1. Spectral partitioning works: Planar graphs and finite element meshes
Daniel A. Spielman, Shang-Hua Teng
Presented By
Yariv Yaari

2. Paper Result
Spectral partitioning on a planar graph of bounded degree will find a cut of ratio
Similar results for k-nearest neighbor graphs in a fixed dimension

3. Outline Introduction Spectral Partitioning
Bound on Fiedler value using Embedding
Bound for Planar graphs
Bisection from low-ratio cut
Questions?

4. Introduction
Our goal is to find good partitioning of graphs. (also called cut)
Partition of G=(V,E) is , we define
Good partition, large and small
The cut ratio is

5. The Laplacian
The Laplacian of a graph G, L(G) is an nxn matrix with entries defined by E.
For G=(V,E), , and
We are interested in eigenvalues, eigenvectors of the Laplacian.

6. Spectral Partitioning
Partition the graph using an eigenvector of the Laplacian.
Choose a number s and split v into: and
There are several approaches to choose s, we will only consider the choice optimizing the cut’s ratio, and only for an eigenvector of a specific eigenvalue, Fiedler value.

7. The Laplacian - properties
For any vector x,
Therefore, L(G) is symmetric positive semidefinite matrix, all eigenvalues are non-negative reals.
0 is always an eigenvalue. If G is connected, its eigenvectors are spanned by (1,1,1,…,1).
The second smallest eigenvalue is called Fiedler value.

8. Fielder Value
Since L(G) is symmetric, the eigenvectors are orthogonal and we get
The minimized quotient is called Rayleigh Quotient, and it can be used to find a low ratio cut.

9. Rayleigh quotient
(Mihail): for a graph G of maximum degree d, for any vector x s.t There is s such that the cut have ratio at most
Therefore, we want to bind Fiedler value.

10. Embedding We will find a bound using an embedding into Rm.
We use: where
This is a direct result of the one-dimensional case.

11. “Kissing Disk” embedding
(Koebe–Andreev–Thurston): For any planar graph G=(V,E) there are disks with pair wise disjoint interiors s.t.

12. Sphere Preserving maps
A sphere preserving map is a map f s.t. the image of any sphere under f is a sphere and similarly the pre-image.
We will use sphere preserving maps between a sphere and a hyperplane.
For our purposes, a hyperplane is a sphere, so a möbius transformation is sphere preserving.

13. Bound for planar graphs
We will soon prove the existence of a sphere preserving map from the plane to the sphere s.t. the centroid of the centers of the disks is the origin.
Then denote the centers, the radii, we then get and for all i,j
Therefore we can split it between i and j and get the bound But the caps are disjoint, so

14. Bound for planar graphs - cont.
Also,
So, summing everything we get
Therefore, and using its eigenvector one can find a cut of ratio

15. Sphere preserving maps - cont
We want a sphere preserving map that will map centers of spheres to a set on a sphere with the centroid at the origin.
First, build a family of sphere preserving maps.
For a sphere and a point denote the stereographic projection of the sphere on the extended hyperplane tangent to the sphere at (extended, with infinity point). This is sphere preserving.

16. Sphere preserving maps - cont
Any möbius transformation is sphere preserving, we will only use dilation, denote a dilation with factor around by
Any composition of Sphere preserving maps is sphere preserving
Our family of sphere preserving functions will be

17. Sphere preserving maps - cont
We now extend the definition for as (this is not continuous)
Now, for a cap C on the unit sphere denote its center by we would like to show that for all there is s.t.

18. Sphere preserving maps - cont
We will need that there is no point shared by the interior of at least half of the caps (in current case, the interiors are disjoint).
We would like to use a mapping from to the centroid of however, this map is not continuous.
Choose s.t. for all , most of the caps are contained within a ball of radius around

19. Sphere preserving maps - cont
Now define a weight function:
And so

20. Sphere preserving maps - cont
is continuous and approach zero where is not continuous, so is continuous.
Now, if the centers are in or and most of them in . Therefore lies on line between the origin and
This implies (by Brewer’s fixed point), and we’re done.

21. Questions?