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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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