Between Two Shapes, Using the Hausdorff Distance Marc van Kreveld, - PowerPoint PPT Presentation

Between Two Shapes, Using the Hausdorff Distance Marc van Kreveld, Till Miltzow, Tim Ophelders Willem Sonke, Jordi Vermeulen

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Directed Hausdorff distance A → B .

Directed Hausdorff distance A → B .

Directed Hausdorff distance A → B .

Directed Hausdorff distance B → A .

Directed Hausdorff distance B → A .

Directed Hausdorff distance B → A .

Undirected Hausdorff distance A ↔ B .

Undirected Hausdorff distance A ↔ B . d H = 1

Directed Hausdorff distance B → A .

Directed Hausdorff distance A → B .

S B A

Find S with minimal Hausdorff distance to A and B . S B A

Find S with minimal Hausdorff distance to A and B . S B A Result: distance 1 / 2 is always possible.

maximal

B A

B A d H = 1

B A A ⊕ D 1 / 2

B A A ⊕ D 1 / 2 B ⊕ D 1 / 2

S B A

B A

S 1 / 8 B A A ⊕ D 1 / 8 B ⊕ D 7 / 8

S 1 / 4 B A A ⊕ D 1 / 4 B ⊕ D 3 / 4

S 1 / 2 B A A ⊕ D 1 / 2 B ⊕ D 1 / 2

S 3 / 4 B A A ⊕ D 3 / 4 B ⊕ D 1 / 4

S 7 / 8 B A B ⊕ D 7 / 8 B ⊕ D 1 / 8

S α B A

Claim: - d H ( A, S ) = α - d H ( B, S ) = 1 − α S α B A

Claim: - d H ( A, S ) = α - d H ( B, S ) = 1 − α S α B A To show: - S ⊆ A ⊕ D α - S ⊆ B ⊕ D 1 − α - A ⊆ S ⊕ D α - B ⊆ S ⊕ D 1 − α

Claim: - d H ( A, S ) = α - d H ( B, S ) = 1 − α S α B A To show: - S ⊆ A ⊕ D α - S ⊆ B ⊕ D 1 − α - A ⊆ S ⊕ D α S ⊕ D α - B ⊆ S ⊕ D 1 − α

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α .

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α �

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α � - A ⊆ S α ⊕ D α

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α � S α - A ⊆ S α ⊕ D α B A

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α � S α - A ⊆ S α ⊕ D α a B A

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α � S α - A ⊆ S α ⊕ D α ≤ 1 a B A b

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α � S α - A ⊆ S α ⊕ D α a ≤ α s B A b ≤ 1 − α

Lemma: S α = ( A ⊕ D α ) ∩ ( B ⊕ D 1 − α ) has d H ( A, S α ) = α and d H ( B, S α ) = 1 − α . - S α ⊆ A ⊕ D α � S α - A ⊆ S α ⊕ D α � a ≤ α s B A b ≤ 1 − α

convex convex

convex convex O ( n + m ) complexity convex

non-convex convex

non-convex convex connected O ( n + m ) complexity

non-convex non-convex

possibly disconnected non-convex O ( nm ) complexity non-convex

maximal S

minimal S

B A

S 1 / 8 B A A ⊕ D 1 / 8 B ⊕ D 7 / 8

S 1 / 4 B A A ⊕ D 1 / 4 B ⊕ D 3 / 4

S 1 / 2 B A A ⊕ D 1 / 2 B ⊕ D 1 / 2

S 3 / 4 B A A ⊕ D 3 / 4 B ⊕ D 1 / 4

S 7 / 8 B A B ⊕ D 7 / 8 B ⊕ D 1 / 8

Conclusion:

Conclusion: - We can compute a shape with d H ( A, S ) = α and d H ( B, S ) = 1 − α

Conclusion: - We can compute a shape with d H ( A, S ) = α and d H ( B, S ) = 1 − α - The shape has linear or quadratic complexity, depending on the convexity of A and B

Conclusion: - We can compute a shape with d H ( A, S ) = α and d H ( B, S ) = 1 − α - The shape has linear or quadratic complexity, depending on the convexity of A and B - The morph obtained by varying α is 1-Lipschitz continuous

Conclusion: - We can compute a shape with d H ( A, S ) = α and d H ( B, S ) = 1 − α - The shape has linear or quadratic complexity, depending on the convexity of A and B - The morph obtained by varying α is 1-Lipschitz continuous Future work: - Extend to more than two input sets

Conclusion: - We can compute a shape with d H ( A, S ) = α and d H ( B, S ) = 1 − α - The shape has linear or quadratic complexity, depending on the convexity of A and B - The morph obtained by varying α is 1-Lipschitz continuous Future work: - Extend to more than two input sets - Minimum required α may be 1

Conclusion: - We can compute a shape with d H ( A, S ) = α and d H ( B, S ) = 1 − α - The shape has linear or quadratic complexity, depending on the convexity of A and B - The morph obtained by varying α is 1-Lipschitz continuous Future work: - Extend to more than two input sets - Minimum required α may be 1 - Not yet clear how to do morphing between three shapes

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Input sets { A 1 , . . . , A k } A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α A 1 A 3 A 2

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α Worst case: smallest α = 1

Input sets { A 1 , . . . , A k } Let S α = � A i ⊕ D α i Find smallest α s.t. A i ⊆ S α ⊕ D α Worst case: smallest α = 1