) = c(−∞) = 0. Denote by disk(+∞) and circle(+∞) the half plane above the strip and the upper boundary of the strip, respectively. Similarly, denote by disk(−∞) and circle(−∞) the half plane below the strip and the lower boundary of the strip, respectively. Consider a sensor s ∈ S + such that circle(s) intersects a vertical line L where circle(s) is the boundary of sensing disk of sensor s. A sensor s  ∈ S + is said to be controlled by s at L, denoted by s  ≺+ L s, if one of the following occurs (Fig. 6.3): • s  does not intersect L. • The lower intersection point of circle(s  )∩L is higher than the lower intersection point of circle(s) ∩ L. • The lower intersection point of circles (s  ) ∩ L is identical to the lower intersection point of circles ∩ L. But, s  is on the right of s. (If s = +∞, then s  is on the right of L.) Similarly, consider a sensor s ∈ S − such that circle(s) intersects a line L. Then a sensor s  ∈ S − is said to be controlled by s, denoted by s  ≺− L s, if one of the following occurs:

6.2 Partition: 28-Approximation

91

Fig. 6.3 s  is controlled by s at line L

• s  does not intersect L. • The upper intersection point of circle(s  )∩L is lower than the upper intersection point of circle(s) ∩ L. • The upper intersection point of circle(s  )∩L is identical to the upper intersection point of circle(s) ∩ L. But, s  is on the right of s. (If s = −∞, then s  is on the right of L.) The following properties follow immediately rom above definition.  Lemma 6.2.5 (Transitive) Let s, s  , s  ∈ S + and L a vertical line. If s  ≺+ L s and + + −  −     −   s ≺L s, then s ≺L s. Similarly, for s, s , s ∈ S , if s ≺L s and s ≺L s, then s  ≺− L s.

Lemma 6.2.6 Let s and s  be two sensors and L a vertical line. Suppose disk(s) ∩ L = ∅. Then +  (a1) if s, s  ∈ S + , then either s  ≺+ L s or s ≺L s ; −  −   (a2) if s, s ∈ S , then either s ≺L s or s ≺− L s .

Next, we prove one more property. Lemma 6.2.7 Let L and L be two vertical lines such that L lies on the right of L . Then the following holds: +   (b1) Let s, s  ∈ S + . If s  ≺+ L s and s ≺L s , then disk(s) ∩ Q(L ) ⊆    disk(s ) ∩ Q(L ) where Q(L ) is the closed lower-left quarter-plane bounded by the upper boundary of the strip and L . −     (b2) Let s, s  ∈ S − . If s  ≺− L s and s ≺L s , then disk(s) ∩ Q (L ) ⊆ disk(s ) ∩     Q (L ) where Q (L ) is the closed upper-left quarter-plane bounded by the lower boundary of the strip and L .

Proof For contradiction, suppose (b1) is not true, that is, there exists a point p ∈ (disk(s)∩Q(L ))\(disk(s  )∩Q(L )). Let L be the vertical line passing through p.   Then s  ≺+ L s. Then L is on the left of L . Let A be the lower intersection point of  circle(s) ∩ L and A the lower intersection point of circle(s) ∩ L . Then disk(s) ∩ L = ∅. Let A be the lower intersection point of circle(s) ∩ L and B  the lower intersection point of circle(s  )∩L . Then A is above B  . They both should lie below

92

6 Weighted Sensor Cover

Fig. 6.4 The proof of Lemma 6.2.5

the line AA (Fig. 6.4). Therefore, disk(s  ) ∩ AA is located inside the segment [A, A ]. Let E and F be two endpoints of disk(s  ) ∩ AA . Clearly, EAF < A A A and |EF | ≤ |AA |. Therefore, radius(disk(s  )) =

|EF | |AA | < = radius(disk(s)), 2 sin EB  F 2 sin A A A

a contradiction. Similarly, (b2) is true.

 

Now, it is ready to show the following. Theorem 6.2.8 (Ambühl et al. [19]) There exists an algorithm with running time O(n4 m) for the minimum weight sensor cover problem on a strip. Proof First, assume every sensor in S has a positive weight since all sensors with zero weight can be removed from S together with target points covered by them at the beginning. Let p1 , .., pn be all target points in A in the ordering from left to right. A disk disk(s) is called a upper disk (lower disk) if s ∈ S + (s ∈ S − ). We design a dynamic programming to find an optimal solution. Let D = {disk(s) | s ∈ S ∪ {+∞, −∞}}. − + −   We extend ≺+ L and ≺L to disks in D, that is, D ≺L D (D ≺L D ) if D = disk(s), +  −    D = disk(s ) and s ≺L s (s ≺L s ). For an upper disk D and a lower disk D  with pi ∈ D ∪ D  , define by Ti (D, D  ) the one with the minimum total weight among disk subsets S  satisfying the following conditions: 1. S  covers p1 , . . . , pi . 2. D, D  ∈ S  .  3. Let Li be the vertical line passing through pi . Then D  ≺+ Li D for any D ∈ −  +     − S ∩ S and D ≺Li D for any D ∈ S ∩ S .

6.2 Partition: 28-Approximation

93

Since disk(+∞) and disk(−∞) have zero weight and cover nothing, for simplicity of the discussion, assume that they cannot appear in Ti (D, D  ) − {D, D  }. In other words, they can play only the role of D or D  . Let c(Ti (D, D  )) be the total weight of disks in Ti (D, D  ). We claim that the following recursive formula holds. c(Ti (D, D  )) = min [c(Ti−1 (D1 , D2 )) + c({D, D  } \ {D1 , D2 })], D1 ,D2

(6.1)

where upper disk D1 and lower disk D2 are overall possible pairs satisfying conditions: (c1) pi−1 ∈ D1 ∪ D2 . −  (c2) Let Li be the vertical line passing through pi . Then D1 ≺+ Li D and D2 ≺Li D . To show this claim, choose D1 to be the upper disk in Ti (D, D  ) which controls every upper disk in Ti (D, D  ) at Li−1 . By Lemma 6.2.5, such a choice must exist. Similarly, choose D2 to be the lower disk in Ti (D, D  ) which controls every lower disk in Ti (D, D  ). By Lemma 6.2.7, D ∩ Q(Li−1 ) ⊆ D1 ∩ Q(Li−1 ) and D  ∩ Q ⊆ D2 ∩ Q . Therefore, (Ti (D, D  ) − {D, D  }) ∪ {D1 , D2 } covers p1 , . . . , pi−1 . Hence, c(Ti (D, D  )) − c({D, D  } \ {D1 , D2 }) ≥ c(Ti−1 (D1 , D2 )), that is, c(Ti (D, D  )) ≥ min [c(Ti−1 (D1 , D2 )) + c({D, D  } \ {D1 , D2 })]. D1 ,D2

On the other hand, for any pair {D1 , D2 } satisfying (c1) and (c2), Ti−1 (D1 , D2 ) ∪ {D, D  } covers p1 , . . . , pi . Moreover, for any upper disk Dˆ in Ti−1 (D1 , D2 ), one + ˆ ˆ must have Dˆ ≺+ Li D. Indeed, for contradiction, suppose D ≺Li D. Then D = D1 + ˆ Note that by and hence Dˆ ∈ D has a positive weight. By Lemma 6.2.5, D1 ≺ D. Li

ˆ the definition of Ti−1 (D1 , D2 ), Dˆ ≺+ Li−1 D1 . By Lemma 6.2.7, then D ∩Q(Li−1 ) ⊆ ˆ D1 ∩Q(Li−1 ), which means that D can be deleted from Ti−1 (D1 , D2 ), contradicting the minimality of Ti−1 (D1 , D2 ).  Similarly, for any lower disk Dˆ in Ti−1 (D1 , D2 ), one must have Dˆ ≺− Li D . Therefore, c(Ti (D, D  )) ≤ Ti−1 (D1 , D2 ) + c({D, D  } \ {D1 , D2 }) for any pair {D1 , D2 } satisfying (c1) and (c2). Therefore, c(Ti (D, D  )) ≤ min [Ti−1 (D1 , D2 ) + c({D, D  } \ {D1 , D2 })] D1 ,D2

for {D1 , D2 } over all pairs satisfying (c1) and (c2). Hence, the equation (6.1) holds.

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6 Weighted Sensor Cover

This recursive formula suggests a dynamic programming for computing an optimal solution for the minimum weight sensor cover on a strip as follows. Dynamic Programming for the minimum weight sensor cover on a strip input:a set A of targets lying a horizontal strip H and a sensor cover S with all sensor lying outside of H ; choose{D, D  } from all pairs satisfying (c1) and (c2) to minimize c(Tn (D, D  )); output: Tn (D, D  ). function c(Tn (D, D  )) begin c(T1 (D, D  )) = c(D) + c(D  ); while i > 1 do c(Ti (D, D  )) = minD1 ,D2 [c(Ti−1 (D1 , D2 )) + c({D, D  } \ {D1 , D2 })] for {D1 , D2 } over all pairs satisfying (c1) and (c2); end. This algorithm runs in time O(n4 m) because there are O(n2 m) possibilities for Ti (D, D  ) and each c(Ti (D, D  )) can be computed recursively in time O(n2 ).   Let A(e) be the set of target points lying inside of the cell e and S(e) a sensor cover for A(e) such that every sensor covers at least one target point in cell e. We will define two subsets of S(e), Sh and Sv , and then partition A(e) into two parts Ah and Av satisfying the following conditions. • S(e) \ e = Sh ∪ Sv , • Ah and Sh form an instance for the problem of minimum weight sensor cover on a horizontal strip. • Av and Sv form an instance for the problem of minimum weight sensor cover on a vertical strip. Let A, B, C, D be four vertices of e. Divide outside of e into eight areas N E (Northeastern), NC (North-Central), NW (Northwestern), ME (Middle-East), MW (Middle-West), SE (Southeastern), SC (South-Central), SW (Southwestern) as shown in Fig. 6.5. Let N = NE ∪ NC ∪ N W , S = SE ∪ SC ∪ SW , E = N E ∪ ME ∪ SE, and W = NW ∪ MW ∪ SW . Define Sv = (N ∪ S) ∩ S and Sh = (E ∪ W ) ∩ S. Fig. 6.5 Outside of e is divided into eight areas

6.2 Partition: 28-Approximation

95

Fig. 6.6 south (p), north (p), west (p), and east (p) Fig. 6.7 The proof of Lemma 6.2.10

Before partition A(e), let us first establish some properties of the optimal solution Opt (e) for the problem of minimum weight sensor cover on cell e. Lemma 6.2.9 If Opt (e) contains a sensor s lying in e, then Opt (e) = {s} and w(s) = mins  ∈e w(s  ). Proof This is because any sensor in e is able to cover all target points in e. Therefore, Opt (e) does not need to contain any sensor lying outside of e.   For any target point p ∈ A(e), let p be a right angle at p such that two edges intersect horizontal line AB each at an angle of π/4. Let south (p) denote the intersection of e and p. Similarly, we can define north (p), east (p), and west (p) as shown in Fig. 6.6. Lemma 6.2.10 If p is covered by a sensor u in area SC, then every target point in south (p) can be covered by u. The similar statement holds for ME and east (p), MW and west (p), and NC and north (p), respectively. Proof Note that south (p) is a convex polygon. It is sufficient to show that the distance from u to every corner of south (p) is at most one. Suppose v is a corner of south (p) on BC (Fig. 6.7). Draw a line L perpendicular to pv and equally divide line segment pv. If u is below L , then d(u, v) ≤ d(u, p) ≤ 1. If u is above L , then uvp < π/2 and hence uvC < 3π/4. Hence, d(u, v) < μ/ cos π/4 = 1. A similar argument can be applied in the case that the vertex v of south (p) is on DA or on AB.   Consider two target points p, p ∈ A(e). Suppose p is on the left of p . Extend the left edge of p and the right edge of p to intersect at point p . Define south (p, p ) to be the intersection of cell e and p (Fig. 6.8). Similarly, we can define north (p.p  ).

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Fig. 6.8 south (p, p  )

Lemma 6.2.11 Let K be a sensor cover for A(e). Suppose p, p ∈ A(e) are covered by some sensors in K ∩ SC, but neither p nor p is covered by any sensor in K ∩ (ME ∪ MW ). Then every target point in south (p, p ) can be covered by sensor in K ∩ (N ∪ S) where N = NE ∪ NC ∪ NW and S = SE ∪ SC ∪ SW . Proof By Lemma 6.2.10, it is sufficient to consider a target point u lying in south (p, p ) \ ( south (p) ∪ south (p ). For contradiction, suppose u is covered by a sensor v in K ∩ (ME ∪ MW ). If v ∈ ME, then east (v) contains p and by Lemma 6.2.10, p is covered by v, a contradiction. A similar contradiction can result from v ∈ MW .   Now, it is ready to give a property of Opt (e) if Opt (e) does not contain any sensor lying in cell e. Lemma 6.2.12 Let Opt (e) be a minimum weight sensor cover for A(e). Suppose no sensor in Opt (e) lies in cell e. Then there exist four nodes p, p , q, q  ∈ A(e) such that Ah = A(e) ∩ ( south (p, p ) ∪ north (q, q  )) is covered by Opth (e) = Opt (e) ∩ (N ∪ S) and Av = A(e) − A1 (e) is covered by Optv (e) = Opt (e) ∩ (E ∪ W ). Proof Let AS be the set of target points in A(e), each of which can be covered by a sensor in SC but not covered by any sensor in ME ∪ MW . Let p be the target point in A(S) such that the left edge of (p) is on the leftmost position among all left edges of (v) for v ∈ VS . Let p be the node in AS such that the right edge of (p ) is on the rightmost position among all right edges of south (v) for v ∈ AS . Clearly, south (p, p ) has the following properties: (p1) Every target point in south (p, p ) can be covered by Opth (e). (p2) AS ⊂ south (p, p ). Similarly, let AN be the set of target points in A(e), each of which can be covered by a sensor in SC but not covered by any sensor in ME ∪ MW . One can find nodes q, q  ∈ AN to meet the following requirement. (q1) Every target point in north (q, q  ) can be covered by Opth (e). (q2) AN ⊂ north (p, p ). It follows from (p1) and (q1) that Ah (e) is covered by Opth (e). It follows from (p2) and (q2) that Av (e) is covered by Optv (e).  

6.2 Partition: 28-Approximation

97

Based on Lemma 6.2.12, we may design a 2-approximation for the minimum weight sensor cover problem on the cell e as follows. 2-Approximation Algorithm for the minimum weight sensor cover in a cell e inputa set A of target points lying in cell e and a sensor cover S for A. Sin ← nil; % define c(nil) = +∞ % Sout ← nil; if S ∩ e = ∅ then u = argminv∈S ∩e c(v) and Sin ← {u}; Sh ← S ∩ (N ∪ S); Sv ← S ∩ (E ∪ W ); if Sh ∪ Sv is a sensor cover then choose {p, p , q, q  } ⊂ A to minimize c(S(p, p , q, q  )) and Sout ← S(p, p , q, q  ); if c(Sin ) ≤ c(Sout ) then S(e) ← Sin else S(e) ← Sout ; output S(e). function S(p, p , q, q  ) begin Ah ← A ∩ ( south (p, p ) ∪ north (q, q  )); Av ← A \ Ah ; find the minimum weight subset Oh of Sh , covering Ah ; find the minimum weight subset Ov of Sv , covering Av ; S(p, p , q, q  ) ← Oh ∪ Ov ; end-function. Clearly, if Opt (e) ∩ e = ∅, then S(e) = Opt (e). If Opt (e) ∩ e = ∅, then for {p, p , q, q  } in Lemma 6.2.12, one has c(Oh ) ≤ c(Opth (e)), c(Ov ) ≤ c(Optv (e)). Therefore, c(A(e)) ≤ c(Oh ∪ Ov ) ≤ c(Opth (e)) + c(Optv (e)) ≤ 2c(Opt (e)). By Theorem 6.2.8, Oh and Ov can be computed in polynomial-time. Therefore, Lemma 6.2.2 is proved. The following lemma will be used frequently later, which can be obtained from above discussion.

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Lemma 6.2.13 Consider a cell e. Let H and V be a horizonal strip and a vertical strip, respectively, such that H ∩ V = e. Let A(e) be a set of targets lying in e and S a sensor cover for A(e). Assume all sensors in S lie outside of e. Then A(e) can be partitioned into two parts Ah (e) and Av (e) such that • Ah (e) is covered by Sh (e) = S \ H , and • Av (e) is covered by Sv (e) = S \ V .

6.3 Double Partition: 6-Approximation Now, we introduce the technique of double partition. Initially, we put all target points into a square. In the first partition, the √ square is partitioned into blocks; each block is a small square with edge length  2/2. In the √ second partition, each block is partitioned into smaller cells with edge length 2/2. The advantage of double partition is on the second one. When  is fixed, each block can be seen to contain a constant-number of cells so that different types of combinations of cells can be enumerated in polynomial-time. By taking this advantage, the minimum weight sensor cover for target points lying in a block may have a better approximation. Suppose that there is a ρ-approximation with running 2 time nO( ) for the problem of minimum weight sensor cover on a block as follows. Problem 6.3.1 (Minimum Weight Sensor Cover on a Block) Consider a block √ B with edge length  2/2 where  is a positive constant integer. Given a set A of target points lying in B and a sensor cover S of sensors with nonnegative weight c : S → R+ , find a sensor cover with minimum total weight. We now show that this ratio ρ can be almost extended to the minimum weight sensor cover problem in general case. Theorem 6.3.2 (Huang et al. [206, 265]) Suppose there exists a polynomial-time ρ-approximation for the problem of minimum weight sensor cover on a block with 2 running time nO( ) . Then for any ε > 0, there exists a polynomial-time (ρ + ε)2 approximation with computation time nO(1/ε ) for the minimum weight sensor cover problem. Proof Choose  = 12 max(1, 1/ε ). Put input target points √ into a grid with each block being an μ × μ square (Fig. 6.9) where μ = 2/2. To make all blocks disjoint, we define each block with boundary open on the left and on the top, but close on the right and on the bottom. For each nonempty block B, compute ρ-approximation for the problem of minimum weight sensor cover on block B. Union those ρ-approximation solutions for all nonempty blocks and denote this union by S(P ) for the partition P induced by this grid. Now, shaft this grid in diagonal direction with distance 4 for /4 times. This would result in /4 partitions P1 , . . . , P/4 . Choose S = S(Pi ) to be the one with

6.3 Double Partition: 6-Approximation

99

Fig. 6.9 √ Double partition (μ = 2/2)

the minimum weight among S(P1 ), . . . , S(P/4 ). Now, we claim that c(S(Pi )) ≤ (ρ + ε)c(Optwsc ) where Optwsc is the minimum weight sensor cover for all target points. For each s ∈ Optwsc , the disk disk(s) may intersect more than one blocks of Pi . Let ζi (s) be the number of blocks in partition Pi , intersecting disk(s). Let O(B) = {s ∈ Optwsc | disk(s) ∩ B = ∅}. Then O(B) is a sensor cover for target points lying in block B. Therefore,   c(O(B)) = c(Optwsc ) + (ζi (s) − 1)c(s). c(S(Pi )) ≤ B∈Pi

s∈Optwsc

Note that disk(s) can intersect at most one horizontal cutline and at most one vertical cutline of Pi for all i. Therefore, ζi (s) has only three possible values. ζi (s) = 1 if disk(s) does not intersect any cutline of Pi , ζi (s) = 2 if disk(s) intersects exactly one cutline of Pi , and ζi (s) = 4 if disk(s) intersects two cutlines of Pi , one horizontal cutline and one vertical cutline. Moreover, √ two vertical cutlines, possibly from two different partition, have distance 2 2 > 2. Thus, overall partitions, each disk disk(s) for s ∈ Optwsc intersects at most one vertical cutline and similarly at most one horizontal cutline. This means that for every s ∈ Optwsc , /4 

(ζi (s) − 1) ≤ 3.

i=1

Therefore, c(S) = min c(S(Pi )) 1≤i≤/4

/4 1  ≤ c(O(B)) /4 i=1 B∈Pi

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6 Weighted Sensor Cover

= c(Optwsc ) +

4 



/4  (ζi (s) − 1)c(s)

s∈Optwsc i=1

12 c(Optwsc )  ≤ (1 + ε)c(Optwsc ). ≤ c(Optwsc ) +

 

Next, we take advantage of the double partition to design a polynomial-time (6 + ε)-approximation for the minimum weight sensor cover problem. By Theorem 6.3.2, it suffices to obtain a polynomial-time 6-approximation for the problem of minimum weight sensor cover on a block B with a constant-size μ × μ. Let S(e) be a sensor cover for A(e), the set of all target points lying in cell e, which is obtained by the 2-Approximation Algorithm for the minimum weight sensor cover on cell e. Let C be the set of those cells e such that A ∩ e = ∅. For each e ∈ C, let (Ah (e), Av (e)) be the partition of A(e) as described in Lemma 6.2.13. Since  is fixed for a fixed ε, we are able to combine Ah (e) along a horizontal strip and combine Av (e) along a vertical strip. With this idea, the approximation performance ratio can be reduced from 28 to 6 on a block as follows. Please note that we define c(nil) = +∞. 6-Approximation for the minimum weight sensor cover on a block B. input a set A of targets lying in block B and a sensor cover S for A; Let C be the set of cells e in B with A(e) = A ∩ e = ∅ and S ∩ e = ∅. Let H1 , . . . , H be horizontal strips and Y1 , . . . , Y vertical strips of B. choose C  ⊆ C and for every e ∈ C  choose se ∈ S ∩ e to minimize c(S(C  , Sin )) where Sin = {se | e ∈ C  }; S(B) ← S(C  , Sin ); output S(B). function S(C  , Sin ) begin A ← A \ ∪s∈Sin disk(s); if ∃e ∈ C − C  : S \ e cannot cover A(e) then S(C  , Sin ) ← nil else S(C  , Sin ) ← Sin ∪ Sout (C  ); end-function function Sout (C  ) begin C  ← C − C  ; for every Hi do begin S(Hi ) ← S \ Hi ; A(Hi ) ← ∪e∈Hi ∩C  Ah (e); compute the minimum weight Opt (Hi ) ⊆ S(Hi ) to cover A(Hi );

6.3 Double Partition: 6-Approximation

101

end-for; for every Vi do begin S(Vi ) ← S \ Vi ; A(Vi ) ← ∪e∈Vi ∩C  Av (e); compute the minimum weight Opt (Vi ) ⊆ S(Vi ) to cover A(Vi ); end-for; Sout (C  ) ← (∪i=1 Opt (Hi )) ∪ (∪i=1 Opt (Vi )); end-function Theorem 6.3.3 (Huang et al. [206, 265]) There exists a polynomial-time 6approximation for the problem of minimum weight sensor cover on block B, with 2 running time nO( ) where n is the number of sensors s such that disk1 (s) ∩ B = ∅ and 2 is the number of cells in block B. 2

Proof First, let us estimate the time for computing S(B). There are O(2 ) possible 2 choices for C and nO( ) possible choices for Sin . Moreover, by Lemma 6.2.13, 2 there are O(n4 ) possible combinations of partitions (Ah (e), Av (e) for all cells in  C − C . For each combination, computing all Opt (Hi ) and all Opt (Yi ) needs time 2 O(nm4 ). Therefore, computing function Sout (C  ) needs time nO( ) mO() where n is the number of sensors and m is the number of targets. Next, we estimate the performance ratio. Let Opt be an optimal solution for the problem of minimum weight sensor cover on block B. Set C  = {e ∈ C | e ∩ Opt = ∅}. Arbitrarily, select se ∈ Opt ∩ e for e ∈ C  . Set Sin = {se | e ∈ C  }. Update A by A ← A \ ∪s∈Sin disk(s). Note that S(C  , Sin ) = Sin ∪ Sout (C  ) and Sout (C  ) = (∪i=1 Opt (Hi )) ∪ (∪i=1 Opt (Vi ). Let O(Hi ) be the set of sensors in {s ∈ (Opt − Sin ) not lying in Hi , each of which covers a target in Hi , not covered by Sin . Then O(Hi ) is a feasible solution for the problem of minimum weight sensor cover on strip Hi with instance (S(Hi ), A(Hi )) in the algorithm. Hence c(Opt (Hi )) ≤ c(O(Hi )). Similarly, Let O(Vi ) be the set of sensors in {s ∈ (Opt − Sin ) not lying in Vi , each of which covers a target in Vi , not covered by Sin . Then c(Opt (Vi )) ≤ c(O(Vi )).

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6 Weighted Sensor Cover

Fig. 6.10 Each disk(v) intersects at most six strips not containing v

Since each disk disk(s) can intersect at most six strips which do not contain s (Fig. 6.10), we have c(S(B)) ≤ c(S(C  , Sin )) ≤ c(Sin ) +

 

c(Opt (Hi )) +

i=1

 

c(Opt (Vi ))

i=1

≤ c(Sin ) + 6 · c(Opt − Sin ) ≤ 6 · c(Opt).

 

6.4 Jigsaw Puzzle: 4-Approximation The design of 6-approximation in Section 6.3 likes playing a jigsaw puzzle. Ah (e) and Av (e) were used for tilling some strips and then take advantage of the polynomial-time solution for the problem of minimum weight sensor cover on a strip. Except strips, could we find other interesting ways to combine Ah (e) and A(e)? The answer is yes. Actually, this is a road going to the improvement of the approximation performance. Dai and Yu [163], Zou et al. [655], and Erlebach and Mihalak [185] all go along this road. In this section, we introduce the work of Zou et al. [655], which is based on the following polynomial-time solvable problem. Problem 6.4.1 (Minimum Weight Chromatic Sensor Cover) Consider  parallel √ horizontal strips H1 , . . . , H with width 2/2 as shown in Fig. 6.11. To have all strips disjoint, assume that each strip has open boundary on the top and close boundary on the bottom. Given a set A of targets points lying in those strips, a set R of red sensors, a set B of blue sensors, and a positive weight function c : R ∪ B → R + , find the minimum weight subset of red sensors and blue sensors such that every target in strip Hi is covered by a chromatic sensor, i.e., by either a red sensor lying above Hi or a blue sensor lying below Hi .

6.4 Jigsaw Puzzle: 4-Approximation

103

Fig. 6.11 Chromatic sensor cover

From definition of the minimum weight chromatic sensor cover problem, we may assume that every red sensor s covers at least one target point lying in a strip below s and every blue sensor s covers at least one target point lying in a strip above s. When  = 1, the minimum weight chromatic sensor cover problem is exactly the problem of minimum weight sensor cover on a strip. The following theorem is a generalization of Theorem 6.2.8. Theorem 6.4.2 (Zou et al. [655]) The minimum weight chromatic sensor cover problem can be solved in time O(nm4 ) where n = |A| and m = |R ∪ B|. Proof Denote Ri = R ∩ Hi−1 for i = 2, . . . ,  and R1 = {s ∈ R | s lies above H1 }. Denote Bi = B ∩ Hi+1 for i = 1, . . . ,  − 1 and B = {s ∈ B | s lies below H }. For simplicity, we introduce 2 dummy sensors +∞i and −∞i for i = 1, . . . ,  all with zero weight and define by disk(+∞i ) the half plane above strip Hi and by disk(−∞i the half plane below strip Hi . Moreover, denote R+ i = {disk(s) | s ∈ R} ∪ {disk(+∞i )} and Bi+ = {disk(s) | s ∈ B} ∪ {disk(−∞i )}. The proof of this theorem is similar to that of Theorem 6.2.8. Therefore, we employ the concept that a disk D is “controlled” by another disk D  at a line L, and −   notations D ≺+ L D and D ≺L D in the proof of Theorem 6.2.8. Consider a 2-dimensional vectors D in + + + P = R+ 1 × · · · × R × B 1 × · · · × B  .

For simplicity, D is also used to denote the set of components of D. Denote by Di + the ith component of D. Then we have Di ∈ R+ i and D+i ∈ Bi for 1 ≤ i ≤ .   Consider D, D ∈ P. For any vertical line L, D is said to be controlled by D, written −  as D ≺L D, if for 1 ≤ i ≤ , Di ≺+ L Di and D+i ≺L D+i . Let L1 , .., Lm be all vertical lines passing through target points in A in the ordering from left to right. Without loss of generality, assume all L1 , .., Lm are distinct. Let Aj be the subset of targets lying on Lj or on the left of Lj . For any D ∈ P, if D does not cover Aj − Aj −1 , then define Tj (D) = nil if there does not exist a disk subset P  satisfying the following conditions: • P  is a chromatic disk cover for Aj . • D ⊆ P .

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6 Weighted Sensor Cover

• For any disk D  ∈ P  ∩ Ri , D  ≺+ Lj Di .

• For any disk D  ∈ P  ∩ Bi , D  ≺− Lj Dm+i .

If such a disk subset P  exists, then define Tj (D) to be the one with the minimum total weight. Since all dummy disks have zero weight and cover nothing, for simplicity of the discussion, one assumes that they cannot appear in Tj (D) \ D. Now, we claim that for D covering Aj − Aj −1 , the following recursive formula holds. c(Tj (D)) = min {c(Tj −1 (D )) + c(D \ D )}.

(6.2)

D ≺Lj D

To show this claim, we first select Di to be the disk in Tj (P) ∩ R+ j which controls every disk in Tj (D) ∩ R+ at L . By Lemma 6.2.5, such a choice must exist. j −1 j  Similarly, one can choose D+i to the disk in Tj (D) which controls every disk in Tj (D) ∩ Bi+ at line Lj −1 . Define D = (Di , 1 ≤ i ≤ 2). By Lemma 6.2.7, Di ∩ Qi (Lj −1 ) ⊆ Di ∩ Qi (Lj −1 ) for 1 ≤ i ≤  where Qi (Lj ) is the close lower-left quarter-plane bounded by Lj −1 and the upper boundary of Hi .  Similarly, D+i ∩ Qi (Lj −1 ) ⊆ D+i ∩ Qi (Lj −1 ) for 1 ≤ i ≤ . This means that if  Tj (D) = nil, then (Tj (D) − (D \ D )) is a chromatic disk cover for Aj −1 . Hence, c(Tj (D)) − c(D \ D ) ≥ c(Tj −1 (D )), that is, c(Tj (D)) ≥ min (c(Tj −1 (D )) + c(D \ D )).

(6.3)

D ≺Lj D

If Tj (D) = nil, then c(Tj (D)) = ∞ and hence (6.3) holds trivially. On the other hand, for any D ≺ D, if Tj −1 (D ) = nil, then c(Tj −1 (D )) = ∞ > c(Tj (D)). Next, assume that Tj −1 (D ) = nil. Then Tj −1 (D ) ∪ D is a chromatic cover of Aj . + ˆ Moreover, for any disk Dˆ in Tj −1 (D ), we must have Dˆ ≺+ Lj Di if D ∈ Ri and Dˆ ≺− D+i if Dˆ ∈ B + . In fact, for contradiction, suppose Dˆ ∈ R+ and Dˆ is not Lj

i

i

ˆ > 0. Moreover, by Lemma 6.2.6, controlled by Di . Thus, Dˆ ∈ D and hence c(D) + +  ˆ ˆ Di ≺Lj D. By Lemma 6.2.5, Di ≺Lj D. By Lemma 6.2.7, Dˆ ∩ Qi (Lj −1 ) ⊆ Di ∩ Qi (Lj −1 ). This means that Dˆ can be deleted from Tj −1 (D ), contradicting the minimality of Tj −1 (D ). Similarly, it is also impossible that Dˆ ∈ Bi+ and Dˆ is not controlled by D+i . From above argument, one can see that Tj −1 (D ) ∪ D satisfies all conditions for above D . Therefore,

6.4 Jigsaw Puzzle: 4-Approximation

105

c(Tj (D)) ≤ c(Tj −1 (D ∪ D)) = c(Tj −1 (D )) + c(D \ D ) for all D ≺Lj D. This completes the proof of (6.2). The recursion (6.2) suggests a dynamic program for computing all Tj (D). There are O(nm2 ) Tj (D)’s and each needs to be computed recursively in time O(m2 ). Therefore, this dynamic program runs in time O(nm4 ). Finally, the minimum weight of subset of disks covering all targets can be computed from minD∈P c(Tm (D)), which requires O(m2 ) time.   Based on above theorem, we can design the following approximation algorithm. Define c(nil) = +∞. 4-Approximation for the minimum weight sensor cover on a block B. input a set A of targets lying in block B and a sensor cover S for A; Let C be the set of cells e in B with A(e) = A ∩ e = ∅ and S ∩ e = ∅. Let H1 , . . . , H be horizontal strips and Y1 , . . . , Y vertical strips of B. choose C  ⊆ C and for every e ∈ C  choose se ∈ S ∩ e to minimize c(S(C  , Sin )) where Sin = {se | e ∈ C  }; S(B) ← S(C  , Sin ); output S(B). function S(C  , Sin ) begin A ← A \ ∪s∈Sin disk(s); if ∃e ∈ C − C  : S \ e cannot cover A(e) then S(C  , Sin ) ← nil; else S(C  , Sin ) ← Sin ∪ Sout (C  ); end-function function Sout (C  ) begin C  ← C − C  ; Ah ← ∪e∈C  Ah (e); Av ← ∪e∈C  Av (e); R ← {s r | s ∈ S − Sin }; B ← {s b | s ∈ S − Sin }; Compute the minimum weight chromatic cover Opth ⊆ R ∪ B for Ah with horizontal strips H1 , . . . , H ; Compute the minimum weight chromatic cover Optv ⊆ R ∪ B for Av with vertical strips V1 , . . . , V ; Sout (C  ) ← {s | s r ∈ Opth ∪ Optv or s b ∈ Opth ∪ Optv }; end-function We now analyze this algorithm.

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Theorem 6.4.3 (Zou et al. [655]) The above algorithm gives a 4-approximation for the problem of minimum weight sensor cover on block B, with running time 2 nO( ) mO() where n is the number of sensors and m is the number of targets. 2

Proof First, let us estimate the time for computing S(B). There are O(2 ) possible 2 subsets of C and nO( ) possible choices of Sin . Moreover, for each (C  , Sin ), the algorithm solves the minimum weight chromatic sensor cover problem twice, which 2 needs time O(nm4 ). Therefore, total computation time is nO( ) mO() . Next, we estimate the performance ratio. Let Opt be an optimal solution for the problem of minimum weight sensor cover on block B. Set C  = {e ∈ C | e ∩ Opt = ∅}. Arbitrarily, select se ∈ Opt ∩ e for e ∈ C  . Set Sin = {se | e ∈ C  }. Update A by A ← A \ ∪s∈Sin disk(s). Note that S(C  , Sin ) = Sin ∪ Sout (C  ) and c(Sout (C  ))| ≤ c(Opth ) + c(Optv ). Let O r = {s r | s ∈ Opt − Sin } and O b = {s b | s ∈ Opt − Sin }. Then O r ∪ O b is a feasible solution for the minimum weight chromatic sensor cover problem with target sets Ah with horizontal strips H1 , . . . , H , and also a feasible solution for the minimum weight chromatic sensor cover problem with target sets Av with vertical strips V1 , . . . , V . Therefore, c(Opth ) ≤ c(O r ) + c(O b ) = 2 · c(Opt − Sin ), and c(Optv ) ≤ c(O r ) + c(O b ) = 2 · c(Opt − Sin ). Therefore c(S(B)) ≤ c(Sin ) + c(Sout (C  )) ≤ c(Sin ) + 4 · c(Opt − Sin ) ≤ 4 · c(Opt).  

6.5 Magic Transformation: 3.63-Approximation

107

6.5 Magic Transformation: 3.63-Approximation Independently, Erlebach and Mihalak [185] also gave a polynomial-time 4approximation for the problem of minimum weight sensor cover on a block. More important than the performance ratio improvement, their work contains a magic transformation, which leads to a further improvement, 3.63-approximation given by Willson [552]. In this section, we first introduce the work of Erlebach and Mihalak [185], and then present the modification given by Willson [552]. Erlebach and Mihalak [185] showed that the following subproblem is polynomial-time solvable. Problem 6.5.1 (Minimum Weight Sensor Cover on Multi-strips) As√shown in Fig. 6.12, consider  parallel horizontal strips H1 , . . . , H with width 2/2 in a block B. To have all strips disjoint, assume that each strip has open boundary on the top and close boundary on the bottom. Given a set S of n sensors with a positive weight function c : D → R + , and a set A of m targets points lying in H1 ∪ H3 ∪ · · · ∪ H2/2 −1 , find the minimum weight sensor cover for A such that every target in strip Hi is covered by a sensor lying outside of Hi . Erlebach and Mihalak [185] transformed this problem to a shortest path problem. To explain this transformation, we study an optimal solution Opt for this problem. Consider a strip Hi for some odd i, 1 ≤ i ≤ . A disk disk(s) for s ∈ Opt is called an upper disk with respect to Hi if s lies above Hi . For simplicity of discussion, we add dummy disks disk(+∞i ) and disk(−∞i ), which are the half plane above Hi and the half plane below Hi , respectively. All upper disks with respect to Hi form an upper area of Hi . The boundary of this area is called the upper envelope of Hi (Fig. 6.13). Similarly, we definite lower disks, the lower area, and the lower envelope of Hi . Fig. 6.12 Multi-strips

Fig. 6.13 The upper envelope

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Lemma 6.5.2 An upper disk disk(s) with respect to Hi appears in the upper envelope of Hi at most once. If an upper disk disk(s  ) is on the left of another upper disk disk(s) on the upper envelope of Hi , then the center s  is on the left of the center s. Similarly, a lower disk disk(s) with respect to Hi appears in the lower envelope of Hi at most once. If a lower disk disk(s  ) is on the left of another lower disk disk(s) on the lower envelope of Hi , then the center s  is on the left of the center s. Proof For contradiction, suppose the upper disk disk(s  ) is on the left of the upper disk disk(s) on the upper envelope of strip Hi , but the center s  is on the right of the center s. Choose a point x from circle(s  ) appearing on the upper envelope and a point y from circle(s) appearing on the upper envelope. Then segments s  x and sy intersect, say at z (Fig. 6.14). Since x and y appear in the upper envelope, one must have |s  x| < |sx| and |sy| < |s  y|. Therefore, |s  x| + |sy| < |sx| + |s  y|. However, |sx| < |sz| + |zx| and |s  y| < |s  z| + |zy|. Hence |sx| + |s  y| < |sz| + |zx| + |s  z| + |zy| = |s  x| + |sy|, a contradiction. Therefore, the second sentence is true. The first sentence is a corollary of the second sentence. In fact, if disk(s) appears twice on the upper envelope, then between two appearances, there must exist another disk disk(s) appearing. This means that s is on the left of s  and also on the right of s  , a contradiction.   A corner of the upper envelope of Hi is an intersection point of two upper disks on the envelope. A corner of the lower envelope of Hi is an intersection point of two lower disks on the lower envelope. A sweep line Li for Hi is a vertical line starts from a position on the left of the leftmost corner and moves to a position on the right of the rightmost corner. Li ’s movement is discrete. Each intermediate position of Li must pass through a corner on either the upper envelope or the lower envelope. Each of such positions is denoted by a quadruple (d1 , d2 ; d3 , d4 ) with either d1 = d2 or d3 = d4 where d1 , d2 are upper disks and d3 , d4 are lower disks. If d1 = d2 , then Li passes through the corner formed by d1 and d2 on the upper envelope. If d3 = d4 , then Li passes through the corner formed by d3 and d4 on the lower envelope. For the initial and the end position of Li , d1 = d2 is the upper dummy disk and d3 = d4 is the lower dummy disk. A move of Li is from its current position (d1 , d2 ; d3 , d4 ) to an adjacent position on the right. If d1 = d2 , this adjacent position is either (d2 , d; d4 , d4 ) or (d1 , d1 ; d4 , d). In this case, we say that the disk d3 leaves Li and the disk d enters. Fig. 6.14 The proof of Lemma 6.5.2

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109

Fig. 6.15 Four possibilities for a sweep line to move to right adjacent position

If d3 = d4 , then the right adjacent position is either (d2 , d; d4 , d4 ) or (d2 , d2 ; d4 , d). In this case, we say that the disk d1 leaves and the disk d enters (6.15). Note that every disk with center in Opt must appear in the upper or lower envelope for some strip Hi (otherwise, this disk can be removed). Hence, the weight of Opt equals the total weight of disks appearing on envelopes. (Here, for simplicity of speaking, we assume that the weight of a disk equals the weight of sensor at the center of the disk.) The sweep line can be used for calculating this total weight, so that we can turn the sweeping process into a shortest path of a graph. There is one trouble if the sweep line is doing counting individually. This trouble is generated from the fact that each disk may appear in the lower envelope of strip Hi and also in the upper envelope of Hi+2 . The consequence of this fact is that if every sweep line does individual counting, then the total weight is not exactly c(Opt). Hence, all sweep lines may be required to do a collaborated action and to employ a technique for avoiding the double counting. A configuration of sweep lines consists of positions of sweep lines L1 , L3 , . . ., L2m/2 −1 at a moment. A legal move from a configuration A to another configuration B consists of exactly one move of one sweep line which is required to satisfy the following constraints: • In this move, if a lower disk d leaves line Li , then the disk d should be already passed by Li+2 . Otherwise, Li has to wait for Li+2 to pass the disk d.

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• In this move, if an upper disk d leaves line Li , then the disk d should be already passed by Li−2 . Otherwise, Li has to wait for Li−2 to pass the disk d. Here, by a disk passed by a sweep line at position (d1 , d2 ; d3 , d4 ), we mean that if d is an upper disk, then the center of d is not on the right of the center of d2 ; if d is a lower disk, then the center of d is not on the right of the center of d4 . Above constraints would eliminate the double counting. This is because we can set counting rule as follows: The counting is performed only on a newly entered disk in each move. This means that when a disk d enters a configuration in a move, the weight of d is counted if the configuration does not contain d, and the weight of d is not counted if the configuration already contains d. Can this constraint stop the moving of configuration? The answer is no because if the moving stops, then every sweep line is waiting for another line to pass a disk. Then there are two cases. Case 1.

There are two sweep lines Li and Li+1 such that Li waits for Li+1 to pass a disk d and Li+2 waits for Li to pass another disk d  . In this case, Li should be in position (d1 , d2 ; d, d4 ) and Li+2 should be in position (d  , d2 ; d3 , d4 ). Since Li waits for Li+2 to pass d, the center of d is on the right of the center of d2 and hence by Lemma 6.5.2, the center of d is on the right of the center of d  . Similarly, the center of d  should be on the right of the center of d, a contradiction. Case 2. Case 1 does not occur. If Li waits for Li+2 , then Li+2 waits for Li+4 , etc. However, since the number of strips is finite, this process cannot go forever, a contradiction. If Li waits for Li−2 , then Li−2 waits for Li−4 . This process cannot go forever, neither. Next, we construct an auxiliary graph G(Opt) to turn the moving of configuration into a shortest path. Each vertex represents a possible configuration. There exists an arc (A, B) from configuration A to configuration B if and only if B can be reached from A through a legal move. The start vertex s is the configuration consisting of all sweep lines on the left of all disks in {disk(s) | s ∈ S}. The ending vertex is the configuration consisting of all sweep lines on the right of all disks in {disk(s) | s ∈ S}. Now, it is ready to show the following. Theorem 6.5.3 (Eriksson and Mihalak [185]) The problem of minimum weight sensor cover on multi-strips can be solved in time O(n6 ). Proof Use S instead of Opt to construct the sweep line positions, configurations, the legal move of configurations, and graph G(D) by following the same rules in the construction of graph G(Opt), except an additional requirement for a sweep line move: during the move, targets between two positions should be covered by two envelopes (Each sweep line position is determined by three disks. The envelope is determined by a move, not determined by S as shown in Fig. 6.16.). Then G(S)contains G(Opt) as a subgraph and the shortest path from configuration s to configuration t would give an optimal solution for the problem of minimum

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111

Fig. 6.16 The envelope is determined by a move

Fig. 6.17 If s is close to the central of Hi , then disk(v) involves only one problem

weight sensor cover on multi-strips. Since each sweep line position is determined by three disks, each sweep line has at most O(n3 ) positions. Hence, the number of configurations is at most O(n3 ). Therefore, computing the shortest path in G(S) takes time O(n6 ).   Based on Theorem 6.5.3, Willson et al. [567] constructed a polynomial-time 3.63-approximation for the problem of minimum weight sensor cover on a block B. Their main idea is motivated from the following observation: To construct an approximation solution, the problem on a block B can be divided into four problems, two on horizontal strips and two on vertical strips. Consider the two on horizontal strips. One is on odd strips H1 , H3 , . . . , H2/2 −1 . The other one is on even strips H2 , H4 , . . . , H2/2 . If the location of s is close to the central of Hi , then disk(s) can be used only for covering targets in Hi−1 and Hi+1 , i.e., it would involve only one problem. This means that in average, disk(s) involves less than two problems. How to take advantage of this average estimation? Shift the partition on the block B. This is a traditional technique (Fig. 6.17).

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3.63-Approximation for the minimum weight sensor cover on a block B. input √ a set A of targets lying in block B and a sensor cover S for A; μ ← 2/2; Set a coordinate system such that the left-lowest corner of block B has coordinate (0, 0). Let P (a, b) be a grid, which partitions block B into cells of size μ × μ, such that the left-lower corner has coordinate (−aμ/q, −bμ/q). choose a and b from {0, 1, . . . , q − 1} to minimize c(A(a, b); S(B) ← A(a, b); output S(B). function A(a, b) begin With grid P (a, b), partition block B into cells of size μ × μ; Let H1 , . . . , H be horizontal strips and Y1 , . . . , Y vertical strips of B. Let C be the set of cells e in B with A(e) = A ∩ e = ∅ and S ∩ e = ∅. choose C  ⊆ C and for every e ∈ C  choose se ∈ S ∩ e to minimize c(S(C  , Sin )) where Sin = {se | e ∈ C  }; A(a, b) ← S(C  , Sin ); end-function function S(C  , Sin ) begin A ← A \ ∪s∈Sin disk(s); if ∃e ∈ C − C  : S \ e cannot cover A(e) then S(C  , Sin ) ← nil; else S(C  , Sin ) ← Sin ∪ Sout (C  ); end-function function Sout (C  ) begin C  ← C − C  ; Hodd ← C  ∩ (H1 ∪ H3 ∪ · · · ∪ H2/2 −1 ); Ah,odd ← ∪e∈Hodd Ah (e); Heven ← C  ∩ (H2 ∪ H4 ∪ · · · ∪ H2/2 ); Ah,even ← ∪e∈Heven Ah (e); Vodd ← C  ∩ (V1 ∪ V3 ∪ · · · ∪ V2/2 −1 ); Av,odd ← ∪e∈Vodd Av (e); Veven ← C  ∩ (V2 ∪ V4 ∪ · · · ∪ V2/2 ); Av,even ← ∪e∈Veven Av (e); S  ← S − Sin ; Compute the minimum weight sensor cover Opth,odd ⊆ S  for Ah,odd ; Compute the minimum weight sensor cover Opth,even ⊆ S  for Ah,even ; Compute the minimum weight sensor cover Optv,odd ⊆ S  for Av,odd ;

6.5 Magic Transformation: 3.63-Approximation

113

Compute the minimum weight sensor cover Optv,even ⊆ S  for Av,even ; Sout (C  ) ← Opth,odd ∪ Opth,even ∪ Optv,odd ∪ Optv,even ; end-function Theorem 6.5.4 (Willson et al. [552]) Above algorithm is a polynomial-time 3.63approximation for the problem of minimum weight sensor cover on a block, with O(2 ) where n is the number of sensors, m is the number of targets, running √ time (mn) and  2/2 is the side length of the block. 2

Proof First, let us estimate the running time of the algorithm. There are O(2 ) 2 possibilities for C  , nO( ) possible choices of Sin for a fixed C  . Moreover, there are 2 O(m4 ) choices of Ah (e) and Av (e) for all cells in C  . For each choice, computing Opth,odd , Opth,even , Optv,odd , and Optv,even needs time O(n6 ). Therefore, total 2 computation time is (nm)O( ) . Next, we study the performance ratio. Let Opt be an optimal solution for the problem of minimum weight sensor cover on block B. Set C  = {e ∈ C | e ∩ Opt = ∅}. For each e ∈ C  , choose a node se ∈ Opt ∩ e. Set Sin = {se | e ∈ C  }. Update A by A ← A \ ∪s∈Sin disk(s). Note that S(C  , Sin ) = Sin ∪ Sout (C  ) and Sout (C  ) = Opth,odd ∪ Opth,even ∪ Optv,odd ∪ Optv,even . Let Oh,odd = {s ∈ Opt − Sin | disk(s) is involved to cover Ah,odd } Oh,even = {s ∈ Opt − Sin | disk(s) is involved to cover Ah,even } Ov,odd = {s ∈ Opt − Sin | disk(s) is involved to cover Av,odd } Ov,even = {s ∈ Opt − Sin | disk(s) is involved to cover Av,even }. Clearly, Oh,odd is a sensor cover for Ah,odd , Oh,even is a sensor cover for Ah,even , Ov,odd is a sensor cover for Av,odd , and Ov,even is a sensor cover for Av,even . Therefore, c(Opth,odd ) ≤ c(Oh,odd ),

(6.4)

c(Opth,even ) ≤ c(Oh,even ),

(6.5)

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c(Optv,odd ) ≤ c(Ov,odd ),

(6.6)

c(Optv,even ) ≤ c(Ov,even ).

(6.7)

For every s ∈ Opt − Sin , define  1 if disk(s) intersects three horizontal strips in partition P (a, b), τh (s; a, b) = 2 otherwise, and



τv (s; a, b) =

1 if disk(s) intersects three vertical strips in partition P (a, b), 2 otherwise.

Then disk(s) involves τh (s; a, b) of two inequalities (6.4) and (6.5), and τv (s; a, b) of two inequalities (6.6) and (6.7), Therefore c(Sout (C  )) ≤ c(Oh,odd ) + c(Oh,even ) + c(Ov,odd ) + c(Ov,even )  ≤ c(s)(τh (s; a, b) + τv (s : a, b)). s∈Opt−Sin

Hence, c(A(a, b)) ≤ c(Sin ) + c(Sout (C  ))  c(s)(τh (s; a, b) + τh (s; a, b)) ≤ c(Sin ) + ≤



s∈Opt−Sin

c(s)(τ1 (s; a, b) + τ2 (s : a, b)),

s∈Opt

where we define that for s ∈ Sin , τh (s; a, b) = τv (s; a, b) = 1. For any s ∈ Opt, note that for any fixed b, there exist at least  3μ−2 μ/q  values of a such that τh (s; a, b) = 1, and for any fixed a, there exist at least  3μ−2 μ/q  values of b such that τv (v; a, b) = 1. Therefore, c(S(B)) ≤

q−1 q−1 1   c(s)(τh (s; a, b) + τv (s; a, b)) q2 a=0 b=0 s∈Opt

√  (3 − 2 2)q  ≤ 4−2 · c(Opt). q √ (3−2 2)q q

√ √ goes to 3 − 2 2. Since 4 − 2(3 − 2 2) < 3.63, there must √     exist a fixed q such that 4 − 2 (3−2q 2)q < 3.63.

As q → ∞,

6.6 PTAS

115

6.6 PTAS After a sequence of efforts, a PTAS finally comes out. Li and Jin [326] showed the following. Theorem 6.6.1 (Li and Jin [326]) For any ε > 0, there exists a (1 + ε)approximation for the problem of minimum weight sensor cover on a block, with 9 running time O(n1/ε ). This is what we have been expecting for a long time. The highest outline of the algorithm is as follows. The algorithm first guesses whether an optimal √ solution OP T contains at most t0 disks, where t0 = L2 /ε2 is a constant (L =  2/2 is the side length of the block). If the answer is yes, then OP T can be found in time O(nt0 ). Suppose the answer is no, then the algorithm continues to guess a set G consisting of t0 heaviest disks in OP T . There are at most O(nt0 ) choices for G. Suppose the lightest disk in G has weight c0 . Reduce the instance to (A , S  ), where A is the set of target points which are not covered by G, and S  is the set of disks obtained from S by removing G as well as those disks whose weights are greater than c0 . Then c(OP T ) ≥ t0 c0 and every disk in S  has weight at most c0 . Next, the algorithm constructs a set H consisting of at most εt0 disks. To cover those remaining target points which are not covered by G ∪ H, an optimal set of disks Q can be obtained by a dynamic programming. Notice that OP T \ G covers all remaining target points. So, c(Q) ≤ c(OP T \ G) = c(OP T ) − c(G). Combining all the above, the total weight of the set of disks computed by the algorithm is at most c(G) + c(H) + c(Q) ≤ c(OP T ) + εt0 c0 ≤ (1 + ε)c(OP T ). The construction of H and the dynamic programming is very complicated. This book only introduces the basic idea behind them. To construct H, the algorithm adds a pair of farthest disks from each cell (such a pair is called a gadget of its cell). Since there are 2 cells in the block, the number of added disks is at most 22 = 4t0 ε2 (which is at most εt0 for ε ≤ 1/4). For the ease of understanding the idea, first consider a cell e. For the region covered by the union of disks whose centers fall into e, call the subregion which is not covered by the gadget disks of e as uncovered region. One has to cover those remaining target points in the uncovered region. Since gadget disks are chosen to be a farthest pair, the uncovered region is composed of two connected subregions (see Fig. 6.18). Intuitively, these two subregions are “strip like” and

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Fig. 6.18 An illustration of uncovered region for a cell. Blackened disks are gadget disks, and shaded region is uncovered region

thus a dynamic programming technique similar to the one in Section 5.2 can be used to find an optimal disk cover for them. However, when considering all cells, uncovered regions belonging to different cells interact with each other, breaking these uncovered regions into smaller pieces, which greatly increases the complexity of the dynamic programming. Furthermore, a suitable ordering of these pieces is needed by a dynamic programming, which requires the pieces to interact in a nice way. Hence a constant-number of disks are further added into H for the purpose of breaking badly interacting pieces, and more involved techniques are implemented. Interested readers may refer to [326]. Let us end this chapter by mentioning what open problems left. Open Problem 8 Is there a PTAS with a short proof for the minimum weight sensor cover problem? Open Problem 9 Is there a polynomial-time constant-approximation for the minimum weight connected sensor cover problem?

Chapter 7

k-Coverage

What would it be like to have not only color vision but culture vision, the ability to see the multiple worlds of others. Mary Catherine Bateson

7.1 Motivation and Overview Consider a set S of n homogeneous sensors and a set A of m targets. A subset S  of S is called a sensor k-cover if every target in A is covered by at least k sensors in S  . In this chapter, we study the following problem. Problem 7.1.1 (Maximum Lifetime k-Coverage) Given a set of sensors with unit lifetime and a set of targets in the Euclidean plane, find a sleep/wakeup schedule for sensors to maximize the lifetime of k-coverage, where the lifetime of coverage is a time period in every moment of which all active sensors form a sensor k-cover. The sensor k-cover satisfies a fault-tolerant requirement that the coverage can be kept in life when at most k − 1 sensors stop working. The k-coverage in wireless sensor networks has been studied extensively in the literature [1, 21– 24, 36, 37, 64, 73, 204, 205, 279, 311, 327, 346, 584, 585, 589, 602, 603, 617, 649]. However, the approximation algorithm with theoretical guaranteed performance for the maximum lifetime k-coverage problem was first given by Erlebach et al. [186]. They extended the claim of Berman et al. [66, 67] from the coverage to k-coverage to show that if the minimum weight sensor k-cover problem (see in Section 6.2) has a polynomial-time ρ-approximation, then for any ε > 0, the maximum lifetime k-coverage problem has a polynomial-time ρ(1 + ε)-approximation. They also presented a polynomial-time (6 + ε)-approximation for the maximum weight sensor 2-cover problem and hence obtained a polynomial-time (6 + ε)-approximation for the maximum lifetime 2-coverage problem. Willson et al. [554] and Zhang et al. [636] made a big progress by showing that there exist polynomial-time (4 + ε)approximation and (3 + ε)-approximation, respectively, for the maximum lifetime k-coverage problem and any ε > 0.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_7

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In this chapter, we would like to discuss those theoretical results given in [186, 636].

7.2 Reduction to Weighted Sensor k-Cover Consider the following weighted k-coverage problem. Problem 7.2.1 (Weighted Sensor k-Cover) Given a set A of target points and a set S of sensors with positive weight c : S → R + in the Euclidean plane, find a sensor k-cover with minimum total weight. Similar to the coverage case, the approximation design for the maximum lifetime k-coverage problem can be reduced to that for the minimum weight sensor cover problem as follows. Theorem 7.2.2 (Erlebach et al. [186]) If the weighted sensor k-cover problem has a polynomial-time ρ-approximation, then the maximum lifetime k-coverage problem has a polynomial-time (1 + ε)ρ-approximation for any ε > 0. Proof Let C be the collection of all sensor k-covers. For any sensor k-cover p, define  1 if s ∈ p, as,p = 0 otherwise. Then the maximum lifetime k-coverage problem can be formulated as the following linear programming: max



(7.1)

xp

p∈C

subject to



as,p xp ≤ 1 for s ∈ S

p∈C

xp ≥ 0

for p ∈ C.

This is the same as the linear programming (5.1) except the different meanings of C. Therefore, the proof of Theorem 7.2.2 can be obtained from the proof of Theorem 5.3.1 by changing the meaning of C, that is, let C be the collection of all sensor k-covers instead of the collection of all sensor covers.  

7.3 Parity Strip Multi-Cover

119

7.3 Parity Strip Multi-Cover Although what we are really interested is the k-cover, it is more convenient to study the multi-cover. Problem 7.3.1 (Weighted Sensor Multi-Cover) Given a set A of target points with cover requirement c : A → {1, . . . , k} and a set S of sensors with positive weight w : S → R + in the Euclidean plane, find a sensor multi-cover with minimum total weight, where a subset S  is called a multi-cover if every target a is covered by at least c(a) sensors. Problem 7.3.2 (Weighted Parity Strip Multi-Cover) Let Q be a square which is divided into√horizontal H (1) , . . . , H (2l+1) for some fixed integer l. Each strip has width μ = 2/2. Suppose all target points lie in even strips H (2) , H (4) , . . . , H (2l) . Every target point a has a covering requirement c(a) ∈ {1, 2, . . . , k}. All sensors lie in Q and every sensor s has a positive weight w(s). A target point a is effectively covered by a sensor s if s is outside of the horizontal strip where a is located, and the sensing disk of s covers a. The problem is to find a minimum total weight subset of sensors such that every target point a is effectively covered by at least c(a) sensors in the subset. Theorem 7.3.3 (Zhang et al. [617]) The weighted parity strip multi-cover problem is polynomial-time solvable. To prove this theorem, we construct an auxiliary digraph and show that the weighted parity strip multi-cover problem is equivalent to the shortest path problem in the auxiliary digraph. For simplicity of speaking, let us first make some assumptions without loss of generality: • No two target points have the same x-coordinate. Otherwise, we may turn around the coordinate system with a proper angle. • Consider each sensor as a disk, the sensing disk of the sensor. Construction of Auxiliary Digraph G Let x(a) denote the x-coordinate of a target point a and a(x) the target point with x-coordinate x. Consider an even horizontal (i) strip H (i) . Let xle be a position (specified by x-coordinate) to the left of all target (i) points and xre a position to the right of all target points. They are considered as the left-end and the right-end, respectively. The set of all positions in H (i) is (i)

(i) }. P (i) = {x(a) | a is a target point in H (i) } ∪ {xle , xre (i)

Let D1 be a set of 2k − 1 disks1 whose centers lie above the upper boundary of (i) H (i) , and D2 a set of 2k − 1 disks whose centers lie below the lower boundary

1 Why

this parameter is set to be 2k − 1? We will explain it later.

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7 k-Coverage (i)

(i)

of H (i) . Denote D (i) = (D1 , D2 ). A configuration is an ordered pair (S, D) where S = (x (2) , x (4) , . . . , x (2l) ) where x (i) ∈ P (i) for i = 2, 4, . . . , 2l, and D = (D (2) , D (4) , . . . , D (2l) ). A configuration is valid if every target a(x (i) ) is covered by at least c(a(p(i) )) disks in D1(i) ∪ D2(i) . (i) (i) It is worth mentioning that D1 (or D2 ) might contain less than 2k − 1 disks. (i) In such a case, we still assume that D1 (or D2(i) ) contains exactly 2k − 1 disks by (i) (i) adding virtual disks. For convenience, we also view D (i) as D1 ∪ D2 and D as (2) (4) (2l) D ∪ D ∪ ···D . Using all valid configurations as vertices, we construct the auxiliary digraph G as follows. For any two valid configurations u = (Su , Du ) and v = (Sv , Dv ), arc (u, v) exists if and only if they satisfy the following two conditions. • Su and Sv have exactly one different component, i.e., there exists i ∈ (j ) (j ) (i) (i) {2, 4, . . . , 2l} such that xu = xv and xu = xv for j = i. (i) (i) • Target point a(xu ) is the immediate predecessor of target point a(xv ) in strip (i) (i) H (i) , i.e., there is no target point between them in H (i) and xu > xv .  Define the weight of arc (u, v) to be z(u, v) = d∈Dv \Du w(d). The definition of auxiliary digraph √ has the following background: Give each even strip a stick with height μ = 2/2 and call them sweeping sticks, denoted by {ss (i) }2,4,...,2l . The first component of a configuration consists of positions of those sweeping sticks. An arc (u, v) exists if and only if Sv can be obtained from Su by moving (sweeping) exactly one stick one step to the right. An illustration is shown in Fig. 7.1. The solid sticks are sweeping sticks of configuration u. In exactly one strip, there is a dashed stick on the right side of the sweeping stick of u, which is a sweeping stick of configuration v. In other strips, sweeping sticks of u and v are identical. Define two special configurations le = (Sle , Dle ) and re = (Sre , Dre ) where (2) (4) (2l) (2) (4) (2l) Sle = (xle , xle , . . . , xle ), Sre = (xre , xre , . . . , xre ), and all disks in Dle and Dre are virtual disks. They are called the starting configuration and the ending configuration, respectively. Next, we establish a relationship between optimal solution of the weighted parity strip multi-cover problem and the shortest path from le to re in the auxiliary digraph.

Fig. 7.1 An illustration of sweeping sticks

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121

Fig. 7.2 An illustration of Claim 1

Lemma 7.3.4 For any le-re path R in the auxiliary digraph G, let D be the set of all real disks appearing in D (i) on the path R. Then D is an effective multi-cover of target set A. Moreover, z(R) = (u,v)∈R z(u, v) ≥ w(D ). Proof From le to re along the path R, each sweeping stick moves from the leftend to the right-end and it must stop at every target point in corresponding strip. This fact implies that D must be an effective multi-cover of A. Moreover, since Dle contains only virtual disks, for any real disk d, there exists an arc (u, v) in R such that d ∈ Dv \ Du and hence w(d) is included in z(u, v). Therefore, z(R) ≥ w(D ).   Clearly, in order to have z(R) = w(D ), for any d ∈ D , we must have exactly one arc (u, v) ∈ R such that d ∈ Dv \ Du . This property will be established for the shortest path from le to re. Especially, we will show that for any optimal solution D∗ of the weighted parity strip multi-cover problem, there exists a shortest path R from le to re such that z(R) = w(D ). Lemma 7.3.5 Let D∗ be an optimal solution for the weighted parity strip multicover problem. Then D∗ is the set of real disks on a shortest le-re path R ∗ in the auxiliary digraph G and moreover, w(D∗ ) = z(R ∗ ). Proof Let us first introduce some notations and terminologies. Let lx denote the vertical line at position x and ∂d the boundary of a disk d. A disk d is said to be lower than another disk d  at position x if the lower intersection point of ∂d and lx is below the lower intersection point of ∂d  and lx . In Fig. 7.2, disk d is lower than d  at position x and d  is lower than d at position x  . If two disks have the same lower intersection points with line lx , then we say that the one with smaller x-coordinate of the center is lower than the other one at position x. Actually, such a case would not incur any technical difficulty in the analysis. For convenience, we will simply ignore it and assume that the case would not occur in the following analysis. Similarly, a disk d is said to be higher than another disk d  at position x if the upper intersection point of ∂d and lx is above the upper intersection point of ∂d  and lx . The following claim can be easily seen from Fig. 7.2. Claim 1 Consider two positions x and x  with x < x  . Suppose that a disk d is lower than another disk d  at position x and d  is lower than d at position x  . Then d is lower than d  at any position y with y < x and d  is lower than d at any position y with y > x  . Similarly, suppose that a disk d is higher than another disk d  at

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Fig. 7.3 An illustration of primary and secondary dominant disk

position x and d  is higher than d at position x  . Then d is higher than d  at any position y with y < x and d  is higher than d at any position y with y > x  . ∗ Let Di,upper be the set of disks of D∗ whose centers lie above the upper boundary of H (i) . Consider a target point a. A disk d is a primary upper dominant disk ∗ at position x(a) if d is among the k lowest disks at position x(a) in Di,upper .d is a secondary upper dominant disk at position x(a) if it is not a primary upper dominant disk at position x(a) and in addition, there are two target points a  and a  in H (i) with x(a  ) < x(a) < x(a  ) such that d is a primary upper dominant disk at both positions x(a  ) and x(a  ). For example, in Fig. 7.3, consider k = 2 and d2 is a secondary upper dominant disk at position x2 because d2 is a primary upper dominant disk in both positions x1 and x3 , but not a primary upper dominant disk at position x2 . ∗ (a) and D ∗ (a) denote the set of primary upper dominant disks and the Let Dpud sud ∗ (a) = set of secondary upper dominant disks at position x(a), respectively. Let Dud ∗ ∗ Dpud (a) ∪ Dsud (a). A disk is an upper dominant disk at position x if it is either a primary upper dominant disk or a secondary upper dominant disk at position x. ∗ (a)| ≤ k for any target point a. If a secondary upper dominant In general, |Dpud ∗ (a)| = k. disk exists at the position x(a), then we must have |Dpud ∗ (a)| ≤ k − 1 for any target point a. Claim 2 |Dsud ∗ (a)| ≥ k. Among k secondary upper dominant For contradiction, suppose |Dsud disks at position x(d), let d have the highest lower intersection point with lx(a) . By the definition, there exist two target points a  and a  with x(a  ) < x(a) < x(a  ) such that d is a primary upper dominant disk at both positions x(a  ) and x(a  ). Note ∗ (a) \ {d} is lower than d at position x(a) and among 2k − 1 that every disk in Dud ∗ \ {d}, there are at most k − 1 lower than d at position x(a  ). Hence disks in Dud there are at least k of them higher than d at position x(a  ). By Claim 1, those k disks are lower than d at position x(a  ), contradicting the fact that d is a primary upper dominant disk at position x(a  ). Claim 2 is proved. ∗ Let Di,lower be the set of all disks whose centers lie below the lower boundary (i) of H . Similarly, we can define the primary lower dominant disk, the secondary ∗ (a), D ∗ (a), lower dominant disk, and the lower dominant disk at position x as Dpld sld ∗ ∗ (a)| ≤ k and and Dld (a) for a target point a, respectively. In general, we have |Dpld ∗ (a)| ≤ k − 1. If |D ∗ (a)| ≥ 1, then |D ∗ (a)| = k. |Dsld sld pld

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123

∗ (a)| = |D ∗ (a)| = k and Next, without loss of generality, we assume |Dpud pld ∗ = |Dsld (a)| = k − 1 by adding virtual disks when necessary. Note that ∗ , then it must be covered by p if a target point a is covered by p disks in Di,upper ∗ ∗ disks in Dud (a) and if a target point a is covered by q disks in Di,lower , then it must ∗ be covered by q disks in Dld (a). Since a is effectively covered by D∗ and hence ∗ (a) ∪ D ∗ (a), in other words, D ∗ (a) = (D ∗ (a), D ∗ (a)). Call D ∗ (a) as an by Dud ld ud ld official bundle. Using official bundles, we can obtain a subset of valid configurations in each of which if the position vector is (x (2) , x (4) , . . . , x (2l) ), then the set D (i) = D ∗ (a(x (i) )) for i = 2, 4, . . . , 2l. Such a valid configuration is called an official configuration. Denote by G the subgraph of auxiliary digraph G induced by official configurations. ∗ (a)| |Dsud

Claim 3 There exists a le-re path R in G such that every disk d ∈ D∗ appears in R consecutively, i.e., all configurations in R, containing disk d, form a subpath of R. This path R will be constructed starting from the left-ending configuration le and moving toward the right-ending configuration re. Recall that moving along an arc is equivalent to moving exactly one sweeping stick ss (i) one step to the right. A sweeping stick is said to enter a disk d if d is not a dominant disk at current position of ss (i) but becomes a dominant disk at next position of ss (i) . ss (i) is said to leave d if d is a dominant disk at current position of ss (i) but not a dominant disk at next position of ss (i) . A disk d ∈ D∗ is a pending disk for strip H (i) if d belongs to some official bundle of strip H (i) and the sweeping stick ss (i) has not entered d. Construction Rule: ss (i) would not leave disk d when d is a pending disk for any strip H (j ) with j = i (see Fig. 7.4).

An illustration of the construction rule is given in Fig. 7.4: Consider k = 1. Sweeping stick ss (i) cannot move on to its next position because sweeping stick ss (i+2) has not entered d. When the construction rule applies, we say that ss (i) is waiting for ss (j ) . Note that every (unit) disk can touch at most two even strips. Therefore, ss (i) can wait only for ss (i+2) or ss (i−2) .

Fig. 7.4 An illustration of the construction rule

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To show Claim 3, we next show the following two facts: 1. Before arriving the right-ending configuration re, the path can always be extended without violating the construction rule. 2. Any disk d appears consecutively in official bundles of strip H (i) for any fixed i. To show fact 1, for contradiction, suppose no sweeping stick can move duo to the construction rule. Then there must exist a cycle, i.e., ss (i) waits for ss (i+2) , ss (i+2) waits for ss (i+4) , . . . , ss (i+2r) waits for ss (i) . Suppose ss (i) waits for ss (i+2) because of disk d1 , ss (i+2) waits for ss (i+4) because of disk d2 , . . . , ss (i+2r) waits for ss (i) because of disk dr+1 . Let xj be the x-coordinate of the center of disk dj . When ss (i+2) moves from current position to next position, d2 would leave the dominant job and d1 may enter the dominant job later. This implies that x1 > x2 . Similarly, we have x1 > x2 > x3 > · · · > xr+1 > x1 , a contradiction. To show fact 2, for contradiction, suppose there exist a disk d ∈ D∗ and a strip (i) H such that d is a dominant disk at positions a1 and a3 , but not a dominant disk at position a2 for x(a1 ) < x(a2 ) < x(a3 ). Without loss of generality, assume d ∈ ∗ ∗ (a ) ∩ Di,upper . Without loss of generality, one may further assume that d ∈ Dpud 1 ∗ ∗ Dpud (a3 ). In fact, if d ∈ Dsud (a1 ), then by the definition of the secondary upper dominant disk, there exists a target point a0 with x(a0 ) < x(a1 ) such that d ∈ ∗ (a ). Hence, we can replace a by a . It is similar to a . Now, since d is not a Dpud 0 1 0 3 ∗ (a ). By the definition, d is the dominant disk at position a2 , we have that d ∈ Dpud 2 secondary upper dominant disk at position d2 , i.e., a dominant disk at position a2 , a contradiction. By fact 1, a le-re path R can be constructed in G satisfying the construction rule. By fact 2, any disk d appears consecutively in official bundles of strip H (i) and moreover, by the construction rule, d appears consecutively along path R. This completes the proof of Claim 3. By Claim 3, for every disk d, there exists at most one arc (u, v) in path R such that d ∈ Dv \ Du . Therefore, z(R) = w(D∗ ). This completes the proof of the lemma.   By Lemmas 7.3.4 and 7.3.5, we have the following conclusion. Theorem 7.3.6 For any shortest le-re path R in auxiliary digraph G, the set of disks appearing in configurations on R form an optimal solution for the weighted parity strip multi-cover problem. Note that |V (G)| = O(nl · m2(2k−1) ) and computing a shortest path in a digraph uses a strongly polynomial-time. Therefore, computing an optimal solution for the weighted parity strip multi-cover problem can also be done in a strongly polynomial-time.

7.4 (4 + ε)-Approximation

125

7.4 (4 + ε)-Approximation To design approximation algorithms for the weighted sensor k-cover problem, we employ the technique of double partition as described in Section 5.3. Initially, we put all target points into a square. In the first partition, the √ square is partitioned into blocks; each block is a small square with edge length  2/2. In the √ second partition, each block is partitioned into smaller cells with edge length 2/2. The advantage of double partition is on the second one. When  is fixed, each block can be seen to contain a constant-number of cells so that different types of combinations of cells can be enumerated in polynomial-time. By taking this advantage, the weighted sensor k-cover problem for target points lying in a block may have a better approximation. Suppose that there is a ρ2 approximation with running time nO( ) for the problem of minimum weight sensor cover on a block as follows. Problem 7.4.1√(Weighted Sensor k-Cover on a Block) Consider a block B with edge length  2/2 where  is a positive constant integer. Given a set A of target points lying in B and a sensor cover S of sensors with nonnegative weight c : S → R+ , find a sensor cover with minimum total weight. This ratio ρ can be almost extended to the weighted sensor k-cover problem in general case as stated in the following. Theorem 7.4.2 Suppose there exists a polynomial-time ρ-approximation for the 2 problem of weighted sensor k cover on a block with running time nO( ) . Then for any ε > 0, there exists a polynomial-time (ρ + ε)-approximation with computation 2 time nO(1/ε ) for the weighted sensor k-cover problem. The proof of Theorem 7.4.2 is the same as that of Theorem 5.3.1. To design a 4-approximation for Problem 7.4.2, we will decompose it into four subproblems each of which is an instance of the weighted parity strip multi-cover problem. To do so, we need first to study some property of the optimal solution for Problem 7.4.2. Let D∗ be an optimal solution for Problem 7.4.2. Denote by Ci the set of all cells each of which contains exactly i centers of disks in D∗ for 0 ≤ i ≤ k − 1. Denote by C≥k the set of all cells each of which contains at least k centers of disks in D∗ . Since every disk has unit radius, any disk with center in a cell e would cover all target points in e. Thus, for any target point a in a cell e ∈ Ci , a requires to be covered by at least k − i disks with centers lying outside of e. For any target point in a cell a ∈ C≥k , such a requirement is zero. For e ∈ Ci , call i as the inner covering requirement e and for e ∈ C≥k , call k as the inner covering requirement for e. Let De∗ be the subset of D∗ , consisting of all disks with centers lying in e. Let ∗ DI be a subset of D∗ containing De∗ for every e ∈ Ci with 0 ≤ i ≤ k and exactly k disks in De∗ for e ∈ C≥k . Then DI∗ fulfils the inner covering requirement for all cells. In addition, each disk in e may cover target points outside e. Therefore, D∗ may fulfil partial outer covering requirement. For each target point a, let αa be the

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Fig. 7.5 Outside of e is divided into eight areas

Fig. 7.6 south (p), north (p), west (p), and east (p)

number of disks in DI∗ , covering a. We call k − αa as the residual outer covering requirement of a. For a cell e, divide the region outside of e into eight areas as shown in Fig. 7.5. Let N = N E ∪ NC ∪ NW , S = SE ∪ SC ∪ SW , E = N E ∪ ME ∪ SE, and W = N W ∪ MW ∪ SW . For simplicity of speaking, a disk is said to be in an area if its center lies in the area. Note that the following lemma has been proved in Section 5.2. Lemma 7.4.3 Suppose that a point p is covered by a disk d in area SC. Let south (p) be the shaded area in Fig. 7.6, formed by two lines with slops 1 and −1 through point p. Then every target point in south (p) can be covered by d. The similar statement holds for ME and east (p), MW and west (p), and N C and north (p), respectively. For e ∈ Ci and three integers 0 ≤ q1 , q2 , q3 ≤ k − i, denote by Ae (q1 , q2 , q3 ) the set of target points in e, covered by exactly q1 disks in N C if q1 ≤ k − i − 1 or at least q1 disks if q1 = k − i, exactly q2 disks in SC if q2 ≤ k − i − 1 or at least q2 disks in SC if q2 = k − i, and at most q3 disks in ME ∪ MW . Let ae1 (q1 , q2 , q3 ), ae2 (q1 , q2 , q3 ), ae3 (q1 , q2 , q3 ), ae4 (q1 , q2 , q3 ) be the upper leftmost, upper rightmost, lower leftmost, lower rightmost target points of Ae (q1 , q2 , q3 ), respectively, and call them anchors of e with respect to (q1 , q2 , q3 ). Note that some lines with slop 1 or −1 through these anchors would form a diamond (the area bounded by wide black lines) and a wedge (the shaded area) as shown in Fig. 7.7. They will be called a (q1 , q2 , q3 )-diamond and a (q1 , q2 , q3 )-wedge, respectively. Moreover, every disk j j in {D∗ (ae (q1 , q2 , q3 ))}4j =1 is called an anchor disk where D∗ (ae (q1 , q2 , q3 )) is the j

set of disks in ME ∪ MW covering anchor ae (q1 , q2 , q3 ). The following lemma states a property of (q1 , q2 , q3 )-wedge.

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127

Fig. 7.7 An illustration of anchors, diamond, and wedge

Lemma 7.4.4 If q1 + q2 ≤ k − i − 1, then every target point a in (q1 , q2 , q3 )-wedge is covered by exactly q1 disks in NC and exactly q2 disks in SC. Proof Note that (q1 , q2 , q3 )-wedge is the union of north (ae1 (q1 , q2 , q3 )) ∩ south (ae3 (q1 , q2 , q3 )) and north (ae3 (q1 , q2 , q3 ))∩ south (ae4 (q1 , q2 , q3 )). Without loss of generality, assume a ∈ north (ae1 (q1 , q2 , q3 )) ∩ south (ae3 (q1 , q2 , q3 )). Suppose a is covered by exactly x disks in NC and exactly y disks in SC. Since q1 + q2 ≤ k − i − 1, every anchor is covered by exactly q1 disks in NC and exactly q2 disks in SC. By Lemma 7.4.3, x ≥ q1 and y ≥ q2 . Note that ae1 (q1 , q2 , q3 ) ∈ north (a) and ae3 (q1 , q2 , q3 ) ∈ south (a). By Lemma 7.4.3, q1 ≥ x and q2 ≥ y. Therefore, x = q1 and y = q2 .   For each target point a in cell e ∈ Ci , the set D∗ \ DI∗ must contain at least k − αa i disks lying outside cell e and covering a. How are these disks distributed? To describe it, we consider three cases as follows. No diamond contains a. In this case, there are at least k − αa disks in ME∪MW covering a. In fact, for contradiction, suppose there are exactly r3 disks in ME ∪ MW covering a and r3 ≤ k − i. Let r1 be the number of disks in NC covering a and r2 the number of disks in SC covering a. Set q1 = min{r1 , k − αa } and q2 = min{r2 , k − αa }. By the definition of diamond, a is contained in (q1 , q2 , r3 )-diamond, a contradiction. Case 2. a lies in some (q1 , q2 , q3 )-wedge, i.e., a ∈ ( north (ae1 (q1 , q2 , q3 )) ∩ south (ae3 (q1 , q2 , q3 )))∪( north (ae3 (q1 , q2 , q3 ))∩ south (ae4 (q1 , q2 , q3 ))). Without loss of generality, assume a ∈ north (ae1 (q1 , q2 , q3 )) ∩ south (ae3 (q1 , q2 , q3 )). Then a is covered by at least q1 disks in N C and at least q2 disks in SC. If q1 +q2 ≥ k−αa , then those q1 +q2 disks already meet the residual outer covering requirement. If q1 + q2 ≤ k − αa − 1, then by Lemma 7.4.4, a is covered by exactly q1 disks in N C and exactly q2 disks in SC, and hence covered by exactly q1 + q2 disks in N C ∪ SC and in addition, at least k − αa − q1 − q2 disks in E ∪ W . Case 3. a lies in some (q1 , q2 , q3 )-diamond, but not in (q1 , q2 , q3 )-wedge. In this case, if a disk d in ME ∪ MW covers a, then d must cover an anchor and hence d is an anchor disk. Therefore, the set of disks in ME ∪ MW covering a is a subset of the anchor disk set. Other disks covering a must lie in N ∪ S.

Case 1.

Next, we use above properties of optimal solution to design a 4-approximation for the weighted sensor k-cover problem on a block. We will employ “guess” in algorithm description. Every guess satisfies two conditions.

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• The guess is from a pool with polynomial size so that enumeration of all cases in the pool would take a polynomial-time. • The pool contains an optimal solution so that the guessed result is compatible with the optimal. For example, for every cell e, we would guess an integer i from {0, 1, 2, . . . , k} and a set De of i disks in e. Let me be the number of disks in e. The pool size for 2  k l2 k this guess is O(k l e∈B me ) = O(k m ) where B is the block. It is compatible with the optimal solution if e ∈ Ci for 1 ≤ i ≤ k − 1 and e ∈ C≥k for i = k. Let DI = ∪e∈B De where B is the block. For each target point a, compute αa which is the number of disks in DI , covering a. For each target point a ∈ e with guessed i for e and guessed set De of i disks in e, the algorithm will assign a covering type (τa , ηa ) where τa + ηa = n − αa . τa indicates that a is required to be covered by τa disks in N ∪ S. ηa indicates that a is required to be covered by ηa disks in E ∪ W . To give such an assignment, the algorithm is first to determine all diamonds, that is, for each e ∈ Ci and 0 ≤ q1 , q2 , q3 ≤ k − i, guess four anchors of e with respect to (q1 , q2 , q3 ). Then for each j = 1, 2, 3, 4, guess at most q3 disks in ME ∪ MW j to form set D(ae (q1 , q2 , q3 )). Now, each target point a ∈ e in Ci or C≥k can be assigned with a covering type (τa , ηa ) as follows, corresponding to previous three cases. a does not belong to any guessed diamond. Assign τa = 0 and ηa = k − αa . Case 2. a is in a guessed (q1 , q2 , q3 )-wedge. If q1 + q2 ≥ k − i, then assign τa = k − i and ηa = 0. If q1 + q2 ≤ k − αa − 1, then assign τa = q1 + q2 and ηa = k − αa − q1 − q2 . Case 3. a is in some (q1 , q2 , q3 )-diamond, but not in (q1 , q2 , q3 )-wedge. Let x be j the number of disks in ∪4j =1 D(ae (q1 , q2 , q3 )) covering a. If x ≥ k − αa , then assign τa = 0 and ηa = k − αa . If x ≤ k − αa − 1, then assign τa = k − αa − x and ηa = x.

Case 1.

With above assigned covering types, the weighted sensor k-cover problem on a block can be decomposed into four instances of the weighted parity strip multicover problem. The first one consists of target points lying in odd horizontal strips and covering requirement τa . The second one consists of target points lying in even horizontal strips and covering requirement τa . The third one consists of target points lying in odd vertical strips and covering requirement ηa . The fourth one consists of target points lying in even vertical strips and covering requirement ηa . The set of disks in every instance is D \ DI where D is the set of all sensing disks of given sensors. For every instance, the optimal solution can be obtained in polynomialtime. Let Dj for j = 1, 2, 3, 4 denote these four optimal solutions for the four instances, respectively. Among all guessed results, select an approximation solution DI ∪ D1 ∪ D2 ∪ D3 ∪ D4 with the minimum weight.

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129

Lemma 7.4.5 In algorithm described as above, the number of possible guessed 2 results is O(k 3+l l 2 n4+k mk ) where n and m are the number of target points and the number of disks, respectively. Proof For each cell e, there are k+1 possibilities for guessed i. For l 2 cells, there are 2 at most (k + 1)l possible guessed results. For any fixed guessed i’s, there are O(nk ) possible DI . For each cell e, there are O(k 3 ) possible (q1 , q2 , q3 ) and for each fixed (q1 , q2 , q3 ), there are O(n4 ) possible groups of anchors and O(m4 ) possible choices j of {D(ae (q1 , q2 , q3 ))}4j =1 . Putting all together, the number of possible guessed 2

results is O(k 3+l l 2 n4+k mk ).

 

Lemma 7.4.6 There exists a polynomial-time 4-approximation for the weighted sensor k-cover problem on a block. Proof Let D∗ be an optimal solution for the weighted sensor k-cover problem. Suppose that the guess agrees with D∗ on every step, that is, for every parameter, guessed result is exactly the value determined by D∗ . For example, for each cell e, i is the number of disks in e and in D∗ , DI = D∗ , (q1 , q2 , q3 ) and corresponding anchors are determined by D∗ , etc. In this case, we have c(Dj ) ≤ c(D∗ \ DI∗ ) for j = 1, 2, 3, 4. Hence, c(DI∗ ∪ D1 ∪ D2 ∪ D3 ∪ D4 ) ≤ c(DI∗ ) + 4c(D∗ \ DI∗ ) ≤ c(D∗ ).

 

Theorem 7.4.7 (Willson et al. [554]) There exists a polynomial-time (4 + ε)approximation for the weighted sensor k-cover problem. Corollary 7.4.8 (Willson et al. [554]) There exists a polynomial-time (4 + ε)approximation for the maximum lifetime k-coverage problem.

7.5 (3 + ε)-Approximation In this section, we modify the 4-approximation algorithm in last section into 3approximation. We still consider the weighted sensor k-cover problem on a block. This modification is based on the following observation: Suppose D∗ is an optimal solution for the weighted sensor k-cover problem on a block. Let DI∗ be the set of disks fulfilling the inner requirement for all cells. When using disks in D \ DI to form feasible solutions for four subproblems (i.e., four instances of the weighted parity strip multi-cover problem), a disk may be involved four times. This is why the performance ratio of this approximation algorithm is 4. In the following, we will show that the number of such disks is quite small so that they can be guessed out like what we did in last section. Then we can get a 3-approximation. To start, we introduce some terminologies and results in graph theory. Consider a digraph G = (V , E). Two nodes u and v are said to be related if (u, v) ∈ E − + or (v, u) ∈ E. Let NG (u) be the set of in-neighbors of u and NG (u) the set of

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− out-neighbors of u. Denote − (G) = max{|NG (u)| | u ∈ V } and + (G) = + max{|NG (u)| | u ∈ V }. A graph G is transitive if (u, v) ∈ E and (v, w) ∈ E imply (u, w) ∈ E. If (u, v) ∈ E, then we say that u dominates v and v is dominated by u.

Lemma 7.5.1 Let G1 and G2 be two transitive acyclic digraphs on the same vertex set V . Suppose that every two nodes u and v are related either in G1 or in G2 . Then − (G1 ) + − (G2 ) ≥ n − 1 where n = |V |. Proof Since G1 and G2 are acyclic, they must have a sink. We prove by induction on n that there exist a sink si in G1 and a sink s2 in G2 such that − − (s1 ) ∪ NG (s2 ). V \ {s1 , s2 } ⊆ NG 1 2

(7.2)

This is trivial for n = 1, 2. Next, consider n ≥ 3 and two cases. Case 1.

Either G1 or G2 has unique sink. Without loss of generality that G1 has unique sink s1 . Then every node v in V has an arc (v, s1 ). Thus, V \{s1 } ⊆ − NG (s1 ). Choose any sink s2 from G2 . Then (7.2) holds. 1 Case 2. Both G1 and G2 have at least two sinks. Let S1 be the set of sinks of G1 . Then the subgraph of G2 induced by S1 has unique sink, denote by s1∗ . Similarly, Let S2 be the set of sinks of G2 . Then the subgraph of G1 induced by S2 has unique sink, denoted by s2∗ . Without loss of generality, assume either s1∗ = s2∗ or s1∗ and s2∗ are related in G2 (i.e., − G2 contains (s1∗ , s2∗ )). Since NG (s ∗ ) contains all sinks of G1 except s1∗ , 2 1 − − ∗  we have NG2 (s1 ) = ∅. Let V = V \ NG (s ∗ ) and Gj the subgraph of Gj 2 1  induced by V for j = 1, 2. By induction hypothesis, there exist a sink s1 of G1 and a sink s2 of G2 such that − −   V  \ {s1 , s2 } ⊆ NG  (s1 ) ∪ NG (s2 ). 1

(7.3)

2

Note that s1∗ is still a sink of G1 . If s1∗ = s1 , then by (7.3), we must have − − −  ∗  either s1∗ = s2 or s1∗ ∈ NG  (s2 ) and hence NG (s1 ) ⊆ NG (s2 ). Let s1 be a 2 2 2

sink reachable by s1 in G1 and s2 a sink reachable by s2 in G2 . Then (7.2) holds. Hence, we may assume s1∗ = s1 . Note that + (s ∗ ). V  \ {s1∗ } ⊆ N − G1 (s1∗ ) ∪ NG 2 1

+ − + (s ∗ ) ∩ NG (s  ) = ∅ or s2 ∈ NG (s ∗ ), then (7.2) holds again by If NG 2 1 2 2 2 1  letting s1 be a sink reachable by s1 in G1 and s2 a sink reachable by s2 in G2 . Therefore, we may assume + − + (s ∗ ) ∩ NG (s  ) = ∅ and s2 ∈ NG (s ∗ ). NG 2 1 2 2 2 1 + − −  ∗  ∗ ∗ By (7.3), NG (s ∗ ) ⊆ NG  (s1 ). Thus, V \ {s1 } ⊆ NG (s ). Let s1 = s 2 1 1 1

and s2 = s2∗ . Then (7.2) holds.

 

7.5 (3 + ε)-Approximation

131

Fig. 7.8 Each disk strides over at most four horizontal strips and at most four vertical strips

Now, we come back to the weighted sensor k-cover problem. Consider an optimal solution D∗ of the weighted sensor k-cover problem on a block. Let DI∗ be a subset of D∗ satisfying the inner covering requirement. Let D1∗ , D2∗ , D3∗ , and D4∗ be minimal feasible solutions in D∗ \ Di∗ for the four instances of the parity strip multi-cover problems, respectively. That is, deleting any disk would destroy the feasibility. In the following, we show that there are at most 4(2k − 1) disks of D∗ involving in all four subsets D1∗ , D2∗ , D3∗ , and D4∗ . Consider a cell e at intersection of a horizontal strip H (i) and a vertical strip V (j ) . Let De∗ be the subset of D∗ \ DI∗ with centers in e. A disk d ∈ De∗ is said to be a far-top dominant disk of cell e if d is a primary lower dominant disk at some position for strip H (i−2) . A disk d ∈ De∗ is said to be a far-bottom dominant disk of cell e if d is a primary lower dominant disk at some position for strip H (i+2) . Note that a disk with unit radius can overlap with at most four horizontal strips (Fig. 7.8). Therefore, a disk in De∗ involving both D1∗ and D2∗ must be either a far-top dominant disk or a far-bottom dominant disk. Similarly, we can introduce the concepts of far-left and far-right dominant disks and a disk in De∗ involving both D3∗ and D4∗ must be either a far-left dominant disk or a far-right dominant disk. Let V = De∗ ∩ D1∗ ∩ D2∗ ∩ D3∗ ∩ D4∗ . Partition V into (Vtl , Vtr , Vbl , Vbr ) where Vtl consists of all disks each of which is a far-top and far-left dominant disk, Vtr consists of all disks each of which is a far-top and far-right dominant disk, Vtr , Vbl consists of all disks each of which is a far-bottom and far-left dominant disk, and Vbr consists of all disks each of which is a far-bottom and far-right dominant disk. First, consider Vtl . Suppose d1 , d2 ∈ Vtl . Then both d1 and d2 are far-top dominant disks and one of the following two possibilities must occur. • ∂d1 and ∂d2 have a crossing point in H (i−1) . • d1 dominates d2 at all positions in H (i−2) or d2 dominates d1 at all positions in H (i−2) . We say that a disk d1 totally dominates d2 in H (i−2) if d1 dominates d2 at all positions in H (i−2) (Fig. 7.9). Construct a digraph G1 with node set Vbl by drawing an arc (d1 , d2 ) if and only if d1 totally dominates d2 . Then G1 is clearly transitive and is acyclic if no two disks in Vbl are identical. To remove this exceptional case, we may delete one of two arcs (d1 , d2 ) and (d2 , d1 ) if both exist. The remainder is still denoted by G1 . Then, we have.

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7 k-Coverage

Fig. 7.9 An illustration of total-domination

Fig. 7.10 An illustration for the proof of Lemma 7.5.3

Lemma 7.5.2 G1 is transitive and acyclic. Similarly, we can introduce the concept of totally domination in V (j −2) and construct a transitive and acyclic digraph G2 with node set Vbl . Lemma 7.5.3 Any two nodes d1 and d2 are related in either G1 or G2 . Proof For contradiction, suppose d1 and d2 are unrelated in both G1 and G2 . Then d1 = d2 , and ∂d1 and ∂d2 have an intersection p1 in H (i−1) and an intersection p2 in V (j −2) (Fig. 7.10). Let o(dh ) be the center of disk dh for h = 1, 2. Then line p1 p2 is the perpendicular bisector of line segment o(d1 )o(d2 ) and hence passes through cell e. Therefore, we must have the Euclidean distance p1 p2  > 2. Actually, the position for reaching the minimum p1 P2  is that the line p1 p2 passes through the top-left corner of e and has slope 1. At this position, p1 p2  = 2. However, since d1 = d2 , the line p1 p2 can approach the position and cannot reach. Therefore, we must have p1 p2  > 2 and hence o(d1 )p1  > 1, contradicting that d has unit radius.   Lemma 7.5.4 |Vbl | ≤ 2k − 1. Proof For contradiction, suppose |Vtl | > 2k − 1. By Lemma 7.5.1, − (G1 ) + − (G2 ) ≥ 2k − 1.

7.5 (3 + ε)-Approximation

133

Thus, − (G1 ) ≥ k or − (G2 ) ≥ k. Without loss of generality, assume − (G1 ) − ≥ k. Then there exists a node d ∈ Vtl such that NG (d) ≥ k, i.e., there are k disks 1 (i−2) so that d cannot be a primal lower dominant disk in Vtl totally dominate d in H at any position in H (i−2) . Hence, d cannot be far-top, contradicting to d ∈ Vbl .   By the similar arguments, we can obtain that |Vtr | ≤ 2k − 1, |Vbl | ≤ 2k − 1, and |Vbr | ≤ 2k − 1. Therefore, we have. Lemma 7.5.5 |V | ≤ 4(2k − 1). Proof |V | = |Vtl | + |Vtr | + |Vbl | + |Vbr | ≤ 4(2k − 1).

 

Theorem 7.5.6 (Zhang et al. [636]) There exists a polynomial-time (3 + ε)approximation for the weighted sensor k-cover problem. Proof Modify the 4-approximation algorithm for the weighted sensor k-cover problem on a block (in Section 6.4) as follows. After guessing DI , for each cell e, guess at most 4(2k − 1) disks; each of them is supposed to be involved in four minimal feasible solutions for four instances of the parity strip multi-cover problem. Suppose that all disks obtained from these new guesses form a set Df our . Then use DI ∩ Df our instead of DI in the rest part of the 4-approximation. We would obtain a feasible solution for the weighted sensor k-cover problem with total weight at most w(DI∗ ) + w(Df our + 3 · w(D∗ \ (DI∗ ∪ Df∗ our ) ≤ 4 · w(D∗ ). Note that the running time is increased by a factor of n4(2k−1) , keeping a polynomialtime.   Corollary 7.5.7 (Zhang et al. [636]) There exists a polynomial-time (3 + ε)approximation for the maximum lifetime k-coverage problem.

Chapter 8

Heterogeneous Sensors

The only difference between suicide and martyrdom is press coverage. Chuck Palahniuk

8.1 Motivation and Overview In previous chapters, the considered sensor system was assumed to be homogeneous almost everywhere. Exceptional two cases are as follows. 1. For the minimum sensor cover problem, a PTAS is designed in Chapter 3, which is held for heterogeneous sensor system. This is because the local search method used there does not require homogeneous property. However, the following is open. Open Problem 10 Could this local search method be extended to weighted case or the problem with connectivity? For example, extend the local search method to design a PTAS for the minimum weight sensor cover problem, or extend the local search method to design a polynomial-time constant-approximation for the minimum connected sensor cover problem. 2. For the minimum connected sensor cover problem, there are three approximation algorithms designed in Chapter 4. The one with the group Steiner tree is suitable in heterogeneous sensor systems while the other two are not, i.e., they can be applied only to homogeneous sensor systems. Let us summarize existing results on approximation solutions for optimization problems about sensor coverage in homogeneous wireless sensor networks.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_8

135

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8 Heterogeneous Sensors

Coverage in homogeneous wireless sensor system Minimum cardinality Sensor cover PTAS [177] Sensor k-cover Connected sensor cover O(r) [567]

Minimum weight PTAS [326] O(1) [636] O(log2 n log log n) [567]

In this chapter, we are interested in heterogeneous wireless sensor networks, which correspond to heterogeneous disk systems. The disk cover is a special case of the set cover. Therefore, the following results on set cover can also be held for the disk cover. Coverage in set system Set cover Set multi-cover Connected set cover

Minimum cardinality 1 + ln n [278] 1 + ln n O(log2 n log log n) [567]

Minimum weight 1 + ln n [155] 1 + ln n O(log2 n log log n) [567]

Since the disk is a special geometric object, the disk cover should admit better results by taking geometric advantage. The following table lists existing better results compared with those in set systems. Coverage in heterogeneous disk system Minimum cardinality Disk cover PTAS [413] Disk multi-cover O(1) [122] Connected disk cover

Minimum weight O(1) [112] O(1) [56]

As mentioned above, the PTAS for the minimum heterogeneous disk cover problem is designed by using the local search technique and this technique has not been extended to weighted case so far. The O(1)-approximation for the problems in weighted case is designed by using another technique, concerning linear program relaxation and randomized rounding. In this chapter, we first include the (1 + ln n)approximation for the minimum weight set multi-cover problem (Section 8.2) and then introduce the latter technique to design O(1)-approximation for the minimum weight disk multi-cover problem (Section 8.3). The second part has used the lecture notes of Chan [109] as a major reference. Again, a purpose of studying the weighted version of disk multi-cover problem is due to its application to the maximum lifetime coverage problem. Note that Garg– Köemann Theorem can be extended to heterogeneous sensor systems. Therefore, we have the following.

8.2 Minimum Weight Set Multi-Cover

137

Theorem 8.1.1 The maximum lifetime k-coverage problem has a polynomial-time O(1)-approximation in heterogeneous wireless sensor networks. Proof Since the minimum weight disk multi-cover problem has a polynomial-time O(1)-approximation, by Garg–Köemann Theorem [217], the maximum lifetime kcoverage problem has a polynomial-time (O(1) + ε)-approximation for any ε > 0. For sufficiently small ε > 0, O(1) + ε = O(1).   Finally, we will indicate that for heterogeneous sensor systems, there exist other models with different optimization objectives, such as energy efficient coverage problem for sensors with variable coverage range [411, 510, 512] and the composite event coverage problem [209, 595]. These two subjects will be explored a little more in Sections 8.4 and 8.5, respectively.

8.2 Minimum Weight Set Multi-Cover In this section, we study the following problem. Problem 8.2.1 (Minimum Weight Set Multi-Cover) Given a collection C of subsets of a base set X, each element e ∈ X with a required covering number re and each subset C ∈ C with a positive weight wC , find a set multi-cover with the minimum total weight, where a set multi-cover is a subcollection A of C such that every element e ∈ X belongs to at least re subsets in A. We show the following. Theorem 8.2.2 The minimum weight set multi-cover problem has a polynomialtime H (n)-approximation where H (·) is the Harmonic function and n is the number of elements. To show this, we define a function f : 2C → N as follows: For every subcollection A ⊆ C,  min(re , ce (A)), f (A) = e∈X

where ce (A) is the number of subsets in A containing e. This function has a close relationship with the set multi-cover.  Lemma 8.2.3 Let R = e∈X re . Then A is a set multi-cover if and only if f (A) = R. Proof By the definition of the set multi-cover, A is a set multi-cover if and only if ce (A) ≥ re for every e ∈ X. This occurs if and only if min(re , ce (A)) = re if and only if f (A) = R.  

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8 Heterogeneous Sensors

By this lemma, we may assume f (C) = R; otherwise, there is no solution for the minimum weight set multi-cover problem. Lemma 8.2.4 If f (A) < R, then there exists C ∈ C \ A such that f (A ∪ {C}) > f (A). Proof Suppose f (A) < R. Then there must exist e ∈ X such that min(re , ce (A)) < re , i.e., ce (A) < re . Since ce (C) ≥ re , there must exist C ∈ C \ A such that ce (A) < ce (A ∪ {C}) and hence f (A) < f (A ∪ {C}).   Above lemmas indicate that the following greedy algorithm would produce a set multi-cover. Greedy Algorithm SMC input a collection C of subsets of X and a weight function w : C → N ; A ← ∅; while f (A) < R do ( A) choose C ∈ C to maximize f (A∪{C})−f w(C) and set A ← A ∪ {C}; output A. Actually, this algorithm produces an approximation solution with performance ratio H (n), which gives a proof of Theorem 8.2.2. To establish this result about performance ratio, let us first introduce a little knowledge on submodular optimization. Consider a function g defined on all subsets of a base set X. g is called a submodular function if for any two subsets A and B, g(A) + g(B) ≥ g(A ∪ B) + g(A ∩ B). g is called a monotone nondecreasing function if A ⊂ B ⇒ g(A) ≤ g(B). The submodularity has a well-known property. Lemma 8.2.5 (Decreasing Marginal Value) g is submodular if and only if for any two sets A and B with A ⊂ B and any x ∈ B, x g(A) ≥ x g(B), where x g(A) = g(A ∪ {x}) − g(A). If g is monotone nondecreasing, submodular and g(∅) = 0, then g is also called a polymatroid function. The following are two properties of polymatroid functions. Lemma 8.2.6 If g1 and g2 are two polymatroid functions, so is g1 + g2 .

8.2 Minimum Weight Set Multi-Cover

139

 

Proof Trivial.

Lemma 8.2.7 Suppose g is a polymatroid function and c is a nonnegative constant. Then ζ (A) = min(c, g(A)) is a polymatroid function. Proof Since g is a polymatroid function, it is clear that ζ (∅) = 0 and ζ is monotone increasing. To show the submodularity of ζ , we would like to use the wellknown Decreasing Marginal Value property of submodular functions and divide our argument into three cases. Consider two subsets A and B with A ⊂ B, and an element x ∈ B. Case 1.

g(A ∪ {x}) > c. In this case, ζ (A ∪ {x}) = c and ζ (B ∪ {x}) = c. Hence x ζ (A) = c − ζ (A) ≥ c − ζ (B) = x ζ (B).

Case 2.

g(A ∪ {x}) ≤ c and g(B) ≤ c. In this case, ζ (A ∪ {x}) = g(A ∪ {x}), ζ (A) = g(A), ζ (B) = g(B). Hence x ζ (A) = g(A ∪ {x}) − g(A) ≥ g(B ∪ {x}) − g(B) ≥ x ζ (B).

Case 3.

g(A ∪ {x}) ≤ c and g(B) > c. In this case, ζ (A ∪ {x}) = g(A ∪ {x}), ζ (A) = g(A), ζ (B) = c, ζ (B ∪ {x}) = c. Thus, x ζ (A) = x g(A) ≥ 0 = c − c = ζ (B ∪ {x}).

 

There is a well-known optimization related to polymatroid. Problem 8.2.8 (Submodular Set Cover) Suppose g is a monotone nondecreasing, submodular function on 2X . Define (g) = {A ⊆ X | g(A) = g(X)}. Let c be a nonnegative cost function on X. Then the submodular set cover problem is a minimization as follows.  min c(A) = c(x) x∈A

subject to A ∈ (g). The minimum weight set multi-cover problem can be seen as a special case of this problem provided that f is a polymatroid function which will be proved later. In general, the submodular set cover problem has a greedy approximation as follows. Greedy Algorithm SSC input a polymatroid function g on 2X and a nonnegative cost function on X; A ← ∅; while g(A) < g(X) do x g(A) choose x ∈ X \ A to maximize c(x)

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8 Heterogeneous Sensors

and set A ← A ∪ {x}; output A. There is a well-known theorem related to this algorithm. Theorem 8.2.9 (L. Wolsey [556] ) If g is an integer-value polymatroid function, then the Greedy Algorithm SSC produces an H (γ )-approximation solution A for the submodular set cover problem, where γ = maxx∈X g({x}) and H (·) is the Harmonic function. Now, we come back to the minimum weight set multi-cover problem. We are going to show that f is a polymatroid function. Lemma 8.2.10 For any e ∈ X, ce (A) is a polymatroid function. Proof It is clear that ce (·) is monotone increasing and ce (∅) = 0. To show its submodularity, consider two subset subcollections A and B. Compare ce (A)+ce (B) with ce (A ∪ B). The difference is that each subset in A ∩ B is counted once in ce (A ∪ B), but twice in ce (A) + ce (B). Therefore, we have ce (A) + ce (B) = ce (A ∪ B) + ce (A ∩ B). This means that ce is a modular function and of course also submodular.

 

Lemma 8.2.11 f is a polymatroid function. Proof It follows immediately from Lemmas 8.2.6, 8.2.7, and 8.2.10.

 

Now, Theorem 8.2.2 follows from Theorem 8.2.9 and Lemma 8.2.11. By Garg– Köemann Theorem, we can also obtain the following corollary. Corollary 8.2.12 The maximum lifetime coverage problem in heterogeneous wireless sensor systems has a polynomial-time (1 + ln n)-approximation where n is the number of sensors.

8.3 Minimum Weight Disk Multi-Cover In this section, we study the following problem. Problem 8.3.1 (Minimum Weight Disk Multi-Cover) Consider a collection C of disks and a set P of target points in the Euclidean plane. Each target point e is associated with a required covering number re and each subset C ∈ C is associated with a positive weight wC . The problem is to find a disk multi-cover with the minimum total weight, where a disk multi-cover is a subcollection A of C such that every target point e ∈ P belongs to at least re disks in A.

8.3 Minimum Weight Disk Multi-Cover

141

Fig. 8.1 Illustration of the depth of a vertex and the proof of Lemma 8.3.4

The best known result in the literature is as follows. Theorem 8.3.2 (Bansal and Pruhs [56]) The minimum weight disk multi-cover problem has a polynomial-time O(1)-approximation. This result is obtained with several advanced techniques. In the following, we introduce them step by step.

8.3.1 Shallow Cell Complexity and ε-Net In [81], Brönnimann and Goodrich reveal an amazing relation between ε-net and geometric coverage problem. Given a set of objects O on the plane, the depth of a point p with respect to O, denoted as depthp (O), is the number of objects containing p, where O ∈ O contains p means that p is in the interior of O, lying on the boundary is not regarded as containment. The intersection points of the boundaries of objects in O are called vertices. We follow the general convention that in a generic position, no three objects have their boundaries intersecting at a common vertex. Figure 8.1 illustrates the depth of vertices. Vertex v1 has depth 0. It is the intersection point of the boundaries of disks d1 and d3 , but belongs to the interior of no disks. Vertex v2 has depth 2, it is the intersection point of the boundaries of disks d1 and d2 , and is contained in the interior of disks d3 and d4 . Denote by O=k (resp. O≤k ) the number of vertices of depth k (resp. at most k) in O. In particular, those vertices of depth 0 are the vertices on the boundary of the union of the objects, and thus are often called vertices of the union. The following result is well-known (see, for example, the book [246]). Lemma 8.3.3 The union of m disks has O(m) vertices. Base on this lemma, we have the following result. Lemma 8.3.4 (Clarkson and Shor [158]) Given a set D of m disks on the plane, there are O(mk) vertices of depth at most k. Proof Let R be a sample of disks obtained by picking each disk with probability 1/k. Then the expected number of disks in R is E[R] = m/k. By Lemma 8.3.3, the expected number of depth-0 vertices in R is

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8 Heterogeneous Sensors

E[|R=0 |] = O(m/k).

(8.1)

For any vertex v ∈ D≤k , let us consider the probability that v becomes a vertex of depth 0 in R. Suppose v is the intersection point of the boundaries of disks d and d  , and is contained in  disks d1 , . . . , d , where  ≤ k. Then v ∈ R=0 if and only if d, d  ∈ R and d1 , . . . , d ∈ R. This event happens with probability     2    1 1  1 1 1− P r[v ∈ D= ∧ v ∈ R=0 ] = = 2 = 2 .    k It follows that E[|R=0 |] ≥

k  =0



 1 P r[v ∈ D= ∧ v ∈ R=0 ] · |D= | =  2 |D≤k | . k

(8.2)

Combining (8.1) and (8.2), |D≤k | = O(mk). The lemma is proved.

 

A set of objects O may divide the plane into regions. A region is said to be a level-k cell if it is the intersection of exactly k objects of O. Lemma 8.3.5 The number of level-≤ k cells in D is O(mk). Proof For a level- cell R, let vR be a vertex on its boundary whose depth is the largest. Then the depth of vR is at most . In this way, we can establish a mapping from the set of level-≤ k cells to the set of vertices of depth at most k. Since in a generic position, every vertex is incident with at most 4 regions, the number of level-≤ k cells is at most four times the number of vertices of depth at most k. The lemma follows from Lemma 8.3.4.   Definition 8.3.6 An object system O is said to have shallow cell complexity f (m, k) if for any subset of objects O ⊆ O with |O | = m, the number of level-≤ k cells of O is f (m, k). As a corollary of Lemma 8.3.5, we have the following result. Corollary 8.3.7 The shallow complexity f (m, k) = O(mk) for disks. An ε-net is a set of objects covering all those points which are sufficiently deep. The formal definition is given as follows. Definition 8.3.8 Given a set of objects O and a set of point P, a subset R ⊆ O is said to be an ε-net if every point p ∈ P with depthp (O) ≥ ε|O| is covered by R. The following result was proved by Matousˇek et al. Theorem 8.3.9 ([406]) For disks on the plane, there is an ε-net of size O(1/ε).

8.3 Minimum Weight Disk Multi-Cover

143

Algorithm 1 Algorithm for minimum disk cover 1: x∗ ← an optimal solution of (8.3).  in which each disk D has 2mx ∗  copies. 2: Create a multiset D D 1  with ε = . 3: Return R ← the set of disks of D in an ε-net of D 2opt f rac

8.3.2 ε-Net and Minimum Cardinality Disk Cover Since an ε-net covers all those points which are sufficiently deep, the idea is to make every point deep enough by duplication. The question is how many copies are needed? In this case, linear program helps. Given a set of disks D and a set of points P on the plane, the following is an LP relaxation of the minimum weight disk cover problem: 

min

wD xD

D∈D

s.t.



xD ≥ 1, ∀p ∈ P

(8.3)

D∈D : p∈D

xD ≥ 0, ∀D ∈ D. In particular, if w ≡ 1, then (8.3) is an LP relaxation for the minimum (cardinality) disk cover problem. Let m = |D|. Denote by opt f rac the optimal value of LP formulation (8.3) and opt the optimal value of the minimum disk cover problem. Then opt f rac ≤ opt. The algorithm is described in Algorithm 1. Theorem 8.3.10 The set R returned by Algorithm 1 is a constant-approximation for the minimum disk cover problem.   ∗ ∗  = Proof Notice that |D| · opt f rac . On D∈D 2mxD  ≤ 2m D∈D xD = 2m  ∗ ≥  = 2mxD the other hand, every point p ∈ P has depth depthp (D) D∈D : p∈D     ∗ − 1) ≥ 2m ∗ − m ≥ m, where the last inequality (2mxD xD

D∈D : p∈D

D∈D : p∈D

 ≥ ε|D|.  So, uses the constraint in (8.3). By the choice of ε, we have depthp (D)  and thus every point is covered by R. every point in P is covered by the ε-net of D, We have proved that R is a feasible solution. By Theorem 8.3.9 and the choice of ε, we have |R| = O(1/ε) = O(opt f rac ) = O(opt), and thus R is a constant-approximation.  

The first realization of a constant-approximation for the minimum disk cover problem using ε-net was given by Brönnimann and Goodrich in [81], which does not rely on the LP. The algorithm first guesses an ε with 4opt ≤ 1ε ≤ 8opt, then recursively doubles the multiplicity of those disks containing a common uncovered

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8 Heterogeneous Sensors

point and finds ε-nets for the uncovered points. The essence of this method is to adaptively deepen the depths of those uncovered points. When the depth of a point is sufficiently large, then it is covered by the ε-net. It was proved that the algorithm 1 can terminate in 2ε log m iterations, which results in an O( 1ε ) = O(opt) set.

8.3.3 Quasi-Uniform Sampling and Minimum Weight Disk Cover A difficulty in generalizing the above method to the weighted case is as follows: the construction of an ε-net starts from a uniform sampling followed by some kind of alteration, the alteration method cannot avoid the possibility that some sets with high weights are picked. In 2010, Varadarajan [506] proposed an elegant method called quasi-uniform sampling to conquer this difficulty, in which disks are picked with a bounded probability and the sampling is not required to be independent. Such a sampling method leads to the following theorem showing the existence of a desired weighted ε-net. Theorem  W  8.3.11 Fora set of disks D on the plane, there is an ε-net of weight O εm , where W = D∈D wD is the total weight of disks and m = |D|. Before proving the theorem, we show that if the theorem is true, then a constantapproximation for the minimum weight disk cover problem can be achieved. Theorem 8.3.12 There is a constant-approximation for the minimum weight disk cover problem. Proof The algorithm is similar to Algorithm 1 with the following two differences: using the method in the proof of Theorem 8.3.11 to construct the ε-net, and taking ε to be  1 2x ∗ . D∈D D  m ∗  ≤ 2m Since |D| D∈D xD < ε , we see that every point p ∈ P has depth  ≥ m ≥ ε|D|,  and thus is covered by an ε net of D.  This proves the depthp (D) feasibility of the algorithm.   ∗ ∗  = Notice that W = w(D) · D∈D 2mxD wD ≤ 2m D∈D xD wD = 2m   m ∗ f rac  opt and |D| ≥ 2m D∈D xD −m = ε −m. So the weight w(R) = O W = f rac O( 2opt 1−ε )

ε|D |

= O(opt). The approximation ratio is proved.

 

The remaining part of this section is to prove Theorem 8.3.11. A subset R is said to be a ρ-sample of O if every object of O is picked with probability ρ independently. A subset of objects R is said to be a quasi-ρ-sample of O if for any 1 object O ∈ O, the probability P r[O ∈ R] ≤ ρ. So, if R is a quasi-O( εm )-sample of disk set D, then R has expected weight

8.3 Minimum Weight Disk Multi-Cover

E[w(R)] =

 D∈D

145

 wD P r[D ∈ R] = O

1 εm

 D∈D

 wD = O

 W . εm

Denote by k = εm. The above argument shows   that to prove Theorem 8.3.11, it suffices to prove the existence of a quasi-O k1 -sample R which is a mk -net of D, that is, R is a sample of D satisfying the following two properties:   (a) P r[D ∈ R] ≤ O k1 for every disk D ∈ D, and (b) R covers all points of depth at least k. Let us first think of the sampling process in an ideal setting. Denote D0 = D. Let D1 be a 1/2-sample of D0 . Then any point p with depthp (D0 ) ≥ k has expected depth at least k/2 in D1 . Repeating the same sampling method, one obtains a sequence of disk sets D0 , D1 , . . . such that every point p with depthp (Di−1 ) ≥ k/2i−1 has expected depth at least k/2i in Di . Suppose in an ideal setting these events happen definitely, that is, suppose every point p with depthp (Di−1 ) ≥ k/2i−1 has depthp (Di ) ≥ k/2i for i = 1, 2, . . .. Then after  = log2 k rounds, every point p with depthp (D0 ) ≥ k has depthp (D ) ≥ k/2 = 1, and thus D satisfies property (b) in the above paragraph. Furthermore, for any disk D ∈ D, the probability P r[D ∈ D ] = (1/2) = 1/k, and thus property (a) is also satisfied. However, in a random sampling, such an ideal setting cannot be guaranteed to happen. In order that the depth of a point is not decreased too much with a high probability, one has to pick every disk with a probability a little larger than 1/2. The following lemma formalizes this idea.    k Lemma 8.3.13 Let R be a 12 + c log -sample of D, where c is some constant. k Then for any point p with depthp (D) ≥ k, the probability   1 k ≤ (c) . P r depthp (R) < 2 k Proof For a point p with depthp (D) ≥ k, suppose the disks containing p are D1 , . . . , Dt with t ≥ k. Define arandom variable Xi = 1 if Di ∈ R and Xi = 0 otherwise. Then depthp (R) = ti=1 Xi is the sum of independent Poisson trials. Notice that    1 c log k k  μ = E[depthp (R)] ≥ k + = + ck log k. 2 k 2 By Chernoff’s bound,     k P r depthp (R) < ≤ P r |depthp (R) − μ| ≥ ck log k 2 ≤ e−(c log k) =

1 . k (c)

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8 Heterogeneous Sensors

 

The lemma is proved.

Although the desired property holds with a high probability, it is still possible that the depths of some points is decreased a lot in the sampling process so that the final sampled disk set cannot cover them. We call such a situation as bad. The following 1 lemma shows that if a bad situation occurs, then one can construct a quasi- k (c) sample covering those points whose depths drop too much.    k Lemma 8.3.14 Let R be a 12 + c log -sample of D, where c is some constant. k 1 One can construct a quasi- k (c) -sample A of D such that for any point p with depthp (D) ≥ k, either depthp (R) ≥ k2 , or p is covered by A.

Proof By Lemma 8.3.5, the number of level-≤ k cells of D is O(mk). Since every level-≤ k cell is contained in at most k disks, there must ∃ a disk D0 intersecting at most O

 mk · k  m

= O(k 2 ) level- ≤ k cells.

(8.4)

1 Let D = D − {D0 }. Suppose by induction that we can construct a quasi- k (c)   sample A of D with the desired property. For any point p with depthp (D) ≥ k, if depthp (D ) ≥ k, then the desired property holds for p by induction. In particular, this holds for p with depthp (D) > k or p ∈ D0 . So, we only need to consider those points in

P  = {p ∈ P : depthp (D) = k and p ∈ D0 }. If every point p ∈ P  satisfies depthp (R) ≥ k/2, then there is nothing more to do. Otherwise, let A = A ∪ {D0 }. By the above construction of A, we have P r[D0 ∈ A] = P r[∃p ∈ P  with depthp (R) < k/2].

(8.5)

Notice that every point in P  belongs to a level-k cell intersecting D0 . By (8.4), there are O(k 2 ) such cells. Furthermore, points in a same level-k cell play the same role for the event in the righthand side of (8.5). Then, it follows from Lemma 8.3.13 that P r[D0 ∈ A] = O(k 2 ) · As to the other disks D ∈ D , P r[D ∈ A ] ≤ 1 -sample of D. k (c)

1 k (c) 1 k (c)

=

1 k (c)

.

by induction. So, A is a quasi 

Now, we are ready to prove Theorem 8.3.11. Proof of Theorem 8.3.11 Let k = εm. For simplicity of statement, assume that k = 2 for some integer . Let R = D. Apply Lemma 8.3.14  times to yield set of

8.3 Minimum Weight Disk Multi-Cover

147

disks R , R . . . , R0  and A , A−1 , . . . , A1 , such that for i = ,  − 1, . . . , 1, −1 ,  c log 2i 1 -sample of Ri and Ai is a quasi- (2i )1(c) -sample of Ri . Ri−1 is a 2 + 2i    Let R = R0 ∪ i=1 Ai . Claim 1 R is an ε-net of D. To prove the claim, we have to show that R covers all points of depth at least k. For any index i = ,  − 1, . . . , 1 and any point p with depthp (Ri ) ≥ 2i , by Lemma 8.3.14, either depthp (Ri−1 ) ≥ 2i−1 or p is covered by Ai . So for any point p with depthp (D) ≥ k = 2 , after  rounds, either depthp (R0 ) ≥ 1 (which  implies that p is covered by R0 ) or p is covered by some disk in i=1 Ai . In any case, p is covered by R. The claim is proved. Claim 2 R is a quasi-O( k1 )-sample of D. For i =  − 1, . . . , 1, 0, set Ri is a any disk D ∈ D, the probability P r[D ∈ Ri ] =

 " j =i+1

≤

1

⎛

#

⎝1 + 2

e 2−i

 ! j =i+1

 1 2

c log 2j 2j

 -sample of D. So for

⎞ ⎛ ⎞ #  c log 2j ⎠ c log 2j ⎠ 1 " ⎝ = −i 1+2 2j 2j 2

√  2 c j =i+1

j =i+1



log 2j 2j

 =

where the last equality uses the observation that consequence, P r[D ∈ R0 ] = we have

+



2i k

 , 



j =i+1

log 2j 2j

( k1 ). Furthermore, for any index i 

P r[D ∈ Ai ] = P r[D ∈ Ri ∧ D ∈ Ai ] = 

2i k



is a constant. As a = , −1, . . . , 1,

  1 1 . =  i (c) k (2 )

Since R0 , A1 , . . . , A are disjoint sets. So any disk D has P r[D ∈ R] = O( k1 ). Combining Claim 1 and Claim 2, Theorem 8.3.11 is proved.  

8.3.4 Minimum Weight Disk Multi-Cover To sum up the above method for the minimum weight disk cover problem: it iteratively takes samples of disks; every disk is sampled with a probability roughly proportional to the optimal LP solution; if the depth of some point is decreased too much, then some disks are forced to be picked; one thing that should be taken care

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8 Heterogeneous Sensors

of is how to guarantee that the probability of forcing is sufficiently small, and this is where the shallow cell complexity comes in. The difficulty in generalizing this method to the multi-cover version is that a point p must be covered by rp distinct disks. Nevertheless, Bansal and Pruhs [56] made a non-trivial extension to obtain a constant-approximation. The idea of their algorithm is as follows: let x ∗ be an optimal solution to the relaxed LP formulation of the minimum weight disk multi-cover problem; for a large constant Q, pick ∗ ≥ 1/Q; then x ∗ < 1/Q for each residual disk D all those disks D with xD D and thus each residual point p must be fractionally covered by Qrp disks by the residual disks, where rp is the residual covering requirement of point p; then in terms of expectation, there will be enough distinct disks to cover the residual points. Notice that the desired property holds in terms of expectation does not guarantee its occurrence in the sampling process. If the number of distinct disks fractionally covering a point is far less than required, a forcing step is needed. Again, one has to guarantee that the probability of forcing is sufficiently small. This is much more involved for the multi-cover problem. Interested readers may refer to [56].

8.4 Minimum Power Cover In this section, we consider the wireless sensor network in which sensors can adjust their power and hence their coverage range can be varied. Denote by p(s) the power of a sensor s and by r(s) the radius of the coverage range of s. Then they have the following relation. p(s) = r(s)α ,

(8.6)

where α is a constant between 1 and 6, determined by environment condition and called the attenuation factor of power. For sensors considered in this section, the coverage range of any sensor s is a disk Disk(s, r(s)) with center s and radius r(s). The aim of this section is to study the problem of minimizing the total power of sensors subject to certain coverage requirement, such as full coverage, partial coverage and multi-coverage. Let us explain them as follows. Problem 8.4.1 (Minimum Power Cover) Consider a set S of m sensors and a set T of n targets on the Euclidean plane. The problem is to determine the power at each sensor such that every target is covered by at least one sensor and the total power is as small as possible. When the service is required to be fault tolerant, each target may need to be covered not only once; the number of times would be determined by its importance. Problem 8.4.2 (Minimum Power Multi-Cover) Consider a set S of m sensors and a set T of n targets on the Euclidean plane. In addition, we are given a covering requirement crt for each t ∈ T . The problem is to determine the power at each

8.4 Minimum Power Cover

149

sensor such that all targets are fully covered (i.e., each target t is covered by at least crt distinct sensors) and the total power is as small as possible. The partial cover requirement means that only a prescribed number of targets should satisfy the coverage requirement. More issues and problems will be discussed in Chapter 10. Here, let us give an example. Problem 8.4.3 (Minimum Power Partial Multi-Cover) Consider a set S of m sensors and a set T of n targets on the Euclidean plane. In addition, we are given a covering requirement crt for each t ∈ T and a positive integer k. The problem is to determine the power at each sensor such that at least k targets are fully covered and the total power is as small as possible. The partial cover requirement arises from the situation in the real world that to accomplish the complete perfection, it requires unbearable amount of effort [360]. The 1-dimensional minimum power cover problem can be solved in polynomialtime, and faster approximation algorithms were also studied [18, 72, 322]. In the Euclidean plane, its NP-hardness was proved by Bilo` et al. [72] for α ≥ 2 and by Alt et al. [18] for α > 1. Charikar and Panigrahy [117] designed a Primal–Dual approximation algorithm with a constant performance ration for the minimum power cover problem. Lev-Tov and Peleg [322] constructed a PTAS for the minimum power cover problem with α = 1. Bilo` et al. [72] obtained a PTAS for general α ≥ 1. For the minimum power multi-cover problem, Abu-Affash et at. [2] considered a special case that α = 2 and the covering requirement is uniform (i.e., crt = k for every target t). They designed an approximation algorithm with performance ratio at most 23.02 + 63.91(k − 1). Bar-Yehuda and Rawitz [63] constructed a localratio algorithm with approximation performance ratio at most 3α · crmax . Note that both approximation performance √ ratios in [2, 63] depend on crmax . Bhowmick et al. [69, 70] constructed a (2d) · (27 d)α -approximation algorithm for the minimum power multi-cover problem in the d-dimensional Euclidean space Rd , which has the performance ratio independent from crmax , but depends on the dimension d. Finally, Bhowmick et al. [71] gave an O(1)-approximation for the minimum power multi-cover problem; the performance ratio is also independent from the dimension d and actually, they studied the problem in a general metric space instead of the Euclidean space. For the minimum power partial cover problem, Freund and Rawitz [199] mentioned a (12 + ε)-approximation for α = 2 and Li et al. [348] constructed a 3α -approximation for α ≥ 1. For α = 2, 2α = 9 < 12 + ε. The minimum power partial multi-cover problem is first studied by Ran et al. [435] for α ≥ 1 under the assumption that crmax is upper bounded by a constant. They generalized the work of [322] and [72] by incorporating partial cover requirement and multi-cover requirement. It also improved the constant-approximation performance ratio for the minimum power multi-cover problem obtained in [69] and [70] to PTAS in the case that the maximum covering requirement is a constant.

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8 Heterogeneous Sensors

The algorithm and analysis in [435] are based on the shifted quadtree technique combined with a dynamic programming. Note that the same idea was used in [322] to obtain a PTAS for the minimum power cover problem with α = 1. Why only α = 1 was considered in [322]? It is because a property called semi-disjointness can be satisfied for α = 1, but not for α ≥ 2. Some insightful ideas are explored in [72] to overcome the difficulty brought by the lack of semi-disjointness. In [435], they further explore geometric properties and the speciality of the power formulation to deal with difficulties brought by the partial requirement and multicover requirement. There are many publications on energy-efficiency or power-saving with various coverage consideration [369, 594], The reader may find a useful survey [411].

8.5 Composite Event Coverage There is a Chinese story, called “blind men touch an elephant” (Fig. 8.2). The story says that a king asked four blind men to touch an elephant and then report what the elephant likes. The first one touched a leg and said: “The elephant seems like a pole.” The second blind man touched the nose and said: “The elephant likes a bamboo.” The third one touched the body and said: “The elephant likes a wall.” The fourth one touched the tail and said: “The elephant likes a rope.” There is also an Indian version, called “six blind men touch an elephant.” When monitoring a composite event with sensors, the situation is similar to this story. Actually, the study of composite event coverage is based on the following two observations. Fig. 8.2 Blind men touch an elephant

8.5 Composite Event Coverage

151

First, each sensor may obtain information on several components, such as temperature, humidity, and ambient light. For example, a production of Crossbow Technology, Inc., MTS400 multi sensor board enables a sensor to sense temperature, humidity, barometric pressure, and ambient light [595]. Secondly, each event may consist of not only one component. For example, a fire may be detected by two components, the temperature reaches a threshold and the light also reaches a threshold. An event is called an atomic event if it consists of only one component. Otherwise, it is called a composite events [209, 337, 509, 595]. Yang et al. [595] formulated an optimization problem as follows. Consider M sensing components x1 , x2 , . . . , xM . Each sensor is able to sense a subset of {x1 , x2 , . . . , xM }. For different components, sensing ranges may be different. Denote by Ri (s) the sensing range for component xi of sensor s. If s cannot sense component i, then Ri (s) is empty. All sensors are battery powered and cannot be recharged. They have two modes, active and sleep. Problem 8.5.1 (Composite Event Detection) Consider a set of sensors deployed in a plane for monitoring a composite event {x1 , x2 , . . . , xM } in an area, design a sensor scheduling mechanism to maximize the lifetime of composite event coverage, i.e., the time length for a period during which the set of active sensors ensures the coverage for all components and connectivity requirements. Yang et al. [595] proposed two heuristics and leave the following open. Open Problem 11 Is there a polynomial-time constant-approximation for the composite event detection problem? Event detection and coverage are closely related issues. The reader may find more from [312, 362, 652] about them.

Chapter 9

Grid-Based Deployment

All the fingerprint paintings are done without a grid. Chuck Close

9.1 Motivation and Overview For applications in agriculture, sensors are often placed at grid points. Which grid (or pattern) minimizes the number of sensors and reach the full (area) coverage? It is not hard to show that the optimal solution is the triangle grid such that each cell is an equilateral triangle [287]. However, when the requirement is the full coverage together with others, such as k-connectivity, the optimal solution would be nontrivial. Actually, there exist a lot of research efforts along this direction in the literature. When consider the connectivity, the ratio rc /rs plays an important role where rc and rs are the communication radius and the sensing radius of the sensor, respectively. For example, the triangle pattern was shown not to be optimal for √ rc /rs 3 [525] and for rc = rs [269]. For all values of rc /rs or within a large range, optimal patterns for full coverage and k-connectivity were determined by Bai et al. [46] for k = 2, by Bai et al. [48] for k = 4, and by Bai et al. [47] and Yun et al. [611] for k ≤ 6. The optimality in above papers is asymptotic, that is, the optimality is reached as the area goes to infinitely large. Yu et al. [606] put their attention in bounded area, such as a rectangle. Optimal patterns have been also studied for coverage with localization [350, 352], k-coverage [29, 53], barrier coverage [27, 28], directed sensors [604, 607], and camera sensors [608]. In the following sections, we will briefly introduce some optimal patterns mentioned above.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_9

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9 Grid-Based Deployment

9.2 Coverage with Connectivity First, let us make clear what means asymptotically optimal. Theorem 9.2.1 (Bai et al. [46]) Consider a square D with edge length L. Let N (rc , rs ) be the minimum number of sensors with communication radius rc and sensing radius rs , covering D such that centers of sensors form a 2-connected network. Then lim π rs2 N(rc , rs ) = K(rc , rs )L2 ,

rs →0

where  K(rc , rs ) =

√ if π/3 ≥ φ, 2π 3/9 π −2φ π/(sin φ + 2 sin 2 ) otherwise,

and φ = arccos 2rrcs . Bai et al. [46] found patterns such that for sensors lying in those patterns, if n(rc , rs ) is the minimum number of sensors covering the square D and forming a 2-connected communication network, then lim π rs2 n(rc , rs ) = K(rc , rs )L2 .

rs →0

Therefore, those patterns are said to be asymptotically optimal. Those patterns are constructed as shown in √ Fig. 9.1, called strip-based deployment. There are two  parameters α = min{rc , √ 3rs } and β = rs + rs2 − α 2 /4. Note that π/3 ≥ φ is equivalent √ to rc /rs ≤√ 3. The construction is divided into two cases based on rc /rs ≤ 3 or rc /rs√> 3. √ In case rc /rs ≤ 3, the pattern is shown in Fig. 9.1a. In case rc /rs < 3, the pattern is shown in Fig. 9.1b. The difference is on the left and right boundaries.

Fig. 9.1 Optimal patterns for coverage with 2-connectivity

9.3 Coverage with Localization

155

Some sensors are added in (b), but not in (a). This is because in (a), all sensors √ form a connected communication network. However, in (b), since rc /rs < 3 those sensors in (a) cannot be connected and hence it needs additional sensors to establish the 2-connectivity. In [47], asymptotically √ optimal pattern is constructed for coverage with 4connectivity and rc /rs > 2. In [47, 611], a complete set of deployment patterns for all values of rc /rs was proposed for coverage with k-connectivity k ≤ 6. However, optimality is established only for certain cases as follows: • For k = 3, 5, optimality is proved among regular deployments for rc /rs ≥ 1 and is conjectured for rc /rs < 1. √ • For k = 4, optimality is conjectured for rc /rs ≤ 2. √ • For k = 6, optimality is proved √ among regular deployment for rc /rs ≥ 3 and is conjectured for rc /rs < 3. There is a bad news for above conjectures: An interesting phenomenon in pattern deployment was discovered in [612], called pattern mutation. Using pattern mutation, a class of new patterns have been constructed and they are optimal for small rc /rs , which are against above conjectures.

9.3 Coverage with Localization In this section, we assume rc /rs ≥ 2 so that the full coverage implies the connectivity. For simplicity of description, we use the sensing disk that represents the sensor. Suppose that n disks are deployed and fully cover an area of interest, . Then  is divided into many small areas. Let us encode each point by the set of disks (sensors) covering the small area. Then each small area is a set of points with the same code. Such a small area is called the basic cross area. The identifying code of a basic cross area is the code of every point in the area. For example, in Fig. 9.2, There are seven basic cross areas with their identifying codes. Now, if a target moves into a basic cross area, then every sensor in the identifying code of the basic cross area would find the target. From information collected from sensors, one may identify the basic cross area where the target is located. In [350, 352], a problem is proposed as follows. Fig. 9.2 Seven basic cross areas with their identifying codes

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9 Grid-Based Deployment

Fig. 9.3 The hexagon pattern

Fig. 9.4 The square pattern deployment

Problem 9.3.1 (Optimal Localization-Deployment) Given a bounded area , a disk radius R, and a localization accuracy requirement d, find a deployment for minimum number of disks to satisfy the following conditions: (a) full coverage for , (b) for every basic cross area A: dia(A) ≤ d where dia(A) is the diameter of A. This problem was studied based on the hexagon pattern (Fig. 9.3) and the square pattern (Fig. 9.4). Actually, the hexagon pattern in [350, 352] is the triangle pattern since all centers of hexagons are also considered as grid points. Each cell is a triangle described by two edges with lengths r and r  , respectively, together with the angle θ between them. When r = r  and θ = π/3, this triangle becomes the equilateral one and the triangle pattern becomes the regular triangle one. In the triangle pattern deployment, all disks of radius R are placed at grid points, that is, the center of each disk is identified with a grid point. The following result is obtained in [350, 352]. √ Theorem 9.3.2 For (2/ 7)R ≤ d ≤ 2R, the optimal solution for the optimal localization-deployment problem among triangle pattern is reached when r/R =  √ ( 3d/R + 16 − (d/R)2 )/4. For the square pattern, each cell is a square with edge length r. In the square pattern deployment, every grid point is deployed with a disk with radius R. When R = 1, the length of segment AB in Fig. 9.4 is determined by r, denoted by f (r). In [350, 352], the following has been proved.

9.3 Coverage with Localization

157

 √ √ Theorem 9.3.3 For 5 2/65 ≤ d/R ≤ f (4 (18 + 4 2)/219), the optimal solution for the optimal localization-deployment problem among the square pattern  2 )/5. is reached at r/R = 2d/R + ( 20 − (d/R)  √ For f (4 (18 + 4 2)/219) < d/R ≤ 2, the optimal solution for the optimal localization-deployment problem among the square pattern is reached at r/R =  √ ( 2d/R + 16 − 2(d/R)2 )/4. In [289, 290], there is another type of work about coverage with localization.

Chapter 10

Barrier Coverage

The barrier between success is not something which exists in the real world; it composed purely and simply of doubts about ability. Franklin D. Roosevelt

10.1 Motivation and Overview A region is a belt if its boundary consists of two parts such that every point in one part has equal distance to the other part. This distance is called width and these two parallel boundary parts are called two banks. (When a belt is considered as river, two boundary parts are two banks.) For example, ring and strip are belts. A belt is closed if it is a closed and bounded region, such as ring. An open belt can be seen as a piece of a closed belt between two parallel lines (Fig. 10.1). In such a case, the boundary on the two lines are considered to be open and called belt-ends. Hence, an open belt keeps its boundary consisting of two banks and two belt-ends. For simplicity, from now on, by a belt, we mean an open belt since the closed belt can be turned to an open belt easily. In fact, use a line to cut a closed belt. Then the closed belt can be turned to an open belt and what we do for an open belt can be extended to a closed belt without any trouble. Usually, a belt is a boundary region used for protecting an area of interest. A belt is separated by a curve if any walk from one bank to the other bank must cross the curve. Such a curve is called a separating curve. A set of sensors is called a barrier cover if it covers a curve separating considered belt. Therefore, a barrier cover is used for protecting an area of interest (Fig. 10.2). For simplicity, we may assume that the considered belt has two vertical belt-ends and the area of interest lies below the belt. Suppose an intruder lying above the belt wants to get in the area of interest. Then he must cross the belt. Therefore a set of sensors is a barrier cover if and only if along any walk, the intruder must be detected by at least one sensor. This explanation can be used for generalization of barrier cover. A set of sensors is a k-barrier cover if along any walk, the intruder must be detected by at least k sensors. © Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_10

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Fig. 10.1 An open belt and a closed belt

Fig. 10.2 A barrier cover is a set of sensors covering a curve connecting two open belt-ends

The notion of barrier cover was proposed by Kumar et al. [313]. Kumar et al. [315] study the maximum lifetime barrier coverage problem. They showed that this problem is polynomial-time solvable and following three statements are equivalent in homogeneous wireless sensor networks. 1. There exists a k-barrier cover. 2. There exists a set of disjoint k barrier cover. 3. There exists a sleep–wakeup schedule with lifetime at least k, i.e., it does not increase the lifetime of barrier coverage by allowing every sensor alternatively to make state change between sleep and wakeup. Therefore, it is enough to study disjoint barrier covers for interest in lifetime maximization and k-barrier cover. Problem 10.1.1 (Maximum Lifetime Barrier Coverage) Suppose a set of homogeneous sensors, S, are randomly deployed into an open belt. Every sensor has a unit lifetime and the same sensing radius Rs . Find the maximum number of disjoint barrier covers. Kim et al. [297, 300] found a security problem when schedule disjoint barrier covers to reach the longer lifetime of barrier coverage. They call this problem as the barrier-breach problem and proposed three approaches to solve this security problem. Consider two barrier covers B1 and B2 for a horizontal belt B (Fig. 10.3). They are not overlapping and B1 lies above B2 . If we schedule B2 works the first and B1 follows, then the intruder can first walk to point a during the period that B1 sleeps and then gets in the area of interest during the period that B1 works and B2 does not, so that the intruder walks into the area of interest, not detected by any sensor. However, if B1 is scheduled to work before B2 , then the barrier-breach would not

10.1 Motivation and Overview

161

Fig. 10.3 Two nonoverlapping barrier covers

Fig. 10.4 Two disjoint barrier covers crossing each other

Fig. 10.5 Three disjoint barrier covers crossing each other, but still have secure schedule

occur. The scheduling without barrier-breach is called the secure scheduling. Does a secure scheduling always exist for a set of disjoint barrier covers? The answer is no. Assume that B1 and B2 cross each other, i.e., there exist two points a1 and a2 such that a1 is below B1 and above B2 , and a2 is below B2 and above B1 (Fig. 10.4). Now, if B1 is used first and B2 is used second, then an intruder can first get at point a2 to pass B2 and then pass B1 , not seen by any sensor, vice versa. A set of disjoint barrier covers in a belt B is said to be non-penetrable if there exist k non-crossing curves, each separating the belt B, such that these k curves are covered by k disjoint barrier covers, respectively. Non-penetrability is a sufficient condition for a set of disjoint barrier covers to have a secure schedule. In fact, a secure schedule can be found by arranging all barrier covers to work in the updown ordering. Given a set of barrier covers, how to find a maximum non-penetrable subset? Kim et al. [297, 300] proposed three methods. The non-penetrability is not necessary for a set of barrier covers to have a secure schedule. In Fig. 10.5, there are three barrier covers B1 , B2 , and B3 . They cross each other. However, they have a secure schedule (B3 , B2 , B1 ). What is the necessary and sufficient condition for a set of barrier covers to have a secure schedule? Zhang et al. [639] solved this problem and also showed that the following two problems are equivalent:

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Problem 10.1.2 (Max Secure Schedule) Given a set of disjoint barrier covers consisting of homogeneous sensors, find a maximum subset with secure scheduling. Problem 10.1.3 (Longest Path) Given a directed graph G = (V , E) with arc length c : E → R+ and two nodes s and t, find a simple path from s to t with the maximum total length. Moreover, Zhang et al. [639] also extended this result from homogeneous sensors to heterogeneous sensors. The quality of a barrier cover may be measured by its width, which is the shortest distance crossing the monitoring area of the barrier cover. Wu et al. [572] formulated several optimization problems on construction of barrier covers with quality requirement. To improve the quality of a barrier cover, mobile sensors are useful. There are more issues regarding improving quality by mobile sensors. For example, an optimization problem can be raised by considering the tradeoff between moving distance and the quality improvement while each mobile sensor has energy limitation. There are many other research issues on the barrier cover. For example, the critical conditions for weak and strong k-barrier coverage with omnidirectional sensors are studied in [313] and [368], respectively. Later, dedicated sensors (e.g., cameras [469], radars [228], UAVs [482], etc.) are taken into consideration and various wireless sensing models including directional sensing model [65] and probabilistic sensing model [345] are considered for barrier coverage problem. The sensor deployment strategy is also a big issue which attracted many interests. In [130], different protocols are designed for one-way barrier coverage under different sensor models. Multi-round sensor deployment is studied in [588]. In [251], He et al. proposed a condition under which line-based deployment is suboptimal, while indicating the advantage of curve-based deployment in terms of vulnerability. With the deployment of mobile sensor networks [171, 250, 328], mobile sensors are introduced to improve the QoS (quality of service) for barrier coverage and reduce the cost of sensor node deployment. In [451], Saipulla et al. indicated how to relocate sensors with limited mobility in order to improve barrier coverage after random sensor deployment. They then study the problem of barrier coverage of a line-based sensor deployment strategy and how to exploit sensor mobility to improve barrier coverage [452]. The mobile sensors are further studied in [543] when sensor nodes have location errors and authors propose a progressive deployment method to improve barrier coverage by deploying mobile sensor nodes after initial deployment. A randomized sleep-wakeup algorithm, Randomized Independent Sleeping (RIS), is proposed in [313]. In this algorithm, time is slotted and in each time slot each sensor independently decides to sleep or stay active with a predefined probability p. The shortcoming of this work is that, their results cannot provide guarantee of barrier coverage, and there is no trivial method to choose a reasonable value for p. A localized algorithm, Localized Barrier Coverage Protocol (LBCP), is proposed in [125] to improve the lifetime of a network deployed for barrier coverage. Although the performance of LBCP is statistically close to optimal for

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some parameter settings, no performance guarantees are provided, neither does it address the case in which sensor nodes are heterogeneous in lifetime. Kumar et al. have shown in [315] that the problem of sleep-wakeup scheduling for lifetime maximizing in k-barrier coverage is polynomial-time solvable. They propose two optimal algorithms, Stint and Prahari, for homogeneous and heterogeneous networks, respectively. This chapter serves as an introduction to this subject. Following sections will contribute to a little detailed discussion on [314, 315, 572, 639]. There are many research topics on barrier coverage [40, 54, 135, 159, 182, 190, 191, 276, 298, 302, 321, 341, 387, 388, 402, 403, 412, 479, 486, 488, 491, 545, 563, 632, 637, 640]. A good survey can be found in [569].

10.2 Disjoint Barrier Covers Consider a set of homogeneous wireless sensors with unit lifetime. Let us construct a coverage digraph as follows (Fig. 10.6) : • Each sensor u is represented by an arc (u+ , u− ) with capacity one. • The node set of G consists of u+ and u− for all sensors u and two additional nodes s and t which represent two belt-ends. • Two sensors u and v have their sensing disks overlapping if and only if there exists an arc (u− , v + ) with capacity +∞. • The belt-end represented by s overlaps with the sensing disk of u if and only if there exists an arc (s, u+ ) with capacity +∞. • The belt-end represented by t overlaps with the sensing disk of u if and only if there exists an arc (u− , t) with capacity +∞. The following theorem states important properties of the coverage digraph. Theorem 10.2.1 Following facts hold:

Fig. 10.6 Coverage digraph

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(a) A subset of sensors is a barrier cover if and only if it induces a set of nodes which together with s and t contains a path from s to t. (b) A subset of sensors is a minimum barrier cover, i.e., a barrier cover with minimum cardinality if and only if it induces a shortest path from s to t. (c) There exists a maximum flow which can be decomposed into arc-disjoint pathflows. Proof (a) is trivial. (b) follows immediately from (a). For (c), note that the capacity of every arc is integer. Hence there exists a maximum flow such that the flow function value at each arc is an integer. Therefore, the value is 0 or 1 on arcs representing sensors. This fact implies that the maximum flow can be decomposed into arc-disjoint path-flows each with flow value one.   Following are corollaries. Corollary 10.2.2 (Kumar et al. [315]) The value of maximum flow from s to t in the coverage digraph is equal to the maximum number of disjoint barrier covers. Proof Since the capacity of every arc representing a sensor is one, there exists a maximum flow such that the flow function value at such an arc is an integer, i.e., 0 or 1. Therefore, it can be decomposed into arc-disjoint flows each corresponding a path from s to t. Since the arc from u+ to u− with capacity one represents sensor u, the arc-disjoint paths would give corresponding sensor-disjoint barrier covers.   The following follows immediately from above proof. Corollary 10.2.3 For a set of sensors in an open belt, a k-barrier cover exists if and only if the maximum lifetime of barrier coverage is at least k. In Chapter 5, we see that when each sensor is allowed to sleep and wakeup alternatively, the lifetime of coverage can get longer. However, such a case does not hold for the barrier coverage as described in following theorem. Theorem 10.2.4 The maximum value of following problem can be achieved by a set of disjoint barrier covers: Given a set of sensors, S in an open belt, find the maximum lifetime of barrier coverage, i.e., the length of longest time period such that at each time moment, all active sensors form a barrier cover under constraint that every sensor has total active time not exceed one. Proof Suppose the sequence of barrier covers each together with time period, (S1 , t1 ), (S2 , t2 ), . . . , (Sk , tk ), is a possible solution where barrier cover Si is activated for time length ti . We construct a flow in the coverage digraph G as follows: Note that the subgraph of G, induced by {s, t} ∪ {u+ , u− | u ∈ Si } must contain a path from s to t. Along this path, we send a flow with value ti from s to t. Then we would obtain a flow from s to t with total value t1 +t2 +· · ·+tk . Therefore, its value cannot exceed the maximum flow value of the flow network G. Conversely, suppose the maximum flow has value k. Then by Theorem 10.2.1 (c), there exists a maximum flow which can be decomposed into k node-disjoint path flow each with value one, which induces k disjoint barrier covers. These k disjoint barrier covers give the barrier coverage with time length k.  

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10.3 Secure Schedule with Homogeneous Sensors A set of scheduled barrier covers has a barrier-breach if the intruder is able to cross the belt without monitored by any sensor. A schedule of barrier covers is secure if it does not have a barrier-breach. Consider a horizontal belt B. Recall that the area of interest is below B and the intruder is located above B. Let B1 and B2 be two barrier covers which cover two separating curves C1 and C2 , respectively. Suppose C1 and C2 are disjoint and there is a point a below the sensing area of B1 and above the sensing area of B2 . If we first use B1 and secondly use B2 , then the intruder cannot cross the belt without discovered by any sensor. This means that schedule (B1 , B2 ) is secure. However, if we first use B2 and secondly use B1 , then the intruder may first get at point a and after B2 stops working, go from point a to protected area. Then no sensor can find the intruder’s walk. This means that the schedule (B2 , B1 ) is not secure. Two barrier covers B1 and B2 are said to cross each other if that there exists a point a1 below the sensing area of B1 and above the sensing area of B2 , and there also exists a point a2 below the sensing area of B2 and above the sensing area of B1 . Theorem 10.3.1 (Kim et al. [300]) For a set of two barrier covers B1 and B2 , there is a secure schedule if and only if B1 and B2 do not cross each other. Proof If B1 and B2 cross each other, then both (B1 , B2 ) and (B2 B1 ) are not secure and hence there does not exist a secure schedule. Next, we assume that B1 and B2 do not have a secure schedule and then prove that B1 and B2 cross each other. Let C1 be the lower boundary of the sensing area of B1 and C2 the upper boundary of sensing area of B2 . If C1 and C2 cross each other, then there exists a point a1 lying below C1 and above C2 . Hence, point a1 lies below B1 and above B2 . If C1 and C2 do not cross each other. In case that C1 is above C2 , (B1 , B2 ) is a secure schedule and in case that C1 is below C2 , (B2 , B1 ) is a secure schedule. Let C1 be the upper boundary of the sensing area of B1 and C2 the lower boundary of sensing area of B2 . By a similar argument, we would see that either there exists a point a2 lying below B2 and above B1 , or there exists a secure schedule. Since B1 and B2 do not have a secure schedule, above argument showed that B1 and B2 cross each other.   For a set of k barrier covers with k ≥ 3, the nonexistence of crossing pair of barrier covers is a sufficient condition for the existence of a secure schedule. Theorem 10.3.2 (Kim et al. [300]) If there exists disjoint k curves C1 , . . . , Ck each separating the belt and covered by barrier covers B1 , . . . , Bk , respectively, then there exists a scheduling to avoid the barrier-beach. Proof Suppose Ci is above Ci+1 for i = 1, 2, . . . , k − 1. Then (B1 , . . . , Bk ) is clearly a secure schedule.  

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How can we remove the minimum number of barrier covers to make remaining barrier covers without crossing? This problem can be reduced to the minimum vertex cover problem in graphs. Problem 10.3.3 (Minimum Vertex Cover) Given a graph, find the minimum subset of vertices such that every edge has at least one endpoint in the subset. (Such a subset of vertices is called a vertex cover.) Consider k barrier covers B1 , . . . , Bk . Construct a graph with those barrier covers as vertices. Put an edge between Bi and Bj if and only if Bi and Bj cross each other. This graph is called the crossing graph of barrier covers B1 , . . . , Bk . Then a subset of vertices is a vertex cover if and only if removal of them results in removal of all edges, that is, no cross-point exists. The minimum vertex cover problem is NP-hard. However, it is well-known that it has a polynomial-time 2-approximation. There exist several constructions. A simple one is to find a maximal matching and collect all endpoints of edges in the maximal matching. This collection is a 2-approximation for the minimum vertex cover problem. Theorem 10.3.4 (Kim et al. [300]) There exists a polynomial-time 2-approximation for the minimum number of barrier covers that the removal of them can result in the existence of a secure schedule for the remaining barrier covers. It is interesting to see directly on k barrier covers B1 , . . . , Bk with the algorithm for finding a maximal matching. 2-Approximation for Minimum Removed Set input k barrier covers B1 , . . . , Bk . C ← {B1 , . . . , Bk }; D ← ∅; while C contains two barrier covers Bi and Bj crossing each other do C ← C − {Bi , Bj } and D ← D ∪ {Bi , Bj }; output D. We may also view the problem from another point, i.e., to maximize the number of remaining barrier covers without crossing, i.e., non-penetrable barrier covers. However, this problem would reduce to the maximum independent set problem, for which the best known polynomial-time approximation algorithm has performance ratio O(k/ log 2 n). Problem 10.3.5 (Maximum Independent Set) Given a graph, find a maximum subset of vertices such that no edge exists between two vertices in the subset, i.e., no pair if vertices in the subset are adjacent. (Such a subset is called an independent set.) Note that the coverage graph of a set of sensors in the Euclidean plane has a geometric structure. Usually, a geometric optimization problem has much better approximation algorithm than general graph version of the problem. This leaves us an open problem: For k barrier covers obtained from the coverage graph of a set of

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Fig. 10.7 Construct barrier covers with a given crossing graph

sensors on the Euclidean plane, is there a good approximation algorithm for finding the maximum subset of non-penetrable barrier covers? Theorem 10.3.6 For any graph G, there exists a set of barrier covers whose crossing graph is isomorphic to G. Proof Suppose G contains n nodes v1 , . . . , vn . We will construct n barrier covers B1 , . . . , Bn corresponding to n nodes, respectively, such that the crossing graph of {B1 , . . . , Bn } is isomorphic to G. Consider an open belt of enough length. Draw a straight separating horizontal line. For each edge (vi , vj ) ∈ E(G), put two points aij and bij on the line. All such points apart each other by a distance at least 3. Cover this separating line by a set B0 of sensors with unit sensing disk (note: a unit sensing disk has unit diameter) such ij ij that for each point aij (or bij ), there exists a sensor sa lying at aij (or sensor sb lying at point bij ). At each point aij , construct four unit disks as shown in Fig. 10.7 such that two disks below point aij and two disks above point aij ; they are sensing disks ij ij ij ij of sensors si1 , si2 , sj 1 , sj 2 , respectively. Similarly, for point bij , we also construct four unit disks as shown in Fig. 10.7 such that two disks below point bij and two ij ij ij ij disks above point bij ; they are sensing disks of sensors sj 3 , sj 4 , si3 , si4 , respectively. Now, for each i = 1, . . . , n, define barrier cover Bi by ij

ij

Bi = [B0 \ ({sa , sb | (vi , vj ) ∈ E(G)}] ij

ij

ij

ij

∪{si1 , si2 , si3 , si4 | (vi , vj ) ∈ E(G)} Then the crossing graph of {B1 , . . . , Bn } is isomorphic to G.

 

Corollary 10.3.7 Finding the maximum subset of non-penetrable barrier covers from a given set of barrier covers is equivalent to the maximum independent set problem for graphs. Moreover, its complement problem, i.e., finding the minimum removed subset of barrier covers from a given set of barrier covers in order to make

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the remaining barrier covers non-penetrable, is equivalent to the minimum vertex cover problem in graphs. Now, let us show a necessary and sufficient condition for a barrier cover schedule (B1 , . . . , Bk ) to be secure. Lemma 10.3.8 (Zhang et al. [639]) A barrier cover schedule (B1 , . . . , Bk ) is not secure, i.e., contains a barrier-breach if and only if there exists two consecutive barrier covers Bi and Bi+1 (1 ≤ i ≤ k − 1) such that there exists a point a lying above Bi and below Bi+1 . Proof If such two barrier covers Bi and Bi+1 together with point a exist, then a barrier-breach occurs as follows: When Bi works and all others are in sleeping, the intruder walks to point a. After Bi does not work and BI +1 wakes up, the intruder walks into the area of interest without getting detected by any sensor. Conversely, suppose a barrier-breach exists, i.e., the intruder stays at a point a when a barrier cover works and can walk into the area of interest after Bi stops working. Then a must lie above Bi and below Bi+1 .   Given a set of barrier covers B1 , . . . , Bk , construct a barrier-breach digraph for those barrier covers as follows. • Take {B1 , . . . , Bk } as vertex set. • No loop exists. • An arc (Bi , Bj ) exists if and only if there exists a point a lying above Bi and below Bj . The complement of the barrier-breach digraph is called the secure digraph. For example, the three barrier covers in Fig. 10.5 have the barrier-breach digraph and the secure digraph as shown in Fig. 10.8. From Lemma 10.3.8, it is easy to know the following. Corollary 10.3.9 A barrier cover schedule (B1 , . . . , Bk ) is secure if and only if it is a path in the secure digraph of {B1 , . . . , Bk }. This corollary reminds us the following problem. Problem 10.3.10 (Longest Path) Given a digraph G = (V , E), find a simple path with maximum number of arcs. Fig. 10.8 Barrier-breach digraph and secure digraph

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Actually, the longest path problem is closely related to the problem that we are studying as follows. Problem 10.3.11 (Longest Secure Schedule) Given a set of barrier covers with homogeneous sensors, find a maximum subset with a secure schedule. Theorem 10.3.12 (Zhang et al. [639]) The longest secure schedule problem is equivalent to the longest path problem. Proof Given a set of barrier covers, B, any secure schedule for a subset of barrier covers corresponding to a path in the secure digraph for B, vice versa. Therefore, the longest secure schedule problem is equivalent to the longest path problem for the secure digraph of the given set of barrier covers. Next, given a digraph G without loop, we show that there exists a set of barrier covers such that their secure digraph is isomorphic to G. To do so, it is sufficient to show that given a digraph G = (V , E) without loop, there exists a set of barrier covers whose barrier-breach digraph is isomorphic to G. Suppose G contains n nodes v1 , . . . , vn . We will construct n barrier covers B1 , . . . , Bn corresponding to n nodes, respectively, such that the barrier-breach digraph of {B1 , . . . , Bn } is isomorphic to G. Consider an open belt of enough length. Draw a straight separating horizontal line. For each arc (vi , vj ) ∈ E(G), put a point (i, j ) on the line. All such points apart each other with a distance at least 6. Cover this separating line by a set B0 of sensor with unit sensing disk (note: a unit sensing disk has unit diameter) such that ij for each point (i, j ), there exists a sensor s0 lying at (i, j ). At each point (i, j ), construct four unit disks as shown in Fig. 10.9 such that two disks below point (i, j ) ij ij ij ij and two disks above point (i, j ); they are sensing disks of sensors si1 , si2 , sj 1 , sj 2 , respectively. Now, for each i = 1, . . . , n, define barrier cover Bi by ij

Bi = [B0 \ ({s0 | (vi , vj ) ∈ E(G)} ji

∪{s0 | (vj , vi ) ∈ E(G)})] ij

ij

ji

ji

∪{si1 , si2 | (vi , vj ) ∈ E(G)} ∪{si1 , si2 | (vj , vi ) ∈ E(G)}. Fig. 10.9 Constructing barrier covers with given barrier-breach digraph

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Fig. 10.10 Three possible configuration at point (h, k). (a) shows the configuration if i ∈ {h, k} or h = i with (i, k) ∈ E(G) or k = i with (h, i) ∈ E(G). (b) shows the configuration if i = h and (i, k) ∈ E(G). (c) shows the configuration if i = k and (h, i) ∈ E(G)

We next show that the barrier-breach digraph of {B1 , . . . , Bn } is isomorphic to G. To do so, it suffices to prove that (vi , vj ) ∈ E(G) if and only if there exists a point a above Bi and below Bj . From the construction of Bi , we see that for any point (h, k) with arc (vh , vk ) ∈ E(G), Bi has three possible positions as shown in Fig. 10.10, corresponding to three cases (a) i ∈ {i, j }, (b) h = i, (c) k = i, respectively. Thus, for (vi , vj ) ∈ E(G), Bi is in position (b) at point (i, j ) and Bj is in position (c) at point (i, j ), that is, point (i, j ) lies above Bi and below Bj . Conversely, if there is a point a lying above Bi and below Bj , then there must exist point (h, k) for (vh , vk ) ∈ E(G) lying above Bi and below Bj , that is, Bi is in case (b) and Bj is in case (c). Therefore, h = i and k = j . Therefore the barrier-breach digraph of {B1 , . . . , Bn } is isomorphic to G.   About the longest path problem, there are several well-known facts. • The longest path problem is NP-hard [456]. • The longest path problem does not have any polynomial-time n1−ε approximation for any ε > 0 unless NP = P [74]. • The longest path problem √ has a polynomial-time approximation with performance ratio n/exp(( log n) [201]. By Theorem 10.3.12, these facts can be extended to the longest secure schedule problem. The longest secure schedule problem is based on given set of disjoint barrier covers with homogeneous sensors. However, from a given set of homogeneous sensors, we may obtain different sets of disjoint barrier covers. Therefore, we still have following research problems left. 1. Given a set of homogeneous sensors, how to find a set of disjoint barrier covers with the longest secure schedule? 2. Given a set of homogeneous sensors, can the longest secure schedule be always achieved by the maximum set of disjoint barrier covers, i.e., the maximum flow in the coverage digraph?

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Fig. 10.11 There are exponential number of maximum flows

Fig. 10.12 Computing the number of maximum flows is #P -complete

3. Given a set of homogeneous sensors, how many maximum flows exist in the coverage digraph? 4. Given a set of homogeneous sensors, what is the computational complexity of computing the number of maximum flows in the coverage digraph? The first problem is the goal. Other three are helpful to find the solution of the first problem. For the third and the fourth problems, we have following solution. Theorem 10.3.13 Given a set of homogeneous sensors, there may exist an exponential number of maximum flows in the coverage digraph and computing the number of maximum flows is #P -complete. Proof An example is given in Fig. 10.11 that there may exist an exponential number of maximum flows. To see the #P -completeness, we choose following #P -complete problem to do reduction. Problem 10.3.14 (Counting Perfect Matchings) Given a bipartite graph G, count how many perfect matchings G has. As shown in Fig. 10.12, a given bipartite graph is reduced to a flow network such that every perfect matching in the bipartite graph corresponds to a maximum flow in the flow network. Since the counting perfect matchings is #P -complete, we see that computing the number of maximum flows in a flow network is #P -hard. Moreover, given a flow network, determining whether the flow network has a maximum flow can be done in polynomial-time. Therefore, computing the number of maximum flows in a flow network is #P -complete.   Theorem 10.3.13 indicates that it is not easy to solve the first problem.

10.4 Secure Schedule with Heterogeneous Sensors Above study on secure schedule with homogeneous sensors can be extended to heterogeneous sensors.

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Fig. 10.13 Coverage digraph of heterogeneous sensors

First, for a set of heterogeneous sensors, the coverage graph can be constructed similarly as follows. • Each sensor u is represented by an arc (u+ , u− ) with capacity equal to the lifetime of sensor u. • The node set of G consists of u+ and u− for all sensors u and two additional nodes s and t which represent two belt-ends. • Two sensors u and v have their sensing disks overlapping if and only if there exists an arc (u− , v + ) with capacity +∞. • The belt-end represented by s overlaps with the sensing disk of u if and only if there exists an arc (s, u+ ) with capacity +∞. • The belt-end represented by t overlaps with the sensing disk of u if and only if there exists an arc (u− , t) with capacity +∞. An example is showed in Fig. 10.13. Lemma 10.4.1 In the coverage digraph of any set of sensors, the maximum flow f can be decomposed into at most n path-flows where n is the number of sensors. This means that the maximum lifetime barrier coverage would be given by at most n barrier covers. Proof Note that each path from s to t must pass through an arc in form (x + , x − ) for some sensor x. Since the capacity of arc (x + , x − ) is equal to the lifetime of sensor x, a flow passing through arc (x + , x − ) must have a finite value. It follows that the maximum flow f has value |f | < +∞. Suppose |f | > 0, i.e., there exists a barrier cover. Consider a path P from s to t in flow f . Since f is the maximum flow, there must exist an arc on P such that the capacity of the arc is equal to the flow value on the arc. This arc must be in form (x + , x − ) for some sensor x because each of other arcs has capacity +∞. Let fP be the path-flow along path P . Construct flow network G from G by reducing |fP | from the capacity of each arc on path P . Then f  = f − fP would be the maximum flow in G . Moreover, for some sensor x, the capacity of (x + , x − ) becomes zero and hence this arc can be deleted. Continue this

10.5 Quality

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Fig. 10.14 Transformation from node weight to arc weight

process until remaining flow has value zero. This process can last at most n steps. Thus f is decomposed into at most n path-flows.   By Lemma 10.4.1, for secure scheduling, we need to study the following problem. Problem 10.4.2 (Weighted Secure Schedule) Given a set of barrier covers each with a positive weight, find a subset of barrier covers with secure schedule to maximize the total weight. Construct the secure digraph with node weight which is equal to the weight of corresponding barrier cover. Then we may reduce the weighted secure schedule problem to the following. Problem 10.4.3 (Node-Weighted Path) Given a digraph with positive node weight, find a simple path with maximum total node weight. Theorem 10.4.4 The weighted secure schedule problem is equivalent to the nodeweighted path problem. Furthermore, the node-weighted path problem is equivalent to the longest path problem. Proof The first equivalence can be proved by a similar argument to the proof of Theorem 10.3.12. The second equivalence can be proved by transformation from node weight to arc weight as shown in Fig. 10.14.  

10.5 Quality The quality of a barrier cover is usually measured by its width, i.e., the shortest distance of walk crossing the sensing area of the barrier cover. A barrier cover is said to be w-simple if all sensors in it have an ordering s1 , . . . , sh satisfying following (Fig. 10.15): • For any i = 1, .., h − 1, sensing disks of si and si+1 have a common chord with length at least w. • The sensing disk of s1 intersects the left belt-end with a chord with length at least w.

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Fig. 10.15 A w-simple barrier cover

Fig. 10.16 Graph Gw 1

• The sensing disk of sh intersects the right belt-end with a chord of length at least w. There are four problems about quality of barrier covers studied in [572]. Problem 10.5.1 Given a set of homogeneous sensors in a belt and a positive width w, compute a minimum-size w-simple barrier cover, where the size of a barrier cover is its cardinality. Problem 10.5.2 Given a set of homogeneous sensors in a belt and a positive width w, compute the maximum number of disjoint w-simple barrier covers. Problem 10.5.3 Given a set of homogeneous sensors in a belt and a positive width w, compute a minimum-size barrier cover with width w. Problem 10.5.4 Given a set of homogeneous sensors in a belt and a positive width w, compute the maximum number of disjoint barrier covers with width w. In this section, we will present following results in [572]. • Problems 10.5.1 and 10.5.2 are polynomial-time solvable. • Problems 10.5.33 and 10.5.4 are NP-hard. Meanwhile, design heuristics for Problems 10.5.3 and 10.5.4. Theorem 10.5.5 (Wu et al. [572]) Problem 10.5.1 can be solved in O(n2 ) time, where n is the number of sensors. Proof Construct a w-coverage graph Gw 1 of sensors in the following (Fig. 10.16). • The node set of Gw 1 consists of all sensors and two additional nodes s and t representing the left and right belt-ends, respectively. • There exists an edge (u, v) between two sensors u and v if and only if sensing disks of u and v have a chord in common with length at least w.

10.5 Quality

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Fig. 10.17 Digraph Gw 2

• There exists an edge (s, u) between s and a sensor u if and only if the left belt-end intersects the sensing disk of u with a chord of length at least w. • There exists an edge (t, u) between t and a sensor u if and only if the right beltend intersects the sensing disk of u with a chord of length at least w. Clearly, any w-simple barrier cover corresponds to a path from s to t in graph Gw 1 and the minimum-size w-simple barrier cover corresponds to a shortest path from s to t in Gw 1 , vice versa. Therefore, Problem 10.5.1 is equivalent to computing the shortest path between s and t in Gw 1 . Note that every edge is assumed to have length one. Therefore, we can have an algorithm for this shortest path problem, with running time O(m) where m is the number of edges in Gw 1 . The proof is completed   by noting m = O(n2 ). Theorem 10.5.6 (Wu et al. [572]) Problem 10.5.2 can be solved in O(n) time. Proof Construct a digraph Gw 2 (Fig. 10.17) as follows. • Put two nodes s and t to represent the left and the right belt-ends, respectively. • For each sensor x, put two nodes x + and x − together with an arc (x + , x − ) of capacity one. • If the sensing disk of sensor x intersects the sensing disk of sensor y with a common chord of length at least w, then add an arc (x − , y + ) with capacity +∞. • If the left belt-end intersects the sensing disk of sensor x with a chord of length at least w, then add an arc (s, x + ) of capacity +∞.

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• If the sensing disk of sensor x intersects the right belt-end, with a chord of length at least w, then add an arc (x − , t) with capacity +∞. It can be seen easily that any k w-simple barrier corresponds to k path-flow each with flow value one, which form a flow with value k from s to t in Gw 2 . Therefore, has a maximum flow with value at least k. Conversely, suppose Gw Gw 2 2 has the w maximum flow f with value k. Since the capacity at each edge in G2 is integer, the maximum flow f has integer value at each edge. Moreover, every path in flow f must pass at least one arc (x + , x − ) for some sensor x. This implies that f can be partitioned into disjoint paths each with flow value one. Hence, the maximum flow f can be decomposed into k path-flows each with value one; they correspond to k disjoint w-simple barrier covers. Therefore, Problem 10.5.2 is reduced to the maximum flow problem in Gw 2 . With Edmonds–Karp algorithm [161], this 2 computation takes time O(m n) = O(n5 ) where m is the number of edges in Gw 2.   Theorem 10.5.7 (Wu et al. [572]) Problem 10.5.3 is NP-hard. Proof A cumbersome proof is sketched as follows. The reduction is from the following well-known NP-complete problem. Problem 10.5.8 (Exact-3-Set-Cover) Given a collection C of 3-subsets of a set X with 3n elements, determine whether C contains n 3-subsets covering X, where a 3-subset means a subset with three elements. The construction of the reduction is as follows: On the left belt-end, put |C| disks (sensors); each represents a 3-subset in C. On the right belt-end, put 3n disks; each represents an element in X. If a 3-set A contains an element x, then connect disk A and disk x with a path with k disks where k is a properly chosen fixed positive integer. Each of such paths has width w/(3n), i.e., two adjacent disks intersect with a chord of length at least w/(3n) and there exist two adjacent disks intersecting with a chord of length exactly w/(3n). Those disk-paths may cross each other. Dealing with cross-points is a quite complicated issue. Finally, we have that C contains n 3-subset covering X if and only if there exists a barrier cover with width w.   Theorem 10.5.9 (Wu et al. [572]) Problem 10.5.4 is NP-hard. Consider the following well-known NP-complete problem. Problem 10.5.10 (Partition) Given n integers a1 , .., an , determine whether or not these n integers can be partitioned into two parts such that their sums are equal. First, we show that following special case is still NP-complete. Problem 10.5.11 (Positive Integer Partition) Given n positive integers a1 , .., an , determine whether or not these n integers can be partitioned into two parts such that their sums are equal. Lemma 10.5.12 The positive integer partition problem is NP-complete.

10.5 Quality

177

Proof The positive integer partition is a decision problem. First, note that this problem belongs to NP. To show its NP-hardness, for n integers a1 , .., an , construct 2n positive integers a1 + m, . . . , an + m, b1 , . . . , bn where m = 1 + max1≤i≤n |ai | and b1 = b2 = (n + 1)m, b3 = · · · = bn = m. Suppose a1 , . . . , an can be partitioned into two disjoint groups G1 and G2 with equal sums. Then partition a1 + m, . . . , an + m, b1 , . . . , bn into groups G1 = {ai + m | ai ∈ G1 } ∪ B1 and G2 = {ai + m | ai ∈ G2 } ∪ B2 where B1 contains b1 and |G2 |−1 elements of {b3 , . . . , bn } and B2 = {b1 , . . . , bn }\ B1 . Clearly, this partition of {a1 + m, . . . , an + m, b1 , . . . , bn } gives equal sums. Conversely, if {a1 + m, . . . , an + m, b1 , . . . , bn } can be partitioned into two groups with equal sums. Note that b1 and b2 must belong to two different groups. Therefore, a1 + m, . . . , a2 + m must belong to different groups, which induces a partition of {a1 , . . . , an } with equal sums. Therefore, the partition problem can be polynomial-time many-one reduced to the positive integer partition problem.   Proof of Theorem 10.5.9 We construct a polynomial-time many-one reduction from the positive integer partition problem to the decision version of Problem 10.5.4. Problem 10.5.13 (Decision Version of Problem 10.5.4) Given a set of homogeneous sensors in a belt, a positive width w and a positive integer k, determine whether there exists or not k disjoint barrier cover with width w.  Let a1 , . . . , an be n positive integers. Set w = 0.5 · ni=1 ai and k = 2. For each ai , construct a ai -simple barrier cover with width exactly ai , and moreover, make those n barrier covers apart each other (Fig. 10.18). Then a1 , . . . , an can be partitioned into two groups with equal sums if and only if those n barrier covers can be partitioned into two groups each of which forms a barrier cover with width w. That is, our construction is actually a polynomial-time many-one reduction from the positive integer partition problem to the decision version of Problem 10.5.4. Hence, Problem 4 is NP-hard.   Above proof also indicates that it is NP-hard to determine whether there exist or not two disjoint barrier covers with width w for a given set of sensors. Fig. 10.18 Disjoint ai -simple barrier covers

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10 Barrier Coverage

Fig. 10.19 Digraph Gw 3

Corollary 10.5.14 (Wu et al. [572]) For any ρ > 0.5, Problem 10.5.4 does not have a polynomial-time ρ-approximation unless NP=P. Proof For contradiction, suppose there is a polynomial-time ρ-approximation algorithm A for Problem 10.5.4 where ρ > 0.5. If a set of sensors has at least two disjoint barrier covers with width w, then algorithm A produces at least 2(= 2ρ ) disjoint barrier covers with width w. If a set of sensors does not have two disjoint barrier covers with width w, then algorithm A produces at most one barrier cover with width w. Thus, algorithm A gives a polynomial-time algorithm to determine whether or not given set of sensors contains two disjoint barrier covers with width w, a contradiction.   There are heuristics for Problems 10.5.3 and 10.5.4 designed in [572]. They use an algorithm to compute the maximum path-flow in following digraph Gw 3 (S), which is (Fig. 10.19) for any set S of sensors lying in a belt. • To represent the left belt-end, place two nodes s + and s − together with arc (s + , s − ) of capacity w. To represent the right belt-end, place nodes t + and t − together with arc (t + , t − ) of capacity w. • For each sensor x, place two nodes x + and x − together with an arc (x + , x − ) of capacity w. • If the sensing disk of sensor x intersects the sensing disk of sensor y with a chord of length c, then place an arc (x − , y + ) with capacity min(w, c). • If the left belt-end intersects the sensing disk of sensor x, with a chord of length c, then place an arc (s − , x + ) with capacity min(c, w).

10.6 Weak Barrier Cover and Local Barrier Cover

179

• If the sensing disk of sensor x intersects the right belt-end, with a chord of length c, then place an arc (x − , t + ) with capacity min(c, w). • For every sensor x, assign arc (x + , x − ) with cost one. For other arcs, assign each with cost 0. Theorem 10.5.15 (Wu et al. [572]) Computing a maximum path-flow from s + to 2 t − in Gw 3 (S) can be done in O(n log w) time. Proof Consider the following algorithm. Algorithm for maximum path-flow a ← 0 and b ← w; while a = b do begin delete all arcs with capacity less than (a + b)/2 from Gw 3 (S); if the remaining digraph contains a path from s + to t − then a ← the smallest arc capacity not less than (a + b)/2 else b ← the largest arc capacity not more than (a + b)/2; end-while find a path f from s + to t − in remainder of Gw 3 after deleting arcs with capacity less than a = b; output path flow f . This algorithm runs in O(n2 log w) time.

 

For details of those heuristics for Problems 10.5.3 and 10.5.4, the reader may find in [572].

10.6 Weak Barrier Cover and Local Barrier Cover Before end of this chapter, let us briefly discuss some variations of barrier covers. Weak Barrier Cover [313, 569] A set of sensors lying in a rectangular belt area is called a weak barrier cover if any intruder going from top to bottom along a vertical line would be detected by a sensor in the set (Fig. 10.20). The network flow approach can also be employed in study of weak barrier covers. To reduce the weak barrier cover problem to a flow problem, we may construct a flow network as follows. Each sensor x is represented by an arc (x + , x − ). For any two sensors x and y such that x lies on the left side of y, if there exists a vertical line passes through both sensing disks of x and y, add an arc (x − , y + ) (Fig. 10.21). The capacity on arcs would be defined based on what problem we are going to study. Local Barrier Cover [125, 127, 569] A set S of sensors lying in a rectangular belt area is called a L-local barrier cover if in every rectangular subarea with horizontal length at most L, S forms a barrier cover, that is, S would detect any intruder crossing from top to bottom along any path in the subarea (Fig. 10.22). For local

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10 Barrier Coverage

Fig. 10.20 Weak barrier cover

Fig. 10.21 Construct flow network for weak barrier cover

Fig. 10.22 Local barrier cover

barrier covers, it would be nice to see if there is a mathematical approach, such as flow networks, to deal with various problems. However, so far we did see such a proper mathematical tool in the literature. Open Problem 12 Find a novel mathematical approach to deal with the local barrier cover. In [49, 126, 128, 579], the quality of a weak barrier cover is defined as the largest L such that the weak barrier cover is also a L-local barrier cover. The worst quality is zero when a vertical line touches one sensor’s sensing disk on the left side and the other one’s sensing disk on the right side (Fig. 10.23).

10.6 Weak Barrier Cover and Local Barrier Cover

181

Fig. 10.23 The quality of a weak barrier cover can be zero

The reader is suggested to find from [569] for more variations of barrier covers.

Chapter 11

Sweep-Coverage

You can’t sweep other people off their feet, if you can’t be swept off your own. Clarence Day

11.1 Motivation and Overview In all previous coverage problems, targets or areas are required to be monitored all the time. But in many applications typically featured as patrolling, target points are only required to be monitored with certain frequency. For example, a set of static sensors are distributed in a forest, surveilling ecological environment. Information collected by those static sensors are gathered by a set of mobile sensors, say, at least once every three hours. One may be wondering what is the minimum number of mobile sensors that are needed to accomplish such a task, and how to design trajectories for those mobile sensors. Motivated by such a consideration, Cheng et al. [144, 340] proposed the following dynamic coverage problem. Definition 11.1.1 (Sweep-Cover) Consider a set of target points in a metric space. Each target point u is required to be monitored at least once in every time period tu (tu is called the sweep-period of u). A set of mobile sensors is called a sweep-cover if they can be scheduled to monitor all target points within required time periods, where a target point gets monitored by a mobile sensor if and only if the mobile sensor moves at the location of the target point. Problem 11.1.2 (Min-Sensor Sweep-Coverage) Find a sweep-cover with the minimum number of mobile sensors. From a dual point of consideration, Gao et al. [210] proposed a min–max version of the sweep-coverage problem. Problem 11.1.3 (Min-Sweep-Period Sweep-Coverage) Given m mobile sensors, find trajectories for these mobile sensors to form a sweep-cover such that the maximum sweep-period of target points is minimized.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_11

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11 Sweep-Coverage

The sweep-coverage problem was first studied under the background of robotic patrolling, and extensively studied nowadays in the field of wireless sensor networks. There are a lot of works on this problem with wide applications in data harvesting, routing for mobile chargers, etc. All existing performance guaranteed algorithms make use of Traveling Salesman Problem (TSP) as a step stone. Under the assumption that all mobile sensors move at the same speed v, Gorain and Mandal [230] proposed a 3-approximation algorithm for Min-Sensor Sweep-Coverage when all target points have the same sweep-period t; and Gao et al. [210] presented a 4-approximation algorithm for Min-SweepPeriod Sweep-Coverage. Ideas of these two algorithms will be discussed in details in the following. Other variations will be briefly introduced in the last section of this chapter.

11.2 Complexity The NP-hardness can be proved by a many-one reduction from the traveling salesman problem (TSP). In a TSP instance, knowing n cities and the traveling distance on every pair of cities, a salesman wants to find out a most economic tour visiting every city exactly once and then back to his office. In the language of graph theory, the goal of a TSP is to find a Hamiltonian cycle with the minimum length. The decision version of TSP is as follows: Given n cities and a distance function defined on all pairs of cities, together with a number L > 0, determine whether there exists a Hamiltonian cycle with length at most L. Theorem 11.2.1 (Cheng et al. [144, 340]) The Min-Sensor Sweep-Coverage problem is NP-hard. Proof For any TSP instance, view each city as a target point with a required sweepperiod t = L/v. Since all mobile sensors move at the same speed v, the traveling distance for a mobile sensor within time period t is at most vt = L. Hence one mobile sensor is enough to monitor all target points if and only if the traveling salesman has a tour visiting all cities with total distance at most L.   In fact, Cheng et al. proved the following lower bound for the inapproximability of Min-Sensor Sweep-Coverage. Corollary 11.2.2 (Cheng et al. [144, 340]) The sweep-coverage problem does not have a polynomial-time (2 − ε)-approximation for any ε > 0 unless N P = P . Proof Using the same reduction in the proof of Theorem 11.2.1, if one mobile sensor is enough, then the number of mobiles sensors found by a (2 − ε)approximation is at most (2 − ε), which must be 1 since the number of sensors is an integer. Hence a (2 − ε)-approximation algorithm can solve TSP, which is impossible unless N P = P .  

11.3 Min-Sensor Sweep-Coverage

185

Notice that TSP can be reduced to Min-Sweep-Period Sweep-Coverage with one mobile sensor. Hence we have the following NP-hardness result. Theorem 11.2.3 Min-Sweep-Period Sweep-Coverage is NP-hard.

11.3 Min-Sensor Sweep-Coverage This section presents a 3-approximation algorithm for the Min-Sensor SweepCoverage problem in which the sweep-period for all target points is a constant t. The length of a graph is the sum of edge-distances for those edges in this graph. From the proof of NP-hardness in the above section, one may obtain a feeling that the sweep-coverage problem is closely related with the traveling salesman problem. In fact, current studies more or less borrow some ideas dealing with TSP. Both the algorithm in this section and the algorithm in next section make use of the following idea embedded in the well-known 2-approximation algorithm for TSP in a metric space (in which the distance function defined on pairs of nodes satisfies triangle inequality) [507]. Remark 11.3.1 Consider a set of nodes V in a metric space with distance function w. Find a minimum length tree T spanning all nodes of V . The graph obtained from T by doubling every edge is an Eulerian graph and thus has an Eulerian tour F . Going along the Eulerian tour and short-cutting repeated nodes results in a Hamiltonian cycle H on V . By triangle inequality, w(H ) ≤ w(F ) = 2w(T ). A natural idea to solve the Min-Sensor Sweep-Coverage problem with a constant sweep-period t is to find a TSP tour, divide it uniformly into segments of length at most vt, and locate mobile sensors at the dividing points. Moving in a same direction along the tour, those mobile sensors form a sweep-cover. However, this idea will not produce satisfactory results when target points are very far from each other. For example, suppose there are two target points at distance M from each other, where M is a large number. The above strategy yields a solution with 2M/v mobile sensors. On the other hand, two sensors stationed at the two target points are sufficient. This example gives us a hint that a crucial problem towards a good solution is how to partition target points into groups. One idea in the following algorithm, which was presented by Gorain and Mandal [230], is to “guess” a good partition. The algorithm consists of two stages. In the first stage, a partition of groups is guessed. In the second stage, mobile sensors are deployed. Every group with only one target point is monitored by one mobile sensor. For each group with more than one target points, find a Hamiltonian cycle. Suppose the Hamiltonian cycle has length L, then L/vt mobile sensors which are uniformly deployed along the cycle and move in the same direction with speed v form a sweep-cover for this group.

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11 Sweep-Coverage

Algorithm for Min-Sensor Sweep-Coverage Input: A set of n target points in a metric space with distance function w. Output: A deployment of mobile sensors forming a sweep-cover. for k = 1, . . . , n Find a minimum length spanning forest Fk with k components. (k) (k) Let T1 , . . . , Tk be the components of Fk . nk ← 0. for i = 1, . . . , k if |V (Ti(k) )| = 1 then nk ← nk + 1 else nk ← nk + 2w(Ti )/vt end-for end-for k0 = arg min{nk : k = 1, . . . , n} for i = 1, . . . , k0 if |V (Ti(k0 ) )| = 1 then (k ) Deploy one mobile sensor on V (Ti 0 ). else (k ) (k ) Find a Hamiltonian cycle Ci 0 on V (Ti 0 ) using Remark 11.3.1. (k ) (k ) Uniformly deploy w(Ci 0 )/vt mobile sensors along Ci 0 . end-for Notice that those Fk ’s can be found by executing Kruskal’s algorithm (for the construction of a minimum length spanning tree) once: the first k edges selected by Kruskal’s algorithm yield Fn−k . Theorem 11.3.2 The above algorithm is a 3-approximation for Min-Sensor SweepCoverage. Proof Suppose an optimal solution uses k ∗ mobile sensors. During time interval [0, t], the trajectories of these mobile sensors form a graph G spanning all target points. Furthermore, G contains at most k ∗ components and thus contains a spanning forest F consisting of exactly k ∗ components. By the choice of Fk ∗ and the assumption that all sensors move at the same speed v, we have w(Fk ∗ ) ≤ w(F ) ≤ w(G) ≤ k ∗ vt. Notice that for each i = 1, . . . , k ∗ , the number of mobile sensors deployed to (k ∗ ) (k ∗ ) (k ∗ ) monitor group V (Ti ) is at most w(Ci )/vt + 1 ≤ 2w(Ti )/vt + 1. Hence the total number of mobile sensors when k ∗ is guessed is at most   k∗ (k ∗ )  2w(Ti ) 2w(Fk ∗ ) +1 = + k ∗ ≤ 3k ∗ . vt vt i=1

The theorem is proved.

 

11.4 Min-Sweep-Period Sweep-Coverage

187

11.4 Min-Sweep-Period Sweep-Coverage A set of cycles the union of which spans all nodes is called a cycle cover. As one can see from previous section, given a cycle cover C = (C1 , . . . , Ct ), an optimal deployment for Min-Sensor Sweep-Coverage with respect to C (which minimizes the total number of mobile sensors) can be easily calculated. But for Min-SweepPeriod Sweep-Coverage, finding an optimal deployment with respect to C (which minimizes the maximum sweep-period) needs more effort, because one does not know in advance how many mobiles sensors should be deployed on each cycle. In fact, we may consider a more general problem: given a family of graphs G = (G1 , . . . , Gt ) and a positive integer m, find an allocation (m1 , . . . , mt ) such that every mi is a positive integer, m1 +· · ·+mt = m, and maxi w(Gi )/mi is minimized. The following algorithm solves this problem optimally. Allocation for a family of graphs. Input: A family of graphs G = (G1 , . . . , Gt ) and an integer m. Output: An optimal allocation (m1 , . . . , mt ). mi ← 1 for i = 1, . . . , t. for l = 1 to m − t k ← arg maxi=1,...,t w(Gi )/mi mk ← mk + 1 end-for Lemma 11.4.1 Given a family of graphs G, the above algorithm finds an optimal allocation for G. Proof Denote by m(l) i the value of mi in the l-th iteration. Suppose (m1 , . . . , mt ) produced by the algorithm is not an optimal allocation. Let (m∗1 , . . . , m∗t ) be an optimal allocation. Then maxi {w(Gi )/m∗i } < maxi {w(Gi )/mi(m−t) }. For the > w(Gk0 )/m∗k0 , index k0 = arg maxi {w(Gi )/mi(m−t) }, we have w(Gk0 )/mk(m−t) 0   (m−t) (m−t) t and thus mk0 < m∗k0 . Since ti=1 m∗i = = m, there exists i=1 mi (m−t)

another index k1 with mk1

> m∗k1 . Suppose in the l-th iteration, the algorithm

increases the value of mk1 from m∗k1 to m∗k1 + 1, that is, mk1

mk1 = m∗k1 + 1. Then w(Gk1 )/m∗k1 = maxi w(Gi )/mi (l)

increases,

(l−1)

(j ) maxi w(Gi )/mi

. Notice that when j

does not increase. Hence

max{w(Gi )/m∗i } ≥ w(Gk1 )/m∗k1 ≥ max w(Gi )/mi

(m−t)

i

a contradiction.

= m∗k1 and

(l−1)

i

,  

The Min-Sweep-Period Sweep-Coverage problem can be viewed as a partition and allocation problem: it partitions target points into groups V1 , . . . , Vt , and for each group Vi , it allocates mi mobile sensors to cooperatively monitor target points in Vi , where two mobile sensors whose trajectories have a common target point are

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11 Sweep-Coverage

viewed as monitoring the target point cooperatively. Hence the trajectories of those mobile sensors cooperatively monitoring Vi form a connected graph. The following algorithm was presented by Gao et al. [210]. Algorithm for Min-Sweep-Period Sweep-Coverage Input: A set of n target points in a metric space with distance function w, and m mobile sensors. Output: A deployment of mobile sensors forming a sweep-cover. for k = 1, . . . , m Find a minimum length spanning forest Fk with k components. (k) (k) Let T1 , . . . , Tk be the components of Fk . (k) (k) Find a cycle cover C (k) = {C1 , . . . , Ck } using Remark 11.3.1. (k) (k) Find an optimal allocation (m(k) 1 , . . . , mk ) for C . (k) (k) For each i, uniformly deploy mi mobile sensors along Ci . end-for Choose the best k with the minimum sweep-period. To analyze the algorithm, we first prove an auxiliary lemma. Lemma 11.4.2 Suppose T is a tree allocated with m sensors. Let l = w(T )/m. If T contains an edge e with w(e) > l, then we can allocate m1 , m2 sensors to the two subtrees T1 , T2 of T − e, respectively, such that m1 + m2 = m and max{w(T1 )/m1 , w(T2 )/m2 } ≤ l. Proof Let m1 be the maximum integer such that w(T1 )/m1 ≤ l. Then w(T1 )/(m1 − 1) > l. Combining this with w(T ) = ml and w(e) > l, we have w(T2 ) ml − (m1 − 1)l − w(e) l − w(e) < =l+ l ∗ . Let Ti,1 i,2 ∗ ∗ ∗ ∗ ∗ two subtrees of Ti − e. Denote Vi,1 = V (Ti,1 ), Vi,2 = V (Ti,2 ). By Lemma 11.4.2, ∗ )/m∗ , w(T ∗ )/m∗ } ≤ l ∗ . one can divide m∗i = m∗i,1 + m∗i,2 such that max{w(Ti,1 i,1 i,2 i,2 ∗ ∗ ∗ ∗ Replacing Vi by Vi,1 and Vi,2 , and replacing mi by m∗i,1 and m∗i,2 , we obtain another optimal solution to Tree Partition and Allocation, but the number of groups is increased, contradicting our choice of the optimal solution. (k ) For an index j ∈ {1, . . . , k0 }, suppose V (Tj 0 ) contains p groups of Vi∗ ’s, say V1∗ , . . . , Vp∗ . Notice that T1∗ , . . . , Tp∗ form a spanning subgraph of V (Tj(k0 ) ), and (k0 )

adding at most p − 1 edges of Tj length at most l ∗ , we have (k ) w(Tj 0 )

≤

p 

w(Ti∗ ) + (p

(k0 )

can connect them. Since every edge in Tj

∗

− 1)l
0 if NP = P . Liu and Liang [361] study a minimum partial cover problem as follows: Problem 12.1.3 (Minimum Connected Partial Cover) Given a set of sensors, a target area, and a number 0 < θ < 1, find a minimum subset of sensors to cover at least θ percentage of the target area and this subset of sensors also induces a connected subgraph in communication network of sensors. This work is followed by some others [363, 523] in the literature. The minimum connected partial cover problem also has several variations, e.g., the target area is replaced by target points and the θ percentage is replaced by a given number of target points or given size of target area. The partial coverage brings a lot of new issues into the study of optimal coverage in wireless sensor networks. A new issue is the way to measure the coverage or the breach. Balister et al. [55] proposed to use the diameter of each hole in the area-coverage, which is the longest distance between two points in the hole. Using this measure, they studied a new type of coverage, trap-coverage. A set of sensors is said to give D-trap coverage if the diameter of every hole is not larger than D. However, there is no solid theoretical research works existing in the literature about the trap-coverage. Another measure is based on probabilistic models. The survey [549] shows a quite large amount of references on probabilistic coverage. Since there are many references on each topic about partial coverage, we have to distribute our discussion on literature review into the following sections when we study them more carefully.

12.2 Geometric Maximum Coverage In area of computing geometry, the geometric maximum coverage problem attracts a lot of research efforts. In this problem, instead of a collection of subsets considered in the maximum coverage problem, a set of geometric objects are considered to cover a given set of target points on the Euclidean plane. To our interest, we would like to consider only disks and hence give the name of the problem as follows.

12.2 Geometric Maximum Coverage

195

Problem 12.2.1 (Max Disk Coverage) Given a set of weighted target points in the Euclidean plane and an integer m ≥ 1, find m unit disks to maximize total weight of covered target points. This problem was first introduced by Drezner [173] for m = 1. In this case, Chazelle and Lee [120] designed an O(n2 )-time algorithm to compute the optimal solution while Agarwal et al. [5] and Aronov and Har-Peled [41] constricted MonteCarlo (1 − ε)-approximation. For m = 2, Cabello et al. [88] studied a special case that the two disks are disjoint. For fixed m ≥ 1, de Berg et al. [165] gave a general result as follows. Theorem 12.2.2 For the maximum disk coverage problem with fixed m, there exists a deterministic (1 − ε)-approximation with running time O(n log n). Later Li et al. [277] improved the running time to linear. Those (1 − ε)approximation algorithms are designed by using two important techniques in computational geometry and approximation algorithm design, respectively. The first technique is about the discrepancy method in computational geometry [118, 119]. This method can replace the given set of weighted target points by a new set of weighted points possibly with different weights, however, less cardinality. However, selected m unit disks give a good approximation for original set of weighted target points. Specifically, let us consider a set P of n weighted target points and a set U of unions each of which is the union of m unit disks. Let A be a subset of P , possibly with different weights on the points. A is called a (1/r)approximation for P if for any U ∈ U , |w(U ∩ A) − w(U ∩ P )| ≤ w(P )/r. The following lemma is proved in [165]. Lemma 12.2.3 (de Berg et al. [165]) Consider a set P of n weighted target points and a number r, 1 ≤ r ≤ n. Then a 1/r-approximation A for P can be constructed in O(nr 12 log6 r) time such that |A| = O(r 2 log r). The second technique √ is the partition. Let us consider the unit disk as a disk with radius 1. Denote s = 2. Let μ = 3ms. Denote by G(a,b) the grid consisting of μ×μ cells such that (as, bs) is one of grid vertices where a, b ∈ {0, 1, . . . , 3m−1}. Let H(a,b) be the set of horizontal lines and V(a,b) the set of vertical lines in grid G(a.,b) , that is, Hb = {(x, y) ∈ R 2 | y = bs + k · μ, k ∈ Z}, and Va = {(x, y) ∈ R 2 | x = as + k · μ, k ∈ Z}, where Z is the set of integers. Furthermore, denote L(a,b) = Hb ∪ Va . Following lemma can be proved by a counting argument. Lemma 12.2.4 Let U be a union of m unit disks. Then there exists a (a, b) ∈ {0, 1, 3m − 1} such that every connected component of U is fully contained in a cell of G(a,b) , that is, L(a,b) does not intersect U .

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In fact, among Va for a = 0, 1, . . . , 3m − 1, there are at most two of them intersecting each unit disk and among Hb for b = 0, 1, . . . , 3m − 1, there are at most two of them intersecting each unit disk. Since there are totally m unit disks, a and b can be selected to make both Va and Hb not intersecting any unit disk. For such a selection, every unit disk is fully contained in a cell of G(a,b) . By Lemmas 12.2.3 and 12.2.4, the problem is reduced to study the maximum disk coverage problem in a cell. Then the following result can be employed. Lemma 12.2.5 The maximum disk coverage problem in a cell can be solved optimally in O(n2m−1 log n) time. An additional technique is about how to distribute m disks into cells. A dynamic programming takes this job. To reduce the running time, an interesting technique is included in the design of this dynamic programming as follows.   Suppose that m disks are distributed into k cells. There are m+k ways. Let k f (m1 , . . . , mk ) be the objective function value for the k cells which receive m1 , . . . , mk disks, respectively. Then the maximum value is given by f ∗ (m) = max{f (m1 , . . . , mk ) | m1 + · · · + mk , m1 ≥ 0, . . . , mk ≥ 0}. Define f ∗ (m1 , . . . , mi , mi+1 ) = max{f (m1 , . . . , mi , mi+1 , . . . , mk ) | mi+1 + · · · + mk = mi+1 , mi+1 ≥ 0, . . . , mk ≥ 0}. Then f ∗ (m) = max{f ∗ (m1 , m2 ) | m1 + m2 = m, m1 ≥ 0, m2 ≥ 0, and f ∗ (m1 , . . . , mi ) = max{f ∗ (m1 , . . . , mi , mi+1 ) | mi + mi+1 = mi , mi ≥ 0, mi+1 ≥ 0}. This recursive formula gives a polynomial-time computation for f ∗ (m).

12.3 Set k-Cover For the set k-cover problem, Abrams et al. [1] designed a very simple randomized algorithm with pretty good performance.

12.3 Set k-Cover

197

Randomized Algorithm for Set k-Cover each sensor chooses a random number i ∈ {1, . . . , k}; assign self to partial cover ci . Theorem 12.3.1 (Abrams et al. [1]) The solution generated by the randomized algorithm is expected to cover at least (1 − 1/e) portion of opt, the total number of times elements covered by an optimal solution. Proof Consider a target point a. Let na be the number of sensors covering a. Since each sensor is assigned to a particular partial cover ci with probability 1/k, it has probability (1 − 1/k)na for a particular partial cover not to cover a. Thus, the expected number of partial covers containing a is k(1 − (1 − 1/k)na ). Hence, the total of times that target points are covered by the partition is  expected number na a k(1 − (1 − 1/k) ). Denote by ta∗ the number of times  a is covered in the optimal solution. Then ∗ ta ≤ min(k, na ). Thus, opt ≤ a min(k, na ). Next, we claim that k(1 − (1 − 1/k)na ) ≥ (1 − 1/e) min(k, na ) and show this claim in the following two cases. Case 1.

k ≤ na . Then we have k(1−(1−1/k)na ) ≥ min(k, na )·(1−(1−1/k)k ) ≥ min(k, na )(1−1/e).

Case 2.

k ≥ na . Denote g(na ) =

k(1 − (1 − 1/k)na ) . na

Then g  (na ) =

k[(1 − 1/k)na (1 − ln(1 − 1/k)na ) − 1] 0.

12.4 Minimum Weight Partial Set Multi-Cover In this section, we study the following problem. Problem 12.4.1 (Minimum Weight Partial Set Multi-Cover) Consider an element set E, a subset collection S ⊆ 2E , a cost wS on each set S ∈ S, and a covering requirement re for each element e ∈ E. Given an integer k > 0, the problem is to find a subcollection F ⊆ S to fully cover at least k elements such that the total cost of F is as small as possible, where an element e is said to be fully covered by F if it belongs to at least re sets in F. This is called the minimum weight partial set cover problem if all covering requirement re are equal to one and are called the minimum partial set cover problem if furthermore, all weights on sets in S are equal to one. The minimum partial set cover problem was first studied by Kearns [286]. In this paper, a greedy algorithm was given with performance ratio at most 2H + 3, (n) n where n = |E| and H (n) is the Harmonic function, i.e., H (n) = i=1 1/i. Slavík et al. [473] improved the algorithm and obtained the performance ratio H (min{ , k)}), where is the maximum size of sets in S. Using primal-dual method, Gandhi et al. [202] presented an approximation algorithm with performance ratio f , where f is the maximum number of sets containing a common element. The same performance ratio f was also obtained by Bar-Yehuda [60] by using local ratio method. A special case is the partial vertex cover problem in which elements are edges in a given graph and each subset is the set of edges covered by a node. In this special case, f = 2. The 2-approximation for the partial vertex cover problem can also be constructed by LP-rounding [82]. There also exist some related research works in [202, 304, 409]. In the following, we show a result for the minimum weight partial set cover problem.

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199

Theorem 12.4.2 The minimum partial set cover problem has a polynomial-time approximation with performance ratio H (min( , k)). Proof For any subcollection F of S, define g(F) =



min(1, |{e ∈ S | S ∈ F}|).

e∈E

In Chapter 7 (Section 7.2), this function has been proved to be monotone nondecreasing and submodular. Therefore, we can obtain an approximation solution by using a greedy algorithm as follows: F ← ∅; while g(F) < k do select S ∈ S to maximize set F ← F ∪ {S}; output F.

S g(F ) w(S) ,

and

By a well-known theorem of Wolsey [556] (see also [177]), this greedy algorithm gives an approximation solution with performance ratio H (γ ), where γ = max min(k, g({S})) ≤ min( , k). S∈S

Actually, when using Theorem 8.2.9 of Wolsey, we could find that above greedy algorithm is not in the standard form for which the theorem can be used. To transform it to the standard form, we need to replace the inequality g(F) < k by min(g(F), k) < k. Therefore, in this standard form, the potential function is min(g(F), k) instead of g(F).   Both the minimum set multi-cover problem and the minimum partial set cover problem have approximation algorithms with the performance ratios, matching the lower bound for the classic set cover problem. However, it is very hard to study the minimum partial set multi-cover problem. Ran et al. [433] were the first to obtain an approximation with guaranteed performance ratio for the minimum partial set multi-cover problem. But, their performance ratio is meaningful only when the covering percentage p = k/n is very close to 1. Later, Ran et al. [432] gave a greedy approximation with performance ratio , where is the size of the maximum set. Since can be as large as n, the performance ratio is still not improved much. Shi et al. [462] and Shi et al. [463] designed bi-criteria approximations. Ran et al. [436] made a significant progress in study of approximation for the minimum partial set multi-cover problem. First, they proved a lower bound for approximation performance ratio by a reduction from the densest l-subgraph problem [400]. The hardness of the densest 1-subgraph problem in [400] implies the following result.

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Theorem 12.4.3 Under the exponential time hypothesis, the minimum weight partial set multi-cover problem cannot be polynomial-time approximated within a 1

factor O(n (log log n)c ) for some constant c. Ran et al. [436] also designed a primal-dual approximation algorithm and obtained the following results. Theorem 12.4.4 Under the assumption that all re for e ∈ E are upper bounded by a constant, the minimum weight partial set multi-cover problem has a polynomial√   time approximation with performance ratio B + B · n, where B = max{ free | e ∈ E} and fe is the number of sets containing element e. Theorem 12.4.5 Under condition that re ≡ 2 and fe ≡ 2 for all e ∈ E, the minimum (cardinality) partial set multi-cover problem has a polynomial-time √ approximation with performance ratio 1 + 2n1/4 . (This restricted version looks the problem on a graph.) In the primal-dual method, design of an integer program is a crucial step. Ran et al. [434, 436] propose a novel integer program such that its relaxation (using Lovász extension [378]) is a convex program. Using the fact that for a submodular function, its Lovász extension coincides with its convex closure, the convex program is modified into a linear program with exponential number of variables. However, the Primal-Dual Algorithm can still be executed in polynomial-time, by making use of an efficient algorithm for minimizing a submodular function divided by a modular function [197]. This algorithm consists of two stages. In the first stage, a Primal-Dual Algorithm is employed to result in the subcollection of sets selected by the last iteration, which may fully cover much more elements than required. In the second stage, this subcollection is refined by iteratively implementing submodular minimization algorithms [197]. Following interesting problem is still open. Open Problem 16 Is there a polynomial-time approximation algorithm with performance ratio nc for the minimum partial set multi-cover problem where c is a constant with 0 < c < 1.

12.5 Minimum Partial Sensor Cover A natural extension of the minimum sensor cover problem is as follows: Problem 12.5.1 (Min Partial Sensor Cover) Given a set P of target points, a set S of sensors in the Euclidean plane, and an integer k ≥ 1, find a minimum subset S  of S to cover at least k target points. This problem has a lot of applications in the real world [133]. For the minimum partial sensor cover problem, there should exist better approximations than those for

12.5 Minimum Partial Sensor Cover

201

the minimum partial set cover problem. For homogeneous wireless sensor networks, we have the following result. Theorem 12.5.2 ([111, 202, 226]) There exists a PTAS for the minimum partial sensor cover problem in homogeneous wireless sensor networks. Proof The proof contains a standard partition technique (see the proof of Theorem 2.5.1) together with a dynamic program technique similar to the one described at the end of Section 10.2. Instead of distributing m disks to k cells, we distribute k target points to finitely many cells.   We may also study the sensor scheduling for the lifetime of partial sensor coverage, which can also be reduced to the minimum weight partial sensor cover problem as follows. Problem 12.5.3 (Min Weight Partial Sensor Cover) Given a set P of target points, a set S of sensors, each with a nonnegative weight, in the Euclidean plane, and an integer k ≥ 1, find a minimum total weight subset S  of S to cover at least k target points. In [359], a polynomial-time O(1)-approximation is designed for this problem. Actually, they pointed out that all O(1)-approximation for the minimum weight sensor cover problem designed by using partition technique can be extended to solve the minimum weight partial sensor cover problem. For partial coverage, we cannot reduce the area-coverage to the target-coverage without weight because simply using a target point to represent a small area would not work. However, each small area may be replaced by a target point with weight equal to the area. This can be a motivation to study the following variation of partial coverage. Problem 12.5.4 (Partial Cover with Weighted Target) Given a positive number k, a set of target points each with a nonnegative weight and a set of sensors, find a minimum set of sensors to cover a target-point subset with total weight at least k. There are some efforts made directly on partial area-coverage in the literature [361, 363, 523]. In study of lifetime of partial sensor coverage [231, 358, 445, 531, 653], the minimum weight partial sensor cover problem may find another application by establishing a similar result of Garg and Könemann [217]. This similar result seems to be held. However, there is record in the literature. To end this section, let us summarize open problems about minimum (weight or unweight) partial sensor networks as follows: Open Problem 17 (1) Is there a polynomial-time O(1)-approximation for the minimum partial sensor cover problem in heterogeneous wireless sensor networks? (2) Is there a PTAS for the minimum partial sensor cover problem with the coverage rate requirement in homogeneous wireless sensor networks?

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12.6 p-Percent Coverage There are different measurements for the quality of partial coverage [347]. Instead of required number of covered target points, the quality is measured by the coverage ratio, that is, the percentage of covered target points or area [45, 100, 342, 343, 367, 540]. Especially, there exists a sequence of efforts [207, 342, 343, 511, 564] made on the following problem. Problem 12.6.1 (p-Percent Coverage) Given a target area  and a set of sensors with unit lifetime, find a sensor active-sleep scheduling to maximize the lifetime of p-percent coverage, i.e., the time length of the period during which at any moment the target area  has at least p percent covered by active sensors. Those efforts contribute several algorithms which perform very well in computer experiments. Especially, Li et al. [343] proposed a framework for transforming the full coverage to percentage-partial coverage. The key idea in this transformation is to adjust sensing radius of sensors. By varying the sensing disk size, they obtain a virtual network, i.e., a network with virtual sensors. Then an algorithm is employed for full coverage to find a subset of virtual sensors fully covering the target area . Finally, they show that the original sensors corresponding to every sensor cover with selected virtual sensors would give a p-percent sensor cover.This technique is quite powerful, which can turn any algorithm for full coverage into an algorithm for ppercent coverage. However, the approximation performance ratio is not established for any algorithm obtained with this technique. Actually, the following is still open. Open Problem 18 Is there a polynomial-time O(1)-approximation for the ppercent coverage problem?

Chapter 13

Probabilistic Coverage

The probability that we may fail in the struggle ought not to deter us from the support of a cause we believe to be just. Abraham Lincoln

13.1 Motivation and Overview Previously, we consider that the state of a sensor covering targets is in binary mode: success (cover the targets) or fail (cannot cover the targets). However in the real world, a sensor covers targets with a certain probability, called the Probabilistic coverage, due to various issues which cause the uncertainty and form various probabilistic models, including the probabilistic network model [131, 152, 296, 371, 458, 566, 596], the probabilistic sensing model [10, 129, 423, 538, 539], the random noise model [310, 530], the random fault model [484, 614], etc. About probabilistic coverage, there is a quite long survey [549]. Here, we introduce this subject briefly. The reader may refer the survey if would like to study more. He et al. [249] consider the fact that a sensor covers targets with users’ satisfied probability and introduced failure probability into target- coverage problems. Liu et al. [366] studied a different scenario. They assume that sensors have no accurate location information. In this case, how to schedule sensor nodes to save energy and meet the requirements for both sensing coverage and network connectivity? They use random scheduling for sensing coverage with guaranteed connectivity, if necessary, turn on extra sensor nodes for network connectivity. Ahmed et al. [8] consider the people-centric sensing. They observed that people with mobile phones can act as mobile sensors, i.e., gather information from environment to fulfill sensing service. Due to uncontrollable mobility of people, the coverage of target area of interest gets the uncertainty and has to be measured by expected coverage degree. They formulated an optimization problem together with two heuristics.

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Chen et al. [129] designed intelligent algorithms to solve the minimum weight sensor probabilistic cover problem, yielding also a solution to the maximum lifetime probabilistic coverage problem. The probabilistic k-coverage problem is studied in [20, 39, 539, 565, 613]. The coverage with other issues, such as connectivity, is also brought into ones’ attention again [241, 380, 501, 565, 573, 614]. Actually, all subjects and issues in the disk model can be reconsidered in the probabilistic sensing model, such as maximum coverage[189, 628], optimal sensor deployment [10, 484, 656], maximum lifetime [167], barrier coverage [131, 458, 587], sweep-coverage [537], area-coverage [529], energy efficiency [152, 255, 461, 465, 596], mobility [371], path coverage [320]. There are also new topics and new issues brought in by probabilistic sensing. One of them is the interference [310] and noise [485, 576]. The interference occurs when addition of signals transmitted by others surrounding sensors. A comparison on interference between different sensor deployments is made by [310]. Another topics are the mobile target-coverage [132, 407] and cooperation between sensors [196, 318, 325]. In the following sections, we select two issues to discuss.

13.2 Random Deployment In [513, 514], Wan and Yi made connection between probabilistic coverage and random deployment of sensors. Under assumption that the sensors are deployed as either a Poisson point process or a uniform point process in a square or disk region, they study how the probability of the k-coverage changes with the sensing radius or the number of sensors. Consider a coordinate system on the Euclidean plane. Let A be the unit-area square centered at the origin and tA the square with area t 2 and center at the origin. Suppose n sensors are deployed in A uniformly at random. They can be represented by n uniform point process {X1 , X2 , . . . , Xn }. Assume n is also a random variable with Poisson distribution P o(λ) where λ > 0 is a Poisson parameter, called mean. Then the stochastic process {X1 , X2 , . . . , XP o(λ) } is called the Poisson point process with mean λ on area A, denoted by Pλ (A). Let k be a fixed nonnegative integer. Let Cn,r denote the even that A is (k + 1)covered by the closed disks √ with radius r and centered at the points in Pλ (A). Let Ks,n denote the event that sA is (k +1)-covered by unit-area closed disks centered at the points in Pλ (A). For simplicity, for any ζ ∈ R, define α(ζ ) =

⎧ ⎨ e−ζ /2 · ⎩ e−ζ /2 ·

√

( 2 +e−ζ /2 )2 √ 16(8√π +e−ζ /2 ) 4 π 2k+6 (k+2)! πn

if k = 0 if k ≥ 1

13.2 Random Deployment

205

and β(ζ ) =

⎧ √ ⎨ 4e−ζ /2 (e−ζ /2 + 2( π + ⎩ 4e−ζ /2 ·

√

√1 ) π

π + √1π

if k = 0 if k ≥ 1.

2k−1 k!

Wan and Yi [513, 514] proved the following. Theorem 13.2.1 (Wan and Yi [513, 514]) Let  rn =

ln n + (2k + 1) ln ln +ζn , πn

If limn→∞ ζn = ζ for some ζ ∈ R, then 1 − β(ζ ) ≤ lim Pr[Cn,rn ] ≤ n→∞

1 . 1 + α(ζ )

If limn→∞ ζn = ∞, then lim Pr[Cn,rn ] = 1.

n→∞

If limn→∞ ζn = −∞, then lim Pr[Cn,rn ] = 0.

n→∞

Theorem 13.2.2 (Wan and Yi [513, 514]) Let μ(s) = ln s + 2(k + 1) ln ln s + ζ (s). If lims→∞ ζ (s) = ζ for some ζ ∈ R, then 1 − β(ζ ) ≤ lim Pr[Ks,μ(s)s ] ≤ s→∞

1 . 1 + α(ζ )

If lims→∞ ζ (s) = ∞, then lim Pr[Ks,μ(s)s ] = 1.

s→∞

If lims→∞ ζ (s) = −∞, then lim Pr[Ks,μ(s)s ] = 0.

s→∞

Random deployment has also been studied in [150, 336, 430, 493, 527].

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13.3 Probabilistic Sensing Model Li et al. [332, 333] solved a problem in [408] and consider a more general sensing model: The sensing ability diminishes as the distance increases. It is possible to motivate from this model to think about extending the SINR (signal-to-interferenceand-noise ratio) model [517] from connectivity to coverage as follows: Let P (x) be the coverage probability of a target position x. Then P (x) = 1 −

"

[1 − ps (x)],

s∈S

where S is the set of sensors and ps (x) denotes the detection probability of sensor s at position x. ps (x) can be calculated according to the sensing model [3, 10, 101, 249, 262, 266, 526, 566, 578, 626, 648]. For example, in the truncated model [7], 

e−αd(s,x) if d(s, x) ≤ Ru , 0 otherwise,

ps (x) =

where α is a decay factor and Ru is the radius of sensor’s upper truncated detection area. In some model, ps (x) is determined in quite complicated way. For example, in Neyman–Peason probabilistic detection model [502], 

 τ − a(s, x) ps (x) = 1 −  , σ where (y) is the normal distribution of the standard Gaussian, that is, $ (y) =

y

−∞

 2 1 t dt, √ exp − 2 2π

and a(s, x) =

θ + ns . d(s, x)α

Moreover, τ , θ , α, and ns are parameters independent from the distance d(s, x). Under those probabilistic sensing model, various coverage problems have been studied extensively. Based on Neyman–Peason probabilistic detection model, Tian et al. [502] studied the minimum sensor γ -cover problem where a sensor set is called a γ -cover if the total detection probability at any target point is at least γ . A distributed algorithm is designed and its performance is evaluated by simulation. Other heuristics are given in [622].

13.3 Probabilistic Sensing Model

207

There are many optimization problems based on probabilistic sensing models [6, 107, 317, 401, 420, 426, 429, 439–441, 453, 459, 500, 542, 624]. Following are examples. Problem 13.3.1 (Minimum Weight γ -Probabilistic Cover [101, 342, 596, 600]) Given a real number γ , 0 < γ < 1, a two-dimensional area A, and set S of sensors, each sensor s ∈ S associated with the nonnegative weight ws , find a γ -probabilistic cover with minimum total weight, where a subset F is called a γ -probabilistic cover if every point in A can be detected by F with probability at least γ . Problem 13.3.2 (Maximum Lifetime γ -Probabilistic Coverage) Given a real number γ , 0 < γ < 1, a two-dimensional area A, and a set S of sensors with unit lifetime, find a sleep-activate schedule to maximum the lifetime of γ -Probabilistic Coverage, which is a time period at any moment of which active sensors form a γ -Probabilistic cover. Problem 13.3.3 (Minimum (γ , T )-Probabilistic Cover) Given a real number γ , 0 < γ < 1, a two-dimensional area A, and set S of sensors with unit lifetime, find the minimum number of sensors which can provide γ -Probabilistic coverage for a time period of length T . This subset of sensors is called a (γ , T )-Probabilistic Cover. Although there are many publications in the literature on those optimization problems, not many solid theoretical results exist. Therefore, they provide us with a lot of research opportunity.

Chapter 14

Mobile Sensors

The mobile phone acts as a cursor to connect the digital and physical. Marissa Mayer

14.1 Motivation and Overview In previous chapters, we already met mobile sensors. Mobile sensors have a lot of applications [138, 212, 301, 546, 550, 633, 634, 647]. In this chapter, we focus on a special application, hole healing. There are many issues [270, 280, 325, 455, 457, 502, 518, 574, 618–621, 627] in controlling coverage and connectivity in wireless sensor networks. One of them is to detect holes [147, 193, 223, 285, 323, 330, 331, 370, 395]. When a sensor stops working, the coverage and the connectivity may be broken, that is, a coverage hole or a connectivity hole may appear. There exist several approaches to repair holes [166, 338, 423, 448–450]. A popular one is to employ mobile sensors, which can recover holes and improve coverage quality [7, 124, 134, 353, 354, 364, 397, 508, 529, 555, 580, 582, 631, 648] To heal the hole, we may replace it by a mobile sensor, that is, move a mobile sensor to the proper location. In this chapter, we study a hybrid sensor network consisting of two types of sensors, static sensors and mobile sensors. For simplicity, we assume that all mobile sensors are homogeneous, that is, they have the same communication radius Rc , the same sensing radius Rs , and the same energy consumption for moving a unit distance; their energy consumption for moving is proportional to the distance. Wang et al. [524] proposed an algorithm for moving mobile sensors to heal holes by minimizing the total energy consumption. Nguyen et al. [417, 418] gave an algorithm to find an optimal solution for hole healing with minimization of the maximum sensor energy consumption over all sensors. Shen et al. [460] studied a dynamic hole healing problem and gave an approximation algorithm with theoretical performance ratio guaranteed.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_14

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Actually, this dynamic hole healing problem should be called the online hole healing problem since after a hole appears, it would not disappear until it gets healed. Usually, a problem is said to be dynamic if its input task could be added or dismissed. For example, in the dynamic Steiner tree problem, input terminals will be dynamically changed, i.e., some new terminals appear and some old terminals disappear. If input task can be only added, then it is usually called an online problem. The analysis of Shen et al. [460] is not a standard way for an online problem. They treated the online problem as a sequence of offline problems and compared their algorithm with the sequence of optimal solutions for the sequence of offline problems. Actually, the sequence of optimal solutions does not give even an online algorithm with good performance. For an online problem, one often compares designed online algorithm with optimal solution of corresponding globally offline problem, i.e., suppose all input tasks are known at the beginning. This compared result is called the competitive ratio. Zhang et al. [641] gave a new algorithm with an analysis on competitive ratio. An important approach of hole healing, which is generally used in the study of hole healing in the literature [224, 460, 524, 641], is called cascade healing. In implementing hole healing, one may need to pay attention to the energy consumption of moving, which is proportional to the moving distance. Since every sensor is powered by batteries with limited energy, each mobile sensor has a limited moving distance. With this restriction, a feasible hole healing schedule may involve a sequence of movings of active mobile sensors, i.e., move those mobile sensors which are already working for the required coverage and connectivity. For example, consider Fig. 14.1. s (b) is a backup sensor lying at the same location as active sensor (a) s3 . Since s (b) is redundant, it is in a sleeping status initially. Suppose at a time slot, an active static sensor s fails, creating a hole h. We may think to use backup sensor s (b) to heal the hole. However, it is too far to move directly to h. Therefore, a sequence of feasible movings may be involved as follows: move active mobile (a) (a) sensor s1 to the hole h, move another active mobile sensor s2 to the original (a) (a) location of s1 , and move backup sensor s (b) to the original location of s2 . This (a) sequence of movements may occur simultaneously, and the path (s (b) , s2 , s1(a) , h) is called a cascade healing path. It is worth mentioning that when multiple holes appear simultaneously, every hole needs a cascade healing path, and Zhang et al. [641] observe that in an optimal solution, these paths must be node-disjoint in a corresponding movement graph constructed based on sensor locations and possible

Fig. 14.1 An illustration of cascade healing

14.2 An Observation on Cascade Healing Paths

211

movements. This is a very important observation which can give a clear and deeper understanding of the cascade hole healing problem and obtained results. For example, what the results obtained in [418, 477] are polynomial-time algorithms for the following two optimization problems. Problem 14.1.1 (Minimum Total Distance Cascade Healing) Consider a set of backup sensors S (b) and a set of active mobile sensors S (a) in a wireless sensor network lying in the Euclidean plane together with a distance upper bound B > 0. Given a set of holes H , find |H | node-disjoint cascade healing paths from S (b) to H , (b) such that for every hole h ∈ H , there exists a unique backup sensor sh ∈ S (b) and (b) a cascade healing path from sh to h, the moving distance of every involved mobile sensor is at most B, and the total length of all cascade healing paths is minimized. Problem 14.1.2 (Bottleneck Cascading Healing) Consider a set of backup sensors S (b) and a set of active mobile sensors S (a) in a wireless sensor network lying in the Euclidean plane together. Given a set of holes H , find |H | node-disjoint cascade healing paths from S (b) to H , such that for every hole h ∈ H , there exists a unique (b) (b) backup sensor sh ∈ S (b) and a cascade healing path from sh to h, and the longest moving distance of an involved mobile sensor is minimized. In the following sections, we start from this observation to introduce above research works.

14.2 An Observation on Cascade Healing Paths Let us first define a movement directed graph. Definition 14.2.1 (Movement Directed Graph) For a set of holes H , a set of backup sensors S (b) , and a set of active mobile sensors S (a) in the Euclidean plane together with a distance upper bound B > 0, the movement directed graph GB (H, S (a) , S (b) ) = (V , A) is defined by 1. the node set V = H ∪ S (a) ∪ S (b) , and 2. the arc set A = {(u, v) | u ∈ S (a) ∪ S (b) , v ∈ H ∪ S (a) , and d(u, v) ≤ B}, where d(u, v) is the distance between nodes u and v. When we study the bottleneck cascade healing problem, assume B = ∞ and denote the movement directed graph by G∞ (H, S (a) , S (b) ). For simplicity of speaking, let us make a convention. Later, by “move u to v,” we mean moving a mobile sensor at location u to location v. For example, as shown in Fig. 14.2a, a movement directed graph GB (H, S (a) , S (b) ) is presented with H = {h1 , h2 }, S (a) = {v3 , v4 }, and S (b) = {v1 , v2 , v5 }. The set of directed paths P1 = {v1 v3 h1 , v5 h2 } requires to move v3 to h1 , v1 to v3 , and v5 to h2 , simultaneously. Could the set of directed paths P2 = {v1 v4 h1 , v2 v4 h2 } in Fig. 14.2a form an optimal healing strategy? The answer is no. Why? Here is an explanation: This set

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Fig. 14.2 (a) GB (H, S (a) , S (b) ) and (b) RB (H, S (a) , S (b) ) with hollow nodes as holes, solid nodes as backup sensors, double cycled nodes as active mobile sensors, and square nodes as source s and sink t

of paths means that move v4 to h1 , v1 to v4 , v2 to v4 , and v4 to h2 . Thus, location v4 has two mobile sensors; either the one coming from v1 moves to h1 or the one coming from v2 moves to h2 . In the former case, a mobile sensor moves from v1 to v4 and then from v4 to h1 , which implies that d(v1 , h1 ) ≤ d(v1 , v4 ) + d(v4 , h1 ) ≤ B, contradicting nonexistence of edge (v1 , h1 ). In the latter case, a contradiction would result from nonexistence of edge (v2 , h2 ). Following result comes from above observation. Lemma 14.2.2 (Zhang et al. [641]) For any set of cascade healing paths, there exists a set of node-disjoint cascade healing paths, (1) using the same subset of backup sensors to heal the same set of holes and (2) keeping the total moving distance and the maximum individual moving distance of mobile sensors not increased. Proof Consider two cascade healing paths with a node u in common, path (s1 , . . . , v, u, x, . . . , h1 ), and path (s2 , . . . , y, u, w, . . . , h2 ). Note that there is only one active sensor at u, which can attend only one cascade healing process, say the first one. Then in cascade healing process of the second path, the mobile sensor at y moves from y to u and has to continue moving from u to w. This means that the distance from y to w is at most B, and hence the arc (y, w) exists. Using (y, w) to replace ((y, u) and (u, w) in the second path. Then common node u is removed and the total length of paths and the longest path length would not be increased. Repeat this process. Finally, we would obtain a set of node-disjoint paths, meeting our requirements.   By this lemma, we will only consider solutions consisting of node-disjoint paths for the cascade healing problems from now on.

14.3 Maximum Flow

213

14.3 Maximum Flow Lemma 14.2.2 enables the cascade healing to be connected to the maximum flow. To see this, let us define an auxiliary directed graph defined as follows. Definition 14.3.1 (Auxiliary Directed Graph) Given a set of holes H , a set of active mobile sensors S (a) , and a set of backup sensors S (b) in the Euclidean plane together with a distance upper bound B > 0, the auxiliary directed graph RB (H, S (a) , S (b) ) is defined to be a flow network with a capacity function u(·) and a cost function c(·) on the arc set, which are constructed from the movement directed graph GB (H, S (a) , S (b) ) in the following way: • For each node x of GB (H, S (a) , S (b) ), replace x by two nodes x + and x − , and an arc (x − , x + ) with cost c(x − , x + ) = 0. • For each arc (x, y) of GB (H, S (a) , S (b) ), replace (x, y) by arc (x + , y − ) with cost c(x + , y − ) = d(x, y). • Add two virtual nodes, the source s and the sink t together with arcs (s, x − ) for all x ∈ S (b) and arcs (h+ , t) for all h ∈ H . Their costs are defined as c(s, x − ) = c(h+ , t) = 0. • Every arc (a, b) in this directed graph has capacity u(a, b) = 1. For example, the auxiliary directed graph of the graph in Fig. 14.2a is shown in Fig. 14.2b. An important property of the auxiliary directed graph is that every two arcdisjoint paths from the source s to the sink t must be node-disjoint. This property is introduced by the construction of replacing each node x by two nodes x + , x − , and an arc (x − , x + ) with capacity one. This property is the basis for establishing the relationship between node-disjoint cascade healing paths and the maximum flow in the auxiliary directed graph. Lemma 14.3.2 There exists a set of |H | node-disjoint directed paths from backup sensors to holes with total distance at most z in the movement directed graph GB (H, S (a) , S (b) ) if and only if there exist maximum flows from s to t with value |H | and total cost at most z in the auxiliary directed graph RB (H, S (a) , S (b) ). Proof First, suppose there exists a set of |H | node-disjoint directed paths from backup sensors to holes with total distance z in GB (H, S (a) , S (b) ). Following the construction of the auxiliary directed graph RB (H, S (a) , S (b) ), we would obtain corresponding |H | node-disjoint directed (s, t)-paths with total cost z in RB (H, S (a) , S (b) ). They give a flow with value |H |, which is a maximum flow. Conversely, suppose that GB (H, S (a) , S (b) ) has a maximum flow with value |H | and total cost z. Since every arc has an integer capacity, such a maximum flow can be assumed to have integer value at each arc, called an integer flow. It is well-known that any network flow can be decomposed into a union of flows along source–sink paths and cycles [161], which are called path-flows and cycle-flows. Moreover, for an integer flow, the decomposition can also result in integer path-flows and integer cycle-flows. For example, the flow in Fig. 14.3a can be decomposed into

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Fig. 14.3 Decomposition of a flow into a union of path-flows and cycle-flows. The numbers indicate the flow values on the corresponding arcs

one cycle-flow and two path-flows shown in Fig. 14.3b, each carrying one unit of flow. Since every arc in the auxiliary directed graph has capacity one, path-flows and cycle-flows obtained by the decomposition should be arc-disjoint and hence by the property of the auxiliary directed graph, they are node-disjoint. Note that each integer path-flow in the decomposition should have value one and each cycle-flow has value zero. Deleting all cycle-flows, we would obtain |H | node-disjoint pathflows each with value one. Deleting s, t and identifying those x + and x − on those path-flows, we would obtain |H | node-disjoint paths from backup sensors to holes in GB (H, S (a) , S (b) ), which can be seen easily to have total distance at most z.   By Lemmas 14.2.2 and 14.3.2, we have the following: Lemma 14.3.3 The minimum total distance cascade healing problem can be reduced to the minimum cost |H |-flow problem in the auxiliary directed graph RB (H, S (a) , S (b) ). There are several polynomial-time algorithms for the minimum cost K-flow problem in the literature. For example, the algorithm in [227] runs in time O(N 3 log(N C)), where N is the number of nodes in the network and C is the maximum absolute value of arc-costs. Using this algorithm, we would obtain the following theorem (note: RB (H, S (a) , S (b) ) has maximum cost at most B). Theorem 14.3.4 (Srinivasan and Chua [477]) The total distance cascade healing problem can be solved in time O(n3 log n) through computing a minimum cost |H |flow in an auxiliary directed graph with the number of nodes n = 2(|S (a) | + |S (b) | + |H | + 1). Using the similar argument as above, the following result can be established. Lemma 14.3.5 The bottleneck cascade healing problem can be reduced to the bottleneck |H |-flow problem in the auxiliary directed graph R∞ (H, S (a) , S (b) ), where the bottleneck |H |-flow problem is defined in the following. Problem 14.3.6 (Bottleneck K-Flow) Given a flow network, find a flow with value K and the longest involved arc is minimized. Note that potential objective function value must be some arc length and hence is discrete. Moreover, the optimal distance must be the minimum B such

14.4 Greedy Is Exponentially Bad

215

Algorithm 2 Offline bottleneck cascade healing 1: Order distances of arcs in R∞ (H, S (a) , S (b) ) as d1 ≤ d2 ≤ · · · ≤ dm where m is the number of arcs in R∞ (H, S (a) , S (b) ). Let d0 ← 0. 2: q ← 0, p ← m. 3: while p − q > 1 do 4: c ← (a + b)/2 5: if Rdc (H, S (a) , S (b) ) has a flow of value |H | then 6: p←c 7: else 8: q←c 9: end if 10: end while 11: Output |H | arc-disjoint path-flows in Rdp (H, S (a) , S (b) ) and dp .

that RB (H, S (a) , S (b) ) contains an |H |-flow. Therefore, the following algorithm (Algorithm 2) solves the bottleneck cascade healing problem in polynomial-time. Theorem 14.3.7 (Nguyen et al.[418]) The bottleneck cascade healing problem can be solved by Algorithm 2 in time O(log m · T f low ), where m = (|S (a) | + |S (b) |)(|H | + |S (a) |) + 2(|H | + |S (b) |) is the number of arcs in R∞ (H, S (a) , S (b) ) and T f low is the running time for solving the maximum flow problem. In next two sections, we study the online version of the minimum total distance cascade healing problem. In the online version, the information about input is not given initially, which would be explored sequentially during the computation. Specifically, for the cascade healing problem, let us consider a time window [0, T ] and suppose holes appear sequentially, that is, at each time slot p of the time window, a set Hp of kp holes appear. For online problems, Greedy is a popular strategy for algorithm design. However, for the online cascade healing, the greedy strategy does not have a good performance. Actually, we will present three results in [641] as follows: • An analysis on greedy algorithm shows that it is exponentially bad. • An online algorithm is designed to have competitive ratio at most 2T − 1 • Show that 2T −1 is the best possible for competitive ratio of an online algorithm.

14.4 Greedy Is Exponentially Bad By the greedy strategy, we mean that when kp new holes appear at time slot p, they would be healed optimally by using minimum cost kp -flow algorithm. For (b) (b) (b) example, consider T backup sensors s0 , s1 , . . . , sT −1 located on the real axis, with coordinates −ε and 2i−1 for i = 1, . . . , T − 1, respectively, where ε is a small positive real number (see Fig. 14.4).

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Fig. 14.4 An example shows that greedy algorithm can be exponentially bad

Suppose holes appear one by one and the distance upper bound B is sufficiently large so that there is no need to have cascade healing since a cascade sequence of movements has the same effect of moving a sensor directly. Thus, when a new hole appears, the greedy algorithm would heal it by using the closest backup sensor. Now, at the first time slot, a hole h1 appears at coordinate 1/2, the greedy algorithm would heal it by using backup sensor s1(b) , and backup sensor s1(b) becomes (a) an active mobile sensor s1 . At the second time slot, a hole h2 appears at coordinate (b) (b) 1, where s1 is originally located, the greedy algorithm would heal it by using s2 , (b) (a) and backup sensor s2 becomes an active mobile sensor s2 . Continuing in this (b) way until the T -th time slot, when hole hT appears at the coordinate where sT −1 is (b)

originally located, the greedy algorithm has to heal it by using s0 . The total moving T −2 i−1 2 + (2T −2 + ε) = 2T −1 − 12 + ε. distance is 12 + i=1 On the other hand, if we know the locations of all holes at the beginning, then an optimal solution is to use s0(b) to heal hole h1 at coordinate 1/2 and keep all the other sensors motionless. The total moving distance is 12 + ε. Therefore, the competitive ratio is asymptotically 2T − 1.

14.5 Online Algorithm with Tight Competitive Ratio Before presenting the algorithm, let us introduce a flow operation, the difference of two flows on the same flow network. Let f and g be two flows on the same flow network. The difference f − g is defined to be the flow whose value on each arc a is (f − g)(a) = f (a) − g(a). (f − g)(a) may be negative for some arc a. In such a case, (f − g)(a) is seen a positive flow on the reverse arc ← a− of a, ← − i.e., (f − g)( a ) = −(f − g)(a). For example, consider flows f (1) and f (2) in Fig. 14.5a,b, respectively. Suppose every arc which is drawn in the figure has flow value 1 and all other arcs have flow value 0. Then (f (2) −f (1) )(s, u2 ) = 0−1 = −1, which says that there is a −1 unit of flow from s to u2 , or it is equivalent to say that there is a unit flow from u2 to s. Thus, we can view arc (u2 , s) the reverse of arc (s, u2 ), to have flow value 1. With such a point of view, a nonnegative flow representation of f (2) − f (1) is depicted in Fig. 14.5c, where the dashed arcs are those reverse arcs. It should be mentioned that although some reverse arcs do not exist in RB (H, S (a) , S (b) ), it is only an intermediate step to help us for finding out the healing paths, and we will see later that all arcs participating in the healing strategy are legal. With the above view, and because the problem in this section

14.5 Online Algorithm with Tight Competitive Ratio

217

Fig. 14.5 An illustration of the online cascade healing problem. For ease of drawing, we contract v − and v + into one node in the figures. (a) to (e) show that the preliminary algorithm may dispatch a sensor which is already used by a previous hole. Figure (e) provides information of how to modify f (2) into f(2) in figure (f). Figure (g) indicates that a mobile sensor at u3 is used to heal the hole in H3 . Figure (h) provides information of how to modify f (3) into f(3) in figure (i). Figure (j) indicates that a mobile sensor at u4 is used to heal the hole in H4

deals with only 0–1 flows, we may regard each flow f as a set of arcs, %B (H, S (a) , S (b) ) : f (u, v) = 1}, where R %B denotes the A(f ) = {(u, v) ∈ R

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symmetric directed graph obtained from RB (H, S (a) , S (b) ) by adding reverse arcs. Since reverse arcs are needed in the analysis, A(f ) consists of all those arcs in %B (H, S (a) , S (b) ) each of which carries a unit flow of f . Note that an s-t flow f R can be decomposed into the union of arc-disjoint path-flows and cycle-flows. This implies that the arc set A(f ) can be decomposed into the union of arc-disjoint directed s-t paths and directed cycles.1 It should also be mentioned that in the later contents of this chapter, symbols S (a) and S (b) always refer to the initial set of active mobile sensors and the initial set of backup sensors, that is, the two symbols will not refer to corresponding updated sets even after some backup sensors are activated. To make a clear explanation of the algorithm, we first introduce a preliminary version of the algorithm, point out a flaw existing in the preliminary version, and then present modified algorithm to fix the flaw. A Preliminary Algorithm At time slot p + 1, a new hole pset Hp+1 appears and hole sets H1 , . . . , Hp are already known. Denote Kp = j =1 kj . Let f (p) be a minimum cost Kp -flow f (p) in the auxiliary directed graph RB (H1 ∪· · ·∪Hp , S (a) , S (b) ), found by the algorithm for offline problem. Similarly, a minimum cost Kp+1 -flow f (p+1) in RB (H1 ∪ · · · ∪ Hp+1 , S (a) , S (b) ) can be found in the same way. Note that by replacing arcs with negative flow value by their reverse arcs, f (p+1) − f (p) is turned to be a kp+1 flow f (p,p+1) with some reverse arcs, which can be decomposed into arc-disjoint union of paths and cycle. Since a cycle-flow contributes 0 to flow value. Deleting all cycle-flows, the remainder is a kp+1 -flow consisting of kp+1 arc-disjoint pathflows which give kp+1 node-disjoint cascade healing paths from S (b) to Hp+1 . An example in Fig. 14.5a–c illustrates this process which finds a cascade healing path (u3 , h1 , u1 , h3 ) for H2 . For simplicity, let us make an assumption that in the rest of this chapter, when we mention a flow, it always means a {0, 1}-flow. If there is an arc with flow value −1, then it should be replaced by its reverse arc with flow value 1. With such an understanding, we will not distinguish f (p,p+1) and f (p+1) −f (p) . Moreover, when there is no danger of confusion, we will not clearly distinguish a location of interest, u, and the sensor located at u. A Flaw with the Preliminary Algorithm Consider the example in Fig. 14.5 again. Suppose H1 = {h1 , h2 }, H2 = {h3 }, and H3 = {h4 }. Constructed f (1) , f (2) , and f (3) as shown in Fig. 14.5a,b,d, respectively. Denote by Sp the set of backup sensors used to heal holes in Hp . The algorithm would yield S1 = {u1 , u2 }, S2 = {u3 }, and S3 = {u2 }, that is, S1 and S3 are conflicted with each other. Actually, in the third time slot, figure (e) indicates that the algorithm should dispatch a mobile sensor of S (b) which is originally located at u2 to heal hole h4 . However, it faces the awkward situation that in the first time slot, 1 For

simplicity we may use a graph to refer to the arc set of the graph. For example, a directed cycle is also used to refer to the arc set of the cycle.

14.5 Online Algorithm with Tight Competitive Ratio

219

this sensor is already activated and moves to h2 , if we retract the sensor and send it to h4 , then the hole h2 is left not healed. This example shows a general problem with the preliminary algorithm. In fact, when a mobile sensor is dispatched to an active location which needs to be served continuously, it would cost a flaw. Again, denote by Sp the set of backup sensors used to heal holes in Hp . Possibly, the sets S1 , S2 , . . . found by the preliminary algorithm are non-disjoint. This means that in time slot p, the algorithm finds a healing strategy which wants to dispatch some mobile sensor in S (b) to heal a hole in Hp , but that sensor has already been activated and moved to heal a hole appeared earlier. This sensor cannot be retracted, since, otherwise, the early-appeared hole would be left unhealed again. A Modified Algorithm To fix such a flaw, the algorithm is modified as follows: in the (p + 1)th time slot, choose the minimum cost Kp+1 -flow f (p+1) satisfying the following condition: A(f (p+1) − f (p) ) has no directed cycle containing s.

(14.1)

An illustration of this strategy is explained by Fig. 14.5c and f. Note that A(f (2) −

f (1) ) in Fig. 14.5c contains a directed cycle (s, u4 , h2 , u2 , s). This cycle can be used for modifying f (2) to satisfy condition (14.1). This can be done by replacing arcs

(s, u4 ), (u4 , h2 ) of flow f (2) by arcs (s, u2 ), (u2 , h2 ). Outcome is the flow f(2) in Fig. 14.5f. Is f(2) still a minimum cost K2 -flow? The answer is yes. The reason is as follows. Note that f(2) is a K2 -flow and f (2) is a K2 -flow with the minimum cost. Thus, c(f(2) ) ≥ c(f (2) ) and c(su2 ) + c(u2 h2 ) ≥ c(su4 ) + c(u4 h2 ). On the other hand, another K1 -flow f(1) results from replacing arcs (s, u2 ), (u2 , h2 ) ∈ A(f (1) ) of the K1 -flow f (1) by arcs (s, u4 ), (u4 , h2 ). Because f (1) has the minimum cost, we have c(f(1) ) ≥ c(f (1) ). Therefore, c(su4 ) + c(u4 h2 ) ≥ c(su2 ) + c(u2 h2 ). This implies that c(su4 ) + c(u4 h2 ) = c(su2 ) + c(u2 h2 ). Thus, f(2) is also a minimum cost K2 -flow. Above argument can be generalized to yield the following. Lemma 14.5.1 At each time slot p, there is a Kp+1 -flow f (p+1) such that condition (14.1) holds. Moreover, such a flow can be computed in polynomial-time. Proof An important observation obtained from above example is as follows: If A(f (p+1) − f (p) ) contains a directed cycle C passing s, then ← − replacing arcs in C ∩ A(f (p+1) ) by arcs in C ∩ A(f (p) ) results in another minimum cost Kp+1 -flow,

(14.2)

← − where C is the cycle C with the reverse direction (note that an arc in C is either an arc in A(f (p+1) ) or the reverse of an arc in A(f (p) )). Let us give a proof for (14.2) in the following. Denote   ← − A = A(f (p+1) ) − (C ∩ A(f (p+1) )) ∪ ( C ∩ A(f (p) )) and

(14.3)

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  ← − A = A(f (p) ) − ( C ∩ A(f (p) )) ∪ (C ∩ A(f (p+1) )).

(14.4)

For a node v and an arc subset A, let dA− (v) and dA+ (v) denote the in-degree and the out-degree of v in A, respectively, i.e., the number of arcs in A which are directed into and directed out of v, respectively. We next show the observation by proving three facts as follows: Fact 1. The arc set A corresponds to a valid flow f  with the same flow value as f (p+1) . Actually, to see the validity of f  , it is sufficient to show the conservation law at internal nodes, that is, dA− (v) = dA+ (v) holds for any node v = s, t.

(14.5)

Because A(f (p+1) ) is a valid flow, we have − + dA(f (p+1) ) (v) = dA(f (p+1) ) (v) holds for any node v = s, t.

(14.6)

Note that − − − dA− (v) = dA(f − (p+1) ) (v) − dC∩A(f (p+1) ) (v) + d←

(v), and

(14.7)

+ + + dA+ (v) = dA(f − (p+1) ) (v) − dC∩A(f (p+1) ) (v) + d←

(v).

(14.8)

C ∩A(f (p) ) C ∩A(f (p) )

There are four cases. Case 1.

− − v is not a node on C. In this case, dC∩A(f − (p+1) ) (v) = d← + dC∩A(f (p+1) ) (v)

Case 2.

C ∩A(f (p) )

= 0.

C ∩A(f

(v) = 0.

(v) =

)

← − v is on C and both of the two arcs of C incident at v belong to A (f (p) ). − + − In this case, d← (v) = d← (v) = 1 and dC∩A(f (p+1) ) (v) = − − (p) (p) C ∩A(f

+ dC∩A(f (p+1) ) (v) = 0.

Case 4.

C ∩A(f (p) )

v is on C and both of the two arcs of C incident at v belong to A(f (p+1) ). − + − In this case, dC∩A(f (v) = − (p+1) ) (v) = dC∩A(f (p+1) ) (v) = 1 and d← (p) + d← −

Case 3.

=

+ d← (v) − C ∩A(f (p) )

)

C ∩A(f

)

One of the two arcs of C incident at v belongs to A(f (p+1) ) and the ← − − other arc belongs to A (f (p) ). In this case, either dC∩A(f (p+1) ) (v) = − d← −

+ + (v) = 1 and dC∩A(f = d← − (p+1) ) (v)

C ∩A(f (p) ) − or dC∩A(f (p+1) ) (v) + d← (v) = 1. − C ∩A(f (p) )

=

− d← (v) − C ∩A(f (p) )

= 0

(v) = 0

C ∩A(f (p) ) + and dC∩A(f (p+1) ) (v)

=

14.5 Online Algorithm with Tight Competitive Ratio

221

Fig. 14.6 An illustration of the proof of Lemma 14.5.1. In (a), the solid arcs belong to A(f (p+1) ) and the dotted arcs belong to A(f (p) ). (b) illustrates A . (c) illustrates A

In any case, by (14.6), (14.7), and (14.8), we have d −ˆ(p+1) (v) = d +ˆ(p+1) (v). (This A A proof is illustrated in Fig. 14.6.) + + Finally, note that dC∩A(f (s) = 1. By applying relation − (p+1) ) (s) = d← (p) C ∩A(f

)

+ (14.8) on node s , we obtain dA+ (s) = dA(f (p+1) ) (s). It follows that the flow value of

f  is the same as f (p+1) . Fact 2.

The arc set A gives a valid flow f  with the same flow value as f (p) .

The proof is similar to that of Fact 1. ← − Fact 3. c( C ∩ A(f (p) )) = c(C ∩ A(f (p+1) )). By combining Fact 1 with the fact that f (p+1) is a minimum cost Kp+1 -flow, we have c(f  ) ≥ c(f (p+1) ). By the definition of A in (14.3), we obtain ← − c( C ∩ A(f (p) )) ≥ c(C ∩ A(f (p+1) )).

(14.9)

Similarly, combining Fact 2, the definition of A in (14.4), and the fact that f (p) is a minimum cost Kp -flow, we obtain c(f  ) ≥ c(f (p) ) and hence ← − c(C ∩ A(f (p+1) )) ≥ c( C ∩ A(f (p) )).

(14.10)

Now, Fact 3 follows from inequalities (14.9) and (14.10). Fact 3 implies that c(f  ) = c(f (p+1) ) and hence the observation (14.2) is proved. Note that after an operation in observation (14.2), the number of common arcs of A(f (p+1) ) ∩ A(f (p) ) is increased by at least one. Thus, a minimum cost Kp+1 -flow satisfying (14.1) can be obtained from f (p+1) through at most |E| operations, where E is the set of arcs in the movement directed graph. Note that a directed cycle in a directed graph can be found through a depth first search and in our case, the directed graph is induced by A(f (p+1) − f (p) ). Thus, the modification takes time O(|E|2 ). Moreover, the maximum flow can be computed in polynomial-time. The flow in the lemma can be computed in polynomial-time.  

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Next, we prove that in different time slots, the sets of backup sensors used to heal holes are mutually disjoint. Lemma 14.5.2 Let Sp denote the set of backup sensors used to heal those holes in Hp for p = 1, . . . , T . If in every iteration, f (p+1) is chosen to satisfy condition (14.1), then the sets S1 , S2 , . . . , ST are mutually disjoint. Proof For p = 1, . . . , T , let Wp denote the set of out-neighbors of source s in f (p) . We claim that for any p = 1, . . . , T , Wp ⊆ Wp+1 .

(14.11)

Actually, since Sp+1 ⊆ Wp+1 , we have that there is no node of Wp \Wp+1 belonging to Sp+1 . Therefore, if Wp \ Wp+1 = ∅, then for any node u ∈ Wp \ Wp+1 , a directed cycle passing arc (u, s) must exist in A(f (p+1) − f (p) ). This contradicts condition (14.1). p p+1 By (14.11) and facts that |Wp | = j =1 kj , |Wp+1 | = j =1 kj , and |Sp+1 | = kp+1 , we have Sp+1 = Wp+1 − Wp . This implies the lemma.

(14.12)  

Consider the example in Fig. 14.5, again. We have S1 = W1 = {u1 , u2 }. After modify f (2) in (b), we obtain a new f(2) in (f). Hence, we have W2 = {u1 , u2 , u3 } and S2 = {u3 }. The minimum cost K3 -flow f (3) is showed in (d) and the flow f (3) − f(2) is showed in (h). which contains a directed cycle su5 h3 u1 h1 u3 s. After modification, we obtain f(3) in (i). Thus, W3 = {u1 , u2 , u3 , u4 } and S3 = {u4 } as showed in (j)). Clearly, S1 , S2 , and S3 are mutually disjoint. The correctness of the modified algorithm follows from the disjointness of Sp ’s. We put this result together with its competitive ratio into the following theorem. It should be mentioned that the solution obtained by the modified algorithm may violate the distance upper bound by a factor of 2, that is, in such a solution, each mobile sensor is allowed to move a distance more than B, but within 2B. This means that the competitive ratio is between a solution of an online algorithm with distance constraint 2B and an optimal solution of offline algorithms with distance constraint B. Theorem 14.5.3 (Zhang et al. [641]) The modified algorithm heals holes online with competitive ratio at most 2T − 1 within time window [0, T ], when the distance constraint is relaxed by at most twice in each time slot. Proof Let opt be the cost of an offline optimal solution for healing all holes in H1 ∪ · · · ∪ HT . Note that f (p) is obtained by an optimal strategy for healing holes in H1 ∪ · · · ∪ Hp . Since f (p+1) also includes the job to heal holes in H1 ∪ · · · ∪ Hp , we obtain c(f (p+1) ) ≥ c(f (p) ). Thus, c(f (1) ) ≤ c(f (2) ) ≤ · · · ≤ c(f (T ) ) = opt. In each time slot p, since all arcs in cascade healing paths are non-zero arcs in

14.5 Online Algorithm with Tight Competitive Ratio

223

Fig. 14.7 The computational process of the modified algorithm on the instance in Fig. 14.4. Numbers indicate corresponding coordinates on the line. (a) f (1) . (b) f (2) . (c) f (2) − f (1) . (d) f (3) . (e) f (3) − f (2)

f (p+1) − f (p) , the total distance moved by mobile sensors in the p-th iteration is upper bounded by c(f (p) ) + c(f (p+1) ) ≤ 2c(f (T ) ) = 2opt. Moreover, in the first time slot, the total moving distance is exactly c(f (1) ) ≤ opt. Therefore, the total distances moved in the online algorithm is at most (2T − 1)opt. Note that there may exist mobile sensors on the healing path which move two hops in the modified algorithm. Consider the example in Fig. 14.5g. Because the backup sensor at u1 is used in the first time slot, it is no longer there in current time slot. Hence, the mobile sensor which moves from h1 to u1 has to continue to move from u1 to h3 . Therefore, the moving distance for this mobile sensor is possibly more than B, however within 2B.   Figure 14.7 shows the computational process of the modified algorithm on the instance in Fig. 14.4. In the first time slot, the backup sensor at coordinate 1 is moved to heal the hole at coordinate 1/2, which is indicated by the minimum 1-flow f (1) in (a). In the second time slot, a minimum cost 2-flow f (2) in (b) and the flow f (2) − f (1) in (c) are computed, and the latter tells us that the active sensor at coordinate 1/2 should be sent back to heal the new hole at coordinate 1 and the backup sensor at coordinate −ε should be moved to coordinate 1/2 to heal the first hole. Starting from the third time slot, once a new hole appears at coordinate 2i−1 for i = 2, . . . , T − 1, the backup sensor at the same coordinate will be activated to heal the hole (see (d) and (e) for an illustration). Therefore, the modified algorithm spends the total distance ε + 3/2. Competitive Ratio 2T − 1 Is Tight Consider a star network consisting of a center u0 and T leaves ui for i = 1, 2, . . . , T . At each leaf, there exists a backup sensor. There are T holes appearing at the center u0 and T −1 leaves. Suppose every edge has distance B in this network. If all holes are given at the beginning, then an offline algorithm needs the total distance B, that is, heal the hole at center u0 by using the backup sensor where no hole appears. Now, we ask holes to appear online and in the first time slot, a hole appears at u0 and then remaining T − 1 holes appear one by one in the other T − 1 time slots. In the first time slot, a backup sensor at a leaf is moved with distance B to heal the hole at center u0 . In the second time slot or later, once a hole at a leaf appears, the

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worst case is that the backup sensor at the same location was moved to heal a hole just in the previous time slot. Hence, the new hole needs to be healed by another active sensor or backup sensor. In the latter case, the backup sensor needs to move with distance 2B. In the former case, the active sensor at u0 moves with distance B and however, the hole at u0 needs to be healed by a backup sensor at a leaf, which needs to move with distance B; hence, the total distance is still 2B. Therefore, in the worst case, any online cascade healing needs a total distance 2T − 1 in the worst case. This example shows that the competitive ratio 2T − 1 is the best possible even if the distance constraint is allowed to be relaxed within a factor of two. It also indicates that such a relaxation is necessary. Otherwise, the online version of the cascade healing problem may have no solution.

14.6 Consider Longest Moving Distance, Again Consider the online bottleneck cascade healing problem. Similarly, the greedy strategy is also exponentially bad. Actually, in a greedy online algorithm, when a set Hp+1 of new holes appear, the greedy strategy is to heal them by computing a kp+1 -flow with min–max individual moving distance. Using the same example in Section 14.4, we would find that the greedy strategy yields a longest moving distance 2T −2 in the latest time slot, while an offline optimal moving distance is ε + 1/2. Therefore, the competitive ratio is roughly 2T −1 . Now, with an idea similar to that in Section 14.5, we design an online algorithm with competitive ratio 2. Again, at time slot p + 1, hole sets H1 , . . . , Hp , Hp+1 p already appear. Suppose |Hi | = ki for i = 1, . . . , p. Denote Kp = j =1 kj . Using Algorithm 2, compute an offline optimal bottleneck Kp -flow f (p) in the auxiliary directed graph Rdb(p) (H1 ∪ · · · ∪ Hp , S (a) , S (b) ), and an offline optimal bottleneck Kp+1 -flow f (p+1) in the auxiliary directed graph Rdb(p) (H1 ∪· · ·∪Hp+1 , S (a) , S (b) ). Again, we can ask f (p+1) to satisfy condition (14.1). Then f (p+1) − f (p) contains kp+1 s-t-paths which form a kp+1 -flow which indicates the way to heal holes in Hp+1 . Let db(p) denote the longest moving distance of a sensor in f (p) . Since f (p + 1) contains the duty that f (p) has, i.e., to heal holes in H1 ∪ · · · ∪ Hp , we have db(p) ≤ db(p+1) . Denote by opt = db(T ) , i.e., the optimal bottleneck objective function value for healing all holes in H1 ∪ · · · ∪ HT . Then we have db(1) ≤ db(2) ≤ · · · ≤ db(T ) = opt. Note that in every time slot p, the flow f which is obtained from f (p+1) − f (p) uses only those arcs with non-zero values in f (p+1) and f (p) and hence the lengths of these arcs are upper bounded by opt. However, it is similar to the proof of Theorem 14.5.3 that there may exist some mobile sensor who moves two hops. Therefore, the moving distance of each mobile sensor is upper bounded by 2opt. Hence, the following theorem holds.

14.7 Remark

225

Fig. 14.8 A tight example for the online bottleneck cascade healing algorithm

Theorem 14.6.1 (Zhang et al. [641]) There exists a polynomial-time online algorithm with competitive ratio 2 for the bottleneck cascade hole healing problem. In [641], Zhang et al. present the example in Fig. 14.8 to show that competitive ratio 2 is tight for the bottleneck cascade hole healing problem. In this example, two (b) (b) backup sensors s1 and s2 , and two holes h1 and h2 lying on a line with distance as shown in the figure. Clearly, the offline solution has objective function value B, i.e., (b) (b) s1 is moved to h1 and s2 is moved to h2 . In online situation, at the first time slot, (b) h1 appears and then s2 is moved to heal h1 . At the second time slot, h2 appears, which requires s2(a) moving to heal h2 and s1(b) moving to heal h1 . Therefore, the longest moving distance is 2 − ε in this time slot. Therefore, the competitive ratio is required to be 2 − ε for any online algorithm. However, this example is ambiguous at the first time slot. In fact, it is unclear why any online algorithm chooses s2(b) to heal h1 . There also exists a question on the objective function. From above example, it can also be seen clearly that the moving distance of a sensor is counted in a time slot, not in entire time period from 0 to T . If the moving distance of a sensor is counted during the entire time period, then the sensor s2 (s2(b) or s2(a) ) has the total moving distance 3 − 2ε. Could above online algorithm still have competitive ratio 2 if the moving distance of each sensor is counted during the whole time period? It seems that this is not easy to be proved. However, in this case, the star network in Section 14.5 can provide a solid example to show that the competitive ratio 2 is the best possible. Actually, we have more open problems on bottleneck consideration. Open Problem 19 If the moving distance of each sensor is calculated during whole time period, is there a online algorithm with competitive ratio 2 for the bottleneck cascade hole healing problem? Open Problem 20 Is there a online algorithm with competitive ratio 2 for the cascade hole healing problem with minimum longest path from backup sensors to holes?

14.7 Remark In the literature, there exist some methods, other than maximum flow, to deal with the cascade hole healing problem. Let us mention one of them as follows.

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Consider the offline cascade hole healing problem. Given a set H of n holes and a set Mb of m backup sensors, with m ≥ n, construct a bipartite graph between H and Mb as follows: For h ∈ H and s ∈ M, assign edge (h, s) with weight w(h, s) = dmax − d(h, s), where dmax is the maximum distance between a hole and a mobile sensor. Then the maximum total weight matching in complete bipartite graph between H and Mb would give a minimum total distance moving schedule for hole healing. In above study, we did not consider a constraint that every sensor has a power limitation. This means that if d(f (h), h) is too long, then sensor f (h) may have no enough power to move to hole h. When this constraint is considered, we may need to consider the cascade hole healing. Let Ma be the set of active mobile sensors. Let d¯ be the maximum distance that a mobile sensor is able to move. Construct a bipartite graph G between Mb ∪ Ma and H ∪ Ma by giving edge set E = {(s, h) | d(s, h) ≤ d¯ for s ∈ Mb ∪ Ma , h ∈ H ∪ Ma }. For each edge (s, h) ∈ E, assign weight w(s, h) = d¯ − d(s, h). Then a feasible moving schedule exists if and only if G has a matching of cardinality n + |Ma | and moreover, the maximum weight matching of G gives a moving schedule with the minimum total distance.

Chapter 15

Camera Sensors

It was lights, camera, inaction. Terry Gilliam

15.1 Motivation and Overview In previous chapters, sensors are all based on omni-sensing model and the sensing area is a disk. In this chapter, we consider the directional antennas or camera sensors and the sensing area is a sector, which can be described by its center s, a sensing angle α, a sensing radius r, and a sensing direction which is a vector f( at s (Fig. 15.1). The vector f( is the bisector of angle α, i.e., divides angle α evenly. The sector may be rotatable, i.e., the sensing direction is adjustable. There are two types of rotations, discrete and continuous. For the discrete rotation, there are a finite number of fixed orientations (Fig. 15.2). For the continuous rotation, the sector can be freely rotated and stops at any orientation. There are a few works on traditional targets (points or areas) covered by directional wireless sensor networks [9, 43, 44, 153, 271, 344, 376, 383, 384, 386, 392, 483, 593, 607]. For example, Ma and Liu [392] studied the coverage rate and Ai and Abouseled [9] and Chow and Lui [153] considered a problem of maximum coverage with minimum number of sensors (min–max sensor coverage). Actually, given a set of directional sensors and a set of targets, it is possibly NP-hard to determine whether all targets can be covered by adjusting sensing directions for all directional sensors. In fact, if ignore the geometric structure, then each directional sensor can be represented by a group of subsets of targets; each subset is covered by the sensor at certain sensing direction. Given a number of groups of subsets of targets, it is NP-complete to determine whether all targets can be covered by selecting at most one subset from each group. Ai and Abouseled [9] designed a greedy algorithm, but was unable to give theoretical analysis. In this chapter, we will show that this greedy algorithm can produce a subset of directional sensors which cover at least a half fraction of maximum coverage. The proof idea is motivated from [121] for the maximum coverage problem with group budget constraint. The

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_15

227

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15 Camera Sensors

Fig. 15.1 Directional sensor Sensing Angle

Sensing direction

Sensing radius

Fig. 15.2 Discrete and continuous rotations

Fig. 15.3 Directional target

better ratio (1 − e−1 ) is achieved by an algorithm obtained from [238]. This result also holds in weighted case in which every subset has a nonnegative weight. The camera sensor is an important type of directional sensors, which is used for collecting the image data. The network consisting of camera sensors is called the visual sensor network (VSN). In the study of coverage of VSNs, not only the sensing area is directional but also the target is associated with a direction. For example, to identify a person, the camera sensor may need to watch the person’s face. Therefore, the face would give a special direction to the person as a target. Such a target can be described by its location t, a facing angle φ, and the facing direction which is a vector g( bisecting the angle φ (Fig. 15.3).

15.1 Motivation and Overview

229

Fig. 15.4 A directional target t is covered by a camera sensor s

A directional target t is covered by a directional sensor s if t lies in the sensing area of s and the line st lies inside the angle α and the angle φ, i.e., (f(, st) ≤ α/2 and (( g , ts) ≤ φ/2 (Fig. 15.4). With this definition, all results about target points can be extended to the case that every directional target has a fixed facing direction. To provide high quality of monitoring, the multiple coverage was considered in many papers [211, 428, 544]. Although the multiple coverage requires a large number of sensors, it increases only the reliability, i.e., the fault-tolerant property, and does not give additional information about the target. Wang and Cao [536] found a better way to use those large amount of sensors. They proposed a concept of fullview on directional targets. Consider a set of camera sensors, S and a set of directional targets, T . A target t ∈ T is called full-viewed by S if for any facing direction of t, there exists a sensor s ∈ S covering t. S is called a sensor full-view cover of T if any target t ∈ T is fullviewed by S. Clearly, a full-view target is equivalent to infinitely many targets with fixed facing direction. Can a full-view target be replaced by finitely many targets with fixed facing direction? The answer is no. Actually, for any finite set of facing directions getting covered, there exists a facing direction uncovered. However, for any ε > 0, we can find a finite set of facing directions such that if this set of facing directions gets covered, then with probability 1 − ε, for any facing direction, the target is covered. In such a case, we may call the target as an ε-full-view target. The full-view coverage has been studied in the literature extensively [213, 253, 263, 324, 398, 608, 635], including the full-view barrier coverage [213, 398, 599, 608] and target area consideration [253, 263, 635]. The generalization of full-view is the β-view [239]. A directional target t is required to have the β-view coverage if there is angle of size β at t such that for every facing direction within angle β, there exists a camera sensor covers the target. Every result in full-view can be generalized to β-view without significant difficulty. Camera sensors have been used in various types of coverage mentioned in previous chapters [233, 581] and generated many research subjects [146, 235, 236, 256–261, 592]. Especially, a mathematical model, the group set cover, is proposed for study of coverage problems of camera sensors [183]. In the following sections, we would select only a few subjects described as above to introduce to the reader. They have a feature in common, that is, they have clear formulation on optimization.

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15.2 Sensing Direction Selection Consider the following problem. Problem 15.2.1 (Sensing Direction Selection) Given a set of directional sensors, S, and a set of target points, A, select a sensing direction for each sensor in S such that all target points in A can be covered. If the geometric structure is ignored, then this problem can be formulated as the following. Problem 15.2.2 (Group Set Cover) Given k groups G1 , . . . , Gk of subsets of finite set X, is it possible to select a subset Si from each group Gi such that collection {S1 , . . . , Sk } covers X. Here, each group Gi corresponds to a directional sensor si . Each subset Si in Ci consists of all target points covered by sensor si with certain sensing direction. Theorem 15.2.3 The group set cover problem is NP-complete. Proof It is easy to construct a polynomial-time many-one reduction from the set cover problem to the group set cover problem. The instance of the former consists of a collection C of subsets of a finite set X, and a positive integer k. The problem is to find whether C has a subcollection of at most k subsets, covering X. Set G1 = · · · = Gk = C. Selection of a subcollection of at most k subsets from C is equivalent to selecting one from each Gi for i = 1, . . . , k. Moreover, the purpose of selection is the same, that is, cover X. Therefore, the reduction is established. It is well-known that the minimum set cover problem is NP-hard and hence the generalized set cover problem is NP-hard. Moreover, the group set cover problem is clearly in NP. Hence, it is NP-complete.   The NP-completeness of the group set cover problem does not imply the NPcompleteness of the sensing direction selection problem. In fact, from the proof of Theorem 15.2.3, we see that the group set cover problem is still NP-hard in the special case that all collections G1 , . . . , Gk are identical. However, the working direction determination problem is polynomial-time solvable when all sensors are located at the same point. This polynomial-time algorithm can be constructed as follows. Since all groups are identical, we may assume that all sensors are homogeneous, that is, they have the same size of sensing radius and the same size of sensing angle. Let us arrange all target points in the counterclockwise ordering a1 , a2 , . . . , an . Also, arrange all sensors in the counterclockwise ordering s1 , s2 , . . . , sk based on their sensing directions. Their sensing directions are said to be in canonical positions if for any i, si and si+1 cover disjoint subsets of targets points. Suppose that there exists a selection of sensing directions for k directional sensors such that all target points are covered. Then we can adjust the sensing directions to reach canonical positions which still cover all target points. Note that each canonical position determines a sequence of subsets of target points. There are at

15.3 Group Set Cover

231

most n different such sequences, starting with subset {a1 , . . .}, or {a2 , . . .}, . . . , {an , . . .}. Call those sequence as canonical sequence. Therefore, we can construct an algorithm which generates all canonical sequences. If one of them contains all target points, then the sensing direction selection problem has a solution; otherwise, the problem has no solution. Actually, in the literature, for many network problem, the NP-hardness was proved only for the problem without geometric structure. If put back the geometric structure, then it is still an open problem, such as the following. Open Problem 21 Is the sensing direction selection problem NP-complete?

15.3 Group Set Cover The group set cover problem is a generalization of the set cover problem. In order to understand the optimization problems on the group set cover, let us first discuss optimization problems on the set cover. There are two classic optimization problems on the set cover, the maximum coverage problem and the minimum set cover problem. Problem 15.3.1 (Maximum Coverage) Given a collection C of subsets of a finite set X and an positive integer k, find k subsets from C to cover the maximum number of elements of X. Problem 15.3.2 (Minimum Set Cover) Given a set cover C of a finite set X, find a minimum set cover C  ⊆ C, where a set cover of X is a collection of subsets of X such that every element of X appears in the collection. Greedy algorithm gives the best approximation for them. Greedy Algorithm for Maximum Coverage input a finite set X, a collection C of subsets of X, and an integer k > 0; A ← ∅; repeat pick the subset A in C which covers the maximum number of uncovered elements and mark elements in selected subset as covered; put A into A and delete A from C; until done ("done" means: k subsets are selected from C.) output A. Theorem 15.3.3 The greedy solution is a polynomial-time (1−e−1 )-approximation for the maximum coverage problem. Proof Let subsets A1 , A2 , . . . , Ak be generated by the greedy algorithm. Denote Ai = {A1 , . . . , Ai }. Let OP T = {S1 , . . . , Sk } be an optimal solution. Denote

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f (A) = | ∪A∈A A|. Then f (S ∗ ) ≤ f (Ai ∪ OP T ) ≤ f (Ai )+ S1 f (Ai )+ S2 f (Ai ∪ {S1 })+ · · · + Sk f (Ai ∪ {S1 , . . . , Sk−1 }) ≤ f (Ai ) + S1 f (Ai ) + S2 f (Ai ) + · · · + Sk f (Ai ) ≤ f (Ai ) + k Ai+1 f (Ai ), where denote S f (A) = f (A ∪ {S}) − f (A). Let us denote αi = opt − f (Ai ), where opt = f (OP T ). Then αi ≤ k(αi − αi+1 ). Therefore, 1 αi+1 ≤ (1 − )αi ≤ e−1/k αi k since 1 + x ≤ ex . Hence, αk ≤ e−1 α0 = e−1 · opt, i.e., f (Ak ) ≥ (1 − e−1 ) · opt.   For the minimum set cover problem, we usually assume that the input collection C covers X. In fact, if any element in X is not covered by C, then we can simply delete it before dealing with the problem. Greedy Algorithm for Min Set Cover input a finite set X and a set cover C of X; A ← ∅; repeat pick the subset A in C which covers the maximum number of uncovered elements and mark elements in selected subset as covered; put A into A and delete A from C; until done ("done" means: no uncovered element exists.) output A. Theorem 15.3.4 The greedy solution is a polynomial-time (1 + ln |X|)approximation for the minimum set cover problem. Proof Let subsets A1 , . . . , Ag be generated by the greedy algorithm. Denote Ai = {A1 , . . . , Ai }. Let OP T = {S1 , . . . , Sopt } be an optimal solution. Define f (A) = | ∪A∈A A|. By the greedy rule, for any i = 1, . . . , g, Ai f (Ai−1 ) ≥ Sk f (Ai−1 ) ∀k = 1, . . . , opt.

15.3 Group Set Cover

233

Hence, δAi f (Ai−1 ) ≥

opt 1  Sk f (Ai−1 ) · opt k=1

≥

1 · ( S1 f (Ai−1 ) + S2 f (Ai−1 ∪ {S1 }) opt + · · · + Sopt f (Ai−1 ∪ {S1 , . . . , Sopt−1 }))

≥

1 · f (Ai−1 ∪ OP T ) − f (Ai−1 ). opt

Denote αi = f (OP T ) − f (Ai ). Note that f (OP T ) = f (Ai−1 ∪ OP T ). Then we have αi−1 − αi ≥

1 · αi−1 . opt

Hence,  1  αi ≤ 1 − αi−1 ≤ e−1/opt αi−1 . opt Note that α0 = f (OP T ) ≥ opt since each subset in OP T would cover at least one element not covered by other subsets in OP T . Moreover, αg = 0. Therefore, there exists 0 < k ≤ g such that αk ≥ opt > αk+1 . Since αi is an integer for any i, we must have αk+1 ≤ opt − 1 and 0 = αg ≤ opt − (g − k). Hence, g ≤ opt + k. Moreover, opt ≤ αk ≤ e−1/opt αk−1 ≤ · · · ≤ e−k/opt α0 = e−k/opt f (OP T ). Therefore, k ≤ opt · ln(f (OP T )/opt) ≤ opt · ln |X|.   In previous chapter, when we deal with optimization problems about sensor covers, we did not mention maximum coverage. Actually, previously considered sensors are often very cheap and they are randomly deployed into the working area. The minimum sensor cover problem and the maximum lifetime coverage problem are both based on this situation. However, camera sensors are not so cheap and they are usually installed at designed locations. Thus, the background is changed and the study of maximum coverage would make sense. There are two real-world problems related to the maximum coverage of camera sensors. Problem 15.3.5 (Direction Adjustment) Given a set of camera sensors and a set of target points, adjust working directions of all camera sensors to maximize the total number of covered target points. Problem 15.3.6 (Camera Sensor Placement) Given a set of target points and a set of camera sensor locations, select k locations for putting camera sensors and adjusting their directions to maximize the total number of covered target points.

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Clearly, Problem 15.3.5 is a special case of Problem 15.3.6 when k is equal to the number of locations. They both can be formulated as the following optimization problem. Problem 15.3.7 (Max Group Set Coverage) Given groups G1 , . . . , G of subsets of a finite set X, and an integer k( ≥ k > 0), select k groups and choose at most one subset from each selected group to form a collection in order to cover the maximum number of elements. Consider a greedy algorithm: Greedy Algorithm for Max Group Set Coverage input a finite set X, an integer k > 0, and (≥ k) groups G1 , . . . , G of subsets of a finite set X; repeat pick the unselected group which contains the subset that covers the maximum number of uncovered elements and mark elements in selected subset as covered; select the group; until done ("done" means: k groups are selected.) output k selected groups. Theorem 15.3.8 The greedy algorithm gives a polynomial-time 0.5-approximation for the maximum group set coverage problem. Proof Let subsets S1 , . . . , Sk be generated by the Greedy Algorithms. Relabel all groups such that S1 ∈ G1 , . . . , Sk ∈ Gk . Let S1∗∗ , . . . , Sk∗∗ be the optimal solution such that 1∗ < · · · < k ∗ and S1∗∗ ∈ G1∗ , . . . , Sk∗∗ ∈ Gk ∗ . Then 1 ≤ 1∗ , . . . , k ≤ k ∗ . Note that for 1 ≤ i ≤ k, |Si \ (S1 ∪ · · · ∪ Si−1 )| ≥ |Si∗∗ \ (S1 ∪ · · · ∪ Si−1 )|. Therefore, we have |S1 ∪ · · · ∪ Sk | = |S1 | + |S2 \ S1 | + · · · + |Sk \ (S1 ∪ · · · ∪ Sk−1 )| ≥ |S1∗∗ | + |S2∗∗ \ S1 | + · · · + |Sk∗∗ \ (S1 ∪ · · · ∪ Sk−1 )| ≥

k 

|Si∗∗ \ (S1 ∪ · · · ∪ Sk )|

i=1

≥ |(S1∗∗ ∪ · · · ∪ Sk∗∗ ) \ (S1 ∪ · · · ∪ Sk )| ≥ |S1∗∗ ∪ · · · ∪ Sk∗∗ | − |S1 ∪ · · · ∪ Sk |.

15.3 Group Set Cover

235

Hence, |S1 ∪ · · · ∪ Sk | ≥ |S1∗∗ ∪ · · · ∪ Sk∗∗ |/2 = opt/2.

 

Actually, 0.5 is the best possible performance ratio that a greedy algorithm can reach for the maximum group set coverage problem. To understand this, let us mention a problem well-studied in the literature. Problem 15.3.9 (Maximum Coverage with Group Budget Constraints) Consider a collection C of subsets of a finite set X and a partition (G1 , . . . , G ) of C. Given an integer k > 0, find a subcollection C  of C to cover the maximum number of elements in X under constraints |C  | ≤ k and |C  ∩ Gi | ≤ 1 for 1 ≤ i ≤ . The maximum group set coverage problem can be transformed to the maximum coverage problem with group budget constraints in the following way: Give each group Gi with an id gi ∈ X and put gi into every subset in Gi . However, for objective function, we count only the number of elements in X. For the maximum coverage problem with group budget constraints, an example is given in [121] to show that the performance ratio of the greedy algorithm cannot be better than 0.5. This example can be easily modified for the maximum group set coverage problem based on above transformation. Guo et al. [238] designed a polynomial-time (1 − e−1 )-approximation with LP relaxation and randomized rounding. Let us introduce a simple version of their algorithm for a special case that k =  which has an application, the direction adjustment problem. First, we formulate the problem into a 0–1 linear programming as follows: Consider k disjoint groups G1 , . . . , Gk which are obtained by transformation from the maximum group set coverage problem with k = . Let |X| = n and G = G1 ∪ · · · ∪ Gk . We use xS to indicate whether subset S is selected or not and yi to indicate whether element i appears in a selected subset or not. The following is the integer linear programming. (I LP ) : max

n 

yi

i=1



s.t. yi ≤ 

xS ∀i = 1, . . . , n,

S:i∈S

xS ≤ 1 ∀j = 1, . . . , k,

S:S∈Gj

yi ∈ {0, 1} ∀i = 1, . . . , n, xS ∈ {0, 1} ∀S ∈ G.

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Its relaxation is as follows: (LP ) : max

n 

yi

i=1



s.t. yi ≤ 

xS ∀i = 1, . . . , n,

S:i∈S

xS ≤ 1 ∀j = 1, . . . , k,

S:S∈Gj

0 ≤ yi ≤ 1 ∀i = 1, . . . , n, 0 ≤ xS ≤ 1 ∀S ∈ G.

Let (yi∗ , xS∗ ) be an optimal solution of linear programming (LP). We do a randomized rounding as follows. Randomized Rounding For each group Gj , select  one subset S with probability xS∗ and not select any subset with probability 1 − S∈Gj xS∗ . Set yi = 1 if element i appears in a selected subset, and yi = 0, otherwise. Theorem 15.3.10 Let (yi , xS ) be approximation solution obtained by above algorithm. Then E[

n 

yi ] ≥ (1 − e−1 )opt,

i=1

where opt is the objective function value of optimal solution for the group set coverage problem with k = . Proof For each i = 1, . . . , n, Prob[y = 1] =

k "

"

(1 − xS∗ )

j =1 S:i∈S∈Gj

=

"

(1 − xS∗ )

S:i∈S

 ≤ 1−

 S:i∈S

Ki

 y ∗ K ≤ 1 − i i. Ki

xS∗ Ki

(Ki = |{S|i ∈ S}|)

15.3 Group Set Cover

237

Fig. 15.5 Function f (z)

Hence,  y ∗ K Prob[yi = 1] ≥ 1 − 1 − i i . Ki  Note that the function f (z) = 1 − 1 − in [0, 1] (Fig. 15.5) since

 z Ki Ki

is monotone increasing and concave

 z Ki −1 ≥ 0, f  (z) = 1 − Ki z Ki −2 Ki − 1  f  (z) = − 1− ≤ 0. Ki Ki Therefore, the function curve is above the line segment between two points (0, f (0)) K  and (1, f (1)). Since f (0) = 0 and f (z) = 1 − 1 − K1i i , we have  1 Ki f (z) ≥ 1 − 1 − z ≥ (1 − e−1 )z. Ki Thus, Prob[yi = 1] ≥ (1 − e−1 )yi∗ . Therefore, E[

n 

yi ] =

i=1

n 

E[yi ]

i=1

=

n 

Prob[yi = 1]

i=1

≥ (1 − e−1 ) ·

n 

yi∗

i=1

≥ (1 − e−1 ) · opt.

 

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For general case that 0 < k ≤ , the algorithm is much more complicated (see [238]). This is because {C | |C ∩ Gj | ≤ 1, 1 ≤ j ≤ } is a matroid. When k = , there is only a matroid constraint. When k ≤ , we have a matroid constraint plus a size-constraint. This additional size-constraint would cost additional techniques to reduce the complexity of running time. A natural generalization of the min set cover problem is as follows: Problem 15.3.11 (Min Group Set Cover) Given groups G1 , . . . , Gk of subsets of a finite set X, select at most one subset from each group to form a minimum set cover for X. It is interesting to study its computational complexity. Consider the decision version of this problem: Given groups G1 , . . . , Gk of subsets of a finite set X, and an integer h > 0, is it possible to select h of them to form a group set cover? A nondeterministic algorithm for this problem may be described as follows: Guess h groups. Check if obtained h groups form a group set cover. Clearly, the check step requires to solve an NP-complete problem. Therefore, the p decision of the minimum group set cover problem belongs to N P (N P ) = 2 . If NP =P, then it is impossible to design a polynomial-time approximation algorithm for the minimum group set cover problem since knowing the feasibility of a solution is NP-complete. A more proper generalization of the min set cover problem may be as follows: Problem 15.3.12 (Min–Max Group Set Cover) Given groups G1 , . . . , Gk of subsets of a finite set X, select at most one subset from each group to form a minimum collection to cover the maximum number of elements in X. Although the greedy algorithm does not give the best approximation solution for the min–max group set cover problem, it gives a pretty good approximation. Greedy Algorithm for Min–Max Group Set Cover input a finite set X and  groups G1 , . . . , G of subsets of X; initially, all groups are marked “unselected”; Repeat pick the unselected group which contains the subset that covers the maximum number of uncovered elements and mark elements in selected subset as covered; mark the group and the subset as selected; until done ("done" means: no uncovered element can be further covered.) output all selected groups together with selected subsets in them.

15.3 Group Set Cover

239

Theorem 15.3.13 Consider groups G1 , . . . , Gk of subsets of a finite set X. Select at most one subset from each group to form a collection. Suppose that the Greedy Algorithm produces g subsets which cover h elements. Then h ≥ emax /2 and g ≤ ρsmin , where that emax is the maximum elements which can be covered by a collection of subsets selected at most one subset from each group, smin is the number of subsets in such a collection which covers at least h elements, and ρ is the approximation performance ratio of a greedy algorithm for a submodular cover problem with a partition matroid constraint. Proof Let subsets S1 , . . . , Sg be generated by the Greedy Algorithms. Relabel all ∗ be the optimal solution groups such that S1 ∈ G1 , . . . , Sg ∈ Gg . Let S1∗∗ , . . . , Sm ∗ ∗ ∗ ∗ ∗ such that 1 < · · · < m and S1∗ ∈ G1∗ , . . . , Sm∗ ∈ Gm∗ . Then 1 ≤ 1∗ , . . . , m ≤ m∗ . If g ≥ m, then for 1 ≤ i ≤ m, we have |Si \ (S1 ∪ · · · ∪ Si−1 )| ≥ |Si∗∗ \ (S1 ∪ · · · ∪ Si−1 )|. If g < m, then we set Sg+1 = · · · = Sm = ∅. Hence, we still have that for 1 ≤ i ≤ m, |Si \ (S1 ∪ · · · ∪ Si−1 )| ≥ |Si∗∗ \ (S1 ∪ · · · ∪ Si−1 )|. Therefore, we have |S1 ∪ · · · ∪ Sm | = |S1 | + |S2 \ S1 | + · · · + |Sm \ (S1 ∪ · · · ∪ Sm−1 )| ∗ ≥ |S1∗∗ | + |S2∗∗ \ S1 | + · · · + |Sm ∗ \ (S1 ∪ · · · ∪ Sm−1 )|

≥

m 

|Si∗∗ \ (S1 ∪ · · · ∪ Sm )|

i=1 ∗ ≥ |(S1∗∗ ∪ · · · ∪ Sm ∗ ) \ (S1 ∪ · · · ∪ Sm )| ∗ ≥ |S1∗∗ ∪ · · · ∪ Sm ∗ | − |S1 ∪ · · · ∪ Sm |.

Hence, ∗ |S1 ∪ · · · ∪ Sm | ≥ |S1∗∗ ∪ · · · ∪ Sm ∗ |/2 = emax /2.

To see the second inequality, let us assume that all groups are disjoints. In this case, M = {C | |C ∩ Gj | ≤ 1 ∀i = 1, . . . , k}

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is a partition matroid. Consider the following submodular cover problem with a matroid constraint. min |C| s.t. | ∪S∈C S| ≥ g, C ∈ M. The greedy algorithm for the min–max group set cover problem can also be seen as a greedy algorithm for this submodular cover problem. Suppose that the approximation performance ratio of this greedy algorithm is ρ. Then we obtain the second inequality.   Open Problem 22 What is the best performance ratio which can be reached by a greedy algorithm for the submodular cover problem with a matroid constraint?

15.4 Directional Targets The aim of using camera sensors is mainly to monitor directional targets, such as cars and persons. For a car, it is important to see its license plate and for a person, it is important to see his/her face. Therefore, each directional target has a facing direction and a facing angle which is divided by the facing direction evenly (Fig.15.3). A directional target t is said to be covered by a camera sensor s if t lies in the sensing area of s and the straight line connecting t and s lies in both the sensing angle of s and the facing angle of t (Fig. 15.4). With this definition of coverage for directional target, if every directional target has a fixed facing direction, then the coverage of camera sensors can still be formulated to problems about the group set cover. A trouble maker is the full-view requirement when the facing direction is not fixed. A directional target t is said to get the full-view covered by a camera sensor network if for any facing direction, there exists a camera sensor covers t (Fig.15.6). Could we replace a target with the full-view requirement by several targets with fixed facing directions? Proposition 15.4.1 If the facing angle is the same, then a target with the full-view requirement cannot be replaced equivalently by a finite set of targets with fixed facing directions. Proof For contradiction, suppose a target t with full-view requirement can be replaced equivalently by a finite number of targets with fixed facing directions. Assume that among those facing directions, the angle between facing directions f( and f( is the smallest one. Put a camera sensor s such that stf = θ/2 − ε, where θ is the facing angle. Then s can cover the target t with facing direction f(. Put another sensor s  such that s  tf  = θ/2 − ε and sts  > θ (choose ε > 0

15.4 Directional Targets

241

Fig. 15.6 A target with full-view requirement

Fig. 15.7 Proof of Proposition 15.4.1

Fig. 15.8 Proof of Proposition 15.4.2

sufficiently small). Let f( be the direction bisecting the angle sts  evenly. Then for t with facing direction f( , there is no camera sensor can cover t, contradicting to the definition of full-view (Fig. 15.7).   Proposition 15.4.2 It is sufficient to replace a target with facing angle α and fullview requirement by 2π/(α − β) targets with fixed facing angle β < α. Proof. Let t be the target with full-view requirement and facing angle α. Replace it by 2π/(α − β) targets at location t with fixed facing direction f(1 , . . . , f(k , where k = 2π/(α − β) such that f1 tf2 = f2 tf3 = · · · = fk tf1 ≤ α − β. Let s1 , . . . , sk be k sensors covering targets t with facing direction f(1 , . . . , f(k , respectively. Then s1 ts2 ≤ β + (α − β) = α, . . . , sk ts1 ≤ α. Therefore, for t with any facing direction and facing angle α, there exists sensor si such that tsi lies in the facing angle, that is, si covers t (Fig. 15.8).

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15.5 Target Area Consider a set S of camera sensors with sensing direction fixed and a target point t. There is a simple way to judge whether x gets the full-view coverage or not. Suppose S  = {s | t lies in the sensing area of s}. Connect t to every s ∈ S  . Let us order S  = {s1 , s2 , . . . , sk } based on counterclockwise ordering of ts1 , ts2 , . . . , tsk (Fig. 15.9). Theorem 15.5.1 t is full-view covered by S  if and only if θ, . . . , sk ts1 ≤ θ , where θ is the facing angle of target t.

s1 ts2

≤

Proof If si tsi+1 > θ , then no camera sensor covers target t with facing direction which divides si tsi+1 evenly. Conversely, if t is not full-view covered by S  , then there is a facing direction tf such that there is no camera sensor s ∈ S  to have segment ts lying in the facing angle with direction tf . Suppose tf is in si tsi+1 . Then we must have si tsi+1 > θ since, otherwise, either tsi or tsi+1 would lie in the facing angle with direction tf .   Consider a set S of camera sensors with sensing direction fixed and a target area A. A is full-view covered by S if every point t in A is full-view covered by S. The boundaries of sensing areas of sensors in S partition A into many small subareas (Fig. 15.10). In each subarea B, all points are covered by the same subset of camera Fig. 15.9 Camera sensors surrounding a target

Fig. 15.10 Target area is partitioned into small subareas

15.5 Target Area

243

Fig. 15.11 Proof of Lemma 15.5.2

Fig. 15.12 The unsafe region of two sensors

sensors in S, denoted by SB . For every point t in B, we can give an ordering of SB in a similar way as above. Lemma 15.5.2 The ordering of SB induced by every point t in B is the same. Proof Consider two points t and t  in B and three camera sensors s1 , s2 , s3 in SB . Suppose ts1 , ts2 , ts3 are in counterclockwise ordering and t  s1 , t  s2 , t  s3 are in clockwise ordering (Fig. 15.11). Then t and t  have to be located in different sides of line passing through s1 and s2 . This is impossible for both t and t  to lie in sensing areas of s1 , s2 , s3 .   Let θ be the facing angle for every target. Consider two camera sensors s and s  (Fig. 15.12). The unsafe region of s and s  is defined by UR(s, s  ) = {x | sxs  > θ }. Theorem 15.5.3 Suppose SB = {s1 , s2 , . . . , sk } be the ordering induced by any point in subarea B. The subarea B is full-view covered by SB if and only if B does not intersect the unreliable area UR(si , si+1 ) for any i = 1, . . . , k (assume sk+1 = s1 ). Proof The subarea B is full-view covered by SB if and only if every point in B is full-view covered by SB . By Theorem 15.5.1, this holds if and only if si tsi+1 ≤ θ for every point t ∈ B and every i = 1, . . . , k, if and only if B does not intersect the unreliable area of si and si+1 for i = 1, . . . , k.  

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Clearly, the target area A is full-view covered by the sensor set S if and only if every subarea B is full-view covered by the sensor set S, which can be judged by Theorem 15.5.3. Usually, the subarea B can also be full-view covered by a subset of SB . For each subarea B, let ΓB be the collection of camera sensor subsets each of which can fullview cover B. In order to find a minimum subset of S which full-view covers target area A, we need to solve the following minimization problem: min | ∪C ∈Γ C| s.t. |Γ ∩ ΓB | = 1 ∀B.

This problem can be expressed as a monotone nondecreasing submodular minimization problem with a matroid base constraint as follows: Problem 15.5.4 Given k groups of subsets in X, select one subset from each group to cover the minimum total number of elements. This problem has been studied in [515].

Chapter 16

Energy-Harvesting Sensors

Don’t judge each day by the harvest you reap but by the seeds that you plant. Robert Louis Stevenson

16.1 Motivation and Overview The sensor is often supplied with electrical power. For sensors studied in previous chapters, the electrical power comes from batteries, which have limited lifetime. Therefore, several technologies for recharging batteries have been developed recently [106, 393, 394, 657, 658]. Such technologies utilize energy sources in environment, such as sunlight [272, 478], electromagnetic radiation, thermal, and vibration [225]. A sensor with such a technology is called the energy-harvesting sensor, that is, an energy-harvesting sensor is able to convert energy of other types, in environment, to electrical energy. Among other energy sources, the sunlight is the most available one outdoor. Solar technologies are developed quite fast and now already reach conversion efficiency of 15% [478]. There exist many research articles about coverage of energy-harvesting sensors [487] or battery-free sensors [140, 466–468] in the literature. However, some mathematical models have no difference from classical sensors. The difference is only on the physical meaning. For example, the lifetime of a sensor depends on the energy level of rechargeable batteries, instead of the real life of batteries. Do there exist optimization problems about energy-harvesting sensors, different from those in previous chapters? Of course, the answer is yes. Here, we list three examples. Mule Scheduling The energy provider of energy-harvesting sensors is possibly a mobile device, called a mule. Suppose all mules are dispatched from a depot to serve target sensors located on an area, e.g., a path. Every target sensor has a handling time, and every mule is required to return to the depot within a time period or before a time. For different objectives, different mule scheduling problems are motivated. Sleep/Wakeup Scheduling Consider a set of small sensors harvesting energy from sunlight. During daytime, the sensor is charged to reach a level which is not enough © Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_16

245

246

16 Energy-Harvesting Sensors

to work for a whole night. Therefore, a sleep/wakeup scheduling is required for sensors in order to reach the coverage lifetime covering up the night time. The problem is to find a schedule to minimize the number of sensors with a coverage lifetime constraint. Hybrid System Consider a hybrid system consisting of a set of wireless rechargeable sensors and a set of larger sunlight-energy harvesting sensors. The later ones are responsible for collecting energy from sunlight and charging the former ones. Each sunlight-energy harvesting sensor may be able to charge a certain number of wireless rechargeable sensors with certain distance. The problem is to find a good placement for sunlight-energy harvesting sensors to minimize the number of sunlight-energy harvesting sensors. In next a few sections, we discuss above three problems.

16.2 Mule Scheduling With fast development of wireless charging technology, the mule scheduling gets more and more attentions. A mule is a mobile device which moves around a target area or a road to assistant with network maintenance, such as battery recharging and/or data collection. The problem is to find the mule routing to achieve certain service cost minimization. Many efforts have been made on this problem in the literature [104, 116, 475, 480]. For example, Sugihara and Gupta [480] study the mule scheduling which aims at minimization of the time latency. They focus on the speed control in a one-dimensional (1D) setting and claim that “Once we choose the path, any 2D/3D data mule scheduling problem can be converted to a 1D problem.” For a similar consideration, Chen et al. [141] also study the mule scheduling on a path with a different objective, i.e., minimization of the number of mules. The exact description of the problem is as follows: Problem 16.2.1 (Mule Scheduling) Consider a path u0 u1 · · · un and mules which move on this path. The traveling time from ui−1 to ui is di . At each node ui for 1 ≤ i ≤ n, there is a sensor waiting for service and the service time is ci ≥ 0. Mules are dispatched simultaneously from the depot u0 and every data mule is required to return to the depot before time span D. The problem is to assign sensors to the minimum number of mules to accomplish the service tasks under the above constraints. This problem is motivated from [116, 547] in which the time-consuming service is considered and hence multiple mules are studied. Actually, there exist many related works in the literature. They may appear in different formats, such as vehicle routing problems [68, 222, 283, 284, 414, 504]. It can be found in [198] more information about mules. In those works, different objectives functions and/or different constraints have been considered. A surprising observation is that although a path is a so simple topology, the mule scheduling problem may also be NP-hard.

16.2 Mule Scheduling

247

Theorem 16.2.2 (Chen et al. [141]) Problem 16.2.1 is NP-hard. Proof We construct a reduction from the bin packing problem as follows. Problem 16.2.3 (Bin Packing) Given n items with sizes s1 , s2 , . . . , sn , respectively, where 0 < si ≤ 1 for 1 ≤ i ≤ n are rational numbers, pack all items into a minimum number of bin of unit-size. Since the sizes of items are rational numbers,  we can find an integer q such that for any index set I ⊆ {1, 2, . . . , n} with i∈I si > 1, ( i∈I si − 1 ≥ 1/q. Moreover, such an integer q can be found in polynomial-time. In fact, q can be the product of all denominators of si for 1 ≤ i ≤ n. We now construct an instance of the mule scheduling problem as follows: Consider a path u0 u1 · · · un as described in Problem 16.2.1. Sensors at u1 , u2 , . . . , un require service times s1 , s2 , . . . , sn , respectively. The traveling time on every edge ui−1 ui is 1/(4qn). The time span is D = 1 + 1/(2q). Next, we show that for the bin packing problem, all items can be packed into k bin of unit-size if and only if for the mule scheduling problem, all services can be accomplished by k mules. First, suppose that all items can be packed into k bins of unit-size. For each bin, we use a mule to serve those sensors corresponding to items in the bin. Then total 1 1 service time is at most one. The traveling time is at most 2n · 4qn = 2q . Therefore, 1 = D time. This is a feasible solution for the mule each mule spends at most 1 + 2q scheduling problem. Conversely, suppose that for the mule scheduling problem, k mules are enough to accomplish the service. For each mule, we put in a bin all items corresponding to sensors served by the mule. We claim that  the total size of those items cannot exceed one. For contradiction, suppose that i∈I si > 1, where I is the set of those  items. Note that i∈I si ≤ D. Therefore,

 i∈I

si − 1 ≤ D − 1 ≤

1 , 2q

contradicting assumption on q.

 

Greedy algorithm is a pretty good choice to provide  solutions or approximate solutions for Problem 16.2.1. For simplicity, denote li = ij =1 dj . Greedy Algorithm for Problem 16.2.1 input: An instance (n; D; d1 , d2 , . . . , dn ; c1 , c2 , . . . , cn ). output: The number of mules, k and a sequence of integers m = i1 > · · · > ik > ik+1 = 0 such that [ij +1 , ij ] is the set of indices of those sensors served by the j th mule. 1: j ← 1, i1 ← n. 2: while ij > 0 do ij 3: Let ij +1 be the smallest index such that 2 · lij + t=i ct ≤ D. j +1 +1

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Fig. 16.1 An illustration for the adjustment of service. The left schedule is an adjustment of the right schedule

4: j ← j + 1. 5: end-while 6: Output k ← j − 1 and i1 , . . . , ik . First, let us consider a special case that all service times ci for 1 ≤ i ≤ n are identical, i.e., ci = c for 1 ≤ i ≤ n. This case is called the uniform case. Theorem 16.2.4 (Chen et al. [141]) In the uniform case, the greedy algorithm for Problem 16.2.1 produces an optimal solution in O(n) time. Proof The proof idea can be explained in Fig. 16.1 that if an optimal schedule does not serve sensors in consecutive manner, then we can adjust it into a consecutive pattern without increasing the number of mules. In fact, suppose a mule ma serves sensors si1 , si2 , . . . , sih , where i1 < i2 < · · · < ih , but i1 + 1 < i2 . Find mule mb which serves sensor si2 −1 . Exchange si1 and si2 −1 . Then ma has no change on service time and traveling time, and mb has no increasing on service time and traveling time. Hence, the feasibility is preserved. After finitely many operations like this, the schedule will be modified into a consecutive manner.   We now study non-uniform case. Let opt denote the minimum number of mules for Problem 16.2.1 and opx the number of mules in the greedy solution computed by the greedy algorithm. Let T ∗ be the total time consumed by those mules in an optimal solution and T A the total time consumed by those mules in the greedy solution. Lemma 16.2.5 T ∗ + 2(opx − opt) · lmax > T A , where lmax = ln . ∗ be mules in an optimal solution in the increasing order of Proof Let v1∗ , . . . , vopt their traveling time (handling time not included). Let pi∗ denote the traveling time for mule vi∗ . (Note that pi∗ = 2li∗ , where li∗ is the traveling time for mule vi∗ to travel from the depot to the farthest sensor it serves.) Similarly, mules in the greedy A , and the traveling time for mule v A is denoted solution are ordered as v1A , . . . , vapx i A ∗ by pi . Suppose that Pi and PiA are the sets of sensors served by vi∗ and viA , respectively. Then,both the optimal solution and the greedy solution have the same total service time ni=1 ci . Therefore, to show the lemma, it suffices to prove that opt  i=1

pi∗ + 2(apx − opt)lmax ≥

apx  i=1

piA .

(16.1)

16.2 Mule Scheduling

249

The right-hand side of (16.1) can be written as apx  i=1

piA =

opt 

apx 

piA +

i=1

piA .

i=opt+1

Note that piA ≤ 2lmax for any i. Thus, 2(apx − opt)lmax ≥

apx 

piA .

i=opt+1

This means that the proof of inequality (16.1) is reduced to show opt  i=1

pi∗ ≥

opt 

piA .

(16.2)

i=1

To this end, we show a stronger claim as follows. Claim There is an optimal solution satisfying pi∗ ≥ piA for any 1 ≤ i ≤ opt. For contradiction, suppose the claim is not true (see Fig.16.2 for an explanation of the following proof). Let i0 be the largest index satisfying pi∗0 < piA0 . Since the farthest sensor which can be served by the first i0 mules is li∗0 away from the depot, 0 we have that ii=1 Pi∗ is a subset of the sensors in the interval [0, li∗0 ]. Moreover, i0 A A i=1 Pi contains all sensors in the interval [0, li0 ] since the greedy solution serves sensors consecutively. Therefore, if we change the schedule of the optimal solution by letting vi∗ serve sensors in set PiA for i = 1, . . . , i0 , and keeping the schedule for the remaining mules, then we have a feasible solution with the same number of mules since [0, pi∗0 ] ⊆ [0, piA0 ]. This feasible solution is an optimal solution satisfying the claim. Inequality (16.2) follows immediately from the claim and hence the lemma is proved.   Theorem 16.2.6 (Chen et al. [141]) The greedy algorithm runs in O(n) time and max produces a solution which uses at most 2D−2l D−2lmax opt + 1 mules, where opt is the number of mules in an optimal solution. Proof The time complexity is O(n) because the algorithm can be executed by scanning only once the sensors in the decreasing order of their distances from the depot. From line 3 of the greedy algorithm, it is easy to see that if the j -th mule serves one more sensors, then the time span constraint will be violated, that is,

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Fig. 16.2 An illustration for the proof of the claim. (a) is an optimal solution, with P1∗ = {u1 }, P2∗ = {u2 , u4 }, P3∗ = {u3 , u5 }. (b) is a greedy solution, with P1A = {u1 , u2 }, P2A = {u3 }, P3A = {u4 }, P4A = {u5 }. In this example, i0 = 1. (c) is an optimal solution satisfying the claim after adjustment

ij 

2lij +

ct + cij +1 > D for j = 1, . . . , apx − 1,

t=ij +1 +1

and 2liapx +

iapx 

ct > 0 for j = opx.

t=1

Summing up j = 1 to j = opx, we have T

A

+

apx−1 

cij +1 > (apx − 1)D,

j =1

since 2

apx−1  j =1

lij + 2liapx +

apx−1 

ij 

j =1 t=ij +1 +1

ct +

iapx  t=1

ct = T A .

16.3 Sleep/Wakeup Scheduling

Note that T ∗ − 2lmax ≥

apx−1 j =1

251

cij +1 . Therefore,

T A + T ∗ − 2lmax > (apx − 1)D. By Lemma 16.2.4 and the observation opt · D ≥ T ∗ , we have 2opt · D + 2(apx − opt)lmax − 2lmax > (apx − 1)D, and hence the theorem is proved.

 

Chen et al. [141] asked several questions on this research work, which may be worth studying in the future. First, note that the approximation performance ratio of the greedy algorithm 2 depends on D and lmax . Actually, it approaches asymptotically to 2 + β−2 , where β = D/ lmax . This means that the ratio is fairly good when D >> 2lmax . However, if D is close to 2lmax , then the ratio is poor. Is there an approximation with performance ratio not involving D and lmax ? This is still open. Secondly, note that this greedy solution has a separate property, i.e., sensors served by each mule are consecutive and hence form an interval; for different mules, their intervals are nonoverlapping. It is easy to show that the greedy algorithm produces an optimal one among feasible solutions with separate property. Actually, for any feasible solution, there is a mule serving the n-th sensor. If this mule does not use its full capacity, then let it make more service so that it serves as many as the greedy algorithm assigns it. By induction argument, we would find the greedy algorithm solution uses mules of the number less than or equal to that in any feasible solution with separable property. An interesting question is: how large does the gap exist between solutions with separate property and allowing cooperation? Thirdly, in Problem 16.2.1, the depot is located at one end. A different location of the depot would give a different problem. For example, if the depot is located at an internal node, then the problem can be transformed into two subproblems consisting of the left side path and the right side path, so that the approximation performance ration gets an extra factor of 2. Similar idea can be applied to the case that the path is replaced by a cycle, which has an application in a fence patrol. What would happen if more depots are involved? This case would be required more ideas.

16.3 Sleep/Wakeup Scheduling We have studied sleep/wakeup scheduling of sensors several times for lifetime maximization in various models on coverage. However, this time, we would not study the lifetime maximization and instead, we are interested in minimization of the number of sensors with lifetime constraint.

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Problem 16.3.1 (Sleep/Wakeup Scheduling) Given a set of sensors with unit lifetime, a target lifetime T , and a set of target points, the problem is to find a sleep/wakeup schedule for a minimum subset of sensors to cover all target points during a time period T . This problem is NP-hard because a special case that T ≤ 1 is actually the minimum sensor cover problem, which is NP-hard. However, its approximation solution has not been studied well. A suggested randomized algorithm is as follows. Algorithm for Problem 16.3.1 input: a set of n sensors, a time span T , and a set of target points. output: a subset of sensors with a sleep/wakeup schedule. 1 a ← 0 and b ← n. 2: while b − a ≥ 2 do 3: k ← (a + b)/2 . 4: Randomly select k sensors and find a sleep/wakeup schedule to maximize lifetime of coverage. 5: if the lifetime of coverage ≥ T then b ← k else a ← k. 6: end-while 7: Output selected sensors with their sleep/wakeup schedule. The random selection may repeat several times to reach certain accuracy of the outcome. Since the lifetime maximization problem is NP-hard, we have to choose certain approximation algorithm as subroutine in this algorithm. Therefore, the performance of this algorithm also depends on the choice of such an approximation. Theoretical analysis of this algorithm has not been done.

16.4 Hybrid Systems A small sensor may not have ability to harvest enough energy during the daytime in order to continue its work in nighttime. Therefore, a hybrid system gets consideration. This system consists of two types of sensors, big sensors and small sensors. The big sensors response to harvest energy from sunlight during daytime and the small sensors are charged by the big sensors. Of course, big sensors are more expensive and we would like to minimize number of them. Since each big sensor has certain capacity to harvest energy and certain capability to charge small sensors, i.e., only charge those small sensors within certain distance. Therefore, we meet the following problem. Problem 16.4.1 (Capacity-Constrained Disk Cover) Given a set of target points in the plan and an integer c > 0, find a minimum number of unit disks to cover all target points under constraint that each unit disk covers at most c target points.

16.5 Energy-Data Dual Coverage

253

This is an NP-hard problem because it is NP-hard when c = n the number of target points. For the minimum disk cover problem without capacity constraint, we can design a PTAS by using partition and shift techniques. Can we extend this design to the capacity-constrained case? Let us give an analysis as follows. To do it, first, we use a square to cover all target points. Then partition the square into small squares, called cells. Suppose the cell is a square with edge length a. Then we can use 2a 2 unit disks to cover it. When no capacity constraint exists, this is an upper bound of the number of unit disks for covering all target points inside the cell. This upper bound is independent from the number of target points inside the cell. This fact is very important to design a PTAS. However, with capacity constraint, this upper bound should be 2a 2 + n0 /c, where n0 is the number of target points lying in the cell. This means that the partition method does not work in the capacityconstrained case. Therefore, following is open. Open Problem 23 Is there a PTAS for Problem 16.4.1? Is there a polynomial-time constant-approximation for Problem 16.4.1?

16.5 Energy-Data Dual Coverage Shi et al. [467] proposed two problems on the energy-data dual coverage in the hybrid sensor system. The aim of classic coverage studied before this section is to collect data from target points or target area. Therefore, we may call it as the data coverage in order to distinct from others, such as energy coverage. The energy coverage aims at deploying power stations (big sensors) and supplies additional energy resources to battery-free nodes (small sensors). The power station scheduling and deployment have been studied by [164, 390, 419]. However, it is in [467] the first time to study the relationship between the power station deployment and the data coverage issues. To describe their problems, let us start with their network setting as follows. Consider a sensor network consisting of a set of sinks, S, a set of battery-free sensors, V = {1, 2, . . . , n}, and a set of power stations, P = {p1 , p2 , . . . , pm }. Assume that all battery-free sensors are uniformly deployed in a target region R and all sinks induce a connected graph. The deployment of power stations will be an issue to be studied. The target region R is partitioned into unit hexagons h1 , h2 , . . . , hH . Let T be the required lifetime of coverage. T is partitioned into multiple sampling periods {T1 , T2 , . . . , TK } such that |T1 | = |T2 | = · · · = |TK | and for each period Tk , the sensory data and ambient energy remain the same. While each battery-free sensor i harvests energy from the ambient environment, it can also be recharged by every power station pj if pj is within a distance rh . As indicated in [164], beyond battery-free node i cannot harvest energy from power station pj and the amount of energy battery-free node i can be obtained from pj is

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16 Energy-Harvesting Sensors

& ei (pj ) =

α (d(i,pj )+β)2

if d(i, pj ) ≤ rh

0

otherwise,

where α and β are hardware parameters. Thus, the total energy of battery-free node i obtained from power stations is p

Ei =



ei (p),

p∈P

and in time period Tk , battery-free node i can harvest energy p

Ei (Tk ) = Ei + Eiα (Tk ), where Eiα (Tk ) is the energy harvested by battery-free node i from the ambient environment. At each time period Tk , every battery-free node i can at either work or sleep status. i can work only if it has energy Bi (Tk ) ≥ Bf , where Bf is the minimum energy that keeps a battery-free being functional. Let ε be the minimum energy consumption for battery-free node i to work. Assume that each battery-free node equips a supper capacitor with capacity B to store energy. Then each battery-free node at beginning of time period Tk+1 keeps energy Bi (Tk + l) = min{B, Bi (Tk ) + Ei (Tk ) − xi (Tk )ε}, where  xi (Tk ) =

1 if i works in Tk , 0 otherwise.

Define, the monitored region of each battery-free node i is Ri = {h ∈ R | d(i, h) ≤ rs }, where rs is the sensing radius of each battery-free node. The energy coverage is E = {h(pj ) ∈ R | pj ∈ P}, where h(pj ) is the location of power station pj . The global data coverage is C(E) = {C1 , C2 , . . . , CK } where each local data coverage Ck satisfies the following conditions:

16.5 Energy-Data Dual Coverage

255

(a) For every 1 ≤ k ≤ K, Ck ⊆ V . (b) For every i ∈ Ck , Bi (Tk ) ≥ Bf . (c) The subgraph induced by Ck ∪ S is connected. The local coverage quality of Ck is Q(Ck ) =

|R(Ck )| , |R|

where R(Ck ) = ∪i∈Ck Ri . The global coverage quality of C(E) is Q(C(E)) =

K 1  · Q(Ck ). |T | k=1

Shi et al. [467] proposed the following two problems: Problem 16.5.1 (Minimum Number of Power Stations) Consider the sensor network described in this section. Given a user input 0 ≤ δ ≤ 1, find an energy coverage E and a global data coverage C(E) such that the number of power stations is minimized and Q(C(E)) ≥ δ. Problem 16.5.2 (Maximum Coverage Quality) Consider the sensor network described in this section. Given m power stations, find an energy coverage E and a global data coverage C(E) such that the global coverage quality Q(C(E)) is maximized. Shi et al. [467] proved that both problems are NP-hard. They also designed approximation algorithms for them together with theoretical analysis. However, following is still open. Open Problem 24 Is there a polynomial-time O(1)-approximation for the maximum coverage quality problem?

Chapter 17

Underwater Sensors

Being able to breathe underwater would be sweet. Cameron Bright

17.1 Motivation and Overview Earth has 70% surface covered by water. There exists a lot of information unknown under water, The underwater sensor is an important tool for collecting such information and it has been an important type of wireless sensor networks. One of the differences between the terrestrial sensor network and the underwater sensor network is the dimension. In a terrestrial sensor network, sensors are often deployed on the surface of the earth and hence they form a two-dimensional (2D) network. However, in an underwater sensor network, underwater sensors may be deployed at different depths of the ocean [11]. Hence, we may need to treat it as a three-dimensional (3D) network. Although many research results in 2D wireless sensor networks can be extended in 3D ones, there still exist many of them which are hard to do such an extension. For example, it is still unknown how to extend those results in Chapter 5 about weighted sensor covers to 3D space. There are many interesting research works in 3D wireless sensor networks with possible applications in underwater sensor networks, regarding deployment designs [12, 14, 15, 25, 26, 50, 51, 623], k-coverage [32, 33, 35], etc.[10, 58, 59, 78, 181, 458, 520, 537, 542, 573, 627]. In the following sections, we choose a few to explore them.

17.2 3D Coverage The study of three-dimensional (3D) coverage was initiated at earlier stage [264] of research in wireless sensor networks. Rather than random deployment [15], Alam and Haas [12, 14] had interest in the grid-based sensor deployment, which leads to interesting problems in combinatorial geometry. © Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_17

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Consider a set of points, S, in the three-dimension Euclidean space. All points closer to a point s ∈ S than to others in S form the interior of a convex polyhedron, called the Voronoi cell of s. All Voronoi cells tessellate the whole space, called the Voronoi tessellation with respect to the set S. Given a 3D-region R, what is the minimum number of sensors to have the full coverage? If each Voronoi cell is identical, then total number of sensors for 3D full coverage can be estimated by the ratio of the volume of R and the volume of one Voronoi cell. Thus, the problem is reduced to finding a space-filling polyhedron with the highest volumetric quotient. Alam and Haas [12] gave a formal definition of the volumetric quotient and made a conjecture as follows. Definition 17.2.1 (Volumetric Quotient) Consider a polyhedron. Suppose that the maximum distance from its center to any vertex is rs and the volume of that polyhedron is V . Then the volumetric quotient of that polyhedron is 3V /4π rs3 . Open Problem 25 (Alam–Haas Conjecture) Consider any two space-filling polyhedrons P1 and P2 . Suppose that P1 has higher isoperimetric quotient than P2 where the isoperimetric quotient of a space-filling polyhedron P with volume V and surface area S is defined to be 36π V 2 /S 3 . Then P1 has higher volumetric quotient than P2 , too. This conjecture leads the problem to an old conjecture about space-filling polyhedrons. In 1887, Lord Kelvin studied the following problem [498]: “What is the optimal way to fill a three-dimensional space with cells of equal volume, so that the surface area (interface area) is minimized?” This is equivalent to the problem of finding a space-filling polyhedrons with the highest isoperimetric quotient. Among known space-filling polyhedrons, Kelven found that the 14-sided truncated octahedron (Fig. 17.1) has the highest isoperimetric quotient 0.753367. However, he cannot prove its optimality. Therefore, he made the following conjecture. Open Problem 26 (Kelven’s Conjecture) Among space-filling polyhedrons, the 14-sided truncated octahedron has the highest isoperimetric quotient. Fig. 17.1 A truncated octahedron

17.3 Underwater Barrier

259

Fig. 17.2 A centered cuboid of size l × w × h

If both Kelven’s conjecture and Alam–Haas conjecture are true, then the minimum number of sensors for full coverage would be reached by placing sensors at centers of the 14-sided truncated octahedrons which fill the space and each lies in a ball with radius rs , where rs is the sensing radius of each sensor. The work of Alam and Haas [12] was followed by a sequence of efforts [33, 145, 551, 586, 642, 651]. Finally, optimal grids for full coverage together with kconnectivity were found by Zhang et al. [623] for k ≤ 4 and by Bai et al. [51] for k + 6, 14. The following result is an example. Theorem 17.2.2 (Zhang et al. [623]) Consider a body-centered grid in which building block is a three-dimensional cuboid of size l × w × h with the center (Fig, 17.2), where l is its length, w is its width, and h is its height. Then the optimal for coverage with 1- or is reached by the following parameters: (a)  2-connectivity  For rc /rs < 4/3, l = (3rs2 − rc2 + rs 9rs2 − 2rc2 )/2, w = (3rs + 9rs2 − 2rc2 )/2, h = 2rc .   √ (b) For 4/3 ≤ rc /rs < 12/ 9 + 32 3, l = w = 4rs2 − rc2 /4, h = 2rc . √ √ √ √ (c) For 12/ l = w = h = 2rc / 3. √ 9√+ 32 3 ≤ rc /rs < 2 3/ 5,√ (d) For 2 3/ 5 ≤ rc /rs , l = w = h = 4rs / 5. Open Problem 27 What is the optimal grid for full coverage and k-connectivity for k ≥ 5 and k = 6, 14 in three-dimensional space?

17.3 Underwater Barrier Barr et al. [57, 58] described an interesting background for studying the barrier coverage with underwater sensors. Using sonar to detect submarines was an effective method. Current technologies, however, have made it possible for submarines to thwart standard (active or passive) sonar mechanisms [305]. Thus, finding alternatives to detect submarines becomes important and timely. One viable alternative is to use magnetic or acoustic sensors in close proximity of possible underwater pathways a submarine may pass through. This approach may require deploying large-scale underwater sensor networks to form strong barriers for coastline protection.

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17 Underwater Sensors

They calculated the stealth distance for random deployment according to a Poisson point process. In a three-dimensional sensor network, assume that an intruder is initially undetected and is moving randomly toward a certain direction. The stealth distance of the intruder is defined as the distance it travels before first detected by any sensor. Barr et al. [57] obtained the following result: Theorem 17.3.1 (Barr et al. [57, 58]) Consider a three-dimensional sensor network where sensors are deployed randomly according to a Poisson point process of density λ. Then the stealth distance of an intruder, X, would follow an exponential distribution with parameter λπ r 2 , that is, X ∼ exp(λπ r 2 ) and P rob(X < x) = 1 − e−λπ r , 2

where r is the sensing radius of the sensor. By Theorem 17.3.1, the expectation of the stealth distance is E[X] =

1 . λπ r 2

This formula indicates how much the density and/or the sensing radius should be decreased if we want to increase the stealth distance. Barr et al. [57] also showed a surprising result as follows. Theorem 17.3.2 (Barr et al. [57, 58]) Consider a three-dimensional cuboid of size l ×w×d (length l, width w, and depth d). Assume that sensors are distributed in this cuboid according to a Poisson point process of density λ. Then there is no (strong) barrier coverage in the direction of depth for finite sensor density λ and depth d as l and w go to infinity. Proof Project the cuboid on the plan along the direction of depth. Then sensor projections would be distributed in a rectangle of size l × w according to a Poisson point process of density λd. Based on a result (Theorem 1 in [363]), when l and w go to infinity, the fraction of area covered by sensor projections approaches to 1 − e−λπ r < 1 2

for finite λ and d. This means that a hole exists in the direction of depth.

 

This result may suggest to use the grid-based deployment. However, Barr et al. [57] showed that adding a little mobility to sensors would also be able to form moving barrier for underwater task. More results about underwater barriers can be found in [59].

Chapter 18

Crowdsensing

We are a stupid species with smart phones. Abhijit Naskar

18.1 Motivation and Overview When you drive on a highway from a city to another city, you may get slow down due to an accident ahead. This accident may cost you at least a half-hour delay. If you have a friend or family member in your car, then he/she may help you by using an app in the smart phone. This app can collect information from cars on the road and then gives you an advice how to go around the accident. This scenario is not a fiction. It is a real experience of many drivers. This app is an application of crowdsensing, which is also called the participatory sensing [288–290]. What is the crowdsensing? It is a service provided by smart phones. The idea is as follows: Nowadays, smart phones have been owned popularly by people. It has multiple functions, not only being a communication tool but also being able to monitor and to collect information from nearby environment. Thus, it is able to play the role of sensors in classic wireless sensor networks when users agree to provide requested service. Of course, for such a service, smart phone users would receive rewards. The mobile crowdsensing, as a novel sensing paradigm, has attracted great attention [203] and developed many applications in various fields, such as driving planning (as described as above), localization, healthcare, etc. [295, 307, 399, 431, 437, 497, 568, 597]. Usually, a general crowdsensing system consists of a platform residing in the cloud and a set of agents with smart phones, registered in the platform. A typical process is as follows: First, the platform sends out a task T with a time period. Each interested agent then reports his/her private information, including cost, available time period, and so on. The platform will make the decision to accept whose service. After success of service, the platform calculates the reward to agents based on certain mechanism.

© Springer Nature Switzerland AG 2020 W. Wu et al., Optimal Coverage in Wireless Sensor Networks, Springer Optimization and Its Applications 162, https://doi.org/10.1007/978-3-030-52824-9_18

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Crowdsensing

Several mechanisms have been proposed in the literature [194, 309, 389, 591, 638, 643]. Here, we have a special interest in one of them, the work of Liu et al. [375], in which they proposed a problem, time coverage. This is completely different from the coverage studied in previous chapters, in which the coverage is about space. Does our interest come from physics theory of Albert Einstein, the time and the space can be closely related? Of course, not. However, the crowdsensing has played a role in more and more classic applications of regular sensors. For example, in the application of transportation planning, it is possible to propose a road coverage problem. In fact, the crowdsensing is already getting a lot of attentions [160, 221, 425, 476, 630]. In next two sections, let us give time coverage a simple introduction.

18.2 Time Coverage Liu et al. [375] proposed two time coverage problems, the single slot coverage and the continuous coverage. Let us start with the simpler one, the single slot coverage. In the single slot coverage mechanism, the platform divides the task time period into several disjoint pieces. Each agent may choose one or more pieces as his/her interested work. The platform can also select one or more time pieces to assign each agent. Therefore, each time piece can be independently dealt with. This means that for simplicity, we may put our study on only one time piece as follows. Initially, the platform releases a task T with a time period [s, f ) and a probability threshold p, that is, the task is expected to complete successfully with probability at least p. Then, there are n agents, N = {1, . . . , n}, responded and each agent i reported his/her private information (ci , pi ), where ci is the cost of agent i to provide this service and pi is the probability for agent i to carry out this service successfully. The platform would select a subset of agents, I ⊆ N , to take the work under constraint that the probability of success should be at least p, that is, 1−

"

(1 − pi ) ≥ p

(18.1)

i∈I

 and meanwhile, to minimize the total cost i∈I ci . Set wi = − ln(1 − pi ) and w = − ln(1 − p). Then inequality (18.1) becomes 

wi ≥ w.

i∈I

Therefore, the platform needs to solve a minimization problem as follows: min

n  i=1

ci xi

18.2

Time Coverage

263

subject to

n 

wi xi ≥ w,

i=1

xi = 0 or 1. This is an NP-hard problem [162]. Actually, its decision version is the same as the decision of well-known NP-hard knapsack problem. Liu et al. [375] proposed the following greedy algorithm. SelectedAgents ([n], w, (c1 , . . . , cn ), (w1 , . . . , wn )) Input: A set of n agents [n] = {1, . . . , n}, threshold weight w, a profile of costs (c1 , . . . , cn ), a profile of weights (w1 , . . . , wn ). Output: A set of selected agents I 1: Sort n agents in ordering c1 /w1 ≤ c2 /w2 ≤ · · · ≤ cn /wn ; 2: I ← ∅; Icur ← ∅; 3: Copt ← ∞; Ccur ← 0; 4: Wcur ← 0; 5: for i ← 1 to n do 6: if Wcur + wi < w then 7: Wcur ← wi + Wcur ; Ccur ← ci + Ccur ; Icur ← Icur ∪ {i}; 8: else 9: if Ccur + ci < Copt then 10: Copt ← Ccur + ci ; I ← Icur ∪ {i}; 11: end if 12: end if 13: end for 14: return I They claim that they proved that this algorithm produces a 2-approximation solution for above 0–1 linear programming. Actually, the claim is incorrect. Let us give a counterexample as follows. Choose w = 1, w1 = 1 − ε, w2 = k, w3 = 1, c1 = 1 − 2ε, c2 = k − ε, c3 = 1, where k ≥ 2 and ε > 0 is a very small number. Clearly, c1 /w1 < c2 /w2 < c3 /w3 . The optimal solution is I ∗ = {3} with objective function value 1. However, the above algorithm would generate I = {1, 2} with objective function value (k + 1) − 3ε. When ε goes to zero, the approximation performance ratio ((k + 1) − 3ε)/1 goes to k + 1. Thus, they leave an open problem. Open Problem 28 Does above 0–1 linear programming have a polynomial-time 2-approximation?

264

18

Crowdsensing

18.3 Reward Mechanism To make crowdsensing successful, it is important to have an attractive reward mechanism. Usually, such a reward mechanism is required to satisfy some properties. Let us assume that each agent’s cost ci is checkable, that is, it is fixed. Then the reward ri for each selected agent i is a function with respect to pi , for i ∈ [n]. For simplicity of description, let us assume I = {1, . . . , k}. Then required properties are as follows: Computational Efficiency For every i ∈ I , ri (p1 , . . . , pn ) can be computed in polynomial-time. Truthfulness (in Expectation) For every i ∈ [n], the profit of the agent i is defined as  r − ci if i ∈ I, ui = i 0 otherwise. Then for every i ∈ [n] and pi = pi , ui (p1 , . . . , pi−1 , pi , pi+1 , . . . , pk ) ≤ ui (p1 , . . . , pi−1 , pi , pi+1 , . . . , pk ). Individual Rationality For every i ∈ I , if the task is completed, then ri (p1 , . . . , pk ) > ci . To satisfy those properties, Liu et al. [375] constructed the following algorithm to compute ri for i ∈ I . For i ∈ I , since agent i is not selected, he/she cannot receive any reward, i.e., ri = 0. However, the cost ci would not occur, that is, the actual cost should be defined as  c if i is selected, cˆi = i 0 otherwise. With cˆi , we would have ui = ri − cˆi . Reward Computation Input A set of n agents [n] = {1, . . . , n}, threshold weight w, a profile of costs (c1 , . . . , cn ), a profile of weights (w1 , . . . , wn ), selected agent set I . Output The reward ri of agent i 1: I  ← SelectedAgents ([n] \ {i}, w, (cj , j ∈ [n] \ {i}), (wj , j ∈ [n] \ {i})); w 2: wˆ i ← minj ∈I  ci · cjj ; 3: pˆ i ← 1 − e−wˆ i ; 4: if the task is completed then 5: ri ← β(1 − pˆ i ) + ci 6: else 7: ri ← −β pˆ i + ci ; 8: end if 9: return ri .

18.4

Continuous Coverage

265

Here, β is pre-determined coefficient. Theorem 18.3.1 The reward ri can be computed efficiently and satisfies the truthfulness and the individual rationality. Proof Clearly, the computational efficiency and the individual rationality hold. First, note that the agent selection algorithm is monotone in pi , that is, as the pi gets larger, agent i has more possibility getting selected. If agent i is selected, then for pi ≥ pˆ i , agent i is still selected and ri does not change and hence ui does not change. If pi is lowered down and becomes below pˆ i , then agent i is risked not to be selected and hence ui becomes 0. For not selected agent i, if reported too high (not truthful) pi , then the task may not be able to complete and ui = ri − cˆi becomes negative.   Finally, let us make a remark on game model. From above discussion, we may find a little tasty of game about crowdsensing. Actually, one often put the crowdsensing into a game formulation. In game model, the profit ui is the private utility of agent i. Suppose V is a social value of the task, then V − i∈I ci is the social utility. The truthfulness condition would be equivalent to say that the reward mechanism makes ui become an equilibrium point. Actually, the crowdsensing is one-round auction, which, based on different mechanisms, can be formulated into various game models [194, 309, 389, 591, 638, 643].

18.4 Continuous Coverage In continuous coverage, initially the platform releases a sensing task T with time period (s, f ) and a probability threshold p and then each agent i who is willing to take the task will report with information (si , fi , ci , pi ). This information means that the agent i is willing to work in time period [si , ti ) (s ≤ si < ti ≤ t) with cost ci and success probability pi . Suppose the platform received reports from n agents, [n] = {1, 2, . . . , n}. Then the decision would be made based on solving the following optimization problem. min

n  i=1

subject to 1 −

ci xi "

(1 − pi )xi ≥ p for s ≤ t ≤ f

t∈[si ,fi )

xi ∈ {0, 1} for i ∈ [n]. Define w = − ln(1 − p) and for any t with s ≤ t ≤ f ,

266

18

 wit =

Crowdsensing

− ln(1 − pi ) if t ∈ [si , fi ) 0 otherwise.

About minimization problem can be simplified to the following: min

n 

ci xi

i=1

subject to

n 

wit xi ≥ w for t ∈ Tˆ ,

i=1

xi ∈ {0, 1} for i ∈ [n], where Tˆ = {si , fi | i ∈ [n]}. This is an NP-hard 0–1 linear programming. Liu et al. [375] gave an approximate solution together with a reward mechanism.

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