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CCCG 2010, Winnipeg MB, August 9–11, 2010

Direction Assignment in Wireless Networks

Boaz Ben-Moshe∗

Paz Carmi‡

Lilach Chaitman‡

Yael Stein‡

Matthew J. Katz‡

Gila Morgenstern‡

Abstract

In this paper we consider a wireless network, where each

transceiver is equipped with a directional antenna, and

study two direction assignment problems, determined

by the type of antennas employed. Given a set S of

transceivers with directional antennas, located in the

plane. We investigate two types of directional antennas

— quadrant antennas and half-strip antennas, and show

how to assign a direction to each antenna, such that the

resulting communication graph is connected.

1Introduction

Wireless ad hoc networks have received much attention

in the last decade due to their role in civilian and mili-

tary applications [3, 4, 6, 7]. A wireless network consists

of numerous devices that are equipped with process-

ing, memory and wireless communication capabilities,

and are linked via short-range ad hoc radio connections.

Each node in such a network has a limited energy re-

source (battery), and each node operates unattended.

Consequently, energy efficiency is an important design

consideration for these networks. One way to conserve

energy is to use directional antennas, whose coverage

area is often modeled by a sector of a given angle and ra-

dius. A direction assignment is the task of aiming each

directional antenna in a certain direction, so that the

induced communication graph has some desired prop-

erties, such as connectivity. The communication graph

G has an edge between two transmitters p and q if and

only if p lies in the region covered by q and q lies in the

region covered by p.

In this paper we consider two types of directional an-

tennas, quadrant and half-strip antennas. For the first

type (quadrant), we give a two approximation on the

required range to ensure connectivity. That is, if there

exists a direction assignment with range r, we find a di-

rection assignment with range at most 2r. For the sec-

ond type (half-strip), we give a direction assignment to

the antennas that induces a connected communication

graph, if one exists. At first glance it is not clear why

∗Department of Computer Science, Ariel University Center.

‡Department of Computer Science, Ben-Gurion University,

Beer-Sheva, Israel. Partially supported by the Lynn and William

Frankel Center for Computer Sciences.

we consider the second type of directional antennas (i.e.,

half-strip antennas). However, consider Figure 1 where

the region covered by a narrow RF antenna is depicted,

see also [9] and [2, 5, 8]. We believe that a half-strip

is a good approximation for the covered region in this

case. Due to space considerations, we restrict our at-

tention to the decision version, where r is given and we

only need to assign a direction to each antenna, such

that the resulting communication graph is connected.

Notice that if r is not given, we can find it using binary

search on the O(|S|2) potential ranges. Moreover, this

can be done efficiently using parametric search [1].

Figure 1: An example of patterns of narrow antennas.

Considering a free space pathloss propagation model,

the coverage region of a directional antenna can be ap-

proximated using a narrow rectangle.

1.1 Notation and problem definitions

Notation 1 Let Hr(p) denote a vertical strip of width

2r, such that the vertical line that passes through the

point p splits the strip into two equal strips of width r.

Notation 2 Let H+

strip Hr(p), having p on its boundary. See Figure 2.

r(p) denote the upper half-strip of

Notation 3 Let H−

strip Hr(p), having p on its boundary. See Figure 2.

r(p) denote the lower half-strip of

In this paper we consider the following problems.

Problem 1 Given a set S of directional antennas in

?2and a range r. The goal is to direct each antenna to

one of the four quadrants, such that the resulting com-

munication graph is connected, where there is an edge

between p and q, if and only if the following conditions

hold.

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22ndCanadian Conference on Computational Geometry, 2010

rr

p

H−

r(p)

H+

r(p)

Figure 2: Strip Hr(p) and the two half strips H+

H−

r(p) and

r(p).

(1) antenna q lies in the quadrant that p is directed to,

(2) antenna p lies in the quadrant that q is directed to,

(3) |pq| ≤ r

Problem 2 Given a set S of directional antennas in

?2and a range r. The goal is to direct each antenna

either upwards or downwards, such that the resulting

communication graph is connected, where there is an

edge between p and q, if and only if one of the following

conditions holds.

(1) antenna p is directed up, q is directed down, and

q ∈ H+

(2) antenna p is directed down, q is directed up, and

q ∈ H−

r(p).

r(p).

2 Quadrant antennas

In this section we consider Problem 1. Let S be a set of n

transceivers of transmission range r, each equipped with

a quadrant antenna. A quadrant antenna can assume

one of four directions: north-east (position I), north-

west (II), south-west (III), and south-east (IV). Thus,

the coverage area of a transceiver is a quarter disk of

radius r that coincides with one of the four quadrants.

The goal is to assign one of the four directions (po-

sitions) to each of the transceivers, such that the re-

sulting communication graph is connected, where there

is an edge between two transceivers, if and only if each

transceiver lies in the assigned quadrant of the other and

the distance between the two transceivers is at most r.

We assume that the transceivers are in general position;

i.e., no two transceivers have the same x-coordinate or

the same y-coordinate.

Lemma 4 In a solution to our problem, if one of the

antennas is in position I (alternatively, position III),

then all antennas are in position I or III.

Proof. Assume by contradiction that there exist two

antennas p and q, such that p is in position I (alterna-

tively, III) and q is in position II or IV. Consider a path

P between p and q. Then, there must be two consecu-

tive antennas p?,q?∈ P, such that p?is in position I or

III and q?is in position II or IV, and there is an edge

between p?and q?. However, since the antennas are in

general position this cannot occur — contradiction.

?

Corollary 5 We may assume w.l.o.g. that all anten-

nas are in positions I or III.

Notation 6 Let Arci

transceiver p of range r when its antenna is in position

i, where i is either I or III.

r(p) denote the region covered by

Assuming there exists a direction assignment with

range r, such that the resulting communication graph

is connected, we show how to find such a direction as-

signment with range at most 2r. I.e., we present a 2-

approximation algorithm (on the range).

Lemma 7 Let SOLr denote a solution with range r.

Let p and q be two antennas in S, such that p ∈

ArcIII

r

(q) and q ∈ ArcI

lution SOL2rwith range 2r, in which p is in position I

and q is in position III.

r(p). Then, there exists a so-

Proof. In order to show this, it is enough to show that

any antenna that is connected in SOLr to p or to q,

can be connected to either p or q when p and q are

in positions I and III, respectively (and the range is

2r). Notice that there are exactly four different ways

to position two antennas. Thus, antennas p and q are

positioned in one of the following four ways in SOLr.

• Both p and q are in position I. Then, all antennas

that are connected to q in SOLr, are connected to p in

SOL2r, and antenna p is positioned as in SOLr.

• p is in position I and q is in position III. In this

case, antennas p and q are positioned as in SOLr.

• Both p and q are in position III. Symmetric to

case 1.

• p is in position III and q is in position I. Then, all

antennas that are connected to q in SOLr, are connected

to p in SOL2r, and all antennas that are connected to p

in SOLr, are connected to q in SOL2r.

?

Lemma 8 Let p,q ∈ S be two antennas with direc-

tion assignment i,j respectively, such that Arci

Arcj

ing s to either p or q will produce a connection between

p, q, and s (allowing range 2r).

r(p) ∩

r(q) ?= ∅. Let s ∈ Arci

r(p) ∩ Arcj

r(q). Then, direct-

Proof. There are two cases to consider.

Case 1: Antennas p and q are in the same position,

w.l.o.g. both are in position I. Then assigning s position

III will clearly connect s to both p and q.

Case 2: Antennas p and q are in different positions,

w.l.o.g. p is in position I and q is in position III. Then,

|pq| ≤ |ps| + |sq| ≤ 2r (and p ∈ ArcIII

ArcI

and q, and we can assign s either position I or III to

obtain connectivity.

2r(q) and q ∈

2r(p)). Therefore, there is a connection between p

?

Corollary 9 From the above lemmas it follows that Al-

gorithm 1 finds a valid direction assignment with range

2r, if there exists a direction assignment with range r.

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CCCG 2010, Winnipeg MB, August 9–11, 2010

Algorithm 1 Quadrant assignment

Input: A set P of n quadrant antennas.

Output: A direction to each antenna of P, such that

the resulting communication graph (with range 2r)

is connected.

1: P?← P, Q ← ∅.

2: while there are two points p,q ∈ P?, such that q ∈

ArcI

r

(q) do

3:

Choose such p and q

4:

Q ← Q ∪ {p,q}.

5:

Assign p position I.

6:

Assign q position III.

7:

P?← P?\ {v ∈ P?|v ∈ ArcI

8: S ← P \ Q.

9: while S ?= ∅ do

10:

Choose s ∈ S and p ∈ Q such that p is in position

i and s ∈ Arci

11:

Direct s to p.

12:

S ← S \ {s}

r(p) and p ∈ ArcIII

r(p) ∪ ArcIII

r

(q)}.

r(p).

3Half-strip antennas

In this section we consider Problem 2. Let S be a set

of n transmitters in general position (i.e., no two have

the same x-coordinate or same y-coordinate), and let r

be a real number. Each transmitter p ∈ S is equipped

with a half-strip antenna that can transmit either up-

wards, covering the region H+

ering H−

or down to each of the antennas, such that the resulting

communication graph is connected. Where there is an

edge between transmitters p and q, if and only if one of

the following cases holds.

• p is directed up, q is directed down, and q ∈ H+

• p is directed down, q is directed up, and q ∈ H−

Lemma 10 If there exists a direction assignment to the

antennas of S, such that in the resulting communica-

tion graph the leftmost and rightmost antennas are con-

nected, then there exists a solution to our problem (i.e.,

a direction assignment, such that the resulting commu-

nication graph is connected).

r(p), or downwards, cov-

r(p). The goal is to assign a direction either up

r(p).

r(p).

Proof. Let p0 be the leftmost antenna, pn the right-

most antenna, and P = (p0,p1,...,pn) a path connect-

ing p0 and pn. In order to show that there exists a

solution to our problem, it is enough to show that each

antenna q ∈ S\P can be connected to an antenna in

the path P, without changing the directions assigned to

the antennas of P. q is either above or bellow an edge

(pi,pi+1) ∈ P; w.l.o.g. assume q is above (pi,pi+1).

Moreover, exactly one of the two antennas piand pi+1

is directed upwards; assume w.l.o.g. that antenna piis

directed upwards. Then, since qxbetween pixand pi+1x,

q is already covered by pi, and by directing q downwards

we establish the edge (q,pi), without changing the di-

rections assigned to the antennas along the path.

?

Among all solutions, consider a solution in which the

shortest path between the two x-extreme points is short-

est (in terms of number of hops), and let OP be such a

shortest path.

Lemma 11 Let OP = (p0,...,pn), then for any two

antennas pi,pj∈ OP, i < j, that are directed upwards

(alternatively, downwards), it holds that piis to the left

of pj.

Proof. Assume to the contrary that there exist two an-

tennas pi,pj ∈ OP, i < j, that are both directed up-

wards and pjx< pix. Consider the first such pair of

antennas pi,pj, i < j (determined by the lexicographic

order over {(i,j)}). We distinguish between two cases.

Case 1: pjy< pi−1y

Notice that both piand pjare to the right of pi−2. Since,

if one of them is leftward to pi−2, then (observing that

pi−2is directed upwards) we reach a contradiction; the

pair pi−2,pjalready violates the claim. It follows that

pi−2x< pjx< pix. But, if so, pi−1 also captures pj

(since pjy< pi−1y), and we can shorten the path OP —

contradiction.

Case 2: pjy> pi−1y

Let j?≥ j be the maximal index such that pj? is directed

upwards and is to the left of pi. Then pj?y> pi−1y,

since, by Case 1, the opposite is impossible. We also

know that pix< pj?+2x, by the maximality of j?. Now,

because piy< pi−1y< pj?y< pj?+1y, and because

pj?x< pix< pj?+2x, we obtain that there is a connec-

tion between piand pj?+1and thus we can shorten the

path OP — contradiction.

?

Notation 12 An edge e = (u,v) ∈ OP is good, where

OP is as in Lemma 11, if for each s ∈ S such that

ux< sx< vx, s is not in OP.

Lemma 13 Let (pi,pi+1) and (pi+1,pi+2) be two con-

secutive edges in OP, such that pix< pi+2x< pi+1x,

then (pi+1,pi+2) is good.

Proof. Assume by contradiction that edge (pi+1,pi+2)

is not good. Assume w.l.o.g. that point piis directed up

and point pi+1is directed down. Let pjbe a point that

lies between pi+2and pi+1(i.e., pi+2x< pjx< pi+1x).

If j < i then by Lemma 11 pj is directed down, and if

j > i then pjis directed up. Let us consider these two

cases.

Case 1: j < i, pj is directed down.

In this case pjy< pi+2y, since otherwise there would

be a shortcut in the path OP between pj and pi+2,

contradicting the minimality of OP. Consider a direc-

tion assignment as in OP with two modifications: pj

is directed up and pi+2 down. Since pj−1x< pixand

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22ndCanadian Conference on Computational Geometry, 2010

pj−1y< pjy< pi+2y, we have a connection between pj−1

and pi+2. Moreover, we have a connection between pj

and pi+3, and between pj and pi+2, contradicting the

minimality of OP.

Case 2: j > i, pj is directed up.

In this case pjy> pi+1y, since otherwise there would be

a shortcut in the path OP between pj and pi+1, con-

tradicting the minimality of OP. Consider a direction

assignment as in OP with two modifications: pj is di-

rected down and pi+1 up.

pjy> pi+1y> pi+2y, we have a connection between

pi+1and pj+1. Moreover, we have a connection between

pj and pi, and between pj and pi+1, contradicting the

minimality of OP.

Since pj+1x> pi+1xand

?

Corollary 14 Let e = (pi,pi+1) and e?= (pi+1,pi+2)

be two consecutive edges in OP, such that pi+1x >

pix,pi+2xor pi+1x< pix,pi+2x, then by Observation 13

either edge e or edge e?is good.

Lemma 15 Consider any three consecutive edges in

OP, then at least one of them is good.

Proof. Assume by contradiction that there are three

consecutive edges (pi−1,pi), (pi,pi+1), (pi+1,pi+2), such

that none of them is good.

pi−1x< pix< pi+1x< pi+2x. If there exists a point

pj such that pix< pjx< pi+1x, then by Lemma 11 it

cannot be directed upwards nor downwards; thus, there

cannot be such a point. We conclude that (pi,pi+1) is

good.

Then, by Corollary 14

?

Observation 16 Consider the ordered sequence of

good edges in OP (by x-coordinate), then it follows from

Lemma 15 that any two consecutive edges in this se-

quence are connected either directly or via a single point

that lies between them (with respect to the x-axis).

Corollary 17 If there exists a solution (with range

r), then there also exists a shortest path OP

(p0,p1,...,pn) as above.Moreover, from Lemma 15

and Observation 16 it follows that we can apply dynamic

programing to find this path, since there always exists an

edge that splits the problem into two independent prob-

lems. Therefore, Algorithm 2 below finds a valid direc-

tion assignment if such exists.

=

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Algorithm 2 Half-strip assignment

Input: A set S of n transmitters.

Output: A direction assignment to each antenna in S,

such that the resulting communication graph is con-

nected.

1: Let L be the set of potential good edges, sorted by

x-coordinate of the left endpoint.

2: Construct a table for each pair of edges ei,ej∈ L.

3: Initialize each entry as false.

4: for each eiand ejin L do

5:

if ei and ej can be connected directly, or via a

point that lies between them (with respect to the

x-axis) then

6:

Set entry (ei,ej) as true. else

7:

if there exists e?in L between eiand ej, such

that the entries (ei,e?) and (e?,ej) are both true

then

8:

Set entry (ei,ej) as true

9: if (there exists an entry (ei,ej) marked true) and

(p0 can be connected to ei) and (pn can be con-

nected to ej) then

10:

Return the direction assignment induced from the

table entries.

11:

Connect each remaining transmitter according to

Lemma 10.

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