Geometric Separators and the

Parabolic Lift

Don Sheehy University of Connecticut

!(work done at INRIA Saclay, France)

Take a Classic Problem (Geometric Separators)

Take a Classic Problem (Geometric Separators)

Apply a Classic Trick (Replace Stereographic Projection with the Parabolic Lift)

Take a Classic Problem (Geometric Separators)

Apply a Classic Trick (Replace Stereographic Projection with the Parabolic Lift)

Simplify a useful algorithm (The Miller-Thurston Geometric Separator Algorithm)

Sphere Separators

Sphere Separators

Sphere Separators

Sphere Separators

Sparse: Intersect only a sublinear number of disks.

Sphere Separators

Sparse: Intersect only a sublinear number of disks.

Balanced: Contains the centers of at least and at most a constant fraction of the disks.

Sphere Separators

Sparse: Intersect only a sublinear number of disks.

Balanced: Contains the centers of at least and at most a constant fraction of the disks.

O(n1� 1d )

Sphere Separators

Sparse: Intersect only a sublinear number of disks.

Balanced: Contains the centers of at least and at most a constant fraction of the disks.

O(n1� 1d )

n

d+ 2

Finding Sparse Sphere Separators is Easy

Finding Sparse Sphere Separators is Easy

Finding Sparse Sphere Separators is Easy

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.Output the stereographic projection.

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.Output the stereographic projection.The output is sparse with constant probability.

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.Output the stereographic projection.The output is sparse with constant probability.Sample until you get a good one.

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.Output the stereographic projection.The output is sparse with constant probability.Sample until you get a good one.

[Miller-Thurston ’90]

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.Output the stereographic projection.The output is sparse with constant probability.Sample until you get a good one.

[Miller-Thurston ’90][Eppstein et al ’95]

Finding Sparse Sphere Separators is Easy

Sample a great circle of a d-sphere in d+1 dimensions uniformly.Output the stereographic projection.The output is sparse with constant probability.Sample until you get a good one.

How to make it balanced?

[Miller-Thurston ’90][Eppstein et al ’95]

Balance from CenterpointsAny halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Balance from Centerpoints

It suffices to find a stereographic map that has centerpoint at the center.

c

Any halfspace containing a centerpoint c contains at least n/d+1 points.

Algorithms

AlgorithmsMiller-Thurston ’90 (Miller-Thurston-Teng-Vavassis ’97)

AlgorithmsMiller-Thurston ’90 (Miller-Thurston-Teng-Vavassis ’97)

Let ⇧ be the stereographic map from Rdto the

unit d-sphere centered at the origin in Rd+1.

First, compute

c 2 Centerpoint(⇧(P )).

Find an orthogonal transformation Q such that

Q(c) =⇥0✓

⇤for some ✓ 2 R.

Let D =

q1�✓1+✓ I, where I is the identity on Rd

.

Choose a random unit vector v 2 Rd+1and let S0

be the d-sphere formed by intersecting the hyperplane

{p | v>p = 0} with the unit d-sphere centered at 0.

Output S = ⇧

�1(Q�1

⇧(D⇧

�1(S0))).

AlgorithmsMiller-Thurston ’90 (Miller-Thurston-Teng-Vavassis ’97) This Paper

Let ⇧ be the stereographic map from Rdto the

unit d-sphere centered at the origin in Rd+1.

First, compute

c 2 Centerpoint(⇧(P )).

Find an orthogonal transformation Q such that

Q(c) =⇥0✓

⇤for some ✓ 2 R.

Let D =

q1�✓1+✓ I, where I is the identity on Rd

.

Choose a random unit vector v 2 Rd+1and let S0

be the d-sphere formed by intersecting the hyperplane

{p | v>p = 0} with the unit d-sphere centered at 0.

Output S = ⇧

�1(Q�1

⇧(D⇧

�1(S0))).

AlgorithmsMiller-Thurston ’90 (Miller-Thurston-Teng-Vavassis ’97) This Paper

Let ⇧ be the stereographic map from Rdto the

unit d-sphere centered at the origin in Rd+1.

First, compute

c 2 Centerpoint(⇧(P )).

Find an orthogonal transformation Q such that

Q(c) =⇥0✓

⇤for some ✓ 2 R.

Let D =

q1�✓1+✓ I, where I is the identity on Rd

.

Choose a random unit vector v 2 Rd+1and let S0

be the d-sphere formed by intersecting the hyperplane

{p | v>p = 0} with the unit d-sphere centered at 0.

Output S = ⇧

�1(Q�1

⇧(D⇧

�1(S0))).

First, compute

c =

⇥ ccd+1

⇤2 Centerpoint(

⇥ p1

kp1k2

⇤, . . . ,

⇥ pn

kpnk2

⇤).

Next, choose a random unit vector v =

⇥ vvd+1

⇤2 Rd+1

.

Let

r =

pcd+1 � kck2|vd+1|

.

Output the sphere S with center (c� rv) and radius r.(In the improbable case that vd+1 = 0, the output

is just the hyperplane {p | v>(p� c) = 0}.)

AlgorithmsMiller-Thurston ’90 (Miller-Thurston-Teng-Vavassis ’97) This Paper

Let ⇧ be the stereographic map from Rdto the

unit d-sphere centered at the origin in Rd+1.

First, compute

c 2 Centerpoint(⇧(P )).

Find an orthogonal transformation Q such that

Q(c) =⇥0✓

⇤for some ✓ 2 R.

Let D =

q1�✓1+✓ I, where I is the identity on Rd

.

Choose a random unit vector v 2 Rd+1and let S0

be the d-sphere formed by intersecting the hyperplane

{p | v>p = 0} with the unit d-sphere centered at 0.

Output S = ⇧

�1(Q�1

⇧(D⇧

�1(S0))).

First, compute

c =

⇥ ccd+1

⇤2 Centerpoint(

⇥ p1

kp1k2

⇤, . . . ,

⇥ pn

kpnk2

⇤).

Next, choose a random unit vector v =

⇥ vvd+1

⇤2 Rd+1

.

Let

r =

pcd+1 � kck2|vd+1|

.

Output the sphere S with center (c� rv) and radius r.(In the improbable case that vd+1 = 0, the output

is just the hyperplane {p | v>(p� c) = 0}.)

AlgorithmsMiller-Thurston ’90 (Miller-Thurston-Teng-Vavassis ’97) This Paper

Let ⇧ be the stereographic map from Rdto the

unit d-sphere centered at the origin in Rd+1.

First, compute

c 2 Centerpoint(⇧(P )).

Find an orthogonal transformation Q such that

Q(c) =⇥0✓

⇤for some ✓ 2 R.

Let D =

q1�✓1+✓ I, where I is the identity on Rd

.

Choose a random unit vector v 2 Rd+1and let S0

be the d-sphere formed by intersecting the hyperplane

{p | v>p = 0} with the unit d-sphere centered at 0.

Output S = ⇧

�1(Q�1

⇧(D⇧

�1(S0))).

First, compute

c =

⇥ ccd+1

⇤2 Centerpoint(

⇥ p1

kp1k2

⇤, . . . ,

⇥ pn

kpnk2

⇤).

Next, choose a random unit vector v =

⇥ vvd+1

⇤2 Rd+1

.

Let

r =

pcd+1 � kck2|vd+1|

.

Output the sphere S with center (c� rv) and radius r.(In the improbable case that vd+1 = 0, the output

is just the hyperplane {p | v>(p� c) = 0}.)

Changing the problem (from spheres to paraboloids)

Changing the problem (from spheres to paraboloids)

The parabola is an ellipse with one focal point at infinity.

Changing the problem (from spheres to paraboloids)

The parabola is an ellipse with one focal point at infinity. As one focal point goes to

infinity, the stereographic projection of the intersection with a plane through the other focal point does not change.

Changing the problem (from spheres to paraboloids)

The parabola is an ellipse with one focal point at infinity. As one focal point goes to

infinity, the stereographic projection of the intersection with a plane through the other focal point does not change.

It suffices to find a stereographic map that has centerpoint at the center.

Changing the problem (from spheres to paraboloids)

The parabola is an ellipse with one focal point at infinity. As one focal point goes to

infinity, the stereographic projection of the intersection with a plane through the other focal point does not change.

It suffices to find a stereographic map that has centerpoint at the center.

parabolic lift

focal point

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Picking a paraboloid

Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.

Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators.

Thank you.

Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators.

Thank you.

Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators.

.

Thank you.

Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators.

.

.