Connections between Theta-Graphs, TD-Delaunay Triangulations, and

Orthogonal Surfaces

WG 2010

Nicolas Bonichon, Cyril GavoilleNicolas Hanusse, David Ilcinkas

LaBRIUniversity of Bordeaux

France

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Let G be a weighted graph, andlet H be a spanning subgraph of G.

H is an s-spanner of G if, for all u,v

dH(u,v) ≤ s dG(u,v)

s is the stretch of H

Ex: dG(b,d)=5, dH(b,d)=7dG(b,e)=4, dH(b,e)=8

H is a 2-spanner of G.

Geometric Spanners

In this talk (E,d) is the Euclidean plane

www.2m40.com

Accidents:

- 26 in 2009

- 10 in 2010

- Last one: June 22nd

Let (E,d) be a metric space.Let S be a set of points of E.G(S) is the complete graph.The length of (u,v) is d(u,v).

Goals:– Small stretch s– Few edges– Small max degree– Routable– Planar– …

Delaunay Triangulation

Voronoï cell:

Delaunay triangulation:

si is a neighbor of sj iff

[Dobkin et al. 90] Delaunay T. is a plane 5.08-spanner[Keil & Gutwin 92] Delaunay T. is a plane 2.42-spanner

Stretch > 1.414 for any plane spanner [Chew 89]Stretch > 1.416 for Delaunay triangulations [Mulzner 04]

Triangular Distance Delaunay Triangulation

Triangular “distance”:

TD(u,v) = size of the smallest equilateral triangle centred at u touching v.

[ TD(u,v) ≠ TD(v,u) in general ]

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[Chew 89] TD-Delaunay is a plane 2-spanner

θk-graph [Clarkson 87][Keil 88]

Vertex set of θk-Graph is S

Space around each vertex of S is split into k cones of angle θk = 2/k.

Edge set of θk-Graph: for each vertex u and each cone C, add an edge toward vertex v in C

with the projection on the bisector that is closest to u.

No bounds on the stretch are known to be tight.

k Stretch

< 9 ???

9 < 8.11

10 < 4.50

… …

14 < 2.14

15 < 1.98

Half-θk-graph

Half-θk-Graph(S):

Like a θk-Graph(S) but one preserves edges from half of the cones only.

Theorem 1: Half-θ6-Graph(S) = TD-Delaunay(S)

Corollary:

- Half-θ6-Graph(S) is a plane 2-spanner

- θ6-Graph(S) is a 2-spanner (optimal stretch)

Theorem 1: Half-θ6-Graph(S) = TD-Delaunay(S)

Corollary:

- Half-θ6-Graph(S) is a plane 2-spanner

- θ6-Graph(S) is a 2-spanner (optimal stretch)

For k=6:

Proof: contact between 2 triangles

Whenever two triangles touch, it’s a tip that touches a side.

v touches north tip of u’s triangle iff v belongs to the north cone of u.

Let v be a vertex in the north cone of u. The time when both triangles touch is y(v)-y(u).

There is an edge between u and v iff v’s triangle is the first to touch the tip of u’s triangle.

QED

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Orthogonal Surface [Miller 02] [Felsner 03] [Felsner & Zickfeld 08]

Coplanar if all points of S are in (P): x+y+z=cste

General position: no two points with same x,y, or z.

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Geodesic Embedding [Miller 02] [Felsner 03] [Felsner & Zickfeld 08]

Properties [Felsner et al.] :

1.The geodesic embedding of every orthogonal surface of coplanar point set S is a plane triangulation.

2.Every plane triangulation is the geodesic embedding of orthogonal surface of some coplanar point set S.

Theorem 2: TD-Delaunay(S) GeoEmbedding(S)

Corollary: Every plane triangulation is TD-Delaunay realizable

Theorem 2: TD-Delaunay(S) GeoEmbedding(S)

Corollary: Every plane triangulation is TD-Delaunay realizable

TD-Voronoï Coplanar Orthogonal Surface

Proof: growing 2D triangles viewed as 3D cones

TD-Delaunay Geodesic Embedding

Delaunay Realizability

• A graph G is Delaunay realizable if there exists S such that G=Delaunay(S).

• [Dillencourt & Smith 96]: some sufficient conditions, and some necessary conditions.

No characterization known.

Decision problem: in PSPACE, NP-hard?

• But, trivial for TD-Delaunay realizability:

Every plane triangulation is TD-Delaunay realizable (S constructible in O(|V(G)|) time).

Graphs that are nonDelaunay realizable

Thank You!