Tuttes Embedding Theorem Reproven and Extended Craig Gotsman - - PowerPoint PPT Presentation

Tutte’s Embedding Theorem Reproven and Extended

Craig Gotsman Center for Graphics and Geometric Computing Technion – Israel Institute of Technology

Joint with Steven Gortler and Dylan Thurston

Tutte’s Theorem

∀i, 0<αi<π αi

planar 3-connected graph

2 ( , )

min

i j i j E

x x

∈

−

∑

straight-line planar embedding

Bad Cases

Non-convex face Non-wheel vertex

Good and Bad Embeddings

double-convex face wheel vertex non-wheel vertex double-wheel vertex convex face non-convex face

Tutte’s Theorem (1963)

If G=<V,E> is a 3-connected planar graph and and the “boundary” of G is constrained to a convex polygon, Then is a straight-line planar embedding – all faces are convex and all vertices are wheels.

n i w

i N j ij

,.., 1 1

) (

= =

∑

∈

⎪ ⎩ ⎪ ⎨ ⎧ ∈ > =

• therwise

E j i wij ) , (

x Wx =

> < E y x V , , ,

y Wy =

Applications

• Planar Graph Drawing
• Texture Mapping
• Remeshing
• Surface Reconstruction
• Morphing
• Compression

Application - Texture Mapping

boundary

good bad

Remeshing

Today

• Some simple results about discrete one-forms on

manifold meshes

• Very simple proof of Tutte’s theorem

– Essentially relies only on Euler’s theorem

• Generalize to case of non-convex boundary
• Generalize to case of higher genus surfaces

One-Forms on Meshes

Definition: A non-vanishing one-form [G,∆z] is an assignment

• f a non-zero real value ∆zuv to each half edge (u,v) of the

mesh G=<V,E,F> such that ∆zuv = -∆zvu. ♦

5 5

• 2
• 2

3.1

• 3.1
• 5.9

5.9 5 2 3.1 5.9

≡

Indices: Topological Sign Changes

(needs faces for edge order)

ind(v) = (2-sc(v))/2

non-singular sc = 2 index = 0 saddle sc > 2 index < 0 source sc = 0 index = 1 non-singular sc = 2 index = 0 saddle sc > 2 index < 0 vortex sc = 0 index = 1

ind(f) = (2-sc(f))/2

Index Theorem

(after Banchoff ‘70, Lazarus and Verroust ‘99, Benjamini and Lovasz ‘02)

Theorem: If G is a closed oriented manifold mesh of genus g, then any one-form [G,∆z] satisfies Proof: Essentially by counting corners and applying Euler’s formula: V+F-E=2-2g. ♦

Corollary:g = 0 → must have at least two sources/sinks/vortices. g ≥ 2 → must have at least one saddle.

( ) ( ) 2 2

v f

ind v ind f g

∈ ∈

+ = −

∑ ∑

V F

Index Theorem

• Natural discretization
• f the “Poincare-Hopf

index theorem”

• Counts types of

singularities in vector fields on surfaces (Hairy ball theorem)

From Tutte Drawing to One-form

• Take straight line Tutte

drawing (may have crossings)

• Pick arbitrary direction: Z
• Project onto Z
• Use Z differences as one-form

3 5 3 2 9 7 9

(0,0) (3,-1) (5,1) (4,-4) (1,-2)

Z=2Y-X

• 3
• 5
• 12

Properties of One-form: Faces (incl. outer)

Closed: sum must be zero → Cannot be vortex → Index ≤ 0

(0,0) (3,-1) (5,1) (4,-4) (1,-2)

Z=2Y-X

• 3
• 5
• 12

3 5 3 2 9 7 9

Properties of One-form: Interior Vertices

In convex hull of its neighbors

→ Co-closed: (weighted) sum must be zero → Cannot be source or sink

→ Index ≤ 0

(0,0) (3,-1) (5,1) (4,-4) (1,-2)

Z=2Y-X

• 3
• 5
• 12

3 5 3 2 9 7 9

Boundary is drawn as convex polygon “upper” vertex is source “lower” vertex is sink “side” vertices non source/sink → All vertices but two have index ≤ 0

Properties of One-form: Boundary Vertices

Now Let’s Count Indices …

• All faces ≤ 0
• All vertices ≤ 0 except for 2
• Planar graph (incl. outer face) is topological

sphere

– 2-2g=2

• Index Theorem: sum of indices must be 2
• → No negative indices are possible
• → No saddles are possible

0, neg + 0, neg + 2 = 2

So Far …

In a one-form obtained as any projection

• f a Tutte drawing,

no faces or interior vertices are saddles.

Properties of Tutte Drawing

• Suppose that there was a flip at a vertex
• Could pick a projection to produce one-form

with a saddle Contradiction !!

non-wheel vertex Y saddle

Possible Neighborhoods

non-convex face X and Y saddles wheel vertex no saddles non-wheel vertex Y saddle double-wheel vertex X and Y saddles convex face no saddles non-convex face Y saddle

Summarizing

• Each face is convex
• Each vertex is a wheel
• Locally an embedding

Local to Global

• Lemma: If each neighborhood is locally an

embedding, and the boundary is simple then the drawing is globally an embedding

Local to Global

• Lemma: If each neighborhood is locally an

embedding, and the boundary is simple then the drawing is globally an embedding

• QED

Non-Convex Boundary

Non-Convex Boundary

• Convex boundary creates

significant distortion

• “Free” boundary is better

3D

Multiple Boundaries

better (non convex boundary)

Main Result

• If the drawing is not an embedding, then you can

detect this at the boundary

• If the method forces the boundary to behave

properly, then the drawing will be an embedding

Bad case: Reflex boundary vertex not in the convex hull of its four neighbors.

Multiple Non-Convex Boundaries

Lemma: If 1. G is an oriented 3-connected mesh of genus 0 having multiple exterior faces. 2. The boundary of the unbounded exterior face is mapped to the plane with positive edge lengths and turning number 2π. 3. The boundaries of the finite exterior faces are mapped to the plane with positive edge lengths and turning number -2π. 4. [G,x,y] is the straight line drawing of G where each interior vertex is positioned as a convex combination of its neighbors. 5. In [G,x,y] each reflex boundary vertex is in the convex hull of its neighbors.

Then for any projection [G,∆z], no vertex or interior face is a

saddle.

Proof: More counting

Like Before

• Each face is convex
• Each vertex is a wheel
• Locally an embedding
• In addition: If the boundary is simple, then

globally an embedding

Theorem difficult to use because cannot tell apriori which vertices should be reflex

Genus 1

Harmonic One-form on Mesh (F,E,V)

• Each face is closed
• Each vertex is co-closed (wrt fixed weights)

Theorem (Mercat ’01): Harmonic one-forms w.r.t. given positive weights on a mesh of a closed surface with genus g form a linear space of dimension 2g.

• Can be determined by computing the nullspace of

a matrix

g = 0

No nullspace → No non-trivial harmonic one-forms So all we can do is …. what we did:

Remove one face to obtain disk Use Tutte embedding

g = 1

• Harmonic → 2D nullspace, all indices ≤ 0
• Index Theorem → sum of indices = 0
• → All indices = 0
• → No saddles

Parameterization for g = 1 (Gu &Yau 03)

• Pick two linearly independent harmonic one-

forms – for x,y coordinates

• Pick one starting vertex, map to origin
• Integrate for x,y, coordinates

– Closed faces: path independent

1 1 2 2 2 3 5 y 5 3 1 2 3 1 1 x

(0,0) (3,-1) (5,1) (4,-4) (1,-2)

g = 1

Stop integration when vertex repeats

g = 1

Theorem: Embedding is always valid Proof: No saddles in either one form → All vertices must be wheels and all faces must be convex in the drawing (otherwise the projection would contain saddles)

g > 1: More Complicated

• Sum of indices = 2-2g < 0
• → Must be saddles in one-forms
• → Must be “badness” in drawing
• Usually “double covers”.

Summary

• Discrete one-forms are useful mathematical tool
• Easy proof of Tutte’s theorem
• Extension to non-convex boundary
• Extension to higher genus

Thank You