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# Algorithms for Sensor Coverage and Sensor Localization Problems in Wireless Sensor Networks

## Material Information

Title: Algorithms for Sensor Coverage and Sensor Localization Problems in Wireless Sensor Networks
Physical Description: 1 online resource (135 p.)
Language: english
Creator: Xu, Xiaochun
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2008

## Subjects

Subjects / Keywords: approximation, computational, sensor, target, wireless
Computer and Information Science and Engineering -- Dissertations, Academic -- UF
Genre: Computer Engineering thesis, Ph.D.
bibliography   ( marcgt )
theses   ( marcgt )
government publication (state, provincial, terriorial, dependent)   ( marcgt )
born-digital   ( sobekcm )
Electronic Thesis or Dissertation

## Notes

Abstract: Recent advances in sensor technology coupled with embedded systems and wireless networking have made it possible to deploy sensors for numerous applications including monitoring microclimates and wildlife habitats, the structural integrity of bridges and buildings, building security, location of valuable assets, traffic, and so on. Among a variety of the essential algorithmic issues that arise in the context of wireless sensor networks, we investigate (1) sensor deployment algorithms for point/region coverage problems, and (2) sensor localization algorithms for the problem of estimating the location of a source in the plane. An integer linear programming (ILP) formulation is developed to find the minimum cost deployment of sensors that provides the desired coverage of a target point set. We also propose a greedy heuristic for this problem. Our formulation permits heterogeneous multimodal sensors and is extended easily to account for nonuniform sensor detection resulting from blockage, noise, fading, and so on. A greedy algorithm for solving the proposed general ILP is developed. Additionally, $\epsilon$-approximation algorithms and a polynomial time approximation scheme are proposed for the case of grid coverage. Experiments demonstrate the superiority of our proposed algorithms over earlier algorithms for point coverage of grids by using heterogeneous sensors. We propose a factor 3 approximation algorithm and a polynomial time approximation scheme for the problem of sensor deployment that provides the desired coverage over the entire region at minimum cost. We further propose an equivalent transformation from region coverage to point coverage, and prove the cost of the optimal point coverage is within a constant factor of that of the optimal region coverage for a network of homogeneous sensors. We also study the performance of deploying sensors in a gridded (equilateral triangle, square, and regular hexagon) layout. We establish several fundamental properties of localization using distance-differences. These properties enable minimalistic realizations of localization systems. We establish conditions for the unique identification of a source in Euclidean plane, and derive minimum number of sensors needed for unique source identification within the Euclidean plane and a polygonal monitoring region. Compared to four possible intersections of two hyperbolas, we show this task leads to at most 2 intersections, which correspond to potential source estimates. We propose a computational geometry method for the source localization problem using measurements of DTOA (Difference of Time-Of-Arrival). Compared to existing solutions to this well-studied problem, our method is (a) computationally more efficient and adaptive in that its precision can be controlled as a function of the number of computational operations, making it suitable to low power devices; and (b) robust with respect to measurement and computational errors, and is not susceptible to numerical instabilities typical of existing linear algebraic or quadratic methods.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Xiaochun Xu.
Thesis: Thesis (Ph.D.)--University of Florida, 2008.

## Record Information

Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2008
System ID: UFE0022099:00001

## Material Information

Title: Algorithms for Sensor Coverage and Sensor Localization Problems in Wireless Sensor Networks
Physical Description: 1 online resource (135 p.)
Language: english
Creator: Xu, Xiaochun
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2008

## Subjects

Subjects / Keywords: approximation, computational, sensor, target, wireless
Computer and Information Science and Engineering -- Dissertations, Academic -- UF
Genre: Computer Engineering thesis, Ph.D.
bibliography   ( marcgt )
theses   ( marcgt )
government publication (state, provincial, terriorial, dependent)   ( marcgt )
born-digital   ( sobekcm )
Electronic Thesis or Dissertation

## Notes

Abstract: Recent advances in sensor technology coupled with embedded systems and wireless networking have made it possible to deploy sensors for numerous applications including monitoring microclimates and wildlife habitats, the structural integrity of bridges and buildings, building security, location of valuable assets, traffic, and so on. Among a variety of the essential algorithmic issues that arise in the context of wireless sensor networks, we investigate (1) sensor deployment algorithms for point/region coverage problems, and (2) sensor localization algorithms for the problem of estimating the location of a source in the plane. An integer linear programming (ILP) formulation is developed to find the minimum cost deployment of sensors that provides the desired coverage of a target point set. We also propose a greedy heuristic for this problem. Our formulation permits heterogeneous multimodal sensors and is extended easily to account for nonuniform sensor detection resulting from blockage, noise, fading, and so on. A greedy algorithm for solving the proposed general ILP is developed. Additionally, $\epsilon$-approximation algorithms and a polynomial time approximation scheme are proposed for the case of grid coverage. Experiments demonstrate the superiority of our proposed algorithms over earlier algorithms for point coverage of grids by using heterogeneous sensors. We propose a factor 3 approximation algorithm and a polynomial time approximation scheme for the problem of sensor deployment that provides the desired coverage over the entire region at minimum cost. We further propose an equivalent transformation from region coverage to point coverage, and prove the cost of the optimal point coverage is within a constant factor of that of the optimal region coverage for a network of homogeneous sensors. We also study the performance of deploying sensors in a gridded (equilateral triangle, square, and regular hexagon) layout. We establish several fundamental properties of localization using distance-differences. These properties enable minimalistic realizations of localization systems. We establish conditions for the unique identification of a source in Euclidean plane, and derive minimum number of sensors needed for unique source identification within the Euclidean plane and a polygonal monitoring region. Compared to four possible intersections of two hyperbolas, we show this task leads to at most 2 intersections, which correspond to potential source estimates. We propose a computational geometry method for the source localization problem using measurements of DTOA (Difference of Time-Of-Arrival). Compared to existing solutions to this well-studied problem, our method is (a) computationally more efficient and adaptive in that its precision can be controlled as a function of the number of computational operations, making it suitable to low power devices; and (b) robust with respect to measurement and computational errors, and is not susceptible to numerical instabilities typical of existing linear algebraic or quadratic methods.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Xiaochun Xu.
Thesis: Thesis (Ph.D.)--University of Florida, 2008.

## Record Information

Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2008
System ID: UFE0022099:00001

Full Text
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ALGORITHMS FOR SENSOR COVERAGE AND SENSOR LOCALIZATION
PROBLEMS IN WIRELESS SENSOR NETWORKS

By

XIAOCHUN XU

A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA

2008

2008 Xiaochun Xu

To my parents

ACKNOWLEDGMENTS

I would like to acknowledge many people for helping me during my Ph.D study. I

am greatly indebted to my advisor, Dr. Sartaj Sahni, for supervising and guiding my

research work. This dissertation would not have been possible without his help. I am

highly grateful for his insightful thoughts and constructive -tI.i: -I i,.1 throughout my

Ph.D study. He continually encouraged me to develop independent thinking and research

skills.

Special thanks go to Dr. N 'i, -. ira S. V. Rao from Oak Ridge National Laboratory.

His research inspired much of the work in this dissertation. He offered many helpful

I am very grateful for having an exceptional committee and wish to thank Dr.

Ravindra Al! ti Dr. Shigang C'!, i1 Dr. S I-,i Iy Ranka, and Dr. Ye Xia for their support

and encouragement. I would like to thank Dr. Zhen Song from Siemens Corporate

Research whose insightful thoughts and -,t.;.; -1 i. ',s have been very helpful in our

numerous discussions.

Finally, I am highly grateful to my parents, my sister, and my girl friend who

have been very instrumental in the successful completion of my Ph.D. Their constant

motivation and encouragement is something that I would alv--,v- cherish in the years to

come.

This work was supported in part by the US National Science Foundation under grant

ITR-0326155. This work was also supported by S. i,- -N. t program at Oak Ridge National

Laboratory managed by UT-Battelle, LLC for U.S. Department of Energy under Contract

No. DE-AC05-00OR22725.

page

ACKNOW LEDGMENTS ................................. 4

LIST OF TABLES ....................... ............. 7

LIST OF FIGURES .................................... 8

ABSTRACT . . . . . . . . . . 11

CHAPTER

1 INTRODUCTION ...................... .......... 13

1.1 Motivation .................................. 13
1.2 Related Work in Sensor Deployment ......... ........ .... 15
1.3 Related Work in Sensor Localization ........ ......... .... 19
1.4 Contribution ............................... 22

2 SENSOR DEPLOYMENT FOR POINT COVERAGE .............. 26

2.1 Integer Linear Programming Formulation ............... 26
2.2 Greedy Algorithm ......... .......... .......... 28
2.3 Grid Coverage .................. . . ...... 29
2.3.1 Problem Definition and Properties .................. .. 29
2.3.2 Asymptotic (1.58+e)-Approximation Algorithm . ... 34
2.3.3 Asymptotic PTAS for 1-Coverage .................. .. 40
2.3.4 Asymptotic PTAS for q-Coverage .................. .. 42
2.4 Experimental Results .................. ........... .. 45
2.4.1 Integer Linear Program . ........... .. 45
2.4.2 Comparison of Approximation Algorithms . . ..... 48
2.4.3 Comparison with Divide-and-Conquer ................ 53

3 SENSOR DEPLOYMENT FOR REGION COVERAGE . . ..... 55

3.1 Exact 3-Approximation Algorithm. ................ ..... 55
3.2 Asymptotic PTAS .................. ............ .. 58
3.3 Region Coverage via Point Coverage ............... .. .61
3.4 Coverage with a Grid of Sensors ......... . .. 66
3.4.1 q = 2 . . . . . .. . . 69
3.4.2 q = 3 ...... ............. .............. .. 71
3.4.3 q = 4 . . . .. . . ... .. .. 72
3.4.4 q = 5 ...... ............. .............. 75
3.5 Experimental Results .................. ........... .. 78

4 PROPERTIES OF DTOA LOCALIZATION ....

Preliminaries and Definitions ........
Properties of Identifying Sensor Sets .
Number of Intersections of L12 and L 13
Indistinguishable Points .. ........
ISSs for Polygonal Regions .. .......

5 COMPUTATIONAL GEOMETRY METHOD FOR TRIANGULATION USING
D T O A . . . . . . . . . ..

5.1 Euclidean and DTOA Spaces . ..
5.2 Geometric DTOA Method . . .
5.3 Correctness and Complexity of the Method
5.4 Monotonicity of Directional Derivative .
5.4.1 Top Left Region ..........
5.4.2 Inside Region . .......
5.4.3 Bottom Right Region . .
5.4.4 Top, Bottom Left, Bottom, and Top
5.5 Simulation Results . ........
5.5.1 F 0 . . . . .
5.5.2 F > 0 . . . ...

6 CONCLUSIONS AND FUTURE WORK ....

REFERENCES .....................

BIOGRAPHICAL SKETCH ..............

102
104
109
111
113
115
116
116
121
123
125

Right Regions

LIST OF TABLES

Table

2-1

2-2

2-3

2-4

2-5

2-6

2-7

2-8

2-9

3-1 dMax, maximum APN,

100 seconds .

1000 seconds

e

Sensor coverage properties . ..........

Performance comparison of ILP of [6] and ours with T

Performance comparison of ILP of [6] and ours with T

Data for 1-coverage . ..............

Data for 2-coverage . ..............

Data for 3-coverage . ....

Data for 4-coverage . ..............

Data for 5-coverage . ..............

Cost/point using [6] together with our ILP with T
1.58 + c algorithm .........

and .i-.mptotic ratio 1 < q < 5 .. .........

5-1 Data for S1

5-2 Data for S1
F0 ....

5-3 Data for S1
F=10/100

5-4 Data for S1
F=5/100

5-5 Data for S1
F=1/100

5-6 Data for S1

5-7 Data for S1

5-8 Data for S1

5-9 Data for S,

5-10 Data for S1

5-11 Data for S,

(0, 0), S2= (0, 50000), S3

(0,0), S2 (0, 50000), S3

(0, 0), S2 (0, 50000), S3

(0, 0), S2 (0, 50000), S3

(0, 0), S2 (0, 50000), S3
(0,0), S2 (0,50000), S3
(0,0), S2 = (0,50000), S3

(0,0), 2 (0, 50000), S3

(0,0), S2 (0,100000), S3
(0,0), S2= (0,100000), S3
(0,0), S2= (0,5100000), S3

(0o, o), 2 (o, l000O), s3

(0o, o), S2= (o, l000O), S3

(0,0), S2= (0,100000), S3

(0.001, 100000), and F0 .

= (0.0000000000000001,

S(0.0000000000000001,

= (0.0000000000000001,

= (0.0000000000000001,

(5000,100000), and F

(5000, 100000), and F

(5000,100000), and F

100000), and

100000), and

100000), and

100000), and

10/100 .

5/100 .

1/100

(100000,0), and F10/100 .

(100000,0), and F5/100 ..

(100000,0), and F1/100 ..

1000 seconds and our

page

. 31

. 46

. 47

. 50

. 50

. 51

. 51

. 52

. 122

. 122

. 122

. 123

. . ...

LIST OF FIGURES
Figure page

1-1 Location determination approaches. A) AOA. B) TOA. C) DTOA. . 21

2-1 Greedy algorithm to deploy sensors .................. ..... .. 29

2-2 A grid with a sensor of range 4 at its center. A) The sensor monitors all dark
locations. B) Largest covering square. C)Largest covering rectangle. ...... ..30

2-3 Asymptotic (1.58+e)-approximation algorithm for 1-coverage . .... 35

2-4 Two regular tiling strategies. A) '+' patterns. B) diamond patterns ...... ..36

3-1 Algorithm 3-approx ............... ............. .. 56

3-2 Sensor at s E S can cover points in at most 3 hexagons . . ..... 57

3-3 Location of sites that can cover points in H ................ 58

3-4 Algorithm AS .................. .................. .. 59

3-5 G(d) for a rectangular region R. The 16 shaded grid points are within a disk of
radius r centered at the location s. .................. .... 62

3-6 The sensor s is located inside the square formed by four grid points Pi, i 1, 2, 3, 4,
and covers MAX(r) grid points, shown in dark color. ............ ..64

3-7 Gridded layout using a geometry size d. A) Equilateral triangle. B) Square. C)
Regular hexagon. .................. ... ......... 66

3-8 Equilateral triangle, q=2 ............... ...... 69

3-9 Square, q=2 .................. ................... .. 70

3-10 Regular hexagon, q=2 ............... ... ........ 71

3-11 Square, q 3 . . . . .. . . . .. .. 72

3-12 Regular hexagon, q=3 ............... ... ........ 72

3-13 Equilateral triangle, q=4 ............... ... ....... 73

3-14 Square, q= 4 . . . . . . . . ..... 74

3-15 Regular hexagon, q=4 ............... ... ........ 75

3-16 Equilateral triangle, q=5 ............... ... ....... 76

3-17 Square, q 5 .. .. .. .. ... .. .. .. .. .. .. .. .. ... .. .. .. 76

3-18 Regular hexagon, q=5 ............... ... ........ 77

3-19 Total sensor cost required v.s. various 6s, where q=l. ... . 80

3-20 Total sensor cost required v.s. various 6s, where q 2. . . 81

3-21 Total sensor cost required v.s. various 6s, where q=3. ............. .82

4-1 Examples of the locus L12 .................. ........... .. 85

4-2 Three non-collinear sensors S1, S2, and S3 form a triangle and two hyperbolas
L12(612) and L13(613) intersect each other at P1 and P2. . . ..... 86

4-3 A hyperbola L that passes through Si (1 < < 4). .............. 88

4-4 Regions of monitoring area: (a) top left, (b) inside, (c) bottom right, (d) top,
(e) bottom left, (f) bottom, and (g) top right. ................. 90

4-5 A hyperbola L = L' U L with focus S and iiiiii i. wr axis y-axis. . 91

4-6 Case 1: S2S3 lies below L12 .................. .......... .. 92

4-7 Case 2: S2S3 intersects L2 .................. ......... ..94

4-8 Case 3: S2S3 intersects L'2 and ZS3SIS2 > 90. ................ ..94

4-9 Case 4: S2S3 intersects L'2 and ZS3S1S2 < 90. ................ 95

4-10 Collinear sensors .................. ................ .. 97

4-11 Sensors Si, S2, and S3 on the boundary of a convex polygon. . .... 99

4-12 A concave polygon, its bounding convex polygon, and three sensors SI, S2, and
S3 placed on the common boundary of the concave and convex polygons . 100

4-13 S1 lies inside a simple polygon while S2 and S3 are on the boundary. Pi in the
top region is a dual point of P2 which lies in the top left region. . ... 101

5-1 Canonical placement of 3 sensors and partitioning of monitoring region ..... 105

5-2 Canonical placement of 3 sensors and L12(612) . . . ........ 106

5-3 P = (x, y) is located in the top left region. ............. .. 113

5-4 P = (x, y) is located inside the triangle. ................ ..... 114

5-5 P = (x, y) is located in the bottom right region. ................ 114

5-6 P = (x, y) is located in the top region. .................. .... 114

5-7 P = (x, y) is located in the bottom left region. ................. 115

5-8 P = (x, y) is located in the bottom region. ............. .. 116

5-9 P = (x, y) is located in the top right region. .................. 117

5-10 Source S = (x, y) is randomly selected, and the sign of the directional derivative
is computed .................. .................. .. 118

5-11 The degenerate case when L12(612) is a vertical ray ............... .118

Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy

ALGORITHMS FOR SENSOR COVERAGE AND SENSOR LOCALIZATION
PROBLEMS IN WIRELESS SENSOR NETWORKS

By

Xiaochun Xu

August 2008

('C! r: Sartaj Sahni
Major: Computer Engineering

Recent advances in sensor technology coupled with embedded systems and wireless

networking have made it possible to deploy sensors for numerous applications including

monitoring microclimates and wildlife habitats, the structural integrity of bridges and

buildings, building security, location of valuable assets, traffic, and so on. Among av ii. I

of the essential algorithmic issues that arise in the context of wireless sensor networks, we

investigate (1) sensor deployment algorithms for point/region coverage problems, and (2)

sensor localization algorithms for the problem of estimating the location of a source in the

plane.

An integer linear programming (ILP) formulation is developed to find the minimum

cost deployment of sensors that provides the desired coverage of a target point set. We

also propose a greedy heuristic for this problem. Our formulation permits heterogeneous

multimodal sensors and is extended easily to account for nonuniform sensor detection

resulting from blockage, noise, f ,lii- and so on. A greedy algorithm for solving the

proposed general ILP is developed. Additionally, c-approximation algorithms and a

polynomial time approximation scheme are proposed for the case of grid coverage.

Experiments demonstrate the superiority of our proposed algorithms over earlier

algorithms for point coverage of grids by using heterogeneous sensors.

We propose a factor 3 approximation algorithm and a polynomial time approximation

scheme for the problem of sensor deployment that provides the desired coverage over the

entire region at minimum cost. We further propose an equivalent transformation from

region coverage to point coverage, and prove the cost of the optimal point coverage

is within a constant factor of that of the optimal region coverage for a network of

homogeneous sensors. We also study the performance of deploying sensors in a gridded

(equilateral triangle, square, and regular hexagon) layout.

We establish several fundamental properties of localization using distance-differences.

These properties enable minimalistic realizations of localization systems. We establish

conditions for the unique identification of a source in Euclidean plane, and derive

minimum number of sensors needed for unique source identification within the Euclidean

plane and a polygonal monitoring region. Compared to four possible intersections of

two hyperbolas, we show this task leads to at most 2 intersections, which correspond to

potential source estimates.

We propose a computational geometry method for the source localization problem

using measurements of DTOA (Difference of Time-Of-Arrival). Compared to existing

solutions to this well-studied problem, our method is (a) computationally more efficient

and adaptive in that its precision can be controlled as a function of the number of

computational operations, making it suitable to low power devices; and (b) robust with

respect to measurement and computational errors, and is not susceptible to numerical

instabilities typical of existing linear algebraic or quadratic methods.

CHAPTER 1
INTRODUCTION

1.1 Motivation

In recent years, the advancement in sensor technology, coupled with embedded

systems and wireless networking, has made it possible to deploy wireless sensors for

numerous applications related to national security, battlefield surveillance, home appliance,

health care, inventory tracking, and environmental monitoring [23, 24, 60, 73], to name

just a few. Wireless sensor networks extend people's capability by establishing an instant

and remote interaction with the physical world. A typical wireless sensor network [11,

55, 65] may comprise thousands of spatially distributed yet cooperative sensor nodes

that continuously monitor a set of prespecified physical or environmental conditions in

the domain of interest [48]. Each sensor node has a sensing capability as well as limited

energy supply, compute power, memory, and communication ability. Nowa,-,-,-- a sensor

node (e.g., mote) usually consists of three parts, that is, a wireless communication device,

a small automatic control unit, and a source of energy (e.g., battery). Iyengar and

Brooks [23, 24] and Culler and Hong [10] provide good overviews of the breadth of sensor

network research topics as well as of applications for sensor networks. Sensor network

algorithms are reviewed in [49].

The purpose of wireless sensor networks is to continuously monitor the domain of

interest (i.e., a region or a set of discrete targets), detect the occurring events and collect

relevant data for further processing/responding. If not enough sensors have been deploy, .1

then there is no way that the sensor network can guarantee full coverage over the domain,

which means events occurring at some locations may not be detectable. Therefore, an

appropriate strategy on sensor deployment that guarantees the domain of interest get

covered at a certain quality of service (QoS) level is a fundamental problem for almost

every sensor network application. M. Cardei and J. Wu [5] reviewed the sensor coverage

problems in the context of static wireless sensor networks. In practice, the coverage

concept may be subject to a wide range of interpretations due to a large Iv ii I of sensors

and applications [33].

In some multi-sensor applications, one is often interested in placing sensors at a

subset of preselected sites so as to minimize the sensor cost while providing a specified

degree of coverage of the domain of interest (i.e. a region or a set of targets). We call it

the Min-Cost Sensor Deployment Problem (MCSDP). A possible application considered

here is in the deployment of chemical and radioactive sensors so as to monitor a portion

of the urban area or some high-risk targets that may be approximated as points. The

cost of a sensor may be from hundreds of dollars to tens of thousands of dollars. The

sensors are mounted on trolleys and may, for example, be powered from wall outlets. The

sensors do not communicate with one another. Rather, each sensor communicates with

a base station. We assume that power and sensor communication range are not design

issues. That is, each of the feasible sensor sites has an abundant energy supply and each

sensor has a sufficiently large transmission range to reach the base station from each of the

preselected sites. Note that these assumptions are in stark contrast to the assumption of

low cost and low power that are made in much of the recent research in sensor networks.

Not only are sensor networks required to provide a certain degree of coverage over

the domain of interest, but they also need certain capabilities to report the accurate

location information of the event. This capability of localizing the event is crucial for

many practical applications. For instance, with the location information of enemy tanks in

a battlefield, it will be easy to deploy the troops very effectively and efficiently. Moreover,

location information can even help on building up some critical system functionalities,

i.e., position-based routing [26, 32, 67] and location-based information querying [16].

Although the cutting-edge GPS equipment can provide the desired location with high

accuracy, it is not practical to have all sensor nodes and objects of interest equipped with

still expensive GPS modules. In addition, the accuracy of GPS measurements degrades

in urban environments as well as inside buildings. Localization, specifically, discovering

spatial relationships among objects under constraints, has been a very important and hot

topic in the sensor network research community. Many algorithms and protocols have been

developed over the years [3, 18, 36-38, 50] Sensor localization is reviewed in [59].

The localization technique based on measurements of Difference of Time-Of-Arrival

(DTOA) has been extensively studied for at least three decades [7, 13, 34, 54, 56].

Recently, it has received a renewed attention [44, 50, 51] due to the increasing proliferation

of wireless sensor networks [30, 73] and embedded networked systems [42]. In some

applications, the wireless nodes are often required to repeat the localization computations,

but they are limited in their computational capabilities and available power. Consequently,

it has become important to study the trade-offs between the number and type of

computations and the quality of localization solution to facilitate power savings by

operations may be limited in certain sensor nodes, and its impact on the localization

solution must be well understood. These factors motivate a closer examination of the

computational aspects of DTOA triangulation methods, and such results could be of more

general interest as well.

1.2 Related Work in Sensor Deployment

In some applications it is possible to select the sites where sensors are placed while in

others (e.g., in hostile environments) we may simply scatter (e.g., air drop) a sufficiently

large number of sensors over the monitoring region with the expectation that the sensors

that survive the air drop will be able to adequately monitor the target region. When site

selection is possible, we use deterministic sensor deployment and when site selection isn't

possible, the deployment is nondeterministic. In some applications, it is desirable that the

deploy, .1 collection of sensors be able to communicate with one another, either directly

or indirectly via multihop communication. So, in addition to covering the region or set of

points to be sensed, we often require the deploy, .1 collection of sensors to form a connected

network. Instead, each sensor communicates directly with a base station that is situated

within the communication range of all sensors.

One practical goal of sensor deployment in the design of distributed sensor systems

is to achieve optimal monitoring and surveillance of a target region. The optimality of a

sensor deployment scheme is a trade-off between implementation cost and coverage quality

levels. Wu et al. [64] have presented a probabilistic sensing model that provides different

sensing capabilities in terms of coverage range and detection quality with a different

cost. They have proven this problem to be NP-complete and an approximate solution is

proposed based on a 2D genetic algorithm.

For a given placement of sensors, it is easy to check whether the collection covers the

target region or point set and also whether the collection is connected. For the coverage

property, we need to know the sensing range of individual sensors. (We assume that a

sensor can sense events that occur within a distance r,, where r8 is the sensors sensing

range.) For the connected property, we need to know the communication range, re, of

a sensor. Zhang and Lou [71] have established the following necessary and sufficient

condition for coverage to imply connectivity.

Theorem 1 (Zhang and Lou [71]). When the sensor 1, ,:;.1:/ (i.e., number of sensors

per unit area) is finite, r, > 2r, is a necessary and sufficient condition for coverage to

:I ', co I.' t. .:i

Xing et al. [66] prove a similar result for the case of q-coverage (each point is covered

by at least q sensors) and q-connectivity (the communication graph for the deploy, 1

sensors is q connected).

Theorem 2 (Xing et al. [66]). When r, > 2rs, q-coverage of a convex region implies

q -, ,' :;

Notice that q-coverage with q > 1 affords some degree of fault tolerance. We are

able to monitor all points so long as no more than q-1 sensors fail. Huang and Tseng [22]

develop algorithms to verify whether a sensor deployment provides q-coverage.

Howard et al. [20, 21] consider the case when the sensors are mobile and self-deploy.

A collection of mobile sensors may be placed into an unknown and potentially hazardous

environment. Following this initial placement, the sensors relocate so as to obtain

maximum coverage of the unknown environment. They communicate the information

they gather to a base station outside of the environment being sensed. A distributed

potential-field-based algorithm to self-deploy mobile sensors under the stated assumptions

is developed in [21], and a greedy and incremental self-deployment algorithm is developed

in [20]. A virtual-force algorithm to redeploy sensors so as to maximize coverage

is developed by Zou and Chakrabarty [74]. Poduri and Sukhatme [41] developed a

distributed self-deployment algorithm that is based on artificial potential fields and that

maximizes coverage while ensuring that each sensor has at least k other sensors within its

communication range.

Kar and Banerjee [25] examine the problem of deploying the fewest number of

homogeneous sensors so as to cover the plane with a connected sensor network. They

assume that the sensing range r, equals the communication range r, (i.e., r, = r,). Kar

and Banerjee [25] have shown that their algorithm has a sensor density that is within 2.'..

of the optimal density. This algorithm may be extended to provide connected coverage for

a set of finite regions [25]. Bai et al [1] extend the result in [25] and prove its optimality

when r,/r, < V3. Commuri and Watfa [9] consider the problem of deploying the fewest

number of limited-energy homogeneous sensors to cover a 3D region.

Kar and Banerjee [25] have proposed an algorithm to deploy a connected sensor

network so as to cover a set of points in Euclidean space. This algorithm, which assumes

that r, = r,, uses at most 7.256 times the minimum number of sensors needed to cover

the given point set. Wang and Zhong [62] consider minimum-cost sensor deployment on

a sensing field that is comprised of discrete points. They relax the ILP that we developed

in [49] and solve the corresponding LP. The obtained noninteger solution to the relaxed

LP is then converted to an integer solution that provides the desired degree of coverage by

using a rounding algorithm. Although the algorithm in [62] runs in polynomial time, it is

not an c-approximation algorithm for any constant c. For the test cases reported in [62],

the developed algorithm had an approximation factor of 3.

Grid coverage is another version of the point coverage problem. In the version of

('!i i1:i iIarty et al. [6], we are given a 2D or 3D grid of points that are to be sensed.

Sensor locations are restricted to these grid points and each grid point is to be covered

by at least q, q > 1, sensors (i.e., we seek q-coverage). For sensing, we have t sensor

types available. A sensor of type i costs ci dollars and has a sensing range ri. At most one

sensor may be placed at a grid point. In this version of the point coverage problem, the

sensors do not communicate with one another and are assumed to have a communication

range large enough to reach the base station from any grid position. Thus, network

connectivity is not an issue. The objective is to find a least cost sensor deployment that

provides q-coverage.

('C!i ,11 I arty et al. [6] formulate this q-coverage deployment problem as an integer

linear program (ILP) with O(tn2) variables and O(tn2) equations, where n is the number

of grid points. For a large n, C'!i i1: hilarty et al. [6] propose a divide-and-conquer

"near-optimal" algorithm in which the base case (a small number of points) is solved

optimally using the ILP formulation. Funke et al. [15] develop a greedy constant-factor

approximation algorithm and a PTAS for 1-coverage of a grid. However, they seek to

minimize the number of deploy, 1 sensors rather than the cost of these sensors. When

optimizing the number of deploy, 1 sensors, only sensors with maximum range need

be considered. The complexity of the greedy algorithm of [15], which accounts for

obscuring obstacles, is O(n log n) and its approximation factor is log(47rR2), where R

is the maximum sensor range. The approximation factor of the PTAS in [15], which

assumes there are no obstacles but which obtains 1-coverage for any specified subset

of grid points, is 4(1 + c) and its complexity is O(n). Wu et al. [64] look at the grid

deployment problem for sensors with probabilistic detection capabilities. They show that

the problem is NP-hard and propose a genetic algorithm.

Xu et al. [68] consider the problem of deploying relay nodes in a heterogeneous sensor

network and Mhatre et al. [35] consider minimum-cost heterogeneous networks with

lifetime constraints. Relay nodes are not deploy, .1 in our model as we assume that each

sensor (or sensor assembly) is capable of single-hop communication with the base station.

Also, lifetime is not an issue for us as our sensors are not energy limited.

1.3 Related Work in Sensor Localization

The goal of sensor localization is to estimate the location of the object or event

of interest through the cooperation of a network of sensors. A variety of approaches

have been proposed for tackling this problem with respective to various assumptions

[3, 18, 32, 37, 38, 40, 50, 51].

There are two basic formulations of the localization problem: (1) estimate the

location of the object, such as origin of a plume; and (2) a device, such as a sensor

node, estimates the location itself, that is self-localization or atomic multilateration.

In a model where each sensor and object of interest is equipped with a GPS receiver,

the process of localization is quite straightforward. In a typical model, however, only a

small portion of sensor nodes, namely beacon nodes, are deterministically deploy, .1 with

location information or equipped with GPS receivers. Other sensor nodes can compute

their locations with the help of beacon nodes. Sensor nodes with known locations can

work together through some localization process to locate an object or event of interest.

The process for locating an object/event works in two steps. Its distances to some nearby

sensor nodes with known location are estimated in the first step. In step two, one sensor

node then collects all these distance estimates and computes the actual location. The

approaches employ, -1 in step two depends on the signal features used in step one, and may

be classified into three groups: (1) Angle of Arrival (AOA), (2) Time of Arrival (TOA),

and (3) Difference of Time-Of-Arrival (DTOA).

Provided that every sensor node is equipped with directional antennas, AOA

estimates, i.e., the directions of arrival of the signals, can be easily calculated by

measuring the phase difference between the directional antennas or other techniques [29].

An estimate of the target location [37], as shown in Figure 1-1-(A), is simply the

intersection of two beams originated in the sensors sl and s2, respectively. AOA methods

do not require clock synchronization between sensor nodes. However, to install directional

antennas inside each sensor node is difficult and expensive. The non-line-of-sight (NLOS)

effect is a ii' r source of error in AOA methods. The time of arrival (TOA) method

takes advantage of the estimates of the propagation time between the source and three

respective sensor nodes [40, 63]. Since the propagation time is directly proportional

to the traversed range, the distance from the source to each sensor node is simply the

multiplication of the corresponding TOA estimate and the known velocity (i.e. the speed

of light is 3 x 08 m/s). In a plane, such three distance measurements, each corresponding

to a circle, can determine a unique location. An example of how TOA methods work

is shown in Figure 1-1-(B). The DTOA based localization method uses time difference

measurements rather than absolute time measurements as TOA does. It is actually

a process solving for the mathematical intersection of multiple hyperbolas as shown

in Figure 1-1-(C). The classic localization problem using DTOA measurements has

been solved using two general approaches: (1) linear algebraic solution which typically

involves matrix inversion and solving a quadratic equation [34, 53], and (2) intersection

of hyperbolic curves [13]. The transmit time of a signal from the source is essential for

the TOA method as sensors need this information to compute their distances from the

source. A small drift on the transmit time will directly lead to an error on the location

estimate. However, this knowledge is not required by the DTOA method because it

cancels out when computing the differences between TOAs. As a result, compared to the

TOA method, the DTOA method does not require strict time synchronization between the

source and sensors. However, Both TOA and DTOA methods require time synchronization

between sensors.
sOurce L12

S2 \

S1 S
(A) AOA (B) TOA (C) DTOA

Figure 1-1. Location determination approaches. A) AOA. B) TOA. C) DTOA.

There are also some localization methods, called r inj--free methods, which do not

try to estimate the distance information based on received signal strength or other

features [18, 38, 50]. Savvides et al. [50] propose a distributed localization algorithm

that recursively calculates the locations of sensors with unknown position from sensors

whose positions are already known, using inter-sensor distance estimates. In [18], He

et al. present a range-free localization algorithm called APIT that performs location

estimation. Every three non-collinear beacon nodes form a triangular region. By putting

a small perturbation on the sensor location, one may determine with very good successful

possibility whether the sensor lies inside or outside a triangular region. By repeating this

procedure upon different triangular regions, the area where the sensor can potentially

reside can be quickly narrowed down. DV-HOP [38] lets beacon nodes flood their locations

throughout the entire network and maintains a running hop-count at each node along

the way. Sensor nodes compute their locations based on the received beacon locations,

the hop-count from the corresponding beacon node, and the average-distance per hop.

Although the range-free methods make the design of hardware greatly simplified, the

estimates by these methods may be far from accurate, and thus not suitable for some

applications (i.e., asset tracking) that require high accuracy.

In general, the quality of the location estimate is a complex function of the precision

with which the underlying numerical operations are implemented, and consequently,

there is no apparent and simple way of relating the computations to the "(I 1 111i of the

location estimate. In particular, it is unclear if devoting more computational operations

would increase the accuracy of these methods, or conversely if it is possible to reduce the

computations while slightly compromising the quality of location estimate. In addition,

sensor errors can have drastic effects on localization methods. For example, under simple

random noise conditions, the quadratic equation of [34, 53] may have imaginary roots

in which case these methods do not return an answer, that is they become incomplete.

More generally, numerical instabilities may arise in computations implemented with low

precision arithmetic operations wherein matrix inversions needed for linear algebraic

methods may become ill-conditioned resulting in large estimation errors.

1.4 Contribution

We have developed a general integer linear programming (ILP) formulation for

the minimum cost sensor deployment problem (MCSDP) for point coverage. This

formulation, when specialized to the case of a grid, uses far fewer variables and constraints

than does the formulation of [6]. The impact of this improvement in ILP formulation is a

reduction of up to 7-'- (based on our test data) in the cost of deploy, ,1 sensors. A greedy

algorithm for general sensor deployment also was developed. This greedy algorithm,

like our ILP, may be specialized to the case of a grid. Fast .i-i-ii!il '1 ic approximation

algorithms as well as PTASs for grids were developed. Although the proofs for these

algorithms assumed square grids, the proofs also apply to rectangular grids (note that

the approximation factors are .i-vmptotic and that, when the rectangle dimensions are

large enough, the approximation factor is not affected). Our experiments indicate that

the 1.58 + c-approximation algorithm is the best of our algorithms for q < 4 and that the

greedy algorithm is usually best for q > 4, where q is the desired degree of coverage.

For the MCSDP problem for region coverage, we have proposed a factor 3

approximation algorithm as well as a polynomial time approximation scheme. In addition,

we further propose an approximation algorithm which is practical on large deployment

instances by conducting an equivalent transformation from region coverage to point

coverage. The proof is given to show the cost of the optimal point coverage is within

a constant factor of that of the optimal region coverage for a network of homogeneous

sensors. We also study the performance of deploying sensors in a gridded (equilateral

triangle, square, and regular hexagon) layout. We show that the triangular grid is the

most efficient for q = 1, 3, 5, while the regular hexagon for q = 2 and the square grid for

q 4.

We study the impact of sensor deployment on the uniqueness of source estimate in

Euclidean plane as well as in a simple ]i" iv.-on. A necessary and sufficient condition is

derived for each case. We provide a tight bound on the size of a minimal identifying sensor

set in Euclidean space R2. We reinvestigate the number of intersections of two hyperbolas

having a common focus, and show it to be at most 2. Specifically, at most one intersection

lies in the union of inside region, top left region, top right region, and bottom region,

while at most one intersection lies in the union of top region, bottom left region, and

bottom right region. Each sensor deployment corresponds to an equivalence relation on

R2. For each identifying sensor set, each equivalence class is of unit cardinality. For each

non-identifying sensor set, at least one equivalence class is of greater than unit cardinality.

We have presented a computational geometric method for DTOA localization based

on a binary search on an algebraic curve defined by a distance-difference function. We

exploit the monotonicity of the directional derivative of the other distance-difference

on it to support the binary search. The computational complexity of this method is

O(log(l/7)), where the computed solution is guaranteed to be within a distance of 7

to the actual location of the source. Alternatively, by fixing the number of operations

to k, one can achieve the precision 7 0 (2-k). This method is robust with respect

to distance measurement errors: (1) 7 is of the same order of magnitude as errors in

distance measurements, such guarantees cannot be made in methods that involve division

operations; and (2) it is complete in that it will ah--bi-i return an answer, even under

random measurement errors. This method is a generalization of the DTOA localization

method in [45] proposed as a part of plume identification, where the source is inside the

acute triangle formed by sensors. In our case, the object can be located anywhere in

the monitoring region. In addition, we also provide a detailed analysis of the underlying

computation and the proof of the required monotonicity property of the underlying

directional derivative.

This dissertation is organized as follows. In C'! lpter 2, we develop an integer linear

programming (ILP) formulation as well as a greedy heuristic for minimum cost sensor

deployment. For the special case of minimum cost sensor deployment to cover a grid of

points, we develop a linear time factor 1.58 approximation algorithm as well as polynomial

time approximation schemes. In Chapter 3, we consider the the placement of sensors for

q-coverage of planar regions. We present two approximation methods with multiplicative

factors of 3 and 1+1/1, where 1 is a tunable parameter tht determines the computational

complexity. We develop a transformation from a region coverage problem to an equivalent

point coverage problem so that known point-coverage algorithms may be used to construct

good region coverage deployments. We also study the performance of deploying sensors

in a gridded layout. In Chapter 4, we establish the properties of sensor sets that uniquely

identify all sources in Euclidean plane using DTOA localization and the bound on the

number of intersections of two DTOA hyperbolas. The minimum number of sensor

needed to uniquely identify all sources in a bounded polygon is also derived. In ('! Ilpter

5, we examine the relationship between proximity in Euclidean space and proximity in

DTOA space. Our analysis shows that the proximity in DTAO space does not guarantee

proximity in Euclidean space. We develop a geometric DTOA triangulation method and

prove its correctness. This method guarantees proximity in both Euclidean and DTOA

spaces. Finally, we conclude this dissertation in C'! lpter 6 with some concluding remarks

and directions for future research.

CHAPTER 2
SENSOR DEPLOYMENT FOR POINT COVERAGE

In this chapter, we first develop an integer linear programming (ILP) formulation

for minimum cost sensor deployment problem (MCSDP) for point coverage. A greedy

heuristic is then proposed. Our ILP formulation permits heterogeneous multimodal sensors

and is easily extended to account for nonuniform sensor detection resulting from blockages,

noise, f lli:- and so on. Thereafter, a special case of minimum cost sensor deployment

to cover a grid of points (i.e., grid coverage) is considered. For this, we propose two

.-i-', i!'il ic approximation algorithms. The first ..-i-~!,illi ic approximation algorithm

is of factor 1.58+c. The second algorithm is a polynomial time approximation scheme

(PTAS). The complexities of both approximation algorithms are linear in the number of

grid points. Although the proofs of these algorithms assumed square grids, the proofs also

apply to rectangular grids (note that the approximation factors are ..-i-ii!l, -l ic and that,

when the rectangle dimensions are large enough, the approximation factor is not affected).

We conclude this chapter by providing extensive experimental results for grid deployment.

2.1 Integer Linear Programming Formulation

We are given a set of locations to be monitored as well as a set of locations where

it is feasible to place sensors. For simplicity, we model these two sets as a single set and

model any differences in the two sets by ILP (integer linear program) constraints. We

assume that at each location we wish to monitor a predefined subset of quantities such

as temperature, sound, levels of different gases, radioactivity and so on. Each quantity

to be monitored is referred to as a i,, .../I.:;,/ The subset of modalities being monitored at

different locations may be different. Let cover(j, 1) be the degree of monitoring coverage

required at location 1 for modality j. For monitoring, we have available sensors of different

types. Each sensor is able to monitor one or more modalities. Let locations(i, j, 1) be

the set of locations with the property that if a sensor of type i is placed at the location

then the sensor provides a unit degree of coverage for modality j at location 1. Notice

that with this formulation it is possible that a sensor, which is capable of monitoring

modalities a and b, placed at some location may cover another location for modality a

but not b (this may, for example, happen because the sensor's range for modality a is

different from that for modality b). Let cap., .:! i(i, 1) be the number of sensors of type i

that may feasibly be placed at location I and let cost(i) be the cost of 1 sensor of type i.

The sensor deployment problem is to determine the number, xi,,, of sensors of type i to

place at each location z so as to achieve the desired degree of coverage, not violate the

capacity constraint at each location, and minimize cost. The problem may be formulated

as the following ILP.

minimize Y (cost(i) xi,")
z I

SXi,z > cover(j, ), Vj, I
i zElocation(i,j,l)
0 < xi,z < i.,,.,. .'/ (i, 1), Vi, z

The total number of variables (i.e., the xi,,s) in the above ILP is sn, where s is the

number of sensor types and n is the number of locations. The total number of constraints

is (s + m)n, where m is the number of modalities. Additional constraints

Y xi,z < totalType(i), Vi

and

xi,z < totalLocation(z), Vz

may be added to the above formulation to limit the total number of sensors of each type

that may be deploi-y 1 as well as the total number of sensors deploy, l1 at a particular

location.

The above ILP formulation may be used to model the grid coverage problem studied

in [6]. In this problem, the locations form a Vn x V/ grid; the number m of modalities is

1; cover(, z) = q for each location z; and totalLocation(z) = 1 for each location z. The

formulation becomes

minimize (cost(i) xi,)
i z

S S i, >q, Vl
i zElocation(i,l,l)

x,,, < 1,VZ

0 < xi,z < 1 Vi, z

The grid ILP is a 0-1 ILP with tn variables and (t + 2)n constraints. In contrast, the

0-1 ILP formulation of [6] has O(tn2) variables and O(tn2) constraints.

2.2 Greedy Algorithm

A very simple and intuitive greedy algorithm for sensor deployment selects and

places one sensor in each round. Rounds are repeated until every location has the desired

degree of coverage or it isn't possible to place a sensor so as to reduce the yet-to-be-met

coverage requirement of any location. In the latter case, the greedy algorithm fails to find

a sensor deployment that provides the desired degree of coverage. The sensor selection and

deployment in each round is done by determining a sensor type and location pair (i, z)

such that

1. The selection of a sensor of type i doesn't violate any bound on the total number of

sensors of this type that may be deploy, 1

2. Placement of a sensor of type i at location z doesn't violate the bound on the

number of sensors of type i that may be placed at location z.

3. Let coverl(j, 1) be the remaining coverage degree to be provided for modality j at

location 1. Initially (i.e., when no sensor has been deploi-, 1), cover(j, 1) = cover(j,1)

for all j and 1. cover (j, 1) is reduced by 1 whenever a sensor is deploy, 1 so as cover

1 for modality j. Let coverll(j, 1) = max{0, coverf(j, 1) 1} for all j and I such that

z c locations(i,j, 1). The incremental coverage cost cost(i)/(Zj (cover/(j, 1) -

covert/(j, 1)) is minimized.

Such an (i, z) is called an optimal sensor-location pair. Figure 2-1 gives our greedy

algorithm for minimum-cost sensor deployment.

Set coverf(j,l) = cover(j,1), Vj,l
while (cover/(j, 1) > 0 for at least one pair (j, 1)){
Let (i, z) be an optimal sensor-location pair.
If there is no such (i, z) or if the incremental coverage cost
is infinity, terminate. // deployment unsuccessful
Deploy a sensor of type i at location z.
Update covert.
}
Figure 2-1. Greedy algorithm to deploy sensors

2.3 Grid Coverage

In this section, we develop approximation algorithms for the grid coverage problem

of ('!i i:.1 ilarty et al. [6]. Before presenting our two proposed .. -mptotoic approximation

algorithms, we first give the detailed definition of this problem and some interesting

properties.

2.3.1 Problem Definition and Properties

Assume the locations to be monitored form a V/ x V/ grid, the distance between

grid neighbors is 1, the number of modalities is 1, each grid location can accommodate 1

sensor, the coverage degree required at each location is q, the number of sensor types is

t, the cost of a sensor of type i is cost(i) and its range is range(i), 1 < i < t. A sensor

whose range is range(i) can monitor/cover the grid location at which it is placed as well

as all grid points within a distance of range(i) of this location. So, for example, when the

range is 1 up to 5 points may be monitored (the sensor location plus the at most 4 grid

neighbors of this location) and when the range is 2, at most 13 points may be monitored.

Figure 2-2 (A) shows all locations monitored by a sensor whose range is 4 and that is

placed at the grid center. Dark locations indicate location monitored by the sensor. The

objective is to find a minimum cost sensor deployment that provides the desired degree

of coverage. Such a sensor deployment is called an optimal deil.- '/1, -I,/ Note that some

instances of the defined grid coverage problem may not have a feasible solution. For

example, when t = 1; range(l) = 1; and q = 4. Now, locations at the corners of the

grid do not have sufficient locations within a distance of 1 to achieve the desired degree of

coverage.

0000 0000 000000000 000000000
00 0 00000 000000000 00 000 00
0 000000 0 00 0 0 0O0 000 0 0 00
0 000000 0 00 0 0 0O0 000 0 0 00
000000000 000000000 00000000
000000000 000 000000 00 0 0 00
0 000000 0 00 000 000 000 00 00
000000000 000000000 00000000
0000 0000 000000000 0000000 0
(A) (B) (C)
Figure 2-2. A grid with a sensor of range 4 at its center. A) The sensor monitors all
dark locations. B) Largest covering square. C)Largest covering rectangle.

Note also that if cost(i) < cost(j) and range(i) > range(j) an optimal deployment

will not use any sensors of type j. So, without loss of generality, we may assume that no

two sensor types have the same range and that range(i) < range(j) => cost(i) < cost(j).

Further, we may assume that cost(j) > 0, 1 < j < t.

Let D(i) be the maximum number of grid points covered by a sensor whose range is i.

Let S(i) be the maximum number of these grid points that fall within a vertically aligned

square and let R(i) be the corresponding number for the case of a vertically aligned

rectangle. Figure 2-2 (B) shows the largest square of covered points when the sensor

range is 4 and Figure 2-2 (C) shows the largest rectangle of covered points for this sensor.

Observe that D(4) = 49, S(4) = 25 and R(4) = 35. Table 2-1 gives D(i), S(i) and R(i)

for 1 < i < 12. This figure gives also the ratios s(i) = S(i)/D(i) and r(i) R(i)/D(i),

1
Lemma 1. s(i) > 0.51 for i > 1.

Proof Let the coordinates of the sensor location be (0,0). If the grid location (x, y) is

covered by the sensor, then -i < x < i and -i < y < i. For any x, -i < x < i, exactly

2L[i x2] + 1 grid locations are covered. Hence,

Table 2-1. Sensor coverage properties
i 1 2 3 4 5 6 7 8 9 10 11 12
D(i) 5 13 29 49 81 113 149 197 253 317 377 441
S(i) 1 9 25 25 49 81 81 121 169 225 225 289
R(i) 1 9 25 35 63 81 99 143 169 225 255 289
s(i) .20 .69 .86 .51 .60 .72 .54 .61 .67 .71 .60 .66
r(i) .20 .69 .86 .71 .78 .72 .66 .73 .67 .71 .68 .66

i
D(i) (2L[V x2 + 1) (2-1)
x=-i
i
< t(2Vi2 -x + 1)
x=-i
i
= 2 (V )+x2i+1
x--i

< 2 /t2 x2dx + 2i + 1

S ri2 + 2i + 1 (2-2)

The coordinates of the largest vertically aligned square of covered locations have the

form (x, +x), where x = [i/v]. So,

S(i) = (2L J + 1)2 (2-3)

> (2( 1)+ 1)2

S(vi 1)2

S2i2- 2vi2 + t (2-4)

From Equations 2-2 and 2-4, we get

s(i) S(i)/D(i)

> (2i2 2/2i + l)/(7-i2 + 2i + 1) (2-5)

Table 2-1 gives s(i) for i < 12 as obtained from Equations 2-1 and 2-3. For i > 1, the

shown values are > 0.51. For i > 12, s(i) > 0.51 follows from Equation 2-5 with i = 13

and the fact that the right side of this equation is a nondecreasing function of i. Note that

the right side of Equation 2-5 has the limiting value of 2/7 as i -0 0. .

Lemma 2. Let d be the width' and height of the /',i, -I ;., ,/. ,;ll; ilt.:ii',, 1 square A

of grid points covered by a sensor. Let w and h, ,'. "i. /,h';. l;' be these quantities for the

1,,, -It., i, i.: ,,/; aligned r. I/-.,'l. B of covered points. Either w < h and w = d and

h E d, d + 2} or w > h and w = d + 2 and h = d.

Proof We prove only that w < h = w = d and h E {d, d + 2}. The remainder of the

proof is symmetric. Assume that w < h. First, we show that h < w + 2.

Since d, w and h are all odd, the coordinates of the corners of B, under the

assumption that the sensor is placed at (0,0), are (" -1 h2 ), ("-1, h-i ), (-"-1, h-1),

and (- "- h-l). Since the grid locations at the corners of B are covered by the sensor at

(0,0), ("-)2 + (h1)2 < i2, where i is the range of the sensor. One may verify that when
S> w + 2, (+1)2 + (3)2 < So, the rectangle Q whose corners are at (+1, h-3),

(w+l h- ), (_ w1, h-), and ( h-3), includes only grid locations that are covered
by the sensor at (0, 0). The width and height of Q are w + 2 and h 2, respectively. The

number of grid locations included in Q is (w + 2)(h 2), which is greater than the number,

wh, of locations included in B when h > w + 2. This contradicts the assumption that B is

the largest rectangle of covered locations. Hence, h < w + 2. Consequently, w < h < w + 2.

Now, if w < d, then w < d 2 as both w and d are odd. So, h < d and the number

of locations covered by B is wh < (d 2) d < d2. So, B is not the largest rectangle of

covered locations. This contradicts the assumption on B. Also, if w > d, w > d + 2 as both

1 Dimensions are measured in terms of the number of included grid locations.
2 I.e., Maximum area, which is equivalent to maximizing the number of covered grid
locations.

w and d are odd. Since w < h, the (d + 2) x (d + 2) square of locations centered at (0,0)

includes only covered locations contradicting the assumption that A is the largest square

of covered locations. So, w = d. From this, w < h < w + 2, and the knowledge that h and

w are odd, it follows that h E {d, d + 2}. m

Lemma 3. r(i) > 0.6366 for i > 1.

Proof Assume that the sensor is located as in Lemma 1. Let B be as in Lemma 2 and

assume that w < h. The case w > h is symmetric. From Lemma 2 and the proof of

Lemma 1, it follows that w = d = 2 [i/v2] + 1, where i is the range of the sensor.

Since B includes only covered locations, the y-coordinate of the top right corner of B is

L[i2- Li/v J2]. Hence, h = 2[ i2- [/ 2] + 1. So,

R(i) (2[i/v/2 + 1)(2[ i2 2] + 1) (2 6)

Table 2-1 gives r(i) for i < 12 as obtained from Equations 2-1 and 2-6. For i > 1, the

shown values are > 0.6366. Using a computer program, we verified that r(i) > 0.6366 for

i < 100, 000. For i > 100, 000, we use r(i) > s(i) and the knowledge that the right side of

Equation 2-5 is more than 0.6366 for i > 100,000 to conclude that r(i) > 0.6366. 0

Let opt(n, q) be the cost of an optimal sensor deployment that achieves q-coverage for

a x/n x /n grid. Let cMin = min
Lemma 4. opt(n, q) > n q cMin

Proof Follows from the observation that the per location, per coverage cost of every

deployment must be at least cMin. m

Let tMax = *I ,, l
tMax and tMin, respectively, give the sensor types with maximum and minimum range.

Let qMax be the maximum coverage degree obtainable by any deployment of the given set

of sensor types.

Lemma 5. qMax < (D(range(tMax)) + 4range(tMax) + 3)/4. The upper bound of

(D(range(tMax))+4range(tMax)+3)/4 on qMax is achievable when /n > range(tMax).

Proof Clearly, coverage degree is maximum when a sensor of type tMax is placed

at every grid point. A sensor of type tMax placed at a grid corner is able to cover

at most (D(range(tMax)) + 4range(tMax) + 3)/4 grid points. From symmetry of

coverage, it follows that there are at most (D(range(tMax)) + 4range(tMax) + 3)/4 grid

points in a Vn x Vn grid from where a sensor of type tMax may cover a grid corner.

Hence a coverage degree larger than (D(range(tMax)) + 4range(tMax) + 3)/4 is not

possible. It is easy to see that when Vn > range(tMax), we obtain a coverage of degree

(D(range(tMax)) + 4range(tMax) + 3)/4 by placing a sensor of type tMax at every grid

location. m

2.3.2 Asymptotic (1.58+e)-Approximation Algorithm

An 'i ;,,I/l/. : : 6-approximation algorithm for an optimization problem finds, for every

instance whose size is more than some constant threshold T, feasible solutions whose value

is within a factor 6 of the value of the optimal solution. 6 is called the approximation

factor of the algorithm. Note that for an ordinary 6-approximation algorithm, T = 0.

Also, note that since instances of size < T may, typically, be solved exactly in 0(1) time

(T is a constant), an ..i-mptotic 6-approximation algorithm may be converted into a

6-approximation algorithm of the same complexity by coupling it with a step to solve

(exactly) instances of size < T in O(1) time. An- ,;imj, ../.. PTAS i..J..,,.'i.l time

approximation scheme) is a family of ..- i-,!ll i ic 6-approximation algorithms, the family

contains an algorithm for each 6 > 1, and the complexity of each algorithm in the family is

polynomial in the instance size (in determining complexity, 6 is regarded as a constant).

Figure 2-3 gives our .. -i-~iill1 ic approximation algorithm for 1-coverage. Later, we

extend this algorithm to 2- and 3-coverage.

The following lemma shows that Step 2 of Figure 2-3 is executed only when

range(dMin) = 1.

Lemma 6. If i = 1 in Figure 2-3, then dMin = rMin.

Step 1: [Initialize]
Let dMin = ,,,iin.:,, jt{cost(j)/D(, ,,,,,. (j))}.
Ties are broken by using the sensor with smaller range.
If ,.,,,. (dMin) = 1 go to Step 2.
If ,.,,,. (dMin) = 2 go to Step 3.
Let rMin= .i, .:,, j Ties are broken by using the sensor with smaller range.
Let i = ,,, ,ii (rMin).
Let w 2 [i/ ] + 1.

Let h = 2 i2 [ i/ 22 +1.
If i > 2 go to Step 4.
If i = 2 go to Step 3.

Step 2: [ ,,,,.. (dMin) = 1, tile with '+'s]
Use the method of Cockayne et al. [8] to cover the /n x /n grid with a minimal number of
"+" patterns (Figure 2-4 (A)).
Place a type rMin sensor at the center of each "+".
If a sensor is placed at a non-grid location, move it to a grid location that is part of its "+"
pattern.
Done.

Step 3: [i = 2 tile with diamonds]
Cover the /nx /n grid with a minimal number of diamond patterns (i.e., 3 x 3 squares rotated
by 45 degrees).
Place a type rMin sensor at the center of each diamond (Figure 2-4 (B)).
If a sensor is placed at a non-grid location, move it to a grid location that is part of its dia-
mond pattern.
Done.

Step 4: [i > 2, tile with w x h rectangles]
Tile the grid with [ /n/h] [/n/wl w x h rectangles.
Note that some rectangles may not fully overlap with grid locations.
Place a type rMin sensor at the center of each rectangle.
Move sensors (if any) that are placed at non-grid locations to the nearest grid location in the
rectangle.

Figure 2-3. Asymptotic (1.58+c)-approximation algorithm for 1-coverage

0
0
0
0

0
0

0
0
O
O

000
0 0

000
000
0 0 0

000

000
*0 0
OO

0
0
0

0 .
0
0
0

0

0
00
00
00
00
00
00
00
S0
00
00
00
00
0
00

(A) (B)

Figure 2-4. Two regular tiling strategies. A) '+' patterns. B) diamond patterns

Proof When i

range(rMin)

1, then w 1= R(range(rMin))

D(range(rMin)) = 5. From the definition of rMin, the assumption that no two sensor

types have the same range, and Lemma 3, we get

cost(rMin)
D(range(rMin))

So, cost(rMin)/D(range(rMin))

1 cost(rMin)
5 R(range(rMin))
1 cost(j)
< -------- 1 5 R(range(j))'
1 cost(j)
< -* ( j < < t, j rMin
5 0.6336D(range(j))'
cost(j)
< D(range) j D(range-j))

cMin. From this, the tie breaker used in

Figure 2-3, and the assumption that no two sensor types have the same range, it follows

that dMin = rMin. m

Theorem 3. The il.- rithm of Figure 2-3 has an ',;,,i,/i.1/:.: approximation factor 1.58 +

for rii C > 0.

1 and

Proof We actually show that when the sensor deployment is done using Step 2, the
.-i, !l 1 -I .1ic approximation factor is 1 + c; the factor is 1.09 + c when Step 3 is used for the

deployment and is 1.58 + c when the deployment is done using Step 4.

Case 1: Sensor deployment is done in Step 2.

For this case, range(dMin) = 1. Cockayne et al. [8] have shown that at most
n+4vn-16 +, patterns are needed to tile a V/ x V/ grid. From this and Lemma 6, it

follows that the cost of the tiling obtained by the algorithm of Figure 2-3 is less than
n+4 cost(dMin). From this, Lemma 4, and the fact that D(1) = 5, it follows that the

approximation ratio is less than

n+4~i cost(dMin) n + 4 /n 4
n cMin n V/

This ratio is < 1 + c for every c > 0 whenever n > 16/C2. Hence, the ..i-mptotic

approximation ratio, for this case, is 1 + e.

Case 2: Sensor deployment is done in Step 3.

Now, either range(dMin) = 2 or i = range(rMin) = 2. When i = range(rMin) = 2,

then D(range(rMin)) = 13 and R(range(rMin)) = 9. Also, note that when Step 3 is

executed, range(dMin) > 1. So,

cost(rMin) 9 cost(rMin)
D(range(rMin)) 13 R(range(rMin))
9 cost(dMin)
13 R(range(dMin))
9 cost(dMin)
13 0.6366D(range(dMin))
< 1.09cMin

Using a method similar to that used in [8], we can show that Step 2 deploys at most

[ + (v/ + 4) sensors of type rMin. From this and Lemma 4, it follows that the

approximation factor is at most

[r+ (V + 4) cost(rMin)
n cMin

n+21n+68 cost(rMin)
D(range(rMin)) *costr
n cMin
21 + 68
< 1.09*(1+ )
12

Since this bound is < 1.09 + c for n > max{52396, 14.24, the ..-mptotic

approximation factor is 1.09 + c. A similar proof shows that the .ii-1',i ll' ic approximation

factor is 1 + c when range(dMin)= 2.

Case 3: Sensor deployment is done in Step 4.

Now, i = range(rMin) > 2. The cost of the tiling obtained by our algorithm is

cost(rMin) [vn I
w h
From Lemma 4, it follows that the approximation ratio is at most

1 1 < ( + 1)( + 1)
n cMin w h n*en cM w h
cost(rMin) n 1 1
( + ^(-+ )+1)
n cMin wh w h
cost(rMin) n 1 1
+ ( n + (- + )+1))
n cmin R(range(rMin)) w h

From Lemma 3, range(dMin) > 1, and the choice of rMin, we get

cost(rMin)
cMin R(range(rMin))

cost(rMin) D(range(dMin))
cost(dMin) R(range(rMin))
cost(rMin) R(range(dMin))
0.6366 cost(dMin) R(range(rMin))
1
0.6366
< 1.58

So, the approximation ratio for our algorithm is less than 1.58 + cost(ri (V W +
-) + 1), which is < 1.58 + for n > max{ 2cost(rMl ), ()2cost( )( + 1))2}
h' '/) -VI I -D L e i C- e v I*Cl w h(T
Hence, the .-i-!,il'I '1ic approximation ratio, for this case, is 1.58 + c. m

For q-coverage, q E {2, 3}, we start with the 1-coverage deployment obtained by

the algorithm of Figure 2-3 and increase the number of sensors at each grid location

that has a sensor from 1 to q. While, we now have q-coverage and a 1.58 + c .,-i-,i!ill ic

approximation factor is guaranteed by Lemma 4 and Theorem 3, the deployment is

infeasible as there are q > 1 sensors deploy, 1 at certain grid locations. To remedy this

infeasibility, we relocate the excess sensors at any location. The sensor deployment of

Figure 2-3 is such that every location L that has a sensor satisfies the following properties

(since we are dealing with .i-i- !II, ill ic properties, we may assume that /n > 2):

1. There is a location either just above or just below L where no sensor is located.

2. There is a location either just to the left or just to the right of L where no sensor is
located.

The relocation of excess sensors is done by moving the first excess sensor at each

location L to the unoccupied location either above or below L; the second excess sensor

(in the case of q = 3) is moved to the unoccupied location that is either to the left or right

of L. While this relocation of sensors fixes the multiple sensors at a location problem, we

may lose q-coverage at some of the boundary locations of the grid. q-coverage is restored

by adding O(Vn) sensors of the same type as used at other locations. Since q E {2, 3} and

every boundary location of the grid has at least 2 neighbors at a distance of 1, it is ah-i--,

possible to restore q-coverage by the deployment of additional sensors on the boundary.

The deployment of these additional boundary sensors doesn't affect the 1.58+e i-mptotic

approximation factor though the value of n at which the bound applies increases slightly.

Note that if we use sensors of type sMin = argmini
tile the grid with squares with side 1/S(range(sMin)), rather than rectangles as is done

in Figure 2-3, the .,-vmptotic approximation factor is 1/0.51 + e < 1.96 + e.

2.3.3 Asymptotic PTAS for 1-Coverage

Before presenting our PTAS for 1-coverage of a grid, we establish two properties

of opt(n, 1). Let optl(n, q) be the cost of an optimal sensor deployment that q-covers a

Vn x V/ grid but in which sensors may be placed at locations outside the grid that is to

be q-covered.

Lemma 7. optf(n, 1) = opt(n, 1).

Proof We prove the lemma by showing how to transform every 1-coverage deployment

that has sensors deploy, -l at locations outside the Vn x Vn grid that is to be covered into

a 1-coverage with no sensors deploy, ,1 outside this grid. The transformation is comprised

of a finite number of iterations. In each iteration, the number of sensors deploy, -l outside

the V/ x V/ grid is decreased by 1, 1-coverage is preserved, and the total cost of the

deploy, ,1 sensors isn't increased. Since the number of such sensors is finite in an optimal

deployment, a finite number of iterations are needed.

Consider a 1-coverage of any instance of the grid coverage problem in which sensors

are permitted to be deploy, ,1 outside the Vn x /n grid. Let St be a sensor deploy, ,1 at a

location Pi outside the /n x /n grid that is to be covered. If there is no such sensor, we

are done with the transformation.

Let P be a grid location in the /n x Vn grid that is closest to S. Clearly,

P is on the boundary of this grid. Suppose there is a sensor S deploy, ,1 at P. If

range(type(S')) > range(type(S)), we remove S from P and move SI from Pi to

P. This preserves 1-coverage, reduces the number of sensors deploy, ,1 outside the

grid by 1, and reduces the total cost of the deploy, ,1 sensors by cost('ifi" (S)). If

range(type(S')) < range('i,," (S)), St covers no location in the n x n grid that is not

covered by S. So, removing St from Pi preserves 1-coverage, reduces the number of sensors

deploy, ,1 outside the grid by 1, and reduces the total cost of the deploy, ,1 sensors by

cost( / (S/)).

If there is no sensor at P, relocating SI from Pi to P preserves 1-coverage as well
as the total cost of the deployment; the number of sensors deploy .1 outside the grid is
reduced by 1. 0
Lemma 8. [V n/ Vd2 opt(d, 1) > opt(n, 1) > Ln/(Vd- + 2R)]2 *opt(d, 1), where Vn and
d are integers, n > d, and R = range(tMax).
Proof The first inequality follows from the observation that a Vn x V/ grid may

be 1-covered by overlaying this grid with [v//V d]2 Vd x Vd grids, using the optimal
1-coverage deployment for each of these Vd x Vd grids, and the strategy in the proof of
Lemma 7 to handle any sensors placed outside the Vn x /n grid.
For the second inequality, we start with a 1-coverage whose cost is opt(n, 1) (all

sensors are deploy. 1 at grid locations) and partition the /n x /n grid into (Vd + 2R) x

(Vd + 2R) subgrids. The number of subgrids is at least L[//(Vd + 2R)J2. Let C be the
minimum cost of the sensors deploy, ,1 in any one of these subgrids. It is easy to see that

the sensors deploy, '1 in each subgrid provide a (not necessarily optimal) 1-coverage of the
Vd x Vd subgrid located at the center of the (Vd + 2R) x (vd + 2R) subgrid. From
Lemma 7, it follows that C > opt(d, 1). Hence, opt(n, 1) > L[n/(vd + 2R)J2 opt(d, 1).

Theorem 4. For every fixed R = range(tMax), there is an ;,,,i/./.:.: PTAS that obtains

1-coverage for a xn x /n grid. The ,.-'impl. ri/ii of this PTAS is linear in the number of

grid points.
Proof For fixed R, a possible .*i-mptotic PTAS proceeds as follows. For any fixed e > 0,

let Vd be the least integer greater than 2R/(v + e 1). Compute the optimal sensor
deployment for a Vd x vrd grid in constant (though, perhaps, impractical), time using the
ILP formulation of Section 2.1 (note that since R and e are constants, the size of the ILP

is constant). Next, tile the Vn x /n grid with v/d x V/d subgrids and use the computed
optimal deployment within each subgrid. Finally, relocate or eliminate sensors that are
deploy, l1 off the /n x /n grid using the strategy used in the proof of Lemma 7.

We show that for Vn > 1 1 c i ), the approximation factor for the
1/+c/ (! i+2R)-1/17d '
preceding algorithm is 1 + c. From the definition of approximation factor, it follows that
the approximation factor is

[vn/vd2 2* opt(d
opt(n, 1)

Since, Vd > 2R/(V/
1/Vd > 0. From this and t

,1) n/ d]2* opt(d, 1)
) < [-/(Vd- 2 opt(d, 1) (from Lemma 8)
Lv/( vd+ 2R)]2 opt(d, 1)
< d+1 2
SV+1- (2
\vd+2R

1), 1/ d< VT+/(V + 2R) and so, V/ +/(/( + 2R)
he stated bound on Vn, we obtain

( V > + V+C
V+ 2R fd

and so,
/+ 1< +( 1)
vd ~vd+ 2R
From this and Equation 2-7, it follows that the approximation factor is 1 + c. Hence the
stated algorithm is a PTAS for 1-coverage of a grid. Its complexity is O(n) as a V/ x V/
grid can be tiled by a Vd x Vd subgrid and the off-grid sensors redeploi-, 1 in this much
time. U
2.3.4 Asymptotic PTAS for q-Coverage

By using a slightly more complex tiling strategy than used in Section 3.2 for
1-coverage, we obtain an .,i-, iiiill ic PTAS for q-coverage, q > 1. Before presenting this
PTAS and its proof of correctness, we establish two lemmas. Let p = D(range(tMax)) *
cost(tMax)/(q cost(tMin)).
Lemma 9. The cost of an optimal sensor 1 /* l..'. ,, ,/. in a /n x /n grid that achieves
q-coverage is no more than p optf(n, q).
Proof Note that if there is a deployment that achieves q-coverage, then q-coverage
is achieved by placing a sensor of type tMax at every grid point. Hence, the cost of

7)

an optimal deployment that achieves q-coverage is at most n cost(tMax). To prove the

lemma, we show that p*optf(n,q) > n*cost(n,q). Let dMin= argmini
Note that cMin = cost(dMin)/D(range(dMin)). From our assumption on sensor types

(i.e., range(i) < range(j) => cost(i) < cost(j)), it follows that sensors of type tMin have

the least cost. Also, from the proof of Lemma 4, it follows that optl(n, q) > n q cMin.

So,

p* opt/(n, q) > D(range(tMa) cost(tMax) (n q cMin)
D(range(tMax))*cost(tMax) ( in
cost(tMin) cMm )
= n cost(tMax) D(range(tMax))cMin
cost (tM in)
> n cost(tMax) D(range(dMin)cMin

n cost(tMax)

Lemma 10. Let R = range(tMax). [(Vn 2R)/d ]2 opt/(d, q) + 4 cost(tMax) R *

(/ R) > opt(n,q) > [ n/(Vd + 2R)J2 optl(d,q), where n and d are integers,

q < qMax, and /n > 2R + 2 Vd.

Proof Since /n > 2R + 2 /, q-coverage is possible (Lemma 5). Let A be the set of all

grid points located on the width R boundary of the grid and let B be the set of remaining

grid points. The size of A is 4R(/n- R) and B is a ( n- 2R) x ( n- 2R) grid. Partition

B into Vd x vd subgrids. This partitioning is done so that when /n- 2R is not a multiple

of Vd, the partitions of size smaller than x x d form a '+' pattern in the center of B.

The number of subgrids is at most [(/ 2R)/d]2.

Consider a sensor deployment of cost optl(d, q) that provides q-coverage for a Vd x /d

subgrid. Note that this deployment may place sensors at grid points outside the subgrid.

Replicate this deployment as needed to q-cover all of the subgrids of B. Since no sensor

has range more than R and subgrids of size less than Vd x Vd are at the grid center,

all deploy, l1 sensors are at grid locations. Note, however, that some locations may have

more than one sensor assigned. Next, place a type tMax sensor at all points in A. Finally,

examine each grid point that has more than one sensor and remove all but the one with
largest range. From Lemma 5, it follows that the remaining sensors provide q-coverage of
all grid points. The cost of this deployment is at most [(V/ 2R)/V]d2 optl(d, q) +
4 cost(tMax) R ( R). Hence, opt(n, q) < [(/ 2R)/ Vd2 optf(d, q) + 4 *
cost(tMax) R (/n R).
Next, start with a q-coverage whose cost is opt(n, q) and partition the V/ x / grid
into (Vd+ 2R) x (Vd+ 2R) subgrids. The number of subgrids is at least [L /( d+2R)]2
Let C be the minimum cost of the sensors deploi-, 1 in any one of these subgrids. It is
easy to see that the sensors deploi-, 1 in each subgrid provide a (not necessarily optimal)
q-coverage of the Vd x Vd subgrid located at the center of the (vd + 2R) x (vd + 2R)
subgrid. So, C > optl(d, q). Hence, opt(n, q) > [L /(vd + 2R) 2 optl(d, q). m
Theorem 5. For ,:1, fixed R = range(tMax), there is an '- ;,,i/,'l. .:' PTAS that obtains
q-coverage for a Vn x Vn grid, where q < qMax. The ,.'iipl. iii; of this PTAS is linear in
the number of grid points.
Proof For fixed R, a possible ..i-mptotic PTAS proceeds as follows. For any fixed C > 0,
let d be the least integer > 2R/( V(1 + e)/(l + c) 1), where 0 < ec < e. Compute
the optimal sensor deployment of cost optl(d, q) for a Vd x Vd grid in constant time using
the ILP formulation of Section 2.1 with locations corresponding to the grid points of a

(vd + 2R) x (vrd + 2R) grid; all points are available for sensor deployment and only
those corresponding to points in the center Vd x Vd subgrid are to be covered. Assume
that Vn > 2R + 2vd and deploy sensors to q-cover the Vn x Vn grid using the strategy
discussed in the first part of Lemma 10. We show that the just stated sensor deployment
algorithm is an ..-i-.. ill ic PTAS for q-coverage.
Let /n- be the minimum integer > 2R + 2Vd such that for any n > ni, we have

((n 2R)/Vd + 1)2 C1 > 4 R (/ R) p/d. The approximation factor for our
q-coverage algorithm is

[(V 2R)/ d]2 optl(d, q) + 4 cost(tMax) R (/ R)
opt(n, q)
< [(Vn- 2R)/ d]2 *opt(d, q) + 4 R (/ R) p optf(d,q)/d
[Lv /(v/d + 2R)J2 opt/(d, q)
[(V 2R)/Vd-2 + 4 R (V ?R) *p/d
Lv /(v + 2R)J2
< ((1 2R)/vd + 1)2 + 4 R (V- R) p/d
S(V/(v + 2R)- 1)2
(1 + 1)((l 2R)/Vd + 1)2, fr n > n (2 8)
< ^for n > al (2-8)
( V/(V2 + 2R) 1)2

From the definition of Vd, it follows that 1 0. From this, it
V 1+(1 ,di+2R Td
follows that whenever V > Vi= (1- 2R + W)/( i* = 1),

1+= 1 1 2R 1+7
( ) >1 +
1 + 1 V+2R d d 1 Vt l+ci

and so
Vr- 2R 1[+7 (V
S2R +1< 1)
d1 + V 1 + 2R )
So, for n > max{ni, n2}, the approximation factor bound of Equation 2-8 is at most
1 + c. Hence the stated algorithm is a PTAS for q-coverage of a grid. Its complexity is
O(n) for the same reasons as the complexity of the PTAS for 1-coverage is O(n).
2.4 Experimental Results
2.4.1 Integer Linear Program
We evaluated the impact of using our ILP formulation of Section 2.1 versus that of [6]
in conjunction with the divide-and-conquer heuristic proposed in [6] for the deployment
of sensors in a grid. Recall that the heuristic of [6] tiles a large grid using an optimal
deployment for the largest square subgrid whose ILP formulation may be solved in a given
amount of time. For this evaluation, we used the ILP solver Ipsolver 5.0 [31] developed by
the Eindhoven University of Technology. This ILP solver was run on a Dell Inspiron PC
with a 1.7 GHz processor and 512 MB memory.

Each test set may be described by a tuple of the form [q, t, ci, rl, C2, r2,..., Ct, rt], where

q is the desired coverage degree, t is the number of sensor types, ci is the cost of a sensor

of type i, and ri is the range of a sensor of type i. Table 2-2 gives the size of the largest

square subgrid whose ILP formulation is solvable in 100 seconds using the formulation

of [6] as well using our formulation of Section 2.1. Also, the total cost of the optimal

deployment divided by the number of points in the subgrid (cost/point) and the reduction

in cost/point achieved using our ILP formulation are given. Table 2-3 gives this data for

the case when Ipsolver is given 1000 seconds to solve the ILP formulation.

Table 2-2. Performance comparison of ILP of [6] and ours with T = 100 seconds
T = 100 seconds

Test case

[1, 2, 3, 3, 1.5, 1]
[2, 2, 3, 3, 1.5, 1]
[3, 2, 3, 3, 1.5, 1]
[4, 2, 3, 3, 1.5, 1]
[5, 2, 3, 3, 1.5, 1]
[1, 2, 7, 4,5, 3]
[2, 2, 7, 4,5, 3]
[3, 2, 7, 4,5, 3]
[4, 2, 7, 4,5, 3]
[5, 2, 7, 4,5, 3]
[1, 2, 6, 4, 5, 3]
[2, 2, 6, 4,5, 3]
[3, 2, 6, 4,5, 3]
[4, 2, 6, 4,5, 3]
[5, 2, 6, 4,5, 3]
[1, 2, 7, 5, 5, 4]
[2, 2, 7, 5,5, 4]
[3, 2, 7, 5,5, 4]
[4, 2, 7, 5,5, 4]
[5,2,7,5,5,4]
[1, 3, 5, 5, 3, 3, 1.5,
[2, 3, 5, 5, 3, 3, 1.5,
[3, 3, 5, 5, 3, 3, 1.5,
[4, 3, 5, 5, 3, 3, 1.5,
[5, 3, 5, 5, 3, 3, 1.5,

size
5*5
4*4
4*4
4*4
3*3
5*5
5*5
4*4
4*4
3*3
5*5
5*5
4*4
4*4
3*3
6*6
5*5
4*4
3*3
3*3
5*5
4*4
4*4
3*3
3*3

[6]
cost/point
0.12
0.375
0.563
0.75
1.67
0.2
0.48
0.938
1.25
2.78
0.2
0.44
0.938
1.25
2.78
0.194
0.4
0.938
2.22
2.78
0.12
0.375
0.563
1.33
1.67

size
25*25
14*14
14*14
14*14
13*13
15*15
14*14
13*13
14*14
13*13
15*15
14*14
14*14
14*14
13*13
18*18
16*16
16*16
16*16
15*15
18*18
17*17
16*16
15*15
15*15

Ours
cost/point
0.12
0.276
0.413
0.551
0.728
0.2
0.429
0.639
0.847
1.07
0.187
0.367
0.551
0.735
0.923
0.130
0.219
0.348
0.478
0.609
0.093
0.175
0.246
0.356
0.444

Reduction( .)

0
26.4
38.5
26.53
56.41
0
10.63
31.88
32.24
61.51
6.5
16.59
41.26
41.2
66.80
32.99
45.25
67.38
78.47
78.09
22.5
53.33
56.31
73.23
73.41

Table 2-3. Performance comparison of ILP of [6] and ours with T = 1000 seconds
T = 1000 seconds

Test case

[1, 2, 3, 3, 1.5, 1]
[2, 2, 3, 3, 1.5, 1]
[3, 2, 3, 3, 1.5, 1]
[4, 2, 3, 3, 1.5, 1]
[5, 2, 3, 3, 1.5, 1]
[1,2,7,4,5,3]
[2,2,7,4,5,3]
[3,2,7,4,5,3]
[4, 2, 7, 4,5, 3]
[5, 2, 7, 4,5, 3]
[1, 2, 6, 4, 5, 3]
[2, 2, 6, 4,5, 3]
[3, 2, 6, 4,5, 3]
[4, 2, 6, 4,5, 3]
[5, 2, 6, 4,5, 3]
[1, 2, 7, 5, 5, 4]
[2, 2, 7, 5,5, 4]
[3, 2, 7, 5,5, 4]
[4, 2, 7, 5,5, 4]
[5, 2, 7, 5,5, 4]
[1, 3, 5, 5, 3, 3, 1.5,
[2, 3, 5, 5, 3, 3, 1.5,
[3, 3, 5, 5, 3, 3, 1.5,
[4, 3, 5, 5, 3, 3, 1.5,
[5,3,5,5,3,3, 1.5,

size
5*5
5*5
4*4
4*4
4*4
6*6
5*5
5*5
4*4
4*4
6*6
5*5
5*5
4*4
4*4
7*7
5*5
5*5
4*4
4*4
6*6
4*4
4*4
4*4
3*3

[6]
cost/point
0.12
0.36
0.563
0.75
1.125
0.333
0.48
0.76
1.25
1.69
0.306
0.44
0.68
1.25
1.63
0.143
0.4
0.6
1.25
1.56
0.139
0.375
0.563
0.75
1.67

size
25*25
14*14
14*14
14*14
14*14
15*15
15*15
14*14
14*14
14*14
15*15
15*15
14*14
14*14
13*13
19*19
18*18
16*16
16*16
16*16
19*19
17*17
17*17
17*17
17*17

Ours
cost/point
0.12
0.276
0.413
0.551
0.689
0.2
0.418
0.638
0.847
1.06
0.187
0.373
0.551
0.735
0.923
0.13
0.256
0.348
0.477
0.605
0.094
0.175
0.260
0.349
0.438

Reduction( .)

0
23.33
26.64
26.53
38.76
39.94
12.92
16.05
32.24
37.28
38.89
15.23
18.97
41.2
43.37
9.09
36
42
61.84
61.22
32.37
53.33
53.82
53.47
73.77

We note that increasing the time available to Ipsolver from 100 seconds to 1000

seconds has little impact on the size of the subgrid whose ILP can be solved. For

the formulation of [6], this 10-fold increase in run time enabled the solution of grids

whose dimension is at most 1 more than what could be solved in 100 seconds. For our

formulation, a 10-fold increase in run time had about the same impact; we were able, in

three of the test cases, to solve for subgrids whose dimension is 2 more. In the remaining

tests cases, the dimension of the solvable subgrid increased by at most 1. On the other

hand, using our ILP formulation versus that of [6] had a dramatic impact on the size of

the solvable subgrid. For example, for our first test case [1,2,3,3,1.5,1] we were able to

solve the ILP of [6] only for a 5 x 5 subgrid, whereas we were able to solve our ILP for

a 25 x 25 subgrid! Although for this particular test case, there was no reduction in the

cost/point in the optimal solution for the larger subgrid, over our set of test cases the

reduction in cost/point in the optimal solution for the largest subgrid solvable using our

ILP formulation versus that of [6] ranged from 0'. to slightly more than 7-'-. The impact

of this reduction on the divide-and-conquer heuristic of [6] is evident-for our test cases

tiling a large grid using the optimal solutions from the larger subgrids solvable using our

ILP formulation reduces the overall cost of the deploy, -l sensors by as much as about 7'-.'

It is interesting to note that the cost/point increased for 3 of our test cases when

the ILP of [6] was given 1000 seconds rather than 100 seconds. In one of these cases, the

increase was as high as 1','.- For our formulation, there was an increase in cost/point for 4

of the 25 test cases; however, in all four cases, the increase was less than The impact

of an increase in cost/point in the optimal deployment for a larger subgrid on the tiling of

a large grid may be reduced by finding the optimal deployment for several small subgrids

and using the subgrid with the smallest cost/point to tile large grids.

2.4.2 Comparison of Approximation Algorithms

We compared the performance of our approximation algorithms against that of the

greedy algorithm. Specifically, we experimented with the following algorithms3

1. The .,i-mptotic (1.58+e)-approximation of Section 2.3.2.

2. The .,i-mptotic (1.96+e)-approximation algorithm described at the end of Section 2.3.2.

3. Iterative versions of both of these algorithms. In the iterative version, Step 4 of the
algorithm was modified to find an optimal deployment for the largest subgrid of size
2w x 2Ah whose ILP could be solved in a specified amount of time T. Here w and h
are as determined in Step 1. Note that the cost/point for this optimal deployment

3 In addition to these algorithms, we experimented with a simulated annealing
algorithm. However, despite the large amount of computing time given to this simulated
annealing algorithm, it was unable to consistently outperform the algorithms listed below

cannot exceed that for the w x h subgrid used in the unmodified approximation
algorithm. This largest subgrid was then used to tile the input grid of points.

4. The greedy algorithm of Section 2.2.

The tiling of a large grid in Step 4 of the first four algorithms stated above was

modified from that stated in the original description of Section 2.3.2. This modification

is for the case when /n is not a multiple of the size of the tiling subgrid. For those

portions of the xn x /n grid that are not covered by whole tiles, the greedy algorithm

of Section 2.2 is used to deploy sensors to achieve the desired coverage in these portions.

This modification is referred to as tiling with greedy filling. We did not experiment with

our PTASs as we do not consider these to be practical from the standpoint of required run

time.

For our experiments, we used 16 test cases, each described by the tuple [t, cl, ri, ..., ct, rt].

For each test case, we sought q-coverage for 1 < q < 5 and the grid size was 300 x 300 (or,

n = 90, 000). The iterative versions of the algorithms were given 3600 seconds to solve the

associated ILP. Although we have not described either a 1.58 + e- or a 1.96 + e-asymptotic

approximation algorithm for the case q e {4, 5}, for our experiments, we proceeded as

for the case q e {2, 3} with the 1-coverage deployment obtained by the algorithm of

Figure 2-3 and increased the number of sensors at each grid location that has a sensor

from 1 to q. Then, we relocated the sensors so as to preserve q-coverage and assure that

no location has more than 1 sensor. Our relocation strategy succeeded in all of our 16

test cases. Tables 2-4-2-8 give the costs of the constructed sensor deployments as well as

the size of the tiling subgrids used. In the case of the 1.58 + e and 1.96 + e algorithms,

the selected sensor type also is given. The percentage reduction in cost obtained by the

1.58 + e-approximate algorithm relative to the cost of the deployment constructed by the

greedy algorithm also is given.

The iterative version of the 1.58 + e-approximation algorithm outperformed the base

1.58 + e algorithm in only 1 of the 80 tests. This occurred in 1 of the tests with q = 4 and

Table 2-4. Data for 1-coverage

Test case

[2, 3, 3, 1.5, 1]
[2, 3, 3, 1, 1]
[2, 3, 3, 1.5, 2]
[2, 7, 4, 5, 3]
[2, 7, 4, 4.5, 3]
[2, 7, 4,4, 3]
[2, 6, 4, 5, 3]
[2,7,5,5,4]
[2, 7, 5,4.5, 4]
[2,7,5,4,4]
[2, 7, 5, 5, 3]
[2,7,5,4,3]
[3, 5, 5,3,3,1.5, 1]
[3, 5, 5, 2.5, 3, 1.5,1]
[3, 5, 5,2.5, 3, 1, 1]
[3,5,5,2,3,1, 1]

q=1, n=90000 and T=3600 seconds
Tiling with greedy filling

(1.96+e)-alg.
10800(1,5*5)
10800(1,5*5)
10800(1,5*5)
18000(2,5*5)
16200(2,5*5)
14400(2,5*5)
18000(2,5*5)
13010(1,7*7)
12942(1,7*7)
12876(1,7*7)
13001(1,7*7)
12985(1,7*7)
9257.5(1,7*7)
9000(2,5*5)
9000(2,5*5)
7200(2,5*5)

Iterative alg.
10800(20*20)
10800(20*20)
10800(10*10)
18000(20*20)
16200(20*20)
14400(20*20)
15975(20*20)
11248(14*14)
10759.5(14*14)
10198(14*14)
11993(14*14)
11512(14*14)
8290(14*14)
9000(10*10)
9000(10*10)
7200(20*20)

(1.58+e)-alg.
10800(1,5*5)
10800(1,5*5)
10800(1,5*5)
18000(2,5*5)
16200(2,5*5)
14400(2,5*5)
15555(1,5*7)
10386(1,7*9)
10296(1,7*9)
10230(1,7*9)
10365(1,7*9)
10401(1,7*9)
7410(1,7*9)
7376(1,7*9)
7321(1,7*9)
7264(1,7*9)

Iterative alg.
10800(20*20)
10800(20*20)
10800(10*10)
18000(20*20)
16200(20*20)
14400(20*20)
15555(10*14)
10446(14*18)
10397(14*18)
10232(14*18)
10515(14*18)
10569(14*18)
7486.5(14*18)
7422.5(14*18)
7378(14*18)
7296(14*18)

Greedy alg.

15259.5
14719
14010
20042
19145
21005
18296
13383
12839
13278
13883
13243
9101
8663
8361.5
7768

Reduction(%)

29.22
26.63
22.91
10.19
15.38
31.44
14.98
22.39
19.81
22.96
25.34
21.46
18.58
14.86
12.44
6.49

Table 2-5. Data for 2-coverage

q=2, n 90000 and T 3600 seconds

Test case

[2, 3, 3, 1.5, 1]
[2, 3, 3,1, 1]
[2, 3, 3,1.5, 2]
[2, 7, 4, 5, 3]
[2, 7, 4,4.5, 3]
[2, 7, 4, 4, 3]
[2, 6, 4, 5, 3]
[2, 7, 5, 5,4]
[2, 7, 5,4.5, 4]
[2, 7, 5,4,4]
[2, 7, 5, 5, 3]
[2, 7, 5,4, 3]
[3, 5, 5,3,3,1.5, 1]
[3, 5, 5, 2.5, 3, 1.5,1]
[3, 5, 5,2.5, 3, 1, 1]
[3,5,5,2,3,1, 1]

(1.96+e)-alg.
21780(1,5*5)
21720(1,5*5)
21691.5(1,5*5)
36205(2,5*5)
32584.5(2,5*5)
28964(2,5*5)
36184(2,5*5)
25728(1,7*7)
25630(1,7*7)
25528(1,7*7)
25852(1,7*7)
25772(1,7*7)
18390(1,7*7)
18101.5(2,5*5)
18120(2,5*5)
14520(2,5*5)

Tiling with
Iterative alg.
25650(10*10)
25200(10*10)
24300(10*10)
39600(10*10)
36900(10*10)
34200(10*10)
37800(10*10)
22202(14*14)
21253.5(14*14)
20204(14*14)
24077(14*14)
23172(14*14)
16677.5(14*14)
20250(10*10)
19800(10*10)
17100(10*10)

greedy filling
(1.58+e)-alg.
21780(1,5*5)
21720(1,5*5)
21691.5(1,5*5)
36205(2,5*5)
32584.5(2,5*5)
28964(2,5*5)
30972(1,5*7)
20431(1,7*9)
20335.5(1,7*9)
20232(1,7*9)
20528(1,7*9)
20619(1,7*9)
14676.5(1,7*9)
14646(1,7*9)
14611.5(1,7*9)
14455(1,7*9)

Iterative alg.
25650(10*10)
25200(10*10)
24300(10*10)
39600(10*10)
36900(10*10)
34200(10*10)
34140(10*14)
22371(14*18)
21792(14*18)
20639(14*18)
22831(14*18)
22430(14*18)
16086(14*18)
15582(14*18)
15517.5(14*18)
17191(7*9)

Greedy alg.
26295
25431
25158
38115
37321.5
36148
33087
23967
23282
22352
24062
23933
16721.5
16645
16350.5
15562

Reduction(%)
17.17
14.59
13.78
5.01
12.69
19.87
6.39
14.75
12.66
9.48
14.69
13.85
12.23
12.01
10.64
7.11

Table 2-6. Data for 3-coverage

q=3, n 90000 and T 3600 seconds

Test case

[2, 3, 3, 1.5, 1]
[2, 3, 3, 1, 1]
[2, 3, 3, 1.5, 2]
[2, 7, 4, 5, 3]
[2, 7, 4, 4.5, 3]
[2, 7, 4, 4, 3]
[2, 6, 4, 5, 3]
[2, 7, 5, 5,4]
[2, 7, 5,4.5, 4]
[2, 7, 5,4,4]
[2, 7, 5, 5, 3]
[2, 7, 5,4, 3]
[3, 5, 5,3,3,1.5, 1]
[3, 5, 5, 2.5, 3, 1.5,1]
[3, 5, 5,2.5, 3, 1, 1]
[3,5,5,2,3,1, 1]

(1.96+e)-alg.
32758.5(1,5*5)
32640(1,5*5)
32583(1,5*5)
54410(2,5*5)
48969(2,5*5)
43528(2,5*5)
54368(2,5*5)
38464(1,7*7)
38343(1,7*7)
38202(1,7*7)
38677(1,7*7)
38645(1,7*7)
27595.5(1,7*7)
27204.5(2,5*5)
27239.5(2,5*5)
21839(2,5*5)

Tiling with
Iterative alg.
40500(10*10)
38700(10*10)
37800(10*10)
59400(10*10)
55350(10*10)
51300(10*10)
56700(10*10)
33260(14*14)
797.5(14*14)
30322(14*14)
35166(14*14)
33435(14*14)
24090.5(14*14)
30600(10*10)
29700(10*10)
26100(10*10)

greedy filling
(1.58+e)-alg.
32758.5(1,5*5)
32640(1,5*5)
32583(1,5*5)
54410(2,5*5)
48969(2,5*5)
43528(2,5*5)
46551(1,5*7)
30683(1,7*9)
30517(1,7*9)
30371(1,7*9)
30739(1,7*9)
30897(1,7*9)
21926.5(1,7*9)
21908.5(1,7*9)
21838(1,7*9)
21656(1,7*9)

Iterative alg.
40500(10*10)
38700(10*10)
37800(10*10)
59400(10*10)
55350(10*10)
51300(10*10)
52033(10*14)
33789(14*18)
32407(14*18)
30581(14*18)
35595(14*18)
34141(14*18)
24668.5(14*18)
23664(14*18)
27333.5(7*9)
271. :i7 'I)

Greedy alg.

36957
36222
36013.5
53099
52285
50337
46012
33271
32771.5
31015
33522
33367
23387
23273.5
22901.5
22313

Reduction(%)

11.36
9.89
9.53
-2.47
6.34
13.53
-1.17
7.78
6.88
2.08
8.30
7.40
6.24
5.87
4.64
2.94

Table 2-7. Data for 4-coverage

q=4, n 90000 and T 3600 seconds

Test case

[2, 3, 3, 1.5, 1]
[2, 3, 3,1, 1]
[2, 3, 3, 1.5, 2]
[2, 7, 4, 5, 3]
[2, 7, 4,4.5, 3]
[2, 7, 4, 4, 3]
[2, 6, 4, 5, 3]
[2, 7, 5, 5,4]
[2, 7, 5,4.5, 4]
[2, 7, 5,4,4]
[2, 7, 5, 5, 3]
[2, 7, 5,4, 3]
[3, 5, 5,3,3,1.5, 1]
[3, 5, 5, 2.5, 3, 1.5, 1]
[3, 5, 5,2.5, 3, 1, 1]
[3,5,5,2,3,1, 1]

(1.96+e)-alg.
43737(1,5*5)
43560(1,5*5)
43474.5(1,5*5)
72615(2,5*5)
65353.5(2,5*5)
58092(2,5*5)
72552(2,5*5)
51337(1,7*7)
51132.5(1,7*7)
50941(1,7*7)
51458(1,7*7)
51511(1,7*7)
36769(1,7*7)
36306(2,5*5)
36359(2,5*5)
29158(2,5*5)

Tiling with
Iterative alg.
54000(10*10)
52200(10*10)
51300(10*10)
79200(10*10)
73800(10*10)
68400(10*10)
75600(10*10)
44324(14*14)
42377(14*14)
40412(14*14)
48028(14*14)
46206(14*14)
33242.5(14*14)
40950(10*10)
39600(10*10)
35100(10*10)

greedy filling
(1.58+e)-alg.
43737(1,5*5)
43560(1,5*5)
43474.5(1,5*5)
72615(2,5*5)
65353.5(2,5*5)
58092(2,5*5)
62158(1,5*7)
41054(1,7*9)
40860.5(1,7*9)
40657(1,7*9)
41036(1,7*9)
41087(1,7*9)
29226.5(1,7*9)
29171.5(1,7*9)
29114(1,7*9)
28857(1,7*9)

Iterative alg.
54000(10*10)
52200(10*10)
51300(10*10)
79200(10*10)
73800(10*10)
68400(10*10)
69916(10*14)
45667(14*18)
52090(7*9)
40265(14*18)
47723(14*18)
46286(14*18)
33010(14*18)
32245.5(14*18)
31177(14*18)
37144(7*9)

Greedy alg.
47373
46522
46477.5
67634
67025.5
64221
58468
42367
41832
39335
42600
42442
29843
29650.5
29248.5
28661

Reduction(%)
7.68
6.37
6.46
-7.36
2.49
9.54
-6.31
3.10
2.32
-3.36
3.67
3.19
2.07
1.62
0.46
-0.68

Table 2-8. Data for 5-coverage

Test case

[2, 3, 3, 1.5, 1]
[2, 3, 3, 1, 1]
[2, 3, 3, 1.5, 2]
[2, 7, 4, 5, 3]
[2, 7, 4, 4.5, 3]
[2, 7, 4, 4, 3]
[2, 6, 4, 5, 3]
[2, 7, 5, 5,4]
[2, 7, 5,4.5, 4]
[2, 7, 5,4,4]
[2, 7, 5, 5, 3]
[2, 7, 5,4, 3]
[3, 5, 5,3,3,1.5, 1]
[3, 5, 5, 2.5, 3, 1.5,1]
[3, 5, 5,2.5, 3, 1, 1]
[3,5,5,2,3,1, 1]

q=5, n=90000 and T=3600 seconds
Tiling with greedy filling

(1.96+e)-alg.
54714(1,5*5)
54480(1,5*5)
54366(1,5*5)
90822(2,5*5)
81742.5(2,5*5)
72660(2,5*5)
90742(2,5*5)
64183(1,7*7)
63884.5(1,7*7)
63648(1,7*7)
64361(1,7*7)
64318(1,7*7)
45910(1,7*7)
45409(2,5*5)
-. 17-(2,5*5)
36476(2,5*5)

Iterative alg.
67500(10*10)
64800(10*10)
63450(10*10)
100800(10*10)
94500(10*10)
88200(10*10)
95400(10*10)
55357(14*14)
53037.5(14*14)
50576(14*14)
60790(14*14)
59025(14*14)
42413(14*14)
51300(10*10)
49950(10*10)
44100(10*10)

(1.58+e)-alg.
54714(1,5*5)
54480(1,5*5)
54366(1,5*5)
90822(2,5*5)
81742.5(2,5*5)
72660(2,5*5)
77668(1,5*7)
51164(1,7*9)
50914.5(1,7*9)
50659(1,7*9)
51446(1,7*9)
51284(1,7*9)
36637(1,7*9)
36482(1,7*9)
36390(1,7*9)
36108(1,7*9)

Iterative alg.
67500(10*10)
64800(10*10)
63450(10*10)
100800(10*10)
94500(10*10)
88200(10*10)
88493(10*14)
56956(14*18)
68565.5(7*9)
64156(7*9)
59701(14*18)
58160(14*18)
41407(14*18)
51576(7*9)
39268.5(14*18)
37340(14*18)

Greedy alg.
57660
56843
56671.5
82374
81442.5
77972
70992
51293
50750.5

51536
51350
36205
35972.5
35525.5
34975

Reduction(%)
5.11
4.16
4.07
-10.26
-0.37
6.81
-9.40
0.25
-0.32
-6.14
0.17
0.13
-1.19
-1.42
-2.43
-3.24

the cost reduction was about 1 .

The iterative version of the 1.96 + e algorithm, however,

outperformed the base version in 31 of the 80 test cases and cost reductions as high as

211'- were observed. In 6 of the 80 tests, the iterative version of the 1.96 + e algorithm

outperformed the base 1.58 + e-algorithm; but, never by more than 1 On the remaining

74 tests, the base 1.58 + e algorithm did as well as or better than the iterative version of

the 1.96 + e algorithm; the cost reductions were as high as 3:l' So, the 1.58 + e algorithm

is generally superior to both the iterative versions from the standpoint of running time

(the time required by each of the base approximation algorithms is less than 1 second per

test case) and of the cost of the deploy, 1 sensors.

Although our greedy algorithm took less than a minute to solve each of our test cases,

it generally produced sensor deployments whose cost exceeded that of the deployment

obtained by our 1.58 + e algorithm. The greedy algorithm produced better solutions on

0, 0, 2, 3, and 9, respectively, of the 16 tests sets for each of q = 1, 2, 3, 4, 5; the 1.58 + e

algorithm did better on the remaining 66 test sets. The cost reduction in the 1.58 + e

solutions relative to the greedy solutions ranged from -1(C t' to 29.2"' We expect that

for larger q values, our greedy algorithm will outperform the remaining 4 algorithms.

2.4.3 Comparison with Divide-and-Conquer

C'!, 11:1 iarty et al. [6] propose tiling an Vn x Vn grid using an optimal solution

for a Vd x V/d subgrid. However, they are not very specific about how to select d. Two

reasonable choices for d are (1) choose the largest d for which the ILP is solvable with the

amount of computational resource we wish to expend and (2) choose d as in (1) but with

the added restriction that Vd be a divisor of Vn. If d is selected as in (1), we need to way

to handle the portions of the Vn x Vn grid that are not covered by whole subgrids.

We compare the performance of the divide-and-conquer method of [6] with that of

our 1.58 + c-approximation algorithm. For this comparison, we use the cost/point of the

optimal solution for the largest ILP solvable using our formulation and the 25 data sets of

Table 2-3. The cost/point for the 1.58 + c-approximation algorithm is obtained by dividing

the total cost of the sensors deploy, 1 on a 300 x 300 grid by the number of points (90,000)

in the grid. This comparison is biased in favor of the divide-and-conquer algorithm as this

comparison assumes that the size of the grid to be covered is a multiple of the size of the

subgrid solved using the ILP. However, for the 1.58 + c-approximation algorithm, we use

the greedy algorithm to cover portions of the 300 x 300 grid not covered by whole subgrids

(i.e., greedy filling). Table 2-9 gives the cost/point for each algorithm and the percent

reduction in cost/point achieved by the 1.58 + c algorithm relative to that achieved by the

divide-and-conquer algorithm of [6] using our ILP formulation; the time allowed to the ILP

solver was 1000 seconds. Both algorithms produced deployments with the same cost/point

on 2 of the 25 data sets. On the remaining 23 data sets, our 1.58 + c-algorithm had a

smaller cost/point. The average reduction in cost/point obtained by our algorithm was

7' the maximum reduction was 12.5'. and the minimum reduction 0' When the time

available to the ILP solver is only 100 seconds, these percentages were 6.9' 23.1.'. and

-3.1..' (on one test set, using our ILP, the divide-and-conquer algorithm of [6] did better

than our 1.58 + c algorithm, on 2 test sets the two algorithms were tied, and our 1.58 + e

algorithm was superior on the remaining 22 data sets). As noted earlier, for the test cases

with q = 5, we expect that our greedy algorithm will provide even a greater cost reduction

than obtained by our 1.58 + c algorithm.

Table 2-9. Cost/point using [6] together with our ILP with T = 1000 seconds and our
1.58 + c algorithm

Test case
[1, 2, 3, 3, 1.5, 1]
[2, 2, 3, 3, 1.5, 1]
[3, 2, 3, 3, 1.5, 1]
[4, 2, 3, 3, 1.5, 1]
[5, 2, 3, 3, 1.5, 1]
[1,2,7,4,5,3]
[2, 2, 7, 4,5, 3]
[3, 2, 7, 4,5, 3]
[4, 2, 7, 4,5, 3]
[5, 2, 7, 4,5, 3]
[1, 2, 6, 4, 5, 3]
[2, 2, 6, 4,5, 3]
[3, 2, 6, 4,5, 3]
[4, 2, 6, 4,5, 3]
[5, 2, 6, 4,5, 3]
[1, 2, 7, 5, 5, 4]
[2, 2, 7, 5,5, 4]
[3, 2, 7, 5,5, 4]
[4, 2, 7, 5,5, 4]
[5, 2, 7, 5,5, 4]
[1, 3, 5, 5, 3, 3, 1.5,
[2, 3, 5, 5, 3, 3, 1.5,
[3, 3, 5, 5, 3, 3, 1.5,
[4, 3, 5, 5, 3, 3, 1.5,
[5, 3, 5, 5, 3, 3, 1.5,

[6] + our ILP
0.12
0.276
0.413
0.551
0.689
0.2
0.418
0.638
0.847
1.06
0.187
0.373
0.551
0.735
0.923
0.13
0.256
0.348
0.477
0.605
0.094
0.175
0.260
0.349
0.438

1.58+c
0.12
0.242
0.364
0.486
0.608
0.2
0.402
0.605
0.807
1.01
0.173
0.344
0.517
0.691
0.863
0.115
0.227
0.341
0.456
0.568
0.0823
0.163
0.244
0.325
0.407

Reduction( )
0
12.31
11.86
11.80
11.76
0
3.83
5.17
4.72
4.72
7.49
7.77
6.17
5.99
6.5
11.54
11.33
2.01
4.4
6.12
12.45
6.86
6.15
6.88
7.08

CHAPTER 3
SENSOR DEPLOYMENT FOR REGION COVERAGE

In this chapter, we consider the MCSDP problem of deploying sensors, at most 1

sensor at each location of S, so as to provide q-coverage, q > 1, for a region R at minimum

cost. Suppose that we have a sufficient supply of sensors of t different types. Let ri > 0

and ci > 0, respectively, be the sensing range and cost of a sensor of type i. Without

loss of generality, we may assume that no two sensor types have the same range or the

same cost and that ri < rj iff ci < cj. Let S be the set of sites where it is feasible to

place a sensor and let R be a region in Euclidean space that is to be monitored. For this

MCSDP for region coverage, we propose three approximation algorithms. The first is

a factor 3 approximation algorithm whose compuational complexity is linear in the area

of the region R. The second is an .,-i-~ ,iil I ic polynomial time approximation scheme

(PTAS). Its complexity is .,-i-,!i,1 ically linear in the multiplication of the area of the

region R and a system parameter 1. The third algorithm tackles the MCSDP problem

from a different perspective. It transforms the region coverage instance into an equivalent

point coverage instance and solves region coverage via point coverage. A special case

where sensors are deploy, ,1 in a gridded fashion is also considered. Finally, experiments are

conducted to evaluate the performance of our algorithms.

3.1 Exact 3-Approximation Algorithm

Let tMax = argmax
positive constant. Our 3-approximation algorithm, 3-approx (Figure 3-1), begins by tiling

R with regular hexagons whose sides have length L; some of the hexagons that overlap the

boundary of R may contain portions of the Euclidean space that are not in R. Next, for

each hexagon, Hi, of the tiling, we find an optimal (i.e., least cost) sensor deployment that

q-covers Hi n R. Finally, the optimal sensor deployments for the hexagons in the tiling

are combined by ensuring that no site in S has two or more sensors. To the algorithm of

Step 1: Tile R with regular hexagons whose side length is L = 2rMax + e.
Step 2: For each hexagon Hi of the tiling find an optimal deployment of sensors to sites
in S so as to q-cover Hi n R.
Step 3: Combine the optimal deployments found in Step 2 for all of the hexagons in the
tiling. In case a site of S has two or more sensors assigned to it, discard all but
the sensor with maximum range.

Figure 3-1. Algorithm 3-approx

Figure 3-1, we may add an optional pruning step in which redundant sensors (i.e., sensors

whose omission doesn't affect the q-coverage property) are eliminated.

It is easy to see that 3-approx constructs a q-cover for R with at most one sensor

per site in S provided such a q-cover exists. In the following, we establish that the cost of

the constructed q-cover is at most 3 times that of an optimal q-cover and we analyze the

complexity of 3-approx.

Lemma 11. A sensor at a site s E S can cover points in at most three of the hexagons of

the tiling of Step 1 of 3-approx.

Proof We consider only the case when s is located within one of the hexagons (-v

H1) of the tiling (see Figure 3-2). The proof for the case when s is in no Hi (this may

happen when S has locations outside of R) is similar. Let d(pi,p2) be the Euclidean

distance between two points pi and p2 and let d(Hi, Hj) be the smallest distance between

two points one of which is in Hi and the other is in Hj. Since, d(pi, s) > L > rMax

for points outside of the 7 hexagons shown in Figure 3-2, the sensor at s cannot cover

points outside of the 7 hexagons. Notice that d(H, Hj) = L for i,j E {2, 4, 6} as well

as for i,j E {3,5,7}. From this observation, L = 2rMax + c > 2rMax, and the

triangle inequality, it follows that for any point pi in Hi and p2 in Hj, i,j e {2, 4, 6}

or i,j e {3, 5, 7}, d(p, s) + d(p2, s) > 2rMax. Hence, either d(p, s) > rMax or

d(p2, s) > rMax or both. So, the sensor at s can cover points in at most one of H2, H4,

and H6 and at most one of H3, H5, and H7. This sensor may cover also points in H1.

Hence, points in at most 3 of the hexagons of the tiling may be covered. m

Theorem 6. 3-approx is a 3-approximation il'.irithm.

Figure 3-2. Sensor at s E S can cover points in at most 3 hexagons

Proof Consider any instance of the q-cover problem for which there is a feasible solution

(i.e., a selection of sensors, at most one per site in S, that q-covers R). Start with an

optimal solution O for this instance. Let cost(O) be the cost of sensors deploy. 1 in O.

To each hexagon Hi in the tiling of Step 1, assign the subset of sensors of O that cover

points in Hi. Let cost(Hi) be the cost of the sensors assigned to Hi. From Lemma 11, it

follows that the assignment of sensors to hexagons, assigns each sensor of 0 to at most 3

hexagons. Hence, E cost(Hi) < 3cost(O). Since the sensors assigned to Hi q-cover Hi n R,

their cost must be at least that of the deployment computed in Step 2. So, the sensor

deployment following Step 3 has a cost that is at most 3cost(O). m

The complexity of algorithm 3-approx is governed by the time it takes to determine

the optimal q-cover for each of the hexagons in the tiling. When computing the optimal

q-cover for a hexagon H, we need consider only those sites in S that line in the shaded

region shown in Figure 3-3. The area A of this region is area(H) + perimeter(H) *

rMax + wrMax2. = 1.50L2 + 6L rMax + wrMax2. Under the assumption that the

site density (i.e., number of sites in any region of area A) is bounded (this is the case,

for instance, when there is a fixed lower bound on the distance between two sites) by

some fixed constant and the number of sensor types t is similarly bounded, the size of the

state space for each Hi is bounded by a (potentially very large) constant. The optimal

deployment for each Hi may, therefore, be found in constant time by simply searching the

state space of Hi. Under these assumptions, the complexity of 3-approx is linear in the

number of hexagons in the tiling.

rMar

I-- 2rMax+r -P-

Figure 3-3. Location of sites that can cover points in H

3.2 Asymptotic PTAS

We employ the shifting strategy of [2, 19] to arrive at an approximation scheme for

the q-cover problem. Let 1 > 1 be an integer shifting parameter. The cost of the sensor

deployments computed by our approximation scheme will be within a multiplicative factor

of 1 + 1/1 of the cost of an optimal deployment. By making 1 suitably large, we can obtain

deployments as close to optimal as desired. Figure 3-4 gives our approximation scheme

AS.

Unlike algorithm 3-approx, which considers a single tiling of the region R that is to

be q-covered, algorithm AS considers a family, T1, *- T1, of tilings. To obtain a tiling in

this family, we begin by determining the smallest bounding rectangle U of R. Next, this

bounding rectangle is tiled using tiles whose height equals that of U but whose width is

1 L, where L = 2rMax + c and c is a positive constant. The first and last tiles in the

tiling are exceptions. In Ti, the width of the first tile is i L. In case tiling with a tile of

Step 1: Let U be the smallest bounding rectangle for the region R. Let Vi be a tile whose
height equals that of U and whose width is i L, 1 < i< 1. Let Ti be the tiling of
U in which the first tile is Vi. This tile is followed by zero or more tiles of type V1.
An additional tile is used at the right end (if necessary) to cover the remainder of
U. The width of this last tile is < 1 L and is chosen so as to cover the remaining
uncovered width of U.
Step 2: Do Steps 3 and 4 for 1 < i < 1.
Step 3: For each tile T in the tiling Ti, find an optimal deployment of sensors to sites in
S so as to q-cover T n R.
Step 4: Combine the optimal deployments found in Step 3 for the tiles in Ti. In case a
site of S has two or more sensors assigned to it, discard all but the sensor with
maximum range.
Step 5: From the constructed deployments for Ti, .. Tl, select the deployment that has
least cost.

Figure 3-4. Algorithm AS

width i L followed by tiles of width I L doesn't exactly cover U, a last tile whose width

is less than I L is used in Ti.

In Steps 2 through 4, we compute an optimal deployment for each of the tilings Ti

using a strategy similar to that used for the hexagonal tiling used in 3-approx (i.e., find

an optimal deployment to q-cover the sub-region of R included in each tile of Ti and then

combine these optimal deployments to obtain a q-cover for R). Finally, in Step 5, the best

of the q-covers over all I tilings TI, T1 is selected as the deployment to use.

As in the case of algorithm 3-approx, we may add an optional pruning step in which

redundant sensors (i.e., sensors whose omission doesn't affect the q-coverage property) are

eliminated.

It is easy to see that AS finds a q-cover for every instance for which there is a q-cover.

The following theorem establishes that AS is, in fact, an approximation scheme for the

q-cover problem.

Theorem 7. Algorithm AS is an approximation scheme for the q-cover problem. Si'.. :7-

, ./l;., the computed q-cover has a cost that is within a multiplicative factor (1 + 1/1) of the

cost of an optimal q-cover.

Proof Consider any instance of the q-cover problem for which there is a feasible solution.

Let O be an optimal solution for this instance, let cost(O) be the cost of sensors deploi-y l

in 0, and let cost(Ti) be the cost of the sensor deployment computed in Step 4 of AS

for the tiling Ti. Since each tile in Ti, except possibly the first and last, has a width

I L > 2rMax, no sensor can cover points in 3 or more consecutive tiles. Let Oi be the

sensors deploi-, 1 in O that cover points in 2 tiles of Ti. Using a distribution scheme similar

to that used in the proof of Theorem 6, we obtain

cost(Ti) < cost(O) + cost(Oi)

Suppose that a sensor of O that lies in the first tile of Ti covers a point in the second

tile of Ti. Since L > 2rMax, this sensor is part of the second tile of Tj, j < i and does

not cover a point in any tile of Tj that is not part of this second tile. For j > i, this sensor

remains in the first tile for Tj and is unable to cover points that are not in the first tile of

Tj. Hence, this sensor, which is in Oi, is not in any Oj, j / i. By reasoning in a similar

fashion, we may show that all Ois are di-I. ii and so

Scost(Oj) < cost(0)
i= 1
Hence,

mm {cost(Ti)}
1 1 I
< Z cost(Ti)
i=1
1
S(cost(O) + cost(O))
i=1

cost(O) + cost(Oi)
i i
< (1 + )cost(O)
I

Under assumptions similar to those made in the analysis of 3-approx, the complexity

of AS is linear in the product of I and the number of rectangles in a tiling Ti.

3.3 Region Coverage via Point Coverage

Since known algorithms [4, 22, 61, 72] to verify that a sensor deployment actually

provides region coverage with the desired coverage degree are rather cumbersome while

similar algorithms for point coverage are rather straightforward, we are motivated to

transform a region coverage instance into an equivalent point coverage instance. We

assume that there is only 1 sensor type and that at most 1 sensor may be placed at

each location in S. Let r be the sensing range of a sensor. We assume that each sensor

can monitor/cover the entire disk of radius r centered at itself. The objective is to find

the minimum cost (since there is only 1 sensor type, minimizing cost is equivalent to

minimizing the number of sensors) sensor deployment that provides the desired degree, q,

of coverage over the region R. Such a sensor deployment, denoted by OPTR(r), is called

an optimal dce/'1',;;'. i/ and its cost is cost(OPTR(r)).

The transformation from region coverage to point coverage may be accomplished

by superimposing a grid of points over R such that every point of R is within a distance

d of at least one grid point that is inside R. Here, d < v2r is the distance between

.,,li i:ent grid points and is an optimization parameter. Let G(d) denote those grid points

that are in (or on the boundary of) R. Note that some or all locations in S may not be

points of G(d) and may not even be in R. Figure 3-5 shows G(d) for the case when R is a

rectangle. G(d) is composed of the small circles (both shaded and unshaded) in this figure;

the outermost grid points are a distance d/2 from the boundary of R. Let ri r d > 0.

Lemma 12. Let D(r/) be a dce/'/.1';,,, I of sensors whose rir,,.. is ri. If D(r/) q-covers

G(d), then D(r) (note that D(r) is -.'.,,i;/, D(r/) with the sensor r,,.' I,J,,,. l from rI to

r) q-covers the region R.

S0 0 0 0 0 0 0 0 0
0 0 0 0 0 O 0 0 0 0
S0 0 0 0

0 *0 0 0

O O ** O

0 0 0 0 0 0 0 0 0 0

Figure 3-5. G(d) for a rectangular region R. The 16 shaded grid points are within a disk
of radius r centered at the location s.

Proof Consider any point P in the region R. Let Pi c G(d) be the grid point closest

to P. From the definition of G(d), it follows that d(P, PI) < Since D(r/) q-covers

G(d), there are at least q sensors in D(r/) that are within a distance ri of PL. From the

triangular inequality, it follows that these q sensors are within a distance ri + d(P, PI) < r

of P. Hence, D(r) q-covers P.

Corollary 1. Let OPTc(d, r) be an optimal de'l..';,i. i,/ of sensors, whose rr,,j. is r/,

that q-covers G(d). cost(OPTG(d, r/)) > cost(OPTR(r)).

Note that when G(d) cannot be q-covered using sensors whose range is r/, cost(OPTc(d))

oc. Note also that sensors may be deploiv, 1 only to locations in S.

Lemma 13. Consider an infinite grid of points with point separation d. Let MAX(r) and

MIN(r), ,. -/,. ,';. ; I, be the maximum and minimum number of grid points covered by a

sensor whose rI.s,. is r. MAX(r) < 7r()2 + 2L2'] + 1 and MIN(r) > 7r(9)2 2 L2']

Proof Without loss of generality, assume that the sensor is located at (0, 0) and that

Pi, i = 1, 2, 3, 4 are the 4 grid points closest to the sensor (Figure 3-6). We may further

assume that P1i (x d, y d) is closer to the sensor than are Pi, i = 2, 3, 4. Clearly,

0 < x < and 0 < y < Label the row in which P1 lies row 0, the row right above

(below) Pi is row 1 (-1), and so on. Let maxrow be the row number of the highest row

that is partially covered by the sensor and let minrow be corresponding lowest row. Note

that L2J] < maxrow minrow + 1 < [2'J + 1.

Let row(i) be the number of grid points in row i that are covered by the sensor.

Clearly, [2 ()2- (Y + i)2J < row(i) < [2 (f)2- (Y + i)2J + 1.

maxrow
MAX(r) = row(i)
i =mnrow
maxrow
S (L2 ( )2 (1 )2J +1)
i=minrow
maxrow
S(L2 ( )2 )2) + L2] +
i=minrow

< (2 ( )2 (y.ii)2)L2']
i=minrow
maxrow
+ E (2 ( )2 (1 + i)2) + L2] + 1
i 1

< 2 ( ()2 y2dy + 2L2 + 1
r d d
d
< ()2 + 2[2> + 1

maxrow
MIN(r) = row(i)
i=minrow
maxrow
> L2 ( )2 (y l+)2j
i=minrow
maxrow
> (2 ( )2 (y +i)2 1)
i=minrow
maxrow
S (2 ( 2 (y +1 )2) [2 ]
i=minrow
> ( ) 2 -2 [L2]
d dxists, costd

Lemma 14. When D(rl) as in Lemma 12 exists, cost(OPTG(d, r/)) < F cost(OPTp(r)),

where F < MAX(r) MIN(r) + 1.

d0 0 0 0 0 0 0 0 0 0

0 0 0 0 O
0 0 S 0Q 0 0 0 0

0 0O00

0 0 0 0 0 0 0

0 0 0 0 00 0 0 0 0

Figure 3-6. The sensor s is located inside the square formed by four grid points Pi,
i = 1, 2, 3, 4, and covers MAX(r) grid points, shown in dark color.

Proof The optimal deployment OPTR(r) q-covers every point in R and hence q-covers

every point in G(d). When the sensing range of the sensors in the deployment OPTR(r)

is reduced to r/, this deployment may no longer q-cover every point in G(d). For each

sensor s in OPTR(r), the difference between the number of grid points covered by s

before and after the sensing range reduction is at most MAX(r) MIN(rl). Note that

even though MAX(r) and MIN(rl) are defined with respect to infinite grids, the bound

MAX(r) MIN(rt) applies to G(d), which is a finite grid. This is so because the stated

bound may overestimate (but not underestimate) the difference. Since the deployment

D(r/) exists, by placing a sensor whose range is rl at at most MAX(r) MIN(rt)

points of S at which there is no sensor, we can cover all points of G(d) covered by s

when its sensing range was r. Repeating this process for each s E OPTR(r), we obtain a

deployment that has at most F size(OPTR(r)) sensors with sensing range rl that q-covers

G(d).

Lemma 15. F < 12.443' 2.399.

Proof

MAX(r) MIN (r)

< () + 2L2]J + 1
d d
d) -

L2)2)+2[2]+t (
d d
< x( )2+2[2 J+l-(x( -
r 1 r

+2(d ) + 2I- v2]

< (8 + ( )2 +41 2+

d 2
< 12.443 3.399
d

d
12

S)2

Hence, F < MAX(r) MIN(rl) + 1 < 12.443' 2.399.

Theorem 8. Let R be a region for which the de~j1. ;;/, -i,/ D(r/) exists. cost(OPTR(r)) <

cost(OPTG(d, r/)) < (12.443 2.399)cost(OPTa(r)).

Proof Follows from Corollary 1 and Lemma 15.

Notice that the use of a finer grid (i.e., smaller d) results in a larger approximation

factor. This would argue in favor of using the maximum possible d; that is d = V2r. For

this value of d, the approximation factor is

(12.443 2.399) 12.443/-- 2.399 6.4

Of course, while the approximation factor is minimized when d = V2r, it is entirely

possible, and intuitively, we expect so, that on some instances, better deployments

are obtained using a smaller d. Our nonintuitive result that the approximation factor

is smaller when d is larger is an artifact of the proof method used to establish the

approximation factor. Further, it may be necessary to use a smaller d to ensure the

existence of the deployment D(r/).

3.4 Coverage with a Grid of Sensors

In this section, we consider q-covering a large rectangular region R by deploying

sensors in a gridded (equilateral triangle, square, and regular hexagon, see Figure 3-7)

layout, exactly one sensor per grid point. We use the symbols T, S, and H, respectively,

to refer to the three grid geometries (equilateral triangle, square, and regular hexagon)

considered by us. We use d to denote the size (i.e., side length) of the geometry being

considered. For the problem considered in the section, there is only 1 sensor type;

the sensor range is r; and we wish to minimize the number of sensors deploy, 1 We

assume that r is much smaller than the height and width of the rectangular region R

that is to be q-covered. So, we neglect boundary effects in our analysis. With these

assumptions, minimizing the number of sensors for each gridded layout is equivalent to

finding dMax(X, q), X E {T, S, H}, the maximum value of d for the geometry X for which

the resulting gridded layout q-covers the plane. We consider the case when 1 < q < 5,

which we believe to be the most practical.

d

AX\ \ \ /

(A) Equilateral tangle (B) Square (C) Regular hexagon

Figure 3-7. Gridded layout using a geometry size d. A) Equilateral triangle. B) Square.
C) Regular hexagon.

Table 3-1. dMax, maximum APN, and .-i-ii.iil. i, ratio 1 < q < 5
q dMax Maximum APN Asymptotic ratio
1 3r 3r .2 2,=1.209
2 9r9
2 r 32r52 t1.814
Equilateral triangle 3 r vr2 2,3 1.21
24 r2 4 1.21
2 2 3 .
r5 r 7r2 745 1.27
1 Vr 2r2 =1.57
2
2 r r2 -1.57
2
r 12
Square 3 -2 4 52 5I 1
4 3v2r 18r2 25- 1.09
5 25 72
5 /r 22 w 1.57
1 r v3r2 473 2.402
4 9
2 r v3r2 2v4 1.209
4 9
Regular hexagon 3 r 3v3r2 ,3 1.41
4 5 75 /2 49" / 1.185
_7 196 V 225
5 v 2 47, 1 .451
4 15

Following [1], we use 7l to denote the maximum area per node (APN) in the gridded

layout1 that results in q-coverage, with geometry X e {T, S, H}. From the well-known

formulas for the area of a geometry and the number of geometries that share a common

vertex in a gridded layout (e.g., each vertex in a hexagonal grid is shared by 3 hexagons),

it follows that

7= 2 dMax(T, q)2

7Yq dMax(S, q)2

7 4 = 3dMax(H, q)2

Since a sensor can cover an area of at most 7rr2, rr2/q is an upper bound on the value

of APN for any q-cover. Hence, the ..- i','.!ll.lic ratio, ratios = rr2/(xq) is a measure

1 APN is the area of the region divided by the number of nodes/vertices in the gridded
layout for that region.

of the efficiency with which geometry X is able to q-cover the plane. This ratio gives an

upper bound on the ratio (number of sensors in the gridded deployment)/(number of

sensors in an optimal deployment). Note that the optimal deployment is not required to

be gridded.

Table 3-1 summarizes the values of dMax, the maximum APN 7~, and the

.-i~,i,!I ,l ic ratio ratioX for the geometries considered in this section. The results for

q = 1 are from Bai et al. [1] and those for 2 < q < 5 are derived later in this section.

Actually, we derive only the value of dMax(X, q), 2 < q < 5 as the values of 7q and

ratioX are simply computed using the equations give earlier and the value of dMax(X, q).

From the ..ii- ,ii1 l i, ratios given in Table 3-1, we see that the triangular grid, is

more efficient than the square and hexagonal grids for 1 coverage. In fact, it achieves

an ..-i-I, l lic ratio of 1.2. This is about 211'. less than that for the square grid and 510'.

less than that for the hexagonal grid. So, for 1-coverage, the triangular grid uses about

211' fewer sensors than does the rectangular grid; the triangular grid uses half the sensors

required by the hexagonal grid. For q = 2, the hexagonal grid is most efficient requiring

about :; fewer sensors than required by the triangular grid (an .,-i-,!l,,l ic ratio of 1.2

for the hexagon versus 1.8 for the triangle) and about 21 fewer sensors than required by

the square grid. For q = 3, the triangular grid is again most efficient (..-i ,1.!.l ic ratio

is 1.2). However, the square and hexagonal grids are quite competitive with ..i -1 11il.l ic

ratios of 1.3 and 1.4, respectively. Not surprisingly, when q = 4, the square grid is most

competitive and comes within 9' of our lower bound on the number of sensors required

for 4-coverage. The triangular and hexagonal grids are within 21 and 19'. respectively,

of this lower bound. When q = 5, the triangular grid is once again most effective and

comes within 27'. of our lower bound. We note that the stated results (though now we use

total sensor cost as our efficiency metric rather than number of sensors) and bounds apply

even when we have a mix of sensor types provided we use only sensors with the least value

of cost/range2.

Ak "B

Figure 3-8. Equilateral triangle, q 2

It is interesting to note that the upper bound of 7rr2/q used by us in the computation

of the .,-vmptotic ratio is rather loose. For example, for the case of 1-coverage (q = 1)

Kershner [27] has shown that placing sensors at the vertices of an equilateral triangle

whose sides are /3r minimizes the number of sensors required to cover the plane (this

result was rediscovered in [71]). So, for 1-coverage, 3-r2 is a tight upper bound on the

achievable APN. Our formula for .,-i-~! .' ic ratio uses the weaker upper bound of 7rr2/q.

In the remainder of this section, we abbreviate dMax(X, q) by dMax as the values of

X and q are evident from the context. We use dk(P) to denote the smallest distance such

that at least k sensors are within this distance of point P. Recall that in the model being

considered, there is a sensor at each vertex of the geometry being used.

3.4.1 q = 2

Equilateral Triangle.

For points P that are geometry vertices, d2(P) d, the length of a side of a triangle.

For example, for the vertex A of Figure 3-8, there is one sensor at a distance of 0 (i.e.,

the sensor at A) and no additional sensor until a distance of d. At distance d from A,

there are several sensors (those at B and C plus another 4 on the 4 distinct vertices of

the 5 other triangles that share A as a vertex). For every other point P, d2(P) < d (see

Figure 3-8). Hence, dMax = r.

Square.

A B
Figure 3-9. Square, q 2

Every point P in the plane lies in2 a triangle with side lengths d, d/ /2, and d//2.

Each vertex at the end of the triangle side whose length is d has a sensor; the remaining

triangle vertex is the center of a square of the gridded layout and has no sensor (see

Figure 3-9). The distance between P and these two sensors is < d. So, d2(P) < d. For

points P that are geometry vertices, however, d2(P) = d. For example, for the vertex A of

Figure 3-9, there is one sensor at a distance of 0 (i.e., the sensor at A) and no additional

sensor until a distance of d. At distance d from A, there are several sensors (those at B

and D plus another 2 on the 3 other squares that share A as a vertex). Hence, dMax = r.

Regular Hexagon.

For points P that are geometry vertices as well as for the center of a hexagon,

d2(P) = d, the length of a side of a hexagon. For example, for the vertex A of Figure 3-10,

there is one sensor at a distance of 0 (i.e., the sensor at A) and no additional sensor until

a distance of d. At distance d from A, there are three sensors (those at B and D plus

another one on the two other hexagons that share A as a vertex). Every point in the plane

is in one of the side d equilateral triangles that partition a regular hexagon into triangles

(see Figure 3-10) and every such triangle has sensors on 2 of its 3 vertices (the third vertex

2 Our use of the word "in" includes the boundary of the geometry referred to.

0 D

d

A

Figure 3-10. Regular hexagon, q 2

is the center of a hexagon and has no sensor). So, for every point P that is not the vertex
of a tiling hexagon, d2(P) < d. Hence, dMax = r.
3.4.2 q = 3
Equilateral Triangle.
A proof similar to that used for the q = 2 case establishes d3(P) = d for the vertices
of the triangle and d3(P) < d for all other points P. So, dMax = r.
Square.
Consider the square of Figure 3-11. P is the mid point of a side of the square. The
sensors at E and F are at a distance of d/2 from P. The remaining sensors are at least
d(A, P) = d away. So, d3(P) = 'd. Next, consider points p (other than B, which
is the mid point of the edge AD) in the shaded triangle ABC. d(p, A) < d(C, A) < d,

d(p,D) < d, and d(p, E) < d(B,E) = d. So, d3(p) < d3(P). By symmetry, d3(p) < d3(P)
for all points other than edge mid points. So, dMax = r. Or, dMax = r.
Regular Hexagon.
Consider the regular hexagon of Figure 3-12. P is the mid point of a side of the
hexagon. The sensors at E and G are at a distance of d/2 from P. The remaining sensors
are at least d(A, P) /d2 + ()2 2dd cos(2,) = "d (from the law of cosines) away. So,

d3(P) 7d. Next, consider points p (other than B, which is the mid point of the edge
AE) in the shaded triangle ABC. d(p, A) < d(C, A) = d, d(p, E) < d(C, E) = d, and

Eo d/2

d

A

d/2 F

C

Figure 3-11. Square, q=3

Figure 3-12. Regular hexagon, q-

D G
-3

d(p F) < d(B,F) -= d. So, d3(p) < d3(P). By symmetry, d3(p) < d3(P) for all points
other than edge mid points. So, $dMax = r. Or, dMax = r. 3.4.3 q = 4 Equilateral Triangle. Figure 3-13 shows 4 of the triangles in a gridded layout. The length of each side of each triangle is d. Let P be the circumcenter of the equilateral triangles BDF and ACE. The circumradius d(A, P) = d(C, P) = d(E, P) of the equilateral triangle ACE is d and that of triangle BDF is The point P has three sensors at a distance These are the B d B C A D F E Figure 3-13. Equilateral triangle, q=4 sensors at B, D, and F. However, the next nearest sensors (those at A, C, and E) are at a distance of -d. So, d4(P) -2d. For any p / P inside triangle BDF, d(p, B) < d, d(p, D) < d, and d(p, F) < d. Further, one of d(p, A), d(p, C), and d(p, E) is less than the circumradius of triangle ACE. So, d4(p) < d. Combining the bounds for the cases P is the circumcenter of a triangle and p / P, we see that no matter which point we consider in the plane, there are 4 sensors within a distance d and this bound is tight. So, dMax = r. Square. Figure 3-14 shows 4 squares of the gridded layout. There is a sensor at each corner of each square (i.e., grid position). The point P is the circumcenter of the isosceles triangle ACE. Note that d(A, C) =d(E, C) = 5d and d(A, E) = 2d. So, the circumradius of triangle ACE is d(A, P) d(C, P) d(E, P) = c = 2d. Let disk(P, c) denote a disk of radius c centered at P. The six sensors A-F are in this disk and every disk centered at P and having smaller radius than c excludes (at least) the sensors at A, C, and E. So, d4(P) = c. Next, consider the triangle FGH, where G is the mid point of the side BF and H is the mid point of the square BCDF. For every point p / P in this triangle, d(p, F) < d d(p,B) < d, and d(p, D) < vd < c. Also, since ', .:,.i,, :(FGH) C disk(A,c) U disk(C, c), either d(p, A) < c or d(p, C) < c. So, there are 4 sensors within a distance c of every point B P nE P i\ 0- Co H C D \isk(C,c) (a) (b) Figure 3-14. Square, q=4 of triangle FGH. By symmetry, it follows that there are 4 sensors within a distance c of every point in the d x d square centered at F. This implies there are 4 sensors within a d x d square centered at each grid point of the layout and hence within a distance c of every point in the plane. So, the maximum d that results in 4-coverage is such that r c 2d. So, dMax = 2vr. Regular Hexagon. Figure 3-15 shows 3 hexagons of the gridded layout. There is a sensor at each corner of each hexagon (i.e., grid position). The point P is the circumcenter of the isosceles triangle ADF. Note that d(A, D) = d(A, F) 7d and d(D, F) = d. So, the circumradius of triangle ADF is d(A, P) d(D, P) = d(F, P) = c = d. Since d(E, P) d and d(J, P) d(K, P) d > c, the sensors at E, J, and K are not in disk(P, c). Only the sensors at A-D, F, and G are in disk(P, c). This together with the observation that every disk centered at P and having smaller radius than c excludes (at least) the sensors at A, D, and F. So, d4(P) = c. Next, consider the triangle BHI, where H is the mid point of the side BG and I is the mid point of the hexagon BCDEFG. For every point p / P in this triangle, d(p, B) < d, d(p,G) < d, and d(p, C) < id < c. Also, since tr:,,l.. (BHI) C disk(A, c) U disk(F,c), (a) (b) Figure 3-15. Regular hexagon, q=4 either d(p, A) < c or d(p, F) < c. So, there are 4 sensors within a distance c of every point of triangle BHI. The hexagon BCDEFG may be partitioned into 12 triangles (including BHI) that are symmetric to BHI. By symmetry, it follows that there are 4 sensors within a distance c of every point in these 12 triangles. In other words, there are 4 sensors within a distance c of every point in the hexagon BCDEFG and hence within a distance c of every point in the plane. So, the maximum d that results in 4-coverage is such that r c d. So, dMax r. 3.4.4 q = 5 Equilateral Triangle. Figure 3-16 shows 4 of the triangles in a gridded layout. The length of each side of each triangle is d. Let P be the mid point of the side DF. The sensors at D and F are d/2 from P while those at B and E are at a distance of$d. The sensors at A are C are

next in proximity to P. From the law of cosines, we obtain d(A, P) = d(C, P) = c

d2+ 2 -2d cos( ) d. So, ds(P) = d.

For any p / P inside triangle FGP, where G is the common circumcenter of the

equilateral triangles ACE and BDF, d(p, B) < d, d(p, D) < d, d(p, E) < 2d < c,

d(p, F) < d < c, and d(p, A) < c. So, the sensors at A, B, D, and F are within c of every
point in triangle FGP. By drawing lines from the vertexes of triangle BDF through the

d B

C G A

D P F

E
Figure 3-16. Equilateral triangle, q=5

A

d /

Figure 3-17. Square, q=5

circumcenter G and up to the opposing edge of the triangle, we partition the triangle into
6 smaller triangles (including triangle FGP) that are symmetric to FGP. From symmetry

it follows that there are 5 sensors within a distance c of every point in triangle BCD and
hence within c of every point in the plane. So, the maximum d that results in 5-coverage is
such that r = c = d. So, dMax r.
Square.
Figure 3-17 shows 4 squares of the gridded layout. The point P is the center of the

square BCDE. The sensors at B, C, D, and E are closest to P; each is a distance
from P. The next closest sensors are at a distance of c = 1-d (e.g., the sensor at A). So,
d(P) = c.

(a) (b)
Figure 3-18. Regular hexagon, q=5

Next, consider the triangle BFP, where F is the mid point of the side BE. For every

point p / P in this triangle, d(p, B) < d, d(p, C) < $d < c, d(p, D) < /2d < c, d(p, E) < d, and d(p, A) < c. By symmetry, it follows that there are 5 sensors within a distance c of every point in the square BCDE and hence within every point in the plane. So, the maximum d that results in 5-coverage is such that r c -d. So, dMax = r r. Regular Hexagon. Figure 3-18 shows 3 hexagons of the gridded layout. The point P is a vertex of a hexagon. There are 4 sensors (i.e., those at A, C, G, and P) that are within a distance d of P. The next nearest sensor is at a distance /3d (e.g., the sensors at D, F, J, and K). So, d5(P) c= V3d. Next, consider the triangle HIP, where H is the mid point of the side GP and I is the mid point of the hexagon CDEFGP. For every point p / P in this triangle, d(p, P) < d, and d(p, C) <$d, d(p, G) < d, and d(p, F) < c. Also, since tr(.'.,l. (HIP) C
disk(A,c) U disk(E, c), either d(p,A) < c or d(p, E) < c. So, there are 5 sensors within a

distance c of every point of triangle HIP. By symmetry, it follows that there are 5 sensors
within a distance c of every point in the hexagon CDEFGP and hence within a distance

c of every point in the plane. So, the maximum d that results in 5-coverage is such that
r c = vd. So, dMax =
,/3

3.5 Experimental Results

In Section 3.1-3.3, we presented several strategies to q-cover a region when the sensor

locations are limited to a specified set S. We believe that all but the region coverage

via point coverage method of Section 3.3 are too compute intensive to be used on large

deployment instances. This is because each of the other methods requires us to find

optimal deployments for several small hexagons or tiles. While an optimal deployment for

a hexagon or tile may be found in 0(1) time, the required constant time is expected to be

large as the optimal solution is found by searching a potentially large though constant size

space of possible solutions. This search will perform many explicit tests for region rather

than point coverage are time consuming and region coverage tests are compute intensive.

As a result, our experiments focus on the method of Section 3.3 adapted to instances with

multiple sensor types.

We programmed the region coverage via point coverage scheme of Section 3.3 on

a Dell Dimension PC with a 2.13 GHz dual-core processor and 2GB memory. To cover

the resulting grid G(d), we used an ILP (integer linear programming) relaxation method

as well the greedy algorithm of Xu and Sahni [69]. For the ILP relaxation, we started

with the ILP formulated in [69] for point coverage, relaxed it to a linear program, solved

the linear program using lp-solver 5.0 [31], and then converted the solution to the linear

program to a feasible integer solution to the original ILP using the rounding method

proposed by Wang and Zhang [62]. Since both the ILP formulation and greedy method

of [69] are for heterogeneous sensor deployment, we experimented with heterogeneous

instances with up to 4 sensor types.

For test data, we used a 35 x 35 region R. To construct G(d) for any given d, we

tiled R with d x d squares and considered the centers of these tiles. Centers that were

outside R were relocated to their nearest point on the boundary of R. The resulting center

locations define G(d). To construct the set S of permissible sensor locations, we tiled R

with 7r,,,/12 x 7r,,,/12 squares and included the center of each square in S, where r,,,

is the maximum range of a sensor. Next, S was augmented by adding [3area(R)/(7r~ ax)]

randomly chosen points from R. For each value of q, 1 < q < 3 (q is the desired coverage

degree), we experimented with the following 9 test cases. The sensor set T specifies the

sensor types as pairs (r, c), where r is the sensor range and c is the cost of the sensor.

1. IsI = 362 T= {(4,5)}

2. ISI = 242 T {(5,6)}

3. ISI = 242 T = {(3.5,4),(5,6)}

4. ISI = 313 T = {(3,4),(4.5,5.5)}

5. ISI = 242 T = {(3,3.5),(4,4),(5,6)}

6. ISI = 207 T = (3.5,4),(4.5,5),(5.5,6)}

7. ISI = 242 T {(3.5,4),(4.5,5),(5,6)}

8. ISI = 242 T = {(3,3.5),(4,4),(4.5,5),(5,6)}

9. ISI = 207 T = {(3,3),(3.5,4),(4,5),(5.5,6)}

For each test case and each q, we experimented with different d values and for each

combination of test case, d, and q, we generated 10 random instances (these differed only

in the set S of allowable locations). Figures 3-19-3-21 plot the average of the cost of the

deployments obtained for these 10 random instances using the ILP and greedy methods.

The figures show also the lower bound for the cost of an optimal deployment for G(d)

as determined by the solution to the linear programming relaxation for the ILP. Note

that when determining a deployment to q-cover G(d), sensor ranges were reduced from

r as specified in the test case to rl. Our experiments show that the relaxed ILP method

generally produces lower cost deployments than does the greedy method. Further, the cost

of the deployment is not well correlated with d. Often, though, a smaller d resulted in a

smaller deployment cost. The use of a smaller d does, however, increase the time needed to

find a deployment. Each of the instances used by us was solved in a few seconds when d

was large and in a little under 15 minutes when d was small.

--LPlow er bound
--- LP rounding
-A-- Greedy

200

400 --LPlower bound
---LP rounding
-A-- Greedy

300 ---LPlowerbound
-- LProunding
-A-- Greedy

j;200

300

1o0 -
o00
02 03 04 05 06 07 08

00 LP lower bound
:--LProunding
-- A Greedy

SLP lower bound
--LProunding
-- Greedy

200 ---

Figure 3-19. Total sensor cost required v.s. various 6s, where q=1.

80

rc~C

01 02 03 04 05 06 07 08

02 03 04 05 06 07 0E

~-p~C~

02 03 04 05 06 07 0

02 03 04 05 06 07 0O

550 --LPlowerbound

Greedy

450

350

250

600 --LPlower bound
--LP rounding
-a-- Greedy

-*- LP low er bound
--LP rounding
-A- Greedy
450

350

250
02 03 04 05 06 07 08

450 -- LProunding
-a-- Greedy

350

Figure 3-20. Total sensor cost required v.s. various 6s, where q=2.

81

500

400

--LPlower bound
---LProundng
-A- Greedy

02 03 04 05 06 07 08
5

01 02 03 04 05 06 07 08

02 03 04 05 06 07 08

~_CXI~C~

ozu LPIow er bound]
rouding
47 ] --A reedy
470

420

370

02 03 04 05 06 07 0;

700 LP lower bound

-A- Greedy

600

500

400

600 --LPlowerbound
-5- LProunding
-A-- Greedy

500

400

02 03 04 05 06 07 08

750 -*-LP lower bound
--LP rounding
-A- Greedy

650

0

550

450 -
02 03 04 05 06 07 08

700 ---LPIowerbound
-- LP rounding
-A- Greedy

600

650 --LPlower bound
-5-LProunding
-A- Greedy

550

450

350
02 03 04 05 06 07 08

Figure 3-21. Total sensor cost required v.s. various 6s, where q=3.

LL 11rb

LP lower bound
950 LProunding
900- --reedy

850
8oo

"'

01 02 03 04 05 06 07 08

m L

G ::

02 03 04 05 06 07 08

400

CHAPTER 4
PROPERTIES OF DTOA LOCALIZATION

Although several methods to localize [7, 13, 17, 34, 45, 46, 53, 70] given a set of

difference of time-of-arrival measurements have been proposed, there appears to be no

studies of fundamental properties of DTOA localization.

In this chapter, we present a number of results that establish fundamental properties

of DTOA localization. We first consider the unique identification of a source and establish

the following:

1. DTOA localization uniquely identifies a source in Euclidean plane R2 iff the sensors
do not lie on a hyperbola1

2. At least four sensors are necessary for unique localization of a source in Euclidean
plane, and it is sufficient to place the four sensors at the corners of a parallelogram to
achieve this.

3. A minimal sensor set to achieve unique source identification (i.e., a sensor set none of
whose proper subsets is also a uniquely identifying sensor set) has between 4 and 6
sensors.

4. Three sensors are sufficient to uniquely identify any source in a monitoring region
bounded by a polygon. These sensors, however, must be placed outside the polygon.

We then consider the computational aspects of DTOA localization that utilizes the

intersection of hyperbolas corresponding to distance-difference measurements. In general,

two hyperbolas may have four intersection points, but we show that two hyperbolas that

correspond to distance-differences to a source that have a common focus may have at

most 2 intersections. We also show that when non-collinear sensors are used, at most 2

points can have the same DTOA values. These results establish that the DTOA problem

is more structured and easier in this sense compared to computing intersection points of

hyperbolas.

1 For convenience, in this dissertation, the term hyperbola is used to refer to one branch
of a hyperbola.

This chapter is organized as follows. In Section 4.1, we present some fundamental

properties and definitions. Properties of sensor sets that uniquely identify all sources

in Euclidean space are developed in Section 4.2. Our detailed analysis of Section 4.3

establishes the bound on the number of intersections of two DTOA hyperbolas. In

Section 4.4 we show that at most 2 points can have the same set of DTOA values. The

minimum number of sensors needed to uniquely identify all sources in a bounding ]p" .Iv.on

is derived in Section 4.5.

4.1 Preliminaries and Definitions

Llet Si = (xi, yi), 1 < i < k, be the locations of k sensors in Euclidean space R2.

These locations are assumed to be distinct. For any point P =(x,y) in R2, the distance,

d(P, Si), between P and Si is /(x x)2 + (y- y)2. A signal originating at P at time 0

arrives at Si at time proportional to d(P, Si). For simplicity, we assume that the arrival

time is d(P, Si). The difference, Aiy, in the time of arrival (DTOA) at Si and Sj is given

by

Ai(P) = d(P, Si) d(P, Sj).

From the triangle inequality, it follows that AI (P)I < d(Si, Sj). Furthermore, the

locus, Lij(6), of points defined by

Lij() = {P|Aij(P) 6}

is a hyperbola2 (see Figure 4-1).

When Aij(P) = Ai(Q) for every i,j e {1,2,..., k}, the points P and Q are

indistinguishable. Actually, since Aij(P) = Aij(P) Ali(P), for all i and j, P and Q

are indistinguishable iff A1j(P) = Aij(Q) for every j E {2,..., k}. So, the set of sensor

2 Strictly -I.'" i1;i- Lij(5) is one branch of a hyperbola and Lyi(-6) is the other branch.
As mentioned earlier, for convenience, in this dissertation, we use the term hyperbola to
refer to one branch of a hyperbola.

L(-2)

Figure 4-1. Examples of the locus L12

locations (also referred to as the sensor set) SS {S1, S2, Sk} can uniquely identify

every source S in Euclidean space R2 iff for every pair P and Q of distinct points in

Euclidean space R2, we have Aij(P) / Aij(Q) for at least one j E {2, 3, .. k}. A sensor

set that can uniquely identify (localize) every possible point in Euclidean space is called an

identif.i:,,.i sensor set, ISS. Two points that are indistinguishable are duals.

The DTOA method localizes the source by determining the common intersections of

the hyperbolas3 L1j(A1j(S)), 2 < j < k. When these hyperbolas have more than one

common intersection, the source is not uniquely localized. Figure 4-2 gives an example of

two hyperbolas L12(512) and L13(513) that intersect at two distinct locations P1 and P2.

So, using L12 and L13 alone, we are unable to uniquely localize the source. We are able

only to assert that the source location is either P1 or P2. To uniquely identify the source

using the DTOA method, the hyperbolas LI, 2 < j < k should have exactly one common

intersection. Alternatively, these hyperbolas should have exactly one common intersection

inside a region in which the source is known to lie.

3 A point in R2 is a common intersection of a set of hyperbolas iff this point is on each
of the hyperbolas

Figure 4-2. Three non-collinear sensors S1, S2, and S3 form a triangle and two hyperbolas
L12( 12) and L13(613) intersect each other at Pi and P2.
4.2 Properties of Identifying Sensor Sets

In this section, we establish, in Theorem 9 a necessary and sufficient condition for a

sensor set SS to be an ISS. Theorem 10 shows that every ISS has at least 4 sensors and

Theorem 12 shows that every ISS with more than 6 sensors has a subset of size at most 6

that is an ISS.

Theorem 9. The sensor set SS = {Si, ,Sk} is an ISS iff no ,',', ,I,. li passes through

all points of SS.

Proof

We first show that if SS is an ISS, then no hyperbola may pass through all points

of SS. By contradiction, suppose there exists a hyperbola, -i L, that passes through all

points of in SS. Let P1 and P2 be the two foci of L. From the definition of a hyperbola,

it follows that d(P1, S) d(P2, Si) = d(P1, Sj) d(P2, Sj), 1 < i < j < k. So, Aij(PI)

d(P1, Si) d(P, Sj) = d(P2, Si) d(P2, Sj) = Ay(P2), 1 < i < j < k. Hence, P1 and P2

are indistinguishable and SS is not an ISS, a contradiction.

Next, we show that if SS is not an ISS, then at least one hyperbola passes through

all points of SS. Let P1 and P2 be two different points that are indistinguishable. So,

Ai(P) P PI, S) d(P1, S) d(P2, S) d(P2, S) AI,(P2), 2 < j < k. Hence,

d(PI, SI) d(P2, SI) d(PI, S) d(P2, Sj), 2 < j < k. Therefore there is a hyperbola with

P1 and P2 as as its foci that passes through all points of SS. m

Theorem 10. If SS is an ISS, then ISSI > 4 and there exist ISSs that have 'i.. r; 4

sensors.

Proof

We first prove that 3 sensors are not sufficient to constitute an ISS and so, \SS| > 4

whenever SS is an ISS. Let SS = {Si, S2, S3}. When S1, 52, and S3 are collinear, the

straight line through these three sensors is a trivial hyperbola through the points of SS.

From Theorem 9, it follows that SS is not an ISS. When S1, 52, and S3 are not collinear,

they define a nontrivial triangle as shown in Figure 4-2. Clearly, there exists a negative

constant, 612, such that the hyperbola L12(612) intersects the line S1S3 at two distinct

points Qi and Q2. Observe that the hyperbola L13(-d(SI, S3)) is actually a ray that

originates at S1 and intersects L12(612) at Q1 only. Let 613 be a negative constant slightly

greater than -d(S1, S3). The hyperbola L13(613) intersects L12(612) at two distinct points

P1 and P2 (see Figure 4-2). So, P1 and P2 are indistinguishable and SS is not an ISS.

Next, we show that whenever SS { S= {S2, S3, S4} are the corners of a parallelogram

with side length > 0, SS is an ISS. We show this by proving that no 4 distinct points of

a hyperbola define the corners of a parallelogram. The result then follows from Theorem 9.

Consider the hyperbola L of Figure 4-3. Let S1, 52, S3, and 54 be 4 points on

this hyperbola. The case shown in Figure 4-3 has S1 and 54 on one part (arm) of the

hyperbola and 52 and S3 on the second part. (There are two other cases for the location

of the 4 points-exactly 3 points on one part of L and 4 points on one part of L.) Let

Q1 and Q2, respectively, be the intersections of the line segments 51S2 and S3S4 with

the x-axis, which is the -, ,,1ii, i r axis of L. If the 4 identified points on L are the

s4 L
S4

Figure 4-3. A hyperbola L that passes through Si (1 < i < 4).
corners of a parallelogram, S1S2 and S3S4 are parallel and of equal length. However,

if these segments are parallel, d(S1, Q1) < d(S4, Q2) and d(S2, Q1) < d(S3, Q2). So,

d(S1, S2) = d(SI, Qi) + d(S2, Q) < d(S4, Q2) + d(S3, Q) d(S3, S4). So, S1S2 and S3S4
cannot be parallel and of equal length. The remaining two cases are similar.

Corollary 2. An infinite number of i,;', ,/..'l.- pass through r,:, 3 non-collinear sensors in

Euclidean space R2.

Corollary 3. Whenever SS contains the corners of a 1,p,'11 l..gram with side length > 0,

SS is an ISS. In particular, whenever 4 sensors of SS are at the 4 corners of a square

with side length > 0, SS is an ISS.

An ISS is a minimal ISS (MISS) iff no proper subset of the ISS is also an ISS.

Theorem 12 establishes an upper bound of 6 on the size of an MISS. To prove this

theorem, we need to use Bezout's bound on the number of intersections of curves in

Euclidean space.

Theorem 11. [Bezout's Theorem [28]]: Let C1 and C2 be curves of degree m and n,

. "i"/.. ,'.; ; in Euclidean space R2. If C1 and C2 have no curves in common, then the

number of intersections of C1 and C2 is at most mn.

Corollary 4. Two Iu;l', ,1.i/..l in Euclidean space R2 have at most 4 intersections.

Lemma 16. At most 1 iuil' ,ti.,ll, In,,; pass through ,:; set of 5 or more distinct points.

Proof Consider any set SS with 5 or more points. If two hyperbolas pass through the

points of SS, then these two hyperbolas intersect at the points of SS and so have more

than 4 intersections. This violates Corollary 4. Hence, at most 1 hyperbola may pass

through the points of SS. m

Theorem 12. Every SS that is an MISS .,-i/:. 4 < ISS <6.

Proof 4 < ISSI follows from Theorem 10 and the fact that a MISS is an ISS. ISSI < 6

may be shown by contradiction. Suppose that ISSI > 6. Let SS be a subset of SS such

that ISS/| = 5. From Lemma 16, SSi has at most 1 hyperbola passing through its 5

points. If no hyperbola passes through these points, then SSi is an ISS (Theorem 9)

and SS cannot be an MISS. So, we may assume that exactly one hyperbola passes

through SSi. Since SS is an ISS, SS contains at least one point Si that does not lie on

this hyperbola. Hence, there is no hyperbola that passes through the 6 points SS/ U{Si}.

From Theorem 9, it follows that SSfU{Si} C SS is an ISS. This contradicts the

assumption that SS is an MISS.

4.3 Number of Intersections of L12 and L13

Although two hyperbolas in Euclidean space may have up to 4 intersections

(Corollary 4), two DTOA hyperbolas L12 and L13 may have no more than 2 intersections

when S1, S2, and S3 are non-collinear. Without loss of generality (w.l.o.g), we choose our

coordinate system as in Figure 4-4. The features of this choice are (a) S1S2 falls on the

y-axis, (b) the midpoint of S1S2 is the origin O of the coordinate system, and (c) S3 lies

on the right side of the y-axis. We see that S1S2, S2S3, and S1S3 partition the Euclidean

space R2 into seven regions (a)-(g). At most one intersection of L12 and L13 lies in the

union of regions (a), (b), (f), and (g) and at most one intersection lies in the union of

regions (c), (d), and (e). To prove these assertions, we need a result from Theorem 13

top left (+)

top right (+)

bottom (+)

Figure 4-4. Regions of monitoring area: (a) top left, (b) inside, (c) bottom right, (d) top,
(e) bottom left, (f) bottom, and (g) top right.
(Refer to the detailed proof from Section 5.4) that establishes the monotonicity of the

directional derivative of A13(P) along the hyperbola L12(A12(P)) within each of the 7

regions of Figure 4-4. The sign of the directional derivative for each region is also given in

Figure 4-4.

Theorem 13. For i.:', point P in Euclidean space R2, the directional derivative of A13(P)

il..,i"j the li,,, ,ii 'l.. L12(A12(P)) is monotone in each of seven regions -.... 7.1 by three

non-collinear sensors, as shown in Figure 4-4. The directional derivative is positive in

regions (a), (b), (f), and (g), and is negative in regions (c), (d), and (e).

In the following, we use LI and L' to refer to the two symmetric parts (arms) of the

hyperbola L (see Figure 4-5). The two parts LI and L' intersect only at the vertex B. 11

and 12 are the two .i-:ii-!11 1.1. of the hyperbola and lI, and 12T are lines that intersect at

the vertex B and are parallel to these .-i- .1iii !l From our choice of coordinate system,

it follows that the .i-i~!iiil, .1 -; intersect at O.

Lemma 17. 1. L (Ll) strictly lies between 11 (12) and 11(121).

2. The shortest Euclidean distance between a point P on Lr (LP) and the i- ;u,,/i -/.: 11

(12) decreases monoto, .:. ll as P gets farther from the vertex B.

Y
12 / 1

Proof Follows from the definition of a hyperbole, its i/and the lines I and
\\ \\ / /

I I \ \
L/ /, L

/ \/ \ \

12//
S \/Pr
/ \/ \ \
\ I

\ iI \ \
/ / \ \

/ / \ \
/ / \ \
/ / \ \
i / \ \

Figure 4-5. A hyperbola, L LUL with fothat 12cus an d -3 Pi axis yxis.
increases monoto liI as P gets farther from the vertex B.
Proof Follows from the definition of a hyperbole, its .\i- and the lines and

In There are 4 possible caseshow that when S1 is closer tofor the sourrelationship between thneS than are SS2S3 and thS3e

hyprestriction on the sore beline is beloser to S(2) than thline remaining two sensors is removed incts
Let (xi, yi), 1 iI m be intersections of L1 and 13 From the definition of a
hyperbole, it follows that A(P) A(P) and A13(P) A13(P) for 1 \ < i m

hyperbola L12-(1) thle line is below L\^, (2) thle line intersects L1^, (3) thle line intersects

Figure 4-6. Case 1: S2S3 lies below L12
Lr2 and ZS3SIS2 > 90, and (4) the line intersects Lr2 and ZS3SIS2 < 90. These 4 cases

are shown in Figures 4-6-4-9, respectively. We show below that L12 and L13 have at most 2

intersections in each of these cases.

Case 1: S2S3 lies below L12

When S2S3 lies below L12, L12 must lie wholly within regions (a) top left, (b) inside,

(d) top, and (g) top right, (Figure 4-6). A13, from Theorem 13, monotonically increases in

regions (a), (b), and (g) and monotonically decreases in (d). So, if no component of L12

is in region (d), then A13 monotonically increases along all of L12 and the value of A13

for each point P on L12 is unique. Hence, L12 and L13 have only 1 intersection. If region

(d) contains a portion of L12, then when one moves the point P from left to right along

L12, (d) is the first region to be visited. So, when moving from left to right along L12, A13

monotonically decreases while we are moving along the portion of L12 that is inside region

(d) and then monotonically increases for the remainder of L12. Hence L12 has at most 2

distinct points for any given value of A13. So, L12 and L13 have at most 2 intersections.

Case 2: S2S3 intersects L12

When S2S3 intersects LI2, ZS3S2S1 > 90 (Figure 4-7). So, L12 cannot have a

component in either of the regions (c) (bottom right) and (f) (bottom). Additionally, L12

cannot have a component in region (d) (top). To see this, observe that L 2 is wholly to the

right of the y-axis while region (d) is wholly to the left of this axis. So, no portion of L'2

is in region (d). To see that no portion of L12 is in region (d) either, note that LI2 is below

12/ (Lemma 17). Since, S2S3 intersects LI2 and 12 is strictly below LI2 (Lemma 17), S2S3
intersects the .i,-'iini, 12. Now, since 121 is parallel to 12, S2S3 also intersects 121. which

implies that the slope of S2S3 is less than that of 121. Hence, the slope of S1S3 is less than

that of 121. From this, the fact that LI2 lies below 121, and the fact that the intersection

(vertex B of L12) of LI2 and 12i is below S1, it follows that no portion of L 2 is inside the
top region (d).

Consequently, as one moves from left to right along L12, the region (e) (i.e., bottom

left) is the first region to be visited. A13 monotonically decreases inside this region and

monotonically increases in the remaining regions that L12 is in. Hence L12 has at most 2

distinct points for any given value of A13. So, L12 and L13 have at most 2 intersections.

Case 3: S2S3 intersects L'2 and ZS3SIS2 > 90

In this case, region (e) (bottom left) lies entirely below L12 (Figure 4-8). Hence,

no portion of L12 is in region (e). Since ZS3SIS2 > 90, 0 < 90 (see Figure 4-8).

Hence, d(P, SI) > d(P, S3) for every point P inside region (c) (bottom right). Since,

by assumption, S1 is closer to the source S than is S3, no portion of L13 is in region (c).

Hence, L12 and L13 have no intersection in region (c).

If L12 has an overlap with region (d) (top), then region (d) is the first region

encountered as we move from left to right along L12 and if L12 overlaps with region

(c) (bottom right), region (c) is the last region encountered as we move from left to right

along L12. A13 monotonically decreases in region (d), L12 and L13 do not intersect in

region (c), and A13 monotonically increases in the remaining regions that L12 may overlap.

So, L12 and L13 have at most 2 intersections.

12

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\ \

\ \
\\\

\\

/ \

/ /
/ /
/ /
/ /
/ /
/ / 2
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /

/
/

/
/ /^
/ /
/ /

/ /
/ /
/ r

/

/
V

Xx\
S\\S +
\\ \ \
+ \\ \\
\ \
\\\ \

\\ \
\ \ \

\\ \\
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12'
2 2

12 \

\ \

\\

S /
/ /
/ /
/ /
+,/ /s
/ / 2
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/ /
//
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/

.II
/1 I

+/
/ ,,, i.,

/ ,1
/ / /

// /I
// /I
+ /- /,

/ //

/7 /

/ \ /
//

/ /

V \ \
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\ \
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\\
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\\
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Figure 4-8. Case 3: S2S3 intersects L'2 and ZS3S1S2 > 90.

Figure 4-7. Case 2: S2S3 intersects L12

y
12' /

L12 // 12

S /

\/ + \L
/\ \ 1

/ / \/ \ \\
/ / \ \ \
\ \ ~\
+ /,

I i \\ \

Figure 4-9. Case 4: S2S3 intersects L'2 and ZS3S1S2 < 90.

Case 4: S2S3 intersects L'2 and ZS3S1S2 < 90

As in Case 3, no portion of L12 is in region (e) (bottom left). Further, L13 may

overlap with either region (c) (bottom right) or region (d) (top) but not both. To see this,

suppose that L13 overlaps with region (c). For this to happen, L'3 must cross S2S3. Using

an argument similar to that used in Case 2, we may show that the slope of S2S3 is greater

than that of L'3. Furthermore, the remaining portion of L'3 once after crossing S2S3 lies

strictly below S2S3. So, no portion of L'3 is in region (d). Since LIM is to the left of SIS3,

no portion of LIM is in region (d) either. So, L13 may overlap only one of the regions (c)

and (d). Therefore, L12 and L13 cannot have an intersection in both region (c) and region

(d). Finally, if a portion of L12 is in region (d), region (d) is the first region encountered as

we move along L12 from left to right and if a portion of L12 is in region (c), then region (c)

is the last region encountered. A13 monotonically decreases as we move from left to right

along L12 inside regions (c) and (d) and monotonically increases in the remaining regions

that L12 overlaps. So, L12 and L13 have at most 2 intersections.

*

Theorem 15. L12 and L13 have at most 2 intersections.

Proof

Since, A23(P) A13(P) A12(P) for every point P, the hyperbola pairs (L12, L13),

(L12, L23), and (L13, L23) have the same set of intersections. Suppose, w.l.o.g., that the

source is closer to S2 than to S and S3. It follows from Theorem 14 that L21 and L23 have

at most 2 intersections. Hence, L12 and L13 have at most 2 intersections. m

4.4 Indistinguishable Points

When SS is not an ISS, there is at least one pair of distinct points that are

indistinguishable. That is, there are distinct points Pi and P2 for which Ayi(PI) A= A(P2),

1 < i < j < k (or equivalently, Alj(P1) = Alj(P2), 2 < j < k). P1 and P2 are dual

points. When SS is an ISS, no point P has a dual. In this section, we first show that the

indistinguishable relation is an equivalence relation. Then, we show that each point P may

have at most 1 dual point.

Theorem 16. The .':,,1.: ./i:,/,.;'.,,1l: relation is an equivalence relation on R2.

Proof

A relation is an equivalence relation iff it is reflexive, symmetric, and transitive.

Reflexivity is immediate as a point is indistinguishable from itself. Also, if P1 and P2

are indistinguishable then so also are P2 and P1. So, the relation is symmetric. For any

three points Pi, P2, and P3 such that Pi and P2 are indistinguishable and P2 and P3 are

indistinguishable, we have Ayi(Pi) = d(P1, S) d(P1, Sj) = d(P2, S) d(P2, Sj) Aij(P2)

and Aij(P2) d(P2, S) d(P2, Sj) = d(P3, S) d(P3, Sj) = A,(P3), 1 < i < j < k. So,

Ai(Pi) = d(PI, S) d~) d(P, 5() d(P3, S)- d(P3, Sj) = Ai(P3), 1 < i < j < k. Hence, the
',/ ', / 1, .//,,,, .1 .: relation is transitive.

Clearly, the .:,.1.:/I.:,, uishable relation partitions Euclidean space R2 into a collection

of di-i Piiil equivalence classes. If SS is an ISS, then each equivalence class is of unit

cardinality; otherwise, the cardinality of at least one equivalence class is more than 1.

S, Si Si+1 Sk /

*p'

Figure 4-10. Collinear sensors

When k = 2, each equivalence class corresponds to a hyperbola with foci S1 and S2

and vice verse. The cardinality of each equivalence class in this case is infinite. When

k > 2 and the sensors are collinear (Figure 4-10), each point on the line segment SISk,

exclusive of S1 and Sk, defines an equivalence class of unit cardinality because no such

point has a dual. All points on the line 1 that runs through the collinear sensors and that

are to the left (right) of SI(Sk), inclusive, form an equivalence class of infinite cardinality.

For each point P not on the line 1, has a single dual point PI that is the reflection of P

with respect to 1. Point P and its dual Pi define an equivalence class of cardinality 2.

When the sensors are not collinear (this can happen only when k > 2), Theorem 17

establishes that the cardinality of each equivalence class is at most 2.

Theorem 17. When the sensors are not collinear, the i ,,;
1, I;,u, 1 by the indistinguishable relation is at most 2.

Proof

We prove this by contradiction. Let SS be the sensor set. Suppose there is an

equivalence class whose cardinality is more than 2. Let PI, P2, and P3 be any three

points in this equivalence class. Since P1 and P2 are indistinguishable, from the proof of

Theorem 9, it follows that there is a hyperbola L12, whose foci are Pi and P2, that passes

through the points of SS. Similarly, there is a hyperbola L13, whose foci are P1 and P3,

that passes through the points of SS. L12 and L13 intersect at at least the points of SS,

which are more than 2 in number. This contradicts Theorem 15, which states that these

two hyperbola may have at most two intersections.

4.5 ISSs for Polygonal Regions

Although 4 properly positioned sensors are required to uniquely identify a source in

Euclidean space (Theorem 10), in many real-world applications, the monitoring region is

bounded by a polygon and 3 sensors suffice. We assume that the sensors are restricted

to be placed on or inside the bounding p" I.v.-on. As an aside, we note that when the

monitoring region is a simple line segment, w SiSj, then two sensors placed at Si and

Sj, respectively, are sufficient to uniquely identify any source on this segment. To see

this, observe that as we move P from Si to Sj along the line segment SiSj, Ayi(P) varies

monotonically from -d(Si, Sj) to d(Si, Sj). Hence, there is no pair of indistinguishable

points on this segment.

Lemma 18. Every non-degenerate simple ...1';;i/,i, has a MISS whose size is 3.

Proof

Case 1: The simple polygon is convex.

Let S1 and S2 be the end points of an edge of the p" I.v.-on. Let S3 be any other point

on this edge. Note that the 3 chosen points are collinear and the entire convex polygon

lies on one side of the edge that these 3 points lie on. From the discussion preceding

Theorem 17, it follows that the dual of every point of the polygon that is not on this edge

is on the other side of this edge. Points on the edge either have no dual or have dual(s)

outside the polygon. Hence every point in or on the polygon is uniquely identifiable and

{S1,S2, S3} is a size 3 MISS for the p" ..v.-on.

An alternative construction for a size 3 MISS is to consider any 3 non-collinear

points S1, S2, and S3 that are on the boundary of the polygon (Figure 4-11). Now, the

entire convex polygon must be contained in the union of four regions: (a) top left, (b)

inside, (f) bottom, and (g) top right. From Theorem 13, the directional derivative of A13

along L12 increases monotonically in each of these four regions. Further, the intersection

of L12 and the convex p" I.v.-on is a continuous curve C that is limited to these four regions

top left (+) ... '"
G

S2
bottom (+) D

bottom left (-) F
E

Figure 4-11. Sensors S', S2, and S3 on the boundary of a convex ]" '..i-on.

(see Theorem 14). Since, A13 is monotonically increasing along C, L12 and L13 have at

most one intersection on C. Hence, every point in or on the convex p" .Iv.-on is uniquely

identifiable.

Case 2: The simple polygon is concave.

We start with a a minimum bounding convex p" ..I -on of the concave p" ..I.-on

(Figure 4-12). Let S1, S2, and S3 be any three points on the intersection of the boundary

of these concave and convex polygons. From Case 1, it follows that every point in and on

the boundary of the convex bounding polygon, and so every point in and on the boundary

of the concave polygon, is uniquely identifiable.

In Lemma 18, we prove that by choosing 3 sensor locations on the boundary of a

simple ]" .Iv.-on, an SS of size 3 uniquely identifies any source S on or inside a simple

l" .v.-on. We show in Lemma 19 when a sensor is placed strictly inside a simple polygon, 3

sensors are not sufficient to uniquely identify every point in or on the p" .Iv.-on.

Lemma 19. Let SS be an ISS set for a non-degenerate simple y 'l, ';i', If at least one

location of SS is inside the p"'';,'i/'' ISS| > 4.

top left (+) ... I k '"
H

S2

bottom left (-) G bottom (+)

Figure 4-12. A concave polygon, its bounding convex 1 .i.-lon, and three sensors S1, S2,
and S3 placed on the common boundary of the concave and convex polygons

Proof

Suppose that SS is an ISS and that ISS| = 3. W.l.o.g, assume S1 lies inside the

simple ]" Ivl..on as shown in Figure 4-13. Note that a portion of the simple polygon must

lie inside the top region. We may choose two negative constants 612 and 613, such that

L12(612) and L13(613) intersect at two distinct points P1 in the top region and P2 in the top

left region. Since both P1 and P2 are inside the simple polygon and P1 is the dual of P2,

SS is not an ISS for the points of the simple polygon.

Theorem 18. 3 sensors can ;";,:'; ;/ ,:il,. ,lfy i,;, source in or on a non-degenerate simple

j' ..'.1;,i- iff the sensors are on the common boundary of the given y .', ..;;,-' and its minimum

bounding convex yI'..;; ,i-,, In case the 3 bo ,,]'..; i sensors are collinear, 2 must be at the

end points of an edge of the 7.. -,,.,.l, convex y','..,;;, and the third at an in-between point.

Proof

Follows from Lemmas 18 and 19.

*

top left (+) -

S2
bottom (+)

bottom left (-) F

E

Figure 4-13. S1 lies inside a simple i" ..v.-on while S2 and S3 are on the boundary. Pi in
the top region is a dual point of P2 which lies in the top left region.

CHAPTER 5
COMPUTATIONAL GEOMETRY METHOD FOR TRIANGULATION USING DTOA

We begin this chapter by examining the relationship between proximity in Euclidean

space and proximity in DTOA space. Although all previous localization methods have

focused on using proximity in DTOA space [45, 46], our analysis shows that this does

not guarantee proximity in Euclidean space. Next, we describe our geometric DTOA

triangulation method that guarantees proximity in both Euclidean and DTOA spaces.

This is followed by a detailed proof of the correctness of our method. We conclude this

chapter by providing simulation results.

5.1 Euclidean and DTOA Spaces

Recall that we let Si = (xi, yi), 1 < i < k, be the locations of k sensors in Euclidean

space R2. These locations are assumed to be distinct. For any point P = (x, y) E R2, the

distance, d(P, Si), between P and Si is V/(x xi)2 + (y yi)2. A signal that originates at

P at time 0 arrives at Si at time proportional to d(P, Si). For simplicity, we assume that

the arrival time is d(P, Si). The difference, Aiy, in the time of arrival (DTOA) at Si and

Sj is given by

A, (P) = d(P, Si) d(P, Sj).

Let SiSj be the line through the points Si and Sj. As we move P from Si to Sj along

the line SiSj, Aij(P) varies monotonically and linearly from -d(Si, Sj) to d(Si, Sj), and

equals 0 at the bisector point. From this observation and the triangle inequality, it follows

that lay (P) I < d(Si, Sj). Furthermore, the locus, Lij(5), of points defined by

Lj() = {P E R2Aij(P) 6

is a hyperbola1 (see Figure 4-1).

The DTOA space of all (k 1)-tuples [A12(P), A13(P),..., Alk(P)] forms a (k -

1)-dimensional vector space denoted by 6k-1. Each point P = (x, y) E R2 has a unique

dual point Pi = (plf,p2t, ...,pk-1') in 6k-1, where p/ Al(j+i)(P), j = 1, 2,..., (k 1).

However, each point Pi/ (pl,p2t, *...,k-1) in 6k-1 may have zero or more dual points

in R2. In fact, the dual points of Pi are those points in R2 that are common to (i.e., the

common intersections) the k 1 hyperbolas LIjj+(P), 1 < j < k.

In this chapter, we consider the DTOA localization problem of estimating the location

of a source S from the measurements of Aij(S), 2 < j < k, SI = [612, ..., 61k] in 6k-1

When there is the possibility of errors in the measurement of the Alj values, existing

DTOA localization algorithms [14, 39, 57, 58], estimate the source location by minimizing

the sum of least squares error in 6k-1 space. We show, in this section, that an estimate

that is close in 6k-1 space may not be close in Euclidean space R2. However, an estimate

that is close to the source in Euclidean space R2, is necessarily close to the source in 6k-1

space (Lemmas 20 and 21, respectively).

Lemma 20. Two points that are close to one another in 6k-1 space I,,in be ,i.:l/,i.:l;; far

apart in R2.

Proof Consider the three sensors S'(1, 0), S2(1, -1), and S3(-, 0) ((1,0) is the location

in R2 of sensor S'). Let 612 = -d(S1, 52) = -1 and 613 = -2e, where e is a small positive.

The hyperbolic equation for L12(612) is x = and y >= 0. In other words, L12(612)

degenerates from a hyperbola to a ray from S1 vertically up to infinity. The hyperbolic

equation for L13(613) is 2/(e2) y2/(1 e2) = 1 where x > 0. The intersection, P, of

L12(612) and L13(613) is (1, (1 e2)/).

1 Strictly -I'" 1:;ii:- Lij(6) is one branch of a hyperbola and Lyi(-6) is the other branch.
As mentioned earlier, for convenience, in this dissertation, we use the term hyperbola to
refer to one branch of a hyperbola.

Now, suppose we change the value of 513 to -4e. The hyperbolic equation for the

new L13(613) is x2/(4e2) y2/(1 4e2) 1 where x > 0 and the new intersection,

Q, between L12 and L13 is (1,1 4e2)/(2e)). The distance between P and Q in 62

space is 2e. However, the distance, d(P, Q), between P and Q in Euclidean space R2 is

|(1 4e2)/(2e) (1 e2)/el e + 1/(2e). As can be seen, d(P, Q) becomes large as the

distance in 52 space approaches zero. m

Lemma 21. Given two points P(xp,yp) and Q(xq,yq) in Euclidean space R2 and their

respective dual points Pt(x,/, y,/) and Qf(xq,, yq,) in 62 space, then we have d(Pi, Qi) <

2 /2 d(P, Q)
Proof From the definition, we have xP = A12(P) = d(P, S1) d(P, S2) and Xq =
A12(Q) d(Q, SI) d(Q, S2).

IXpl Xql| = I(d(P, Si) d(P, S2)) (d(Q, Si) d(Q, S2))|
I(d(P, Si) d(Q, Si)) (d(P, S2) d(Q, S2))l
< Id(P, Q) (-d(P, Q))I
S2 d(P, Q)

Similarly, we have lYp yql < 2 d(P, Q) as well. So,

d(PI, Ql) = /(x Xq)2 + (ypf- Yq)2 < 22 d(P, Q)

Corollary 5. Given two points P and Q in Euclidean space R2 and their respective dual

points PI and QI in DTOA space 6k, d(Pi, QI) < 2vk d(P, Q).

5.2 Geometric DTOA Method

In the remainder of this chapter, we consider only the case when we have k = 3

sensors, S1, S2, and S3. Without loss of generality (w.l.o.g.), we choose our coordinate

system so that the line S1S2 falls on the y-axis and so that the midpoint of this line is the

origin O as shown in Figure 5-1. The [Dx1, Dx2] x [Dy1, DY2] box shown in Figure 5-1 is

the monitoring region within which the source S is to be localized. The lines S1S2, S2S3,

and S1S3 partition the monitoring region as shown in Figure 5-1. Although this figure has

Figure 5-1. Canonical placement of 3 sensors and partitioning of monitoring region
all sensors within the monitoring region, our development of the geometric localization

method does not require this. In fact, the method works even when some or all of the

sensors are outside the monitoring region.

Figure 5-2 shows our three sensors together with the locus L12(612). This locus

may be partitioned into segments that lie wholly within a region of the partitioning of

Figure 5-1. The segment end points are designated Sj, where j is a lowercase letter.

So, SaSb and SbSc are two of the segments that L12 is partitioned into in Figure 5-2.

Notice that because of our choice of coordinate system, as we move a point P along any

segment of L12(612), the x- and y-coordinates of the point vary monotonically. This is

a consequence of the vertical orientation of L12, which, in turn, is assured by the chosen

coordinate system.

Let (xi, yi) and (xj, yj), xi < xj be the end points of an L12 segment and let P

(x, y) be any point on this segment. From Lemma 22 (Section 5.3), it follows that xi <
x < xj and min{y, yj} < y < max{yj, y}. Also, as we move P along a segment

of L12(612), A13(P) varies monotonically (Section 5.4). In particular, it monotonically

Figure 5-2. Canonical placement of 3 sensors and L12(612)

decreases with x for the segments in the top, bottom left, and bottom right regions and

monotonically increases for the remaining segments. Based on these key observations,

our overall strategy to estimate the source S is to utilize the monotonicity of A13(P) to

perform a binary search within each segment of L12 to determine a set, U, of points such

that U has at least one point within a specified accuracy 7 of each intersection between

L12(612) and L13(613) that is in the monitoring region. Further, the number of points in

U is at most equal to the number of such intersections. Recall that the number of such

intersections is at most 2 as proved in Theorem 15. It follows that the true source location

is within a distance 7 (in R2) of one of the points in U. The details are presented in

algorithm geometric_DTOA(612, 613).

Algorithm geometricDTOA first determines the segments of L12. The end points

of these segments are just the intersections of the curve L12(612) with each of the three

lines S1S2, S1S3, and S2S3. Although a line and a hyperbola may intersect twice (except

in the degenerate case when the hyperbola is a vertical ray), our choice of coordinate

system ensures that L12 intersects S1S2 exactly once, except when L12 is a ray. We ignore

this case when L12 is a vertical ray for now. So, the number of intersections is at most

il.. ':thm geometric_DTOA((12, 613);
begin
(x12, Y12) -- intersection point of L12 with S1S2;
Ixi <- set of x-coordinates of intersections of L12 with S1S3;
Ix2 s- set of x-coordinates of intersections of L12 with S2S3;
Ix {Dxl, x12, Dx2}U IxU Ix2;
Ix Ix {xlx c Ix && (x < Dxi II x > Dx2)};
Sort sort (Ix);
let I sort {X(1), X(2), ... X |, l }
let {Y(1), y(2), Y* Ysort, } be the corresponding y-coordinates;
U <-- 0;
for i= 1,... Lsor.t 1 do
U -- U U{ locateL13 x(i), y (i) (i+l), Y(i+l));
return U;
end

5 and, in the worst case, we need to consider 6 segments of the hyperbola L12(612). The

computation of (x12, Y12), Ixi and IX2 may be carried out either by binary searches on

the lines S1S2, S1S3, and S2S3 with L12 as objective function (as in [46]) or by a method

similar to that used in il.., .:thm locate_L12(x, YL, YR) 2 Note that intersections outside

the monitoring region may be ignored.

Next, a binary search is performed within each segment, as shown in il.>rithm

locate_L13 (L, YL, XR, YR). If 613 is not in the range [Ami, A,,x], the algorithm concludes,

from the monotonicity property, that there is no point P with A13(P) 613 on the

segment currently being searched. Otherwise, the continuity of the directional derivative of

A13 implies that there is a point P on the segment for which A13(P) 613 and a binary

search, as described in the do-until loop of il, .rithm locate_L13(xLL, L,R, YR), to locate

a point on L12 that is within 7 of P in R2. In each iteration, either the x- or y-range to be

2 Algorithm locate_L12 can be augmented to detect the case when the quadratic
equation in step 2 does not yield a real solution and resort to the binary search of [46].
Such an extension ensures the theoretical completeness of the method albeit at an
additional complexity of O [(log(1/7))2]. However, in all of our simulations of Section 5.5,
the quadratic equation of step 2 yielded a real solution. Hence, we retain the unaugmented
version for simplicity of presentation.

considered is halved by appropriately updating Pi or P2. As proved in Theorem 19, our

algorithm guarantees to return a point that is within a distance 7, in R2 space, of the true

source location.

li.. 'rithm locate_L13 (XL, YL, XR, YR);
begin
Xl XL;
Yi YL;
PI =(XL, YL);
X2 <-- XR;
Y2 --R;
P2 = (XR, YR);
Amin = min{A13(P), A13(P2)};
Amax = max {A13(P), A13(P2)};
if (Amin > 613) or (Aax < 613) then
return(null);
do{
if IX1 x21 > ly Y2 then
x -- (xl + x2)/2;
y -- locate_L12(x, YL, YR);
else
y ~ (y + y2)/2;
x -- locateL2(y, XL, xR);
P (x, y);
if (A13(Pi) 613) (A3(P) 13) > 0 then
P,= -P;
else
P2 P;
} until dist(P, P2) <7
return(P);
end

When L12 is a vertical ray, the source lies on the line S1S2 but outside the segment

S1S2, whose end points are S1 and S2. In this case, we may do a binary search on the

il,.>,rithm locate_L12(x, YL, YR);
begin
substitute x into the hyperbolic equation for L12(612);
solve the quadratic equation for y;
return the solution that is in the range [min{yL, YR}, max{yL, YR}].
end

relevant segment of the y-axis that is contained in the monitoring region and excludes

either the segment from S1 to -oo or the segment from S2 to oo. As shown in Section 5.4,

L13 is monotone on both these vertical segments.

5.3 Correctness and Complexity of the Method

In this section, we establish the correctness of our geometric DTOA method subject

to the monotonicity of L13 on each segment of L12. This monotonicity property is

established in Section 5.4. The following assumes that L12 is not a vertical ray. The

correctness proof for the case when L12 is a vertical ray (note that this case, which is not

included in the statement of algorithm geometricDTOA, is handled by a binary search on

a segment of the y-axis) is similar and simpler.

Lemma 22. As ;,. ;, move i1.. 'u each segment of L12(12), the x-coordinate (y-coordinate)

monoto,:d. ,ll:; increases or decreases.

Proof Follows from the definition of a segment and our choice of coordinate system. *

Lemma 23. For ,:,' point P on a segment SSj of L12(612),

max{d(Si, P), d(Sj, P)} < d(Si, Sj).

Proof Let Si = (xi, y), Sj = (xj,yj), and P = (x,y). W.l.o.g., we may assume that the

segment is oriented so that xi < xj. From Lemma 22, we have xi < x < xj and yi < y < yj

(or y < y < yi). So, max{xi- x|, Ix- Xj} < Ix,- Xjl and max{|ly, y, ly- yj} < lyi yj.
Hence, max{d(S, P),d(Sj, P)} < \/max{|x x1, x Xj\}2 + max{ly yl, y yj\}2 <

d(Si, Sj). m

Lemma 24. Let P = (x, y) be a point on a segment SiSj of L12(612) such that A13(P)

613. The search of this -i* uit, using i,'l',.:thm locate_L13 returns a point P on L12(612)
such that d(P, P) < 7, where 7 is the desired '.. ;,.'. ,;

Proof (XL, YL) and (xR, yR) are the end points of the segment SiSj. Since A13 is

monotone on this segment and P is on the segment, 613 is in the range [Amin,Amax]. So,

the binary search described in the algorithm is performed. The original search rectangle

is determined by point Pi = (XL, yL) and P2 (xR, R). In each iteration, we chop the x-

or y-range, whichever is larger, of the search rectangle into half and choose the half that

contains P as the new search rectangle by updating P1 or P2 accordingly. This basic step

is repeated until the Euclidean distance between P1 and P2 is no more than 7. From this

and Lemmas 22 and 23, it follows that d(P1, P) < 7 and d(P2, P) < 7. The lemma now

follows from the observation that the point P returned by the algorithm is either P1 or P2.

Theorem 19. The set of points U returned by il/.'rithm geometricDTOA contains at

least one point that is within 7 of each intersection between L12(012) and L13(613) that is

in the monitoring region and the number of points in U is at most equal to the number of

such intersections in the monitoring region. Hence, at least one point of U is within 7 of

the true source location provided this location is in the monitoring region.

Proof The theorem follows from Lemma 24 and the observations (a) every segment (or

segment portion) of L12(012) in the monitoring region is searched, (b) every intersection

within the monitoring region is on exactly one of the segments, of L12, and (c) algorithm

locate_L13 returns at most one point per intersection. U

Note that the points in the set U returned by algorithm geometricDTOA are on

the locus L12(612). So, for each point P E U, A12(P) 612. Since each returned point

P E U is within 7, in R2 space, of an intersection of L12(612) and L13(613), it follows

that A13(P) < 2v/27 (Lemma 21). By changing the condition on the binary search loop

of algorithm locate_L13, we can ensure that the returned points are within a specified

tolerance of intersection points in 62 space or within specified tolerances in both R2 and 62

spaces.

The set Isot may be computed in 0(1) time. Let 1 = max{Dx2 Dx1, Dy2 Dy1}.

In computing U, there are altogether up to 6 calls to locate_L13(xL, YL, XR, YR). Since the

number of intersection points is at most 2 as shown in Theorem 15, at most 2 such calls

make O(log(l/7)) calls to locate_L12(x, L, YR), which in turn can be done in 0(1) time.

Thus the complexity of algorithm geometricDTOA is O(log(1/7)), which can be adapted

by suitably specifying 7. If the number of basic computational operations is fixed at c,

then we have 7 < 0(1 2-c). We note that the inclusion of the case when L12 is a vertical

ray does not change the .,-i-1,1i, il ic complexity of our algorithm.

5.4 Monotonicity of Directional Derivative

In this section, we establish the monotonicity of the directional derivative of A13 on

each segment of L12(612) 3 We do this first for the case when L12 is not a vertical ray.

For this case, we consider explicitly each of the seven regions: (a) top left, (b) inside, (c)

bottom right, (d) top, (e) bottom left, (f) bottom, and (g) top right as shown in Figure

5-1. We show that the directional derivative of A13(.) along the curve L12(.) is monotone

in each of these regions: it is positive in regions (a), (b), (f), and (g) and is negative in

regions (c), (d), and (e).

We have for i = 1, 2, 3,

9d(P, S) (x xj) 9d(P, S) (y yi)
dx d(P, S) ya d(P, S)'

Also, the tangent vector to L12(612) at P = (x, y) is given by

Az12 (P)
ay
6a12 (P)
ax

3 This method is more direct than using the directional derivative of A13 on the
gradient of L12(612) as in [46].

So, the directional derivative of A13(P) at P on the locus L12(612) {= 1P 2(P)

612}, for any 612, is given by

-T
dA13(P) dA12(P)
ax 0 y
A13 (P) DAi2(P)
jy ax
T
x-x1 x-x3 Y-y1 Y-Y2
d(P,Si) d(P,S3) d(P,Si) d(P,S2)
Y--1 Y--Y3 x-x x-x2
d(P,Si) d(P,S3) d(P,Si) d(P,S2)

We note that some authors define the directional derivative by doing an inner product

with a unit tangent vector rather than with any tangent vector. If we wish to conform to

this definition, we must divide the directional derivative as given by the above expression

by the quantity ( + (d (P)) Since we are interested only in the sign of the
S a Dx ay t

directional derivative, it doesn't matter which of the two definitions we use. We continue

with the simpler definition that does not require the use of a unit tangent vector.

We use the following three basic identities extensively in our derivations:

sin a sin = 2 sin cos ( --

sin a + sin = 2 sin a + cos -
S2 (2

cos a + cos = 2cos cos
2 2

5.4.1 Top Left Region

In this case, we have 0 < 71 + 72 < 7, 0 < 71 + 73 < 7, and 73 > 72 as shown in Figure

5-3. The directional derivative is given by

-T
X--X21 X3 Y--Yi Y-Y2
d(P,Si) d(P,S3) d(P,S) d(P,S2)
Y-Y1 Y--Y3 x--1 x-x2
d(P,S) ,3) dP,) d(P,S) P,S2)
S (- sin 7i + sin 73)(cos 7i + cos 72)

+(cos 7i + cos 73) (sin 71 sin 72)

- sin(7i + 72) + sin(7i + 73) + sin(73 72)

71 72 +273 71+72 .
2 sin 71-72+273 cos -7172 sin(7i + 72)

2 cos 71 72

sin. 71 -72+273 71+721
2 2 1
4 C(71 7+72 7Y3 2 1 +73
4 cos sm 2- cos
2 2 CO 1+Y2

We have 0 < 71 +72 < 7 which makes the first cos term positive. We have 73 > 72 and

0 < 73 < 7. Thus 0 < -3 2 < 7/2, which makes the second sin term positive. We have

0 < 71 + 73 < 7, which makes the third cos term positive. Hence the directional derivative

is positive.

bottom

Figure 5-3. P = (x, y) is located in the top left region.

Figure 5-4. P

Figure 5-5. P

bottom

(x, y) is located inside the triangle.

bottom

(x, y) is located in the bottom right region.

bottom

(x, y) is located in the top region.

Figure 5-6. P

5.4.2 Inside Region

In this case, we have 0 < 72 + 73 < 7T, and 73 > 71 as shown in Figure 5-4. The

directional derivative of A(SI, S3) on the locus {(x, y) |A( SI, S2) 12}, for any 612, is

given by

T
X--21 X-3 Y-Y1 Y-Y2
d(P,Si) d(P,S3) d(P,SI) d(P,S2)
Y-Y1_ Y-Y3 x--1 x--2
d(P,S1) d(P,S3) d(P,S1) d(P,S2)
S (sin 71 + sin 73)(cos 71 + cos 72)

+(- cos 7i cos 73)(sin 71 sin 72)

Ssin(in + 72) + sin(3 71) + sin2 + 73)

n. 72 +73 cos 271 + 72 73 os71 +72
= 2 sm cos + cos ^ )
2 2 2
4sin7273 COS7172 C 71 73
2 2 2

We have 0 < 72 + 73 < 7r, which makes the first sin term positive. Since 73 > 71, we have

0 < 71 + 72 < 7r, which makes the second cos term positive. Since 73 > 71 and 0 < 73 < r,

we have -r/2 < 712 < 0, which makes the third cos term positive. Hence the directional

derivative is positive.

bottom

Figure 5-7. P = (x, y) is located in the bottom left region.

Figure 5-8. P = (x, y) is located in the bottom region.

5.4.3 Bottom Right Region

In this case, we have 0 < 71 + 73 < 7T and 73 > 72 as shown in Figure 5-5. The

directional derivative of A(SI, S3) on the locus {(x, y) lA(SI, S2) =12}, for any 612, is

given by
T
x--x1 2x-3 Y--Y Y--Y2
d(P,S1) d(P,S3) d(P,S) d(P,S2)
Y-Yi_ Y--Y3 x-x x-x-2
d(P,Si) d(P,S3) d(P,Si) d(P,S2)
S(sin 71 sin 73) (cos 71 + COS 72)

+(- ccos cos 73) (sin 71 sin 72)

Sssin(7i + 72) in(1 3)+ sin(72 73)

sin72 73) cos( 27 + 72 73 +co 72 -73
-2 22 2 )1

S4 sin 2-73 COS 1 72 COS 1 73
2 2 2

Since 73 > 72 and 0 < 73 < 7, we have -T/2 < 223 < 0, which makes the sin term

negative. We have 0 < 71 + 72 < 7 and 0 < 71 + 73 < 7r, which makes the last two cos

terms positive. Hence the directional derivative is negative.

5.4.4 Top, Bottom Left, Bottom, and Top Right Regions

In the following, we briefly show the directional derivative is monotone in each of the

remaining four regions:

bottom

Figure 5-9. P = (x, y) is located in the top right region.

* Top: The case of top is identical to the top left region except that 7r < 71 + 73 < 27
as shown in Figure 5-6, which makes the third cos term negative, and hence the
directional derivative is negative.

* Bottom Left: The case of bottom left is identical to the top left region except that
72 > 73 as shown in Figure 5-7, which makes the sin term negative, and hence the
directional derivative is negative.

* Bottom Region: For bottom region, the derivation is identical to the case of inside
region except that 7 < 72 + 73 < 27 as shown in Figure 5-8, which keeps the first sin
term still positive, and hence the directional derivative is positive.

* Top Right: The case of top right region, as shown in Figure 5-9, is identical to
inside region except that 73 < 71. Thus we have 0 < 713 < /2, which makes the
third cos term still positive, and hence the directional derivative is positive.

Computational results indicating the signs of the directional derivative of randomly

generated sources are shown in Figure 5-10.

When L12 is a vertical ray, we need to consider the portion of the segments (a) from

Si to oo and (b) from S2 to -oo that lie within the monitoring region. We consider only

(a). The proof for (b) is similar. Let P1 and P2 be two points on the segment (a). W.l.o.g.,

assume that Pi is closer to S1 than is P2 (see Figure 5-11). We see that

negative

O S3 negative

positive

E S2

negative

-10 -5 0 5 10

Figure 5-10. Source S = (x, y) is randomly selected, and the sign of the directional
derivative is computed.

Figure 5-11. The degenerate case when L12(612) is a vertical ray

Table 5-1. Data for S1 = (0,0), S2 (0,50000), S3 = (0.001,100000), and F=0
S = (0, 0), S2 = (0, 50000), S3 (0.001, 100000), F 0, and N=12635
space R?2 space
Method #fail ratio space R
Y1 #count ratio 72 #count ratio
100 1328 0.1051 100 722 0.0571
500 4683 0.3706 500 2384 0.1887
1000 7698 0.6093 1000 4004 0.3169
Mellen 161 0.0127
2500 11503 0.9104 2500 7487 0.5926
5000 12420 0.9830 5000 10512 0.8320
10000 12473 0.9872 10000 12249 0.9694
0.00000001 2619 0.2073 0.00000001 246 0.0195
Ours 0 0.0 0.0000001 12011 0.9506 0.0000001 2273 0.1799
0.000001 12635 1.0 0.000001 12635 1.0

Table 5-2. Data for S1 = (0, 0), S2 = (0, 50000), S3 = (0.0000000000000001, 100000), and
F=0
S1 (0,0), S2 (0, 50000), S3 (0.0000000000000001, 100000), F 0, and N 12345
S6 space R2 space
Method #fail ratio space s
71 #count ratio 72 #count ratio
100 60 0.0049 100 3 0.00024301
500 134 0.0109 500 16 0.0013
1000 210 0.0170 1000 30 0.0024
2500 378 0.0306 2500 88 0.0071
Mellen 3426 0.2775
5000 572 0.0463 5000 200 0.0162
10000 935 0.0757 10000 421 0.0341
50000 6652 0.5388 50000 2070 0.1677
100000 8919 0.7225 100000 4037 0.3299
0.00000001 1437 0.1164 0.00000001 225 0.0182
Ours 0 0.0 0.0000001 10546 0.8543 0.0000001 2185 0.1770
0.000001 12345 1.0 0.000001 12345 1.0

A13(P) 13(P2) (d(Pl, S)-d(P1,S3))

-(d(P2, S) d(P2,S3))

(d(Pi, S) d(P2, S1))

-(d(P1,S3) d(P2,S3))

-d(P1, P2) d(P, S3) + d(P2, S3)

< 0 (from the triangle inequality)

Hence, the directional derivative of L13 on segment (a) is monotone.

Table 5-3. Data for S1 = (0,0), S2 (0,50000), S3 = (0.0000000000000001, 100000), and
F=10/100
S1 =(0,0), S2 = (0,50000), S3 = (0.0000000000000001,100000), F 10/100, and N=12598
6 space R2 space
Method #fail ratio space
71 #count ratio 72 #count ratio
10000 907 0.0720 10000 276 0.0219
25000 2455 0.1949 25000 861 0.0683
Mellen 3944 0.3131
50000 8034 0.6377 50000 1922 0.1526
100000 11873 0.9425 100000 3992 0.3169
100 25 0.0020 100 0 0.0
1000 1183 0.0939 1000 258 0.0205
Ours 2149 0.1706 2500 4495 0.3568 2500 1067 0.0847
5000 9273 0.7361 5000 2483 0.1971
10000 12481 0.9907 10000 5097 0.4046

Table 5-4. Data for S = (0,0), S2 (0,50000), 3 S (0.0000000000000001, 100000), and
F 5/100
S1 =(0,0), S2 = (0, 50000), S3 = (0.0000000000000001,100000), F-5/100, and N=12341
6 space R2 space
Method #fail ratio space
71 #count ratio 72 #count ratio
10000 901 0.0730 10000 304 0.0246
25000 2230 0.1807 25000 877 0.0711
Mellen 3708 0.3005
50000 6382 0.5171 50000 1881 0.1524
100000 8721 0.7067 100000 3978 0.3223
100 78 0.0063 100 9 0.00072928
1000 3048 0.2470 1000 672 0.0545
Ours 1484 0.1202 2500 8945 0.7248 2500 2129 0.1725
5000 12236 0.9915 5000 4544 0.3682
10000 12341 1.0 10000 8029 0.6506

Table 5-5. Data for S1 = (0,0), S2 = (0,50000), S3 = (0.0000000000000001, 100000), and
F=1/100
S1 = (0,0), S2 = (0, 50000), S3 = (0.0000000000000001,100000), F 1/100, and N=12599
6 space R' space
Method #fail ratio space s
I7 #count ratio 72 #count ratio
10000 951 0.0755 10000 409 0.0325
25000 2185 0.1734 25000 1063 0.0844
Mellen 3513 0.2788
50000 6829 0.5420 50000 2092 0.1660
100000 9109 0.7230 100000 4140 0.3286
100 840 0.0667 100 156 0.0124
250 3693 0.2931 250 760 0.0603
500 8594 0.6821 500 1911 0.1517
Ours 650 0.0516 1000 12480 0.9906 1000 4164 0.3305
2500 12599 1.0 2500 8846 0.7021
5000 12599 1.0 5000 11691 0.9279
10000 12599 1.0 10000 12577 0.9983

Table 5-6. Data for S1 = (0, 0), S2 = (0,50000), S3 = (5000, 100000), and F-

S(0, 50000), S3

Method #fail ratio
'1
250
500
1000
Mellen 1649 0.1322 00
2500
5000
10000
250
500

Ours 1647

1000
0.1320 2500
5000
10000

:(5000,100000), F-

6 space
#count
52
189
725
3404
7604
10628
78
287
1026
4237
9093
12389

ratio
0.0042
0.0152
0.0581
0.2729
0.6096
0.8521
0.0063
0.0230
0.0823
0.3397
0.7290
0.9933

10/100, and N=12473

72
250
500
1000
2500
5000
10000
250
500
1000
2500
5000
10000

R2 space
#count
19
58
215
932
2258
4804
19
58
215
932
2364
5378

ratio
0.0015
0.0047
0.0172
0.0747
0.1810
0.3852
0.0015
0.0047
0.0172
0.0747
0.1895
0.4312

Table 5-7. Data
S1= (0, 0), S2

for S= (0,0),
S(0, 50000), S3

S2 (0,50000), S3 = (5000,100000), and F-
= (5000, 100000), F-5/100, and N-12591

Method #fail ratio
'1
250
500
1000
Mellen 1007 0.0800 00
2500
5000
10000
250
500
1000
Ours 1007
0.0800 2500
5000
10000

6 space
#count
228
798
2536
8183
11502
11579
258
921

9045
12514
12591

ratio
0.0181
0.0634
0.2014
0.6499
0.9135
0.9196
0.0205
0.0731
0.2292
0.7184
0.9939
1.0

72
250
500
1000
2500
5000
10000
250
500
1000
2500
5000
10000

R2 space
#count
51
238
726
2468
5009
2"'".,
51
238
726
2468
5116
8776

ratio
0.0041
0.0189
0.0577
0.1960
0.3978

0.0041
0.0189
0.0577
0.1960
0.4063
0.6970

5.5 Simulation Results

We compared the performance of our binary search algorithm of Section 5.2 versus

the linear algebra method of [34, 53], which requires a solution to a quadratic equation

as well as the inversion of matrices. Both algorithms were implemented in Matlab on a

Dell Dimension PC with a 2.13 GHz dual-core processor and 2 GB memory. The typical

execution times of both methods are only several milliseconds.

S= (0, 0), S2

-5/100

10/100

Table 5-8. Data for S1 = (0,0), S2 (0,50000), S3 = (5000,100000), and F-

(0,50000), S3

ratio

<71
100
250
500
Mellen 267 0.0211 00
1000
2500
5000
100
250
Ours 267 0.0211 500
1000
2500
5000

Table 5-9.

Data for S1

(5000,100000), F=1/100, and N=12683

SI (0,0), S2

Method #fail

ratio
0.0594
0.2824
0.6670
0.9727
0.9784
0.9784
0.0599
0.2879
0.6837
0 -1-1 1 ;
1.0
1.0

72
100
250
500
1000
2500
5000
100
250
500
1000
2500
5000

R2 space
#count ratio
219 0.0173
965 0.0761
2374 0.1872
5030 0.3966
9636 0.7598
11926 0.9403
219 0.0173
965 0.0761
2374 0.1872
5030 0.3966
9636 0.7598
12008 0.9468

(0,0), S2 (0,100000), 3 (100000,0), and F

S1 (0,0), S2 = (0,100000), S3 = (100000,0), F=10/100, and N=12518

Method #fail ratio
71
500
1000
Mellen 20 0.0016 2500
5000
10000
250
500
1000
Ours 0 0.0
2500
5000

8 space

#count
12
39
182
624
1875
119
193
725
3460
8504

ratio
0.00095862
0.0031
0.0145
0.0498
0.1498
0.0095
0.0154
0.0579
0.2764
0.6793

10000 12511 0.9994

72
500
1000
2500
5000
10000
250
500
1000
2500
5000

R' space
#count
435
1520
6270
10909
12449
119
435
1520
6272
10917

ratio

0.1214
0.5009
0.8715
0.9945
0.0095

0.1214
0.5010
0.8721

10000 12465 0.9958

Table 5-10. Data for S1 = (0,0), S2 -
S'1 (0,0), S2 = (0,100000), S3

S(0,100000), S3
(100000,0), F

6 space
Method #fail ratio space
71 #count
500 44
1000 112
Mellen 4 0.00032
2500 542
5000 1542
250 183
500 680
Ours 0 0.0 1000 2392
1000 2392
2500 8494
5000 12478

ratio
0.0035
0.0090
0.0434
0.1235
0.0147
0.0545
0.1916
0.6804
0.9996

= (100000,0), and F-
5/100, and N-12483

72
500
1000
2500
5000
250
500
1000
2500
5000

=5/100

R2 space
#count ratio
1516 0.1214
4729 0.3788
11042 0.8846
12459 0.9981
415 0.0332
1516 0.1214
4729 0.3788
11043 0.8846
12463 0.9984

6 space
#count
753
3582
8459
12337
12409
12409
760
3652
8672
12611
12683
12683

o10/100

1/100

Table 5-11. Data for S1 = (0,0), S2 (0,100000), S3 = (100000,0), and F=1/100
S = (0,0), S2 (0,100000), S3 = (100000, 0), F=1/100, and N=12398
6 space R2 space
Method #fail ratio 6 spae
71 #count ratio 72 #count ratio
250 149 0.0120 250 6337 0.5111
500 500 0.0403 500 11029 0.8896
Mellen 0 0.0
1000 1382 0.1115 1000 12394 0.9997
2500 1393 0.1124 2500 112:'' 1.0
100 708 0.0571 100 1542 0.1244
250 3432 0.2768 250 6337 0.5111
Ours 0 0.0 500 i :. 0.6804 500 11029 0.8896
1000 12394 0.9997 1000 12394 0.9997
2500 11:2;' 1.0 2500 11:2;' 1.0

Each sensor measurement corresponds to (1 + f)r where r is the actual distance

from sensor to source, and f is uniformly randomly generated in the interval [0, F] for

a fixed multiplicative factor F. While f values are generated independently, sensor

error magnitude is proportional to the distance from the sensor to plume origin. Also,

the sensor errors are correlated due to the spatial relationships between the sensor

locations, a source close to one sensor generates a small error there and larger errors at

other sensors, which are located farther away. From these measurements, we computed

distance-differences and tested DTOA localization methods. In our experiments, We

considered two different scenarios: (1) sensor errors are zero (i.e., F = 0), and (2) sensor

errors are greater than zero (i.e., F > 0). On a related note, the method of [7] accounts for

random errors that are independent Gaussian, and hence is not directly applicable to this

case.

Our simulation was conducted in a network of three sensors on a [0, 100000] x

[0, 100000] grid, where location of sources are randomly generated based on the uniform

distribution.

5.5.1 F 0

We compare the performance of both methods in case that all sensor measurements

are accurate. When three sensors form a "good triangle", the method of [34, 53] may

accurately estimate the source location as shown in [46]. By good triangle, we mean its

smallest (largest) angle is not close to 0 (180) degree. However, when three sensors lie in

an almost collinear manner, the method of [34] may fail to find a solution either because

the quadratic equation has imaginary roots or because the matrices being inverted are

close to singular. Although our method also could potentially fail because the roots of

the quadratic being solved by Algorithm locateL12 are imaginary, this did not happen

in any of our simulations (Tables 5-1 and 5-2). For our experiments, each test case may

be described by a tuple of [S1, S2, S3, F, N], where S1, S2, and S3 are coordinates of

three respective sensors, F is the sensor error, and N is the number of randomly generated

sources. Note that we ahv-,-i keep S1 closest to the source. Table 5-1 gives the number of

sources such that [34] returns imaginary roots as well as our method fails to find a solution

where S1 = (0, 0), S2 = (0, 50000), S3 = (0.001, 100000), and N = 12635. The ratio of the

number of such sources against the total number of sources is given. For each test case,

we consider various 71 and 72, where 71 and 72 are the desired errors acceptable in 6 space

and R2 space, respectively. For each 71(72), Table 5-1 gives the number of sources whose

estimate returned by [34] as well as by our method is within the desired error l1(72) of

the actual source. The ratio of the number of such sources to the total number of sources

also is given. Table 5-2 gives this data for the case where S1 = (0, 0), S2 = (0,50000),

S3 = (0.0000000000000001, 100000), and N = 12345.

We note that using our binary search based method versus that of [34] had a great

impact on the number of sources that could be estimated. For example, for the two test

cases shown in Tables 5-1 and 5-2 the percentage of sources for which the method of [34]

failed to return an estimate (either because the quadratic being solved had imaginary

roots or because of failure to invert matrices) is 1."7'. and 27.7.'. respectively, whereas

our method never failed to estimate the source. Note that almost 2-'. of sources can't be

estimated by the method of [34] in the second test case. Further, the source estimate given

by our method also shows much better accuracy in both 6 space and R2 space than that

of the method of [34]. As shown in Table 5-1, to get the ratio of successful estimates to be

more than I'- in 6 space, the method of [34] needs to set 71 to be almost 5000, whereas

our method alv--, gives the successful estimate when 71 is as small as 0.000001! A similar

phenomenon is observed in R2 space as well. When 72 is set to 10000, the success ratio

of the method of [34] is still slightly less than 97' whereas our method achieves 1011' .

success even if we reduce 72 by as much as about 1010 times! This improvement is even

more impressive for the second test case shown in Table 5-2, where the estimating quality

is improved by more than 1011 times!

Another observation is that an estimate that is accurate in 6 space may not be

accurate in R2 space. For example, in Table 5-1, when 71 and 72 are both 1000, the ratio

of successful estimates in 6 space and R2 space is 60.9 ;' and 31.1.'1., respectively, which

implies that more than 2' i' of the estimates that are close to the source in 6 space are

distant from the source in R2 space.

5.5.2 F > 0

When sensor measurements are inaccurate, our method may fail to estimate the

source because the L12 and L13 curves described by the inaccurate measurements do

not intersect. This cause for failure is in addition to the possibility that the quadratic

being solved by Algorithm locate_L12 has imaginary roots. To overcome this additional

cause, a finalization step is added to the end of the original description of our method of

Section 5.2. When U is empty, in other words, algorithm locateL13 returns null each time

it is invoked, this finalization step chooses as the source estimate the point P of Isort, for

which A13(P) 6131 is minimized. This modification is referred to as binary search with

finalization.

For our experiments, we used 9 test cases, each described by the tuple [S1, S2, S3,

F, N]. We choose F from {10/100, 5/100, 1/100}. For each test case, we used various 7ls

and 72s. Tables 5-3 5-11 give the simulation results. Specifically, the value listed under

#fail for our method gives the number of sources that could not be estimated without

the finalization step. Since, in all of our simulations, the quadratic solved by Algorithm

locate_L12 had real roots, our method with finalization was able to estimate the source

1011' of the time. So, for binary search with finalization, the entries #fail and ratio are

0. Even without finalization, our method outperforms that of [34] in terms of the number

of sources that could be estimated. In all 9 of our test cases, the number of sources that

could not be estimated by our method without finalization is less than or equal to that of

the method of [34]; the reductions were as high as 22'"

As noted earlier (Section 5.2), should the quadratic in Algorithm locateL12 have

imaginary roots for some instance, the algorithm may be augmented to with a binary

search to make it complete at an additional complexity.

For all tested 71s and 72s in each test case, the estimate given by our method

consistently shows as good as or much better accuracy in both 6 space and R2 space

than that of the method of [34]. In particular, when three sensors are almost collinear, the

improvement made by our method is significant. For example, when 72 is 10000 as shown

in Tables 5-3 -5-5, the increment of the ratio of successful estimate by our method versus

the method of [34] is more than 3 -'. (,-'., and ',,'.- respectively.

CHAPTER 6
CONCLUSIONS AND FUTURE WORK

We studied two fundamental problems, sensor coverage and sensor localization,

arising in wireless sensor networks. Sensor coverage is the process that deploys a network

of sensors to provide a certain degree of coverage over the domain of interest (i.e., a

region or a set of discrete targets). Sensor localization is the process that given a set of

measurements (i.e. Difference of Time-Of-Arrival) estimates the accurate location of a

source in the plane.

We developed a general ILP formulation to minimize the cost of deploy, -l sensors

while providing the desired degree of coverage over a set of discrete targets. A greedy

algorithm for solving the general ILP was also developed. For the case of a grid,

linear-time .,i-~,i! I .I i, approximation algorithms and PTASs were developed. Experiments

demonstrate the superiority of our proposed algorithms over earlier algorithms for point

coverage of grids by using heterogeneous sensors. We presented two approximate solutions

to the problem of minimizing the cost of placing heterogeneous sensors at certain locations

to ensure q-coverage of a planar region. A transformation was developed from region

coverage to point coverage. We also studied the performance of deploying homogeneous

sensors in a gridded layout.

We studied the impact of sensor deployment on the uniqueness of source estimate

in Euclidean plane as well as in a simple polygon. A necessary and sufficient condition

was derived for each case. We gave a tight bound on the size of a minimal identifying

sensor set in Euclidean plane. We reinvestigated the number of intersections of two

hyperbolas having a common focus, and showed that number is at most 2. Each sensor

deployment corresponds to an equivalence relation on Euclidean plane. Specifically, for

each non-identifying sensor set, at least one equivalence class is of greater than unit

cardinality. Given measurements of distance-difference, we presented a computational

geometric method for the problem of estimating the source location in the plane.

This method is particularly suited for deployment in sensor nodes that adapt their

computations in response to power budgets.

Naturally, there also exist some unexplored areas. Many of them are clearly

interesting topics for future research work.

For sensor coverage, it would be interesting to incorporate costs associated with

the region such as population within sensor regions in addition to the sensor costs. It

would also be interesting to consider non-circular and probabilistically specified sensor

regions. For grid coverage, the problem whether it is NP-hard is still open. In general,

the linear-time approximation algorithm we developed only works for q-covering a square

grid where q < 3. It would be interesting to come up with a general method that is

able to provide sensor deployments that satisfy the desired degree of coverage up to

the theoretical upperbound(See Lemma 5). Approximation algorithms for different grid

layouts (i.e., regular hexagon, equilaterial triangle) are also worth exploring.

This work is only a step towards utilizing computational geometry methods for

solving sensor localization problems. It would be of future interest to consider extensions

of this method for cases where more than three sensors are deploy, .1 and multiple

measurement sets are provided [47]. It would also be interesting to see if the proposed

method can be extended under random noise models, particularly when sensor errors are

correlated and the noise model is unknown. For the special case when Si, S2 and S3 form

an acute triangle, a training method was proposed in [45] wherein the localization method

can be trained in-situ to account for sensor correlations. The current method can be

similarly employ, .1 but the training procedure is likely to be more involved. It would be of

future interest to explore the tracking ability of this method by repeatedly executing it on

a stream of distance-difference measurements corresponding to a moving object. It would

be interesting to investigate the effects of randomness in distance-differences on both

uniqueness and minimality results presented in this work. Applications of these results to

practical radiation detection systems would be of future interest.

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BIOGRAPHICAL SKETCH

Xiaochun Xu was born in C('! 1; ,. !nu, a midsize city in the eastern coastal area of

C!ii, i He studied in the field of computer software at the Department of Computer

Science and Technology, N ,'iiii-; University, C'liii from 1995 to 2002, and obtained his

master's degree in computer software. He started his Ph.D. study at the Department of

Computer and Information Science and Engineering at the University of Florida in August

2002. His research areas are approximation algorithms, computational geometry, wireless

sensor networks, and wireless ad hoc networks.

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page ACKNOWLEDGMENTS ................................. 4 LISTOFTABLES ..................................... 7 LISTOFFIGURES .................................... 8 ABSTRACT ........................................ 11 CHAPTER 1INTRODUCTION .................................. 13 1.1Motivation .................................... 13 1.2RelatedWorkinSensorDeployment ..................... 15 1.3RelatedWorkinSensorLocalization ..................... 19 1.4Contribution ................................... 22 2SENSORDEPLOYMENTFORPOINTCOVERAGE .............. 26 2.1IntegerLinearProgrammingFormulation ................... 26 2.2GreedyAlgorithm ................................ 28 2.3GridCoverage .................................. 29 2.3.1ProblemDenitionandProperties ................... 29 2.3.2Asymptotic(1.58+)-ApproximationAlgorithm ........... 34 2.3.3AsymptoticPTASfor1-Coverage ................... 40 2.3.4AsymptoticPTASforq-Coverage ................... 42 2.4ExperimentalResults .............................. 45 2.4.1IntegerLinearProgram ......................... 45 2.4.2ComparisonofApproximationAlgorithms .............. 48 2.4.3ComparisonwithDivide-and-Conquer ................. 53 3SENSORDEPLOYMENTFORREGIONCOVERAGE ............. 55 3.1Exact3-ApproximationAlgorithm ....................... 55 3.2AsymptoticPTAS ............................... 58 3.3RegionCoverageviaPointCoverage ..................... 61 3.4CoveragewithaGridofSensors ........................ 66 3.4.1q=2 ................................... 69 3.4.2q=3 ................................... 71 3.4.3q=4 ................................... 72 3.4.4q=5 ................................... 75 3.5ExperimentalResults .............................. 78 5

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.................... 83 4.1PreliminariesandDenitions .......................... 84 4.2PropertiesofIdentifyingSensorSets ..................... 86 4.3NumberofIntersectionsofL12andL13 89 4.4IndistinguishablePoints ............................ 96 4.5ISSsforPolygonalRegions ........................... 98 5COMPUTATIONALGEOMETRYMETHODFORTRIANGULATIONUSINGDTOA ......................................... 102 5.1EuclideanandDTOASpaces ......................... 102 5.2GeometricDTOAMethod ........................... 104 5.3CorrectnessandComplexityoftheMethod .................. 109 5.4MonotonicityofDirectionalDerivative .................... 111 5.4.1TopLeftRegion ............................. 113 5.4.2InsideRegion .............................. 115 5.4.3BottomRightRegion .......................... 116 5.4.4Top,BottomLeft,Bottom,andTopRightRegions ......... 116 5.5SimulationResults ............................... 121 5.5.1F=0 ................................... 123 5.5.2F>0 ................................... 125 6CONCLUSIONSANDFUTUREWORK ...................... 127 REFERENCES ....................................... 129 BIOGRAPHICALSKETCH ................................ 135 6

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Table page 2-1Sensorcoverageproperties .............................. 31 2-2PerformancecomparisonofILPof[ 6 ]andourswithT=100seconds ...... 46 2-3PerformancecomparisonofILPof[ 6 ]andourswithT=1000seconds ..... 47 2-4Datafor1-coverage .................................. 50 2-5Datafor2-coverage .................................. 50 2-6Datafor3-coverage .................................. 51 2-7Datafor4-coverage .................................. 51 2-8Datafor5-coverage .................................. 52 2-9Cost/pointusing[ 6 ]togetherwithourILPwithT=1000secondsandour1:58+algorithm .................................. 54 3-1dMax,maximumAPN,andasymptoticratio1q5 ............. 67 5-1DataforS1=(0;0),S2=(0;50000),S3=(0:001;100000),andF=0 ....... 119 5-2DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=0 .......................................... 119 5-3DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=10/100 ....................................... 120 5-4DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=5/100 ....................................... 120 5-5DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=1/100 ....................................... 120 5-6DataforS1=(0;0),S2=(0;50000),S3=(5000;100000),andF=10/100 .... 121 5-7DataforS1=(0;0),S2=(0;50000),S3=(5000;100000),andF=5/100 .... 121 5-8DataforS1=(0;0),S2=(0;50000),S3=(5000;100000),andF=1/100 .... 122 5-9DataforS1=(0;0),S2=(0;100000),S3=(100000;0),andF=10/100 ..... 122 5-10DataforS1=(0;0),S2=(0;100000),S3=(100000;0),andF=5/100 ...... 122 5-11DataforS1=(0;0),S2=(0;100000),S3=(100000;0),andF=1/100 ...... 123 7

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Figure page 1-1Locationdeterminationapproaches.A)AOA.B)TOA.C)DTOA. ....... 21 2-1Greedyalgorithmtodeploysensors ......................... 29 2-2Agridwithasensorofrange4atitscenter.A)Thesensormonintorsalldarklocations.B)Largestcoveringsquare.C)Largestcoveringrectangle. ....... 30 2-3Asymptotic(1.58+)-approximationalgorithmfor1-coverage ........... 35 2-4Tworegulartilingstrategies.A)+'patterns.B)diamondpatterns ....... 36 3-1Algorithm3-approx .................................. 56 3-2Sensorats2Scancoverpointsinatmost3hexagons .............. 57 3-3LocationofsitesthatcancoverpointsinH 58 3-4AlgorithmAS ..................................... 59 3-5G(d)forarectangularregionR.The16shadedgridpointsarewithinadiskofradiusrcenteredatthelocations. ......................... 62 3-6ThesensorsislocatedinsidethesquareformedbyfourgridpointsPi,i=1;2;3;4,andcoversMAX(r)gridpoints,shownindarkcolor. .............. 64 3-7Griddedlayoutusingageometrysized.A)Equilateraltriangle.B)Square.C)Regularhexagon. .................................. 66 3-8Equilateraltriangle,q=2 ............................... 69 3-9Square,q=2 ...................................... 70 3-10Regularhexagon,q=2 ................................ 71 3-11Square,q=3 ...................................... 72 3-12Regularhexagon,q=3 ................................ 72 3-13Equilateraltriangle,q=4 ............................... 73 3-14Square,q=4 ...................................... 74 3-15Regularhexagon,q=4 ................................ 75 3-16Equilateraltriangle,q=5 ............................... 76 3-17Square,q=5 ...................................... 76 3-18Regularhexagon,q=5 ................................ 77 8

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.............. 80 3-20Totalsensorcostrequiredv.s.variouss,whereq=2. .............. 81 3-21Totalsensorcostrequiredv.s.variouss,whereq=3. .............. 82 4-1ExamplesofthelocusL12 85 4-2Threenon-collinearsensorsS1,S2,andS3formatriangleandtwohyperbolasL12(12)andL13(13)intersecteachotheratP1andP2. ............. 86 4-3AhyperbolaLthatpassesthroughSi(1i4). ................ 88 4-4Regionsofmonitoringarea:(a)topleft,(b)inside,(c)bottomright,(d)top,(e)bottomleft,(f)bottom,and(g)topright. ................... 90 4-5AhyperbolaL=LlSLrwithfocusSandsemimajoraxisy-axis. ....... 91 4-6Case1: 92 4-7Case2: 94 4-8Case3: .................. 94 4-9Case4: .................. 95 4-10Collinearsensors ................................... 97 4-11SensorsS1,S2,andS3ontheboundaryofaconvexpolygon. ........... 99 4-12Aconcavepolygon,itsboundingconvexpolygon,andthreesensorsS1,S2,andS3placedonthecommonboundaryoftheconcaveandconvexpolygons .... 100 4-13S1liesinsideasimplepolygonwhileS2andS3areontheboundary.P1inthetopregionisadualpointofP2whichliesinthetopleftregion. ......... 101 5-1Canonicalplacementof3sensorsandpartitioningofmonitoringregion ..... 105 5-2Canonicalplacementof3sensorsandL12(12) ................... 106 5-3P=(x;y)islocatedinthetopleftregion. ..................... 113 5-4P=(x;y)islocatedinsidethetriangle. ...................... 114 5-5P=(x;y)islocatedinthebottomrightregion. .................. 114 5-6P=(x;y)islocatedinthetopregion. ....................... 114 5-7P=(x;y)islocatedinthebottomleftregion. ................... 115 5-8P=(x;y)islocatedinthebottomregion. ..................... 116 5-9P=(x;y)islocatedinthetoprightregion. .................... 117 9

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..................................... 118 5-11ThedegeneratecasewhenL12(12)isaverticalray ................ 118 10

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23 24 60 73 ],tonamejustafew.Wirelesssensornetworksextendpeople'scapabilitybyestablishinganinstantandremoteinteractionwiththephysicalworld.Atypicalwirelesssensornetwork[ 11 55 65 ]maycomprisethousandsofspatiallydistributedyetcooperativesensornodesthatcontinuouslymonitorasetofprespeciedphysicalorenvironmentalconditionsinthedomainofinterest[ 48 ].Eachsensornodehasasensingcapabilityaswellaslimitedenergysupply,computepower,memory,andcommunicationability.Nowadays,asensornode(e.g.,mote)usuallyconsistsofthreeparts,thatis,awirelesscommunicationdevice,asmallautomaticcontrolunit,andasourceofenergy(e.g.,battery).IyengarandBrooks[ 23 24 ]andCullerandHong[ 10 ]providegoodoverviewsofthebreadthofsensornetworkresearchtopicsaswellasofapplicationsforsensornetworks.Sensornetworkalgorithmsarereviewedin[ 49 ].Thepurposeofwirelesssensornetworksistocontinuouslymonitorthedomainofinterest(i.e.,aregionorasetofdiscretetargets),detecttheoccuringeventsandcollectrelevantdataforfurtherprocessing/responding.Ifnotenoughsensorshavebeendeployed,thenthereisnowaythatthesensornetworkcanguaranteefullcoverageoverthedomain,whichmeanseventsoccuringatsomelocationsmaynotbedetectable.Therefore,anappropriatestrategyonsensordeploymentthatguaranteesthedomainofinterestgetcoveredatacertainqualityofservice(QoS)levelisafundamentalproblemforalmosteverysensornetworkapplication.M.CardeiandJ.Wu[ 5 ]reviewedthesensorcoverageproblemsinthecontextofstaticwirelesssensornetworks.Inpractice,thecoverage 13

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62 ]runsinpolynomialtime,itisnotan-approximationalgorithmforanyconstant.Forthetestcasesreportedin[ 62 ],thedevelopedalgorithmhadanapproximationfactorof3.Gridcoverageisanotherversionofthepointcoverageproblem.IntheversionofChakrabartyetal.[ 6 ],wearegivena2Dor3Dgridofpointsthataretobesensed.Sensorlocationsarerestrictedtothesegridpointsandeachgridpointistobecoveredbyatleastq,q1,sensors(i.e.,weseekq-coverage).Forsensing,wehavetsensortypesavailable.Asensoroftypeicostscidollarsandhasasensingrangeri.Atmostonesensormaybeplacedatagridpoint.Inthisversionofthepointcoverageproblem,thesensorsdonotcommunicatewithoneanotherandareassumedtohaveacommunicationrangelargeenoughtoreachthebasestationfromanygridposition.Thus,networkconnectivityisnotanissue.Theobjectiveistondaleastcostsensordeploymentthatprovidesq-coverage.Chakrabartyetal.[ 6 ]formulatethisq-coveragedeploymentproblemasanintegerlinearprogram(ILP)withO(tn2)variablesandO(tn2)equations,wherenisthenumberofgridpoints.Foralargen,Chakrabartyetal.[ 6 ]proposeadivide-and-conquer\near-optimal"algorithminwhichthebasecase(asmallnumberofpoints)issolvedoptimallyusingtheILPformulation.Funkeetal.[ 15 ]developagreedyconstant-factorapproximationalgorithmandaPTASfor1-coverageofagrid.However,theyseektominimizethenumberofdeployedsensorsratherthanthecostofthesesensors.Whenoptimizingthenumberofdeployedsensors,onlysensorswithmaximumrangeneedbeconsidered.Thecomplexityofthegreedyalgorithmof[ 15 ],whichaccountsforobscuringobstacles,isO(nlogn)anditsapproximationfactorislog(4R2),whereRisthemaximumsensorrange.TheapproximationfactorofthePTASin[ 15 ],whichassumestherearenoobstaclesbutwhichobtains1-coverageforanyspeciedsubsetofgridpoints,is4(1+)anditscomplexityisO(n).Wuetal.[ 64 ]lookatthegrid 18

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34 53 ]mayhaveimaginaryrootsinwhichcasethesemethodsdonotreturnananswer,thatistheybecomeincomplete.Moregenerally,numericalinstabilitiesmayariseincomputationsimplementedwithlowprecisionarithmeticoperationswhereinmatrixinversionsneededforlinearalgebraicmethodsmaybecomeill-conditionedresultinginlargeestimationerrors. 6 ].TheimpactofthisimprovementinILPformulationisareductionofupto78%(basedonourtestdata)inthecostofdeployedsensors.Agreedyalgorithmforgeneralsensordeploymentalsowasdeveloped.Thisgreedyalgorithm,likeourILP,maybespecializedtothecaseofagrid.FastasymptoticapproximationalgorithmsaswellasPTASsforgridsweredeveloped.Althoughtheproofsforthesealgorithmsassumedsquaregrids,theproofsalsoapplytorectangulargrids(notethattheapproximationfactorsareasymptoticandthat,whentherectangledimensionsarelargeenough,theapproximationfactorisnotaected).Ourexperimentsindicatethatthe1:58+-approximationalgorithmisthebestofouralgorithmsforq4andthatthegreedyalgorithmisusuallybestforq>4,whereqisthedesireddegreeofcoverage. 22

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6 ].Inthisproblem,thelocationsformap 27

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6 ]hasO(tn2)variablesandO(tn2)constraints. 1. Theselectionofasensoroftypeidoesn'tviolateanyboundonthetotalnumberofsensorsofthistypethatmaybedeployed. 2. Placementofasensoroftypeiatlocationzdoesn'tviolatetheboundonthenumberofsensorsoftypeithatmaybeplacedatlocationz. 3. Letcover0(j;l)betheremainingcoveragedegreetobeprovidedformodalityjatlocationl.Initially(i.e.,whennosensorhasbeendeployed),cover0(j;l)=cover(j;l)foralljandl.cover0(j;l)isreducedby1wheneverasensorisdeployedsoascoverlformodalityj.Letcover00(j;l)=maxf0;cover0(j;l)1gforalljandlsuchthatz2locations(i;j;l).Theincrementalcoveragecostcost(i)=(PjPl(cover0(j;l)cover00(j;l))isminimized. 28

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2-1 givesourgreedyalgorithmforminimum-costsensordeployment. Setcover0(j;l)=cover(j;l);8j;lwhile(cover0(j;l)>0foratleastonepair(j;l))fLet(i;z)beanoptimalsensor-locationpair.Ifthereisnosuch(i;z)oriftheincrementalcoveragecostisinnity,terminate.//deploymentunsuccessfulDeployasensoroftypeiatlocationz.Updatecover0.gFigure2-1. Greedyalgorithmtodeploysensors 6 ].Beforepresentingourtwoproposedasymptotoicapproximationalgorithms,werstgivethedetaileddenitionofthisproblemandsomeinterestingproperties. 2-2 (A)showsalllocationsmonitoredbyasensorwhoserangeis4andthatisplacedatthegridcenter.Darklocationsindicatelocationmonitoredbythesensor.Theobjectiveistondaminimumcostsensordeploymentthatprovidesthedesireddegreeofcoverage.Suchasensordeploymentiscalledanoptimaldeployment.Notethatsome 29

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Agridwithasensorofrange4atitscenter.A)Thesensormonintorsalldarklocations.B)Largestcoveringsquare.C)Largestcoveringrectangle. Notealsothatifcost(i)0,1jt.LetD(i)bethemaximumnumberofgridpointscoveredbyasensorwhoserangeisi.LetS(i)bethemaximumnumberofthesegridpointsthatfallwithinaverticallyalignedsquareandletR(i)bethecorrespondingnumberforthecaseofaverticallyalignedrectangle.Figure 2-2 (B)showsthelargestsquareofcoveredpointswhenthesensorrangeis4andFigure 2-2 (C)showsthelargestrectangleofcoveredpointsforthissensor.ObservethatD(4)=49,S(4)=25andR(4)=35.Table 2-1 givesD(i),S(i)andR(i)for1i12.Thisguregivesalsotheratioss(i)=S(i)=D(i)andr(i)=R(i)=D(i),1i12. 30

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Sensorcoverageproperties (2{1) (2{2) Thecoordinatesofthelargestverticallyalignedsquareofcoveredlocationshavetheform(x;x),wherex=bi=p (2{4) FromEquations 2{2 and 2{4 ,weget (2{5) 31

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2-1 givess(i)fori12asobtainedfromEquations 2{1 and 2{3 .Fori>1,theshownvaluesare0:51.Fori>12,s(i)0:51followsfromEquation 2{5 withi=13andthefactthattherightsideofthisequationisanondecreasingfunctionofi.NotethattherightsideofEquation 2{5 hasthelimitingvalueof2=asi!1. 2;h1 2),(w1 2;h1 2),(w1 2;h1 2),and(w1 2;h1 2).SincethegridlocationsatthecornersofBarecoveredbythesensorat(0,0),(w1 2)2+(h1 2)2i2,whereiistherangeofthesensor.Onemayverifythatwhenh>w+2,(w+1 2)2+(h3 2)2i2.So,therectangleQwhosecornersareat(w+1 2;h3 2),(w+1 2;h1 2),(w+1 2;h3 2),and(w+1 2;h3 2),includesonlygridlocationsthatarecoveredbythesensorat(0;0).ThewidthandheightofQarew+2andh2,respectively.ThenumberofgridlocationsincludedinQis(w+2)(h2),whichisgreaterthanthenumber,wh,oflocationsincludedinBwhenh>w+2.ThiscontradictstheassumptionthatBisthelargestrectangleofcoveredlocations.Hence,hw+2.Consequently,whw+2.Now,ifwd,wd+2asboth 32

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1 .LetBbeasinLemma 2 andassumethatwh.Thecasew>hissymmetric.FromLemma 2 andtheproofofLemma 1 ,itfollowsthatw=d=2bi=p 2-1 givesr(i)fori12asobtainedfromEquations 2{1 and 2{6 .Fori>1,theshownvaluesare>0:6366.Usingacomputerprogram,weveriedthatr(i)>0:6366fori100;000.Fori>100;000,weuser(i)s(i)andtheknowledgethattherightsideofEquation 2{5 ismorethan0.6366fori>100;000toconcludethatr(i)>0:6366. Letopt(n;q)bethecostofanoptimalsensordeploymentthatachievesq-coverageforap LettMax=argmax1jtfrange(j)gandtMin=argmin1jtfrange(j)g.NotethattMaxandtMin,respectively,givethesensortypeswithmaximumandminimumrange.LetqMaxbethemaximumcoveragedegreeobtainablebyanydeploymentofthegivensetofsensortypes.

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2-3 givesourasymptoticapproximationalgorithmfor1-coverage.Later,weextendthisalgorithmto2-and3-coverage.ThefollowinglemmashowsthatStep2ofFigure 2-3 isexecutedonlywhenrange(dMin)=1. 2-3 ,thendMin=rMin.

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8 ]tocoverthep 2-4 (A)).PlaceatyperMinsensoratthecenterofeach"+".Ifasensorisplacedatanon-gridlocation,moveittoagridlocationthatispartofits"+"pattern.Done. 2-4 (B)).Ifasensorisplacedatanon-gridlocation,moveittoagridlocationthatispartofitsdia-mondpattern.Done. Asymptotic(1.58+)-approximationalgorithmfor1-coverage

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6 ?ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddttttttttttttttttttttdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddtttttttttttttttttt 6 ?IR(A)(B)Figure2-4. Tworegulartilingstrategies.A)+'patterns.B)diamondpatterns 3 ,weget 5cost(rMin) 5cost(j) 5cost(j) 0:6336D(range(j));1jt;j6=rMin0.

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8 ]haveshownthatatmostn+4p 5`+'patternsareneededtotileap 6 ,itfollowsthatthecostofthetilingobtainedbythealgorithmofFigure 2-3 islessthann+4p 4 ,andthefactthatD(1)=5,itfollowsthattheapproximationratioislessthann+4p n=1+4 13cost(rMin) 13cost(dMin) 13cost(dMin) 0:6366D(range(dMin))<1:09cMinUsingamethodsimilartothatusedin[ 8 ],wecanshowthatStep2deploysatmostdp 13e(p 4 ,itfollowsthatthe 37

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13e(p wedp heFromLemma 4 ,itfollowsthattheapproximationratioisatmostcost(rMin) wedp he1,andthechoiceofrMin,wegetcost(rMin) 0:6366cost(dMin)R(range(rMin))1 0:6366<1:58 38

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Forq-coverage,q2f2;3g,westartwiththe1-coveragedeploymentobtainedbythealgorithmofFigure 2-3 andincreasethenumberofsensorsateachgridlocationthathasasensorfrom1toq.While,wenowhaveq-coverageanda1:58+asymptoticapproximationfactorisguaranteedbyLemma 4 andTheorem 3 ,thedeploymentisinfeasibleasthereareq>1sensorsdeployedatcertaingridlocations.Toremedythisinfeasibility,werelocatetheexcesssensorsatanylocation.ThesensordeploymentofFigure 2-3 issuchthateverylocationLthathasasensorsatisesthefollowingproperties(sincewearedealingwithasymptoticproperties,wemayassumethatp 2-3 ,theasymptoticapproximationfactoris1=0:51+<1:96+. 39

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7 tohandleanysensorsplacedoutsidethep 7 ,itfollowsthatCopt(d;1).Hence,opt(n;1)bp 2.1 (notethatsinceRandareconstants,thesizeoftheILPisconstant).Next,tilethep 7 41

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8 )

1.BeforepresentingthisPTASanditsproofofcorrectness,weestablishtwolemmas.Letp=D(range(tMax))cost(tMax)=(qcost(tMin)). 42

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4 ,itfollowsthatopt0(n;q)nqcMin.So,popt0(n;q)D(range(tMax))cost(tMax) cost(tMin)ncost(tMax)D(range(dMin))cMin cost(dMin)=ncost(tMax) 5 ).LetAbethesetofallgridpointslocatedonthewidthRboundaryofthegridandletBbethesetofremaininggridpoints.ThesizeofAis4R(p 43

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5 ,itfollowsthattheremainingsensorsprovideq-coverageofallgridpoints.Thecostofthisdeploymentisatmostd(p 2.1 withlocationscorrespondingtothegridpointsofa(p 10 .WeshowthatthejuststatedsensordeploymentalgorithmisanasymptoticPTASforq-coverage.Letp 44

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Fromthedenitionofp 2{8 isatmost1+.HencethestatedalgorithmisaPTASforq-coverageofagrid.ItscomplexityisO(n)forthesamereasonsasthecomplexityofthePTASfor1-coverageisO(n). 2.4.1IntegerLinearProgramWeevaluatedtheimpactofusingourILPformulationofSection 2.1 versusthatof[ 6 ]inconjunctionwiththedivide-and-conquerheuristicproposedin[ 6 ]forthedeploymentofsensorsinagrid.Recallthattheheuristicof[ 6 ]tilesalargegridusinganoptimaldeploymentforthelargestsquaresubgridwhoseILPformulationmaybesolvedinagivenamountoftime.Forthisevaluation,weusedtheILPsolverlpsolver5.0[ 31 ]developedbytheEindhovenUniversityofTechnology.ThisILPsolverwasrunonaDellInspironPCwitha1.7GHzprocessorand512MBmemory. 45

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2-2 givesthesizeofthelargestsquaresubgridwhoseILPformulationissolvablein100secondsusingtheformulationof[ 6 ]aswellusingourformulationofSection 2.1 .Also,thetotalcostoftheoptimaldeploymentdividedbythenumberofpointsinthesubgrid(cost/point)andthereductionincost/pointachievedusingourILPformulationaregiven.Table 2-3 givesthisdataforthecasewhenlpsolverisgiven1000secondstosolvetheILPformulation. Table2-2. PerformancecomparisonofILPof[ 6 ]andourswithT=100seconds T=100seconds Testcase[ 6 ]OursReduction(%) sizecost/pointsizecost/point [1;2;3;3;1:5;1]5*50.1225*250.120[2;2;3;3;1:5;1]4*40.37514*140.27626.4[3;2;3;3;1:5;1]4*40.56314*140.41338.5[4;2;3;3;1:5;1]4*40.7514*140.55126.53[5;2;3;3;1:5;1]3*31.6713*130.72856.41[1;2;7;4;5;3]5*50.215*150.20[2;2;7;4;5;3]5*50.4814*140.42910.63[3;2;7;4;5;3]4*40.93813*130.63931.88[4;2;7;4;5;3]4*41.2514*140.84732.24[5;2;7;4;5;3]3*32.7813*131.0761.51[1;2;6;4;5;3]5*50.215*150.1876.5[2;2;6;4;5;3]5*50.4414*140.36716.59[3;2;6;4;5;3]4*40.93814*140.55141.26[4;2;6;4;5;3]4*41.2514*140.73541.2[5;2;6;4;5;3]3*32.7813*130.92366.80[1;2;7;5;5;4]6*60.19418*180.13032.99[2;2;7;5;5;4]5*50.416*160.21945.25[3;2;7;5;5;4]4*40.93816*160.34867.38[4;2;7;5;5;4]3*32.2216*160.47878.47[5;2;7;5;5;4]3*32.7815*150.60978.09[1;3;5;5;3;3;1:5;1]5*50.1218*180.09322.5[2;3;5;5;3;3;1:5;1]4*40.37517*170.17553.33[3;3;5;5;3;3;1:5;1]4*40.56316*160.24656.31[4;3;5;5;3;3;1:5;1]3*31.3315*150.35673.23[5;3;5;5;3;3;1:5;1]3*31.6715*150.44473.41 46

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6 ]onlyfora55subgrid,whereaswewereabletosolveourILPfora2525subgrid!Althoughforthisparticulartestcase,therewasnoreductioninthecost/pointintheoptimalsolutionforthelargersubgrid,overoursetoftestcasesthereductionincost/pointintheoptimalsolutionforthelargestsubgridsolvableusingourILPformulationversusthatof[ 6 ]rangedfrom0%toslightlymorethan78%.Theimpactofthisreductiononthedivide-and-conquerheuristicof[ 6 ]isevident{forourtestcasestilingalargegridusingtheoptimalsolutionsfromthelargersubgridssolvableusingourILPformulationreducestheoverallcostofthedeployedsensorsbyasmuchasabout78%!Itisinterestingtonotethatthecost/pointincreasedfor3ofourtestcaseswhentheILPof[ 6 ]wasgiven1000secondsratherthan100seconds.Inoneofthesecases,theincreasewasashighas65%.Forourformulation,therewasanincreaseincost/pointfor4ofthe25testcases;however,inallfourcases,theincreasewaslessthan6%.Theimpactofanincreaseincost/pointintheoptimaldeploymentforalargersubgridonthetilingofalargegridmaybereducedbyndingtheoptimaldeploymentforseveralsmallsubgridsandusingthesubgridwiththesmallestcost/pointtotilelargegrids. 2.3.2 .2.Theasymptotic(1.96+)-approximationalgorithmdescribedattheendofSection 2.3.2 .3.Iterativeversionsofbothofthesealgorithms.Intheiterativeversion,Step4ofthealgorithmwasmodiedtondanoptimaldeploymentforthelargestsubgridofsize2iw2ihwhoseILPcouldbesolvedinaspeciedamountoftimeT.HerewandhareasdeterminedinStep1.Notethatthecost/pointforthisoptimaldeployment 48

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2.2 .ThetilingofalargegridinStep4oftherstfouralgorithmsstatedabovewasmodiedfromthatstatedintheoriginaldescriptionofSection 2.3.2 .Thismodicationisforthecasewhenp 2.2 isusedtodeploysensorstoachievethedesiredcoverageintheseportions.Thismodicationisreferredtoastilingwithgreedylling.WedidnotexperimentwithourPTASsaswedonotconsiderthesetobepracticalfromthestandpointofrequiredruntime.Forourexperiments,weused16testcases,eachdescribedbythetuple[t;c1;r1;:::;ct;rt].Foreachtestcase,wesoughtq-coveragefor1q5andthegridsizewas300300(or,n=90;000).Theiterativeversionsofthealgorithmsweregiven3600secondstosolvetheassociatedILP.Althoughwehavenotdescribedeithera1:58+-ora1:96+-asymptoticapproximationalgorithmforthecaseq2f4;5g,forourexperiments,weproceededasforthecaseq2f2;3gwiththe1-coveragedeploymentobtainedbythealgorithmofFigure 2-3 andincreasedthenumberofsensorsateachgridlocationthathasasensorfrom1toq.Then,werelocatedthesensorssoastopreserveq-coverageandassurethatnolocationhasmorethan1sensor.Ourrelocationstrategysucceededinallofour16testcases.Tables 2-4 2-8 givethecostsoftheconstructedsensordeploymentsaswellasthesizeofthetilingsubgridsused.Inthecaseofthe1:58+and1:96+algorithms,theselectedsensortypealsoisgiven.Thepercentagereductionincostobtainedbythe1:58+-approximatealgorithmrelativetothecostofthedeploymentconstructedbythegreedyalgorithmalsoisgiven.Theiterativeversionofthe1:58+-approximationalgorithmoutperformedthebase1:58+algorithminonly1ofthe80tests.Thisoccurredin1ofthetestswithq=4and 49

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Datafor1-coverage TestcaseTilingwithgreedyllingGreedyalg.Reduction(%) (1.96+)-alg.Iterativealg.(1.58+)-alg.Iterativealg. [2;3;3;1:5;1]10800(1,5*5)10800(20*20)10800(1,5*5)10800(20*20)15259.529.22[2;3;3;1;1]10800(1,5*5)10800(20*20)10800(1,5*5)10800(20*20)1471926.63[2;3;3;1:5;2]10800(1,5*5)10800(10*10)10800(1,5*5)10800(10*10)1401022.91[2;7;4;5;3]18000(2,5*5)18000(20*20)18000(2,5*5)18000(20*20)2004210.19[2;7;4;4:5;3]16200(2,5*5)16200(20*20)16200(2,5*5)16200(20*20)1914515.38[2;7;4;4;3]14400(2,5*5)14400(20*20)14400(2,5*5)14400(20*20)2100531.44[2;6;4;5;3]18000(2,5*5)15975(20*20)15555(1,5*7)15555(10*14)1829614.98[2;7;5;5;4]13010(1,7*7)11248(14*14)10386(1,7*9)10446(14*18)1338322.39[2;7;5;4:5;4]12942(1,7*7)10759.5(14*14)10296(1,7*9)10397(14*18)1283919.81[2;7;5;4;4]12876(1,7*7)10198(14*14)10230(1,7*9)10232(14*18)1327822.96[2;7;5;5;3]13001(1,7*7)11993(14*14)10365(1,7*9)10515(14*18)1388325.34[2;7;5;4;3]12985(1,7*7)11512(14*14)10401(1,7*9)10569(14*18)1324321.46[3;5;5;3;3;1:5;1]9257.5(1,7*7)8290(14*14)7410(1,7*9)7486.5(14*18)910118.58[3;5;5;2:5;3;1:5;1]9000(2,5*5)9000(10*10)7376(1,7*9)7422.5(14*18)866314.86[3;5;5;2:5;3;1;1]9000(2,5*5)9000(10*10)7321(1,7*9)7378(14*18)8361.512.44[3;5;5;2;3;1;1]7200(2,5*5)7200(20*20)7264(1,7*9)7296(14*18)77686.49 Datafor2-coverage TestcaseTilingwithgreedyllingGreedyalg.Reduction(%) (1.96+)-alg.Iterativealg.(1.58+)-alg.Iterativealg. [2;3;3;1:5;1]21780(1,5*5)25650(10*10)21780(1,5*5)25650(10*10)2629517.17[2;3;3;1;1]21720(1,5*5)25200(10*10)21720(1,5*5)25200(10*10)2543114.59[2;3;3;1:5;2]21691.5(1,5*5)24300(10*10)21691.5(1,5*5)24300(10*10)2515813.78[2;7;4;5;3]36205(2,5*5)39600(10*10)36205(2,5*5)39600(10*10)381155.01[2;7;4;4:5;3]32584.5(2,5*5)36900(10*10)32584.5(2,5*5)36900(10*10)37321.512.69[2;7;4;4;3]28964(2,5*5)34200(10*10)28964(2,5*5)34200(10*10)3614819.87[2;6;4;5;3]36184(2,5*5)37800(10*10)30972(1,5*7)34140(10*14)330876.39[2;7;5;5;4]25728(1,7*7)22202(14*14)20431(1,7*9)22371(14*18)2396714.75[2;7;5;4:5;4]25630(1,7*7)21253.5(14*14)20335.5(1,7*9)21792(14*18)2328212.66[2;7;5;4;4]25528(1,7*7)20204(14*14)20232(1,7*9)20639(14*18)223529.48[2;7;5;5;3]25852(1,7*7)24077(14*14)20528(1,7*9)22831(14*18)2406214.69[2;7;5;4;3]25772(1,7*7)23172(14*14)20619(1,7*9)22430(14*18)2393313.85[3;5;5;3;3;1:5;1]18390(1,7*7)16677.5(14*14)14676.5(1,7*9)16086(14*18)16721.512.23[3;5;5;2:5;3;1:5;1]18101.5(2,5*5)20250(10*10)14646(1,7*9)15582(14*18)1664512.01[3;5;5;2:5;3;1;1]18120(2,5*5)19800(10*10)14611.5(1,7*9)15517.5(14*18)16350.510.64[3;5;5;2;3;1;1]14520(2,5*5)17100(10*10)14455(1,7*9)17191(7*9)155627.11

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Datafor3-coverage TestcaseTilingwithgreedyllingGreedyalg.Reduction(%) (1.96+)-alg.Iterativealg.(1.58+)-alg.Iterativealg. [2;3;3;1:5;1]32758.5(1,5*5)40500(10*10)32758.5(1,5*5)40500(10*10)3695711.36[2;3;3;1;1]32640(1,5*5)38700(10*10)32640(1,5*5)38700(10*10)362229.89[2;3;3;1:5;2]32583(1,5*5)37800(10*10)32583(1,5*5)37800(10*10)36013.59.53[2;7;4;5;3]54410(2,5*5)59400(10*10)54410(2,5*5)59400(10*10)53099-2.47[2;7;4;4:5;3]48969(2,5*5)55350(10*10)48969(2,5*5)55350(10*10)522856.34[2;7;4;4;3]43528(2,5*5)51300(10*10)43528(2,5*5)51300(10*10)5033713.53[2;6;4;5;3]54368(2,5*5)56700(10*10)46551(1,5*7)52033(10*14)46012-1.17[2;7;5;5;4]38464(1,7*7)33260(14*14)30683(1,7*9)33789(14*18)332717.78[2;7;5;4:5;4]38343(1,7*7)797.5(14*14)30517(1,7*9)32407(14*18)32771.56.88[2;7;5;4;4]38202(1,7*7)30322(14*14)30371(1,7*9)30581(14*18)310152.08[2;7;5;5;3]38677(1,7*7)35166(14*14)30739(1,7*9)35595(14*18)335228.30[2;7;5;4;3]38645(1,7*7)33435(14*14)30897(1,7*9)34141(14*18)333677.40[3;5;5;3;3;1:5;1]27595.5(1,7*7)24090.5(14*14)21926.5(1,7*9)24668.5(14*18)233876.24[3;5;5;2:5;3;1:5;1]27204.5(2,5*5)30600(10*10)21908.5(1,7*9)23664(14*18)23273.55.87[3;5;5;2:5;3;1;1]27239.5(2,5*5)29700(10*10)21838(1,7*9)27333.5(7*9)22901.54.64[3;5;5;2;3;1;1]21839(2,5*5)26100(10*10)21656(1,7*9)27183(7*9)223132.94 Datafor4-coverage TestcaseTilingwithgreedyllingGreedyalg.Reduction(%) (1.96+)-alg.Iterativealg.(1.58+)-alg.Iterativealg. [2;3;3;1:5;1]43737(1,5*5)54000(10*10)43737(1,5*5)54000(10*10)473737.68[2;3;3;1;1]43560(1,5*5)52200(10*10)43560(1,5*5)52200(10*10)465226.37[2;3;3;1:5;2]43474.5(1,5*5)51300(10*10)43474.5(1,5*5)51300(10*10)46477.56.46[2;7;4;5;3]72615(2,5*5)79200(10*10)72615(2,5*5)79200(10*10)67634-7.36[2;7;4;4:5;3]65353.5(2,5*5)73800(10*10)65353.5(2,5*5)73800(10*10)67025.52.49[2;7;4;4;3]58092(2,5*5)68400(10*10)58092(2,5*5)68400(10*10)642219.54[2;6;4;5;3]72552(2,5*5)75600(10*10)62158(1,5*7)69916(10*14)58468-6.31[2;7;5;5;4]51337(1,7*7)44324(14*14)41054(1,7*9)45667(14*18)423673.10[2;7;5;4:5;4]51132.5(1,7*7)42377(14*14)40860.5(1,7*9)52090(7*9)418322.32[2;7;5;4;4]50941(1,7*7)40412(14*14)40657(1,7*9)40265(14*18)39335-3.36[2;7;5;5;3]51458(1,7*7)48028(14*14)41036(1,7*9)47723(14*18)426003.67[2;7;5;4;3]51511(1,7*7)46206(14*14)41087(1,7*9)46286(14*18)424423.19[3;5;5;3;3;1:5;1]36769(1,7*7)33242.5(14*14)29226.5(1,7*9)33010(14*18)298432.07[3;5;5;2:5;3;1:5;1]36306(2,5*5)40950(10*10)29171.5(1,7*9)32245.5(14*18)29650.51.62[3;5;5;2:5;3;1;1]36359(2,5*5)39600(10*10)29114(1,7*9)31177(14*18)29248.50.46[3;5;5;2;3;1;1]29158(2,5*5)35100(10*10)28857(1,7*9)37144(7*9)28661-0.68

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Datafor5-coverage TestcaseTilingwithgreedyllingGreedyalg.Reduction(%) (1.96+)-alg.Iterativealg.(1.58+)-alg.Iterativealg. [2;3;3;1:5;1]54714(1,5*5)67500(10*10)54714(1,5*5)67500(10*10)576605.11[2;3;3;1;1]54480(1,5*5)64800(10*10)54480(1,5*5)64800(10*10)568434.16[2;3;3;1:5;2]54366(1,5*5)63450(10*10)54366(1,5*5)63450(10*10)56671.54.07[2;7;4;5;3]90822(2,5*5)100800(10*10)90822(2,5*5)100800(10*10)82374-10.26[2;7;4;4:5;3]81742.5(2,5*5)94500(10*10)81742.5(2,5*5)94500(10*10)81442.5-0.37[2;7;4;4;3]72660(2,5*5)88200(10*10)72660(2,5*5)88200(10*10)779726.81[2;6;4;5;3]90742(2,5*5)95400(10*10)77668(1,5*7)88493(10*14)70992-9.40[2;7;5;5;4]64183(1,7*7)55357(14*14)51164(1,7*9)56956(14*18)512930.25[2;7;5;4:5;4]63884.5(1,7*7)53037.5(14*14)50914.5(1,7*9)68565.5(7*9)50750.5-0.32[2;7;5;4;4]63648(1,7*7)50576(14*14)50659(1,7*9)64156(7*9)47729-6.14[2;7;5;5;3]64361(1,7*7)60790(14*14)51446(1,7*9)59701(14*18)515360.17[2;7;5;4;3]64318(1,7*7)59025(14*14)51284(1,7*9)58160(14*18)513500.13[3;5;5;3;3;1:5;1]45910(1,7*7)42413(14*14)36637(1,7*9)41407(14*18)36205-1.19[3;5;5;2:5;3;1:5;1]45409(2,5*5)51300(10*10)36482(1,7*9)51576(7*9)35972.5-1.42[3;5;5;2:5;3;1;1]45478(2,5*5)49950(10*10)36390(1,7*9)39268.5(14*18)35525.5-2.43[3;5;5;2;3;1;1]36476(2,5*5)44100(10*10)36108(1,7*9)37340(14*18)34975-3.24 52

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6 ]proposetilinganp 6 ]withthatofour1:58+-approximationalgorithm.Forthiscomparison,weusethecost/pointoftheoptimalsolutionforthelargestILPsolvableusingourformulationandthe25datasetsofTable 2-3 .Thecost/pointforthe1:58+-approximationalgorithmisobtainedbydividingthetotalcostofthesensorsdeployedona300300gridbythenumberofpoints(90,000)inthegrid.Thiscomparisonisbiasedinfavorofthedivide-and-conqueralgorithmasthiscomparisonassumesthatthesizeofthegridtobecoveredisamultipleofthesizeofthesubgridsolvedusingtheILP.However,forthe1:58+-approximationalgorithm,weusethegreedyalgorithmtocoverportionsofthe300300gridnotcoveredbywholesubgrids(i.e.,greedylling).Table 2-9 givesthecost/pointforeachalgorithmandthepercentreductionincost/pointachievedbythe1:58+algorithmrelativetothatachievedbythedivide-and-conqueralgorithmof[ 6 ]usingourILPformulation;thetimeallotedtotheILPsolverwas1000seconds.Bothalgorithmsproduceddeploymentswiththesamecost/pointon2ofthe25datasets.Ontheremaining23datasets,our1:58+-algorithmhadasmallercost/point.Theaveragereductionincost/pointobtainedbyouralgorithmwas7%;themaximumreductionwas12.5%andtheminimumreduction0%.WhenthetimeavailabletotheILPsolverisonly100seconds,thesepercentageswere6.9%,23.6%and3:65%(ononetestset,usingourILP,thedivide-and-conqueralgorithmof[ 6 ]didbetterthanour1:58+algorithm,on2testsetsthetwoalgorithmsweretied,andour1:58+algorithmwassuperiorontheremaining22datasets).Asnotedearlier,forthetestcases 53

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Table2-9. Cost/pointusing[ 6 ]togetherwithourILPwithT=1000secondsandour1:58+algorithm Testcase[ 6 ]+ourILP1.58+Reduction(%) [1;2;3;3;1:5;1]0.120.120[2;2;3;3;1:5;1]0.2760.24212.31[3;2;3;3;1:5;1]0.4130.36411.86[4;2;3;3;1:5;1]0.5510.48611.80[5;2;3;3;1:5;1]0.6890.60811.76[1;2;7;4;5;3]0.20.20[2;2;7;4;5;3]0.4180.4023.83[3;2;7;4;5;3]0.6380.6055.17[4;2;7;4;5;3]0.8470.8074.72[5;2;7;4;5;3]1.061.014.72[1;2;6;4;5;3]0.1870.1737.49[2;2;6;4;5;3]0.3730.3447.77[3;2;6;4;5;3]0.5510.5176.17[4;2;6;4;5;3]0.7350.6915.99[5;2;6;4;5;3]0.9230.8636.5[1;2;7;5;5;4]0.130.11511.54[2;2;7;5;5;4]0.2560.22711.33[3;2;7;5;5;4]0.3480.3412.01[4;2;7;5;5;4]0.4770.4564.4[5;2;7;5;5;4]0.6050.5686.12[1;3;5;5;3;3;1:5;1]0.0940.082312.45[2;3;5;5;3;3;1:5;1]0.1750.1636.86[3;3;5;5;3;3;1:5;1]0.2600.2446.15[4;3;5;5;3;3;1:5;1]0.3490.3256.88[5;3;5;5;3;3;1:5;1]0.4380.4077.08 54

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3-1 ),beginsbytilingRwithregularhexagonswhosesideshavelengthL;someofthehexagonsthatoverlaptheboundaryofRmaycontainportionsoftheEuclideanspacethatarenotinR.Next,foreachhexagon,Hi,ofthetiling,wendanoptimal(i.e.,leastcost)sensordeploymentthatq-coversHi\R.Finally,theoptimalsensordeploymentsforthehexagonsinthetilingarecombinedbyensuringthatnositeinShastwoormoresensors.Tothealgorithmof 55

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Algorithm3-approx Figure 3-1 ,wemayaddanoptionalpruningstepinwhichredundantsensors(i.e.,sensorswhoseomissiondoesn'taecttheq-coverageproperty)areeliminated.Itiseasytoseethat3-approxconstructsaq-coverforRwithatmostonesensorpersiteinSprovidedsuchaq-coverexists.Inthefollowing,weestablishthatthecostoftheconstructedq-coverisatmost3timesthatofanoptimalq-coverandweanalyzethecomplexityof3-approx. 3-2 ).TheproofforthecasewhensisinnoHi(thismayhappenwhenShaslocationsoutsideofR)issimilar.Letd(p1;p2)betheEuclideandistancebetweentwopointsp1andp2andletd(Hi;Hj)bethesmallestdistancebetweentwopointsoneofwhichisinHiandtheotherisinHj.Since,d(p1;s)L>rMaxforpointsoutsideofthe7hexagonsshowninFigure 3-2 ,thesensoratscannotcoverpointsoutsideofthe7hexagons.Noticethatd(Hi;Hj)=Lfori;j2f2;4;6gaswellasfori;j2f3;5;7g.Fromthisobservation,L=2rMax+>2rMax,andthetriangleinequality,itfollowsthatforanypointp1inHiandp2inHj,i;j2f2;4;6gori;j2f3;5;7g,d(p1;s)+d(p2;s)>2rMax.Hence,eitherd(p1;s)>rMaxord(p2;s)>rMaxorboth.So,thesensoratscancoverpointsinatmostoneofH2,H4,andH6andatmostoneofH3,H5,andH7.ThissensormaycoveralsopointsinH1.Hence,pointsinatmost3ofthehexagonsofthetilingmaybecovered.

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Sensorats2Scancoverpointsinatmost3hexagons 11 ,itfollowsthattheassignmentofsensorstohexagons,assignseachsensorofOtoatmost3hexagons.Hence,Pcost(Hi)3cost(O).SincethesensorsassignedtoHiq-coverHi\R,theircostmustbeatleastthatofthedeploymentcomputedinStep2.So,thesensordeploymentfollowingStep3hasacostthatisatmost3cost(O). Thecomplexityofalgorithm3-approxisgovernedbythetimeittakestodeterminetheoptimalq-coverforeachofthehexagonsinthetiling.Whencomputingtheoptimalq-coverforahexagonH,weneedconsideronlythosesitesinSthatlineintheshadedregionshowninFigure 3-3 .TheareaAofthisregionisarea(H)+perimeter(H)rMax+rMax2.=1:5p 57

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LocationofsitesthatcancoverpointsinH 2 19 ]toarriveatanapproximationschemefortheq-coverproblem.Letl1beanintegershiftingparameter.Thecostofthesensordeploymentscomputedbyourapproximationschemewillbewithinamultiplicativefactorof1+1=lofthecostofanoptimaldeployment.Bymakinglsuitablylarge,wecanobtaindeploymentsasclosetooptimalasdesired.Figure 3-4 givesourapproximationschemeAS.Unlikealgorithm3-approx,whichconsidersasingletilingoftheregionRthatistobeq-covered,algorithmASconsidersafamily,T1,,Tl,oftilings.Toobtainatilinginthisfamily,webeginbydeterminingthesmallestboundingrectangleUofR.Next,thisboundingrectangleistiledusingtileswhoseheightequalsthatofUbutwhosewidthislL,whereL=2rMax+andisapositiveconstant.Therstandlasttilesinthetilingareexceptions.InTi,thewidthofthersttileisiL.Incasetilingwithatileof 58

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6 ,weobtaincost(Ti)cost(O)+cost(Oi)SupposethatasensorofOthatliesinthersttileofTicoversapointinthesecondtileofTi.SinceL>2rMax,thissensorispartofthesecondtileofTj,ji,thissensorremainsinthersttileforTjandisunabletocoverpointsthatarenotinthersttileofTj.Hence,thissensor,whichisinOi,isnotinanyOj,j6=i.Byreasoninginasimilarfashion,wemayshowthatallOisaredisjointandsolXi=1cost(Oi)cost(O)Hence,min1ilfcost(Ti)g1 60

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d)2+2b2r dc+1andMIN(r)(r d)22r db2r dc.ProofWithoutlossofgenerality,assumethatthesensorislocatedat(0;0)andthatPi,i=1;2;3;4arethe4gridpointsclosesttothesensor(Figure 3-6 ).WemayfurtherassumethatP1=(xd;yd)isclosertothesensorthanarePi,i=2;3;4.Clearly,0x1 2and0y1 2.LabeltherowinwhichP1liesrow0,therowrightabove(below)P1isrow1(1),andsoon.Letmaxrowbetherownumberofthehighestrowthatispartiallycoveredbythesensorandletminrowbecorrespondinglowestrow.Notethatb2r dcmaxrowminrow+1b2r dc+1. 62

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d)2(y1+i)2crow(i)b2p d)2(y1+i)2c+1.MAX(r)=maxrowXi=minrowrow(i)maxrowXi=minrow(b2r d)2(y1+i)2c+1)maxrowXi=minrow(b2r d)2(y1+i)2c)+b2r dc+11Xi=minrow(2r d)2(y1+i)2)+b2r dc+maxrowXi=1(2r d)2(y1+i)2)+b2r dc+12Zr dr dr d)2y2dy+2b2r dc+1(r d)2+2b2r dc+1MIN(r)=maxrowXi=minrowrow(i)maxrowXi=minrowb2r d)2(y1+i)2c>maxrowXi=minrow(2r d)2(y1+i)21)maxrowXi=minrow(2r d)2(y1+i)2)b2r dc(r d)22r db2r dc 12 exists,cost(OPTG(d;r0))Fcost(OPTR(r)),whereFMAX(r)MIN(r0)+1.

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ThesensorsislocatedinsidethesquareformedbyfourgridpointsPi,i=1;2;3;4,andcoversMAX(r)gridpoints,shownindarkcolor. d2:399.Proof

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d)2+2b2r dc+1((rd d)2+2b2r dc+1((r d1 d1 dp d)2+4r d+1(r d1 d1 dp d+12p d3:399Hence,FMAX(r)MIN(r0)+112:443r d2:399. d2:399)cost(OPTR(r)).ProofFollowsfromCorollary 1 andLemma 15 Noticethattheuseofanergrid(i.e.,smallerd)resultsinalargerapproximationfactor.Thiswouldargueinfavorofusingthemaximumpossibled;thatisd=p d2:399)=12:443=p 65

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3-7 )layout,exactlyonesensorpergridpoint.WeusethesymbolsT,S,andH,respectively,torefertothethreegridgeometries(equilateraltriangle,square,andregularhexagon)consideredbyus.Weusedtodenotethesize(i.e.,sidelength)ofthegeometrybeingconsidered.Fortheproblemconsideredinthesection,thereisonly1sensortype;thesensorrangeisr;andwewishtominimizethenumberofsensorsdeployed.WeassumethatrismuchsmallerthantheheightandwidthoftherectangularregionRthatistobeq-covered.So,weneglectboundaryeectsinouranalysis.Withtheseassumptions,minimizingthenumberofsensorsforeachgriddedlayoutisequivalenttondingdMax(X;q),X2fT;S;Hg,themaximumvalueofdforthegeometryXforwhichtheresultinggriddedlayoutq-coverstheplane.Weconsiderthecasewhen1q5,whichwebelievetobethemostpractical. Griddedlayoutusingageometrysized.A)Equilateraltriangle.B)Square.C)Regularhexagon. 66

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Equilateraltriangle1p 2r22 2r2 2r22 2r3 8p 7p Square1p 5r25 5p 25r225 5r2 5r2 Regularhexagon1r3 4p 4p 7p 7r75 196p 4r24 Following[ 1 ],weuseqXtodenotethemaximumareapernode(APN)inthegriddedlayout 2dMax(T;q)2qS=dMax(S;q)2qH=3 4p 67

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Regularhexagon,q=2 isthecenterofahexagonandhasnosensor).So,foreverypointPthatisnotthevertexofatilinghexagon,d2(P)
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Square,q=3 Regularhexagon,q=3 2d.So,d3(p)
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Equilateraltriangle,q=4 sensorsatB,D,andF.However,thenextnearestsensors(thoseatA,C,andE)areatadistanceof2 2r. 3-14 shows4squaresofthegriddedlayout.Thereisasensorateachcornerofeachsquare(i.e.,gridposition).ThepointPisthecircumcenteroftheisoscelestriangleACE.Notethatd(A;C)=d(E;C)=p 6p 2d
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Regularhexagon,q=5 Next,considerthetriangleBFP,whereFisthemidpointofthesideBE.Foreverypointp6=Pinthistriangle,d(p;B)
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3.1 3.3 ,wepresentedseveralstrategiestoq-coveraregionwhenthesensorlocationsarelimitedtoaspeciedsetS.WebelievethatallbuttheregioncoverageviapointcoveragemethodofSection 3.3 aretoocomputeintensivetobeusedonlargedeploymentinstances.Thisisbecauseeachoftheothermethodsrequiresustondoptimaldeploymentsforseveralsmallhexagonsortiles.WhileanoptimaldeploymentforahexagonortilemaybefoundinO(1)time,therequiredconstanttimeisexpectedtobelargeastheoptimalsolutionisfoundbysearchingapotentiallylargethoughconstantsizespaceofpossiblesolutions.Thissearchwillperformmanyexplicittestsforregionratherthanpointcoveragearetimeconsumingandregioncoveragetestsarecomputeintensive.Asaresult,ourexperimentsfocusonthemethodofSection 3.3 adaptedtoinstanceswithmultiplesensortypes.WeprogrammedtheregioncoverageviapointcoverageschemeofSection 3.3 onaDellDimensionPCwitha2.13GHzdual-coreprocessorand2GBmemory.TocovertheresultinggridG(d),weusedanILP(integerlinearprogramming)relaxationmethodaswellthegreedyalgorithmofXuandSahni[ 69 ].FortheILPrelaxation,westartedwiththeILPformulatedin[ 69 ]forpointcoverage,relaxedittoalinearprogram,solvedthelinearprogramusinglp-solver5.0[ 31 ],andthenconvertedthesolutiontothelinearprogramtoafeasibleintegersolutiontotheoriginalILPusingtheroundingmethodproposedbyWangandZhang[ 62 ].SinceboththeILPformulationandgreedymethodof[ 69 ]areforheterogeneoussensordeployment,weexperimentedwithheterogeneousinstanceswithupto4sensortypes.Fortestdata,weuseda3535regionR.ToconstructG(d)foranygivend,wetiledRwithddsquaresandconsideredthecentersofthesetiles.CentersthatwereoutsideRwererelocatedtotheirnearestpointontheboundaryofR.TheresultingcenterlocationsdeneG(d).ToconstructthesetSofpermissiblesensorlocations,wetiledRwith7rmax=127rmax=12squaresandincludedthecenterofeachsquareinS,wherermax

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Totalsensorcostrequiredv.s.variouss,whereq=1. 80

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Totalsensorcostrequiredv.s.variouss,whereq=2. 81

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Totalsensorcostrequiredv.s.variouss,whereq=3. 82

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7 13 17 34 45 46 53 70 ]givenasetofdierenceoftime-of-arrivalmeasurementshavebeenproposed,thereappearstobenostudiesoffundamentalpropertiesofDTOAlocalization.Inthischapter,wepresentanumberofresultsthatestablishfundamentalpropertiesofDTOAlocalization.Werstconsidertheuniqueidenticationofasourceandestablishthefollowing:1.DTOAlocalizationuniquelyidentiesasourceinEuclideanplaneR2ithesensorsdonotlieonahyperbola 83

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4.1 ,wepresentsomefundamentalpropertiesanddenitions.PropertiesofsensorsetsthatuniquelyidentifyallsourcesinEuclideanspacearedevelopedinSection 4.2 .OurdetailedanalysisofSection 4.3 establishestheboundonthenumberofintersectionsoftwoDTOAhyperbolas.InSection 4.4 weshowthatatmost2pointscanhavethesamesetofDTOAvalues.TheminimumnumberofsensorsneededtouniquelyidentifyallsourcesinaboundingpolygonisderivedinSection 4.5 4-1 ).Whenij(P)=ij(Q)foreveryi;j2f1;2;:::;kg,thepointsPandQareindistinguishable.Actually,sinceij(P)=1j(P)1i(P),foralliandj,PandQareindistinguishablei1j(P)=1j(Q)foreveryj2f2;:::;kg.So,thesetofsensor 84

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ExamplesofthelocusL12 4-2 givesanexampleoftwohyperbolasL12(12)andL13(13)thatintersectattwodistinctlocationsP1andP2.So,usingL12andL13alone,weareunabletouniquelylocalizethesource.WeareableonlytoassertthatthesourcelocationiseitherP1orP2.TouniquelyidentifythesourceusingtheDTOAmethod,thehyperbolasL1j,2jkshouldhaveexactlyonecommonintersection.Alternatively,thesehyperbolasshouldhaveexactlyonecommonintersectioninsidearegioninwhichthesourceisknowntolie. 85

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Threenon-collinearsensorsS1,S2,andS3formatriangleandtwohyperbolasL12(12)andL13(13)intersecteachotheratP1andP2. 9 anecessaryandsucientconditionforasensorsetSStobeanISS.Theorem 10 showsthateveryISShasatleast4sensorsandTheorem 12 showsthateveryISSwithmorethan6sensorshasasubsetofsizeatmost6thatisanISS. 86

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9 ,itfollowsthatSSisnotanISS.WhenS1,S2,andS3arenotcollinear,theydeneanontrivialtriangleasshowninFigure 4-2 .Clearly,thereexistsanegativeconstant,12,suchthatthehyperbolaL12(12)intersectsthelineS1S3attwodistinctpointsQ1andQ2.ObservethatthehyperbolaL13(d(S1;S3))isactuallyaraythatoriginatesatS1andintersectsL12(12)atQ1only.Let13beanegativeconstantslightlygreaterthand(S1;S3).ThehyperbolaL13(13)intersectsL12(12)attwodistinctpointsP1andP2(seeFigure 4-2 ).So,P1andP2areindistinguishableandSSisnotanISS.Next,weshowthatwheneverSS=fS1;S2;S3;S4garethecornersofaparallelogramwithsidelength>0,SSisanISS.Weshowthisbyprovingthatno4distinctpointsofahyperboladenethecornersofaparallelogram.TheresultthenfollowsfromTheorem 9 .ConsiderthehyperbolaLofFigure 4-3 .LetS1,S2,S3,andS4be4pointsonthishyperbola.ThecaseshowninFigure 4-3 hasS1andS4ononepart(arm)ofthehyperbolaandS2andS3onthesecondpart.(Therearetwoothercasesforthelocationofthe4points{exactly3pointsononepartofLand4pointsononepartofL.)LetQ1andQ2,respectively,betheintersectionsofthelinesegments 87

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AhyperbolaLthatpassesthroughSi(1i4). cornersofaparallelogram, 12 establishesanupperboundof6onthesizeofanMISS.Toprovethistheorem,weneedtouseBezout'sboundonthenumberofintersectionsofcurvesinEuclideanspace. [Bezout'sTheorem[ 28 ]]:LetC1andC2becurvesofdegreemandn,respectively,inEuclideanspaceR2.IfC1andC2havenocurvesincommon,thenthenumberofintersectionsofC1andC2isatmostmn.

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4 .Hence,atmost1hyperbolamaypassthroughthepointsofSS. 10 andthefactthataMISSisanISS.jSSj6maybeshownbycontradiction.SupposethatjSSj>6.LetSS0beasubsetofSSsuchthatjSS0j=5.FromLemma 16 ,SS0hasatmost1hyperbolapassingthroughits5points.Ifnohyperbolapassesthroughthesepoints,thenSS0isanISS(Theorem 9 )andSScannotbeanMISS.So,wemayassumethatexactlyonehyperbolapassesthroughSS0.SinceSSisanISS,SScontainsatleastonepointSithatdoesnotlieonthishyperbola.Hence,thereisnohyperbolathatpassesthroughthe6pointsSS0SfSig.FromTheorem 9 ,itfollowsthatSS0SfSigSSisanISS.ThiscontradictstheassumptionthatSSisanMISS. 4 ),twoDTOAhyperbolasL12andL13mayhavenomorethan2intersectionswhenS1,S2,andS3arenon-collinear.Withoutlossofgenerality(w.l.o.g),wechooseourcoordinatesystemasinFigure 4-4 .Thefeaturesofthischoiceare(a) 13 89

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Regionsofmonitoringarea:(a)topleft,(b)inside,(c)bottomright,(d)top,(e)bottomleft,(f)bottom,and(g)topright. (RefertothedetailedprooffromSection 5.4 )thatestablishesthemonotonicityofthedirectionalderivativeof13(P)alongthehyperbolaL12(12(P))withineachofthe7regionsofFigure 4-4 .ThesignofthedirectionalderivativeforeachregionisalsogiveninFigure 4-4 4-4 .Thedirectionalderivativeispositiveinregions(a),(b),(f),and(g),andisnegativeinregions(c),(d),and(e).Inthefollowing,weuseLlandLrtorefertothetwosymmetricparts(arms)ofthehyperbolaL(seeFigure 4-5 ).ThetwopartsLlandLrintersectonlyatthevertexB.l1andl2arethetwoasymptotesofthehyperbolaandl10andl20arelinesthatintersectatthevertexBandareparalleltotheseasymptotes.Fromourchoiceofcoordinatesystem,itfollowsthattheasymptotesintersectatO. 2. TheshortestEuclideandistancebetweenapointPonLr(Ll)andtheasymptotel1(l2)decreasesmonotonicallyasPgetsfartherfromthevertexB.

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AhyperbolaL=LlSLrwithfocusSandsemimajoraxisy-axis. TheshortestEuclideandistancebetweenapointPonLr(Ll)andthelinel10(l20)increasesmonotonicallyasPgetsfartherfromthevertexB.ProofFollowsfromthedenitionofahyperbola,itsasymptotes,andthelinesl10andl20. InTheorem 14 ,weshowthatwhenS1isclosertothesourceSthanareS2andS3,L12(12(S))andL13(13(S))haveatmost2intersectionsincludingthesourceS.ThisrestrictiononthesourcebeingclosertoS1thantheremainingtwosensorsisremovedinTheorem 15 .WeoftenuseLijasanabbreviationforLij(ij(S)). 91

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Case1: 4-6 4-9 ,respectively.WeshowbelowthatL12andL13haveatmost2intersectionsineachofthesecases.Case1: 4-6 ).13,fromTheorem 13 ,monotonicallyincreasesinregions(a),(b),and(g)andmonotonicallydecreasesin(d).So,ifnocomponentofL12isinregion(d),then13monotonicallyincreasesalongallofL12andthevalueof13foreachpointPonL12isunique.Hence,L12andL13haveonly1intersection.Ifregion(d)containsaportionofL12,thenwhenonemovesthepointPfromlefttorightalongL12,(d)istherstregiontobevisited.So,whenmovingfromlefttorightalongL12,13monotonicallydecreaseswhilewearemovingalongtheportionofL12thatisinsideregion(d)andthenmonotonicallyincreasesfortheremainderofL12.HenceL12hasatmost2distinctpointsforanygivenvalueof13.So,L12andL13haveatmost2intersections.Case2:

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4-7 ).So,L12cannothaveacomponentineitheroftheregions(c)(bottomright)and(f)(bottom).Additionally,L12cannothaveacomponentinregion(d)(top).Toseethis,observethatLr12iswhollytotherightofthey-axiswhileregion(d)iswhollytotheleftofthisaxis.So,noportionofLr12isinregion(d).ToseethatnoportionofLl12isinregion(d)either,notethatLl12isbelowl20(Lemma 17 ).Since, 17 ), 4-8 ).Hence,noportionofL12isinregion(e).Since\S3S1S290,<90(seeFigure 4-8 ).Hence,d(P;S1)>d(P;S3)foreverypointPinsideregion(c)(bottomright).Since,byassumption,S1isclosertothesourceSthanisS3,noportionofL13isinregion(c).Hence,L12andL13havenointersectioninregion(c).IfL12hasanoverlapwithregion(d)(top),thenregion(d)istherstregionencounteredaswemovefromlefttorightalongL12andifL12overlapswithregion(c)(bottomright),region(c)isthelastregionencounteredaswemovefromlefttorightalongL12.13monotonicallydecreasesinregion(d),L12andL13donotintersectinregion(c),and13monotonicallyincreasesintheremainingregionsthatL12mayoverlap.So,L12andL13haveatmost2intersections. 93

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Case2: Case3: 94

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Case4: Case4: 95

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14 thatL21andL23haveatmost2intersections.Hence,L12andL13haveatmost2intersections. Clearly,theindistinguishablerelationpartitionsEuclideanspaceR2intoacollectionofdisjointequivalenceclasses.IfSSisanISS,theneachequivalenceclassisofunitcardinality;otherwise,thecardinalityofatleastoneequivalenceclassismorethan1. 96

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Collinearsensors Whenk=2,eachequivalenceclasscorrespondstoahyperbolawithfociS1andS2andviceverse.Thecardinalityofeachequivalenceclassinthiscaseisinnite.Whenk>2andthesensorsarecollinear(Figure 4-10 ),eachpointonthelinesegment 17 establishesthatthecardinalityofeachequivalenceclassisatmost2. 9 ,itfollowsthatthereisahyperbolaL12,whosefociareP1andP2,thatpassesthroughthepointsofSS.Similarly,thereisahyperbolaL13,whosefociareP1andP3,thatpassesthroughthepointsofSS.L12andL13intersectatatleastthepointsofSS,whicharemorethan2innumber.ThiscontradictsTheorem 15 ,whichstatesthatthesetwohyperbolamayhaveatmosttwointersections. 97

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10 ),inmanyreal-worldapplications,themonitoringregionisboundedbyapolygonand3sensorssuce.Weassumethatthesensorsarerestrictedtobeplacedonorinsidetheboundingpolygon.Asanaside,wenotethatwhenthemonitoringregionisasimplelinesegment,say 17 ,itfollowsthatthedualofeverypointofthepolygonthatisnotonthisedgeisontheothersideofthisedge.Pointsontheedgeeitherhavenodualorhavedual(s)outsidethepolygon.HenceeverypointinoronthepolygonisuniquelyidentiableandfS1;S2;S3gisasize3MISSforthepolygon.Analternativeconstructionforasize3MISSistoconsiderany3non-collinearpointsS1,S2,andS3thatareontheboundaryofthepolygon(Figure 4-11 ).Now,theentireconvexpolygonmustbecontainedintheunionoffourregions:(a)topleft,(b)inside,(f)bottom,and(g)topright.FromTheorem 13 ,thedirectionalderivativeof13alongL12increasesmonotonicallyineachofthesefourregions.Further,theintersectionofL12andtheconvexpolygonisacontinuouscurveCthatislimitedtothesefourregions 98

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SensorsS1,S2,andS3ontheboundaryofaconvexpolygon. (seeTheorem 14 ).Since,13ismonotonicallyincreasingalongC,L12andL13haveatmostoneintersectiononC.Hence,everypointinorontheconvexpolygonisuniquelyidentiable.Case2:Thesimplepolygonisconcave.Westartwithaaminimumboundingconvexpolygonoftheconcavepolygon(Figure 4-12 ).LetS1,S2,andS3beanythreepointsontheintersectionoftheboundaryoftheseconcaveandconvexpolygons.FromCase1,itfollowsthateverypointinandontheboundaryoftheconvexboundingpolygon,andsoeverypointinandontheboundaryoftheconcavepolygon,isuniquelyidentiable. InLemma 18 ,weprovethatbychoosing3sensorlocationsontheboundaryofasimplepolygon,anSSofsize3uniquelyidentiesanysourceSonorinsideasimplepolygon.WeshowinLemma 19 whenasensorisplacedstrictlyinsideasimplepolygon,3sensorsarenotsucienttouniquelyidentifyeverypointinoronthepolygon.

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Aconcavepolygon,itsboundingconvexpolygon,andthreesensorsS1,S2,andS3placedonthecommonboundaryoftheconcaveandconvexpolygons 4-13 .Notethataportionofthesimplepolygonmustlieinsidethetopregion.Wemaychoosetwonegativeconstants12and13,suchthatL12(12)andL13(13)intersectattwodistinctpointsP1inthetopregionandP2inthetopleftregion.SincebothP1andP2areinsidethesimplepolygonandP1isthedualofP2,SSisnotanISSforthepointsofthesimplepolygon. 18 and 19 100

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101

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4-1 ).TheDTOAspaceofall(k1)-tuples[12(P),13(P),...,1k(P)]formsa(k1)-dimensionalvectorspacedenotedbyk1.EachpointP=(x;y)2R2hasauniquedualpointP0=(p10;p20;:::;pk10)ink1,wherepj0=1(j+1)(P),j=1;2;:::;(k1).However,eachpointP0=(p10;p20;:::;pk10)ink1mayhavezeroormoredualpointsinR2.Infact,thedualpointsofP0arethosepointsinR2thatarecommonto(i.e.,thecommonintersections)thek1hyperbolasL1;j+1(P),1j
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5-1 .The[DX1;DX2][DY1;DY2]boxshowninFigure 5-1 isthemonitoringregionwithinwhichthesourceSistobelocalized.Thelines 5-1 .Althoughthisgurehas 104

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Canonicalplacementof3sensorsandpartitioningofmonitoringregion allsensorswithinthemonitoringregion,ourdevelopmentofthegeometriclocalizationmethoddoesnotrequirethis.Infact,themethodworksevenwhensomeorallofthesensorsareoutsidethemonitoringregion.Figure 5-2 showsourthreesensorstogetherwiththelocusL12(12).ThislocusmaybepartitionedintosegmentsthatliewhollywithinaregionofthepartitioningofFigure 5-1 .ThesegmentendpointsaredesignatedSj,wherejisalowercaseletter.So, 5-2 .Noticethatbecauseofourchoiceofcoordinatesystem,aswemoveapointPalonganysegmentofL12(12),thexandycoordinatesofthepointvarymonotonically.ThisisaconsequenceoftheverticalorientationofL12,which,inturn,isassuredbythechosencoordinatesystem.Let(xi;yi)and(xj;yj),xixjbetheendpointsofanL12segmentandletP=(x;y)beanypointonthissegment.FromLemma 22 (Section 5.3 ),itfollowsthatxixxjandminfyi;yjgymaxfyi;yjg.Also,aswemovePalongasegmentofL12(12),13(P)variesmonotonically(Section 5.4 ).Inparticular,itmonotonically 105

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-xyOS1S2S3SaSbSc,,,,,,,,,,,,,,,,HHHHHHHHHHHHHHHHHHHH Canonicalplacementof3sensorsandL12(12) decreaseswithxforthesegmentsinthetop,bottomleft,andbottomrightregionsandmonotonicallyincreasesfortheremainingsegments.Basedonthesekeyobservations,ouroverallstrategytoestimatethesourceSistoutilizethemonotonicityof13(P)toperformabinarysearchwithineachsegmentofL12todetermineaset,U,ofpointssuchthatUhasatleastonepointwithinaspeciedaccuracyofeachintersectionbetweenL12(12)andL13(13)thatisinthemonitoringregion.Further,thenumberofpointsinUisatmostequaltothenumberofsuchintersections.Recallthatthenumberofsuchintersectionsisatmost2asprovedinTheorem 15 .Itfollowsthatthetruesourcelocationiswithinadistance(inR2)ofoneofthepointsinU.Thedetailsarepresentedinalgorithmgeometric DTOA(12;13).Algorithmgeometric DTOArstdeterminesthesegmentsofL12.TheendpointsofthesesegmentsarejusttheintersectionsofthecurveL12(12)witheachofthethreelines 106

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5.4 ,L13ismonotoneonboththeseverticalsegments. 5.4 .ThefollowingassumesthatL12isnotaverticalray.ThecorrectnessproofforthecasewhenL12isaverticalray(notethatthiscase,whichisnotincludedinthestatementofalgorithmgeometric DTOA,ishandledbyabinarysearchonasegmentofthey-axis)issimilarandsimpler. 22 ,wehavexixxjandyiyyj(oryjyyi).So,maxfjxixj;jxxjjgjxixjjandmaxfjyiyj;jyyjjgjyiyjj.Hence,maxfd(Si;P);d(Sj;P)gp 109

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22 and 23 ,itfollowsthatd(P1;P)andd(P2;P).Thelemmanowfollowsfromtheobservationthatthepoint^PreturnedbythealgorithmiseitherP1orP2. DTOAcontainsatleastonepointthatiswithinofeachintersectionbetweenL12(12)andL13(13)thatisinthemonitoringregionandthenumberofpointsinUisatmostequaltothenumberofsuchintersectionsinthemonitoringregion.Hence,atleastonepointofUiswithinofthetruesourcelocationprovidedthislocationisinthemonitoringregion.ProofThetheoremfollowsfromLemma 24 andtheobservations(a)everysegment(orsegmentportion)ofL12(12)inthemonitoringregionissearched,(b)everyintersectionwithinthemonitoringregionisonexactlyoneofthesegments,ofL12,and(c)algorithmlocate NotethatthepointsinthesetUreturnedbyalgorithmgeometric DTOAareonthelocusL12(12).So,foreachpointP2U,12(P)=12.SinceeachreturnedpointP2Uiswithin,inR2space,ofanintersectionofL12(12)andL13(13),itfollowsthat13(P)2p 21 ).Bychangingtheconditiononthebinarysearchloopofalgorithmlocate 15 ,atmost2suchcallsmakeO(log(l=))callstolocate DTOAisO(log(l=)),whichcanbeadapted 110

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5-1 .Weshowthatthedirectionalderivativeof13(:)alongthecurveL12(:)ismonotoneineachoftheseregions:itispositiveinregions(a),(b),(f),and(g)andisnegativeinregions(c),(d),and(e).Wehavefori=1;2;3,@d(P;Si) 46 ]. 111

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@12(P)

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5-3 .Thedirectionalderivativeisgivenby264xx1 113

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114

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5-4 .Thedirectionalderivativeof(S1;S3)onthelocusf(x;y)j(S1;S2)=12g,forany12,isgivenby264xx1 115

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5-5 .Thedirectionalderivativeof(S1;S3)onthelocusf(x;y)j(S1;S2)=12g,forany12,isgivenby264xx1 116

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5-6 ,whichmakesthethirdcostermnegative,andhencethedirectionalderivativeisnegative.BottomLeft:Thecaseofbottomleftisidenticaltothetopleftregionexceptthat2>3asshowninFigure 5-7 ,whichmakesthesintermnegative,andhencethedirectionalderivativeisnegative.BottomRegion:Forbottomregion,thederivationisidenticaltothecaseofinsideregionexceptthat<2+3<2asshowninFigure 5-8 ,whichkeepstherstsintermstillpositive,andhencethedirectionalderivativeispositive.TopRight:Thecaseoftoprightregion,asshowninFigure 5-9 ,isidenticaltoinsideregionexceptthat3<1.Thuswehave0<13 5-10 .WhenL12isaverticalray,weneedtoconsidertheportionofthesegments(a)fromS1to1and(b)fromS2tothatliewithinthemonitoringregion.Weconsideronly(a).Theprooffor(b)issimilar.LetP1andP2betwopointsonthesegment(a).W.l.o.g.,assumethatP1isclosertoS1thanisP2(seeFigure 5-11 ).Weseethat 117

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SourceS=(x;y)israndomlyselected,andthesignofthedirectionalderivativeiscomputed. -xyOS1S2S3P1P2,,,,,,,,,,,,,,,,HHHHHHHHHHHHHHHHHHHHFigure5-11. ThedegeneratecasewhenL12(12)isaverticalray 118

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DataforS1=(0;0),S2=(0;50000),S3=(0:001;100000),andF=0 Method#failratiospaceR2space Mellen1610.012710013280.10511007220.057150046830.370650023840.1887100076980.6093100040040.31692500115030.9104250074870.59265000124200.98305000105120.832010000124730.987210000122490.9694 Ours00.00.0000000126190.20730.000000012460.01950.0000001120110.95060.000000122730.17990.000001126351.00.000001126351.0 DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=0 Method#failratiospaceR2space Mellen34260.2775100600.004910030.000243015001340.0109500160.001310002100.01701000300.002425003780.03062500880.007150005720.046350002000.0162100009350.0757100004210.03415000066520.53885000020700.167710000089190.722510000040370.3299 Ours00.00.0000000114370.11640.000000012250.01820.0000001105460.85430.000000121850.17700.000001123451.00.000001123451.0 119

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DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=10/100 Method#failratiospaceR2space Mellen39440.3131100009070.0720100002760.02192500024550.1949250008610.06835000080340.63775000019220.1526100000118730.942510000039920.3169 Ours21490.1706100250.002010000.0100011830.093910002580.0205250044950.3568250010670.0847500092730.7361500024830.197110000124810.99071000050970.4046 DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=5/100 Method#failratiospaceR2space Mellen37080.3005100009010.0730100003040.02462500022300.1807250008770.07115000063820.51715000018810.152410000087210.706710000039780.3223 Ours14840.1202100780.006310090.00072928100030480.247010006720.0545250089450.7248250021290.17255000122360.9915500045440.368210000123411.01000080290.6506 DataforS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andF=1/100 Method#failratiospaceR2space Mellen35130.2788100009510.0755100004090.03252500021850.17342500010630.08445000068290.54205000020920.166010000091090.723010000041400.3286 Ours6500.05161008400.06671001560.012425036930.29312507600.060350085940.682150019110.15171000124800.9906100041640.33052500125991.0250088460.70215000125991.05000116910.927910000125991.010000125770.9983

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DataforS1=(0;0),S2=(0;50000),S3=(5000;100000),andF=10/100 Method#failratiospaceR2space Mellen16490.1322250520.0042250190.00155001890.0152500580.004710007250.058110002150.0172250034040.272925009320.0747500076040.6096500022580.181010000106280.85211000048040.3852 Ours16470.1320250780.0063250190.00155002870.0230500580.0047100010260.082310002150.0172250042370.339725009320.0747500090930.7290500023640.189510000123890.99331000053780.4312 DataforS1=(0;0),S2=(0;50000),S3=(5000;100000),andF=5/100 Method#failratiospaceR2space Mellen10070.08002502280.0181250510.00415007980.06345002380.0189100025360.201410007260.0577250081830.6499250024680.19605000115020.9135500050090.397810000115790.91961000082950.6588 Ours10070.08002502580.0205250510.00415009210.07315002380.0189100028860.229210007260.0577250090450.7184250024680.19605000125140.9939500051160.406310000125911.01000087760.6970 5.2 versusthelinearalgebramethodof[ 34 53 ],whichrequiresasolutiontoaquadraticequationaswellastheinversionofmatrices.BothalgorithmswereimplementedinMatlabonaDellDimensionPCwitha2.13GHzdual-coreprocessorand2GBmemory.Thetypicalexecutiontimesofbothmethodsareonlyseveralmilliseconds. 121

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DataforS1=(0;0),S2=(0;50000),S3=(5000;100000),andF=1/100 Method#failratiospaceR2space Mellen2670.02111007530.05941002190.017325035820.28242509650.076150084590.667050023740.18721000123370.9727100050300.39662500124090.9784250096360.75985000124090.97845000119260.9403 Ours2670.02111007600.05991002190.017325036520.28792509650.076150086720.683750023740.18721000126110.9943100050300.39662500126831.0250096360.75985000126831.05000120080.9468 Table5-9. DataforS1=(0;0),S2=(0;100000),S3=(100000;0),andF=10/100 Method#failratiospaceR2space Mellen200.0016500120.000958625004350.03471000390.0031100015200.121425001820.0145250062700.500950006240.04985000109090.87151000018750.149810000124490.9945 Ours00.02501190.00952501190.00955001930.01545004350.034710007250.0579100015200.1214250034600.2764250062720.5010500085040.67935000109170.872110000125110.999410000124650.9958 DataforS1=(0;0),S2=(0;100000),S3=(100000;0),andF=5/100 Method#failratiospaceR2space Mellen40.00032500440.003550015160.121410001120.0090100047290.378825005420.04342500110420.8846500015420.12355000124590.9981 Ours00.02501830.01472504150.03325006800.054550015160.1214100023920.1916100047290.3788250084940.68042500110430.88465000124780.99965000124630.9984 122

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DataforS1=(0;0),S2=(0;100000),S3=(100000;0),andF=1/100 Method#failratiospaceR2space Mellen00.02501490.012025063370.51115005000.0403500110290.8896100013820.11151000123940.9997250013930.11242500123981.0 Ours00.01007080.057110015420.124425034320.276825063370.511150084360.6804500110290.88961000123940.99971000123940.99972500123981.02500123981.0 Eachsensormeasurementcorrespondsto(1+f)rwhereristheactualdistancefromsensortosource,andfisuniformlyrandomlygeneratedintheinterval[0;F]foraxedmultiplicativefactorF.Whilefvaluesaregeneratedindependently,sensorerrormagnitudeisproportionaltothedistancefromthesensortoplumeorigin.Also,thesensorerrorsarecorrelatedduetothespatialrelationshipsbetweenthesensorlocations.asourceclosetoonesensorgeneratesasmallerrorthereandlargererrorsatothersensors,whicharelocatedfartheraway.Fromthesemeasurements,wecomputeddistance-dierencesandtestedDTOAlocalizationmethods.Inourexperiments,Weconsideredtwodierentscenarios:(1)sensorerrorsarezero(i.e.,F=0),and(2)sensorerrorsaregreaterthanzero(i.e.,F>0).Onarelatednote,themethodof[ 7 ]accountsforrandomerrorsthatareindependentGaussian,andhenceisnotdirectlyapplicabletothiscase.Oursimulationwasconductedinanetworkofthreesensorsona[0;100000][0;100000]grid,wherelocationofsourcesarerandomlygeneratedbasedontheuniformdistribution. 34 53 ]may 123

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46 ].Bygoodtriangle,wemeanitssmallest(largest)angleisnotcloseto0(180)degree.However,whenthreesensorslieinanalmostcollinearmanner,themethodof[ 34 ]mayfailtondasolutioneitherbecausethequadraticequationhasimaginaryrootsorbecausethematricesbeinginvertedareclosetosingular.AlthoughourmethodalsocouldpotentiallyfailbecausetherootsofthequadraticbeingsolvedbyAlgorithmlocate 5-1 and 5-2 ).Forourexperiments,eachtestcasemaybedescribedbyatupleof[S1;S2;S3;F;N],whereS1,S2,andS3arecoordinatesofthreerespectivesensors,Fisthesensorerror,andNisthenumberofrandomlygeneratedsources.NotethatwealwayskeepS1closesttothesource.Table 5-1 givesthenumberofsourcessuchthat[ 34 ]returnsimaginaryrootsaswellasourmethodfailstondasolutionwhereS1=(0;0),S2=(0;50000),S3=(0:001;100000),andN=12635.Theratioofthenumberofsuchsourcesagainstthetotalnumberofsourcesisgiven.Foreachtestcase,weconsidervarious1and2,where1and2arethedesirederrorsacceptableinspaceandR2space,respectively.Foreach1(2),Table 5-1 givesthenumberofsourceswhoseestimatereturnedby[ 34 ]aswellasbyourmethodiswithinthedesirederror1(2)oftheactualsource.Theratioofthenumberofsuchsourcestothetotalnumberofsourcesalsoisgiven.Table 5-2 givesthisdataforthecasewhereS1=(0;0),S2=(0;50000),S3=(0:0000000000000001;100000),andN=12345.Wenotethatusingourbinarysearchbasedmethodversusthatof[ 34 ]hadagreatimpactonthenumberofsourcesthatcouldbeestimated.Forexample,forthetwotestcasesshowninTables 5-1 and 5-2 thepercentageofsourcesforwhichthemethodof[ 34 ]failedtoreturnanestimate(eitherbecausethequadraticbeingsolvedhadimaginaryrootsorbecauseoffailuretoinvertmatrices)is1.27%and27.75%,respectively,whereasourmethodneverfailedtoestimatethesource.Notethatalmost28%ofsourcescan'tbeestimatedbythemethodof[ 34 ]inthesecondtestcase.Further,thesourceestimategivenbyourmethodalsoshowsmuchbetteraccuracyinbothspaceandR2spacethanthat 124

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34 ].AsshowninTable 5-1 ,togettheratioofsuccessfulestimatestobemorethan98%inspace,themethodof[ 34 ]needstoset1tobealmost5000,whereasourmethodalwaysgivesthesuccessfulestimatewhen1isassmallas0.000001!AsimilarphenomenonisobservedinR2spaceaswell.When2issetto10000,thesuccessratioofthemethodof[ 34 ]isstillslightlylessthan97%,whereasourmethodachieves100%successevenifwereduce2byasmuchasabout1010times!ThisimprovementisevenmoreimpressiveforthesecondtestcaseshowninTable 5-2 ,wheretheestimatingqualityisimprovedbymorethan1011times!AnotherobservationisthatanestimatethatisaccurateinspacemaynotbeaccurateinR2space.Forexample,inTable 5-1 ,when1and2areboth1000,theratioofsuccessfulestimatesinspaceandR2spaceis60.93%and31.69%,respectively,whichimpliesthatmorethan29%oftheestimatesthatareclosetothesourceinspacearedistantfromthesourceinR2space. 5.2 .WhenUisempty,inotherwords,algorithmlocate 5-3 5-11 givethesimulationresults.Specically,thevaluelistedunder#failforourmethodgivesthenumberofsourcesthatcouldnotbeestimatedwithout 125

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34 ]intermsofthenumberofsourcesthatcouldbeestimated.Inall9ofourtestcases,thenumberofsourcesthatcouldnotbeestimatedbyourmethodwithoutnalizationislessthanorequaltothatofthemethodof[ 34 ];thereductionswereashighas22%.Asnotedearlier(Section 5.2 ),shouldthequadraticinAlgorithmlocate 34 ].Inparticular,whenthreesensorsarealmostcollinear,theimprovementmadebyourmethodissignicant.Forexample,when2is10000asshowninTables 5-3 5-5 ,theincrementoftheratioofsuccessfulestimatebyourmethodversusthemethodof[ 34 ]ismorethan38%,62%,and96%,respectively. 126

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5 ).Approximationalgorithmsfordierentgridlayouts(i.e.,regularhexagon,equilaterialtriangle)arealsoworthexploring.Thisworkisonlyasteptowardsutilizingcomputationalgeometrymethodsforsolvingsensorlocalizationproblems.Itwouldbeoffutureinteresttoconsiderextensionsofthismethodforcaseswheremorethanthreesensorsaredeployedandmultiplemeasurementsetsareprovided[ 47 ].Itwouldalsobeinterestingtoseeiftheproposedmethodcanbeextendedunderrandomnoisemodels,particularlywhensensorerrorsarecorrelatedandthenoisemodelisunknown.ForthespecialcasewhenS1,S2andS3formanacutetriangle,atrainingmethodwasproposedin[ 45 ]whereinthelocalizationmethodcanbetrainedin-situtoaccountforsensorcorrelations.Thecurrentmethodcanbesimilarlyemployedbutthetrainingprocedureislikelytobemoreinvolved.Itwouldbeoffutureinteresttoexplorethetrackingabilityofthismethodbyrepeatedlyexecutingitonastreamofdistance-dierencemeasurementscorrespondingtoamovingobject.Itwouldbeinterestingtoinvestigatetheeectsofrandomnessindistance-dierencesonbothuniquenessandminimalityresultspresentedinthiswork.Applicationsoftheseresultstopracticalradiationdetectionsystemswouldbeoffutureinterest. 128

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