Efficiently navigating a random Delaunay triangulation Nicolas - - PowerPoint PPT Presentation

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Efficiently navigating a random Delaunay triangulation Nicolas - - PowerPoint PPT Presentation

Apr 07, 2024 148 likes •796 views

Efficiently navigating a random Delaunay triangulation Nicolas Broutin, Olivier Devillers, Ross Hemsley The Delaunay Triangulation Take some points (Assume no 4 points on circle, or 3 points on line) The Delaunay Triangulation Add all

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Efficiently navigating a random Delaunay triangulation

Nicolas Broutin, Olivier Devillers, Ross Hemsley

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The Delaunay Triangulation

Take some points

(Assume no 4 points on circle, or 3 points on line)

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The Delaunay Triangulation

Add all triangles having empty circumcircles

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The Delaunay Triangulation

Result: Delaunay Triangulation

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The Delaunay Triangulation

Useful for Interpolation Surface Reconstruction Routing ... Easily extends to higher dimensions

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The Problem: Point Location

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Input: Delaunay Triangulation

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Input: Delaunay Triangulation, Point

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Output: Triangle containing point

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A Possible Solution

(1) Choose arbitrary start vertex

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A Possible Solution

(2) Move to neighbour that is closest to destination

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A Possible Solution

(2) Move to neighbour that is closest to destination...

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A Possible Solution

(2) Move to neighbour that is closest to destination...

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A Possible Solution

(2) Move to neighbour that is closest to destination...

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A Possible Solution

(3) Terminate when we cannot advance

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  • No (non-trivial) analysis
  • Very effective in practice

Idea: Analyse for n random points in unit disc?

Vertex Walk Call this:

Conjecture: Θ(|zq|√n ) (expectation for random z, q)

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Why is it tricky to analyse?

Next step depends non-trivially on previous steps.

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A similar algorithm can give the properties we want

Our Contribution:

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Introducing Cone Walk

“Search Cone” “Search Disc”

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Zi q Ri Far from destination ⇒ cones do not intersect ⇒ progress for each step is iid r.v

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Zi q Ri How does this relate to Delaunay... ?

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Lemma Always a path from Zi to Zi+1 in DT ∩ search disc.

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Lemma Always a path from Zi to Zi+1 in DT ∩ search disc. Proof By ‘growing circles’

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Lemma Always a path from Zi to Zi+1 in DT ∩ search disc. Proof By ‘growing circles’

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Lemma Always a path from Zi to Zi+1 in DT ∩ search disc. Proof By ‘growing circles’

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Lemma Always a path from Zi to Zi+1 in DT ∩ search disc. Proof By ‘growing circles’

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q Cone Walk on the Delaunay Triangulation

Algorithm gives a path, like vertex walk

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Vertex Walk Cone Walk

Competitve Worst case complexity Memoryless Worst Complexity ∀z, q No Yes (< 3.7) Yes No⋆ O(n) Op(|zq|√n log log n + 5

  • |zq| log6 n)

O(n) Op(√n log log n)

All bounds are best known bounds For n uniform points in a unit disc

[see paper] Vertices on Path ∀z, q O(n) Op(|zq|√n + 5

  • |zq| log6 n)
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Vertex Walk Cone Walk

Competitve Worst case complexity Memoryless Worst Complexity ∀z, q No Yes (< 3.7) Yes No⋆ O(n) Op(|zq|√n log log n + 5

  • |zq| log6 n)

O(n) Op(√n log log n)

All bounds are best known bounds For n uniform points in a unit disc

[see paper] Vertices on Path ∀z, q O(n) Op(|zq|√n + 5

  • |zq| log6 n)
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Bounding Vertices Accessed

  • Know behaviour within discs
  • Pointset has no border
  • Planar Poisson process (intensity 1)

Assume: [random points with average density 1 on R2]

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Bounding Vertices Accessed q z Consider some walk

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Bounding Vertices Accessed q z Consider some walk It accesses all points in discs

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Bounding Vertices Accessed q z Consider some walk It accesses all points in discs + 1-neighbourhood

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Bounding Vertices Accessed q z Consider some walk It accesses all points in discs + 1-neighbourhood Outputs path

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Bounding Vertices Accessed q z

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Bounding Vertices Accessed q z

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Bounding Vertices Accessed

Accessed ≤ Blue Crossing Edges +

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Bounding Vertices Accessed

How many crossing edges?

x

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Bounding Vertices Accessed

How many crossing edges? Wheel [Devroye et al.]

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Bounding Vertices Accessed

How many crossing edges? in 1-neighbourhood ⇒ ≥ 1 ×

x

empty

(because of Delaunay property)

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

E[#border points] =

  • R2\W

P(x a border point) λ(dx)

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

E[#border points] =

  • R2\W

P(x a border point) λ(dx) Slivnyak-Mecke Formula

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

E[#border points] =

  • R2\W

24 exp

  • − π

32|xW|2 λ(dx)

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

E[#border points] =

  • R2\W

24 exp

  • − π

32|xW|2 λ(dx) ≤

∞

  • i=0
  • Ui

24 exp

  • − π

32i2 λ(dx) U1 U2

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

E[#border points] =

  • R2\W

24 exp

  • − π

32|xW|2 λ(dx) ≤

∞

  • i=0
  • Ui

24 exp

  • − π

32i2 λ(dx) U1 U2 ≤

∞

  • i=0

λ(Ui) 24 exp

  • − π

32i2

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Bounding Vertices Accessed

How many crossing edges?

P(x a border point) = 24 exp

  • − π

32|xW|2

[we assume input is R2 for now]

|xW|

E[#border points] =

  • R2\W

24 exp

  • − π

32|xW|2 λ(dx) ≤

∞

  • i=0
  • Ui

24 exp

  • − π

32i2 λ(dx) U1 U2 ≤

∞

  • i=0

λ(Ui) 24 exp

  • − π

32i2 ≤ 24

∞

  • i=0

κ−1

  • j=0

π(2(Rj + i) + 1) exp

  • − π

32i2

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Bounding Vertices Accessed

How many crossing edges?

E[#border points] ≤ C · max

  • |zq|, log6 n
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Bounding Vertices Accessed

How many crossing edges?

E[#border points] ≤ C · max

  • |zq|, log6 n
  • z

q

Recall

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Bounding Vertices Accessed

How many crossing edges?

E[#border points] ≤ C · max

  • |zq|, log6 n
  • z

q

Recall

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Bounding Vertices Accessed

How many crossing edges?

E[#border points] ≤ C · max

  • |zq|, log6 n
  • z

q

Recall Euler Relation: crossing 3 ×

  • border + blue

# ≤

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Bounding Vertices Accessed From expectation to concentration

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Bounding Vertices Accessed

From expectation to concentration Recall: # border =

  • x∈X

1{x a border point}

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Bounding Vertices Accessed

From expectation to concentration Recall: # border =

  • x∈X

1{x a border point} Sum of dependent r.v’s

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Bounding Vertices Accessed

From expectation to concentration Recall: # border =

  • x∈X

1{x a border point} Sum of dependent r.v’s

P(#border points ≥ E[#border points] + t) ≤ E

  • exp
  • −

2t2 χ · |X ∩ W ∂|

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Bounding Vertices Accessed

From expectation to concentration Recall: # border =

  • x∈X

1{x a border point} Sum of dependent r.v’s Janson

P(#border points ≥ E[#border points] + t) ≤ E

  • exp
  • −

2t2 χ · |X ∩ W ∂|

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Bounding Vertices Accessed

From expectation to concentration Recall: # border =

  • x∈X

1{x a border point} Sum of dependent r.v’s Janson

P(#border points ≥ E[#border points] + t) ≤ E

  • exp
  • −

2t2 χ · |X ∩ W ∂|

  • max. # overlapping ‘wheels’
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Bounding Vertices Accessed

From expectation to concentration Recall: # border =

  • x∈X

1{x a border point} Sum of dependent r.v’s Janson

P(#border points ≥ E[#border points] + t) ≤ E

  • exp
  • −

2t2 χ · |X ∩ W ∂|

  • max. # overlapping ‘wheels’

Special region with ‘all’ border points

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Bounding Vertices Accessed

Bounding the worst case (in brief)

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Bounding Vertices Accessed

Bounding the worst case (in brief) 1) At most n4 walks on n points

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Bounding Vertices Accessed

Bounding the worst case (in brief) 1) At most n4 walks on n points 2) Generate ‘small’ sample containing them all

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Bounding Vertices Accessed

Bounding the worst case (in brief) 1) At most n4 walks on n points 2) Generate ‘small’ sample containing them all 2) Concentration inequalities overwhelm sample size

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Thanks

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D 1 p q |pq| Contributes O(|pq|√n )