5.5.1 Definition
Algorithm II is an enhanced version of Algorithm I. Compared to Algorithm I, Al- gorithm II further increases nodes’ coverage radii so that for any two nodes, if Algo- rithm I would create a directed edge between them, then Algorithm II will create a bi-directional edge between them. The reason for doing this is that in engineering, it’s often desirable to use bi-directional links for the ease of medium-access control and 2-way communication (although in principal, having short loops instead may also work).
Algorithm II has two steps, and it is defined as follows: “step 1, use Algorithm I to determine the coverage radius of every node; step 2, for every node v, if its coverage radius was r and the maximum length of its incoming edges was t in step 1, then v makes its coverage radius to be max{r, t}.”
5.5.2 Connectivity and Routing Property
All the good routing and connectivity properties that Algorithm I can achieve is also achieved by Algorithm II. Moreover, Algorithm II guarantees the strong connectivity of the network:
Theorem 5.5 A network constructed using Algorithm II is strongly connected.
Proof: For a fixed set of nodes on the 2-dimensional plane, let’s use Algorithm I (Algorithm II) to construct a network which we shall call Network I (Network II).
Clearly for any two nodespand q, if there is a directed path from ptoq or fromqto p in Network I, then in Network II, all the edges in that path become bi-directional
— so pcan q can reach each other. Now consider two arbitrary nodes u1 and u2. By Theorem 5.4, in Network I there is a strongly-connected subgraph thatu1 andu2 can both reach. Then it’s not hard to see that in Network II, u1 and u2 can reach each other. So Network II is strongly connected. 2
5.5.3 Coverage Radius and Node Degree
The distribution of a node’s coverage radius is hard to compute for Algorithm II.
Nevertheless, we can show that its expectation and variance are upper bounded by small constants.
Lemma 5.1 In a network constructed using Algorithm I, randomly and uniformly select an edge, and denote the edge’s length byL. Then the probability density function of L is fL(x) = 2π5 x(e−πx2 +πx2e−πx22), for x≥0.
Proof: Without loss of generality, we assume there is a node uat the origin point of the 2-dimensional plane, and consider its outgoing edges. Let’s use ∆(x) to denote the probability that “uhas an outgoing edge whose length is betweenxandx+dx”, where dx→ 0. Then ∆(x) =P r{the cone-angles ofu caused by the nodes within distance xare not all smaller than π}P r{there exists a node whose distance tou is betweenx andx+dx}=P r{the cone-angles ofucaused by the nodes within distancexare not
all smaller thanπ}[π(x+dx)2−πx2][1+o(1)] =P r{the cone-angles ofucaused by the nodes within distancexare not all smaller thanπ}[2πxdx+o(dx)]. From Theorem 5.3, we can see thatP r{the cone-angles ofucaused by the nodes within distancexare not all smaller thanπ}=e−πx2+πx2e−πx22. So ∆(x) = (e−πx2+πx2e−πx22)[2πxdx+o(dx)].
It is not hard to see that fL(x) = [limdx→0 ∆(x)dx ]·C = (e−πx2 +πx2e−πx22)·2πx·C, where C is some appropriate scaling factor. From R∞
0 fL(x)dx = 1, we find that C = 15. So fL(x) = 2π5 x(e−πx2 +πx2e−πx22), forx≥0.
2
Theorem 5.6 Let R denote the coverage radius of a node in a network constructed using Algorithm II. Then E(R)≤1.6585, and V ar(R)≤1.338.
Proof: Imagine we have a very large bag, and let’s play the following game: “generate the nodes following the Poisson point process and use Algorithm I to construct a network; corresponding to each node u of the network, we put a red stick whose length equals the coverage radius of u into the bag; corresponding to each directed edge of the network, we put a green stick whose length equals the length of the edge into the bag; repeat all the above steps.”
Clearly there will be infinitely many sticks in the bag, because each network has infinite nodes and we generate the network infinitely many times. Nevertheless, if we use LR to denote the length of a red stick and useLG to denote the length of a green stick, we know the probability density functions of LR andLG: fLR(x) = 2πxe−πx2+ π2x3e−πx22 −2πxe−πx22 (by Theorem 5.3), and fLG(x) = 2π5 x(e−πx2 +πx2e−πx22) (by Lemma 5.1). Also, Theorem 5.2 tells us that the average degree of a node is 5, so the number of green sticks is 5 times the number of red sticks. Therefore, if we use Lto denote the length of a stick in the bag (regardless of its color), then the probability density function of L is fL(x) = [fLR(x) + 5fLG(x)]· 16 = 2π3 xe−πx2 + π22x3e−πx22 −
π
3xe−πx22.
For Algorithm II, we see (from its definition) that a node determines its coverage radius through these two steps: “firstly, Algorithm I is used to construct a network;
next, every node sets its coverage radius to be the larger value of its old coverage
radius and the length of its longest old incoming edge.” Equivalently, we can also see the nodes as using the following method to determine their coverage radii: “every node picks a red stick and a green stick (not necessarily uniformly) from the bag, and sets its coverage radius to be the maximum length of these two sticks.”. Clearly, no two nodes need to, or should, pick the same red stick or green stick.
We can see that the average coverage radius for Algorithm II cannot exceed the average coverage radius for the following scheme: “pick just one stick out of the bag for each node (one by one), and set the node’s coverage radius to be equal to that stick’s length; for every node, we always pick the longest stick that is still available (regardless of its color).” What is the average length of the sticks picked out in the above scheme? The number of sticks in the bag is 6 times the number of nodes. So if we are picking out exactly those sticks longer than a valuez, thenR∞
z fL(x)dx= 16
— so we find z ≈ 1.4115. Then the average length of the sticks picked out in the above scheme is [R∞
z xfL(x)dx]·6 = 1.6585. So E(R)≤1.6585.
With the same method we can prove that E(R2)≤2.7948. The average coverage radius for Algorithm II is no less than that of algorithm I, so E(R) ≥ √2+12 ≈ 1.207 (by Theorem 5.3). So V ar(R) =E(R2)−E2(R)≤2.7948−1.2072 = 1.338.
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From the proof of Theorem 5.6, it is easy to realize that, in fact, all the moments of the coverage radius (not just its expectation and second moment) in a network constructed using Algorithm II are finite constants (instead of ∞). As to the node degree, we can see that in such a network, the neighbors within a node’s coverage range do not follow a Poisson point process, because the node’s coverage radius is dependent on the positions of the neighbors. Nevertheless, it is not hard to see that the expectation of the node degree is also a finite constant.