Establishing symmetric connectivity in directional wireless sensor networks equipped with $2\pi /3$ antennas

Abstract

In this paper, we study the antenna orientation problem concerning symmetric connectivity in directional wireless sensor networks. We are given a set of nodes each of which is equipped with one directional antenna with beam-width $\theta = 2\pi /3$ and is initially assigned a transmission range 1 that yields a connected unit disk graph spanning all nodes. The objective of the problem is to compute an orientation of the antennas and to find a minimum transmission power range $r=O(1)$ such that the induced symmetric communication graph is connected. We propose two algorithms that orient the antennas to yield symmetric connected communication graphs where the transmission power ranges are bounded by 6 and 5, which are currently the best results for this problem. We also study the performance of our algorithms through simulations.

Appendix: Proof of Lemma 2

Claim Let A and B be two groups, each contains three sensor nodes equipped with antennas of degree $2\pi /3$ whose range is $\infty$. If A and B are separated by a line l , then there exists an orientation of the antennas such that the six nodes form a symmetric connected graph (See Fig. 16).

Fig. 16

Two groups of sensors separated by line l

The idea is to show that if we orient the antennas of each group independently using the method of Lemma 1, then there is a bi-directional edge between a node in group A and a node in group B. We first make the following observation.

Observation Suppose that there are 3 nodes whose antennas cover the entire plane. Let p be one of these nodes. Then the antenna at p covers one third of the plane called p’s sector. The remaining two thirds are divided into two areas (1) and (2), of which each is one third of the plane as depicted in Fig. 17. The other two nodes must cover the areas (1) and (2). Let q be the node that covers (1) and $l_1$ the side of p’s sector that separates area (1) and p’s sector. We claim that q must lie in the half plane defined by $l_1$ that does not contain area (1) as depicted in Fig. 17. To show this, let us look at Fig. 18. If q lies in the half plane containing area (1), then there exists an area in (1) that is not covered by q and p, and as the third point must cover area (2), it cannot cover that area either.

Fig. 17

Three sensors cover all the plane

Fig. 18

Position of sensor q

Fig. 19

Plane B is covered by 2 nodes of group A

Fig. 20

Case 1.c

Fig. 21

$a_2$ must reside in area (*)

Fig. 22

$a_1$ lies above $b_2$’s sector

Fig. 23

$a_1$ lies below $b_2$’s sector

Fig. 24

Plane B is covered by 3 nodes of group A

Fig. 25

Case 2.c

Fig. 26

Case 2.c.1

Fig. 27

Case 2.c.2

We now return to the proof of Lemma 2 which will be shown by considering various cases concerning the coverage of plane B by nodes in group A. Clearly, plane B is covered by at least two nodes in group A. Thus we have two main cases: either plane B is covered by two nodes of group A or by three nodes.

• Case 1: Plane B is covered by 2 nodes of group A, say $a_1$ and $a_2$ (Fig. 19).

  Since the nodes in group B cover the entire plane, let $b_1$ be a node of B that covers $a_1$.

  1.a: If $b_1$ is in $a_1$’s sector, then we have

  

  1.b: $b_1$ is in $a_2$’s sector and $b_1$ covers $a_1$ and $a_2$

  

  1.c: $b_1$ is in $a_2$’s sector and $b_1$ covers $a_1$ but $b_1$ does not cover $a_2$. Let (*) be the sector adjacent with $b_1$ sector that intersects plane A. Let $\alpha$ be the angle formed by line l and one edge of $a_1$’s sector and $\beta$ be the angle formed by line l and one edge of $b_1$’s sector as depicted in Fig. 21. Note that $\frac{\pi }{3}\le \alpha \le \frac{2\pi }{3}$ because $a_1$ and $a_2$ cover the entire plane B. Therefore $\beta \le \frac{2\pi }{3}$ which implies that $a_2$ must lie within area (*). There are two possible scenarios as depicted in Fig. 20.

  In case 1.c.1, there must be a node $b_2$ that covers area (*). Hence, $b_2$ also covers $a_2$ since node $a_2$ lies in area (*). Since $a_2$ covers $b_2$, there is an edge $(a_2, b_2)$.

  For case 1.c.2, let $b_2$ be the node that covers area (*) in Fig. 20. By a similar reasoning, $a_2$ must be covered by $b_2$.

  If $b_2$ is in $a_2$’s sector, then $a_2$ and $b_2$ cover each other yielding edge $(a_2,b_2)$.

  If $b_2$ is in $a_1$’s section, we claim that it must also cover $a_1$ yielding edge $(a_1,b_2)$. Suppose by way of contradiction that $b_2$ does not cover $a_1$. There are 2 possible cases for the position of $a_1$ in regard to $b_2$’s sector.

  • $a_1$ lies above $b_2$’s sector as depicted in Fig. 22.

    Based on the orientation of the antennas of group B as done in Lemma 1, if $b_2$ does not cover $a_1$, it cannot cover $b_1$. Thus, $b_2$ must be the node that holds the biggest angle of $\triangle b_1b_2b_3$ and the third node $b_3$ must be the node that holds the smallest angle of $\triangle b_1b_2b_3$. Therefore, $b_3$ must lie on one of the edges of $b_1$’s sector and is not covered by $b_2$. Consequently, $b_1,b_2,b_3$ do not form a symmetric connected graph. A contradiction!

  • $a_1$ lies below $b_2$’s sector as depicted in Fig. 23.

    We have that $b_1$ is in $a_2$’s sector and $b_2$ is in $a_1$’s sector. In this case, $b_1$ must cover all the area below line $l_1$ that contains $a_2$. This contradicts our assumption that $b_1$ is in $a_2$’s sector but it does not cover $a_2$.

• Case 2: Plane B is covered by all 3 nodes of group A as depicted in Fig. 24. Let $b_1$ be the node that covers node $a_1$. For the position of $b_1$ there are three cases.

  2.a: $b_1$ is in $a_1$’s sector, then there is edge $(a_1,b_1)$.

  2.b: $b_1$ is not in $a_1$’s sector. In this case, $b_1$ may lie in $a_2$’s sector or $a_3$’s sector. Without loss of generality, let $b_1$ be in $a_3$’s sector. If $b_1$ covers $a_3$, then there is edge $(a_3,b_1)$.

  2.c: $b_1$ is in $a_3$’s sector but it does not cover $a_3$. Then there are two possible configurations as depicted in Fig. 25.

  2.c.1. $b_1$ cover $a_1$ but it does not cover $a_3$ as shown in Fig. 26:

  We can see that $a_3$ must lie in area (*) adjacent with $b_1$’s sector. Let $b_3$ be the node of group B that covers area (*). From the previous observation, $b_3$ must lie in the half plane above $l_1$. Then $b_3$ is covered by $a_3$ and $a_3$ is also covered by $b_3$, yielding edge $(a_3,b_3)$.

  2.c.2. $b_1$ covers $a_1$ but doesn’t cover $a_3$ as shown in Fig. 27.

  If $b_3$ is in $a_3$’s sector, then $b_3$ covers $a_3$ and $a_3$ covers $b_3$, yielding edge $(a_3,b_3)$. Otherwise, if $b_3$ is in $a_1$’s sector, then we can use the same argument as in Case 1.c.2 to prove that $b_3$ must cover $a_1$. Thus, $b_3$ covers $a_1$ and $a_1$ covers $b_3$, yielding edge $(a_1,b_3)$.