Geometric Separators
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1 Geometric Separators and the Parabolic Lift Don Sheehy University of Connecticut! (work done at INRIA Saclay, France)
2
3 Take a Classic Problem (Geometric Separators)
4 Take a Classic Problem (Geometric Separators) Apply a Classic Trick (Replace Stereographic Projection with the Parabolic Lift)
5 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)
6 Sphere Separators
7 Sphere Separators
8 Sphere Separators
9 Sphere Separators Sparse: Intersect only a sublinear number of disks.
10 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.
11 Sphere Separators Sparse: Intersect only a sublinear number of disks. O(n 1 1 d ) Balanced: Contains the centers of at least and at most a constant fraction of the disks.
12 Sphere Separators Sparse: Intersect only a sublinear number of disks. O(n 1 1 d ) Balanced: Contains the centers of at least and at most a constant fraction of the disks. n d +2
13 Finding Sparse Sphere Separators is Easy
14 Finding Sparse Sphere Separators is Easy
15 Finding Sparse Sphere Separators is Easy
16 Finding Sparse Sphere Separators is Easy Sample a great circle of a d-sphere in d+1 dimensions uniformly.
17 Finding Sparse Sphere Separators is Easy Sample a great circle of a d-sphere in d+1 dimensions uniformly. Output the stereographic projection.
18 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.
19 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.
20 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]
21 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]
22 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] How to make it balanced?
23 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points.
24 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c
25 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c
26 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c
27 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c
28 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c
29 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c
30 Balance from Centerpoints Any halfspace containing a centerpoint c contains at least n/d+1 points. c It suffices to find a stereographic map that has centerpoint at the center.
31 Algorithms
32 Miller-Thurston 90 (Miller-Thurston-Teng-Vavassis 97) Algorithms
33 Miller-Thurston 90 (Miller-Thurston-Teng-Vavassis 97) Let be the stereographic map from R d to the unit d-sphere centered at the origin in R d+1. First, compute c 2 Centerpoint( (P )). Find an orthogonal transformation Q such that Q(c) = 0 for some 2 R. q 1 Let D = 1+ I,whereI is the identity on Rd. Choose a random unit vector v 2 R d+1 and let S 0 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 (S 0 ))). Algorithms
34 Algorithms Miller-Thurston 90 (Miller-Thurston-Teng-Vavassis 97) This Paper Let be the stereographic map from R d to the unit d-sphere centered at the origin in R d+1. First, compute c 2 Centerpoint( (P )). Find an orthogonal transformation Q such that Q(c) = 0 for some 2 R. q 1 Let D = 1+ I,whereI is the identity on Rd. Choose a random unit vector v 2 R d+1 and let S 0 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 (S 0 ))).
35 Algorithms Miller-Thurston 90 (Miller-Thurston-Teng-Vavassis 97) Let be the stereographic map from R d to the unit d-sphere centered at the origin in R d+1. First, compute c 2 Centerpoint( (P )). Find an orthogonal transformation Q such that Q(c) = 0 for some 2 R. q 1 Let D = 1+ I,whereI is the identity on Rd. Choose a random unit vector v 2 R d+1 and let S 0 be the d-sphere formed by intersecting the hyperplane {p v > p =0} with the unit d-sphere centered at 0. This Paper First, compute c = c p1 pn c d+1 2 Centerpoint( kp 1 k 2,..., kp n k 2 ). Next, choose a random unit vector v = v v d+1 2 R d+1. Let p cd+1 kck r = 2. v d+1 Output the sphere S with center (c rv) and radius r. (In the improbable case that v d+1 = 0, the output is just the hyperplane {p v > (p c) =0}.) Output S = 1 (Q 1 (D 1 (S 0 ))).
36 Algorithms Miller-Thurston 90 (Miller-Thurston-Teng-Vavassis 97) Let be the stereographic map from R d to the unit d-sphere centered at the origin in R d+1. First, compute c 2 Centerpoint( (P )). Find an orthogonal transformation Q such that Q(c) = 0 for some 2 R. q 1 Let D = 1+ I,whereI is the identity on Rd. Choose a random unit vector v 2 R d+1 and let S 0 be the d-sphere formed by intersecting the hyperplane {p v > p =0} with the unit d-sphere centered at 0. This Paper First, compute c = c p1 pn c d+1 2 Centerpoint( kp 1 k 2,..., kp n k 2 ). Next, choose a random unit vector v = v v d+1 2 R d+1. Let p cd+1 kck r = 2. v d+1 Output the sphere S with center (c rv) and radius r. (In the improbable case that v d+1 = 0, the output is just the hyperplane {p v > (p c) =0}.) Output S = 1 (Q 1 (D 1 (S 0 ))).
37 Algorithms Miller-Thurston 90 (Miller-Thurston-Teng-Vavassis 97) Let be the stereographic map from R d to the unit d-sphere centered at the origin in R d+1. First, compute c 2 Centerpoint( (P )). Find an orthogonal transformation Q such that Q(c) = 0 for some 2 R. q 1 Let D = 1+ I,whereI is the identity on Rd. Choose a random unit vector v 2 R d+1 and let S 0 be the d-sphere formed by intersecting the hyperplane {p v > p =0} with the unit d-sphere centered at 0. This Paper First, compute c = c p1 pn c d+1 2 Centerpoint( kp 1 k 2,..., kp n k 2 ). Next, choose a random unit vector v = v v d+1 2 R d+1. Let p cd+1 kck r = 2. v d+1 Output the sphere S with center (c rv) and radius r. (In the improbable case that v d+1 = 0, the output is just the hyperplane {p v > (p c) =0}.) Output S = 1 (Q 1 (D 1 (S 0 ))).
38 Changing the problem (from spheres to paraboloids)
39 Changing the problem (from spheres to paraboloids) The parabola is an ellipse with one focal point at infinity.
40 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.
41 Changing the problem (from spheres to paraboloids) The parabola is an ellipse with one focal point at infinity. It suffices to find a stereographic map As one focal point goes to infinity, the stereographic projection of the intersection with a plane through the other focal point does not change. that has centerpoint at the center.
42 Changing the problem (from spheres to paraboloids) The parabola is an ellipse with one focal point at infinity. parabolic lift It suffices to find a stereographic map As one focal point goes to infinity, the stereographic projection focal of the point intersection with a plane through the other focal point does not change. that has centerpoint at the center.
43 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
44 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
45 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
46 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
47 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
48 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
49 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
50 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
51 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
52 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
53 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
54 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
55 Picking a paraboloid Goal: Find a paraboloid so that the lifted points have a centerpoint at the focal point.
56
57 Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators.
58 Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators. Thank you.
59 Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators. Thank you..
60 Take away: Use the parabolic lifting rather than the stereographic map to compute geometric separators. Thank you...
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