Lecture Notes CMSC M
CMSC M Computer Graphics
Spring
DaveMount
Lecture Intro duction
Tuesday Sep
What is Computational Geometry Computational geometry is a term claimed bya number
of di erent groups The term was coined p erhaps rst by Marvin Minsky in his b o ok Percep
trons whichwas ab out pattern recognition and has b een used often to describ e algorithms
solid mo deling But its most widely recognized use is to describ e the sub eld of algorithm
theory that involves the design and analysis of e cient algorithms for problems involving geo
metric inputs primarily in or p erhaps consant dimensional spaces It primarily involves
straight or at ob jects lines line segments p olygons planes and p olyhedra as opp osed to
curves and surfaces It is this latter sense of the term that we will b e covering in this course
The eld develop ed rapidly in the late s and through the s and s and it still continues
to develop Because of the area from which it grew discrete algorithm design the eld of
computational geometry has always emphasized problems of a discrete mathematic nature For
most problems in computational geometry the input is a nite set of p oints or other geometric
ob jects and the output is a typically some sort of structure consisting of a nite set of p oints
or line segments
Here is an example of a typical problem called the shortest path problem Given a set p olygonal
obstacles in the plane nd the shortest obstacle avoiding path from some given start p oint
toagiven goal p oint Although it is p ossible to reduce this to a shortest path problem on a
graph called the visibility graph whichwe will discuss later this semester and then apply a
nongeometric algorithm such as Dijkstra s algorithm it seems that by solving the problem in
its geometric domain it should b e p ossible to devise more e cient solutions This is one of the
main reasons for the growth of interest in geometric algorithms
s t s t
Figure Shortest path problem
The measure of the quality of an algorithm in computational geometry has traditionally b een
its asymptotic worst case running time Thus an algorithm running in O n time is b etter than
one running in O n log n time which is b etter than one running in O n time This particular
problem can b e solved in O n log n timeby a fairly simple algorithm and in O n log n by
avery complex algorithm In some cases average case running time is considered instead
Copyright David M Mount Dept of Computer Science University of Maryland College Park MD
These lecture notes were prepared byDavid Mount for the course CMSC Computational Geometry at
the University of Maryland College Park Permission to use copy modify and distribute these notes for educational
purp oses and without fee is hereby granted provided that this copyright notice app ear in all copies
Lecture Notes CMSC M
However for manytyp es of geometric inputs it is di cult to de ne input distributions that
are b oth easy to analyze and representativeoftypical inputs
There are many elds of computer science that deal with solving problems of a geometric
nature These include computer graphics computer vision and image pro cessing rob otics
computer aided design and manufacturing computational uid dynamics and geographic in
formation systems to name a few One of the goals of computational geometry is to provide
the basic geometric to ols needed from which application areas can then build their programs
There has b een signi cant progress made towards this goal but it is still far from b eing fully
realized
Limitations of Computational Geometry There are some fairly natural reasons why compu
tational geometry maynever fully address the needs of all these applications areas and these
limitations should b e understo o d b efore undertaking this course One is the discrete naturea
of computational geometry In some sense any problem that is solved on digital computers
must b e expressed in a discrete form but many applications areas deal with discrete approx
imations to continuous phenomenon For example in image pro cessing the image maybea
discretization of a continuous dimensional gray scale function and in rob otics issues of vibra
tion oscillation in dynamic control systems are of a continuous nature Nonetheless there are
many applications in which ob jects are of a very discrete nature For example in geographic
information systems road networks are discretized into collections of line segments
The other limitation is the fact that computational geometry deals primarily with straightor
at ob jects To a large extent this is a result of the fact that computational geometers were
not trained in geometry but in discrete algorithm design So they chose problems for which
geometry and numerical computation plays a fairly small role Much of solid mo deling uid
dynamics and rob otics deals with ob jects that are mo deled with curved surfaces However it
is p ossible to approximate curved ob jects with piecewise planar p olygons or p olyhedra This
assumption has freed computational geometry to deal with the combinatorial elements of most
of the problems as opp osed to dealing with numerical issues This is one of the things that
makes computational geometry fun to study y ou do not have to learn a lot of analytic or
di erential geometry to do it But it do es limit the applications of computational geometry
One more limitation is that computational geometry has fo cused primarily on dimensional
problems and dimensional problems to a limited extent The nice thing ab out dimensional
problems is that they are easy to visualize and easy to understand But many of the daunting
applications problems reside in dimensional and higher dimensional spaces Furthmore issues
related to top ology are much cleaner in and dimensional spaces than in higher dimensional
spaces
Trends in CG in the s and s In spite of these limitations there is still a remarkable array
of interesting problems that computational geometry has succeeded in addressing Throughout
the s the eld develop ed many techniques for the design of e cient geometric algorithms
These include well known metho ds such as divide and conquer and dynamic programming
along with a numb er of newly discovered metho ds that seem to b e particularly well suited
to geometric algorithm These include plane sweep randomized incremental constructions
duality transformations and fractional cascading
One of the ma jor fo cuses of this course will b e on understanding technique for designing
e cient geometric algorithms A ma jor part of the assignments in this class will consist of
designing and or analyzing the e ciency of problems of a discrete geometric nature
However throughout the s there a nagging gap was growing b etween the theory and
practice of designing geometric algorithms The s and early s sawmany of the op en
problems of computational geometry solved in the sense that theretically optimal algorithms
Lecture Notes CMSC M
were develop ed for them However many of these algorithms were nightmares to implement
b ecause of the complexity of the algorithms and the data structures that they required Fur
thermore implementations that did exist were often sensitive to geometric degeneracies that
caused them to pro duce erroneous results or ab ort For example a programmer designing
an algorithm that computes the intersections of a set of line segments may not consider the
situation when three line segments intersect in a single p oint In this rare situation the data
structure b eing used may b e corrupted and the algorithm ab orts
Much of the recentwork in computational geometry has dealt with trying to make the the
oretical results of computational geometry accessible to practioners This has b een done by
simplifying existing algorithms dealing with geometric degeneracies and pro ducing libraries
of geometric pro cedures This pro cess is still underway Whenever p ossible we will discuss the
simplest known algorithm for solving a problem Often these algorithms will b e randomized
algorithms We will also discuss hop efully without getting to o b ogged down in details some
of the techniques for dealing with degenerate situations in order to pro duce clean and yet
robust geometric software
A Grand Overview Here are some of the topics that we will discuss this semester
Convex Hulls Convexityisavery imp ortant geometric prop erty A geometric set is convex
if for every two p oints in the set the line segment joining them is also in the set One of
the rst problems identi ed in the eld of computational geometry is that of computing
the smallest convex shap e called the convex hul l that encloses a set of p oints
Convex hull Polygon triangulation
Figure Convex hulls and p olygon triangulation
Intersections One of the most basic geometric problems is that of determining when two sets
of ob jects intersect one another Determining whether complex ob jects intersect often
reduces to determining which individual pairs of primitiveentities e g line segments
intersect We will discuss e cient algorithms for computing the intersections of a set of
line segments
Triangulation and Partitioning Triangulation is a catchword for the more general prob
lem of sub dividing a complex domain into a disjoint collection of simple ob jects The
simplest region into which one can decomp ose a planar ob ject is a triangle a tetrahedron
in d and simplex in general We will discuss how to sub divide a p olygon into triangles
and later in the semester discuss more general sub divisions into trap ezoids
Low dimensional Linear Programming Many optimization problems in computational
geometry can b e stated in the form of a linear programming problem namely nd the
extreme p oints e g highest or lowest that satis es a collection of linear inequalities
Linear programming is an imp ortant problem in the combinatorial optimization and
p eople often need to solvesuch problems in hundred to p erhaps thousand dimensional
spaces However there are manyinteresting problems e g nd the smallest disc enclosing
a set of p oints that can b e p osed as low dimensional linear programming problems In
low dimensional spaces very simple e cient solutions exist
Lecture Notes CMSC M
Line arrangements and duality Perhaps one of the most imp ortant mathematical struc
tures in computational geometry is that of an arrangement of lines or generally the
arrangement of curves and surfaces Given n lines in the plane an arrangement is just
the graph formed by considering the intersection p oints as vertices and line segments join
ing them as edges We will show that such a structure can b e constructed in O n time
These reason that this structure is so imp ortantisthatmany problems involving p oints
can b e transformed into problems involving lines by a metho d of duality For example
supp ose that you want to determine whether any three p oints of a planar p oint set are
collinear This could b e determines in O n time by brute force checking of each triple
However if the p oints are dualized into lines then as we will see later this semester
this reduces to the question of whether there is a vertex of degree greater than in the
arrangement
Voronoi Diagrams and DelaunayTriangulations Given a set S of p oints in space one
of the most imp ortant problems is the nearest neighb or problem Given a p ointthatisnot
in S whic p ointofS is closest to it One of the techniques used for solving this problem
is to sub divide space into regions according to whichpoint is closest This gives rise to a
geometric partition of space called a Voronoi diagram This geometric structure arises in
many applications of geometry The dual structure called a Delaunay triangulation also
has manyinteresting prop erties
Figure Voronoi diagram and Delaunay triangulation
Search Geometric search problems are of the following general form Given a data set e g
points lines p olygons which will not change prepro cess this data set into a data struc
ture so that some typ e of query can b e answered as e ciently as p ossible For example
a nearest neighbor search query is determine the p oint of the data set that is closest to a
given query p oint A range query is determine the set of p oints or countthenumber of
points from the data set that lie within a given region The region may b e a rectangle
disc or p olygonal shap e like a triangle
Course Texts The text for this course is the excellent new b o ok Computational Geometry
Algorithms and Applications bydeBerg van Kreveld Overmars and Schwarzkopf Many
of the algorithms notations de nitions and gures will b e b orrowed from their presen
tation We ll refer to their b o ok by their initials BKOS We ll also refer to the excellent
algorithms textb o ok Introduction to Algorithms by Cormen Leiserson and Rivest for
basic algorithms and data structures
Lecture Geometric Background and Convex Hulls
Thursday Sep
Up dated pm Sep
Reading Chapter in BKOS deBerg vanKreveld Overmars and Schwarzkopf
Lecture Notes CMSC M
A ne space We will usually b e rather informal in our presentation of geometry but it is go o d to
start things o on a somewhat formal fo oting We b egin with some background on Euclidean
geometry which is not presented in the text
There are a numb er of di erent geometric systems that one can work under This semester
wewillbeworking almost exclusively with Euclidean geometry although wemayintro duce a
few concepts from pro jective geometry later in the semester We b egin with a short review of
geometry Our approach will b e di erent from what is done in most math texts and most com
putational geometry texts The standard approach in math texts is to b egin by assuming that
everything is a tuple of real numb ers But this approach do es lend itself to ob ject oriented
programming b ecause it blurs the distinction b etween di erent geometric elements The ap
proach in computational geometry texts is to say nothing ab out geometric representations but
to rely up on an intuitive understanding of geometric concepts
Rather than de ning Euclidean geometry we will rst de ne a somewhat more basic geome
try called a ne geometry Later we will add one op eration called an inner pro duct which
di erentiates Euclidean from a ne geometry
An a ne geometry consists of a set of scalars the real numb ers a set of points and a set
of freevectors or simply vectors Points are used to sp ecify p osition Free vectors are used
to sp ecify direction and magnitude but have no xed p osition in space This is in contrast
to linear algebra where there is no real distinction b etween p oints and vectors However this
distinction is useful since the two are really quite di erent
The following are the op erations that can b e p erformed on scalars p oints and vectors Vector
op erations are just the familiar ones from linear algebra It is p ossible to subtract two p oints
The di erence p q of twopoints results in a free vector directed from q to p It is also
p ossible to add a p ointtoavector In p oint vector addition p v results in the p ointwhich
is translated by v from p The following are the legal op erations
S V V scalar vector multiplication
V V V vector addition
P P V p oint subtraction
P V P point vector addition
u+v q u p+v p-q p v p v
vector addition point subtraction point-vector addition
Figure A ne op erations
Anumb er of op erations can b e derived from these For example we can de ne the subtraction
of twovectors u v as u v or scalar vector division v as v provided There
is one sp ecial vector called the zerovector which has no magnitude suchthat v v
Note that it is not p ossible to multiply a p oint times a scalar or to add twopoints together
However there is a sp ecial op eration that combines these two elements called an a ne combi
nation Given any scalar and twopoints p and p de ne the a ne combination A p p
Lecture Notes CMSC M
to b e
p p p p p
Note that the left hand side of this equation is not legal given our list of op erations But
this is how the a ne combination is typically expressed namely as the weighted average of
two p oints The right hand side which is easily seen to b e algebraically equivalent is legal
An imp ortant observation is that if p p then the p ointA p p lies on the line
connecting p and p and generally as varies from to it traces out all the p oints
on this line Aff(p,q,-1/2) Aff(p,q,0) Aff(p,q,1/2) p Aff(p,q,1)
q
Figure A ne combination
In the sp ecial case where A p p isapoint that sub divides the line segment
p p into two subsegments of relative sizes to When is in this range the op eration is
called a convex combination and the set of all convex combinations traces out the line segment
p p
Homogeneous Co ordinates In order to assign co ordinates to p oints and vectors we assume that
there is a designated standardcoordinate system whichisspeci edbyanoriginpoint and d
orthogonal unit vectors In many applications such as computer graphics and solid mo deling
it is often convenient to de ne other lo cal co ordinate systems Wewillnothavemuch need of
this though
To representvectors and p oints in a ne space we use homogeneous coordinates Supp ose that
we are working in d dimensional a ne space It is common to represent b oth free vectors and
p oints as d tuples of real numbers To represent a free vector wetake its standard d tuple
of co ordinates an prep end an additional to the b eginning To representapointwetake its
d tuple and prep end an additional In many application areas graphics and solid mo deling
in particular it is common to app end the or to the end of the tuple In principal there
is no reason that one representation is b etter than the other but b eware that the concept of
orientation de ned b elowisreversed in o dd dimensions in this case
This may sound rather ad ho c but it has some very nice algebraic prop erties For example
if you take the di erence of twopoints which results in a free vector observethatby simply
subtracting their homogeneous co ordinate tuples comp onentby comp onen t you will get the
prop er representation of this free vector since the rst two co ordinates will cancel each other
giving We will sometimes b e sloppy When it is clear that we are dealing with p oints we
may drop the homogenizing co ordinate
Orientation So far all of the op erations map numerical geometric entities into other numerical
entities In order to make discrete decisions we need a geometric op eration that is somewhat
analogous to the relational op erations with numb ers There do es not seem to b e
any natural intrinsic waytocomparetwo p oints in d dimensional space but there is a natural
relation b etween ordered d tuples of p oints in d space which extends the notion of
relations in space called orientation
Lecture Notes CMSC M
v: (0, 2, 1) v p p : (1, 2, 2) p-q q : (1, 3, 0) p-q : (1-1, 2-3, 2-0) = (0, -1, 2)
q
Figure Homogeneous co ordinates
Given an ordered triple of three p oints hp q r i in the plane wesay that they have positive
orientation if they de ne a counterclo ckwise oriented triangle negative orientation if they
de ne a clo ckwise oriented triangle and zero orientation if they are collinear Note that
orientation dep ends on the order in whichthepoints are given In general given an ordered
tuple p oints in space we can also de ne their orientation as b eing either p ositive forming a
right handed screw negative a left handed screw or zero coplanar This can b e generalized
to any ordered d tuple p oints in d space r q q q r
p p p r
positive zero negative
Figure Orientations of the ordered triple p q r
Orientation is formally de ned as the sign of the determinant of the p oints given in homoge
neous co ordinates For example in the plane wehave
p p
x y
A
Orient p q r det q q
x y
r r
x y
Observe that in the dimensional case Orient p q isjust q p Hence it is p ositiveifp q
zero if p q and negativeifp q Thus orientation generalized in dimensional
space
Convexity Now that wehave discussed some of the basics let us consider an initial problem The
computation of convex hulls is among the most basic problems in computational geometry An
O n log n algorithm for computing convex hulls was one of the earliest results in computational
geometry due to Ron Graham
The convex hull can b e de ned intuitively by surrounding a collection of p oints with a rubb er
band and letting the rubb er band snap tightly around the p oints There are a n umber of
reasons that the convex hull of a p oint set is an imp ortant geometric structure One is that it
is one of the simplest shap e approximations for a set of p oints Also many algorithms compute
the convex hull as an initial stage in their execution b ecause convex p olygons are often easier
to deal with than p oint sets For example in order to nd the smallest rectangle or triangle
that encloses a set of p oints it su ces to rst compute the convex hull of the p oints and then
nd the smallest rectangle or triangle enclosing the hull
Lecture Notes CMSC M
Convexity Aset S is convex if given anypoints p q S anyconvex combination of p and q
is in S or equivalently the line segment pq S
Convex hull The convex hul l of any set S is the intersection of all convex sets that contains
S ormoreintuitively the smallest convex set that contains S Following our b o ok s
notation we will denote this CH S
An equivalent de nition of convex hull is the set of p oints that can b e expressed as convex
combinations of the p oints in S A pro of can b e found in any book on convexitytheory A
convex combination of three or more p oints is an linear combination of the p oints in which the
co e cients sum to and all the co e cients are in the interval
In general convex sets mayhave either straight or curved b oundaries may b e b ounded or
unb ounded e g an in nite cone is convex and may b e top ologically op en or closed that
is they mayormay not contain their b oundary The convex hull of a nite set of p oints is
necessarily a b ounded closed convex p olygon
Convex hull problem The planar convex hul l problem is given a set of n points P in the plane
output the vertices of the convex hull Normally p olygons are presented in counterclo ckise
order For some reason our b o ok outputs the hull in clo ckwise order but obviouslyitisa
trivial matter to convert from one to the other The hull should consist only of extreme points
in the sense that if three p oints lie on an edge of the convex hull then the middle p oint should
not b e output as part of the hull
The b o ok intro duces a simple O n convex hull algorithm whichoperatesby considering each
ordered pair of p oints p q and the determining whether all the remaining p oints of the set
lie within the half plane lying to the right of the directed line from p to q Observe that this
can b e tested using the orientation test The question is can we do b etter
Graham s scan We will presentanO n log n algorithm for convex hulls It is a simple variation
of a famous algorithm for convex hulls called Graham s scan This algorithm dates backtothe
early s The algorithm is based on an approach called incremental construction inwhich
p oints are added one at a time and the hull is up dated with each new insertion If wewere to
add p oints in some arbitrary order wewould need some metho d of testing whether p oints are
inside the existing hull or not Toavoid the need for this test we will add p oints in increasing
order of x co ordinate thus guaranteeing that each newly added p oint is outside the current
hull It is not clear that this is necessarily a go o d thing to do After all if a p oint is inside
the convex hull then no up date need b e made Perhaps a b etter algorithm should seek to add
p oints in suchaway that most of the p oints that are not on the hull are eliminated quickly
from consideration Nonetheless Graham s scan is easy to implement and is optimal in the
worst case so it is not a bad algorithm byany means
Since weareworking from left to right it would b e convenient if the convex hull vertices were
also ordered from left to right The convex hull is a cyclically ordered sets Cyclically ordered
sets are somewhat messier to work with than simple linearly ordered sets so we will break the
hull into twohulls an upper hul l and lower hul l The break p oints common to b oth hulls will
b e the leftmost and rightmost vertices of the convex hull After building b oth the twohulls
are merged into a single cyclic list
Let us consider the upp er hull since the lower hullissymmetric Let hp p p i denote
n
the set of p oints sorted by increase x co ordinates As wewalk around the upp er hull from left
to right observe that each consecutive triple along the hull makes a right hand turn That
is if p q r are consecutivepoints along the upp er hull then Orient p q r When a new
p oint p is added to the currenthull this may violate the right hand turn invariant So we
i
check the last three p oints on the upp er hull including p They fail to form a right hand turn
i
Lecture Notes CMSC M
Upper hull p i
p n p 1
Lower hull
Figure Convex hulls and Graham s scan
then we delete the p oint prior to p This is rep eated until the numberofpoints on the upp er
i
hull including p is less than three or the right hand turn condition is reestablished See the
i
text for a complete description of the co de Wehave ignored a numb er of sp ecial cases We
will consider these next time
Analysis Let us prove the main result ab out the running time of Graham s scan
Theorem Graham s scan runs in O n log n time
Pro of Sorting the p oints according to x co ordinates can b e done byany e cient sorting
algorithm in O n log n time For each p oint added the amount of time sp ent is clearly
O D where D is the numberofpoints deleted in the pro cess of adding p oint p
i i i
The reason is that each orientation test takes constant time and wemust p erform one
orientation test for eachpoint deleted p erhaps along with one extra one for the last p oint
P
n
which is not deleted Thus the total running time is prop ortional to D
i
i
P P
n
D How large is D Observe that once a p oint is deleted it can never b e n
i i
i i
P
deleted again Since eachofn points can b e deleted at most once D n Thus after
i
i
sorting the total running time is O n Since this is true for the lower hull as well the
total time is O n O n
Lecture More Convex Hulls
Tuesday Sep
Reading Chapter in BKOS
Degeneracies Last time we presented the Graham s scan algorithm for computing convex hulls
Recall that the algorithm computes just the upp er and lower hulls of a set of n points each
in O n log n time The metho d is an incremental algorithm that adds p oints to the hull in
increasing order of x co ordinate
One of the issues that we neglected to deal with was that of degenerate inputs A degeneracy
is somewhat hard to de ne formally buta working de nition is that a degeneracy is any
sp ecial con guration of p oints lines etc that will b e broken with very high probability
if the ob jects de ning co ordinates are randomly p erturb ed F or example three p oints form
a degeneracy in all dimensions strictly higher than since any p erturbation of any their
co ordinates will cause the orientation test to b ecome nonzero
Of course we are only interested in degeneracies that a ect the algorithm of interest A
functional de nition of degeneracy is any con guration of p oints that causes a sign test based
on real values such as the orientation test to return a zero value
Let us b egin by assuming that we p erform all sign tests using exact arithmetic There are two
places in Graham s algorithm where degeneracies might arise The rst is that in sorting the
Lecture Notes CMSC M
p oints by their x co ordinates there maybetwopoints with the same x co ordinate The second
is when adding a p oint p and testing the last three consecutive p oints for the right hand turn
i
condition the three p oints may b e collinear
For the case of equal x co ordinates there is a very simple metho d for dealing with this that is
well worth rememb ering We will imagine that an in nitesimal p erturbation has b een applied
to the p oints Such a p erturbation should have the prop erty of destroying degeneracies but
creating no new degeneracies For example to break ties in x co ordinates we could apply a small rotation of the plane
Before perturbation After perturbation
Figure Perturbation to eliminate degeneracies
If we rotate to o much we might create a new tie among x co ordinates How small is small
enough Howmuchnumerical precision is needed The trick is that we do not actually
compute the p erturbation numerically Instead we p erform the computation on the original
exact inputs but make all decisions as if this rotation had b een applied This technique of
simulating a p erturbation is called symbolic perturbation
For example supp ose that we rotate all the p oints in nitesimally clo ckwise ab out the origin
of the co ordinate system If twopoints had the same x co ordinate prior to the rotation then
the one with the larger y co ordinate will have the larger x co ordinate after the rotation Since
the rotation is in nitesimal in nitely small then no new ties amongst x co ordinates will b e
created Tosimulate the p erturbation we simply sort the p oints lexicographical ly rst by
x co ordinate and then amongst p oints with the same x co ordinate by y co ordinate They
key is that the p oints have not actually b een rotated numerically but the algorithm b ehaves
as if they had b een
We could have p erturb ed the p oints byacounterclo ckwise rotation as well either would b e
ne This p erturbation also takes care of the question of where the upp er hull ends and the
lo wer hull b egins if there are ties for the leftmost and rightmost p oints
To deal with three collinear p oints observe that since the p oints are sorted by x co ordinate
as we visit them the middle p ointofanysuch triple should not b e considered as lying on
the hull So when implementing the right hand turn condition we should require that it is a
strict right hand turn otherwise the middle p oint is deleted
If wewere to use oating p oint arithmetic the b est we could hop e for is an approximation
to the convex hull in the sense that if that the hull weoutputwould b e the correct hull for
some small p erturbation of the input When dealing with oating p oint computation the
most imp ortant goal is that numerical errors should never cause the program to pro duce an
output that is structurally unsound For convex hulls this is not to o complex However for
other geometric problems it may b e quite hard to devise an algorithm that insensitive to small
numerical errors
Lecture Notes CMSC M
Quickhull Toshow that there is not just one way to compute planar convex hulls let us consider
some other approaches The rst is called quickhul l and can b e viewed as a dimensional
generalization of quicksort
Likequicksort this algorithm runs in O n log n time for favorable inputs but can take as long
as O n time for unfavorable inputs However unlikequicksort there is no obvious wayto
convert it into a randomized algorithm with O n log n exp ected running time Nonetheless
quickhull tends to p erform very well in practice
The intuition is that in many applications most of the p oints lie in the interiorofthehull For
example if the p oints are uniformly distributed in a unit square then the exp ected number of
p oints on the convex hull is O log n Here is a pro of
Theorem Supp ose that n points are uniformly distributed in a unit square The exp ected
number of points on the convex hull is O log n
Pro of First we will break the hull into four parts the upp er left upp er right lower left and
lower righthulls This is done by breaking the hull at its leftmost rightmost topmost
and b ottommost p oints We will show that eachhasO log n p oints By symmetry we
may consider one say the upp er righthull
Rather than b ounding the numberofpoints on the hull we will b ound a larger quantity
A p oint p x y issaidtobe dominated by another p oint p if x x and y y
i i i j j i j i
The maxima of a p ointset P are the p oints that are not dominated byany other p oint
in P Intuitively the maxima form a staircase along the upp er right side of the p oint set
Observe that the set of maxima is a sup erset of the p oints on the upp er righthull so an
upp er b ound on the numb er of maxima is an upp er b ound on the numb er of p oints in
the upp er righthull
Upper-right hull
Maxima
Figure Upp er righthull and maxima
Supp ose that the p oints are sorted in decreasing order of x co ordinate We will assume
that all p oints have distinct co ordinates to simplify things Clearly p is a maximum if
i
and only if p has the largest y co ordinate among the subset fp p p g What is
i i
the probability that this is true The y co ordinates of these p oints are indep endent from
each other Therefore the probability that p has the largest y co ordinate is just i
i
Thus each p oint p is a maxima with probability i The exp ected numb er of maxima
i
is
n
X
ln n O log n E
n
i
i
The sum is the harmonic series see Cormen Leiserson and Rivest if you haven t seen
this From the comments made earlier it follows that the exp ected number of points on
all the hulls is at most times this quantity whichisO log n
Lecture Notes CMSC M
The idea b ehind quickhull is to discard p oints that are not on the hull as quickly as p ossi
ble QuickHull b egins by computing the p oints with the maximum and minimum x and
y co ordinates Clearly these p oints must b e on the hull Horizontal and vertical lines passing
through these p oints are supp ort lines for the hull and so de ne a b ounding rectangle within
which the hull is contained Furthermore the convex quadrilateral de ned by these four p oints
lies within the convex hull so the p oints lying within this quadrilateral can b e eliminated from
further consideration All of this can b e done in O n time
discard these
Figure Quickhull s initial quadrilateral
To continue the algorithm we classify the remaining p oints into the four corner triangles
that remain In general as this algorithm executes we will have an inner convex p olygon
and asso ciated with eachedgewehave a set of p oints that lie outside of that edge More
formally these p oints are witnesses to the fact that this edge is not on the convex hull b ecause
they lie outside the half plane de ned by this edge When this set of p oints is empty the edge
isa naledgeofthehull Consider some edge ab Assume that the p oints that lie outside
of this hull edge have b een placed in a bucket that is asso ciated with ab Our job is to nd
apoint c among these p oints that lies on the hull discard the p oints in the triangle abc and
split the remaining p oints into two subsets those that lie outside ac and those than lie outside
of cb We can classify each p ointby making two orientation tests c b b
discard these
a a
Figure Quickhull elimination pro cedure
How should c b e selected There are a numb er of p ossible selection criteria that one might
think of The suggested metho d is that c be the point that maximizes the p erp endicular
distance from the line ab Another p ossible choice mightbethepoint that maximizes the
angle cba or cab It turns out that these are p o or choices b ecause they do not pro duce even
splits of the remaining p oints We replace the edge ab with the two edges ac and cb and
classify the p oints as lying in one of three groups those that lie in the triangle abc which are
discarded those that lie outside of ac and those that lie outside of cb We put these p oints in
buckets for these edges and recurse We claim that it is not hard to classify each p oint p by
computing the orientations of the triples acp and cbp
The running time of quickhull as with quicksort dep ends on howevenly the p oints are split
at each stage Let T n denote the running time on the algorithm assuming that n p oints
remain outside of some edge In O n time we can select a candidate splitting p oint c and
classify the p oints in the bucket in O n time Let n and n denote the numb er of remaining
Lecture Notes CMSC M
p oints Then the running time is given by the recurrence
T
T n T n T n n
To solve the recurrence it would b e necessary to determine the exp ected values of n n
whichwould dep end on the p oint distribution If we assume that the p oints are evenly dis
tributed in the sense that max n n n for some constant then by applying the
same analysis as that used in quicksort see Cormen Leiserson Rivest the running time will
solveto O n log n where the constant factor dep ends on
Do es quickhull outp erform Graham s scan This dep ends to a great extent on the distribution
of the p oint set There are variations of quickhull that are designed for sp eci c p oint distribu
tions e g p oints uniformly distributed in a square and their authors claim that they manage
to eliminate almost all of the p oints in a matter of only a few iterations
Other convex hull algorithms There are still other convex hull algorithms A natural question
In fact it is p ossible to show that to ask is whether it is p ossible to improveon O n log n
any algorithm for convex hulls must at least sort the p oints along the hull We ll leave this as
an exercise In the worst case the hull has n p oints and so this implies an n log n lower
b ound
Butifyou know or exp ect fewer p oints on the hull then you maybeabletodomuch b etter
For example if your inputs consist of p oints uniformly distributed in a square then you exp ect
the numb er of p oints on the hull to b e only O log n Is it p ossible to nd a b etter algorithm
in this case An algorithm that takes into consideration b oth the input and output size is
called an output sensitive algorithm Let n denote the number of input points and h denote
the numb er of edges on the convex hull
There is an algorithm called Jarvis s march which builds the hull in O nh time by a pro cess
called gift wrapping In O n time it is p ossible to nd the next edge on the hull and
this algorithm op erates bywalking around the hull one edge at a time But this gives no
improvement for the case of uniformly distributed p oints in the square Nonetheless manyof
the most practical higher dimensional convex hull algorithms are based on this approac h
There exists an O n log h time algorithm for the convex hull problem and it can b e shown
that this algorithm is optimal with resp ect to output sensitive algorithms The st algorithm
with this complexitywas discovered back in the mid s But recently in amuch
simpler algorithm was discovered
Lecture Line SegmentIntersection
Thursday Sep
Revised Sept Fixed the analysis
Reading Chapter in BKOS
Geometric intersections One of the most basic problems in computational geometry is that of
computing intersections Intersection computation is basic to manydi erent application areas
In solid mo deling p eople often build up complex shap es by applying various b o olean
op erations intersection union and di erence to simple primitive shap es The pro cess
in called constructive solid geometry CSG In order to p erform these op erations the most
basic step is determining the p oints where the b oundaries of the two ob jects intersect
Lecture Notes CMSC M
In rob otics and motion planning it is imp ortanttoknowwhentwo ob jects intersect for
collision detection and collision avoidance
In geographic information systems it is often useful to overlaytwo sub divisions e g a
road network and county b oundaries to determine where road maintenance resp onsibilities
lie Since these networks are formed from collections of line segments this generates a
problem of determining intersections of line segments
Another example comes from computer graphics The clipping problem involves deter
mining what parts of a set of p olygons are visible through a rectangular window This
involves determining the intersection of the p olygons and the window and then clipping
the p olygons at these p oints
Line segmentintersection The problem that we will consider is given n line segments in the
plane rep ort all p oints where a pair of line segments intersect We assume that each line
segment is represented by giving the co ordinates of its twoendpoints
n
Observe that n line segments can intersect in as few as and as manyas O n
di erentintersection p oints We could settle for an O n algorithm claiming that it is worst
case asymptotically optimal but it would not b e v ery useful in practice since in many instances
of intersection problems intersections may b e rare Therefore it seems reasonable to lo ok for
an output sensitive algorithm that is one whose running time should b e e cient b oth with
resp ect to input and output size
We will let I denote the number of intersections How should wecount I in the degenerate
con guration where three or more lines intersect in a single p oint If k lines intersect in a
single p oint one might think of this as a single intersection p oint or one mightinterpret
k
k k pairwise intersections The former interpretation would lead to this as
faster running times and so would b e the goal of a careful implementation And indeed the
implementation in the book achieves this On the other hand one might argue that such
degeneracies are rare in practice and so it is not worth worrying ab out Wewilltake the
latter sloppier attitude and refer to the text for a more careful implementation
Complexity We will present a not quite optimal O n log n I log n time algorithm for the line
segmentintersection problem A natural question is whether this is optimal
time The b est that one might hop e for is O n log n I time algorithm Clearly we need O I
to output the intersection p oints What is not so obvious is that O n log n time is needed
This results from the fact that the following problem is known to require n log n time
Element uniqueness Given a list of n numb ers do es anynumb er app ear at least twice in
the list
Note that element uniqueness can b e solved in O n log n time by sorting and checking whether
adjacent elements are equal Given this lower b ound even determining whether the same
endp ointisusedintwo di erent line segments would require n log n time and hence this
is a lower b ound for determining even the existence of a single intersection
Note that this lower b ound result assumes the algebraic decision treemodel of computation
in which all decisions are made by comparisons made based on exact algebraic op erations
applied to numeric inputs Although this encompasses most of what one would
consider to b e normal geometric computations there are alternative mo dels of computation
For example by taking mo ds o ors or ceilings you would b e able to implement hashing
This is a data access metho d taught in most data structures and algorithms courses See
Cormen Leiserson and Rivest for example With hashing it is p ossible to solv e the element
uniqueness in exp ected O n time Unfortunately even in this more p owerful computational
Lecture Notes CMSC M
mo del no one knowhow to determine whether anytwo line segments intersect in faster than
O n log n I time in the worst case
Later in the semester we will discuss an optimal O n log n I time algorithm for this problem
Line segmentintersection In rather typical computational geometry fashion our b o ok do es not
discuss the issue of how to determine the intersection p ointoftwo line segments Let ab and
cd be two line segments given by their endp oints It is an easy exercise to determine whether
these line segments intersect simply by applying an appropriate combination of orientation
tests
To determine the co ordinates of the intersect p ointinvolves solving a small system of equations
The most natural way to set up this computation is to intro duce the notion of a parametric
representation of the line segment Recall that anypoint on the line segment ab can b e written
as a convex combination involving a real parameter s
p s s a sb for s
Similarly for cd wemayintro duct a parameter t
q t t c td for s
An intersection o ccurs if and only if we can nd s and t suchthatp s p t Thus we get
the two equations
s a sb t c td
x x x x
s a sb t c td
y y y y
The co ordinates of the p oints are all known so it is just a simple exercise in linear algebra
to solve for s and t The computation of s and t will involve a division If the divisor is
this corresp onds to the case where the line segments are parallel and p ossibly collinear
These sp ecial cases should b e dealt with carefully If the divisor is nonzero then we get values
for s and t as rational numb ers the ratio of twointegers We can approximate them as
oating p ointnumbers or if wewant to p erform exact computations it is p ossible to simulate
rational numb er algebra exactly using high precision integers and multiplying through by
least common multiples Once the values of s and t have b een computed all that is needed is
to check that b oth are in the interval
Plane Sweep Algorithm Let S fs s s g denote the line segments whose intersections
n
we wish to compute The metho d is called plane sweep Here are the main elements of any
plane sweep algorithm and howwe will apply them to this problem
Sweep line We will simulate the sweeping of a vertical line called the sweep line from left
to right Our text uses a horizontal line but there is obviously no signi cant di erence
We will maintain the line segments that intersect the sweep line in sorted order say from
top to b ottom
Even ts Although we might think of the sweep line as moving continuously we only need to
up date data structures at p oints of some signi cantchange in the sweep line contents
called event points
Di erent applications of plane sweep will have di erent notions of what event p oints
are For this application eventpoints will corresp ond to instances where the sweep line
encounters an endp oints of a line segment which are all known in advance and when the
sweep line encounters an intersection p ointoftwo line segments which will b e discovered
as the algorithm executes
Lecture Notes CMSC M
Event up dates When an eventisencountered wemust up date the data structures asso ci
ated with the event It is a go o d idea to b e careful in sp ecifying exactly what invariants
you intend to maintain For example when we encounter an intersect p oint wemust
interchange the order of the intersecting line segments along the sweep line
There are a great numb er of nasty sp ecial cases that complicate the algorithm and obscure the
main p oints We will makeanumb er of simplifying assumptions for now
No line segmentisvertical Easily xed through a symb olic p erturbation
If two segments intersect then they intersect in a single p oint that is they are not
collinear
No three lines intersect in a common p oints
future event points
intersections detected
l
Figure Plane sweep
Detecting intersections Wementioned that endp ointevents are all known in advance But how
do we detect intersection events It is imp ortantthateachevent b e detected b efore the actual
event o ccurs Our strategy will b e as follows Whenever two line segments b ecome adjacent
along the sweep line wewillcheck whether they haveanintersection o ccuring to the rightof
the sweep line If so we will add this new event
A natural question is whether this is su cient In particular if two line segments do intersect
is there necessarily some prior placementofthesweep line such that they are adjacent Happily
this is the case but it requires a pro of
Lemma Given two segments s and s whichintersect in a single p oint p and assuming no
i j
other line segment passes through this p oint there is a placementofthesweep line prior
to this event such that s and s are adjacent along the sweep line and hence will b e
i j
tested for intersection
Pro of Consider such a pair of segments and assume that prior to this all intersections
have b een correctly handled by the algorithm thus the sweep line order prior to their
intersection is correct Consider the contents of the sweep line immediately prior to this
event We claim that at this time there could b e no segment s between s and s along
k i j
the sweep line If there were then either there would have to b e endp ointofs prior to
k
the intersection implying this was not the event immediately prior to the intersection or
that s intersects either s or s again implying the existence of an intermediate event
k i j
Data structures In order to p erform the sweep we will need two data structures
Lecture Notes CMSC M
si
sk
sj
l
Figure Correctness of plane sweep
Event queue This holds the set of future events sorted according to increasing x co ordinate
Eacheventcontains the auxiliary information of what typ e of eventthisis segment
endp ointorintersection and which segment s are involved The op erations that this
data structure should supp ort are inserting an event if it is not already present in the
queue and extracting the minimum event
It seems like a heap data structure would b e ideal for this since it supp orts insertions
and extract min in O log M time where M is the number of entries in the queue See
Cormen Leiserson and Rivest for details However a heap cannot supp ort the op eration
of checking for duplicate events
There are twoways to handle this One is to use a more sophisticated data structure such
as a balanced binary tree or skip list This adds a small constant factor but can check
that there are no duplicates easily The second is use the heap but when an extraction
is p erformed you mayhave to p erform many extractions to deal with multiple instances
of the same event Our b o ok recommends the prior solution
If events have the same x co ordinate then we can handle this easily through a symb olic
p erturbation by sorting them lexicographically by x y This has the same e ect as
imagining that the sweep line is rotated in nitesimally counterclo ckwise
Sweep line status To store the sweep line status wemaintain a balanced binary tree or
p erhaps a skiplist whose entries are p ointers to the line segments stored in decreasing
order of y co ordinate along the currentsweep line
Normally when storing items in a tree the key values are constants Since the sweep
line varies we need variable keys To do this let us assume that each line segment
computes a line equation y mx b as part of its representation The key value in
each no de of the tree is a p ointer to a line segment To compute the y co ordinate of some
segment at the lo cation of the currentsweep line we simply take the current x co ordinate
of the sweep line and plug it into the line equation for this line
The op erations that we need to supp ort are those of deleting a line segment inserting a
line segment swapping the p osition of two line segments and determining the immediate
predecessor and successor of any item Assuming any balanced binary tree or a skiplist
these op erations can b e p erformed in O log n time each
The Complete Algorithm We can now present the complete plane sweep algorithm
Initially we insert all of the line segment endp oints into the event queue The initial
sweep status is empty
While the event queue is nonempty extract the next event in the queue There are three
yp e of event cases dep ending on the t
Lecture Notes CMSC M
Segment left endp oint Insert this line segmentinto the sweep line status based on the
y co ordinate of this endp ointandthey co ordinates of the other segments currently
along the sweep line Test for intersections with the segment immediately ab ove and
b elow
Segment right endp oint Delete this line segmentfromthesweep line status For the
entries immediately preceding and succeeding this entry test them for intersections
Intersection p oint Swap the two line segments in order along the sweep line For the
new upp er segment test it against its predecessor for an intersection For the new
lower segment test it against its successor for an intersection
Analysis The work done by the algorithm is dominated by the time sp ent up dating the various
data structures since otherwise we sp end only constant time p er sweep event We need to
counttwo things the numb er of op erations applied to each data structure and the amountof
time needed to pro cess each op eration
For the sweep line status there are at most n elements intersecting the sweep line at any time
and therefore the time needed to p erform any single op eration is O log n from standard
results on balanced binary trees
Since we do not allo w duplicate events to exist in the event queue the total numb er of elements
in the queue at any time is at most n I Since we use a balanced binary tree to store the
event queue each op eration takes time at most logarithmic in the size of the queue whichis
O log n I Since I n this is at most O log n O log n O log n time
Eacheventinvolves a constantnumb er of accesses or op erations to the sweep status or the
event queue and since each such op eration takes O log n time from the previous paragraph
it follows that the total time sp ent pro cessing all the events from the sweep line is
O n I log n O n I logn O n log n I log n
Thus this is the total running time of the plane sweep algorithm
Lecture DCEL s and Sub division Intersection
Tuesday Sep
Revised Sep Augmented Figure with alternative view
Reading Chapter in BKOS
Top ological Information In most applications of segmentintersection problems we are not in
terested in just a listing of the segmentintersections but wanttoknowhow the segments are
connected together Typically the plane has b een sub divided into regions and wewantto
store these regions in a way that allows us to reason ab out their prop erties e ciently
This leads to the concept of a planar subdivision or what might b e called a cel l complex in
top ology A planar sub division is de ned by a graph with straight line edges emb edded in
the plane so that no two edges intersect except p ossibly at their endp oints The condition
that the edges b e straight line segments may b e relaxed to allow curved segments but we
will assume line segments here This naturally sub divides the plane into regions The
dimensional vertices dimensional edges and dimensional faces We consider these three
typ es of ob jects to b e disjoint implying each edge is top ologically op en it do es not include it
endp oints and that eachfaceisopen itdoesnotincludeitsboundary Thereisalways one
unb ounded face that stretches to in nity Note that the underlying planar graph need not b e
a connected graph In particular faces maycontain holes and these holes maycon tain other
holes
Lecture Notes CMSC M
vertex edge
face
Figure Planar straight line sub division
Planar sub divisions will form the basic ob jects of many di erent structures that we will discuss
later this semester triangulations and Voronoi diagrams in particular so this is a go o d time to
consider them in greater detail The rst question is how should we represent such structures
so that they are easy to manipulate and reason ab out For example at a minimum wewould
like to b e able to list the edges that b ound each face of the sub division in cyclic order and we
would like to b e able to list the edges that surround eachvertex
Planar graphs There are a numberofimportant facts ab out planar graphs that we should discuss
Generally sp eaking an undirected graph is just a nite set of vertices and collection of
unordered pairs of distinct vertices called edges A graph is planar if it can b e drawn in the
plane the edges need not b e straight lines so that no two distinct edges cross eachother An
embedding of a planar graph is anysuchdrawing In fact in sp ecifying an emb edding it is
su cient just to sp ecify the counterclo ckwise cyclic list of the edges that are incident to each
vertex Since we are interested in geometric graphs our emb eddings will contain complete
geometric information co ordinates of vertices in particular
There is an imp ortant relationship b etween the number of vertices edges and faces in a planar
graph or more generally an emb edding of any graph on a top ological manifold but we will
stick to the plane Let V denote the number of vertices E the numb er of edgs F the number
of faces in a connected planar graph Euler s form ula states that
V E F
If weallow the graph to b e disconnected and let C denote the numb er of connected comp o
nents then wehave the more general formula
V E F C
In our example ab ovewehave V E F and C which clearly satis es this
formula An imp ortant fact ab out planar graphs follows from this
Theorem Astraight line planar graph with n vertices has O n edges and O n faces
Pro of Since the graph is a straight line graph there cannot b e multiple edges b etween the
same pair of vertices and there cannot b e self lo op edges Wewillprove the theorem in
the more general setting of planar graphs without these typ es of edges
First we will mo dify the graph in a way that it is still planar but no more edges can
b e added This mo di cation will only increase the numb er of edges and faces If the
graph is disconnected then add an edge b etween vertices in two di erent comp onents
thus reducing the numb er of connected comp onents by one and p ossibly increasing the
b er faces eachby one Rep eat until the graph is numb er of edges and p ossibly the num
connected
Lecture Notes CMSC M
Next if any face has more than three edges then add an edge through this face thus
increasing the numb er of edges and faces eachby one This cannot generally b e done
without adding curved lines unless if there are more than three vertices on the convex
hull of the vertex set But since we are not requiring that this b e a straight line planar
graph this is not an issue
Figure Mo di ed planar graph
Let E E and F F denote the numb er edges and faces in the mo di ed graph
The resulting graph has the prop erty that it has one connected comp onent every face is
b ounded by exactly three edges and each edge has a di erent face on either side of it
The last claim involves a little thought
If we countthenumb er of faces and multiply by then every edge will b e counted exactly
twice once by the face on either side of the edge Thus F E Euler s formula states
that V E F Using the fact that E F wehave
F
F F F V V
and using the face that F E wehave
E
V E E E V
Thus the numb er of faces is at most V and the numb er of edges is at most V
There are a numb er of reasonable representations that are used in practice The most widely
used on is the winged edge data structure Unfortunately it is probably also the messiest
There is another called the quad edge data structure which is quite elegant and has the nice
prop erty of b eing self dual We will discuss duality later in the semester We discuss a simple
and relatively elegant data structure called a doubly connectededge list or DCEL
Doubly connected Edge List Like most representations for planar graphs the DCEL is an edge
based representation but vertex and face information is also included for whatever geometric
application is using the data structure There are three sets of records whichmayberep
resented either as arrays or linked lists as you prefer a set of vertex records a set of edge
records and a set of facerecords Each undirected edge is represented bytwo directed edges
whic h our book calls half edges Here is what each record contains for the data structure
We will make a simplifying assumption that faces do not have holes inside of them This
assumption can b e satis ed byintro ducing some number of dummy edge joining each hole
either to the outer b oundary of the face or to some other hole that has b een connected to
the outer b oundary in this way Our text do es not make this assumption and so presents a
somewhat more general data structure How to add these dummy edges is left as an exercise
With this assumption it may b e assumed that the edges b ounding each face form a single
cyclic list
Lecture Notes CMSC M
Vertex Eachvertex stores its co ordinates along with a p ointer to any incident directed edge
that has this vertex as its origin v inc edge
Edge Each undirected edge is represented as two directed edges Each edge has a p ointer
to the opp ositely directed edge called its twin Each directed edge has an origin and
destination vertex Each directed edge is asso ciate with two faces one to its left and one
to its right
We store a p ointer to the origin vertex e org we can access the destination as the origin
of the twin edge We store a p ointer to the face to the left of the edge e left we can
access the face to the rightfromthetwin edge This is called the incident face We also
store the next and previous directed edges in counterclo ckwise order ab out the incident
face e next and e prev resp ectively
Face Each face f stores a p ointer to a single edge for which this face is the incident face
f inc edge See the text for the more general case of dealing with holes
e.twin e e.org
e.next e.prev e.left
DCEL Alternative view
Figure Doubly connected edge list
The gure shows twoways of visualizing the DCEL One is in terms of a collection of doubled
up directed edges An alternativeway of viewing the data structure that gives a b etter sense
of the connectivity structure is based on covering eachedgewithatwoelementblock one for
e and the other for its twin The next and prev p ointers provide links around each face of
the p olygon The next p ointers are directed counterclo ckwise around each face and the prev
p ointers are directed clo ckwise
Of course in addition the data structure may b e enhanced with whatever application data is
relevant In some applications it is not necessary to know either the face or vertex information
or b oth at all and if so these records may b e deleted See the b o ok for a complete example
For example supp ose that wewanted to enumerate the vertices that lie on some face f Here
is the co de
Vertex enumeration using DCEL
enumerate vertices Face f
Edge start f inc edge
Edge e start
do
output e org
e e next
while e start
Lecture Notes CMSC M
Merging sub divisions Let us return to the applications problem that lead to the segmentinter
section problem Supp ose that wehavetwo planar sub divisions S and S andwewantto
compute their overlay In particular this is a sub division whose vertices are the union of the
vertices of each sub division and the p oints of intersection of the line segments in the sub divi
sion Because we assume that each sub division is a planar graph the only new vertices that
could arise will arise from the intersection of two edges one from S and the other from S
Supp ose that each sub division is represented using a DCEL Can we adapt the plane sweep
algorithm to generate the DCEL of the overlaid sub division
The answer is yes Furthermore if we will not b e needing the original sub divisions it is
p ossible to do this by mo difying the existing sub divisions If not then the pro cess b egins by
making a copy of each The rst part of the problem is straightforward but p erhaps a little
tedious This part consists of building the edge and vertex records for the new sub division The
second part involves building the face records It is more complicated b ecause it is generally
not p ossible to know the face structure at the momentthatthesweep is advancing without
lo oking into the future of the sweep to see whether regions will merge You mighttryto
convince yourself of this The entire sub division is built rst and then the face information
is constructed and added later We will skip the part of up dating the face information see
the text
For the rst part the most illustrative case arises when the sweep is pro cessing an intersection
event In this case the segments come from the tw o di erent sub divisions The pro cess involves
the following steps and is illustrated in the gure
Create a new vertex v at the intersection p oint
Let a and b denote the edges that intersect oriented from left to right across the sweep
line We split eachofthetwointersecting edges byaddingavertex at the intersection
point Let a and b b e the new edge pieces They are created by the calls Split a
a and Split b b where the pro cedure is given b elow This pro cedure creates the
new edge links it into place and then returns the newly created edge After this the
edges have b een split but they are not linked to each other The edge constructor is
given the origin and destination of the new edge We will not give the details Also note
that the destination of a that is the origin of a s twin must b e up dated whichwehave
omitted
This pro cedure creates b oth the new edge and its twin The pro cedure a dest returns
the destination of a which is equivalentto a twin org
Split edge a edge a a is returned
a new edge v a dest create edge v a dest
a next a next a next prev a
a next a a prev a
a t a twin a t a twin the twins
a t prev a t prev a t prev next a t
a t prev a t a t next a t
Link the four edges together
Splice edge a edge a edge b edge b
a t a twin a t a twin find the twins
b t b twin b t b twin
a next b b prev a link the edges together
b t next a a prev b t
a t next b t b t prev a t
Lecture Notes CMSC M
b next a t a t prev b
a1 a1 b2
a1t split a1t b2t a2 b1 b1
b1t b1t a2t
splice
a1 b2
a1t b2t a2 b1
b1t a2t
Figure Up dating the DCEL
Lecture Polygon Triangulation
Thursday Sep
Revised Sep Moved last section to Lecture
Reading Chapter in BKOS
Simple Polygons Todaywe b egin study of the problem of triangulating p olygons Weintro duce
this problem byway of a cute example in the eld of combinatorial geometry
We b egin with some de nitions A polygonal curve is a nite sequence of line segments
called edges joined end to end The endp oints of the edges are vertices For example let
v v v denote the set of n vertices and let e e e denote a sequence of n
n n
edges where e v v A p olygonal curveisclosed if the last endp oint equals the rst
i i i
v v A p olygonal curveissimple if it is not self intersecting More precisely this means
n
that each edge e do es not intersect any other edge except for the endp oints it shares with its
i
adjacent edges
The famous Jordan curve theorem states that every simple closed plane curve divides the plane
into two regions the interior and the exterior Although the theorem seems intuitively
obvious it is quite di cult to prove We de ne a polygon to b e the region of the plane
b ounded by a simple closed p olygonal curve The term simple polygon is also often used to
emphasize the simplicity of the p olygonal curve We will assume that the vertices are listed in
counterclo ckwise order around the b oundary of the p olygon
Lecture Notes CMSC M
v6 v2 v0 v3
v1 v4 v5
polygonal curve closed but not simple simple polygon
Figure Polygonal curves
Art Gallery Problem Wesaythattwopoints x and y in a simple p olygon can see each other
or x and y are visible if the op en line segment xy lies entirely within the interior of P
If we think of a p olygon as the o or plan of an art gallery consider the problem of where to
place guards and howmany guards to place so that every p oint of the gallery can b e seen
by some guard Victor Klee p osed the following question Supp ose wehave an art gallery
whose o or plan can b e mo deled as a p olygon with n vertices As a function of n what is
the minimum numb er of guards that su ce to guard such a gallery Observe that are you are
told ab out the p olygon is the numb er of sides not its actual structure Wewanttoknow the
fewest numb er of guards that su ce to guard al l p olygons with n sides
A guarding set A polygon requiring n/3 guards.
Figure Guarding sets
Before getting into a solution let s consider some basic facts Could there b e p olygons for
which no nite numb er of guards su ce It turns out that the answer is no but the pro of is
not immediately obvious You might consider placing a guard at eachofthevertices Sucha
set of guards will su ce in the plane But to showhow counterintuitive geometry can b e it is
interesting to not that there are simple nonconvex p olyhedra in space suchthateven if you
place a guard at every vertex there would still b e p oints in the p olygon that are not visible to
any guard As a challenge try to come up with one with the fewest numberofvertices
An interesting question in combinatorial geometry is howdoesthenumb er of guards needed
to guard any simple p olygon with n sides grow as a function of n Ifyou play around with
the problem for a while trying p olygons with n sides for example you will
eventually come to the conclusion that bn c is the rightvalue The gure ab ove shows
aworst case example where bn c guards are required A cute result from combinatorial
geometry is that this number always su ces The pro of is based on three concepts p olygon
triangulation dual graphs and graph coloring The remarkably clever and simple pro of was
discovered by Fisk
Theorem The Art Gallery Theorem Given a simple p olygon with n vertices there exists
a guarding set with at most bn c guards
Lecture Notes CMSC M
Before giving the pro of we explore some asp ects of p olygon triangulations We b egin by
intro ducing a triangulation of P Atriangulation of a simple p olygon is a planar sub division
of the interior of P whose vertices are the vertices of P and whose faces are all triangles An
imp ortant concept in p olygon triangulation is the notion of a diagonal that is a line segment
between twovertices of P that are visible to one another A triangulation can b e viewed as
the union of the edges of P and a maximal set of noncrossing diagonals
Lemma Every simple p olygon with n vertices has a triangulation consisting of n diagonals
and n triangles
We ll refer you to the text for a pro of The pro of is based on the fact that given any n
vertex p olygon with n it has a diagonal This may seem utterly trivial but actually
takes a little bit of work to prove For example it fails to hold in space The addition
of the diagonal breaks the p olygon into two p olygons of say m and m vertices such that
m m n since b oth share the vertices of the diagonal Thus by induction there are
m m n n triangles total A similar argument holds for the case
of diagonals
It is a well known fact from graph theory that any planar graph can b e colored with colors
The famous color theorem This means that we can assign a color to eachofthevertices
of the graph from a collection of di erent colors so that no two adjacentvertices have the
same color However we can do even b etter for the graph wehave just describ ed
3 1 3 1 2 1 2 3 2
1 2 1 1 3 3 1
2 3 2
Figure Polygon triangulation and a coloring
Lemma Let T b e the triangulation graph of a triangulation of a simple p olygon Then T is
colorable
Pro of For every planar graph G there is another planar graph G called its dual The dual
G is the graph whose vertices are the faces of G andtwovertices of G are connected
by an edge if the two corresp onding faces of G share a common edge
Since a triangulation is a planar graph it has a dual shown in the gure b elow We
do not include the external face in the dual Because each diagonal of the triangulation
splits the p olygon into two it follows that each edge of the dual graph is a cut edge
meaning that its deletion would disconnect the graph As a result it is easy to see that
the dual graph is a freetree that is a connected acyclic graph and its maximum degree
is This would not b e true if the p olygon had holes
The coloring will b e p erformed inductively If the p olygon consists of a single triangle
then just assign any colors to its vertices An imp ortant fact ab out any free tree is that
it has at least one leaf in fact it has at least two Remove this leaf from the tree This
corresp onds to removing a triangle that is connected to the rest triangulation by a single
edge Such a triangle is called an ear By induction color the remaining triangulation
Lecture Notes CMSC M
an ear
Figure Dual graph of triangulation
When you add back the deleted triangle twoofitsvertices have already b een colored
and the remaining vertex is adjacent to only these twovertices Give it the remaining
color In this waytheentire triangulation will b e colored
We can now give the simple pro of of the guarding theorem
Pro of of the Art Gallery Theorem Consider any coloring of the vertices of the p olygon
At least one color o ccurs at most bn c time Otherwise we immediately get there are
more than n vertices a contradiction Place a guard at eachvertex with this color We
use at most b n c guards Observethatevery triangle has at least one vertex of each
of the tree colors since you cannot use the same color twice on a triangle Thus every
pointintheinterior of this triangle is guarded implying that the interior of P is guarded
A somewhat messy detail is whether you allow guards placed at a vertex to see along the
wall However it is not a di cult matter to push each guard in nitesimally out from his
vertex and so guard the entire p olygon
Lecture More Polygon Triangulation
Tuesday Sep
Reading Chapter in BKOS
The Polygon Triangulation Problem The art gallery exercise was intended as motivation for
the problem of triangulating a simple p olygon This op eration is used in many other applica
tions where complex shap es are to b e decomp osed into a set of disjoint simpler shap es There
are many applications in which the shap es of the triangles is an imp ortant issue e g skinny
triangles should b e avoided but there are equally manyinwhich the shap e of the triangle is
unimp ortant We will consider the problem of given an arbitrary simple p olygon compute
any triangulation for the p olygon
This problem has b een the fo cus of computational geometry for manyyears There is a simple
naive p olynomial time algorithm based on adding successive diagonals but it is not partic
ularly e cient There are very simple O n log n algorithms for this problem that have b een
known for manyyears A longstanding op en problem was whether there exists an O n time
algorithm The problem was solved by Chazelle in but the algorithm is so amazingly
intricate it could never comp ete with the practical but asymptotically slower O n log n al
gorithms In fact there is no known algorithm that runs in less than O n log n time that is
really practical enough to replace the standard O n log n algorithm whichwe will discuss
We will present one of many known O n log n algorithms The approachwe presenttoday
isatwo step pro cess although with a little cleverness b oth steps can b e combined into one
algorithm The rst is to consider the sp ecial case of triangulating a monotone polygon After
Lecture Notes CMSC M
this we consider howtoconvert an arbitrary p olygon into a collection of disjoint monotone
p olygons Then we will apply the rst algorithm to each of the monotone pieces The former
algorithm runs in O n time The latter algorithm runs in O n log n time
Monotone Polygons A p olygonal chain C is said to b e strictly monotone with resp ect to a given
line L ifany line that is orthogonal to L intersects C in at most one p oint Achain C
is monotone with resp ect to L if each line that is orthogonal to L intersects C in a single
connected comp onent Thus it mayintersect not at all at a single p oint or along a single
line segment A p olygon P is said to b e monotone with resp ect to a line L if its b oundary
P can b e split into twochains each of which is monotone with resp ect to L
x-monotone polygon Monotone decomposition
Figure Monotonicity
Henceforth let us consider monotonicity with resp ect to the x axis We will call these p olygons
horizontal ly monotone It is easy to test whether a p olygon is horizontally monotne How
a Find the leftmost and rightmost vertices min and max x co ordinate in O n time
b These vertices split the p olygon s b oundary into twochains an upper chain and a lower
chain Walk from left to right along eachchain verifying that the x co ordinates are
nondecreasing This takes O n time
As a challenge consider the problem of determining whether a p olygon is monotone in any
unsp eci ed direction This can b e done in O n time but is quite a bit harder
Triangulation of Monotone Polygons We can triangulate a monotone p olygon by a simple
variation of plane sweep metho d We b egin with the assumption that the vertices of the p oly
gon have b een sorted in increasing order of their x co ordinates For simplicitywe assume no
duplicate x co ordinates Otherwise break ties b etween the upp er and lower chains arbitrarily
and within a chain break ties so that the chain order is preserved Observe that this do es not
require sorting We can simply extract the upp er and lower chain and merge them as done
in mergesort in O n time
erything The idea b ehind the triangulation algorithm is quite simple Try to triangulate ev
you can to the left the currentvertex by adding diagonals and then remove the triangulated
region from further consideration
In the example there is obviously nothing to do until wehaveatleast vertices With vertex
it is p ossible to add the diagonal to vertex and so we do this In adding vertex we can
add the diagonal to vertex However vertices and are not visible to any other nonadjacent
vertices so no new diagonals can b e added When we get to vertex it can b e connected to
and The pro cess continues until reaching the nal vertex
The imp ortant thing that makes the algorithm e cient is the fact that when wearriveata
vertex the untriangulatedregion that lies to the left of this vertex alwayshasavery simple
structure This structure allows us to determine in constant time whether it is p ossible to
add another diagonal And in general we can add each additional diagonal in constant time
Lecture Notes CMSC M
2 2
1 3 1 3 4
7 7 2 2
1 3 4 1 3 4 5 5 8 6 13 6
12 12 11 11 7 7 2 2 10 10 1 3 4 1 3 4 9 9 5 8 5 8
6 6 13
Figure Triangulating a monotone p olygon
Since any triangulation consists of n diagonals the pro cess runs in O n total time This
structure is describ ed in the lemma b elow
Lemma Main Invariant For i let v be the vertex just pro cessed by the triangulation
i
algorithm The untriangulated region lying to the left of v consists of two x monotone
i
chains a lower chain and an upp er chain eachcontaining at least one edge If the chain
from v to u has two or more edges then these edges form a re ex chain that is a sequence
i
of vertices with interior angles all at least degrees The other chain consists of a
single edge whose left endp ointisu and whose right endp oint lies to the rightofv
i
We will provetheinvariantby induction As the basis case consider the case of v Here
u v and one chain consists of the single edge v v and the other chain consists of the other
edge adjacenttov
To prove the main invariant we will give a case analysis of how to handle the next event
involving v assuming that the invariant holds at v and see that the invariant is satis ed
i i
after eachevent has b een pro cessed There are the following cases that the algorithm needs
to deal with
Case v lies on the opp osite chain from v In this case we add diagonals joining v to all
i i i
the vertices on the re ex chain from v back to but not including u Now u v
i i
and the re ex chain consists of the single edge v v
i i
Case v is on the same chain as v Wewalk back along the re ex chain adding diagonals
i
joining v to prior vertices until we nd the rst that is not visible to v whichmay mean
i i
that we add no diagonals
As can b e seen in the gure this mayin volve connecting v to one or more vertices a
i
or it mayinvolve connecting v to no additional vertices b dep ending on whether the
i
Lecture Notes CMSC M
u
vi-1 Initial invariant
vi
u u u vi vi-1 v i-1 vi-1
Case 1 Case 2a Case 2b vi
Figure Triangulation cases
rst angle is less or greater than degrees In either case the vertices that were cut o
by diagonals are no longer in the chain and v b ecomes the new endp ointtothechain
i
Note that when we are done analogous to Graham s scan the remaining chain from v
i
to u isare exchain
How is this implemented The vertices on the re ex chain can b e stored in a stack Wekeep
a ag indicating whether the stackisontheupperchain or lower chain and assume that
with each new vertex we knowwhichchain of the p olygon it is on Note that decisions ab out
visibility can b e based simply on orientation tests involving v andthetoptwoentries on the
i
stack When we connect v by a diagonal we just p op the stack
i
Analysis We claim that this algorithm runs in O n time As wementioned earlier the sorted list
of vertices can b e constructed in O n time through merging The re ex chain is stored on a
stack In O time p er diagonal we can p erform an orientation test to determine whether to
add the diagonal and assuming a DCEL the diagonal can b e added in constant time Since
the numb er of diagonals is n the total time is O n
Monotone Sub division In order to run the ab ove triangulation algorithm we rst need to sub di
vide an arbitrary simple p olygon into monotone p olygons This is also done by a plane sweep
approach We will add a set of nonintersecting diagonals that partition the p olygon into
monotone pieces
Observe that the absence of x monotonicity o ccurs only at vertices in which in which the
interior angle is greater than degrees and b oth edges lie either to the left of the vertex or
b oth to the right Following the b o oks notation we call the rst typ e a merge vertex since
as the sweep passes over this vertex the edges seem to b e merging and the latter typ e a split
vertex
Let s discuss split vertices rst b oth edges lie to the right of the vertex When a split vertex
is encountered in the sweep there will b e an edge e of the p olygon lying ab ove and an edge
j
e lying b elow We might consider attaching the split vertex to left endp oint of one of these
k
two edges but it might b e that neither endp oint is visible to the split vertex But this would
imply that there is a closer vertex lying b etween e and e We will attach the split vertex to
j k
the closest vertex to the left of the sweep line whichliesbetween e and e Call this vertex
j k
the helper e We could have just as easily asso ciated the help er with e it do esn t really
j k
Lecture Notes CMSC M
helper(e1)
e1
e2
e3 helper(e3)
e4 e5
e6
helper(e5)
Figure Split vertices merge vertices and help ers
matter If there is no vertex b etween these edges then helper e is de ned to b e the left
j
endp ointofe or e that lies closer to the sweep line See the gure
j k
Note that helper e is de ned with resp ect to the current lo cation of the sweep line As the
j
sweep line moves its value changes Also it is only de ned when the sweep line intersects e
j
One way to visualize helper e is to imagine a trap ezoid with vertical sides and b ounded ab ove
j
and b elowby e and e sweeping to the left of the currentsweep line The rst vertex this
j k
sweeping trap ezoid hits is the help er These trap ezoids are illustrated in the gure ab ove
Here are basic elements of the plane sweep algorithm to x the split vertices We consider
merge vertices later
Events The endp oints of the edges of the p olygon These are sorted by increasing order of
x co ordinates Since no new events are generated the events may b e stored in a simple
sorted list i e no priority queue is needed
Sweep status The sweep line status consists of the list of edges that intersect the sweep line
sorted from top to b ottom Our b o ok notes that we actually only need to store edges
such that the p olygon lies just b elow this edge since these are the only edges that we
evaluate helper from
These edges are stored in a dictionary e g a balanced binary tree or a skip list so
that the op erations of insert delete nd predecessor and successor can b e evaluated in
O log n time each
Event pro cessing There are eventtyp es based on a case analysis of the lo cal structure of
edges around eachvertex Let v b e the currentvertex encountered bythesweep
Split vertex Searchthesweep line status to nd the edge e lying immediately ab ove
v Add a diagonal connecting v to helper e Add the two edges incidenttov in the
sweep line status and make v the help er of the lower of these two edges and make v
the new help er of e
Merge vertex Find the two edges incidenttothisvertex in the sweep line status they
must b e adjacent Delete them b oth Let e b e the edge lying immediately ab ove
them Make v the new help er of e
Lecture Notes CMSC M
Start vertex Both edges lie to the rightofv but the interior angle is less than
degrees Insert this vertex and its edges into the sweep line status Set the help er
of the upp er edge to v
End vertex Both edges lie to the left of v but the interior angle is less than
degrees Delete b oth edges from the sweep line status
Upp er chain vertex One edge is to the left and one to the right and the p olygon
interior is b elow Replace the left edge with the right edge in the sweep line status
Make v the help er of the new edge
Lower chain vertex One edge is to the left and one to the right and the p olygon
interior is ab ove Replace the left edge with the right edge in the sweep line status
Let e b e the edge lying ab ove here Make v the help er of e
e e e v v v v v v
Split Merge Start End Upper Lower
Figure Plane sweep cases
This only inserts diagonals to x the split vertices What ab out the merge vertices This
could b e handled by applying essentially the same algorithm using a reverse right to left
sweep It can b e shown that this will never intro duce crossing diagonals but it might attempt
to insert the same diagonal twice However the b o ok suggests a simpler approach Whenever
wechange a help er vertex check whether the original help er vertex is a merge vertex If so
the new help er vertex is then connected to the merge vertex by a new diagonal It is not hard
to show that this essentially has the same e ect as a reverse sweep and it is easier to detect
the p ossibility of a duplicate insertion in case the new vertex happ ens to b e a split vertex
There are many sp ecial cases what a pain but each one is fairly easy to deal with so the
algorithm is quite e cient As with previous plane sweep algorithms it is not hard to show
that the running time is O log n times the number of events In this case there is one event
per vertex so the total time is O n log n This gives us an O n log n algorithm for p olygon
triangulation
Lecture Intersection of Halfplanes
Thursday Sep
Revised Sept xed the clo ckwise bug in the last lemma
Reading Chapter in BKOS with some elements from Section and Section
Halfplane Intersection Todaywe b egin studying another very fundamental topic in geometric
computing and along the waywe will show a rather surprising connection b etween this topic
the topic of convex hulls whichwe discussed earlier
Any line in the plane splits the plane into two regions called halfplane one lying on either side
of the line Wemay refer to a halfplane as b eing either closed or open dep ending on whether
it contains the line itself For this lecture wewillbeinterested in closed halfplanes
Lecture Notes CMSC M
Howdowe represent lines and halfplanes Wemay assume each halfplane is expressed byan
inequality of the form
ax by c
and the halfplane is represented by the three co e cients a b c The line b ounding the
halfplane is ax by c Notethatifwemultiply a b andc by the same nonzero scalar
value then the line equation do es not change Thus we can think of this triple as a form of
homogeneous co ordinates for lines For example if c thenwe can divide the equation
through by c yielding an equation of the form a c x b c y which can b e expressed by
the homogeneous co ordinates a c b c
If the scalar multiple is negative then the sense of the inequalitywillbereversed Thus we
do not need a separate inequality of the form ax by c since we can just negate all three
co e cients to get the same e ect
Halfplane intersection problem The halfplane intersection problem is given a set of n closed
halfplanes H fh h h g compute their intersection A halfplane closed or op en is
n
a convex set and hence the intersection of anynumb er of halfplanes is also a convex set
Unlike the convex hull problem the intersection of n halfplanes may generally b e emptyor
even unb ounded A reasonable output representation might b e to list the lines b ounding the
intersection in counterclo ckwise order p erhaps along with some annotation as to whether the
nal gure is b ounded or unb ounded
Figure Halfplane intersection
How many sides can b ound the intersection of n halfplanes in the worst case Observe that
byconvexity each of the halfplanes can app ear only once as a side and hence the maximum
numb er of sides is n How fast can we compute the intersection of halfspaces As with
the convex hull problem it can b e shown through a suitable reduction from sorting that the
problem has a lower b ound of n log n
Who cares ab out this problem Our b o oks discusses a rather fanciful application in the
area of casting More realistically halfplane intersection and halfspace intersection in higher
dimensions are used as a metho d for generating convex shap e approximations In computer
graphics for example a b ounding b ox is often used to approximate a complex multi sided
p olyhedral shap e If the b ounding b ox is not visible from a given viewp oint then the ob ject
within it is certainly not visible Testing the visibility of a sided b ounding b oxismuch
easier than a multi sided nonconvex p olyhedron and so this can b e used as a lter for a
more costly test A b ounding b ox is just the intersection of axis aligned halfspace in
space If more accurate but still convex approximations are desired then wemay compute
the intersection of a larger numb er of tight b ounding halfspaces in various orientations as the
nal approximation
Solving the halfspace intersection problem in higher dimensions is quite a bit more challenging
than in the plane For example just storing the output as a cyclic sequence of b ounding planes
is not su cient In general some sort of adjacency structure ala DCEL s is needed
Lecture Notes CMSC M
We will discuss two algorithms for the halfplane intersection problem The rst is given in
the text For the other we will consider somewhat simpler problem of computing something
called the lower envelope of a set of lines and show that it is closely related to the convex hull
problem
Divide and Conquer Algorithm We b egin bysketching a divide and conquer algorithm for
computing the intersection of halfplanes The basic approachisvery simple
If n then just return this halfplane as the answer
Split the n halfplanes of H into subsets H and H of sizes b n c and d n e resp ectively
Compute the intersection of H and H eachby calling this pro cedure recursively Let
C and C b e the results
Intersect the convex p olygons C and C which mightbeunb ounded into a single convex
p olygon C and return C
The running time of the resulting algorithm is most easily describ ed using a recurrence that
is a recursively de ned equation If we ignore constant factors and assume for simplicity that
n is a p ower of then the running time can b e describ ed as
if n
T n
T n S n if n
where S n is the time required to compute the intersection of twoconvex p olygons whose total
complexityis n Ifwe can showthatS n O n then by standard results in recurrence s
it will follow that the overall running time T n isO n log n See Cormen Leiserson and
Rivest for a pro of
Intersecting Two Convex Polygons The only nontrivial part of the pro cess is implementing an
algorithm that intersects two convex p olygons C and C into a single convex p olygon Note
that these are somewhat sp ecial con vex p olygons b ecause they maybeemptyorunb ounded
We know that it is p ossible to compute the intersection of line segments in O n I log n
time where I is the number of intersecting pairs Twoconvex p olygons cannot intersect in
more than I O n pairs This follows from the observation that each edge of one p olygon
can intersect at most two edges of the other p olygon byconvexity This would given O n log n
algorithm for computing the intersection and an O n log n solution for T n which is not as
good as wewould like
However there is a very simple plane sweep approach to solving this problem Supp ose that
we p erform a left to right plane sweep to compute the intersection Observethatbyconvexity
the sweep line intersects each C in at most two p oints Therefore there are at most four
i
ts in the sweep line status at any time Thus we do not need a fancy dictionary for storing p oin
the sweep line status All the op erations can b e p erformed in constant time Furthermore you
do not need a priority queue for the events The only twoevents that are imp ortant are the
leftmost p oints of C and C note that they might stretchbackto To determine the
next event p oint you just need to check the four current edges for intersections and check the
four edges for their next endp oints Since there are only a constantnumb er of p ossibilities
eachevent can b e handled in O time
Lower Envelop es and Duality Next we consider a slightvariant of this problem to demonstrate
some connections with convex hulls These connections are very imp ortant to an understanding
of computational geometry andwe see more ab out them in the future These connections have
to do with a convex called point line duality Inanutshell there is a remarkable similarity
Lecture Notes CMSC M
potential next event
Figure Convex p olygon intersection
between how p oints interact with each other an how lines interact with each other Sometimes
it is p ossible to take a problem involving p oints and map it to an equivalent problem involving
lines and vice versa In the pro cess new insights to the problem may b ecome apparent
The problem to consider is called the lower envelope problem and it is a sp ecial case of the
halfplane intersection problem We are given a set of n lines L f g where is of
n i
the form y a x b Think of these lines as de ning n halfplanes y a x b each lying
i i i i
below one of the lines The lower envelope of L is the b oundary of the intersection of these
half planes There is also an upp er envelop e formed by considering the intersection of the
halfplanes lying ab ove the lines
Upper envelope
Lower envelope
Figure Lower and upp er envelop es
The lower envelop e problem is a restriction of the halfplane intersection problem but it an
interesting restriction Notice that any halfplane intersection problem that do es not involve
anyvertical lines can b e rephrased as the intersection of twoenvelop es a lower envelop e de ned
bythelower halfplanes and an upp er envelop e de ned bytheupward halfplanes
I will show that solving the lower envelop e problem is essentially equivalent to solving the
upp er convex hull problem In fact they are so equivalent that exactly the same algorithm
will solve b oth problems without changing even a single character of co de All that changes
is the wayinwhichyou view the two problems
Duality Let us b egin by considering lines in the plane Each line can b e represented in a number
of ways but for now let us assume the representation y ax b for some scalar values a and
b We cannot representvertical lines in this way and for nowwe will just ignore them Later
Lecture Notes CMSC M
in the semester we will x this up Why did we subtract b We ll see later that this is just a
convenience
Therefore in order to describ e a line in the plane you need only give its two co ordinates a b
In some sense lines in the plane can b e thoughtofaspoints in a new plane in which the
co ordinate axes are lab eled a b rather than x y Thus the line y x corresp onds
to the p oint in this new plane Eachpoint in this new plane of lines corresp onds to
a nonvertical line in the original plane We will call the original x y plane the primal plane
and the new a b plane the dual plane
What is the equation of a line in the dual plane Since the co ordinate system uses a and b
wemight write a line in a symmetrical form for example b a where the values and
could b e replaced byany scalar values
Consider a particular p oint p p p in the primal plane and consider the set of all
x y
nonvertical lines passing through this p oint Anysuch line must satisfy the equation p
y
ap b The images of all these lines in the dual plane is a set of p oints
x
j p ap bg L f a b
y x
f a b j b p a p g
x y
Notice that this set is just the set of p oints that lie on a line in the dual a b plane And
this is whywe negated b Thus not only do lines in the primal plane map to p oints in the
dual plane but there is a sense in whicha point in the primal plane corresp onds to a line in
the dual plane
Tomake this all more formal we can de ne a function that maps p oints in the primal plane
to lines in the dual plane and lines in the primal plane to p oints in the dual plane We denote
it using a asterisk as a sup erscript Thus given p oint p p p and line y ax b
x y
in the primal plane wede ne and p to b e a p oint and line resp ectively in the dual plane
de ned by
a b
p b p a p
x y
r*:(b=-2a+1)
y b p:(-2,2) s*:(b=a+1) q*:(b=-a-1) q:(-1,1) L*:(0,1) x a
L:(y=0x-1) r:(-2,-1) s:(1,-1)
p*:(b=-2a-2)
Primal plane Dual plane
Figure Dual transformation
We can de ne the same mapping from dual to primal as well Dualityhasa number of
interesting prop erties each of which is easy to verify by substituting the de nition and a little
algebra
Lecture Notes CMSC M
Self Inverse p p
Order reversing Point p lies ab ove on b elow line in the primal plane if and only if line
p passes b elow on ab ove p oint in the dual plane resp ectively
Intersection preserving Lines and intersect at p oint p if and only if line p passes
through p oints and in the dual plane
Collinearity Coincidence Three p oints are collinear in the primal plane if and only if their
dual lines intersect in a common p oint
To nish things up weneedtomake the connection b etween the upp er convex hull of a set of
p oints and the lower envelop e of a set of lines
Lemma Let P b e a set of p oints in the plane The counterclo ckwise order of the p oints along
ex hull of P is equal to the left to right order of the sequence of the upp er lower conv
lines on the lower upp er envelop e of the dual P
Pro of We will prove the result just for the upp er hull and lower envelop e since the other case
is symmetrical For simplicity let us assume that no three p oints are collinear Observe
that a necessary and su cient condition for a pair of p oints p p to form an edge on the
i j
upp er convex hull is that the line that passes through b oth of these p oints has every
ij
other p ointinP strictly b eneath it
A necessary and su cient condition that these lines and p Consider the dual lines p
j i
are adjacentonthelower envelop e is that the dual p ointatwhich they meet lies
ij
b eneath all of the other dual lines in P
The order reversing condition of duality assures us that the primal condition o ccurs if and
only if the dual condition o ccurs Therefore the sequence of edges on the upp er convex
hull is identical to the sequence of vertices along the lower envelop e
As wemovecounterclo ckwise along the upp er hull observe that the slop es of the edges
increase monotonically Since the slop e of a line in the primal plane is the a co ordinate
of the dual p oint it follows that as wemove counterclo ckwise along the upp er hull we
visit the lower envelop e from left to right
One rather cryptical feature of this pro of is that although the upp er and lower hulls app ear
to b e connected the upp er and lower envelop es of a set of lines app ears to consist of two
disconnected sets To make sense of this we should interpret the primal and dual planes from
the p ersp ective of pro jective geometry and think of the rightmost line of the lower envelop e as
wrapping around to the leftmost line of the upp er envelop e and vice versa We will discuss
pro jective geometry later in the semester
Another interesting question is that of orientation We know the the orientation of three p oints
is p ositive if the p oints have a counterclo ckwise orientation What do es it mean for three lines
to have a p ositiveorientation The de nition of line orientation is exactly the same in terms
of a determinant of the co e cients of the lines
Lecture Linear Programming
Tuesday Sep
Reading Chapter in BKOS
Linear Programming Last time we considered the problem of computing the intersection of n
halfplanes and presented in optimal O n log n algorithm for this problem In many applica
tions it is not imp ortant to knowtheentire p olygon or generally the entire p olytop e in higher
dimensions but only to nd the p oin t that that is extreme in some given direction
Lecture Notes CMSC M
One particularly imp ortant application is that of linear programming In linear programming
LP we are given a set of linear inequalities or contraints whichwemay think of as de ning
a p ossibly empty p ossibly unb ounded p olyhedron in space called the feasible region and
we are given a linear objective function whichistobeminimized or maximized sub ject to the
given constraints A typical description of a d dimensional linear programming problem might
be
Maximize c x c x c x
d d
Sub ject to a x a x b
d d
a x a x b
d d
x a x b a
n d d n n
where a c and b are given real numbers
i j i i
From a geometric p ersp ective the feasible region is the intersection of halfspaces and hence
is a convex p olyhedron We can think of the ob jective function as a vector c and the problem
is to nd the p oint of the feasible region that is furthest in the direction c called the optimal
vertex Inmany of our examples we will imagine that the vector c is p ointing down and hence
the problem is just that of nding the lowest p oint of the feasible region
Linear programming is a very imp ortant technique used in solving large optimization problems
Typical instances mayinvolvehundreds to thousands of contraints in very high dimensional
space It is without doubt one of the most imp ortantformulations of general optimization
problems
We will restrict ourselves to low dimensional instances of linear programming There are
anumber of interesting optimization problems that can p osed as a low dimensional linear
programming problem or as closely related optimization problems One whichwe will see
later is the problem of nding a minimum radius circle that encloses a given set of n points
Normally this extreme p ointwillbeavertex of the feasible region p olyhedron but there
some other p ossibilities as well The feasible region ma ybeempty in which case the linear
programming problem is said to b e infeasible and there is no solution It maybeunb ounded
and if c p oints in the direction of the unb ounded part of the p olyhedron then there may
solutions with in nitely large values of the ob jective function In this case there is no nite
solution and the LP problem is said to b e unbounded Finally observe that in degenerate
situations it is p ossible to have an in nite numb er of nite optimum solutions b ecause an
edge or face of the feasible region is p erp endicular to the ob jective function vector In such
instances it is common to break ties by requiring that the solution b e lexicographically maximal
e g among all maximal solutions take the one with the lexicographically maximum vector
This is exactly analogous to a applying a symb olic p erturbation by rotating space
dimensional LP The principal metho ds used for solving high dimensional linear programming
problems are the simplex algorithm and various interior point methods The simplex algo
rithm works by nding a vertex on the feasible p olyhedron then then walking edge by edge
downwards until reaching a lo cal minimum Byconvexity thelocalminimum is the global
minimum It has b een long known that there are instances where the simplex algorithm runs
in exp onential time The question of whether linear programming was even solvable in p oly
nomial time was op en until a little over years ago when Karamarkar showed a p olynomial
time algorithm based on moving through the interior of the feasible region It should b e men
tioned that the metho d is p olynomial in the number of numberofconstraints the dimension
and the numb er of bits of precision in the numb ers Although this is go o d enough for prac
tice one of the ma jor op en problems left in the area is whether there is a strongly polynomial
time algorithm that is p olynomial without the assumption that the numb ers are of b ounded
precision
Lecture Notes CMSC M
Let us consider the problem just in the plane Here weknow that there is an O n log n
algorithm based on just computing the feasible p olyhedron and nding its lowest vertex
However since we are only interested in one p oint on the p olyhedron it seems that we might
hop e to do b etter We will show that dimensional LP problems can b e solved in O n time
and in fact more generally any xed dimensional linear programming can b e solved in O n
time The problem with using such an algorithm in practice is that the constant factors grow
very rapidly with the dimension faster than p olynomial time We will discuss one simple
one later whose running time is O d n Thus these algorithms are only acceptable for relative
small dimensional problems
Incremental Construction The algorithms that we will discuss for linear programming are very
simple and are based on a metho d called incremental construction Plane sweep and incre
mental construction are the two pillars of computational geometry and so this is another
interesting reason for studying the linear programming problem
Assume that we are given a set of n linear inequalities halfplanes h h of the form
n
a x a y b
i x i y i
and a nonzero ob jective function giv en byvector c c c The problem is to nd p
x y
p p that is feasible and maximizes the dot pro duct c p c p In our illustrations we
x y x x y y
will assume that c is p ointing downwards e g c This is our b o ok s convention
It is a bit confusing b ecause note that as you decrease the y co ordinate of a p oint you
increase the ob jective function
Let us supp ose for simplicity that the LP problem is b ounded and furthermore we can nd
two halfplanes whose intersection a cone is b ounded with resp ect to the ob jective function
Let us assume that the halfplanes are renumb ered so that these are h and h and let v
denote this rst optimum vertex
We will then add halfplanes one byone h h and with each addition we will up date
the current optimum vertex After the addition of fh h h g let C denote the current
i i
feasible region their common intersection and let v denote the current feasible vertex Our
i
job is to up date v with each new addition Note that we will not compute the C s explicitly
i i
since doing so would require O n log n time but it is handy to think ab out their existence
for the purp oses of proving things Notice that with each new constraint the feasible region
b ecomes smaller and hence the value of the ob jective function at optimum vertex can only
decrease
There are two cases that can arise when h is added The rst is that v is already satis es
i i
constraint h that is it lies within this halfplane If so then it is easy to see that the optimum
i
vertex do es not change that is v v On the other hand if v violates constraint h
i i i i
then we need to nd a new optimum vertex Where shall we lo ok for it
The imp ortant observation is that assuming that the feasible region is not empty the new
optimum vertex must lie on the line that b ounds h Call this line The b o ok proves this
i i
formally Intuitively if the new optimum v ertex did not lie on thendraw a line segment
i
from v to the new optimum Observe that as you walk along this segment the value
i
of the ob jective function is decreasing monotonically by linearity and that this segment
must cross b ecause it go es from b eing infeasible with resp ect to h to b eing feasible Thus
i i
it is maximized at the crossing p oint which lies on Convexity a linearity are b oth very
i
imp ortant for the pro of
So this leaves the question Howdowe nd the optimum vertex lying on line Thisturns
i
out to b e a dimensional LP problem Simply intersect each of the halfplanes with this line
Eachintersection will take the form of a ray that lies on the line We canthinkofeachray
Lecture Notes CMSC M
as representing an interval unb ounded to either the left or to the right All we need to do is
to intersect these intervals and nd the p oint that maximizes the ob jective function that is
the lowest p oint Computing the intersection of a collection of intervals is very easy and can
b e solved in linear time We just need to nd the smallest upp er b ound and the largest lower
b ound
Lecture More Linear Programming
Thursday Oct
Reading Chapter in BKOS
Recap Last time we presented a simple incremental algorithm for linear programming in the plane
We are given a set of n halfspaces h h h and an ob jective function describ ed bya
n
vector c The goal is to nd the p oint p that has the maximum dot pro duct with c that is is
furthest in direction c sub ject to the constraint that it lies in the intersection of the halfspaces
the convex feasible region
Let us recall the algorithm last time from the p ersp ective of linear programming in d
dimensional space We are given a set of halfspaces and wewant to nd the p oint that
lies in the intersection of the halfspaces and is most extreme in some direction We start by
nding an initial set of d halfspaces that de ne an initial b ounded solution We renumb er the
halfspaces so that these are the rst d halfspaces Let v b e the initial optimum vertex Then
d
we add halfspaces one at a time On adding the h we test whether v is feasible with resp ect
i i
to h Ifso we set v v Otherwise we claim that the new optimum must lie on the
i i i
hyp erplane that b ounds h Weintersect all the previous i halfspaces with this hyp erplane
i
thus reducing the dimension of the problem by pro ject the ob jective function vector onto
this hyp erplane and then recursively solve the resulting d dimensional LP problem The
recursion b ottoms out when we get down to a dimensional LP problem which is just the
problem of intersecting a collection of intervals Thus wesolvead dimensional problem by
rep eatedly reducing it to a d dimensional problem until wegetdown to a dimensional
problem which is trivial to solve
Analysis What is the worst case running time of this algorithm in the planar case There are
roughly n halfspace insertions In step i wemay either nd that the current optimum vertex
is feasible in which case the pro cessing time is constant On the other hand if the current
optimum vertex is infeasible then wem ust solve a dimensional linear program with i
constraints In the worst case this second step o ccurs all the time and the overall running
time is given by the summation
n n
X X
n n
i O n i
i i
from standard results on summations see Cormen et al
Randomized Algorithm The O n time is not very encouraging considering that we could
compute the entire feasible region in O n log n time But we presented it b ecause it leads to
avery elegant randomized O n time algorithm The algorithm op erates in exactly the same
way but we insert the halfplanes in random order This is called a randomizedincremental
algorithm For example just prior to calling the algorithm given last time we call a pro cedure
for randomly p ermuting the initial input list Our text gives an example
We analyze the running time in the exp ected case where weaverage over all n p ossible p erm u
tations It is imp ortant to observe that the analysis makes no assumptions ab out the nature
Lecture Notes CMSC M
of the input only on the random p ermutation used In other words there are no bad inputs
only bad random p ermutations The fact that the rst two halfplanes were chosen sp ecially
adds a confusing element to this To b e correct we should think of them as b eing xed and
we randomize over the n p erm utations of the remaining halfplanes However we ll try
to ignore this complicating factor b elow
To analyze the algorithm we use an interesting technique called backward analysis Before
discussing backward analysis think of howa typical probabilistic analysis e g the one used
for quicksort works At each stage the algorithm has a set of random choices that might
b e made For example in quicksort the choice is which element to select as the pivot
Eachchoice has a certain probability of o ccuring Typically all choices are equally likely
We analyze the e ect of eachchoice on the running time and then multiply each times it
probability of o ccuring and sum everything up to get the exp ected running time
The di culty of applying this to a problem like the linear programming problem is that at the
next stage it is very di cult to predict what is going to happ en since it dep ends heavily on
what has gone on up to this p oint In a backwards analysis rather than asking what is the
e ect of the next random choice on the running time we ask what was the e ect of the previous
random choice This may sound just like like a semantical distinction but the remarkable fact
is that it really makes things much easier to analyze The reason is that wehavefullknowledge
of the past
Consider the example shown b elow Wehave i random halfplanes that have b een added
so far and v v is the current optimum vertex see the gure Rather than consider which
i
halfplane will b e added next let s consider whichonewas added last For example if h was
the last to b e added then imagine the picture without h Prior to this v would have b een the
optimum vertex Therefore if h was added last then it would have b een added with O
cost This applies to all the other halfplanes as well except those that de ne v namelyh and
h For example if h was added last then prior to its addition wewould have had a di erent
optimum vertex namely v Thus if h or h were added last then it would incur the higher
running time of O i to solve the dimensional LP problem
h5
h13 c
h4
v h8 v’
h15 h
11 h3
Figure Backward analysis
What is the probabilityofeach of these events o ccuring The imp ortant observation here is
that the order in which halfplanes are inserted in completely indep endent of the geometric
structure Among the halfplanes in the diagram so far fh h h h h h h g there are
an equal number of permutations of these ending with h as there are ending with h as there
Lecture Notes CMSC M
are with any of them in particular Here w e are ignoring the two initial halfplanes which
are xed Imagine that they lie outside of the p ortion shown in the gure Thus each halfplane
has an equal probabilityof i of b eing the last halfplane to b e added Among the i halfplanes
present two of them the two that b ound the currentoptimum vertex would incur a cost of
O i on the running time and the other i would incur a cost of O Thus there is
a i chance of incuring time O i and a i i chance of incuring time O Thus the
exp ected time for the i th stage is
i
O i O O
i i
If we sum this up over all n stages wejustgetO n as the total exp ected running time by
the linearity of exp ectation
This may all seem like a bit of probabilistic magic If you nd it unb elievable you are
encouraged to draw a set of halfplanes and exp eriment with di erent orders to convince yourself
at least at an intuitivelevel of the soundness of this technique We will b e applying it on
other randomized algorithms this semester
Unb ounded LP s We b egan by assuming that we could always nd two or generally d initial
halfplanes to provide us with an initial b ounded optimal vertex Wenever dealt with this
issue but we claim that it is always p ossible to determine either that the LP is unb ounded
or to determine a pair of halfplanes that b ound the feasible region in O n time Our b o ok
presents an algorithm for doing this in the plane This algorithm has the nice feature that it
generalizes to higher dimensions although they leave the generalization as an exercise
In the planar case there is a much simpler algorithm than this Consider the outward p oint
ing normal vectors for the various halfplanes imagine they are scaled to unit length They
partition the unit circle into n angular sectors Now consider the angular sector that contains
the ob jectivevector c Consider the two halfplanes lying immediates clo ckwise and counter
clo ckwise from c If the angle b etween these two normals is less than degrees then the no
matter how far apart they are these two halfplanes will converge at a p oint p and the feasible
set if it is nonempty will lie in the cone b ounded by these halfplanes
c
Figure Boundedness test
Intersecting with a hyp erplane Our b o ok do es not talk ab out howwegoaboutintersecting
given set of halfspaces with a hyp erplane Su ce it to say that it is a geometric primitive
that can b e evaluated in O d time p er hyp erplane For completeness let s consider how this
might b e done
Supp ose that the halfspace that we just added was given by the inequality
h a x a x a x b
i i i i d d i
The corresp onding hyp erplane is given by the equation
a x a x a x b
i i i i d d i
Lecture Notes CMSC M
We can express this more succinctly in matrix notation Let A denote the d vector
i
consisting of the i th rowoftheA matrix A a a a The inequalitymaybe
i i i i d
written A x b where x is a d vector and b is a scalar
i i i
Wewanttointersect the other halfspaces with this hyp erplane Furthermore wewould like
to represent the result as a LP in d dimensional problem Observe that after intersection
the hyp erplane still resides in d space
The idea is to apply one step of Gauss elimination using the equation of to eliminate a
i
variable from all the other inequalities Let us assume that a if not select a dimension
i
k such that a andswap with the rst Consider an arbitrary constraint h that we
i k j
wish to intersect with
i
h A x b
j j j
To eliminate the rst dimension from h wemultiply A by a a and subtract from A
j i j i j
do the same for b and b
i j
a
j
A A A
i j
j
a
i
a
j
b b b
j i
j
a
i
To see that this works supp ose that x is a p oint on the hyp erplane implying that A x b
i i i
Supp ose that x satis ed constraint h Thenwehave
j
a
j
A x x A A
i j
j
a
i
a
j
A x A x
j i
a
i
a
j
b b b
i j
j
a
i
Thus every p ointonthehyp erplane satis es the mo di ed constraint if and only if it satis es
the original constraint A similar elimination can b e p erformed to the ob jectivevector c It
is also easy to show that as is standard in Guass elimination the rst term of each equation
vanishes so we are left with a d dimensional problem Reversing the pro cess allows us to
pro ject the d dimensional solution backinto d space
Lecture Orthogonal Range Searching
Tuesday Oct
Chapter in BKOS
Range Queries We shift our fo cus from algorithm problems to data structures for the next few
lectures In general we consider the question given a collection of ob jects prepro cess them
storing the results in a data structure of some variety so that queries of a particular form
can b e answered e ciently Generally we measure data structures in terms of two quantities
the time needed to answer a query and the amount of space needed by the data structure
Often there is a tradeo b etween these twoquantities but most of the structures that we will
be interested in will have either linear or near linear space Prepro cessing time is an issue of
secondary imp ortance but most of the algorithms we will consider will have either linear of
O n log n prepro cessing time
Lecture Notes CMSC M
Range queries are queries of the general form given a set P of p oints list or count or compute
some commutative function of the subset of P lying within a given region Regions maybe
rectangles triangles halfspaces circles etc There are many data structures for pro cessing
range queries dep ending on the typ e of region
An imp ortant concept b ehind all geometric range searching is that the subsets that can b e
formed by simple geometric ranges is much smaller than the set of p ossible subsets called the
power set of P Given a particular class of ranges a range space can b e describ ed abstractly as
apair P R consisting of the p oints P and the collection R of all subsets of P that b e formed
by ranges of this class For example the following gure shows the range space assuming
rectangular ranges for a set of p oints in the plane In particular note that the sets f g and
f g cannot b e formed by rectangular ranges
R = { {}, {1}, {2}, {3}, {4}, 2 {1,2}, {1,3}, {2,3},{2,4},{3,4}, 1 {1,2,3}, {1,3,4}, {2,3,4} , {1,2,3,4} } 3
4
Figure Rectangular range space
Todaywe consider orthogonal rectangular range queries that is ranges de ned by rectangles
whose sides are aligned with the co ordinate axes One of the nice thing ab out rectangular
ranges is that they can b e decomp osed into a collection dimensional searches
One dimensional range queries Before consider how to solve general range queries let us con
sider how to answer dimension range queries or interval queries We are given a set of
p oints P fp p p g on the line and given an interval x x rep ort all the p oints
n lo hi
lying within the interval Our goal is to nd a data structure that can answer these queries
in O log n k time where k is the numb er of p oints rep orted an output sensitive result
Range counting queries can b e answered in O log n time with minor mo di cations
Clearly one way to do this is to simply sort the p oints and apply binary search to nd the
rst p ointofP that is greater than or equal to x and less than or equal to x and then
lo hi
list all the p oints b etween However this will not generalize to higher dimensions
A basic approach to solving almost all range queries is to represent P as a collection of canonical
subsets fS S S g each S S where k is generally a function of n and the typ e of
k i
ranges such that the answer to any query can b e expressed as a disjoint union of a small
numb er of canonical subsets Note that these subsets mayoverlap each other The trickto
solving a range searching problem is to identify a collection of canonical subsets having these
prop erties and then nding the prop er subsets for a given range
Supp ose that the p oints of P are sorted in increasing order and then stored in the leaves of a
balanced binary tree We can asso ciate each no de of this tree with the canonical subset of
p oint stored in the leaves that are descendents of this no de This gives rise to O n canonical
subsets Further we claim that the canonical subsets corresp onding to any range can b e
computed in O log n time
Given anyinterval x x we search the tree to nd the leaf whose key is greater than or
lo hi
equal to x and the leaf whose key is less than or equal to x Then wetake all of the
lo hi
Lecture Notes CMSC M
maximal subtrees lying b etween these two search paths In particular the two search paths
will generally travel along some common subpath until reaching a no de v where the two
split
paths split After this each search path go es its own way When the left path to x travels
lo
along a left edge of the tree then the right subtree lies entirely within the interval Similarly
when the right path to x travels along a right edge then the left subtree lies entirely within
hi
the interval The leaves of these maximal subtrees are the desired canonical subsets whose
disjoint union is the answer to the range query Toanswer a range rep orting query we simply
traverse these subtrees rep orting the p oints of their leaves Toanswer a range counting query
we store the total number of points in each subtree and sum all of these counts
Since the search paths are of length at most log n itfollows that there are O log n total
canonical subsets Thus the range counting query can b e answered in O log n time For
rep orting queries since the leaves of each subtree can b e listed in time that is prop ortional to
the number of leaves in the tree a basic fact ab out binary trees it follows that the total time
in the searchisO log n k where k is the number of points rep orted
1,31
17,31 1,15 {9,12,14,15} {17,20,22,23}
9,15 17,23 25,31 1,7 {4,5} {25,27}
4,7 9,12 14,15 17,20 22,23 25,27 29,31 1,3 {3}
13 4 7 9 1214 1517 2022 232527 2931
xlo=2 xhi=28
Figure Canonical sets for interval queries
Thus dimensional range queries can b e answered in O log n time using O n storage How
can we extend this to higher dimensional range queries
Kd trees The natural question is how to extend dimensional range searching to higher dimen
sions First we will consider kd trees This data structure is easy to implement and quite
practical and useful for many di erenttyp es of searching problems nearest neighb or search
ing for example However it is not the asymptotically most e cient solution for the orthogonal
range searching as we will see later
The idea b ehind a kd tree is to extend the notion of a one dimension tree but alternate is
using the x ory co ordinates to split on In general dimension the kd tree cycles among the
various p ossible splitting dimensions
Eachinternal no de of the kd tree is asso ciated with twoquantities a splitting dimension either
x or y and a splitting value s Ithastwo subtrees Since the splitting dimension alternates
between x and y some implementations do not store this explicitly but keep track of it while
traversing the tree If the splitting dimension is x then all p oints whose x co ordinates are less
than or equal to s are stored in the left subtree and p oints whose x co ordinates are greater
than or equal to s are stored in the right subtree If a p oint s co ordinate is equal to s thenwe
reserve the right to store it on either side This is done to allow us to balance the number of
p oints in the left and right subtrees When a single p oint or more generally a small constant
numberofpoints remains we store it in a leaf
How is the splitting value chosen To guarantee that the tree is balanced the most common
metho d is to let the splitting value b e the median splitting co ordinate The resulting tree will
Lecture Notes CMSC M
have O log n height
It is p ossible to build a kd tree in O n log n time by a simple recursive pro cedure The most
costly step of the pro cess is determining the median co ordinate However if we presort the
p oints into two cross referenced lists one sorted by x and the other sorted by y thenitis
an easy matter to nd the median at each step The two lists can then b e split in O n
time each where n is the numb er of remaining p oints This leads to a recurrence of the form
T n T n n which solves to O n log n
Searching the kd tree Toanswer an orthogonal range query we pro ceed as follows Let R denote
the query rectangle Assume that the current splitting line is vertical The horizontal case
is similar Let v denote the current no de in the tree Observe that each no de of the tree is
naturally asso ciated with a rectangular region of space Call this reg v The search pro ceeds
as follows If v is a leaf then wecheckthepoin t s stored in v as to whether it lies in R If
so we rep ort count them If v is an internal no de then we rst consider the left subtree If
its region lies entirely within R then we call a di erent routine to enumerate all the p oints in
this subtree or for a counting query we return a precomputed count of the numb er of p oints
in this subtree If the left child s region partially overlaps R then we search it recursively If
the left child s region do es not overlap R at all then we ignore it We do the same for the
rightchild of v
p
n Howmany no des do es this metho d visit altogether We claim that the total number is O
k where k is the numb er of p oints rep orted
Theorem The times needed to answer an orthogonal range query using kd tree with n p oints
p
in O n k where k is the numb er of rep orted p oints
Pro of First observe that for each subtree with m points we can enumerate the p oints of this
subtree in O m time by a simple traversal Thus it su ces to b ound the number of
no des visited in the search not counting subtree enumeration
Tocount the numb er of no des visited rst observe that if we visit a no de then its
asso ciated region must intersect one of the four sides of the rectangle range Otherwise
it lies entirely inside or outside and so is not visited We will do this separately for each
of the four sides and multiply the result byfour
Rather than consider the line segment de ning a side of the range rectangle we consider
the numb er of no des visited if we had applied an orthogonal range query to the entire
line on which this side lies This will only overestimate the numb er of no des visited
Because the kd tree pro cesses even and o dd levels di erently itwillbeimportantto
analyze twolevels at a time or generally d levels at a time for dimension d Let Q n
denote the query time for visiting a no de v that has n p oints descended from it Consider
any orthogonal line that intersects the region asso ciated with v Thekey observation
tersect at most two of the four subregions is that any horizontal or vertical line can in
asso ciated with the grandchildren of v The other two will either b e entirely contained
within the halfspace or lie entirely outside the halfspace In either case the algorithm will
not make a recursive call on them Since eachchild has half as manypoints b ecause
we split at the median the numberofpoints in each grandchild is roughly n Thus
we make at most two recursive calls on subtrees of size n This gives the following
recurrence
if n
Q n
Q n otherwise
p
n It is an easy pro cess to expand this recursion and derive the fact that it solves to O
p
Including the factor of four we still haveanO n b ound on the total number of nodes
p
visited and adding the enumeration time the total time is O n k as desired
Lecture Notes CMSC M
Lecture More Orthogonal Range Searching
Thursday Oct
Read Chapter in BKOS
Orthogonal Range Trees Last time wesaw that kd trees could b e used to answer orthogonal
p
range queries in the plane in O n k time Todaywe consider a b etter data structure
called orthogonal range trees
d
An orthogonal range tree is a data structure which in all dimensions d uses O n log n
d
space and can answer orthogonal rectangular range queries in O log n k time where k
is the numb er of p oints rep orted Prepro cessing time is the same as the space b ound Thus in
the plane we can answer range queries in time O log n and space O n log n We will present
the data structure in two parts the rst is a version that can answer queries in O log n time
in the plane and then we will showhowtoimprove this in order to strip o a factor of log n
from the query time
The data structure is based on the concept of leveling a complex multi dimensional searchinto
a constantnumb er of simpler range searches In this case we will reduce a d dimensional range
search to a series of dimensional range searches
Supp ose you hav e a query which can b e stated as the intersection of a small numb er of simpler
queries For example a range query in the plane can b e stated as two range queries Find all
the p oints whose x co ordinates are in the range x x and all the p oints whose y co ordinates
lo hi
are in the range y y Let us consider how to do this for dimensional range queries and
lo hi
then consider how to generalize the pro cess First build a range tree for the rst range query
which in this case is just a dimensional range tree for the x range Recall that this is just
a balanced binary tree on these p oints sorted by x co ordinates Also recall that eachnodeof
this binary tree is implicitly asso ciated with a canonical subset of p oints These are the p oints
lying in the subtree ro oted at this no de The answer to any dimensional range query can
b e represented as the disjoint union of a small collection of m O log n canonical subsets
fS S S g where each subset corresp onds to a no de in the search This constitutes the
m
rst level of the search tree For the second level for eachnode v in this x range tree we build
an auxiliary tree each of whichisay co ordinate range tree whichcontains all the p oints in
the canonical subset asso ciated with v
h that each no de p oints to auxiliary Thus the data structure consists of a x range tree suc
y range tree This notion of a tree of trees is basic to solving range queries by leveling For
example for d dimensional range trees wewillhave d levels of trees
To answer a query we determine the canonical sets that satisfy the rst query there will b e
O log n of them Each of these sets is just represented as a no de in the range tree Weknow
that that these sets are disjoint and that every p oint in these sets lies within the range of
x co ordinates Thus to answer the query we just need to nd out which p oints from each
canonical subset lies within the range of y co ordinates Todothis foreach canonical subset
we access the auxiliary tree for this no de and p erform a dimensional range search on the
y range This pro cess is illustrated in the following gure
What is the query time for a range tree Recall that it takes O log n timetolocatethenodes
representing the canonical subsets for the dimensional range query For each weinvokea
dimensional range search Thus there O log n canonical sets eachinvoking an O log n
range search for a total time of O log n As b efore listing the elements of these sets can b e
p erformed in additional k time by just traversing the trees Counting queries can b e answered
b y precomputing the subtree sizes and then just adding them up
d
The space used by the data structure is O n log n in the plane and O n log n indi
mension d The reason comes by summing the sizes of the two data structures The tree for
Lecture Notes CMSC M
x-range tree
xlo xhi y hi Level 2 canonical sets.
ylo
Level 1 canonical sets.
Figure Range tree search
the x co ordinates requires only O n storage But we claim that the total storage in all the
auxiliary trees is O n log n Wewanttocount the total sizes of all these trees The number of
no des in a tree is prop ortional to the numberofleaves and hence the numb er of p oints stored
in this tree Rather than countthenumberofpoints in each tree separately instead let us
countthenumb er of trees in whicheachpoint app ears This will give the same total Observe
that a p oint app ears in the auxiliary trees of each of its ancestors Since the tree is balanced
eachpointhasO log n ancestors and hence eachpoint app ears in O log n auxiliary trees
It follows that the total size of all the auxiliary trees is O n log n
By the way observe that b ecause the auxiliary trees are just dimensional trees we could
just store them as a sorted array
We claim that it is p ossible to construct a dimensional range tree in O n log n time Con
structing the dimensional range tree for the x co ordinates is easy to do in O n log n time
However w e need to b e careful in constructing the auxiliary trees b ecause if wewere to sort
each list of y co ordinates separately the running time would b e O n log n Instead the trick
is to construct the auxiliary trees in a b ottom up manner The leaves whichcontain a single
p oint are trivially sorted Then we simply merge the two sorted lists for eachchild to form
the sorted list for the parent Since sorted lists can b e merged in linear time the set of all
auxiliary trees can b e constructed in time that is linear in their total since or O n log n Once
the lists have b een sorted then building a tree from the sorted list can b e done in linear time
Summarizing here is the basic idea to this and many other query problems based on leveled
searches Let S R denote the range space consisting of p oints S and range sets R Con
struct a data structure which represents S by a collection of canonical sets fS S S g
m
For each canonical subset construct a data structure for answering the second typ e of range
query and so on The main prop erty of the canonical subsets is that for any range query
we can e ciently determine a small numb er of canonical sets whose disjoint union is equal to
the answer to the query in our case this was O log n subsets Furthermore this collection
of canonical subsets can b e determined e ciently in our case in O log n time Toansw er a
range query wesolve the rst range query resulting in a collection of canonical subsets whose
Lecture Notes CMSC M
union is the answer to this query We then invoke the second range query problem on eachof
these subsets and so on Finally we take the union of all the answers to all these queries
Fractional Cascading Can we improveonthe O log n query time Wewould like to reduce
the query time to O log n As we descend the searchthe x interval tree for eachnodewe
visit weneedtosearch the corresp onding y interval tree It is this combination that leads
to the squaring of the logarithms If we could searcheach y interval in O time then we
could eliminate this second log factor The trick to doing this is used in a numb er of places in
computational geometry and is generally a nice idea to remember We are rep eatedly searching
di erent lists but always with the same key The idea is to merge all the di erent lists into
a single massive list do one search in this list in O log n time and then use the information
ab out the lo cation of the key to answer all the remaining queries in O time each This is a
simpli cation of a more general searchtechnique called fractional cascading
In our case the massive list on whichwe will do one searchistheentire of p oints sorted by
y co ordinate In particular rather than store these p oints in a balanced binary tree let us
assume that they are just stored as sorted arrays The idea works for either trees or arrays
but the arrays are a little easier to visualize Call these the auxiliary lists We will do one
exp ensive search on the auxiliary list for the ro ot which takes O log n time However after
this we claim that wecankeep track of the p osition of the y range in each auxiliary list in
constant time as we descend the tree of x co ordinates
Here is howwe do this Let v b e an arbitrary internal no de in the range tree of x co ordinates
and let v and v b e its left and rightchildren Let A b e the sorted auxiliary list for v and
L R v
let A and A b e the sorted auxiliary lists for its resp ectivechildren ObservethatA is the
L R v
disjoint union of A and A assuming no duplicate y co ordinates For each elementinA
L R v
we store twopointers one to the item of equal or larger value in A and the other to the item
L
of equal or larger value in A If there is no larger item the p ointer is null Observe that
R
once weknow the p osition of an item in A then we can determine its p osition in either A
v L
or A in O additional time
R
Here is a quick illustration of the general idea Let v denote a no de of the x tree and let v
L
and v denote its left and rightchildren Supp ose that in b ottom to top order the asso ciated
R
no des within this range are hp p p p p p i and supp ose that in v we store the p oints
L
hp p p i and in v we store hp p p i Thisisshown b elow For each p oint in the auxiliary
R
list for v we store a p ointer to the lists v and v to the p osition this elementwould b e
L R
y values That is we store a p ointer to the inserted in the other list assuming sorted by
largest element whose y value is less than or equal to this p oint
p6 v p5 p4 p3 p1 p2 p3 p4 p5 p6 vLRv p2
p2 p3 p5 p1 p4 p6
p1
Figure Cascaded search in range trees
At the ro ot of the tree we need to p erform a binary search against all the y values to determine
which p oints lie within this interval for all subsequentlevels once weknow where the y interval
falls with resp ect to the order p oints here we can drop down to the next level in O time
Lecture Notes CMSC M
Thus as with fractional cascading the running time is O log n rather than O log n It
turns out that this trick can only b e applied to the last level of the search structure b ecause
all other levels need the full tree search to compute canonical sets
d
Theorem Given a set of n p oints in R orthogonal rectangular range queries can b e answered
d d
in O log n k time from a data structure of size O n log n whichcanbe
d
constructed in O n log n time
Lecture Planar Point Lo cation
Tuesday Oct
Today s material is not covered in our text
Point Lo cation Todaywe consider is the point location problem The problem in space is
given a p olygonal sub division of the plane with n vertices prepro cess this sub division so that
h face of the sub division contains q given a query p oint q we can e ciently determine whic
Wemay assume that each face has some identifying lab el which is to b e returned We also
assume that the sub division is represented in any reasonable form e g as a DCEL
In general q may coincide with an edge or vertex To simplify matters we will assume that q
do es not lie on an edge or vertex but these sp ecial cases are not hard to handle Our b o ok
discusses how to handle degeneracies As usual this is done bycho osing a careful way to break
ties
It is remarkable that although this seems likesuch a simple and natural problem it to ok quite
a long time to discover an query time space optimal solution It has long b een known that
there are data structures that can p erform these searches reasonably well e g quad trees
and kd trees but for which no go o d theoretical b ounds could b e proved There were data
structures of with O log n query time but O n log n space and O n space but O log n
query time
The rst construction to achievebothO n space and O log n query time was a remarkably
clever construction due to Kirkpatrick It turns out that Kirkpatrick s idea has some large
emb edded constant factors that make it less attractive practically but the idea is so clever
that it is worth discussing nonetheless Later we will discuss a more practical randomized
metho d that is presented in our text
Kirkpatrick s Algorithm Kirkpatrick s idea starts with the assumption that the planar sub divi
sion is a triangulation and further that the outer face is a triangle If this assumption is not
met then we b egin by triangulating all the faces of the sub division The lab el asso ciated with
each triangular face is the same as a lab el for the original face that contained it For the outer
face is not a triangle rst compute the convex hull of the p olygonal sub division triangulate
everything inside the convex hull Then surround this convex p olygon with a large triangle
call the vertices a b andc and then add edges from the convex hull to the vertices of the
convex hull It may sound likewe are adding a lot of new edges to the sub division but recall
from earlier in the semester that the numb er of edges and faces in any straight line planar
sub division is prop ortional to n the number of vertices Thus the addition only increases the
size of the structure by a constant factor
Note that once we nd the triangle containing the query p oint in the augmented graph then
we will know the original face that contains the query p oint The triangulation pro cess can b e
p erformed in O n log n time byaplanesweep of the graph or in O n time if you wantto
use sophisticated metho ds like the linear time p olygon triangulation algorithm In practice
Lecture Notes CMSC M
a b
c
Figure Triangulation of a planar sub division
many straight line sub divisions may already haveconvex faces and these can b e triangulated
easily in O n time
Let T denote the initial triangulation What Kirkpatrick s metho d do es is to pro duce a
sequence of triangulations T T T T where k O log n suchthatT consists only of
k k
a single triangle the exterior face of T and each triangle in T overlaps a constantnumber
i
of triangles in T
i
We will see how to use such a structure for p oint lo cation queries later but for now let us
concentrate on how to build such a sequence of triangulations Assuming that wehave T we
i
wish to compute T In order to guarantee that this pro cess will terminate after O log n
i
stages we will wanttomake sure that the number of vertices in T decreases by some
i
constant factor from the n umber of vertices in T In particular this will b e done by carefully
i
selecting a subset of vertices of T and deleting them and along with them all the edges
i
attached to them After these vertices have b een deleted we need retriangulate the resulting
graph to form T The question is Howdowe select the vertices of T to delete so that
i i
each triangle of T overlaps only a constantnumb er of triangles in T
i i
There are two things that Kirkpatrick observed at this p oint that make the whole scheme
work
Constant degree We will make sure that eachofthevertices that we delete have constant
d degree that is each is adjacenttoatmostd edges Note that the when we
delete suchavertex the resulting hole will consist of at most d triangles When we
retriangulate each of the new triangles can overlap at most d triangles in the previous
triangulation
Indep endent set We will makesurethatnotwoofthevertices that are deleted are adjacent
to each other that is the vertices to b e deleted form an independent set in the current
planar graph T This will make retriangulation easier b ecause when we remove m
i
indep endentvertices and their incident edges we create m indep endent holes non
triangular faces in the sub division whichwewillhave to retriangulate However each
of these holes can b e triangulated indep endently of one another Since each hole contains
a constantnumber of vertices we can use any stupid triangulation algorithm since the
running time will b e O anyway
The big question is whether such a subset exists That is given any triangulation can wealways
nd a indep endentsubsetofvertices of b ounded degree whose size is at least a constant fraction
of the size of the whole vertex set Fortunately the answer is yes and in fact it is quite
easy to nd such a subset Part of the trickistopickthevalue of d to b e large enough to o
small and there may not b e enough of them It turns out that d is go o d enough
Theorem Given a planar graph with n vertices there is an indep endent set consisting of
vertices of degree at most with at least n vertices This indep endent set can b e
constructed in O n time
Lecture Notes CMSC M
Wow The numb er is probably smaller in practice but this is the b est b ound that this
pro of generates However the size of this constant is one of the reasons that Kirkpatrick s
algorithm is not used in practice But the construction is quite clever nonetheless and once
a simple solution is known to a problem it is often not long b efore a practical solution
follows
Before proving this theorem we rst complete the description of the p oint lo cation algorithm
We start with T and rep eatedly select an indep endent set of vertices of degree at most
However never include the three vertices a b andc forming the outer face in suchan
indep endent set We delete these vertices from the graph and retriangulate the resulting
holes Observe that each triangle in the new triangulation can overlap at most triangles in
the old triangulation Since we can eliminate a constant fraction of vertices with each stage
after O log n stages wewillbedowntothelast vertices
The constant factors here are not so great With each stage the numberofvertices falls bya
k
factor of To reduce to the nal vertices implies that n or that
k log n lg n
It can b e shown that byalways selecting the vertex of smallest degree this can b e reduced to
a more palatable lgn
Point lo cation algorithm Now to p erform p oint lo cation we create a tree The ro ot of the tree
is the single triangle of T The no des at the next lower level are the triangles of T followed
k k
by T until wereach the leaves which are the triangles of our initial triangulation T Each
k
no de for a triangle in triangulation T stores p ointers to all the triangles it overlaps in T
i i
there are at most of these
Tolocateapoint we start with the ro ot T If the query p oint do es not lie within this single
k
triangle then we are done it lies in the exterior face Otherwise we searcheach of the at
most triangles in T that overlap this triangle When we nd the correct one we search
k
each of the triangles in T that overlap this triangles and so forth Eventually we will nd
k
the triangle containing the query p oint in the last triangulation T and this is the desired
output
es a constant amountoftimeto The tree has O log n levels one for each triangulation it tak
move from one level to the next at most p oint in triangle tests thus the total query time
is O log n The size of the data structure is the sum of sizes of the triangulations Since the
numb er of triangles in a triangulation is prop ortional to the numberofvertices it follows that
the size is prop ortional to
n n
using standard formulas for geometric series Thus the data structure size is O n again
with a pretty hefty constant
Construction and Analysis The last thing that remains is to showhow to construct the inde
p endent set of the appropriate size We rst present the algorithm for nding the indep endent
set and then prove the b ound on its size
Mark all no des of degree
While there exists an unmarked no de do the following
a Cho ose an unmarked vertex v
b Add v to the indep endent set
c Mark v and all of its neighb ors
Lecture Notes CMSC M
ko z n y m x T j h g T v t 0 l i 1 w u a cd p f q r s b e
G J F T3 T2 E D H I A B C
T4 (not shown) Triangulation Refinement.
K T0
HI J T1
AB C D EFG T2
p qrstuvwxyzT3
abc def gh ijklmno T4
Final Point Location Tree.
Figure Kirkpatrick s p oint lo cation algorithm
Lecture Notes CMSC M
It is easy to see that the algorithm runs in O n time e g bykeeping unmarked vertices in
astack and representing the triangulation so that neighb ors can b e found quickly
Intuitively the argument that there exists a large indep endent set of low degree is based on
the following simple observations First b ecause the average degree in a planar graph is less
than there must b e a lot of vertices of degree at most otherwise the average would b e
unattainable Second whenever we add one of these vertices to our indep endent set only
other vertices b ecome ineligible for inclusion in the indep endent set
Here is the rigorous argument Recall from Euler s formula that if a planar graph is fully
triangulated then the numb er of edges e satis es e n If we sum the degrees of all the
vertices then each edge is counted twice Thus the average degree of the graph is
X
deg v e n n
v
Next we claim that there must b e at least n vertices of degree or less To see why suppose
to the contrary that there were more than n vertices of degree or greater The remaining
vertices must have degree at least with the p ossible exception of the vertices on the outer
face and thus the sum of all degrees in the graph w ould have to b e at least as large as
n n
n
whichcontradicts the equation ab ove
Now when the ab ove algorithm starts execution at least n vertices are initially unmarked
Whenever we select suchavertex b ecause its degree is or fewer we mark at most new
vertices this no de and at most of its neighb ors Thus this step can b e rep eated at least
n n times b efore we run out of unmarked vertices This completes the pro of
Lecture More Planar PointLocation
Thursday Oct
Read Chapt of BKOS
More Point Lo cation Last time we presented Kirkpatrick s p oint lo cation algorithm Todaywe
will consider another asymptotically optimal algorithm for p oint lo cation but the interesting
element of this algorithm is that it is randomized and in particular it is based on a randomized
incremental construction We will show that the size prepro cessing time and query time are
O n O n log n and O log n resp ectively in the exp ected case Here the exp ectation is
indep endent of the choice of the p olygonal sub division or the query p oint It dep ends only on
the order in which the ob jects are inserted
Trap ezoidal Map The algorithm is based on a construction called a trapezoidal map which also
go es under many other names in the computational geometry literature Although we nor
mally think of the input to a p oint lo cation algorithm as b eing a planar sub division we will
de ne the algorithm under the assumption that the input is just a collection of line segments
S fs s s g such that these line segments do not intersect except p ossibly at their
n
endp oints To construct a trap ezoidal map imagine sho oting a bullet vertically upwards and
downwards from eachvertex in the p olygonal sub division For simplicity we will assume that
there are no vertical segments in the initial sub division and no twosegments have the same
x co ordinate Both of these are easy to handle with an appropriate symb olic p erturbation
The bullet travels until it hits another line segmentofS The resulting bullet paths together
with the initial line segments de ne the trap ezoidal map Toavoid in nite bullet paths at the
Lecture Notes CMSC M
Figure Trap ezoidal map
top and b ottom of the sub division wemay assume that the initial sub division is contained
entirely within a large b ounding rectangle An example is shown in the gure b elow
First observe that all the faces of the resulting sub division are trap ezoids with vertical sides
The left or right side might degenerate to a line segment of length zero implying that the
resulting trap ezoid degenerates to a triangle We claim that the pro cess of converting an
arbitrary p olygonal sub division into a trap ezoidal decomp osition increases its size byatmost
a constant factor The nal trap ezoidal map will b e given as a sub division represented say
using a DCEL
Claim Given a p olygonal sub division with n segments the resulting trap ezoidal map has at
most n vertices and n trap ezoids
Pro of To prove the b ound on the number of vertices observethateachvertex sho ots two
bullet paths each of which will result in the creation of a new vertex Thus each original
vertex gives rise to three vertices in the nal map Since each segment has twovertices
this implies at most n vertices
To b ound the numb er of trap ezoids observe that for each trap ezoid in the nal map its
left side and its rightaswell is b ounded byavertex of the original p olygonal sub division
The left endp ointofeach line segment can serve as the left b ounding vertex for two
trap ezoids one ab ove the line segment and the other b elow and the right endp ointofa
line segment can serve as the left b ounding vertex for one trap ezoid Thus each segment
of the original sub division gives rise to at most three trap ezoids for a total of n The
last trap ezoid is the one b ounded by the left side of the b ounding b ox
An imp ortant fact to observe ab out each trap ezoid is that it is de ned by exactly four entities
from the original sub division a segmentontop asegment on the b ottom a b ounding vertex
on the left and a b ounding vertex on the right This simple observation will play an imp ortant
role in the analysis
Trap ezoidal decomp ositions like triangulations are interesting data structures in their own
right It is another example of the idea of converting a complex shap e into a disjoint collection
of simpler ob jects The fact that the sides are vertical makes trap ezoids simpler than arbitrary
quadrilaterals Finally observe that the trap ezoidal decomp osition is a re nement of the orig
inal p olygonal sub division and so once we knowwhich face of the trap ezoidal map a query
p oint lies in we will knowwhich face of the original sub division it lies in either implicitly or
b ecause we lab el each face of the trap ezoidal map in this way
Construction We could construct the trap ezoidal map easily by plane sweep Hop efully this
is an easy exercise by this p oint but think ab out howyou would do it We will build the
trap ezoidal map by a randomized incremental algorithm b ecause the p oint lo cation algorithm
is based on this construction In fact historically this algorithm arose as a metho d for
Lecture Notes CMSC M
computing the trap ezoidal decomp osition of a collection of intersecting line segments and the
p oint lo cation algorithm arose as an artifact that was needed in the construction
The incremental algorithm starts with the initial b ounding rectangle that is one trap ezoid
and then we add the segments of the p olygonal sub division one by one in random order As
each segment is added we up date the trap ezoidal map Let S denote the subset consisting of
i
the rst i random segments and let T denote the resulting trap ezoidal map
i
To p erform the up date this we need to knowwhich trap ezoid the left endp oint of the segment
lies in We will let this question go until later since it will b e answered by the p oint lo ca
tion algorithm itself Then we trace the line segment from left to right determining which
trap ezoids it intersects Finally we go back to these trap ezoids and x them up There are
two things that are involved in xing First the left and right endp oints of the new segment
need to have bullets red from them Second one of the earlier bullet paths might hit this
line segment When that happ ens the bullet path must b e trimmed back Weknowwhich
vertices are from the original sub division vertices so weknow which side of the bullet path to
trim The pro cess is illustrated in the gure b elow
Figure Incremental up date
Observe that the structure of the trap ezoidal decomp osition do es not dep end on the order
in which the segments are added This observation will b e imp ortant for the probabilistic
analysis The following is also imp ortant to the analysis
Claim Ignoring the time sp ent to lo cate the left endp oint of an segment the time that it
takes to insert the ith segment and up date the trap ezoidal map is O k where k is the
i i
numb er of newly created trap ezoids
Pro of Consider the insertion of the ith segment and let K denote the numb er of bullet paths
that this segmentintersects We need to sho ot four bullets two from eachendpoint and
then trim eachofthe K bullet paths for a total of K op erations that need to b e
p erformed If the new segment did not cross any of the bullet paths then wewould get
exactly four new trap ezoids For eachoftheK bullet paths we cross we add one more to
the numb er of newly created trap ezoids for a total of K Thus letting k K be
i
the numb er of trap ezoids created the numb er of up date op erations is exactly k Each
i
of these op erations can b e p erformed in O time given any reasonable representation of
the trap ezoidal map e g a DCEL
Analysis We left one detail out whichishowwe lo cate the left endp oin tofeach new segment that
we add But ignoring the time for this whichwe will see will b e O log n time on average
we will show that the exp ected time to add each segmentisO Since there are n insertions
this will lead to a total exp ected time complexityof O n log n O n log n
We know that the size of the nal trap ezoidal map is O n It turns out that the total size of
the p oint lo cation data structure will actually b e prop ortional to the numb er of new trap ezoids
that are created with each insertion In the worst case when we add the ith segment it might
Lecture Notes CMSC M
cut through a large fraction of the existing O i trap ezoids and this would lead to a total
P
n
i n However the magic of the incremental construction is that size prop ortional to
i
this do es not happ en We will showthatonaverage each insertion results in only a constant
numb er of trap ezoids b eing created
You might stop to think ab out this for a moment b ecause it is rather surprising at rst
Clearly if the segments are short then each segment might not intersect very many trap ezoids
But what if all the segments are long It seems as though it might b e p ossible to construct a
counterexample Give it a try b efore you read this
Lemma Consider the randomized incremental construction of a trap ezoidal map and let
k denote the numb er of new trap ezoids created when the ith segment is added Then
i
E k O where the exp ectation is taken over all p ermutations of the segments
i
Pro of The analysis will b e based on a backwards analysis Let T denote the trap ezoidal map
i
after the insertion of the ith segment Because we are averaging over all p ermutations
among the i segments that are presentinT each one has an equal probability i of
i
b eing the last one to have b een added For each of the segments s wewant to count the
number of trapezoids that would have b een created had s been the last segment to be
added Let s say that a trap ezoid depends on an segment s ifs would have caused
to b e created had s b een added last Wewant to countthenumb er of trap ezoids
that dep end on eachsegment and then compute the average over all segments If we let
s if segment s dep ends on and otherwise then the exp ected complexityis
X X
E k s
i
i
T s S
i i
Some segments mighthave resulted in the creation of lots of trap ezoids and other very few
Howdowe get a handle on this quantity The trick is rather than countthenumber
of trap ezoids that dep end on each segment we count the numb er segments that each
trap ezoid dep ends on The old combinatorial trick of reversing the order of summation
In other words wewant to compute
X X
E k s
i
i
T s S
i i
s
The segments that the trapezoid
The trapezoids that depend on s depends on.
Figure Trap ezoid segment dep endencies
This is much easier to determine In particular each trap ezoid is b ounded by four sides
The top and b ottom sides are each determined byasegmentofS and clearly if either of
i
these was the last to b e added then this trap ezoid would have come into existence as a
Lecture Notes CMSC M
result The left and right sides are each determined by a endp ointofasegmentinS and
i
clearly if either of these was the last to b e added then this trap ezoid would have come
into existence Thus each trap ezoid is dep endent onatmostfoursegments implying
P
that s Since T consists of O i trap ezoids wehave
i
s S
i
X
jT j O i O E k
i i
i i i
T
i
Lecture Point Lo cation in Trap ezoidal Maps
Tuesday Oct
Read Chapt of BKOS
Point Lo cation Last time we presented a randomized incremental algorithm for constructing a
trap ezoidal map in the plane Todaywe consider how to mo dify this algorithm to answer
p oint lo cation queries The prepro cessing time will b e O n log n in the exp ected case as was
the time to construct the trap ezoidal map and the space and query time will b e O n and
O log n resp ectively in the exp ected case
Recall from last time that we are treating the input as a set of segments S fs s g
n
p ermuted randomly that S denotes the subset consisting of the rst i segments of S and T
i i
denotes the trap ezoidal map of S One imp ortantelement of the analysis to rememb er from
i
last time is that each time we add a new line segment it may result in the creation of the
collection of new trap ezoids whichwere said to depend on this line segment We presented
b er of new trap ezoids that are created with each stage is abackwards analysis that the num
exp ected to b e O This will play an imp ortant role in to day s analysis
Point Lo cation Data Structure As in Kirkpatrick s algorithm the p oint lo cation data struc
ture will b e based on a ro oted directed acyclic graph In particular to the query pro cessor
it will lo ok like a binary tree but there may b e sharing of subtrees There are twotyp es of
no des x nodes and y nodes Each x no de contains the x co ordinate of an endp oint of one of
the segments Its twochildren corresp ond to the p oints lying to the left and to the right of this
co ordinate Each y no de contains a p ointer to a segment The left and rightchildren corre
sp ond to whether the query p ointisaboveorbelowlinecontaining the segment resp ectively
We will visit a y no de only if we already know that the x co ordinate of the query p oint lies
between the left and right endp oints of the segment
Our construction of the p oint lo cation data structure mirrors the incremental construction of
the trap ezoidal map In particular if we freeze the construction just after the insertion of any
segment the existing structure will b e a p oint lo cation structure for the existing trap ezoidal
map In the gure b elowweshow a simple example of what the data structure lo oks likefor
two line segments from our textb o ok There is one leaf for each trap ezoid The y no des are
shown as hexagons For example if the query p oin t is in trap ezoid D wewould rst detect
that it is to the rightofp thenleftofq then b elow s the rightchild then rightofp
then ab ove s the left child
The question is howdowe build this data structure incrementally First observe that when
a new line segment is added we only need to adjust the p ortion of the tree that involves the
trap ezoids that have b een deleted as a result of this new addition This will b e set of leaves
in the current tree Eachsuch leaf will b e replaced with a new subtree which determines
which one of the new trap ezoids contains the query p oint In Kirkpatrick s algorithm we just
said check each of the new triangles that overlapp ed one of the old triangles and we will do
essentially the same thing here As in Kirkpatrick s algorithm it will turn out that each old
Lecture Notes CMSC M
p1 q B 1 q q1 A 2 p s 1 s1 1 B s2 G C E G p A D 2 E p C 2 s s2 2
F q2 DF
Figure Trap ezoidal map p oint lo cation data structure
trap ezoid is overlapp ed by a constantnumb er of new trap ezoids However let us consider the
pro cess in a little more detail here
Supp ose that we add a line segment This results in the replacement of an existing set of
trap ezoids with a set of new trap ezoids If the segment passes entirely through an existing
trap ezoid then there will b e twooverlapping trap ezoids in the new trap ezoidal map and
thus we just need to compare against the newly added segment one y no de If the existing
trap ezoid contains one or b oth endp oints then we need to test on which sides of these endp oints
we lie one or two x no des and if we lie b etween them then we need to test whether we lie
ab oveorbelow the line segment oney no de If weaddalinesegment to the example ab ove
resulting in the replacementofC D E and G with new trap ezoids we will replace these
leaves of the subtree as shown in the gure b elow It is imp ortant to notice that through
sharing each trap ezoid app ears exactly once as a leaf in the resulting structure
p1 q B 1 K q q1 A 2 p s 1 s1 1 I q q s 3 B s2 3 3 p A p N 2 s 3 p 3 p L M 3 s3 N 2 s s2 2 H F q s3 M H J 2 s3 F
I J K L
Figure Line segment insertion
Analysis We claim that the size of the p oint lo cation data structure is O n and the query time
is O log n b oth in the exp ected case As usual the exp ectation dep ends only on the order of
insertion not on the line segments or the lo cation of the query p oint
To prove the space b ound of O n observe that the numb er of new no des added to the
structure with each new segment is prop ortional to the numb er of newly created trap ezoids
This follows by observing that we create a constantnumb er of new no des for each existing
trap ezoid that was destroyed and the numb er of new trap ezoids is prop ortional to the number
Lecture Notes CMSC M
of old trap ezoids that were destroyed Last time weshowed that with each new insertion the
exp ected numb er of trap ezoids that were created was O Therefore weaddO new no des
with each insertion in the exp ected case implying that the total size of the data structure is
O n
Analyzing the query time is a little subtler In a normal probabilistic analysis of data structures
we think of the data structure as b eing xed and then compute exp ectations over random
queries Here the approach will b e to imagine that wehave exactly one query to handle
Imagine an adversary that tries to select the worst p ossible query p oint but do es not know
what random choices the algorithm made in building the data structure Wewillshow that
no matter how the query p oint is selected most random orderings of the line segments will
lead to a search path of length O log n in the resulting tree Since our analysis will b e made
without any assumptions on the nal trap ezoid in which q lies it will apply equally well to all
query p oints
Rather than consider the search path for q in the nal search structure we will consider how
q moves incrementally through the structure with the addition of each new line segment Let
denote the trap ezoid of the map that q lies in after the insertion of the rst i segments
i
Observe that if then insertion of the ith segment did not a ect the trap ezoid that
i i
q was in and therefore q will stay where it is relative to the current search structure For
example if q was in trap ezoid B prior to adding s in the gure ab ove then the addition of s
do es not incur any additional cost to lo cating q However if then the insertion
i i
of the ith segment caused q s trap ezoid to b e deleted As a result q must lo cate itself with
resp ect to the newly created trap ezoids that overlap Since there are a constantnumber
i
of such trap ezoids at most four there will b e O work needed to lo cate q with resp ect to
these In particular q mayfallasmuch as three levels in the search tree For example if q
was in trap ezoid C b efore the addition of s in the gure ab ove and q was in trap ezoid J
afterwards then q would have to pass through two new levels of the structure as a result of
this insertion
To compute the exp ected length of the search path it su ces to compute the probability
that the trap ezoid that contains q changes as a result of the ith insertion Let P denote this
i
probability Since q could fall through up to three levels in the search tree as a result of each
the insertion the exp ected length of q s search path in the nal structure is at most
n
X
P
i
i
Wewillshow that P i From this it will follow that the exp ected path length is at most
i
n n
X X
i i
i i
which is roughly ln n O log n by the Harmonic series
To showthat P i we apply a backwards analysis Recall from last time that the trap ezoid
i
that contains q is dep endent on at most four trap ezoids which de ne the top and b ottom edges
and the left and right sides of the trap ezoid Since each segment is equally likely to b e the
last segment to b e added the probability that the last insertion caused q to b elong to a new
trap ezoid is at most i This completes the pro of
Guarantees on SearchTime The only problem with this result is that even though the search
time is provably small in the exp ected case for a given query p oint it might still b e the case
that once the data structure has b een constructed there is a single very long path in the search
Lecture Notes CMSC M
structure and the user rep eatedly p erforms queries along this path The result provides no
guarantees on the running time of all queries
Although we will not prove it the b o ok presents a stronger result namely that the length of
the maximum search path is also O log n with high probability In particular they prove the
following
Lemma Given a set of n non crossing line segments in the plane and a parameter the
probability that the total depth of the randomized search structure exceeds ln n
ln
is at most n
For example for the probability that the search path exceeds ln n isatmost
n The constant factors here are rather weak but a more careful analysis leads to
a b etter b ound
Nonetheless this itself is enough to lead to variant of the algorith for which O log n time is
guaranteed Rather than just running the algorithm once and taking what it gives instead
keep running it and checking the structure s depth As so on as the depth is at most c log n for
some suitably chosen c then stop here Dep ending on c and n the ab ove lemma indicates how
long you may need to exp ect to rep eat this pro cess until the nal structure has the desired
depth For su ciently large c the probability of nding a tree of the desired depth will b e
b ounded away from by some constant factor and therefore after a constantnumb er of trials
dep ending on this probability you will eventually succeed in nding a p oint lo cation structure
of the desired depth A similar argument can b e applied to the space b ounds
Theorem Given a set of n non crossing line segments in the plane in exp ected O n log n
time it is p ossible to construct a p oint lo cation data structure of worst case size O n
that can answer p oint lo cation queries in worst case time O log n
Lecture Review for the Midterm
Thursday Oct
Midterm Exam Oct The exam will b e closed b o ok closed notes but you are allowed one
cheat sheet front and back of notes
What we vecovered so far
Geometric Background A ne geometry homogeneous co ordinates orientation test sym
b olici p erturbation
Convex hulls Graham s scan an O n log n algorithm
Line SegmentIntersection Plane sweep algorithm O n k log n where k is the number
of intersections
Representing planar sub divisions Euler s formula DCEL s dual graphs
Polygon Triangulation The art gallery problem monotone decomp osition by plane sweep
in O n log n time triangulating monotone p olygons in O n time There exists an O n
time triangulation but it is quite complicated
Intersection of Halfplanes Divide and conquer solution that runs in O n log n time
Point line duality The lower and upp er envelop es of a set of lines can b e computed in
O n log n time by applying p oint line duality and computing the convex hull of the dual
points
Lecture Notes CMSC M
Linear Programming Randomized incremental algorithm for low dimensional linear pro
gramming RunsinO d n exp ected time in dimension d
Orthogonal Range Searching kd trees and range trees Both can b e constructed in O n log n
p
time Showed that kd trees could b e used to answer orthogonal range queries in O n
time and that range trees could answer them in O log n time Both can b e generalized
d
to higher dimensions with range trees taking O log n time
Planar PointLocation Kirkpatrick s algorithm O n log n prepro cessing time O n space
and O log n query time Based on decimation of a triangulation
Trap ezoidal Maps Randomized incremental construction and p ointlocation O n log n
exp ected prepro cessing time and O n space and O log n query time
Techniques
Plane sweep
Divide and conquer
Randomized incremental construction
Decomp osing search problems into layers of trees
Point line duality
Fractional cascading
Backwards analysis
Lecture Midterm
Tuesday Oct
Midterm Exam to day No lecture
Lecture Voronoi diagrams
Thursday Oct
Reading BKOS Chapt
Euclidean Geometry Wenow will make a subtle but imp ortant shift Up to know virtually
everything that wehave done has not needed the notion of angles lengths or distances except
for our work on circles All geometric tests were made on the basis of orientation tests a
purely a ne construct
But there are imp ortant geometric algorithms which dep end on nona ne quantities suchas
distances and angles Let us b egin by de ning the Euclidean length ofavector v v v
x y
q
p
in the plane to b e jv j The v v and in general dimension it is jv j v v
x y
d
distance b etween two p oints p and q denoted dist p q is de ned to b e jp q j
Voronoi Diagrams Voronoi diagrams like convex hulls are among the most imp ortant structures
in computational geometry AVoronoi diagram records information ab out what is close to
what
Let P fp p p g b e a set of p oints in the plane or in any dimensional space which
n
wecallsites De ne V p the Voronoi cel l for p to b e the set of p oints q in the plane such
i i
Lecture Notes CMSC M
that dist q p dist q p That is the Voronoi cell for p consists of the set of p oints for
i j i
which p is the unique nearest neighbor to q
i
V p fq j dist p q dist p q j ig
i i j
Another way to de ne V p is in terms of the intersection of halfplanes Given two sites p
i i
and p the set of p oints that are strictly closer to p than to p is just the open halfplane
j i j
whose b ounding line is the p erp endicular bisector b etween p and p Denote this halfplane
i j
h p p It is easy to see that a p oint q lies in V p if and only if q lies within the intersection
i j i
of h p p for all j i In other words
i j
V p h p p
i j i i j
Since the intersection of halfplanes is a p ossibly unb ounded convex p olygon it is easy to see
that V p is a p ossibly unb ounded convex p olygon Finally de ne the Voronoi diagram of
i
P denoted Vor P to b e what is left of the plane after weremove all the op en Voronoi cells
It is not hard to prove see the text that the Voronoi diagram consists of a collection of line
segments whichmaybeunb ounded either at one end or b oth
Figure Voronoi diagram
Voronoi diagrams haveanumb er of imp ortant applications These include
Nearest neighb or queries One of the most imp ortant data structures problems in com
putational geometry is solving nearest neighb or queries Given a p ointset P and given
a query p oint q determine the closest p ointinP to q This can b e answered by rst
computing a Voronoi diagram and then lo cating the cell of the diagram that contains q
Wehave already discussed p oint lo cation algorithms
Computational morphology Some of the most imp ortant op erations in morphology used
very much in computer vision is that of growing and shrinking or thinning
ob jects If wegrow a collection of p oints by imagining a grass re starting simultaneously
from each p oint then the places where the grass res meet will b e along the Voronoi
diagram The medial axis of a shap e used in computer vision is just a Voronoi diagram
of its b oundary
Facility lo cation Wewant to op en a new Blo ckbuster video It should b e placed as far as
p ossible from any existing video stores Where should it b e placed It turns out that the
vertices of the Voronoi diagram are the p oints that lo cally at maximum distances from
any other p oint in the set
High clearance path planning Arobotwants to move around a set of obstacles To mini
mize the p ossibility of collisions it should stay as far away from the obstacles as p ossible
To do this it should walk along the edges of the Voronoi diagram
Lecture Notes CMSC M
Prop erties of the Voronoi diagram Some theoretical observations ab out the Voronoi diagram
are warranted at this p oint
Voronoi edges Each p ointonanedgeoftheVoronoi diagram is equidistant from its two
nearest neighbors p and p Thus there is a circle centered at suchapoint such that p
i j i
and p lie on this circle and no other site is interior to the circle
j
Voronoi vertices It follows that vertex at which three Voronoi cells V p V p and V p
i j k
intersect is equidistant from all sites Thus it is the center of the circle passing through
these sites and this circle contains no other sites in its interior
Degree If we assume that no four p oints are co circular then the vertices of the Voronoi
diagram all have degree three
Convex hull A cell of the Voronoi diagram is unb ounded if and only if the corresp onding
site lies on the convex hull Observe that a p ointisontheconvex hull if and only if it
is the closest p oint from some p oint at in nity
Size If n denotes the numb er of sites then the Voronoi diagram is a planar graph if we
imagine all the unb ounded edges as going to a common vertex in nity with exactly n
faces It follows from Euler s formula that the numberofVoronoi verticesisatmost
n and the numb er of edges is at most n See the text for details
DelaunayTriangulation Since the Voronoi diagram is a planar graph wemay naturally ask
what is the corresp onding dual graph The vertices for this dual graph can b e taken to b e the
sites themselves Since assuming general p osition the vertices of the Voronoi diagram are of
degree three it follows that the faces of the dual graph excluding the exterior face will b e
triangles The resulting dual graph is a triangulation Among the many triangulations of a
set of p oints this one has a numb er of nice geometric prop erties and is called the Delaunay
triangulation
Figure Delaunay triangulation
Delaunay triangulations haveanumber of interesting prop erties that are consequences of the
structure of the Voronoi diagram
Convex hull The exterior face of the Delaunay triangulation is the convex hull of the p oint
set
Circumcircle prop erty The circumcircle of any triangle in the Delaunay triangulation is
empty containsnopoints of P
Empty circle prop erty Two sites p and p are connected by an edge in the Delaunay
i j
triangulation if and only if there is an empty circle passing through p and p One
i j
direction of the pro of is trivial from the circumcircle prop erty In general if there is an
empty circumcircle passing through p and p then the center c of this circle is a p oint
i j
Lecture Notes CMSC M
on the edge of the Voronoi diagram b etween p and p b ecause c is equidistant from each
i j
of these p oints and there is no closer p oint
Closest pair prop erty The closest pair of p oints in P are Delaunay neighb ors
There are a numb er of other interesting prop erties of Delaunay triangulations that are not so
easy to prove
Max Min Angle criterion Among all triangulations of the p ointset P the Delaunay trian
gulation maximizes the minimum angle in the triangulation This only holds in dimension
MST prop erty The minimum spanning tree of a set of p oints in the plane is a subgraph of
the Delaunay triangulation
This provides a go o d algorithm for computing MST s of p oints in the plane First compute
the Delaunay triangulation of the p oint set wewillshow this can b e done in O n log n time
and then in O n log n time you can compute the MST of this sparse graph
Lecture Fortune s Voronoi Diagram Algorithm
Tuesday Nov
Reading BKOS Chapt
Computing Voronoi Diagrams Last time weintro duced the Voronoi diagram of a set of n p oints
P fp p g in the plane called sites Recall that this is a planar straight line sub division
n
with n faces each a p ossibly unb ounded convex p olygon The face or Voronoi cel l asso ciated
with site p denoted V p consists of all the p oints in the plane that are strictly closer to
i i
p thantoany other site Recall that the edge joining cells for p and p is the p erp endicular
i i j
bisector b etween these p oints and so a circle grown ab out any p ointonaVoronoi edge will
touch these two sites and contain no other site in its interior Also recall that the Voronoi
vertices are of degree three assuming general p osition and a cicle grown ab out this p oint will
touch all three sites an contain no other site in its center
There are a numb er of algorithms for computing Voronoi diagrams Of course there is a
naive O n log n time algorithm which op erates by computing V p byintersecting the n
i
bisector halfplanes h p p for j i However there are much more e cientways which
i j
run in O n log n time Since the convex hull can b e extracted from the Voronoi diagram in
O n time it follows that this is asymptotically optimal in the worst case
Note from the comments made last time ab out the Delaunay triangulation that the Voronoi
diagram can either b e computed directly or else it can b e built by rst constructing the
Delaunay triangulation of the sites and then taking its dual graph with some care as to
where the Voronoi vertices are to b e placed Historically the rst algorithm for computing
Voronoi diagrams is the simple incremental algorithm that was used for computing Delaunay
triangulations It was known for manyyears that this is not asymptotically optimal and
in fact ran in O n timeintheworst case but this did not deter the practitioners who
were quite happy with it When computational geometry came along a more complex but
asymptotically sup erior O n log n algorithm was discovered This algorithm was based on
divide and conquer But it was rather complex and somewhat di cult to understand Later
SteveFortune invented a plane sweep algorithm for the problem whichprovided a simpler
O n log n solution to the problem It is his algorithm that we will discuss Somewhat later
still it was discovered that the incremental algorithm is actually quite e cient if it is run as
Lecture Notes CMSC M
a randomized incremental algorithm We will discuss this algorithm later but in the form of
aDelaunay triangulation algorithm
Before discussing Fortune s algorithm it is interesting to consider why this algorithm wasn t
invented much earlier In fact it is quite a bit trickier than any plane sweep algorithm wehave
seen so far The key to any plane sweep algorithm is the ability to discover all up coming
events in an e cient manner For example in the line segmentintersection algorithm we
considered all pairs of line segments that were adjacent in the sweep line status and inserted
their intersection p oint in the queue of up coming events The problem with the Voronoi
diagram is that of predicting where the up coming events will o ccur The reason is that a site
that lies ahead of the sweep line may generate a Voronoi vertex that lies b ehind the sweep line
It is these unanticipated events that make the design of a plane sweep algorithm challenging
Beach line Unanticipated Sweep line
events
Figure Plane sweep Voronoi diagrams
Fortune s Algorithm Fortune made the clever observation of rather than computing the Voronoi
diagram through plane sweep in its nal form instead to compute a distorted but top olog
ically equivalentversion of the diagram This distorted version of the diagram was based on
a transformation that alters the way that distances are measured in the plane The resulting
diagram had the same structure as the Voronoi diagram but its edges were parab olic arcs
rather than straight line segments Once this distorted diagram was generated it was an easy
matter to undistort it to pro duce the correct Voronoi diagram
Our presentation will b e di erentfromFortune s Rather than distort the diagram we can
think of this algorithm as distorting the sweep line Actually we will think of two ob jects
that control the sweeping pro cess First there will b e a horizontal sweep line moving from
top to b ottom We will also maintain a x monotonic curve called a beach line I guess b ecause
it lo oks likewaves rolling up on a b each The b each line is formed from parab olic arcs As
the sweep line moves downward the b eachlinemoves downward as well but the b each line s
shap e dep ends on the lo cations of the sites We will see that the Voronoi diagram grows out
of the b each line
In order to make these ideas more concrete recall that the problem with ordinary plane sweep
is that sites that lie b elowthesweep line may a ect the diagram that lies ab ovethesweep
tain only the p ortion of the diagram that cannot b e line Toavoid this problem we will main
a ected byanything that lies b elowthesweep line To do this we will cut the halfplane
lying ab ovethesweep line into two regions those p oints that are closer to some site p ab ove
the sweep line than they are to the sweep line itself and those p oints that are closer to the
sweep line than anysiteabove the sweep line The set of p oints q that are equidistant from
the sweep line to their nearest site ab ovethesweep line is called the beach line Observe that
for anypoint q ab ovethebeach line we know that its closest p oint cannot b e a ected byany
Lecture Notes CMSC M
site that lies b elow the sweep line Hence the p ortion of the Voronoi diagram that lies ab ove
the b each line is safe in the sense that wehave all the information that we need in order to
compute it without knowing ab outs what p oints are still to app ear b elowthesweep line
What do es the b each line lo ok like Weknow from high scho ol geometry that the set of
p oints that are equidistant from a site lying ab ove a horizontal line and the line itself forms
a parab ola that is op en on top With a little analytic geometry itiseasytoshow that the
parab ola b ecomes skinnier as the site b ecomes closer to the line the parab ola degenerates
into a vertical ray emanating from the site You should work out the equations to see why
this is so
Thus the b each line consists of the lower envelop e of these parab olas By the way ifwewere
interested in computing the Voronoi diagram of line segments the b each line is the b oundary
of the Voronoi cell of the sweep line itself Because the parab olas are x monotone so is the
b each line Also observe that the vertex where two arcs of the b each line intersect whichwe
call a breakpoint isapoint that is equidistant from two sites and the sweep line and hence
must lie on some Voronoi edge In particular if the b each line arcs corresp onding to p oints p
i
and p share a common breakp oint on the b each line then this breakp ointliesontheVoronoi
j
edge b etween p and p From this wehave the following imp ortantcharacterization
i j
Lemma The b eachlineisanx monotone curve made up of parab olic arcs The breakp oints
of the b each line lie on Voronoi edges of the nal diagram
Fortune s algorithm consists of simulating the growth of the b each line as the sweep line moves
downward and in particular tracing the paths of the breakp oints as they travel along the edges
of the Voronoi diagram Of course as the sweep line moves the parab olas forming the b each
line change their shap es continuously As with all plane sweep algorithms we will b e interested
in simulating the discrete event p oints where there is a signi cantevent that is anyevent
that changes the top ological structure of the b each line It turns out these signi cantevents
will b e of twovarieties
Site events When the sweep line passes over a new site a new arc will b e inserted into the
beach line
Vertex events What our text calls circle events When the length of a parab olic arc
shrinks to zero the arc disapp ears and a new Voronoi vertex will b e created at this p oint
The algorithm consists of pro cessing these tw otyp es of events As the Voronoi vertices are
b eing discovered byvertex events it will b e an easy matter to up date a DCEL for the diagram
as we go and so to link the entire diagram together For the rest of the lecture we fo cus on
the nature of the b each line and how these events are discovered and pro cessed
Site events A site event is generated whenever the sweep line passes over a site As wementioned
b efore at the instant that the sweep line touches the p oint its asso ciated parab olic arc will
degenerate to a vertical ray sho oting up from the p oint to the currentbeach line As the sweep
line pro ceeds downwards this ray will widen into an arc along the b each line To pro cess a site
eventwe will determine the arc of the sweep line that lies directly ab ove the new site Let us
make the general p osition assumption that it do es not fall immediately b elowavertex of the
b each line We then split this arc of the b each line in twoby inserting a new in nitesimally
small arc at this p oint As the sweep pro ceeds this arc will start to widen and eventually will
join up with other edges in the diagram See the gure b elow
It is imp ortant to consider whether this is the only way that new arcs can b e intro duced into
the sweep line In fact it is We will not prove it but a careful pro of is given in the text As a
consequence of this pro of it follows that the maximum numb er of arcs on the b each line can
Lecture Notes CMSC M
Figure Site events
b e at most n since each new p oint can result in creating one new arc and splitting an
existing arc for a net increase of two arcs p er p oint except the rst
The nice thing ab out site events is that they are all known in advance Thus after sorting
the p oints by y co ordinate all these events are known However the other typeofevent are
somewhat harder to predict
Vertex events In contrast to site events vertex events are generated dynamically as the algorithm
runs As with the line segment plane sweep algorithm the imp ortant idea is that each such
event is generated by ob jects that are neighbors on the b each line However unlike the seg
mentintersection where pairs of consecutivesegments generated events here triples of p oints
generate the events
In particular consider any three consecutive p oints p p andp whose arcs app ear consec
i j k
utively on the b each line from left to right See the gure b elow Further supp ose that the
circumcircle for these three sites lies at least partially b elow the currentsweep line meaning
that the Voronoi vertex has not yet b een generated and that this circumcircle con tains no
p oints lying b elowthesweep line meaning that no future p oint will blo ck the creation of the
vertex
Consider the momentatwhichthesweep line falls to a p oint where it is tangenttothelowest
p oint of this circle At this instant the circumcenter of the circle is equidistant from all three
sites and from the sweep line Thus all three parab olic arcs pass through this center p oint
implying that the contribution of the arc from p has disapp eared from the b each line In
j
terms of the Voronoi diagram the bisectors p p and p p have met each other at the
i j j k
Voronoi vertex and a single bisector p p remains
i k
pj pj pj
pi pi
pk pkpi pk
Figure Vertex events
Sweep line algorithm We can now present the algorithm is greater detail The main structures
that we will maintain are the following
Lecture Notes CMSC M
Partial Voronoi diagram The partial Voronoi diagram that has b een constructed so far
will b e stored in a DCEL There is one technical di culty caused by the fact that the
diagram contains unb ounded edges To handle this we will assume that the entire diagram
is to b e stored within a large b ounding b ox This b ox should b e chosen large enough
that all of the Voronoi vertices t within the b ox
Beach line The b each line is represented using a dictionary e g a balanced binary tree or
skip list An imp ortant fact of the construction is that we do not explicitly store the
parabolic arcs They are just their for the purp oses of deriving the algorithm Instead
for each parab olic arc on the current b each line we store the site that gives rise to this
arc Notice that a site mayappearmultiple times on the b each line in fact linearly many
times in n But the total length of the b each line will never exceed n
Between each consecutive pairofsites p and p there is a breakp oint Although the
i j
breakp ointmoves as a function of the sweep line observe that it is p ossible to compute
the exact lo cation of the breakp oint as a function of p p and the current y co ordinate of
i j
the sweep line Thus as with b each lines we do not explicitly storebreakpoints Rather
we compute them only when we need them
The imp ortant op erations that wewillhave to supp ort on the b each line are
Given a xed lo cation of the sweep line determine the arc of the b each line that in
tersects a given vertical line This can b e done by a binary search on the breakp oints
which are computed on the y Think ab out this
Compute predecessors and successors on the b each line
Insert an new arc p within a given arc p thus splitting the arc for p into two This
i j j
creates three arcs p p and p
j i j
Delete an arc from the b eachline
It is not di cult to mo dify a standard dictionary data structure to p erform these op era
tions in O log n time each
Event queue The event queue is a priority queue with the ability b oth to insert and delete
new events Also the event with the largest y co ordinate can b e extracted For each site
we store its y co ordinate in the queue
For each consecutive triple p p p on the b each line we compute the circumcircle of
i j k
these p oints We ll leave the messy algebraic details as an exercise but this can b e done
t of the circle the minimum y co ordinate on the in O time If the lower endp oin
circle lies b elowthesweep line then we create a vertex event whose y co ordinate is the
y co ordinate of the b ottom endp oint of the circumcircle We store this in the priority
queue Eachsuchevent in the priority queue has a cross link back to the triple of sites
that generated it and each consecutive triple of sites has a cross link to the event that it
generated in the priority queue
The algorithm pro ceeds likeany plane sweep algorithm We extract an event pro cess it and
go on to the next event Eacheventmay result in a mo di cation of the Voronoi diagram and
the b each line and may result in the creation or deletion of existing events
Here is howthetwotyp es of events are handled
Site event Let p b e the current site Weshootavertical ray up to determine the arc that
i
lies immediately ab ove this p ointinthebeach line Let p b e the corresp onding site
j
We split this arc replacing it with the triple of arcs p p p whichwe insert into the
j i j
beach line Also we create new dangling edge for the Voronoi diagram which lies on the
bisector b etween p and p Any old triple that involved p as its center arc is deleted
i j j
from the priority queue and we generate new events for each of the p ossible three new
triples that result from this insertion
Lecture Notes CMSC M
Vertex event Let p p andp b e the three sites that generate this event from left to
i j k
right We delete the arc for for p from the b each line We create a new vertex in the
j
Voronoi diagram and tie the edges for the bisectors p p p p to it and start a new
i j j k
edge for the bisector p p that starts growing down b elow Finally we delete anyevents
i k
that arose from triples involving this arc of p and generate new events corresp onding to
j
consecutive triples involving p and p there are two of them
i k
The analysis follows a typical analysis for plane sweep Eacheventinvolves O pro cessing
time plus a constantnumb er accesses to the various data structures Each of these accesses
takes O log n time and the data structures are all of size O n Thus the total time is
O n log n and the total space is O n
Lecture DelaunayTriangulations
Thursday Nov
Sp ecial Announcement Next Tuesday Nov class will meet in AVW to attend Prof
De oriani s lecture on multiresolution surface representations You will b e resp onsible for material
presented in De oriani s talk
Reading BKOS Chapt
The DelaunayTriangulation Last time wegave an algorithm for computing Voronoi diagrams
Todaywe consider the related structure called a Delaunay triangulation Recall that a trian
gulation of a p oint set P is a straight line planar sub division whose vertices are the p oints of
P and which is maximal in the sense that no edge may b e added without violating planarity
This implies that very internal face of this sub division is a triangle and the external face is
b ounded by the b oundary of the convex hull of the p oint set
There are many p ossible triangulations of a given p oint set but one stands out as having the
nicest geometric prop erties This is the Delaunay triangulation The Delaunay triangulation
can b e de ned in a number of equivalentways The de nition that we use is that it is the
straight line dual of the Voronoi diagram of the sites In particular the vertices are the sites
t in the triangulation if and only if their Voronoi cells themselves and two sites are adjacen
share a common edge
Figure DelaunayTriangulation
If the sites are not in general p osition in the sense that four or more are co circular then the
Delaunay triangulation may not b e a triangulation at all but just a planar graph since the
Voronoi vertex that is incidenttofourormoreVoronoi cells will induce a face whose degree is
equal to the number of such cells In this case the more appropriate term would b e Delaunay
graph However it is common to either assume the sites are in general p osition or to enforce
it through some sort of symb olic p erturbation or else to simply triangulate the faces of degree
Lecture Notes CMSC M
four or more in any arbitrary way Henceforth we will assume that sites are in general p osition
so we do not have to deal with these messy situations
Given a p ointset P with n sites where there are h sites on the convex hull it is not hard to
prove that the Delaunay triangulation has n h triangles as wesaw on the exam and
n h edges by a similar pro of Recall that in all planar graphs the average degree of a
vertex is a constant at most
Also it is not hard to prove that the graph is planar Weknow that the graph is top ologically
planar since it is the dual of a planar graph but this only implies that there is some wayto
draw the edges so they do not cross It would still need to b e proven that if the edges are
drawn as straight line segments then they still do not cross See the text for details
An equivalent condition that we mentioned earlier is that every triangle in the Delaunay
triangulation has the prop erty called the empty circle property that the circumcircle of the
three vertices of the triangle do es not contain any other site of P in its interior
Maximizing Angles and Edge Flipping One of the interesting prop erties of Delaunay triangu
lations is that among all triangulations the Delaunay triangulation maximizes the minimum
angle In fact a much stronger statement holds as well Among all triangulations with the
same smallest angle the Delaunay triangulation maximizes the second smallest angle and so
on In particular any triangulation can b e asso ciated with a sorted angle sequence thatis the
increasing sequence of angles app earing in the triangles of the triangulation
m
Note that the length of the sequence will b e the same for all triangulations of the same p oint
set since the numb er dep ends only on n and h
Theorem Among all triangulations of a given p oint set the Delaunay triangulation has the
lexicographically maximum angle sequence
Before getting into the pro of we should recall a few basic facts ab out angles from basic
geometry First recall that if we consider the circumcircle of three p oints then each angle of
the resulting triangle is exactly half the angle of the minor arc subtended by the opp osite two
p oints along the circumcircle It follows as well that if a p oint is inside this circle then it will
subtend a larger angle and a p oint that is outside will subtend a smaller angle This in the
gure part a b elow wehave
θ θ 3 d d 2 ϕ a θ a ab cd ϕbc θ1 θbc ϕda
θda ϕcd θab b c b c
(a) (b) (c)
Figure Angles and edge ips
We will not give a formal pro of of the theorem One app ears in the text The main idea
is to showthatforany triangulation that fails to satisfy the empty circle prop erty itis
p ossible to p erform a lo cal op eration called an edge ip which increases the lexicographical
sequence of angles An edge ip is an imp ortant fundamental op eration on triangulations
Lecture Notes CMSC M
in the plane Given two adjacent triangles abc and cda such that their union forms a
convex quadrilateral abcd the edge ip op eration replaces the diagonal ac with bd Note
that it is only p ossible when the quadrilateral is convex Supp ose that the initial triangle
pair violates the empty circle condition in that p oint d lies inside the circumcircle of abc
implying equivalently that b lies inside the circumcircle of cda then if we ip the edge
it will follows that the two circumcircles of the two resulting triangles abd and bcd are
nowempty relative to these four p oints and the observation ab ove ab out circles and angles
proves that the minimum angle increases at the same time In particular in the gure ab ove
wehave
ab ab bc bc cd cd da da
There are two other angles that need to b e compared as well can you sp ot them but even
though their values might decrease it can b e seen that neither of them is the minimum angle
ou see why after the swap can y
Since there are only a nite numb er of triangulations this pro cess must eventually terminate
with the lexicographically maximum triangulation and this triangulation must satisfy the
empty circle condition and hence is the Delaunay triangulation
Randomized Algorithm This observation provides a basis for a simple but p otentially very slow
algorithm Simply build any initial triangulation e g the plane sweep triangulation presented
in the exam and then start p erforming edge ips until every pair of triangles satis es the empty
circle prop erty However there are no go o d b ounds on the numb er of edge ips required or
what a go o d strategy would b e for selecting them Instead we will present a simple randomized
O n log n exp ected time algorithm for constructing Delaunay triangulations for n sites in the
plane The algorithm is remarkably similar in spirit to the trap ezoidal map algorithm in
that builds its own p oint lo cation data structure as a side e ect We will not discuss the
p oint lo cation data structure in detail but the details are easy to ll in
As with any randomized incremental algorithm the idea is to insert sites in random order
one at a time and up date the triangulation with each new addition The issues involved with
the analysis will b e showing that the numb er of structural changes in the diagram is not very
large it will b e O in the exp ected case for each insertion As with other incremental
algorithm we need some wayofkeeping track of where newly inserted sites are to b e placed
in the diagram Unlike the trap ezoidal map algorithm we will not create a data structure
but will instead use a simpler metho d that puts each of the uninserted p oints into buckets
according to the triangle that it overlaps in the current triangulation In this case we will
need to argue that the n umb er of times that a site is rebucketed on average is not to o large
it will b e O log n in the exp ected case
Incremental up date The basic issue in the design of the algorithm is how to up date the trian
gulation when a new site is added In order to do this we rst investigate the basic prop erties
of a Delaunay triangulation Recall that a triangle abc is in the Delaunay triangulation if
and only if the circumcircle of this triangle contains no other site in its interior Recall that
we make the general p osition assumption that no four sites are co circular Howtowe test
whether a site d lies within the interior of the circumcircle of abc It turns out that this can
b e reduced to a determinant computation It can b e shown that a site d lies in the circumcircle
determined by the counterclo ckwise triangle abc if and only if the following determinantis
p ositive This is called the incircle test We will assume that this primitiveisavailable to us
a a a a
x y
x y
C B
b b b b
x y
x y
C B
in a b c d det
A
c c c c
x y
x y
d d d d
x y
x y
Lecture Notes CMSC M
When we add the next site p the problem is to convert the current Delaunay triangulation
i
into a new Delaunay triangulation This will b e done by creating an illegal nonDelaunay
triangulation containing the new p oint and then incrementally xing this triangulation to
restore the Delaunay prop erties The fundamental changes will b e adding a site to the
middle of a triangle and creating three new edges and p erforming an edge ip Both of
these op erations can b e p erformed in O time assuming that the triangulation is maintained
say as a DCEL
Figure Basic triangulation changes
Here is how the algorithm works We start with an initial triangulation Guibas Knuth and
Sharir suggest starting with the triangle of three sites at in nity Not just any enclosing
triangle will work since we need to b e sure that the newly added vertices will not a ect the
structure of the circumcircles and hence the interior structure of the triangulation This
guarantees that all sites to b e added will lie within some triangle of the existing triangulation
The sites are added in random order When a new site p is added we nd the triangle abc
of the current triangulation that contains this site insert the site in this triangle and join
this site to the three surrounding vertices This creates three new triangles pab pbc and
pca each of whichmayormay not satisfy the empty circle condition Howdowe test this
For each of the triangles that have b een added wecheck the vertex of the triangle that lies
on the other side of the edge that do es not include p Ifthisvertex fails the incircle test then
weswap the edge creating two new triangles that are adjacenttop and rep eat the same test
with these triangles An example is shown b elow
b b b
d d p p p
ca c a c a
Triangle pab is illegal. Triangles pdb and pbc are illegal. Final triangulation.
Figure Point insertion
The following is a description of the algorithm Guibas Knuth and Sharir give a nonrecursive
version The current triangulation is kept in a global data structure The edges in the
following algorithm are actually p ointers to the quad edge data structure
Insert p f
Find the triangle abc containing p
Insert edges pa pb and pc into triangulation
SwapTest ab
Lecture Notes CMSC M
SwapTest bc
SwapTest ca
g
SwapTest ab f
if ab is an edge on the exterior face return
Let d be the vertex to the right of edge ab
if in p a b d f
Replace edge ab with pd
SwaptTest ad
SwaptTest db
g
g
As you can see the algorithm is very simple The only things that have to b e implemented
are the DCEL or other data structure to store the triangulation the incircle test and
lo cating the triangle that contains p Tasks and are straightforward
Task can b e accomplished by one of two means Our text discusses the idea of building a
p oint lo cation data structure for dealing with this The other somewhat simpler approachis
based on a simple bucketing idea whichinvolves the same total amountofwork but it a bit
easier to implement Think of each triangle of the current triangulation as a bucket that holds
the future sites that lie within this triangle and are yet to b e inserted Whenever an edge
is ipp ed or when a triangle is split into three triangles through p oint insertion then some
triangles are destroyed and are replaced by a constantnumb er of new triangles When this
happ ens all the sites in the buckets asso ciated with the old triangles are rebucketed according
to the new triangles that they lie in
There is only one ma jor issue in establishing the correctness of the algorithm When we
p erformed empty circle tests we only tested triangles containing the site p and only
sites that lay on the opp osite side of an edge of such a triangle To establish observe that
it su ces to consider only triangles containing p b ecause since p is the only newly added site
it is the only site that can cause a violation of the empty circle prop erty
To establish we argue that if for every site d which is opp osite from p along some edge ab
lies outside the circumcircle of pab then all these circumcircles are empty A complete pro of
takes some e ort but here is a simple justi cation What could go wrong It might b e that d
lies outside the circumcircle but there is some other site e g a vertex e of a triangle adjacent
to d that lies inside the circumcircle This is illustrated in the following gure We claim that
this cannot happ en It can b e shown that if e lies within the circumcircle of pab then a
must lie within the circumcircle of bde The argument is a exercise in high scho ol geometry
However this violates the assumption that the initial triangulation b efore the insertion of p
was a Delaunay triangulation
Analysis To analyze the algorithm we need to b ound two things howmanychanges are made
in the triangulation on average with the addition of each new site and howmuch e ort is
sp ent in rebuck eting sites
We argue that the exp ected numb er of edge changes with each insertion is O by a simple
application of backwards analysis Observe that whenever the triangulation p erforms an edge
swap italways adds a new edge to p Therefore the total number of changes made in the
triangulation for the insertion of p is equal to the degree of p after the insertion is completed
By backwards analysis weknow that each of the sites that are currently in the triangulation
are equally likely to b e deleted With the exception of the three initial sites at in nity but
they only slightly bias this argument so we ignore them
Lecture Notes CMSC M
d e b a
p
Figure Pro of of su ciency of testing neighb oring sites
By the argument just given weknow that the total number of edge changes needed if p were
the last site added is just the degree of p in the resulting triangulation However we also know
that the average degree of a vertex in a planar graph is constant Therefore the exp ected
numberofchanges to the structure is O p er insertion
Next we argue that the exp ected numb er of times that a site is rebucketed as to which
triangle it lies in is O log n Again this is a standard application of backwards analysis
Consider the i th stage of the algorithm after i sites have b een inserted into the triangulation
Among the remaining n i sites we claim that the probability that this site changes triangles
is at most i Observe that when we insert a new site p all of the newly created triangles
include the site p Each triangle in the current triangulation is determined by exactly sites If
none of these was the last site to b e inserted then no uninserted site in this triangle needs
to b e rebucketed Thus for each of the remaining n i sites the probability that this site
needs to b e rebucketed is at most i The probability is indep endent of the site distribution
and dep ends only on the sequence with which the sites are inserted Thus the total exp ected
numb er of rebuc ketings is given by the sum
n n n
X X X
n i n n n ln n
i i i
i i i
Lecture DeFloriani s Lecture
Tuesday Nov
Sp ecial Announcement Today s class will meet in AVW to attend Prof De oriani s
lecture on multiresolution surface representations You will b e resp onsible for material presented in
De oriani s talk
Lecture Line Arrangements
Thursday Nov
Reading BKOS Chapt
Arrangements So far wehave studied a few of the most imp ortant in my opinion geometric
structures convex hulls and Voronoi diagrams and Delaunay triangulations The next most
imp ortant structure again in my opinion in computational geometry is that of a line ar
rangement Aswithhulls and Voronoi diagrams it is p ossible to de ne arrangements in any
dimension but we will concentrate on the plane As with Voronoi diagrams a line arrangement
is a p olygonal sub division of the plane Unlike most of the structures wehaveseenuptonow
a line arrangement is not de ned in terms of a set of p oints but rather in terms of a set L of
Lecture Notes CMSC M
lines However line arrangements are used mostly for solving problems on p oint sets The con
nection is that the arrangements are typically constructed in the dual plane We will b egin by
de ning arrangements discussing their combinatorial prop erties and how to construct them
and nally discuss applications of arrangements to other problems in computational geometry
Before discussing algorithms for computing arrangements and applications we rstprovide
de nitions and some imp ortant theorems that will b e used in the construction A nite set L
of lines in the plane sub divides the plane The resulting sub division is called an arrangement
denoted A L Arrangements can b e de ned for curves as well as lines and can also b e de ned
for d dimensional hyp erplanes in dimension d Butwe will only consider the case of lines
in the plane here
In the plane the arrangement de nes a planar graph whose vertices are the p oints where two
or more lines intersect edges are the intersection free segments or rays of the lines and faces
are p ossibly unb ounded convex regions containing no line An example is shown b elow vertex
face
edge
bounding box
Figure Arrangementoflines
An arrangementissaidtobe simple if no three lines intersect at a common p oint We will
make the usual general p osition assumptions that no three lines intersect in a single p oint
This assumption is easy to overcome by some sort of symb olic p erturbation
An arrangement is not formally a planar graph b ecause it has unb ounded edges We can
x this top ologically by imagining that a vertex is added at in nity and all the unb ounded
edges are attached to this vertex A somewhat more geometric way to x this is to imagine
that there is a b ounding b ox which is large enough to contain all the vertices and we tie all
the unb ounded edges o at this b ox Rather than computing the co ordinates of this huge
box which is p ossible in O n time it is p ossible to treat the sides of the b ox as existing
at in nity and handle all comparisons symb olically For example the lines that intersect the
right side of the b ox at in nity have slop es b etween and and the order in which they
intersect this side from top to b ottom is in decreasing order of slop e If you don t see this
rightaway think ab out it
The combinatorial complexity of an arrangement is the total number of vertices edges and
faces in the arrangement The following shows that all of these quantities are O n
Theorem Give a set of n lines L in the plane in general p osition
n
i the number of vertices in A L is
ii the number of edges in A L isn and
n
n iii the numb er of faces in A L is
n
is clear from the fact that each pair of lines Pro of The fact that the number of vertices is
we use induction intersects in a single p oint Toprovethatthenumb er of edges is n
The basis case is trivial line and edge When we add a new line to an arrangement
of n lines having n edges by the induction hyp othesis we split n existing
edges thus creating n new edges and weaddn new edges from the n intersections
Lecture Notes CMSC M
with the new line This gives a total of n n n n Thenumb er of faces
comes from Euler s formula v e f with the little trickthatwe need to create
one extra vertex to attach all the unb ounded edges to
Incremental Construction Arrangements are used for solving many problems in computational
geometry But in order to use arrangements we rst must b e able to construct arrangements
We will present a simple incremental algorithm which builds an arrangementby adding lines
one at a time Unlike most of the other incremental algorithms wehave seen so far this one
will not require randomization In fact its asymptotic running time will b e the same O n
no matter what order we insert the lines This is asymptotically optimal since there this is
the size of the arrangement The algorithm will also require O n since this is the amount
of storage needed to store the nal result Later we will consider the question of whether it
is p ossible to traverse an arrangement without actually building it
As usual we will assume that the arrangement is represented in any reasonable data structure
for planar graphs a DCEL for example Let L f g denote the lines We will
n
simply add lines one by one to the arrangement The order do es not matter We will show
that the i th line can b e inserted in O i time If wesumover i this will give O n total
time
Supp ose that the rst i lines have already b een added and consider the e ort involved
in adding Recall our assumption that the arrangement is assumed to lie within a large
i
bounding box Since each line intersects this b oxtwice the rst i lines have sub divided
the b oundary of the b oxinto i edges We determine where intersects the b ox and
i
which of these edge it crosses intersects This will tell us which face of the arrangement rst
i
enters
Next we trace the line through the arrangement from one face to the next Whenever weenter
a face the main question is where do es the line exit the face Weanswer the question bya
very simple strategy Wewalk along the edges face say in a counterclo ckwise direction recall
that a DCEL allows us to do this and as we visit each edge we ask whether intersects this
i
edge When we nd the edge through which exits the face we jump to the face on the other
i
side of this edge again the DCEL supp orts this and continue the trace This is illustrated in
the gure b elow
l l
traversing the arrangement the zone
Figure Traversing an arrangement and zones
Once weknowthepoints of entry and exit into each face the last matter is to up date the
DCEL by inserting edges for eachentry exit pair This is easy to handle in constant time by
adjusting the appropriate p ointers in the DCEL Details are left as an exercise
Clearly the time that it takes to p erform the insertion is prop ortional to the total number
of edges that havebeentraversed in this tracing pro cess Anaive argumentsays that we
encounter i lines and hence pass through i faces assuming general p osition Since each
Lecture Notes CMSC M
face is b ounded byatmosti lines each facial traversal will take O i time and this gives a
total O i Hey whatwent wrong Ab ovewe said that wewould do this in O i time The
claim is that the traversal do es indeed traverse only O i edges but to understand why we
need to delve more deeply into a concept of a zone of an arrangement
Zone Theorem The most imp ortant combinatorial prop erty of arrangements which is critical to
their e cient construction is a rather surprising result called the zone theorem Given an
arrangement A of a set L of n lines and given a line that is not in L the zone of in A
denoted Z is the set of faces whose closure intersects The gure ab ove illustrates a
A
ve construction we are only interested in the zone for the line For the purp oses of the ab o
edges of the zone that lie b elow but if we b ound the total complexity of the zone then this
i
will b e an upp er b ound on the numb er of edges traversed in the ab ove algorithm
The combinatorial complexity of a zone as argued ab ove is O n The Zone theorem states
that the complexity is actually much smaller only O n
Theorem The Zone Theorem Given an arrangement A L ofn lines in the plane and given
any line in the plane the total numb er of edges in all the cells of the zone Z isat
A
most n
Pro of As with most combinatorial pro ofs the key is to organize everything so that the
counting can b e done in an easy way Note that this is not trivial b ecause it is easy to
see that any one line of L mightcontribute manysegments to the zone of The key in
the pro of is nding a way to add up the edges so that each line app ears to induce only a
constantnumberofedgesinto the zone
The pro of is based on a simple inductive argument We will rst imagine through a
suitable rotation that is horizontal and further that none of the lines of L is horizon tal
through an in nitesimal rotation We split the edges of the zone into two groups those
that b ound some face from the left side and those that b ound some face from the right
side More formally since each face is convex if we split it at its topmost and b ottommost
vertices we get twoconvex chains of edges The left bounding edges are on the left chain
and the right bounding edges are on the rightchain We will show that there are at most
n lines that b ounded faces from the left Note that an edge of the zone that crosses
itself contributes only once to the complexity of the zone In the b o ok s pro of they seem
to count this edge twice and hence their b ound they get a b ound of n instead We will
also ignore the edges of the b ounding b ox
For the basis case when n then there is exactly one left b ounding edge in s zone
and n Assume that the hyp othesis is true for anysetofn lines Consider the
rightmost line of the arrangementtointersect Call this Selecting this particular
line is very imp ortant for the pro of Supp ose that we consider the arrangementofthe
other n lines By the induction hyp othesis there will b e at most n left b ounding
edges in the zone for
Now let us add back and see howmany more left b ounding edges result Consider the
n
rightmost face of the arrangementofn lines Note that all of its edges are left b ounding
edges Line will intersect within this face Let e and e denote the two edges of
a b
this that intersects one ab ove and the other b elow The insertion of creates a
new left b ounding edge along itself and splits the left b ounding edges e and e into
a b
two new left b ounding edges for a net increase of three edges Observethat cannot
contribute any other left b ounding edges to the zone b ecause dep ending on slop e either
the line supp orting e or the line supp orting e blo cks s visibilityfromfrom Note
a b
that it might provide right b ounding edges but we are not counting them here Thus
the total numb er of left b ounding edges on the zone is at most n n and
hence the total numb er of edges is at most n as desired
Lecture Notes CMSC M
ea
l1
l
eb
Figure Pro of of the Zone Theorem
Lecture Applications of Arrangements
Tuesday Nov
Reading This material is not covered in the text
Revised Nov Added material ab out angular sorting
Sweeping Arrangements Last time weshowed how to construct an arrangement of lines in the
plane in O n time Since an arrangement is of size O n this is optimal Usually it is not
su cient just to build the arrangement but necessary to traverse the arrangementaswell In
some instances any graph traversal will su ce For other problems it is nice to p erform the
traversal in some order for example as a plane sweep from left to right
If an arrangement is to b e built just so it can b e swept then mayb e you didn t need to construct
the entire arrangement after all You can just p erform the plane sweep on the lines exactly as
we did for the line segmentintersection algorithm Assuming that we are sweeping from left to
right the initial p osition of the sweep line is at x computed symb olically The sweep
line status maintains the lines in top to b ottom order according to their intersection with the
sweep line The initial order is according to increasing order of slop e Then the sweep line
pro ceeds exactly as it did in the algorithm for determining line segmentintersections but the
only events are intersection events since there are no endp oints The sweep ends when the
last intersection event is pro cessed Sweeping an arrangement in this w aytakes O n log n
time but only O n space since we need only maintain the currentsweep line For many
applications involving arrangements this o ers a reasonable tradeo b etween time and space
But is the O log n factor necessary In many applications involving plane sweep it is not
necessary to sweep in strictly left to right order as long as the vertices lying on each line of
the arrangement are swept from left to right In this case wecansavetheO log n factor by
constructing the arrangement then thinking of the edges as b eing directed from left to right
p erform a topological sort of the vertices A top ological sort of an acyclic directed graph is
an ordering of the vertices of a directed graph such that for eachedge u v in the graph u
app ears b efore v in the order It is well known that the vertices of a graph can b e top ologically
sorted in time that is prop ortional to the size of the graph or O n in this case See CLR
for details Given the top ological ordering of the vertices of the arrangement wecansweep
the arrangementby simply extracting the vertices in this order This gives an O n time and
space algorithm for sweeping arrangements O n
Is it p ossible to sweep arrangements in O n time but with only O n space It turns out
that the answer is yes There is an algorithm called topological plane sweep which incorp orates
Lecture Notes CMSC M
5 9 1 4 6
7 12 2 14 13 16 15 11 8
3 10
Figure One p ossible top ological ordering of the vertices of an arrangement
the top ological nature of the ab ovesweeping algorithm with the incremental nature of plane
sweep
Applications of Arrangements and Duality The computational and mathematical to ols that
wehavedevelop ed with geometric duality and arrangements allow a large numb er of problems
to b e solved Here are some examples Unless otherwise stated all these problems can b e
solved in O n time and O n spaceby constructing a line arrangement or in O n time
and O n space through top ological plane sweep
General p osition test Given a set of n points in the plane determine whether any three
are collinear
Minimum area triangle Given a set of n points in the plane determine the minimum area
triangle whose vertices are selected from these p oints
Minimum k corridor Given a set of n p oints and an integer k determine the narrowest
pair of parallel lines that enclose at least k points of the set The distance b etween the
lines can b e de ned either as the vertical distance b etween the lines or the p erp endicular
distance b etween the lines
Visibility graph Given line segments in the plane wesay that twopoints are visible if the
interior of the line segment joining them intersects none of the segments Given a set
of n non intersecting line segments compute the visibility graph whose vertices are the
endp oints of the segments and whose edges a pairs of visible endp oints
Maximum stabbing line Given a set of n line segments in the plane determine whether
the line that stabs intersects the maximum numb er of these line segments
Hidden surface removal Given a set of n non intersecting p olygons in space imagine
pro jecting these p olygon onto a plane either orthogonally or using p ersp ective Deter
mine which p ortions of the p olygons are visible from the viewp oint under this pro jection
Note that in the worst case the complexity of the nal visible scene may b e as high as
O n so this is asymptotically optimal However since such complex scenes rarely o ccur
in practice this algorithm is really only of theoretical interest
Ham SandwichCut Given n red p oints and m blue p oints nd a single line that simul
taneously bisects these p oint sets It is a famous fact from mathematics called the
Ham Sandwich Theorem thatsuch a line always exists If the p oint sets are separated
O n m by a line then this can b e done in time O n m space
Sorting all angular sequences Here is a natural application of duality and arrangements that
turns out to b e imp ortant for the problem of computing visibility graphs Consider a set of n
Lecture Notes CMSC M
p oints in the plane For eachpoint p in this set wewant to p erform an angular sweep sayin
counterclo ckwise order visiting the other n points of the set For each p oint it is p ossible
to compute the angles b etween this p oint and the remaining n points and then sort these
angles This would take O n log n time p er p oint and O n log n timeoverall
With arrangements we can sp eed this up to O n total time getting rid of the extra O log n
factor Here is how Recall the duality transformation describ ed in Lecture Apoint
p p p and line y ax b in the primal plane are mapp ed through duality to dual
x y
p oint and dual line p by the following relationship
a b
p b p a p
x y
Observe that the a co ordinate in the dual plane corresp onds to the slop e of a line in the primal
plane Supp ose that p is the p oint that wewant to sort around and let p p p b e the
n
p oints in nal angular order ab out p
p p 4 7 p p p 5 6 8 p p p 5 2 p 3 6 p p 2 p 3
p p 4 7 p p 1 p
8 p 1
Figure Arrangements and angular sequences
Consider the arrangement de ned by the dual lines p How is this order revealed in the
i
arrangement Consider the dual line p and its intersection p oints with each of the dual lines
p These form a sequence of vertices in the arrangementalongp Consider this sequence
i
ordered from left to right It would b e nice if this order were the desired circular order but
this is not quite correct The a co ordinate of each of these vertices in the dual arrangementis
the slop e of some line of the form pp Thus the sequence in whichthevertices app ear on the
i
line is a slopeordering of the p oints ab out p not an angular ordering
i
However given this slop e ordering we can simply test which primal p oints lie to the left of p
that is have a smaller x co ordinate in the primal plane and separate them from the p oints
that lie to the rightofp having a larger x co ordinate We partition the vertices into two
sorted sequences and then an concatenate these two sequences with the p oints on the right
side rst and the p oints on the left side later The resulting is an angular sequence starting
with the angle degrees and pro ceeding up to degrees
Thus once the arrangement has b een constructed we can reconstruct each of the angular
orderings in O n time for a total of O n time
Lecture More Applications of Arrangements
Thursday Nov
Reading The material on discrepancies is in Chapt of BKOS The other material is not covered
in the text
Lecture Notes CMSC M
Applications of Arrangements Todaywe will discuss a few of applications of line arrangements
in the plane This is to give a feel for how one converts problems into dual form and then uses
arrangements to solve these problems Recall the duality op erator from Lecture We de ned
a function that maps p oints in the primal plane to lines in the dual plane and lines in the
primal plane to p oints in the dual plane We denoted it using a asterisk as a sup erscript
Thus given p oint p p p and line y ax b in the primal plane wede ne and
x y
p to b e a p oint and line resp ectively in the dual plane de ned by
a b
p b p a p
x y
The main prop erty of this transformation was the following order reversing prop erty a p oint
p lies ab ove on b elow line in the primal plane if and only if line p passes b elow on ab ove
p oint in the dual plane resp ectively
Narrowest corridor As mentioned ab ove in this problem we are given a set P of n p oints
in the plane and an integer k k n and we wish to determine the narrowest pair of
parallel lines that enclose at least k p oints of the set In this case we will de ne the vertical
distance b etween the lines as the distance to minimize It is easy but a bit tricky to adapt
the algorithm for p erp endicular distance
To simplify the presentation we assume that k The generalization to general k is an
exercise We will assume that no three p oints of P are collinear This will allow us to assume
that the narrowest corridor contains exactly three p oints and has width strictly greater than
zero Observe that if three p oints were collinear then they would corresp ond to three lines
that intersect at a common vertex in the arrangement which could b e tested as we build the
arrangement We will also assume that no two p oints have the same x co ordinate The dual
transformation that we consider cannot representvertical lines but this assumption will imply
that the solution is not a vertical line We could x this by using homogeneous co ordinates
but we will not worry ab out it now
If we dualize the p oints of P then in dual space wehave a set L of n lines f g The
n
slop e of each dual line is the x co ordinate of the corresp onding p ointofP and its y intercept
is the negation of the p oint s y co ordinate
Anarrowest corridor in the primal plane consists of two parallel lines and in primal
a b
space Their duals and are dual p oints whichhave the same x co ordinates since the
a
b
lines are parallel and the vertical distance b etween these p oints is the di erence in the y
intercepts of the two primal lines Thus the vertical width of the corridor is the vertical length
of the line segment
In the primal plane there are exactly three p oints lying in the corridor that is there are three
p oints that are b oth ab ove and b elow Thus by the order reversing prop erty in the dual
b a
Combining all plane there are three dual lines that pass b oth b elowpoint and ab ove
a
b
these observations it follows that the dual formulation of the narrowest corridor problem is
the following
Shortest vertical stabb er Given an arrangementofn lines determine the shortest ver
tical segment that stabs three lines of the arrangement
It is easy to show by a simple p erturbation argument that the shortest vertical stabb er
may b e assumed to have one of its endp oints on a vertex of the arrangement implying that the
other endp oint lies on the line of the arrangement lying immediately ab oveorbelowthisvertex
In the primal plane the signi cance is that we can assume that the minimum corridor one
Lecture Notes CMSC M
l p* a * lb r q* p lb d q d r* la*
Primal Plane Dual Plane
Figure Narrowest corridor primal and dual form
of the lines passes through of the p oints and the other passes through a third p oint and
there are no p oints within the interior of the corridor This is shown in the gure b elow
We can compute the minimum stabb er in an arrangement by a simple plane sweep of the
arrangement using a vertical sweep line Whenever we encounter a vertex of the arrangement
we consider the distance to the edge of the arrangement lying immediately ab ove this vertex
and the edge lying immediately b elow We can solve this problem by top ological plane sweep
in O n timeandO n space
We can also solve this by constructing the arrangement and the computing the vertical
trap ezoidal map Each trap ezoidal edge will corresp ond to a corridor The shortest such edge
is the nal answer This leads to an O n time and space solution
Figure Narrowest corridor and trap ezoidal maps
Maximum Discrepancy Next we consider a problem derived from computer graphics and sam
pling Supp ose that we are given a collection of n points S lying in a unit square U
Wewant to use these p oints for random sampling purp oses In particular the prop erty that
wewould like these p oints to have is that for any halfplane h wewould like the size of the
fraction of p oints of P that lie within h should b e roughly equal to the area of intersection of
h with U That is if we de ne h to b e the area of h U and h jS hj jS j then
S
wewould like h h for all h This prop erty is imp ortant when p oint sets are used for
S
things like sampling and Monte Carlo integration
To this end w e de ne the discrepancy of S with resp ect to a halfplane h to b e
h j h h j
S S
For example in the gure the area of h U is h and there are out of p oints
in h thus h Thus the discrepancy of h is j j De ne
S
the halfplane discrepancy of S to b e the maximum or more prop erly the supremum or least
Lecture Notes CMSC M
h
Figure Discrepancy of a p oint set
upp er b ound of this quantityover all halfplanes
S sup h
S
h
Since there are an uncountably in nite numb er of halfplanes it is imp ortant to derive some
sort of niteness criterion on the set of halfplanes that might pro duce the greatest discrepancy
Lemma Let h denote the halfplane that generates the maximum discrepancy with resp ect
to S and let denote the line that b ounds h Then either i passes through at least
two p oints of S or ii passes through one p ointofS and this p oint is the midp ointof
the line segment U
Remark If a line passes through one or more p oints of S then should this p oint b e included
in h For the purp oses of computing the maximum discrepancy the answer is to
S
either include or omit the p oint whichever will generate the larger discrepancy The
justi cation is that it is p ossible to p erturb h in nitesimally so that it includes none or
all of these p oints without alterning h
Pro of If do es not pass through anypointofS then dep ending on which is larger h
or h wecanmove the line up or down without changing h and increasing or
S S
decreasing h to increase their di erence If passes through a p oint p S but is not
the midp oint of the line segment U thenwe can rotate this line ab out p and hence
increase or decrease h without altering h to increase their di erence
S
Since for eachpoint p S there are only a constantnumber of lines at most two I think
through this p ointsuch that p is the midp ointof U it follows that there are at most O n
lines of typ e i ab ove and hence the discrepancy of all of these lines can b e tested in O n
time
To compute the discrepancies of the other lines we can dualized the problem In the primal
p S is mapp ed in the dual plane to a plane a line that passes through two p oints p
i j
p oint at which the lines p and p intersect This is just a vertex in the arrangement of the
i j
dual lines for S So if wehave computed the arrangement then all we need to do is to visit
eachvertex and compute the discrepancy for the corresp onding primal line
It is easy to see that the area U of each corresp onding line in the primal plane can b e
computed in O time So all that is needed is to compute the number of points of S lying
beloweach such line In the dual plane the corresp onds to determining the numb er of dual
lines that lie b eloworabove eachvertex in the arrangement If weknowthenumb er of dual
lines that lie strictly ab oveeachvertex in the arrangement then it is trivial to compute the
numb er of lines that lie b elowby subtraction
We de ne a p ointtobeat level k L in an arrangement if there are at most k lines ab ove
k
el of an arrangementisan this p oint and at most n k lines b elow this p oint The k th lev
x monotone p olygonal curve as shown b elow For example the upp er envelop e of the lines is
Lecture Notes CMSC M
level of the arrangement and the lower envelop e is level n of the arrangement Note that a
vertex of the arrangement can b e on multiple levels Also note that our de nition of level is exactly one greater than our text s de nition
L1
L3
L5
Figure Levels in an arrangement
We claim that it is an easy matter to compute the level of eachvertex of the arrangement e g
by plane sweep The initial levels at x are determined by the slop e order of the lines
As the plane sweep pro ceeds the index of a line in the sweep line status is its level Thus by
using top ological plane sweep in O n time we can compute the minimum and maximum level
number of eachvertex in the arrangement From the order reversing prop erty for eachvertex
of the dual arrangement the minimum level number minus indicates the numb er of primal
p oints that lie strictly b elow the corresp onding primal line and the maximum level number is
the numb er of dual p oints that lie on or b elow this line Thus given the level numb ers and
the fact that areas can b e computed in O time we can compute the discrepancy in O n
time and O n space through top ological plane sweep
Lecture Shortest Paths and Visibility Graphs
Tuesday Nov
Reading The material on visibility graphs is tak en roughly from Chapter but we will present
slightly more e cientvariant of the one that app ears in this chapter
Shortest paths We are given a set of n disjoint p olygonal obstacles in the plane and two p oints
s and t that lie outside of the obstacles The problem is to determine the shortest path from s
to t that avoids the interiors of the obstacles It maytravel along the edges or pass through
the vertices of the obstacles The complement of the interior of the obstacles is called free
space Wewant to nd the the shortest path that is constrained to lie entirely in free space
Todaywe consider a simple but p erhaps not the most e cient waytosolve this problem We
assume that we measure lengths in terms of Euclidean distances Howdowe measure paths
lengths for curved paths Luckily wedonothave to b ecause we claim that the shortest path
will always b e a p olygonal curve
Claim The shortest path b etween anytwopoints that avoids a set of p olygonal obstacles is
a p olygonal curve whose vertices are either vertices of the obstacles or the p oints s and
t
Pro of Weshowthatanypath that violates these conditions can b e replaced b y a slightly
shorter path from s to t Since the obstacles are p olygonal if the path were not a p olygonal
curve then there must b e some p oint p in the interior of free space such that the path
passing through p is not lo cally a line segment If we consider any small neighborhood
Lecture Notes CMSC M
ab out p small enough to not contain s or t or anypartofany obstacle then since the
shortest path is not lo cally straight we can shorten it slightly by replacing this curved
segmentby a straight line segment jointing one end to the other Thus is not shortest
acontradiction
Thus is a p olygonal path Supp ose that it contained a vertex v that was not an obstacle
vertex Again we consider a small neighb or ho o d ab out v that contains no part of any
obstacle We can shorten the path as ab ove implying that is not a shortest path
From this it follows that the edges that constitute the shortest path must travel b etween s and
t and vertices of the obstacles Each of these edges must have the prop erty thatitdoesnot
intersect the interior of any obstacle implying that the endp oints must b e mutually visible
De nition The visibility graph of s and t and the obstacle set is a graph whose vertices are s
and t the obstacle vertices and vertices v and w are joined by an edge if the line segment
v w lies entirely in free space
s
t
Figure Visibility graph
It follows from the ab ove claim that the shortest path can b e computed by rst computing
the visibility graph and lab eling each edge with its Euclidean length and then computing the
shortest path by say Dijkstra s algorithm see CLR Note that the visibility graph is not
planar and hence may consist of n edges Also note that even if the input p oints have
integer co ordinates in order to compute distances we need to compute square ro ots and then
sums of square ro ots This can b e approximated by oating p oint computations If you are
a stickler for exactness this can really b e a problem b ecause there is no known p olynomial
time pro cedure for p erforming arithmetic with arbitrary square ro ots of integers
Computing the Visibility Graph Next we givean O n pro cedure for constructing the visi
bility graph of n line segments in the plane The more general task of computing the visibility
graph of an arbitrary set of p olygonal obstacles is a very easy generalization In this context
twovertices are visible if the line segment joining them do es not intersect any of the obstacle
line segments However we allow each line segmenttocontribute itself as an edge in the visi
bility graph We will make the general purp ose assumption that no three vertices are colinear
but this is not hard to handle with some care The algorithm is not output sensitive An
O n log n k algorithm do es exist but it quite complicated
The text gives an O n log n time algorithm W e will givean O n time algorithm Both
algorithms are based on the same concept namely that of p erforming an angular sweep around
eachvertex The text s algorithm op erates by doing this sweep one vertex at a time Our
algorithm do es the sweep for all vertices simultaneously Using the fact that angular sorting
can b e implemented through top ologically sweeping the a line arrangementin O n time we
manage to shaveo theextraO log n factor in running time
Here is a high level intuitive view of the algorithm First recall the algorithm for computing
trap ezoidal maps We sho ot a bullet up and down from every vertex until it hits its rst
line segment This implicitly gives us the vertical visibility relationships b etween vertices
Lecture Notes CMSC M
and segments Now we imagine that angle continuously sweeps out all slop es from
to Imagine that all the bullet lines attached to all the vertices b egin to turn slowly
counterclo ckwise If we play the mind exp eriment of visualizing the rotation of these bullet
paths the question is what are the signi cantevent p oints and what happ ens with each
event As the sweep pro ceeds we will eventually determine everything that is visible from
every vertex in every direction Thus it should b e an easy matter to piece together the edges
of the visibility graph as we go
Figure Visibility graph bymultiple angular sweep
Let us consider this multiple angular sweep in greater detail Observe that a signi cant
event o ccurs whenever a bullet path jumps from one line segment to another This o ccurs
when reaches the slop e of the line joining two visible vertices v and w Unfortunately itis
somewhat complicated to keep trackofwhichvertices are visible and which are not although
if we could do so it would lead to a more e cient algorithm Instead we will takeevents to
be al l angles between twovertices whether they are visible or not By duality the slop e of
suchanevent will corresp ond to the a value of the intersection of lines v and w in the dual
arrangement Convince yourself of this b efore going on Thus bysweeping the arrangement
we will generate all these events
Next let s consider what happ ens at eacheventpoint Consider the state of the angular sweep
algorithm for some slop e For eachvertex v therearetwo bullet paths emanating from v
along the line with slop e Call one the forward bul let path and the other the backward bul let
path Letf v andb v denote the line segments that these bullet paths hit resp ectively If
either path do es not hit any segmentthenwe store a sp ecial null value As varies the following
events can o ccur Assuming through symb olic p erturbation that each slop e is determined by
exactly two lines whenever we arriveatanevents slop e there are exactly twovertices v and
w that are involved Here are the p ossible scenarios
Same segment If v and w are endp oints of the same segment then they are visible and we
add the edge v w to the visibility graph
Invisible Consider the distance from v to w First determine whether w lies on the same
side as f v orb v For the remainder assume that it is f v The case of b v is
symmetrical
Lecture Notes CMSC M
f(v) f(v) (new) w w (new) w w f(v)(old) v v v v f(v)(old) f(v)
Same segment Invisible Entry Exit
Figure Possible events
Compute the contact p oint of the bullet path shot from v in direction with segment
f v If this path hits f v strictly b efore w thenwe knowthat w is not visible to v and
so this is a nonevent
Segmententry Consider the segmentthatisincidenttow Either the sweep is just ab out
to enter this segment or is just leaving it If we are entering the segment then we set
f v to this segment
Segment exit If we are just leaving this segment then then the bullet path will need to
sho ot out and nd the next segmentthatithits Normally this would require some
searching In particular this is one of the reasons that the text s algorithm has the extra
O log n factor to p erform this search However we claim that the answer is available
to us in O time
In particular since we are sweeping over w at the same time that we are sweeping over
v Thus weknow that the bullet extension from w hits f w All we need to do is to set
f v f w
This is a pretty simple algorithm although there are a numb er of cases The only information
that we need to keep track of is a priority queue for the events and the f v andb v
p ointers for eachvertex v The priority queue is not stored explicitly Instead it is available
from the line arrangement of the duals of the line segmentvertices By p erforming a top ological
sweep of the arrangement we can pro cess all of these events in O n time
There is one subtlety ab out top ological sweep that we should b e careful ab out though Re
memb er that top ological sweep and regular plane sweep are not identical in that a top ological
sweep do es pro cess events in strictly left to right order However we are assured that along
any dual line the sweep will pro cess events from left to right along the line
Here is our fear In the Segment Exit event we accessed f w But since the sweeps along the
dual lines for v and w are pro ceeding indep endently in general mightitbethat f w is lagging
b ehind or has moved ahead of the sweep along v If so this could b e disastrous b ecause f w
would b e set for some other angle To see whythisisnotadanger we need to lo ok at what
is happ ening in the arrangement At the momentofthisevent we are pro cessing a line that
passes through b oth v and w In the dual plane this line corresp onds to the intersection p oint
of lines v and w in the dual plane The top ological plane sweep sweeps eachvertex of the
arrangement exactly once implying that these events will b e pro cessed at the same time in
the top ological plane sweep and so there is no danger of one b eing ahead or lagging b ehind
the other
There are examples of algorithms where the natural events to b e pro cessed in top ological plane
sweep do not corresp ond to single vertices but rather to groups of events in di erent parts
of the arrangement that are supp osed to o ccur at the same time but might b e out of phase
Lecture Notes CMSC M
b ecause of the top ological nature of the sweep Do es this mean that top ological sweep cannot
b e used Sometimes yes and sometimes no Recall that the top ological sweep arises from a
top ological sort of a DAG the directed edges of the arrangement Sometimes it is p ossible
to add additional edges i e additional ordering constraints thus forcing the sweep to b ehave
itself
Lecture Motion Planning
Tuesday Dec
Reading Chapt in BKOS
Motion planning Last time we considered the problem of computing the shortest path of a p oint
in space around a set of obstacles Todaywe will study a very general approach to the more
general problem of how to plan the motion of one or more rob ots each with p otentially many
degrees of freedom in terms of its movement and p erhaps having articulated joints
Work Space and Con guration Space The environment in which the rob ot op erates is called
its work space which consists of a set of obstacles that the rob ot is not allowed to intersect We
assume that the work space is static that is the obstacles do not move We also assume that
a complete geometric description of the work space is available to us There are interesting
variants of motion planning in environments where the obstacles either unknown or are known
but move In unknown environments the rob ot must have some way of sensing its environment
either by sightortouch In environments with moving ob jects the rob ot m ust b e capable of
detecting the presense of moving ob jects early enough that it can avoid colliding with them
Henceforth let S denote the set of obstacles de ning the environment
For our purp oses a robot will b e mo deled bytwo main elements The rst is a con guration
which is a nite sequence of values that fully sp eci es the p osition of the rob ot The second
element is the rob ot s geometric shap e description Combined these two element fully de ne
the rob ot s exact p osition and shap e in space For example supp ose that the rob ot is a
dimensional p olygon that can translate and rotate in the plane Its con guration maybe
describ ed by the x y co ordinates of some reference p oint for the rob ot and an angle
that describ es its its orientation Its geometric information would include its shap e sayat
some canonical p osition given say as a simple p olygon Given its geometric description and
a con guration x y this uniquely determines the exact p osition R x y of this rob ot
in the plane Thus the p osition of the rob ot can b e identi ed with a p oint in the rob ot s
con guration space
R(4,3,45)
R(0,0,0)
Figure Con gurations of a translating and rotating rob ot
A more complex example would b e an articulatedarmconsisting of a set of links connected
to one another by a set of revolute joints The con guration of sucharobotwould consist
ofavector of joint angles The geometric description would probably consist of a geometric
Lecture Notes CMSC M
representation of the links Given a sequence of joint angles the exact shap e of the rob ot
could b e derived bycombining this con guration information with its geometric description
For example a typical dimensional industrial rob ot has six joints and hence its con guration
can b e thoughtofasapoint in a dimensional space Why six Generally there are three
degrees of freedom needed to sp ecify a lo cation in space and more degrees of freedom
needed to sp ecify the direction and orientation of the rob ot s end manipulator
Given a p oint p in the rob ot s con guration space let R p denote the placement of the rob ot
at this con guration The gure b elow illustrates this in the case of the planar rob ot de ned
ab ove
Work space Configuration space
Figure Work space and con guration space
Because of limitations on the rob ot s physical structure and the obstacles not every p ointin
con guration space corresp onds to a legal placement of the rob ot Any con guration whichis
illegal in that it causes the rob ot to intersect one of the obstacles is called a forbidden con gu
ration The set of all forbidden con gurations is denoted C R S and all other placements
forb
are called freecon gurations and the set of these con gurations is denoted C R S
free
Now consider the motion planning problem in rob otics Given a rob ot R anwork space S and
initial con guration s and nal con guration t b oth p oints in the rob ot s free con guration
space determine if p ossible a waytomove the rob ot from one con guration to the other
without intersecting any of the obstacles This reduces to the problem of determining whether
there is a path from s to t that lies entirely within the rob ot s free con guration space Thus
we map the task of computing a rob ot s motion to the problem of nding a path for a single
p oint through a collection of obstacles
Con guration spaces are typically higher dimensional spaces and are typically b ounded by
curved surfaces esp ecially when rotational elements are involved Perhaps the simplest case
to visualize is that of translating a convex p olygonal rob ot in the plane amidst a collection
of p olygonal obstacles In this cased b oth the work space and con guration space are two
dimensional Consider a reference p oint placed in the center of the rob ot As shown in the
gure ab ove the pro cess of mapping to con guration space involves replacing the rob ot with
t its reference p oint and growing the obstacles by a comp ensating amount a single p oin
These grown obstacles are called con guration obstacles or C obstacles
This approach while very general ignores many imp ortant practical issues It assumes that we
have complete knowledge of the rob ot s environment and have p erfect knowledge and control
of its placement As stated we place no requirements on the nature of the path but in reality
physical ob jects can not b e broughttomove and stop instantaneously Nonetheless this
abstract view is very p owerful since it allows us to abstract the motion planning problem into
avery general framework
Lecture Notes CMSC M
For the rest of the lecture we will consider a very simple case of a convex p olygonal rob ot that
is translating among a convex set of obstacles Even this very simple problem has a number
of interesting algorithmic issues
Planning the Motion of a Point Rob ot As mentioned ab ove we can reduce complex motion
planning problems to the problem of planning the motion of a p oint in free con guration
space First we will consider the question of how to plan the motion of a p oint amidst a set of
p olygonal obstacles in the plane and then we will consider the question of how to construct
con guration spaces
To determine whether there is a path from one p oint to another of free con guration space we
will sub divide free space into simple convex regions In the plane we already knowhowtodo
this by computing a trap ezoidal map We can construct a trap ezoidal map for all of the line
segments b ounding the obstacles then throwawayany faces that lie in the forbidden space
We also assume that wehaveapoint lo cation data structure for the trap ezoidal map
Next we create a planar graph called a road map based on the trap ezoidal map To do this
we create a vertex in the center of each trap ezoid and a vertex at the midp ointofeachvertical
edge We create edges joining eachcenter vertex to the vertices on its at most four edges
t
s
Figure Motion planning using road maps
Now to answer the motion planning problem weassumewe are given the start p oint s and
destination p oint t We lo cate the trap ezoids containing these twopoints and connect them
to the corresp onding center vertices We can join them by a straight line segment b ecause the
cells of the sub division are convex Then we determine whether there is a path in the road
map graph b etween these twovertices sayby breadth rst search Note that this will not
necessarily pro duce the shortest path but if there is a path from one p osition to the other it
will nd it
This description ignores many practical issues that arise in motion planning but it is the
basis for many practical motion planning problems More realistic con guration spaces will
contain more information for example enco dings of the currentjoint rotation velo cities
and will usually re ne the road map to a much nerextent so that short paths can b e
approximated well as well as handling other elements suchasguaranteeing minimal clearances
around obstacles
Con guration Obstacles and Minkowski Sums Nowthatwe knowhow to nd paths in con
guration space let us consider how to build con guration space for a set of p olygonal obsta
cles We again consider the simplest case of translating a convex p olygonal rob ot amidst a
collection of convex obstacles If the obstacles are not convex then wemay sub divide them
into convex pieces
Lecture Notes CMSC M
Consider a rob ot R whose placementisde nedby a translation x y so that R x y denotes
the placement of the rob ot Given an obstacle P the con guration obstacle is de ned as all
the placements of R that intersect P that is
CP f x y jR x y P g
One way to visualize CP is to imagine scraping R along the b oundary of P and seeing the
region traced out by R s reference p oint
The problem we consider next is given R and P compute the con guration obstacle CP To
do this we rst intro duce the notion of a Minkowski sum Let us violate our notions of a ne
geometry for a while and think of p oints x y in the plane as vectors Given anytwo sets S
and S in the plane de ne their Minkowski sum to b e the set of all pairwise sums of p oints
taken from eachset
p S q S g S S fp q j
Also de ne S f p j p S g Observe that for the case of a translating rob ot we could
de ne R x y asR f x y g
CP P+(-R(0,0))
P P
R(0,0) -R(0,0)
Figure Con guration obstacles and Minkowski sums
Claim Given a planar translating rob ot R and an obstacle P then the C obstacle of P is
P R
Pro of We show that R x y intersects P if and only if x y P R First if
R x y intersects P then there must b e a p oint q q q such that q P R x y
x y
This implies that q x q y R and hence x q y q R Since
x y x y
ve x y P R q P by adding q q tothisweha
x y
Conversely if x y P R then there must b e p oints p p P and
x y
r r R such that p r p r x y Thus wehave p p r
x y x x y y x y x
x r y which implies that P intersects R x y
y
Since it is an easy matter to compute R in linear time by simply negating all of its
vertices the problem of computing the C obstacle CP reduces to the problem of computing a
ex p olygons We claim that this can b e done in O m n time Minkowski sum of twoconv
where m is the number of vertices in R and n is the number of vertices in P The algorithm
is based on the following observation Given a vector d Wesay that a p oint p is extreme in
direction d if it maximizes the dot pro duct p d
Observation Given two p olygons P and R then the set of extreme p oints of P R in
direction d is the set of sums of p oints p and r that are extreme in direction d for P and
R resp ectively
Lecture Notes CMSC M
The b o ok leaves the pro of as an exercise It follows easily by the linearity of the dot pro duct
From this observation it follows that there is a simple rotating calip ers algorithm for computing
P R when b oth are convex p olygons In particular we p erform an angular sweep by
considering a unit vector d rotating counterclo ckwise around a circle As d rotates it is an
easy matter to keep trackofthevertex or edge of P and R that is extreme in this direction
Whenever d is p erp endicular to an edge of either P or R we add this edge to the vertex of
the other p olygon The algorithm is given in the text and is illustrated in the gure b elow
P e P+R
d e+r
d R r
d
Figure Computing Minkowski sums
Assuming P and R are convex observe that eachedgeofP and eachedgeofR contributes
exactly one edge to P R If two edges are parallel and on the same side of the p olygons
then these edges will b e combined into one edge which is as long as their sum Thus wehave
the following
Claim Given twoconvex p olygons P and R withn and m edges resp ectively their Minkowski
sum P R can b e computed in O n m time and consist of at most n m edges
Lecture More Motion Planning
Thursday Dec
Reading Chapt in BKOS
Motion planning Recall that the motion planning problem is that of determining whether there is
a path from one placement or con guration of a rob ot to another sub ject to the constraint
that the rob ot do es not intersect any of a set of p olygonal obstacles Weshowed that this
problem could b e solved by transforming the obstacles into con guration space where each
p oint in this space corresp onds to a particular con guration or p osition of the rob ot Then the
problem reduces to determining whether it is p ossible to get from the starting con guration
p oint to the ending con guration p oint The transformed obstacles are called con guration
obstacles or C obstacles
We also showed that in the sp ecial case where the rob ot R can b e translated but not rotated
i e the con guration space is two dimensional then given a work space obstacle P the
corresp onding con guration obstacle CP is the Minkowski sum P R Wealsoshowed
that if P and R are convex then it is p ossible to compute this Minkowski sum by a rotating
calip ers algorithm in O n m time where n and m are the numb er of sides in P and R
resp ectively
Complexity of Minkowski Sums Todaywecontinue to consider the case of translation only
b ecause it is the simplest nontrivial case and consider the questions of howwe can compute
Lecture Notes CMSC M
a complete description of the con guration space and what the combinatorial complexityof
this description mightbe
To b egin with let s see just how bad things might b e Supp ose you are given a rob ot R with
m sides and a set of work space obstacle P with n sides Howmany sides might the Minkowski
sum P R have in the worst case O n m O nm even more The complexity generally
dep ends on what sp ecial prop erties if any P and R have
Nonconvex Rob ot and Nonconvex Obstacles Supp ose that b oth R and P are nonconvex sim
ple p olygons Let m be the numb er of sides of R and n b e the numb er of sides of P How
many sides might there b e in the Minkowski sum P R in the worst case We can derivea
to the quick upp er b ound as follows First observe that if we triangulate P we can break it in
union of at most n triangles That is
n
P T
i
i
m
R S
j
j
It follows that
n m
P R T S
i j
i j
Thus the Minkowski sum is the union of O nm p olygons each of constant complexity Thus
there are O nm sides in all of these p olygons The arrangement of all of these line segments
can have at most O n m intersection p oints if eachsideintersects with each other and
hence this is an upp er b ound on the number of vertices in the nal result
Could things really b e this bad Yes they could Consider the two p olygons in the gure b elow
left There are O n m ways that these two p olygons can b e do cked as shown on the right
The Minkowski sum P R is shown in the text Notice that the large size is caused by the
numb er of holes It might b e argued that this is not fair since we are not really interested
in the entire Minkowski sum just a single face of the Minkowski sum Proving b ounds on the
complexity of a single face is an interesting problem and the analysis is quite complex
P
R
Figure Minkowski sum of O n m complexity
As a nal observation notice that the upp er b ound holds even if P and R for that matter is
not a single simple p olygon but any union of n triangles
Convex Rob ot and Nonconvex Obstacles Wehave seen that the worst case complexity of the
Minkowski sum might range from O n m to as high as O n m which is quite a gap Let
us consider an intermediate but realistic situation Supp ose that we assume that P is an
arbitrary n sided simple p olygon and R is a convex m sided p olygon Typically m is much
Lecture Notes CMSC M
smaller than n What is the combinatorial complexityof P R in the worst case As b efore
we can observethatP can b e decomp osed into the union of n triangles T implying that
i
n
P R T R
i
i
Each Minkowski sum in the union is of complexity m So the question is howmany sides
might there b e in the union of O n convex p olygons each with O m sides We could derive
a b ound on this quantity but it will give a rather p o or b ound on the worst case complexity
To see why consider the limiting case of m Wehave the union of n convex ob jects
each of complexity O This could have complexityashighas n as seen by generating
ou try to construct suchacounter a criss crossing pattern of very skinny triangles But if y
example you won t b e able to do it
To see why suchacounterexample is imp ossible supp ose that you start with nonintersecting
triangles and then take the Minkowski sum with some convex p olygon The claim is that it is
imp ossible to generate this sort of criss cross arrangement So how complex an arrangement
can you construct We will show the following
Theorem Let R be a convex m gon and P and simple n gon then the Minkowski sum P R
has total complexity O nm
Is O nm an attainable b ound The idea is to go back to our analogy of scraping R around
the b oundary of P Canwe arrange P such that most of the edges of R scrap e over most of
the n vertices of P Supp ose that R is a regular convex p olygon with m sides and that P has
the comb structure shown in the gure b elow where the teeth of the comb are separated by
a distance at least as large as the diameter of R In this case R will havemany sides scrap e
across each of the p oint y ends of the teeth implying that the nal Minkowski sum will have
total complexity nm
P+R
R
P
Figure Minkowski sum of O nm complexity
The Union of Pseudo disks Toshow that O nm is an upp er b ound we need some way of ex
tracting the sp ecial geometric structure of the union of Minkowski sums Recall that we are
computing the union of T R where the T s have disjointinteriors The con guration that
i i
wewanttoavoid is the criss cross pattern shown ab ove Howdoweprove that such a pattern
cannot b e created The key is in the way the Minkowski sums of disjoint ob jects can intersect
We call a pair of convex ob jects o and o are a pair of pseudodisks if the b oth of the di erences
o no and o no are connected In particular if the ob jects intersect then they do not cross
through one another
Note that the b eing a pseudodisk is not a prop erty of a single ob ject but is the prop ertyof
pairs of ob jects A collection of ob jects is said to b e a col lection of pseudodisks if each pair is
a pair of pseudo disks The theorem ab o ve will follow from the next two lemmas
Lecture Notes CMSC M
Pseudodisks Not pseudodisks
Figure Pseudo disks
Lemma Given a set convex ob jects T T T with disjointinteriors and convex R
n
the set
fT R j i ng
i
is a collection of pseudo disks
Lemma Given a collection of pseudo disks with a total of n vertices the complexityof
their union is O n In our case the total number of vertices is O nm
First weprove Lemma Consider two p olygons T and T with disjointinteriors Wewant
to show that T R and T R do not cross over one another There is a very easy wayto
visualize why this is true but it is not a rigorous pro of Recall that the Minkowski sum
T R is related to scraping the negation R around the b oundary of T Supp ose we scrap e
R around the union of T and T Iftheinteriors of these two p olygons do not intersect then
either these two scrapings do not intersect each other at all or else they intersect in exactly
two p oints But from convexity it is plausible that there should not b e more than two
intersections
Here is a careful pro of Recall from yesterday s lecture that given any directional unit vector d
the most extreme pointofR in direction d is the p oint r R that maximizes the dot pro duct
d r Recall that we treat the p oints of the p olygons as if they were vectors The p oint
of T R that is most extreme in direction d is the sum of the p oints t and r that are most
extreme for T and R resp ectively
Given two convex p olygons T and T with disjointinteriors they de ne two outer tangents
as shown in the gure b elow Let d and d b e the outward p ointing p erp endicular vectors for
these tangents Because these p olygons do not intersect it follows easily that as the directional
vector rotates from d to d T will b e the more extreme p olygon and from d to d T will
b e the more extreme See the gure b elow
Now if to the contrary T R and T R had a crossing intersection then observe that we
can nd p oints p p p andp in cyclic order around the b oundary of the convex hull of
T R T R suchthatp p T R and p p T R First consider p Because
it is on the convex hull consider the direction d p erp endicular to the supp orting line here
Let r t andt b e the extreme p oints of R T and T in direction d resp ectively From our
basic fact ab out Minkowski sums wehave
p r t p r t
Since p is on the convex hull it follows that t is more extreme than t in direction d that
is T is more extreme than T in direction d By applying this same argument we nd that
is more extreme than T in directions d and d but that T is more extreme than T in T
directions d and d But this is imp ossible since from the observation ab ove there can b e
Lecture Notes CMSC M
d d2 T1 2 T1 extreme
d d 1 1 T2 extreme
T2 Figure Alternation of extremes
T2 extreme d2 d2 d3 d3 T1ext T1+R
T2+R d1 d1 d4 T1 extreme T2 extreme
d4
Figure Pro of of Lemma
Lecture Notes CMSC M
at most one alternation in extreme p oints for nonintersecting convex p olygons See the gure
below
Next we prove Lemma This is a rather cute combinatorial lemma Wearegiven some
collection of pseudo disks and told that altogether they have n vertices We claim that their
entire union has complexity O n Recall that in general the union of n convex p olygons can
have complexity O n by criss crossing The pro of is based on a clever charging scheme
Eachvertex in the union will b e charged to a vertex among the original psuedo disks such that
no vertex is charged more than twice This will imply that the total complexityisatmost n
There are twotyp es of vertices that may app ear on the b oundary The rst are vertices from
the original p olygons that app ear on the union There can b e at most n suchvertices and
eachischarged to itself The more troublesome vertices are those that arise when two edges of
two pseudo disks intersect each other Supp ose that twoedges e and e of pseudo disks P and
P intersect along the union Followedge e into the interior of the pseudo disk e Two things
might happ en First wemight hit the endp oint v of this e b efore leaving the interior P In
this case charge the intersection to v Note that v can get at most twosuchcharges one from
either incident edge If e passes all the waythrough P b efore coming to the endp oint then
try to do the same with edge e Again if it hits its endp oint b efore coming out of P then
charge to this endp oint See the gure b elow
e2 e1 e2 e1 e2 e1
v v u v u
Charge v Charge u Cannot happen
Figure Pro of of Lemma
But what do we do if b oth e sho ots straightthrough P and e sho ots straight through P
Nowwehavenovertex to charge This is okay b ecause the pseudo disk prop erty implies that
this cannot happ en If b oth edges sho ot completely through then the two p olygons must cross
over each other
Lecture Final Review
Tuesday Dec
Announcements Class is cancelled this Thursday O ce hours on Wednesday will b e moved up
to pm I ll b e holding usual o ce hours on Monday from pm on the day b efore
the nal
Final Exam Tues Dec am The exam will b e closed b o ok closed notes but you
are allowed twocheat sheets front and back
Before the Midterm Convex hulls line segmentintersection planar graphs and DCEL s p oly
gon triangulation intersection of halfspaces linear programming in low dimensions orthogonal
range searching planar p oint lo cation trap ezoidal maps You should b e aware of the results
de nitions and general techniques but I will not ask you for detailed information e g simu
lating Kirkpatrick s algorithm on a data set
Lecture Notes CMSC M
Voronoi diagrams Recall the de nition of Voronoi diagrams We presented Fortune s algorithm
which op erated bysweeping a distorted sweep line Voronoi diagrams havemany applica
tions in problems dealing with distances More generally Voronoi diagrams are an example
of the notion of sub dividing the plane into regions according to some criterion In this case
we sub divide the plane into regions according to who the nearest neighb or is It is p ossible to
de ne other sorts of diagrams For example the nd order Voronoi diagram is a sub division
of the plane according to who your rst two nearest neighb ors are and the k th order diagram
is de ned similarly for the k nearest neighb ors Also the furthest p ointVoronoi diagram is
de ned according to whichpoint is the furthest from this p oint Voronoi diagrams can b e
computed for line segments as well and are often used in motion planning applications
DelaunayTriangulations This is the dual of the Voronoi diagram We discussed prop erties such
as the fact that the Delaunay triangulation maximized the minimum angle We presented a
randomized incremental algorithm for the Delaunay triangulation Generally triangulations
play an imp ortant role in computational geometry since they allow us to reduce complex
domains to a collection of simpler domains triangles DeFloriani describ ed howany algorithm
for incrementally up dating a triangulation could b e used for generating a hierarchical surface
representation similar to Kirkpatrick s algorithm You are not resp onsible for the details of
her presentation but for understanding how up date rules for triangulations can b e used to
generate hierarchical surface representations
Line Arrangements Weshowed that line arrangements together with dualit y could b e used for
answering many problems having to do with lines and p oints in the plane We presented a
simple incremental construction algorithm for arrangements in the plane proved its O n
running time with the zone theorem We discussed top ological plane sweep as a metho d for
traversing an planar arrangement Duality and arrangements can also b e generalized to higher
dimensions
We presented a numb er of applications of arrangements These included sorting angular se
quences computing the narrowest corridor computing the maximum discrepancy and dis
cussed the notion of a level in an arrangement On Homework Problem we also saw
that there are problems that can b e solved by using the arrangement to conceptualize a
solution even though the actual solution may not involve constructing the arrangement You
should probably exp ect at least one problem on the exam whose solution will require use of
arrangements
Shortest Paths and Visibility Graphs We discussed the problem of computing shortest paths
in the plane and how visibility graphs could b e used to reduce the shortest path problem to
a problem on graphs Notice that using visibility graphs do es not extend to three dimensional
shortest path problems since a shortest path maybendanywhere along the interior of an edge
Nonetheless b ecause graph problems are often much easier to solve the technique of extracting
a graph from a geometric setting is an imp ortant concept Wesaw in Homework Problem
that some sp ecial shortest path problems can b e solved without explicitly constructing a
visibility graph at all For example shortest paths in simple p olygons can b e solved in O n
time once the p olygon has b een triangulated The solution is somewhat more complicated
but still similar to the homework solution
Motion Planning Weintro duced con guration spaces and discussed the idea of mapping work
space obstacles in to obstacles in con guration space We discussed Minkowski sums as a
metho d for computing con guration obstacles where translation is involved Weshowed how
to compute the Minkowski sum of twoconvex p olygons in linear time We also discussed
the combinatorial complexity of the union of a geometric ob jects when these ob jects satis ed
sp ecial conditions Although the general complexity could b e as high as O n weshowed
that if the ob jects are convex pseudo disks then the complexity only has complexity O n
Lecture Notes CMSC M
Lecture X Randomized Trap ezoidal Decomp osition
Supplemental
Randomized Incremental Algorithms We consider the following problem Given a set of p os
sibly intersecting line segments in the plane sub divide the plane into a collection of trap ezoids
formed by sho oting a bullet to the left and rightofeachvertex and eachintersection p oint
until it hits the rst ob ject Following Mulmuley s presentatio
The running time of this algorithm will b e O k n log n with high probability where n is the
numb er of segments and k is the number of intersection p oints b etween the segments More
succinctly the running time is O k n log n Mulmuley uses the notation f n O g n
if for all su ciently large n f n cg n for some constant c with probability p n
where p n is a p olynomial whose degree increases with c Thus we can reduce the probability
that the running time fails to b e O g n as lowaswe like up to the recipro cal of high degree
p olynomial function of n Observe that this running time is asymptotically b etter than the
previous in the sense that wehaveremoved the log n multiplicative factor from in front of the
k The price wehave paid is that the running time is now randomized and so not guaranteed
Randomized Incremental Algorithm The basic idea of the algorithm is describ ed b elow
i
Randomly permute the set of segments Let N denote the rst i segments In general
i
H N will denote the trap ezoidal decomp osition of the rst i segments
Initialize the structure to a trivial empty trap ezoidal decomp osition e g a large empty
enclosing b ox H N
One by one add the segments in random order For each segment do the following
i
a Locate the trap ezoid of H n containing the left endp oint of the segmen t There
areacoupleofways to do this More on this later
b Trace the segment from one trap ezoid to the next At the endp oints of the segment
split the current trap ezoid by adding a vertical wall For each trap ezoid determine
the p ointofentry and p oint of exit of the segment If the segment crosses the top
or b ottom then create new intersection p oints and split the trap ezoid by adding a
vertical wall
i
c After tracing the segment determine the vertical walls of H N thatwere stabb ed
by the segment For each suchwall trim it backtotheportioncontaining its sup
p orting vertex This results in the merger of two adjacent trap ezoids
The implementation of the algorithm relies on the ability to p erform the necessary lo cal ma
nipulations of the planar graph representing the trap ezoidal decomp osition In particular we
assume that we can trace a segment through a face of the decomp osition split a face
by the addition of a vertical wall and merge two adjacent faces when a vertical wall is
trimmed in time prop ortional to the complexity of the face that is the number of vertices
equivalently edges on the face We let f denote a trap ezoidal face of the decomp osition
f denote the complexity of this face Observe that the running time of the ab ove and let c
algorithm is O c f for each face that is traced by this pro cedure b ecause there can only b e
a constantnumb er at most split p erformed for each trap ezoid and a constantnumb er at
most merges for each trap ezoid
The other question weleftopenwas how to lo cate the left endp oint from which to start the
tracing We do this using a simple bucketing strategy For eachofthen left endp oints we
store which trap ezoid of the current sub division contains this p oint and with each trap ezoid f
we asso ciate a list L f whichcontains the segments whose left endp oint lies within this face
When we wish to trace a segment we can determine the trap ezoid containing this segment
Lecture Notes CMSC M
Initial decomposition.
Trace and split.
Trim and merge.
Figure Randomized incremental trap ezoidal decomp osition
Lecture Notes CMSC M
in constant time When a segmentintersects a trap ezoid f itmay split a trap ezoid f into
a constantnumb er of new trap ezoids We can walk through the list L f and determine
which of the new trap ezoids contains a given p oint in constant time each We form new left
endp oint lists for each of the new faces L f L f etc When we merge two trap ezoids
we simply concatenate their lists Observe that b oth of these op erations can b e p erformed in
time O l f where l f jL f j
f 3
f f 1 f 2
c(f) = 9 l(f) = 5 c(f1) = 7 l(f1) = 2 c(f2) = 5 l(f2) = 2
c(f3) = 4 l(f3) = 1
Figure Splitting a face
i
Lemma Consider the insertion of segmentinto H N whichintersects faces F ff f f g
k
in the decomp osition The time to insert this segmentis
X
A
O c f l f
f F
Analysis Analyzing the exp ected case running time of this algorithm is quite a tricky task We
want to show that no matter what n segments are given initially ifwe randomize of all p ossible
insertion orders the total exp ected time to build the decomp osition is O k n log n We will
only prove that this is an exp ected b ound The pro of that this holds with high probability
takes some additional work In particular we seem to need to b e able to determine the
P
exp ected value of c f l f over all segments that might b e inserted next
f
k called backward This task would b e quite daunting if it were not for a very clever analysis tric
analysis Here is the idea we imagine that we are running time algorithm backwards deleting
segments one at a time Whether wecountbackwards or forwards makes no di erence The
running time of a given stage is still given by the lemma ab ove Observe that if all insertion
orders are equally likely then the last segment to b e deleted in the reversed algorithm is equally
i
likely to b e any one of the segments that already exists in the decomp osition H N Since
i
every segmentofN is equally likely to b e deleted to determine the exp ected time for the i th
stage it su ces to average over all p ossible segments i to b e deleted and for each determine
the complexity if this were the segmentchosen
i
When we delete some segment from the H N observe that the only trap ezoids that would b e
i
a ected from its deletion consist of the trap ezoids of H N that are adjacent to this segment
An example is shown in the gure b elow
Lemma For each j j i let F denote the faces that the j th segmentintersects in
j
i i
H N The exp ected time for the last stage in the construction of H N is on the order
Lecture Notes CMSC M
Regions affected by deletion.
Final decomposition.
Figure A ected trap ezoids
of
i
X X
E i c f l f
i
j
f F
j
We seem to b e no closer to our goal since there do es not app ear to b e an easy way to analyze the
sums of complexities of all the trap ezoids that a given segmentintersects However the crucial
trick is not to count segmentby segment but to count trap ezoid by trap ezoid Assuming that
the segments are in general p osition observe the following imp ortant fact
Lemma Bounded degree prop erty Each trap ezoid in a trap ezoidal decomp osition is adja
cent to at most segments Namely the segments immediately ab ove and b elow and
the segments which touch the left and rightwalls of the trap ezoid either b ecause of an
endp ointoranintersection p oint
Thus if wetake the complexityofeach trap ezoid and multiply by we will have an upp er
b ound on the sum of complexities of all the trap ezoids that are intersected by all segments
X
c f l f E i
i
i
f H N
X X
A
c f l f
i
i i
f H N f H N
P
Now this is something we can analyze Weknowthevalue of c f is the sum of all the
f
edges summed over all the faces in the decomp osition However this counts every edge in
the decomp osition twice Since the decomp osition is a planar graph this is prop ortional to
P
the number of vertices in the decomp osition currently Thevalue of l f is just equal to
f
n i since it includes all the left endp oints of segments that haveyet to b e added If we
let k denote the numberofintersection p oints at the current stage of the decomp osition the
i
i
numberofvertices in H N is just i k so the ab oveformula simpli es to
i
E i i k n i
i
i
Lecture Notes CMSC M
c n k
i
i
for some constant c The interesting thing at this p oint is that although our analysis was
i
conditional on the structure of H N the resulting b ound is almost entirely indep endentof
this structure Only the value k is dep endent
i
Atthispointwe can see where the analysis is going If we ignore the k term in the ab ove we
i
see that E i cn i To get the total time we sum these up getting at total exp ected time
of
n
X
TE n E i
i
n
X
cn
i
i
n
X
cn
i
i
cn ln n O n log n
P
The last line uses the well known fact that the Harmonic series i tends to ln n
i n
We need to get a handle on what k is exp ected to b e Obviously the value of k should
i i
eventually approach k the total numberofintersections as i approaches n The interesting
fact is that this quantity approaches k quadratically not linearly
i
Lemma For xed i the exp ected value of k assuming that N is a random sample of
i
N of size i is O ki n
Pro of For eachintersection p oint v between twosegments s and s let I denote the
v
i
random variable that this if this intersection is part of the decomp osition H N and
P
otherwise Observe that the exp ected value of k is I However v o ccurs within
i v
v
I if and only if b oth s and s o ccur within the rst i randomly selected segments The
v
probability that each one has b een selected alone is i n The probabilitythatbothhave
b een selected is roughly i n for i and n large Thus the exp ected value of k is k times
i
this quantity or O ki n
P
To complete the analysis wemake use of another basic fact from summations i
i n
n
n
X
E i TE n
i
n
X
c n k
i
i
i
n n
X X
cn ck
i
i i
i i
n n
X X
ki
c cn
i n i
i i
n
X
ck
cn ln n i
n
i
cn ln n ck
O n log n k
Lecture Notes CMSC M
Lecture X Voronoi Diagrams
Supplemental
Read O Rourke Chapt Mulmuley Sect
Planar Voronoi Diagrams Recall that given n p oints P fp p p g in the plane the
n
Voronoi p olygon of a p oint p V p is de ned to b e the set of all p oints q in the plane for
i i
which p is among the closest p oints to q in P Thatis
i
V p fq jp q j jp q j j ig
i i j
The union of the b oundaries of the Voronoi p olygons is called the Voronoi diagram of P
denoted VD P The dual of the Voronoi diagram is a triangulation of the p oint set called
the Delaunay triangulation Recall from our discussion of quad edge data structure that given
a go o d representation of any planar graph the dual is easy to construct Hence it su ces
to showhow to compute either one of these structures from which the other can b e derived
easily in O n time
There are four fairly well known algorithms for computing Voronoi diagrams and Delaunay
triangulations in the plane They are
Divide and Conquer For b oth VD and DT The rst O n log n algorithm for this prob
lem Not widely used b ecause it is somewhat hard to implement Can b e generalized to
higher dimensions with some di culty Can b e generalized to computing Vornoi diagrams
of line segments with some di culty
Randomized Incremental For DT and VD The simplest O n log n time with high
probability Can b e generalized to higher dimensions as with the randomized algorithm
for convex hulls Can b e generalized to computing Voronoi diagrams of line segments
fairly easily
Fortune s Plane Sweep For VD A very clever and fairly simple algorithm It computes
a deformed Voronoi diagram by plane sweep in O n log n time from which the true
diagram can b e extracted easily Can b e generalized to computing Voronoi diagrams of
line segments fairly easily
Reduction to convex hulls For DT Computing a Delaunay triangulation of n points in
dimension d can b e reduced to computing a convex hull of n p oints in dimension d
Use your favorite convex hull algorithm Unclear how to generalize to compute Voronoi
diagrams of line segments
We will cover all of these approaches except Fortune s algorithm O Rourkedoesnotgive
detailed explanations of any of these algorithms but he do es discuss the idea b ehind For
tune s algorithm Todaywe will discuss the divide and conquer algorithm This algorithm is
presented in Mulmuley Section
Divide and conquer algorithm The divide and conquer approachworks like most standard ge
ometric divide and conquer algorithms We split the p oints according to x co ordinates into
roughly equal sized groups e g by presorting the p oints by x co ordinate and selecting medi
ans We compute the Voronoi diagram of the left side and the Voronoi diagram of the right
side Note that since each diagram alone covers the entire plane these twodiagramsoverlap
We then merge the resulting diagrams into a single diagram
The merging step is where all the work is done Observethatevery p oint in the the plane
lies within twoVoronoi p olygons one in VD L andoneinVD R We need to resolve this
overlap by separating overlapping p olygons Let V l betheVoronoi p olygon for a p oint
Lecture Notes CMSC M
from the left side and let V r betheVoronoi p olygon for a p oint on the right side and
supp ose these p olygons overlap one another Observethatifwe insert the bisector b etween
l and r and through away the p ortions of the p olygons that lie on the wrong side of the
bisector we resolve the overlap If we do this for every pair of overlapping Voronoi p olygons
we get the nal Voronoi diagram This is illustrated in the gure b elow
Left/Right Diagrams and Contour
Final Voronoi Diagram
Figure Merging Voronoi diagrams
The union of these bisectors that separate the left Voronoi diagram from the rightVoronoi
diagram is called the contour Apoint is on the contour if and only if it is equidistant from
p oints in S one in L and one in R
Presort the p oints by x co ordinate this is done once
Split the p ointset S bya vertical line into twosubsets L and R of roughly equal size
Compute VD L andVD R recursively These diagrams overlap one another
Merge the two diagrams into a single diagram by computing the contour and discarding
the p ortion of the VD L that is to the right of the contour and the p ortion of VD R
that is to the left of the contour
Assuming we can implement step in O n time where n is the size of the remaining p oint
set then the running time will b e de ned by the familiar recurrence
T n T n n
whichwe know solves to O n log n
Lecture Notes CMSC M
Computing the contour What makes the divide and conquer algorithm somewhat tricky is the
task of computing the contour Before giving an algorithm to compute the contour let us
make some observations ab out its geomtetric structure Let us make the usual simplifying
assumptions that no p oints are co circular
Lemma The contour consists of a single p olygonal curve whose rst and last edges are
semiin nite which is monotone with resp ect to the y axis
Pro of A detailed pro of is a real hassle Here are the main ideas though The contour
separates the plane into two regions those p oints whose nearest neighbor lies in L from
those p oints whose nearest neighb or lies in R Because the contour lo cally consists of
points that are equidistantfrom points it is formed from pieces that are p erp endicular
bisectors with one p oint from L and the other p oint from R Thus it is a piecewise
p olygonal curve Because no p oints are co circular it follows that all vertices in the
Voronoi diagram can have degree at most However b ecause the contour separates the
plane into only typ es of regions it can contain only vertices of degree Thus it can
consist only of the disjoint union of closed curves actually this never happ ens as we
will see and unb ounded curves Observe thatifwe orientthecontour counterclo ckwise
with resp ect to each p ointin R clo ckwise with resp ect to each p ointinL then each
segmentmust b e directed in the y directions b ecause L and R are separated bya
vertical line Thus the contour contains no horizontal cusps This implies that the
contour cannot contain any closed curves and hence contains only vertically monotone
unb ounded curves Also this orientability also implies that there is only one such curve
Lemma The topmost b ottommost edge of the contour is the p erp endicular bisector for the
twopoints forming the upp er lower tangent of the left hull and the righthull
Pro of This follows from the fact that the vertices of the hull corresp ond to unb ounded
Voronoi p olygons and hence upp er and lower tangents corresp ond to unb ounded edges
of the contour
These last two theorem suggest the general approach We start by computing the upp er
tangent whichweknow can b e done in linear time once weknow the left and righthulls or
by prune and search Then we start tracing the contour from top to b ottom When we are
in Voronoi p olygons V l and V r we trace the bisector b etween l and r in the negative
y direction until its rst contact with the b oundaries of one of these p olygons Supp ose that
we hit the b oundary of V l rst Assuming that we use a go o d data structure for the Voronoi
diagram e g quad edge data structure we can determine the p oint l lying on the other side
of this edge in the left Voronoi diagram Wecontinue following the contour by tracing the
bisector of l and r
However in order to insure e ciency wemust b e careful in howwe determine where the
bisector hits the edge of the p olygon Consider the gure shown b elow We start tracing the
contour b etween l and r Bywalking along the b oundary of V l we can determine the edge
that the contour would hit rst This can b e done in time prop ortional to the number of edges
in V l which can b e as large as O n However we discover that b efore the contour hits the
b oundary of V l it hits the b oundary of V r W e nd the new p oint r and now trace the
bisector b etween l and r Again we can compute the intersection with the b oundary of V l
in time prop ortional to its size However the contour hits the b oundary of V r rst and so
wegoontor As can b e seen if we are not smart we can rescan the b oundary of V l over
and over again until the contour nally hits the b oundary IfwedothisO n times and the
b oundary of V l isO n then we are stuck with O n timetotracethecontour
Wehavetoavoid this rep eated rescanning However there is a way to scan the b oundary of
eachVoronoi p olygon at most once Observethataswewalk along the contour each time we
Lecture Notes CMSC M
Contour
r0
l 0 r1
r2
Figure Tracing the contour
stay in the same p olygon V l we are adding another edge onto its Voronoi p olygon Because
the Voronoi p olygon is convex weknow that the edges we are creating turn consistently in
the same direction clo ckwise for p oints on the left and counterclo ckwise for p oints on the
right To test for intersections b etween the contour and the currentVoronoi p olygon we trace
the b oundary of the p olygon clo ckwise for p olygons on the left side and counterclo ckwise for
p olygons on the right side Whenever the contour changes direction wecontinue the scan
from the p ointthatwe left o In this way we knowthatwe will never need to rescan the
same edge of anyVoronoi p olygon more than once
Lecture X DelaunayTriangulations and Convex Hulls
Supplemental
Read O Rourke and
DelaunayTriangulations and Convex Hulls At rst Delaunay triangulations and convex hulls
app ear to b e quite di erent structures one is based on metric prop erties distances and the
other on a ne prop erties collinearity coplanarity Todayweshow that it is p ossible to con
vert the problem of computing a Delaunay triangulation in dimension d to that of computing
aconv ex hull in dimension d Thus there is a remarkable relationship b etween these two
structures
We will demonstrate the connection in dimension by computing a convex hull in dimension
Some of this may b e hard to visualize but see O Rourke for illustrations You can also
reason by analogy in one lower dimension of Delaunay triangulations in d and convex hulls
in d but the real complexities of the structures are not really apparent in this case
The connection b etween the two structures is the paraboloid z x y Observe that this
equation de nes a surface whose vertical cross sections constant x or constant y are parab olas
and whose horizontal cross sections constant z are circles For eachpoint in the plane x y
the vertical projection of this p ointonto this parab oloid is x y x y in space Given a set
of p oints S in the plane let S denote the pro jection of every p ointin S onto this parab oloid
Consider the lower convex hul l of S This is the p ortion of the con vex hull of S whichis
visible to a viewer standing at z We claim that if we take the lower convex hull of S
and pro ject it backonto the plane then we get the Delaunay triangulation of S In particular
let p q r S and let p q r denote the pro jections of these p oints onto the parab oloid
Then p q r de ne a face of the lower convex hull of S if and only if pq r is a triangle of the
Delaunay triangulation of S The pro cess is illustrated in the following gure
Lecture Notes CMSC M
Project onto paraboloid. Compute convex hull. Project hull faces back to plane.
Figure Delaunay triangulations and convex hull
The question is whydoesthiswork To see why we need to establish the connection b etween
the triangles of the Delaunay triangulation and the faces of the convex hull of transformed
p oints In particular recall that
Delaunay condition Three p oints p q r S form a Delaunay triangle if and only if the
circumcircle of these p oints contains no other p ointofS
Convex hull condition Three p oints p q r S form a face of the convex hull of S if
and only if the plane passing through p q and r has all the p oints of S lying to one
side
Clearly the connection we need to establish is b etween the emptiness of circumcircles in the
plane and the emptiness of halfspaces in space Wewillprove the following claim
Lemma Consider distinct p oints p q r s in the plane and let p q r s b e their resp ective
pro jections onto the parab oloid z x y The p oint s lies within the circumcircle of
p q r if and only if s lies on the lower side of the plane passing through p q r
To prove the lemma rst consider an arbitrary nonvertical plane in space whichwe assume
is tangent to the parab oloid ab ove some p oint a b intheplane To determine the equation
of this tangentplane we take derivatives of the equation z x y with resp ect to x and y
giving
z z
x y
x y
At the p oint a b a b these evaluate to a and b It follows that the plane passing through
these p oint has the form
z ax by
Tosolvefor we know that the plane passes through a b a b sowe solve giving
a b a a b b
Implying that a b Thus the plane equation is
z ax by a b
If we shift the plane upwards by some p ositiveamount r we get the plane
z ax by a b r
Lecture Notes CMSC M
How do es this plane intersect the parab oloid Since the parab oloid is de ned by z x y
we can eliminate z giving
x y ax by a b r
which after some simple rearrangements is equal to
x a y b r
This is just a circle Thus wehaveshown that the intersection of a plane with the parab oloid
pro duces a space curve which turns out to b e an ellipse which when pro jected backonto the
x y co ordinate plane is a circle centered at a b
Thus we conclude that the intersection of an arbitrary lower halfspace with the parab oloid
when pro jected onto the x y plane is the interior of a circle Going back to the lemma when
we pro ject the p oints p q r onto the parab oloid the pro jected p oints p q and r de ne a
plane Since p q andr lie at the intersection of the plane and parab oloid the original
p oints p q r lie on the pro jected circle Thus this circle is the unique circumcircle passing
through these p q andr Thus the p oint s lies within this circumcircle if and only if its
pro jection s onto the parab oloid lies within the lower halfspace of the plane passing through
p q r
p’ r’ s’ q’
q s
r p
Figure Planes and circles
Nowwe can prove the main result
Theorem Given a set of p oints S in the plane assume no are co circular and given
points p q r S the triangle pq r is a triangle of the Delaunay triangulation of S if
and only if triangle p q r is a face of the lower convex hull of the pro jected set S
From the de nition of Delaunay triangulations we know that pq r is in the Delaunay trian
gulation if and only if there is no p oint s S that lies within the circumcircle of pq r From
the previous lemma this is equivalenttosaying that there is no p oint s that lies in the lower
convex hull of S which is equivalenttosaying that p q r is a face of the lower convex hull
This completes the pro of
In order to test whether a p oint s lies within the circumcircle de ned by p q r it su ces to
lies within the lower halfspace of the plane passing through p q r Ifwe test whether s
assume that p q r are oriented counterclo ckwise in the plane this this reduces to determining
whether the quadruple p q r s is p ositively oriented or equivalently whether s lies to the
left of the oriented circle passing through p q r
Lecture Notes CMSC M
This leads to the incircle test we presented last time
p p p p
x y
x y
B C
q q q q
x y
x y
B C
in p q r s det
A
r r r r
x y
x y
s s s s
x y
x y
Voronoi Diagrams and Upp er Envelop es Weknow that Voronoi diagrams and Delaunay tri
angulations are dual geometric structures Wehave also seen informally that there is a dual
relationship b etween p oints and lines in the plane and in general p oints and planes in space
From this latter connection we argued that the problems of computing convex hulls of p oint
sets and computing the intersection of halfspaces are somehow dual to one another It turns
are not surprisingly interrelated In particular in out that these two notions of duality
the same way that the Delaunay triangulation of p oints in the plane can b e transformed to
computing a convex hull in space it turns out that the Voronoi diagram of p oints in the
plane can b e transformed into computing the intersection of halfspaces in space
Here is howwe do this For each p oint p a b in the plane recall the tangent plane to the
parab oloid ab ovethispoint given by the equation
z ax by a b
De ne H p to b e the set of p oints that are ab ove this halfplane that is H p f x y z j z
ax by a b g LetS fp p p g b e a set of p oints Consider the intersection of
n
the halfspaces H p This is also called the upper envelope of these halfspaces The upp er
i
envelop e is an unb ounded convex p olyhedron If you pro ject the edges of this upp er envelop e
down into the plane it turns out that you get the Voronoi diagram of the p oints
Theorem Given a set of p oints S in the plane assume no are co circular let H denote
the set of upp er halfspaces de ned by the previous transformation Then the Voronoi
diagram of H is equal to the pro jection onto the x y plane of the skeleton of the
convex p olyhedron which is formed from the intersection of halfspaces of S
q’ p’
p
q
Figure Intersection of halfspaces
It is hard to visualized this surface but it is not hard to showwhy this is so Supp ose wehave
p oints in the plane p a b and q c d The corresp onding planes are
z ax by a b and z cx dy c d
Lecture Notes CMSC M
If we determine the intersection of the corresp onding planes and pro ject onto the x y
co ordinate plane by eliminating z from these equations weget
x a c y b d a c b d
We claim that this is the p erp endicular bisector b etween a b and c d To see this observe
that it passes through the midp oint a c b d b etween the twopoints since
b d a c
a c b d a c b d
and its slop e is a c b d which is the negative recipro cal of the line segment from
a b to c d From this it can b e shown that the intersection of the upp er halfspaces de nes
a p olyhedron whose edges pro ject onto the Voronoi diagram edges
Lecture X Top ological Plane Sweep
Supplemental
Read The material on top ological plane sweep is not discussed in any of our readings The
algorithm for top ological plane sweep can b e found in the pap er Top ologically sweeping an ar
rangement by H Edelsbrunner and L J Guibas J Comput Syst Sci with
Corrigendum in the same journal volume
Top ological Plane Sweep In the last two lectures wehaveintro duced arrangements of lines and
geometric duality as imp ortant to ols in solving geometric problems on lines and p oints Today
give an e cient algorithm for sweeping an arrangement of lines
As we will see many problems in computational geometry can b e solved by applying line
sweep to an arrangement of lines Since the arrangement has size O n and since there are
O n events to b e pro cessed eachinvolving an O log n heap deletion this typically leads to
algorithms running in O n log n time using O n space It is natural to ask whether we
can disp ense with the additional O log n factor in running time and whether we need all of
O n space since in theory we only need access to the current O n contents of the sweep
line
We discuss a variation of plane sweep called topological plane sweep This metho d runs in O n
time and uses only O n space byessentially constructing only the p ortion of the arrangement
that we need at any p oint Although it may app ear to b e somewhat sophisticated it can
b e implemented quite e ciently and is claimed to outp erform conventional plane sweep on
arrangements of any signi cant size e g over lines
plane sweep b ecause we do not Cuts and top ological lines The algorithm is called topological
sweep a straightvertical line through the arrangement but rather wesweep a curved topological
line that has the essential prop erties of a vertical sweep line in the sense that this line intersects
each line of the arrangement exactly once The notion of a top ological line is an intuitive one
but it can b e made formal in the form of something called a cut Recall that the faces of
the arrangement are convex p olygons p ossibly unb ounded Assuming no vertical lines the
edges incidenttoeach face can naturally b e partitioned into the edges that are above the face
and those that are below the face De ne a cut in an arrangement to b e a sequence of edges
c c c in the arrangement one taken from each line of the arrangement suchthatfor
n
i n c and c are incident to the same face of the arrangement and c is ab ove
i i i
the face and c is b elow the face An example of a top ological line and the asso ciated cut is
i
shown b elow
Lecture Notes CMSC M
c1
c2
c3 c4
c5
Figure Top ological line and asso ciated cut
The top ological plane sweep starts at the leftmost cut of the arrangement This consists of
all the left unb ounded edges of the arrangement Observe that this cut can b e computed in
O n log n time b ecause the lines intersect the cut in inverse order of slop e The top ological
sweep line will sweep to the rightuntil wecometotherightmost cut which consists all of the
right unb ounded edges of the arrangement The sweep line advances by a series of what are
called elementary steps In an elementary steps we ndtwo consecutive edges on the cut that
meet at a vertex of the arrangement we will discuss later how to determine this and push
the top ological sweep line through this vertex Observe that on doing so these two lines swap
in their order along the sweep line This is shown b elow
Figure Elementary step
It is not hard to show that an elementary step is always p ossible since for any cut other than
the rightmost cut there must b e two consecutive edges with a common right endp oint In
particular consider the edge of the cut whose rightendpoint has the smallest x co ordinate It
is not hard to show that this endp oint will always allow an elementary step Unfortunately
determining this vertex would require at least O log n time if we stored these endp oints in
a heap sorted by x co ordinate and wewant to p erform eachelementary step in O time
Hence we will need to nd some other metho d for nding elementary steps
Upp er and Lower Horizon Trees To nd elementary steps weintro duce two simple data struc
tures the upper horizon tree UHT and the lower horizon tree LHT To construct the upp er
horizon tree trace each edge of the cut to the right When two edges meet keep only the one
with the higher slop e and continue tracing it to the right The lower horizon tree is de ned
symmetrically There is one little problem in these de nitions in the sense that these trees
need not b e connected forming a forest of trees but this can b e xed conceptually at least by
the addition of a vertical line at x For the upp er horizon we think of its slop e as b eing
and for the lo wer horizon we think of its slop e as b eing Notethatwe consider the
left endp oints of the edges of the cut as not b elonging to the trees since otherwise they would
not b e trees It is not hard to show that with these mo di cations these are indeed trees Each
edge of the cut de nes exactly one line segmentineach tree An example is shown b elow
The imp ortant things ab out the UHT and LHT is that they giveusaneasyway to determine
Lecture Notes CMSC M
Upper Horizon Tree Lower Horizon Tree
Figure Upp er and lower horizon trees
the right endp oints of the edges on the cut Observe that for each edge in the cut its right
endp oint results from a line of smaller slop e intersecting it from ab ove aswe trace it from left
to right or from a line of larger slop e intersecting it from b elow It is easy to verify that the
UHT and LHT determine the rst suchintersecting line of eachtyp e resp ectively It follows
that if weintersect the two trees then the segments they share in common corresp ond exactly
to the edges of the cut Thus byknowing the UHT and LHT we know where are the right
endp oints are and from this we can determine easily which pairs of consecutive edges share
a common right endp oint and from this we can determine all the elementary steps that are
legal We store all the legal steps in a stack or queue or any list is ne and extract them
one by one
The sweep algorithm Here is an overview of the top ological plane sweep
Input the lines and sort by slop e Let C b e the initial leftmost cut a list of lines in
decreasing order of slop e
Create the initial UHT incrementally by inserting lines in decreasing order of slop e Create
the initial LHT incrementally by inserting line in increasing order of slop e More on this
later
By consulting the LHT and UHT determine the right endp oints of all the edges of the
initial cut and for all pairs of consecutive lines l l sharing a common right endp oint
i i
store this pair in stack S
Rep eat the following elementary step until the stack is empty implying that wehave
arrived at the rightmost cut
a Pop the pair l l from the top of the stack S
i i
b Swap these lines within C the cut we assume that each line keeps track of its p osition
in the cut
c Up date the horizon trees More on this later
d Consulting the changed entries in the horizon tree determine whether there are any
new cut edges sharing rightendpoints and if so push them on the stack S
The imp ortant un nished business is to showthatwe can build the initial UHT and LHT in
O n time and to show that for eachelementary step we can up date these trees and all other
relevant information in O amortizedtime Byamortized time we mean that even though
a single elementary step can take more than O time the total time needed to p erform all
O n elementary steps is O n and hence the average time for each step is O
Thisisdoneby an adaptation of the same incremental face walking technique weusedinthe
incremental construction of line arrangements Let s consider just the UHT since the LHT is
Lecture Notes CMSC M
symmetric To create the initial leftmost UHT we insert the lines one by one in decreasing
order of slop e Observe that as each new line is inserted it will start ab ove all of the current
lines The upp ermost face of the current UHT consists of a convex p olygonal chain see the
gure b elow left As we trace the newly inserted line from left to right there will b e some
p ointatwhich it rst hits this upp er chain of the current UHT By walking along the chain
from left to right we can determine this intersection p oint Eachsegment that is walked over
is never visited again by this initialization pro cess b ecause it is no longer part of the upp er
chain and since the initial UHT has a total of O n segments this implies that the total time
sp entinwalking is O n Thus after the O n log n time for sorting the segments the initial
UHT tree can b e built in O n additional time
new line vv
Initial UHT construction. Updating the UHT.
Figure Constructing and up dating the UHT
Next weshowhow to up date the UHT after an elementary step The pro cess is quite similar
as shown in the gure right Let v be the vertex of the arrangement which is passed over in
the sweep step As we pass over v thetwoedgesswap p ositions along the sweep line The new
lower edge call it l which had b een cut o of the UHT by the previous lower edge nowmust
b e reentered into the tree We extend l to the left until it contacts an edge of the UHT At its
rst contact it will terminate and this is the only change to b e made to the UHT In order
to ndthiscontact we start with the edge immediately b elow l the current cut Wetraverse
the face of the UHT in counterclo ckwise order until nding the edge that this line intersects
Observethatwemust eventually nd such an edge b ecause l has a lower slop e than the other
edge intersecting at v and this edge lies in the same face
Analysis A careful analysis of the running time can b e p erformed using the same amortization
pro of based on p ebble counting that was used in the analysis of the incremental algorithm
We will not give the pro of in full detail Observe that b ecause wemaintain the set of legal
elementary steps in a stack as opp osed to a heap as would b e needed for standard plane sweep
tary step in O time The only part of the elementary we can advance to the next elemen
step that requires more than constant time is the up date op erations for the UHT and LHT
However we claim that the total time sp ent up dating these trees is O n The argument
is that when we are tracing the edges as shown in the previous gure we are essentially
traversing the edges in the zone for L in the arrangement This is not quite true b ecause
there are edges ab ove l in the arrangement whichhave b een cut out of the upp er tree but
the claim is that their absence cannot increase the complexity of this op eration only decrease
it However a careful pro of needs to takethisinto account Since the zone of each lineinthe
arrangement has complexity O n all n zones have total complexity O n Thus the total
time sp ent in up dating the UHT and LHT trees is O n
Lecture Notes CMSC M
Lecture X Ham Sandwich Cuts
Supplemental
Ham Sandwich Cuts of Linearly Separated Point Sets We are given n red p oints A and m
blue p oints B and wewant to compute a single line that simultaneously bisects b oth sets If
the cardinality of either set is o dd then the line passes through one of the p oints of the set
We make the simplifying assumption that the sets are separated by a line This assumption
makes the problem much simpler to solve but the general case can still b e solved in O n
time using arrangements
To make matters even simpler we assume that the p oints have b een translated and rotated
so this line is the y axis Thus all the red p oints set A have p ositive x co ordinates and
hence their dual lines have p ositive slop es whereas all the blue p oints set B have negative
x co ordinates and hence their dual lines have negative slop es As long as we are simplifying
things let s make one last simpli cation that b oth sets haveanoddnumberofpoints This
is not di cult to get around but makes the pictures a little easier to understand
Consider one of the sets say A Observe that for each slop e there exists one way to bisect the
p oin ts In particular if we start a line with this slop e at p ositive in nity so that all the p oints
lie b eneath it and drop in downwards eventually we will arrive at a unique placement where
there are exactly n points ab ove the line one p oint lying on the line and n
p oints b elow the line assuming no two p oints share this slop e This line is called the median
line for this slop e
What is the dual of this median line If we dualize the p oints using the standard dual trans
formation D a b y ax b thenwegetn lines in the plane By starting a line with
a given slop e ab ovethepoints and translating it downwards in the dual plane wemoving a
p ointfrom upwards in a vertical line Each time the line passes a p oint in the primal
plane the vertically moving p oint crosses a line in the dual plane When the translating line
hits the median p oint in the dual plane the moving p oint will hit a dual line such that there
e are exactly n dual lines ab ove this p ointand n dual lines b elow this p oint W
de ne a p ointtobeat level k L in an arrangement if there are at most k lines ab ove
k
this p oint and at most n k lines b elow this p oint The median level in an arrangementofn
lines is de ned to b e the d n e th level in the arrangement This is shown as M A in
the following gure on the left
M(B)
M(A) M(A) Ham sandwich point
Dual arrangement of A. Overlay of A and B’s median levels.
Figure Ham sandwich Dual formulation
Thus the set of bisecting lines for set A in dual form consists of a p olygonal curve Because
this curve is formed from edges of the dual lines in A and b ecause all lines in A have p ositive
slop e this curve is monotonically increasing Similarly the median for B M B is a p olygonal
curvewhich is monotonically decreasing It follows that A and B must intersect at a unique
p oint The dual of this p oint is a line that bisects b oth sets
Lecture Notes CMSC M
We could compute the intersection of these twocurves byasimultaneous top ological plane
sweep of b oth arrangements However it turns out that it is p ossible to do much b etter and in
fact the problem can b e solved in O n m time Since the algorithm is rather complicated I
will not describ e the details but here are the essential ideas The algorithm op erates by prune
and search In O n m timewe will generate a hyp othesis for where the ham sandwichpoint
is in the dual plane and if we are wrong we will succeed in throwing away a constant fraction
of the lines from future consideration
First observe that for anyvertical line in the dual plane it is p ossible to determine in O n m
time whether this line lies to the left or the rightoftheintersection p oint of the median levels
M A andM B This can b e done by computing the intersection of the dual lines of A with
this line and computing their median in O n time and computing the intersection of the
dual lines of B with this line and computing their median in O m time If A s median lies
below B s median then we are to the left of the ham sandwich dual p oint and otherwise we
are to the right of the ham sandwichdualpoint It turns out that with a little more work it
is p ossible to determine in O n m time whether the ham sandwichpoint lies to the right
or left of a line of arbitrary slop e The trick is to use prune and search We ndtwo lines L
and L in the dual plane by a careful pro cedure that I will not describ e These two lines
de ne four quadrants in the plane By determining which side of each line the ham sandwich
p oint lies weknowthatwe can throwawayany line that do es not intersect this quadrant from
further consideration It turns out that by a judicious choice of L and L we can guarantee
that a fraction of at least n m lines can b e thrown awayby this pro cess We recurse
on the remaining lines By the same sort of analysis we made in the Kirkpatrick and Seidel
prune and search algorithm for upp er tangents it follows that in O n m time we will nd
the ham sandwichpoint