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
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