lecture 4 (cgal)

Lecture 4 (CGAL)

Panos Giannopoulos, Dror AtariahAG TI

WS 2012/13

É Exercise 3?

Outline

Note: Random points

Delaunay Triangulations and Voronoi diagrams

Exercise 4

Project ideas

,FU Berlin, Lecture 4 (CGAL), WS 2012/13 3

Note: Random points

Choosing random points in a convex polygon


Triangulation of a point set P

Section 9.1TRIANGULATIONS OF PLANAR POINT

SETS

mined by two points that are relatively far away. This happens because q lies inthe middle of an edge of two long and sharp triangles. The skinniness of thesetriangles causes the trouble. So it seems that a triangulation that contains smallangles is bad. Therefore we will rank triangulations by comparing their smallestangle. If the minimum angles of two triangulations are identical, then we canlook at the second smallest angle, and so on. Since there is only a finite numberof different triangulations of a given point set P, this implies that there must bean optimal triangulation, one that maximizes the minimum angle. This will bethe triangulation we are looking for.

9.1 Triangulations of Planar Point Sets

Let P := {p1, p2, . . . , pn} be a set of points in the plane. To be able to formallydefine a triangulation of P, we first define a maximal planar subdivision asa subdivision S such that no edge connecting two vertices can be added toS without destroying its planarity. In other words, any edge that is not in S

intersects one of the existing edges. A triangulation of P is now defined as amaximal planar subdivision whose vertex set is P.

With this definition it is obvious that a triangulation exists. But does itconsist of triangles? Yes, every face except the unbounded one must be atriangle: a bounded face is a polygon, and we have seen in Chapter 3 that anypolygon can be triangulated. What about the unbounded face? It is not difficultto see that any segment connecting two consecutive points on the boundary ofthe convex hull of P is an edge in any triangulation T. This implies that the

convex hull boundary

union of the bounded faces of T is always the convex hull of P, and that theunbounded face is always the complement of the convex hull. (In our applicationthis means that if the domain is a rectangular area, say, we have to make surethat the corners of the domain are included in the set of sample points, so thatthe triangles in the triangulation cover the domain of the terrain.) The numberof triangles is the same in any triangulation of P. This also holds for the numberof edges. The exact numbers depend on the number of points in P that are onthe boundary of the convex hull of P. (Here we also count points in the interiorof convex hull edges. Hence, the number of points on the convex hull boundaryis not necessarily the same as the number of convex hull vertices.) This is madeprecise in the following theorem.

Theorem 9.1 Let P be a set of n points in the plane, not all collinear, and let kdenote the number of points in P that lie on the boundary of the convex hullof P. Then any triangulation of P has 2n!2! k triangles and 3n!3! k edges.

Proof. Let T be a triangulation of P, and let m denote the number of trianglesof T. Note that the number of faces of the triangulation, which we denote byn f , is m + 1. Every triangle has three edges, and the unbounded face has kedges. Furthermore, every edge is incident to exactly two faces. Hence, thetotal number of edges of T is ne := (3m+ k)/2. Euler’s formula tells us that

n!ne +n f = 2. 193

Collection of trianglesÉ triangle corners are the

points of PÉ union of triangles =

convex hull of PÉ any two triangles are

É either disjointÉ or intersect in a

corner or edge

Note: the number triangles is the same any triangulation of P

Delaunay Triangulation: max min angle (avoids ’skinny triangles’)


Delaunay Triangulation

(Some) motivation: (Polyhedral) Terrains

9 Delaunay TriangulationsHeight Interpolation

When we talked about maps of a piece of the earth’s surface in previous chapters,we implicitly assumed there is no relief. This may be reasonable for a countrylike the Netherlands, but it is a bad assumption for Switzerland. In this chapterwe set out to remedy this situation.

We can model a piece of the earth’s surface as a terrain. A terrain is a2-dimensional surface in 3-dimensional space with a special property: everyvertical line intersects it in a point, if it intersects it at all. In other words, itis the graph of a function f : A ! R2 " R that assigns a height f (p) to everypoint p in the domain, A, of the terrain. (The earth is round, so on a globalscale terrains defined in this manner are not a good model of the earth. Buton a more local scale terrains provide a fairly good model.) A terrain can bevisualized with a perspective drawing like the one in Figure 9.1, or with contourlines—lines of equal height—like on a topographic map.

Figure 9.1A perspective view of a terrain

Of course, we don’t know the height of every point on earth; we only know itwhere we’ve measured it. This means that when we talk about some terrain, weonly know the value of the function f at a finite set P ! A of sample points. Fromthe height of the sample points we somehow have to approximate the heightat the other points in the domain. A naive approach assigns to every p # A theheight of the nearest sample point. However, this gives a discrete terrain, which 191



Terrain from sample points

Chapter 9DELAUNAY TRIANGULATIONS

doesn’t look very natural. Therefore our approach for approximating a terrainis as follows. We first determine a triangulation of P: a planar subdivisionwhose bounded faces are triangles and whose vertices are the points of P. (Weassume that the sample points are such that we can make the triangles coverthe domain of the terrain.) We then lift each sample point to its correct height,thereby mapping every triangle in the triangulation to a triangle in 3-space.Figure 9.2 illustrates this. What we get is a polyhedral terrain, the graph of acontinuous function that is piecewise linear. We can use the polyhedral terrainas an approximation of the original terrain.

Figure 9.2Obtaining a polyhedral terrain from a

set of sample points

The question remains: how do we triangulate the set of sample points? Ingeneral, this can be done in many different ways. But which triangulation is themost appropriate one for our purpose, namely to approximate a terrain? Thereis no definitive answer to this question. We do not know the original terrain, weonly know its height at the sample points. Since we have no other information,and the height at the sample points is the correct height for any triangulation, alltriangulations of P seem equally good. Nevertheless, some triangulations lookmore natural than others. For example, have a look at Figure 9.3, which showstwo triangulations of the same point set. From the heights of the sample pointswe get the impression that the sample points were taken from a mountain ridge.Triangulation (a) reflects this intuition. Triangulation (b), however, where onesingle edge has been “flipped,” has introduced a narrow valley cutting throughthe mountain ridge. Intuitively, this looks wrong. Can we turn this intuition intoa criterion that tells us that triangulation (a) is better than triangulation (b)?

Figure 9.3Flipping one edge can make a big

difference (a) (b)height = 985 height = 23

0

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The problem with triangulation (b) is that the height of the point q is deter-192

Approximation of a real terrain(earth’s surface)É measure height at sample

points PÉ approximate the height of

other points (real terrain)É triangulate PÉ lift each sample point to

its correct height



Terrain from sample points: which triangulation?

To flip or not to flip?


doesn’t look very natural. Therefore our approach for approximating a terrainis as follows. We first determine a triangulation of P: a planar subdivisionwhose bounded faces are triangles and whose vertices are the points of P. (Weassume that the sample points are such that we can make the triangles coverthe domain of the terrain.) We then lift each sample point to its correct height,thereby mapping every triangle in the triangulation to a triangle in 3-space.Figure 9.2 illustrates this. What we get is a polyhedral terrain, the graph of acontinuous function that is piecewise linear. We can use the polyhedral terrainas an approximation of the original terrain.

Figure 9.2Obtaining a polyhedral terrain from a

set of sample points

The question remains: how do we triangulate the set of sample points? Ingeneral, this can be done in many different ways. But which triangulation is themost appropriate one for our purpose, namely to approximate a terrain? Thereis no definitive answer to this question. We do not know the original terrain, weonly know its height at the sample points. Since we have no other information,and the height at the sample points is the correct height for any triangulation, alltriangulations of P seem equally good. Nevertheless, some triangulations lookmore natural than others. For example, have a look at Figure 9.3, which showstwo triangulations of the same point set. From the heights of the sample pointswe get the impression that the sample points were taken from a mountain ridge.Triangulation (a) reflects this intuition. Triangulation (b), however, where onesingle edge has been “flipped,” has introduced a narrow valley cutting throughthe mountain ridge. Intuitively, this looks wrong. Can we turn this intuition intoa criterion that tells us that triangulation (a) is better than triangulation (b)?

Figure 9.3Flipping one edge can make a big

difference (a) (b)height = 985 height = 23

0

10

6

20

36

28

1240

1000

980

990

1008

890

q

0

4 23

190

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The problem with triangulation (b) is that the height of the point q is deter-192 ’Skinny’ triangles (-> small angles) give counter-intuitive results(e.g., cutting a mountain ridge into a narrow valey)



Delaunay Triangulation: max min angle (over all triangles) over alltriangulations

Note: there is only a finite number of different triangulations


Delaunay Triangulation(Given some initial triangulation)

Flip ’illegal’ edges to increase min angle (if possible)


Plugging the values for ne and n f into the formula, we get m = 2n!2!k, whichin turn implies ne = 3n!3! k.

Let T be a triangulation of P, and suppose it has m triangles. Consider the3m angles of the triangles of T, sorted by increasing value. Let !1,!2, . . . ,!3mbe the resulting sequence of angles; hence, !i ! ! j, for i < j. We call A(T) :=(!1,!2, . . . ,!3m) the angle-vector of T. Let T" be another triangulation of thesame point set P, and let A(T") := (! "

1,! "2, . . . ,! "

3m) be its angle-vector. Wesay that the angle-vector of T is larger than the angle-vector of T" if A(T) islexicographically larger than A(T"), or, in other words, if there exists an index iwith 1 ! i ! 3m such that

! j = ! "j for all j < i, and !i > ! "

i .

We denote this as A(T) > A(T"). A triangulation T is called angle-optimal ifA(T) " A(T") for all triangulations T" of P. Angle-optimal triangulations areinteresting because, as we have seen in the introduction to this chapter, they aregood triangulations if we want to construct a polyhedral terrain from a set ofsample points.

Below we will study when a triangulation is angle-optimal. To do this it isuseful to know the following theorem, often called Thales’s Theorem. Denotethe smaller angle defined by three points p, q, r by #pqr.

! C

p

q

r

s

a

b

Theorem 9.2 Let C be a circle, ! a line intersecting C in points a and b, and p,q, r, and s points lying on the same side of !. Suppose that p and q lie on C, thatr lies inside C, and that s lies outside C. Then

#arb > #apb = #aqb > #asb.

Now consider an edge e = pi p j of a triangulation T of P. If e is not an edgeof the unbounded face, it is incident to two triangles pi p j pk and pi p j pl . If thesetwo triangles form a convex quadrilateral, we can obtain a new triangulationT" by removing pi p j from T and inserting pk pl instead. We call this operationan edge flip. The only difference in the angle-vector of T and T" are the six

Figure 9.4Flipping an edge

! "1

! "4

! "3

! "5

! "2

! "6

pi

p j

pk

pl

pi!4

!1

!3

!5

!2

!6

p j

pk

pl

edge flip

angles !1, . . . ,!6 in A(T), which are replaced by ! "1, . . . ,! "

6 in A(T"). Figure 9.4illustrates this. We call the edge e = pi p j an illegal edge if

min1!i!6

!i < min1!i!6

! "i .194

’Illegal’ edge criterion:

Section 9.1TRIANGULATIONS OF PLANAR POINT

SETS

In other words, an edge is illegal if we can locally increase the smallest angleby flipping that edge. The following observation immediately follows from thedefinition of an illegal edge.

Observation 9.3 Let T be a triangulation with an illegal edge e. Let T! be thetriangulation obtained from T by flipping e. Then A(T!) > A(T).

It turns out that it is not necessary to compute the angles !1, . . . ,!6,! !1, . . . ,! !

6to check whether a given edge is legal. Instead, we can use the simple criterionstated in the next lemma. The correctness of this criterion follows from Thales’sTheorem.

pi

p j

pk

pl

illegal

Lemma 9.4 Let edge pi p j be incident to triangles pi p j pk and pi p j pl , and let Cbe the circle through pi, p j, and pk. The edge pi p j is illegal if and only if thepoint pl lies in the interior of C. Furthermore, if the points pi, p j, pk, pl forma convex quadrilateral and do not lie on a common circle, then exactly one ofpi p j and pk pl is an illegal edge.

Observe that the criterion is symmetric in pk and pl : pl lies inside the circlethrough pi, p j, pk if and only if pk lies inside the circle through pi, p j, pl . Whenall four points lie on a circle, both pi p j and pk pl are legal. Note that the twotriangles incident to an illegal edge must form a convex quadrilateral, so that itis always possible to flip an illegal edge.

We define a legal triangulation to be a triangulation that does not containany illegal edge. From the observation above it follows that any angle-optimaltriangulation is legal. Computing a legal triangulation is quite simple, once weare given an initial triangulation. We simply flip illegal edges until all edges arelegal.

Algorithm LEGALTRIANGULATION(T)Input. Some triangulation T of a point set P.Output. A legal triangulation of P.1. while T contains an illegal edge pi p j2. do (" Flip pi p j ")3. Let pi p j pk and pi p j pl be the two triangles adjacent to pi p j.4. Remove pi p j from T, and add pk pl instead.5. return T

Why does this algorithm terminate? It follows from Observation 9.3 that theangle-vector of T increases in every iteration of the loop. Since there is onlya finite number of different triangulations of P, this proves termination of thealgorithm. Once it terminates, the result is a legal triangulation. Although thealgorithm is guaranteed to terminate, it is too slow to be interesting. We havegiven the algorithm anyway, because later we shall need a similar procedure.But first we will look at something completely different—or so it seems. 195


Voronoi diagram (’dual’ of DT)

Given a set P of n points (sites) in the plane

Section 7.1DEFINITION AND BASIC PROPERTIES

say that a Voronoi diagram is connected we mean that the union of its edges andvertices forms a connected set. The cell of Vor(P) that corresponds to a site piis denoted V(pi); we call it the Voronoi cell of pi. (In the terminology of theintroduction to this chapter: V(pi) is the trading area of site pi.)

We now take a closer look at the Voronoi diagram. First we study thestructure of a single Voronoi cell. For two points p and q in the plane we definethe bisector of p and q as the perpendicular bisector of the line segment pq. Thisbisector splits the plane into two half-planes. We denote the open half-planethat contains p by h(p,q) and the open half-plane that contains q by h(q, p).Notice that r ! h(p,q) if and only if dist(r, p) < dist(r,q). From this we obtainthe following observation.

Observation 7.1 V(pi) =!

1! j!n, j "=i h(pi, p j).

Thus V(pi) is the intersection of n#1 half-planes and, hence, a (possiblyunbounded) open convex polygonal region bounded by at most n#1 verticesand at most n#1 edges.

What does the complete Voronoi diagram look like? We just saw that eachcell of the diagram is the intersection of a number of half-planes, so the Voronoidiagram is a planar subdivision whose edges are straight. Some edges are linesegments and others are half-lines. Unless all sites are collinear there will be noedges that are full lines:

Theorem 7.2 Let P be a set of n point sites in the plane. If all the sites arecollinear then Vor(P) consists of n # 1 parallel lines. Otherwise, Vor(P) isconnected and its edges are either segments or half-lines.

Proof. The first part of the theorem is easy to prove, so assume that not all sitesin P are collinear.

We first show that the edges of Vor(P) are either segments or half-lines. Wealready know that the edges of Vor(P) are parts of straight lines, namely parts ofthe bisectors between pairs of sites. Now suppose for a contradiction that thereis an edge e of Vor(P) that is a full line. Let e be on the boundary of the Voronoicells V(pi) and V(p j). Let pk ! P be a point that is not collinear with pi and p j.The bisector of p j and pk is not parallel to e and, hence, it intersects e. But then

pi p j

pke

the part of e that lies in the interior of h(pk, p j) cannot be on the boundary ofV(p j), because it is closer to pk than to p j, a contradiction.

It remains to prove that Vor(P) is connected. If this were not the casethen there would be a Voronoi cell V(pi) splitting the plane into two. BecauseVoronoi cells are convex, V(pi) would consist of a strip bounded by two parallelfull lines. But we just proved that the edges of the Voronoi diagram cannot befull lines, a contradiction.

Now that we understand the structure of the Voronoi diagram we investigateits complexity, that is, the total number of its vertices and edges. Since there aren sites and each Voronoi cell has at most n#1 vertices and edges, the complexityof Vor(P) is at most quadratic. It is not clear, however, whether Vor(P) canactually have quadratic complexity: it is easy to construct an example wherea single Voronoi cell has linear complexity, but can it happen that many cells 149

Voronoi diagram V(P): subdivision of the plane into n cells (one foreach site of P), such that a point q lies in the cell corresponding toa site pi ∈ P iff distance(q,pi) < distance(q,pi)



How are the cells formed?

Intersection of (open) half-planes formed by bisectors betweensites

Section 7.1DEFINITION AND BASIC PROPERTIES

say that a Voronoi diagram is connected we mean that the union of its edges andvertices forms a connected set. The cell of Vor(P) that corresponds to a site piis denoted V(pi); we call it the Voronoi cell of pi. (In the terminology of theintroduction to this chapter: V(pi) is the trading area of site pi.)

We now take a closer look at the Voronoi diagram. First we study thestructure of a single Voronoi cell. For two points p and q in the plane we definethe bisector of p and q as the perpendicular bisector of the line segment pq. Thisbisector splits the plane into two half-planes. We denote the open half-planethat contains p by h(p,q) and the open half-plane that contains q by h(q, p).Notice that r ! h(p,q) if and only if dist(r, p) < dist(r,q). From this we obtainthe following observation.

Observation 7.1 V(pi) =!

1! j!n, j "=i h(pi, p j).

Thus V(pi) is the intersection of n#1 half-planes and, hence, a (possiblyunbounded) open convex polygonal region bounded by at most n#1 verticesand at most n#1 edges.

What does the complete Voronoi diagram look like? We just saw that eachcell of the diagram is the intersection of a number of half-planes, so the Voronoidiagram is a planar subdivision whose edges are straight. Some edges are linesegments and others are half-lines. Unless all sites are collinear there will be noedges that are full lines:

Theorem 7.2 Let P be a set of n point sites in the plane. If all the sites arecollinear then Vor(P) consists of n # 1 parallel lines. Otherwise, Vor(P) isconnected and its edges are either segments or half-lines.

Proof. The first part of the theorem is easy to prove, so assume that not all sitesin P are collinear.

We first show that the edges of Vor(P) are either segments or half-lines. Wealready know that the edges of Vor(P) are parts of straight lines, namely parts ofthe bisectors between pairs of sites. Now suppose for a contradiction that thereis an edge e of Vor(P) that is a full line. Let e be on the boundary of the Voronoicells V(pi) and V(p j). Let pk ! P be a point that is not collinear with pi and p j.The bisector of p j and pk is not parallel to e and, hence, it intersects e. But then

pi p j

pke

the part of e that lies in the interior of h(pk, p j) cannot be on the boundary ofV(p j), because it is closer to pk than to p j, a contradiction.

It remains to prove that Vor(P) is connected. If this were not the casethen there would be a Voronoi cell V(pi) splitting the plane into two. BecauseVoronoi cells are convex, V(pi) would consist of a strip bounded by two parallelfull lines. But we just proved that the edges of the Voronoi diagram cannot befull lines, a contradiction.

Now that we understand the structure of the Voronoi diagram we investigateits complexity, that is, the total number of its vertices and edges. Since there aren sites and each Voronoi cell has at most n#1 vertices and edges, the complexityof Vor(P) is at most quadratic. It is not clear, however, whether Vor(P) canactually have quadratic complexity: it is easy to construct an example wherea single Voronoi cell has linear complexity, but can it happen that many cells 149



Characterization of edges and vertices of a Voronoi diagramSection 7.2COMPUTING THE VORONOI DIAGRAM

Proof. (i) Suppose there is a point q such that CP(q) contains three or more siteson its boundary. Let pi, p j, and pk be three of those sites. Since the interiorof CP(q) is empty q must be on the boundary of each of V(pi), V(p j), andV(pk), and q must be a vertex of Vor(P).

On the other hand, every vertex q of Vor(P) is incident to at least threeedges and, hence, to at least three Voronoi cells V(pi), V(p j), and V(pk).Vertex q must be equidistant to pi, p j, and pk and there cannot be anothersite closer to q, since otherwise V(pi), V(p j), and V(pk) would not meet at q.Hence, the interior of the circle with pi, p j, and pk on its boundary does notcontain any site.

(ii) Suppose there is a point q with the property stated in the theorem. SinceCP(q) does not contain any sites in its interior and pi and p j are on itsboundary, we have dist(q, pi) = dist(q, p j) ! dist(q, pk) for all 1 ! k ! n.It follows that q lies on an edge or vertex of Vor(P). The first part of thetheorem implies that q cannot be a vertex of Vor(P). Hence, q lies on an edgeof Vor(P), which is defined by the bisector of pi and p j.

Conversely, let the bisector of pi and p j define a Voronoi edge. Thelargest empty circle of any point q in the interior of this edge must contain piand p j on its boundary and no other sites.

7.2 Computing the Voronoi Diagram

In the previous section we studied the structure of the Voronoi diagram. Wenow set out to compute it. Observation 7.1 suggests a simple way to do this:for each site pi, compute the common intersection of the half-planes h(pi, p j),with j != i, using the algorithm presented in Chapter 4. This way we spendO(n logn) time per Voronoi cell, leading to an O(n2 logn) algorithm to computethe whole Voronoi diagram. Can’t we do better? After all, the total complexityof the Voronoi diagram is only linear. The answer is yes: the plane sweepalgorithm described below—commonly known as Fortune’s algorithm afterits inventor—computes the Voronoi diagram in O(n logn) time. You may betempted to look for an even faster algorithm, for example one that runs in lineartime. This turns out to be too much to ask: the problem of sorting n real numbersis reducible to the problem of computing Voronoi diagrams, so any algorithmfor computing Voronoi diagrams must take !(n logn) time in the worst case.Hence, Fortune’s algorithm is optimal.

The strategy in a plane sweep algorithm is to sweep a horizontal line—thesweep line—from top to bottom over the plane. While the sweep is performedinformation is maintained regarding the structure that one wants to compute.More precisely, information is maintained about the intersection of the structurewith the sweep line. While the sweep line moves downwards the informationdoes not change, except at certain special points—the event points.

Let’s try to apply this general strategy to the computation of the Voronoi diagramof a set P = {p1, p2, . . . , pn} of point sites in the plane. According to the plane 151

É a point is a vertex ofV(P) iff its largest emptycircle contains three ormore sites on itsboundary

É a bisector between twosites defines an edge iffthere is a point on thebisector such that itslargest empty circlecontains both sites onits boundary but noother site

The Voronoi diagram can be computed in O(n logn) time



What’s the connection?

Dual graph G of V(P)


9.2 The Delaunay Triangulation

Let P be a set of n points—or sites, as we shall sometimes call them—in theplane. Recall from Chapter 7 that the Voronoi diagram of P is the subdivisionof the plane into n regions, one for each site in P, such that the region of asite p ! P contains all points in the plane for which p is the closest site. TheVoronoi diagram of P is denoted by Vor(P). The region of a site p is called

Figure 9.5The dual graph of Vor(P)

Vor(P)

G

the Voronoi cell of p; it is denoted by V(p). In this section we will study thedual graph of the Voronoi diagram. This graph G has a node for every Voronoicell—equivalently, for every site—and it has an arc between two nodes if thecorresponding cells share an edge. Note that this means that G has an arc forevery edge of Vor(P). As you can see in Figure 9.5, there is a one-to-onecorrespondence between the bounded faces of G and the vertices of Vor(P).

Figure 9.6The Delaunay graph DG(P)

Consider the straight-line embedding of G, where the node correspondingto the Voronoi cell V(p) is the point p, and the arc connecting the nodes ofV(p) and V(q) is the segment pq—see Figure 9.6. We call this embedding theDelaunay graph of P, and we denote it by DG(P). (Although the name soundsFrench, Delaunay graphs have nothing to do with the French painter. They196

É a node for everyVoronoi cell (equiv. site)

É an arc between twonodes if the corr. cellsshare and edge



What’s the connection?

Delaunay graph DG(P)


9.2 The Delaunay Triangulation

Let P be a set of n points—or sites, as we shall sometimes call them—in theplane. Recall from Chapter 7 that the Voronoi diagram of P is the subdivisionof the plane into n regions, one for each site in P, such that the region of asite p ! P contains all points in the plane for which p is the closest site. TheVoronoi diagram of P is denoted by Vor(P). The region of a site p is called

Figure 9.5The dual graph of Vor(P)

Vor(P)

G

the Voronoi cell of p; it is denoted by V(p). In this section we will study thedual graph of the Voronoi diagram. This graph G has a node for every Voronoicell—equivalently, for every site—and it has an arc between two nodes if thecorresponding cells share an edge. Note that this means that G has an arc forevery edge of Vor(P). As you can see in Figure 9.5, there is a one-to-onecorrespondence between the bounded faces of G and the vertices of Vor(P).

Figure 9.6The Delaunay graph DG(P)

Consider the straight-line embedding of G, where the node correspondingto the Voronoi cell V(p) is the point p, and the arc connecting the nodes ofV(p) and V(q) is the segment pq—see Figure 9.6. We call this embedding theDelaunay graph of P, and we denote it by DG(P). (Although the name soundsFrench, Delaunay graphs have nothing to do with the French painter. They196

The Delaunay triangulationis any triangulationobtained from DG(P) byadding edges to it (ifneeded)

Sraight-line embedding ofthe dual graph GIt’s a plane graph

It has a face for every vetexof V(P)

Section 9.2THE DELAUNAY TRIANGULATION

are named after the Russian mathematician Boris Nikolaevich Delone, whowrote his own name as “Boris Nikolaeviq Delone,” which would betransliterated into English as “Delone.” However, since his work was publishedin French—at his time, the languages of science were French and German—hisname is better known in the French transliteration.) The Delaunay graph of apoint set turns out to have a number of surprising properties. The first is that itis always a plane graph: no two edges in the embedding cross.

Theorem 9.5 The Delaunay graph of a planar point set is a plane graph.

Proof. To prove this, we need a property of the edges in the Voronoi diagramstated in Theorem 7.4(ii). For completeness we repeat the property, phrasedhere in terms of Delaunay graphs.

The edge pi p j is in the Delaunay graph DG(P) if and only if thereis a closed disc Ci j with pi and p j on its boundary and no other siteof P contained in it. (The center of such a disc lies on the commonedge of V(pi) and V(p j).)

Ci j

pi

p j

contained in V(pi)

contained in V(p j)

Define ti j to be the triangle whose vertices are pi, p j, and the center of Ci j.Note that the edge of ti j connecting pi to the center of Ci j is contained in V(pi);a similar observation holds for p j. Now let pk pl be another edge of DG(P),and define the circle Ckl and the triangle tkl similar to the way Ci j and ti j weredefined.

Suppose for a contradiction that pi p j and pk pl intersect. Both pk and plmust lie outside Ci j and so they also lie outside ti j. This implies that pk pl mustintersect one of the edges of ti j incident to the center of Ci j. Similarly, pi p jmust intersect one of the edges of tkl incident to the center of Ckl . It followsthat one of the edges of ti j incident to the center of Ci j must intersect one of theedges of tkl incident to the center of Ckl . But this contradicts that these edgesare contained in disjoint Voronoi cells.

The Delaunay graph of P is an embedding of the dual graph of the Voronoidiagram. As observed earlier, it has a face for every vertex of Vor(P). The edges

vf

around a face correspond to the Voronoi edges incident to the correspondingVoronoi vertex. In particular, if a vertex v of Vor(P) is a vertex of the Voronoicells for the sites p1, p2, p3, . . . , pk, then the corresponding face f in DG(P) hasp1, p2, p3, . . . , pk as its vertices. Theorem 7.4(i) tells us that in this situation thepoints p1, p2, p3, . . . , pk lie on a circle around v, so we not only know that f is ak-gon, but even that it is convex.

If the points of P are distributed at random, the chance that four pointshappen to lie on a circle is very small. We will—in this chapter—say that a setof points is in general position if it contains no four points on a circle. If P isin general position, then all vertices of the Voronoi diagram have degree three,and consequently all bounded faces of DG(P) are triangles. This explains whyDG(P) is often called the Delaunay triangulation of P. We shall be a bit morecareful, and will call DG(P) the Delaunay graph of P. We define a Delaunaytriangulation to be any triangulation obtained by adding edges to the Delaunaygraph. Since all faces of DG(P) are convex, obtaining such a triangulation 197

If P contains no 4 points ona cirlce then all vertices ofV(P) have degree 3 -> all(bounded) faces of DG(P)are triangles



Both Delaunay triangulation and Voronoi diagram of an n point setcan be computed in O(n logn) time


Exercise 4

É Read CGAL 2D Triangulations (Ch. 37.1–37.5)É Compute the Delaunay triangulation T and its dual Voronoi

diagram V of your favorite n point set P (2d). Draw both usingQt.

É Given a point p (not necessarily in P), find the triangle of Tand the cell of V that p lies in.

É Given two points s and t (not necessarily in P), find a shortestpath (measured in number of triangles) from s to t in T.

É Given two points s and t compute the set of triangles of Tinterected by the straight-line segment st.


Project ideas (so far...)

É Porting Arrangments package demo (currently in Qt3) to Qt4É Random point generator on a circleÉ Convex hull (2d, 3d) in O(nlogh) time (T. Chan)É Triangulating simple polygons (non-convex) in O(n) time

(Amato et al.)É Shape matching: minimizing the Hausdorff distance of two

point sets under translationÉ Guarding art galleries, guarding terrains (visibility polygons,

regions)É Shortest paths among obstacles (visibility graphs)É Shortest paths in terrainsÉ Motion planning in robotics: can a robot reach its distination?

(minkowski sums, visibility-voronoi complex)É Surface reconstruction from point setsÉ Binding CGAL to other (scripting) languages..É Look at the packages of CGAL to find something that you like..


lecture 4 (cgal)

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