以最小區域圓覆蓋集合為基礎用於異質無線隨意網路之廣播

國

立

交

通

大

學

網路工程研究所

碩

士

論

文

以 最 小 區 域 圓 覆 蓋 集 合 為 基 礎 用 於

異 質 無 線 隨 意 網 路 之 廣 播

Minimum Local Disk Cover Sets for Broadcasting

in Heterogeneous Wireless Ad Hoc Networks

研 究 生：劉芳君

以最小區域圓覆蓋集合為基礎用於異質無線隨意網路之廣播

Minimum Local Disk Cover Sets for Broadcasting

in Heterogeneous Wireless Ad Hoc Networks

研 究 生：劉芳君 Student：Fang-Chun Liu

指導教授：易志偉 Advisor：Chih-Wei Y

i

國 立 交 通 大 學

網 路 工 程 研 究 所

碩 士 論 文

A Thesis

Submitted to Institute of Network Engineering College of Computer Science

National Chiao Tung University in partial Fulfillment of the Requirements

for the Degree of Master

in

Computer Science

June 2007

Hsinchu, Taiwan, Republic of China

以最小區域圓覆蓋集合為基礎用於

異質無線隨意網路之廣播

學生：劉芳君 指導教授：易志偉

國立交通大學網路工程研究所

摘要

在無線隨意網路中，廣播是其中一個最基礎的網路操作方法。他被廣泛使用 在發現網路拓蹼、傳輸路徑與監視網路的完整性。而其中一個簡單的機制，稱為 氾濫式演算法，它的方法是當節點收到第一份訊息時，這個節點會傳送這個訊息 給它所有的鄰居。儘管它很簡單，氾濫式演算法傾向去產生大量多餘的重傳。因 此氾濫式演算法可能會引起廣播風暴問題，而且在電源和頻寬上也沒有效率。為 了減輕廣播風暴問題，當某個節點需要廣播時，這個節點只選擇部分的鄰居，稱 為前傳集合來幫它重送訊息，而不是所有鄰居都來傳送訊息。在前傳集合內的所 有節點會覆蓋住所有 2-hop 鄰居，所以可以確定所有網路上的節點都會收到這廣 播訊息。 在同質性網路中，根據 1-hop 鄰居的覆蓋範圍，可以計算出前傳集合。這個 節點的前傳集合的覆蓋範圍跟所有鄰居的覆蓋範圍是一樣的。而被提出的演算法 是區域性且分散式的演算法，而且具有最佳的時間複雜度。 在這篇論文中，我們提出一個機制以最小區域圓覆蓋集作為前傳集合在異質 性網路中(各個節點有不同的傳輸半徑)。對每個節點來說，如果它的鄰居的子集 合有最少的個數而且它們的覆蓋範圍和所有鄰居的覆蓋範圍是一樣的，則這個子 集合稱作最小區域圓覆蓋集合。首先我們證明找它的最小區域性的圓盤覆蓋集合 與找它的輪廓線集合是相等的。我們提出一個各個擊破的演算法來局部地計算此 輪廓線集合，而證明它有最佳的時間複雜度 O(nlogn)。 關鍵字：最小區域圓覆蓋集合、廣播、無線隨意網路、前傳集合

Minimum Local Disk Cover Sets for Broadcasting

in Heterogeneous Wireless Ad Hoc Networks

Student: Fang-Chun Liu Advisor: Prof. Chih-Wei Yi

Department of Computer Science National Chiao-Tung University

ABSTRACT

In wireless ad hoc networks, broadcasting is one of the fundamental networking operations. It is widely and frequently used to explore the network topology, find routing paths, and monitor network integrity. A simple broadcasting mechanism, known as flooding, is to let every node relay messages to all its 1-hop neighbors when it receives the messages the first time. Despite its simplicity, flooding tends to

generate too many redundant retransmissions. It may cause the broadcast storm problem and is neither power nor bandwidth efficient. To relieve the problem, when a node does broadcasting, it selects a subset of neighbors called forwarding set to relay messages instead of all neighbors. Nodes in the forwarding set of a node would cover all its 2-hop neighbors, so it ensures that messages can reach all nodes in the network.

In homogeneous networks, it has been proposed computing the forwarding set based on the coverage area of 1-hop neighbors. The nodes in the forwarding set of a node can cover the same area as its all 1-hop neighbors. The proposed algorithm is localized, distributed, and with the optimal time complexity O(nlogn). In this paper, we propose to use the minimum local disk cover set as forwarding set in

heterogeneous networks, where nodes may have different transmission radii. A

minimum local disk cover set of a node is a subset of 1-hop neighbors and the number of set is smallest. The nodes in the minimum local disk cover set cover the same area of all 1-hop neighbors. We show that the minimum local disk cover set of a node is equivalent to its skyline set. We propose a divide-and-conquer algorithm with the optimal time complexity O(nlogn) to compute the skyline set locally.

Keywords: minimum local disk cover sets, broadcasting, wireless ad hoc networks, forwarding set

誌謝

進入研究所這兩年，從指導教授與系上老師們學到許多寶貴的經

驗與知識，無論是做研究的方法與態度，還是做人處事的道理與如何

與其他人相處，我都受益良多，對提攜我的老師與長輩們心懷感激。

最後再次感謝所有指導、協助與鼓勵我的教授、同學與家人。

Contents

中文摘要

i

英文摘要

ii

誌謝

iii

Contents

iv

List of Figures

vi

1 Introduction

1

1.1 Wireless Ad hoc Networks ………

1

1.2 Broadcast Storm Problem ..………

2

1.3 Motivations and Research Methods …..………

3

2 Related Works

5

2.1 Multipoint relaying ………

5

2.2 Selecting Forwarding Sets ………….………

6

2.3 Minimum Disk Cover Sets ………

7

3 Minimum Local Disk Cover Sets

8

3.1 Network Model ……….………

8

3.2 Problem Description ……….………

9

3.3 The Geometry of Skyline Sets …………..…………

11

3.4 A Divide-and-Conquer Algorithm ………

15

4 Time Complexity Analysis

₁₉

4.1 Geometric Property of Disks ………

₂₀

4.2 Time Complexity of Algorithm ………

29

5 Simulations and Conclusions

₃₁

5.1 Simulations

₃₁

5.1.1 Homogeneous Networks ..………

₃₂

5.1.2 Heterogeneous Networks ..………

35

5.2 Conclusions

₃₈

Bibliography

39

List of Figures

3.1 The configuration of a node u_i. . . 9

3.2 The source node o relays the message to MLDCS. . . . 11

3.3 oa is contained in B(u_i, r_i). . . . 12

3.4 An arc ab is represented by 4 parameters α_i, u_j, r_uj, α_i+1. . . 13

3.5 B (a, r)∩ B (u_i, r_i) is exclusively covered by B(u_i, r_i). . . 14

3.6 Merge two arcs with the same corresponding angles. . . 18

4.1 The configuration B(u₄, r₄) contributes 3 arcs in the skyline. . 20

4.2 Proof of Lemma 4. . . 22

4.3 The structure in Lemma 5. . . 23

4.4 A extreme case when ∂B₁ and ∂B₂ are tangent. . . 24

4.5 Rotate B₁ and B₂ by a, and then b− c ≥ b − c. . . . . 24

4.6 The configuration of Lemma 6. . . 25

4.7 n≤ 3, the skyline contains 2n arcs at most. . . . 27

4.8 3 disks have a common intersection point. . . 27

4.9 In the configuration, B_k+1 doesn’t contribute 3 arcs. . . 29

5.1 The algorithms are compared in homogeneous networks. . . . 33

5.2 The distribution of forward nodes with 1-hop neighbors = 10. 34 5.3 The distribution of forward nodes with 1-hop neighbors = 20. 35 5.4 The algorithms are compared in heterogeneous networks. . . . 36

LIST OF FIGURES iv

Chapter 1

Introduction

1.1

Wireless Ad Hoc Networks

Wireless ad hoc networks consist of a collection of wireless devices sharing a common channel without the need of centralized controls or fixed infrastruc-tures. Each node is equipped with an omnidirectional antenna to access the common channel and usually applies the CSMA/CA(carrier sense multiple ac-cess with collision avoidance) [20] protocol. A communication session is built either through a single-hop radio transmission if the communication parties are close enough, or through relaying by intermediate nodes. Wireless ad hoc networks can be flexibly and quickly deployed for many applications such as personal area networks, smart home environment, environmental monitoring, battlefield surveillance, and emergency disaster relief.

Wireless ad hoc networks still have some challenges, including [18]

(1) Energy conservation: Because the devices in wireless ad hoc networks are typically battery equipped, one of the design goals is to use the limited energy as efficiently as possible.

are limited in wireless ad hoc network. Protocols must provide the good per-formance although the few available resource.

1.2

Broadcast Storm Problem

Broadcasting is one of the fundamental operations in wireless ad hoc networks. A source node sends messages to their 1-hop neighbors, and 1-hop neighbors relay messages to their neighbors, and so on so forth, until all nodes receive one copy of messages. It is widely and frequently used to explore the network topology, find routing paths, and monitor network integrity. Flooding, the most known broadcasting mechanism, is to let every node relay messages to all its 1-hop neighbors when it receives the messages the first time.

Despite its simplicity, flooding has a serious drawback, known as the broad-cast storm problem [1]. First, because the radio propagation is omnidirec-tional and a physical location may be covered by the transmission ranges

of several hosts, many rebroadcasts are considered to be redundant.

Sec-ond, heavy contention could exist because rebroadcasting hosts are probably close to each other. Third, collisions are more likely to occur because the RTS/CTS dialogue is inapplicable and the timing of rebroadcasts is highly cor-related. Since every node retransmits messages without considering whether their neighbors have received messages, unnecessary transmission occurs fre-quently. That results in shortening lifetime of battery-driven devices, and boosting network traffic and bandwidth contention. So, flooding is neither power-efficient nor bandwidth-efficient. Instead, many alternative broadcast-ing algorithms [1] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] have been discussed in literature.

1.3

Motivations and Research Methods

In most works, to relieve the broadcast storm problem, when a node receives a broadcast, it selects a subset of neighbors to relay messages instead of all neighbors. Such subset is referred as forwarding set [7] or multipoint relays (MPRs) [15]. Basically, each node needs to maintain a local network topology of its 1-hop or even 2-hop neighborhood. To ensure that a broadcast can reach all nodes in the network, nodes in the forwarding set of a node should cover all its 2-hop neighbors. At the same time, to better relieve the broadcast storm problem, the forwarding set should be kept as small as possible.

In [5], Sun et al. proposed constructing the forwarding set based on the coverage area of 1-hop neighbors. The idea is to ensure nodes in the forward-ing set of a node to cover the same area as all its 1-hop neighbors. However, the algorithm works only when all nodes in the network have the same trans-mission radius. In this paper, we extend the work to heterogeneous networks in which nodes may have different transmission radii. The network topology is represented by disk graphs with bidirectional link. In other words, each node is associated with a transmission radius. A link exists between two nodes if the distance between them is not larger than the minimum radius of them. We propose to use the minimum local disk cover set as forwarding set. The minimum local disk cover set of a node is a subset of 1-hop neighbors, and its coverage is the same as the coverage of all 1-hop neighbors and cardinality is smallest. Thus, it would better alleviate the broadcast storm problem. We propose a divide-and-conquer algorithm to construct the minimum local disk cover set. Our algorithm has the advantage of needing 1-hop information and being an optimal solution with time complexity O(nlogn).

The remaining of this paper is organized as follows. The related works are introduced in Chapter 2. In Chapter 3, network model is defined and the

forwarding set problem is formulated as the minimum local disk cover set problem. Then the equivalence between minimum local disk cover sets and skyline sets is build. A divide-and-conquer algorithm is given to find skyline sets. In Chapter 4, the time complexity of proposed algorithm is derived. In Chapter 5, simulations are given to compare with other forwarding set algorithms and concludes this paper.

Chapter 2

Related Works

Many localized algorithms have been proposed to alleviate the broadcast storm problem in wireless ad hoc networks. Localized algorithm refers that the nodes build their view of the network topology by using local information only, that is, the information regarding up to h-neighbors in the graph, where h is a small constant (2-3 at most) [18]. We introduce some localized forwarding schemes, including the multipoint relays (MPRs) [15], selecting forwarding sets [6] and minimum disk cover sets [5] in the following sections.

2.1

Multipoint Relays

Qayyum et al. [15] introduced the scheme of multipoint relays (MPRs) to for-ward broadcasting messages in the wireless networks. The idea of multipoint relays is to reduce the number of redundant retransmissions while broadcast-ing messages. The mechanism reduces the forwardbroadcast-ing set to a small set of neighbors instead of all neighbors like in flooding.

Qayyum et al. proposed a scheme to select these MPRs in a wireless environment. The heuristic is adapted from the Chv˘atal’s greedy algorithm [3]

for Set Cover, and gives an approximation ratio of O(log Δ), where Δ is the maximum number of 1-hop neighbors. The heuristic iteratively chooses a 1-hop neighbor covering the maximum number of 2-hop neighbors not yet covered as multipoint relays, and completes when all 2-hop neighbors have been covered. The information requiring for the multipoint relays is the 1-hop neighbors and 2-hop neighbors’ information of each node. Multipoint relays scheme works in a distributed manner. Each node calculates its own MPRs, which is completely independent of other nodes’ selection of their MPRs. In [2], multipoint relays are used to transmit control messages from a mesh point into the network.

2.2

Selecting Forwarding Sets

C˘alinescu et al. [6] solved the Minimum Forwarding Set problem for the geo-metric aspect. The problem is formulated as that given a source node and the sets of 1-hop and 2-hop neighbors. The idea is to select a minimum subset of 1-hop neighbors which dominates all 2-hop neighbors. The Minimum For-warding Set problem is s special case of the Unit-Disk Cover problem known as NP-hard. The complexity of the Minimum Forwarding Set problem is not known.

C˘alinescu et al. proposed several heuristics with constant approximation ratios based on partition the plane into quadrant and established basic geomet-ric properties of the 1-hop and 2-hop neighboring sets in wireless networks. A 2-approximation O(nlogn) time algorithm is proposed, when all 2-hop neigh-bors are in the same quadrant with respect to the source node. First, the algo-rithm computes the skyline and numbers the skyline disks in counter-clockwise order. Then, each 2-hop neighbors are constructed the intervals covered by the skyline disks. Finally, using a simple greedy algorithm picks the disks.

The work needs to collect 1-hop and 2-hop information. The algorithms work only when the network model is homogeneous.

2.3

Minimum Disk Cover Sets

Sun et al. [5] show that the concept of the disk cover set. Given the set of disks S, the problem of disk cover set is to find a minimum subset of S, say

S, such that the union of the disks in S is equal to the union of the disks in S. All nodes in wireless ad hoc networks have the same transmission range

in two-dimensional. Constructing the minimum disk cover set is based on the coverage area of 1-hop neighbors. The nodes in minimum disk cover set cover the same area as all one-hop neighbors. A node can simply request the neighbors in the minimum disk cover set to rebroadcast messages.

Sun et al. proved that finding the minimum disk cover set for disks is equivalent to finding the arcs of disks that make up the boundary. Then, a divide-and-conquer algorithm is proposed to construct the minimum disk cover set efficiently in homogeneous networks. The time complexity of construction the minimum disk cover set is O(nlogn), where n is the number of 1-hop neighbors. Any algorithm that solves the minimum disk cover set problem needs at least O(nlogn) time in the worst case. Therefore, based on 1-hop information, the algorithm is optimal in the aspect of time complexity.

Chapter 3

Minimum Local Disk Cover Sets

3.1

Network Model

To alleviate the broadcast storm problem, the size of the forwarding set needs to be reduced. However, at the same time, we need to make sure that messages may reach all nodes in the network. The selection of the forwarding set at each node should guarantee that the message will reach all its 2-hop neighbors.

We assume that wireless nodes are distributed in a two-dimensional plane

R2. In what follows, x is the Euclidean norm of a point x ∈ R2, and |A| is shorthand for the cardinality of a countable set A. For any two points x,

y ∈ R2, x − y is the Euclidean distance between x and y; xy denotes a line segment between x and y; and −xy denotes a ray (or called a half line) from x→

to y.

The topology of a heterogeneous ad hoc network is modeled by disk graphs with bidirectional links. Each node u_i is associated with a transmission radius

r_i. See Figure 3.1(a). A link exists between two nodes if the distance between them is not larger than their transmission radii. Two nodes u_i, u_j are said to be neighbors if and only if the Euclidean distanceu_i− u_j <= min(r_i, r_j).See

(a)

i ri _{ui uj}

(b)

u

Figure 3.1: The configuration of a node u_i.

Figure 3.1(b). For a node u_i, its coverage is modeled as a disk with center u_i and radius r_i, which is denoted as B(u_i, r_i). A node u_j is said to be covered by u_i if u_j ∈ B(u_i, r_i). The topological boundary of a set S ⊂ R2 is denoted by ∂S, and thus, ∂B(u_i, r_i) is the circle centered at u_i and with radius r_i.

We assume each node knows the information of their 1-hop neighbors, es-pecially locations and transmission radii. Due to the bidirectional link model, for any two nodes u, v, if u is the one-hop neighbor of v, then it implies v is the one-hop neighbor of u. So, the intersection of coverage of 1-hop neighbors

is not empty, i.e. 

ui∈S

B(u_i, r_i)= ∅, where S is 1-hop neighbors.

3.2

Problem Description

We propose a strategy for each node to choose its forwarding set locally and statelessly based on the coverage of 1-hop neighbors. To make sure all 2-hop neighbors may receive messages, each node selects its forwarding set that the coverage of the forwarding set is the same as the coverage of all 1-hop neighbors.

We define the disk cover set of a set of nodes V as S ⊆ V such that



ui∈S

B(u_i, r_i) = 

ui∈V

B(u_i, r_i). For a set of nodes V , if there exists a node in V such that all other nodes in V are neighbors to that node, then V is called the

local set and the corresponding disk set, {B (u₁, r₁) , ..., B (u_n, r_n)}, is called the local disk set. For a local set, its cover set is called a local disk cover set. Obviously, the closed neighbor set of a relay node n₀ is a local set.

To alleviate the broadcast storm problem and ensure the reachability of broadcast messages, we propose to use the minimum local disk cover set (MLDCS) as the forwarding set. The problem of the minimum local disk cover set is formally stated as follows.

Minimum Local Disk Cover Set (MLDCS) Problems Input: A local disk set {B (u₀, r₀) , B (u₁, r₁) , ..., B (u_n, r_n)}

such that for all 1≤ i ≤ n, u₀− u_i ≤ min (r₀, r_i), i.e. u_i ∈ B(u₀, r₀) and u₀ ∈ B(u_i, r_i).

Let V ={u₀, u_1,..., u_n} be the set of disk centers.

Output: A subset S of V such that 

ui∈S

B(u_i, r_i) = 

ui∈V

B(u_i, r_i).

Measure: |S| is minimal.

Figure 3.2 illustrates the minimum local disk cover set. Assume source node is o. Node o has five one-hop neighbors u₁, u_2,...u₅ and five two-hop neighbors

u₆, u_7,..., u₁₀. We find the union of B(u₁, r₁), B(u₂, r₂), B(u₄, r₄), B(u₅, r₅) = (5

i=0

B(u_i, r_i)). B(u₃, r₃) is covered by (5

i=0

B(u_i, r_i)), so B(u₃, r₃) isn’t in the minimum local disk cover set. The collection of these disks is minimum local disk cover set. The coverage of MLDCS is equal to the coverage of all one-hop neighbors, and the cardinality of MLDCS is 4.

We assume that each node learns the locations and radii of its neighbors through the exchanges of beacons. In addition, we define the skyline for a disk set as the boundary of the union of disks in the set. Hence, the skyline of the local disk set {B(u₀, r₀), B(u₁, r₁), . . . , B(u_n, r_n)} is ∂(n

i=0

B(u_i, r_i)). Ob-viously, a skyline is composed of arcs of disks. The collection of origins of disks that contribute arcs to a skyline is called the skyline set. In the next

o 7 u₈ u₁₀ u6 u2 u₅ u₁ u3 u9 u₄ u

Figure 3.2: The source node o relays the message to MLDCS.

section, we will show that the MLDCS of a local set is the skyline set of the corresponding local disk set, and thus, we can solve the MLDCS problem for a given local disk set by computing the corresponding skyline.

3.3

The Geometry of Skyline Sets

In this section, we give properties of skylines and build the relation between the MLDCS for a given local set and the skyline set for the corresponding local disk set. We then propose a divide-and-conquer algorithm to compute the skyline set.

The following geometric lemma and corollary are used to build the relation between MLDCS for a local set V = {u₀, u₁, . . . , u_n} and the skyline set for

the corresponding local disk set{B(u₀, r₀), B(u₁, r₁), . . . , B(u_n, r_n)}. Without loss of generality, in the following discussion we assume n₀ = o.

o

i

ri

a

u

Figure 3.3: oa is contained in B(u_i, r_i).

Proof. Note that o ∈ B (u_i, r_i) sinceo − u_i ≤ r_i. Since B (u_i, r_i) is convex and o, a ∈ B (u_i, r_i), the line segment oa ⊂ B (u_i, r_i). See Figure 3.3.

Then, we have the following corollary.

Corollary 2 For any ray from o, it intersects the skyline at exactly one point. Proof. Obviously, any ray from o intersects the skyline. We only need to show the uniqueness of the intersection point. We can prove this by contradiction. Assume there exists a ray that intersects the skyline at two points a and b. Without loss of generality, we also assume a is farther from o than b. Since

a is in the skyline, a is in ∂B (u_i, r_i) for some i. According to Lemma 1, we have oa ⊂ B (u_i, r_i). This implies that b is inside of B (u_i, r_i), and b can’t be in the skyline. It is a contradiction, and thus the corollary is proved.

According to the corollary, we can know that a skyline is composed of a sequence of arcs surrounding the origin from 0 to 2π. An arc can be repre-sented by four parameters α_i, u_j, r_uj, α_i+1. Here u_j and r_j respectively are the center and radius of the disk contributing the arc, and α_i and α_i+1 with

α_i < α_i+1are two angles corresponding to two endpoints of the arc measured at o in counterclockwise direction from the x-axis. See Figure 3.4. Note that the

j αi αi+1 b o Y X u a

Figure 3.4: An arc ab is represented by 4 parameters α_i, u_j, r_uj, α_i+1. reference point to measure α_i and α_i+1 is o, not u_i. For convenience, if an arc crosses the x-axis in the positive direction then it is splitting by the x-axis into two arcs, i.e. an arcα_i, u_j, r_uj, α_i+1with α_i < 360◦ and α_i+1> 360◦, then we split this arc into 2 arcs (α_i, u_j, r_uj, 360◦) and0◦, u_j, r_uj, α_i+1. Without loss of generality, we could assume that there are no arcs exceeding 360◦. Thus, a sky-line consisting n arcs can be represented as (α₀, u_s0, r_s0, α₁, u_s1, r_s1, α₂, ..., α_n), where 0 = α₀ < α₁ < ... < α_n = 2π and for any 0≤ i ≤ n−1, (α_i, u_si, r_si, α_i+1) is an arc in the skyline.

Now, we give the following theorem that builds the relation between a skyline set and the corresponding minimum local disk cover set.

Theorem 3 For a given local set V = {u₀, u_1,..., u_n}, its minimum

lo-cal disk cover set is the skyline set for the corresponding lolo-cal disk set {B (u0, r0) , B (u1, r1) , ..., B (un, rn)}.

Proof. To prove that the skyline set is the minimum local disk cover

set of a local set, we first prove that a skyline set is a local disk cover set. Assume the skyline is composed of arcs a₁b₁, a₂b₂, . . . , a_kb_k belonging to B (u_i1, r_i1) , B (u_i2, r_i2) , . . . , B (u_ik, r_ik), respectively. Let oa_jb_j denote the

i uj ri rj o u x a

Figure 3.5: B (a, r)∩ B (u_i, r_i) is exclusively covered by B(u_i, r_i). sector area surrounded by line segments oa_j, ob_j, and arc a_jb_j. Each sector has no intersection with others. The covered area n_i=0B (u_i, r_i) is equal to the union of sectorsk_j=1oa_jb_j. According to Lemma 1, for each skyline arc

a_jb_j, we have oa_jb_j ⊆ Bu_ij, r_ij. Thus, n_i=0B (u_i, r_i) ⊆ k_j=1Bu_ij, r_ij. This means {B (u_i1, r_i1) , B (u_i2, r_i2) , . . . , B (u_ik, r_ik)} is a disk cover set.

Next, we prove that if B (u_i, r_i) is in the skyline set, B (u_i, r_i) must be in any disk cover set. To prove this, we are going to show there exists some area belonging to B (u_i, r_i) but not belonging to any other disk. Assume a is a point in the internal of the skyline arc contributed by B (u_i, r_i). For any

B (u_i, r_i), j = i, we have the distance between u_j and a is larger than r_j, i.e.

uj − a > rj. Let r = 1₂(minj=iuj − a − rj). We may draw a disk which

is with center located at a and whose radius is r. For any point x ∈ B (a, r) and j = i, see Figure 3.5.

uj − x ≥ uj− a − x − a ≥ uj− a − 1₂  min j=i uj − a − rj  >u_j − a − (u_j− a − r_j) = r_j.

Then we have u_j− x ≥ r_j. Thus, for any j = i, B (a, r) ∩ B (u_j, r_j) = ∅. This implies that B (a, r)∩ B (u_i, r_i) belongs to B (u_i, r_i) but does not belong to any other disk, thus B (a, r)∩ B (u_i, r_i) is exclusively covered by B(u_i, r_i). So the theorem is proved.

3.4

A Divide-and-Conquer Algorithm

According to Theorem 3, computing the MLDCS of a local set is the same as finding the skyline set of the corresponding local disk set. In this subsection, a divide-and-conquer algorithm is proposed to find the skyline set.

In the algorithm, the disk set is divided into 2 subsets of disks DS1 and

DS2 recursively. If the disk set is small enough (i.e. the size of disk set is 1),

then the skyline returns the skyline of the disk. The skyline of each subset is discovered by recursive techniques, and then two skylines are merged to find the skyline of all disks. Eventually, Skyline returns a new skyline list. Without loss of generality, we may assume that the position of n₀ (i.e., o) and the value r₀ are stored in the algorithms.

1 procedure Skyline (DS ={(u₁, r₁) , ..., (u_n, r_n)}) 2 // (u_i, r_i) represents the center and radius of a disk.

3 begin

4 if |DS| = 1 return the skyline of {B (u₁, r₁)}

5 else // if|DS| > 1 6 begin 7 DS1 =  (u₁, r_u1) , ...,  u_n 2, run₂  8 DS2 =  un 2 +1, run2+1  , ..,u_n, r_un 9 Skyline1 = Skyline (DS1) 10 Skyline2 = Skyline (DS2)

11 return M erge (Skyline1, Skyline2)

12 end

13 end

1 procedure M erge (SL1, SL2)

2 // SL1 and SL2 are skylines.

3 begin

4 Step 1: Refine SL1 and SL2 to align arcs in skylines.

5 Then, we may assume SL1 = (α₀, u₁, r_u1, α₁, ..., α_m) 6 and SL2 = (α₀, v₁, r_v1, α₁, ..., α_m)

7 Step 2: For each 0≤ i ≤ m, decide new skyline arcs from

8 (α_i, u_i, r_ui, α_i+1) and (α_i, v_i, r_vi, α_i+1)

9 with the same angle span.

10 Step 3: Combine neighboring skyline arcs from the same disk.

11 Return the new skyline.

12 end

Skyline is a classical divide-and-conquer procedure, and most works

First, two skylines are aligned by splitting arcs such that two

sky-lines have the same angle sequences. For example, assume SL1 =



β₀, u₀, r_u₀, β₁, u₁, r_u₁, β₂, ..., β_k and SL2 =γ₀, v₀, r_v0, γ₁, v₁, r_v₁, γ₂, ..., γ_lare two skylines. Let (α₀, α₁,· · · , α_m) be a monotonic sequence of angles such that {α₀, α₁,· · · , α_m} = {β₀, β₁, ..., β_k} ∪ {γ₀, γ₁, ..., γ_l}. Then, SL1 and SL2

are refined according to angles α₀, α₁,· · · , α_m. After refinement, both lists should have the same angles sequence and the same number of arcs, and we may assume SL1 = (α₀, u₁, r_u1, α₁, ..., α_m) and SL2 = (α₀, v₁, r_v1, α₁, ..., α_m).

For instance, there are two skyline lists, ⎧

⎨ ⎩

SL1 =0◦, u₀, r_u₀, 40◦, u₁, r_u1, 100◦, u₂, r_u₂, 150◦ SL2 =0◦, v₀, r_v₀, 60◦, v₁, r_v1, 120◦

We split the skyline lists into smaller arcs. After the step 1 in procedure

M erge, they will be refined to

⎧ ⎨ ⎩ SL1 =0◦, u₀, r_u0, 40◦, u₁, r_u₁, 60◦, u₁, r_u₁, 100◦, u₂, r_u2, 120◦, u₂, r_u₂, 150◦ SL2 = 0◦, v₀, r_v 0, 40 ◦ , v₀, r_v 0, 60 ◦ , v₁, r_v 1, 100 ◦ , v₁, r_v 1, 120 ◦ , v₁, r_v 1, 150 ◦

In the second step, for each 0 ≤ i ≤ m, new skyline arcs are decided

from (α_i, u_i, r_ui, α_i+1) and (α_i, v_i, r_vi, α_i+1). Given two arcs (α, u, r_u, β) and

(α, v, r_v, β) with the same angle span, we have following three cases to decide

the new skyline arc.

Case 1: arc(α, u, r_u, β) and arc(α, v, r_v, β) have no intersection. One arc

is closer to o than the other, and the arc closer to o will be removed from the skyline. The outer arc remains on the skyline. For instance, in Figure 3.6(a), arc ab and cd have no intersection point. Arc ab are the new skyline arc of arcs ab and cd.

Case 2: arc(α, u, r_u, β) and arc(α, v, r_v, β) intersect at one point e. Let

f (a) (b) (c) d b c a b d c a e o o o c a e d b

Figure 3.6: Merge two arcs with the same corresponding angles.

used in Case 1, new skyline arcs can be decided from arc(α, u, r_u, γ) and

arc(α, v, r_v, γ); and arc(γ, u, r_u, β) and arc(γ, v, r_v, β). For instance, in Figure

3.6(b), e is the intersection point of arcs ab and cd. Arcs ae and ed are the new skyline arcs of arcs ab and cd.

Case 3: arc(α, u, r_u, β) and arc(α, v, r_v, β) intersect at two points e, f .

Let γ₁ and γ₂(γ₁ < γ₂) be the angles corresponding to the intersection points. Applying the principle used in Case 1, new skyline arcs can be decided from

arc(α, u, r_u, γ₁) and arc(α, v, r_v, γ₁); arc(γ₁, u, r_u, γ₂) and arc(γ₁, v, r_v, γ₂); and

arc(γ₂, u, r_u, β) and arc(γ₂, v, r_v, β). For instance, in Figure 3.6(c), e, f is the

intersection points of arcs ab and cd. Arcs ae, ef and f b are the new skyline arcs of arcs ab and cd.

In the first and second steps, one arc may be split into several pieces. So, in the last step, before returning the new skyline, we try to combine neighboring skyline arcs if they are from the same disk. This step could reduce the overhead in splitting skyline lists, if there are many small arcs in the refined skyline lists.

Chapter 4

Time Complexity Analysis

In this chapter, we show that the time complexity of the proposed algorithm is O (n ln n). In [5], it has been shown that the time complexity of the al-gorithm that computes the minimum local disk cover set for homogeneous networks (i.e., all nodes have the same radius) is bounded by O (n ln n). Since homogeneous networks are special cases of heterogeneous networks, the time complexity for the algorithm that computes the minimum local disk cover set for heterogeneous networks is at least O (n ln n). Hence, the proposed algo-rithm is with the optimal time complexity.

Skyline is a divide-and-conquer algorithm. The time complexity can be

formulated by a recursive equation ⎧

⎨ ⎩

T (n) = O (1) if n = 1,

T (n) = 2Tn₂+ f (n) otherwise.

Here T (n) is the time complexity of Skyline. f (n) is the time complexity of M erge. f (n) is linear with respect to the number of arcs. If we may have

f (n) = O (n), according to the master theorem [22], then we have T (n) =

O (n log n). In Lemma 8, we will show that the number of arcs in a skyline is

(b) u i u ui+1 p x o (a) 1 u 2 u 3 u o 4

Figure 4.1: The configuration B(u₄, r₄) contributes 3 arcs in the skyline. First we present the geometric arguments and related lemmas to support the proof in the following section. Then we give the time complexity of the algorithm.

4.1

Geometric Property of Disks

We would like to show that the number of arcs in a skyline can not be more than 2n. This can be proved that if we can show that adding a disk into a disk set increases the number of skyline arcs by at most 2. However, this is not true in the general case. For a positive integer k ≥ 3, let u_i =1₂cos2πi_k ,1₂sin 2πi_k for any i = 0, 1,· · · , k−1, the center of these disks averagely located on ∂Bo,1₂. In Figure 4.1(a), p is an intersection point of ∂B (u_i, 1) and ∂B (u_i+1, 1) outside B (o, 1), and x is the intersection point of u_iu_i+1 and op. We have u_i− x =

1

2sin πk, x − p =



1− (1₂sinπ_k)2, and o − p = 1₂ cosπ_k + 

1− (1₂ sinπ_k)2. Let r ∈ o − p , 1 + 1₂ be a constant. If we add B (o, r) to the disk set

{B (o, 1) , B(u1, r2), B(u2, r2), ..., B(uk, rk)}, then B (o, r) contributes k

dis-joint arcs to the skyline and doesn’t fully cover any previous skyline arc. Thus, adding B (o, r) increases the number of skyline arcs by k. Figure 4.1(b) illus-trates the configuration, for k = 3, B(u₄, r₄) contributes 3 arcs in the skyline,

where u₄ = o and r₄ ∈1₂cos π₃ + 

1− (1₂sinπ₃)2 , 1 + 1₂ 

.

However, if the disk is added into the disk set follows the decreasing order of radius, the claim will be true. Since the order to add disks into the disk set doesn’t change the final skyline, we change the arrangement of disks added. Therefore, in what follows, we assume disks are added according to their radii in the decreasing order of radius.

In the following, we introduce some lemmas about the geometry of disks.

Lemma 4 In the skyline of B₁, B₂, . . . , B_n, if B_n contributes at least 3 arcs, then we can pick B_i, B_j, B_k from B₁, B₂, . . . , B_n−1 such that B_n contributes 3 arcs in the skyline of B_i, B_j, B_k, B_n.

Proof. First, we claim that if B_n contributes at least 3 arcs in the skyline of

B_i, B_j, B_k, B_n, then B_ncontributes exactly 3 arcs in the skyline. Assume there are more than 3 arcs contributed by B_n. Since each arc is with 2 endpoints, there are at least 8 intersection points in ∂B_n. But ∂B_i, ∂B_j, ∂B_k have at most 6 intersection points with ∂B_n. This is a contradiction. Thus, our claim is true. So, to prove the lemma, it is enough to show that B_n contributes at least 3 arcs in the skyline of B_i, B_j, B_k, B_n.

Choose 3 arcs contributed by B_n in the skyline of B₁, B₂, . . . , B_n, and consider those disks whose boundaries intersect with ∂B_n at endpoints of the 3 arcs. These 6 endpoints are from intersection points of B_n with another 3, 4, 5, or 6 disks since the boundary of each disk can contribute either 1 or 2 intersection points with ∂B_n in the skyline. If there are 3 disks, the lemma is proved. So, we consider the case that there are more than 3 disks. By counting the endpoints, if there are more than 3 disks, at lease one intersects with ∂B_n exactly at 1 point in the skyline. If we remove a disk which intersects

does not decrease. Thus, in the skyline of remaining disks, B_n contributes at least 3 arcs, but the number of disks decrease by 1. Figure 4.2(a) illustrates the configuration in which the skyline contains 4 disks B₁, B₂, B₃, B₄ and B_n contributes 3 arcs. The arcs of B_nare presented in dotted line. We remove the

(b) B 3 B 2 B 4 B (a) 1 B 3 B 2 B 1

Figure 4.2: Proof of Lemma 4.

disk B₄ which intersects ∂B_n at exact 1 point in the skyline, and the number of arcs contributed by B_n doesn’t decrease, see Figure 4.2(b). Then, we repeat the process until there remains 3 disks, and then the lemma is proved.

Lemma 5 Assume two circle ∂B₁ and ∂B₂ have two intersection points a and

d. Let ac (and ab, respectively) be a diameters of B₁ (and B₂, respectively), and c (and b, respectively) is a point in arc cd (bd respectively). See Figure 4.3. If the angle ∠cab is obtuse, we have

b − c > 2 min (r1, r2) .

Proof. First, we explore an extreme case in which ∂B₁ and ∂B₂ are tangent,

i.e. c, a, b are in a line and a, d are overlapping and cab is π₂. See Figure 4.4. Since ac (ab respectively) is the diameter of B₁ (B₂ respectively), cca

c’ B _a B2 c d b’ b 1

Figure 4.3: The structure in Lemma 5. since cca and bba are also right angles, we have

b − c2 >a − c2+a − b2 =a − c2+  r₂ r₁ c _{− c} ₂ =a − c2+c− c2+  r₂ r₁ ₂ − 1  c_{− c}2 = (2r₁)2+  r₂ r₁ ₂ − 1  c_{− c}2 _{≥ (2r} 1)2 = (2 min (r1, r2))2.

Similarly, if r₂ ≤ r₁, we also have

b − c2 >a − c2+a − b2 =a − b2+  r₁ r₂ b _{− b} ₂ =a − b2+b − b2+  r₁ r₂ ₂ − 1  b_{− b}2 = (2r₂)2+  r₁ r₂ ₂ − 1  b _{− b}2 _{≥ (2r} 2)2 = (2 min (r1, r2))2.

Thus b − c > 2 min (r₁, r₂) . The lemma is correct for this extreme case. The inequality can be extended for general cases by the following simple

observation. Rotate B₁ and/or B₂ by a to let cab become smaller but

don’t let ac across ac and ab across ab. Let c denote the intersection of the ray ac and ∂B₁ and b denote the intersection of the ray ab and ∂B₂.

a 2 1 B 2 r 1 r c’ b’ b c B

Figure 4.4: A extreme case when ∂B₁ and ∂B₂ are tangent.

See Figure 4.5. We have a − b ≥ a − b and a − c ≥ a − c. Thus,

b_{− c}_{ ≥ b − c. So, the proof is complete.}

b’’ 2 1 B a c’ c b b’ c’’ B

Figure 4.5: Rotate B₁ and B₂ by a, and thenb− c ≥ b − c.

Lemma 6 Given an acute triangle, for each edge of the triangle, draw a circle

with the edge as a chord whose center is outside the triangle and radius is equal to the circumradius. Then, three circles intersect at the orthocenter.

Proof. Let Δabc be an acute triangle, C₁ be the circumcircle of Δabc, and C₂

(respectively, C₃ and C₄) be a circle with the circumradius of Δabc as its radius, and edge ab (respectively, bc and ac) as a chord, and its center outside Δabc. See Figure 4.6(a). To prove this lemma, it is enough to show that C₂,

f 1 C₃ C₂ C₁ C₄ C₂ e (b) a b c h d (a) a c d b g C

Figure 4.6: The configuration of Lemma 6.

C₃, C₄ contain the orthocenter of Δabc. Figure 4.6(b) illustrates the relation between C₁ and C₂, and we want to prove C₂ contains the orthocenter.

Let f denote the perpendicular foot on ab. Let h denote the intersection point of C₁ and line cf , and e(respectively, d) denote the intersection points of C₂ and line cf outside (respectively, inside) Δabc. Let g be the intersection point of lines bd and ac. If we can show g is the perpendicular foot on ac, then

d is the orthocenter. Since C₁ and C₂ have the same radius, and ab and ce are

perpendicular, acbe is a rhombus. In the rhombus, ∠aef = ∠acf. Since ∠aed

and ∠abd correspond to the same arc ad and they are inscribed angles on C₂, ∠aed = ∠abd. Thus, ∠abd = ∠acf. In Δdbf and Δdcg, ∠dbf = ∠dcg and

∠bdf = ∠cdg, so ∠bfd = ∠cgd. Since ∠bfd = 90◦_{, ∠cgd = 90}◦_{. bg⊥ac and}

cf⊥ab, therefore d is the orthocenter of Δabc.

Similarly, we can prove C₃, C₄ also contain the orthocenter. So, the lemma is proved.

Then, we have the following corollary.

Corollary 7 Given an acute or right triangle, for each edge of the triangle,

draw a circle with the edge as a chord whose center is outside the triangle and radius is larger than the circumradius. Then, three circles have no intersection.

We apply simple counting techniques. Since each arc has two endpoints, each endpoint is an intersection point of two disks, and two disks have at most 2 intersection points.

Lemma 8 The number of arcs in a skyline of n disks is upper bounded by 2n. Proof. We prove this by mathematical induction on n. Without loss of gen-erality, we may assume all n nodes have arcs in the skyline and B₁, B₂,· · · , B_n

have been sorted according to their radii in decreasing ordering.

If n = 1, there is only one disk, and thus, the skyline consists of one arc. If n = 2, two disks intersect at 2 points. There are at most 2 arcs in the skyline. See Figure 4.7(a). If n = 3, the topology can be categorized into 2 configurations, like Figure 4.7(b) and Figure 4.7(c). Figure 4.7(b) illustrates one configuration in which each disk contains one intersection point of the other two disks, and the skyline is composed of 3 arcs. Figure 4.7(c) illustrates the other configuration in which one disk contains two intersection points of the other two disks. In addition, 3 disks is allowed to have a common intersection point like Figure 4.8. The skyline is composed of 3 or 4 arcs. No matter how, the number of arcs is no more than 2n.

Now, assume that as n = k, the skyline contains at most 2k arcs. If we can show that after a disk B_k+1 is added into the set, the number of arcs in the skyline increases at most by two, then the new skyline contains no more than 2 (k + 1) arcs, and the proof is complete. Without loss of generality, we may assume B_k+1 is the disk with the smallest radius among B₀, B₁, B₂, . . . , B_k+1

since it doesn’t change the skyline. We denote the number of arcs in the sky-line of B₁, B₂, . . . , B_k as Sky(B₁, B₂, . . . , B_k). Now, we are going to prove this

by contradiction. Assume B_k+1 can contribute at least 3 arcs. According to

arcs in the skyline of B₁, B₂, B₃, B_k+1. According to the two possible configu-rations of B₁, B₂, B₃ like Figure 4.7(b) or Figure 4.7(c), we discuss the problem in the two cases.

(c)

(a) (b)

Figure 4.7: n≤ 3, the skyline contains 2n arcs at most.

Figure 4.8: 3 disks have a common intersection point.

First, we consider the configuration like Figure 4.7(b). Let a be the inter-section point of B₁ and B₂ not in B₃; b be the intersection point of B₁ and B₃ not in B₂; and c be the intersecting point of B₂ and B₃ not in B₁. In order to contribute 3 arcs, B_k+1 must intersect with 3 disks and contain a, b, c. Now, the problem is discussed by following two cases: (1) Δabc is an acute or right triangle; and (2) Δabc is an obtuse triangle.

Case 1: Δabc is an acute or right triangle. Let r_c be the circumradius

of Δabc. Since Δabc is an acute or right triangle and B_k+1 contains a, b, c, we have r_k+1 is larger than r_c. In addition, since B_k+1 is the smallest among

Corollary 7, if r₁, r₂, r₃ are larger than r_c, B₁, B₂, B₃ have no intersection. This is a contradict to the fact that the intersection of B₁, B₂, B₃ is not empty.

Case 2: Δabc is an obtuse triangle. It doesn’t affect the correctness of

following argument if we assume ∠cab is obtuse. Since B_k+1 contribute three arcs in the skyline of B₁, B₂, B₃, B_k+1 and r_k+1 ≤ r₁, r₂, r₃, degree of each arc in the skyline of B₁, B₂, B₃ must be larger than π. If ac is a diameter of B₂ and ab is a diameter of B₁, c and b are in the skyline of B₁, B₂, B₃. c is in the arc ac intersected by arc ab and not contained in B₁ and b is in the arc

ab intersected by arc ac and not contained in B₂. According to Lemma 5, if ∠cab is obtuse, we have b − c > 2 min (r1, r2). On the other hand, since Bk+1

contains Δabc, we have r_k+1 ≥ 1₂b − c, and 2r_k+1 ≥ b − c > 2 min (r₁, r₂). Thus, we have a contradiction.

Next, we consider the configuration like Figure 4.7(c). Without affecting

the correctness of following argument, we assume B₃ is the one containing

two intersection point of the boundary of other two disk. Let b, e denote intersection points of B₁ and B₃, and c, f denote intersection points of B₂ and B₃. To contribute three arcs to the skyline of B₁, B₂, B₃, B_k+1, B_k+1 should cover at least 3 intersection points. According to the number of covered intersection points, there are two cases.

Case 3: If B_k+1 cover exactly 3 intersection points, like Figure 4.9(a), then

B_k+1 must have 3 intersection points with B₃. This is not possible to happen, since 2 disks have at most 2 intersection points.

Case 4: If B_k+1 cover exactly 4 intersection points, like Fig. 4.9(b), let

e, f denote the two intersection points covered by the same arc of B_k+1 and

a is a intersection point of B₁ and B₂ that is closer to the arc covering e, f . Since B_k+1 is smaller than B₁, B₂, B₃, arcs bc of B₃ outside B₁, B₂, arc be of

B₁ outside B₃, and arc cf of B₂ outside B₃ are larger than π. So, the angle bac > π/2, the diameter of B1 with one endpoint at a is outside of B2, and

c 1 B₁ B2 B₃ B2 B₃ b e f _{e f} (a) (b) b c a B

Figure 4.9: In the configuration, B_k+1 doesn’t contribute 3 arcs.

the diameter of B₂ with one endpoint at a is outside of B₁. According to

Lemma 5, just like Case 2, we have 2r_k+1 >b − c > 2 min (r₁, r₂). This is a contradiction.

According to previous discussion, B_k+1 can’t contribute 3 arcs to the sky-line of B₁, B₂,· · · , B_k+1, and therefore, the number of arcs in the skyline

B₁, B₂,· · · , B_k+1 is at most 2(k + 1). By mathematical induction, we con-clude that the number of arcs in the skyline of n disks is upper bounded by 2n.

Now we show that our algorithm has time complexity O (n log n) .

4.2

Time Complexity of Algorithm

Theorem 9 The time complexity of Skyline is O (n log n), where n is the

number of disks.

Proof. The running time T (n), of Skyline has the recursive equation ⎧

⎨ ⎩

T (n) = O (1) if n = 1,

If n = 1, we have T (n) = O (1). If n ≥ 2, Skyline executes Skyline twice

but with half problem size and then executes M erge once. According to

Lemma 8, the time complexity of M erge is O (n). Hence, according the master theorem [22], T (n) = O (n). The proposed algorithm has the time complexity

Chapter 5

Simulations and Conclusions

5.1

Simulations

Simulations are done to compare the performance of forwarding set algorithms, including the blind flooding, skyline algorithm, selecting forwarding set [6], greedy algorithm, and optimal algorithm. In the simulation, nodes are de-ployed over a 12.5×12.5 square. A source node u is placed at the center of the deployment region. If n is the average number of neighbors, 12.5_πr22n nodes are

generated with uniform distribution over the deployment region, where r is the transmission radius of a node. Two types of networks, homogeneous net-works and heterogeneous netnet-works, are considered. In homogeneous netnet-works, all nodes are with the same transmission radius 1. In heterogeneous networks, every node may have different transmission radius that is generated by a func-tion which produces randomly a real number between 1 and 2, including the source node. Bidirectional links are considered. 200 random point sets are generated in the simulation. For each random point set, the forwarding sets of node u are calculated by each algorithm. For each algorithm, we calculate the average size of forwarding set and the distribution of the size of forwarding sets.

Since it is still open that whether the minimum forwarding set problem of disk graphs is NP-Complete or not, we use brute force algorithm to find the minimum forwarding set. In the blind flooding, all 1-hop neighbors relay messages, so the number of forwarding nodes is equal to the number of 1-hop neighbors. In the skyline algorithm, each node chooses a subset of one-hop neighbors that contribute arcs to the skyline as its forwarding set. The selecting forwarding set algorithm introduced in [6] for homogeneous networks is with constant approximation ratios. In the greedy algorithm is that each node iteratively chooses a 1-hop neighbor covering the maximum number of 2-hop neighbors not yet covered, and completes when all 2-hop neighbors have been covered.

Excepting the skyline algorithm that only needs 1-hop information, the optimal algorithm, greedy algorithm, and selecting forwarding set algorithm [6], need one-hop and two-hop information to calculate the forwarding set. To obtain the information of 1-hop neighbors, each node periodically sent HELLO messages to report its current status. To obtain information of 2-hop information, the HELLO messages should contain the sender’s 1-hop neighbor list are sent periodically.

5.1.1

Homogeneous Networks

In homogeneous networks, all nodes have the same transmission radii 1. We run simulation for the blind flooding, skyline algorithm, selecting forwarding set algorithm [6], greedy algorithm, and optimal algorithm. The average num-ber of forwarding nodes of five algorithms is illustrated in Figure 5.1. The

x-axis denotes the average number of 1-hop neighbors of node u. The y-axis

denotes the average number of forward nodes. The five curves in the figure form top to down are corresponding to the blind flooding, skyline algorithm,

selecting forwarding set [6], greedy algorithm and optimal algorithm, respec-tively. ˛̂̀̂˺˸́˸̂̈̆ʳˡ˸̇̊̂̅˾̆ ˃ ˈ ˄˃ ˄ˈ ˅˃ ˅ˈ ˃ ˈ ˄˃ ˄ˈ ˅˃ ˅ˈ ˔̉˸̅˴˺˸ʳ˄ˀ˻̂̃ʳ́˸˼˺˻˵̂̅̆ ˹̂̅̊˴̅˷˼́˺ʳ̆˸̇ ˹˿̂̂˷˼́˺ ̆˾̌˿˼́˸ ̆˸˿˸˶̇ʳ˙˦ ˺̅˸˸˷̌ ̂̃̇˼̀˴˿

Figure 5.1: The algorithms are compared in homogeneous networks.

The selecting forwarding set algorithm and greedy algorithm usually gen-erate smaller forwarding sets than the skyline algorithm. But instead of 2-hop information, the skyline algorithm needs only 1-hop information. So, the se-lecting forwarding set algorithm and greedy algorithm demand more resource than the skyline algorithm. In addition, if nodes have mobility, more efforts are needed to maintain 2-hop information then the nodes cost a lot of space and time in collecting two-hop information. Thus, the skyline algorithm is more easily to implement than the other two algorithms in wireless ad hoc networks.

Figure 5.2 and Figure 5.3 illustrate the distribution of the number of for-warding nodes. Figure 5.2 is of the network in which nodes have 10 1-hop

neighbors in average, and Figure 5.3 is of the network in which nodes have 20 1-hop neighbors in average. The x-axis is corresponding to the number of forwarding nodes, and y-axis is corresponding to the number of random point sets. ˛̂̀̂˺˸́˸̂̈̆ʳˡ˸̇̊̂̅˾̆ʿʳ˔̉˸̅˴˺˸ʳ˄ˀ˻̂̃ʳːʳ˄˃ ˃ ˅˃ ˇ˃ ˉ˃ ˋ˃ ˄˃˃ ˄˅˃ ˃ ˄ ˅ ˆ ˇ ˈ ˉ ˊ ˋ ˌ ˄˃ ˄˄ ˄˅ ˄ˆ ˄ˇ ˄ˈ ̇˻˸ʳ́̈̀˵˸̅ʳ̂˹ʳ˹̂̅̊˴̅˷˼́˺ʳ́̂˷˸̆ ́̈̀˵˸̅ʳ̂˹ʳ́̂˷˸̆ ˹˿̂̂˷˼́˺ ̆˾̌˿˼́˸ ̆˸˿˸˶̇˼́˺ʳ˙˦ ˺̅˸˸˷̌ ̂̃̇˼̀˴˿

Figure 5.2: The distribution of forward nodes with 1-hop neighbors = 10.

If an algorithm is with better performance, in most cases, it generates forwarding sets with small size. So, the distribution is located at the left hand side in the figures. Without too much surprise, the greedy algorithm and selecting forwarding set algorithm generate smaller forwarding sets than the skyline algorithm. Note that in Figure 5.3, since the main part of forwarding nodes is distributed below 25, for simplicity, we show the main part instead of all, and there are some nodes with forwarding set larger than 25 nodes generated by the blind flooding algorithm.

˛̂̀̂˺˸́˸̂̈̆ʳˡ˸̇̊̂̅˾̆ʿʳ˔̉˸̅˴˺˸ʳ˄ˀ˻̂̃ʳːʳ˅˃ ˃ ˄˃ ˅˃ ˆ˃ ˇ˃ ˈ˃ ˉ˃ ˊ˃ ˃ ˄ ˅ ˆ ˇ ˈ ˉ ˊ ˋ ˌ ˄˃ ˄˄ ˄˅ ˄ˆ ˄ˇ ˄ˈ ˄ˉ ˄ˊ ˄ˋ ˄ˌ ˅˃ ˅˄ ˅˅ ˅ˆ ˅ˇ ˅ˈ ̇˻˸ʳ́̈̀˵˸̅ʳ̂˹ʳ˹̂̅̊˴̅˷˼́˺ʳ́̂˷˸̆ ́̈̀˵˸̅ʳ̂˹ʳ́̂˷˸̆ ˹˿̂̂˷˼́˺ ̆˾̌˿˼́˸ ̆˸˿˸˶̇˼́˺ʳ˙˦ ˺̅˸˸˷̌ ̂̃̇˼̀˴˿

Figure 5.3: The distribution of forward nodes with 1-hop neighbors = 20.

5.1.2

Heterogeneous Networks

Since the selecting forwarding set algorithm [6] doesn’t work for heterogeneous networks, we don’t consider it in the simulation for heterogeneous networks. We run simulation for the blind flooding algorithm, skyline algorithm, greedy algorithm, and optimal algorithm. In heterogeneous networks, the transmis-sion range of each node is randomly assigned between 1 and 2.

Figure 5.4 illustrated the average number of forward nodes. The x-axis denotes the average number of 1-hop neighbors of node u. The y-axis denotes the average number of forward nodes. In the figure, there are four curves, form top to down, corresponding to the blind flooding algorithm, skyline algorithm, greedy algorithm, and optimal algorithm.

Because we use the bidirectional link, the number of 1-hop of source node

u may less than the average 1-hop neighbors.

˛˸̇˸̅̂˺˸́˸̂̈̆ʳˡ˸̇̊̂̅˾̆ ˃ ˅ ˇ ˉ ˋ ˄˃ ˄˅ ˄ˇ ˄ˉ ˄ˋ ˃ ˈ ˄˃ ˄ˈ ˅˃ ˅ˈ ˔̉˸̅˴˺˸ʳ˄ˀ˻̂̃ʳ́˸˼˺˻˵̂̅̆ ˹̂̅̊˴̅˷˼́˺ʳ̆˸̇ ˹˿̂̂˷˼́˺ ̆˾̌˿˼́˸ ˺̅˸˸˷̌ ̂̃̇˼̀˴˿

Figure 5.4: The algorithms are compared in heterogeneous networks. networks in which nodes with 10 1-hop neighbors in average. The x-axis is corresponding to the number of forwarding nodes, and y-axis is corresponding to the number of random point sets.

Last, we want to point out a drawback of the skyline algorithm that is due to the directional links and bidirectional links. In Figure 5.6, node u has three 1-hop neighbors u₁, u₂, u₃ and u₄, u₅ are 2-hop neighbors. u₄ is a neighbor of

u₁, and u₅ is a neighbor of u₂. The transmission range of u₃ can cover u₄ and

u₅, but the transmission range of u₄ or u₅ can not cover u₃. So, u₄ and u₅ are not neighbors of u₃. The optimal forwarding set under the bidirectional link model is {u₁, u₂}, but the skyline set is {u₃}. Since the skyline algorithm

utilizes only 1-hop information, it can’t know the information about 2-hop neighbors. We leave this problem as our one of our future works.