arXiv:1106.0708v1  [math.OC]  3 Jun 2011
Optimal Sensor Configurations
for Rectangular Target Dectection
Franc¸ois-Alex Bourque and Bao U. Nguyen
Abstract— Optimal search strategies where targets are ob-
served at several different angles are found. Targets are
assumed to exhibit rectangular symmetry and have a uniformly-
distributed orientation. By rectangular symmetry, it is meant
that one side of a target is the mirror image of its opposite
side. Finding an optimal solution is generally a hard problem.
Fortunately, symmetry principles allow analytical and intuitive
solutions to be found. One such optimal search strategy consists
of choosing n angles evenly separated on the half-circle and
leads to a lower bound of the probability of not detecting
targets. As no prior knowledge of the target orientation is
required, such search strategies are also robust, a desirable
feature in search and detection missions.
I. INTRODUCTION
In mine hunting operations it is known that the detection
performance improves when a target is observed many times
at different aspect angles [1]–[6]. Similarly, classification
algorithms
[7]–[9] and sensor-arrays deployed for target
localization and tracking [10]–[13] benefit from multi-aspect
observations. This fact is, however, often overlooked. For
example, the formula for the probability of detecting a target
in a random search derived by Koopman is widely used yet
it assumes no angular dependence [14].
In this paper, search strategies that minimize the overall
probability of not detecting a target observed at several
different angles are identified. Finding an optimal angular
configuration is a priori intractable as it is multi-dimensional
in the sense that each observation is independent of one
another and hence each observation angle must be considered
as a separate dimension. What is more, the explicit expres-
sion for the overall probability of detection can be hopelessly
complicated even when the probability of detection for a
single observation is simple and the number of observations
is small.
The novelty of our approach lies in the fact that this
normally intractable problem is solved using an elegant
symmetry argument. Specifically, targets are assumed to
exhibit rectangular symmetry. That is, the left-hand side of a
target is the mirror image of its right-hand side, and its rear
end is the mirror image of its front end. Many targets can be
approximated with this class of symmetry including canoes,
ships, submarines, mines and human bodies.
Optimal angles are then shown to be evenly distributed
on multiples of the half-circle as in Ref. [13]. However, this
constitutes a departure from the current literature on sensor
F-.A. Bourque and B. U. Nguyen are scientists with Defence Research
and Development Canada Centre for Operational Research and Analysis,
National Defence Headquarters, 101 Colonel By drive, Ottawa, Canada.
Email enquiries should be sent to alex.bourque@drdc-rddc.gc.ca.
x
Fig. 1.
Rectangular target observed at angle x.
geometry [10]–[13] as our result is derived for finite-extent
targets rather than for point targets. The simplicity of the
solution implies that no complicated calculations are required
prior to a search as long as the target has the assumed
approximate symmetry. This fact should improve the task
of planning the path of mobile sensors, such as unmanned
vehicles, to search for fixed targets, as well as of deploying
a fixed sensor array to monitor traffic through choke points.
Assumptions and the minimization problem are stated in
Section II, while Section III presents a set of search strategies
that minimizes the probability of no detection. Section IV
provides a lower bound of not detecting targets achievable
with these strategies, which is illustrated with a specific
example in Section V. Conclusions including future work
are discussed in Section VI. Supplementary lemmata used in
Sections III and IV are found in the Appendix.
II. PROBLEM STATEMENT
The dependence of detection process on angle occurs
often in search and detection operations. In general, the
effectiveness of such an operation also depends on the
distance between the sensor and the target. However, here,
the probability of detection is assumed constant as a function
of range and, hence, the focus is only on the angular
dependence. For more details on the range dependence, refer
to Ref. [6].1
1Note that the probability of detection as a function of range is primarily
a characteristic of the sensor, while the probability of detection as a function
of angle is primarily a characteristic of the target.
x
�x
x +


Fig. 2.
Symmetries of the target: Reflection through the short axis of
the target (dashed line) and reflection through the center of the target (dot-
dashed line).
As shown in Fig. 1, the problem is modeled on a two-
dimensional plan and the observation angle, x, is defined
as the counter-clockwise angle measured in radian between
the sensor beam and the short axis of a rectangular (positive
horizontal axis) target. An observation angle of zero degree
corresponds to the observation of the long side of the target,
while an observation angle of π/2 degrees corresponds to the
observation of the short side of the target. Targets considered
will have approximate rectangular symmetries as shown in
Fig. 2. That is, they possess a reflection axis through their
short axis (left-right mirror symmetry) and a point reflection
through their center.2
In what follows, the probability of no detection rather
than the probability of detection is considered; one being
the complement of the other. Define the single probability
of no detection as the probability of not detecting the target
at angle x and denote this single-value real function as g(x).
Note that the single probability of no detection is even due to
the reflection symmetry through the short axis of the target
and periodic due to the reflection through the center of the
target. Specifically,
g(x) = g(−x),
g(x) = g(x + π).
Next, define the multiple probability of no detection as
the probability of not detecting a target after n observations.
Let x be the orientation of the target. Let ui be the i-
th angle at which the target is observed relative to x and
⃗µ = (µ0, . . . , µn−1) be the vector of the n relative ob-
servation angles. Assume the multiple observation detection
process is a Bernoulli process, i.e., all observations are
independent. Then, the multi-observation probability of no
detection is modeled as the product of single probabilities
2Note that the composition of a reflection through the short axis followed
by a reflection through the center of the target is equivalent to a reflection
through the long axis of the target (forward/backward mirror symmetry).
of no detection. In general, however, the exact value of x,
i.e., the orientation of the target is unknown. To circumvent
this problem, assume that the target’s orientation is uniformly
distributed and evaluate the average multiple probability of
no detection by G(⃗µ). Then,
G(⃗µ) = 1
π
Z
π
2
−π
2
dx
n−1
Y
i=0
g(x + µi).
(1)
Therefore, the problem amounts to finding search strate-
gies that minimize G(⃗µ). For simplicity, the probability of no
detection is taken to mean the average multiple probability
of no detection in what follows.
III. A SET OF OPTIMAL SEARCH STRATEGIES
In this section, a condition ensuring that all partial deriva-
tives of the probability of no detection are equal to zero
is derived. From this condition, a set of optimal search
strategies is then identified and the probability of no detection
is recast into a form used in the subsequent section.
Let us first introduce some useful notation and definitions.
Let i ∈{0, . . . , n −1}, Ni = {0, . . . , n −1} \{i} and j ∈
Ni. Define ∂i = ∂/∂i. Denote ⃗µ∗as an optimal point of
G (⃗µ) . Let a be an integer and b be a positive integer. Define
the modulo operation as
a mod b = a −
ja
b
k
b.
Let m be a non-negative integer and µ = π/n. Define
˜G (m, n) = 1
π
Z
π
2
−π
2
dx
n−1
Y
i=0
g(x + miµ).
Then the following holds.
Lemma 3.1: The partial derivative can be written as
∂iG (⃗µ) = 1
2π
Z
π
2
−π
2
dx g′ (x)
 Y
j∈Ni
g (x + µj −µi)
−
Y
j∈Ni
g (x −µj + µi)

.
Proof:
Apply the partial derivative to the expression
for G (⃗µ) given by (1). Then
∂iG(⃗µ) =
1
π
Z
π
2
−π
2
dx g′ (x + µj)
Y
j∈Ni
g(x + µj).
Let x →x−µi and note that Lemma A.1 applies and dictates
that the integral is invariant under this change of variable.
Thus,
∂iG(⃗µ) =
1
π
Z
π
2
−π
2
dx g′ (x)
Y
j∈Ni
g(x + µj −µi).
(2)
Because g(x) is even,
∂iG (⃗µ) = 1
π
Z
π
2
−π
2
dx g′ (x)
Y
j∈Ni
g(−x −µj + µi).
Let x →−x and remark that g′(x) is odd. Then
∂iG (⃗µ) = −1
π
Z
π
2
−π
2
dx g′ (x)
Y
j∈Ni
g(x −µj + µi).
(3)
And the result follows from the average of (2) and (3).
A condition for optimizing G (⃗µ) is thus for the integrands
of all partial derivatives to be equal to zero. The next Lemma
identifies a set of search strategies for which this condition
holds, i.e., optimizes the probably of no detection.
Lemma 3.2: Define (m, n) to be the search strategy such
that the separation between two consecutive observations is
a constant and equal to mπ/n with m a positive integer.
Then the search strategy (m, n) is an optimum of G (⃗µ) and
defines a subset of all possible optimal search strategies.
Proof: Evaluate the product in (2) at point ⃗µ∗giving
Y
j∈Ni
g(x + µ∗
j −µ∗
i ).
Assume that (m, n) is an optimal search strategy. Then, the
definition of the strategy (m, n) implies that
µ∗
j −µ∗
i = m [−(2i −j) + i] µ.
From definition of the modulo note that
2i −j =
2i −j
n

n + (2i −j) mod n.
Define σi (j) = (2i −j) mod n and recall that g(x) is
periodical. Then,
g(x + µ∗
j −µ∗
i ) = g (x −mσi (j) µ + miµ) .
Lemma A.2 implies that the map σi (j) is a bijection from
the set of j to itself. Therefore,
Y
j∈Ni
g
�x + µ∗
j −µ∗
i


=
Y
j∈Ni
g (x −mσi (j) µ + miµ)
=
Y
j∈Ni
g (x −mjµ + miµ)
=
Y
j∈Ni
g
�x −µ∗
j + µ∗
i


.
And Lemma 3.1 then entails that ∂iG (⃗µ∗) = 0 for all i.
For an optimal search strategy (m, n), it will be useful in
what follows to cast the integral in the following form.
Lemma 3.3: Let the search strategy be (m, n). Then the
probability of no detection is equal to ˜G (m, n).
Proof: First, remark that only the difference between
two consecutive observations is specified in Lemma 3.2.
Thus, liberty exists in the choice of the absolute reference
angle. Without loss of generality, take µ∗
0 to be this absolute
reference angle. Then,
µ∗
i = µ∗
0 + miµ.
Substituting this expression into (1) yields
G (⃗µ∗) = 1
π
Z
π
2
−π
2
dx
n−1
Y
i=0
g (x + µ∗
0 + miµ).
Next, let x →x −µ∗
0. Then Lemma A.1 entails that the
domain of integration is invariant under such a shift of the
integration variable.
IV. A LOWER BOUND OF THE
PROBABILITY OF NO DETECTION
In the previous section, a set of optimal search strategies
was identified, the (m, n) search strategies. In this section,
a lower bound of the probability of no detection achievable
with these strategies is proven.
Let us first introduce some useful notation and definitions.
Let gcd (·, ·) be the greatest common divider. Let r, q and p
be strictly positive integers such that m = pq, n = pr and
p = gcd(m, n). Let i ∈{0, . . . , n −1}, j ∈{0, . . . , r −1}
and k ∈{1, . . . , p}. Define
h (x) = g (x) g (x + mµ) . . . g (x + (r −1) mµ) .
Lemma 4.1: The following identities hold:
p
Y
k=1
h (x + (k −1) µ) =
n−1
Y
i=0
g(x + iµ),
(4)
n−1
Y
i=0
g(x + miµ) = h(x)p.
(5)
Proof: Consider the first identity. Using the definition
of h(x), write the left-hand side of (4) as
p
Y
k=1
r−1
Y
j=0
g (x + mjµ + (k −1) µ) .
The definitions of m and n and of the modulo operation
imply that
mj = n
qj
r

+ p (qj mod r) .
Using this expression for mj and the periodicity of g (x),
the right-hand side further becomes equal to
p
Y
k=1
r−1
Y
j=0
g (x + p(qj mod r)µ + (k −1) µ).
Next, from Lemma A.3, the map σ (j) = qj mod r is known
to be a bijection from the set of j to itself. Therefore, the
product can also be written as
p
Y
k=1
r−1
Y
j=0
g (x + pjµ + (k −1) µ).
Finally, from Lemma A.4, the set of (j, k) pairs can be
mapped to the set of i using the bijection σ(k −1, j) =
(k −1) + pj. Therefore,
p
Y
k=1
r−1
Y
j=0
g (x + pjµ + (k −1) µ) =
n−1
Y
i=0
g (x + iµ) .
Now, consider the second identity. From Lemma A.4, the
set of i can be mapped to the set of (j, k) pairs using the
bijection σ(j, k −1) = j + r(k −1). Therefore, the left-hand
side of (5) becomes
p
Y
k=1
r−1
Y
j=0
g (x + mjµ + m (k −1) rµ) .
Next, note that the definitions of m and n imply mr = nq
and that the definition of µ implies nµ = π, from which
follows that m (k −1) rµ = (k −1)qπ. This equality and
the periodicity of g (x), then imply that
g (x + mjµ + m (k −1) rµ) = g (x + mjµ) .
Finally, from the definition of h (x),
p
Y
k=1
r−1
Y
j=0
g (x + mjµ) =
p
Y
k=1
h(x) = h(x)p.
Theorem 4.2 (Lower Bound Estimate): For
any
(m, n)
search strategies,
˜G (m, n) ≥˜G (1, n) .
Proof: Consider a given (m, n) pair. Then, Lemma 4.1
allows to write
˜G (1, n) = 1
π
Z
π
2
−π
2
dx
p
Y
k=1
h (x + (k −1) µ),
˜G (m, n) = 1
π
Z
π
2
−π
2
dx h(x)p .
Next, let that l ∈{0, . . ., p −1} and define
λl(x) =
p−l
Y
k=1
h (x + (k −1) µ).
Assume that
Z
π
2
−π
2
dx λl(x)h(x)l ≤
Z
π
2
−π
2
dx λl+1 (x) h(x)l+1.
Then
˜G (1, n) = 1
π
Z
π
2
−π
2
dx λ0 (x) ≤1
π
Z
π
2
−π
2
dx λ1 (x) h(x) ≤. . .
≤1
π
Z
π
2
−π
2
dx λl (x) h(x)l ≤. . .
≤1
π
Z
π
2
−π
2
dx λp−1 (x) h(x)p−1 = 1
π
Z
π
2
−π
2
dx h(x)p
= ˜G (m, n) .
To show that the assumption holds true, proceed by
generating a partially ordered set. Let δl = (p −l −1) µ and
note that λl (x) = λl+1 (x) h (x + δ). Then, the first element
of the partially ordered set is
Z
π
2
−π
2
dx λl+1 (x) h(x + δ)h(x)l.
And the last element is
Z
π
2
−π
2
dx λl+1 (x) h(x)l+1.
For greater clarity, the indice of λl+1 (x) and δl are
suppressed in what follows. Next, apply recursively t
times h(x)ah(x + δ)b ≤
1
2h(x)a−b h
h(x)2a + h(x + δ)2ai
from the first element. After summing the resulting geometric
series, the element generated is
Z
π
2
−π
2
dx λ (x)
 "
1 −
1
2
t#
h(x)l+1
+
1
2
t
h(x)l+1−2t
h(x + δ)2l
.
Remark however that beyond a recursion step defined such
that 2t+1 > l + 1 ≥2t, the power of h(x) becomes negative
in the second term. This is problematic because the last
element of the partially ordered set has only positive powers
of h(x). To circumvent this problem, let x →x −δ in
the the second term of the element. Since the domain of
integration remains unchanged (Lemma A.1), h(x) is even
(Lemma A.5), and λ(x) = λ(x −δ) (Corollary A.6), then
the second term is also equal to
Z
π
2
−π
2
dx λ (−x) h(−x)l+1−2th(−x + δ)2l
Finally, letting x →−x, the t-element becomes
Z
π
2
−π
2
dx λ (x)
 "
1 −
1
2
t#
h(x)l+1
+
1
2
t
h(x + δ)l+1−2t
h(x)2l
And the power of h(x) is now greater than or equal to that
of h (x + δ) in the second term allowing again the recursive
application of the inequality.
The algorithm thus loops through successive recursion
steps and changes of variable. Note that for l + 1 = 2t,
the last element is generated after the first iteration. Lemma
A.7 states that this is the only case for which the algorithm
terminates in a finite number of iterations. For all other cases,
the last element arises by letting the number of iterations go
to infinity.
V. AN ANALYTICAL EXAMPLE
In the previous section, a lower bound of the probability of
no detection was found for an arbitrary g(x). In this section,
the probability of no detection is evaluated and the inequality
of Theorem 4.2 is explicitly verified when g (x) = sin (x)2
Lemma 5.1: Let g (x) = sin (x)2. Then
˜G (m, n) = 2p (2p −1)!!
4np!
.
(6)
Proof: Recall that
˜G (m, n) =
1
π
Z
π
2
−π
2
dx
p
Y
k=1
r−1
Y
j=0
g (x + jmµ).
Then Lemma A.8 implies that
˜G (m, n) = 1
π
Z
π
2
−π
2
dx
p
Y
k=1
r−1
Y
j=0
g

x + jπ
r

.
Let g (x) = sin (x) 2 and use the identity [15]:
sin (rx) = 2r−1
r−1
Y
j=0
sin

x + jπ
r

.
Then,
˜G (m, n) = 1
π
Z
π
2
−π
2
dx
p
Y
k=1
1
4r−1 sin (rx)2
= 1
π
Z
π
2
−π
2
dx
1
4p(r−1) sin (rx)2p
Finally, carrying out the integral [16] and recalling that n =
pr,
˜G (m, n) =
1
4n−p
(2p −1)!!
2p!!
.
Since 2p!! = 2pp!, (6) follows.
Corollary 5.2: For p = gcd (1, n) = 1,
˜G (1, n) = 2
4n .
And Theorem 4.2 follows since (2p −1)!! = 1 × 3 × · · · ×
(2p −1) ≥1 × 2 × · · · × p = p!.
VI. CONCLUSIONS AND FUTURE WORK
In this paper, the angular dependence of the detection
process which is often overlooked for search and detection
mission is explicitly accounted for by assuming that the tar-
get possesses rectangular symmetry. One major consequence
of this approximate symmetry is that the long side of a target
is endowed with the largest cross section, which results in
the highest probability of detection given that the target is
observed only once. However, since the orientation of the
target is in general unknown, there is likelihood that it will
be imaged on the short side, i.e., the smallest cross-section.
Therefore, the probability of not detecting the target may not
be zero even if the search area is entirely covered. Making
several observations of the target in order to increase the
change of observing its long side is one way to address this
problem.
Assuming that the observations are independent, an opti-
mal search strategy is then to observe the target such that
the separation between two consecutive observations is a
constant and equal to a multiple of 180 degrees divided by
the number of observations. The resulting tactic is simple,
intuitive and robust (as no prior knowledge of the target
orientation is required). For example, two observations sep-
arated by 90 degrees or three observations separated by 60
degrees will minimize the probability of no detection.
Having shown that one of these search strategies leads
to a lower bound of the probability of no detection, work
is currently underway to prove that it is also a globally
minimal search strategy. Another logical extension of this
work is to relax the assumption that information provided
by subsequent observations is uncorrelated (i.e., follows
a Bernoulli process) as it entails that the probability of
not detecting a target decreases with increasing number of
observations even if the observations are co-linear. This is
questionable as no additional information is gained.
VII. ACKNOWLEDGMENTS
One of the authors (A. Bourque) would like to acknowl-
edge A. Percival from Defence R&D Canada - Atlantic for
bringing to his attention the issue of correlation effects in
the detection process.
APPENDIX
SUPPLEMENTARY LEMMATA
Lemma A.1 (Rotational Invariance): Let ω
∈
R and
h(x) = h(x + π). Then
Z
π
2
−π
2
dx h(x −ω) =
Z
π
2
−π
2
dx h(x).
Proof:
Let x →x + ω on the RHS and break the
integration interval of the resulting integral into [−π
2 −ω, −π
2 [
and [−π
2 , π
2 −ω]. Then let x →x−π in the integral over the
first interval and recall that by assumption h(x) is periodic.
Lemma A.2: Let
i
∈
{0, . . . , n −1},
Ni
=
{0, . . . , n −1} \{i}
and
j
∈
Ni.
Then
σi (j)
=
(2i −j) mod n is a bijection from Ni
to itself and
its own inverse.
Proof: Composition gives
σi (σi (j)) = (2i −(2i −j) mod n) mod n.
Use the definition of the modulo twice to give
σi (σi (j)) = j +
(2i −j)
n

n −

j +
j
(2i−j)
n
k
n
n
n.
Recall that j
∈
Ni and note that
j
n
<
n. Then
j
j
n +
j
(2i−j)
n
k
n
k
=
j
(2i−j)
n
k
n and σi (σi (j)) = j.
Lemma A.3: Let r, q be positive integers such that
gcd (r, q)
=
1, i.e., r and q are co-primes. Let i
∈
{0, . . . , r −1} . Then the map σ(i)
=
qi mod r is a
bijection of the set of i to itself.
Proof: Proceed with a proof by contradiction. Assume
this map is not a bijection. Then there exists a pair u, v ∈
{0, . . . , r −1} such that u ̸= v and qu mod r = qv mod r.
Next, assume that u > v then q (u −v) mod r = qw mod
r = 0 where w = u −v. This implies that qw = ry with y a
positive integer, as w, q > 0. Because q and r are co-primes,
i.e., gcd (r, q) = 1 then w = rz with z > 0, which leads to
a contraction as w = u −v < r −1.
Lemma A.4: Let u ∈{0, . . . a −1}, v ∈{0, . . . , b −1},
and w ∈{0, . . . , n −1} where n = ab. Then σ (u, v) =
u + av and σ−1 (w) =
�w mod a,
 w
a


are bijections.
Proof: Composition gives
σ
�σ−1(w)


=
w mod a + a
jw
a
k
= w,
where the last equality follows from the definition of the
modulo operation. Similarly,
σ−1 (σ (u, v)) =

(u + av) mod a,
u + av
a

= (u, v) ,
where the last equality follows from the definition of the
modulo operation and since u < a. Therefore, σ(u, v) and
σ−1 (w) are both one-to-one, onto, and inverse of each other.
Lemma A.5: h (x) = h (−x).
Proof:
Because of the periodicity of g (x) and nq =
mr, g (x + mjµ) = g(x+m(j−r)µ). Let j →−j+r. Then
h (x) = g (x) . . . g (x −(r −1) jµ) and the proof follows
since by definition g (x) is even.
Corollary A.6: Consider λl(x) and δl
=
(p −k −
1)µ. Then λl (x −δl) = h (x −(p −l −1) µ) . . . h (x) and
λl (x −δl) = λl (−x) since h(x) is even by Lemma A.5.
Lemma A.7: Let i and c be positive integers. Let {ti} be
the set of non-negative integers such that 2ti+1 > c ≥2ti.
And let {ui} be the set of non-negative integer such that
ui = c −2tiui−1 and u0 = 1. Then, a fix point exists if and
only if c = 2l1 .
Proof: After i iterations,
ui = c −2tiui−1 = c−2ti �c −2ti−1ui−2


=
�c −2ti �. . .
�c −21

. . .



= c

1 −2ti �. . .
�1−2t2

. . .


+ (−1)i2ti+···+t1.
Then ui = 0 implies that the prime factorization of (k + 1)
must be 2a where a is a non-negative integer. Because
2t1+1 > (l + 1) = 2a ≥2t1, then a = t1 and ui = 0
for i > 0.
Lemma A.8:
r−1
Y
j=0
g

x + mjπ
n

=
r−1
Y
j=0
g

x + jπ
r

.
Proof: Recall that r, q and p are positive integers such
that m = pq, n = pr, and p = gcd(m, n). Then
g

x + mjπ
n

= g

x + qjπ
r

.
Use the periodicity of g (x) and the definition of the modulo
to further write the right-hand side of the equality as
g

x + qjπ
r
−
qj
r

π

= g

x + (qj mod r) π
r

.
Lemma A.3 implies that map σ (j) = qj mod r is a bijection
from the set of j to itself.
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