ONLINE PURSUIT ALGORITHMS AND OPTIMAL STRATEGIES FOR
HETEROGENEOUS ROBOTS

By

Shivam Bajaj

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

Electrical and Computer Engineering—Doctor of Philosophy

2023

ABSTRACT

The advancement in technology, especially Unmanned Aerial Vehicles (UAVs) or drones, has

helped mankind in many aspects of everyday life, such as environmental monitoring and

in surveillance. However, an easy access to UAV technology has spurred its malicious use,

leading to numerous attempts of flying UAVs into restricted areas or in public places. One

possible way to counter against such adversarially intruding UAVs is to tag or disable them

before they reach a specified location by using superior drones. But the problem of how to

plan the motion of these drones, i.e., designing algorithms that have provable guarantees on

the numbers of adversarial UAVs that can be disabled has remained an open problem.

This dissertation addresses design of control strategies and online algorithms, i.e., algo-

rithms that do not have information about the intruders a priori, for drones to pursue and

disable one or many intruders and is divided into two parts. The first part involves many,

possibly infinite, intruders that move directly towards a region of interest. For this scenario,

we design decentralized as well as cooperative online algorithms with provable worst-case

guarantees for 1) a single drone defender, 2) a team of homogeneous defenders, and 3) a

team of heterogeneous defenders. The aim of such defender drones is to capture as many

intruders as possible that arrive in the environment. To quantify how well the algorithms

perform in the worst-case, we adopt a competitive analysis technique.

In particular, the

algorithms designed in this dissertation exhibit a finite competitive ratio, meaning that the

performance of an online algorithm is no worse than a finite value determined in this dis-

sertation. We also determine fundamental limits on the existence of online algorithms with

finite competitive ratios.

In terms of heterogeneity, the first part addresses drones of different capabilities as well

as motion models, such as a drone and a turret operating in the same environment. The

second part of this dissertation considers coupling between the motion. Specifically, this

part considers a laser attached to a fixed wing aircraft. The aircraft is modelled as a planar

Dubins vehicle and the laser with a finite range can rotate clockwise or anti-clockwise. We

design an optimal control strategy for both the Dubins vehicle and the laser such that a static

target, located in the environment, is tagged in minimum time. By applying the minimum

principle, we establish cooperative properties between the laser and the Dubins vehicle. We

further establish that the shortest path must lie in a family of 13 candidate paths and

characterize the solutions to all of these types. Finally, this part also addresses a scenario

when the location of the perimeter is not precisely known to the adversary requiring the

adversary to release surveilling agents to estimate the location of the perimeter. Specifically,

a pursuit problem of surveilling agents is considered wherein the objective of the pursuer

is to capture any one of the surveilling agents whereas the surveilling agents aim to jointly

maximize a weighted combination of the determinant of the Fisher Information Matrix and

the distance to the pursuer from the nearest tracker. We establish the optimal strategy

for the pursuer and show that the optimization problem for the surveilling agents can be

converted to a Quadratically Constrained Quadratic Program. We further establish that the

optimal strategies obtained for the pursuer and the trackers form a Nash equilibrium of this

game.

Copyright by
SHIVAM BAJAJ
2023

Dedicated to my family.

v

ACKNOWLEDGEMENTS

I would like to express my deepest appreciation to my advisor, Dr. Shaunak D. Bopardikar.

He has not only taught me research but has also provided unwavering support and construc-

tive criticism throughout my research projects. I would also like to thank my committee

members, Dr. Eric Torng, Dr. Xiaobo Tan, and Dr. Vaibhav Srivastava for their time,

support, and valuable feedback.

I would also like to express my gratitude towards my collaborators, Dr. Alexander Von

Moll, Dr. Isaac Weintraub, Dr. Eloy Garcia, and Dr. David Casbeer at the Control Science

Center in Air Force Research Laboratory for numerous insightful discussions. Special thanks

to Dr. Bhargav Jha for a wonderful joint work.

I am very grateful to all my lab mates and friends, especially Sandeep Banik, for all the

discussions over coffee breaks as well as the treasured memories. I would also like to thank

Dr. Shardula Gawankar for continuously motivating me as well as supporting me throughout

this journey. Finally and most importantly, I would like to thank my family, without whom

none of this would be possible, for their immense support, love, sacrifice, and continuously

motivating me all this while.

vi

TABLE OF CONTENTS

CHAPTER 1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . .

1

CHAPTER 2

COMPETITIVE PERIMETER DEFENSE IN LINEAR ENVI-
RONMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

CHAPTER 3

RANDOMIZED COMPETITIVE PERIMETER DEFENSE IN
LINEAR ENVIRONMENTS . . . . . . . . . . . . . . . . . . . . .

CHAPTER 4

COMPETITIVE PERIMETER DEFENSE IN CONICAL EN-
VIRONMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

49

63

CHAPTER 5

COMPETITIVE PERIMETER DEFENSE WITH A TURRET
AND A MOBILE DEFENDER . . . . . . . . . . . . . . . . . . . 115

CHAPTER 6

OPTIMAL CONTROL OF A DUBINS-LASER SYSTEM . . . . 150

CHAPTER 7

OPTIMAL PURSUIT OF SURVEILLING AGENTS NEAR A
HIGH VALUE TARGET . . . . . . . . . . . . . . . . . . . . . . . 176

CHAPTER 8

CONCLUSIONS AND FUTURE DIRECTIONS . . . . . . . . . . 196

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

APPENDIX A: PROOFS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

APPENDIX B: EXPRESSIONS FOR MATRICES . . . . . . . . . . . . . . . . . . . 215

vii

CHAPTER 1

INTRODUCTION

The advancement in technology, especially Unmanned Aerial Vehicles (UAVs) or drones, has

helped mankind in many aspects of everyday life, such as monitoring [43, 21] and surveil-

lance [53, 67]. However, the malicious use of such technology has also increased; recent

instances include flying UAVs into restricted areas or crash landing [71] and using drones for

surveillance in an ongoing armed conflict [42]. Although these instances use a single drone,

a scenario involving hundreds or thousands of such drones is also feasible. An excerpt from

[29] substantiates this statement.

"Swarms of low-cost, autonomous air and space systems can provide adaptability, rapid

upgradability, and the capacity to absorb losses that manned systems cannot. By lever-

aging advances in artificial intelligence, low-cost sensors, and networked communications,

low-end systems can restore the agility to attack adversary weaknesses in unexpected

ways by exploiting numbers and complexity."

Simply put, by using many robotic drones (referred to as intruders from here on), or

equivalently a robotic swarm, an adversary can severely damage critical infrastructures. To

counter such intruders, superior drones (referred to as defenders from this point onward) may

be utilized for defense applications, including patrolling, surveillance, or perimeter defense.

Deploying these defenders for such applications requires designing motion plans with certain

provable guarantees. However, as expected, the information possessed by defenders about

the intruders or vice-versa plays a crucial role in designing the motion plans for the defenders.

This dissertation sheds light on perimeter defense and surveillance scenarios under infor-

mation asymmetry. Specifically, this work is in two parts. First, we consider a perimeter

defense scenario in which the defenders do not have complete information about the in-

truders and thus, must take decisions as and when new information is released. Then we

consider two surveillance scenarios in which 1) the defenders are used to gain information

about critical locations and 2) deter intruders from gaining critical information.

1

When defenders do not have complete information about the intruders, the defenders

must take decisions as and when new information is released to them while ensuring certain

guarantees for the application in consideration. For instance, consider a scenario in which a

single intruder, A, arrives near a high value facility which is being guarded by a defender,

D. If D does not attack A, by say pursuing A, then the facility faces a risk. So D, given the

information of A, must plan its motion to pursue A.

Now, while D is in pursuit of A, consider that a swarm S of intruders arrive. In fact, it

may be the case that A deliberately lured D away from the facility so that S can damage the

facility. In such a scenario, was pursuing A a valid motion plan for D? Could D have done

better by chasing A sufficiently away from the facility and then returning to the facility?

Clearly, if the arrival of S was known a priori, one may have designed a better motion plan

for D. However, in this scenario, D wasted valuable time pursuing A and, as a consequence,

the facility faces a risk.

This example highlights an online optimization problem or an online problem. In simple

words, the defender faces incomplete information about the arrival of intruders and must

plan its motion as new information is released over time. Thus a critical question arises:

Question 1 (Perimeter Defense): How to design motion plans with certain guar-

antees for the defenders that seek to defend a perimeter against many intruders and without

any information about the intruders arrival process?

Similarly, in surveillance applications, the defenders may be used either to 1) deter intrud-

ers from gaining information (maintaining informational advantage) or 2) gather information

(gaining informational advantage) leading to the following question:

Question 2 (Surveillance): How to design motion plans with certain guarantees for

the defenders to maintain or gain informational advantage?

This work addresses both Question 1 and Question 2.

In particular, we address the

first question by focusing on non-maneuverable intruders that seek to damage a facility.1

1We will see in Chapter 2 that some of the results can be applied to maneuverable intruders with slight

modifications.

2

More precisely, we consider perimeter defense problems in which defenders seek to defend a

perimeter from an arbitrary number of intruders that arrive in the environment at arbitrary

time instants and locations. As the defenders do not have the information about the number,

arrival locations, and arrival times of the intruders, we design online algorithms with provable

guarantees as well as fundamental limits for the defenders for the worst-case scenarios. In

simpler terms, irrespective of how and when the intruders arrive in the environment, the

algorithms must guarantee that certain number of intruders are always captured. Further,

we address the second question by focusing on designing optimal control strategies for the

defenders for both, deterring intruders as well as gathering information.

We now provide some background information and review concepts that will be used in

the subsequent chapters which address Question 1.

1.1 Competitive Analysis: A Brief Overview

Recall Question 1 that considers that the defenders do not have information about the

arrival of the intruders. Such type of problems in which the information is not known a

priori and is revealed over time are called online problems.

The information revealed to the defenders is modelled as a sequence, known as input

sequence I, which can be defined as a set of 3-tuples comprising: (i) an arbitrary time

instant t ≤ T , where T denotes the final time instant, (ii) the number of intruders N (t) that

are released at time instant t, and (iii) the release location of each of the N (t) intruders.

Formally, I = {t, N (t), LN (t)}T

t=0, where LN (t) denotes the set of locations of N (t) intruders
that arrive at time instant t. An input instance I(t) ⊆ I is a part of the input sequence that

has been revealed until the current time instant t. Since the information is not known to the

defenders in advance, we design online algorithms that only uses the information I(t) ⊆ I

that is revealed to it until time t. Formally, an online algorithm A assigns velocity(ies) to

the defender(s) at time t as a function of the input instance I(t) ⊆ I revealed until time t.

For any online problem, we can assume the existence of an optimal offline algorithm, de-

fined as a non-causal algorithm which has complete information of the entire input sequence

3

I to assign velocity(ies) to the the defender(s) at any time t. Such an algorithm can be used

to measure the quality of an online algorithm. The idea of using an offline optimal algorithm

to compare with an online algorithm was first introduced in [69] and is known as competitive

analysis. Before we describe this technique, we briefly describe an instance of the perimeter

defense problem.

A problem instance P for a perimeter defense problem (Question 1) is characterized by

the following parameters:

• The speed of the intruders which is normalized by the speed of the defenders, unless

otherwise stated.

• The size of the environment as well as the perimeter

Note that a problem instance P can also be characterized by additional properties of the

defenders such as the capture radius or service times of the defenders. A more precise and

formal definition of a problem instance is provided in each of the subsequent chapters.

As we consider the worst-case scenarios, an online algorithm A’s performance is measured

using the notion of competitive ratio: the ratio of the optimal offline algorithms performance

and algorithm As performance for a worst-case input sequence for algorithm A. An algorithm

is c-competitive if its competitive ratio is no larger than c which means its performance is

guaranteed to be within a factor c of the optimal, for all input sequences. More formally,

the definition of competitive ratio and a c-competitive algorithm is as follows:

Definition 1 (Competitive Ratio for Deterministic Algorithms). Given a problem

instance P, an input sequence I, and an online algorithm A, let nA(I, P) (resp. nO(I, P))

denote the performance of an online algorithm A (resp. optimal offline algorithm O) on an

input sequence I. Then,

i) the competitive ratio of A on I is defined as CA(I, P) = nO(I,P)
nA(I,P)

≥ 1,

ii) the competitive ratio of A for the problem instance P is cA(P) = supI CA(I, P),

4

Figure 1.1 Description of Example 1. The green line segment denotes the perimeter and the blue
and the red dots denotes the defender and the intruder, respectively.

iii) the competitive ratio for the problem instance P is c(P) = inf A cA(P)

Finally, an algorithm is c-competitive for the problem instance P if cA(P) ≤ c, where c ≥ 1

is a constant.

In this work, the performance of an online or optimal offline algorithm is the number of

intruders captured by the defenders following the respective algorithm.

Competitive analysis falls under a general framework of Request-Answer games and thus,

can be viewed as a two player zero-sum game between an online player and an adversary

[16]. An adversary is defined as a pair (Q, O), where Q is the input component responsible

for generating the input sequences I and O is an optimal offline algorithm which maximizes

nO(I, P). Thus, the adversary, with the complete information of the online algorithm,

constructs a worst-case input sequence so as to maximize the competitive ratio, i.e., it

minimizes the number of intruders captured by the online algorithm and simultaneously

maximizes the number of intruders captured by an optimal offline algorithm. On the other

hand, the online player operates an online algorithm on an input sequence created by the

adversary. Thus, in this work, we restrict the choice of inputs I to those for which there

exists an optimal offline algorithm O such that nO(I, P) ≥ 1. Clearly, nO(I, P) ≥ nA(I, P).

However, if for some I, nA(I, P) = 0, then we say that A is not c-competitive for any

finite c. We now provide an example for a better understanding of competitive ratio and the

information structure. We say that a defender captures an intruder if it is collocated with

the intruder, unless otherwise stated.

Example 1. Consider a linear environment as shown in Figure 1.1 in which intruders arrive

at the two endpoints L and −L. Upon arrival, the intruders move towards the perimeter,

5

defined as a set of two locations ρ and −ρ, with a fixed speed v. A single defender that can

move with unit speed seeks to capture maximum number of intruders before they breach the

perimeter. Suppose the defender follows an algorithm which is defined as follows:

Simple Algorithm: Move towards endpoint +1 and stay at the endpoint for all time.

Recall, that an adversary has complete information of the entire algorithm a priori

whereas the Simple Algorithm does not have any knowledge of how and when the intruders

arrive in the environment. Given the information about Simple Algorithm, suppose that the

adversary releases intruders based on the input sequence I1 described as follows. Release all

intruders at endpoint +1 at time 1. Against such an input sequence, the Simple Algorithm

can capture all intruders. Similarly, an optimal offline algorithm, having the complete infor-

mation of I1, would also move its defender towards endpoint +1 and capture all intruders.

Thus, from Definition 1(i) and against input sequence I1, Simple Algorithm has a competi-

tive ratio of 1. However, this does not mean that Algorithm Simple is 1-competitive. This is

explained as follows. There exists an input sequence, I2, which has intruders arriving only

at endpoint −1. Against such an input sequence, Simple Algorithm does not capture any

intruder. On the other hand, since an optimal offline algorithm has the information of entire

I2 a priori, it can move its defender towards −1 and capture all intruders. This implies

that in the worst-case, Simple Algorithm does not capture any intruder, meaning that there

exists a constant greater than 1 that bounds the competitive ratio of Simple Algorithm and

thus, it not 1-competitive (refer to Definition 1 (ii)). In fact, as Simple Algorithm does not

capture any intruder, it is not c-competitive for any finite c.

Thus far, we have explored the concept of competitive ratio within the context of de-

terministic online algorithms. Now, let us turn our attention to the various adversarial

models and the definition of competitive ratio for randomized online algorithms. We begin

by defining a randomized online algorithm.

Let Adet = {Ai : i ∈ N} denote the set of all online deterministic algorithms Ai and let

nA,i(I, P) (resp. nO(I, P)) denote the performance, i.e., the number of intruders captured,

6

of the vehicle when following a deterministic algorithm Ai ∈ Adet (resp. O) on an input

sequence I. Then, a randomized algorithm A is defined as a probability distribution over

the set of deterministic online algorithms Adet.

Unlike the case of deterministic algorithms, there are two different models of the ad-

versary; oblivious and adaptive [16]. These models differ in the information they have in

generating the input sequence. In this work, we consider the oblivious adversary which is

defined as follows:

Definition 2 (Oblivious Adversary [16]). For any online randomized algorithm, an obliv-

ious adversary generates the entire input sequence in advance.

Although an oblivious adversary has prior information of the randomized algorithm, it

does not know the outcome of the random choices made by the randomized algorithm. Ir-

respective of the model of the adversary, an adversary, for a given randomized algorithm,

generates input sequence such that the number of intruders captured by the optimal offline

algorithm is maximized and the number of intruders captured, in expectation, by the ran-

domized algorithm is minimized. We now define the competitive ratio for a randomized

online algorithm.

Given a problem instance P, an input sequence I, and an optimal offline algorithm O,

the competitive ratio for randomized algorithms is defined as:

Definition 3 (Competitive Ratio for Randomized Algorithms). Given a problem

instance P, an input sequence I, and a randomized online algorithm A, let nA(I, P) (resp.

nO(I, P)) denote the performance of a randomized online algorithm A (resp. optimal offline

algorithm O) on an input sequence I. Then,

i) the competitive ratio of A on I is defined as CA(I, P) = nO(I,P)
E[nA(I,P)]

≥ 1, where the

expectation E[.] is with respect to the distribution over Adet.

ii) The competitive ratio of A for the problem instance P is cA(P) = supI CA(I, P),

7

iii) the competitive ratio for the problem instance P is c(P) = inf A cA(P)

Finally, an algorithm is c-competitive for the problem instance P if cA(P) ≤ c, where c ≥ 1

is a constant.

We now summarize relevant prior works on perimeter defense problems.

1.2 Perimeter Defense and Dynamic Vehicle Routing: A Brief History

We start with the existing and relevant literature on perimeter defense problems followed

by the relevant literature on dynamic vehicle routing problems.

1.2.1 Perimeter Defense Problems

Introduced in [41] as a target guarding problem, perimeter defense problem is a variant of

pursuit evasion problems in which the aim is to determine optimal policies for the pursuers

(or defenders) and evaders (or intruders) by formulating it as a differential game. Since

then, several extensions have been studied such as Target-Attacker-Defender (TAD) games

[47, 28] or reach-avoid games[34]. Versions of the same with multiple vehicles and intruders

have been studied extensively as [24, 81, 82, 48, 36, 35]. The classic approach either requires

computing solutions to the Hamilton-Jacobi-Bellman-Isaacs equation which, due to the curse

of dimensionality, is applicable only for low dimensional state spaces and simple environments

[51] or determining the value function of the game. Another work [46] addresses a class of

perimeter defense problems, called perimeter defense games, which require the defenders to

be constrained on the perimeter. We refer the reader to [68] for a review of perimeter defense

games. More recently, [62, 63] considered a target guarding problem in which targets arrive

sequentially according to some random process.

Considering a turret, which is a different type of defender compared to a mobile de-

fender, [1] introduced a differential game between a turret and a mobile intruder with an

instantaneous cost based on the angular separation between the two. A similar problem with

possibility of retreat was considered in [77, 78]. Further, [79] and [80] considered a scenario

in which the turret seeks to align its angle to that of the intruders in order to neutralize an

8

attacker. Other recent works include approach based on control barrier functions [37] and

task assignment problems [75, 76]. All of these works require that some information, such

as location, of the intruders is known a priori.

1.2.2 Dynamic Vehicle Routing Problems

Online problems which require that the route of the vehicle be re-planned as information is

revealed gradually over time are known as dynamic vehicle routing problems [64, 12, 20]. The

most relevant works in this area are TRP-TW problems in which most of the works consider

either zero or stochastic service times [56, 31, 59, 11, 38]. In these problems, the input (also

known as demands) is static and therefore, the problem is to find the shortest route through

the demands in order to minimize (maximize) the cost (reward). Examples of such metrics

would be the total service time or the number of inputs serviced.

In perimeter defense

scenarios, the input (intruders) are not static. Instead, they are moving towards a specified

region, making this problem more challenging than the former. With the assumption that

the arrival process of the intruders is stochastic, [3, 70, 49] consider the perimeter defense

problem as a vehicle routing problem. Other related works that do not make any assumptions

on the intruders and focus on design of must-win algorithms, i.e., algorithms that detect every

intruder in an environment, are [30, 54].

Most prior work on perimeter defense has either focused on determining an optimal

strategy for few intruders or intruders generated by a stochastic process. The optimal strat-

egy approaches do not scale with an arbitrary number of intruders released online. While

stochastic approaches yield important insights into the average-case performance of defense

strategies, they do not account for the worst-case in which intruders may coordinate their

arrival to overcome the defense.

1.3 Organization

The remainder of this dissertation is organized as follows. Chapter 2 and Chapter 3

considers a perimeter defense problem in linear environments with a single defender, focusing

on deterministic and randomized algorithms, respectively. Chapter 4 considers a perimeter

9

defense problem in conical environments with multiple mobile defenders, and Chapter 5

addresses a perimeter defense problem with heterogeneous defenders. Chapter 6 considers

a novel motion planning problem of a Dubins-Laser System and Chapter 7 considers a

surveillance problem with two trackers and a single defender. Finally, Chapter 8 summarizes

this work and discusses some possible extensions.

10

CHAPTER 2

COMPETITIVE PERIMETER DEFENSE IN LINEAR ENVIRONMENTS

This chapter addresses a problem in which a robotic defender, having a maximum unit

speed, is tasked with defending a given set of points, termed as the perimeter from mobile

intruders. The intruders are released in an arbitrary manner at specified locations. Upon

release, each intruder moves with a fixed speed v < 1 towards the perimeter. The defender

seeks to intercept maximum number of intruders before the intruders reach the perimeter.

We consider two possible scenarios: (i) the intruders are released at the endpoints of the

line segment and move inwards, i.e., towards the midpoint of the environment, or (ii) the

intruders are released in the interior of the line segment and move outwards, i.e., away from

the midpoint of the environment. We focus on design and analysis of online algorithms to

route the defender as the input, consisting of release times, number of intruders and locations

of intruders, is unknown a priori. This work is motivated by the potential use of a single

drone to relay real-time information about a road hazard (perimeter) to vehicles driving on

both sides of a highway. Military applications include situational awareness using a drone

around defense assets in narrow environments, e.g., boats traversing a river, or convoys

moving in a valley.

This chapter presents a competitive analysis approach to perimeter defense accounting

for worst case inputs. We do not impose any assumption on the number or the release

process of the intruders. The contributions of this chapter are three-fold:

1. Perimeter defense model with two scenarios: This chapter considers a perimeter

defense problem in two scenarios: 1) Inward moving intruders and 2) Outward moving

intruders.

In particular, both scenarios can be specified using two parameters: (i)

the speed ratio v ∈ (0, 1] between the intruders and the defender and, (ii) the ratio

ρ > 0, which denotes the size of the perimeter (distance between the points of interest)

relative to the size of the environment.

11

(a) Incoming intruderss (P1): The intruders
move inwards with speed v.

(b) Escaping intruders (P2): The intruders
move outwards with speed v.

Figure 2.1 Description of the environment E(ρ). The vehicle is shown as a blue dot and the
red dots depict the intruders.

2. Necessary conditions: For each scenario, we prove necessary conditions on the

competitiveness of any algorithm. Specifically, we characterize a parameter regime of

v and ρ in which no algorithm can be c-competitive for any constant c ≥ 1, and a

second parameter regime in which every algorithm is at best 2-competitive.

3. Algorithm design and analysis: For incoming moving intruders, we design and

analyze three classes of algorithms and establish their competitive ratios. Specifically,

the first two algorithms are provably 1 and 2-competitive, respectively. The third is a

family of algorithms, each of which covers a specific parameter regime with a specified

competitive ratio in that parameter regime. Finally, for outward moving intruders,

we propose and analyze two algorithms and characterize parameter regimes for which

they are provably 1 and 4-competitive, respectively.

This chapter, content of which is based on [9, 8], is organized as follows.

In Section

2.1, we formally define our problem. Section 2.2 presents the algorithms and analysis for

inward moving intruders. Section 2.3 extends our algorithms and analysis to outward moving

intruders. Section 2.4 presents and discusses the parameter regime plots for the two scenarios.

Finally, section 2.5 summarizes this chapter.

2.1 Problem Formulation

In this work, we consider a perimeter defense problem in a linear environment E(ρ) in

which intruders arrive over time at either position −1 or 1 and move with fixed speed v

towards the perimeter, defined by the two points −ρ and ρ, respectively. To avoid a trivial

problem formulation, we will suppose that ρ ̸= 1. There are two possibilities for E(ρ). If

12

ρ < 1, then E(ρ) = [−1, 1], and we define the incoming intruders problem variant P1 (Figure

2.1a) where the intruders move inwards to reach the perimeter. If ρ > 1, then E(ρ) = [−ρ, ρ],

and we define the escaping intruders problem variant P2 (Figure 2.1b) where the intruders

move outwards to reach the perimeter. In both variants, the intruders are located between ρ

and 1 or −ρ and −1. In both variants, a single defender with simple motion and a maximum

speed of unity is used to defend the perimeter. The defender captures an intruder when

its location is the same as that of the intruder. We assume that the defender captures an

intruder instantaneously, so it can capture an intruder while moving. An intruder is said to

be lost by the defender if the intruder reaches the perimeter without being captured. Finally,

we assume that the defender is located at the origin at time 0, unless otherwise specified.

Although the online and the optimal offline algorithms compute the velocity of the de-

fender, it turns out that specifying the direction in which the defender must move is more

important than the speed of the defender. In fact, we will formally establish in Lemma 4

that it is not advantageous for the defender to move with speed less than 1 in any interval

of time; that is, the defender either does not move or it moves with speed 1.

Problem Statement: Our aim is to design fundamental limits on achieving finite c-

competitiveness and deterministic c-competitive algorithms for problems P1 and P2.

Let extreme speed algorithms A′ be the algorithms that always move the defender with

unit speed or keep it stationary. Further, let non-adaptive input instances be the inputs in

which the arrival of intruders is not based on the movement of the defender. Then, due to

the following result, we restrict our attention to extreme speed algorithms in this work.

Lemma 4 (Extreme speed algorithms). Given any problem instance P(ρ, v), for any non-

adaptive input instances, extreme speed algorithms are no worse than general algorithms

which can move with any speed in the range [−1, 1].

Proof. Let A denote any general speed algorithm and I denote an arbitrary non-adaptive

input instance in which there are n distinct time points where A captures at least one

intruder of I. We define the capture profile of A’s execution on I as the set of pairs (xi, ki)

13

for 1 ≤ i ≤ n where ki is the time of the ith capture and xi is the location where the ith

capture occurred. Let I(xi, ki) denote the intruders in I captured by A at location xi at time

ki. We add to this capture profile the pair (x0, k0) which denotes the starting location for

the defender and the starting time instant. Our goal is to show that there exists an extreme

speed algorithm A′ that can match A’s capture profile for I: namely having the defender at

location xi at time ki for 0 ≤ i ≤ n. If we can show this, then since I is non-adaptive, A′

will capture the intruders in I(xi, ki) for 1 ≤ i ≤ n at xi at time ki and the result follows.

We now show that there exists an A′ that can match A’s capture profile for I. We prove

this by induction on i; namely, that there exists an A′ that will have the defender at point xj

at exactly time kj for 0 ≤ j ≤ i for 0 ≤ i ≤ n. The base case where i = 0 is straightforward

as the defender starts at location x0 at time k0 for both algorithms. We now show that if

this holds up to i, then it also holds for i + 1. Given the induction hypothesis, we know that

there exists an A′ that will have the defender at point xj at exactly time kj for 0 ≤ j ≤ i.

We extend that by showing that A′ can also have the defender at xi+1 at time ki+1.

Let VA ∈ [−1, 1] denote the speed with which A moves its defender. Then, at time ki,

A′ keeps the defender stationary at xi for (xi+1−xi)(1−VA)

VA

amount of time. A′ then moves the

defender with unit speed to location xi+1 arriving exactly at time ki+1. This completes the

inductive step and, in turn, completes the proof.

We now provide an example for problem P1 with ρ = v = 0.5 to elucidate the use of the

competitive analysis technique for this work. An analogous example can be constructed for

problem P2.

2.1.1 Simple Example

For this example, we use a natural algorithm, First-Come-First-Served (FCFS), defined

as follows:

FCFS Algorithm: FCFS moves the defender with unit speed towards the earliest

intruder that is released in the environment which is not lost or captured by the defender.

If intruders arrive at the same time instant, then FCFS breaks ties arbitrarily.

14

Let’s first consider an input instance I1 consisting of a single intruder released at endpoint

+1 a time 0. FCFS can capture the single intruder of I1 at location 2/3 ≈ 0.66. Similarly, an

optimal offline algorithm O also captures the single intruder. Although nO(I1) = nFCFS(I1),

this does not mean that FCFS has a competitive ratio of 1. Such a result requires FCFS to

be 1-competitive for all input instances.

The adversary, which knows the online algorithm a priori and can design the input

instance based on this knowledge, would choose a more aggressive input instance I2, which

has an intruder arriving at endpoint +1 at time instant 0 and a burst of x + 1 intruders

arriving at endpoint −1 at time ϵ, where ϵ > 0 is a very small number. Against I2, FCFS

captures only the first intruder which arrives at endpoint +1. This is because the total time

required by the defender following FCFS to reach −ρ after capturing the first intruder is

2/3 ∗ 2 + 0.5 ≈ 1.833. On the other hand, the x + 1 burst intruders reach −ρ at time 1 + ϵ.

Meanwhile, optimal offline algorithm O, knowing I2 at time 0, will move its defender towards

endpoint −1 at time 0 and capture the x + 1 burst intruders sacrificing the first intruder

that is released at endpoint +1. Thus, nO(I2) = (x + 1)nFCFS(I2). The existence of input

instance I2 proves that FCFS is not c-competitive for any c > 1 when ρ = v = 0.5 as the

adversary can set x = c. This concludes the example and leads to the following remark,

which is established formally in [9].

Remark 1. For any problem instance P for which 2

v+1 + ρ > 1−ρ

v + ϵ for some small ϵ > 0,

FCFS is not c-competitive for any constant c.

We now briefly explain our motivation for designing multiple algorithms. The effective-

ness of each of our algorithm is characterized by (i) its competitive ratio c and (ii) the

parameter regime, i.e., the set of values of parameters ρ and v, in which the algorithm is

c-competitive. For each algorithm that we present, there exists a parameter regime where

it provides the smallest competitive ratio. For example, in Section 2.2, we present algo-

rithms Sweep that is 1-competitive and CaC that is 2-competitive. Clearly, for parameter

15

regimes where both are effective, Sweep should be used since it has a lower competitive

ratio. However, there exist parameter regimes where CaC is 2-competitive and Sweep is

not c-competitive, so CaC should be used in such situations. Note that in Section 2.2,

the parameter regimes are described mathematically as they correspond to equations of the

problem parameters. To better visualize the regimes, we provide numerical plots in Section

2.4. Similar explanations hold for algorithms designed in Section 2.3.

2.2 Incoming Intruders

This section addresses problem P1 in which ρ < 1, i.e., the physical region R(ρ) = [−1, 1]

and the intruders move inwards with fixed speed 0 < v < 1. We first establish two necessary

conditions on achieving a finite competitive ratio and then design three classes of online

algorithms with provably finite competitive ratios.

2.2.1 Fundamental Limits

We start with a necessary condition for achieving a c-competitive ratio for any constant

c followed by a necessary condition for achieving a strictly better than 2-competitive ratio.

Theorem 5 (Necessary condition for c-competitiveness). For any problem instance P(ρ, v)

such that v > 1−ρ

2ρ , there does not exist a c-competitive algorithm for any finite c ≥ 1.

Proof. Recall from Definition 1 that an online algorithm A is c-competitive if inf γ≥1{γ :

nO(I, P) ≤ γ nA(I, P), ∀I}} = c. Thus, the idea behind the proof is to construct an input

sequence I such that for any constant γ ≥ 1, the condition nO(I, P) ≤ γnA(I, P) does not

hold for any online algorithm A.

We start by constructing an input sequence for any online algorithm A and then compare

the total number of intruders captured to that of the optimal offline algorithm O. We assume

that the defender starts at the origin for both A and O.

The input sequence has two phases: a stream of intruders that arrive at endpoint 1, 2

time units apart starting at time 1 and a burst of c + 1 intruders that arrive at endpoint −1

at time t. Time instant t corresponds to the first time the defender moves to ρ according

16

to A. If the defender never moves to location ρ, the stream never ends and the burst never

arrives, so A will not be c-competitive for any constant c, and the first result follows. Let

i be the number of stream intruders released up to and including time t; note that i might

be 0 if the defender reaches ρ before time 1 when the first stream intruder is released. The

defender, following A, will capture at most intruder i. In particular, because the stream

intruders arrive 2 time units apart, all i − 1 stream intruders have reached ρ before time

t. Furthermore, since v > 1−ρ

2ρ , the defender will not be able to capture the burst of c + 1
intruders that arrive at time t. We now determine the number of intruders O can capture.

Two cases arise: i = 0 and i > 0.

Case 1: If i = 0, then this means that the defender reached location ρ before time 1, so t < 1.

O can move the defender to location −ρ by time t and thus capture all c + 1 burst intruders,

and the result follows. Case 2: If i > 0, then this means that the defender, following A,

reached location ρ no earlier than time 1, so 1 ≤ t. In this case, O can capture the first i − 1

stream intruders immediately upon arrival by moving the defender to location 1 at time 1

and staying there until the first i − 1 stream intruders have been captured. O can then move

the defender to position −ρ before the burst intruders have been released since the stream

intruders arrive 2 time units apart. Note that if i = 1, O moves the defender immediately

to −ρ. Therefore, from Definition 1, we obtain the result.

In either case, we show that O can capture all the burst intruders and at least all but

the last stream intruder whereas A will capture at most 1 stream intruder, and the result

follows.

Corollary 1. For any problem instance P(ρ, v) such that v > 1−ρ

2ρ , no algorithm (online or

offline) can capture all intruders.

Proof. The result follows by observing that in the input sequence I, constructed for the proof

of Theorem 5, there are choices for t including t = 1 such that no algorithm can capture all

i stream intruders and all c + 1 burst intruders, where t and i are analogously defined as in

proof of Theorem 5. This concludes the proof.

17

While the previous result provides a fundamental limit on the existence of a c-competitive

algorithm, the following result provides a necessary condition to achieve a strictly better than

2-competitive ratio for an ecosystem.

Theorem 6 (Necessary condition for c(P) < 2). Given any problem instance P, c(P) < 2

only if v < 1−ρ
1+ρ .

Proof. In this proof, all of our input sequences consist of two intruders, a and b, where a

arrives at location -1 and b arrives at location 1. We assume that the defender starts at the

origin at time 0. We first consider the case where v = 1−ρ

1+ρ . Consider the following input
instance I1 where both a and b arrive at time 1. For input sequence I1, we describe one

of the two symmetric solutions below. At time 0, the defender moves toward location 1

capturing intruder b at location 1 at exactly time 1. The defender then moves to location

−ρ to capture intruder a at location −ρ at exactly time 2 + ρ. The defender has just enough

time to do this given the condition that v = 1−ρ

1+ρ which is equivalent to 1 + ρ = 1−ρ

v . Thus,

any algorithm that hopes to be better than 2-competitive must capture both intruders in

I1, and the only way to do so is to move immediately to location -1 or 1 reaching at exactly

time 1.

Now consider input sequences I2 and I3 which also consists of two intruders.

In I2,

intruder a arrives at time 1 and intruder b arrives at time 1 + ϵ where ϵ < 2ρv.

In I3,

intruder b arrives at time 1 and intruder a arrives at time 1 + ϵ where ϵ < 2ρv. Online

algorithms which have the defender arriving at location 1 at time 1 can capture at most

one of the two intruders in input instance I2. This follows as the defender can only capture

intruder a if it moves immediately to arrive at location −ρ at time 2 + ρ. However, because

ϵ < 2ρv and v = 1−ρ

1+ρ , the defender cannot intercept intruder b before b passes location ρ.

Similarly, algorithms which have the defender arrive at location -1 at time 1 can capture

at most one of the two intruders in input instance I3. At the same time, there exists an

optimal offline algorithm having future information of the input sequences which captures

both intruders in input instance I2 and a different algorithm which captures both intruders

18

in I3 by simply moving to the correct location at time 1 and then intercept the other intruder

before it reaches +ρ or −ρ.

We now consider the case where v > 1−ρ

1+ρ . Now we only need two input sequences I4 and I5.
In I4, intruder a arrives at time 1 and intruder b arrives at time 1 + ϵ where ϵ = 1 + ρ − 1−ρ
v .
In I5, intruder b arrives at time 1 and intruder a arrives at time 1 + ϵ where ϵ = 1 + ρ − 1−ρ
v .
Because v > 1−ρ
1+ρ , ϵ > 0. Using similar arguments to those for I2 and I3, it follows that no

single online algorithm can capture both intruders in both I4 and I5.

In summary, even restricting the set of possible input sequences to {I1, . . . , I5}, no single

online algorithm can capture both intruders from all five input sequences, but since there

exist an optimal offline algorithms which capture both intruders for all five input sequences,

it follows that c(P) ≥ 2 when v ≥ 1−ρ

1+ρ and the result is established.

First-Come-First-Served (FCFS) is defined as the algorithm which moves the defender

at unit speed towards the earliest intruder to arrive that is not lost (guaranteed to reach its

goal before the defender) or already captured breaking ties arbitrarily. We now show that

FCFS is not a c-competitive algorithm.

Lemma 7. For any P where

competitive for any constant c.

2

v+1 + ρ > 1−ρ

v + ϵ for some small ϵ > 0, FCFS is not c-

Proof. We prove this result by constructing an adversarial input sequence against FCFS.

Consider the following input sequence I(c). Let the first intruder be released at time 0 at

point 1. Let c + 1 intruders be released at time ϵ at point -1. FCFS will move the defender

the condition 2

immediately towards the first intruder and intercept it at point

v+1 . Because of
v + ϵ, FCFS cannot get the defender to position −ρ before the
c + 1 intruders released at time ϵ reach position −ρ. It follows that nF CF S(I(c), P) = 1. On

v+1 + ρ > 1−ρ

v+1 at time 1

1

the other hand, if the defender had moved toward position −ρ immediately, it would capture

those c + 1 intruders. Thus, nO(I(c), P) = c + 1, and the result follows.

19

This result generalizes to any variation of FCFS which services the first arriving intruder, if

possible, before any later arriving intruder.

With these necessary conditions in place, we now turn our focus to the design of algo-

rithms with provable guarantees on the competitive ratio.

2.2.2 Algorithms

We now describe and analyze three algorithms, characterizing parameter regimes with

provably finite competitive ratios for incoming intruders.

2.2.2.1 Sweeping Algorithm

We define the Sweeping algorithm (Sweep) as follows. At initial time, Sweep moves the

defender with maximum speed towards location 1. Hereafter, the defender changes direction

only when it reaches an endpoint 1 or −1. Once the defender reaches an endpoint, it moves

with maximum speed towards the opposite endpoint. Sweep is an open-loop algorithm;

meaning, Sweep does not require any information about intruders. One logical modification

is to stop moving in a given direction if there are no intruders in that direction. We establish

that, even with this modification, Sweep achieves the same competitive ratio.

Theorem 8. For any problem instance P, Sweep is 1-competitive if v ≤ 1−ρ

3+ρ . If not, Sweep

is not c-competitive for any constant c.

Proof. Consider that v ≤ 1−ρ

3+ρ holds. Any intruder will take 1−ρ
location to the perimeter. Without loss of generality, consider an intruder i that arrives at

time to get from its arrival

v

location 1 and thus, takes 1−ρ
v

time to reach ρ. In the worst case, the defender will have left

1 just before intruder i arrived. In the worst case, the defender will take 3 + ρ time to reach

ρ moving towards intruder i. Since 3 + ρ ≤ 1−ρ

v , Sweep’s defender will reach ρ before the

intruder i and thus, intruder i will be captured. This concludes the proof for the first result.

For v > 1−ρ

3+ρ there exists an input sequence where intruders only arrive at endpoint 1 just

after the defender has left 1. As v > 1−ρ

3+ρ , all intruders will be lost and the second result

follows.

20

Note that the first result holds for the modified Sweep where it stops moving in a given

direction if there are no intruders in that direction. For the second result, we introduce some

intruders at −1 to ensure that the modified Sweep will move the defender towards −1. The

result still holds, even though the intruders released at location −1 might be captured, by

releasing more intruders at location 1.

The following definition will be helprooful for the analysis of the subsequent algorithms.

A set of intruders S is said to be on the same side as the defender if the defender is located at

ρ (resp. −ρ) and S ⊂ (ρ, 1] (resp. [−1, −ρ)). Similarly, we say that S is on the opposite side

of the defender if the defender is located at ρ (resp. −ρ) and S ⊂ [−1, −ρ] (resp. [ρ, 1]). For

i ≥ 1, Si

opp and Si

same denotes the set of intruders in an input instance I that arrive in the

ith time interval that are on the opposite side and same side of the defender, respectively

and |S| denotes the cardinality of S.

2.2.2.2 Compare and Capture algorithm

We now present a Compare and Capture (CaC) algorithm that is provably 2-competitive

beyond the parameter regime of the Sweep algorithm. CaC is not open-loop, but is mem-

oryless, i.e., the actions of CaC depend only on the present state of the defender and the

intruders.

An epoch k for the CaC algorithm is the time interval when the defender moves from

location xk to location xk+1 and is about to move from xk+1, capturing some intruders on

the way. Location xk is always either ρ or −ρ. We denote the beginning of epoch k by

kS. For epoch k, we define Sk

same as the set of intruders on the same side as the defender

at time kS, and Sk
between ρ + 2ρv and ρ + 2vρ + 2v(1−ρ)

1+v

opp as the set of intruders on the opposite side of the defender that are

away from the origin at time kS. More precisely, if

the defender is located at xk = ρ, then Sk
[−(ρ + 2ρv + 2v(1−ρ)

1+v ), −(ρ + 2ρv)].

opp is defined as the set of intruders contained in

The CaC algorithm, summarized in Algorithm 1, works as follows: At epoch k, for any

21

intruders in [ρ + 3ρv, 1] ≤ intruders in [−1, −(ρ + 3ρv)] then
Move to −ρ

time units at the center

v

Algorithm 1: Compare-and-Capture Algorithm
1 Defender waits for 1−ρ−3ρv
2 if
3
4 else
5
6 end
7 for each epoch k ≥ 1 do
if at epoch k, |Sk
8

Move to ρ.

| < |Sk

| then

same

opp

9

10

11

12

13

Move to capture all intruders in Sk
Move to xk+1 = −xk

opp

else

Move to capture intruders in Sk
Move to xk+1 = xk

same.

end

14
15 end

k ≥ 1, CaC computes the total number of intruders located in Sk

same and Sk

opp. If the total

number of intruders in Sk
defender moves, in the direction of the origin, for at most 2ρ + 4v(1−ρ)

opp is more than the total number of intruders in Sk

(1+v)2 time to capture all
opp and then returns to xk+1 = −xk. Otherwise, the defender moves

intruders located in Sk

same, then the

away from the origin for at most 1−ρ

1+v time to capture all intruders from the set Sk

same and

then returns to xk+1 = xk.

For the initial case, CaC waits at the origin until the first intruder that arrives is 3ρv

distance away from the perimeter. If the total number of intruders located in [ρ + 3ρv, 1]

is more than the total number of intruders located in [−1, −(ρ + 3ρv)], then the defender

moves to x1 = ρ. If not, the defender moves to x1 = −ρ. The first epoch starts when the

defender reaches x1.

To establish 2-competitiveness of CaC, we first establish that any intruder not belonging

to Sk

same or Sk

say that an intruder lies outside of the set Sk
and the intruder is contained in [−1, −(ρ + 2ρv + 2v(1−ρ)

opp in an epoch k will not be lost during epoch k. At the start of epoch k, we
opp, if the defender is located at ρ (resp. −ρ)
1+v )) (resp. (ρ + 2ρv + 2v(1−ρ)

, 1]).

1+v

Lemma 9. In every epoch k, any intruder that lies outside of the set Sk

same and Sk

opp is not

22

lost if ρv

1−ρ + v2

(1+v)2

≤ 1
4 .

Proof. Without loss of generality, consider that the defender is located at xk = ρ at time kS.

Two cases arise:

Case 1: |Sk

same

| ≥ |Sk

opp

|: In this case, the defender moves away from the origin to capture

intruders located in Sk

same. The defender takes at most 1−ρ

1+v time to capture these intruders.

The total time taken by the defender to capture the intruders and return to xk+1 = xk in
epoch k is at most 2(1−ρ)
1+v ), −(ρ+
opp will be at least 2ρv

1+v . Since the intruders in Sk
2ρv)], any intruder in the opposite side which was outside the set Sk

opp are located in [−(ρ+2ρv+ 2v(1−ρ)

distance away from ρ and thus, will be contained in Sk+1
opp .

Case 2:

located in Sk

same

| < |Sk

|Sk
opp and return back to xk+1 = −xk is at most 2ρ + 4v(1−ρ)

|: The total time taken by the defender to capture intruders

(1+v)2 . In order to ensure

opp

that any intruder that did not belong to Sk

same is at least 2ρv distance away from ρ at the

end of epoch k, we require

2ρ +

4v(1 − ρ)
(1 + v)2

≤ 1 − ρ − 2ρv
v

⇒ ρv
1 − ρ

+

v2
(1 + v)2

≤ 1
4

.

This concludes the proof.

Lemma 10. In each epoch k of CaC, Sk

opp is well defined if ρ + 2ρv + 2v(1−ρ)

1+v

≤ 1.

Proof. In order to ensure that Sk

environment, i.e., Sk
to ρ + 2ρv + 2v(1−ρ)

1+v

opp is well defined, we require that Sk

opp is contained in the
opp does not extend beyond 1 or −1. Mathematically, this corresponds
≤ 1, and the result follows.

Theorem 11. Compare and Capture algorithm defined in Algorithm 1 is 2-competitive for

any problem instance P in parameter regime for which both Lemma 9 and Lemma 10 hold.

Proof. First, observe that CaC is well-defined as, from Lemma 10, Sk

opp is well-defined in

every epoch k. Lemma 9 ensures that every intruder will belong to Sk

same or Sk

opp for some

epoch k. Due to the comparison of the number of intruders in Algorithm 1, in each epoch,

the number of intruders lost is at most the number of intruders captured. Recall that an

23

| + · · · + |S

| + · · · + |S
Move to x2 = −ρ, same side becomes left of origin

Algorithm 2: Capture with Patience Algorithm
1 Input: Parameter i such that 2i is a positive integer
2 Assumption: Defender stays at origin until time 2ρ, same side is right of origin
3 At time 2ρ, if |S1
4
5 else
6
7 end
8 Wait until time z = 1−ρ
v
9 for each time instant z + 2j′ρ + 2ρj/i, j, j′ ≥ 0 do

Move to x2 = ρ, same side unchanged

⌈i⌉
same| then

⌈i⌉
opp| > |S1

same

opp

10

11

12

13

14

15

16

if |Sj+⌊i⌋+1

| > |Sj+1
same

| + · · · + |Sj+2i+1

| then
opp
Move to xj+1 = −xj, same side changes
Stay at xj+1 and capture Sj+⌊i⌋+1
j′ = j′ + 1, j = j + 1

same

opp

else

Stay at xj and capture interval Sj+1
same
j = j + 1

end

17
18 end

(a) Intervals of length 2ρv (i = 1)

(b) Intervals of length 2ρv/1.5 (i = 1.5)

Figure 2.2 Breakdown of the physical region into regions of length (a) 2ρv (i = 1) and (b) 2ρv/1.5
(i = 1.5) or equivalently time into intervals of length 2ρ (i = 1) and 2ρ/1.5 (i = 1.5) by the CAP
algorithm.

intruder is not considered lost in epoch k if it is contained in Sk+1

same or Sk+1

opp in the next epoch

k + 1. Thus, CaC captures at least half of the total number of intruders. This concludes the

proof.

2.2.2.3 Capture with Patience Algorithm

We now describe a family of algorithms that we call Capture with Patience (CAP) which

covers a larger parameter regime than the previous algorithms at the expense of a higher

competitive ratio. CAP is formally defined in Algorithm 2 and is described as follows.

24

The basic idea behind CAP is to label intruders by the time interval that they arrive

in as depicted in Figure 2.2. CAP defines the initial time to be the time instant when

the first intruder arrives. CAP has the defender stay at ρ or −ρ and compares a specific

interval on the opposite side of the origin with some number of intervals on the same side

of the origin. If the opposite side interval contains more intruders than all of the same side

intervals collectively, then CAP moves the defender to the opposite side and captures that

interval. Otherwise, CAP keeps the defender where it is and captures the next interval on

the same side. After capturing an interval, it makes the same comparison to decide which

interval to capture next. If an interval represents t units of time, then its length in Figure

2.2 is tv since that is how far an intruder will move in t time units.

We can make CAP have a lower competitive ratio by choosing bigger intervals. CAP

specifies the interval size using a parameter i ≥ 1/2 such that 2i is an integer. Specifically,

the time intervals span for 2ρ/i time units and have length 2ρv/i in Figure 2.2. Clearly,

a smaller i leads to a larger interval. Furthermore, i denotes the number of intervals the

defender will lose if it moves from ρ to −ρ or vice versa. If 2i is an even integer (i is an

integer), then the intervals are symmetric on both sides of the origin as depicted in Figure

2.2a. If 2i is an odd integer, then the intervals are offset by half an interval on one side of

the origin as depicted in Figure 2.2b.

A key feature of the problem is the quantity z = 1−ρ

v which represents the time required
for any intruder originating at 1 or −1 to reach the corresponding location ρ or −ρ. In order

to guarantee a competitive ratio, we require that z ≥ 2ρ(2i+1)

i

which means that at least 2i+1

intervals exist on each side of the origin. For the smallest i = 1/2, z ≥ 8ρ. This condition is

easier to satisfy as i increases asymptotically approaching the condition z ≥ 4ρ.

After the initial time z, CAP operates as follows: At any time instant 2j′ρ + 2jρ/i + z

for j, j′ ≥ 0, the defender is stationed at either −ρ or ρ. Here j′ is used to characterize the

amount of time the defender has spent moving from ρ to −ρ or vice versa and j is used to

characterize the amount of time the defender has spent capturing intervals. Without loss of

25

generality, consider that the defender is located at ρ.

First, note that for any j ≥ 0, the intruders in Sj+1

same are located between ρ and ρ + 2ρv/i,

the intruders in Sj+2

same are located between ρ + 2ρv/i and ρ + 4ρv/i, until the intruders in

Sj+2i+1
same

are located between ρ + 4iρv/i and ρ + 2(2i + 1)ρv/i. Further, because z ≥ 2ρ(2i+1)

i

,

all the intruders in Sj+2i+1

same

have arrived by time 2j′ρ + 2jρ/i + z. Analogous conclusions

can be drawn for the intruders in Sj+1

opp, Sj+1
opp ,
when 2i is odd. Denote ⌈x⌉ and ⌊x⌋ as the ceil and the floor functions on

when 2i is even and Sj

opp , . . . , Sj+2i+1

opp , Sj+2

opp

. . . , Sj+2i+1

opp

a real number x, respectively.

If |Sj+⌊i⌋+1

opp

| > |Sj+1
same

| + |Sj+2
same

| + · · · + |Sj+2i+1

same

|, then the

defender moves to −ρ arriving at time 2(j′ + 1)ρ + 2jρ/i + z, which is just in time to
capture all the intruders in Sj+⌊i⌋+1
Sj+⌊i⌋+1
|Sj+⌊i⌋+1

. The defender then reevaluates at time 2(j′ + 1)ρ + 2(j + 1)ρ/i + z. Otherwise (if

|), the defender stays at ρ and captures all the intruders

and stays at −ρ to capture all of the intruders in

| + · · · + |Sj+2i+1

opp

opp

opp

| ≤ |Sj+1
same

same

in Sj+1

same and reevaluates at time 2j′ρ + 2(j + 1)ρ/i + z. The key idea is that the defender
moves from ρ to −ρ only when it sees sufficient benefit in terms of the number of intruders
in Sj+⌊i⌋+1

to sacrifice all the intruders in Sj+1

same, . . . , and Sj+2i+1

same, Sj+2

same

opp

.

For the initial case, the defender stays at the origin until time 2ρ. At time 2ρ, if |S1

opp

| +

· · · + |S

⌈i⌉
opp| > |S1

same

| + · · · + |S

⌈i⌉
same| (intruders to the left of the origin are considered on the

opposite side while intruders to the right of the origin are considered on the same side for

this special case), then the defender moves to −ρ. Otherwise, the defender moves to ρ. In

either case, the defender then stays at either −ρ or ρ until time z.

Lemma 12. Algorithm CAP never moves the defender from ρ to −ρ and then back to ρ (or

vice versa) without capturing at least one interval of intruders.

Proof. For the defender to move from ρ to −ρ at time 2j′ρ + 2jρ/i + z, the condition
|Sj+⌊i⌋+1

| must hold. This implies that |Sj+⌊i⌋+1

| + · · · + |Sj+2i+1

| >

| > |Sj+1
same

| + |Sj+2
same

same

opp

opp

|. In order to move directly back to ρ without capturing the intruders in the interval

, it must be true that |Sj+2i+1

same

| > |Sj+⌊i⌋+1

opp

|+|Sj+⌊i⌋+2

opp

|+· · ·+|Sj+⌊3i⌋+1

opp

|. This cannot

|Sj+2i+1
same
Sj+⌊i⌋+1
be true because |Sj+⌊i⌋+1

opp

opp

| > |Sj+2i+1

same

| and the claim is established.

26

Figure 2.3 Labeling of intervals for an integer i.

Corollary 2. For any j, j′ ≥ 1, Algorithm CAP will capture all intruders from one of Sj
or Sj+⌊i⌋
opp .

same

Proof. At time z + 2(j′ − 1)ρ + 2ρ(j − 1)/i, the defender is in position to capture Sj

same by

definition of CAP. If it captures Sj
capture point and by Lemma 12 will capture Sj+⌊i⌋
opp .

same, the claim follows. If not, it moves to the opposite

Theorem 13. The CAP algorithm, which takes an input i and is defined in Algorithm 2 is
2i + 2-competitive for any problem instance P with v ≤ i(1−ρ)

2ρ(2i+1) .

Proof. We prove this claim by using a charging scheme where we “charge” lost intervals of

intruders to captured intervals of intruders, or equivalently, captured intervals of intruders

“pay” for the lost intervals of intruders. We prove this result in two parts. First, we establish

that each captured interval is charged at most 2i + 1 times. Second, we establish that each

lost interval is paid for by the captured intervals.

Without loss of generality, consider that the defender captures its first and last interval

of intruders at ρ and ρ, respectively (our proof easily generalizes to an infinite number

of intruders). We summarize an execution of CAP with a trace which records how many

intervals are captured at each location before switching to the opposite location. Formally,

we define a trace as a list [k1

ρ, k2

−ρ, . . . , km

ρ ] wherein we use superscripts to index the elements

in the list. From Corollary 2, every element in the trace must be greater than zero with

the exception of k1
ρ

≥ 0. A trace fully defines the movement of CAP’s defender; namely, it

initially moves from 0 to ρ, stays at ρ to capture kl

ρ intervals, then moves to −ρ, stays at −ρ

to capture k2

−ρ, intervals, and so on, until it stays at ρ to capture the final km

ρ intervals.

27

We now describe how intervals of captured intruders pay for intervals of lost intruders.

Each captured interval can be classified into one of two types: type (a) where the defender

did not move to capture it, meaning that the defender also captured the previous interval on

the same side (or the interval corresponds to the first interval S1

same) or type (b), where the

defender did move to capture it, i.e., the defender spent the last 2ρ time units moving from

ρ to −ρ or vice versa. We establish that both type (a) and type (b) captured intervals are

charged at most 2i+1 times. We first consider type (a) captured intervals. For any arbitrary

lth, 1 ≤ l ≤ m, element in the trace with kl
ρ

≥ 1 (resp. kl

−ρ

≥ 1), note that the last kl
ρ

− 1

−ρ

− 1) intervals are classified as type (a) intervals. A captured interval Sj+q

(resp. kl
1 ≤ q ≤ 2i is charged q times to pay for the lost intervals Sj+1+⌊i⌋

, . . . , Sj+q+⌊i⌋

opp

(Fig. 2.3).

same, where

opp

Further, each captured interval Sj+q
times to pay for each of the lost intervals Sj+q−⌈i⌉

opp

same, where 2i + 1 ≤ q ≤ kl

ρ (resp. kl

−ρ) is charged 2i + 1

, . . . , Sj+q+⌊i⌋

opp

. Note that if kl
ρ

≤ 2i (resp.

kl
−ρ

≤ 2i), then each captured interval is charged at most 2i times.
A type (b) captured interval Sj+1+⌊i⌋

opp

(Fig. 2.3) will be charged at most 2i + 1 times –

a unit charge to pay all at once for lost intervals Sj+1
help pay for each of the lost intervals Sj+1−⌈i⌉

, . . . , Sj+⌊i⌋
opp .

opp

same, . . . , Sj+1+2i

same

, and up to 2i times to

To establish the second part of the proof, we decompose a CAP trace into a sequence of

atomic traces which we define as follows. As we noted earlier, a trace defines the movement of

CAP’s defender. An atomic trace is an interval characterizing the movement of the defender

between ρ and −ρ as well as capture of intruder intervals at ρ and −ρ. Specifically, for any

lth, 2 ≤ l ≤ m − 1, element of CAP’s trace, we define atomic trace Al

ρ (resp. Al

−ρ) as an

interval that has the following components:

i An initial component where the defender moves from −ρ (resp. ρ) to ρ (resp. −ρ).

ii A middle component where the defender captures kl

ρ (resp. kl

−ρ) intervals at ρ (resp.

−ρ).

iii A final component where the defender returns to −ρ (resp. ρ) and completes capturing

28

the first type (b) interval, denoted as Cb, in kl+1

−ρ (resp. kl+1

ρ

).

Atomic trace A1

ρ differs in the initial component as the defender moves from 0 to ρ and

the middle component as it may possibly not capture any intervals at ρ if CAP moves the

defender to −ρ without capturing an interval at ρ. Atomic trace Am

ρ has no final component

because the defender does not return to −ρ. Note that component (iii) for an atomic trace

Al

ρ (resp. Al

−ρ) is included in the next atomic trace Al+1

ρ

(resp. Al+1

−ρ ) (components (i) and

(ii)). We include component (iii) in the next atomic trace because we separate how this

captured type (b) interval, denoted by Cb, is charged to pay for lost intervals. Specifically,

in component (iii), Cb is used to pay for up to 2i + 1 lost intervals at the same location as

Cb whereas in components (i) and (ii), Cb is used to pay for 2i + 1 lost intervals at the other

location. We explain this further below.

We now prove the second part of the proof that the lost intruder intervals are fully paid

for by the captured intruder intervals. Specifically, we prove that all the intervals lost at

−ρ between those captured in atomic trace Al−1

−ρ and atomic trace Al+1

−ρ are paid for by the

captured intervals in atomic trace Al

ρ. Likewise, we prove that all the intervals lost at ρ

between those captured in atomic trace Al−1

ρ

and atomic trace Al+1

ρ

are paid for by the

captured intervals in atomic trace Al

−ρ. We separately handle the boundary cases for the

intervals lost at −ρ in atomic traces A1

ρ and Am

ρ . Since the arguments for atomic traces Al
ρ

and Al

−ρ are analogous due to symmetry, we only give the proof for atomic trace Al
ρ.

We first focus on atomic trace Al

loses kl

ρ + 2i intervals at −ρ. At the start of Al

ρ where 1 < l < m. During atomic trace Al

ρ, the defender
ρ, the defender moves from −ρ to ρ which

implies that the first interval that the defender captures at ρ in Al

ρ pays for the first 2i+1 lost

intervals at −ρ. If kl

ρ = 1, then the captured interval at ρ accounts for all 2i + 1 = 2i + kl

ρ lost

intervals at −ρ. Thus, for the rest of this proof, we assume that kl
ρ

≥ 2 and show that the

remaining kl
ρ

− 1 lost intervals at −ρ are accounted for by the remaining captured intervals

in the atomic trace Al
ρ.

The defender spends (kl
ρ

− 1)2ρ/i amount of time staying at ρ capturing kl
ρ

− 1 type (a)

29

intervals. Let t = z + 2ρ(j+kl

ρ

i

−1)

+2ρj′, where j, j′ > 0, be the time when the defender finishes

capturing the last interval at ρ within Al

ρ. First, because the defender stays at ρ from time

t − (kl
ρ

− 1)2ρ/i to time t − 2ρ

i , the following conditions hold:

|S

j+kl
ρ
opp

−1+⌊i⌋

| ≤ |S

j+kl
ρ
same

−1

| + |S

j+kl
same | + · · · + |S
ρ

j+kl
ρ
same

−1+2i

|, at time t − 2ρ
i

|S

j+kl
ρ
opp

−2+⌊i⌋

| ≤ |S
...

−2

j+kl
ρ
same

| + |S

j+kl
ρ
same

−1

| + · · · + |S

ρ+2i−2

j+kl
same

|, at time t − 4ρ
i

|Sj+1+⌊i⌋
opp

| ≤ |Sj+1
same

| + |Sj+2
same

| + · · · + |Sj+2i+1

same

|, at time t − (kl

ρ

−1)2ρ
i

.

For ease of presentation, we do not update j after each 2ρ/i time interval in the above

recursive conditions. As the defender moves from location −ρ to ρ and captures the first

interval out of kl

ρ before making its first comparison, Sj+1

same denotes the second interval out
ρ that the defender captured at ρ. Next, because the defender goes from ρ to −ρ at time

of kl

t, the following condition holds:

|S

ρ+⌊i⌋

j+kl
opp

| > |S

j+kl
same | + |S
ρ

ρ+1

j+kl
same

| + · · · + |S

ρ+2i

j+kl
same

|.

(2.1)

We first consider that kl
ρ
Sj+1+⌊i⌋

opp

− 1 ≥ 2i + 1. Since kl
ρ

− 1 ≥ 2i + 1, the first lost interval

is accounted for by the 2i + 1 captured intervals at ρ as the condition |Sj+1+⌊i⌋

opp

| ≤

|Sj+1
same

| + |Sj+2
same

| + · · · + |Sj+2i+1

same

| holds. Then, every subsequent captured interval at ρ will

account for one lost interval at −ρ as the following conditions hold:

|Sj+2+⌊i⌋
opp

| ≤ |Sj+2
same

| + |Sj+3
same

| + · · · + |Sj+2i+2

same

|

|Sj+3+⌊i⌋
opp

| ≤ |Sj+3
same
...

| + |Sj+4
same

| + · · · + |Sj+2i+3

same

|

|S

j+kl
ρ
opp

−1−2i+⌊i⌋

| ≤ |S

j+kl
ρ
same

−1−2i

| + |S

j+kl
ρ
same

−2i

| + · · · + |S

−1

j+kl
ρ
same

|.

We now account for the last 2i lost intervals, S

j+kl
ρ
opp

−2i+⌊i⌋

, . . . , S

j+kl
ρ
opp

−1+⌊i⌋

, lost at −ρ.

Consider any arbitrary lost interval Sj+q+⌊i⌋

opp

, where kl
ρ

− 2i ≤ q ≤ kl
ρ

− 1. To account

for this lost interval, CAP was supposed to capture the 2i + 1 intervals Sj+q

same, . . . , Sj+q+2i

same

.

30

CAP actually captures Sj+q

same, . . . , S

−1

j+kl
ρ
same

but does not capture S

j+kl
same , . . . , Sj+q+2i
ρ

same

. CAP

makes up for these intervals by capturing the interval S

ρ+⌊i⌋

j+kl
opp

which, due to equation (2.1),

contains more intruders than all of the intervals S

j+kl
same , . . . , Sj+q+2i
ρ

same

combined. By redefining

the bounds on q as 1 ≤ q ≤ 2i + 1, the proof for the case when kl
ρ

− 1 < 2i + 1 is analogous

to when we consider the remaining 2i intervals out of kl
ρ

− 1, so we omit this case for brevity.

Earlier, we noted that type (b) captured intervals would be charged up to 2i times to pay

for lost intervals on the same side as the captured intervals. Here we see that the last type

(b) interval in atomic interval Al

ρ is charged exactly max{2i, kl

ρ

− 1} times for lost intruders

at −ρ.

Thus, we conclude that in an atomic trace of the type [kl

ρ, 1], all lost intervals have

been fully accounted for by the captured intervals. After the defender finishes capturing

the interval S

ρ+⌊i⌋

j+kl
opp

, note that the next atomic trace Al+1

−ρ includes the captured interval

ρ+⌊i⌋

j+kl
opp

S

. At the start of the next atomic trace Al+1

−ρ , a new steady state is achieved as all

j+kl
ρ
same , . . . , S

S

ρ+2i

j+kl
same

intervals are accounted for since equation (2.1) holds. Further, following

the same argument as for atomic trace Al

ρ, it can be shown that all of the lost intervals at ρ

from the start of Al+1

−ρ until the start of Al+2

ρ

are accounted for by the captured intervals of

Al+1
−ρ .

Now it remains to be shown that the lost intervals are fully accounted for in the first

and the last atomic trace. We first consider the last atomic trace, i.e., Am

ρ . As the defender

moves from −ρ to ρ, the first type (b) interval that the defender captures at ρ pays for 2i + 1

lost intervals at ρ. If km

ρ = 1, then the captured interval accounts for all lost intervals at

−ρ. Thus we now consider that km

ρ > 1. As the defender does not switch sides, starting

from the time instant when the defender finished capturing the first of km
ρ

intervals, after

every 2ρ

i time units, the total number of intruders that are on the same, i.e., right side of

the defender will be no less than the total number of the intruders on the opposite, i.e., left

side of the defender. Thus, all the intruder intervals that are lost at −ρ are accounted for.

We now consider the first atomic trace A1

ρ. As the defender decides to move to ρ at

31

time 2ρ, the condition |S1

same

| + · · · + |S

⌈i⌉
same| ≥ |S1

opp

| + · · · + |S

⌈i⌉
opp| must hold. If k1

ρ = 0

then this means that at time z, the defender starts to move towards −ρ as the condition
|S1+⌈i⌉
⌈i⌉ intervals at −ρ are accounted for by capturing S1+⌈i⌉

| holds. Thus, the 2i + 1 lost intervals in A1

ρ as well as the first

opp . The proof for k1

| + · · · + |S2i+1
same

opp | > |S1

ρ > 0 is analogous

same

to the proof for Al

ρ, 1 < l < m − 1 and has been omitted for brevity.

Since we have shown our charging scheme pays for all lost intruder intervals and each

captured intruder interval pays for at most 2i+1 lost intruder intervals, the result follows.

We now establish two lower bounds on the competitive ratio for the CAP algorithm

including showing that the bound is tight for i = 0.5 and i = 1.

Lemma 14. Algorithm CAP is no better than 2i+1-competitive for v ≤ i(1−ρ)

(resp. i = 1), Algorithm CAP is no better than 3-competitive for v ≤ min{ 1+ρ

2ρ(2i+1) . For i = 0.5
2ρ , 1−ρ
} (resp.

8ρ

v ≤ min{ 1

3 , 1−ρ

6ρ

}).

Proof. We prove this result using an input sequence I consisting of two streams of intruders,

one stream each arriving at opposite endpoints. Specifically, at location 1, one intruder

arrives at every time instant 2ρ(2i+1)k

i

, 0 ≤ k ≤ K for some very large K. At location −1,

one intruder arrives at every time instant (2i+1)ρ

i + ρ
As the first intruder at location 1 arrives (2i+1)ρ

i , 0 ≤ k ≤ K for the same K.
iv time units before the first intruder
arrives at location −1, CAP moves the defender to ρ from the origin, reaching ρ at time

i + ρ

iv + k2ρ

3ρ. As the intruders at location 1 arrive every 2ρ(2i + 1)/i time units and the intruders at

location −1 arrive every 2ρ/i time units, from this moment on until the end, the defender
remains at ρ because there will always be at most one intruder in Sj+⌊i+1⌋

and exactly one

opp

intruder in Sj+1

same, Sj+2

same until Sj+2i+1

same

combined. Thus, except at the beginning and at the end

of the streams, every 2ρ(2i + 1)/i time units, the defender captures 1 intruder that arrives

at +1 but loses 2i + 1 intruders that arrive at −1.

For v ≤ i(1−ρ)

captures all the intruders that arrive at location −1 captures 2i+1

2ρ(2i+1) , a simple optimal offline algorithm that moves the defender to −ρ and
2i+2 of the intruders. Thus,

32

for the specified parameter settings, CAP cannot be better than 2i + 1-competitive.

We now consider the case when i = 0.5 (resp. i = 1) and show that an optimal offline

algorithm O can capture all intruders for specific parameter settings. First consider ρ+ ρ
i

≤ 1.

Let p1 and p2 denote two distinct points which are ρ and −ρ − ρ/i distance away from the

origin respectively.

O moves the defender towards p1 at time instant 1−ρ

v

− ρ reaching p1 just in time to

capture the first intruder that arrived at endpoint 1. Since the first intruder at endpoint

−1 arrives (2i+1)ρ

i + ρ

iv time after the arrival of first intruder at endpoint +1, the defender

captures the second intruder at p2 by moving immediately towards p2. This is because the

intruder will be located at a distance of (2i + 1)ρv/i + ρ/i from −ρ at the time when the

defender captured the intruder at p1. This means that the distance between the defender and

the intruder will be (2i+1)ρ(1+v)

i

implying that the defender captures this intruder in 2ρ + ρ
i

time or equivalently at location p2. The defender takes 2(2i + 1)ρ/i time units to move

from p1 to p2 and back to p1. Thus, the defender can capture every intruder, that arrives at
endpoint 1 at time instant (2i + 1)2kρ/i, at location p1 at time instant 1−ρ

v + (2i + 1)2kρ/i.
Now consider the intruders that arrive at −1. They arrive 2ρ/i time units apart, and we

already showed that the first intruder that arrived at −1 can be captured at p2. This means

will take 2ρ/i time to reach p2 and 2ρ

that from the moment the defender leaves p2 after capturing an intruder, the next intruder
iv time units to reach −ρ, whereas the defender will
i time to move from p2 to p1 and then to −ρ. Thus, in order to ensure that the
next intruder is not lost while the defender moves from p2 to p1 and then to −ρ, we need

take 4ρ + ρ

i + ρ

4ρ + ρ
i

≤ 2ρ

i + ρ

iv implying v ≤ 1

4i−1 , which always holds for i = 0.5 and yields v ≤ 1

3 for

i = 1. In summary, after capturing an intruder at p1, the defender moves to p2, capturing

an intruder that arrives at −1 at time (2i+1)ρ

i + ρ

iv + 2ρk(2i+1)

i

at location p2.

We now consider ρ + ρ

i > 1. Since ρ + ρ

of −ρ − ρ

i > 1, we cannot set the point p2 at a distance
i ; instead, we set p2 to be −1 and have the defender idle at −1 for (2ρ − 2) + 2ρ
i
time. Note that the total time that the defender takes to move from p1 to −1 combined

33

with the waiting time at −1 is 3ρ − 1 + 2ρ

i . This time is sufficient for the defender to capture

the intruders that arrives at time (2i+1)ρ

i + ρ
iv + 2kρ(2i+1)
as the defender leaves −1 at time 3ρ − 1 + 2ρ

however, arrives at time (2i+1)ρ

i + ρ

i

iv + 2kρ(2i+1)
+ 2ρ

i

i + 2kρ(2i+1)

i

immediately. The next intruder,

i which is after the defender leaves −1
. This is equivalent to the intruder

arriving at −1, 1 + ρ

iv + ρ(2i+1)

i

− 3ρ time after the defender has left −1. The total time the

defender takes to move from −1 to ρ and then to −ρ is 1 + 3ρ and the intruder will take

− 3ρ time. Thus, in order to ensure that the next intruder is not lost,

1 + ρ

iv + 1−ρ
we need 3ρ ≤ ρ

i

v + ρ(2i+1)
iv + 1−ρ
2ρ and substituting i = 1 yields v ≤ 1

v + ρ(2i+1)

i

implies v ≤ 1+ρ

− 3ρ. Substituting i = 0.5 yields 3ρ ≤ 2ρ

3ρ . As, for i = 1, v ≤ 1

v + ρ, which

v + 1−ρ
3 holds, then this

implies that v ≤ 1

3ρ always holds. This concludes the proof.

Remark 2 (Evasive motion of intruders). We say that an intruder evades when the intruder

moves toward the endpoint that generated the intruder or remains stationary. Since the

analysis in Theorems 5, 6 and the Sweep algorithm are independent of the nature of the

motion of intruders, the two fundamental results as well as Theorem 8 hold for evasive

intruders. Furthermore, by modifying the set Sk

opp in Algorithm 1 to [−1, −(ρ + 2vρ)] (resp.

[1, (ρ + 2vρ)]), it can be shown that 1 is 2-competitive for evasive intruders as well.

2.3 Escaping Intruders

This section addresses problem P2 with the physical region being R(ρ) = [−ρ, ρ] with

ρ > 1, where intruders arrive at either −1 or 1 and they move at a fixed speed 0 < v < 1

towards −ρ and ρ, respectively, as depicted in Figure 2.1b. Recall that if they pass their

destination before the defender, moving at speed 1, reaches them, they are not captured.

If the defender reaches them before they pass the target point (−ρ or ρ), then they are

captured.

2.3.1 Fundamental Limits

Similar to Section 2.2, we start by deriving a fundamental limit for c-competitiveness of

any algorithm.

34

Theorem 15 (Necessary condition for c-competitiveness). For any problem instance P with

v > ρ−1

ρ+1 , there does not exist a c-competitive algorithm.

Proof. Assume that the defender starts at the origin for both online algorithm and the

optimal offline algorithm. Similar to the proof of Theorem 5, we start by constructing an

input sequence I. The input sequence consists of two stages: a stream of intruders that

arrive at location 1, max{2, ρ−1
v

} time units apart starting at time 1 and a burst of c + 1

intruders which arrive at location −1 at time t. Time instant t is the time instant the

defender moves to 1 according to any online algorithm for the first time. Note that if the

defender never moves to location 1, the stream never ends, and the burst never arrives, so

the algorithm will not be c-competitive for any constant c, and the result follows.

Let i be the number of stream intruders released up to and including time t; note that i

will be at least 1 if the defender reaches 1 exactly at time 1 when the first stream intruder

is released. We first observe that the defender will capture at most intruder i. In particular,

because the stream intruders are released max{2, ρ−1
v

} time units apart, all stream intruders

before i have reached ρ before time t. Two cases arise:

Case 1 : If t = 1, this means the defender reached location 1 exactly at time 1 capturing

the intruder that arrived at location 1 at time instant 1. Since v > 1−ρ

1+ρ , the defender will not

be able to capture the burst of c + 1 intruders that arrive at time t = 1. We now consider

how many intruders the optimal offline algorithm could have captured. For the same input

instance as for the online algorithm, the optimal offline algorithm can move the defender to

location −1 by time t and will thus capture all c + 1 burst intruders, and the result follows.

Case 2 : If t > 1, this means the defender reached location 1 after time 1. In this case, the

optimal offline algorithm can capture the first i−1 stream intruders immediately upon arrival

by moving the defender to location 1 at time 1 and staying there until the first i − 1 stream

intruders have been captured. This algorithm can then move the defender to position −1

before the burst intruders have been released since the stream intruders arrive max{2, ρ−1
v

}

time units apart. Note that if i = 1, the algorithm moves the defender immediately to −1.

35

In either case, we show that the optimal offline algorithm can capture all the burst

intruders and at least all but the last stream intruder whereas the online algorithm will

capture at most 1 stream intruder, and the result follows.

Corollary 3. For any problem instance P with v > ρ−1

ρ+1 , no algorithm (online or offline)

can capture all intruders.

Proof. Consider the same input instance as described in the proof of Theorem 15. The result

then follows by observing there are choices for t including t = 1 such that no algorithm can

capture all i stream intruders and all c + 1 burst intruders, for same t and i as defined in the

proof of Theorem 15. This concludes the proof.

2.3.2 Algorithms

We now analyze two algorithms for the escaping intruders scenario. Note that CAP is

not effective for escaping intruders since the perimeter locations −ρ and ρ are far apart.

2.3.2.1 Sweeping Algorithm

Sweep is fundamentally the same as described for incoming intruders in Section 2.2; the

only change is that the endpoints are now −ρ and ρ rather than −1 and 1. That is, the

defender starts at the origin and moves toward ρ. Thereafter, the defender does not change

speed or direction until it reaches an endpoint in which case it moves at unit speed towards

the other endpoint.

Theorem 16. Sweep is 1-competitive for any problem instance P for v ≤ ρ−1

1+3ρ . Otherwise,

Sweep is not c-competitive for any constant c.

Proof. Similar to the proof of Theorem 8 assume that v ≤ ρ−1

1+3ρ . Any intruder that arrives
time to reach the perimeter after arriving in R(ρ).

in the physical region R(ρ) takes ρ−1
v

Without loss of generality, consider an arbitrary intruder i that arrives at location 1. In the

worst case for Sweep, which is that intruder i arrived just after the defender left location 1,

the defender will take 3ρ + 1 time to reach ρ. This follows as the defender first moves to

36

location −1 from ρ and then moves in the direction of the intruder, i.e., towards ρ. Since

3ρ + 1 ≤ ρ−1

v holds, then Sweep’s defender will get to ρ first and intruder i will be captured,

and the first result follows.

Now suppose that v > ρ−1

1+3ρ . Then there is an input instance where intruders only arrive

at location 1 just after Sweep’s defender has left location +1. Given that v > ρ−1

1+3ρ , all

intruders will be lost and the second result follows.

Next we describe a 4-competitive algorithm. For this algorithm, we make an assumption

on the finiteness of the number of intruders that can arrive in the physical region R(ρ).

This is because, unlike the previous algorithms, the defender now keeps an account of the

intruders at every time instant. Thus, if the arrival of the intruders does not end in finite

duration, it is possible that the defender does not capture any intruder and so the algorithm

will not be competitive for any constant c.

2.3.2.2 Instantaneous Compare and Capture Algorithm

Earlier, for incoming intruders, Algorithm CaC could compare the intruders that arrived

in a particular time interval T from each entry point, capture the intruders that arrived in

T from one entry point, and still be able to capture any intruders that arrived after T . In

the escaping intruders case, this is not possible. Specifically, the defender may not be able

to capture its currently targeted intruders and still capture intruders that arrived at the

other arrival location after making its current capture plan. Thus, Instantaneous Compare

and Capture algorithm (iCaC) constantly monitors for new arrivals and considers changing

direction. We make the following assumption:

Assumption A1: Finite number of intruders arrive in the physical region R(ρ).

Note that this assumption implies that the last intruder arrives at a finite time instant.

To determine if the defender should start chasing intruders at a time instant k, iCaC tracks

three quantities: the total number of intruders it has captured nk

c and lost nk

l until instant

k, as well as the total number of intruders that exist in the physical region nk

e . iCaC tracks

these quantities from the most recent time instant these quantities were set to zero which

37

may be time 0. iCaC should begin a chase at time k if the chase condition C1, defined as

2nk

c < nk

l + nk

e , holds. While the defender is chasing intruders at some time instant k, iCaC

monitors the balance of intruders on the left and right; more precisely, it tracks two sets of

intruders: Skl

left which is the set of intruders that arrived at location −1 no earlier than time

instant kl ≤ k and Skr

right which is the set of intruders that arrived at location 1 no earlier
than time instant kr ≤ k. The time instants kl and kr might not be equal in order to track

the intruders in Skl

left and Skr
iCaC, the defender either chases intruders in Skl

right appropriately. We will formally define kl and kr shortly. In

left or in Skr

right, and it switches direction when

the other set of intruders is at least double the current set of intruders it is chasing. We

describe this more precisely when we define the two chase modes.

A key parameter for iCaC is the capture limit; the point of no return such that if the

defender crosses it, then it is impossible for the defender to capture intruders escaping on

the other side. Some problem instances do not have a capture limit meaning the sweeping

algorithm is guaranteed to capture all intruders.

Definition 17 (Capture limit). If ρ(1−v)−1−v
has two capture limits, namely ρ(1−v)−1−v

2v

< ρ, then the physical region R(ρ) for ρ > 1

and − ρ(1−v)−1−v

2v

2v

, which we denote as pc and −pc,

respectively.

We outline the iCaC algorithm in Algorithm 3 and describe the three modes, namely rest

mode, chase mode and locked mode, that iCaC operates in, below.

Rest mode: The defender is located at either 1 or −1 and C1 does not hold. This means

that the defender has captured sufficiently many intruders and can stay at the same location

at time instant k. The defender resets nk

c , nk

e and nk

l to zero (line 9-11 in Algorithm 3).

With the reset of nk

c , nk

e and nk

l to zero, a single intruder arriving on the opposite side is

sufficient to make the defender leave rest mode thus guaranteeing that iCaC will not rest on

many early captures to allow many later arriving intruders to escape. iCaC exits rest mode

and enters either chase mode (line 5) or locked mode if C1 holds.

38

Figure 2.4 Snapshot at time instant k for pc > 1 when the defender switches from chase-right mode
to chase-left mode as C2 holds.

Chase mode: We have two modes under chase mode, chase-right and chase-left, where

the defender is chasing intruders but can change direction instantaneously; these are the

most complex modes and represent the key new modes for escaping intruders. iCaC can be

in either of the two chase modes when the defender is in the range [−pc, pc]. In particular,

when pc ≥ 1, iCaC is in chase-left (resp. chase-right) mode when the defender is in the range

[−pc, pc] and is moving towards −ρ (resp. ρ). When pc < 1, iCaC is in chase-left (resp.

chase-right) mode when the defender is located in [−pc, 1) (resp. (−1, pc]) and is moving

towards −ρ (resp. ρ). For pc ≥ 1, we define the chase region as the range [−pc, pc] and for

pc < 1, we define chase region as the range [−pc, 1) (resp. (−1, pc]) when the defender is in

chase-left (resp. chase-right) mode. During chase mode, the defender updates nk

c , nk

l and nk
e

at every time instant.

iCaC monitors a switch condition C2 during either of the two chase modes at every

time instant. In the chase-right mode, C2 corresponds to |Skl
left

| ≥ 2|Skr

right

|. In the chase-

left mode, C2 corresponds to |Skr

right

| ≥ 2|Skl
left

|. As the name implies, if the corresponding

switch condition applies, iCaC toggles its chase mode (lines 14-17) and reverses defender

direction (cf Fig: 2.4). The switch condition ensures that the payoff for changing direction is

large enough to warrant the cost of switching directions. Otherwise, the defender continues

to move in the same direction (lines 18-20). If the defender captures all intruders it was

chasing before reaching the capture limit, the defender enters locked mode (defined later)

where it will return directly to 1 or −1.

We now define how Algorithm iCaC sets time instants kl and kr and updates nk

e , nk

c , and

nk
l when a switch occurs. When Algorithm iCaC first enters a chase mode from either rest

mode or locked mode, assuming there are a finite number of intruders, it will go through

39

Algorithm 3: Instantaneous Compare-and-Capture Algorithm
1 Defender is at origin
2 for each time instant k do
3

c , nk
Update nk
l .
if defender is in rest mode then

e , nk

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

if chase condition holds then

Set kl = kr = t1.
Enter chase-left (resp. chase-right) mode. Move to capture intruders that
arrive at −1 (resp. 1).

else

Stay at 1 (resp. −1) and reset nk

e , nk

c , nk

l to 0.

end

else if defender is in chase-right (resp. chase-left) mode then

if switch condition holds then

Add |Skr
Move to capture |Skl
left

| (resp. |Skl
|) to nk
left
| (resp. |Skr

right

right

|).

l . Set kr = k (resp. kl = k).

else

Move to capture |Skr

right

| (resp. |Skl
left

|).

end
if

right

|Skr
| (resp. |Skl
left
Enter locked mode and reset nk

e , nk

c , nk

l to 0.

|) is captured or defender exits chase region then

end

else

if defender is located in chase region then

Return to nearest location out of 1 or −1.

else

Move to capture all intruders on same side.
Return to nearest location out of 1 or −1 after capture.

end
if chase condition holds after return to 1 or −1 then

Set kl = kr = tl

e−1 and enter chase mode.

else

Enter rest mode reset nk

e , nk

c , nk

l to 0.

end

end

33
34 end

a sequence of n consecutive chase modes for some n ≥ 1 before ending in locked mode.

We assume without loss of generality that the first chase mode is chase-right. The defender

begins this first chase mode at position −1 at some time t1. For this first chase mode, kl = kr

and are set equal to t1 if the previous mode was rest mode or t1 = 0. Otherwise, kl = kr

40

(a) Defender is in chase region and has cap-
tured intruders it was pursuing at time in-
stant k.

(b) Defender is at location pc and C2 does
not hold.

Figure 2.5 Snapshots at time instant k when defender enters locked mode for pc > 1.

and are set equal to the start of the previous locked mode. For the ith chase mode in this

sequence for 2 ≤ i ≤ n, we define ti to be the time when the switch condition caused iCaC

to change direction for the (i − 1)th time. If i is even, then kl = ti−1 and kr = ti; if i is odd,

then kr = ti − 1 and kl = ti. This assignment of kl and kr is required to ensure an accurate

comparison of intruders on either side for the switch condition. In either case, Algorithm

iCaC adds the intruders that the defender was pursuing prior to reversing the direction to

l i.e., the defender will not include these intruders in either Skl
nk

left or Skr

right again and will not

count such intruders as captured even if the defender does capture them.

Locked mode: In this mode, the defender’s route is fixed until it returns to the nearest

location out of 1 or −1. At the start of locked mode, Algorithm iCaC resets nk

c , nk

e , and nk

l to

0 and then updates these quantities as appropriate but will not change mode until it returns

back to −1 or +1 after capturing intruders. When the defender returns to −1 or +1, C1

will dictate whether iCaC enters rest mode or one of the chase modes. If the defender is in

the chase region, then the defender enters locked mode only when the defender has captured

all intruders it is pursuing (lines 21-24, cf Fig. 2.5a). In this case, the defender returns to

the nearest location out of 1 or −1 (lines 26-28). For pc > 1, the defender is guaranteed to

be in locked mode if it is in the region [−ρ, −pc) ∪ (pc, ρ]. In particular, Algorithm iCaC

transitions from chase-left mode (resp. chase-right mode) to locked mode at time k if the

defender is at −pc (resp. pc) and C2 does not hold (cf Fig. 2.5b). In this case, the defender’s

route is locked into capturing all the intruders it can that are to its left (resp. right). Once it

has done so, it returns to −1 (resp. 1) (lines 29-31). Recall that for pc < 1, Algorithm iCaC

is in chase-left (resp. chase-right) mode in [−pc, 1) (resp. (−1, pc]). So, Algorithm iCaC is

41

in locked mode when the defender is in [−ρ, −pc) ∪ [1, ρ] (resp.

[−ρ, −1] ∪ (pc, ρ]). In this

case, the defender crosses −1 (resp. 1) when pursuing intruders after entering locked mode

from chase-left (resp. chase-right) mode and then returns to −1 (resp. 1) after capture.

We now prove two properties of algorithm iCaC.

Lemma 18. If the defender leaves 1 (resp. −1) at time t and goes directly to −pc (resp. pc)

and then turns around and goes to ρ (resp. −ρ), it will arrive at ρ (resp. −ρ) before any

intruder that arrives at 1 (resp. −1) strictly after time t.

Proof. The time it takes for the defender to leave 1 and go to −pc and then go to ρ is
exactly 2pc + 1 + ρ. The time it would take an intruder to go from 1 directly to ρ is ρ−1
v .

Since the defender will leave 1 before the intruder in question arrives, we need to show that

2pc + 1 + ρ ≤ ρ−1

v which is true given our definition of pc. This concludes the proof.

Note that if pc ≥ ρ, then Sweep is guaranteed to capture all intruders and iCaC is not

necessary, so we assume that pc < ρ for the remainder of this algorithm description.

In order for Algorithm iCaC to be effective, we assume that v ≤ 3(ρ−1)

7ρ+1 . This ensures

that if the defender crosses either capture limit at time t, it has sufficient time to capture

any intruders it was pursuing plus any intruders that arrive after time t. We formalize this

condition below.

Lemma 19. Let the defender be in the locked mode at time t. Then, the defender can capture
all intruders that arrive in the physical region R(ρ) at time t′ ≥ t for v ≤ 3(ρ−1)
7ρ+1 .

Proof. We first consider the case when the defender is in [−ρ, −pc) ∪ (pc, ρ]. In the worst

case, the time it takes the defender to leave pc (resp. −pc) and go to ρ (resp. −ρ) and then

go to −ρ (resp. ρ) is exactly 3ρ − pc. The time it would take an intruder to go from −1
v . Thus, we simply need to show that 3ρ − pc ≤ ρ−1
(resp. 1) directly to −ρ (resp. ρ) is ρ−1
which is true given v ≤ 3(ρ−1)
7ρ+1 . For the case when the defender enters the locked mode in

v

[−pc, pc], Lemma 18 ensures the capture of the intruders. This concludes the proof.

42

A consequence of Lemma 19 is that there is no benefit for intruders to arrive at either

location 1 or −1 once the defender enters the locked mode. We now establish that Algorithm

3 is 4-competitive.

Theorem 20. For input instances with a finite number of intruders, the Instantaneous

Compare and Capture (iCaC) algorithm defined in Algorithm 3 is 4-competitive for any
problem instance P for which v ≤ 3(ρ−1)
7ρ+1 .

Proof. We first observe that Algorithm iCaC would not be competitive against some input

instances with an infinite number of intruders as Algorithm iCaC can be forced to keep

changing directions without capturing any intruders. Our assumption A1 of a finite number

of intruders implies that the last intruder must arrive at a finite time instant t. If the chase

condition never holds which means the defender is always in rest mode, it is obvious that

nc ≥ nl/2. We thus assume for the rest of the proof that the chase condition holds at least

once.

We start by defining an epoch τe which starts at the time instant te when the defender

leaves locked mode and ends at the time instant te+1 when the defender leaves the next

consecutive locked mode. The first epoch τ1 starts at time 0, i.e., we treat time 0 as the end

of locked mode. As the number of intruders is finite, the defender will be in locked mode at

least once, so one epoch must exist.

In any epoch τe, the defender will enter a sequence of n chase modes for some finite

n ≥ 1, starting at times tc,i
e

for 1 ≤ i ≤ n, and then locked mode, starting at time tl

e. Recall

that, for condition C2, Algorithm iCaC sets kl and kr to the time instances tc,i

e depending

on the type of chase mode. The defender may skip rest mode completely meaning te = tc,1
e ,

or the defender starts in rest mode at te before transitioning to chase mode at tc,1

e > te, i.e.,

the defender is in rest mode in the interval [te, tc,1

e ).

During epoch τe, we account for all the intruders that arrived from tl

e−1 (the start of

the previous epoch’s locked mode) or time 0 for τ1 to time tl

e. Suppose Algorithm iCaC

enters rest mode at time te. This means the defender caught at least 1/3 of the intruders

43

that arrived during the interval [tl

e−1, te] so the chase condition does not hold at time te.

Likewise, during rest mode, the defender resets nk

c , nk

l and nk

e to zero until time tc,1

e when the

number of intruders that arrive at the other entry point are more than double the number

of intruders that arrive at the entry point the defender is at. Thus, in summary, Algorithm

iCaC captures at least 1/3 of all the intruders that arrived in the time interval [tl

e−1, tc,1

e ).

We then only need to account for the intruders that arrive from [tc,1

e , tl

e) when analyzing the

sequence of chase modes and locked mode. On the other hand, if Algorithm iCaC skips rest

mode, then we must account for all the intruders that arrived since tl

e−1 when analyzing the

sequence of chase modes and locked mode.

We now analyze the sequence of n chase modes, n ≥ 1. One crucial observation is that

because of Lemma 18 and Lemma 19 and the fact that Algorithm iCaC adds any intruders

it was chasing before switching directions to nk

l , all the intruders the defender is chasing

during a given chase mode will be captured if the defender does not switch directions again

in the chase mode. We focus on the final chase mode which starts at time tc,n

e . Without loss

of generality, we assume this nth chase mode is chase-right mode. Let an be the number of

intruders in Skr

right, kr = tc,n−1

e

, at time tc,n

e

that caused Algorithm iCaC to move towards ρ.

Specifically, the intruders an are all of the intruders that arrived at location 1 in the interval

[tc,n−1
e

, tc,n

e ]. We now characterize the number of intruders lost that arrived before time tc,n
e .

These correspond to the number of intruders that were pursued but never captured in the

earlier n − 1 chase modes (which might be 0 if there were no such chase modes). Recall

that Algorithm iCaC adds these intruders to nk

l after switching direction in the chase mode.

Because the switch condition requires the number of intruders in the opposite direction to

be twice the number of intruders in the direction the defender was pursuing, this number of

intruders is at most an/2 + an/4 + · · · + an/2n−1 which is strictly less than an.

We now characterize the number of intruders bn that arrived during [tc,n

e , tl

−1 and thus might be lost. Note that these bn intruders correspond to |Skl
left

e] at entry point
| where kl = tc,n
e .

Let a′

n be the number of intruders that arrived at entry point 1 that were captured as well

44

Figure 2.6 Parameter regimes for incoming intruders (P1). Solid lines, extend to the left and lower
bounds (dashed lines), extend to the right.

as pursued during [tc,n

e , tl

e] which might be larger than an since new intruders might have

arrived at 1 after tc,n

e . Because of the switch condition, we know that bn < 2a′

n. Finally, from

Lemma 18, Algorithm iCaC will capture all a′

n intruders during the nth chase mode or the

ensuing locked mode. This means that the number of intruders lost in epoch τe, an + bn, is

strictly less than 3a′

n, the number of intruders captured during the sequence of chase modes

and locked modes of epoch τe.

This analysis applies to all epochs meaning Algorithm iCaC captures at least 1/4 of all

intruders that arrive before the stock of locked mode for the final epoch. The final epoch

must end with a transition to rest mode meaning that iCaC captures at least 1/3 of the

intruders that arrive after the start of locked mode for the final epoch. Thus, the result

follows.

We now provide two parameter regime plots highlighting the parameter regime in which

our algorithms are effective as well as the parameter regime for the established necessary

conditions for both problem variants P1 and P2.

45

2.4 Results and Discussion

Figure 2.6 and Figure 2.7 show v-ρ plots summarizing our results for incoming (P1) and

escaping (P2) intruders, respectively.

For incoming intruders (Figure 2.6), we see that CAP has relatively small area of utility.

This is the region between the curves for CAP and CaC which increases as the input i to CAP

increases. Note that CAP does not extend beyond the dark red curve (i → ∞) in Figure

2.6. For ρ ≥ 0.9, the parameter curve of CAP for i → ∞ overlaps that of the CaC algorithm

implying that for any ρ < 0.9, there exists a CAP algorithm whose parameter regime is

completely above that of the CaC. Although CAP covers higher parameter regime for higher

values of i, the increase in the parameter regime decreases as i increases implying that for

high values of i, slight increase in the parameter regime may increases the competitiveness of

CAP considerably. Interestingly, CAP’s regions extend beyond the curve for Theorem 6 and

for small values of ρ, CAP’s parameter regime covers all values of intruder speed v. Finally,

we observe that for ρ > 0.95, the parameter curve for CaC and CAP (i → ∞) overlaps to

that of the Sweep. Key open problems include closing the 2-competitive gap between the

CaC curve and the curve for Theorem 6 and closing the c-competitive gap between the CAP

(i → ∞) curve and the curve for Theorem 5.

For escaping intruders, we plot the lower bound and the two algorithms. We observe

that the region of utility for Sweep is completely contained within the region of utility for

Algorithm iCaC. The key open problem is to close the gap between the curves for Theorem

15 and Algorithm iCaC.

We now briefly comment on the choice of algorithms and the applicability of this work.

The choice of which algorithm to use depends on the problem parameters, i.e., the size of the

perimeter and the speed ratio, and the acceptable bound on competitiveness. Note that the

acceptable competitive ratio is governed by the application. For instance, given the problem

parameters, a military application may require algorithms with the least competitive ratio1

1Designing a 1-competitive algorithm may not always be feasible.

46

Figure 2.7 Parameter regimes for escaping intruders (P2). Solid lines, extend to the right and
lower bounds (dashed lines), extend to the left.

whereas an application focused on relaying information may not have this requirement.

Finally, we remark that the sufficient conditions on the algorithms are derived in the worst-

case.

In some applications, these worst-case inputs may not be realizable and thus the

algorithms will perform better for those applications.

2.5 Conclusions

We study a problem where a single defender defends a linear perimeter from mobile

intruders. A key contribution is the combination of concepts and techniques from competitive

analysis of online algorithms with pursuit of multiple mobile intruders. We considered two

scenarios: incoming intruders and escaping intruders. We designed and analyzed three classes

of algorithms for incoming intruders. The first two algorithms are demonstrated to be 1, and

2-competitive, respectively. We generalized our third algorithm to a family of algorithms and

demonstrated that it is 2i + 2 competitive, where i is an input for the generalized algorithm.

For escaping intruders, we designed and analyzed two algorithms and proved that they are

1 and 4-competitive. We also derived fundamental limits on the c-competitiveness for any

47

constant c for both scenarios.

48

CHAPTER 3

RANDOMIZED COMPETITIVE PERIMETER DEFENSE IN LINEAR
ENVIRONMENTS

In Chapter 2, our main focus was on the design and analysis of deterministic algorithms with

provable competitive ratios. Recall that the inputs designed by the adversary heavily rely on

the information about the online algorithms. In simpler terms, the adversary takes advantage

of the information asymmetry to construct worst-case inputs against an online algorithm,

prompting the question, "How can online algorithms limit the information available to the

adversary?".

To address the aforementioned question, in this chapter, we design randomized algo-

rithms to withhold some information from the adversary. Specifically, we address a problem

in which a robotic defender, having a maximum unit speed, is tasked with defending a given

set of points (perimeter) from mobile intruders. The intruders are released in an arbitrary

manner at the endpoints of the line segment and move inwards, i.e., towards the midpoint

of the environment. Upon release, each intruder moves with a fixed speed v < 1 towards

the perimeter. The defender seeks to intercept maximum number of intruders before the

intruders reach the perimeter. We focus on design and analysis of randomized online algo-

rithms to route the defender as the input, consisting of release times, number of intruders,

and release location of every intruder, is unknown a priori. The contributions of this chapter

are as follows:

1. Necessary conditions: We establish necessary conditions on the competitiveness of

any randomized online algorithm. Specifically, we characterize a parameter regime in

the v − ρ space in which no randomized algorithm can be better than 2-competitive,

and a second parameter regime in which no algorithm is better than 1.33-competitive.

We also characterize a parameter regime on the existence of 2-competitive algorithms1

for a special class of randomized online algorithms that do not allow the vehicle to

1in the limit when an adversary can release arbitrarily many intruders

49

remain stationary. Such algorithms are called zealous algorithms.

2. Algorithm design and analysis: We design and analyze three randomized algo-

rithms and establish their competitive ratios. Specifically, our first algorithm is prov-

ably 2-competitive for any value of v and ρ while the competitive ratio of our second

and third algorithms depends on the problem parameters.

This chapter is organized as follows.

In Section 3.1, we formally define our problem.

Section 3.2 establishes the fundamental limits of the problem and Section 3.3 presents the

randomized algorithms and their analysis. Section 3.4 presents and discusses the parameter

regime plots and finally, Section 3.5 summarizes this chapter.

3.1 Problem Formulation

Analogous to the environment in Chapter 2, consider a linear environment E = [−1, 1]

which consists of a region R(ρ) = {y ∈ R : |y| ≤ ρ < 1}. The perimeter ∂R(ρ) is defined as

the set of two locations {ρ, −ρ}, which needs to be defended from intruders. An adversary

releases mobile intruders over time at either location −1 or 1. Each intruder moves with fixed

speed v < 1 towards the nearest location in ∂R(ρ). The defense consists of a single vehicle

with motion modeled as a single integrator. The vehicle is said to capture an intruder when

the vehicle’s location coincides with it. An intruder is said to be lost if the intruder reaches

the perimeter without being captured. An intruder is said to be capturable if the vehicle can

capture it before the intruder reaches the perimeter. Let n(t) denote the number of intruders

that arrive at time instant t at endpoint +1. In this work we consider that n(t) ≤ 2N , i.e.,

the adversary can release at most 2N, N ∈ Z>0, intruders at each time instant. Further, we

will also suppose that the adversary requires at least ∆ ∈ R>0 time to release intruders in

the environment once it has released some intruders in the environment. More precisely, if

n(t) ≤ 2N intruders are released at time instant t, then the next intruders are released no

earlier than time t + ∆. Note that if ∆ ≥ 2 holds, then the problem is trivial as the defender

can cover the environment before the next intruder arrives. Thus, we will focus on the case

50

that 0 < ∆ < 2. Finally, we denote the n(t) intruders released at +1 at time t as in

t,+1.

Intruders released at endpoint −1 are denoted analogously. Since our problem is defined by

four parameters, we denote a problem instance as P(ρ, v, N, ∆) or P for brevity. With the

c-competitive randomized algorithms defined in Chapter 1 (Definition 3), the objective of

this chapter is the following.

Problem Statement: The aim of this chapter is to establish fundamental guarantees

and design c-competitive randomized algorithms for the vehicle.

3.2 Fundamental Limits

We start with establishing the necessary conditions in the (v, ρ) parameter space for any

randomized algorithm. Our approach is based on Yao’s mini-max principle, stated below,

to lower bound the expected performance of any randomized algorithm.

Lemma 21 (Yao’s mini-max principle[45]). Let ¯X denote the probability distribution over

a set of deterministic input instances Ix, x ∈ X . Suppose that there exists a constant c ≥ 1

such that

inf
Ai∈Adet

{E ¯X[nO(Ix)] − cE ¯X[nA,i]} ≥ 0,

then the constant c is a lower bound on the competitive ratio of any randomized algorithm

against an oblivious adversary, where Adet is the set of deterministic online algorithms as

defined in Chapter 1.

Yao’s principle is a valuable tool that allows us to convert the supremum over determinis-

tic inputs (cf. Definition 3) into a search using the expected performance over a finite number

of inputs and then optimize the choice of the probabilities used to compute the expected val-

ues. We demonstrate this technique in establishing the fundamental limits for the problem.

In particular, we first establish fundamental limits for a particular class of algorithms known

as zealous algorithms and then establish fundamental limits for any randomized algorithms.

We start with the definition of zealous algorithms.

51

Definition 22. An algorithm Az is called zealous if it satisfies the following:

1. If there are capturable intruders, then direction of vehicle as per Az changes only if

one of the conditions hold: i) a new intruder is released at an endpoint, ii) the vehicle

is at the origin, or iii) the vehicle has just captured an intruder.

2. At any time if there are capturable intruders, Az’s vehicle moves towards a capturable

intruder or the origin (if the vehicle is not at the origin).

Note that in all of the proofs for the fundamental limits, we will start by describing a set

of input sequences I and a probability distribution on I. The set I = {I∞, I∈}, i.e., I consists

of two input sequences I∞ and I∈, each of which will be described in the proof. We then

determine the performance of the optimal offline algorithm as well as the best deterministic

algorithm and apply Yao’s principle to establish the result.

Theorem 23. Let ϵ := 1 + ρ − 1−ρ

v . For any problem instance P, inf Az cAz (P) ≥ 4N

2N +1 if

v ≥ 1−ρ

1+ρ and





∆ <

min{1, 2ρ}, if v = 1−ρ
1+ρ

min{1, ϵ}, otherwise.

Proof. In this proof, the set of input sequences I consists of two input sequence, i.e., I =

{I1, I2}. Both input sequences in I start at time 0 and are described as follows. In input

sequence I1, a single intruder is released at +1 at time 0. We denote this intruder as i0

1,+1.

At time 1, 2N intruders are released at endpoint −1 (denoted as i1

2N,−1) and at time instant

v = 1−ρ

t, 2N intruders are released at +1 (denoted as it

2N,+1). The time instant t = 1 + ∆ for
1+ρ and t = 1 + ϵ, otherwise. Similarly, in input sequence I2, a single intruder is released
at −1 at time 0. At time 1, 2N intruders are released at endpoint +1 and at time instant t,

2N intruders are released at endpoint −1. Both I1 and I2 have equal probability of 1/2 of

being selected. We now describe the performance of an optimal offline algorithm O followed

52

by the performance of the best deterministic zealous algorithm Az. For both O and Az, we

assume that the vehicle starts at the origin at time 0.

As O has the entire information of the selected input sequence, O can move the vehicle

to the endpoint at which the first 2N intruders arrive. Suppose that I1 is selected. Then, O

moves the vehicle towards −1 and captures i2N

1,−1 immediately upon release. Upon capture,

O then moves the vehicle directly towards +ρ, capturing the 2N intruders that were released

at endpoint 1 at ρ. The capture of these intruders is ensured as these intruders arrive at

time t. The performance of O against I2 is analogous. Thus, E[nO] = 4N .

From zealousness of Az, it follows that the Az’s vehicle will move towards the single

intruder that arrives at time 0. Further, after time 0, the vehicle can only change direction

once it captures the single intruder. Suppose that input sequence I1 is selected. Then, since

vehicle moves towards +1 at time 0, the vehicle will be

1
1+v distance away from the origin

when it captures the first intruder. At this moment, the vehicle can either continue to move

towards +1 or towards the origin.

Suppose that the vehicle moves towards +1 and consider that v > 1−ρ

1+ρ holds. Since 2N
intruders arrive at −1, the vehicle will not be able to capture these intruders if v > 1−ρ
1+ρ .
Thus, in this case and against I1, the vehicle captures at most 2N + 1 intruders. Now
consider that v = 1−ρ

1+ρ holds. Then, the vehicle can capture the i1

2N,−1 intruders that arrive

at endpoint −1 exactly at location −ρ by moving towards −ρ as soon as these intruders are

released. Further, as it

2N,+1 intruders arrive at +1 at time t = 1 + ∆, the vehicle cannot

capture these intruders since ∆ < 2ρ. Finally, if the vehicle changes direction when it

2N,+1

intruders are released, it cannot capture it

2N,+1 and i1

vehicle captures at most 2N + 1 intruders for v ≥ 1−ρ

2N,−1 as v < 1. Thus, against I1, the
1+ρ . The performance against I2 is

analogous and has been omitted for brevity.

Now suppose that the vehicle decides to move towards the origin after capturing i0

1,+1
against I1. In this case, it is guaranteed that the vehicle will not be at −1 by time 1 and

thus, it follows that the vehicle will capture at most 2N + 1 intruders. As the performance

53

against I2 is analogous, E[nAz ] = 2N + 1. The result then follows from Lemma 21.

Theorem 24 (Necessary condition for c(P) ≥ 2). For any problem instance P for which

v > 1−ρ

ρ , c(P) ≥ 2 against an oblivious adversary.

Proof. The set of input sequences I consists of two input sequences; I1 and I2. Both input

sequences start at time instant ρ and involves a single intruder i. Specifically, in input

sequence I1, intruder i arrives at endpoint −1 at time t ≥ ρ. On the other hand, in input

sequence I2 intruder i arrives at endpoint +1 at the same time t. Both I1 and I2 have equal

probability of 1/2 of being selected. Thus, our set of input sequences is I = {I1, I2} with a

probability distribution ¯X on I. We now describe how the optimal offline algorithm and the

best deterministic algorithm captures intruder i. We assume that the vehicle starts at the

origin for both optimal offline and the deterministic algorithm.

Recall that once the input sequence Is, s ∈ {1, 2} is selected, the optimal offline algorithm

O has the entire information of the selected input sequence. As intruder i arrives after time

ρ in both input sequence, O can move the vehicle in the direction of the endpoint where the

intruder i arrives. Given that the vehicle has covered at least ρ distance moving towards

the end point where i arrives, capture of i is ensured in both input sequences. This yields

E ¯X[nO] = 1

2 + 1

2 = 1.

For any deterministic online algorithm A, if the vehicle is at the origin at time t then,

since 1−ρ

v > ρ, the vehicle cannot capture intruder i from any input sequence in I. Thus,
the vehicle must be either located in (0, 1] or [−1, 0) by time t. Without loss of generality,

assume that the vehicle is located in (0, t] when intruder i arrives.

Intruder i arrives at

endpoint +1 with probability 1/2, in which case, A captures i. Otherwise, i.e., intruder i

arrives at endpoint −1 with probability 1

2 , the vehicle will not be able to capture intruder i

as 1−ρ

v > ρ. Thus, E ¯X[nA] = 1

2 and the result follows from Lemma 21.

The previous result establishes a necessary condition for the existence of online random-

ized algorithms with competitive ratio of at least 2. Then the next result establishes a

54

necessary condition for the existence of online randomized algorithms with competitive ratio

of at least 1.33 under specific parameter regime.

Theorem 25 (Necessary condition for c(P) ≥ 1.33). Let ϵ :=






2ρv, if v = 1−ρ
1+ρ .

.

1 + ρ − 1−ρ

v , if v > 1−ρ
1+ρ .
3 against an

Then, for any problem instance P that satisfies v ≥ 1−ρ

1+ρ and ∆ < ϵ, c(P) ≥ 4

oblivious adversary.

Proof. The set of input sequence consists of two input sequences; I1 and I2.

In input

sequence I1, a single intruder i1

1,+1 arrives at endpoint +1 at time instant 1 and a single

intruder i1+ϵ

1,−1 arrives at endpoint −1 at time 1 + ϵ, where




2ρv, if v = 1−ρ
1+ρ .

ϵ =

1 + ρ − 1−ρ

v , if v > 1−ρ
1+ρ .

In input sequence I2, a single intruder i1

1,−1 arrives at endpoint −1 at time instant 1 and

a single intruder i1+ϵ

1,+1 arrives at endpoint +1 at time 1 + ϵ, for the same ϵ. Both input

sequence have equal probability of 1/2 of being selected. Thus, we have set of input sequence

I = {Ix : x ∈ {1, 2}}, with a probability distribution ¯X on I.

We first describe how the optimal offline algorithm O captures both intruders. As O

knows the entire input sequence at time 0, O can have the vehicle move to endpoint +1

arriving endpoint +1 at exactly time +1, if input sequence I1 is selected. Otherwise, O

can move the vehicle to endpoint −1, arriving endpoint −1 at exactly time 1. Either way,

the vehicle captures the intruder that arrives at the endpoint upon arrival. To capture the

second intruder, O moves the vehicle for 1+ρ time units, arriving at the perimeter at exactly

time 2 + ρ. The vehicle can capture the second intruder given the definition of ϵ yielding

E ¯X[nO] = 2.

We now describe the performance, in expectation, of a deterministic algorithm A against

the same probability distribution ¯X on I. Observe that the best way to capture both

intruders is to move the vehicle to one of the endpoint. However, as A operates online, A

55

does not know the arrival time and location of the intruders. Thus, given that the input

sequence is random, the probability that A will have the vehicle at the correct endpoint at

time 1 is 1/2. More precisely, assume that A moves the vehicle to endpoint +1 by time +1.

Then, against input I1, the vehicle can capture both i1

1,+1 and i1+ϵ

1,−1 by moving towards −ρ
1,−1 if v = 1−ρ

1+ρ

after capturing i1

holds. Otherwise, the vehicle captures only i1+ϵ

1,+1. However, against input I2, the vehicle can capture only it
2 = 3

1,+1. Thus, E ¯X[nA] = 1 + 1

2 . Thus, from

Lemma 21 we obtain the result.

3.3 Randomized Algorithms

In this section, we design and analyze three randomized algorithms. Our first algorithm

is an open loop and memoryless algorithm and highlights the advantage of using randomized

algorithms. Specifically, we establish that this algorithm is 2-competitive for all values of v

and ρ.

3.3.1 Algorithm Stay

Algorithm Stay is summarized as follows. At time 0, assuming that the vehicle is located

at the origin, the vehicle randomly selects a location out of the two choices {−ρ, ρ}, each

having equal probability of 0.5. Once a location is picked, the vehicle moves to that location

and stays at that location until the arrival of the intruders end.

Theorem 26. Algorithm Stay is 2-competitive against an oblivious adversary for any prob-

lem instance P.

Proof. Let R ≥ 0 and L ≥ 0 denote the total number of intruders preplanned to be released

at endpoints +1 and −1, respectively, in the time interval [0, T ], where T > 0 denotes the

time instant until which the intruders are released. Then, the expected number of intruders

captured by the vehicle is 0.5 R

R+L + 0.5 L

R+L = 0.5. Thus, from Definition 1 and by assuming

that the optimal offline algorithm can capture all, we obtain the claim that Algorithm Stay

is 2-competitive.

56

We now provide an input sequence I to establish that the bound is tight. Input sequence

I begins at time ρ with 0 < R ≤ 2N intruders arriving at endpoint +1 and no intruders

arriving at endpoint −1. The number of intruders captured by Algorithm Stay, in expec-

tation, is R

2 . On the other hand, since an optimal offline algorithm has prior information

about the input sequence, it captures all R intruders by moving the vehicle to ρ. Thus, from

Definition 3, the competitive ratio is exactly 2 and hence the bound obtained is tight.

Recall that in Chapter 1, we introduced an analogous algorithm called Simple Algorithm

in Example 1 with the key difference that Simple Algorithm was deterministic. By selecting

a location randomly instead of deterministically, Algorithm Stay achieves a competitive ratio

of 2 whereas Simple Algorithm was not c-competitive. This highlights the effect of using

randomization to design online algorithms. We now describe our second algorithm that

randomly selects the initial location of the vehicle from a set of locations.

3.3.2 Randomized Stay Algorithm

At time 0 or at the start of the first epoch, the vehicle is either located at endpoint +1 or

−1. The choice of the endpoint is random, i.e., with probability 0.5, the vehicle is located at

endpoint +1 and with probability 0.5, the vehicle is located at −1. The vehicle then waits

until the first intruder arrives in the environment. Denote the time instant when the first

intruder arrives as t ≥ 0. At time instant t, with probability 0.5, the vehicle either stays at

the endpoint until time t + 2 or, with probability 0.5, moves to the other endpoint. Note

that the vehicle reaches the other endpoint by exactly time t + 2. At time t + 2, the vehicle

then repeats, i.e., the first epoch ends and the second epoch begins.

Lemma 27. Suppose that the vehicle is located at endpoint +1 (resp. −1) at time instant t

and lt (resp. rt) intruders arrive at endpoint −1 (resp. +1) at time t. Then the vehicle can

capture lt (resp. rt) intruder if

v ≤ 1 − ρ
1 + ρ

.

57

Proof. Suppose that the vehicle is located at endpoint +1 and lt intruders arrive at −1 at
time t. These intruders will reach the perimeter −ρ at time t + 1−ρ

v . The vehicle can reach
the perimeter −ρ at time t + 1 + ρ as it first moves the origin and then to location −ρ. Thus,

the vehicle can capture the intruder released at time t if 1 + ρ ≤ 1−ρ
v .

Theorem 28. For any problem instance P for which v ≤ 1−ρ

1+ρ holds,





cRand−Stay =

⌋−1)2N

⌋−1)N

1+(⌊ 2
∆
3
+(⌊ 2
4
∆
1+⌊ 2
⌋2N
∆
+⌊ 2
3
∆
4

⌋N

, if 2
∆

∈ Z+

, otherwise.

Proof. Suppose that the vehicle selects endpoint +1 at time instant 0 and rt (resp.

lt)

intruders arrive at endpoint +1 (resp. −1) at time t ≥ 0. First consider that the vehicle

decides to move towards endpoint −1 at time instant t. As the vehicle waits until the first

intruder is released, at least one out of rt and lt must be strictly greater than 0. Further, the

vehicle captures rt intruders upon arrival. Let p = ⌊ 2
∆

⌋ − 1 if 2

∆ is an integer and p = ⌊ 2

∆

⌋,

otherwise. Denote r1, r2, . . . , rp (resp.

l1, l2, . . . , lp) as the intruders that are released at

endpoint +1 (resp. −1) during the time interval [t + ∆, t + 2). From Lemma 27, the vehicle

can capture all the intruders that are released at endpoint −1 in the time interval [t+∆, t+2).

Thus, in this case, the vehicle captures 1

4 is the probability
that the vehicle selects endpoint +1 and decides to move towards endpoint −1. We now

4 [rt + lt + l1 + · · · + lp], where 1

consider that the vehicle decides to stay at endpoint +1 until time t + 2. In this case, the

vehicle captures 1

4 [rt + r1 + · · · + rp]. The number of intruders captured when the vehicle
selects endpoint −1 at time 0 can be determined analogously. Since the vehicle selects an

endpoint with probability 0.5 and then decides to stay or move to the other endpoint with

probability 0.5, it follows that the expected number of intruders that the vehicle captures in

the time interval [t, t + 2) is

3
4

(rt + lt) +

[ p∑

]

rq + lq

.

q=1

1
2

58

Algorithm 4: Sleep-Switch Algorithm
1 Select endpoint +1 or −1 with probability 0.5.
2 Select sleep time s from {0, ∆, . . . , p∆} with probability 1
3 Wait until the first intruder arrives.
4 Reset time to 0 and sleep until time s at endpoint.
5 Move to other endpoint.
6 Repeat from Step 3.

p+1 .

At time t + 2, the probability of the vehicle to be at endpoint +1 and −1 is 0.5 each.

Thus, following analogous steps as before, it can be shown that for any epoch k ≥ 1, the

expected number of intruders captured by the vehicle is 3

4 (rt + lt) + 1
2 [

∑

p
q=1 rq + lq]. Finally,

assuming that there exists an optimal offline algorithm that can capture all intruders yields
∑
p
q=1 rq + lq
[ ∑

cRand−Stay =

.

rt + lt +
4 (rt + lt) + 1

2

3

] ≤ 1 + 2pN
3
4 + pN

p
q=1 rq + lq

This concludes the proof.

Observe that Algorithm Rand-Stay achieves the same performance as Algorithm Stay if

pN tends to infinity and achieves the fundamental limit if pN tends to zero.

We now present an algorithm that has a competitive ratio strictly lower than that of

Algorithm Rand-Stay but requires the information of parameter ∆.

3.3.3 Algorithm Sleep-Switch

Contrary to the previous algorithms, Algorithm Sleep-Switch requires the information

of the parameter ∆. Algorithm Sleep-Switch is defined in Algorithm 4 and summarized as

follows.

Algorithm Sleep-Switch first selects an endpoint randomly with equal probability of 0.5.

Then, with the same p as defined for in Algorithm Rand-Stay, Algorithm Sleep-Switch selects

a sleep time s from a set of sleep times S := {0, ∆, 2∆, . . . , p∆} with equal probability of

1
p+1 and stays at the selected endpoint until time s. Suppose that the vehicle is located at
endpoint +1. Then, at time s, the vehicle moves to endpoint −1. The vehicle then sleeps

again for the same amount s of time units and repeats the process.

59

Theorem 29. For any problem instance P, Algorithm Sleep-Switch is 2p+2

p+2 -competitive if

v ≤ 1−ρ

1+ρ holds.

Proof. Suppose that the vehicle is at endpoint +1 at time ks, where ks denotes the time when

an epoch k begins. Let r0, . . . , rp (resp. l0, . . . , lp) denote the intruders that arrive at +1 (resp.

−1) at time instants 0, ∆, . . . , p∆, respectively. Similarly, let rp+1, rp+2, . . . , r2p+1 (resp.

lp+1, lp+2, . . . , l2p+1) denote the intruders that arrive at time (p+1)∆, (p+2)∆, . . . , (2p+1)∆,

respectively. Note that (p + 1)∆ ≥ 2. For any sleep time s ∈ {0, . . . , p∆}, the expected

number of intruders captured by the vehicle in time interval [ks, ks + s + 2) is

(

p∑

q=0

1
2

p+s∑

)

rq + lq + ls + rs +

lq + rq

,

q=p+1

where we used the fact that the probability of vehicle being at endpoint +1 and endpoint

−1 is 0.5. Observe that the

∑

2p+1
q=p+s+1 lq + rq intruders are not lost by time ks + s + 2 and

can be captured in the next epoch. Thus, the number of intruders captured by the vehicle

in time interval [ks, ks + s + 2) is

(

p∑

p+s∑

2p+1∑

)

rq + lq + ls + rs +

lq + rq +

lq + rq

=

q=p+1

q=p+s+1

(

2p+1∑

q=0

1
2

)

rq + lq + rs + ls

,

1
2

q=0
∑

where

2p+1
q=p+s+1 lq + rq denotes the number of intruders captured by the vehicle in the next

epoch. Since the sleep time s is chosen with probability 1

p+1 , the expected number of intruders

captured in an epoch is

∑

(p + 2)(

2p+1
q=0 rq + lq)

2(p + 1)

.

Finally, assuming that there exists an optimal offline algorithm that can capture all intruders

yields the result.

We now provide a parameter regime plot highlighting the parameter regime in which

our algorithms are effective as well as the parameter regime for the established necessary

conditions.

60

Figure 3.1 Parameter regime plot highlighting the fundamental limits.

3.4 Results and Discussions

Figure 3.1 shows the (v, ρ) plot summarizing our results. The area above the green

dashed curve denotes the fundamental limit established in Theorem 24. Similarly, the area

above the purple dashed curve denotes the fundamental limit established in Theorem 25 as

well as Theorem 23. Although the parameter space for Theorem 25 and Theorem 23 is the

same, the lower bound on the achievable competitive ratio is different.

The area below the purple dashed curve denotes the effective parameter space for Ran-

domized Stay and Sleep Switch algorithms. Specifically, for any values of perimeter size ρ

and speed ratio v that is below the purple curve, Algorithm Randomized Stay and Algorithm

Sleep Switch have a finite competitive ratio as established in Theorem 28 and Theorem 29,

respectively. Note that algorithm Stay is 2-competitive in the entire parameter space, i.e.,

all values of v and ρ in Figure 3.1.

3.5 Conclusion

We study a problem where a single defender defends a linear perimeter from mobile

intruders that move inwards towards the perimeter upon arrival. A key contribution is

the combination of concepts and techniques from competitive analysis of randomized on-

line algorithms with pursuit of multiple mobile intruders. We designed and analyzed three

61

randomized algorithms for incoming intruders. The first algorithm is demonstrated to be

2-competitive in the entire parameter space whereas the competitive ratio of the other two

algorithms vary with the problem parameters. We also derived a fundamental limits on

the existence of randomized algorithms having competitive ratio of at least 2 and another

fundamental limit on the existence of randomized algorithms having competitive ratio of at

least 1.33. Finally, we derived a fundamental limit on the existence of zealous algorithms

having competitive ratio of at least 2, when arbitrary many intruders can be released in the

environment.

62

CHAPTER 4

COMPETITIVE PERIMETER DEFENSE IN CONICAL ENVIRONMENTS

In this chapter, we extend the perimeter defense problem considered in Chapter 2 to planar

environments and multiple defenders. The environment contains M ≥ 1 defenders which

seek to defend a perimeter by intercepting mobile intruders (cf. Fig. 4.1). The intruders

are released at the boundary of the environment and move radially inwards with fixed speed

toward the perimeter. The defenders, each having a finite capture radius and having access

to intruder locations only after they are released, move with unit speed and with the aim

of capturing as many intruders as possible before they reach the perimeter. This is an

online problem as the input, i.e., the release times, numbers of intruders and their release

locations, is gradually revealed over time. Thus, we focus on the design and analysis of online

algorithms to route the defenders. Aside from military applications, this scenario models

relaying critical information to targets and defending wildlife against poachers [22].

The key question that is addressed in this chapter is how does the analytical bound on

the competitiveness scale with multiple defenders? The main contributions of this work are

as follows:

1. Perimeter defense problem with M defenders: We address a perimeter defense

problem in a conical environment with M ≥ 1 identical defenders tasked to defend a

perimeter. Each defender has a finite capture radius and moves with unit speed. We do

not impose any assumption on the arrival process of the intruders. More precisely, an

arbitrary number of intruders can be released in the environment at arbitrary locations

and time instances. Upon release, the intruders move with fixed speed v towards the

perimeter. Thus, the perimeter defense problem is characterized by five parameters:

(i) angle θ of the conical environment, (ii) the speed ratio v between the intruders and

the defenders, (iii) the perimeter radius ρ, (iv) the number of defenders M , and (v)

the capture radius of the defenders r.

63

Figure 4.1 Illustration of the problem considered in this work. Multiple defenders (light blue disc)
seek to defend a perimeter (green arc) from arbitrary many intruders (light red disc) that arrive
over time in the conical environment. The capture radius of the defenders is shown by the dark
blue circle surrounding the defenders.

2. Necessary condition: We establish a necessary condition on competitiveness of any

online algorithm. Specifically, we establish a necessary condition on the existence of

any c-competitive algorithm for any finite c. We also characterize the minimum number

of defenders required to capture all intruders.

3. Algorithm Design and Analysis: We first design and analyze three classes of decen-

tralized algorithms. We then design and analyze two classes of cooperative algorithms.

Specifically, the first two decentralized algorithms are provably 1 and 2-competitive,

respectively and the third decentralized algorithm exhibits a finite competitive ratio

which depends on the problem parameters. Similarly, the first cooperative algorithm is

provably 1.5-competitive and the second cooperative algorithm exhibits a finite com-

petitive ratio which depends on the problem parameters.

Additionally, through multiple parameter regime plots, we shed light into the relative com-

parison and the effectiveness of our algorithms. We also provide an insight into how the

results in this work can be extended when the defenders have different capture radii.

This chapter, content of which is based on [10, 6], is organized as follows. In Section 4.1,

we formally define the problem. Section 4.2 establishes the necessary conditions, Section 4.3

presents the algorithms and their analysis. Section 4.4 provides several numerical insights and

64

Figure 4.2 Problem description for M = 3 defenders. The red dots denote the intruders whereas the
blue dots and blue circles denote the defenders and the capture circle, respectively. The perimeter
is depicted by the green curve. The first (m = 1) partition contains two intruders, second (m = 2)
partition contains three intruders and the third (m = 3) partition contains one intruder. The
common boundary of V1 and V2 is denoted as the blue dot line and the common boundary of V2
and V3 is denoted as the blue dashed-dot line, respectively.

outlines an extension to the case of different capture radii. Finally, Section 4.5 summarizes

the chapter.

4.1 Problem Formulation

Consider a conical environment of E(θ) = {(y, α) : 0 < y ≤ 1, −θ ≤ α ≤ θ} which

contains a conical region R(ρ, θ) = {(z, α) : 0 < z ≤ ρ < 1, −θ ≤ α ≤ θ}, where θ is

measured with respect to y−axis. An arbitrary number of intruders are released at arbitrary

time instants at the circumference of the environment, i.e., y = 1. Each intruder moves

radially with a fixed speed v < 1 towards the origin in order to breach the perimeter. The

perimeter is defined as the boundary of R(ρ, θ), i.e., ∂R(ρ, θ) = {(y, α) : y = ρ, −θ ≤ α ≤ θ}

(cf Fig. 4.2). We consider M ≥ 1 identical defenders having their motion modeled as a first

order integrator1, tasked to defend the perimeter. Each defender can move with unit speed2

and has a finite capture radius r < ρ as the problem is trivial for any r ≥ ρ. Later, in Lemma

32, we will characterize even lower values of r for which problem is trivial. A capture circle of

a defender is defined as a circle of radius r, centered at the defender’s location. An intruder

is captured and subsequently removed from E(θ) if it lies within or on the capture circle. An

1This work can be extended to other models such as double integrator and will be considered in the

future.

2The speed is set to unity to normalize the speed of the intruders. Thus we use the two terms, speed of

the intruders and speed ratio, interchangeably.

65

intruder is lost if it reaches the perimeter without being captured by the defender. Thus,

a problem instance P(θ, ρ, v, r, M ) for this problem is characterized by five parameters: the

speed of the intruders (or speed ratio), v < 1, the perimeter’s radius 0 < ρ < 1, the angle

that defines the size of the environment as well as the perimeter, 0 < θ ≤ π, the capture

radius of each of the defender r < ρ, and the number of defenders M > 1. For compactness,

we will denote a problem instance as P.

Problem Statement: The aim is to understand how the number of defenders effects

the competitive ratio and the parameter regimes. Specifically, the key objectives of this work

are the following:

1. Establish a necessary condition for the existence of online algorithms with a finite

competitive ratio.

2. Design and analyze algorithms with provable guarantees on the competitive ratio.

3. Characterize the dependence of the competitive ratio on parameter regimes, especially

M .

In the next section, we establish a necessary condition on the existence of a c-competitive

algorithm for any finite c ≥ 1.

4.2 Fundamental Limits

As our analysis relies heavily upon partitioning the environment, we start by formally

defining a partition of the environment.

Definition 30 (Partition of E(θ)). A partition of E(θ) is a collection of q ≥ 1 cones W =

{W1, W2, . . . , Wq} of unit radius with vertex at the origin, disjoint interiors, and whose union

is E(θ). We refer to a cone Wm, 1 ≤ m ≤ q as the m-th dominance region.

We first observe that for any set of initial locations of the M vehicles in which no two

vehicles have the same angular coordinate, there exists a partition of the environment with M

dominance regions such that each dominance region can be assigned to a particular vehicle.

66

Figure 4.3 Blowout image of the perimeter contained in the m-th partition for the description
of proof of Lemma 31 for m-th partition.

Given M dominance regions {W1, . . . , WM }, with each dominance region corresponding to

a vehicle, we refer to an individual vehicle Vm, 1 ≤ m ≤ M based on the dominance region

it occupies, unless stated otherwise. More precisely, we denote a vehicle as Vm, 1 ≤ m ≤ M ,

associated with the m-th dominance region. We assume that the dominance regions are

numbered from left to right when θ < π and in clockwise order if θ = π (cf. Fig. 4.2).

Further, two vehicles Vm and Vm+1 for any 1 ≤ m ≤ M − 1 are called neighbors if their

dominance regions share a boundary. A common boundary for neighbors Vm and Vm+1 is the

rightmost (resp. leftmost) boundary of the m-th (resp. m + 1-th) dominance region (cf. Fig.

4.2). Finally, we denote the portion of the perimeter ∂R(ρ, θ) corresponding to dominance

region m, 1 ≤ m ≤ M as ∂Rm.

We now characterize the locations for each vehicle Vm, 1 ≤ m ≤ M such that the distance

from that location to any location on ∂Rm is the least.

Lemma 31. For a problem instance P, positioning each defender Vm, 1 ≤ m ≤ M, at

locations (ρ cos( θ

M ), −(M − 2m + 1) θ

M ) minimizes the distance, and equivalently the time

taken, from the defender’s location to any location on ∂Rm.

Proof. Let (xm, αm) ∈ E(θ) denote the location for vehicle Vm, ∀m ∈ {1, . . . , M − 1}, such

that the distance from location (xm, αm) to any location on ∂Rm is the least.

Consider the m-th dominance region of angle 2βm that contains the vehicle Vm and let L

be the line joining the two points (ρ, ψ1) and (ρ, ψ2) such that ψ2 − ψ1 = 2βm and ψ2 > ψ1,

which correspond to the two endpoints of ∂Rm (cf. Fig. 4.3). To minimize the distance

to any location on ∂Rm from the location (xm, αm), the vehicle Vm must be located at the

67

midpoint of line L, i.e., at the angle bisector of angle 2βm. Further, since at least one out

of the two locations (ρ, ψ1) and (ρ, ψ2) is shared by the neighbors of Vm, we require that the

distance to the location (ρ, ψ1) (resp. (ρ, ψ2)) for both Vm and Vm−1 (resp. Vm+1) vehicles

must be equal. Without loss of generality, we assume that (ρ, ψ2) is shared by vehicle Vm

and Vm+1. Then,

ρ sin(βm) − r = ρ sin(βm+1) − r ⇒ βm = βm+1,

(4.1)

where 2βm+1 is the angle of the (m + 1)-th dominance region. Equation (4.1) must hold for

obtain βm = θ

each 1 ≤ m ≤ M − 1, which yields β1 = β2 = · · · = βM . Observing that

M
i=1 2βi = 2θ we
M for every 1 ≤ m ≤ M . In other words, this means that every dominance
M . Combining this fact with the fact that

region m ∈ {1, . . . , M } must have an angle of 2θ

∑

each vehicle Vm must be located at the angle bisector of the m-th dominance region, it

follows that αm = −(M − 2m + 1) θ

M . Finally, using basic trigonometric identities, we obtain

xm = ρ cos( θ

M ).

Lemma 3 formally established the intuitive conclusion that E must be equally partitioned

between the vehicles. The next result characterizes a value for the capture radius for which

all intruders released in the environment can be captured by simply keeping the vehicles

stationary at specific locations.

Our next result characterizes the minimum capture radius required, given M defenders,

such that all intruders that arrive in the environment are captured. Equivalently, the result

characterizes the minimum number of defenders required, given the capture radius of each

defender, such that all intruders that arrive in the environment are captured.

Lemma 32. For any problem instance P with r ≥ ρ sin( θ

by keeping vehicle Vm stationary at location

ρ cos( θ

(

M ), all intruders can be captured
)
, ∀1 ≤ m ≤ M .

M ), −(M − 2m + 1) θ

M

Proof. Recall from Lemma 31 that the perimeter can be partitioned into M partitions, each

of angle 2θ

M . The idea is to determine the location (xm, αm) ∈ E(θ) for defender Vm ∀m ∈
{1, . . . , M } such that when Vm is located at (xm, αm), ∂Rm is completely contained within

68

Figure 4.4 Description of proof of Lemma 33. The blue dashed circles C and C′ are centered at
(ρ, θ) and (ρ, −θ) respectively. The defender, denoted by the blue dot, is located at (x, α). The
line L is denoted by the black dashed line.

the capture circle of Vm. To ensure this, the two end points of ∂Rm must be equidistant

from (xm, αm), which implies that the defender Vm must be located at the angle bisector of

the dominance region. This yields αm = −(M − 2m + 1) θ

M . Finally, applying trigonometric

identities yields r = ρ sin( θ

M ). This concludes the proof.

In light of Lemma 32, we assume that r < ρ sin(θ/M ) in the remainder of this chapter.

Our next result will be useful in establishing the necessary condition for the special case,

M = 1.

Lemma 33. Suppose that M = 1. Then the minimum time required by the defender to move

from a location such that the capture circle contains one end of the perimeter, (ρ, θ), to a

location such that the capture circle contains the opposite end of the perimeter, (ρ, −θ), is

2(ρ sin(θ) − r) if θ < π

2 and 2(ρ − r), otherwise.

Proof. Consider two circles C and C′, each of radius r and with centers coinciding with the

points (ρ, θ) and (ρ, −θ), respectively (Fig. 4.4). Observe that, if the defender is located

at any point within the intersection of the circle C (resp. C′) and the perimeter, then the

location (ρ, θ) (resp. (ρ, −θ)) will be contained within the capture circle.

Let (x, α) ∈ E(θ) (resp. (x, −α)) denote a point on the circumference of circle C (resp.

C′). Our aim is to determine the closest possible pair of locations (x, α) and (x, −α); that

69

is, the pair of positions that minimizes the distance the defender needs to travel to go from

one point to the other. We consider two cases: (i) θ ≤ π

2 and (ii) θ > π
2 .

For case (i), finding the closest pair of points corresponds to determining the shortest

distance along the line L between the two circles C and C′. The shortest line that connects

any two non intersecting circles must pass through the center of the circles, so L must pass

through (ρ, θ) and (ρ, −θ), and the points (x, α) and (x, −α) are where L intersects C and C′,

respectively. We compute the distance between the two points as follows. We first find the

angle bisector which bisects the angle 2θ of the environment. Note that the angle bisector

is also the perpendicular bisector of line L as the triangle formed by joining points (ρ, θ),

(ρ, −θ) and the origin is an isosceles triangle. By constructing a triangle that joins the points

(x, α), origin, and the midpoint of line L and using trigonometric identities, we determine
that x = ρ sin(θ)−r

ρ cos(θ) ). From geometry, it follows that the length of the
line segment joining the points (x, α) and (x, −α) is 2(ρ sin(θ) − r). Note that when θ = π
2 ,
L runs through the origin simplifying this expression to 2(ρ − r) matching the expression

and α = tan−1( ρ sin(θ)−r

sin(α)

from case (ii).

For case (ii), the corresponding line L is not contained in the environment which means

the defender cannot travel on L to move from (x, α) to (x, −α). Instead, the shortest path

is to first move to the origin from (x, α) and then to the location (x, −α) from the origin.

This gives us x = ρ − r and α = θ and a minimum distance of 2(ρ − r). This concludes the

proof.

We now establish a necessary condition for the existence of c-competitive algorithms. We

first establish the result for M = 1 and the for any M > 1.

Theorem 34. (Necessary condition for finite c(P)) For any problem instance P with

M = 1 and parameters satisfying

2(ρ sin(θ) − r) >

2(ρ − r) >

1 − ρ
v
1 − ρ
v

π
, if θ <
2
, if θ ≥ π
2

,

,

70

there does not exist a c-competitive algorithm for any constant c and no algorithm, either

online or offline, can capture all intruders.

Proof. Recall from Definition 1 that any online algorithm A is c-competitive if the condition

nO(I, P) ≤ cnA(I, P) holds for every input sequence I. Thus, the aim is to construct

an input sequence I such that the condition nO(I, P) ≤ cnA(I, P) does not hold for any

constant c ≥ 1 and for every online algorithm.

For both A and optimal offline O algorithms, assume that the defender starts at the

origin at time 0. The input sequence starts at time instant 1 with a stream of intruders,

i.e., a single intruder being released every 1−ρ
v

time units apart, at location (1, θ). If A never

captures any stream intruders, the stream never ends meaning the algorithm A will not be

c-competitive for any constant c ≥ 1, and the first result follows as O can move to (ρ, θ)

and capture all the stream intruders. We thus assume A does capture at least one stream

intruder, say the ith one, at time t. The input instance ends with the release of a burst of

c + 1 intruders that arrive at location (1, −θ) at the same time instant t.

We now identify how many intruders A can capture. First, it cannot capture stream

intruders 1 through i − 1 because the stream intruders arrive 1−ρ
v

time units apart meaning

the previous intruder reaches the perimeter and thus is lost just as the next stream intruder

arrives. We now show that the defender cannot capture any of the c + 1 burst intruders. At

time t, the defender must be at most r distance away from the ith stream intruder in order

to capture it. Likewise, it has only 1−ρ
v

time to move to capture the c + 1 burst intruders

that arrived at time t. Given the conditions, 2(ρ sin(θ) − r) > 1−ρ
v

(resp. 2(ρ − r) > 1−ρ

v ) for

θ < π

2 (resp. θ ≥ π
On the other hand, the optimal offline algorithm O can move the defender to location

2 ), the defender is ensured to lose the burst intruders.

(x, α), as defined in Lemma 33, until the first i − 1 intruders have been captured and then

move the defender to (x, −α) capturing the burst intruders, losing only the ith intruder. This

concludes the proof.

71

Theorem 35 (Necessary condition for finite c(P)). For any problem instance P with

M > 1, let α = α∗ denote a solution to the equation

Further, suppose that

2 sin(θ − (M − 1)α) − sin(α) − r
ρ

= 0.





v >

1−ρ
2(ρ sin(θ−(M −1)α∗)−r) ,

if 0 < α∗ < π
2 ,

1−ρ
ρ−r ,

otherwise.

Then, there does not exist a c-competitive algorithm, for any finite c.

Proof. Recall from Definition 1 that any online algorithm A is c-competitive if the condition

nO(I) ≤ cnA(I) holds for every input sequence I. Thus, the aim is to construct an input

sequence I such that the condition nO(I) ≤ cnA(I) does not hold for any constant c ≥ 1

and for every online algorithm. The proof is divided into three parts. First, we describe a

family of M input sequences. We will show that any one input sequence from these M inputs

yields the same result. Then, we characterize the best locations for the vehicles against these

input sequence. Finally, we determine the performance of the best online algorithm as well

as an optimal offline algorithm to establish the result. We start by constructing the input

sequences.

Let I = {I1, I2, . . . , IM } denote a set or family of input sequences. Each input sequence

Il ∈ I, where l ∈ {1, . . . , M }, differs in the release location of intruders. Without loss of

generality, we assume that both A and O have all of their vehicles at the origin at time

instant 0.

i.e., a sequence of a single intruder arriving at location (1, θ) at every time instant 1+k 1−ρ

Every input sequence Il ∈ I starts at time instant 1 and consists of a stream of intruders,
v , k ∈
N ∪ {0} until time instant t. The time instant t ≥ 0 corresponds to the time instant when

a vehicle Vm, for any 1 ≤ m ≤ M , captures an intruder from the stream. A burst of c + 1

intruders is then released at time instant t. The location where the burst of intruder is

72

Figure 4.5 Snapshot of proof of Theorem 35 at time instant t. Vehicle V3 has just captured i-th
stream intruder. Note that V3 has to move 2ρ sin(β)−2r distance to capture intruders at (ρ, θ −2β).

released is different for each input sequence Il. Since there are M dominance regions in the

environment, there are exactly M locations where the burst of intruders can arrive. These M

locations will have the same angular coordinate as the endpoints of each ∂Rm, ∀1 ≤ m < M,

excluding θ and including −θ. Without loss of generality, the burst of intruders are released

at location (1, −θ) for I1.

Note that for any input sequence Il, if no vehicle captures the stream intruder, the

stream never ends and the result follows as the optimal offline algorithm O can have one of

its vehicles move to location (ρ, θ) starting at time 0 and thus, capture all stream intruders.

Thus, without loss of generality we assume that vehicle VM captures the stream intruders

at time instant t according to any online algorithm A. Note that since the stream intruders

arrive every 1−ρ
v

time units apart and stops when an intruder from the stream is captured,

no online algorithm can capture more than 1 intruder from the stream. For some i > 0, let

the i-th intruder be the stream intruder that was captured by VM at time t.

We first determine the best location for VM to capture the i-th stream intruder as well as

the best location for the remaining M − 1 vehicles at time instant t. Recall from Lemma 31

that for any 1 ≤ m ≤ M , positioning vehicle Vm at the midpoint of the line joining the two

endpoints of ∂Rm minimizes the distance to any location on ∂Rm. Thus, the best location

for all M vehicles at time instant t is determined based on the following three observations

(cf. Fig. 4.5):

73

• O1: Vehicle VM must be located on the line joining the two endpoints of the perimeter

within its dominance region and at a distance r from location (ρ, θ).

• O2: The time taken by VM to reach the other endpoint of the perimeter must be equal

to the time taken by VM −1 to reach the same location.

• O3: All other remaining M − 1 vehicles must be located at the midpoints of the line

joining the two endpoints of the perimeter contained in their dominance region.

Let 2β be the angle of M -th dominance region and, from observation O3, let 2α be the

angle of each of the M − 1 dominance regions. From observations O1 and O2, we obtain

2ρ sin(β) − 2r = ρ sin(α) − r ⇒ r
ρ

= 2 sin(β) − sin(α).

Using the fact that 2(M − 1)α + 2β = 2θ yields

f (α) := 2 sin(θ − (M − 1)α) − sin(α) − r
ρ

= 0.

(4.2)

Let α∗ denote a solution of (4.2). We now establish that there exists an α∗ that satisfies

equation (4.2).

Notice that equation (4.2) is a continuous function of α. Suppose that α = ϵ, where

ϵ > 0 is a very small number. Then, f (ϵ) = 2 sin(θ − (M − 1)ϵ) − r

ρ > 0 as 2 sin θ > r
−ϵ)− r

ρ . Now

consider that α = θ

ρ < 0,
for a sufficiently small ϵ. This means that for a sufficiently small ϵ > 0, f (·) changes its sign

−ϵ for the same ϵ. Then, f ( θ

−ϵ) = 2 sin ϵ−sin(

θ
(M −1)

M −1

M −1

in the interval [ϵ,

θ
M −1

− ϵ]. Thus, from Intermediate Value Theorem, it follows that there

must exist an α∗ such that f (α∗) = 0. Further, as there may be more than one solution to

equation (4.2), α∗ corresponds to the smallest value among those multiple solutions.

Let A1 denote any online algorithm that positions its vehicles as described above. We

now describe the number of intruders captured by A1 on Il ∈ I. Two cases arise based on

the solution of equation (4.2); either α∗ < π

2 or α∗ ≥ π

2 . We first consider α∗ < π
2 .

Case 1 (α∗ < π

2 ): Observe that the solution of equation (4.2) yields angle α∗ from
which we compute the time taken, i.e., 2ρ sin(θ − (M − 1)α∗) − 2r, by vehicle VM to reach

74

the location (ρ, θ − 2β). Equivalently, from observations O2 and O3, this expression also

represents the time taken by any vehicle Vm, 1 ≤ m < M , to reach an endpoint of ∂Rm.

Thus, for any input sequence Il ∈ I releasing the c + 1 intruders at time instant t ensures
that no vehicle can capture the burst intruders if 2ρ sin(θ − (M − 1)α∗) − 2r > 1−ρ
v . For I1,

this location corresponds to (1, −θ).

Before we consider Case 2, we briefly remark on online algorithms that positions Vm, 1 ≤

m < M, vehicles at location other than the ones identified in Case 1, i.e., all online al-

gorithms except A1. Recall that there are exactly M locations for the burst to arrive in

the environment. Of these M locations, there exists at least one corresponding dominance

region, say ∂R1, having angle 2α1 > 2α∗. This implies that positioning the vehicle at the

midpoint of the line joining the two endpoints ∂R1, the vehicle will not be able to capture

the burst intruders as ρ sin(α1) − r > ρ sin(α∗) − r holds. Thus, for any online algorithm

that positions the M vehicles at locations other than the location determined from the three

observations, there exists an input sequence Il ∈ I for which the result still holds.

Case 2 (α∗ ≥ π

2 ): Given that α∗ > 0 and M > 1, this case is possible only when M = 2.
2 , the condition (M − 1)α∗ + β = θ does not
hold since θ ≤ π. Thus for this case, we assume that M = 2. In this case, the line joining

This is because for any M > 2, β > 0, and α∗ ≥ π

the two endpoints of the perimeter in the (M − 1)-th dominance region either coincides with

that of M -th dominance region (when θ = π) or lies outside the environment (when θ < π).

Thus, it follows that the best position for VM −1 is to be at the origin. Given the condition
that ρ − r > 1−ρ

v , vehicle VM −1 cannot capture the c + 1 intruders.

We now show that the optimal offline algorithm O captures all of the intruders on the

input sequence Il. Recall that O has complete information of when, where, and how many

intruders will arrive at time instant 0. Thus, O moves one of its vehicles, say VM , to location

(ρ, θ) to capture all i intruders and at the same time O moves any one of the remaining

M − 1 vehicles, say V1 to location (ρ, τl) capturing the c + 1 intruders, where τl denotes

the angular coordinate where the burst intruders is released in input sequence Il. Thus,

75

(a) The defender is located at (t1, α1) at
time t1. Intruder b is at (1, −θ).

(b) The defender is located at (t2, α2). In-
truder b is captured but intruder a is lost.

Figure 4.6 Description of the proof of Theorem 36 for I3. The green curve denotes the perimeter.
The circles C and C′ are denoted by blue dashed circles and the line segment L′ is denoted by the
black dashed line.

nO(I) = i + c + 1 and nA(I) = 1 and the result follows.

We highlight that the input sequence constructed for the proof of Theorem 50 aims to

constrain one out of the total M defenders at a location so as to increase the separation

between the defenders. Observe that releasing multiple streams of intruders does not further

increase the separation. The explanation is as follows. Suppose that two streams of intruders

are released in the environment. First, note that the burst intruders must arrive after the

intruders from both the streams are captured. Otherwise, the input sequence is the same

as in the proof of Theorem 50. For such an input sequence there exists an online algorithm

that captures intruders from one of the stream intruders. This implies that the burst never

arrives and thus, the online algorithm will have a finite (may be arbitrarily high) competitive

ratio.

The next result characterizes a necessary condition for the existence of online algorithms

having a competitive ratio of at least 2 for when M = 1. This result easily extends to M > 1.

We first characterize locations (t1, α1) ∈ E(θ) and (t2, α2) ∈ E(θ) for the defender (Fig.

76

4.6), where

√

1 + r2 − 2r(1−ρ cos(2θ))
√
1+ρ2−2ρ cos(2θ)
√
√

sin(θ)

(

,

1+ρ2−2ρ cos(2θ)−r(1+ρ) sin(θ)
1+ρ2−2ρ cos(2θ)−r(1−ρ) cos(θ)

cos(θ)

)

,

t1 =

α1 = tan−1
√

t2 =

ρ2 + r2 + 2rρ(cos(2θ)−ρ)
√
1+ρ2−2ρ cos(2θ)
√
√

−ρ sin(θ)

(

,

α2 = tan−1

1+ρ2−2ρ cos(2θ)+r(1+ρ) sin(θ)
1+ρ2−2ρ cos(2θ)+r(1−ρ) cos(θ)

ρ cos(θ)

)

.

The locations (t1, α1) and (t2, α2) are determined analogously to the proof of Lemma 33, and

so, we only give an outline for it. Construct two circles C and C′, each of radius r, centered

at (1, θ) and (ρ, −θ) and consider a line segment L′ that joins the centers of the two circles

(Fig. 4.6). Then, location (t1, α1) (resp. (t2, α2)) corresponds to the intersection points of

the line segment L′ with circles C and C′, respectively.

Theorem 36 (Necessary condition for c(P) ≥ 2). For any problem instance P and

M = 1, c(P) ≥ 2 if

√

1 + ρ2 − 2ρ cos(2θ) − 2r,

≤

≤ 1 + ρ − 2r,

1 − ρ
v
1 − ρ
v

if θ ≤ π
2
π
2

if θ >

.

Proof. The key idea is to construct input sequences for which any online algorithm is guaran-

teed to lose half the intruders while proving that an offline algorithm exists that can intercept

all intruders. All of our input instances consists of two intruders denoted by a and b that

arrive at location (1, θ) and (1, −θ), respectively, and we assume that the defender starts at

the origin.

Two cases arise; (i) θ ≤ π
√

Consider that 1−ρ

2 and (ii) θ > π
1 + ρ2 − 2ρ cos(2θ) − 2r and consider an input sequence I1 in

2 . We first consider case (i), i.e., θ ≤ π
2 .

v =

which both intruders a and b arrive at time instant t1. This is the time that the defender

takes to move from the origin directly to location (t1, α1). We claim that the best way for

any algorithm to capture both intruders is to capture either intruder a or b exactly at time

77

t1, i.e., as soon as it arrives and then move to capture the second intruder in minimum time.

The explanation is as follows.

The total time taken by the defender to capture both the intruders in the worst case is

i + ρ2 − 2xiρ cos(2θ)−2r, where ρ ≤ xi ≤ 1 is the radial component of the location
x2

√

1−xi
v +

of the first of the two intruders at the time of capture. The expression of the total time is

determined as follows: The term 1−xi
v

is the intercept time for the first intruder. For the

second term, construct a line segment L joining two points (xi, θ) (resp. (xi, −θ)) and (ρ, −θ)

(resp. (ρ, θ)). Then, the length of the line segment L is given by

i + ρ2 − 2xiρ cos(2θ)
x2

√

from which we subtract 2r to account for the capture radius, to obtain the second term. As

√

1−xi
v +

i + ρ2 − 2xiρ cos(2θ) − 2r is monotonically decreasing function of xi, its minimum
x2

is achieved at xi = 1. This establishes our claim that the minimum time any algorithm can

take is to capture one intruder exactly when it arrives followed by the second intruder at
√

1 + ρ2 − 2ρ cos(2θ) − 2r.

We now describe how an offline algorithm can capture both the intruders in the input

sequence I1. At time 0, the defender starts at the origin and moves towards location (t1, α1)

capturing the intruder at location (1, θ) exactly at time t1. Then the defender moves directly

to location (t2, α2) exactly at time t1 +

1 + ρ2 − 2ρ cos(2θ) − 2r capturing the second

√

intruder at (ρ, −θ). Note that placing the defender at (t1, α1) (resp. (t2, α2)) ensures that

the location (1, θ) (resp. (ρ, −θ)) is on the circumference of the capture circle of the defender.

Therefore, an algorithm that hopes to be better than 2-competitive must capture both the

intruders in this input sequence and the only way to do so is to move to either location

(t1, α1) or (t1, −α1) arriving exactly at time t1.

Now consider input sequences I2 and I3. In I2, intruder a arrives at time t1 and intruder

(

)

b arrives at time t1 + ϵ, where ϵ < L = 2 sin(θ)

1 −

√

r(1+ρ)

1+ρ2−2ρ cos(2θ)

and L denotes the

minimum time required by the defender to move from (t2, α2) to (t2, −α2). In I3, intruder

b arrives at time t1 and intruder a arrives at time t1 + ϵ.

Input sequence I2 (resp I3)

are constructed for algorithms that have the defender arriving at location (t1, −α1) (resp.

78

(t1, α1)) at time t1. Any algorithm that has the defender arriving at location (t1, −α) (resp.

(t1, α1)) at time t1 can capture only one intruder from I2 (resp.

(I3)). As the solution

is symmetric, we only provide the explanation for input sequence I3. This follows as the

defender can capture intruder b if it moves directly to location (t2, α2) (Fig. 4.6a). However,

as intruder a arrives in at most ϵ < L time units, the defender will not be able to capture

intruder a (Fig. 4.6b). An optimal offline algorithm can capture both the intruders by simply

moving to (t1, −α1) at time t1, capturing intruder a upon arrival and then to (t2, −α2) to

capture intruder b.

We now consider the case when 1−ρ

v <

√

1 + ρ2 − 2ρ cos(2θ) − 2r. Consider input se-

quences I4 and I5. In I4, intruder a arrives at time t1 and intruder b arrives at time t1 + ϵ,

where ϵ =

1 + ρ2 − 2ρ cos(2θ) − 2r − 1−ρ

v . In I5, intruder b arrives at time t1 and intruder

√

a arrives at time t1 + ϵ. Following similar reasoning as input instance I2 and I3, it follows

that no online algorithm can capture both intruders from input instance I4 or I5.

We now consider case (ii), i.e., θ > π

2 . Except for when θ = π, as the line segment L′ will
not be contained completely, the defender must move first to the origin and then to the next

intercept point. Note that, the defender will do the same when θ = π. Thus, in this case,

the location (t1, α1) is (1 − r, θ) and location (t2, α2) is (1 + ρ − 2r, −θ). Following similar

steps as case (i), we construct input sequences I1, . . . , I5 (omitted for brevity) and show that

no online algorithm can capture both the intruders from those input sequences.

In summary, even restricting our input sequences to {I1, . . . , I5}, no online algorithm can

capture both intruders whereas an optimal offline algorithm can capture both the intruders.

This concludes the proof.

We now turn our attention to design of algorithms with guaranteed upper bounds on

their competitive ratios. In the next section, we design and analyze decentralized as well as

cooperative algorithms, characterizing the parameter regimes in which they have provably

finite competitive ratios.

79

(a) Defender Vm is located at (xm
2m) θ

M ).

S , −(M −

(b) Defender Vm is located at (xm
2m + 2) θ

M ).

S , −(M −

Figure 4.7 Snapshot of Dec-Sweep for M = 3. The light green dashed curve denotes the angular
path.

4.3 Algorithms

We start by defining an angular path for defender Vm, 1 ≤ m ≤ M . Let defender Vm be

located at (xm, αm) ∈ E(θ) for any 0 < xm ≤ 1 and αm ∈ [−θ, θ]. An angular path is a

circular arc centered at the origin and radius equal to the radial location of defender Vm, i.e.,

xm. Formally, the angular path for defender Vm is defined as Tm(xm, β

m

β

m

≤ β ≤ βm

} for any β

, βm

m

∈ [−θ, θ] such that β

m

≤ αm ≤ βm and β

, βm) := {(xm, β) :
̸= βm (cf Fig.

m

4.7). We say that the defender completes its motion on the angular path when the defender

returns to its starting location after moving along all of the points in Tm twice – once to

move from the starting location (xm, αm) to (xm, βm) (resp. (xm, β

)), and the second time,

m

to move from location (xm, βm) (resp. (x, β

)) to location (xm, β

) (resp. (xm, βm)) and

m

m

then back to the starting location (xm, αm).

We start with designing decentralized algorithms followed by cooperative algorithms.

4.3.1 Decentralized Algorithms

The idea for the decentralized algorithms is to first partition the environment into M

dominance regions and assign a single defender into each dominance region. Then, since each

partition contains a single vehicle, every partition can be viewed as a conical environment

with a single vehicle.

80

4.3.1.1 Decentralized Sweep Algorithm (Dec-Sweep)

Dec-Sweep algorithm is an open loop as well as memoryless algorithm and is described

as follows.

Dec-Sweep partitions the environment E(θ) into M dominance regions, each of angle 2θ
M .

Each defender Vm, 1 ≤ m ≤ M starts at location (xm

S , −(M − 2m + 1) θ

M ), where

xm
S

∈ [ρ − r, min{1 − r, ρ + r}].

The choice of xm

angular path Tm(xm

S will be justified shortly. Each defender Vm, 1 ≤ m ≤ M , then follows the
S , −(M −
M ) at time 0. More precisely, at time 0, each defender Vm, 1 ≤ m ≤ M, picks a velocity
with unit magnitude and direction tangent to the angular path Tm, oriented to the right

M ) moving towards location (xm

S , −(M − 2m + 2) θ

M , −(M − 2m) θ

2m) θ

until it reaches (xm

S , −(M − 2m) θ

M ). Note that the angular coordinate −(M − 2m) θ

M and

−(M − 2m + 2) θ

M correspond to the angular coordinates of the two endpoints of ∂Rm. Once

it reaches (xm

S , −(M − 2m) θ

M ) (cf Fig. 4.7a), the defender switches direction and moves

towards the other endpoint of ∂Rm, i.e., (xm

S , −(M − 2m + 2) θ

M ) (cf Fig. 4.7b). From this

moment on, the defender only switches direction after it reaches an endpoint of ∂Rm. More

precisely, upon reaching location (xm

M ), the defender keeps on moving on an

angular path Tm(xm

S , −(M − 2m) θ

S , −(M − 2m) θ
M ), −(M − 2m + 2) θ
M .

The following three aspects jointly justify the choice of location xm

S , ∀m ∈ {1, . . . , M }.

1. There is no benefit for defender Vm to be located below a distance of ρ − r from the

origin as the capture circle will be completely inside the perimeter and the defender Vm

cannot capture any intruder using the angular sweep algorithm. This yields xm
S

≥ ρ−r.

2. For any radial location xm

S > ρ + r, the defender Vm will take exactly 4 θ

M xm

S time units

to complete one angular path. On the other hand, in the worst case, the intruders

will require exactly 2r

v time to cross the region swept by the defender (cf. Figure 4.7).

Since the time taken 4 θ

M xm

S by defender Vm increases with xm

S , while the time taken

81

2r

v by the intruders remains the same for any xm
S > ρ + r. Equivalently, xm
benefit for xm
S

≤ ρ + r must hold.

S > ρ + r, it follows that there is no

3. Finally, to ensure that xm

S + r is contained within the environment, we require xm
S

≤

1 − r.

Since ρ − r < ρ + r and ρ − r < 1 − r, xm

S is well defined.

Theorem 37. Algorithm Decentralized Sweep is 1-competitive, i.e., cDec-Sweep = 1, for any

problem instance P which satisfies

{

M (xm

S + r − ρ)
4θxm
S

}

,

v ≤ min

m∈{1,...,M }

]

[

for any xm
S

∈

ρ − r, min{1 − r, ρ + r}

and ∀m ∈ {1, . . . , M }. Otherwise, Decentralized

Sweep is not c-competitive for any constant c.

Proof. Without loss of generality, we assume that defender Vm, 1 ≤ m ≤ M , has just left

location (xm

S , −(M − 2m) θ

2m) θ

M ). Defender Vm takes exactly 4 θ
M xm
M ) whereas intruder i takes exactly xm
S +r−ρ)
4θxm
S

2m) θ
the condition v ≤ M (xm

M ) at time instant t and an intruder i is located at (xm
S time units to reach location (xm
S +r−ρ
v

S + r, −(M −
S + r, −(M −

time units to reach the perimeter. Given

which is equivalent to 4 θ

M xm

S

≤ xm

S +r−ρ
v

, it follows that intruder

i will not be lost. Since this explanation holds for every defender Vm, 1 ≤ m ≤ M , the

first result is established. We now establish the second result. Without loss of generality
consider that v > M (xM

. Then we can construct an input sequence with intruders only

S +r−ρ)
4θxM
S

arriving at (1, θ) such that at the time instant defender VM leaves location (xM

S , θ), arbitrary

number of intruders are located at location (xM
v > M (xM

, all intruders will be lost and the result follows.

S +r−ρ)
4θxM
S

S + r, θ) at the same time instant. Since,

Corollary 4. Placing each defender Vm at a radial distance of min{1 − r, ρ + r} maxi-

mizes the speed ratio for which Theorem 37 holds. In other words, cDec-Sweep = 1 if v ≤

min{ 2rM

4θ(ρ+r) , M (1−ρ)

4θ(1−r)

}.

82

This corollary follows from the fact that M (xm

S +r−ρ)
4θxm
S

is a monotonically increasing function

of xm

S . We now establish that any modification to the partitioning in Dec-Sweep Algorithm

does not improve the parameter space. Note that since Dec-Sweep is already 1-competitive,

one cannot further improve the competitive ratio.

Corollary 5. Partitioning the environment into dominance regions of equal angles maxi-

mizes the speed ratio for which Theorem 37.

Proof. We establish this claim by arriving at a contradiction. Suppose that there exist two

dominance regions that have different angles 2γ1 and 2γ2 such that 2γ1 < 2γ2.

Now, all defenders must be placed so that they take the same time to complete a sweep

of their respective dominance regions. Otherwise, the inputs will be concentrated in the

dominance region of the slowest such defender, which will govern the upper bound on v for

1-competitiveness. Therefore, we must have

4γ2x2

S = 4γ1x1
S

⇒ x1

S = x2
S

γ2
γ1

,

where x1

S and x2

S denote the radial locations of the corresponding defenders in the two

dominance regions under consideration. Thus, in the worst-case, for 1-competitiveness we

require

{

v ≤ min

S + r − ρ
x2
4γ2x2
S

,

S + r − ρ
x1
4γ1x1
S

}

=

S + r − ρ
x2
4γ2x2
S

.

From Corollary 4 and since x2

S < x1

S, it follows that x2

S < min{1 − r, ρ − r} as x1

S =

min{1 − r, ρ − r}. In other words, the requirement on v for 1-competitiveness can be relaxed

by increasing γ1 until γ1 = γ2. This concludes the proof.

Although Dec-Sweep is a 1-competitive algorithm, it is effective only for small values of

speed ratio v. Thus, the motivation behind our second algorithm is to capture intruders

with higher speeds at the expense of losing some intruders.

83

Figure 4.8 Description of Dec-CaC Algorithm for M = 2. The black dashed line depicts partition
of the environment. The region between the yellow-dashed and yellow-dot lines denote the set Rk
m
and Lk

| = 0.

m. |Rk
1

| = |Lk
2

| = 1 and |Lk
1

| = |Rk
2

4.3.1.2 Decentralized Compare and Capture (Dec-CaC)

The intuition behind this algorithm is to move the defender within a portion of its

assigned dominance region so as to obtain a higher parameter regime. In other words, the

algorithm first partitions the environment into M dominance regions and positions a single

defender at the center of each dominance region. The algorithm then compares the number

of intruders arriving on each side of every defender. After comparison, the defenders in every

dominance region move on an angular path towards the side which has higher number of

intruders. Hence, our second algorithm is not open-loop. However, we will see that this is a

memoryless algorithm when we formally describe the algorithm.

We first provide some definitions that will be useful in describing and analyzing the Dec-

CaC algorithm. An epoch km in a dominance region m is defined as the time interval in

which defender Vm completes its motion on angular path Tm(xm

C , β

, βm) with a specified

m

distance xm

C such that

[

xm
C

∈

ρ − r, min{ρ + r,

M (1 − r)
M + vθ

]
}

.

The distance xm

C , ∀m ∈ {1, . . . , M }, for defender Vm in dominance region m is same in all

epochs whereas β

m

and βm are set at the start of every epoch k for every dominance region

m. The choice of xm

C and the values that β

and βm take will be described shortly. Denote

|Rk
m

| and |Lk
m

m
| as the total number of intruders contained in the sets Rk

m and Lk

m (cf. Figure

84

Algorithm 5: Decentralized Compare-and-Capture (Dec-CaC) Algorithm
1 for each defender 1 ≤ m ≤ M do

2

3

4

5

6

7

8

9

10

11

12

M +vθ ]}.

Select xC ∈ [ρ − r, min{ρ + r, M (1−r)
for each epoch k ≥ 1 do
| then
| ≤ |Rk
m
= −(M − 2m + 1) θ

if |Lk
m
Set β
Move on angular path to location (xm
Move on angular path to return to location (xm

M , βm = −(M − 2m) θ

C , βm).

m

M

C , β

).

m

else

m

= −(M − 2m + 2) θ

Set β
Move on angular path to location (xm
Move on angular path to return to location (xC, βm).

M , βm = −(M − 2m + 1) θ
C , β

m

)

M

end

end

13
14 end

4.8), respectively, defined as

Rk

m :={(y, ϕ) : ρ + ϕxm

C v < y ≤

min{1, xm

C + r + (2 θ
M

− ϕ)vxm
C

}∀ϕ ∈ [0, θ

M ] and

m :={(y, ϕ) : ρ − ϕxm
Lk

C v < y ≤

min{1, xm

C + r + (2 θ

M + ϕ)vxm

C

}∀ϕ ∈ [− θ

M , 0).

Dec-CaC algorithm is defined in Algorithm 5 and is summarized as follows. At the start

of every epoch km, each defender Vm compares the total number of intruders in the set Rk
m

m

β

and Lk

| < |Lk
m

m. If |Rk

|, then defender Vm moves on the angular path, Tm(xm
M and βm = −(M − 2m + 1) θ
and βm = −(M − 2m) θ
M .

= −(M − 2m + 2) θ

M . Otherwise β

m

m

C , β

, βm) with

m
= −(M − 2m + 1) θ
M

For the initial case, we assume time 0 as the time when the first intruder is released.

Each defender Vm starts at location (xm

C , −(M − 2m + 1) θ

M ) and begins its first epoch at

time 0.

We now justify the range of values for xm

C . The explanation for xm
C

≥ ρ−r and xm
C

≤ ρ+r

is analogous to that for the choice of xm

S (points 1 and 2) for the Dec-Sweep algorithm.

85

Further, we require xm

C + r + vxm
C

θ
M

≤ 1 ⇒ xm
C

≤ M (1−r)
M +vθ

for the algorithm to have a finite

competitive ratio. This yields

[

{

xm
C

∈

ρ − r, min

ρ + r,

M (1 − r)
M + vθ

}]

.

Note that xm

C is well defined only if

ρ − r ≤ M (1 − r)
M + vθ

⇒ v ≤ M (1 − ρ)
θ(ρ − r)

.

(4.3)

Recall that once a value of xm

C is selected for a defender Vm, it remains fixed for all epochs

and two defenders may not have the same value of xm

C . We now establish two results that

will be useful in establishing that Algorithm Dec-CaC is 2-competitive.

Lemma 38. Any intruder that lies beyond3 the location (xm

C + r + (2 θ
M

− ϕ)vxm

C , ϕ), ∀ϕ ∈

[−(M − 2m + 1) θ

M , −(M − 2m) θ
1 ≤ m ≤ M , will either be contained in the set Rk+1

M ] in epoch km and in a dominance region m, for any

m or in Lk+1

m in epoch km + 1 and is not

lost at the start of epoch km + 1 if

v ≤ M (xm

C + r − ρ)
2θxm
C

.

Proof. Without loss of generality, assume that |Lk
m

| < |Rk
m

| at epoch km holds in dominance

region m ∈ {1, . . . , M }. The total time taken by defender Vm to capture intruders in Rk
m

and return back to its starting location (xm

C , −(M − 2m + 1) θ

M ) is 2 θ

M xm

C . In the worst-case,

in order for any intruder i to be not considered in the start of epoch km, the intruder i must

be located just above (xm

M vxm
time the defender reaches location (xm

C + r + θ

C , −(M − 2m + 2) θ
C , −(M − 2m + 2) θ

M ) at the start of epoch km. By the

M ), intruder i will be located just

above the location (xm

If v ≤ M (xm

C +r−ρ)
2θxm
C

C + r, −(M + 2m)θ) and will not be captured.
holds for dominance region m, intruder i will be at least θ

M vxm

C distance

away from the perimeter at the end of epoch km. Thus, intruder i will be considered in the

start of epoch km +1 as per the definitions of the sets Lk+1

m and Rk+1

m . Clearly, as the intruder

3intruders with radial coordinate more than xm

C + r + (2 θ
M

− ϕ)vxm
C

86

i will be considered for comparison in epoch km + 1, it is not lost unless the defender decides

to move to Lk+1

m in epoch k + 1. This concludes the proof.

Theorem 39 (Dec-CaC competitiveness). For any problem instance P, suppose that

v ≤ min

m∈{1,...,M }

{

M (xm

C + r − ρ)
2θxm
C

}

(4.4)

and equation (4.3) holds. Then, for all m ∈ {1, . . . , M }, with the choice of an

[

xm
C

∈

ρ − r, min{ρ + r,

M (1 − r)
M + vθ

]
}

,

cDec−CaC = 2, i.e., Algorithm Dec-CaC algorithm is 2-competitive.

Proof. The idea behind this proof is to show that in any dominance region 1 ≤ m ≤ M ,

defender Vm always captures at least the same number of intruders it loses. The result then

follows as the above claim then holds for every dominance region. Lemma 38 ensures that

every intruder will belong to either set Lk

m or Rk

m in every epoch km. In every epoch km,

defender Vm compares the total number of intruders on either side contained in the set Lk
m

and Rk

m and moves to the side where the number of intruders is higher. Thus, it is guaranteed

that the defender will capture at least half of the total number of intruders that arrive in

the environment, assuming that an optimal offline algorithm can capture all intruders.

C +r−ρ)
Remark 3. As M (xm
2θxm
C
defender Vm, 1 ≤ m ≤ M at min{ρ + r, M (1−r)

M +vθ

is a monotonically increasing function of xm

} yields v ≤ min{ M r

C , positioning each
(2−3r+ρ)θ , M (1−ρ)

θ(ρ−r)

}.

θ(ρ+r) , M (1−ρ)

Similar to Dec-Sweep Algorithm, the next result establishes that there is no benefit

obtained in the parameter regime of Dec-CaC algorithm if the environment is not partitioned

into equal dominance region. As the proof is analogous to that of Corollary 4, we omit the

proof for brevity.

Corollary 6. Partitioning the environment into equal dominance region of equal angles

yields the maximum speed ratio for which Dec-CaC Algorithm is 2-competitive.

87

Figure 4.9 Description of Dec-SNP Algorithm for m-th dominance region and ns = 4. The three
intervals of length Dv are represented by the black dashed curves and the resting point for defender
Vm are denoted by blue dashed circles. The defender Vm is located at (xm

4 , αm

4 ).

As explained in more detail in Section 4.4, Dec-Sweep and Dec-CaC are not very effective

for small values of capture radius r and high values of perimeter size ρ. Our next algorithm

addresses this issue by further dividing each of the M partition into sectors.

4.3.1.3 Decentralized Stay Near Perimeter Algorithm (Dec-SNP)

Unlike the previous two algorithms, in this algorithm, the defenders do not follow an

angular path. Instead, the idea is to further divide each of the M dominance regions into

sectors and position the defenders close to the perimeter in a specific sector of the dominance

region.

In algorithm Dec-SNP, we divide each dominance region m, 1 ≤ m ≤ M , into ns = ⌈ θ
M θs

⌉

sectors, where 2θs = 2 arctan( r

ρ ) denotes the angle of each sector. Let Nl,m, l ∈ {1, . . . , ns},
denote the l-th sector of the m-th dominance region, where N1,m denotes the leftmost sector

of the m-th dominance region (cf. Fig. 4.9). Further, let ∂Rm

perimeter contained in sector Nl,m. Then, a resting point (xm

l denote the portion of the
l ) ∈ E(θ) of sector Nl,m is

l , αm

defined as a location for defender Vm such that when positioned at that location, all of ∂Rm
l

is contained completely within the capture radius of the defender. Mathematically,

(xm

l , αm

l ) :=

(

ρ
cos(θs)

, −(M − 2m + 2) θ

M + (2l − 1)θs

)

.

Lastly, we let D denote the distance between the two resting points that are furthest apart

88

Algorithm 6: Decentralized Stay Near Perimeter (Dec-SNP) Algorithm
1 for each defender 1 ≤ m ≤ M do
1 , αm
2

1 ) until time 2D.

Stay at (xm
for each j ≥ 1 do

3

4

5

6

7

8

9

10

k∗ = arg maxk∈{1,...,ns}{η1
No,m = Nk∗,m
if No,m ̸= Ni,m and |Sj+2
o,m

i (m), . . . , ηns

i (m)}

| ≥ |Sj+1
i,m

| then

Move to (xm

o , αm

o ) and then capture |Sj+2

o,m

|

else

Stay at (xm

i , αm

i ) and capture |Sj+1

i,m

|

end

end

11
12 end

in a dominance region (cf. Fig. 4.9). Formally,





2

2

D =

ρ

cos(θs) sin((ns − 1)θs), if (ns − 1)θs < π

2

ρ

cos(θs) , otherwise.

(4.5)

Note that each of the M dominance regions will have equal number of ns sectors and

therefore, the same value of D. After dividing each dominance region m in the environment

into ns sectors, Dec-SNP algorithm radially divides the environment E(θ) into three intervals

of length Dv, corresponding to time intervals of time length D each. Specifically, the jth

time interval for any j > 0 is defined as the time interval [(j − 1)D, jD]. In order to ensure a

finite competitiveness for this algorithm, we require 1−ρ
v

≥ 3D, i.e., the intruders require at

least 3D time units to reach the perimeter after they are released in the environment. For

any j ≥ 1, let Sj

l,m be the set of intruders present in sector Nl,m in the jth time interval.

The Dec-SNP algorithm (cf. Algorithm 6) is based on the following two steps. Each

defender Vm, 1 ≤ m ≤ M , first selects the sector (within dominance region m) with the

maximum number of intruders, and then determines whether it is beneficial to move to the

resting location of that sector. These two steps are achieved by the two comparisons C1

(lines 7-8 in Algorithm 6) and C2 (lines 9-13 in Algorithm 6) elaborated below.

Comparison C1: For each dominance region 1 ≤ m ≤ M , Dec-SNP determines that

89

sector which has the most number of intruders in the last two intervals as compared to

the total number of intruders in the entire sector in which the defender Vm is located. In

particular, suppose that defender Vm is located at the resting point of sector Ni,m, 1 ≤ i ≤ ns

at the jth iteration. Corresponding to any sector Nl,m, 1 ≤ l ≤ ns in dominance region m,

we define ηl

i(m) as





ηl
i(m) =

|Sj+2
l,m

| + |Sj+3
l,m

|, if l ̸= i

|Sj+1
i,m

| + |Sj+2
i,m

| + |Sj+3
i,m

|, if l = i

Then, Dec-SNP selects the sector Nk∗,m for each dominance region m, where

k∗ = arg max

k∈{1,...ns}

{η1

i (m), . . . , ηns

i (m)}.

If there are multiple sectors in dominance region m, m ∈ {1, . . . , M } with same number of

intruders, then Dec-SNP breaks the tie as follows. If the tie includes the sector Ni,m, then

Dec-SNP selects Ni,m. Otherwise, Dec-SNP selects the sector with the maximum number

of intruders in the interval j + 2. If this results in another tie, then this second tie can be

resolved by picking the sector with the least index. Let the sector chosen for dominance

region m as the outcome of comparison C1 be No,m, o ∈ {1, . . . , ns}.

Comparison C2: If the sector chosen is No,m, o ̸= i, and the total number of intruders

in the set Sj+2

o,m is greater than or equal to the total number of intruders in the set Sj+1

i,m , then

Dec-SNP moves defender Vm to location (xm

o , αm

o ) arriving in at most D time units. Then,

defender Vm waits at that location for another D time units to capture all intruders in Sj+2
o,m .

Otherwise (i.e., if o = i or |Sj+2
o,m

| < |Sj+1
i,m

|), defender Vm stays at its current location (xm

i , αm
i )

for D time units and captures Sj+1

i,m . After capture of either Sj+2

o,m or Sj+1

i,m , the defender then

reevaluates at time D or 2D, respectively.

At time 0, each defender Vm, 1 ≤ m ≤ M waits at the resting point of the first sector

of their corresponding dominance regions, i.e., the resting point of N1,m, for 2D time units.

The first epoch begins at time 2D.

90

Lemma 40. For each m ∈ {1, . . . , M }, let defender Vm be located at (xm

i , αm

i ) of a sector

Ni,m, 1 ≤ i ≤ ns. Then, for any j ≥ 1, Vm always captures either Sj+1

i,m or Sj+2

o,m , where Sj+2
o,m

denotes the set of intruders in sector No,m selected after C1.

Proof. Suppose that after comparison C1, No,m = Ni,m. Then the result follows from Algo-

rithm 6 (line 9) because the defender stays at its current location, i.e., (xm

i , αm

i ) and captures

Sj+1
i,m .

Now suppose that after comparison C1, No,m ̸= Ni,m. Then there are two cases: (i)

Either the defender decides to stay at its current location for D time interval implying that

|Sj+1
i,m

| > |Sj+2
o,m

| (line 9) or (ii) the defender decides to move to the resting point corresponding

to the sector No,m selected from C1 implying that |Sj+1
i,m

| ≤ |Sj+2
o,m

| (line 7). In case (i), the

defender stays at its current location and captures Sj+1

i,m . In case (ii), the defender spends

at most D time units to moves to the resting point of the sector No,m and then captures

intruders in the set Sj+2

o,m . This concludes the proof.

To establish the competitive ratio of Algorithm Dec-SNP, we use an accounting analysis

in which captured intervals pay for the lost intervals or equivalently, captured intervals are

charged for the intervals lost. The following two lemmas, proofs of which are contained in

Appendix, will jointly establish the competitive ratio of Dec-SNP algorithm. A single charge

to a particular captured interval consists of all the lost intervals that were considered in a

comparison step C1 involving that particular captured interval. Note that we consider the

first interval of every sector in which the defender is not contained as a lost interval because

that interval can never be captured by that defender.

Lemma 41. Using Algorithm Dec-SNP, for every defender Vm, m ∈ {1, . . . , M }, any two

consecutive captured intervals pay for a total of 3(ns − 1) lost intervals.

Proof. Please see Appendix (12)

We now establish that each lost interval is fully accounted for by the captured intervals.

91

Lemma 42. Each lost interval is accounted for by the captured intervals in the Dec-SNP

algorithm.

Proof. Please see Appendix (12)

Together, these two lemmas lead to the following main result.

Theorem 43 (Dec-SNP competitiveness). For any problem instance P that satisfies 3D ≤
1−ρ
v , cDec-SNP = 3ns−1

, where ns = ⌈θ/(M θs)⌉, θs = arctan(r/ρ) and D is defined in (4.5).

2

Proof. From Lemma 41 and Lemma 42 it follows that for any given trace of SNP algorithm,

every two consecutively captured intervals pay for 3ns−3 lost intervals and every lost interval

is accounted by two consecutive captured intervals. Thus, the claim follows.

In all of the above algorithms, the central idea was to partition the environment E(θ) into

M dominance regions and then assign a single defender to each dominance region. Next, we

design two cooperative algorithms with provably finite competitive ratio.

4.3.2 Cooperative Algorithms

4.3.2.1 Cooperative Compare and Capture Algorithm (Coop-CaC)

Recall that in Algorithm Dec-CaC, every defender captures the intruders from only one

out of the two sets, Rk

m and Lk

m. The idea behind Coop-CaC algorithm is to have an overlap

of the set Rk

m and Lk

m+1 of the dominance region m and m + 1 so that the defenders do not

lose the same number of intruders that they capture. Since there is an overlap of the sets,

we require that M is an even number for Algorithm Coop-CaC.

Algorithm Coop-CaC first partitions the environment into M

2 dominance regions, each
}. Each

of angle 4θ

M and assigns two defenders to each dominance region m ∈ {1, . . . , M

2

dominance region is then further divided into three equal sectors, each of angle 4θ

3M (Fig.

4.10). As will be described shortly, the motion of the defenders in any dominance region

m ∈ {1, . . . , M
2
∀m′ ∈ {1, . . . , M
2

} is independent of the motion of the defenders in dominance region m′,

} \ {m}. However, the two defenders within dominance region m cooperate

92

Figure 4.10 Description of Coop-CaC for a dominance region m with three sectors. The two
defenders (shown in yellow and orange color) share the sector in the middle. The region between
the yellow dashed lines and the yellow dots and on the left (resp.
right) of the first (moving
clockwise) black dashed line is the set Lk
1,m). The region between the orange dashed
lines and the orange dots and on the right (resp.
left) of the second black dashed line is the set
Rk
2,m). The region between the intersection of the yellow and orange dashed lines and
dots denotes the set I k
2,m. The green dashed line denotes the angular paths for the
two defenders when |Lk

∩ Lk
| and |Rk

1,m (resp. Rk

2,m (resp. Lk

| holds.

m = Rk
1,m

1,m
| > |I k
m

| > |I k
m

2,m

with each other. The cooperation is achieved by assigning the two defenders a common

sector. In particular, let V m

s denote the defender assigned to sector s and s + 1 in dominance

region m, where s ∈ {1, 2}. Then the two defenders V m

1 and V m
2

share the second sector of

the m-th dominance region (cf. Fig. 4.10).

Before we summarize the algorithm, we first determine the values of the parameters

and βs,m which characterize the angular paths Ts(xs,m

xs,m
, βs,m) of defender V m
C , β
s
for all 1 ≤ s ≤ 2 in dominance region m. Let an epoch km of dominance region m be defined

C , β

s,m

s,m

as the time the two defenders in dominance region m take to complete their motion on their

respective angular paths. Note that this requires that the time taken by the two defenders

must be the same. Since the angle of each of the sector in a dominance region m is equal, it

follows that x1,m

C = x2,m

C must hold in every dominance region m. The distance xs,m

C

is fixed

for a defender in all epochs. However, unlike Algorithm Dec-CaC, β

and βs,m are set at

s,m

the start of every epoch k for every sector s based on the number of intruders as well as the

angular path of the other defender in the same dominance region m.

We now determine the range of values that the parameter xs,m

C

can take. From points 1

93

and 2 of the justification for xm

S in Algorithm Dec-Sweep, it follows that xs,m

C

≤ ρ + r and

xs,m
C

≥ ρ − r. Further, as we require that xs,m

C + r + 4θvxs,m

C
3M

≤ 1, we obtain xs,m

C

≤ 3M (1−r)
3M +4θv .

Thus,

[

xs,m
C

∈

ρ − r, min{ρ + r, 3M (1−r)

3M +4θv

]
}

.

Note that the set xs,m

C

is defined when

ρ − r <

3M (1 − r)
3M + 4θv

⇒ v <

3M (1 − ρ)
4θ(ρ − r)

.

(4.6)

| and |Lk

s,m

|

For all s ∈ {1, 2} and every dominance region m ∈ {1, . . . , M
2
s,m and Lk

as the total number of intruders contained in the sets Rk

}, denote |Rk

s,m

s,m (cf. Figure 4.10),

respectively. Mathematically, the sets are defined as

Rk

s,m := {(y, ϕ) : ρ + ϕxs,m

C v < y ≤

min{1, xs,m

C + r + ( 8θ
3M

− ϕ)vxs,m

C

}∀ϕ ∈ [0, 4θ
3M ]

s,m := {(y, ϕ) : ρ − ϕxs,m
Lk

C v < y ≤

min{1, xs,m

C + r + ( 8θ

3M + ϕ)vxs,m

C

}∀ϕ ∈ [− 4θ

3M , 0).

Note that |Rk

1,m

| may not be equal to |Lk

2,m

| for any m ∈ {1, . . . , M
2

}. Finally, let |I k
m

| denote

the number of intruders contained in the intersection of the set I k
m

≜ Rk

1,m

∩ Lk

2,m( cf Fig.

4.10). Then, Algorithm Coop-CaC which is defined in Algorithm 7, is summarized as follows.

For every dominance region m, Algorithm Coop-CaC computes the total number of

intruders in the set Lk

1,m, Rk

2,m and I k

m. As that there are three sectors and two defenders

in a particular dominance region, the idea is to capture the two out of the three sets that

contain the most number of intruders. This requires that the two defenders must be assigned

the appropriate sets. Once the set is assigned, the defenders can then capture the intruders

by moving on the angular paths specified by the sets assigned to them. Thus, in what follows,

we describe the assignment of the sets to the defenders, which can be summarized into two

cases.

94

Case 1: Both |I k
m

| ≥ |Rk

2,m

| and |I k
m

| ≥ |Lk

1,m

| holds. This means that the set I k

m has the

most number of intruders. Thus, by determining which out of the remaining two sets, i.e.,

Rk

2,m and Lk

1,m, has more number of intruders, Algorithm Coop-CaC assigns the sets to the

two defenders. Mathematically, if both |I k
m

| ≥ |Rk

2,m

| and |I k
m

| ≥ |Lk

1,m

| hold, then

• If |Lk

| ≥ |Rk

2,m

|, then defender V m

1 and V m

2 moves on the angular paths T1(xs,m

C , β

, β1,m)

1,m

1,m
and T2(xs,m

C , β

, β2,m), respectively, with

2,m

β

s,m

= ϕ1 − 4θ

3M , βs,m = ϕ1, ∀s ∈ {1, 2},

where ϕ1 = −( 3

2 (M − 4(m − 1)) − 2s) 2θ

3M and is measured from the positive y-axis.

In other words, both defenders move counter clockwise by moving on their respective

angular paths.

• Otherwise, i.e., |Lk

1,m

| < |Rk

2,m

| holds then defenders V m

1 and V m

2 move on the angular

paths T1(xs,m

C , β

1,m

, β1,m) and T2(xs,m

C , β

2,m

, β2,m), respectively, with

β

s,m

= ϕ1, βs,m = ϕ1 + 4θ

3M , ∀s ∈ {1, 2}.

This means that both defenders move clockwise by moving on their respective angular

paths.

Case 2: Both |I k
m

| ≥ |Rk

2,m

| and |I k
m

| ≥ |Lk

1,m

| do not hold. This implies that one set

out of Rk

2,m and Lk

1,m has the maximum number of intruders. In this scenario, the algorithm

compares the two sets associated with each defender individually. In particular, the algorithm

compares the number of intruders in the sets Lk

1,m (resp. Rk

2,m) and I k

m for defender V m
1

(resp.

2 ) and then assigns the set with more number of intruders to defender V m
V m
1

(resp. V m

2 ). If

set Lk

1,m (resp. Rk

2,m) has more number of intruders as compared to the set I k

m, then defender

V m
1

(resp. V m

2 ) moves counter clockwise (resp. clockwise). Otherwise, defender V m
1

(resp.

V m
2 ) moves clockwise (resp. counter clockwise). This is mathematically described as follows.

95

Algorithm 7: Cooperative Compare-and-Capture (Coop-CaC) Algorithm
1 Select x1,m
2 for each epoch k ≥ 1 do
3

for all s ∈ {1, 2} and every dominance region m ∈ {1, . . . , M
2

∈ [ρ − r, min{ρ + r, 3M (1−r)

C = x2,m

} do

}]}.

3M +4θv

C

Determine β

, βs,m depending on Case 1 or Case 2 from the description.

s,m

4

5

6

7
8 end

end
Move every defender V m
Move every defender V m

s on Ts(xs,m
C , β
s back to starting position.

, βs,m).

s,m

• For defender V m
1

• For defender V m
2

β

β

1,m

1,m

β

β

2,m

2,m

= ϕ1 − 4θ

3M , β1,m = ϕ1 if |I k

m

| < |Lk

1,m

|

= ϕ1, β1,m = ϕ1 + 4θ

3M , otherwise.

= ϕ1, β2,m = ϕ1 + 4θ

3M if |I k

m

| < |Rk

2,m

|

= ϕ1 − 4θ

3M , β2,m = ϕ1, otherwise.

Once the parameters of the angular paths of the defenders are set, the defenders move on

the angular paths and then return to their starting locations. The next epoch begins once

the defenders reaches their starting locations.

For the initial case, we assume time 0 as the time when the first intruder is released.

Each defender Vs,m, 1 ≤ s ≤ 2 in dominance region m, m ∈ {1, . . . , M
2

} starts at location

(xs,m

C , ϕ1) and begins its first epoch at time 0.

The following result characterizes the parameter regime for Algorithm Coop-CaC.

Lemma 44. Any intruder in dominance region m at a radial location exceeding xs,m

( 8θ
3M

− ϕ)vxs,m

C , ∀ϕ ∈ [0, 4θ

either in the set Rk+1

s,m or in Lk+1

C + r +
3M ] and for any s ∈ {1, 2}, at the start of epoch k is contained
s,m in epoch k + 1 and is not lost at the start of epoch k + 1 if

v ≤ 3M (xs,m

C + r − ρ)
8θxs,m
C

.

96

Proof. Without loss of generality, let s = 2 and assume that |I k
m

| < |Rk

2,m

| holds in dominance

region m ∈ {1, . . . , M
2

}. The total time taken by defender V m
2

to capture intruders in Rk

2,m

and return back to its starting location (x2,m

C , ϕ1) is 8θ

3M x2,m

C . In the worst-case, in order for

any intruder i to be not considered in the start of epoch k, the intruder i must be located just

above (x2,m

C + r + 4θ

3M vx2,m

C , ϕ1 + 4θ

3M ) at the start of epoch k. By the time the defender reaches

location (x2,m

C , ϕ1 + 4θ

3M ), intruder i will be located just above the location (x2,m

C + r, ϕ1 + 4θ

3M )

and will not be captured.

If v ≤ 3M (xs,m
C +r−ρ)
8θxs,m
C

3M vx2,m
distance away from the perimeter at the end of epoch k. Thus, intruder i gets considered

holds for dominance region m, intruder i will be at least 4θ

C

at the start of epoch k + 1 as per the definitions of the sets Lk+1

2,m and Rk+1

2,m implying that

intruder i is not lost at the start of epoch k + 1. This concludes the proof.

Corollary 7. Any intruder in dominance region m which lies beyond the set I k

m in epoch k

will be contained in the set I k+1

m in epoch k + 1 if Lemma 44 holds.

Proof. The proof directly follows from the fact that Lemma 44 holds for both defenders V m
1

and V m
2

in dominance region m.

Theorem 45. The competitive ratio cCoop−CaC = 1.5 for any problem instance P that satis-

fies

v ≤ min

m∈{1,...,M }

{

3M (xs,m

C + r − ρ)
8θxs,m
C

}

and equation (4.6) for any xs,m

C

∈ [ρ − r, min{ρ + r, 3M (1−r)

3M +4θv

}].

Proof. The idea is to prove the result for any dominance region m ∈ {1, . . . , M
2

}. The

result then will follow from the fact that the motion of defenders in a dominance region is

independent of the motion of the defenders in other dominance regions.

Let there be r2, l1 and l2 + r1 intruders in the sets Rk

2,m, Lk

1,m and I k

m, respectively, in a

dominance region m. Then based on the number of intruders in the sets Rk

2,m, Lk

1,m and I k
m

in an epoch k, three cases arise.

97

Case 1: |I k
m

| ≥ |Lk

1,m

| and |I k
m

| ≥ |Rk

2,m

|. This implies that the following two conditions

must hold in epoch k.

r1 + l2 ≥ l1 and r1 + l2 ≥ r2.

Further, without loss of generality, assume that l1 ≥ r2 holds, i.e., the condition |Lk

1,m

| ≥

|Rk

2,m

| holds. Then, it follows that the two defenders jointly captured r1 + l2 + l1 intruders

out of r1 + l2 + l1 + r2 in epoch k. Assuming that an optimal offline algorithm can capture

all intruders in the worst-case and using the aforementioned conditions, we obtain

cCoop−CaC =

r1 + l2 + l1 + r2
r1 + l2 + l1

≤ 3r2
2r2

= 1.5.

Case 2:

|I k
m

| < |Lk

1,m

| and |I k
m

| ≥ Rk

2,m holds. This implies that the following two

conditions must hold in epoch k.

r1 + l2 < l1 and r1 + l2 ≥ r2.

In this case, the defenders jointly capture r1 +l2 +l1 intruders out of r1 +l2 +l1 +r2 intruders.

Thus, assuming that the optimal offline algorithm can capture all intruders and using the

aforementioned conditions yields

cCoop−CaC =

r1 + l2 + l1 + r2
r1 + l2 + l1

≤ 3r2
2r2

= 1.5.

The proof for the case when |I k
m

| ≥ |Lk

1,m

| and |Im| < Rk

2,m holds is omitted for brevity

as it is analogous to Case 2.

Case 3: |Lk

1,m

| > |I k
m

| and |R2,m| > |I k
m

|. This implies that the following two conditions

must hold in epoch k.

r1 + l2 < l1 and r1 + l2 < r2.

In this case, the defenders jointly capture l1 + r2 intruders out of r1 + l2 + l1 + r2 intruders.

Following similar steps as in previous two cases yields

cCoop−CaC =

r1 + l2 + l1 + r2
r2 + l1

≤ 3(r1 + l2)
2(r1 + l2)

= 1.5.

98

Thus, in any epoch k and in a dominance region m ∈ {1, . . . , M
2

}, the two defenders

contained in dominance region m jointly capture at least 2
3

rd of the total intruders that

arrive in that particular dominance region. Since, the number of intruders captured by the

defenders in any dominance region m ∈ {1, . . . , M
2

} does not depend on any other dominance

regions, it follows that in every dominance region, the defenders capture at least 2
3

rd of the

total intruders that arrive in their respective dominance region. This implies that Algorithm

Coop-CaC captures at least 2
3

rd of the total intruders that arrive in the environment. Finally,

Lemma 44 and Corollary 7 ensures that every intruder is considered in the sets Lk

1,m, Rk

2,m

and I k

m. Since every intruder is considered and the defenders capture at least 2

3 -rd of the

total number of intruders it follows that cCoop−CaC = 1.5 and the result is established.

Remark 4. As 3M (xs,m
C +r−ρ)
8θxs,m
C

is a monotonically increasing function of xs,m

defender V m

s , 1 ≤ s ≤ 2 in every dominance region m ∈ {1, . . . , M

{

C , positioning each
} at min{ρ + r, 3M (1−r)
}

3M +4vθ

2

yields v ≤ min

3M r

4θ(ρ+r) , 3M (1−ρ)

4θ(2−3r+ρ) , 3M (1−ρ)

4θ(ρ−r)

}

.

Remark 5. The maximum speed ratio for which Algorithm Coop-CaC is effective is 3/4-th

of the maximum speed ratio for which Algorithm Dec-CaC is effective.

Recall that in Coop-CaC algorithm, the motion of the defenders in a dominance region

is independent of the motion of the defenders contained in another dominance region. Due

to this, Coop-CaC Algorithm requires an even number of defenders. Next, we propose a

variation of Algorithm Coop-CaC that relaxes this requirement, i.e., the next algorithm

is effective for any value of M > 1. An important distinction to note between the two

algorithms is that in Algorithm Coop-CaC, every defender shares at most one sector whereas

in the Modified Coop-CaC algorithm, every defender shares at least one and at most two

sectors. As Modified Coop-CaC algorithm is similar to Coop-CaC algorithm, we briefly

describe the algorithm and its competitive ratio in the following remark and omit establishing

the result formally due to space constraints.

99

Figure 4.11 Description of Modified Coop-CaC Algorithm for an epoch k and M = 3. Set
2 (resp. I k
I k

3 ) denote the intersection of the sets Rk

2 (resp. Rk

2 and Lk

1 and Lk

3).

Remark 6 (Modified Coop-CaC). Modified Coop-CaC algorithm partitions the environment

into M + 1 sectors, each of angle 2θ

m and
m+1 overlap, for all m ∈ {1, . . . , M − 1} (cf. Fig. 4.11). Then, following similar steps
Lk

M +1 and positions the defenders such that the sets Rk

as for Algorithm Coop-CaC, it can be shown that the Modified Coop-CaC algorithm is M +1
M -

competitive. Further, it can be shown by following the steps of Algorithm Coop-CaC that the

parameter regime of the modified algorithm is lower than that of the Coop-CaC algorithm.

Although the parameter regime of Modified Coop-CaC is lower than that of Algo-

rithm Coop-CaC, note that cM odif ied−CaC < cCoop−CaC for any M > 2 and cM odif ied−CaC =

cCoop−CaC for M = 2. We now present our second cooperative algorithm.

4.3.2.2 Cooperative Stay Near Perimeter (Coop-SNP) Algorithm

The idea behind this algorithm is to divide the entire environment E(θ) into N = ⌈ θ
θs

⌉

sectors, each of angle 2θs = 2 arctan( r

ρ ). We highlight an important distinction between

Algorithm Coop-SNP and Algorithm Dec-SNP. In Algorithm Dec-SNP, the environment is

first partitioned into M dominance regions and then each dominance region is further divided

into ns sectors. On the other hand, in Algorithm Coop-SNP, we directly divide the entire

environment into N sectors. We assume that the number of defenders is strictly less than

the number of sectors, i.e., M < N . If this does not hold, then by assigning a defender to

every sector, Coop-SNP will capture all intruders, yielding the competitive ratio as 1. Let

100

Algorithm 8: Cooperative Stay Near Perimeter Algorithm
1 Assumes each defender Vm, 1 ≤ m ≤ M is at (xm, αm) in sector Nm
2 for each j ≥ 1 do
3

Determine sets O, ˆO and ˜O.
for every sector δg ∈ ˜O do

Assign the defender located in δg to sector δg for 2 ˜D time units.

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

end
for every sector δq ∈ ˆO do
| then

| ≥ |Sj+1

if |Sj+2
q

p

Assign Vp to sector δq for 2 ˜D time units.
Remove Vp from set ˆV and add to set V2.

else

Assign δp for ˜D time units
Assign δq for ˜D time units.
Remove Vp from set ˆV add to set V1.

end

end
Defenders in ˜V stay at their location for 2 ˜D time.
for every Vq ∈ V2 do

Move to the resting point of assigned sector.
Stay at the resting point until time 2(j + 2) ˜D.

end

21
22 end
23 for every Vq ∈ V1 do
24

25
26 end

Stay location until time 2(j + 1) ˜D.
Move to resting point of the second sector assigned.

δl, l ∈ {1, . . . , N }, denote the lth sector, where δ1 denotes the leftmost sector and we assume

that the sectors are numbered from left to right. Further, let ∂Rl denote the portion of the

perimeter contained in sector δl. Then, a resting point (xl, αl) ∈ E(θ) of a sector δl is defined

as a location for defender Vm such that when positioned at that location, the entire ∂Rl is

contained completely within the capture radius of the defender Vm. Mathematically,

(

(xl, αl) :=

ρ cos(θs),

(

l − N + 1
2

)

)

2θs

.

101

Figure 4.12 Description of Coop-SNP Algorithm. Blue dashed circles denote the resting points.
Defender V1, V2 and V3 are located in δ3, δ4, and δ5, respectively. O = {δ1, δ2, δ3}, ˜O = {δ3},
ˆO = {δ2, δ1}, ˜V = {V1} and ˆV = {V3, V2}.

Lastly, let ˜D denote the distance between the two resting points that are furthest apart (cf.

Fig. 4.12). Formally,






˜D =

2ρ cos(θs) sin((N − 1)θs), if (N − 1)θs < π
2

2ρ cos(θs), otherwise.

(4.7)

After dividing the environment into N sectors, analogous to Dec-SNP algorithm, Coop-

SNP algorithm radially divides the environment E(θ) into three intervals of length ˜Dv or

equivalently, time intervals of time length ˜D each. Specifically, the jth time interval for any

j > 0 is defined as the time interval [(j − 1) ˜D, j ˜D]. The requirement to ensure a finite

competitiveness for Coop-SNP algorithm is the same is Dec-SNP, i.e., we require 1−ρ
v

≥ 3 ˜D.

For any j ≥ 1, let Sj

l be the set of intruders that arrive in a sector δl in the jth time interval.

The Coop-SNP algorithm, defined in Algorithm 8, is based on the following two steps:

1) For every j ≥ 0, at time instant 3 ˜D + 2j ˜D, Coop-SNP selects M sectors with highest

number of intruders and, 2) assigns M defenders to the resting locations of the M sectors

selected in the first step. We describe each step in more detail below.

Comparison (S1): Coop-SNP determines M sectors which has the most number of

intruders. In particular, suppose that defenders V1, . . . , VM are located at the resting points

of sectors δ1, . . . , δM , respectively, at the jth iteration. Let L denote the set of those corre-

102

sponding indices. Then, for every i ∈ {1, . . . , N }, let




ηi
L =



|Sj+2
i

| + |Sj+3

i

|,

if i /∈ L,

|Sj+1
i

| + |Sj+2

i

| + |Sj+3

i

|,

otherwise.

Coop-SNP then selects the top M sectors which have the highest number of intruders.

Specifically, Coop-SNP sorts the set {η1

L, . . . , ηN
L

} in a decreasing order and then selects

the sectors corresponding to the first M elements obtained from the ordered set. If there

are multiple sectors with the same number of intruders, then Coop-SNP breaks the tie

analogously to Dec-SNP algorithm.

We introduce the following notations for the second step. Let O be a set of sectors chosen

as the outcome of comparison S1 and let ˜O ⊆ O denote the set of those sectors which have

defenders already positioned at their respective resting points (cf. Fig. 4.12). Let ˜V denote

the set of defenders that are located at the resting points of the sectors in ˜O. Note that

| ˜V| = | ˜O|. For the sectors in ˆO = O \ ˜O, we assume that the sectors in the set ˆO are

ordered with respect to the number of intruders in their second, i.e., (j + 2)th, interval (cf.

Fig. 4.12). Specifically, the first (resp. last) element of ˆO) has the most (resp. least) number

of intruders in the (j + 2)th interval. Similarly, let ˆV = V \ ˜V be the set of defenders not

located in the sectors contained in ˜O. We assume that the set ˆV is in an increasing order

according to the number of intruders in the first, i.e., (j + 1)th interval of the sectors the

defenders are located in (cf. Fig. 4.12).

Assignment (S2): The aim of this step is to ensure that every sector in the set O is

assigned a defender. For every element δq ∈ O, Algorithm Coop-SNP does the following:

i For all δq ∈ ˜O, Coop-SNP assigns the defender that is already located in δq to sector δq

for 2 ˜D time units.

ii For every sector δq ∈ ˆO,

a) Let Vp be the first element of the set ˆV and suppose that Vp is located in sector
δp. Then Coop-SNP assigns defender Vp to sector δq ∈ ˆO for 2 ˜D time units if

103

| > |Sj+1

p

| holds. After assigning Vp to δq, Coop-SNP removes Vp from the set

|Sj+2
q
ˆV.

b) Otherwise, Coop-SNP first assigns the sector δp to defender Vp for ˜D time units and
then sector δq for the next ˜D time units. Coop-SNP then removes Vp from the set
ˆV.

Once the assignment S2 is complete, each defender in the set ˜V captures the first two

intervals in their respective sectors by staying at their locations for 2 ˜D time units. The

defenders that are assigned sectors based on point (ii-a) of S2, move to the resting points of

their assigned sectors, reaching the location in at most ˜D time units, and then stay at that

location until all of the intruders in the (j + 2)th interval are captured. Finally, the defenders

that are assigned sectors based on point (ii-b) of S2, first stay at their locations until all of

the intruders in the (j + 1)th interval are captured. Then, they move to the resting point of

the assigned sectors.

At time 0, each defender Vi, i ∈ {1, . . . , M } is located at the resting point of sector δi

and waits at its location until ˜D time units. Then, setting |Sj+1

i

| = |Sj+2

i

| = 0, the steps S1

and S2 are applied. The first iteration then begins at time 3 ˜D.

Lemma 46. For any j ≥ 1, every defender Vm ∈ V, assumed to be located in sector δm and

assigned sector δi, captures at least one intruder interval, i.e., either Sj+1

m or Sj+2

i

, in the

time interval [j ˜D, (j + 2) ˜D].

Proof. Algorithm Coop-SNP assigns a sector to a defender in only three possible ways. First,

if δm = δi (point (i) of S2), then the result follows as the defender stays at its current location
m of sector δi. Second, sector δi is assigned for 2 ˜D time

and captures interval Sj+1

m and Sj+2

(point (ii-a) of S2), then the defender moves to sector δi and then captures Sj+2

i

. Finally, if

sector δm is assigned followed by sector δi (point (ii-b) of S2), then the defender stays at its

location for ˜D time and captures Sj+1

m . Thus, in all three cases, the defender captures at least

one interval. Since, this proof holds for every defender in V, the result is established.

104

Similar to Dec-SNP algorithm, to establish the competitive ratio of Algorithm Coop-

SNP, we use an accounting analysis in which captured intervals pay for the lost intervals or

equivalently, captured intervals are charged for the intervals lost. A charge to the captured

interval(s) consists of all the lost intervals that are considered in a single comparison step

S1 involving that particular captured interval(s).

The following two lemmas will jointly establish the competitive ratio of Coop-SNP algo-

rithm, proofs of which are provided in the Appendix.

Lemma 47. Under Algorithm Coop-SNP, for each defender, every two consecutive captured

intervals are charged at most 3(N − M ) times.

Proof. Please see Appendix 12.

We now establish that each lost interval is fully accounted for by the captured intervals.

Lemma 48. Every lost interval is accounted for by the captured intervals of Algorithm Coop-

SNP.

Proof. Please see Appendix 12.

Theorem 49. Algorithm Coop-SNP is 3N −M

satisfies 3 ˜D ≤ 1−ρ

v , where N = ⌈ θ

θs

2M -competitive for any problem instance P which
ρ ) and ˜D is defined in equation (4.7).

⌉, θs = arcsin( r

Proof. From Lemma 47, every two consecutive captured intervals for each defender account

for 3(N − M ) lost intervals implying that every two captured intruders pay for at most

3(N − M ) intruders. Since there are total of M defenders, the total number of intervals are

3(N − M ) + 2M = 3N − M . Lemma 48 ensures that every lost interval is fully accounted

by the captured intervals. Thus, assuming that the optimal offline captures all intruder

intervals, we obtain the result.

Corollary 8. Given a problem instance P, cCoop-SNP ≤ cDec-SNP.

105

Proof. Recall from Theorem 43 and Theorem 49, the competitive ratio of Algorithm Dec-

2
≥ 3N −M

SNP and Algorithm Coop-SNP is 3ns−1

and 3N −M

2M , respectively. The idea is to show that

for any problem instance P, 3ns−1

2

Since N > M is an integer, let ⌈ θ
θs

2M or equivalently, ns ≥ N
M .
⌉ = M + p, where p ≥ 1 ∈ Z+. This yields

(

=

1 +

)

.

p
M

N
M

Two cases arise:

Case 1: If θ
θs

= M + p, i.e., θ
θs

is an integer. Then

ns = ⌈1 +

p
M

⌉ ≥ 1 +

p
M

,

and we obtain the result.

Case 2: If θ
θs

is not an integer. Then, for an 0 < ϵ < 1,

ns = ⌈1 +

p − 1 + ϵ
M

⌉.

If p ≤ M , it follows that ⌈1 + p−1+ϵ

M

⌉ ≥ 1 + p

result follows as p

M can be expressed as q + l

M and the result holds. For any p > M , the
M , where q ∈ Z+ and 0 < l < M yielding a

similar expression as in Case 2. This concludes the proof.

Remark 7 (Practical Considerations). While Algorithm Coop-SNP can be applied as is

to UAVs flying at different altitudes, collision avoidance algorithms may be required to be

implemented in other cases when the vehicles move from their respective sectors to some

other sectors. In such cases, the parameter ˜D must be suitably scaled to account for the extra

time required by the vehicle to move to a sector. Doing so may lower the upper bound on

parameter v characterized for the algorithm but the bound on the competitive ratio still holds

as long as 3 ˜D ≤ (1 − ρ)/v holds, as the competitive ratio of Algorithm Coop-SNP does not

depend on the value of ˜D.

Remark 8. Although the competitive ratio of Coop-SNP as well as Coop-CaC algorithm is

lower than that of Dec-SNP and Dec-CaC, respectively, the cooperative algorithms require

106

Figure 4.13 (v, M ) parameter regime plot for fixed ρ = 0.3, r = 0.01, and θ = π
2 . For these values
of parameters, a c-competitive algorithm may exist for all values of v ≤ 1 (Theorem 35). The
algorithms designed in this work have a finite competitive ratio for values of v that lie below their
corresponding markers.
In particular, Algorithm Dec-Sweep, Dec-CaC and Coop-CaC are 1, 2
and 1.5-competitive, respectively, for all values of M and for values of v below their corresponding
markers. The competitive ratio of both Algorithm Dec-SNP and Coop-SNP varies with the param-
eter M . The parameter regime of Algorithm Coop-SNP does not depend on M .

intruders to be slower than the decentralized algorithms. Equivalently, the cooperative algo-

rithms require higher number of vehicles to cover the same (v, ρ) parameter regime as the

decentralized algorithms but achieve a lower competitive ratio.

4.4 Numerical Observations and Discussions

We now provide a numerical visualization of the analytical bounds derived in this chapter.

We focus on the parameter regimes (v, M ) and (v, ρ). We also compare the competitive ratios

of the decentralized and the centralized SNP algorithms and provide insights into the case

of heterogeneous capture radii and perimeters of arbitrary shapes.

4.4.1

(v, M ) parameter regime

Figure 4.13 and Figure 4.14 shows the (v, M ) parameter regime plot for the fixed value

of ρ, r, and θ.

Observe that as the number M of defenders increases, the curves for Algorithms Dec-

107

Figure 4.14 (v, M ) parameter regime plot for fixed ρ = 0.85, r = 0.05, and θ = π
2 . For any value of v
that lies beyond the green markers, there does not exist a c-competitive algorithm. The algorithms
designed in this work have a finite competitive ratio for values of v that lie below their corresponding
markers. Specifically, Algorithm Dec-Sweep, Dec-CaC and Coop-CaC are 1, 2 and 1.5-competitive,
respectively, for all values of M and values of v that lie below their respective markers. The
competitive ratio of both Algorithm Dec-SNP and Coop-SNP varies with the parameter M . The
parameter regime of Algorithm Coop-SNP does not depend on M .

Sweep, Dec-CaC, Coop-CaC and Dec-SNP approach the curve depicting the condition of

Theorem 35. This implies that it is possible to capture faster intruders with a greater

number of defenders. Further, while the curve for Algorithm Dec-Sweep, Algorithm Dec-

CaC and Algorithm Coop-CaC are linear, the curve for Algorithm Dec-SNP is in regions,

with each region having a specific competitive ratio. Since the condition in Theorem 49

does not depend on M , the curve for the conditions of Algorithm Coop-SNP is a straight

line. Given the values of the parameters ρ, r, θ for Figure 4.13, note that the green curve is

constant v = 1 for all values of M , implying that there may exist a c−competitive algorithm

in this parameter regime. Interestingly, the curve for Dec-SNP algorithm reaches the green

curve. Finally, note that the competitive ratio of Algorithm Coop-SNP is strictly less than

4 for M > 16 whereas the competitive ratio of Algorithm Dec-SNP is greater than 4 for all

2 ≤ M ≤ 20.

In Figure 4.14, the curve for Algorithm Dec-SNP is never above the curve for Algorithm

108

Figure 4.15 (v, ρ) parameter regime plot for fixed r = 0.02, θ = π
2 and M = 4. No c-competitive
algorithm exists beyond the green curve. For values of v below the yellow (resp. blue) curve,
Algorithm Dec-Sweep (resp. Dec-CaC) is 1 (resp. 2)-competitive. For values of v below the pink
curve, Algorithm Coop-CaC is 1.5-competitive. Competitive ratio of Algorithms Dec-SNP and
Coop-SNP varies with parameter ρ.

Dec-CaC implying that Dec-CaC is more effective for high values of ρ unless Algorithm Dec-

SNP has a competitive ratio less than 2. Specifically, for M < 27, Algorithm Dec-SNP has a

competitive ratio strictly greater than 2 and its curve lies below that of Algorithm Dec-CaC

and thus, Algorithm Dec-CaC is more effective. For M > 27, Algorithm Dec-SNP as well as

Algorithm Coop-SNP are 1-competitive. Thus, for high values of M , Algorithm Dec-SNP

is more effective. However, since the curve of Algorithm Coop-SNP lies below the curve of

Dec-Sweep, it is not effective.

4.4.2

(v, ρ) parameter regime

Figure 4.15 and Figure 4.16 shows the (v, ρ) parameter regime for fixed values of r, M

and θ.

From both Figure 4.15 and Figure 4.16, since the competitiveness of Algorithm Coop-SNP

and Algorithm Dec-SNP is based on the number of sectors, the curve for both algorithms is

divided into regions, where each region corresponds to a constant value of competitiveness.

As the capture radius r increases, the number of regions decreases. Similarly, the number

109

Figure 4.16 (v, ρ) parameter regime plot for fixed r = 0.1, θ = π
2 and M = 2. No c-competitive
algorithm exists beyond the green curve. For values of v below the yellow (resp. blue) curve,
Algorithm Dec-Sweep (resp. Dec-CaC) is 1 (resp. 2)-competitive. For values of v below the pink
curve, Algorithm Coop-CaC is 1.5-competitive. Competitive ratio of Algorithms Dec-SNP and
Coop-SNP varies with parameter ρ.

of regions increases as θ increases. Therefore, an important feature of Algorithm Coop-SNP

and Algorithm Dec-SNP is that they can be used to determine the tradeoff between the

competitiveness and the required parameter regime for the problem instance.

In Figure 4.15 (resp. Figure 4.16), Algorithm Dec-CaC is more effective than Algorithm

Dec-SNP and Algorithm Coop-SNP as the curves for Algorithm Dec-SNP and Algorithm

Coop-SNP are below the curve for Algorithm Dec-CaC for ρ > 0.76 (resp. ρ > 0.38)

and ρ > 0.88 (resp. ρ > 0.57), respectively. A similar observation is made for Algorithm

Coop-CaC, Algorithm Dec-SNP and Algorithm Coop-SNP. Further, compared to that in

Figure 4.16, both Algorithm Dec-SNP and Algorithm Coop-SNP have a larger area of utility

suggesting that with a smaller capture radius, Dec-SNP and Coop-SNP can capture faster

intruders, and it covers a larger area in the parameter regime but at the cost of higher

competitive ratio.

In Figure 4.16, Algorithm Dec-SNP is 2.5-competitive for 0.15 ≤ ρ ≤ 0.26 whereas

Algorithm Coop-SNP is 1.75-competitive for ρ ≤ 0.2 and 2.5-competitive for 0.2 < ρ ≤ 0.26.

110

Figure 4.17 Comparison of competitive ratios for Algorithm Dec-SNP and Algorithm Coop-SNP
for θ = π, r = 0.13 and ρ = 0.7. Competitive ratio of Algorithm Coop-SNP is always lower than
that of Algorithm Dec-SNP.

Both Dec-SNP and Coop-SNP are 1-competitive for ρ < 0.15. Similarly, in Figure 4.15, both

Algorithm Dec-SNP and Algorithm Coop-SNP are 1-competitive for ρ ≤ 0.05 suggesting that

all intruders can be captured by increasing the number of vehicles.

4.4.3 Comparison of competitive ratios of Dec-SNP and Coop-SNP

Figure 4.17 shows the different values of competitive ratios for Algorithm Dec-SNP and

Algorithm Coop-SNP for a fixed θ = π, r = 0.13 and ρ = 0.7. Note that the competitive

ratio of Algorithm Dec-SNP is constant for some values of M whereas the competitive ratio

of Algorithm Coop-SNP decreases as M increases. For instance, Algorithm Dec-SNP is 2.5

competitive for 9 ≤ M ≤ 16 whereas Algorithm Coop-SNP has a competitive ratio strictly

less than 2.5 for the same values of M .

4.4.4 Extension to defenders with heterogeneous capture radii

We now describe how this work can be generalized to heterogeneous defenders having

different capture radii r1, r2, . . . , rM .

Recall from the proof of Lemma 31, to determine the dominance region for the M de-

fenders, we use the fact that the time taken by the neighboring defenders must be the same

111

to reach the endpoint of the common boundary. When the defenders have different capture

radii, then following similar steps, one can obtain M equations. Solving the M equations

will yield the angles for the M dominance regions. If 2β1, 2β2, . . . , 2βM denote the angles of
M in Lemma 31 and Lemma 32 by βi, 1 ≤ i ≤ M

the M dominance regions, then replacing θ

yields the position for defender Vi.

We now describe how Theorem 35 extends to defenders having different capture radii.

The key idea behind the proof of Theorem 35 is to ensure that no defender can capture the

burst intruders. From the partition angles 2βi, ∀i ∈ {1, . . . , M }, one can compute the time

taken by any defender Vi to reach the endpoint of its dominance region. Since this time is

same for all defenders V1, . . . , VM , the time taken by any defender Vi, i ∈ {1, . . . , M }, except

V1, to reach the endpoint of its dominance region is ρ sin(βi) − ri. Here, without loss of

generality, we assumed that defender V1 captures an intruder from the stream of intruders

and thus, the time taken by V1 is 2ρ sin(β1) − 2r1 and thus, the result is obtained.

Finally, once the dominance region for M defenders is obtained, then all of our decen-

tralized algorithms can be directly applied. Their respective analytical lower bounds can

also be obtained analogously. Our conjecture is that even when the defenders have different

capture radii, Algorithm Dec-Sweep and Algorithm Dec-CaC are still 1 and 2-competitive,

respectively.

We now briefly comment on the choice of algorithms and the applicability of this work.

The choice of which algorithm to use depends on the problem parameters, i.e., the size of the

perimeter and the speed ratio, and the acceptable bound on competitiveness. For instance,

given the speed of the intruders and an acceptable bound on the competitive ratio, one

may select a larger or smaller perimeter size. Note that it may not always be feasible to

select a smaller perimeter size. In that case, faster defenders or algorithms with higher value

of competitive ratio may be chosen. Further, the acceptable competitive ratio is governed

by the application. For instance, defense applications may need c = 1 while agricultural

applications may tolerate higher values of c. Finally, we remark that the sufficient conditions

112

on the algorithms are derived in the worst-case. In some applications, these worst-case inputs

may not be realizable and thus, the algorithms will perform better for those applications.

We now briefly describe how this work extends to perimeters with arbitrary shapes.

Recall that the two important quantities that are used in the analysis are the time taken

by the intruders to reach the perimeter and the time taken by a vehicle to move from one

point to the other, while capturing intruders. For successful capture of an intruder, the time

taken by the vehicle must be at most the time taken by the intruder. This work extends to

perimeters with arbitrary shapes as these two quantities can be determined for any given

(closed) shape of a perimeter. If the shape of the perimeter is such that the dominance region

for each vehicle ensures the same time to move from one point to another for every vehicle,

then the upper bound on v is characterized by the maximum time taken by a vehicle to

move within its dominance region (for instance angular motion in Dec-Sweep or Coop-CaC

algorithm). Otherwise, the upper bound on v is characterized by the vehicle in a particular

dominance region which requires the maximum time for a vehicle. This is because in the

worst-case, all of the intruders can be released in that vehicle’s dominance region.

Through this work, we aim to study how an adversary can overwhelm the defense by

using many (possibly infinite) attackers. This is useful to obtain insights about the problem

including different adversarial inputs such as the one designed in the proof of Theorem 35.

The insights gained on the worst-case inputs can be used for perimeter defense problems

with many responsive intruders, i.e., intruders that can move in any direction. Note that the

some of our results, including the fundamental limit and the Decentralized Sweep algorithm

applies as is when responsive intruders are considered.

4.5 Conclusions

This work analyzed the problem wherein M defenders, each having a finite capture

radius, are tasked to defend a perimeter in a conical environment from intruders that are

released arbitrarily in the environment. Our approach was based on competitive analysis

that first yielded a fundamental limit on the problem parameters for finite competitiveness

113

of any online algorithm. We then designed and analyzed five algorithms, i.e., Decentralized

Sweep, Decentralized CaC, Decentralized SNP, Cooperative CaC and Cooperative SNP and

established sufficient conditions that guaranteed a finite competitive ratio for each algorithm.

In particular, we establish that Decentralized Sweep, Cooperative CaC and Decentralized

CaC are 1, 1.5, and 2-competitive, respectively. Further, their competitive ratio increases

linearly with M . The competitive ratio of Decentralized SNP and Cooperative SNP changes

with the function of the parameters and does not always extend beyond that of Decentralized

CaC algorithm in specific parameter regimes.

114

CHAPTER 5

COMPETITIVE PERIMETER DEFENSE WITH A TURRET AND A
MOBILE DEFENDER

Until now, we focused on perimeter defense problems with homogeneous defenders. However,

there are many scenarios when there are different types of defenders, i.e., heterogeneous

defenders that must cooperate with each other to capture intruders. In this chapter, contents

of which are based on [5], we look at two such heterogeneous defenders.

This chapter addresses a perimeter defense problem in a conical environment. The envi-

ronment contains two heterogeneous defenders, namely a turret and a mobile vehicle, which

seek to defend a perimeter by capturing mobile intruders. The intruders are released at

the boundary of the environment and move radially inwards with fixed speed toward the

perimeter. Defenders have access to intruder locations only after they are released in the

environment. Further, the defenders have distinct motion and capture capability and thus,

are heterogeneous in nature. Specifically, the vehicle, having a finite capture radius, moves

with unit speed in the environment whereas the turret has a finite range and can only turn

clockwise or anti-clockwise with a fixed angular rate. Jointly, the defenders aim to capture

as many intruders as possible before they reach the perimeter. This is an online problem

as the input, which consists of the total number of intruders, their release locations, as well

as their release times, is gradually revealed over time to the defenders. Thus, we focus on

the design and analysis of online algorithms to route the defenders. Aside from military

applications, this work is also motivated by monitoring applications wherein a drone and a

camera jointly monitor the crowd entering a stadium. The main contributions of this work

are as follows:

1. Perimeter defense problem with heterogeneous defenders: We address a perime-

ter defense problem in a conical environment with two cooperative heterogeneous de-

fenders, i.e., a vehicle and a turret, tasked to defend a perimeter. The vehicle has a

finite capture radius and moves with unit speed, whereas the turret has a finite engage-

115

ment range and turns in the environment with a fixed angular rate. We do not impose

any assumption on the arrival process of the intruders. More precisely, an arbitrary

number of intruders can be released in the environment at arbitrary locations and time

instances. Upon release, the intruders move with fixed speed v towards the perimeter.

Thus, the perimeter defense problem is characterized by six parameters: (i) angle θ of

the conical environment, (ii) the speed v of the intruders, (iii) the perimeter radius ρ,

(iv) the engagement range of the turret rt, (v) the angular rate of the turret ω, and

(vi) the capture radius of the vehicle rc.

2. Necessary condition: We establish a necessary condition on the existence of any

c-competitive algorithm for any finite c. This condition serves as a fundamental limit

to this problem and identifies regimes for the six problem parameters in which this

problem does not admit an effective online algorithm.

3. Algorithm Design and Analysis: We design and analyze four classes of cooperative

algorithms with provably finite competitive ratios under specific parameter regimes.

Specifically, the first two cooperative algorithms are provably 1-competitive, the third

cooperative algorithm exhibits a finite competitive ratio which depends on the problem

parameters and finally, the fourth algorithm is 1.5-competitive.

Additionally, through multiple parameter regime plots, we shed light into the relative com-

parison and the effectiveness of our algorithms. We also provide a brief discussion on the

time complexity of our algorithms and how this work can be extended to other models of

the vehicle.

This chapter is organized as follows. In Section 5.1, we formally define the competitive

ratio and our problem. Section 5.2 establishes the necessary conditions, Section 5.3 presents

the algorithms and their analysis. Section 5.4 provides several numerical insights and Section

5.5 discusses the time complexity of our algorithms and provides insights into how this work

116

Figure 5.1 Problem Description. The vehicle is depicted by a blue dot and the blue circle
around the vehicle depicts the capture circle. The direction of the vehicle is shown by the
blue arrow. The yellow arrow depicts the turret (located at the origin of E(θ)) and the
blue dashed curve denotes the engagement range of the turret. The green curve denotes the
perimeter and the red dots denote the intruders. Note that the intruder that the turret is
pointing to (black dashed line) is not captured unless it is within the engagement range of
the turret (blue dashed curve).

extends to defenders with different motion models. Finally, Section 5.6 summarizes the

chapter and outlines future directions.

5.1 Problem Formulation

Consider a planar conical environment (Figure 5.1) described by E(θ) = {(y, α) : 0 <

y ≤ 1, −θ ≤ α ≤ θ}, where (y, α) denotes a location in polar coordinates. The environment

has two endpoints, (1, θ) and (1, −θ). The environment contains a concentric and coaxial

region, R, described by a set of points (z, α) in polar coordinates, where 0 < z ≤ ρ and

α ∈ [−θ, θ]. Mathematically, R(ρ, θ) = {(z, α) : 0 < z ≤ ρ < 1, −θ ≤ α ≤ θ} for some

ρ ∈ (0, 1). Analogous to the environment, R’s endpoints are (ρ, θ) and (ρ, −θ). Arbitrary

numbers of intruders are released at the circumference of the environment, i.e., y = 1, at

arbitrary time instants. Upon release, each intruder moves radially inward with a fixed speed

v > 01 toward the perimeter ∂R(θ) = {(ρ, α) : −θ ≤ α ≤ θ}. Mathematically, if the ith

intruder is released at time ti, then its location is represented by a constant angle θi and

its distance zt

i = 1 − v(t − ti), ∀t ∈ [ti, ti + (1 − ρ)/v]. Two
defenders are employed to defend the perimeter, ∂R of the region R(ρ, θ): a turret located

i from the origin satisfying zt

1As the speed of the intruders is normalized by the speed of the vehicle, we use the speed of intruders

and speed ratio interchangeably.

117

at the origin of E(θ) and a vehicle, both with simple motion dynamics. The vehicle has a

finite capture radius, rc > 0 and can either move with unit speed or remain stationary. The

turret has a finite engagement range, rt such that ρ ≤ rt ≤ 1, and can either turn clockwise

or anti-clockwise with an angular speed of at most ω or remain stationary. We consider

that the vehicle’s capture radius is sufficiently small, in particular, rc < min{ρ, ρ tan(θ)}.

Otherwise, this problem becomes trivial (refer to [10]).

Intruder i, located at (zt

i , θi), is said to be captured at time instant t if either one of the

following holds:

• intruder i is inside or on the capture circle of the mobile vehicle at time t, or

• intruder i is at most rt distance away from the origin and γt = θi holds, where γt

denotes the heading of the turret at time instant t.

The intruder is said to be lost by the defenders if it reaches the perimeter without getting

captured. The intruder is removed from E(θ) if it is either captured or lost. We assume that

the turret and the vehicle neutralize an intruder instantaneously, i.e., they do not require

any additional service time. This implies that the defenders do not need to stop to complete

the capture of an intruder. We further assume that the turret can start and stop firing

instantaneously. This implies that the turret does not neutralize the vehicle in case the

turret’s heading angle is the same as the vehicle’s angular coordinate at a particular time

instant. It is worth highlighting that this work extends, with minor modifications, to when

turret requires a finite amount of spool-up time. We refer to [7] where a turret with finite

service time is considered.

A problem instance P is characterized by six parameters: (1) the speed of the intruders

v > 0, (2) the perimeter radius 0 < ρ < 1, (3) the angle 0 < θ ≤ π that defines the size

of the environment as well as the perimeter, (4) the capture radius of the vehicle 0 < rc <

min{ρ, ρ tan(θ)}, (5) the angular speed of the turret ω > 0, and (6) the range of the turret

ρ ≤ rt ≤ 1. We now formally define the objective of this work.

118

Problem Statement: Design online deterministic cooperative algorithms with finite

competitive ratios for the defenders and establish fundamental guarantees on the existence of

online algorithms with finite competitive ratio.

We start by determining a fundamental limit on the existence of c-competitive algorithms

followed by designing online cooperative algorithms.

5.2 Fundamental Limits

We start by defining a partition of the environment. A partition of E(θ) is a collection

of q ≥ 1 cones W = {W1, W2, . . . , Wq} with disjoint interiors and whose union is E(θ).

Additionally, each cone is of unit radius having a finite positive angle and is concentric with

the environment. We refer to a cone Wm, 1 ≤ i ≤ q as the mth dominance region. Further, an

endpoint of a dominance region is defined analogously as the endpoints of the environment.

Given any set of initial locations of the defenders with distinct angular coordinates, the

environment E(θ) can be partitioned into two dominance regions such that each dominance

region corresponds to a particular defender. We denote the portion of the perimeter contained

in dominance region m, 1 ≤ m ≤ 2, as ∂Rm. Without loss of generality, we assume that ∂R1

(resp. ∂R2) corresponds to the leftmost (resp. rightmost) dominance region.

Let f1(α) = 2ω(ρ sin(α)−rc)+α−θ and f2(α) = 0.5ω(ρ sin(α)−rc)+α−θ. Let ¯α1 denote

the solution to the equation f1(α) = 0 and let ¯α2 be the solution to the equation f2(α) = 0.

Then, following result establishes a necessary condition on the existence of c-competitive

algorithms.

Theorem 50 (Necessary Condition for finite cA(P)). Let




θ − 2ω(ρ − rc),

if θ ≥ π

α∗
1 =

otherwise,

¯α1,

2 + 2ω(ρ − rc),

and let





α∗
2 =

θ − 0.5ω(ρ − rc),

if θ ≥ π

2 + ω(ρ − rc),

¯α2,

otherwise.

119

Then, for any problem instance P such that ρ sin(θ) > rc and v

ω >

there does not exist a c-competitive algorithm for any finite c.

min{θ−α∗

1−ρ
1,2(θ−α∗

2)} holds,

Proof. Recall from Definition 1 that any online algorithm A is c-competitive if the condition

nO(I, P) ≤ cnA(I, P) holds for every input sequence I. Thus, the aim is to construct

an input sequence I such that the condition nO(I, P) ≤ cnA(I, P) does not hold for any

constant c ≥ 1 regardless of which online algorithm is used. The proof is in three parts. First,

we construct an input sequence I. Then, we determine the best locations for the defenders

against such an input sequence. Finally, we evaluate the performance of any online algorithm

A on the input I as well as the performance of the optimal offline algorithm O on the same

input sequence I, to establish the result. Without loss of generality, we assume that both

A and O have the vehicle at the origin at time instant 0 and the turret at angle γ0 = 0.

Let I = {I1, I2} denote a set of two input sequences. Each input sequence Il ∈ I, where

l ∈ {1, 2}, differs in the location of the arrival of intruders. Both input sequences I1 and I2

start at time instant max{1, θ
ω

} and consists of a stream of intruders, i.e., a sequence of a

single intruder arriving at location (1, θ) at every time instant max{1, θ
ω

v , k ∈ N ∪ {0}
until time instant t. The time instant t ≥ 0 corresponds to the time instant when either

} + k 1−ρ

the vehicle or the turret, following any online algorithm A, captures an intruder from the

stream. A burst of c + 1 intruders are then released at time instant t. The location where the

burst of intruder arrives is different for each input sequence Il, l ∈ {1, 2}. Given the location

of the turret and the vehicle at time instant t, there can be at most two dominance regions

of the environment and thus, at most two locations where the burst of intruders can arrive.

These locations have the same angular coordinate as the endpoints of each ∂Rm, ∀m ∈ {1, 2}

excluding θ and including −θ. Without loss of generality, the burst of intruders are released

at location (1, −θ) for I1. Further, if the heading angle of the turret is the same as the

angular coordinate of the vehicle at time t, i.e., γt = θ and the vehicle’s angular coordinate

is θ at time instant t, then the burst intruders arrive at (1, −θ) for both I1 and I2 (In this

case, I1 is same as I2).

120

(a) Vehicle captures the ith intruder (Case 1).

(b) Turret captures the ith intruder (Case 2).

Figure 5.2 Depiction the perimeter depicting the two cases under which one of the defenders
captures the ith intruder.

If neither the vehicle nor the turret captures the stream intruder, the stream never ends

and the result follows as the optimal offline algorithm O can have its vehicle move, at time

instant 0, to location (ρ, θ) and capture all stream intruders. Thus, in the remainder of

the proof, we only consider online algorithms A for which either the vehicle or the turret

captures at least one stream intruder at time instant t. Since the stream intruders arrive

every 1−ρ
v

time units apart and stops when an intruder from the stream is captured, it follows

that no online algorithm can capture more than one intruder from the stream. Thus, we

assume that the ith stream intruder was captured at time instant t, for some i ∈ Z+, where

Z+ denotes the set of positive integers.

We now determine the best locations, or equivalently the dominance regions of the envi-

ronment, for the turret and the vehicle at time instant t. Note that the heading angle of the

turret must not be equal to the angular coordinate of the vehicle at time instant t. This is

because in such case, the burst arrives at (1, −θ) and thus, there always exist a location closer

to angle −θ such that the vehicle or the turret can reach angular location −θ in less time.

This implies that at time instant t, the environment consists of two dominance regions, each

of which contains a defender. We denote the dominance region which contains the vehicle

(resp. turret) as WVeh (resp. WTur) and determine them in the following two cases. These

two cases arise based on whether the vehicle or the turret captures the ith intruder, each of

which is considered below.

121

Case 1 (Vehicle captures the ith intruder): Let 2α1 and 2β1 be the angles of WVeh

and WTur, respectively. The best location for the vehicle and the turret in this case can be

summarized as follows. The vehicle must be located on the line joining the two endpoints

of the perimeter within its WVeh (∂R2) only if 2α1 < π. Otherwise, vehicle must be located

on the line joining the origin to the location (1, θ). In both cases, it must be at a distance

r from location (ρ, θ). The angle of the turret must be equal to the angle bisector of 2β1

(Figure 5.2a). Finally, the time taken by the vehicle to reach the other endpoint of ∂R2

must be equal to the time taken by the turret to turn to the same angle corresponding to

that location. This is denoted mathematically as




θ − α1
ω

2(ρ sin(α1) − rc) if α1 < π
2

2(ρ − rc), otherwise

(5.1)

=



where by definition 2θ = 2α1 + 2β1 and the time taken by the vehicle to capture intruders

at the other endpoint of the perimeter contained in its dominance region is 2(ρ sin(α1) − rc)

(resp. 2(ρ − rc)) when α1 < π

2 (resp. α1 ≥ π

2 ). As θ−α1

ω = 2(ρ − rc) only holds if α1 ≥ π

2 , it

follows that α∗

1 is determined by
solving the equation f1(α) = 0, where f1(α) = 2ω(ρ sin(α) − rc) + α − θ where α1 ∈ [0, π/2).

2 + 2ω(ρ − rc) holds. Otherwise, α∗

1 = θ − 2ω(ρ − rc) if θ ≥ π

We now show that the solution to f1(α) = 0 always exist if rc < ρ sin(θ).

Suppose that α1 = ϵ, where ϵ > 0 is a very small number. Then, f (ϵ) = 2ωρ sin(ϵ) +

ϵ − θ − 2rcω < 0. Now consider that α1 = θ − ϵ for the same ϵ. Then, as ρ sin(θ) > rc, it

follows that f1(θ − ϵ) = 2ω(ρ sin(θ − ϵ) − rc) − ϵ > 0, for a sufficiently small ϵ. This means

that for a sufficiently small ϵ > 0, f1(·) changes its sign in the interval [ϵ, θ − ϵ]. Thus, from

Intermediate Value Theorem and using the fact that f1(α1) is continuous function of α1, it

follows that there must exist an α∗

1 such that f1(α∗

1) = 0 if rc < ρ sin(θ). Further, since f1(α)

is a continuous function and its derivative is strictly increasing for α ∈ [0, π

2 ) and hence,

there exists a unique ¯α1 ∈ [0, π

2 ) which satisfies f1(α).

Case 2 (Turret captures the ith intruder): Similar to Case 1, let 2α2 and 2β2 be

the angles of WVeh and WTur, respectively (Figure 5.2b). As the turret captures the stream

122

intruder, it follows that γt = θ. Further, the vehicle must be located at the midpoint of the

line joining the two endpoints of the perimeter within its dominance region. Finally, the time

taken by the vehicle to reach any endpoint of the perimeter must be equal to the time taken

by the turret to turn to the same angle corresponding to that location. Mathematically, this

yields

2(θ − α2)
ω

=






ρ sin(α2) − rc, if α2 < π
2

(ρ − rc), otherwise

(5.2)

where we used the fact that 2θ = 2α2 +2β2 and ρ sin(α2)−rc (resp. (ρ−rc)) denotes the time

taken by the vehicle to capture intruders at the other endpoint of the perimeter contained

in its dominance region when α2 < π

2 ). As 2 θ−α2
ω = (ρ − rc) only holds if
2 + 0.5ω(ρ − rc) holds. Otherwise, α∗
2
is determined by solving the equation f2(α) = 0, where f2(α) = 0.5ω(ρ sin(α) − rc) + α − θ.

2 (resp. α2 ≥ π
2 = θ − 0.5ω(ρ − rc) if θ ≥ π

2 , it follows that α∗

α2 ≥ π

By following similar steps as in Case 1, it can be shown that a unique solution to f2(α) = 0

always exists if rc < ρ sin(θ) and thus, has been omitted for brevity.
1)−rc), ρ sin(α∗

v < min{2(ρ sin(α∗

As 1−ρ

holds for α∗

1 < π
I1. Further, as 2(ρ sin(α∗

2)−rc} or equivalently v >
2)−rc}
2 , it follows that the vehicle cannot capture the burst intruders from I2 or
holds at time instant t, it

or ρ sin(α∗

2) − rc = 2(θ−α∗
2)

1) − rc) = θ−α∗

1)−rc),ρ sin(α∗

min{2(ρ sin(α∗

1

1−ρ

ω

ω

follows that the turret can also not capture the burst intruders from both I1 and I2. There-

fore, the turret and the vehicle jointly captures at most one intruder from input instance I1

as well as I2. A similar conclusion holds when α∗
1

≥ π

2 or when α∗

2

≥ π
2 .

Thus, we have established that for any online algorithm A, the vehicle and the turret

can jointly capture at most a single intruder from input instance Il, 1 ≤ l ≤ 2. We now

show that the optimal offline algorithm O captures all of the intruders on the input sequence

Il, 1 ≤ l ≤ 2.

Recall that O has complete information at time 0, of when, where, and how many

intruders will arrive. Thus, at time 0, O moves its vehicle to location (ρ, θ) and the turret

to angle −θ. The defenders of O have sufficient time to reach these locations as the first

123

intruder arrives at time instant max{1, θ
ω

} and thus, the capture of all i stream intruders

as well as the burst intruders is guaranteed. Thus, nO(I, P) = i + c + 1 and nA(I, P) = 1
which yields that nO(I,P)

nA(I,P) = i + c + 1. As i + c + 1 > c for any constant c, it follows that

nO(I) ≤ cnA(I) does not hold for any c. This concludes the proof.

Remark 9. Since we do not impose any restriction on the number of intruders that can

arrive in the environment, an adversary can repeat the input sequence designed in the proof

of Theorem 50 any number of times. Thus, the lower bound derived in Theorem 50 holds

asymptotically for when ¯c > 0 in the definition of c-competitive algorithms as well.

We now turn our attention to designing online algorithms and deriving upper bounds on

their competitive ratios. In the next section, we design and analyze four online algorithms,

each with a provably finite competitive ratio in a specified parameter regime.

5.3 Online Algorithms

In this section, we design and analyze online algorithms and characterize the parameter

space in which these algorithms have finite competitive ratios. The parameter space of all

of our algorithms is characterized by two main quantities:

• The time taken by the intruders to reach the perimeter in the worst-case and

• the time taken by the defenders to complete their respective motions.

Intuitively, the parameter space is obtained by comparing the aforementioned quantities2.

Since, the time taken by the intruders is inversely proportional to v, the (v, ρ) parameter

space of our algorithms can be increased by designing the algorithms such that the time

taken by the defenders to complete their motions is the least. We characterize the time

taken by the defenders to complete their respective motions as epochs, which is formally

defined as follows.

An epoch k for an algorithm is defined as the time interval which begins when at least

one of the two defenders moves from its initial position and ends when both defenders return

2the time taken by the defenders must be at most the time taken by the intruders to reach the perimeter

124

(a) Configuration of the defenders for Algo-
rithm SiR for Case 1 (α = θ
). The blue
dot curve depicts the path taken by the ve-
hicle for angular motion.

1+ωzv

(b) Configuration of the defenders for Algo-
rithm SiR for Case 2 (α = arctan(rc/ρ)).

Figure 5.3 Partition (shown by the orange line) of the environment and the configuration of
the defenders for the two cases of Algorithm SiR.

back to their respective initial positions. We denote the time instant when an epoch k begins

as ks.

We now describe our first algorithm that has the best possible competitive ratio.

5.3.1 Sweep within Dominance Region (SiR)

Algorithm SiR is an open-loop and memoryless algorithm in which we constrain the

vehicle to move in an angular motion, i.e., either clockwise or anti-clockwise. This can be

achieved by moving the vehicle with unit speed in the direction perpendicular to its position

vector (see Figure 5.3a). By doing so, we aim to understand the worst-case scenarios for

heterogeneous defenders and gain insights into the effect of the heterogeneity that arises due

to the capture range of the defenders, i.e., the capture circle and the engagement range of the

turret. We say that a defender sweeps in its dominance region if it turns, from its starting

location, either clockwise or anti-clockwise to a specified location and then turns back to

its initial location. Algorithm SiR is formally defined in Algorithm 9 and is summarized as

follows.

Algorithm SiR first partitions the environment E(θ) into two dominance regions and

assigns a single defender to a dominance region. Let 2α denote the angle of WVeh. Then, the

vehicle takes exactly 4αzv to sweep within its dominance region, where zv denotes the radial

125

Algorithm 9: Sweep within Dominance Region Algorithm
1 Turret is at angle −θ
2 if α = arctan( rc

ρ ) then

Vehicle is located at (
for each epoch k do

ρ

cos(α) , θ − α)

Turn the turret clockwise to angle θ − 2α
Turn the turret anti-clockwise to angle −θ

end

7
8 else
9

Vehicle is located at (zv, θ)
for each epoch k do

3

4

5

6

10

11

12

Turn vehicle (resp. turret) anti-clockwise (resp. clockwise) until it reaches
location (resp. angle) (zv, θ − 2α) (resp. θ − 2α)
Turn vehicle (resp. turret) clockwise (resp. anti-clockwise) until it reaches
location (resp. angle) (zv, θ) (resp. −θ)

end

13
14 end

location of the vehicle and will be determined shortly (see Case 1 below). Similarly, as the

turret can only turn either clockwise or anti-clockwise with at most angular speed ω, the
turret takes exactly 4(θ−α)

to sweep in its respective dominance region WTur. Observe that

ω

the environment must be partitioned such that the time taken by the defenders to complete

their motion in their respective dominance region is equal. Otherwise, in the worst-case, all

of the intruders will be concentrated in the dominance region of that defender that takes
more time to sweep its dominance region. Mathematically, this means 4αzv = 4(θ−α)

ω must
. Observe that as ω → ∞, α → 0. This means that

hold which yields that α = θ

1+ωzv

the turret sweeps the entire environment, in time 4θ

ω , if ω is sufficiently high. Recall that

the (v, ρ) parameter space is characterized by the time taken by the defenders to complete

their motion and can be improved by reducing the time taken by the turret, for high values

of ω. In case of very high ω, this can be achieved by having the vehicle remain static at

a specific location while the turret sweeps the remaining environment as opposed to the

entire environment. This means that although angle α = θ

1+ωzv

characterizes the vehicle’s

dominance region, there exists an angle ˆα ≥ α for some values of problem parameters for

126

which we can obtain an improved parameter regime by assigning a dominance region of

angle 2ˆα to the vehicle. Thus, in Algorithm SiR, there are two cases based on the values

of the problem parameters. First, as described above, the defenders sweep the environment

in their respective dominance regions and second, the vehicle remains static at a specific

location while the turret sweeps its dominance region. In what follows, we determine the

location at which the vehicle must remain static for the second case, followed by formally

describing the two cases.

The vehicle’s location must be such that its capture circle covers the perimeter contained

in its dominance region entirely, ensuring that any intruder that arrives in that dominance

region is guaranteed to be captured. To achieve this, the boundary of the dominance re-

gion assigned to the vehicle must be tangent to its capture circle (see Figure 5.3b) which,

through geometry, yields that ˆα = arctan rc

ρ and the location for the vehicle as (

Therefore, the angle of the vehicle’s dominance region is defined as 2α = 2 max{ θ

ρ

cos( ˆα) , θ − ˆα).
, ˆα},

1+ωzv

where ˆα = arctan rc

ρ , and angle α determines if the vehicle sweeps in its dominance region or

remains stationary. We first describe the motion of the turret followed by formally describing

the motion of the vehicle in the two cases.

At time instant 0, the turret is at an angle −θ. The turret turns clockwise, with angular

speed ω towards angle θ − 2α. Upon reaching angle θ − 2α, the turret turns anti-clockwise
towards angle −θ. Note that the turret takes exactly 4(θ−α)

time to complete its motion in

ω

a particular epoch. We now describe the motion of the vehicle which can be summarized in

two cases described as follows:

Case 1 (α = θ

1+ωzv

): At time instant 0, the vehicle is located at (zv, θ), where zv =

min{ρ + rc, 1 − rc} and was determined in [10] and was proved to be optimal in Chapter 4.

The vehicle then moves anti-clockwise with unit speed in the direction perpendicular to its

position vector until it reaches the location (zv, θ − 2α). Then, the vehicle moves clockwise,

with direction perpendicular to its position vector, until it reaches location (zv, θ). Note that

the vehicle takes exactly 4αzv time to complete its motion in a particular epoch. Since α

127

is chosen so that 4αzv = 4(θ−α)

ω

, the vehicle and the turret return to their respective initial

locations at the same time instant, at which the next epoch begins.

Case 2 (α = arctan( rc

cos(α) , θ − α)
and remains stationary at this location for the entire duration. In this case, the next epoch

ρ )): At time instant 0, the vehicle is located at (zv = ρ

begins once the turret turns back to angle −θ.

The following result characterizes the parameter regime in which Algorithm SiR is 1-

competitive.

Theorem 51. Algorithm SiR is 1-competitive for a set of problem parameters that satisfy





v ≤

{

min

(rt−ρ)(1+ωzv)
4θzv

, (zv+rc−ρ)(1+ωzv)
4θzv

}

if

θ
1+ωzv

> arctan(rc/ρ)

(rt−ρ)ω
4(θ−arctan(rc/ρ)) ,

otherwise.

Otherwise, Algorithm SiR is not c-competitive for any constant c.

Proof. Suppose that α = θ

1+ωzv

. At the start of any epoch k, i.e., at time instant ks,

we assume that, in the worst-case, intruders i1 and i2 are located at (zv + rc + ϵ1, θ) and

(rt + ϵ2, −θ), respectively, where ϵ1 and ϵ2 are arbitrary small positive numbers (see Figure

5.3a). To ensure that the vehicle (resp. turret) does not lose any intruder, we require that the

time taken by the vehicle (resp. turret) to return to location (resp. angle) (zv, θ) (resp. −θ)

must be less than the time taken by intruder i1 (resp. i2) to reach the perimeter. Formally,

rt+ϵ2−ρ
v

≥ 4(θ−α)
ω

and zv+rc+ϵ1−ρ

v

≥ 4αzv must hold. Given the first condition on v, these two

conditions always hold, so any intruder that arrives in the environment is guaranteed to be

captured.

If v > min{ (rt−ρ)(1+ωzv)

4θzv

, (zv+rc−ρ)(1+ωzv)
4θzv

}, then there exists an input instance with intrud-

ers arriving only at (1, −θ) such that these intruders are located at (rt + ϵ, −θ) at the time
instant the turret turns from angle −θ. As v > min{ (rt−ρ)(1+ωzv)

} all of these

, (zv+rc−ρ)(1+ωzv)
4θzv

4θzv

intruders will be lost and thus, from Definition 1, Algorithm SiR will not be c-competitive.

Now consider that α = arctan( rc

ρ ). As the vehicle remains stationary in its dominance re-

gion and the location of the vehicle is such that no intruder that is released in that dominance

128

region can reach the perimeter, we only focus on the turret. Assume that, in the worst-case,

intruder i1 is located at (rt +ϵ, −θ) where ϵ is an arbitrary small positive numbers. To ensure

that the turret does not lose any intruder, we require that the time taken by the turret to

return to angle −θ must be less than the time taken by intruder i1 perimeter. Given the

second condition on v, i.e., v ≤

(rt−ρ)ω

4(θ−arctan(rc/ρ)) holds, it is ensured that intruder i1 will be

captured. The proof for Algorithm SiR not being c-competitive when v >

(rt−ρ)ω
4(θ−arctan(rc/ρ))

is analogous to the previous case and has been omitted for brevity. This concludes the

proof.

Although Algorithm SiR is 1-competitive, note that for rt = ρ, the algorithm is not

effective as Theorem 51 yields v ≤ 0. However, by allowing the vehicle to sweep the entire

environment, it is still possible to capture all intruders for some small v > 0. This is

addressed in a similar algorithm below.

5.3.2 Sweep in Conjunction (SiCon)

At time instant 0, the turret is at angle θ and the vehicle is located at location (min{rt +

rc, 1}, θ). The idea is to move the two defenders together in angular motion. Thus, the

vehicle moves anti-clockwise with unit speed in the direction perpendicular to its position

vector until it reaches the location (min{rt + rc, 1}, −θ). Similarly, the turret turns anti-

clockwise, in conjunction with the vehicle, to angle −θ. Upon reaching −θ, the vehicle and

the turret move clockwise until they reach angle θ. The defenders then begin the next epoch.

As the two defenders move in conjunction, 2θ

ω = 2θ min{rt + rc, 1} ⇒ ω =

1

min{rt+rc,1} must

hold. Thus, this algorithm is effective for ω ≥

1

min{rt+rc,1} by turning the turret exactly with

angular speed

1
min{rt+rc,1} .

The following result establishes that Algorithm SiCon is 1-competitive for specific pa-

rameter regimes.

Theorem 52. Algorithm SiCon is 1-competitive for a set of problem parameters which satisfy

ω ≥

min{rt+rc,1} and v ≤ min{rt+2rc,1}−ρ

1

4θ min{rt+rc,1} . Otherwise, it is not c-competitive for any constant

129

c.

Proof. As the proof is analogous to the proof of Theorem 51, we only provide an outline of
this proof for brevity. In the worst-case, an intruder requires exactly min{rt+2rc,1}−ρ

time to

v

reach the perimeter whereas, the defenders synchronously require 4θ min{rt + rc, 1} time to

complete their motion in any epoch. Thus, as the time taken by the defenders must be at

most the time taken by the intruders we obtain the competitive ratio. If the condition on v

does not hold, then by constructing an input analogous to the input in the proof of Theorem

51, it can be shown that Algorithm SiCon is not c-competitive.

Remark 10 (Maneuvering Intruders). As the analysis of the fundamental limit (Theorem

50), Algorithm SiR, and Algorithm SiCon are independent of the nature of motion of the

intruders, the results of Theorem 50, Algorithm SiR, and Algorithm SiCon apply directly to

the case of maneuvering or evading intruders.

Recall that in Algorithm SiR, the idea was to partition the environment and assign a

single defender in each dominance region. By doing so, we obtain valuable insight into the

parameter regime wherein we are guaranteed to capture all intruders. However, we refrain

from designing such algorithms in this work due to the following two reasons. First, such

an algorithm requires that ratio of intruders captured by a defender to the total number of

intruders that arrived in that corresponding defender’s dominance region is equal for both

defenders. Otherwise, since the adversary has the information of the entire algorithm, it

will release more intruders in the dominance region of the defender that has the lower ratio,

which determines the competitive ratio of such an algorithm. Second, such algorithms are

not cooperative and thus fall out of the scope of this chapter. The objective of this work

is to study how heterogeneous defenders can be used to improve the competitive ratio of a

single defender. Thus, in the next algorithm, although we partition the environment, the

defenders are not restricted to remain within their own dominance region.

130

(a) Description of sets P k
left
and P k
right. The grey dashed
line denotes the splitting of
the
into two
halves. The turret’s heading
angle is 0 at time instant ks.

environment

(b) Splitting the environment
into N = 3 sectors. The blue
dashed circles denote the rest-
ing points for the vehicle.

(c) Final representation of the
for Algorithm
environment
Split.
The black dashed
curves denotes splitting of the
radially corre-
environment
sponding to the three time in-
tervals, each of length D. The
vehicle is located at (x1, ϕ1).

Figure 5.4 Description of Algorithm Split. The light grey dashed line denotes the splitting
of the environment. The region between the yellow dashed curve and the yellow dot curve
on the left (resp. right) side of the grey line denotes P k
right). The blue dashed
straight lines denote the sectors (N = 3) of the environment and the black dashed curves
denotes the three radial intervals of length Dv each. The vehicle is located at (x1, ϕ1).

left (resp. P k

5.3.3 Split and Capture (Split)

The motivation for this algorithm is to utilize the vehicle’s ability to move in any direction

while the turret rotates either clockwise or anti-clockwise. Since the turret can only turn

either clockwise or anti-clockwise, the idea is to first partition the environment into two

halves and turn the turret towards the side which has higher number of intruders. By

doing so, we hope to capture at least half of the intruders by the turret, assuming they

are sufficiently slow, that arrive in every epoch. Further, while the turret moves to capture

intruders on one side, the vehicle moves to the other side to capture intruders, ensuring that

the defenders jointly capture more than half of the intruders that arrive in the environment

in every epoch. Algorithm Split is formally defined in Algorithm 10 and is summarized as

follows where we first describe the motion of the turret in every epoch followed by that of

the vehicle.

The heading angle of the turret is always γks = 0 at the start of every epoch k. To

determine whether the turret turns clockwise or anti-clockwise at time instant ks, we first

131

Algorithm 10: Split and Capture Algorithm
1 Turret is at angle 0
2 Vehicle is located at (x1, ϕ1)
3 for each epoch k do
Compute P k
4
if |P k

left, Np∗
| then

right, P k

| ≥ |P k
left

right
Turn the turret clockwise to angle θ
Turn the turret back to angle 0

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

else

Turn the turret anti-clockwise to angle −θ
Turn the turret back to angle 0

end
if Np∗ ̸= Ni then
p∗
Move vehicle to (xp∗, ϕp∗) and then capture intruders in Sj+2
p∗

| ≥ |Sj+1

if |Sj+2

| then

i

else

Keep vehicle at (xi, ϕi) and capture intruders in Sj+1
Move vehicle to (xp∗, ϕp∗)

i

end

else

Keep vehicle at (xi, ϕi) and capture intruders in Sj+1

i

and Sj+2

i

end

21
22 end

describe two sets Pleft and Pright. These sets characterize a region on the left and right side

of the y-axis, respectively, and are determined once at time instant 0.

βv
Pright(ρ, v) := {(y, β) : ρ +
ω
Pleft(ρ, v) := {(y, β) : ρ − βv
ω

< y ≤ min{1, rt +

< y ≤ min{1, rt +

(2θ − β)v
ω
(2θ + β)v
ω

},

},

∀β ∈ [0, θ]},

∀β ∈ (0, −θ]}.

Let P k

right and P k

left denote the set of intruders contained in Pright and Pleft (see Figure

5.4a), respectively, at the start of an epoch k and let |S| denote the cardinality of a set S

of intruders. Then, at time instant ks, the turret compares the total number of intruders

in P k

left to the total number of intruders in P k

has higher number of intruders. More formally, if |P k

right and turns in the direction of the set which
| holds at time instant ks,

| < |P k
left

right

then the turret turns anti-clockwise towards angle −θ. Upon turning to angle −θ, the turret

turns to angle 0. Otherwise, i.e., if |P k

right

| ≥ |P k
left

| holds at time instant ks, then the turret

132

turns clockwise towards angle θ. Upon turning to angle θ, the turret turns to angle 0. As the

turret’s dominance region is determined at the start of every epoch k, we denote the turret’s

dominance region as W k

Tur, and consequently, the other dominance region as the vehicle’s

dominance region denoted as W k

Veh. We now characterize the motion for the vehicle which

builds upon the SNP algorithm designed in [10].

Algorithm Split further divides the environment E(θ) into N = ⌈ θ
θs

⌉ sectors, where 2θs =

2 arctan(rc/ρ) denotes the angle of each sector (see Figure 5.4b). The value 2 arctan(rc/ρ)

of the angle of each of the sectors is to ensure that the portion of the perimeter in a sector

can be completely contained in the capture circle of the vehicle. This can be achieved by

positioning the vehicle at resting points (see Figure 5.4b), which is a specific location in every

sector and is formally defined as follows.

Definition 53 (Resting points). Let Nl denote the lth sector, for every l ∈ {1, . . . , N }

where N1 corresponds to the leftmost sector. Then, a resting point (xl, ϕl) ∈ E(θ) for sector

Nl, is the location for the vehicle such that when positioned at (xl, ϕl), the portion of the

perimeter inside sector Nl is contained completely within the capture circle of the vehicle.

Mathematically, this is equivalent to

(

(xl, ϕl) =

ρ

cos(θs) , (l − N +1

2 )2θs

)

.

Now, let D denote the distance between the two resting points that are furthest apart in

the environment (see Figure 5.4b). Formally,





D =

2

ρ

cos(θs) sin((N − 1)θs), if (N − 1)θs < π
2 ,

2

ρ

cos(θs) , otherwise.

(5.3)

Observe that when N = 1, D = 0. This means that the vehicle captures all intruders that

arrive in the environment by positioning itself to the unique resting point of the single sector.

Next, Algorithm Split radially divides the environment E(θ) into three intervals of length

Dv, corresponding to time intervals of time length D each (see Figure 5.4c). Specifically, the

133

jth time interval for any j > 0 is defined as the time interval [(j − 1)D, jD]. Note that this

time interval is different than the epoch of the algorithm. Let Sj

l denote the set of intruders
that are contained in the lth, l ∈ {1, . . . , N }, sector and were released in the jth interval.

Then, at the start of each epoch, the motion of the vehicle is based on the following two

steps: First, select a sector with the maximum number of intruders. Second, determine if

it is beneficial to switch over to that sector. These two steps are achieved by two simple

comparisons; C1 and C2 detailed below.

Suppose that the vehicle is located at the resting point of sector Ni at the start of the

jth epoch and let ˜N denote the set of sectors in the vehicle’s dominance region. Then, to

specify the first comparison C1, we associate each sector Nl ∈ ˜N with the quantity





≜

ηl
i

|Sj+2
l

| + |Sj+3

l

|,

if l ̸= i,

|Sj+1
i

| + |Sj+2

i

| + |Sj+3

i

|,

if l = i.

Note that ηl

i is not defined for every sector in the environment. Instead, it is only defined

for the sectors in W j

Veh, which may contain Ni. Then, as the outcome of C1, Algorithm

Split selects the sector Np∗, where p∗ = arg maxp∈ ˜N ηp

i . In case there are multiple sectors

with same number of intruders, then Algorithm Split breaks the tie as follows. If the tie

includes the current sector Ni (which is only possible if Ni ∈ ˜N holds3), then Algorithm Split
selects Ni. Otherwise, Algorithm Split selects the sector contained in ˜N with the maximum
|, where ˆN denotes the
number of intruders in the interval j + 2, i.e., p∗ = arg maxp∈ ˆN

|Sj+2
p

set of sectors that have the same number of intruders. If this results in another tie, then this

second tie is resolved by selecting the sector with the least index. Let the sector chosen as

the outcome of C1 be No.

We now describe comparison C2 jointly with the motion of the vehicle in the following

two points:

3This case arises when the Algorithm Split moves the turret in the same direction for at least two

consecutive epochs.

134

• If the sector chosen as the outcome of C1 is No, o ̸= i, and the total number of intruders

in the set Sj+2

o

is no less than the total number of intruders in Sj+1

i

, then Algorithm

Split moves the vehicle to (xo, ϕo) arriving in at most D time units. Then the vehicle

waits at that location to capture all intruders in Sj+2

o

. Otherwise, i.e., the total number

of intruders in Sj+2

o

is less than Sj+1

i

, then the vehicle stays at (xi, ϕi) until it captures

all of the intruders in Sj+1

i

and then moves to (xo, ϕo) arriving in at most D time units.

• If the sector chosen is Ni, then the vehicle stays at its current location (xi, ϕi) for 2D

time units capturing intruders in Sj+1

i

and Sj+2

i

.

Note that the vehicle takes at most 2D time units in every case above. The vehicle then re-

evaluates after 2D time. At time instant 0, the turret’s heading angle is 0 and the vehicle is

located at (x1, ϕ1). The first epoch begins when the first intruder arrives in the environment.

We now describe two key requirements for the algorithm. The first requirement is to

ensure that the defenders start their individual motion in an epoch at the same time instant.

Recall that the turret requires exactly 2θ

ω time to turn from its initial heading angle to either
θ or −θ, at time instant ks, and turn back to its initial heading angle. On the other hand,

the vehicle requires 2D time units to capture intruders in at least one interval. Thus, to

ensure that the defenders begin their motion at the same time instant, 2θ

ω = 2D must hold,

i.e., the speed of the turret must be at least θ

D . The second requirement is to ensure that

the algorithm has a finite competitive ratio. This is achieved by ensuring that any intruder

that was not accounted for comparison by the defenders ( for instance intruders that are

not in P k

left or P k

right) in an epoch k, are accounted in epoch k + 1. Our next result formally

characterizes this requirement for the turret.

Lemma 54. Any intruder with radial coordinate greater than min{1, rt+ (2θ−β)v
(resp. min{1, rt + (2θ+β)v

}, ∀β ∈ (0, −θ]) at time instant ks will be contained in the set P k+1
right

}, ∀β ∈ [0, θ]

ω

ω

(resp. P k+1

left ) at time instant (k + 1)s if v ≤ ω(rt−ρ)

2θ

holds.

135

Proof. Without loss of generality, suppose that |P k
left

| ≤ |P k

right

| holds at time instant ks.

Then, the total time taken by the turret to move towards θ and turn back to angle 0 is 2θ
ω .

In order for any intruder i to not be captured in epoch k, in the worst-case, the intruder i

must be located at (min{1, rt} + ϵ, θ), where ϵ is a very small positive number, by the time

the turret reaches angle θ. Note that 1 + ϵ here means that the intruder is released after ϵ

time at location (1, θ) after the turret’s heading angle is θ. Thus, in order to ensure that

i can be captured in epoch k + 1, the condition 2θ
ω

≤ rt−ρ

v must hold, where we used the

fact that rt ≤ 1 and ϵ is a very small positive number. From the definition of P k

right, if the

intruder i can be captured in epoch k + 1, then it follows that the intruder i was contained

in the set P k+1

right at the start of epoch (k + 1). This concludes the proof.

The proof of Lemma 54 is established in the worst-case scenario which is that an intruder,

with angular coordinate θ (resp. −θ), is located just above the range of the turret at the

time instant when the turret’s heading angle is θ (resp. −θ) in an epoch k. This is because

in an epoch k, the angle θ is the only heading angle that is visited by the turret once as the

turret turns to all angles β ∀β ∈ [0, θ) (resp. [0, −θ)) twice; once when the turret turns to

angle θ or −θ and second, when turret turns back to angle 0. This results in the following

corollary, the proof of which is analogous to the proof of Lemma 54.

Corollary 9. Suppose that the turret moves to capture intruders in P k

left (resp. P k

right) in

epoch k. Then, the intruders contained in P k

right (resp. P k

left) with radial coordinate strictly

greater than rt + vθ
(k + 1) if v ≤ ω(rt−ρ)

.

2θ

ω at time instant ks will be considered for comparison at the start of epoch

Recall that the adversary selects the release times and the locations of the intruders in

our setup. Thus, with the information of the online algorithm, the adversary can release

intruders such that all the intruders have their radial coordinates at most rt + θv

ω and at
angular location θ of −θ at the start of every epoch k, which is considered to be the worst-

case scenario. This ensures that if the turret selects to turn towards −θ (resp. θ) at time

136

ks, then the turret cannot capture any intruder that was contained in P k

right (resp. P k

left) in

epoch k + 1. As the idea is to have the vehicle capture these intruders, we require that the

intruders must be sufficiently slow. This is explained in greater detail as follows.

For the vehicle, the requirement is that the intruders take at least 3D time units to reach

the perimeter. This is to ensure that the vehicle can account for intruders that are very

close to the perimeter at the start of an epoch. From Corollary 9, as the intruders with

radial location greater than rt + θv

ω are counted for comparison in next epoch by the turret,

we require that these intruders must also be counted by the vehicle in the next epoch. This
yields that 3D ≤ min{1,rt+

}−ρ

θv
ω

2D must hold,

which implies that either v ≤ 1−ρ

3D or v ≤ rt−ρ
ω = 2D. Finally, as Lemma 54 requires that v ≤ rt−ρ

where we used the fact that 2θ

v

the second requirement for Algorithm Split is that v ≤ min{ 1−ρ

3D , rt−ρ

2D

2D must hold,
}. We now establish

the competitive ratio of Algorithm Split.

Theorem 55. Let θs = arctan(rc/ρ) and N = ⌈ θ
θs

⌉. Then, for any problem instance P with

the turret’s angular velocity ω ≥ θ
3⌊0.5N ⌋+2 -competitive if v ≤ min{ 1−ρ
3N −1

D , where D is defined in equation 5.3, Algorithm Split is
3D , rt−ρ

}.

2D

Proof. First observe that although the turret can capture intruders from one half of the

environment, the vehicle only captures at most two intervals out of all intervals that are in

W k

Veh (the total number of intervals in W k

Veh will be determined shortly). Thus, in the worst-

case, the intruders are released in the environment such that there are as many intruders

possible in the vehicle’s dominance region. Since W k

Veh is selected based on W k

Tur, there

cannot be more number of intruders in the vehicle’s dominance region as than those in the

turret’s dominance region. This implies that there are equal number of intruders in each

dominance region in every epoch in the worst-case. We now characterize the total number

of intervals in the vehicle’s dominance region.

If N is even then, the vehicle’s dominance region contains N

2 sectors and 3 N

2 intervals

due to the three intervals of length Dv each. Otherwise, the total number of intervals in the

vehicle’s dominance region is 3⌈ N
2

⌉. The explanation is as follows. Observe that, for odd

137

N , the sector in the middle is contained in the turret’s as well as the vehicle’s dominance

region. As the portion of the middle sector which is contained in the vehicle’s dominance

region may contain intruders and from the fact that the number of intervals must be an

integer, we obtain that there are 3⌈ N
2

⌉ intervals in the vehicle’s dominance region. Since the

total number of intervals in the environment is 3N , this implies that the turret’s dominance

region has 3N − 3⌈ N
2

⌉ = 3⌊ N
2

⌋ intervals and not 3⌈ N
2

⌉ intervals as we already accounted for

the portion of the middle sector contained in the turret’s dominance region in the vehicle’s

dominance region by using the ceil function. Intuitively, this means that there is no benefit

for the adversary to release intruders in the portion of the middle sector contained in the

turret’s dominance region as the turret captures all intruders in its dominance region in an

epoch. Thus, the adversary can have all intruders in a single interval within the turret’s

dominance region and the number of intruders that the turret capture remain the same,

which is not the case in the vehicle’s dominance region. We now account for the number of

intruders jointly captured by the defenders in any epoch k.

Since at the start of every epoch k, the turret selects a dominance region based on the

number of intruders on either side of the turret, it follows that the turret captures at least

half of the total number of intruders that arrive in epoch k. This means that the turret

captures intruders in all 3⌊ N
2

⌋ intervals. The number of intruders captured by the vehicle

in an epoch k is determined as follows. Recall that in Algorithm Split, the vehicle’s motion

is independent of the turret’s motion. The only information exchange that is required is

the dominance region selected by the turret at the start of each epoch, which governs the

number of sectors that the vehicle must account intruders in. Hence, this part of the analysis

of accounting the number of intruders captured by the vehicle is identical to the proof of

Lemma IV.5 in [10], so we only give an outline of the proof. From the fact that the vehicle’s

dominance region can have at most 3⌈ N
2

⌉ intervals and by following similar steps as in proof

of Lemma IV.5 from [10], it follows that for every two consecutively captured intervals, the

vehicle loses at most 3⌈0.5N ⌉−3 intervals. Further, from Lemma 54 and by following similar

138

steps as in the proof of Lemma IV.6 from [10], it follows that every lost interval is accounted

for by the captured intervals of the turret and the vehicle. Thus, we obtain that the turret

and the vehicle jointly capture at least 2 + 3⌊0.5N ⌋ intervals of intruders and lose at most

3⌈0.5N ⌉ − 3 intruders in every epoch of Algorithm Split. Therefore, by assuming that there

exists an optimal offline algorithm that can capture all 2 + 3⌊0.5N ⌋ + 3⌈0.5N ⌉ − 3 = 3N − 1

intruder intervals in every epoch establishes that Algorithm Split is

3N −1

3⌊0.5N ⌋+2 -competitive.

This concludes the proof.

Recall that the motion of the vehicle in Algorithm Split builds upon Algorithm SNP

designed in [10], which was shown to be 3N −1

2

-competitive. A major drawback of Algorithm

SNP was that its competitive ratio increases linearly with the number of sectors N . The

following remark highlights that Algorithm Split does not suffer from this drawback and is

effective under the same parameter regime as Algorithm SNP.

Remark 11 (Heterogeneity improves competitive ratio of Algorithm Split). The competitive

ratio of Algorithm Split is at most 2, achieved when N → ∞. Further, for rt = 1, the
parameter regime that required by Algorithm Split (v ≤ 1−ρ

3D )) is the same as that of Algorithm

SNP in [10].

Further note that if N is odd and N ̸= 1, then the competitive ratio of Algorithm Split

is higher than that for N + 1. This is because when N is odd, the adversary can exploit

the fact that there are higher number of intervals that the vehicle can lose as compared to

that in the turret’s dominance region. Finally, for N = 2, Algorithm Split is 1-competitive.

The explanation is as follows. For N = 2, the two sectors of the environment overlap the

two dominance region. Thus, in this case, the turret captures all intruders in one dominance

region while the vehicle remains stationary at the resting point of the second dominance

region, ensuring that all intruders that are released in the environment are captured.

Given that the turret can only move clockwise or anti-clockwise and from the requirement

that the defenders must start their motions at the same time instant, the parameter regime of

139

Algorithm 11: Partition and Capture Algorithm
1 Turret’s heading angle is θ
3 .
2 Vehicle is located at (zv, − θ
3 ).
3 for each epoch k ≥ 1 do
Compute |V k
4
left
left and |I k| ≥ T k
if |I k| ≥ V k
| then
| ≥ |T k
if |V k
left
Assign V k

right holds then

|, and |I k|.

|, |T k

right

5

7

6

right
left to the vehicle and I k to the turret.

else

Assign T k

right to the turret and I k to the vehicle.

end

else

| < |I k| (resp. |T k

if |V k
| < |I k|) then
left
Assign I k to the vehicle (resp. turret).

right

else

Assign V k

left (resp. T k

right) to the vehicle (resp. turret).

end

8

9

10

11

12

13

14

15

16

17

18

19
20 end

end
Turn the defenders in an angular motion to the respective endpoint of the
assigned set.
Turn the defenders back to the initial position.

Algorithm Split is primarily defined by the time taken by the turret to sweep its dominance

region. This means that by reducing the time taken by the turret to complete its motion, it

is possible to achieve an algorithm with higher parameter regime. This is exploited in our

next algorithm which is provably 1.5-competitive.

5.3.4 Partition and Capture (Part)

Algorithm Part, formally defined in Algorithm 11, partitions the environment into three

equal dominance region, each of angle 2θ

3 . We denote these dominance regions as W1, W2,

and W3, where W1 denotes the leftmost dominance region. The idea is to move the vehicle

and the turret similar to the motion of the turret in Algorithm Split and capture all intruders

from two out of the three total dominance region in each epoch. The dominance region are

determined as follows.

At the start of every epoch k, the turret’s heading angle, measured from the y-axis, is

140

Figure 5.5 Description of Algorithm Part. The black dashed line denotes the partitioning
of the environment, each of angle 2θ
3 . The region between the orange (resp. yellow) dashed
curve and the orange (resp. yellow) dot curve on the left (resp. right) side of the vehicle
(resp. turret) denotes the Vleft (resp. Tright).

set to θ

3 . Similar to Algorithm Split, we describe two sets for the turret that characterize

specific regions in the two dominance regions that surround the turret, i.e., W2 and W3.

Intuitively, these sets corresponds to the locations in the environment that the turret can

capture intruders at during its sweep motion.

Tright(ρ, v) := {(y, β) : ρ +

Tleft(ρ, v) := {(y, β) : ρ −

3 )v

3 )v

(β − θ
ω
(β − θ
ω

≤ y ≤ min{1, rt +

≤ y ≤ min{1, rt +

( 5θ
3

− β)v
ω
(θ + β)v
ω

}∀β ∈ [

θ
3
}∀β ∈ [− θ
3

, θ]},

,

θ
3

)}.

Similarly, at the start of every epoch k, the vehicle is assumed to be located at (zv, − θ

3 ),

where the angle is measured from the y-axis and zv is as defined for Algorithm SiR. Next,

we define two sets that characterize a specific region in W1 and W2, respectively.
3 + β)zvv ≤ y ≤ min{1, zv + rc + (θ − β)zvv} ∀β ∈ [− θ
3

Vright(ρ, v) := {(y, β) : ρ + ( θ

θ
3

Vleft(ρ, v) := {(y, β) : ρ − (β +

θ
3

)zvv ≤ y ≤ min{1, zv + rc + (

5θ
3

,

]},
+ β)vzv} ∀β ∈ [−θ, − θ
3

)}.

Let T k

right, T k

left,V k

right, and V k

left denote the set of intruders contained in Tright, Tleft, Vright,

and Vleft, respectively, at the start of an epoch k. Finally, denote I k as the set of intruders

contained in V k

right

∩ T k

left (see Figure 5.5).

We now describe the motion of the defenders. The objective is to move the defenders

such that intruders from any two sets out of V k

left, T k

right, and I k are captured. This requires

141

assigning the defenders to the sets containing maximum number of intruders, which can be

summarized into two cases.

Case 1: The set I k contains maximum number of intruders, i.e., |I k| ≥ V k

left and |I k| ≥

T k
right hold at the start of epoch k. This means that one of the defenders must be assigned

to the set I k. By determining which set has more intruders out of V k

left and T k

right, Algorithm

Part performs an assignment of the sets to the defenders. Mathematically, if |I k| ≥ |V k
left

|

and |I k| ≥ |T k

right

|, then

• If |V k
left

| ≥ |T k

right

|, then the vehicle is assigned the set V k

left and the turret is assigned

the set I k.

• Otherwise, the vehicle is assigned the set I k and the turret is assigned the set T k

right.

Case 2: |V k
left

| < |I k| or |T k

right

| < |I k| holds at the start of epoch k. This implies that

at least one set out of V k

left and T k

right has the maximum number of intruders out of the three

left, I k, and T k
V k

right sets. Then, the sets are assigned as follows:

• If |V k
left

| < |I k|, then the vehicle is assigned the set I k. Otherwise, the vehicle is assigned

the set V k

left.

• Similarly, if |T k

right

| < |I k|, then the turret is assigned the set I k. Otherwise, the

turret is assigned the set T k

that means that |I k| ≥ |V k
left

right. Note that if the vehicle is already assigned set I k then
| holds. Given the condition in Case 2, this implies that

|T k

right

| < |I k| holds and the turret is assigned T k

right. Thus, in Case 2, both defenders

are never assigned the set I k.

Once the sets are assigned, the vehicle turns as follows. If the set assigned to the vehicle

is I k, then the vehicle moves clockwise with unit speed in the direction perpendicular to its

position vector until it reaches location (zv, θ

3 ). Upon reaching the location, the vehicle moves

anti-clockwise with unit speed in the direction perpendicular to its position vector until it

returns to location (zv, − θ

3 ). Otherwise (if the vehicle is assigned the set V k

left), the vehicle

142

moves anti-clockwise with unit speed in the direction perpendicular to its position vector

until it reaches location (zv, −θ). Upon reaching that location, the vehicle moves clockwise

with unit speed in the direction perpendicular to its position vector until it returns to location

(zv, − θ

3 ).

Before we describe the motion of the turret, we determine its angular speed to ensure

that the defenders start an epoch at the same time instant. As we require that the defenders

take the same amount of time to return to their starting locations in an epoch, we require

that 4θ

3ω = 4θ

3 zv ⇒ ω = 1

zv

, which means that the angular speed of the turret must be at least

1
zv

.

We now describe the turret’s motion in an epoch. Similar to the motion of the vehicle,

if the set assigned to the turret is I k, then the turret turns to angle − θ

3 and then turns back

to the initial heading angle θ

3 with angular speed 1
zv

. Otherwise, the turret turns to angle θ

and then back to angle θ

3 with angular speed 1
zv

.

Analogous to Lemma 54, we have the following lemma which ensures that any intruder

that was not considered for comparison at the start of epoch k is considered for comparison

at the start of epoch (k + 1).

Lemma 56. Any intruder which lies beyond the sets V k

right and T k

left at the start

of epoch k will be contained in the sets V k+1

left , V k+1

right, T k+1

left , respectively, at the start

right, T k

left, V k
right and T k+1

of epoch (k + 1) if

{

v ≤ min

3(min{1, zv + rc} − ρ)
4θzv

,

3(rt − ρ)
4θzv

}

Proof. The proof is analogous to the proof of Lemma 54 and has been omitted for brevity.

Corollary 10. Any intruder that lies beyond the set I k at time instant ks will be contained

in the set I k+1 at the start of epoch (k + 1) if the conditions of Lemma 56 hold.

Proof. The proof directly follows from the fact that Lemma 56 holds for both the defenders

and I k represents the intersection of V k

right and T k

left.

143

(a) (ω, ρ) plot for θ = π
v = 0.1.

4 , rt = 1, rc = 0.05, and

(b) (ω, ρ) plot for θ = π
v = 0.3.

4 , rt = 1, rc = 0.1, and

Figure 5.6 (ω, ρ) plot for different values of θ, rc, rt, and v. Markers represent a lower bound
on the turret’s angular speed.

Theorem 57. Algorithm Part is 1.5-competitive for any problem instance P with ω ≥ 1
zv

that satisfies

{

v ≤ min

3(min{1, zv + rc} − ρ)
4θzv

,

3(rt − ρ)
4θzv

}

Proof. Observe that from Lemma 56 and Corollary 10, every intruder is accounted for and

no intruder that is not considered for comparison in a particular epoch is lost under the

condition on v. Now, from the definition of Algorithm Part, the defenders are assigned two

sets out of the total three in every epoch. Further, the assignment is carried out in a way that

the sets with maximum number of intruders are assigned to the defenders in every epoch.

Assuming that there exists an optimal offline algorithm that captures all intruders from all

three sets then, from Definition 1, competitive ratio of Algorithm Part is at most 3
2 .

5.4 Numerical Observations

We now provide numerical visualization of the bounds derived in this work and emphasize

on the (v, ρ) and (ω, ρ) parameter regime. These plots allow the defenders to choose an

appropriate online algorithm out of the four proposed, based on the values of the problem

parameters.

144

5.4.1

(ω, ρ) Parameter Regime

Figure 5.6 shows the (ω, ρ) parameter regime plot for fixed values of rt, rc, θ, and v and

provides insights into the requirement of the angular speed ω for different values of the ρ.

Note that the markers represent a lower bound on the angular speed of the turret.

In Figure 5.6a, the condition in Theorem 50 for the existence of c-competitive algorithms

is represented by the green circles. For all values of 0.1 ≤ ρ ≤ 0.9, as the green circles are at

ω = 0.1, it implies that there exists a c-competitive algorithm for all values of ω ≥ 0.1 and the

values of v, rc, rt, and θ selected for this figure. We now provide insights into the requirement

on ω for our algorithms. Algorithm SiR, represented by the yellow triangle, requires higher

angular speed for the turret as the radius of the perimeter increases. However, Algorithm

Split and Algorithm Part, represented by the red star and blue square respectively, require

lower angular speed for the turret when the radius of the perimeter is sufficiently large.

Although counter intuitive, this can be explained as follows. Recall that Algorithm Part

and Algorithm Split require the two defenders to be synchronous and the vehicle moves

with a fixed unit speed. As the perimeter size increases, the time taken by the vehicle to

complete its motion increases, which in turn requires lower values of ω to ensure synchronicity.

Observe that for ρ ≥ 0.8, there are no markers for Algorithm Split. This is because for

ρ ≥ 0.8 and the values of θ, rt, rc and v considered for this figure, the condition defined for

Algorithm Split in Theorem 55 is not satisfied for any 0 < ω ≤ 7, implying that Algorithm

Split is not c-competitive. Analogous conclusions can be drawn for Algorithms SiCon (resp.

Part), represented by orange diamond (resp. blue square), for values of ρ ≥ 0.6 (resp.

ρ ≥ 0.9). Finally, note that Algorithm SiCon requires ω ≥ 1 for all values of ρ ≤ 0.6. This is

because in this algorithm, the turret is required to move with unit angular speed to maintain

synchronicity with the vehicle.

Analogous observations can be drawn in Figure 5.6b. For instance, when ρ = 0.9 and ω ≥

1.5, there always exists a c-competitive algorithm with a finite constant c. Equivalently, there

does not exist a c-competitive algorithm for ω < 1.5, ρ = 0.9 and for the values of rt, rc, θ,

145

(a) (v, ρ) plot for θ = π
0.05.

3 , rt = 0.8, and rc =

(b) (v, ρ) plot for θ = π
0.1.

2 , rt = 0.6, and rc =

Figure 5.7 (v, ρ) plot for different values of θ, rc and rt.

and v selected. Similarly, as ρ increases, Algorithm SiR requires a faster turret whereas

Algorithm Part and Algorithm Split can work with a slower turret to ensure synchronicity.

Note that Algorithm Part and Split do not have markers beyond ρ = 0.6 and ρ = 0.5,

respectively, which is lower than the values of ρ in Figure 5.6a. This implies that, although

the values of rc are slightly higher than those in Figure 5.6a, it is more difficult to capture

intruders given the higher value of v. Finally, there are no markers for Algorithm SiCon as

it is not c-competitive for the values of parameters selected for this figure.

5.4.2

(v, ρ) Parameter Regime

Figure 5.7 shows the (v, ρ) parameter regime for fixed values of θ, rc, and rt. Since Algo-

rithms Split, Part, SiCon require fixed but different values of ω, we set ω = max{

1

min{1,rt+rc} , 1

zv

, θ
D

}.

Note that the value of ω for this figure depends on the value of ρ as zv is a function of ρ.

Figure 5.7a shows the (v, ρ) parameter regime plot with θ, rt, and rc set to π

3 , 0.8, and

0.05, respectively. For any value of parameters ρ and v, for instance 0.7 and 1, respectively,

that lie beyond the green curve, there does not exists a c-competitive algorithm. For any

value of parameters ρ and v that lie below the yellow curve, Algorithm SiR is 1-competitive.

Similarly, for any value of parameters ρ and v that lie below the blue curve, Algorithm Part

is 1.5-competitive. Analogous observations can be made for Algorithm Sicon and Algorithm

146

Split. Note that for parameter regime that lies below the yellow curve, Algorithm Part is

not effective as there exists Algorithm SiR with a better competitive ratio. For instance, for

ρ = 0.2 and v = 0.2, it is better to use Algorithm SiR as it has a lower competitive ratio.

Observe that for very high values of ρ, Algorithm SiCon is the most effective as it has the

highest parameter regime curve. Finally, the light red curve of Algorithm Split is divided

into regions where each region corresponds to a specific competitiveness. An important

characteristic for Algorithm Split is that it can be used to determine the tradeoff between

the competitiveness and the desired parameter regime for a specific problem instance.

Figure 5.7b shows the (v, ρ) parameter regime plot with θ, rt, and rc set to π

2 , 0.6, and

0.1, respectively. Note that the green curve, which represents the curve for Theorem 50,

is shifted slightly upwards as compared to in Figure 5.7a. This follows from the two cases

considered in the proof which is based on the capture capability of the defenders (vehicle is

now more capable and Theorem 50 is independent of rt). As the angle of the environment

increases and the engagement range of the turret decreases, it is harder to capture intruders.

This is visualized in Figure 5.7b as the curves for all the algorithms have shifted downward

compared to those in Figure 5.7a. Finally, for values of ρ > 0.3, Algorithm Part is more

effective than Algorithm Split only if the competitive ratio of Algorithm Split is less than 1.5

for the chosen values of parameters. Similar to the curve of Algorithm Split, note that the

curve for Algorithm SiR is also divided into regions. This is because of the different values

of ω for different perimeter sizes.

5.5 Discussion

In this section, we provide a brief discussion on the time complexity of our algorithms and

how this work extends to different models of the vehicle. We start with the time complexity

of our algorithms.

5.5.1 Time Complexity

We now establish the time complexity of each our algorithms and show that they can

be implemented in real time if the information about the total number of intruders in every

147

epoch is provided to the defenders.

Algorithm SiR and SiCon : Since Algorithm SiR and SiCon are open loop algorithms,

the time complexity is O(1).

Algorithm Split : There are three quantities that must be computed at the start of every

epoch of Algorithm Split, i.e., |P k

right

|, |P k
left

|, and Np∗. Since Np∗ is determined using a max()

function over N sectors, its time complexity is O(N ). Similarly, determining the sets P k

right

and P k

left also have a time complexity of O(n), where n is the number of intruders in an

epoch. This yields that the time complexity of Algorithm Split is O(max{n, N }). Recall

that N is finite as rc > 0. Thus, in the case when n → ∞, if the information about the

number of intruders in P k

right and P k

left is provided to the defenders (through some external

sensors), then this algorithm can be implemented in real time.

Algorithm Part : Similar to Algorithm Split, Algorithm Part computes |T k

right

|, |V k
left

|, and

|I k| at the start of every epoch which yields that the time complexity of Algorithm Part is

O(n). This requires that the information about the total number of intruders in each of

these sets must be provided to the defenders, for n → ∞, to implement this algorithm in

real time.

We now discuss how this work extends to different motion model of the vehicle.

5.5.2 Different Motion Models for the Vehicle

Observe that the analysis in this work is based upon two quantities; first, the time

taken by the intruders to reach the perimeter and second, the time taken by the defenders

to complete the motion. This work can be extended to other models for the vehicle, for

instance double integrator, by suitably modifying the time taken by the vehicle to complete

its motion. By doing so, it may be that the parameter regimes may be lower than in

Figure 5.7 but the bounds on the competitive ratios will remain the same. The reason that

the parameter regimes will be lower is as follows. Note that the parameter regimes are

characterized by the conditions determined for each of the algorithms. Essentially, these

conditions are determined by requiring the intruders to be sufficiently slow such that they

148

take more time to reach the perimeter than the time taken by the vehicle to complete its

motion. For a different model of the vehicle, such as the Dubins model, the path and the

time taken by the vehicle to complete its motion can be determined by suitably incorporating

the turn radius. Precise dependence of the competitiveness of such realistic models will be

a topic of a future investigation.

5.6 Conclusions

This work analyzed a perimeter defense problem in which two cooperative heterogeneous

defenders, a mobile vehicle with finite capture range and a turret with finite engagement

range, are tasked to defend a perimeter against mobile intruders that arrive in the environ-

ment. Our approach was based on a competitive analysis that first yielded a fundamental

limit on the problem parameters for finite competitiveness of any online algorithm. We then

designed and analyzed four algorithms and established sufficient conditions that guaranteed

a finite competitive ratio for each algorithm under specific parameter regimes.

In this chapter, we considered a turret whose motion is not coupled with the vehicle. In

the next chapter, we consider a scenario in which a turret is attached to a mobile defender.

149

CHAPTER 6

OPTIMAL CONTROL OF A DUBINS-LASER SYSTEM

In the previous chapter, we considered a team of heterogeneous defenders whose motion was

independent of each other. However, in some scenarios, it is possible that the motion of one

defender depends on the motion of another defender. Examples of such scenarios include a

UAV with a gimbaled camera having an ability to rotate clockwise or anti-clockwise. Note

that the angular displacement of the camera is affected differently when the UAV turns in

the same or the opposite direction. This chapter delves into systems of this nature.

This chapter considers an optimal control problem in the plane for a Dubins-Laser System

which consists of a Dubins vehicle, that moves with unit speed, and a laser which is attached

to the vehicle. The laser has a finite range and can rotate either clockwise or anti-clockwise

with a bounded angular speed. The environment consists of a single static target. The

objective of this work is to determine a time optimal control strategy for the Dubins-Laser

System such that the following two conditions jointly hold at final time.

• The target is within the range of the laser and

• the laser is oriented towards the target.

The contributions of this chapter is as follows:

1. Optimal Control of Dubins-Laser System: We formulate a novel optimal control

problem of a Dubins vehicle with a gimbaled laser system. The laser, having a finite

range, is modelled as a single integrator and rotates either clockwise and anti-clockwise

in the environment. The environment consists of a single static target. The aim is

to determine an optimal control strategy of the Dubins-Laser System such that the

target is within the range of the laser and the laser is oriented towards the target in

minimum time.

150

2. Set of Candidate Paths: We establish multiple, including cooperative, properties of

the Dubins-Laser System. We also determine the set of candidate paths for the same.

In particular, we establish that the set of candidate paths consists of 13 paths, i.e., the

shortest path for the Dubins-Laser System is of type {CSC, CC} or a subsegment of

it.

3. Semi-Analytical Solution of the Optimal Path: Finally, we establish that the

optimal path can be efficiently determined by solving a set of non-linear polynomial

equations, thus, providing a semi-analytical solution to the problem.

This chapter is organized as follows.

In Section 6.1, we provide relevant background

information and in Section 6.2, we formally define the problem. Section 6.3 characterizes the

optimal control for the Dubins-Laser system, Section 6.4 characterizes multiple properties of

the shortest path, and Section 6.5 characterizes the solution of the shortest path. In Section

6.6, we provide numerous numerical results and finally, Section 6.7 summarizes this chapter.

6.1 Preliminaries

In this section, we provide some relevant background information that will be instrumen-

tal in understanding the content of this chapter.

6.1.1 The Minimum Principle

[

]

For a given a positive integer n, let x(t) =

x1(t), . . . , xn(t)

denote the state of a

dynamical system given by

˙x = f (x(t), u(t)),

where u(t) : R → Rn is a control signal and u ∈ U where U denotes the set of admissible

control signals. Then, given the boundary conditions (initial and/or final state), a minimum

time optimal control problem is to find controls u∗(t) ∈ U such that the cost J(u(t)) :=
∫

tf
0 dt = tf is minimized, where tf denotes the final time. This means that we want to

minimize the time to go from the initial to the final state.

151

Let p0, . . . , pn be the dual variables, also known as the costates, and p = [p0, . . . , pn] :

R → Rn+1 be the adjoint vector. Then, for a given control u(t) and an initial condition, a

Hamiltonian is defined by H(p(t), x(t), u(t)) : R3n+1 → R

H(p(t), x(t), u(t)) =

n∑

j=0

pjfj(x(t), u(t)),

where f0 = 1 and fj(x(t), u(t)), for j ≥ 1, denotes the jth element of f (x(t), u(t)). The

costates are a solution of the adjoint system

˙pj = −

n∑

i=0

∂fj
∂xi

(x(t), u(t)),

j ∈ {0, . . . , n}.

Then, the minimum principle states the following.

Theorem 58 (Minimum Principle [61]). If u∗(t) is an admissible optimal control, then there

exists a nonzero adjoint vector p(t) such that

1. H(p(t), x(t), u∗(t)) = minu∈U H(p(t), x(t), u(t)), ∀t ∈ [0, tf ].

2. H(p(t), x(t), u∗(t)) = 0 and p0 ≥ 0 ∀t ∈ [0, tf ].

6.1.2 Optimal Control of a Dubins vehicle

A Dubins vehicle is a vehicle that moves with a constant speed (usually normalized to

unity) and has a minimum turn constraint. Formally, a Dubins vehicle has the following

kinematic model:

˙x = cos(θ),

˙y = sin(θ),

˙θ =

u
ρ

,

where (x, y) ∈ R2 denotes the position of the Dubins vehicle in the plane, θ ∈ S1 denotes the

orientation of the vehicle, ρ > 0 denotes the minimum turn radius, and u ∈ [−1, 1] denotes

the control.

The classical problem of finding the shortest path for a Dubins vehicle from a given ini-

tial state to a given final state was studied by Dubins in [27]. Dubins, through geometry,

established that for given endpoints and orientation of the vehicle at these endpoints, the

152

(a) An illustration of a RSR type path.

(b) An illustration of a RLR type path.

Figure 6.1 Illustration of a CSC and a CCC type path for a given initial state [x0, y0, θ0] and
final state [x, y, θ]. The blue dot represents the Dubins vehicle and the blue arrow represents
its orientation. The light blue circles denote the circle of radius ρ subtended by the circular
arcs.

shortest path is one out of 15 candidates paths. He further established that these candidate

paths consists of only straight line segments and circular arcs. Later, authors in [15] estab-

lished the same result by applying the minimum principle (also known as the Pontryagin’s

maximum principle). We now describe the set of candidate paths for the Dubins vehicle

determined in [15] and [27]. We will first describe the classical result for the Dubins vehicle

and then provide a brief explanation about the set of candidate paths.

Theorem 59 (Shortest path for a Dubins vehicle [15]). Any shortest path in the plane

between given endpoints and orientations is either of type CSC or of type CCC, or a sub-

segment of it, where C denotes a circular arc of radius ρ and S denotes a straight line

segment.

The aforementioned result states that the shortest path for a Dubins vehicle is formed

of a sequence of at most three segments, joined at isolated points called as inflexion points.

Further, the shortest path can have at most one straight line segment which implies that

the shortest path is of type CSC or CCC or a subsegment of it. The subsegments of a path

of type CSC are CS, SC, S. Similarly, the subsegments of a path of type CCC are CC, C.

Thus, the set of candidate paths is {CSC, CS, SC, S, CCC, CC, C} or compactly denoted as

{CSC, CCC}. Note that a circular arc can be a clockwise (or right) turn or an anticlockwise

(or left) turn. A clockwise turn is denoted as R and an anticlockwise turn is denoted as L.

153

θ0(x0,y0)ρ(x,y)θ−πθ0(x0,y0)ρ(x,y)θFigure 6.2 Problem Description. The red dot depicts the static target located at the origin
and the blue dot represents the Dubins vehicle. The laser is depicted by the yellow (longer)
arrow and the blue (shorter) arrow depicts the orientation of the Dubins vehicle.

Considering all possible combinations of R and L turns yield the total number of candidate

paths. For instance, a CSC path can be any one out of the set {RSR, RSL, LSL, LSR}.

Figure 6.1a and Figure 6.1b illustrates a CSC and a CCC type path, respectively.

When the orientation of the Dubins vehicle is not specified at the final location, then the

optimal control problem of determining the shortest path for the Dubins vehicle for given

endpoints and orientation at the initial location is known as a Relaxed Dubins problem. In

this case, the set of candidate paths reduces to {CS, CC} [19].

6.2 Problem Description

We consider an optimal control problem of a Dubins-Laser (Dub-L) System in the XY

plane that consists of a Dubins vehicle [27] and a laser which is attached to the vehicle (Fig.

6.2). The laser has a finite range R > 0. The vehicle, with a minimum turn radius ρ > 0,

moves with a constant unit speed and the laser, modelled as a single integrator, has the

ability to rotate either clockwise or anti-clockwise with speed ω ≤ ωM , where ωM > 0. The

environment consists of a static target, assumed to be located at (xT = 0, yT = 0). The state
[

]⊤

vector of the Dub-L system is denoted by x =

x y θ ψ

, where (x, y) ∈ R2, θ ∈ S1, and

ψ ∈ S1 denotes the position of the vehicle, the heading direction of the vehicle measured in the

anticlockwise direction from the positive X-axis, and the orientation of the laser measured in

the anticlockwise direction from the positive x-axis, respectively. The kinematic equations

154

xyθ0ψ0Target(x0,y0)ρthat describe the motion of the Dub-L System is

˙x = cos(θ), ˙y = sin(θ), ˙θ =

, ˙ψ =

u
ρ

u
ρ

+ ω,

where u ∈ [−1, 1] denotes the control input of the Dubins vehicle, and ω ∈ [−ωM , ωM ]

[

]⊤

denotes the angular speed of the laser. We denote the control vector as u =

u ω

and is

defined as u : [0, ∞) → R2.

The target is said to be captured at time t if the distance between the target and the

Dub-L system is at most R and the laser is oriented towards the target. Mathematically, a

target is captured at time t if the following two conditions jointly hold:

x2(t) + y2(t) ≤ R2,

(

π + arctan

[

)

yT − y(t)
xT − x(t)

− ψ(t) = 0.

]⊤

(6.1)

Given an initial state x(0) := x0 =

x0 y0 θ0 ψ0

of the Dub-L System, the objective

of this work is to determine an optimal trajectory for the Dub-L System such that the target

is captured in minimum time, i.e., equations (6.1) hold at final time tf .

The proposed optimal control problem may have non-unique solutions for some initial

conditions and problem parameters. For instance, let x∗(t), y∗(t), θ∗(t) be the time optimal

trajectory for the Dub-L system for all t ∈ [0, tf ], where tf denotes the final time. Further,

suppose that the initial condition is such that the optimal control for the laser is



ωM , if t ∈ [0, t′].

ω∗(t) =



0, otherwise

for some time t′ < tf . Then, for some very small ϵ > 0, there exists another optimal control

for the laser






˜ω∗(t) =

ωM , if t ∈ [ϵ, t′ + ϵ].

0, otherwise

for which the final time at which the target is captured remains the same. Thus, to charac-

terize one trajectory of the laser out of infinitely many, we make the following assumption.

155

Assumption 1. At time instant 0, the laser is switched off and is switched on at time

0 ≤ tl ≤ tf . Once the laser is switched on, it remains switched on until the target is

captured.

The time instant tl is such that the laser remains switched off for maximum amount of

time. To ensure that the target is within the range of the laser, the laser must be switched

on at some time instant 0 ≤ tl ≤ tf . Hence, the time instant at which the laser must be

switched on becomes an optimization variable that needs to be determined to characterize

the trajectory of the Dub-L System.

Let i(t) denote whether the laser is switched on or off. Formally,






0, if t < tl,

1, otherwise.

i(t) =

Then, based on Assumption 1, we model this problem as a two stage optimal control problem

with the following kinematic model.

where

2∑

˙x =

i(t)fk(x, u),

k=1

f1(x, u) =

[

[

cos(θ)

sin(θ)

f2(x, u) =

cos(θ)

sin(θ)

]⊤

u
ρ

]⊤

u
ρ + ω

.

u
ρ

u
ρ

We now formally describe the objective of this work.

Problem Statement: Given an initial state vector x0, the aim is to determine an

[

]⊤

optimal control u∗ =

u∗ ω∗

and time instant t∗
l

≥ 0, such that equation (6.1) holds in

minimum time.

We now apply the necessary conditions and characterize a cooperative nature between

the Dubins vehicle and the laser in the next section.

156

6.3 Necessary Conditions
]

[

Let p =

px py pθ pψ

∈ R4 denote the adjoint vector corresponding to x, y, θ, and

ψ, respectively. Then, given that the aim is to minimize the time before and after time tl,

we can write the Hamiltonian Hk, ∀k ∈ {1, 2}, as

Hk(x, u, p, p0) = p0 + p · fk(x, u),

where p0 ≥ 0 is the abnormal multiplier. Although the costates may be discontinuous at

time tl, they must satisfy the adjoint equations in the time interval t ∈ [t0, tf ], t ̸= tl which

yields

˙px(t) = ˙py(t) = ˙pψ(t) = 0,

˙pθ(t) = px(t) sin(θ) − py(t) cos(θ).

This implies that px(t), py(t), and pψ(t) are constant in the time interval [t0, tl) and (tl, tf ].

Let t−

l (resp. t+

l ) denote the time just before (resp. after) time tl. Then, at time tl, since there

is no cost associated to switching from f1(x, u) to f2(x, u) and the state does not exhibit
jumps at time instant tl, we obtain px(t−
l ), and H1(t−
pθ(t−

l ). Thus, integrating px, py, pθ, and pψ yields

l ) = pψ(t+

l ) = px(t+

l ) = py(t+

l ) = H2(t+

l ) = pθ(t+

l ), pψ(t−

l ), py(t−

l ),

px(t) = cx, py(t) = cy, pψ(t) = cψ and

pθ(t) = cxy(t) − cyx(t) + c0, ∀t ∈ [t0, tf ]

where cx, cy, cψ, and c0 are some constants.

The following lemma will be useful in characterizing the singular control for the Dub-L

System.

Lemma 60. Suppose that the path of the Dub-L System consists of straight line segments.

Then, pθ + pψ = 0 at the inflexion points and on the line segments.

Proof. If pθ + pψ = cxy(t) + cyx(t) = 0 for any non-zero interval of time, then it follows that

the path is a straight line in that interval. Finally, as u changes sign at the inflexion point

157

between two arcs and since pθ + pψ is continuous, it follows that pθ + pψ = 0 at the inflexion

point between two arcs.

From the minimum principle [61] and Lemma 60, we obtain






−1, pθ + pψ > 0

+1, pθ + pψ < 0

0, pθ + pψ = 0,

arg min
u

H1 = arg min

u

H2 ⇒ u∗ =

arg min
ω

H2 ⇒ ω∗ =





−ωM , pψ > 0

ωM , pψ < 0.

(6.2)

(6.3)

where we used the fact that pψ(t−

l ) = pψ(t+

l ), and pθ(t−

l ) = pθ(t+

l ). From equation (6.2),

it follows that the path of the Dub-L System comprises of arcs of radius ρ (when u = +1

or u = −1) and straight line segments (when u = 0). Note that equation (6.3) does not

characterize the control ω∗ when pψ = 0. We characterize the control when pψ = 0 in Section

6.4.

Let S denote a straight line segment and C denote a circular arc of radius ρ. As a circular

arc can be a right turn or a left turn, in the sequel, a right turn circular arc will be denoted

as R and a left turn circular arc will be denoted as L. The following two results follow

immediately from Lemma 60.

Corollary 11. On any optimal path, all inflexion points and all of the line segments are

collinear with the target location.

Proof. The proof is analogous to that of the proof of Lemma 7 of [18] and has been omitted

for brevity.

Corollary 12. The segment of the optimal path before and after time instant tl is one out

of {R|R, L|L, S|S}.

Proof. The result follows from the fact that pθ and pψ is continuous at time instant tl.

158

The transversality conditions yield

px(tf ) = λ1x(tf ) − λ2

py(tf ) = λ1y(tf ) + λ2

y(tf )
x2(tf ) + y2(tf )
x(tf )
x2(tf ) + y2(tf )

pθ(tf ) = 0, pψ(tf ) = −λ2,

(6.4)

where λ1 and λ2 are Lagrange multipliers. Since px(tf ) = cx and py(tf ) = cy, by substituting

px(tf ) and py(tf ) in the equation of pθ(t) and using the fact that pθ(tf ) = 0 yields λ2 = c0 ⇒

cψ = −c0. Thus, equation (6.2) and (6.3) becomes

u∗(t) =






−1, cxy(t) − cyx(t) > 0

+1, cxy(t) − cyx(t) < 0

0, cxy(t) − cyx(t) = 0,

ω∗(t) =






−ωM , c0 < 0

ωM , c0 > 0.

(6.5)

(6.6)

We now characterize a cooperative nature between the laser and the Dubins vehicle.

Lemma 61. Suppose that the optimal trajectory of the Dub-L system ends with a circular

arc, i.e., the final segment of the optimal path is of type C. Then, ω∗(t) and u∗(t) have the

same sign in the final C type segment.

Proof. Consider that at time tf , u∗(tf ) = −1 and ω∗(tf ) = ωM . As pθ(tf ) = 0, it follows

that cxy(tf ) − cyx(tf ) = −c0. As u∗(tf ) = −1 ⇒ cxy(tf ) − cyx(tf ) > 0 which yields that

c0 < 0. This implies that ω∗(tf ) = −ωM which is a contradiction. Thus, u∗ and ω∗ have the

same sign at time instant tf . Finally, since cψ = −c0 is constant in [t0, tf ], and u∗ does not

change sign in the final C segment, the result follows. The case when u∗ = 1 and ω∗ = −ωM

is analogous and has been omitted for brevity. This concludes the proof.

159

6.4 Characterization of Shortest Path

In this section, we characterize the shortest path of the trajectories. We start with the

case when pψ = 0.

6.4.1 Characterization of Shortest Path of Trajectories when pψ = 0

In this section, we establish that when pψ = 0, the problem is equivalent to solving the

Relaxed Dubins problem to a circle. We start with a sufficient condition for pψ = 0.

Lemma 62. The costate pψ = 0 if t∗

l > t0.

Proof. Since tl > t0 and there is no cost associated to switching from f1(x, u) to f2(x, u) at
time instant tl, it follows that H1(t−

l ) which yields

l ) = H2(t+

p0 + px cos θ(t−

l ) + py sin θ(t−

l ) +

(

pθ(t−

l ) + pψ

) u(t−
l )
ρ

=

p0 + px cos θ(t+

l ) + py sin θ(t+

l ) + pθ(t+
l )

u(t+
l )
ρ

+

(

pψ

u(t+
l )
ρ

)

+ ω(t+
l )

⇒ pψω(t+

l ) = 0,

(6.7)

where we used px(t−

l ) = px(t+

l ), py(t−

l ) = py(t+

l ), pψ(t−

l ) = pψ(t+

l ), pθ(t−

l ) = pθ(t+

l ). Since

ω = 0 is non-optimal, it follows that pψ = 0. This concludes the proof.

The next result characterizes the set of candidate paths for the Dub-L System when

pψ = 0.

Lemma 63. If pψ = 0, then the optimal path of the Dub-L System must be of type {CS, CC}

or a subsegment of it.

Proof. If pψ = 0, then Lemma 60, Corollary 11, and Corollary 12 still holds. Further, from

transversality conditions, λ2 = 0 meaning that the constraint that the laser must be oriented

towards the target at final time is inactive. In other words, this means that the laser has

sufficient time to orient itself towards the target while the Dubins vehicle moves towards

the final location. Thus, it follows that when pψ = 0, the problem is equivalent to solving

the Relaxed Dubins problem to a circle. Finally, analogous to Lemma 2.1 in [19] it can be

160

shown that the CSC and the CCC path are not optimal for the Relaxed Dubins problem

to a circle. Thus, the optimal path of the Dub-L System when pψ = 0 is of type {CS, CC}

or a subsegment of it. This concludes the proof.

Corollary 13. x2(tf ) + y2(tf ) = R2 holds at time tf if pψ = 0.

Proof. The result directly follows from Lemma 63 as the problem is equivalent to the Relaxed

Dubins problem to a circle.

We now determine the control ω when pψ = 0.

Lemma 64. If pψ = 0, then

ω∗ =






ωM , if ψ(t∗

l ) − ψ(tf ) <= π

−ωM , otherwise.

Proof. The proof is straightforward and follows from the fact that λ2 = pψ = 0 which implies

that the laser has sufficient time to turn towards the final orientation while the vehicle moves

to the final location.

In this subsection, we characterized the shortest path for the Dub-L System when the

speed of the laser is such that it has sufficient time to turn towards the final orientation. We

now characterize trajectories when the laser must be switched on at time t0.

6.4.2 Characterization of Shortest Path of Trajectories when pψ ̸= 0

This subsection characterizes the main result of this work that the shortest path of the

Dub-L System is of type {CSC, CC} or a subsegment of it. As a consequence of Lemma 62,

we consider that tl = t0 in this subsection. We start with the following remark.

Remark 12. No path of type {CCCC} taken by the Dub-L System can be optimal.

The explanation is as follows. Suppose that at time tf , the vehicle reaches location

(x(tf ), y(tf )) with an orientation θ(tf ). If ωM = 0, then from Lemma 11 in [18], it follows

that a {CCCC} path is not optimal. Let LωM =0 (resp. LωM >0) denote the length of the

161

path when ωM = 0 (resp. ωM > 0). From Lemma 61, it follows that that LωM >0 < LωM =0.
This implies that a path of type {CCCC} cannot be optimal in the interval [t∗

l , tf ] for any

ωM > 0.

Since a path of type CCCC is not optimal, we turn our attention towards characterizing

properties of path of type CCC which will be instrumental is establishing the main result

of this work.

Lemma 65. Suppose that a path of type CCC for the Dub-L system exists and ωM ̸= u
ρ .

Then, 0 < ∆ψ2 < π for any ωM ̸= 1

ρ , where ∆ψ2 denotes the angle that the laser rotates in

the middle C segment of the CCC path.

Proof. Denote the path of type CCC as τ and suppose that ∆ψ2 ≥ π. Without loss of

generality, we assume that the laser is turning clockwise in trajectory τ . Then, the time

taken by the turret to rotate in the middle C segment is T = ∆ψ2
ωM − 1
ρ

ωM > 1

ρ . The proof when ωM < 1

ρ is analogous.

. Here, we assume that

Consider another path τ ′ of type CCC with the same initial and the final location as well

as the same inflexion points as τ . The only difference is that the laser changes its direction

in the middle C segment, i.e., the laser turns anticlockwise in the middle C segment and

clockwise in the first and the last C segment in trajectory τ ′. Let ∆ψ′

2 denote the the angle
that the laser rotates in the middle C segment in trajectory τ ′. As the initial state and

the inflexion points are same in both τ and τ ′, the time taken by the Dub-L System in the

second C segment must be the same. This implies

∆ψ′

2 =

∆ψ2(ωM + 1
ρ )
ωM − 1
ρ

⇒ ∆ψ′

2 > ∆ψ2.

As ∆ψ2 ≥ π, it follows that by changing the direction in which the laser rotates in the middle

C segment, the same orientation can be achieved in less time; meaning that by switching off

the laser over some non-trivial interval of time in the middle C segment and then rotating it

anti-clockwise, the same orientation of the laser is achieved as in trajectory τ at the second

inflexion point. This implies that, for trajectory τ ′, t∗

l > t0. From Lemma 63, this further

162

implies that there exists a trajectory of type {CS, CC} that requires shorter time than

trajectory τ ′. Finally, as the time taken by the Dub-L System in trajectory τ and τ ′ is the

same, it follows that there exists a trajectory of type {CS, CC} that requires shorter time

than trajectory τ . Thus, τ is not optimal. Finally, ∆ψ2 > 0 as ωM ̸= u

ρ . This concludes the

proof.

Lemma 66. Suppose that a path of type CCC for the Dub-L system exists and ωM = 1
ρ .

Then, 0 < β < π, where β denotes the angles subtended by the middle C segment of the

CCC path.

Proof. Denote the path of type CCC as τ and suppose that β ≥ π. Since ωM = u

ρ and

the vehicle and the laser turn in opposite direction in the middle C segment, it follows that

the orientation of the laser, with respect to the positive x-axis, at the first and the second

inflexion point is equal. Mathematically, ψ(t1

inf ) = ψ(t2

inf ), where ψ(t1

inf ) (resp. ψ(t2

inf ))

denotes the orientation of the laser at the first and the second inflexion point.

Similar to the proof of Lemma 65, consider another path τ ′ of type CCC with the same

initial and the final location as well as the same inflexion points as τ . The only difference is

that the laser changes its direction in the middle C segment. Let ∆ψ′

2 denote the angle that
the laser turns in the middle C segment in trajectory τ ′. Then, equating the time taken by

the vehicle in the second segment and the time taken by the turret to rotate and using that

ωM = 1

ρ yields

ρβ =

ρ∆ψ′
2
2

⇒ 2β = ∆ψ′
2.

Since β ≥ π, ∆ψ′
2

≥ 2π meaning that by having the laser switched off over some non-zero

interval of time in τ ′, the same orientation of the laser can be achieved as in trajectory τ at

final time. This implies that there exists a shorter trajectory than τ ′ and τ of type {CS, CC}

and thus, τ is not optimal. This concludes the proof.

Lemma 67. No path of type CCC taken by the Dub-L System is optimal.

163

Proof. The proof is in two parts based on whether ωM = 1

ρ or not. We start with when

ωM ̸= 1
ρ .

Case 1 (ωM ̸= 1

ρ ): Let τ denote a CCC type path for the Dub-L System with angles of

the three C segments as α0, β0, and γ0, respectively, and without loss of generality, suppose

that the laser turns clockwise. Then, the time taken by the Dub-L System or equivalently

the length of τ is Lτ = ρ(α0 + β0 + γ0) = ∆ψ1
1
ωM +
ρ
denotes the angle that the laser rotates in the first, second, and third C segment of τ . Let

, where ∆ψ1, ∆ψ2, and ∆ψ3

+ ∆ψ2
ωM − 1
ρ

+ ∆ψ3
1
ωM +
ρ

ϵ1 > 0 and ϵ2 > 0 denote an infinitesimally small positive real number. We can deform this

path into a similar one with same endpoints and angles α = α0 + ϵ1, β = β0 − ϵ2, and γ0 so

that

cos(∆ψ′

1) + cos(∆ψ′

1

− ∆ψ′

2) = cos(∆ψ1)+

cos(∆ψ1 − ∆ψ2) and

(6.8)

sin(∆ψ′

1) + sin(∆ψ′

1

− ∆ψ′

2) = sin(∆ψ1)+

sin(∆ψ1 − ∆ψ2),

where ∆ψ′

1 = ∆ψ1+ρϵ1(ωM + 1

ρ ), ∆ψ′

2 = ∆ψ2−ρϵ2(ωM − 1

ρ ) and are obtained by using the fact

that the time taken by the vehicle in the first (resp. second) segment of the deformed path
is ρ(α0 + ϵ1) = ∆ψ′

). The determinant of the Jacobian matrix

2

1
1
ωM +
ρ

(resp. ρ(β0 − ϵ2) = ∆ψ′
ωM − 1
ρ

obtained from the set of equations in (6.8) with respect to ϵ1 and ϵ2 is ρ2ω2

M sin(∆ψ2 − ϵ2).

From Lemma 65, it follows that the determinant of the Jacobian is not zero near ∆ψ2, and

thus, such a deformation is possible.

Now, consider one such deformation with ϵ1 = ϵ2 = ϵ, where ϵ > 0 is a very small positive

real number and let τ ′ denote the trajectory obtained though this deformation. Although

τ ′ may have the laser’s orientation different than that of trajectory τ , note that Lτ = Lτ ′,

where Lτ ′ denotes the length of trajectory τ ′. This implies that both trajectories end at the

164

same time tf . Further, as the laser turns clockwise, we have

ψ′
2

− ψ2 = ψ′

1

− ψ1 + ρϵ(ωM − 1
ρ )

⇒ ψ′
2

− ψ2 = −2ϵ,

where ψ′

1 and ψ′

2 (resp. ψ1 and ψ2) denote the orientation of the laser at the first and the
second inflexion point, respectively, in trajectory τ ′ (resp. τ ). Let ψτ (tf ) and ψτ ′(tf ) denote

the orientation of the laser at final time tf in trajectory τ and τ ′, respectively. Let T1 = ∆ψ1
1
ωM +
ρ
. Since Lτ = Lτ ′ and using the fact that the laser turns clockwise, we obtain

and T2 = ∆ψ2
ωM − 1
ρ

T1 + T2 +

ψ2 − ψ(tf )
ωM + 1
ρ

= T1 + T2 +

− ψ′(tf )
ψ′
2
ωM + 1
ρ

⇒ ψ′(tf ) − ψ(tf ) = ψ′

2

− ψ2 < 0,

which is a contradiction as ψτ ′(tf ) ≥ ψτ (tf ) must hold if trajectory τ is optimal, given that

the laser is turning clockwise. This is because as ψ′(tf ) − ψ(tf ) < 0, ψ′(tf ) = ψ(tf ) can be

achieved by switching the laser off over some non-trivial interval of time. From Lemma 63,

this implies that there exists a CS or a CC type trajectory from the same initial location

which requires less time than trajectory τ ′ ( and equivalently τ ). Thus, τ is not optimal and

the result is established.

Case 2 (ωM = 1

ρ ): As the proof of this case is analogous to Case 1, we only provide an
outline of the proof for brevity. Consider the same trajectories τ and τ ′ as in Case 1 and let

ϵ1 and ϵ2 > 0 denote an infinitesimally small positive real number so that

cos(α) + cos(α − β) = cos(α0) + cos(α0 − β0),

sin(α) + sin(α − β) = sin(α0) + sin(α0 − β0).

(6.9)

Then, the determinant of the Jacobian matrix obtained from the set of equations in (6.9)

with respect to ϵ1 and ϵ2 is sin(β0 − ϵ2) which, from Lemma 66 is not equal to zero near β0.

Thus, such a deformation is possible.

Since ωM = 1

ρ and the vehicle and the laser turns in the opposite direction in the middle

C segment in both τ and τ ′, it follows that ψ2 = ψ1 and ψ′

2 = ψ′

1 holds, respectively. Using

165

Figure 6.3 Illustration of Case 1 of proof of Lemma 69. The blue line denotes trajectory τ
and the cyan curve denotes trajectory τ ′.

Lτ = Lτ ′ and that the laser turns clockwise, we obtain

T1 + T2 +

ψ2 − ψ(tf )
ωM + 1
ρ

= T1 + T2 +

ψ′
− ψ′(tf )
2
ωM + 1
ρ

⇒ ψ′(tf ) − ψ(tf ) = ψ′

1

− ψ1 < 0.

Analogous to Case 1, this is a contradiction and thus, the result is established.

We now establish the main result of this work.

Theorem 68. The optimal path of the Dub-L system is of type {CSC, CC} or a subsegment

of it.

Proof. If an optimal trajectory of the Dub-L System contains a straight line segment, it

follows from Corollary 11 that the trajectory must be of type CSC or a degenerate form

if it. Further, from Lemma 67, since a path of type CCC is not optimal, it follows that

the optimal path of the Dub-L System is of type {CSC, CC} or a subsegment of it. This

concludes the proof.

Through the next two results, we establish that an optimal path, except a path of type

C, ends at a distance R from the target.

Lemma 69. For any optimal trajectory of the Dub-L System that consists of a straight line,

x2(tf ) + y2(tf ) = R2 holds at time tf .

166

ψ(t)ψ(tf)−πRψ′(tf)Figure 6.4 Illustration of Case 2 of proof of Lemma 69. The blue line denotes trajectory τ
and the cyan curve denotes trajectory τ ′.

Proof. Since the trajectory of the Dub-L system may end with an arc or a straight line, we

consider two cases based on whether the trajectory ends with an S or a C. For both cases,

we assume that the laser is turning anti-clockwise. The proof when the laser turns clockwise

is analogous.

Case 1 (Trajectory is of S type at tf ): Consider a trajectory τ , as shown in Figure

6.3, in which the Dubins vehicle moves on a straight line until time instant tf and for which

x2(tf ) + y2(tf ) < R2 holds. Consider another trajectory τ ′ that starts from the same initial

condition as τ and has the vehicle turn anti-clockwise at time t = tf − ϵ, where ϵ > 0 is a

very small positive real number, such that the length of both τ and τ ′ is equal. As the laser

is turning anti-clockwise, from Lemma 61, the vehicle must turn anti-clockwise in τ ′ in the

final segment. Let ψ(t) be the orientation of the laser at time instant t. Since both τ and τ ′

have the same initial condition, it follows that the orientation of the laser is same for both

trajectories at time t. Let ψ(tf ) (resp. ψ′(t′

f )) denote the orientation of the laser at time tf
f ) in trajectory τ (resp. τ ′). Analogous to the proof of Lemma 67, as the length, or

(resp. t′

167

ψ(t)Rψ(tf)ψ′(tf)ǫequivalently the time taken by both trajectories is equal,

ψ′(t′

f ) − ψ(t)
ωM + 1
ρ

=

ψ(tf ) − ψ(t)
ωM
ψ(tf ) − ψ(t)
ρ

⇒ ψ′(t′

f ) − ψ(tf ) =

> 0.

This is a contradiction as this implies that ψ′(t′

f ) = ψ(tf ) can be achieved by having the laser

switched off over some non-trivial interval of time implying that there exists a trajectory that

requires length shorter than τ ′ which turns the vehicle until some time strictly less than tf .

Case 2 (Trajectory is of C type at tf ): Consider a trajectory τ (cf. Fig. 6.4) which

ends with a circular arc at time instant tf and for which x2(tf )+y2(tf ) < R2 holds. Consider

another trajectory τ ′ that starts from the same initial condition as τ and that has the vehicle

turn anti-clockwise at time t = t2

inf

− ϵ, where t2

inf denotes the time the vehicle turns anti-

clockwise in trajectory τ , i.e., the time instant when the vehicle reaches the second inflexion

point in τ . As the time taken by the vehicle is equal in both trajectories, we obtain

ψ′(t′

f ) − ψ(t)
ωM + 1
ρ

= ϵ +

ψ(tf ) − ψ(t) − ϵωM
ωM + 1
ρ

⇒ ψ′(t′

f ) − ψ(tf ) =

ϵ
ρ

> 0

which, analogous to Case 1, implies that there exists a trajectory of length shorter than τ ′

and τ . This concludes the proof.

Lemma 70. If the trajectory of the Dub-L System is of type CC, then x2(tf ) + y2(tf ) = R2

holds at time tf .

Proof. The proof is omitted for brevity as it is analogous to that of Lemma 69.

6.5 Solution for Optimal Trajectory

In this section, we characterize the trajectories of the Dub-L System. We start by char-

acterizing the trajectories when t∗

l > t0.

168

Figure 6.5 An illustration of RS type trajectory when t∗

l > t0 with the laser turning clockwise.

6.5.1 Solution for trajectories when pψ = 0

Recall, from Lemma 63 and Corollary 13, that the optimal trajectory ends at a distance

R from the target and is of type {CS, CC} or a subsegment of it. Since the location and

the orientation of the Dubins vehicle at final time can be determined through elementary

geometry, we only describe the solution of CS type trajectory as the solution for the CC

type trajectory is analogous.

Given the initial state of the Dub-L System (refer to Fig. 6.5), the center (cx1, cy1) of the

circle formed by the C segment of the CS trajectory is (x0 + ρ sin(θ0), y0 − ρ cos(θ0)). Note

that Figure 6.5 is only an illustration and the description for an LS trajectory is analogous.

Then, from Corollary 11, as the S segment passes through the target location, it follows that

the equation of the S segment is y = tan(θ(tf ))x. As the orientation of the vehicle does

not change in the S segment, θ(tf ) can be determined by determining the orientation of the

vehicle at the inflexion point. In particular, θ(tf ) is determined by finding the distance from

the center (cx1, cy1) to the S segment at the inflexion point. Mathematically, this yields

|cy1 − tan(θ(tf ))cx1|
√
tan2(θ(tf )) + 1

= ρ,

√

⇒ tan(θ(tf )) =

cy1cx1 ± ρ
c2
x1

c2
y1 + c2
x1
− ρ2

− 1

.

Note that, for a specified θ0, one out of the two values of θf can be eliminated. Finally, by

determining the intersection between the S segment and a circle of radius R, centered at the

169

R(x0,y0)θ0ψ0ψf−πρ(cx1,cy1)(x1,y1)target location, we obtain (x(tf ), y(tf )) as (

√

±R

1+tan2(θ(tf ))

,

± tan(θ(tf ))R
√
1+tan2(θ(tf ))

). We now determine

t∗
l .

Let TD denote the time taken by the Dub-L System to move from (x0, y0) with orientation

θ0 to the location (x(tf ), y(tf )) with orientation θ(tf ). The time required by the laser to rotate

from ψ0 to ψ(tf ) = θ(tf ) is TL =

, where ψ1

tinf

denotes the orientation

of the laser at the inflexion point. Then, t∗

ψ0−ψ(t1
inf )
1
ρ

ωM +

+

inf )

ψf −ψ(t1
ωM
l = TD − TL.

We now characterize the solution of trajectories when pψ ̸= 0.

6.5.2 Solution for trajectories when t∗

l = t0

In this section, we restrict to the case when p0 > 0 under the conjecture, similar to the

work in [44], that the class of trajectories obtained when p0 = 0 is a subset of {CSC, CC}.

Thus, we normalize all other costates by p0. Further, for brevity, we only describe the

solution for CC and CSC trajectories and omit the description of the subsegments as the

process is analogous.

From Lemma 69 and Lemma 70, as the condition x2(tf )+y2(tf ) = R2 holds for both CSC

and CC trajectory, we have that x(tf ) = R cos(η) and y(tf ) = R sin(η), where η denotes the

angle to the final location of the Dub-L System measured from the positive x-axis. Further,

from the transversality conditions, we obtain
cx = λ1R cos(η) − c0 sin(η)

R

, cy = λ1R sin(η) +

c0 cos(η)
R

.

(6.10)

We now establish the solution for the CC type trajectory followed by the CSC type trajec-

tory.

6.5.2.1 CC type trajectory

Let the location and the orientation of the Dub-L System at the inflexion point be

x(t1

inf ), y(t1

inf ), and θ(t1

inf ), respectively. Further, let (cx1, cy1) and (cx2, cy2) denote the cen-

ters of the two circles of the CC segment (Fig. 6.6). Then, as θ(t1

inf ) is perpendicular to the

line joining the two centers (cx1, cy1) and (cx2, cy2), we obtain

tan(θ1) = − cx2 − cx1
cy2 − cy1

,

170

Figure 6.6 An illustration of RL type trajectory when t∗

l = t0.

where (cx1, cy1) is (x0 + ρ sin(θ0), y0 − ρ cos(θ0)), (cx2, cy2) is (x(tf ) − ρ sin(θ(tf )), y(tf ) +

ρ cos(θ(tf ))). Further, (x(t1

inf ), y(t1

inf )) can be written as (cx2+ρ sin(θ(t1

inf )), cy2−ρ cos(θ(t1

inf ))).

Next, as the Hamiltonian must be zero at the initial location, inflexion point, and the final

location, we obtain

(

H0 = 1 + cx

cos(θ0) +

)

(

+ cy

y0u∗
i
ρ

sin(θ0) − x0u∗
ρ

i

)

+

cψω∗

M = 0.

(

)

H 1

inf = 1 + cx

cos(θ(t1

inf )) +

+ cψω∗

M = 0.

y(t1

inf ) sin(θ(t1
x(t1

inf )

inf ))

Hf = 1 + cx cos(θ(tf )) + cy sin(θ(tf )) + cψ(

u∗
f
ρ

+ ω∗

M ) = 0,

(6.11)

(6.12)

(6.13)

where we have used that pθ(tf ) = 0 and pθ + pψ = 0 at the inflexion points. Next, from Fig.

6.6 and Lemma 61, as the vehicle turns anti-clockwise in the final C segment, the laser must

turn anti-clockwise during the entire trajectory. Finally, let TD = ρ(α + β) be the time taken

by the Dub-L system to move from the initial location and orientation to the final location

and orientation, where α and β denotes the angle subtended by the C segments. Further, let

TL =

inf )

ψ0−ψ(t1
ωM − 1
ρ

+

ψ(t1

inf )−ψ(tf )
1
ωM +
ρ

be the time taken by the laser to orient to the final orientation

ψ(tf ) = η + π. Then, as the time taken by the laser and the Dubins vehicle must be the

171

R(x0,y0)θ0ψ0ρ(cx1,cy1)ψf(cx2,cy2)ρθ1−πηFigure 6.7 An illustration of RSR type trajectory when t∗

l = t0.

same, we obtain

TD − TL = 0.

(6.14)

Note that the angles α and β can be determined and expressed in terms of η and θ(tf ) by

using Law of Cosines. Thus, by substituting cx and cy from equation (6.10) in equations

(6.11)-(6.14) yields a set of four equations with four unknowns η, λ1, θf , c0. Solving these

equations yield θf , ψf , xf , yf .

6.5.2.2 CSC type trajectory

We now characterize the solution of the CSC trajectory of the Dub-L System.

Analogous to the CS trajectory is subsection 6.5.1, the first inflexion point and the

orientation of the Dub-L System at the first inflexion point can be determined through

elementary geometry. Then, as the Hamiltonian must be 0 for all time t ∈ [t0, tf ], equating

the Hamiltonian at the initial H0, first inflexion point H 1

inf , second inflexion point H 2

inf , and

the final location Hf yields the following set of equations.

(

H0 = 1 + cx

cos(θ0) +

)

(

+ cy

y0u∗
i
ρ

sin(θ0) − x0u∗
ρ

i

)

+

(6.15)

cψω∗

M = 0

172

R(x0,y0)θ0ψ0ρρ(x2,y2)(x1,y1)θf−π(

H 1

inf = 1 + cx

cos(θ(t1

inf )) +

+ cψω∗

M = 0.

(

H 2

inf = 1 + cx

cos(θ(t1

inf )) +

+ cψω∗

M = 0.

y(t1

inf ) sin(θ(t1
x(t1

inf )

inf ))

)

)

y(t2

inf ) sin(θ(t1
x(t2

inf )

inf ))

Hf = 1 + cx cos(θ(tf )) + cy sin(θ(tf )) + cψ(

u∗
f
ρ

+ ω∗

M ) = 0

(6.16)

(6.17)

(6.18)

where we have pθ(tf ) = 0 and pθ + pψ = 0 at the inflexion points and θ1 = θ2 (Fig. 6.7).

Further, location (x(t2

inf )) can be expressed in terms of (x(tf ), y(tf )) as (R cos(η) −

ρ sin(θ(tf )) + ρ sin(θ(t1

inf ), y(t2
inf )), R sin(η) + ρ cos(θ(tf )) − ρ cos(θ(t1

inf ))). Finally, substituting cx

and cy from equation (6.10) in equations (6.15)-(6.18) yields a system of four equations with

four unknowns, solving which yields the final location for the vehicle. If multiple solutions

of x(tf ), y(tf ), and θ(tf ) are obtained by solving equations (6.15)-(6.18) then, by using the

terminal constraint that the laser must be oriented towards the target (equation (6.1)), the

optimal location at time tf can be determined.

6.6 Numerical Simulations

In this section, we present some numerical simulations to illustrate the properties char-

acterized in this work. For all of our simulations, R = ρ = 1.

Figure 6.8 depicts the shortest path of a CS trajectory when t∗

l > t0, where the final

location and the orientation of the Dub-L System is determined according to 6.5.1. The

initial state, i.e., x0, y0, θ0, ψ0, and ωM are set to 2, 2, π

2 , π, and 0.3 respectively. It is evident
that the condition x2(tf ) + y2(tf ) = R2 holds and the inflexion point and the straight line

segment are collinear with the target location.

Figure 6.9 shows a C type trajectory with t∗

l = t0 with x0, y0, θ0, ψ0, and ωM as 0.5, 0.5, π

2 , π,

and 0.01 respectively. Recall that a C type trajectory is the only trajectory for which the

condition x2(tf ) + y2(tf ) = R2 may not hold when t∗

l = t0. To determine the final location

173

Figure 6.8 A CS type trajectory when t∗
2 , π, and
0.3 respectively. The black dashed circle denotes the circle of radius R around the target.
The initial location is denoted as a green dot and the final location is denoted as the blue
dot. The yellow line denotes the laser.

l > t0 with x0, y0, θ0, ψ0, and ωM as 2, 2, π

and the orientation of the Dub-L System, we use fsolve function in MATLAB to obtain the

solution of equations (6.11),(6.13), and (6.14) characterized in Subsection 6.5.2.

Finally, Figure 6.10 shows a RSR trajectory when t∗

l = t0 with x0, y0, θ0, ψ0, and ωM as

2, 2, π

2 , 4π

3 , and 0.01 respectively. It is evident that the condition x2(tf ) + y2(tf ) = R2 holds

and the inflexion point and the straight line segment are collinear with the target location.

6.7 Conclusions

We considered a joint motion planning problem of a Dubins-Laser System which consists

of a Dubins vehicle and a laser that is attached to the vehicle. The laser has a finite

range and can rotate either clockwise or anti-clockwise with a bounded angular speed. The

environment consists of a single static target. We characterized various geometric properties

of the shortest path of the Dubins-Laser System, including the cooperation between the laser

and the Dubins vehicle. We established that the shortest path for the Dubins-Laser System

is of type {CSC, CC} or a subsegment of it and characterized the solution of each of the 13

candidate trajectories for the shortest path.

174

Figure 6.9 A C type trajectory when t∗
2 , π, and
0.01 respectively. The black dashed circle denotes the circle of radius R around the target.
The initial location is denoted as a green dot and the final location is denoted as the blue
dot. The yellow line denotes the laser.

l = t0 with x0, y0, θ0, ψ0, and ωM as 0.5, 0.5, π

Figure 6.10 A CSC type trajectory with x0, y0, θ0, ψ0, and ωM as 2, 2, π
3 , and 0.01 respec-
tively. The black dashed circle denotes the circle of radius R around the target. The initial
location is denoted as a green dot and the final location is denoted as the blue dot. The
yellow line denotes the laser.

2 , 4π

175

CHAPTER 7

OPTIMAL PURSUIT OF SURVEILLING AGENTS NEAR A HIGH VALUE
TARGET

In most of the works on perimeter defense, it is assumed that the location of the perimeter

is known to the adversary, which may not be true in all applications. For instance, the

location of a high value defense/research facility (target/perimeter) is not precisely known

to the adversary. In such scenarios, prior to deploying intruders to breach the perimeter, the

adversary will typically obtain the estimates of the location of the facility by deploying ad-

versarial UAVs (or trackers) equipped with some low cost range sensor [14]. To counter these

UAVs, the defense system can release mobile pursuers (or defenders) that have the ability to

intercept and disable the UAVs or to corrupt the information gathered by obstructing the

line-of-sight (cf. Fig. 7.1). These scenarios raise an important question which has not yet

been fully explored in the literature – how does the motion strategy of adversarial trackers

change in the presence of one or many pursuers?

This work is a first step towards formulating an adversarial information gathering problem

in presence of a mobile pursuer. Specifically, we introduce a tracking pursuit game in a planar

environment comprising one single static high value target, a single mobile pursuer and two

mobile trackers (adversarial UAVs). We assume that the trackers are slower than the pursuer

and can only measure their individual distances from the target. The trackers jointly seek to

maximize the tracking performance while simultaneously evading the pursuer at every time

instant. On the other hand, the pursuer seeks to capture one of the trackers to hinder the

tracking objective of the trackers. Although we considers a planar environment and range-

only measurements, we also show how this work can be extended to other sensing models

such as bearing measurements.

7.0.1 Related Works

General target tracking problems involve a static/moving target whose state (e.g., the

position and velocity) needs to be estimated by trackers using measurements based on the

176

Figure 7.1 Problem Description. Adversarial UAVs (trackers) move to maximize the information
gathered using the distance measurements to the facility (target) while simultaneously evading from
the pursuer.

distance or bearing or both [25, 13, 55, 60]. A generic approach in these works is to optimize

a measure (e.g., trace or determinant) of the estimation error covariance matrix obtained

from an Extended Kalman Filter (EKF) used to estimate the state. We refer to [23] and the

references therein for the application of state estimation to various target tracking scenarios.

Since the relative geometry of the target and the trackers plays an important role in the

tracking process, many works [52, 13, 85, 86, 72] have focused on identifying such geometries

and motion strategies that optimize the tracking performance such as the determinant of

the Fisher Information Matrix (FIM) or the trace of the estimation error covariance of the

EKF. Tracking based on metrics of observability have also been considered [26, 65, 66].

All of the above mentioned works only focus on determining optimal trajectories for

the trackers to optimize a certain tracking performance, but do not consider the presence

of a pursuer. Authors in [40] design strategies for the pursuers to optimize the tracking

performance while maintaining a desired formation. In [73], an adaptive sampling approach

is considered to track mobile targets and maintain them in the field of view. Authors in [2]

propose an algorithm based on rapidly exploring random trees for pursuers to detect and

track a target.

Pursuit of mobile agents (or evaders) in the presence of a target has been extensively

studied as a differential game known as Target-Attacker-Defender (TAD) games [84, 33, 28,

177

32]. In these works, the attacker tries to capture a target while simultaneously evading a

defender. The objective in these works is to determine optimal cooperative strategies for

the the target and the defender to delay the time taken to capture or evade the attacker.

This work differs from the aforementioned TAD games as the trackers do not seek to capture

the target. Instead, through the measurements obtained, the trackers aim to maximize the

information gathered about the target while, evading the pursuer. Another variant of pursuit

evasion games is pursuit tracking [74, 57, 87] where the objective of the pursuer is to track

the evader by maintaining a fixed distance or Line of Sight to it. In contrast, in this work,

the pursuer seeks to capture the trackers which are tracking a static target. The content of

this chapter is based on [4].

7.0.2 Preliminaries and Contributions

Recall that one of the objectives of the trackers is to maximize the information obtained

from the set of range measurements to the target. This motivates the use of Fisher Infor-

mation Matrix (FIM). The FIM is a symmetric, positive definite matrix that characterizes

the amount of information provided by the measurements for the position of the target that

is to be estimated. In other words, by moving to locations that provide the highest infor-

mation, the trackers aim to improve the outcome of the estimation process. Maximizing

the FIM can be achieved by maximizing a real-valued scalar function (or a metric) of the

FIM. The most commonly used metrics are the trace, determinant and the eigenvalues of the

FIM, also known as the A-optimality, D-optimality and E-optimality criteria, respectively

[83]. Although the trace of the FIM is easy to compute, we consider the determinant as a

metric to be maximized by the trackers. This is because the trace of FIM may be non zero

even when the FIM is singular, implying that optimizing the trace of the FIM can result in

singular configurations.

In this chapter, we seek to understand the role of a pursuer in tracking problems. Equiv-

alently, we aim to understand how the cost of evasion combined with the tracking cost affects

the trajectories, and consequently, the payoff of the trackers. In particular, we consider an

178

instantaneous two player zero sum game between the pursuer and the trackers wherein the

pursuer seeks to minimize the (square of) distance to one of the trackers at every time instant

whereas the trackers aim to jointly maximize a weighted combination of the determinant of

the FIM at every time instant and the distance from the pursuer. Our main contributions

are as follows:

1. Tracking-Pursuit Game with a target: We introduce a tracking-pursuit problem,

modelled as a zero sum game, in a planar environment which consists of a single mobile

pursuer, two mobile trackers and a single static target. For ease of presentation, we

assume that the tracking agents can only measure the distance to the target and are

assumed to be slower than the pursuer. At every time instant, the pursuer aims to

minimize the square of the distance to a tracker, whereas the trackers aim to jointly

maximize a weighted combination of the determinant of the FIM and the square of

the distance to the pursuer from the nearest tracker. The game terminates when the

pursuer captures a tracker.

2. Computing Nash Equilibrium Strategies: We first establish the optimal strategy

for the pursuer. Although the payoff for the trackers is a non-convex function, we

show that the optimization problem can be converted to a Quadratically Constrained

Quadratic Program (QCQP). We further establish that the optimal strategies obtained

for the pursuer and the trackers form a Nash equilibrium of this game.

3. Numerical Insights: We provide several numerical examples highlighting the trajec-

tories of the mobile agents and the affect on the instantaneous payoff. In particular,

we show that due to the presence of pursuer the determinant of the FIM achieves a

lower value. We also show, through one of the examples, that the pursuer can capture

a tracker even when the tracker is faster than the pursuer.

4. Extension to multiple trackers and targets: Finally, we thoroughly describe how

this work extends to the scenarios when there are multiple trackers or multiple targets.

179

Figure 7.2 Problem Description with α = 1. The trackers and the pursuer is denoted by the
blue and the red circles, respectively. The (static) high value target is denoted by a green
square.

This chapter is organized as follows. Section 7.1 comprises the formal problem definition.

In section 7.2, we derive optimal strategies for the pursuer and the trackers, Section 7.3

provides several numerical insights into the problem and Section 7.4 describes the extension

of this work to the case of multiple targets and trackers. Finally, Section 7.5 summarizes

this work and outlines future directions for this work.

7.1 Problem Description

We consider a tracking evasion problem in a planar environment which consists of a

single static target, a single mobile pursuer and two mobile trackers. We denote the two

trackers as E1 and E2, respectively (cf. Fig. 7.2). Each mobile agent is modelled as a

single order integrator with bounded maximum speed. The pursuer is assumed to be faster

than the trackers and can move with a maximum speed normalized to unity. At every time

instant t, each tracker i, where i ∈ {1, 2}, has access to the range measurements, zt

i , to the
target located at s ≜ [sx sy]′ ∈ R2, where r′ denotes the transpose of some vector r. Let
]′

]′

[

[

≜

et
i

et
x,i

et
y,i

∈ R2 (resp., pt ≜

pt
x

pt
y

∈ R2) denote the position of the ith tracker

(resp. the pursuer) and let vt

i (resp. ut) denote the ith tracker’s (pursuer’s) control. Then,

180

the motion model and the measurements are given by

et+1
i = et

i + vt

i + wt
i,

∥vt
i

∥ ≤ µi < 1, ∀i ∈ {1, 2},

pt+1 = pt + ut + wt
p,

∥ut∥ ≤ 1,

i = ∥s − et
zt

i

∥ + V t

i , ∀i ∈ {1, 2},

(7.1)

where, V t
i

∼ N (0, σ2

V ), ∀i ∈ {1, 2} denotes the measurement noise, assumed to be mutually

independent, wt
i

∼ N (0, σwe) as well as wt
p

∼ N (0, σwp) denotes the process noise.

A tracker is said to be captured by the pursuer at time instant t, if its location is within

a unit distance from the pursuer at time t. Note that since the pursuer is faster than both

trackers, the pursuer can always capture both trackers successively. However, we will see

that the game ends when the pursuer captures any one of the trackers. A strategy for the

pursuer is defined as ut : R2 × R2 → R2. Similarly, a strategy for an ith tracker is defined as

i : R2 × R2 × R2 → R2, ∀i ∈ {1, 2}.
vt

We now determine the expression for the determinant of the FIM. Let

h(s, et

1, et

[ ∂
∂sx

∂
∂sy

2) ≜ [∥s − et
]′. Then, for a given model (7.1) and a measurement vector h(s, et

∥]′ denote the measurement vector at time instant t and ∇s ≜

∥ ∥s − et
2

2), the FIM at

1, et

1

time instant t + 1 is [52]

f (s, et

1, et

2, vt

1, vt

2) =

=

1
σ2
V

1
σ2
V

(∇sh(s, et


1, et

2))′(∇sh(s, et

1, et

2)),

2∑

i=1




1
S2
i

(X t
i

− vt

x,i)2

(X t
i

(X t
i

− vt

x,i)(Y t
i

− vt

y,i)

− vt

y,i)




 ,

− vt

(Y t
i

x,i)(Y t
i
− vt

y,i)2

where X t

i = sx −et

x,i (resp. Y t

i = sy −et

y,i) denotes the difference in the x (resp. y)-coordinate

√

of the target and the ith tracker (cf. Fig. 7.2) and Si =

(X t
i

− vt

x,i)2 + (Y t
i

− vt

y,i)2. Since

the trackers do not know the location of the target, we replace s by its estimate ˆs which is

181

obtained from a centralized EKF. Thus, we obtain the determinant of the FIM as

det(f (ˆs, et

1, et

2, vt

1, vt

2)) =

1
σ2
V

[ 2∑

i=1
( 2∑

( ˆX t
i

− vt

x,i)2

2∑

i=1

1
ˆS2
i

( ˆX t
i

− vt

x,i)( ˆY t

i

− vt

y,i)

]

)

2

( ˆY t
i

− vt

y,i)2−

1
ˆS2
i

1
ˆS2
i

i=1
1
ˆS2
1

ˆS2
2

=

σ2
V

(( ˆX t
1

− vt

x,1)( ˆY t

2

− vt

y,2) − ( ˆY t

1

− vt

y,1)( ˆX t

2

− vt

x,2))2,

(7.2)

where ˆX t

i = ˆsx − et

x,i and ˆY t

i = ˆsy − et

y,i for all i ∈ {1, 2}. Note that the determinant of the

FIM for a single tracker is equal to zero implying that the configuration is always singular

in the case of one tracker.

At the first time instant (t = 1), the pursuer selects the tracker which is closest to the

pursuer. This selection is characterized by α ∈ {0, 1}. Specifically, if tracker E1 is closest to

the pursuer, then α = 1. Otherwise, α = 0. The pursuer, then selects its control, ut, such

that the square of the distance to the selected tracker is minimized at every time instant

t ≥ 1. On the other hand, the trackers jointly select their control at every time instant t ≥ 1

to maximize a weighted combination of the determinant of the FIM and the square of the

distance between the selected tracker and the pursuer. We assume that the trackers have

information of the location of the pursuer and thus, the choice of α is known to the trackers.

Since the two trackers jointly maximize the payoff, we model the interaction between the

trackers and the pursuer as a two player zero sum game with the payoff, at time instant

t + 1, defined as

J(vt

1, vt

2, ut) = det(f (ˆs, et

1, et

2, vt

1, vt

2)) + δ(α∥et

1 + vt
1

− pt − ut∥2+

(7.3)

(1 − α)∥et

2 + vt
2

− pt − ut∥2),

where δ ∈ R is a fixed weight associated with the evasion cost (distance between the pursuer

and the selected tracker) and is assumed to be known by all agents. The game terminates

when the pursuer captures the selected tracker since the determinant of the FIM is always

zero for one tracker. We use tf to denote the time instant when the game terminates.

182

We now provide two definitions that will be helprooful in establishing our main result in

Section 7.2.

Definition 71 (Best Response). For a two player zero sum game with the payoff defined as

J(γ, σ), the strategy γ∗ ∈ Γ1 for player 1 (minimizer) is called the best response to player

2’s (maximizer) strategy σ ∈ Γ2 if the following holds

J(γ∗, σ∗) ≤ J(γ, σ∗), ∀γ ∈ Γ1.

Note that the best response for the maximizer can be analogously defined.

Definition 72 (Nash Equilibrium). Given a strategy γ ∈ Γ1 for player 1 and a strategy

σ ∈ Γ2 for player 2 in a two player zero sum game with the payoff J(γ, σ), the pair of

strategies (γ∗, σ∗) is said to be a saddle-point equilibrium strategy if the following holds.

J(γ∗, σ) ≤ J(γ∗, σ∗) ≤ J(γ, σ∗), ∀γ ∈ Γ1, σ ∈ Γ2.

(7.4)

Observe that equation (7.4) in Definition 72 can be rewritten as [39]

J(γ∗, σ∗) = min
γ∈Γ1

J(γ, σ∗), and J(γ∗, σ∗) = max
σ∈Γ2

J(γ∗, σ),

implying that the pair of strategies (γ∗, σ∗) form a Nash equilibrium if γ∗ (resp. σ∗) is the

best response to σ∗ (resp. γ∗).

We now formally state our objective for the above model.

Problem Statement: The aim of this work is to determine saddle-point strategies

ut∗ ∈ R2 and {vt∗

1 , vt∗

2

} ∈ R2 × R2 at every time instant t < tf such that

max
1,vt
vt
2

min
ut

J(vt

1, vt

2, ut) = min
ut

J(vt

1, vt

2, ut)

max
1,vt
vt
2

holds subject to the individual agents maximum speed constraints, i.e.,

∥ut∥ ≤ 1, ∥vt
1

∥ ≤ µ1 < 1, ∥vt
2

∥ ≤ µ2 < 1.

183

For the problem to be well-posed, we make the following assumption.

Assumption 1 (A1)[EKF Convergence]: There exists a time instant te < tf at

which the estimates obtained by the trackers are equal to the true location of the target,

i.e., ||ˆs − s|| = 0, ∀t ≥ te.

7.2 Optimal Strategies

In this section, we determine the optimal strategies for the pursuer and the trackers. We

start with the optimal strategy of the pursuer followed by that of the trackers.

7.2.1 Optimal Strategy of the Pursuer

Without loss of generality, let α = 1 at the first time instant and suppose that vt

1 was

known to the pursuer. Then, the pursuer solves the following optimization problem.

min
ut

δ(∥et

1 + vt
1

− pt − ut∥)2,

subject to ∥ut∥ ≤ 1

(7.5)

where we used the fact that the term det(f (ˆs, et

1, et

2, vt

1, vt

2)) is not a function of the pursuer’s

control ut. It follows directly that the solution to the optimization problem (7.5) is

ut∗

(vt

1) =

1 + vt
et
1
∥et
1 + vt
1

− pt
− pt∥,

meaning that the pursuer moves directly towards the first tracker, E1, with unit speed as long

as the tracker’s position at time instant t + 1, i.e., et

1 + vt

1, is not within a unit distance from

the current position of the pursuer. Otherwise, the optimal pursuer strategy is et

1 + vt
1

− pt,

implying that the tracker is guaranteed to be captured (evasion cost is zero) at time instant

t+1 = tf . This further implies that the trackers will move only to maximize the determinant

of FIM at time instant t = tf − 1 as tracker E1 is guaranteed to be captured at time t + 1.

Thus, the optimal strategy for the pursuer at every time instant t < tf − 1 is

184

ut∗

(vt

1, vt

2) =






et
1+vt
1
∥et
1+vt
1

−pt
−pt∥ , if α = 1,

et
2+vt
2
∥et
2+vt
2

−pt
−pt∥ , otherwise.

Further, the optimal strategy for the pursuer at time instant t = tf − 1 is

ut∗

(vt

1, vt

2) =





1 + vt
et
1

− pt, if α = 1,

2 + vt
et
2

− pt, otherwise

(7.6)

(7.7)

if ∥et

1 + vt
1

− pt∥ < 1 (resp. ∥et

2 + vt
2

− pt∥ < 1) for α = 1 (resp. α = 0). Otherwise, the

optimal strategy for the pursuer at time instant tf − 1 is given by equation (7.6). Note that

although we assumed that vt

i, ∀i ∈ {1, 2} was known to the pursuer, in reality, the pursuer

does not have this information. This means that the optimal strategy defined in (7.6) is an

anticipatory strategy of the pursuer based on the belief of the trackers’ strategy. As will be

clear from the next subsection, we use the optimal strategy, ut∗

determine the optimal state-feedback strategies of the trackers vt∗

1, vt
(vt
1 and vt∗

2), of the pursuer to
2 . Substituting vt∗

1

and vt∗

2 into (7.6) implies that ut∗

is a state-feedback strategy.

In the next section, we determine the optimal strategies of the trackers. Since maximizing

only the determinant of the FIM has been extensively studied [52], we only focus on the case

that the position of the tracker being pursued at time instant tf − 1 is more than a unit

distance from the current position of the pursuer.

In other words, the pursuer’s optimal

strategy is given by equation (7.6) at time tf − 1. Note that the optimal strategy of the

pursuer for any time instant t < tf − 1 is given by equation (7.6). Further, for ease of

presentation, we drop the dependency on time from the notations in the next subsection.

7.2.2 Optimal Strategies of the Trackers

For a given value of α, the trackers jointly solve the following optimization problem.

max
v1,v2

det(f (ˆs, e1, e2, v1, v2)) + δ(∥e1 + v1 − p − u∥)2

subject to ∥v1∥ ≤ µ1, ∥v2∥ ≤ µ2,

185

where, without loss of generality, we assumed that α = 1. Substituting u∗(v1, v2) from

equation 7.6 as well as the expression of the determinant yields

(

max
v1,v2

1
ˆS2
1

ˆS2
2

σ2
V

( ˆX1 − vx,1)( ˆY2 − vy,2)−( ˆY1 − vy,1)( ˆX2 − vx,2)

)

2

+

δ(∥e1 + v1 − p∥ − 1)2,

(7.8)

subject to ∥v1∥ ≤ µ1, ∥v2∥ ≤ µ2.

Although the constraints are convex, the objective is a non-convex function of vi, ∀i ∈ {1, 2}

and thus, computing a global maximizer is difficult.

In what follows, we show that this

optimization problem is equivalent to solving a quadratically constrained quadratic program

(QCQP) [17].

For ease of presentation, we use the following notation in the next result.

[

]′

Let V =

vx,1

vy,1

vx,2

vy,2

vx,1vy,2

vx,2vy,1

∈ R6 and let

z2 =

(

1
ˆS2
ˆS2
2
1

( ˆX1 − vx,1)( ˆY2 − vy,2) − ( ˆY1 − vy,1)( ˆX2 − vx,2)

)

2.

Lemma 73. Suppose α = 1 and let m = ∥e1 + v1 − p∥. Then, the optimization problem

defined in (7.8) is equivalent to solving a QCQP given by

˜V ′P ˜V

max
˜V

subject to

˜V ′Qj

˜V ≤ 0, ∀j ∈ {1, 2}

˜V ′F ˜V = 0,

˜V ′M1

˜V = 0,

˜V ′Lg

˜V = 0, ∀g ∈ {1, . . . , 10}

˜V8 = 1,

]′

[

(7.9)

where ˜V ≜

∈ R17, ˜Vk denotes the kth entry of
vector ˜V and the matrices P, M1, Qj, ∀j ∈ {1, 2} and Lg, ∀g ∈ {1, . . . , 10} are as defined in

V ′ m 1

zvx,1vx,2

zvy,1vy,2

zV ′

z

the Appendix.

186

Proof. By replacing ∥e1 + v1 − p∥ by m, the optimization problem defined in (7.8) can be

rewritten as

max
v1,v2,m,z

1
σ2
V

z2 + δ(m − 1)2

subject to ∥v1∥ ≤ µ1, ∥v2∥ ≤ µ2, m2 = ∥e1 + v1 − p∥2,
(

)

( ˆX1 − vx,1)( ˆY2 − vy,2) − ( ˆY1 − vy,1)( ˆX2 − vx,2)

2

− z2 ˆS2
1

ˆS2
2 = 0.

Observe that the optimization problem is now a polynomial in the original optimization

variables and the additional variables z and m. By adding some extra variables corresponding

to the terms that are polynomial in the optimization variables vx,i and vy,i, ∀i ∈ {1, 2}, we

now convert the aforementioned optimization problem into a QCQP.

[

]′

Let V =

vx,1
optimization variables ˜V ∈ R17 as

vx,2

vy,1

vy,2

vx,1vy,2

vx,2vy,1

∈ R6. Then, we define a vector of

[

˜V =

V ′ m 1

zV ′

zvx,1vx,2

zvy,1vy,2

]′

z

.

Taking the square on both sides of the norm constraints, the above optimization problem

yields the QCQP form as defined in (7.9). Note that the constraint ˜V ′M1
m2. Further, the set of constraints ˜V ′Lg
between the elements of ˜V . As described in [50], the equality constraints in optimization

˜V ≡ ∥e1 + v1 − p∥2−
˜V = 0, ∀g ∈ {1, . . . , 10} characterize the relationship

problem (7.9) can be replaced with two inequality constraints ,thus, reducing the optimiza-

tion problem in the standard QCQP form. This concludes the proof.

187

Following similar steps, an analogous optimization problem when α = 0 is

˜V ′P ˜V

max
˜V

subject to

˜V ′Qj

˜V ≤ 0, ∀j ∈ {1, 2}

˜V ′M0

˜V = 0,

˜V ′F ˜V = 0,

˜V ′Lg

˜V = 0, ∀g ∈ {1, . . . , 10}

˜V8 = 1,

(7.10)

where matrix M0 is as defined in the Appendix. Note that all of the matrices Qj, ∀j ∈ {1, 2},

P, M0, M1, F and Lg, ∀g ∈ {1, . . . , 10} are sparse matrices.

We now establish the main result, i.e., that the pair of strategies (ut∗

, {vt∗

1 , vt∗

2

}) form a

pair of Nash equilibrium. Note that if the optimal strategies of the trackers and the pursuer

form a pair of Nash equilibrium, there is no incentive for the trackers to deviate from their

optimal strategy (see Definition 72). This means that the pursuer has the correct belief of

the trackers strategy and can determine its state-feedback strategy, ut∗

by first solving the

QCQP to determine vt∗

1 and vt∗

2 and then substituting these into (7.6). However, to determine

the strategy of the trackers, the pursuer needs the information of the estimates that the

trackers have of the target’s location. Since the pursuer does not have this information, we

propose that the pursuer uses the true value of the target’s location to solve the QCQP and

consequently determine ut∗

.

Theorem 74. At every instant te ≤ t < tf , the pair of strategies (ut∗

, {vt∗

1 , vt∗

2

}) defined in

(7.6) and obtained by solving the optimization problem (7.8), form a pair of Nash equilibrium

strategies for the payoff function J(ˆs, et

i, pt, vt

i, ut) as defined in (7.3).

Proof. Observe that once the estimates about the location of the target converges to the true

value, all of the mobile agents use the same value of the target’s location to solve the QCQP.

188

Further, at every time instant te ≤ t < tf , the optimal strategy of the pursuer defined in

equation (7.6) is the best-response of the pursuer to the trackers strategy. Similarly, the

optimal strategy of the trackers obtained by solving the optimization problem defined (7.9)

(if α = 1) and (7.10) (otherwise) is the best response of the trackers to an optimal pursuer

strategy. The result then follows directly from Definition 72. This concludes the proof.

We now briefly describe how the game is solved. At every time instant, depending on the

value of α, the pursuer solves the optimization problem 7.9 or 7.10 using the true location of

the target (s) to obtain vt∗

1 and vt∗

2 . The pursuer then moves to the location by determining

its control via 7.6. On the other hand, the trackers jointly solve the same optimization

problem using the estimates of the target (ˆs) and move to the next location using vt∗

1 and

vt∗
2 .

Remark 13 (Bearing Measurements). If the trackers use a sensor that measures the bear-

ing (angle) of the target relative to their positions instead of range measurements, then the

determinant of the FIM is given by [60]

f (ˆs, et

1, et

2, vt

1, vt

2) =

(

( ˆX t
1

1
ˆS4
1

ˆS4
2

σ2
V

− vt

x,1)( ˆY t

2

− vt

y,2) − ( ˆY t

1

− vt

y,1)( ˆX t

2

)

2

.

− vt

x,2)

As the pursuer’s optimal strategy does not change, by following similar steps as in Section

7.2.2, the optimization problem for the trackers can similarly be expressed as a QCQP and

thus, this work easily extends to scenarios when trackers have access to bearing measurements.

7.3 Numerical Observations

We now present numerical simulations of the optimal strategies defined in Section 7.2

and highlight the trajectories of the mobile agents. In all of our simulations, the parameter

σ2
V was kept fixed to 0.03 and the target’s location was chosen to be (0, 0). Due to the

number and size of the sparse matrices in the proposed QCQP optimization problem (7.9),

generating the trajectories was time consuming. Thus, we use fmincon function in MATLAB

to determine the optimal strategies of the trackers which was verified to be consistent with

the strategies obtained by solving optimization problem in (7.9).

189

(a) Trajectories of the mobile
agents for δ = 0. tf = 44.

(b) Trajectories of the mobile
agents for δ = 0.2. tf = 68.

(c) Trajectories of the mobile
agents for δ = 5. tf = 71.

Figure 7.3 Trajectories of the pursuer and the trackers for different values of δ. The cross
represents the starting locations of the mobile agents. The target is denoted by the green
square.

(a) δ = 0

(b) δ = 0.2

(c) δ = 5

Figure 7.4 Determinant of FIM vs Time plots for different values of δ.

7.3.1 Example 1 (α = 1)

Our first numerical simulation (cf. Fig. 7.3) focuses on the trajectories of the mobile

agents when the pursuer moves to capture the first tracker, E1. Specifically, we select the

initial locations such that α = 1. To highlight the role of evasion by the trackers, we

provide a numerical plot with δ = 0 in Fig. 7.3, i.e., the evaders move to maximize only

the determinant of the FIM. Note that the time taken by the pursuer to capture E1 is

mentioned in the description of each sub-figure in Fig. 7.3. The initial locations for all of the

simulations presented in Fig. 7.3 were kept the same and selected to be (−10, −10), (20, −1)

and (−35, −15) for E1, E2 and the pursuer, respectively. Further, the parameters µ1 and µ2

were set to be 0.65 and 0.5, respectively.

Observe that in Fig. 7.3a, the trackers move to position themselves such that the an-

190

gle subtended at the target by the position of the trackers is π

2 . This is consistent with

trajectories that maximize only the FIM as reported in [13]. Upon reaching that position,

the trackers remain at that position until tracker E1 is captured by the pursuer. Based on

the value of δ in Fig. 7.3b as well as in Fig. 7.3c, observe that the tracker E1 first moves

away from the pursuer and then it moves away from the target, maximizing both the time

to capture as well as the determinant of the FIM. Fig. 7.3b shows the cooperative behaviour

of E2. In particular, although the pursuer does not move towards E2, tracker E2 first moves

downwards and then changes its direction in order maximize the determinant of the FIM

by moving to a location such that the position of the trackers subtend an angle of π

2 at

the target. Once the angle between the position of the trackers is π

2 , tracker E2 remains

stationary at its location while E1 evades

Finally, in Fig. 7.4, observe that the determinant of the FIM monotonically increases

in Fig. 7.4a and then converges to 33.33. Although in Fig. 7.4b the determinant of FIM

reaches the value 33.3, the value then decreases as the trackers cannot stay at that position

due to the evasion cost. Note that at time t = 60, the cost converges to 33.3 highlighting

the fact that the angle subtended by the position of the trackers to the target is now at

π
2 , and thus, tracker E2 remains at its position whereas tracker E1 moves in a straight line

maintaining the same angle. Similar trend is observed in Fig. 7.4c. However, tracker E1

is captured before the angle subtended by the trackers to the target is π

2 . This is due to

the higher value of δ as compared to that in Fig. 7.4b because of which tracker E1 moves

directly away from the pursuer. Thus, the trackers require more time to reach the positions

from which the angle subtended to the target is π
2 .

7.3.2 Example 2 (Faster trackers)

This numerical simulation considers a scenario that at one tracker is faster than the

pursuer. The initial locations of the trackers and the pursuer was set to (18, −1), (−15, −15)

and (−13, −20), respectively. Finally the parameter δ was kept fixed to 0.14 and from the

initial locations, α = 1.

191

(a) µ1 = 1.1 and µ2 = 0.5.
tf = 17.

(b) µ1 = 1.2 and µ2 = 0.5.

(c) µ1 = 1.1 and µ2 = 0.65.

Figure 7.5 Trajectories of the pursuer and the trackers for δ = 0.14 and different values of
µi, ∀i ∈ {1, 2}.

In Fig. 7.5a, although tracker E1 is faster (µ1 = 1.1 and µ2 = 0.5) than the pursuer,

the pursuer is able to capture tracker E1. However, for the same initial locations of all of

the mobile agents, the pursuer is unable to capture tracker E1 when µ1 = 1.2 and µ2 = 0.5

(cf. Fig. 7.5b), implying that for faster trackers, there may exist winning regions for the

pursuer as well as the trackers. Specifically, it may be possible to partition the environment

into a winning region (ΩP ) for the pursuer, i.e., the pursuer can always capture a tracker if

the initial locations of all of the mobile agents lie inside ΩP . Similarly, it may be possible

to characterize the winning region (ΩT ) for the trackers, i.e., the trackers can always evade

the pursuer if the initial locations of all of the mobile agents lie inside ΩT . Finally, observe

that for the same initial locations and µ1 = 1.1 (cf. Fig. 7.5c), the pursuer cannot capture

tracker E1 if the speed of the tracker E2 is increased from 0.5 (Fig. 7.5a) to 0.65.

We now describe how this work extends to two different scenarios. We start with a

scenario with multiple targets followed by a scenario with multiple trackers.

7.4 Extensions

In this section, we describe how our analysis extends to the case of multiple targets

and multiple trackers. We also show that in both scenarios the pursuer’s optimal strategy

remains the same as established in Section 7.2. We further establish that the optimization

problem for the trackers can be converted to a QCQP.

192

7.4.1 Multiple targets

In this scenario, we consider that there are N > 1 targets, two mobile trackers and

a single mobile pursuer. Each tracker has access only to range measurements from each

of the N targets. Thus, in this case, the measurement vector is h(s1, . . . , sN , et
[

1, et

2) =

]′

∥s1 − et
1

∥ ∥s1 − et
2

∥ . . . ∥sN − et
1

∥ ∥sN − et
2

∥

, where s1, . . . , sN denote the fixed loca-

tions of the N targets. By taking the partial derivatives with respect to the locations of the

targets and replacing sj∀j ∈ {1, . . . , N } with its estimate ˆsj, the FIM at time instant t + 1

becomes a block diagonal matrix given by

F (ˆs1, . . . , ˆsN , et

i, vt

i) =












f (ˆs1, et

i, vt
i)

02×2

i, vt
i)

f (ˆs2, et
...

02×2
...

02×2

. . .

. . .
. . .

02×2

02×2
...












. . .

. . . f (ˆsN , et

i, vt
i)

where f (ˆsj, et

i, vt

i), ∀1 ≤ j ≤ N is the FIM defined analogously as f (ˆs1, et

i, vt

i) (see Section

3.1). Using the fact that determinant of a block diagonal matrix is the product of the

determinant of its blocks yields

det(F (ˆs1, . . . , ˆsN , et

i, vt

i)) =

N∏

j=1

det(f (ˆsj, et

i, vt

i)).

Thus, the expression for the payoff is given by

J(ˆs1, . . . , ˆsN ,et

i, pt, vt

i, ut) = det(F (ˆs1, . . . , ˆsN , et

i, vt

i))+

δ(α∥et

1 + vt
1

− pt − ut∥2 + (1 − α)∥et

2 + vt
2

− pt − ut∥2).

Since the determinant of the FIM is not a function of the pursuer’s control, it follows that the

pursuer’s strategy remains the same as defined in (7.6). Observe that det(F (ˆs1, . . . , ˆsN , et

i))
y,i for all i ∈ {1, 2}. Therefore, following similar steps

i, vt

is a polynomial function of vt

x,i and vt

as in Section 7.2 and from the fact that any polynomial can be expressed into the standard

QCQP form [50], it follows that the optimization problem obtained for the trackers after

193

substituting ut∗

(vt

1, vt

2) can also be converted into a QCQP of the same form as defined in

Lemma 73. Finally, given that the pair of strategies (ut∗

, {vt∗

1 , vt∗

2

}) are best responses to

each other, it follows that the pair of strategies forms a Nash equilibrium.

7.4.2 Multiple trackers

We now consider the scenario with a single target, M > 2 trackers and a single mobile

pursuer.

[

]′

Let at time instant t < tf , α ≜

M
j=1 αj = 1 and
αj ∈ {0, 1}, ∀j ∈ {1, . . . , M }. Let D ∈ RM denote a vector consisting of the distance

∈ RM such that

. . . αM

α1

[

between the pursuer and the trackers, i.e.,

∥pt − et
1

∥ . . . ∥p − et
M

. Then, the payoff is

∑

]′
∥

given by

where

J(ˆs, et

1, . . . , et

M , vt

1, . . . , vt

M , pt) = det(f (ˆs, et

1, . . . , et

M , vt

1, . . . , vt

M )) + δα′

tD,

det(f (ˆs, et

1, . . . , et

M , vt

1, . . . , vt

M )) =

1
σ2
V

M∑

M∑

j=1

l=j+1

(

( ˆX t
j

1
ˆS2
ˆS2
j
l

− vt

x,j)( ˆY t

l

− vt

y,l)−
)

2

( ˆY t
j

− vt

y,j)( ˆX t

l

− vt

x,l)

.

For a given vector α at the first time instant, the strategy of the pursuer is the same as defined

in Section 7.2 and thus, following similar steps, the payoff for the trackers can be expressed

as a polynomial function in the optimization variables vt

x,i and vt

y,i for all i ∈ {1, . . . , M }.

Hence, following similar steps as in Section 7.2 and given the fact that any polynomial can

be expressed into the standard QCQP form [50], it follows that the optimization problem

obtained for the trackers after substituting ut∗

(vt

1, vt

2) can also be converted into a QCQP of

the same form as defined in Lemma 73.

Finally, given that the pair of strategies (ut∗

, {vt∗

1 , . . . , vt∗

M

}) are best responses to each

other, it follows that the pair of strategies forms a Nash equilibrium.

194

7.5 Conclusions

This chapter introduced a tracking-evasion game consisting of a single pursuer, two track-

ers and a single target. The pursuer seeks to deter the tracking performance of the trackers

by minimizing the square of the distance to the closest tracker, whereas, the trackers aim

to jointly maximize a weighted combination of the determinant of the Fisher Information

Matrix and the square of the distance between the pursuer to the tracker being pursued.

We determined optimal strategies of the pursuer and and showed that the optimal strategies

of the trackers can be obtained by solving a Quadratically Constrained Quadratic Program.

We then established that the pair of strategies form a Nash equilibrium and provided several

numerical observations highlighting the trajectories and the payoff. Finally, we discussed the

extension of this work to multiple trackers and multiple targets.

195

CHAPTER 8

CONCLUSIONS AND FUTURE DIRECTIONS

In this work, we considered perimeter defense problems in linear and conical environments

consisting of 1) a single defender, 2) multiple homogeneous defenders, and 3) heterogeneous

defenders and designed online algorithms with finite competitive ratios and established nec-

essary conditions for the existence of algorithms with finite competitive ratios. We also

addressed two surveilling problems. Specifically, we first address a joint motion planning

for a Dubins-Laser system which consists of a laser attached to a Dubins vehicle. Apart

from establishing multiple properties of the shortest path including the cooperative nature

of the Dubins vehicle and the laser, we establish that the shortest path is {CSC, CC} or a

subsegment of it. Second, we consider a scenario in which intruders need to estimate the

location of a perimeter while the defenders move to deter the intruders from obtaining the

measurements and determined Nash equilibrium strategies for both the defenders and the

intruders.

There are numerous possible extensions of these works, including defenders with sensing

constraints or perimeter defense in higher dimensional environments. We outline some of

the extensions of this work for future direction in the next two subsections.

8.0.1 Competitive Perimeter Defense Problems

An immediate future direction for our perimeter defense works (Chapters 2-5) is to close

the gap between the established fundamental limits and the online algorithms. Another

future direction is to consider the perimeter defense problems in graphical environments with

single or multiple defenders. In Chapter 2 and Chapter 3, we addressed perimeter defense

for linear environments which can serve as a starting point for general graphs. Investigating

how the competitive ratio of online algorithms vary with the properties of various graphical

environments, such as the degree and the diameter of the graph will provide valuable insights.

One thing to note is that a graphical environment can have multiple nodes to defend which

may lead to completely different algorithms than the ones designed in this work.

196

Another possible direction is to explore learning based approaches for these problems.

Reinforcement learning is a valuable tool that can be used to design algorithms for the

defenders that learn to maximize the number of intruders captured over time. Similar to

competitive analysis, reinforcement learning provides a notion of regret, i.e., the difference

between the performance of a static optimal offline algorithm and an online algorithm. This

further leads to the question of designing algorithms that minimize regret as well as com-

petitive ratios.

8.0.2 Surveillance Problems

For the Dubins-Laser System, there are numerous future directions. One such direction

is to consider multiple static targets in the environment that the Dubins-Laser System need

to sequentially capture. Finding the shortest tour that require a Dubins vehicle to visit

multiple static locations are known as Dubins TSP or D-TSP problems and are known to

be NP-Hard [58]. We believe that a similar result can be established for the Dubins-Laser

System as well. Designing approximation algorithms as well as bounds on the shortest path

or designing online algorithms with finite competitive ratios are possible ways to approach

the Dubins-Laser System TSP problem.

Another future direction of these works is to consider a single or multiple moving intruders

that the Dubins-Laser System must capture. The intruders may move on a known trajectory

or may move adversarially to evade the vehicle. Another possible related future direction is

when the Dubins-Laser System needs to move to keep a moving intruder within the range of

the laser for some finite amount of time. Simultaneous capture of intruders via multiple such

defenders or multiple defenders and multiple intruders are also possible future directions.

Finally, for Chapter 7, apart from leveraging the sparse-structure of the matrices for

the optimization problem, a key future direction includes a generalized setup with multiple

pursuers and trackers with motion and energy constraints.

197

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APPENDIX A: PROOFS

We now provide proofs for Lemma 41 and Lemma 42 established for Algorithm Dec-SNP

and proofs for Lemma 47 and Lemma 48 for Algorithm Coop-SNP.

Dec-SNP Algorithm

For the proof of Lemma 41 and as a consequence of Lemma 40, we observe that any

two consecutive captured intervals of defender Vm, 1 ≤ m ≤ M , assumed be located at the

resting point of sector Ni,m, can be classified into four types;

1. stay at the current location and capture the first two intervals in sector Ni,m (cf. Figure

1a).

2. stay at the current location and capture the first interval in Ni,m and then move to the

resting point of sector No,m and capture the third interval (cf. Figure 1b), where No,m

is the sector that is chosen after C1.

3. move to the resting point of No,m and capture both (second and third) intervals (cf.

Figure 1c).

4. move to the resting point of sector No,m and capture the second interval and then move

to the resting point of another sector, No′,m, to capture its second interval (cf. Figure

1d). Here, sector No′,m, o′ ∈ {1, . . . , ns} \ {o}, denotes the sector chosen after C1 once

the defender finishes capturing an interval from the sector No,m.

Proof of Lemma 41. The aim of this proof is to establish that for each of the four types of

captured intervals identified above, the defender Vm, for any m ∈ {1, . . . , ns}, is charged

3(ns − 1) times. We first consider type (1) captured intervals Sj+1

i,m and Sj+2
i,m .

Since defender Vm decides at time instant jD and (j + 1)D to capture Sj+1

i,m and Sj+2
i,m ,

respectively, it loses Sj+2

l,m , respectively, from the other remaining sectors of partition
m, i.e., l ∈ {1, . . . , ns} \ {i}. Thus, the two captured intervals are charged 2ns − 2 times.

l,m and Sj+3

The remaining ns − 1 charge is as follows. Since the defender Vm is currently located at

205

(a) Type (1) captured intervals.

(b) Type (2) captured intervals.

(c) Type (3) captured intervals.

(d) Type (4) captured intervals.

Figure 1 Description of captured intervals for Coop-SNP (resp. Dec-SNP) for a single defender
located at the resting point of sector δ3 (resp. N3,m). The captured intervals are highlighted with
intruders in yellow color.

(xm

i , αm

i ), it follows that the defender captured Sj

i,m. This implies that the comparison C1
must have yielded sector Ni,m at either time instant (j − 2)D (if the defender was located

at (xm

l , αm
requires at least Sj

l ), l ̸= i)) or (j − 1)D (if the defender was located at (xm

i , αm

i ). Recall that C1

i,m and Sj+1

i,m for comparison. Since, defender Vm captured Sj

i,m followed by

i,m , the captured intervals are charged ns − 1 times for all Sj+1
Sj+1
charge is 3(ns − 1).

l,m intervals. Thus, the total

Analogously, type (2) captured intervals Sj+1

i,m and Sj+3

o,m are also charged ns − 1 times for

lost intervals Sj+1

l,m , l ∈ {1, . . . , ns} \ {i} and another ns − 1 times for lost intervals Sj+2

remaining ns − 1 charge is as follows. One charge for all of the intervals Sj+2

i,m , Sj+3
l′,m, l′ ∈ {1, . . . , ns} \ {i, o}, combined.
We now consider type (3) captured intervals. Type (3) captured intervals Sj+2

combined and ns − 2 charge for Sj+3

l′,m and Sj+4

are charged once for the lost intervals Sj+1

lost intervals Sj+2

l,m and Sj+3

ns − 1 times for the lost intervals Sj+4

i,m , Sj+2

o,m and Sj+3
o,m
i,m combined and ns − 2 times for the
l,m , l ∈ {1, ns} \ {i, o}. The captured intervals are charged another
l,m , l ∈ {1, . . . , ns} \ {o}. Finally, the last ns − 1 charge

i,m and Sj+3

l,m . The
i,m and Sj+4

i,m

206

is for the lost intervals Sj

l,m and Sj+1

l,m , l ∈ {1, . . . , ns} \ {i} as defender Vm captured Sj+2

o,m and

not Sj+1
i,m .

Finally, we consider type (4) captured intervals Sj+2

o′,m. For ease of presentation
and without loss of generality, we assume o′ = i. In other words, the defender moves from

o,m and Sj+4

sector Ni,m to sector No,m to capture Sj+2

capture Sj+4

i,m . Type (4) captured intervals are charged once for the lost intervals Sj+1

and Sj+3

i,m combined and ns − 2 times for the lost intervals Sj+2

Similarly, captured intervals are charged once for the lost intervals Sj+3

o,m . Then, defender moves back to sector Ni,m to
i,m , Sj+2
l,m , l ∈ {1, ns} \ {i, o}.
o,m and Sj+5
o,m , Sj+4
o,m
l,m , l ∈ {1, . . . , ns} \ {i, o}.
l,m , l ∈ {1, . . . , ns} \ {i} as

l,m and Sj+5

l,m and Sj+3

l,m and Sj+1

i,m

combined and ns − 2 times for the lost intervals Sj+4

Finally, the last ns − 1 charge is for the lost intervals Sj

defender Vm captured Sj+2

o,m and not Sj+1
i,m .

Since each type of captured intervals are charged 3ns − 3 times, and this result holds for

every defender Vm, the result is established.

Proof for Lemma 42. Since Dec-SNP directs every defender Vm to stay at a resting point

of any sector, in a partition m, for some time interval, it can be viewed as a sequence of

traces, in which the defender Vm spends some number of intervals at one resting point and

some number of intervals at another. Each trace, of a partition m, is defined by a set

{k1, k2, . . . , kns

}, where each element kl, l ∈ {1, . . . , ns} denotes the number of intervals that

the defender Vm decides to capture by staying at the corresponding resting point of the sector

Nl,m.

Note that for any partition m, any realization of Dec-SNP can be achieved by the com-

bination of one or more traces as described in the following cases. Case (i) ki = 3 and

kl = 0 ∀l ∈ {1, . . . , ns} \ {i}, Case (ii) 0 ≤ ki < 3 and ko = 2 and Case (iii) ki = 0, ko = 1

and ko′ = 1, ∀o ∈ {1, . . . , ns} \ {i} and ∀o′ ∈ {1, . . . , ns} \ {o}. The idea is to identify all of

the lost and captured intervals in each case and show that each lost interval is accounted by

the captured intervals.

Case (i): Due to comparison steps C1 and C2 at time jD, the captured intervals Sj+1
i,m ,

207

i,m and Sj+3
Sj+2

i,m account for all of the lost intervals Sj+2

l,m and Sj+3

l,m , ∀l ∈ {1, . . . , ns} \ {i}.

There are two sub-cases; sub-case (a) No = Ni at time instant jD and sub-case (b), there

exists a sector No ̸= Ni at time instant jD (comparison C1) such that |Sj+2
o,m

| < |Sj+1
i,m

|

(comparison C2). We first consider sub-case (a). Sub-case (a) implies that at time instant

jD, the total number of intruders in sector Ni,m is more than in any other sector in the

environment. Thus, captured intervals Sj+1

i,m , Sj+2

i,m and Sj+3
In sub-case (b), we account for lost intervals Sj+2

i,m account for all of the lost intervals
l,m , ∀l ∈

l,m , Sj+3

l,m , ∀l ̸= i.

l,m and Sj+3
Sj+2
{1, . . . , ns} \ {i, o} and Sj+2

o,m , Sj+3

o

, separately. Lost intervals Sj+2

l,m are accounted for

because |Sj+2
l,m

| + |Sj+3
l,m

| ≤ |Sj+1
i,m

| + |Sj+2
i,m

| + |Sj+3
i,m

| or equivalently ηl

i(m) (comparison

C1). Now it remains to account for lost intervals Sj+2

o,m and Sj+3

o,m . Observe that if there exists

a sector No,m ̸= Ni at time instant jD such that |Sj+2
o,m

| < |Sj+1
i,m

|, then there cannot exist

the same No,m at time instant (j + 1)D (from comparison C1). Thus, even if No,m ̸= Ni,m

exists, then the lost interval Sj+2

o,m is accounted by Sj+1

i,m as |Sj+2

o,m

| < |Sj+1

i

| (comparison C2).

Since, at time (j + 1)D, sector No,m cannot be selected again, it follows that ηo

i (m) < ηi

i(m)

at time (j + 1)D and thus, Sj+3

o,m is accounted for.

l,m and Sj+1+ki

Case (ii): To account for the lost intervals Sj+ki

, ∀l ∈ {1, . . . , ns} \ {i}, from
comparison C1 and C2 at time (j + ki)D, the defender was supposed to capture all Sj−2+ki
intervals. While the defender captured Sj−2+ki
Sj−1+ki

i,m intervals, it
i(m) at time instant (j + ki)D, lost intervals Sj+ki
l,m
, ∀l ∈ {1, . . . , ns} \ {i} are fully accounted for. The remaining lost intervals

did not capture Sj+1+ki

, . . . , Sj+1+ki

and Sj+1+ki

i (m) > ηl

, . . . , Sj+ki

. As ηo

i,m

i,m

i,m

i,m

i,m

l,m

,

l,m and Sj+3
i(m) ≤ ηi

l,m
, Sj+2+ki
i,m

Sj+1+ki
i,m

, Sj+3+ki
i,m

Sj+2+ki
l,m

, and Sj+3+ki

l,m

∀l ∈ {1, . . . , ns} \ {o} are fully accounted

by the captured intervals Sj+2+ki

o,m

and Sj+3+ki

o,m

because the conditions ηo

i (m) > ηi

i(m) and

i (m) > ηl
ηo

i(m) are satisfied at time instant (j + ki)D (comparison C1).

Case (iii): To account for lost intervals Sj+1

i,m , Sj+2

i,m , Sj+3

i,m , Sj+2

l,m , and Sj+3

l,m

∀l ∈ {1, . . . , ns}\

{i, o}, the defender was supposed to capture Sj+2

o,m and Sj+3

o,m . This follows because at time

instant jD, ηo

i (m) > ηi

defender captured Sj+2

i(m) (comparison C1) and |Sj+2
i,m as |Sj+2

o,m which accounts for Sj+1

o,m

o,m

| ≥ |Sj+1
i,m

| (comparison C2). The

| ≥ |Sj+1
i,m

|. As the defender moved

208

to capture Sj+4

o′,m at time (j + 2)D, it implies that |Sj+4
o′,m
i,m , Sj+2
i,m , Sj+3
o,m , Sj+2
Sj+3
the lost intervals |Sj+4
l,m

l,m are all accounted by the captured interval |Sj+4
o′,m
|, ∀l ∈ {1, . . . , ns} \ {o′} are accounted for as follows: If the defender

| ≥ |Sj+3
o,m

| (comparison C2) and thus,

l,m , and Sj+3

|. Finally,

also captures Sj+5

o′,m, then lost intervals Sj+4

l,m are accounted for by per case (ii) (ki = 1).

Otherwise (i.e., the defender moved to another sector N˜o,m, ˜o ̸= o to capture Sj+6

˜o,m ), Sj+4

l,m is

accounted for as per case (iii) as now the lost intervals will be Sj+3

i,m , Sj+4

i,m , Sj+5

i,m , Sj+4

l,m , and

Sj+5
l,m

∀l ∈ {1, . . . , ns} \ {i, o}.

Finally, note that the boundary cases of the first and the last intervals fall into these

cases by adding dummy intervals S0

i,m, ∀i ∈ {1, . . . , ns} and SY +1

i,m , where Y denotes the last

interval that consists of intruders in any sector. We assume that the defender captures all

of the dummy intervals. This concludes the proof.

Coop-SNP Algorithm

Similar to that of Algorithm Dec-SNP, any two consecutive captured intervals of any

defender, say Vi ∈ V and assumed be located at the resting point of sector δi, can be

classified into four types;

1. stay at the current location and capture the first two intervals in sector δi ∈ ˜O (cf Fig.

1a).

2. stay at the current location and capture the first interval in δi and then move to the

resting point of sector δo ∈ ˆO and capture the third interval (cf. Fig. 1b).

3. move to the resting point of δo ∈ ˆO and capture both (second and third) intervals of

δo (cf. Fig. 1c).

4. move to the resting point of sector δo ∈ ˆO and capture the second interval of δo and
then move to the resting point of another sector, δp ∈ ˆO, to capture its second interval

(cf. Fig. 1d). Here, sector δp denotes the sector chosen after S1 once the defender

finishes capturing an interval from the sector δo.

209

Proof of Lemma 47. The aim of this proof is to establish that for each of the four types of

captured intervals identified above, every defender Vm ∈ V is charged at most 3(N − M )

times. Without loss of generality, suppose that Vm is located at the resting point of sector
δi at the jth interval. Let L′ denote the set of sectors that do not contain a defender, i.e.,
the complement of the set L. Note that |L′| = N − M .

We first consider type (1) captured intervals Sj+1

i

and Sj+2

i

. Since defender Vm decides

at time instant j ˜D to capture both Sj+1

i

and Sj+2

i

, it loses Sj+2

l

and Sj+3

l

, respectively, from

the other remaining sectors which do not contain a defender, i.e., δl ∈ L′. Thus, the two

captured intervals are charged N − M times. The next N − M charge is as follows. Since

defender Vm is currently located at (xi, αi), it follows that the comparison S1 must have
yielded sector δi at time instant (j − 2) ˜D. Recall that S1 requires at least Sj

i and Sj+1

for

i

comparison. Since defender Vm captured Sj+1

i

, the captured intervals are charged N − M

times for all Sj+1

l

intervals, where δl ∈ L′. Thus, the total charge is 2(N − M ).

Analogously in type (2), captured interval Sj+1

i

is charged N − M times for lost intervals

Sj+1
l

for all sectors δl ∈ L′. Two cases arise for the remaining charge:

Case 1: The captured intervals are Sj+1

i

and Sj+3

o

. Then, one charge for lost intervals

Sj+2
o

, Sj+2
i

and Sj+3

i

combined as the defender captured Sj+1

i

instead of Sj+2

o

and N − M − 1

charge for all Sj+2

l

and Sj+3

l

combined for all sectors δl ∈ L′ \ {o}. Further, the captured

intervals are also charged N − M times for the lost intervals Sj+4

l

for all sectors δl ∈ L′.

Case 2: The captured intervals are Sj+1

i

and Sj+5

p

. This case arises when the defender

moves to sector δo, after capturing Sj+1

i

(point (ii-b) of S2) and then moves to another

sector δp and captures Sj+5

p

(point (ii-a) of S2). Without loss of generality, we assume that

p = i. In this case, one charge for lost intervals Sj+2

o

, Sj+2
i

and Sj+3

i

combined as the defender

captured Sj+1

i

instead of Sj+2

o

and N − M − 1 charge for all Sj+2

l

and Sj+3

l

combined for all

sectors δl ∈ L′ \ {o}. Finally, one charge is for lost intervals Sj+3

o

, Sj+4
o

and Sj+5

o

combined

and N − M − 1 charge for the lost intervals Sj+4

l

and Sj+5

l

combined for sectors δl ∈ L′.

Thus, the total charge for type (2) captured intervals is 3(N − M ).

210

We now consider type (3) captured intervals. Type (3) captured intervals Sj+2

o

and Sj+3

o

are charged once for the lost intervals Sj+1

i

, Sj+2
i

and Sj+3

i

combined and N − M − 1 times

for the lost intervals Sj+2

l

and Sj+3

l

of all sectors δl ∈ L′. The captured intervals are charged

another N − M times for the lost intervals Sj+4

l

of sectors δl ∈ L. Finally, the last N − M

charge is for the lost intervals Sj

l and Sj+1

l

as defender Vm captured Sj+2

o

and not Sj+1

i

.

Finally, we consider type (4) captured intervals Sj+2

o

and Sj+4

p

. For ease of presentation

and without loss of generality, we assume p = i. In other words, the defender moves from

sector δi to sector δo to capture Sj+2

o

. Then, defender moves back to sector δi to capture

Sj+4
i

. Type (4) captured intervals are charged once for the lost intervals Sj+1

i

, Sj+2
i

and Sj+3

i

combined and N −M −1 times for the lost intervals Sj+2

l

and Sj+3

l

of sectors δl ∈ L′. Similarly,

captured intervals are charged once for the lost intervals Sj+3

o

, Sj+4
o

and Sj+5

o

combined and

N − 1 − M times for the lost intervals Sj+4

l

and Sj+5

l

of sectors δl ∈ L′. Finally, the last

N − M charge is for the lost intervals Sj

l and Sj+1

l

combined as defender Vm captured Sj+2

o

and not Sj+1

i

.

Since we have shown that each type of captured intervals are charged 3(N − M ) times,

the result is established.

Proof of Lemma 48. Since Coop-SNP directs every defender Vm to stay at a resting point

of a sector for some time interval, it can be viewed as a sequence of traces, in which each

defender Vm spends some number of intervals at one resting point and some number of

intervals at another. Observe that a trace of Coop-SNP is composed of M vehicular traces,

i.e., a trace for an individual defender. A vehicular trace for defender Vm is defined by a

set {k1, k2, . . . , kN }, where each element kl, l ∈ {1, . . . , N } denotes the number of intervals

that the defender Vm decides to capture by staying at the corresponding resting point of the

sector δl.

At any time instant 2j ˜D, j > 0, consider that Vm is located at the resting point of sector

δi. Observe that any realization of a vehicular trace for defender vm can be achieved by the

combination of one or more atomic traces as described in the following cases. Case (i) ki = 3

211

and kl = 0 ∀l ∈ {1, . . . , N } which satisfy δl ∈ L′, Case (ii) ki = 3 and ko = 1, Case (iii)
ki = 2, ko = 1, case (iv) ki = 1 and ko = 1, and Case (v) ki = 0 and ko = 1. Here, δo ∈ ˆO

denotes the sector that was selected after S1 and assigned to defender Vm after S2.

Case (i): ki = 3 and kl = 0. This case means that the defender stays at its current

location for three consecutive intervals and does not change its sector after capturing the

three intervals. The only way defender Vm does so is if δi ∈ ˜O. This implies that, at time
L holds for all p such that δp ∈ O′.
instant 2j ˜D as well as at 2(j + 1) ˜D, the condition ηi
L
L holds at time instant 2j ˜D and given that the defender captures
for all sectors δp ∈ O′

Since the condition ηi
L
all three intervals Sj+1

, the lost intervals Sj+2

and Sj+3

and Sj+3

≥ ηp

≥ ηp

, Sj+2
i

p

p

i

i

are accounted for. Further, as explained in Lemma 46, lost intervals Sj+1

p

are all accounted

as the defender captured Sj+1

i

.

Case (ii): ki = 3 and ko = 1. This means that at time 2j ˜D, the condition ηi

L holds
(point (i) of S2), for all p such that δp ∈ O′ and, since the defender moved to sector δo ∈ ˆO
after capturing the third interval in sector δi, it follows that at time instant 2(j + 1) ˜D, the

≥ ηp

L

≥ ηp

conditions ηo
L
defender captured Sj+1

i

L, for all p such that δp ∈ O′, and Sj+4

o < Sj+3

i

hold (point (ii-b) of S2). As

given that the condition ηo
L

≥ ηp

, the lost intervals Sj+1

of sectors δp ∈ O′, are accounted for. Second,
L holds at time instant (j + 2) ˜D, if the defender captured

p

Sj+4
o

and Sj+5

o

, then the lost intervals Sj+4

p

, Sj+5
p

, Sj+3
i

, Sj+4
i

and Sj+5

i

would have been

accounted for. However, note that the defender captured Sj+3

i

and Sj+5

o

. Since Sj+4

o < Sj+3

i

,

the captured intervals account for all these lost intervals. Finally, since the defender captured

, Sj+4
i

and Sj+5

Sj+3
i
holds at time instant 2j ˜D.

i

, the lost intervals Sj+2

p

and Sj+3

p

are accounted as the condition ηi
L

≥ ηp
L

Case (iii): ki = 2, ko = 1. This means that at time instant 2j ˜D, the condition ηi
holds (point (i) of S2), for all p such that δp ∈ O′, and at time instant 2(j + 1) ˜D, ηo

L

L

≥ ηp
L
≥ ηp
L

hold (point (ii-a) of S2). To account for the lost intervals Sj+2

l

and Sj+3

l

, the defender was

supposed to capture Sj+1

i

, Sj+2
i

and Sj+3

i

. While the defender captured Sj+1

i

and Sj+2

i

, it did

not capture Sj+3

i

. Given that ηo
L

≥ ηp

L and Sj+4

o

≥ Sj+3
i

holds at time 2(j + 1) ˜D, capturing

212

, Sj+2
i

and Sj+4

Sj+1
i
defender, at time instant 2(j + 2) ˜D, decides to stay and capture Sj+5

accounts for all the lost intervals Sj+1

, Sj+2
l

o

i

and Sj+3

l

. Now, if the

, then lost intervals

o

Sj+4
p

and Sj+5

p

are accounted for by either case (i) or case (ii). Otherwise, they are accounted

for by case (iv) which we consider next.

Case (iv): ki = 1 and ko = 1. There are two sub-cases that arise.

Sub-case (a): This sub-case corresponds to when the defender moved from a sector, say

δq, to sector δi and captured an interval after arrival (point (ii-a) of S2). Then, the defender

moved to sector δo and captured an interval (point (ii-a) of S2). This means that at time
L holds, for all p such that δp ∈ O′.
instant 2j ˜D, S2 assigned sector δi, i.e., condition ηi

≥ ηp

L

To account for lost intervals Sj+1

q

, Sj+2
q

, Sj+3
q

, Sj+2
p

and Sj+3

p

, the defender was supposed to

capture Sj+2

i

and Sj+3

i

. Note that the defender captured Sj+2

i

but not Sj+3

i

as ki = 1. Since

defender moved to sector δo, it follows that at time 2(j + 1) ˜D, the condition ηo

L

≥ ηp

L and

Sj+4
o

≥ Sj+3
i

holds. As defender captured Sj+4

o

instead of Sj+3

i

, the lost intervals Sj+1

q

, Sj+2
q

,

Sj+3
q

, Sj+2
p

and Sj+3

p

are accounted for. Further, as the defender captured Sj+2

i

instead of

Sj+1
q

, all lost intervals Sj+1

p

are accounted.

L, for all p such that δp ∈ O′,
Sub-case (b): At time instant 2j ˜D, the condition ηo
L
holds (point (ii-b) of S2). In this case, the defender first captures Sj+1

≥ ηp

and Sj+1

i > Sj+2

o

i

and then moves to sector δo. As ko = 1, it follows that the condition ηo
L

≥ ηp

L holds at time

instant 2(j + 1) ˜D. As condition ηo
L

≥ ηp

o would have accounted for lost intervals Sj+1
Sj+3

L holds at time instant 2j ˜D, capturing Sj+2
, Sj+2
p

and Sj+3

, Sj+2
i

, Sj+3
i

p

o

i

. Although the

and

defender captured Sj+3

o

, it did not capture Sj+2

o

. Given that Sj+1

i > Sj+2

o

holds and that the

defender captured Sj+1

i

, the lost intervals Sj+2

o

, Sj+2
i

, Sj+3
i

, Sj+2
p

and Sj+3

p

are accounted for.

Further, capturing Sj+1

i

also accounts for the lost intervals Sj+1

p

.

Case (v): ki = 0 and ko = 1. At time instant 2(j − 1) ˜D let the defender be located
in sector δq. Then, this case is only possible if, at time instant 2(j − 1) ˜D, the conditions

≥ ηp

L, for all p such that δp ∈ O′, and Sj−1

ηi
L
time instant 2j ˜D, the conditions ηo
L

≥ ηp

L and Sj+2

q > Sj

o

i both hold (point (ii-b) of S2) and at

≥ Sj+1
i

hold (point (ii-a) of S2), for all p

213

such that δp ∈ O′. Thus, the defender does not capture any interval in sector δi. We account

for the lost intervals as follows.

Given that the condition ηi
L

would have accounted for intervals Sj−1

≥ ηp

L holds at time instant 2(j − 1) ˜D, capturing Sj
p and Sj+1

q , Sj+1
q

, Sj

, Sj

p

q

. Further, since conditions

i and Sj+1

i

≥ ηp

L and Sj+2
ηo
L
account for Sj+1

o

≥ Sj+1
i

both hold at time instant 2j ˜D, capturing Sj+2

and Sj+3

o would

o
. Given that conditions Sj−1

q

≥ Sj

i and Sj+2

o

≥

, Sj+2
i

, Sj+3
p
i
hold and the defender captures Sj−1

and Sj+3

, Sj+2
p

i

and Sj+2

o

, the lost intervals Sj

i , Sj

q , Sj+1
q

, Sj

p and

q

are accounted. Further, lost intervals Sj+1

i

, Sj+2
i

, Sj+3
i

, Sj+2
p

and Sj+3

p

are accounted

Sj+1
i

Sj+1
p

based on whether the defender decides to capture Sj+3

o

(case (i), case (ii) or case (iii)) or if

it moves to some other sector (case (iv)).

Finally, note that the boundary cases of the first and the last intervals fall into these

cases by adding dummy intervals of zero cardinality S0

i , ∀i ∈ {1, . . . , N } and SY +1

i

, where

Y denotes the last interval that consists of intruders in any sector. We assume that the

defender captures all of the dummy intervals. This concludes the proof.

214

APPENDIX B: EXPRESSIONS FOR MATRICES

We now provide the expression for the matrices P, Qj, M and L, respectively. For ease of
notation, denote ai = ˆX t

i . Further, let In×p (resp. 0n×p) denote the identity (resp.

i , bi = ˆY t

zero) matrix of dimension n × p. Then,

P =

×

1
σ2
V















06×6

06×1

06×1

06×8

06×1

01×6

δσ2
V
01×6 − δσ2
V

− δσ2
V

δσ2
V

01×8

01×8

0

0

08×6

01×6

08×1

08×1

08×8

08×1

0

0

01×8

1




















, F =




 ,

F1

08×9

09×8

F2

where F1 =

−a2b2

−b1b2

2a1b2 − a2b1 −b2

b2

0

a2b1b2 − a1b2
2
























b1 −b1

a2 −a2 0 a1a2b2 − b1a2
2
a1b1b2 − a2b2
1
0 a1a2b1 − b2a2
1

a1

0

−a1

1

−1

0

0

−1

1

0

0

0

0

0

0

0

0

0 (a1b2 − a2b1)2
























b2
2

−a2b2

−b1b2

a2
2
2a2b1 − a1b2

2a1b2 − a2b1

−a1a2

−b2

b2

0

a2

−a2

0

2a2b1 − a1b2

b2
1
−a1b1

b1

−b1

0

−a1a2

−a1b1

a2
1
−a1

a1

0

a2b1b2 − a1b2

2 a1a2b2 − b1a2

2 a1b1b2 − a2b2

1 a1a2b1 − b2a2
1

215

and F2 =


−(a2

2 + b2
2)

0

0

−2a1a2

−2a1b2

−(a2

2 + b2
2)
−2a2b1

−2b1b2

b2

0

a2

0

0

a2

0

b2

























−2a1a2

−2a2b1

−2a1b2

−2b1b2

−(a2

1 + b2
1)

0

b2

0

0

0

a2

a2

0

b1 a1

0

0

b1

a1

0

−(a2

1 + b2

1) a1

a1

0

0

b1

b2(a2

1 + b2
1)

1

0

0

0

0

0

0

1

0

0

0

0

0

0

1

0

0

0

b2

0

b1

0

0

0

1

a1(a2

2 + b2
2)

b1(a2

2 + b2
2)

a2(a2

1 + b2
1)

b2(a2

1 + b2
1)

0

0

0

0



























.

0 −(a2

1 + b2

1)(a2

2 + b2
2)

a1(a2

2 + b2
2)

Moreover,

b1(a2

2 + b2

2) a2(a2

1 + b2
1)












I2×2

02×6

02×9

05×4

05×4

05×9

01×7 − µ2
1

01×9

09×4

09×4

09×9












Q1 =

, Q2 =















02×2

02×2

02×4

02×9

02×2

I2×2

02×4

02×9

03×2

03×2

03×4

03×9

01×2

01×5 − µ2
2

01×9

09×2

09×5

09×1

09×9















,















M1 =



















02×2

02×2

02×2

01×2

01×2

09×2

M0 =

I2×2

04×2

01×6

02×5

04×4

−1

[et
1

− pt]

04×2

0

02×9

04×9

01×9

[et
1

− pt]′ 01×5 ∥et
1

− pt∥2 01×9

09×2

09×2

09×2

09×11

02×2

I2×2

02×2

01×4

02×2

02×3

02×2

−1

02×2

[et
2

− pt]

02×2

0

02×9

02×9

02×9

01×9

[et
2

− pt]′ 01×3 ∥et
2

− pt∥2 01×9

09×2

09×2

09×2

09×9

































,

.

216

We now define the matrices Lg ∈ R17×17, ∀g ∈ {1, . . . , 10}. Let Lg(k, l) denote an element

at the kth row and the lth column of the matrix Lg, g ∈ {1, . . . , 10}. Then,






0.5, if k = 1, l = 4,

0.5, if k = 4, l = 1,

−0.5, if k = 5, l = 8,

, L2(k, l) =

−0.5, if k = 8, l = 5,

0 otherwise






0.5, if k = 1, l = 17,

0.5, if k = 17, l = 1,

−0.5, if k = 9, l = 8,

, L4(k, l) =

−0.5, if k = 8, l = 9,

0 otherwise

L1(k, l) =

L3(k, l) =






0.5, if k = 3, l = 17,

0.5, if k = 17, l = 3,

−0.5, if k = 11, l = 8,

, L6(k, l) =

−0.5, if k = 8, l = 11,

0 otherwise

L5(k, l) =






0.5, if k = 2, l = 3,

0.5, if k = 3, l = 2,

−0.5, if k = 6, l = 8,

,

−0.5, if k = 8, l = 6,

0 otherwise






0.5, if k = 2, l = 17,

0.5, if k = 17, l = 2,

−0.5, if k = 10, l = 8,

,

−0.5, if k = 8, l = 10,

0 otherwise






0.5, if k = 4, l = 17,

0.5, if k = 17, l = 4,

−0.5, if k = 12, l = 8,

,

−0.5, if k = 8, l = 12,

0 otherwise

217






0.5, if k = 5, l = 17,

0.5, if k = 17, l = 5,

−0.5, if k = 13, l = 8,

, L8(k, l) =

−0.5, if k = 8, l = 13,

0 otherwise






0.5, if k = 3, l = 9,

0.5, if k = 9, l = 3,

−0.5, if k = 15, l = 8,

, L10(k, l) =

−0.5, if k = 8, l = 15,

0 otherwise






0.5, if k = 6, l = 17,

0.5, if k = 17, l = 6,

−0.5, if k = 14, l = 8,

,

−0.5, if k = 8, l = 14,

0 otherwise






0.5, if k = 4, l = 10,

0.5, if k = 10, l = 4,

−0.5, if k = 16, l = 8,

,

−0.5, if k = 8, l = 16,

0 otherwise

L7(k, l) =

L9(k, l) =

218