On solving coverage problems in a wireless sensor networks using diagrams

On Solving Coverage Problems in a Wireless
Sensor Network Using Voronoi Diagrams
Anthony Man–Cho So1
and Yinyu Ye2
1
Department of Computer Science, Stanford University, Stanford, CA 94305, USA.
manchoso@cs.stanford.edu
2
Department of Management Science and Engineering and, by courtesy, Electrical
Engineering, Stanford University, Stanford, CA 94305, USA.
yinyu-ye@stanford.edu
Abstract. Owing to numerous potential applications, wireless sensor
networks have been the focus of a lot of research efforts lately. In this
note we study one fundamental issue in such networks, namely the cov-
erage problem, in which we would like to determine whether a region
of interest is sufficiently covered by a given set of sensors. This problem
is motivated by monitoring applications in sensor networks, as well as
robustness concerns and protocol requirements. We show that the cover-
age problem and some of its variants can be treated in a unified manner
using suitable generalizations of the Voronoi diagram. As a result, we are
able to give algorithms that have better runtimes than those proposed
in previous works (see, e.g., [5, 6]). Our approach also yields efficient al-
gorithms for coverage problems where the sensing region of a sensor is
an ellipse or an Lp–ball, where p ≥ 1.
1 Introduction
1.1 Background
Due to recent advances in wireless communication and embedded micro–sensing
MEMS technologies, inexpensive wireless nodes capable of sensing, storing, pro-
cessing, and communicating data are becoming increasingly common and readily
available. Such devices provide the means for building large–scale wireless sensor
networks for various applications, such as distributed data storage [11], target
tracking [12], and habitat monitoring [13], just to name a few. In these appli-
cations, sensors collect information about their surroundings, and the data so
obtained are then aggregated to give a complete picture of the region of interest.
Hence, a fundamental issue in the design of wireless sensor networks is coverage,
i.e. how well are the sensors covering the region of interest. This issue has been
tackled by various researchers in the sensor networks community. Typically, a
sensor s is modelled as a point ps in space, and its sensing region is modelled as
an Euclidean ball centered around ps. From this simple model, many different
coverage problems have been proposed. For example, the authors of [10] consid-
ered the problems of finding the maximal breach path and the maximal support

path in a network, which informally can be viewed as the paths that are least and
best monitored in the network, respectively. The maximal support path problem
is further investigated by the authors of [9] in which energy consumptions of the
sensors are also being taken into account. The authors of [14, 15], also motivated
by energy consumption considerations, studied the problem of finding a subset
of sensors C so that the region of interest S is covered by those sensors (i.e. each
point in S is within the sensing range of some sensor in C ). This problem is later
generalized by the authors of [16], in which they considered finding a subset of
sensors C of minimal cardinality such that (i) the communication graph formed
by the sensors in C is connected, and (ii) the region of interest S is k–covered by
those sensors (i.e. each point in S is within the sensing ranges of k distinct sen-
sors in C ). They proposed an algorithm that yields a subset whose cardinality is
at most O(log n) times larger than the optimal and has the required properties.
In this note, we consider the following two coverage problems: (i) given a
set of sensors C with their sensing regions and a region of interest S, determine
whether every point in S is k–covered by the sensors in C , where k ≥ 1 is
a given constant; (ii) given a set of sensors C with their sensing regions and
a region of interest S, determine the largest k such that S is k–covered by the
sensors in C . These two problems are motivated by robustness concerns as well as
protocol requirements. For example, triangulation–based localization protocols
require at least three sensors to localize an object, and hence we would like
every point in S to be at least 3–covered. Regarding the first problem, Huang
and co–authors [5, 6] presented an O(nm log m) (resp. O(nm2
log m)) algorithm
for determining whether a region in R2
(resp. R3
) is k–covered by the given n
sensors, where m is the maximum number of sensing regions that can intersect
the sensing region of a particular sensor. Since m can be of order Θ(n) (see
Figure 1), their algorithms have worst–case runtimes of O(n2
log n) (for the 2–d
case) and O(n3
log n) (for the 3–d case), respectively. Also, it is curious that
the runtimes of both algorithms are independent of k. Thus, it is natural to ask
whether there exists an algorithm that can exploit the tradeoff between runtime
and the parameter k. Regarding the second problem, a na¨ıve approach would be
to run the above algorithm k(≤ n) times, resulting in a worst–case O(n3
log n)
(resp. O(n4
log n)) algorithm for the 2–d case (resp. 3–d case). However, to the
best of our knowledge, this problem has not been addressed directly in the sensor
networks community.
1.2 Our Contribution
We propose to treat the aforementioned coverage problems in a unified manner
using suitable generalizations of the Voronoi diagram. This provides a rigorous
mathematical framework for tackling those problems and enables us to design
algorithms that have much more favorable worst–case runtimes than those pro-
posed in previous works. Specifically, we show that for the 2–d case, if the sensors
have identical sensing ranges, then problem (i) can be solved in O(n log n+nk2
)
time by computing the so–called k–th order Voronoi diagram [4, 8]. Thus, our
algorithm is independent of the configuration of the sensors and depends only on

S
Fig. 1. A scenario where m = Θ(n): the n + 1 circles are centered at (1 − 2i/n, 0),
where i = 0, 1, . . . , n, and each of them has radius 1 + for some > 0.
the number of sensors, n, and the parameter k. Moreover, for k = o(
√
n log n),
our algorithm has a better runtime than that proposed in [5]. In the case where
the sensing ranges are different, we are able to obtain an O(n log n) algorithm
to determine whether a region S ⊂ R2
is 1–covered. Our framework also allows
us to study coverage problems in which the sensing region is not a ball. In par-
ticular, we give efficient algorithms for determining whether a region S ⊂ R2
is
1–covered when the sensing regions are modelled as ellipses or Lp–balls, where
p ≥ 1. Both of these algorithms are faster than the straightforward adapta-
tion of the techniques presented in [5]. For problem (ii), we obtain an O(n3
)
(resp. O(n4
)) algorithm for the 2–d case (resp. for the 3–d case) by using a well–
known relationship between power diagrams and arrangements [2]. This is again
faster than a direct application of the algorithms described in [5, 6].
2 Problem Statement
We formulate the aforementioned coverage problems as follows. For simplicity
of discussion, let the region of interest S be the cube {x ∈ Rd
: x ∞ ≤ 1}. Let
ai ∈ Rd
be the position of the i–th sensor, and let Bi ≡ B(ai, ri) = {x ∈ Rd
:
x − ai
2
≤ r2
i } be the sensing region of sensor i, where 1 ≤ i ≤ n. In other
words, the sensing region of sensor i is a ball centered at ai with radius ri. In
the sequel, we shall mainly concern ourselves with the cases d = 2 and 3. For
any given integer k ≥ 1, we say that x ∈ S is k–covered if there exist indices
1 ≤ i1 < i2 < · · · < ik ≤ n such that x ∈ ∩k
j=1Bij
. We would then like to answer
the following queries in an efficient manner:
Query 1: (k–Coverage; k–Cov) Is every point in S k–covered by the balls
B1, . . . , Bn? In other words, is it true that for every x ∈ S, there exist indices
1 ≤ i1 < · · · < ik ≤ n such that x ∈ ∩k
j=1Bij ?
Query 2: (Max–k–Coverage; Max–k–Cov) Determine the largest k such
that every point x ∈ S is k–covered by the balls B1, . . . , Bn.

3 The Proposed Solutions
In the above problems, we are required to certify that every point in S is k–
covered by the given balls. Thus, in order to obtain efficient algorithms for both
problems, we could attempt to find a finite and small set of points in S such
that the queries can be answered by examining only these points. Indeed, our
techniques are motivated by this idea, and they depend on the notion of power
defined as follows. Given the ball B(a, r), we define the power of a point x ∈ Rd
with respect to B(a, r) by pow(x, B(a, r)) = x − a 2
− r2
. Note that:
pow(x, B(a, r))



< 0 if x ∈ int(B(a, r))
= 0 if x ∈ ∂B(a, r)
> 0 if x ∈ Rd
B(a, r)
As we shall see, the notion of power allows us to treat the aforementioned prob-
lems in a unified manner.
3.1 The k–Cov Problem in R2
with Identical Disks
To begin, let us consider a version of the k–Cov problem where d = 2 and
ri ≡ r for all i = 1, 2, . . . , n. In other words, we are interested in the two–
dimensional case of the k–Cov problem, with the sensing ranges of all sensors
being identical. Let B1 ≡ B(a1, r), . . . , Bn ≡ B(an, r) be the given balls, and let
C = {a1, . . . , an} be the set of centers. For any subset U ⊂ {1, 2, . . . , n} with
|U| = k, define:
cell(U) = x ∈ R2
: pow(x, Bu) < pow(x, Bv) ∀u ∈ U, v ∈ {1, 2, . . . , n}U
= x ∈ R2
: x − au
2
< x − av
2
∀u ∈ U, v ∈ {1, 2, . . . , n}U
The set cell(U) is the so–called Voronoi region of U. Intuitively, cell(U) is the
set of points in R2
closer to all points in U than to any point in C U. It is
well–known (see, e.g., [4, 8]) that the Voronoi regions of all subsets of cardinality
k induce a subdivision of R2
, called the k–th order Voronoi diagram of C and
denoted by Vk(C ). Moreover, such a subdivision is polyhedral. Now, observe that
for x ∈ cell(U) ∩ S, we have x being k–covered iff pow(x, Bu) < 0 for all u ∈ U.
This suggests the following algorithm for determining whether S is k–covered or
not. First, we compute the k–th order Voronoi diagram Vk(C ) of the point set
C . Then, as we will demonstrate below, it suffices to check whether the following
points are k–covered: (i) the vertices of Vk(C ) in S, (ii) the intersections between
the sides of S and Vk(C ), and (iii) the corners of S. We say that x ∈ S is a critical
point if it belongs to one of the above three categories. Thus, in order to correctly
answer the query, it suffices to check for the critical points. We summarize our
algorithm below (see Algorithm 1).

Algorithm 1 Testing k–Coverage by Identical Disks in R2
Given: A set of n disks B1 ≡ B(a1, r), . . . , Bn ≡ B(an, r) in R2
, and an integer k ≥ 1.
1: Compute the k–th order Voronoi diagram Vk(C ) for the set of points C =
{a1, . . . , an}.
2: for each vertex v of Vk(C ) that lies within S do
3: Let {ai1 (v), . . . , aik (v)} be the k closest neighbors of v in C .
4: if ∃j such that pow(v, aij (v)) > 0 then
5: Return No
6: end if
7: end for
8: for each side l of the square S do
9: Let {pj}M
j=1 be the union of the two endpoints of l and the intersections between
l and Vk(C ).
10: for each j = 1, 2, . . . , M do
11: Let ai1 (pj), . . . , aik (pj) be the k closest neighbors of pj in C .
12: if ∃t such that pow(pj, ait (pj)) > 0 then
13: Return No
14: end if
15: end for
16: end for
17: Return Yes
Theorem 1 Algorithm 1 correctly answers the query.
Proof. For each x ∈ S, let a(x) be its k–th closest neighbor in C . Clearly, we have
pow(x, a(x)) > 0 if x is not k–covered. Let ˆx ∈ S be such that pow(ˆx, a(ˆx)) is
maximized. Note that such an ˆx exists since S is compact and the power distance
to the k–th closest point in C is a continuous function. We now consider two
cases:
Case 1: ˆx ∈ int(S)
We claim that ˆx is a vertex of Vk(C ). Suppose that this is not the case. Then,
we have the following possibilities:
Case 1.1: ˆx ∈ cell(U) for some U ⊂ {1, 2, . . . , n} such that |U| = k. Then,
the gradient of pow(·, a(ˆx)) at ˆx is d(ˆx) = 2(ˆx − a(ˆx)). In particular, an ascent
direction is given by d(ˆx) (if ˆx = a(ˆx), then all directions are ascent directions).
Now, let x = ˆx + αd(ˆx), where α > 0 is suﬃciently small so that x ∈ cell(U).
Then, we have x −a(ˆx) 2
> ˆx−a(ˆx) 2
. Moreover, since every x ∈ cell(U) has
the same k closest neighbors in C , we have x − a(x ) 2
≥ x − a(ˆx) 2
. This
shows that ˆx is not a maximizer, which is a contradiction.
Case 1.2: ˆx lies on the interior of an edge e in Vk(C ) deﬁned by the intersection
of the closure of two cells, say e ≡ int cell(U) ∩ cell(V ) , where |U| = |V | = k.
Let ai, aj be two distinct k–th closest neighbors of ˆx in C (cf. [8]). Note that
the equation of the line Le containing e is given by x − ai
2
= x − aj
2
, or
equivalently,
Le : 2(aj − ai)T
x = aj
2
− ai
2

Define u = (aj − ai)/ aj − ai , and let v = 0 be such that uT
v = 0. Note that
v is a vector along the direction of Le. Now, consider the point x = ˆx + αv for
some α ∈ R to be determined shortly. Note that for sufficiently small |α| > 0,
we have x ∈ e. Moreover, we compute:
x − aj
2
= ˆx − aj
2
+ α2
v 2
+ 2αvT
(ˆx − aj) (1)
Thus, if α = 0 is chosen with a suitable sign with |α| small enough, then we
have 2αvT
(ˆx − aj) ≥ 0 and x ∈ e. This implies that x − aj
2
> ˆx − aj
2
.
Moreover, x has the same k closest neighbors as ˆx in C (since x ∈ e), and for
all points q on the edge e, we have q − ai
2
= q − aj
2
≤ q − ak
2
for all
k ∈ U ∪ V . This contradicts the definition of ˆx.
Case 2: ˆx ∈ bd(S)
Let l be the side of S such that ˆx ∈ l. Suppose that ˆx ∈ int(l) ∩ cell(U) for some
U. We claim that ˆx is not a maximizer. To see this, let v = 0 be a vector along
the direction of l. Now, consider the point x = ˆx + αv for some α ∈ R to be
determined shortly. As before, we have x ∈ int(l) ∩ cell(U) if |α| > 0 is chosen
to be sufficiently small. Moreover, using (1), we see that α can be chosen such
that x − a(ˆx) 2
> ˆx − a(ˆx) 2
(recall that a(ˆx) is the k–th closest neighbor of
ˆx in C ). Since every x ∈ cell(U) has the same k closest neighbors in C , we have
x − a(x ) 2
≥ x − a(ˆx) 2
. This again contradicts the definition of ˆx.
Theorem 2 Algorithm 1 runs in time O(n log n + nk2
).
Proof. The k–th order Voronoi diagram Vk(C ) of the n–point set C can be
computed in O(n log n + nk2
) time using the algorithm described in [1]. As is
shown in [8], the total number of vertices, edges and cells in Vk(C ) is bounded
by O(k(n − k)). Thus, the remaining steps can be done in O(k2
(n − k)) time.
This completes the proof.
The algorithm above possesses some interesting features that deserve further
discussion. First, observe that we have reduced the coverage problem to an opti-
mization problem, in which we are trying to find an x ∈ S such that the distance
to its k–th closest neighbor in C is maximized. Such a viewpoint enables us to
answer the coverage query by examining only a set of critical points, and the
efficiency of our algorithm comes from the facts that (i) the k–th order Voronoi
diagram can be computed efficiently, and (ii) there exists a small set of criti-
cal points. As we shall see in subsequent sections, these ideas can be used to
obtain efficient algorithms for other coverage problems. Secondly, note that the
same analysis would go through if the region S is, say, a convex polygon with a
constant number of sides. All we need is that the number of candidate maxima
on the boundary of S is small. Thirdly, we observe that the k–th order Voronoi
diagram depends only on the locations of the centers a1, . . . , an but not on the
common sensing range r. Thus, it is not necessary to re–compute Vk(C ) every
time we change the value of r. It suffices to check the critical points described in
the algorithm, and this requires only O(nk2
) time. This opens up the possibility
of a binary search strategy to determine the smallest r such that S is k–covered.

3.2 The 1–Cov Problem with Various Sensing Regions
In most previous works, the sensing region of a sensor is modelled as an Euclidean
ball. In this section, we show how our techniques can be extended to yield efficient
algorithms for the 1–Cov problem with various sensing regions. As before, we
shall assume that S is the cube {x ∈ Rd
: x ∞ ≤ 1}.
Covering with Non–Uniform Disks Suppose that we are given a set of balls
B1 ≡ B(a1, r1), . . . , Bn ≡ B(an, rn), and we would like to determine whether
S is covered by these balls. To answer this query, we first compute the power
diagram of the collection B = {B1, . . . , Bn} [2], which is a collection of cells of
the form:
cell(i) = x ∈ R2
: pow(x, Bi) < pow(x, Bj) ∀j = i
where pow(x, Bi) = x − ai
2
− r2
i for x ∈ R2
and i = 1, 2, . . . , n. Clearly, a
point x ∈ S is not covered iff min1≤i≤n pow(x, Bi) > 0. Thus, we see that S is
not covered iff the optimal value of the optimization problem
maximize min
1≤i≤n
pow(x, Bi)
subject to x2
j ≤ 1 j = 1, 2
(2)
is positive. Using the arguments in the proof of Theorem 1, it can be shown that
the candidate maxima of (2) are precisely the critical points introduced before,
i.e. (i) the vertices of the power diagram P(B), (ii) the intersections between the
sides of S and P(B), and (iii) the corners of S. Thus, to determine the optimal
value of (2), we can proceed as in Algorithm 1. Since the power diagram P(B)
of the collection B can be computed in O(n log n) time and the total number of
vertices, edges and cells in P(B) is bounded by O(n) [2], we obtain the following
theorem:
Theorem 3 The 1–Cov problem in R2
can be solved in O(n log n) time.
We remark that the above approach works in the three–dimensional case as
well. Specifically, by following the arguments in the proof of Theorem 1, one can
show that it is enough to check whether the following points are covered: (i) the
vertices of the power diagram P(B) that lie in S, (ii) the 0–flats that arise from
the intersections between the sides of S and P(B), and (iii) the corners of S.
Since the power diagram P(B) in R3
can be computed in O(n2
) time and the
total number of features (i.e. vertices, edges, ...) in P(B) is bounded by O(n2
)
[2], we have the following theorem:
can be solved in O(n2
) time.
Note that in both cases, our algorithms have much better worst–case runtimes
than those proposed in [5, 6].

Covering with Ellipses Suppose that we are given a collection of ellipses
E1, . . . , En, where Ei is the ellipse defined by (x − ai)T
Qi(x − ai) ≤ r2
i , with Qi
symmetric positive definite, and we would like to answer the query with the Bi’s
replaced by the Ei’s. As before, we may define pow(x, Ei) = (x−ai)T
Qi(x−ai)−
r2
i for x ∈ R2
. To answer the query, we first compute the “power diagram” of the
collection {E1, . . . , En}. Although this “power diagram” is no longer polyhedral,
we can still show that the optimum must occur at a critical point as defined
before. First, by using a similar argument as before, it is easy to see that ˆx
cannot lie in the interior of a cell. Now, the intersection σ of the closure of two
cells is a portion of a quadratic curve Cσ, whose equation is given by:
Cσ : (x − ai)T
Qi(x − ai) − r2
i = (x − aj)T
Qj(x − aj) − r2
j
Let ˆx ∈ int(σ), and consider the two vectors v1 = Qi(ˆx − ai)/ Qi(ˆx − ai) and
v2 = Qj(ˆx−aj)/ Qj(ˆx−aj) . Note that vi is the gradient vector of the function
pow(·, Ei) at ˆx, where i = 1, 2. Thus, if the angle spanned by v1 and v2 is strictly
less than π, then we can increase the objective value by moving along the curve
Cσ in the direction that lies within the cone spanned by v1 and v2, which is a
contradiction. On the other hand, if the angle spanned by v1 and v2 is exactly
π, then we can find an v = 0 such that vT
v1 = vT
v2 = 0. Now, for any α = 0,
we have:
(ˆx + αv − ai)T
Qi(ˆx + αv − ai) − r2
i
= (ˆx − ai)T
Qi(ˆx − ai) − r2
i + 2αvT
Qi(ˆx − ai) + α2
vT
Qiv
= (ˆx − ai)T
Qi(ˆx − ai) − r2
i + α2
vT
Qiv
and similarly,
(ˆx + αv − aj)T
Qj(ˆx + αv − aj) − r2
j = (ˆx − aj)T
Qj(ˆx − aj) − r2
j + α2
vT
Qjv
Since Qi and Qj are positive definite, we see that (ˆx+αv−ai)T
Qi(ˆx+αv−ai) >
(ˆx−ai)T
Qi(ˆx−ai) and (ˆx+αv −aj)T
Qj(ˆx+αv −aj) > (ˆx−aj)T
Qj(ˆx−aj) for
all α = 0. It follows that by moving along Cσ, we can increase the objective value,
which again contradicts the definition of ˆx. Finally, the case where ˆx ∈ bd(S)
can be handled as before.
The “power diagram” for ellipses can have complexity Θ(n2+
) for any > 0,
and it can be computed in O(n2+
) time as well [3]. Thus, we have the following
theorem:
with ellipsoidal sensing regions can be
solved in O(n2+
) time, for any > 0.
Covering with Lp–Balls Let p ≥ 1 and r > 0 be fixed, and suppose now
that we are given a collection of Lp–balls B1, . . . , Bn, where Bi is defined by
x − ai
p
p ≤ rp
(i.e. all the Bi’s have the same radius). As before, we would
like to answer the query of whether S is covered by the Bi’s. For this purpose,

we define pow(x, Bi) = x − ai
p
p − rp
for x ∈ R2
and formulate an optimiza-
tion problem similar to (2). The optimal value and the optimal solution can
then be determined by first computing the Lp–Voronoi diagram of the point set
{a1, . . . , an} (see [7]) and then checking the objective values at the corresponding
critical points. We omit the details and summarize the results as follows:
with identical Lp sensing regions can be
solved in O(n log n) time.
3.3 The Max–k–Cov Problem
We now turn our attention to the Max–k–Cov problem, which is to determine
the largest k such that S is k–covered by the given balls B1, . . . , Bn (here, we
consider the Euclidean metric). To do this, we first use the algorithm in [2] to
compute all the k–th order power diagrams (where 1 ≤ k ≤ n − 1). This takes
O(n3
) time for the 2–d case and O(n4
) time for the 3–d case. Then, we can check
the critical points in each of these diagrams as before and determine the largest
k such that S is k–covered by the balls B1, . . . , Bn. Since the total number of
features in all the k–th order power diagrams is bounded by O(n3
) for the 2–d
case and by O(n4
) for the 3–d case [2], we obtain the following theorem:
Theorem 7 The Max–k–Cov problem in R2
(resp. R3
) can be solved in O(n3
)
(resp. O(n4
)) time.
Notice that this is more efficient than a direct application of the algorithms in
[5, 6].
4 Conclusion
In this note we have proposed to use suitable generalizations of the Voronoi di-
agram to treat various coverage problems that arise from the design of wireless
sensor networks. We have shown that in many cases, the runtimes of our algo-
rithms are better than those proposed in earlier works. Moreover, our approach
allows us to handle sensing regions whose shapes cannot be conveniently mod-
elled as Euclidean balls. We remark, however, that the proposed algorithms are
not distributed in nature. Since it is desirable to have decentralized computa-
tions in a wireless sensor network, an interesting future direction would be to
find efficient distributed algorithms for the coverage problems discussed in this
note.
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