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Between Two Shapes, Using the Hausdorff Distance Marc van Kreveld, - - PowerPoint PPT Presentation
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 ? Directed Hausdorff distance A B . Directed Hausdorff distance A B . Directed Hausdorff distance A B .
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?
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Directed Hausdorff distance A → B.
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Directed Hausdorff distance A → B.
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Directed Hausdorff distance A → B.
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Directed Hausdorff distance B → A.
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Directed Hausdorff distance B → A.
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Directed Hausdorff distance B → A.
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Undirected Hausdorff distance A ↔ B.
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dH = 1 Undirected Hausdorff distance A ↔ B.
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Directed Hausdorff distance B → A.
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Directed Hausdorff distance A → B.
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A B S
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A B S Find S with minimal Hausdorff distance to A and B.
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A B S Find S with minimal Hausdorff distance to A and B. Result: distance 1/2 is always possible.
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maximal
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B A
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dH = 1 B A
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B A A ⊕ D1/2
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B B ⊕ D1/2 A A ⊕ D1/2
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B A S
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B A
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A ⊕ D1/8 B ⊕ D7/8 B A S1/8
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A ⊕ D1/4 B ⊕ D3/4 S1/4 B A
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A ⊕ D1/2 B ⊕ D1/2 S1/2 B A
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A ⊕ D3/4 B ⊕ D1/4 B A S3/4
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B ⊕ D7/8 B ⊕ D1/8 B A S7/8
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B A Sα
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B A Claim:
- dH(A, S) = α
- dH(B, S) = 1 − α
Sα
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B A Claim:
- dH(A, S) = α
- dH(B, S) = 1 − α
Sα To show:
- S ⊆ A ⊕ Dα
- S ⊆ B ⊕ D1−α
- A ⊆ S ⊕ Dα
- B ⊆ S ⊕ D1−α
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B A Claim:
- dH(A, S) = α
- dH(B, S) = 1 − α
Sα To show:
- S ⊆ A ⊕ Dα
- S ⊆ B ⊕ D1−α
- A ⊆ S ⊕ Dα
- B ⊆ S ⊕ D1−α
S ⊕ Dα
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
- A ⊆ Sα ⊕ Dα
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
- A ⊆ Sα ⊕ Dα
B A Sα
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
- A ⊆ Sα ⊕ Dα
B A Sα a
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
- A ⊆ Sα ⊕ Dα
B A Sα a ≤ 1 b
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
- A ⊆ Sα ⊕ Dα
B A Sα a b s ≤ α ≤ 1 − α
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Lemma: Sα = (A ⊕ Dα) ∩ (B ⊕ D1−α) has dH(A, Sα) = α and dH(B, Sα) = 1 − α.
- Sα ⊆ A ⊕ Dα
- A ⊆ Sα ⊕ Dα
B A Sα a b s ≤ α ≤ 1 − α
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convex convex
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convex convex convex O(n + m) complexity
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convex non-convex
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convex non-convex connected O(n + m) complexity
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non-convex non-convex
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non-convex non-convex possibly disconnected O(nm) complexity
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maximal S
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minimal S
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B A
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A ⊕ D1/8 B ⊕ D7/8 B A S1/8
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A ⊕ D1/4 B ⊕ D3/4 S1/4 B A
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A ⊕ D1/2 B ⊕ D1/2 S1/2 B A
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A ⊕ D3/4 B ⊕ D1/4 B A S3/4
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B ⊕ D7/8 B ⊕ D1/8 B A S7/8
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Conclusion:
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Conclusion:
- We can compute a shape with dH(A, S) = α and dH(B, S) = 1 − α
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Conclusion:
- We can compute a shape with dH(A, S) = α and dH(B, S) = 1 − α
- The shape has linear or quadratic complexity, depending on the
convexity of A and B
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Conclusion:
- We can compute a shape with dH(A, S) = α and dH(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
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Conclusion:
- We can compute a shape with dH(A, S) = α and dH(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
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Conclusion:
- We can compute a shape with dH(A, S) = α and dH(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
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Conclusion:
- We can compute a shape with dH(A, S) = α and dH(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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?
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Input sets {A1, . . . , Ak} A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα A1 A2 A3
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Input sets {A1, . . . , Ak} Let Sα =
i
Ai ⊕ Dα Find smallest α s.t. Ai ⊆ Sα ⊕ Dα Worst case: smallest α = 1
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