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Smoothed Motion Complexity⋆
Valentina Damerow1, Friedhelm Meyer auf der Heide2, Harald R¨ acke2,
Christian Scheideler3, and Christian Sohler2
1PaSCo Graduate School and
2Heinz Nixdorf Institute, Paderborn University, D-33102 Paderborn, Germany
vio, fmadh, harry, csohler@upb.de
3Dept. of Computer Science, Johns Hopkins University, Baltimore, MD 21218, USA,
scheideler@cs.jhu.edu
Abstract. We propose a new complexity measure for movement of objects, the
smoothed motion complexity. Many applications are based on algorithms dealing
with moving objects, but usually data of moving objects is inherently noisy due
to measurement errors. Smoothed motion complexity considers this imprecise
information and uses smoothed analysis [13] to model noisy data. The input is
object to slight random perturbation and the smoothed complexity is the worst
case expected complexity over all inputs w.r.t. the random noise. We think that
the usuallyapplied worst caseanalysis of algorithms dealing withmoving objects,
e.g., kinetic data structures, often does not reflect the real world behavior and that
smoothed motion complexity is much better suited to estimate dynamics.
Weillustratethisapproach on theproblemof maintaining an orthogonal bounding
box of a set of n points in Rdunder linear motion. We assume speed vectors
and initial positions from [−1,1]d. The motion complexity is then the number of
combinatorial changes tothe description of the bounding box. Under perturbation
with Gaussian normal noise of deviation σ the smoothed motion complexity is
only polylogarithmic: O(d · (1 + 1/σ) · logn3/2) and Ω(d ·√logn). We also
consider the case when only very little information about the noise distribution is
known. We assume that the density function is monotonically increasing on R≤0
and monotonically decreasing on R≥0and bounded by some value C. Then the
motion complexity is O(√nlogn · C + logn) and Ω(d · min{5√n/σ,n}).
Keywords: Randomization, Kinetic Data Structures, Smoothed Analysis
1Introduction
The task to process a set of continuously moving objects arises in a broad vari-
ety of applications, e.g., in mobile ad-hoc networks, traffic control systems, and
computer graphics (rendering moving objects). Therefore, researchers investi-
gated data structures that can be efficiently maintained under continuous motion,
e.g., to answer proximity queries [5], maintain a clustering [8], a convex hull [4],
⋆The third and the fifth author are partially supported by DFG-Sonderforschungsbereich 376,
DFG grant 872/8-1, and the Future and Emerging Technologies program of the EU under
contract number IST-1999-14186 (ALCOM-FT).
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or some connectivity information of the moving point set [9]. Within the frame-
work of kinetic data structures the efficiency of such a data structure is analyzed
w.r.t. to the worst case number of combinatorial changes in the description of
the maintained structure that occur during linear (or low degree algebraic) mo-
tion. These changes are called (external) events. For example, to maintain the
smallest orthogonal bounding box of a point set in Rdhas a unique description
at a certain point of time consisting of the 2d points that attain the minimum and
maximum value in each of the d coordinates. If any such minimum/maximum
point changes then an event occurs. We call the worst case number of events
w.r.t. the maintainance of a certain structure under linear motion the worst case
motion complexity.
We introduce an alternative measure for the dynamics of moving data called
the smoothed motion complexity. Our measure is based on smoothed analysis, a
hybrid between worst case analysis and average case analysis. Smoothed anal-
ysis has been introduced by Spielman and Teng [13] in order to explain the
typically good performance of the simplex algorithm on almost every input. It
asks for the worst case expected performance over all inputs where the expec-
tation is taken w.r.t. small random noise added to the input. In the context of
mobile data this means that both the speed value and the starting position of an
input configuration are slightly perturbed by random noise. Thus the smoothed
motion complexity is the worst case expected motion complexity over all inputs
perturbed in such a way. Smoothed motion complexity is a very natural measure
for the dynamics of mobile data since in many applications the exact position
of mobile data cannot be determined due to errors caused by physical measure-
ments or fixed precision arithmetic. This is, e.g., the case when the positions
of the moving objects are determined via GPS, sensors, and basically in any
application involving ’real life’ data.
We illustrate our approach on the problem to maintain the smallest orthogo-
nal bounding box of a point set moving in Rd. The bounding box is a fundamen-
tal measure for the extend of a point set and it is useful in many applications,
e.g., to estimate the sample size in sublinear clustering algorithms [3], in the
construction of R-trees, for collision detection, and visibility culling.
1.1The Problem Statement
We are given a set P of n points in Rd. The position posi(t) of the ith point at
time t is given by a linear function of t. Thus we have posi(t) = si·t+piwhere
piis the initial position and sithe speed. We normalize the speed vectors and
initial positions such that pi,si∈ [−1,1]d.
Themotion complexity oftheproblem isthe numberofcombinatorial changes
to the set of 2d extreme points defining the bounding box. Clearly this motion
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complexity is O(d·n) in the worst case, 0 in the best case, and O(d·logn) in the
average case. When we consider smoothed motion complexity we add to each
coordinate of the speed vector and each coordinate of the initial position an i.i.d.
random variable from a certain probability distribution, e.g., Gaussian normal
distribution. Then the smoothed motion complexity is the worst case expected
complexity over all choices of piand si.
1.2 Related Work
In [4] Basch et al. introduced kinetic data structures (KDS) which is a frame-
work for data structures for moving objects. In KDS the (near) future motion of
all objects is known and can be specified by so-called pseudo-algebraic func-
tions of time specified by linear functions or low-degree polynomials. This
specification is called a flight plan. The goal is to maintain the description of
a combinatorial structure as the objects move according to this flight plan. The
flight plan may change from time to time and these updates are reported to the
KDS. The efficiency of a KDS is analyzed by comparing the worst case num-
ber of internal (events needed to maintain auxiliary data structures) and exter-
nal events it processed against the worst case number of external events. Using
this framework many interesting kinetic data structures have been developed,
e.g., for connectivity of discs [7] and rectangles [9], convex hulls [4], proximity
problems [5], and collision detection for simple polygons [10]. In [4] the au-
thors developed a KDS to maintain a bounding box of a moving point set in Rd.
The number of events these data structures process is O(nlogn) which is close
to the worst case motion complexity of Θ(n). In [1] the authors showed that it
is possible to maintain an (1 + ǫ)-approximation of such a bounding box. The
advantage of this approach is that the motion complexity of this approximation
is only O(1/√ǫ). The average case motion complexity has also been considered
in the past. If n particles are drawn independently from the unit square then it
has been shown that the expected number of combinatorial changes in the con-
vex hull is Θ(log2(n)), in the Voronoi diagram Θ(n3/2) and in the closest pair
Θ(n) [15].
Smoothed analysis has been introduced by Spielman and Teng [13] to ex-
plain the polynomial run time of the simplex algorithm on inputs arising in
applications. They showed that the smoothed run time of the shadow-vertex
simplex algorithm is polynomial in the input size and 1/σ. In many follow-up
papers other algorithms and values have been analyzed via smoothed analysis,
e.g., the perceptron algorithm [6], condition numbers of matrices [12], quick-
sort, left-to-right maxima, and shortest paths [2]. Recently, smoothed analysis
has been used to show that many existing property testing algorithms can be
viewed as sublinear decision algorithms with low smoothed error probability
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[14]. In [2] the authors analyzed the smoothed number of left-to-right maxima
of a sequence of n numbers. We will use the left-to-right maxima problem as
an auxiliary problem but we will use a perturbation scheme that fundamentally
differs from that analyzed in [2].
1.3 Our Results
Typically, measurement errors are modelled by the Gaussian normal distribu-
tion and so we analyze the smoothed complexity w.r.t. Gaussian normally dis-
tributed noise with deviation σ. We show that the smoothed motion complexity
of a bounding box under Gaussian noise is O(d·(1+1/σ)·log n3/2) and Ω(d·
√logn). In order to get a more general result we consider monotone probability
distributions, i.e., distributions where the density function f is bounded by some
constant C and monotonically increasing on R≤0and monotonically decreasing
on R≥0. Then the smoothed motion complexity is O(d·(√nlogn · C+logn)).
Polynomial smoothed motion complexity is, e.g., attained by the uniform distri-
bution where we obtain a lower bound of Ω(d · min{5√n/σ,n}).
Note that in the case of speed vectors from some arbitrary range [−S,S]d
instead of [−1,1]dthe above upper bounds hold if we replace σ by σ/S.
These results make it very unlikely, that in a typical application the worst
case bound of Θ(d · n) is attained. As a consequence, it seems reasonable to
analyze KDS’s w.r.t. the smoothed motion complexity rather than the worst case
motion complexity.
Our upper bounds are obtained by analyzing a related auxiliary problem: the
smoothed number of left-to-right maxima in a sequence of n numbers. For this
problem we also obtained lower bounds which only can be stated here: in the
case of uniform noise we have Ω(?n/σ)and in the case of normally distributed
noise we can apply the average case bound of Ω(logn). These bounds differ
only by a factor of√logn from the corresponding upper bounds. In the second
case the bounds are even tight for constant σ. Therefore, we can conclude that
our analysis is tight w.r.t. the number of left-to-right maxima. To obtain better
results a different approach that does not use left-to-right maxima as an auxiliary
problem is necessary.
2Upper Bounds
To show upper bounds for the number of external events while maintaining the
bounding box for a set of moving points we make the following simplifications.
We only consider the 1D problem. Since all dimensions are independently from
each other an upper or lower bound for the 1D problem can be multiplied by d
to yield a bound for the problem in d dimensions.
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Further, we assume that the points are ordered by their increasing initial
positions and that they are all moving to the left with absolute speed values be-
tween 0 and 1. We only count events that occur because the leftmost point of the
1D bounding box changes. Note that these simplifications do not asymptotically
affect the results in this paper.
A necessary condition for the jth point to cause an external event is that all
its preceding points have smallerabsolute speed values, i.e.that si< sj, ∀i < j.
If this is the case we call sja left-to-right maximum. Since we are interested in
an upper bound we can neglect the initial positions of the points and need only
to focus on the sequence of absolute speed values S = (s1,...,sn) and count
the left-to-right maxima in this sequence.
Thegeneral concept forestimating thenumberofleft-to-right maximawithin
the sequence is as follows. Let f and F denote the density function and distribu-
tion function, respectively, of the noise that is added to the initial speed values.
(This means ? si= si+ φiwhere φiis chosen according to density function f.)
can write this probability as
?∞
−∞
i=1
This holds since F(x − si) is the probability that the ith element is not greater
than x after the pertubation. Since all pertubations are independently from each
other,?j−1
probablity that the jth element reaches x and is a left-to-right maximum. Hence,
integration over x gives the probability Pr[LTRj].
In the following we describe how to derive a bound on the above integral.
First suppose that all siare equal, i.e., si = s for all i. Then Pr[LTRj] =
?∞
z := F(x − s). (Note that this result only reveals the fact that the probability
for the jth element to be the largest is 1/j.)
Now, suppose that the speed values are not equal but come from some inter-
val [smin,smax]. In this case Pr[LTRj] can be estimated by
Let Pr[LTRj] denote the probability that ? sjis a left-to-right maximum. We
Pr[LTRj] =
F(x − si) · f(x − sj)dx .
j−1
?
(1)
i=1F(x − si) is the probability that all elements preceding ? sjare be-
low x. Consequently,?j−1
i=1F(x − si) · f(x − sj)dx can be interpreted as the
−∞F(x − s)j−1· f(x − s)dx =
?1
0zj−1dz = 1/j, where we substituted
Pr[LTRj] =
?∞
−∞
?∞
−∞
?∞
−∞
j−1
?
i=1
F(x − si) · f(x − sj)dx
≤F(x − smin)j−1· f(x − smax)dx
=F(z + δ)j−1f(z)dz ,
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where we use δ to denote smax− smin. Let Zf
δ) ≥ r} denote the subset of R that contains all elements z for which the ratio
f(z)/f(z + δ) is larger than r. Using this notation we get
?
R\Zf
?
R\Zf
?
R\Zf
≤ r ·1
Zf
δ,r
δ,r:= {z ∈ R | f(z)/f(z +
Pr[LTRj] ≤
δ,r
F(z + δ)j−1f(z)dz +
?
Zf
δ,r
F(z + δ)j−1f(z)dz
≤
δ,r
F(z + δ)j−1
f(z)
f(z + δ)f(z + δ)dz +
?
Zf
δ,r
f(z)dz
≤ r ·
δ,r
?
F(z + δ)j−1f(z + δ)dz +
?
Zf
δ,r
f(z)dz
j+
f(z)dz .
(2)
Now, we can formulate the following lemma.
Lemma 1. Let f denote the density function of the noise distribution and define
for positive parameters δ and r the set Zf
δ) ≥ r}. Further, let Z denote the probability of the set Zf
i.e., Z :=?
of n elements that are perturbed with noise distribution F is at most
δ,r⊆ R asZf
δ,r:= {z ∈ R | f(z)/f(z+
δ,rwith respect to f,
Zf
δ,rf(z)dz. Then the number of left-to-right maxima in a sequence
r · ⌈1/δ⌉ · logn + n · Z .
Proof. We are given an input sequence S of n speed values from (0,1]. Let
L(S) denote the expected number of left-to-right maxima in the corresponding
sequence of speed values perturbed with noise distribution f. We are interested
in an upper bound on this value. The following claim shows that we only need
to consider input sequences of monotonically increasing speed values.
Claim. The maximum expected number of left-to-right maxima in a sequence
of n perturbed speed values is obtained for an input sequence S of initial speed
values that is monotonically increasing.
⊓ ⊔
From now on we assume that S is a sequence of monotonically increasing
speed values. We split S into ⌈1/δ⌉ subsequences such that the ℓth subsequence
Sℓ, ℓ ∈ {1,...,⌈1/δ⌉} contains all speed values between (ℓ − 1)δ and ℓδ, i.e.,
Sℓ := (s ∈ S : (ℓ − 1) · δ < s ≤ ℓ · δ). Note that each subsequence is
monotonically increasing.
Let L(Sℓ) denote the expected number of left-to-right maxima in subse-
quence Sℓ. Now we first derive a bound on each L(Sℓ) and then we utilize
L(S) ≤?
ℓL(Sℓ) to get an upper bound on L(S).
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Fix ℓ ∈ {1,...,⌈1/δ⌉}. Let kℓdenote the number of elements in subse-
quence Sℓ. We have
L(Sℓ) =
j=1
kℓ
?
Pr[LTRj] ,
where Pr[LTRj] is the probability that the jth element of subsequence Sℓis a
left-to-right maximum within this subsequence. We can utilize Inequality 2 for
Pr[LTRj] because the initial speed values in a subsequence differ at most by δ.
This gives
kℓ
?
j=1
Hence, L(S) ≤?
2.1Normally distributed noise
L(Sℓ) ≤
(r ·1
j+ Z) ≤ r · logn + kℓ· Z .
ℓL(Sℓ) ≤ r · ⌈1/δ⌉ · logn + n · Z, as desired.
⊓ ⊔
In this section we show how to apply the above lemma to the case of normally
distributed noise. We prove the following theorem.
Theorem 1. The expected number of left-to-right maxima in a sequence of n
speed values perturbed by random noise from the standard normal distribution
N(0,σ) is O(1
σ· (logn)3/2+ logn).
Proof. Let ϕ(z) :=
with expectation 0 and variance σ2. In order to utilize Lemma 1 we choose
δ :=
1
√2πσe−z2
2σ2denote the standard normal density function
σ
√logn. For z ≤ 2σ√logn it holds that
ϕ(z)/ϕ(z + δ) = e(δ/σ2)·z+δ2/(2σ2)= ez/(σ√logn)+1/(2logn)≤ e3.
Therefore, if we choose r := e3we have Zϕ
derive a bound on?
for the normal density function with expectation 0 and variance σ2it holds that
?∞
?
Zϕ
δ,r
Altogether we can apply Lemma 1 with δ = σ/√logn, r = e3and Z = 1/n.
This gives that the number of left-to-right maxima is at most O(1
log(n)), as desired.
δ,r⊂ [2σ√logn,∞). Now, we
Zϕ
δ,rϕ(z)dz. It is well known from probability theory that
kσϕ(z)dz ≤ e−k2/4. Hence,
ϕ(z)dz ≤
?∞
2σ√logn
ϕ(z)dz ≤1
n
.
σ· log(n)3/2+
⊓ ⊔
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2.2 Monotonic noise distributions
In this section we investigate upper bounds for general noise distributions. We
call anoise distribution monotonic ifthecorresponding density function ismono-
tonically increasing on R≤0and monotonically decreasing on R≥0. The follow-
ing theorem gives an upper bound on the number of left-to-right maxima for
arbitrary monotonic noise distributions.
Theorem 2. The expected number of left-to-right maxima in a sequence of n
speed values perturbed by random noise from a monotonic noise distribution is
O(?nlogn · f(0) + logn).
Proof. Let f denote the density function of the noise distribution and let f(0)
denote the maximum of f. We choose r := 2 whereas δ will be chosen later. In
ordertoapply Lemma Lemma1weonly need toderive abound on?
Therefore, we first define sets Zi, i ∈ N such that ∪iZi⊇ Zf
how to estimate?
First note that for z + δ < 0 we have f(z) < f(z + δ) because of the
monotonicity of f. Hence Zf
of the form [(ℓ − 1) · δ,ℓ · δ] for ℓ ∈ N0. Now, we define Zito be the ith
interval that has a non-empty intersection with Zf
a non-empty intersection then Ziis the empty set.) By this definition we have
∪iZi⊇ Zf
We can derive a bound on?
Ziis an interval of length δ and the maximum density within this interval is
f(ˆ zi). Furthermore it holds that f(ˆ zi+2) ≤1
consider some zi∈ Zi∩Zf
where we utilized that zi∈ Zf
?
by
?
∪iZi
i∈N
≤
i∈N
≤ 2δf(ˆ z1) + 2δf(ˆ z2) + δ · f(0) ≤ 5δ · f(0) .
Lemma1 yields that the number of left-to-right maxima is at most 2·⌈1
n · 5δ · f(0). Now, choosing δ :=
Zf
δ,rf(z)dz.
δ,rand then we show
∪iZif(z)dz.
δ,r⊆ [−δ,∞). We partition [−δ,∞) into intervals
δ,r. (If less than i intervals have
δ,ras desired.
∪iZif(z)dz as follows. Suppoe that all Zi⊂
Zif(z)dz ≤ δ·f(ˆ zi) because
R≥0. Let ˆ zidenote the start of interval Zi. Then?
2f(ˆ zi) for every i ∈ N. To see this
δ,r.Wehavef(ˆ zi) ≥ f(zi) > 2·f(zi+δ) ≥ 2·f(ˆ zi+2),
δ,rand that zi+δ ≤ ˆ zi+2. If Z1= [−δ,0] we have
Z1f(z)dz ≤ δ · f(0) for similar reasons. Now we can estimate?
?
Z2i−1
?
∪iZif(z)dz
f(z)dz ≤
?
f(z)dz +
?
i∈N
?
i∈N
?
Z2i
1
f(z)dz +
?
[−δ,0]
f(z)dz
1
2i−1δ · f(ˆ z1) +
2i−1δ · f(ˆ z2) + δ · f(0)
δ⌉·logn+
?logn/(f(0) · n) gives the theorem.
⊓ ⊔
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a)
V3
V0
V1
V2
V4
E1
γ1
α
ǫσ
b)
Vi
Vi+1
Ei
δi
R
ǫσ
Fig.1. (a) The partitioning of the plane into different regions. If the extreme
point Eiof a boundary region i falls into the shaded area the corresponding
boundary region is not valid.(b) The situation where the intersection between
a boundary region i and the corresponding range square Riis minimal.
3Lower Bounds
For showing lower bounds we consider the 1D problem and map each point
with initial position piand speed sito a point Pi= (pi,si) in 2D. We utilize
that the number of external events when maintaining the bounding box in 1D is
strongly related to the number of vertices of the convex hull of the Pi’s. If we
can arrange the points in the 2D plane such that after perturbation L points lie
on the convex hull on expectation, we can deduce a lower bound of L/2 on the
number of external events.
Bythismethod the results of[11]directly implyalowerbound of Ω(√logn)
for the case of normally distributed noise. Forthe case of monotonic noise distri-
butions we show that the number of vertices on the convex hull is significantly
larger than for the case of normally distributed noise.
We choose the uniform distribution with expectation 0 and variance σ2. The
density function f of this distribution is
?1/ǫσ
0
else
We construct an input of n points that has a large expected number of ver-
tices on the convex hull after perturbation. For this we partition the plane into
different regions. We inscribe an ℓ-sided regular polygon into a unit circle cen-
tered at the origin. The interior of the polygon belongs to the inner region while
everything outside the unit circle belongs to the outer region. Let V0,...,Vℓ−1
denote the vertices of the polygon. The ith boundary region is the segment of
the unit circle defined by the chord ViVi+1where the indices are modulo ℓ,
c.f. Figure 1a). An important property of these regions is expressed in the fol-
lowing observation.
f(x) =
|x| ≤ ǫσ/2
, where ǫσ=
√12σ.
Page 10
Observation 1 If no point lies in the outer region then every non-empty bound-
ary region contains at least one point that is a vertex of the convex hull.
⊓ ⊔
In the following, we select the initial positions of the input points such that
it is guaranteed that after the perturbation the outer region is empty and the
expected number of non-empty boundary regions is large.
We need the following notations and definitions. For an input point j we
define the range square R to be the axis-parallel square with side length ǫσ
centered at position (pj,sj). Note that for the uniform distribution with standard
deviation σ the perturbed position of j will lie in R. Further, the intersection
between the circle boundary and the perpendicular bisector of the chord ViVi+1
is called the extremal point of boundary region i and is denoted with Ei. The line
segment from the midpoint of the chord to Eiis denoted with δi, c.f. Figure 1b).
The general outline for the proof is as follows. We try for a boundary region
i to place a bunch ofn
common range square R lies in the extremal point Eiof the boundary region.
Furthermore we require that no point of R lies in the outer region. If this is
possible it can be shown that the range square and the boundary region have
a large intersection. Therefore it will be likely that one of then
corresponding to the square lies in the boundary region after perturbation. Then,
we can derive a bound on the number of vertices in the convex hull by exploiting
Observation 1, because we can guarantee that no perturbed point lies in the outer
region.
Now, we formalize this proof. We call a boundary region i valid if we can
place input points in the described way, i.e., such that their range square Riis
contained in the unit circle and a vertex of it lies in Ei. Then Riis called the
range square corresponding to boundary region i.
ℓinput points in the plane such that a vertex of their
ℓinput points
Lemma 2. If σ ≤ 1/8 and ℓ ≥ 23 then there are at least ℓ/2 valid boundary
regions.
Proof. If σ ≤ 1/8 then the relationship between ǫσand σ gives ǫσ= 2√3σ ≤
1/2. Let γidenote the angle of vector Eiwith respect to the positive x-axis. A
boundary region is valid iff sin(γi) ≥ ǫσ/2 and cos(γi) ≥ ǫσ/2. The invalid
regions are depicted in Figure 1a). If ǫσ≤ 1/2 these regions are small. To see
this let β denote the central angle of each region. Then 2sin(β/2) = ǫσ≤ 1/2
and β ≤ 2 · arcsin(1/4) ≤ 0.51. At most
their extreme point in a single invalid region. Hence the total number of invalid
boundary regions is at most 4(
β
2π/ℓ+ 1 boundary regions can have
β
2π/ℓ+ 1) ≤ ℓ/2.
⊓ ⊔
The next lemma shows that a valid boundary region has a large intersection with
the corresponding range square.
Page 11
Lemma 3. Let Ridenote the range square corresponding to boundary region i.
Then the area of the intersection between Riand the ith boundary region is at
least min{(4
Proof. Letα denote the central angle of the polygon. Then α =2π
cos(α
for α ≤ 2. Plugging in the value for α this gives δi≥ (4
The intersection between the range square and the boundary region is mini-
mal when one diagonal of the square is parallel to δi, c.f. Figure 1b). Therefore,
the area of the intersection is at least δ2
if δi≥√2ǫσ.
Lemma 4. If ℓ ≤ min{5?n/ǫ2
empty with probability at least 1 − 1/e, after perturbation.
Proof. We placen
ability that none of these points lies in the boundary region after perturbation
is
ℓ)4,ǫ2
σ/2} if ℓ ≥ 4.
ℓand δi= 1−
2). By utilizing the inequality cos(φ) ≤ 1−1
2φ2+1
24φ4we get δi≥11
ℓ)2for ℓ ≥ 4.
96α2
i≥ (4
ℓ)4if δi≤√2ǫσand at least ǫ2
σ/2
⊓ ⊔
σ,n/2} then every valid boundary region is non-
ℓinput points on the center of a valid range square. The prob-
Pr[boundary region is empty] ≤
?
1 −min{δ2
i,ǫ2
ǫ2
σ
σ/2}
?n
ℓ
,
because the area of the intersection is at least min{δ2
of the range square is ǫ2
i,ǫ2
σ/2} and the whole area
σ/2} the result follows since
σ. If δ2
i= min{δ2
σ/2}≤ǫ2
i≥ 1/ℓ4which follows from the proof of Lemma 3. In
σ/2 = min{δ2
Theorem 3. If σ ≤ 1/8 the smoothed worst case number of vertices on the
convex hull is Ω(min{5√n/σ,n}).
Proof. By combining Lemmas 2 and 4 with Observation 1 the theorem follows
immediatly if we choose ℓ = Θ(min{5√n/ǫσ,n}).
i,ǫ2
ǫ2
i,ǫ2
σ
min{δ2
σ
δ2
i
≤ ǫ2
σ· ℓ4= ǫ2
σ· ℓ5/ℓ ≤ n/ℓ .
Here we utilized that δ2
the case that ǫ2
i,ǫ2
σ/2} the result follows sincen
ℓ≥ 2.
⊓ ⊔
⊓ ⊔
4 Conclusions
We introduced smoothed motion complexity as a measure for the complexity
of maintaining combinatorial structures of moving data. We showed that for the
problem of maintaining the bounding box of a set of points the smoothed motion
complexity differs significantly from the worst case motion complexity which
makes it unlikely that the worst case is attained in typical applications.
Page 12
A remarkable property of our results is that they heavily depend on the
probability distribution of the random noise. In particular, our upper and lower
bounds show that there is an exponential gap in the number of external events
between the cases of uniformly and normally distributed noise. Therefore we
have identified an important sub-task when applying smoothed analysis. It is
mandatory to precisely analyze the exact distribution of the random noise for a
given problem since the results may vary drastically for different distributions.
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Maintaining approximate extent measures of
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