Maintaining
Connectivity in Sensor
Networks Using
Directional Antennae
Evangelos
Kranakis
,
Danny
Krizanc
,
Oscar Morales
Presented by Tal Beja
Outline
Introduction.
The main problem.
Definitions.
Solving the main problem.
Introduction
Connectivity in wireless sensor networks may be
established using either omnidirectional or
directional antennae.
Each antennae has a transmission range (radios).
Directional antennae has also a spread (angle).
Omnidirectional
Antennae
Directional
Antennae
r
r
𝛼
Introduction
Given a set of sensors positioned in the plane
with antennae (directional or omnidirectional),
a directed network is formed as follows:
The edge (u, v) is in the network if v lies within u’s
sector.
It is easy to see that with omnidirectional
antennae with the same radius all network is
unidirectional, because u in the range of v
iff
v in the range of u.
Introduction
Examples:
Introduction
Why to use directional antennae?
The coverage area of omnidirectional antennae with
range r is:
𝜋
2
The coverage area of directional antennae with range
r and spread
𝛼
is:
𝛼
𝑟
2
2
The energy is proportional to the coverage area.
With the same energy E the omnidirectional antennae
can reach a range of
𝜋
, while the directional
antennae can reach a range of
2
𝛼
.
The main problem
Consider a set S of n point in the plane that can be
identified with sensors having a range
>
0
.
For a given angel
0
≤
𝛼
≤
2
𝜋
and an integer
≥
1
, each sensor is allowed to use at most
directional antennae each of angle at most
𝛼
.
Determine the minimum range
required so that
by appropriately rotating the antennae, a directed,
strongly connected network on S is formed.
Definitions and notations
𝑘
(
,
𝛼
)
–
the minimum range of directed antennae of angular
spread at most
𝛼
so that every sensor in S uses at most
such
antennae (under appropriate rotation) a strongly connected network
on S results.
𝑘
=
𝑘
,
0
𝐷
𝑘
(
)
–
the set of all strongly connected graphs on S with out

degree at most k.
𝑘
,
∈
𝐷
𝑘
(
)
–
the maximum length of an edge in G.
𝑘
=
min
𝐺
∈
𝑘
𝑘
.
(
)
–
a set of all MSTs on S.
,
∈
(
)

the maximum length of an edge in T.
𝑀
=
min
∈
𝑀
(
)
.
𝑀
≤
𝑘
(
,
𝛼
)
.
MST and out degrees
If the set S of n points is on two

dimensional plane there is an MST with a
maximum degree six.
This can be improved to an MST with max
degree of five.
Definition: for
≥
2
,
𝐷
−
is a spanning
tree with max degree of k.
Solving the main problem
When k =
1
, and
𝛼
=
0
We have a problem of finding a Hamiltonian cycle than
minimize the maximum length of an edge.
This known as the bottleneck traveling agent problem (BTSP).
Parker and Rardin study the BTSP where the lengths satisfy
the triangle inequality.
They found
2

approximation algorithm for this problem.
They also proved that there is no better approximation
algorithm with polynomial time to the problem, unless P = NP.
Solving the main problem
Sensors on a line, k =
1
:
Where
𝛼
≥
𝜋
we have the same problem as
omnidirectional antennae. And r is the maximum
length between a pair of two adjacent sensors.
When
0
≤
𝛼
<
𝜋
and k =
0
, there exist an orientation
of the antennae where the graph is strongly
connected if and only if the distance between points I
and i+
2
is at most r, for any
1
≤
𝑖
≤
−
2
.
Solving the main problem
Proof:
Assume
𝑖
,
𝑖
+
2
>
, for some
𝑖
≤
−
2
.
Consider the antenna at
𝑖
+
1

there are two
cases to consider:
The antenna directed left
–
then the left side of the
graph cannot be connected to the rest.
The antenna directed right
–
then the right side of the
graph cannot be connected to the rest.
If we direct the odd label antennae to the right
and the even label antennae to the left we will get
a strongly connected graph.
Solving the main problem
Sensors on a plane, k=
1
.
Where
𝛼
=
0
we have the BTSP problem.
When
𝜋
≤
𝛼
<
8
𝜋
5
there exist a polynomial
algorithm that when given MST on S
computes an orientation of the antennae
with radius of
2
sin
(
𝜋
−
𝛼
2
)
∙
1
,
𝛼
Solving the problem
The algorithm:
←
∅
,
Λ
←
∅
←
∈
𝑉
=
⋃
,
𝑖
.
ℎ
𝑖
←
𝐵
𝑖
𝑖
𝑖
Λ
←
Λ
⋃
𝑖
,
𝜋
∉
=
⋃
,
𝑖
ℎ
𝑖
ℎ
𝑖
.
ℎ
,
∈
𝑖
ℎ
ℎ
𝑖
𝑖
.
ℎ
∈
Λ
𝑖
ℎ
𝜋
.
Solving the problem
Proof:
Lemma
–
for each
,
∈
and for each
neighbor
of either
or
, G (the
transmission graph) contains two opposite
directed edge between
and
, and it
contains a directed edge between
either
or
to
.
Solving the problem
Proof (of the lemma):
𝛼
=
2
𝑖
(
𝜋
−
𝛼
2
)
–
the range of the antennae.
Since
𝜋
≤
𝛼
<
8
𝜋
5
,
d
𝛼
=
2
𝑖
𝜋
−
𝛼
2
≥
2
𝑖
𝜋
5
>
2
𝑖
𝜋
6
=
Because
and
are neighbors
≤
and their
antennae directed at each other they both in each
other sector.
So we have in G two directed edges between
and
.
Solving the problem
Proof (of the lemma):
Assuming
is a neighbor of
(if he is a neighbor of
we
have a symmetrical case).
If the counter clockwise angel
≤
𝛼
then
is in the
sector of
, since
≤
.
If
the counter clockwise angel
>
𝛼
then by the low of
cosines in the triangle define by
,
and
:
=
2
+
2
−
2
≤
2
−
2
=
2
𝑖
𝑥
2
≤
2
𝑖
𝜋
−
𝛼
2
=
(
𝛼
)
So
is in the sector of
.
So we have in G and edge between either
or
to
.
Solving the problem
Proof :
We need to prove now that for any edge
(
,
)
in T we have a direct
path from
to
and from
to
in G.
Without loss of generality, assume that
is closer to the root of T (the
first node we selected in the algorithm).
If the edge
,
∈
than we have two directed edges between them in
G(from the lemma).
Otherwise let
1
be the node where
1
,
∈
.
Since
is a neighbor of
there are a directed edge form
or from
1
to
and a directed edge from
to
1
in G (from the Lemma)
–
so we have
a directed path from
to
in G.
If
∈
Λ
(lead) there are a directed edge between
and
in G (from the
algorithm).
Otherwise let
2
be the node where
2
,
∈
.
Since
is a neighbor of
there are a directed edge form
or from
2
to
and a directed edge from
to
2
in G (from the Lemma)
–
so we have
a directed path from
to
in G.
Solving the main problem
Theorem: Consider a set S of n sensors in the
plane and suppose each sensor has
,
1
≤
≤
5
,
directional antennae. Then the antennae can be
oriented at each sensor so that the resulting
spanning graph is strongly connected and the
range is at most
2
sin
𝜋
𝑘
+
1
∙
𝑘
(
,
𝛼
)
.
Moreover, given an MST on S the spanner can be
compute with additional
Ο
(
)
overhead.
Solving the main problem
When k =
1
,
𝛼
<
2
𝜋
3
and r >
0
.
Determining whether there exists an
orientation of the antennae so the
transmission graph is strongly connected
is NP

complete.
Proof: by reduction from an NP

hard
problem of finding Hamiltonian cycles in
degree three planar graphs.
Solving the main problem
When k =
2
, and the angular sum of the
antennae is
𝛼
then it is NP

hard to
approximate the optimal radius to within a
factor of x, where
=
2
sin
(
𝛼
)
.
Proof: this is also proven with reduction
from finding Hamiltonian cycles in degree
three planar graphs.
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