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 planar 3-connected graph straight-line planar embedding α i ∑ 2 − ∀ i, 0< α i < π min x x i j ∈ ( , ) i j E

Non-wheel vertex Bad Cases Non-convex face

Good and Bad Embeddings wheel vertex non-wheel vertex double-wheel vertex � � � non-convex face convex face double-convex face � � �

Tutte’s Theorem (1963) If G =< V , E > is a 3-connected planar graph > ∈ ⎧ 0 ( i , j ) E Wx = ∑ ⎪ = = x w 1 i 1 ,.., n = ⎨ ij and w ij Wy = ∈ j N ( i ) y ⎪ ⎩ 0 otherwise and the “boundary” of G is constrained to a convex polygon, < > Then is a straight-line planar V , x , y , E embedding – all faces are convex and all vertices are wheels .

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

Application - Texture Mapping boundary

bad good

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 of a non-zero real value ∆ z uv to each half edge ( u , v ) of the mesh G=< V , E , F > such that ∆ z uv = - ∆ z vu . ♦ -5.9 5 3.1 5.9 5.9 3.1 5 ≡ - 5 -3.1 - 2 2 2

Indices: Topological Sign Changes (needs faces for edge order) ind(v) = (2-sc(v))/2 non-singular source saddle sc = 2 sc = 0 sc > 2 index = 0 index = 1 index < 0 ind(f) = (2-sc(f))/2 vortex non-singular saddle sc = 0 sc = 2 sc > 2 index = 1 index = 0 index < 0

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 ∑ ∑ + = − ind v ( ) ind f ( ) 2 2 g ∈ ∈ V F v f Proof: Essentially by counting corners and applying Euler’s formula: V + F - E =2-2 g . ♦ Corollary: g = 0 → must have at least two sources/sinks/vortices. g ≥ 2 → must have at least one saddle.

Index Theorem • Natural discretization of 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 Z=2Y-X (5,1) (0,0) 3 0 (3,-1) 2 (1,-2) 9 5 3 -3 -5 7 (4,-4) 9 -12

Properties of One-form: Faces (incl. outer) Closed: sum must be zero → Cannot be vortex → Index ≤ 0 Z=2Y-X (5,1) (0,0) 3 0 (3,-1) 2 (1,-2) 9 5 3 -3 -5 (4,-4) 7 9 -12

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 Z=2Y-X (5,1) (0,0) 3 0 (3,-1) 2 (1,-2) 9 5 3 -3 -5 (4,-4) 7 9 -12

Properties of One-form: Boundary Vertices 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

Now Let’s Count Indices … • All faces ≤ 0 0, neg • All vertices ≤ 0 except for 2 + 0, neg + 2 • Planar graph (incl. outer face) is topological sphere – 2-2g=2 = 2 • Index Theorem: sum of indices must be 2 → No negative indices are possible • → No saddles are possible •

So Far … In a one-form obtained as any projection of a Tutte drawing, no faces or interior vertices are saddles.

Properties of Tutte Drawing • Suppose that there was a flip at a vertex non-wheel vertex Y saddle • Could pick a projection to produce one-form with a saddle Contradiction !!

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

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 3D • Convex boundary creates significant distortion • “Free” boundary is better

Multiple Boundaries (non convex boundary) better

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 2 π . 3. The boundaries of the finite exterior faces are mapped to the plane with positive edge lengths and turning number -2 π . 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 : M ore 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 2 g . • 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 (5,1) y x (0,0) 5 1 2 2 1 5 (3,-1) 3 1 (1,-2) 1 2 1 3 3 2 (4,-4)

when vertex repeats Stop integration g = 1

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