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 Embedding Theorem Reproven and Extended
Craig Gotsman
Center for Graphics and Geometric ComputingTechnion – 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 Ex 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 planarembedding – all faces are convex and all vertices are wheels.
niwiNj
ij ,..,11)(
==∑∈
⎪⎩
⎪⎨
⎧ ∈>=
otherwise
Ejiwij
),(
0
0xWx =
>< EyxV ,,,
yWy =
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 of 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
5.95
2
3.15.9
≡
Indices: Topological Sign Changes(needs faces for edge order)
ind(v) = (2-sc(v))/2
non-singularsc = 2
index = 0
saddlesc > 2
index < 0
sourcesc = 0
index = 1
non-singularsc = 2
index = 0
saddlesc > 2
index < 0
vortexsc = 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 2v f
ind v ind f g∈ ∈
+ = −∑ ∑V F
Index Theorem
• Natural discretizationof the “Poincare-Hopfindex 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
53
2
97
9
(0,0)
(3,-1)
(5,1)
(4,-4)
(1,-2)
Z=2Y-X
0
-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
0
-3-5
-12
3
53
2
97
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
0
-3-5
-12
3
53
2
97
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 of 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 vertexY saddle
Possible Neighborhoods
non-convex faceX and Y saddles
wheel vertexno saddles
non-wheel vertexY saddle
double-wheel vertexX and Y saddles
convex faceno saddles
non-convex faceY 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 BoundariesLemma: If1. 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 diskUse 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
12
2
2 3
5
y5
312
31
1
x (0,0)
(3,-1)
(5,1)
(4,-4)
(1,-2)
g = 1Stop 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