Geometric Algorithms - Faculteit Wiskunde en...

Geometric AlgorithmsDelaunay Triangulations and Voronoi Diagrams

Terrain Modeling

heightmeasurementsp = (xp, yp, zp)

Terrain Modeling


projectionπ(p) = (px, py, 0)

Terrain Modeling



Interpolation: triangulate & interpolate

Terrain Modeling



Interpolation: triangulate & interpolate within triangles

Terrain Modeling



Interpolation: triangulate & interpolate within triangles

What is a ‘good’ triangulation?

Triangulation of a Point Set

Def.: A triangulation of a point set P ⊂ R2 is a maximalplanar subdivision with vertex set P .



Obs.:



Obs.: • all inner faces are triangles




• outer face is complement of convex hull





CH(P )





CH(P )

Thm 1: Let P be a set of n non-collinear points and let h bethe number of vertices of CH(P ).

Then every triangulation of P has t(n, h) triangles ande(n, h) edges.





CH(P )



Compute t and e!





CH(P )



Compute t and e!

Euler’s formula for connected plane graphs:# faces−# edges + # vertices = 2,also counting the outer face.





CH(P )


Then every triangulation of P has (2n− 2− h)triangles and (3n− 3− h) edges.

Back to Height Interpolation

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Lets look at the interpolation along an edge:


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intuition: avoid ‘thin’ triangles!



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intuition: avoid ‘thin’ triangles!

or: maximize the smallest angle within triangles!


Angle-Optimal Triangulations

Def.:

T A(T ) = (60◦, 60◦, 60◦, 60◦, 60◦, 60◦)

Let T be a triangulation of P with m triangles and 3mvertices. Its angle vector is A(T ) = (α1, . . . , α3m)where α1, . . . , α3m are the angles of T sorted byincreasing value.


Def.:

T A(T ) = (60◦, 60◦, 60◦, 60◦, 60◦, 60◦) T ′

A(T ′) = (30◦, 30◦, 30◦, 30◦, 120◦, 120◦)


For two triangulations T and T ′ of P define orderA(T ) > A(T ′) as the lexicographical order.


Def.:

T A(T ) = (60◦, 60◦, 60◦, 60◦, 60◦, 60◦) T ′

A(T ′) = (30◦, 30◦, 30◦, 30◦, 120◦, 120◦)


For two triangulations T and T ′ of P define orderA(T ) > A(T ′) as the lexicographical order.

T is angle optimal, if A(T ) ≥ A(T ′) for alltriangulations T ′ of P .

Edge Flips

Def.: Let T be a triangulation.An edge e of T is illegal,if thesmallest angle of the triangles incident to e can beincreased by flipping e.

eT

Edge Flips


mini αi = 30◦

eT

Edge Flips


mini αi = 30◦

eT

flip(T , e)

Edge Flips


mini αi = 30◦

eT ′ T

flip(T , e)e′

Edge Flips


mini αi = 30◦mini αi = 60◦

eT ′ T

flip(T , e)e′

Edge Flips


mini αi = 30◦mini αi = 60◦

e

Obs.: Let e be an illegal edge in T and let T ′ = flip(T , e).Then A(T ′) > A(T ).

T ′ Tflip(T , e)

e′

Thales Theorem

a

Thm 2: if ab is a diameter, then the angle atany third point on the circle c is 90◦.

b

Thales Theorem

a

Thm 2: if ab is a diameter, then the angle atany third point on the circle c is 90◦.

bThm 2′: Let C be a circle, ` a line intersecting C in points a

and b, and p, q, r, s points lying on the same side of `.Suppose that p, q lie on C, r lies inside C, and s liesoutside C. Then ]arb > ]apb = ]aqb > ]asb,where ]abc denotes the smaller angle defined bythree points a, b, c.

ab

c c′e

d

∠aeb < ∠acb = ∠ac′b < ∠adb

c

Legal Triangulations

Lemma 1: Let ∆prq and ∆pqs be two adjacent triangles in Tand C the circumcircle of ∆prq. Then:

pq illegal ⇔ s ∈ int(C).If p, q, r, s form a convex quadrilateral and s 6∈ ∂C,then either pq or rs is illegal.

proof sketch:

p

q

r

s C




proof sketch:

p

q

r

s

ϕpr

ϕps

ϕrq

ϕsq

θpr

θpsθrq

θsq

ϕpr > θpr

ϕps > θps

ϕrq > θrq

ϕsq > θsq

C




proof sketch:

p

q

r

s

ϕpr

ϕps

ϕrq

ϕsq

θpr

θpsθrq

θsq

ϕpr > θpr

ϕps > θps

ϕrq > θrq

ϕsq > θsqϕp

ϕq

ϕp = θrq + θsq

ϕq = θpr + θps

C




Def.: A triangulation without illegal edges is called legaltriangulation





Are there legal triangulations?





while T has illegal edge e doflip(T , e)

return T






return Tterminates?






return Tterminates, since A(T ) increases and#triangulations finite

Are we stuck?

We know: Every angle-optimal triangulation is legal.

But is every legal triangulation angle-optimal?

. . . now for something completely different . . .

Voronoi Diagrams

Motivation

?

It is a beautiful sunny day. You know where all of the locationsof ice cream parlors in town and would like to know which oneis closest. How can we determine the closest ice cream parlorfor every point on the map?

Motivation

?


!

Motivation

?


!

The solution is a subdivision of the plane, called Voronoidiagram.It has many applications, e.g., in geography, physics, robotics

Voronoi Diagrams

Definition: The Voronoi diagram of n points {p1, . . . , pn} in the plane isthe subdivision of the plane into n cells, such that a point q is in the cellof pi if and only if dist(q, pi) < dist(q, pj) for all i6=j.

Voronoi Diagrams

long history• Descartes 1644• Dirichlet 1850• Voronoi 1907

Definition: The Voronoi diagram of n points {p1, . . . , pn} in the plane isthe subdivision of the plane into n cells, such that a point q is in the cellof pi if and only if dist(q, pi) < dist(q, pj) for all i6=j.

Voronoi Diagrams

for 2 points

Voronoi Diagrams

p

q

b(p, q) = {x ∈ R2 : |xp| = |xq|}bisector of p and q

for 2 points

Voronoi Diagrams

p

q

b(p, q)

h(p, q)

= {x ∈ R2 : |xp| = |xq|}

= {x : |xp| < |xq|}

bisector of p and q

open half plane

for 2 points

Voronoi Diagrams

p

q

b(p, q)

h(p, q) h(q, p)

= {x ∈ R2 : |xp| = |xq|}

= {x : |xp| < |xq|} = {x : |xq| < |xp|}

bisector of p and q

open half plane open half plane

for 2 points

Voronoi Diagrams

P = {p1, p2, . . . , pn}for n points

Voronoi Diagrams

for n points

Voronoi Diagrams

Vor(P ) =

Voronoi diagramof P

for n points

Voronoi Diagrams

Questions: 1) What are Voronoi cells, edges and vertices?

Voronoi Diagrams

Questions: 1) What are Voronoi cells, edges and vertices?

2) Are Voronoi cells convex?

Voronoi Diagrams

Let P be a set of points in th plane and p, p′, p′′ ∈ P .

Voronoi Diagrams


Vor(P )Voronoi diagramp

p′

p′′

Voronoi Diagrams

V({p}) =


Vor(P )Voronoi diagram

• Voronoi cell

p

p′

p′′

Voronoi Diagrams

V({p}) =


Vor(P )Voronoi diagram

• Voronoi cell

p

p′

p′′

=V(p)

Voronoi Diagrams

V({p}) =


Vor(P )

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}Voronoi diagram

• Voronoi cell

p

p′

p′′

=V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}Voronoi diagram

• Voronoi cell

p

p′

p′′

=V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

V({p, p′})

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}• Voronoi edge

Voronoi diagram

• Voronoi cell

p

p′

p′′

=

=V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

V({p, p′})

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}

{x : |xp| = |xp′| and |xp| < |xq| ∀q 6= p, p′}• Voronoi edge

Voronoi diagram

• Voronoi cell

p

p′

p′′

=

=V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

V({p, p′})∂V(p) ∩ ∂V(p′)=

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}


Voronoi diagram

• Voronoi cell

p

p′

p′′

=

=V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

V({p, p′})∂V(p) ∩ ∂V(p′)= rel-int( )

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}


Voronoi diagram

• Voronoi cell

p

p′

p′′

=

=

i.e. without endpoints,

V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

V({p, p′})∂V(p) ∩ ∂V(p′)= rel-int( )

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}

{x : |xp| = |xp′| and |xp| < |xq| ∀q 6= p, p′}

• Voronoi vertexV({p, p′, p′′})

• Voronoi edge

Voronoi diagram

• Voronoi cell

p

p′

p′′

=

=


V(p)

Voronoi Diagrams

V({p}) =

=⋂

q 6=p h(p, q)


Vor(P )

V({p, p′})∂V(p) ∩ ∂V(p′)= rel-int( )

{x ∈ R2 : |xp| < |xq| ∀q ∈ P \ {p}

}

{x : |xp| = |xp′| and |xp| < |xq| ∀q 6= p, p′}

• Voronoi vertex= ∂V(p) ∩ ∂V(p′) ∩ ∂V(p′′)V({p, p′, p′′})

• Voronoi edge

Voronoi diagram

• Voronoi cell

p

p′

p′′

=

=


V(p)

Characterization

Definition: Let q be a point. Define CP (q) as the largest diskwith center q containing no points of P in itsinterior.

q

Characterization


q

CP (q)

Characterization


q

CP (q)

Obs.: • A point q is a Voronoi vertex⇔ |CP (q) ∩ P | ≥ 3,• the bisector b(pi, pj) defines a Voronoi edge⇔∃q ∈ b(pi, pj) with CP (q) ∩ P = {pi, pj}.

Delaunay Triangulation

Def.: The graph G = (P,E) withE = {pq | V(p) und V(q) are adjacent}is called the dual graph of Vor(P ).

Let Vor(P ) be the Voronoi diagram of P .




G




Def.: The straight-line drawing of G is called Delaunay graphDG(P ).

DG(P )

Properties

Thm 3: DG(P ) has no crossing edges.

Properties


proof sketch:the bisector b(p, q) defines a Voronoi edge⇔∃r ∈ b(p, q) with CP (r) ∩ P = {p, q}.

the edge pq is inDG(P )⇔ there is an empty circle Cp,q with p and q on its boundary.

i.e.

C i j

pi

p j

in V ( pi)

in V ( p j)

pk

pl

⇒ pq can’t intersect another edge

Suppose pipj is crossed by pkpl. Every circlewith pk, pl has to contain either pi or pj .Contradiction.

Properties




i.e.

Obs.: A voronoi vertex v in Vor(P ) of degree k correspondsto a convex k-gon inDG(P ).

Properties




i.e.


If P is in general position (no 4 points ona circle), then all (inner) faces ofDG(P )are triangles.

Properties




i.e.


If P is in general position (no 4 points ona circle), then all (inner) faces ofDG(P )are triangles.

A triangulation ofDG(P ) is called Delaunay triangulation.

Empty-Circle Property

Theorem about Voronoi diagrams:• a point q is a Voronoi vertex⇔ |CP (q) ∩ P | ≥ 3,• the bisector b(pi, pj) defines a Voronoi edge⇔∃q ∈ b(pi, pj) with CP (q) ∩ P = {pi, pj}.



Thm 4: Let P be a set of points.• points p, q, r are vertices of the same face inDG(P )⇔

circle through p, q, r is empty• edge pq is inDG(P )⇔ there is an empty circle Cp,q through p and q



Thm 4: Let P be a set of points.• points p, q, r are vertices of the same face inDG(P )⇔

circle through p, q, r is empty• edge pq is inDG(P )⇔ there is an empty circle Cp,q through p and q

Corollary: Let P be a set of points and T a triangulation of P . Tis a Delaunay triangulation⇔ circumcircle of every triangle is empty.

Legal vs Delaunay

Thm 5: Let P be a set of points in R2. A triangulation T of Pis legal if and only if T is a Delaunay triangulation.

proof sketch:

"‘⇐"’ obvious, use

Lemma 1: Let ∆prq and ∆pqs be two adjacent triangles in T and C thecircumcircle of ∆prq. Then: pq illegal ⇔ s ∈ int(C).

Legal vs Delaunay


proof sketch:

p

q

r

s

"‘⇒"’

• suppose s in the interior of circle of pqr• let pr be the edge maximizing ∠psr

Legal vs Delaunay


t

proof sketch:

p

q

r

s

"‘⇒"’


Legal vs Delaunay


t

proof sketch:

p

q

r

s

"‘⇒"’


• consider t adjacent to p and r• s also lies in the circle of prt

Legal vs Delaunay


t

proof sketch:

p

q

r

s

"‘⇒"’


• consider t adjacent to p and r• s also lies in the circle of prt

Thales theorem: ∠tsr > ∠psrContradiction to choice of pr

Legal vs Delaunay


Obs.: If P is in general position, then the Delaunaytriangulation of P is unique.

Legal vs Delaunay


Obs.: If P is in general position, then the Delaunaytriangulation of P is unique.⇒ legal triangulation unique

Legal vs Delaunay


Obs.: If P is in general position, then the Delaunaytriangulation of P is unique.⇒ legal triangulation uniquewe know: T angle optimal⇒T legal

Legal vs Delaunay


Obs.: If P is in general position, then the Delaunaytriangulation of P is unique.⇒ legal triangulation uniquewe know: T angle optimal⇒T legal⇒ angle-optimal triangulation isDG(P )!

Legal vs Delaunay


Obs.: If P is in general position, then the Delaunaytriangulation of P is unique.⇒ legal triangulation uniquewe know: T angle optimal⇒T legal⇒ angle-optimal triangulation isDG(P )!

Is P not in general position, then the smallest anglein every triangulation of the "‘large"’ faces inDG(P )is the same.(proof uses Thales theorem)

Randomized Incremental Construction


DelaunayTriangulation(P )Initialize T as a large triangle ∆(p0, p−1, p−2) containing all

points from PCompute a random permutation of p1, . . . , pnfor r ← 1 to n do

Insert(pr, T )

Discard p0, p−1 and p−2 with all their incident edgesreturn T

Pp−1 p−2

p0


Insert(pr, T )Find triangle T containing prif pr lies in triangle pipjpk then

add edges from pr to pi, pj , pkLegalizeEdge(pr, pipj , T )LegalizeEdge(pr, pjpk, T )LegalizeEdge(pr, pkpi, T )

elseadd edges from pr to pl, pkLegalizeEdge(pr, pipl, T )LegalizeEdge(pr, plpj , T )LegalizeEdge(pr, pjpk, T )LegalizeEdge(pr, pkpi, T )

// pr lies of edge pipj

pi

pr

pk

pj

pi

pl pkpr

pj


LegalizeEdge(pr, pipj , T )if pipj is illegal then

let pipjph be the adjacent trianglereplace pipj by prphLegalizeEdge(pr, pjph, T )LegalizeEdge(pr, phpi, T )

// Edge flip ph

pj

pi

pr

Correctness

1. All edges inserted are legal• All edges inserted are adjacent to pr.

Correctness

1. All edges inserted are legal• All edges inserted are adjacent to pr.

• All edges inserted are Delaunay edges.

pr pr

pj

pipr

pl

Correctness

1. All edges inserted are legal

2. All other edges are legalfollows from Lemma 1: an edge can only be illegal if it is incident to anew triangle.

• All edges inserted are adjacent to pr.

• All edges inserted are Delaunay edges.

pr pr

pj

pipr

pl

Initialization

Choose p0, p−1, p−2 far enough away from P , such that theylie in none of the circles of P and such that P lies in theirtriangle.

Pp−1 p−2

p0

Initialization

better:Treat p0, p−1, p−2 symbolically by modifying tests/predicatesused for point location and testing illegal edges.

Pp−1 p−2

p0

Search Structure

build search structure for point location:directed acyclic graph with• leaves: current triangles• inner nodes: deleted triangles

∆1

∆2

∆3

∆1 ∆2 ∆3

Search Structure


∆1

∆2

∆3

∆1 ∆2 ∆3

split

pr

pi

p j

∆2

∆3

∆1

∆1 ∆2 ∆3

Search Structure


∆1

∆2

∆3

∆1 ∆2 ∆3

split

pr

pi

p j

∆2

∆3

∆1

∆1 ∆2 ∆3

flip pi p j

pi

pk

∆4∆5

∆5∆4

∆1 ∆2 ∆3

Search Structure


∆1

∆2

∆3

∆1 ∆2 ∆3

split

pr

pi

p j

∆2

∆3

∆1

∆1 ∆2 ∆3

flip pi p j

pi

pk

∆4∆5

∆5∆4

∆1 ∆2 ∆3

flip pi pk

∆6

∆7

∆7∆6

∆5∆4

∆1 ∆2 ∆3

Analysis

Lemma: The expected number of triangles created is at most9n+ 1 = O(n).

Analysis


How many triangles are created when pr is inserted?proof:

Analysis



• Every point in p1, . . . , pr has the same probability 1/r to be the lastpoint.

backwards analysis:

Analysis




backwards analysis:

• expected degree of pr is≤ 6

Analysis




backwards analysis:

• expected degree of pr is≤ 6• number of triangles created at pr is≤ 2(degree(pr))− 3

Analysis




backwards analysis:

• expected degree of pr is≤ 6• number of triangles created at pr is≤ 2(degree(pr))− 3

• overall≤ 2 ∗ 6− 3 = 9; plus 1 for the outer triangle

Analysis


Lemma: The expected number of triangles which are visited inthe search structure during the construction is O(n log n).proof in the book.

Analysis


Lemma: The expected number of triangles which are visited inthe search structure during the construction is O(n log n).

Theorem: The Delaunay triangulation of n points can becomputed in O(n log n) expected time using randomizedincremental construction.

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