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

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

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

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

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.

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.

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.

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.

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.

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

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

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.

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

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

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.

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

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.

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

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

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.

Questions?