Sharp Thresholds for Relative Neighborhood
Graphs in Wireless Ad Hoc Networks
Chih-Wei Yi, Member, IEEE, Peng-Jun Wan, Lixin Wang, and Chao-Min Su
Abstract—In wireless ad hoc networks, relative neighborhood graphs (RNGs) are widely used for topology control. If every node
has the same transmission radius, then an RNG can be locally constructed by using only one hop information if the transmission radius is set no less than the largest edge length of the RNG. The largest RNG edge length is called the critical transmission radius for the RNG. In this paper, we consider the RNG over a Poisson point process with mean density𝑛 in a unit-area disk. Let𝛽0= √ 1/(2 3− √ 3 2𝜋 )
≈ 1.6. We show that the largest RNG edge length is asymptotically almost surely at most𝛽√ln 𝑛
𝜋𝑛 for any fixed 𝛽 > 𝛽0 and at least 𝛽
√ ln 𝑛
𝜋𝑛 for any fixed 𝛽 < 𝛽0. This implies that the threshold width of the critical transmission radius is 𝑜(√ln 𝑛
𝑛
)
. In addition, we also prove that for any constant𝜉, the expected number of RNG edges whose lengths are not less than𝛽0
√ ln 𝑛+𝜉
𝜋𝑛 is asymptotically equal to 𝛽0 2
2 𝑒−𝜉.
Index Terms—Wireless ad hoc networks, relative
neighbor-hood graphs, critical transmission radii, Poisson point processes, thresholds.
I. INTRODUCTION
A
wireless ad hoc network is a collection of radio devices located in a geographic region. Each node is equipped with an omni-directional antenna with limited transmission power. A communication session is established either through a single-hop radio transmission if the communication parties are close enough, or through relaying by intermediate de-vices otherwise. Because of no need for fixed infrastructures, wireless ad hoc networks can be flexibly deployed at low cost for varying missions such as battlefield decision making, emergency disaster relief and environmental monitoring.In slow fading channel models, signals transmitted at power level 𝑝 will be received at distance 𝑑 with strength 𝑐𝑑−𝜅𝑝 where𝜅 is an environment-related constant, called path loss
factors, and 𝑐 is an antenna-related constant. A received
signal can be decoded only if its strength is not less than a threshold. In a homogeneous network, due to the same (maximal) transmission power and similar environments, we
Manuscript received September 22, 2006; revised October 1, 2007; accepted June 7, 2009. The associate editors coordinating the review of this paper and approving it for publication were R. Negi and V. K. Bhargava.
C.-W. Yi (corresponding author) and C.-M. Su are with the Department of Computer Science, National Chiao Tung University, Hsinchu City 30010, Taiwan (e-mail: yi@cs.nctu.edu.tw, adjaumin@et4.thit.edu.tw).
P.-J. Wan and L. Wang are with the Department of Computer Science, Illinois Institute of Technology, Chicago, IL 60616 (e-mail: wan@cs.iit.edu, wanglix@iit.edu).
Part of Dr. Wan’s work was conducted during his visit at the Department of Computer Science, City University of Hong Kong, Hong Kong.
Digital Object Identifier 10.1109/TWC.2010.02.060736
assume all nodes have the same (maximal) transmission radius
𝑟. The induced network topology is called 𝑟-disk graphs in
which two nodes have an edge if and only if the distance between them is at most 𝑟. In many applications, a large
number of ad hoc wireless devices are randomly deployed in a (finite) deployment region. Consequently, it is natural to represent the vertex set by a random point process in a bounded region, and the induced 𝑟-disk graphs are called
random geometric graphs [1] which are a bounded version of the network model proposed by Gilbert (1961) [2].
Without fixed infrastructures, virtual backbones are con-structed and maintained for routing packets. In [3] and [4], relative neighborhood graphs (RNGs) [5] and Gabriel graphs (GGs) [2] are used to construct planar virtual backbones. However, most previous works assumed underlying networks are connected, or implicitly assumed the transmission radius can be arbitrarily increased if necessary. In addition, in order to locally construct virtual backbones, only part of structures will be constructed. This could increase the dilation factor and cause more energy consumption. In this work, by investigating the largest RNG edge length, we show that "complete" RNGs can be locally constructed with high probability by using only 1-hop information if the transmission radius is properly set.
A graph property is said increasing if a graph has this property, then all its supergraphs are with this property. For a given set of nodes and an increasing property, the smallest transmission radius such that the induced network topology has the specified property is called the critical transmission radius for this property. The property of an𝑟-disk graph
with-out isolated nodes is increasing, and the corresponding critical transmission radius is equal to the largest nearest-neighbor link length. The largest nearest-neighbor link problem had been studied by Dette and Henze (1989) [6]. Connectivity is another increasing property. Philips et al. (1989) [7] proved that if the average number of neighbors is less than𝛽 ln 𝑛 for some
constant 𝛽 < 1, the network is almost surely disconnected.
So, there is no (finite) magic number for connectivity. They also conjectured that 𝛽 > 1 is a sufficient condition for
connectivity. Penrose (1997) [8] proved that the two critical transmission radii for connectivity and for no isolated nodes are asymptotically equal. Santi (2003) [9] studied the connec-tivity problem on a more general network model in which nodes are deployed in a cube [0, 𝑙]𝑑 for𝑑 = 1, 2, 3 and the
transmission radius is a function of 𝑙 and 𝑛. Kozma et al.
(2004) [10] proved that the largest Delaunay triangulation edge length is𝑂 ( 3 √ ln 𝑛 𝑛 )
. Goel et al. (2004) [11] showed that all monotonic properties have sharp thresholds and the threshold
u v
Fig. 1. The shaded lunar area in the intersection of two disks with radii
∥𝑢 − 𝑣∥ and centered at 𝑢 and 𝑣 respectively. 𝑢𝑣 is an RNG edge if and
only if no other node is in the shaded area.
width is upper bounded by𝑂(ln3/4𝑛
√𝑛 )in the plane.
In RNGs, two nodes have an edge if and only if there are no other nodes in the intersection of two disks centered respectively at the two nodes and with the distance between them as radii. In Fig. 1,𝑢 and 𝑣 have an RNG edge between
them if and only if the shaded area, called a lens, doesn’t contain any other node. Assume all nodes have the same transmission radius 𝑟. If 𝑟 is no less than the largest RNG
edge length, the RNG is a subgraph of the 𝑟-disk graph.
So, only edges of the 𝑟-disk graph are candidates of RNG
edges. For any two neighboring nodes, all nodes in the corresponding lens are neighbors of these two nodes. Thus, 1-hop information is enough to decide RNG edges. On the other hand, if𝑟 is less than the largest RNG edge length, it is
easy to construct a counter example in which RNG edge can not be decided according to only 1-hop information. Hence, the largest edge length of an RNG is the critical transmission radius to construct the RNG by using only 1-hop information. In this paper, we assume a wireless ad hoc network is represented by a Poisson point process in a unit-area disk with mean density𝑛, denoted by 𝒫𝑛, and all nodes have the same maximal transmission radius𝑟𝑛which is a function of𝑛. We use𝒢𝑛 to denote the RNG over𝒫𝑛 and𝜆𝑛 to denote the maximal edge length of𝒢𝑛. Let 𝛽0 =
√ 1/(2 3− √ 3 2𝜋 ) . We prove that for any constant 𝛽, 𝜆𝑛 is asymptotically almost surely at most 𝛽√ln 𝑛
𝜋𝑛 if 𝛽 > 𝛽0, and at least 𝛽 √
ln 𝑛 𝜋𝑛 if 𝛽 < 𝛽0. This result implies that the threshold width of
the critical transmission radius is 𝑜(√ln 𝑛 𝑛 ) . Furthermore, let 𝑙𝑛 = 𝛽0 √ ln 𝑛+𝜉
𝜋𝑛 for some constant 𝜉 and 𝒩𝑛 denote the number of RNG edges whose lengths are at least 𝑙𝑛. For convenience, these edges are called long (RNG) edges. We prove that the expected number of long RNG edges is asymptotically equal to 𝛽022𝑒−𝜉.
In what follows,∥𝑥∥ denotes the Euclidean norm of a point 𝑥 ∈ ℝ2.∣𝐴∣ is shorthand for 2-dimensional Lebesgue measure
(or area) of a measurable set𝐴 ⊂ ℝ2. All integrals considered
will be Lebesgue integrals. The topological boundary of a set
𝐴 ⊂ ℝ2is denoted by∂𝐴. The disk of radius 𝑟 centered at 𝑥 is
denoted by𝐵 (𝑥, 𝑟). The unit-area disk centered at the origin
is denoted by𝔻, and 𝑅0 = 1/√𝜋 is the radius of 𝔻. Let 𝑛
denote the node density of the Poisson point process. An event is said to be asymptotic almost sure (abbreviated by a.a.s.) if it occurs with a probability converges to one as 𝑛 → ∞.
The symbols 𝑂, 𝑜, ∼ always refer to the limit 𝑛 → ∞. To
avoid trivialities, we tacitly assume 𝑛 to be sufficiently large
if necessary. For simplicity of notation, the dependence of sets and random variables on𝑛 will be frequently suppressed.
The rest of this paper is organized as follows. In Section II, we give a brief of our main results. In Section III, we introduce some terminologies and present several useful geometric and integral lemmas. In Section IV, we derive the asymptotic length of the longest RNG edge. In Section V, we derive the asymptotic expected number of long RNG edges. We give simulation results in Section VI and summarize this paper in Section VII.
II. MAINRESULTS
Recall that 𝒫𝑛 denotes a Poisson point process in a unit-area disk with mean density 𝑛, 𝒢𝑛 denotes the RNG over 𝒫𝑛, 𝜆𝑛 denotes the largest edge length of 𝒢𝑛, and
𝛽0 = √ 1/(2 3− √ 3 2𝜋 )
≈ 1.6. One of our main results is the
following theorem.
Theorem 1: For any constant𝜀 > 0, as 𝑛 → ∞, we have
Pr [ (1 − 𝜀) 𝛽0 √ ln 𝑛 𝜋𝑛 ≤ 𝜆𝑛 ≤ (1 + 𝜀) 𝛽0 √ ln 𝑛 𝜋𝑛 ] → 1. Let 𝑟𝑛 = 𝛽 √ ln 𝑛
𝜋𝑛. According to Theorem 1, the 𝑟𝑛-disk graph over𝒫𝑛 a.a.s. contains𝒢𝑛 if 𝛽 > 𝛽0, and on the other
hand, the 𝑟𝑛-disk graph a.a.s. does not contain 𝒢𝑛 if𝛽 < 𝛽0.
Therefore,𝛽0 is corresponding to the threshold of the critical
transmission radius for RNGs. For reference, we remark that
𝛽 = 1 is corresponding to the threshold for connectivity [12]
[13]. The threshold width of RNGs implied by Theorem 1 is
𝑜(√ln 𝑛 𝑛
)
that is far smaller than𝑂(ln3/4𝑛 √
𝑛 )
given in [11]. The next theorem gives the asymptotic expected number of long RNG edges. Recall that 𝑙𝑛 = 𝛽0
√
ln 𝑛+𝜉
𝜋𝑛 for some constant 𝜉, and 𝒩𝑛 denotes the number of long RNG edges whose lengths are not less than 𝑙𝑛.
Theorem 2: For the expected number of long RNG edges,
we have
lim
𝑛→∞E [𝒩𝑛] =
𝛽02
2 𝑒−𝜉.
Since Pr [𝑋 = 0] = 1 − Pr [𝑋 ≥ 1] ≥ 1 − E [𝑋] for any non-negative integer value RV𝑋,
Pr [ 𝜆𝑛 < 𝛽0 √ ln 𝑛 + 𝜉 𝜋𝑛 ] = Pr [𝒩𝑛= 0] ≥ 1 − E [𝒩𝑛] ∼ 1 − 𝛽202𝑒−𝜉. Therefore, lim 𝜉→∞𝑛→∞lim Pr [ 𝜆𝑛 < 𝛽0 √ ln 𝑛 + 𝜉 𝜋𝑛 ] = 1,
and thus 𝜉 → ∞ is an a.a.s. sufficient condition for 𝜆𝑛 <
𝛽0
√
ln 𝑛+𝜉 𝜋𝑛 .
π 1 (1) Dr Dr(2) (0) Dr r r
Fig. 2. The partition of the unit-area disk𝔻.
III. PRELIMINARIES
In this section, we introduce some terminologies and lem-mas that will be used to prove our main results. For a given real number𝑟 ∈ (0, 𝑅0), the unit-area disk 𝔻 is partitioned
into 𝔻𝑟(0), 𝔻𝑟(1) and 𝔻𝑟(2) as shown in Fig. 2: 𝔻𝑟(0) is the disk of radius1/√𝜋 − 𝑟 centered at the origin; 𝔻𝑟(1) is the annulus of radii 1/√𝜋 − 𝑟 and √1/𝜋 − 𝑟2 centered at
the origin; and𝔻𝑟(2) is the annulus of radii√1/𝜋 − 𝑟2 and 1/√𝜋 centered at the origin. Then,
∣𝔻𝑟(0)∣ =(1 −√𝜋𝑟)2,
∣𝔻𝑟(1)∣ = 2𝜋𝑟(1/√𝜋 − 𝑟),
∣𝔻𝑟(2)∣ = 𝜋𝑟2.
For any two points𝑢 and 𝑣, let 𝐿𝑢𝑣 denote the lens𝐿𝑢𝑣 =
𝐵 (𝑢, ∥𝑢 − 𝑣∥) ∩ 𝐵 (𝑣, ∥𝑢 − 𝑣∥), i.e. the intersection of two
disks centered at𝑢 and 𝑣, respectively, and with radii ∥𝑢 − 𝑣∥.
We have ∣𝐿𝑢𝑣∣ = 𝛽1 02𝜋 ∥𝑢 − 𝑣∥ 2 where𝛽0= √ 1/(2 3− √ 3 2𝜋 )
. For a lens𝐿𝑢𝑣, the middle point of𝑢 and 𝑣 is called its center, the line segment 𝑢𝑣 is called its
waist, and the two intersection points of∂𝐵 (𝑢, ∥𝑢 − 𝑣∥) and ∂𝐵 (𝑣, ∥𝑢 − 𝑣∥) are called its vertices. A lens is called an
𝑟-lens if the length of its waist is2𝑟. A lens is feasible if its waist is contained in𝔻. In what follows, we only consider feasible lenses. Assume∥𝑢 − 𝑣∥ = 2𝑟. Then, if the center of 𝐿𝑢𝑣is in 𝔻√
3𝑟(0), the lens is contained in 𝔻 and ∣𝐿𝑢𝑣∩ 𝔻∣ = ∣𝐿𝑢𝑣∣;
if the center of 𝐿𝑢𝑣 is in 𝔻╲𝔻𝑟(2), at least a half lens is contained in 𝔻 and ∣𝐿𝑢𝑣∩ 𝔻∣ ≥ 12∣𝐿𝑢𝑣∣. The next lemma giving a lower bound for the area of the union of two lenses is from [14] (Lemma 2).
Lemma 3: Assume 𝑐 = 0.039, 𝑅 > 0, and 𝑎1, 𝑏1, 𝑎2, 𝑏2∈
ℝ2. Let𝑧1= 1
2(𝑎1+ 𝑏1), 𝑟1= ∥𝑎1− 𝑏1∥, 𝑧2= 12(𝑎2+ 𝑏2),
and𝑟2= ∥𝑎2− 𝑏2∥. If 𝑟1, 𝑟2∈[12𝑅, 𝑅],∥𝑧1− 𝑧2∥ ≤√3𝑅, 𝑎1, 𝑏1 /∈ 𝐿𝑎2𝑏2, and𝑎2, 𝑏2 /∈ 𝐿𝑎1𝑏1, then
∣𝐿𝑎1𝑏1∪ 𝐿𝑎2𝑏2∣ − ∣𝐿𝑎1𝑏1∣ ≥ 𝑐𝑅 ∥𝑧1− 𝑧2∥ .
An 𝜀-tessellation is to divide the plane by vertical and
horizontal lines into grid cells with width 𝜀. Without loss
of generality, we assume the 𝑥-axis and 𝑦-axis are one of
the horizontal lines and vertical lines, and the origin is at a
Fig. 3. A polyquadrate is a collection of grids that intersect with a convex polygon.
corner of a grid cell. A collection of grid cells intersecting with a convex polygon (or a compact convex set) is called a polyquadrate. For example, in Fig. 3, the shaded grid cells form a polyquadrate. The horizontal span of a polyquadrate is the horizontal distance measured by the number of grid cells from the leftmost cells to the rightmost cells. The vertical span of a polyquadrate is with a similar definition but for the vertical direction.
Lemma 4: If 𝑆 is a collection of 𝑚 grid cells and 𝑛 is a
fixed positive integer, the number of polyquadrates with span less than𝑛 and intersecting with 𝑆 is Θ (𝑚).
Proof: Since𝑛 is fixed, the number of polyquadrates that
have spans less than 𝑛 and contain a specified grid cell is
bounded. Since𝑆 consists of 𝑚 grid cells, the lemma follows.
The next lemma gives an a.a.s. upper bound and lower bound of a collection of Poisson RVs, and we leave its proof in Appendix.
Lemma 5: Assume𝑐 > 0 and 𝛽 > 0 are constant and 𝐼𝑛= Θ(( 𝑛
ln 𝑛
)𝑐). Let 𝑌
𝑖 be a Poisson RV with rate 𝜇𝑖 for 𝑖 = 1, ⋅ ⋅ ⋅ , 𝐼𝑛. 1) If 𝜇𝑖≥ 𝛽 ln 𝑛 and 𝛽 ≥ 𝑐, we have lim 𝑛→∞Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖> 0 ] = 1.
2) If𝑌1, 𝑌2, ⋅ ⋅ ⋅ , 𝑌𝐼𝑛are independent,𝛽 ∈ (0, 𝑐), and 𝜇𝑖≤ 𝛽 ln 𝑛, we have lim 𝑛→∞Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖= 0 ] = 1.
At the end of this section, we give a lemma about the limits of integrals, and similarly, its proof can be found in Appendix.
Lemma 6: Let 𝑟𝜉 = 𝛽0
√
ln 𝑛+𝜉
𝜋𝑛 for some constant 𝜉, and either𝑅𝑛 = 3 √ ln 𝑛 𝜋𝑛 or𝑅𝑛= 𝛽0 √ ln 𝑛+𝜉𝑛 𝜋𝑛 with𝜉𝑛= 𝑜 (ln 𝑛) and𝜉𝑛 → ∞. Then 𝑛2 2 ∫ ∫ 𝑢,𝑣∈𝔻 𝑟𝜉≤∥𝑢−𝑣∥<𝑅𝑛 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣 ∼ 𝛽0 2 2 𝑒−𝜉. IV. ASYMPTOTICLENGTH OFTHELONGESTEDGE
This section is dedicated to the proof of Theorem 1. The proof is divided into two parts. In Subsection IV-A, we give Lemma 7 which provides an upper bound for the largest edge
length. In Subsection IV-B, we give Lemma 8 which provides a lower bound for the largest edge length. Since two bounds are tight, Theorem 1 follows Lemma 7 and 8.
A. Upper Bound for the Longest Edge Length
Lemma 7 says that if𝛽 > 𝛽0, there a.a.s. do not exist RNG
edges whose lengths are not less𝛽√ln 𝑛
𝜋𝑛. In the proof, we are going to show that any lens whose waist is fully contained in 𝔻 and not less than 𝛽√ln 𝑛
𝜋𝑛 a.a.s. contains some other nodes. Therefore, we can conclude that all RNG edges are a.a.s. less than𝛽√ln 𝑛
𝜋𝑛.
Lemma 7: For any constant𝛽 > 𝛽0, we have
lim 𝑛→∞Pr [ 𝜆𝑛≥ 𝛽 √ ln 𝑛 𝜋𝑛 ] = 0. Proof: Let 𝑑 = 𝛽√ln 𝑛
𝜋𝑛 and 𝑟 = 𝑑2. Pick a constant 𝛽1 ∈ (𝛽0, 𝛽), and let 𝑑′ = 𝛽1 √ ln 𝑛 𝜋𝑛, 𝑟′ = 𝑑 ′ 2 and 𝜀 = 𝑟−𝑟 ′ √ 2 .
Let C𝑟 be the collect of all feasible 𝑟-lenses, whose waists are contained in𝔻. If 𝑢𝑣 is an RNG edge, there are no other nodes in𝐿𝑢𝑣. So, 𝜆𝑛 ≥ 𝑑 implies that there exists a lens in C𝑟 that does not contain nodes of𝒫𝑛. Therefore,
Pr [ 𝜆𝑛≥ 𝛽 √ ln 𝑛 𝜋𝑛 ] ≤ Pr [ min 𝐶∈C𝑟∣𝐶 ∩ 𝒫𝑛∣ = 0 ] . (1)
Divide the plane by an 𝜀-tessellation. The distance of any
two points within a grid cell is at most 𝑟 − 𝑟′. If 𝐴 and 𝐵 respectively are an𝑟-lens and 𝑟′-lens with the same center and overlapping waists, grid cells intersected with𝐵 are contained
in𝐴 since any point of 𝐵 is apart from ∂𝐴 by at least 𝑟 − 𝑟′. So, the polyquadrate induced by 𝐵 is contained in 𝐴. Let {𝑃1, ⋅ ⋅ ⋅ , 𝑃𝐼𝑛} denote the set of polyquadrates induced by the 𝑟′-lenses inC𝑟′, and𝑌𝑖denote the number of nodes of𝒫𝑛 in
𝑃𝑖. Since any𝑟-lens in C𝑟 contains at least one𝑃𝑖, we have Pr [ min 𝐶∈C𝑟∣𝐶 ∩ 𝒫𝑛∣ = 0 ] ≤ Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖 = 0 ] . (2) Note that 𝑌𝑖 is a Poisson RV with rate 𝑛 ∣𝑃𝑖∣. Assume 𝐸 is the collection of polyquadrates induced by 𝑟′-lenses with center on𝔻√
3𝑟′(0) and 𝐹 is the collection of polyquadrates
induced by feasible 𝑟′-lenses with center on 𝔻 ∖ 𝔻√ 3𝑟′(0). Since𝐸 ∪ 𝐹 = {𝑃1, ⋅ ⋅ ⋅ , 𝑃𝐼𝑛}, Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖= 0 ] = Pr [ min 𝑃𝑖∈𝐸𝑌𝑖= 0 or min𝑃𝑖∈𝐹𝑌𝑖= 0 ] (3) ≤ Pr [ min 𝑃𝑖∈𝐸𝑌𝑖= 0 ] + Pr [ min 𝑃𝑖∈𝐹𝑌𝑖= 0 ] .
For any𝑃𝑖 ∈ 𝐸, we have
𝑛 ∣𝑃𝑖∣ ≥ 𝑛 1 𝛽02𝜋𝑑 ′2=(𝛽1 𝛽0 )2 ln 𝑛 > ln 𝑛. Applying Lemma 4, we also have∣𝐸∣ = Θ(1
𝜀2 ) = Θ( 𝑛 ln 𝑛 ) . Therefore, by Lemma 5, Pr [ min 𝑃𝑖∈𝐸𝑌𝑖= 0 ] = 1 − Pr [ min 𝑃𝑖∈𝐸𝑌𝑖> 0 ] ∼ 0. (4)
For any 𝑃𝑖∈ 𝐹 , we have
𝑛 ∣𝑃𝑖∣ ≥ 𝑛 ( 1 2𝛽02𝜋𝑑 ′2 ) >12ln 𝑛. Applying Lemma 4, we also have ∣𝐹 ∣ = Θ(𝑟′
𝜀2 ) = Θ(√ 𝑛 ln 𝑛 ) . Therefore, by Lemma 5, Pr [ min 𝑃𝑖∈𝐹𝑌𝑖= 0 ] = 1 − Pr [ min 𝑃𝑖∈𝐹𝑌𝑖> 0 ] ∼ 0. (5) Put Eq. 1, 2, 3, 4, and 5 together, and the lemma is proved.
B. Lower Bounds for the Longest Edge Length
Lemma 8 says that if𝛽 < 𝛽0, there a.a.s. exist RNG edges
whose lengths are not less than 𝛽√ln 𝑛
𝜋𝑛. To prove this, the plane is tessellated into equal-size square cells. For each cell, an event that implies the existence of such RNG edges is introduced, and a lower bound for the probability of the event is derived. Since these events are identical and independent among cells, we can easily find a probability lower bound that is asymptotically equal to 1.
Lemma 8: For any constant𝛽 ∈ (0, 𝛽0),
lim 𝑛→∞Pr [ 𝜆𝑛≥ 𝛽 √ ln 𝑛 𝜋𝑛 ] = 1.
Proof: Assume 𝛽1 and 𝛽2 are positive constants, and 𝑅1 and 𝑅2 respectively are given by 𝑛𝜋𝑅21 = 𝛽1ln 𝑛 and 𝑛𝜋𝑅2
2= 𝛽2ln 𝑛. Choose 𝛽1, 𝛽2 such that max(14𝛽02, 𝛽2)< 𝛽1 < 𝛽2 < 𝛽02 and 𝑐𝜋2 ( 1 −𝑅1 𝑅2 ) < 1. Here 𝑐 is given by Lemma 3. We have 1 2𝑅2 ≤ 𝑅1 ≤ 𝑅2. Divide 𝔻 by ( 4√ln 𝑛 𝑛𝜋 )
-tessellation. Let𝐼𝑛 denote the number of grid cells fully contained in𝔻. Here 𝐼𝑛= 𝑂(ln 𝑛𝑛 ). For each such cell, we draw a disk with radius 1
2
√
ln 𝑛
𝑛𝜋 at the center of the cell. For1 ≤ 𝑖 ≤ 𝐼𝑛, let𝐸𝑖be the event that there exist two nodes
𝑋, 𝑌 ∈ 𝒫𝑛 such that𝑋𝑌 is a RNG edge, their midpoint is in the 𝑖-th disk, and their distance is between 𝑅1 and 𝑅2.
Therefore, Pr [ 𝜆𝑛≥ 𝛽 √ ln 𝑛 𝜋𝑛 ]
≥ Pr [at least one 𝐸𝑖 occur] = 1 − Pr [none of 𝐸𝑖 occurs] .
𝐸1, ⋅ ⋅ ⋅ , 𝐸𝐼𝑖 are identical and independent. Thus,
Pr [none of 𝐸𝑖 occurs] = (1 − Pr [𝐸1])𝐼𝑛 ≤ 𝑒−𝐼𝑛Pr[𝐸1].
We can prove𝐼𝑛Pr [𝐸1] → ∞. (See Appendix B for details.)
So, Pr [ 𝜆𝑛≥ 𝛽 √ ln 𝑛 𝜋𝑛 ] → 1.
V. EXPECTEDNUMBER OFLONGEDGES
We have proved that the ratio of the largest RNG edge length to√ln 𝑛
𝑛𝜋 is a.a.s. equal to𝛽0. Let 𝑙𝑛= 𝛽0 √
ln 𝑛+𝜉 𝑛𝜋 for a constant𝜉. In this section, we are going to prove Theorem
2 that gives the asymptotic expected number of long RNG edges.
Proof of Theorem 2: Assume 𝑌 and 𝑉 are point sets
and 𝑌 ⊆ 𝑉 . Let ℎ𝑟(𝑌, 𝑉 ) denote a function such that
ℎ𝑟(𝑌 = {𝑢, 𝑣} , 𝑉 ) = 1 only if ∥𝑢 − 𝑣∥ ≥ 𝑟 and there is no other node of 𝑉 in the disk area 𝐿𝑢𝑣; otherwise,
ℎ𝑟(𝑌, 𝑉 ) = 0. Let 𝑋1 and 𝑋2 denote independent random
points with uniform distribution over 𝔻 and independent of
𝒫𝑛. According to the Palm theory, E [𝒩𝑛] = E ⎡ ⎢ ⎣ ∑ {𝑋′ 1,𝑋2′}⊆𝒫𝑛 ℎ𝑟({𝑋1′, 𝑋2′} , 𝒫𝑛) ⎤ ⎥ ⎦ =𝑛2!2E [ℎ𝑟({𝑋1, 𝑋2} , {𝑋1, 𝑋2} ∪𝒫𝑛)] . Let𝐹 (𝑟) be the probability of the event that 𝑋1𝑋2is a RNG
edge and∥𝑋1− 𝑋2∥ ≥ 𝑟. Then,
𝐹 (𝑟) = E [ℎ𝑟({𝑋1, 𝑋2} , {𝑋1, 𝑋2} ∪𝒫𝑛)] , and 𝐹 (𝑟) = ∫ ∫ 𝑢,𝑣∈𝔻 ∥𝑢−𝑣∥≥𝑟 Pr [ 𝑋1𝑋2 is a RNG edge 𝑋1= 𝑢 𝑋2= 𝑣 ] 𝑑𝑢𝑑𝑣 = ∫ ∫ 𝑢,𝑣∈𝔻 ∥𝑢−𝑣∥≥𝑟 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣. According to Theorem 1, 𝐹 ( 𝛽0 √ ln 𝑛 + 𝜉 𝑛𝜋 ) ∼ 𝐹 ( 𝛽0 √ ln 𝑛 + 𝜉 𝑛𝜋 ) − 𝐹 ( 3 √ ln 𝑛 𝑛𝜋 ) = ∫ ∫ 𝑢,𝑣∈𝔻 3√ln 𝑛 𝑛𝜋>∥𝑢−𝑣∥≥𝛽0 √ln 𝑛+𝜉 𝑛𝜋 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣.
Hence, by Lemma 6, we have
E [𝒩𝑛] = 𝑛 2 2!𝐹 ( 𝛽0 √ ln 𝑛 + 𝜉 𝑛𝜋 ) ∼𝛽202𝑒−𝜉. Thus, the theorem is proved.
VI. SIMULATIONS
We run simulations to validate our asymptotic results. In the simulation,400 random point sets over a unit-area square with uniform distribution are generated for each node density
𝑛 = 100, 200, 400, 800, and 1600. The cumulative distribution
functions of the largest RNG edge length are illustrated in Fig. 4. The curves from right to left are corresponding to the CDFs with respect to𝑛 = 100, 200, 400, 800, and 1600. Without
too much surprise, as the node density becomes larger, the largest RNG edge length becomes smaller, and the threshold
CDF of the largest RNG edge length
0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 Length Prob. n=100 n=200 n=400 n=800 n=1600
Fig. 4. CDF of the largest RNG edge length. TABLE I DILATION FACTORS. 𝑛 = 100 𝑛 = 200 𝑛 = 400 𝑛 = 800 𝑛 = 1600 MAX 5.039 6.021 5.795 5.651 6.977 MIN 2.375 2.752 2.895 3.279 3.463 AVG 3.337 3.659 3.934 4.177 4.425
width, that is corresponding to the width of the CDF curve, also becomes more narrow.
To investigate the energy efficiency issue, for each pair of nodes, we calculate the ratio of the length of the shortest path in the RNG between them and the Euclidean distance between them. The maximal ratio over all pairs of nodes in a random point set is the dilation factor of the RNG. In Table I, we list the maximal, minimal, and average dilation factors among 400 random point sets for each node density. The dilation factor is a lower bound of the stretch factor. Although Bose (2002) [15] proved that the stretch factor of RNGs is Θ (𝑛) in the worst case, we can see in Table I the dilation factor doesn’t increase significantly. So, the power efficiency of RNG is not too bad in the average case.
VII. CONCLUSIONS
The relative neighborhood graphs are a geometric structure used in topology control for wireless ad hoc networks and can be constructed by distributed and localized algorithms. Motivated by constructing geometric structures using only 1-hop information, we studies the maximal length of RNG edges which is the smallest transmission radius for constructing the RNG by only 1-hop information if all nodes have the same transmission radius. In this paper, we assume a wireless ad hoc networks is represented by a Poisson point process with mean 𝑛 on a unit-area disk. We first showed that the ratio of
the maximal length of RNG edges to√ln 𝑛𝜋𝑛 is a.a.s. equal to √ 1/(2 3− √ 3 2𝜋 )
≈ 1.6. Next, we proved that for a constant 𝜉, the expected number of long RNG edges, whose lengths
are at least𝛽0
√
ln 𝑛+𝜉
𝜋𝑛 , are a.a.s. equal to 𝛽0 2
2 𝑒−𝜉. This imply
that if𝜉 → ∞, it is a.a.s. that the maximal length is less than 𝛽0
√
ln 𝑛+𝜉 𝜋𝑛 .
APPENDIXA
PROOFS OFLEMMAS INSECTIONIII
First, we give the proof of Lemma 5.
Proof of Lemma 5: Consider the first case. If𝜇𝑖≥ 𝛽 ln 𝑛, Pr [𝑌𝑖 = 0] = 𝑒−𝜇𝑖 ≤ 𝑒−𝛽 ln 𝑛. So, Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖> 0 ] = 1 − Pr [∃𝑖 such that 𝑌𝑖= 0] = 1 − Pr [𝐼 𝑛 ⋁ 𝑖=1 (𝑌𝑖= 0) ] ≥ 1 −∑𝐼𝑛 𝑖=1 Pr [𝑌𝑖= 0] ≥ 1 −∑𝐼𝑛 𝑖=1 𝑒−𝛽 ln 𝑛= 1 − Θ(( 𝑛 ln 𝑛 )𝑐) 𝑒−𝛽 ln 𝑛 = 1 − Θ ( 𝑛𝑐−𝛽 ln𝑐𝑛 ) . Therefore, if𝛽 ≥ 𝑐, we have Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖> 0 ] ∼ 1.
Consider the second case. If 𝑌1, 𝑌2, ⋅ ⋅ ⋅ , 𝑌𝐼𝑛 are
indepen-dent, Pr [ 𝐼𝑛 min 𝑖=1𝑌𝑖> 0 ] = Pr [𝐼 𝑛 ⋀ 𝑖=1 (𝑌𝑖> 0) ] = 𝐼𝑛 ∏ 𝑖=1 Pr [𝑌𝑖> 0] = 𝐼𝑛 ∏ 𝑖=1 (1 − Pr [𝑌𝑖= 0]) ≤ 𝐼𝑛 ∏ 𝑖=1 𝑒− Pr[𝑌𝑖=0] = 𝑒−∑𝐼𝑛𝑖=1Pr[𝑌𝑖=0]. If𝜇𝑖≤ 𝛽 ln 𝑛, Pr [𝑌𝑖 = 0] = 𝑒−𝜇𝑖 ≥ 𝑒−𝛽 ln 𝑛. Put the two inequalities together. Then,
Pr [ 𝐼 𝑛 min 𝑖=1 𝑌𝑖 = 0 ] = 1 − Pr [ 𝐼 𝑛 min 𝑖=1𝑌𝑖> 0 ] ≥ 1 − 𝑒−∑𝐼𝑛 𝑖=1Pr[𝑌𝑖=0]≥ 1 − 𝑒−∑𝐼𝑛𝑖=1𝑒−𝛽 ln 𝑛 = 1 − 𝑒−Θ(( 𝑛 ln 𝑛)𝑐)𝑒−𝛽 ln 𝑛 = 1 − 𝑒−Θ ( 𝑛𝑐−𝛽 ln𝑐 𝑛 ) . Therefore, if𝛽 ∈ (0, 𝑐), we have Pr ( min 1≤𝑖≤𝐼𝑛𝑌𝑖= 0 ) ∼ 1.
Next, we are going to prove Lemma 6. But before that, we need a lemma that gives a tighter lower bound for∣𝐿𝑢𝑣∩ 𝔻∣ as the center of𝐿𝑢𝑣 is near the boundary of𝔻 and a tool for variable transformation.
Lemma 9: For any𝑢, 𝑣 ∈ 𝔻, let 𝑟 = 1
2∥𝑢 − 𝑣∥, 𝑧 = 𝑢+𝑣2 , and𝑡 = √1 𝜋 − ∥𝑧∥. If 𝑧 ∈ 𝔻╲ ( 𝔻√ 3𝑟(0) ∪ 𝔻𝑟(2)), we have ∣𝐿𝑢𝑣∩ 𝔻∣ ≥ 12∣𝐿𝑢𝑣∣ + 𝑟𝑡.
0000000
0000000
0000000
0000000
0000000
1111111
1111111
1111111
1111111
1111111
u
b
a
c
v
z
Fig. 5. A lens near the boundary of𝔻.
t(z) z b a r c d Fig. 6. If𝑧 ∈ 𝔻𝑟(1), then 𝜃 (𝑧, 𝑟) = 4∠𝑎𝑧𝑏.
Proof: Let 𝑎, 𝑏 with ∥𝑎∥ ≥ ∥𝑏∥ denote the two vertices
of𝐿𝑢𝑣. If the segment𝑎𝑧 does not intersect ∂𝔻, the triangle
𝑎𝑢𝑣 is contained in 𝔻 and with area 1
2∥𝑢 − 𝑣∥ ∥𝑧 − 𝑎∥ = 1
2(2𝑟)
(√
3𝑟) = 𝑟(√3𝑟) ≥ 𝑟𝑡. Otherwise, let 𝑐 denote the
intersection point of𝑎𝑧 and ∂𝔻. See Fig. 5. Note that 𝑡 is the
shortest distance from𝑧 to ∂𝔻. The triangle 𝑐𝑢𝑣 is contained
in 𝔻 and with area 1
2∥𝑢 − 𝑣∥ ∥𝑧 − 𝑐∥ = 12(2𝑟) ∥𝑧 − 𝑐∥ = 𝑟 ∥𝑧 − 𝑐∥ ≥ 𝑟𝑡. In addition, the half lens surrounded by the
segment𝑢𝑣 and arcs 𝑢𝑏 and 𝑣𝑏 is always contained in 𝔻. So,
we have
∣𝐿𝑢𝑣∩ 𝔻∣ ≥ 12∣𝐿𝑢𝑣∣ + 𝑟𝑡, and the lemma is proved.
For 𝑧 ∈ 𝔻 and 𝑟 ∈ ℝ, let 𝜃 (𝑧, 𝑟) denote the (total)
central angle corresponding to the portion of ∂𝐵 (𝑧, 𝑟) in
which if a diameter of𝐵 (𝑧, 𝑟) has an endpoint, the diameter
is fully contained in𝔻. For example, in Figure (6), 𝑏 , 𝑐 are the intersection points of ∂𝐵 (𝑧, 𝑟) and ∂𝔻, and the segment 𝑏𝑑
is a diameter of𝐵 (𝑧, 𝑟). Then, 𝜃 (𝑧, 𝑟) = 4∠𝑎𝑧𝑏. In addition,
if 𝑧 ∈ 𝔻𝑟(1), we use 𝑡 (𝑧) to denote the distance between 𝑧 and∂𝔻. We have
𝜃 (𝑧, 𝑟) = 2𝜋, if 𝑧 ∈ 𝔻𝑟(0) ;
𝜃 (𝑧, 𝑟) ≤ 4 arcsin𝑡 (𝑧)𝑟 ≤ 4𝑡 (𝑧)𝑟 , if𝑧 ∈ 𝔻𝑟(1) ; (6)
𝜃 (𝑧, 𝑟) = 0, if 𝑧 ∈ 𝔻𝑟(2) .
Proof of Lemma 6: Let 𝑧 = 𝑢+𝑣
2 and 𝑟 = ∥𝑢−𝑣∥2 . If 𝑧 ∈ 𝔻√
3𝑟(0), 𝐿𝑢𝑣 is fully contained in𝔻 and
∣𝐿𝑢𝑣∩ 𝔻∣ = 1
𝛽02𝜋 ∥𝑢 − 𝑣∥ 2= 4
𝛽02𝜋𝑟 2.
First, we calculate the integration over𝑧 ∈ 𝔻√ 3𝑟(0). 𝑛2 2 ∫ ∫ 𝑢,𝑣∈𝔻 𝑢+𝑣 2 ∈𝔻√3𝑟(0) 𝑟𝜉≤∥𝑢−𝑣∥<𝑅𝑛 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣 =𝑛2 2 ∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 ∫ 𝑧∈𝔻√ 3𝑟(0) 𝑒−𝛽024 𝑛𝜋𝑟28𝜋𝑟𝑑𝑧𝑑𝑟 ∼ 2𝑛2∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 𝑒 − 4 𝛽02𝑛𝜋𝑟22𝜋𝑟𝑑𝑟 = 2𝑛2∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 𝑒−𝛽024𝑛𝜋𝑟2𝑑(𝜋𝑟2) = 2𝑛2 ⎛ ⎝−𝛽02 4𝑛𝑒−𝑛𝜋𝑟 2 𝑅𝑛 2 𝑟=𝑟𝜉2 ⎞ ⎠ ∼𝛽202𝑒−𝜉.
Next, we calculate the integration over𝑧 ∈ 𝔻√
3𝑟(1) ∖𝔻𝑟(2).
Let𝑡 denote the distance from 𝑧 to ∂𝔻. By Lemma 9 and Eq.
(6), we have∣𝐿𝑢𝑣∩ 𝔻∣ ≥ 𝛽022𝜋𝑟2+ 𝑐1𝑟𝑡 and 𝑟𝜃 (𝑧, 𝑟) ≤ 𝑐2𝑡. Thus, 𝑛2 2 ∫ ∫ 𝑢,𝑣∈𝔻 𝑢+𝑣 2 ∈𝔻√3𝑟(1)∖𝔻𝑟(2) 𝑟𝜉≤∥𝑢−𝑣∥<𝑅𝑛 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣 ≤ 𝑛22 ∫ ∫ 𝑢,𝑣∈𝔻 𝑢+𝑣 2 ∈𝔻√3𝑟(1)∖𝔻𝑟(2) 𝑟𝜉≤∥𝑢−𝑣∥<𝑅𝑛 𝑒−𝑛 ( 2 𝛽02𝜋𝑟2+𝑐1𝑟𝑡 ) 𝑑𝑢𝑑𝑣 ≤ 𝑛22 ∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 ∫ 𝑧∈𝔻√ 3𝑟(1)∖𝔻𝑟(2) 𝑒−𝑛 ( 2 𝛽02𝜋𝑟2+𝑐1𝑟𝑡 ) 4𝑟𝜃 (𝑧, 𝑟) ⋅ 𝑑𝑧𝑑𝑟 ≤ 𝑂 (1) 𝑛2∫ 𝑅𝑛2 𝑟=𝑟𝜉2 ∫ √ 3𝑟 𝑡=0 𝑒 −𝑛( 2 𝛽02𝜋𝑟2+𝑐1𝑟𝑡 ) 𝑡𝑑𝑡𝑑𝑟 ≤ 𝑂 (1) 𝑛2𝑒−2𝛽021 𝑛𝜋𝑟2𝜉 ∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 ∫ √ 3𝑟 𝑡=0 𝑒 −𝑐1𝑛𝑟𝑡𝑡𝑑𝑡𝑑𝑟 ≤ 𝑂 (1) 𝑛2𝑒−12(ln 𝑛+𝜉) ∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 ∫ ∞ 𝑡=0𝑒 −𝑐1𝑛𝑟𝑡𝑡𝑑𝑡𝑑𝑟 = 𝑂 (1) 𝑛2𝑒−1 2(ln 𝑛+𝜉) ∫ 𝑅𝑛 2 𝑟=𝑟𝜉2 (𝑛𝑟) −2𝑑𝑟 ≤ 𝑂 (1) 𝑛2𝑒−1 2(ln 𝑛+𝜉)(𝑛𝑟𝜉)−2𝑅𝑛 ≤ 𝑂 (1) 𝑒−1 2(ln 𝑛+𝜉) (√ ln 𝑛 𝑛 )−1 = 𝑂 (1) (ln 𝑛)−1/2 = 𝑜 (1) .
Last, we calculate the integration over𝑧 ∈ 𝔻𝑟(2). Since the measure of the set{(𝑢, 𝑣) ∣ 𝑢, 𝑣 ∈ 𝔻,𝑢+𝑣
2 ∈ 𝔻𝑟(2) } is zero, 𝑛2 2 ∫ ∫ 𝑢,𝑣∈𝔻 𝑢+𝑣 2 ∈𝔻𝑟(2) 𝑟𝜉≤∥𝑢−𝑣∥<𝑅𝑛 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣 = 0. Therefore, 𝑛2 2 ∫ ∫ 𝑢,𝑣∈𝔻 𝑟𝜉≤∥𝑢−𝑣∥<𝑅𝑛 𝑒−𝑛∣𝐿𝑢𝑣∩𝔻∣𝑑𝑢𝑑𝑣 ∼ 𝛽0 2 2 𝑒−𝜉. APPENDIXB
SUPPLEMENTS TO THEPROOF OFLEMMA8
To prove Lemma 8, we introduce several relevant events and derive their probabilities. Let𝐴 denote the disk centered
at the origin with area ln 𝑛
4𝑛, i.e. with radius 12
√
ln 𝑛
𝑛𝜋. Let 𝛽1 and𝛽2denote two positive constants, and𝑅1and𝑅2be given
by𝑛𝜋𝑅2
1= 𝛽1ln 𝑛 and 𝑛𝜋𝑅22= 𝛽2ln 𝑛. Choose 𝛽1, 𝛽2 such
thatmax(1 4𝛽02, 𝛽2 ) < 𝛽1< 𝛽2< 𝛽02and𝑐𝜋2 ( 1 −𝑅1 𝑅2 ) < 1.
Here 𝑐 is given by Lemma 3. We have 1
2𝑅2≤ 𝑅1≤ 𝑅2.
Assume 𝑉 is a point set and 𝑌 ⊂ 𝑉 . Let ℎ1(𝑌, 𝑉 )
denote a function such that ℎ1(𝑌 = {𝑢, 𝑣} , 𝑉 ) = 1 only if 1
2(𝑢 + 𝑣) ∈ 𝐴, 𝑅1 ≤ ∥𝑢 − 𝑣∥ ≤ 𝑅2, and there is no other
node of 𝑉 in 𝐿𝑢𝑣; otherwise, ℎ1(𝑌, 𝑉 ) = 0. Then, 𝐸1 is
the event that there exist two nodes 𝑋, 𝑌 ∈ 𝒫𝑛 such that
ℎ1({𝑋, 𝑌 } , 𝒫𝑛) = 1. In what follows, we use 𝑋1,𝑋2, 𝑋3
and 𝑋4 to denote independent random points with uniform
distribution over 𝔻 and independent of 𝒫𝑛. Let 𝐹1 be the
event that
ℎ1({𝑋1, 𝑋2} , {𝑋1, 𝑋2} ∪ 𝒫𝑛) = 1,
𝐹2 be the event that
ℎ1({𝑋1, 𝑋2} , {𝑋1, 𝑋2, 𝑋3} ∪ 𝒫𝑛)
⋅ ℎ1({𝑋1, 𝑋3} , {𝑋1, 𝑋2, 𝑋3} ∪ 𝒫𝑛) = 1, and𝐹3 be the event that
ℎ1({𝑋1, 𝑋2} , {𝑋1, 𝑋2, 𝑋3, 𝑋4} ∪ 𝒫𝑛) ⋅ ℎ1({𝑋3, 𝑋4} , {𝑋1, 𝑋2, 𝑋3, 𝑋4} ∪ 𝒫𝑛) = 1. We claim that Pr [𝐸1] ≥ 𝑛 2 2 Pr [𝐹1] − 𝑛3 2 Pr [𝐹2] − 𝑛4 8 Pr [𝐹3] . (7) We shall prove this claim by the Palm theory and Boole’s inequalities. For clarity, we use𝑋′
1,𝑋2′,𝑋3′ and𝑋4′ to denote
elements of𝒫𝑛. For any{𝑥1, 𝑥2, 𝑥3} ⊆ 𝑉 , let
ℎ2({𝑥1, 𝑥2, 𝑥3} , 𝑉 ) = ℎ1({𝑥1, 𝑥2} , 𝑉 ) ⋅ ℎ1({𝑥1, 𝑥3} , 𝑉 )
+ ℎ1({𝑥2, 𝑥1} , 𝑉 ) ⋅ ℎ1({𝑥2, 𝑥3} , 𝑉 )
+ ℎ1({𝑥3, 𝑥1} , 𝑉 ) ⋅ ℎ1({𝑥3, 𝑥2} , 𝑉 ) .
For any {𝑥1, 𝑥2, 𝑥3, 𝑥4} ⊆ 𝑉 , let ℎ3({𝑥1, 𝑥2, 𝑥3, 𝑥4} , 𝑉 )
= ℎ1({𝑥1, 𝑥2} , 𝑉 ) ⋅ ℎ1({𝑥3, 𝑥4} , 𝑉 )
+ ℎ1({𝑥1, 𝑥3} , 𝑉 ) ⋅ ℎ1({𝑥2, 𝑥4} , 𝑉 )
+ ℎ1({𝑥1, 𝑥4} , 𝑉 ) ⋅ ℎ1({𝑥2, 𝑥3} , 𝑉 ) .
Note that the addition and multiplication in ℎ2 and ℎ3 are
Boolean operators. Let 𝐹′
1({𝑋1′, 𝑋2′}) be the events that ℎ1({𝑋1′, 𝑋2′} , 𝒫𝑛) = 1, 𝐹2′({𝑋1′, 𝑋2′, 𝑋3′}) be the event that ℎ2({𝑋1′, 𝑋2′, 𝑋3′} , 𝒫𝑛) = 1, and 𝐹3′({𝑋1′, 𝑋2′, 𝑋3′, 𝑋4′}) be
the Palm theory (refer to Theorem 7 in [14]), we have ∑ {𝑋′ 1,𝑋2′}⊆𝒫𝑛 Pr [𝐹′ 1({𝑋1′, 𝑋2′})] = E ⎡ ⎢ ⎣ ∑ {𝑋′ 1,𝑋2′}⊆𝒫𝑛 ℎ1({𝑋1′, 𝑋2′} , 𝒫𝑛) ⎤ ⎥ ⎦ = 𝑛2!2E [ℎ1({𝑋1, 𝑋2} , {𝑋1, 𝑋2} ∪ 𝒫𝑛)] = 𝑛22Pr [𝐹1] ; (8) ∑ {𝑋′ 1,𝑋2′,𝑋′3}⊆𝒫𝑛 Pr [𝐹′ 2({𝑋1′, 𝑋2′, 𝑋3′})] = E ⎡ ⎢ ⎣ ∑ {𝑋′ 1,𝑋′2,𝑋3′}⊆𝒫𝑛 ℎ2({𝑋1′, 𝑋2′, 𝑋3′} , 𝒫𝑛) ⎤ ⎥ ⎦ =𝑛3!3E [ℎ2({𝑋1, 𝑋2, 𝑋3} , {𝑋1, 𝑋2, 𝑋3} ∪ 𝒫𝑛)] = 3𝑛3!3Pr [𝐹2] =𝑛 3 2 Pr [𝐹2] ; (9) and ∑ {𝑋′ 1,𝑋′2,𝑋3′,𝑋4′}⊆𝒫𝑛 Pr [𝐹3′({𝑋1′, 𝑋2′, 𝑋3′, 𝑋4′})] = E ⎡ ⎢ ⎣ ∑ {𝑋′ 1,𝑋2′,𝑋3′,𝑋4′}⊆𝒫𝑛 ℎ3({𝑋1′, 𝑋2′, 𝑋3′, 𝑋4′} , 𝒫𝑛) ⎤ ⎥ ⎦ =𝑛4!4E [ℎ3({𝑋1, 𝑋2, 𝑋3, 𝑋4} , {𝑋1, 𝑋2, 𝑋3, 𝑋4} ∪ 𝒫𝑛)] = 3𝑛4!4Pr [𝐹3] = 𝑛 4 8 Pr [𝐹3] . (10)
Applying Boole’s inequalities and Eq. (8), (9), and (10), we have Pr [𝐸1] ≥ ∑ {𝑋′ 1,𝑋′2}⊆𝒫𝑛 Pr [𝐹′ 1({𝑋1′, 𝑋2′})] − ∑ {𝑋′ 1,𝑋′2,𝑋3′}⊆𝒫𝑛 Pr [𝐹′ 2({𝑋1′, 𝑋2′, 𝑋3′})] − ∑ {𝑋′ 1,𝑋′2,𝑋3′,𝑋4′}⊆𝒫𝑛 Pr [𝐹′ 3({𝑋1′, 𝑋2′, 𝑋3′, 𝑋4′})] = 𝑛22Pr [𝐹1] −𝑛 3 2 Pr [𝐹2] − 𝑛4 8 Pr [𝐹3] . Hence, our claim is true.
In the next, we derive the probabilities of𝐹1,𝐹2, and𝐹3.
Let𝑆1 denote the set
{ (𝑥1, 𝑥2)12(𝑥1+ 𝑥2) ∈ 𝐴, 𝑅1≤ ∥𝑥1− 𝑥2∥ ≤ 𝑅2 } . We have Pr [𝐹1] = ∫ ∫ 𝑆1Pr [𝐹1∣ 𝑋1= 𝑥1, 𝑋2= 𝑥2] 𝑑𝑥1𝑑𝑥2 = ∫ ∫ 𝑆1𝑒 −𝑛∣𝐿𝑥1𝑥2∣𝑑𝑥 1𝑑𝑥2 = ∫ ∫ 𝑆1𝑒 −𝑛 1 𝛽02𝜋∥𝑥1−𝑥2∥2𝑑𝑥1𝑑𝑥2. Let𝑧 = 𝑥1+𝑥2 2 and𝑟 = 12∥𝑥1− 𝑥2∥. Then, Pr [𝐹1] = ∫ 𝑧∈𝐴 ∫ 𝑅2 2 𝑟=𝑅12 𝑒−𝛽024 𝑛𝜋𝑟28𝜋𝑟𝑑𝑟𝑑𝑧 = 4 ∫ 𝑧∈𝐴 ∫ 𝑅2 2 𝑟=𝑅12 𝑒−𝛽024 𝑛𝜋𝑟22𝜋𝑟𝑑𝑟𝑑𝑧 = 4 ∫ 𝑧∈𝐴 ∫ 𝑅2 2 𝑟=𝑅12 𝑒 − 4 𝛽02𝑛𝜋𝑟2𝑑(𝜋𝑟2)𝑑𝑧 = − ⎛ ⎝ 𝛽02 𝑛 𝑒 − 4 𝛽02𝑛𝜋𝑟2 𝑅2 2 𝑟=𝑅12 ⎞ ⎠ ∣𝐴∣ = 𝛽4𝑛022 ( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 ) ln 𝑛. (11)
Let𝑆2denote the set
⎧ ⎨ ⎩(𝑥1, 𝑥2, 𝑥3) 𝑥1+𝑥2 2 ,𝑥1+𝑥2 3 ∈ 𝐴; 𝑥1, 𝑥2 /∈ 𝐿𝑥1𝑥3; 𝑥1, 𝑥3 /∈ 𝐿𝑥1𝑥2; 𝑅1≤ ∥𝑥1− 𝑥2∥ ≤ 𝑅2; 𝑅1≤ ∥𝑥1− 𝑥3∥ ≤ 𝑅2 ⎫ ⎬ ⎭. Applying Lemma 3, if (𝑥1, 𝑥2, 𝑥3) ∈ 𝑆2, we have
Pr [𝐹2∣𝑋1= 𝑥1, 𝑋2= 𝑥2, 𝑋3= 𝑥3] ≤ 𝑒−𝑛∣𝐿𝑥1𝑥2∪𝐿𝑥1𝑥3∣ ≤ 𝑒−𝑛 ( 1 𝛽0𝜋∥𝑥1−𝑥2∥2+𝑐𝑅2∥𝑥1+𝑥22 −𝑥1+𝑥32 ∥ ) . Therefore, Pr [𝐹2] = ∫ ∫ ∫ 𝑆2Pr [𝐹2∣𝑋1= 𝑥1, 𝑋2= 𝑥2, 𝑋3= 𝑥3] ⋅ 𝑑𝑥1𝑑𝑥2𝑑𝑥3 ≤ ∫ ∫ ∫ 𝑆2𝑒 −𝑛( 1 𝛽02𝜋∥𝑥1−𝑥2∥2+𝑐𝑅2∥𝑥1+𝑥22 −𝑥1+𝑥32 ∥ ) ⋅ 𝑑𝑥1𝑑𝑥2𝑑𝑥3. Let 𝑧1 = 𝑥1+𝑥2 2, 𝑟1 = 21∥𝑥1− 𝑥2∥, 𝑧2 = 𝑥1+𝑥2 3, and 𝜌 = ∥𝑧1− 𝑧2∥. Then, Pr [𝐹2] ≤ 16 ∫ 𝑧1∈𝐴 ∫ 𝑅2 2 𝑟1=𝑅1 2 ∫ 𝑧2∈𝐴𝑒 −𝑛( 4 𝛽02𝜋𝑟21+𝑐𝑅2∥𝑧1−𝑧2∥ ) 2𝜋𝑟1 ⋅ 𝑑𝑟1𝑑𝑧1𝑑𝑧2 ≤ 16 ∫ 𝑧1∈𝐴 ∫ 𝑅2 2 𝑟1=𝑅12 𝑒 − 4 𝛽02𝑛𝜋𝑟212𝜋𝑟1𝑑𝑟1𝑑𝑧1 ⋅ ∫ 𝑧2∈𝐴𝑒 −𝑐𝑛𝑅2∥𝑧1−𝑧2∥𝑑𝑧 2 ≤ 16 ∫ 𝑧1∈𝐴 ∫ 𝑅2 2 𝑟1=𝑅12 𝑒−𝛽024 𝑛𝜋𝑟21𝑑(𝜋𝑟2 1 ) 𝑑𝑧1
⋅ ∫ ∞ 𝜌=0𝑒 −𝑐𝑛𝑅2𝜌2𝜋𝜌𝑑𝜌 = − ⎛ ⎝ 4𝛽02 𝑛 𝑒 − 4 𝛽02𝑛𝜋𝑟2 𝑅2 2 𝑟=𝑅12 ⎞ ⎠ ∣𝐴∣ 2𝜋 (𝑐𝑛𝑅2)2 =𝑐2(𝑛𝑅2𝜋𝛽022 2) 𝑛3 ( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 ) ln 𝑛. (12)
Let𝑆3 denote the set
⎧ ⎨ ⎩(𝑥1, 𝑥2, 𝑥3, 𝑥4) 𝑥1+𝑥2 2 ,𝑥3+𝑥2 4 ∈ 𝐴; 𝑥1, 𝑥2 /∈ 𝐿𝑥3𝑥4; 𝑥3, 𝑥4 /∈ 𝐿𝑥1𝑥2; 𝑅1≤ ∥𝑥1− 𝑥2∥ ≤ 𝑅2; 𝑅1≤ ∥𝑥3− 𝑥4∥ ≤ 𝑅2 ⎫ ⎬ ⎭. Applying Lemma 3, if(𝑥1, 𝑥2, 𝑥3, 𝑥4) ∈ 𝑆3, we have
Pr [𝐹3∣𝑋1= 𝑥1, 𝑋2= 𝑥2, 𝑋3= 𝑥3, 𝑋4= 𝑥4] ≤ 𝑒−𝑛∣𝐿𝑥1𝑥2∪𝐿𝑥3𝑥4∣ ≤ 𝑒−𝑛 ( 1 𝛽02𝜋∥𝑥1−𝑥2∥2+𝑐𝑅2∥𝑥1+𝑥22 −𝑥3+𝑥42 ∥ ) . Therefore, Pr [𝐹3] = ∫ ∫ ∫ ∫ 𝑆3Pr [ 𝐹3 𝑋𝑋1= 𝑥1, 𝑋2= 𝑥2, 3= 𝑥3, 𝑋4= 𝑥4 ] ⋅ 𝑑𝑥1𝑑𝑥2𝑑𝑥3𝑑𝑥4 ≤ ∫ ∫ ∫ ∫ 𝑆3𝑒 −𝑛( 1 𝛽02𝜋∥𝑥1−𝑥2∥2+𝑐𝑅2∥𝑥1+𝑥22 −𝑥3+𝑥42 ∥ ) ⋅ 𝑑𝑥1𝑑𝑥2𝑑𝑥3𝑑𝑥4. Let 𝑧1 = 𝑥1+𝑥2 2, 𝑟1 = 12∥𝑥1− 𝑥2∥, 𝑧2 = 𝑥3+𝑥2 4, 𝑟2 = 1 2∥𝑥3− 𝑥4∥, and 𝜌 = ∥𝑧1− 𝑧2∥. Then, Pr [𝐹3] ≤ ∫ 𝑧1∈𝐴 ∫ 𝑅2 2 𝑟1=𝑅12 ∫ 𝑧2∈𝐴 ∫ 𝑅2 2 𝑟2=𝑅12 𝑒−𝑛 ( 4 𝛽02𝜋𝑟21+𝑐𝑅2∥𝑧1−𝑧2∥ ) ⋅ (8𝜋𝑟1𝑑𝑟1𝑑𝑧1) (8𝜋𝑟2𝑑𝑟2𝑑𝑧2) ≤ ( 4 ∫ 𝑧1∈𝐴 ∫ 𝑅2 2 𝑟1=𝑅12 𝑒 − 4 𝛽02𝑛𝜋𝑟122𝜋𝑑𝑟1𝑑𝑧1 ) ⋅ ( 8𝜋𝑅2 2 ( 𝑅2 2 − 𝑅1 2 ) ∫ 𝑧2∈𝐴𝑒 −𝑐𝑛𝑅2∥𝑧1−𝑧2∥𝑑𝑧 2 ) ≤ ( 4 ∫ 𝑧1∈𝐴 ∫ 𝑅2 2 𝑟1=𝑅12 𝑒−𝛽024 𝑛𝜋𝑟12𝑑(𝜋𝑟2 1 ) 𝑑𝑧1 ) ⋅ ( 8𝜋𝑅22 ( 𝑅2 2 − 𝑅1 2 ) ∫ ∞ 𝜌=0𝑒 −𝑐𝑛𝑅2𝜌2𝜋𝜌𝑑𝜌) = ( 𝛽02ln 𝑛 4𝑛2 ( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 )) ⋅ ( 4𝜋2 (𝑐𝑛𝑅2)2𝑅2(𝑅2− 𝑅1) ) =𝜋𝑐22𝛽𝑛042 ( 1 − 𝑅𝑅1 2 ) ( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 ) ln 𝑛. (13)
Put Eq. (7), (11), (12) and (13) together. We have Pr [𝐸1] ≥ ( 𝛽02 8 − 𝜋𝛽02 𝑐2(𝑛𝑅2 2)− 𝜋2𝛽02 8𝑐2 ( 1 − 𝑅1 𝑅2 )) ⋅ ( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 ) ln 𝑛 ∼ 𝛽802 ( 1 −𝜋𝑐22 ( 1 − 𝑅𝑅1 2 )) ( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 ) ln 𝑛. Since 𝜋𝑐22 ( 1 −𝑅1 𝑅2 ) < 1 and 𝐼𝑛= Ω(ln 𝑛𝑛 ), we have Pr [𝐸1] = Ω (( 𝑛−𝛽02𝛽1 − 𝑛−𝛽02𝛽2 ) ln 𝑛 ) , and 𝐼𝑛Pr [𝐸1] = Ω ( 𝑛1−𝛽02𝛽1 ) → ∞. (14) ACKNOWLEDGMENT
This work of Dr. Yi described in this paper was partially supported by NSC under Grant No. NSC97-2221-E-009-052-MY3, by MoEA under Grant No. 98-EC-17-A-02-S2-0048, by ITRI under Grant No. 8352BA4124, and by the MoE ATU plan.
The work of Dr. Wan described in this paper was partially supported by NSF of USA under grant CNS-0831831.
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Chih-Wei Yi Dr. Chih-Wei Yi received his PhD degree from the Illinois Institute of Technology, and MS and BS degrees from the National Taiwan University. He is currently an Assistant Professor in Computer Science at the National Chiao Tung University. He is a member of the IEEE and the ACM. He had been a Senior Research Fellow of the Department of Computer Science, City University of Hong Kong. He was bestowed the Outstanding Young Engineer Award by the Chinese Institute of Engineers in 2009. His research focuses on wireless ad hoc and sensor networks, vehicular ad hoc networks, network coding, and algorithm design and analysis.
Peng-Jun Wan Dr. Peng-Jun Wan received his PhD degree from University of Minnesota, MS degree from The Chinese Academy of Science, and BS degree from Tsinghua University. He is currently an Associate Professor in Computer Science at Illinois Institute of Technology, and at City University of Hong Kong. His research interests include wireless networks, optical networks, and algorithm design and analysis.
Lixin Wang Mr. Lixin Wang received the M.S. degree in CS from the University of Houston at Clear Lake, the M.S. degree in Applied Math from the University of Houston and the M.S. degree in Math from the Fudan University, Shanghai, China. He is currently a Ph.D. student in Computer Science at the Illinois Institute of Technology, Chicago. His research is on wireless networks, and algorithm design and analysis.
Chao-Min Su Mr. Chao-Min Su received his MS and BS degree from the National Taiwan University. Currently, he is a Ph.D. candidate in Computer Science at the National Chiao Tung University. His research interests are in wireless ad hoc and sensor networks, and sensor applications.
