1. Boaz Ben-Moshe Binay Bhattacharya Qiaosheng Shi
  Farthest Neighbor Voronoi Diagram in the Presence of Rectangular Obstacles  CCCG 2005
Boaz Ben-Moshe
Binay Bhattacharya
Qiaosheng Shi
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2. Model O: a set of m disjoint Obstacles
§        Input: {O,P}
O: a set of m disjoint Obstacles
(axis-aligned rectangles) in R2.
P: a set of n points in F=R2 - O.
§        Distance:
d(a,b): the shortest L1 obstacle-
avoiding path between a and b.
§        Motivation:
VLSI systems design.
Plotter movement.
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3. Problem Farthest neighbor query (FVD representation).
Farthest Voronoi Diagram: (n*m) complexity
Farthest Neighbor Query:
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4. Applications:
Diameter
All Farthest pairs
Center point
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5. Previous Related Results
|P| = n , |O| = m , N = m+n.
Nearest neighbor queries: O(N log N). preprocessing, O(log N ) query [CKV87], [M89].
Farthest neighbor queries: O(m*n log N) preprocessing, O(log N) query [BKM01].
All farthest pairs & diameter: O(m*n log N).
Our results:
Farthest neighbor queries: O(N3/2 log2 N) preprocessing, O(N1/2 log N) query.
All farthest pairs & diameter: O(nN1/2 log N).
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6. Preliminaries
Lemma 1: For any two points p, q, the shortest path (between p and q) is either X-monotone or Y-monotone (or both).
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7. Preliminaries
Lemma 1: If there exists an X-monotone path (between p and q) and a Y-monotone path, then there exists an XY-monotone path.
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8. Preliminaries
Corollary: If there exists an X-monotone path between p and q, then all shortest paths between p and q are X-monotone. (Alternatively, Y-monotone).
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9. Preliminaries The farthest neighbor from p on the left (p.left):
choose the farthest among
the points to the left of p
with an X-monotone
path to p.
(right, above, below).
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10. Preliminaries Lemma 2: The farthest neighbor from p is
max (p.left, p.right, p.above , p.below).
One can divide the DS for finding the farthest point from p into four independent DSs.
We consider the DS for farthest neighbor on the left.
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11. Theorem 3 Given 3 points p1, p2, q: p2.x < p1.x < q.x
The Path(p1,q) is X monotone
dx=p1.x–p2.x, dy=|p1.y-p2.y|
d(p2,q)  d(p1,q)+dx–dy
Note: If dx  dy  d(p2,q)  d(p1,q)
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12. Constructing the Filter
For each point in the input scene check whether it is a dominant point
Define: X-, X+, Y-, Y+.
Focus on X-.
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13. Filtering example
In the ‘average’ case the size of the remaining set of points is O(n1/2).
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14. Farthest order theorem
Let q1,q2 be two points on
a vertical line where q1.y>p2.y
If q1 (q2) is the farthest
point from p1 (p2):
p1.y < p2.y
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15. Farthest range on a vertical line.
The upper envelope of the distance functions.
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16. Two source points distance query [EM94]
Given an input P,O
One can construct a DS of size O(N3/2), in O(N3/2 log N) time.
To answer distance queries d(p,q) in O(N1/2) time.
Note: p,q belong to P.
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17. FVD infinite cell theorem
If a point q is farthest prom p.
On any vertical line l to the right of q, there is a point q’ that is farthest from p
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18. Computing the FVD(X-) on a vertical line
Given a vertical line l(x) we want to divided to farthest ranges.
O(m) ‘portals’
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19. Computing the FVD(X-) on a vertical line
Use the order claim to perform a divide & concur algorithm on l(x).
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20. Computing the FVD(X-)
Goal: attach each FVD cell to the left most vertical line it starts at!
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21. Computing the FVD(X-)
Use the infinite cell theorem to perform a divide & concur algorithm on the cells starting point.
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22. Computing an implicit FVD(X-)
Use a persistent tree DS [DSST89] to store the cells y-order for each vertical line.
The y-‘middle’ cell can be found in O(log n) time, for any vertical qurey line l(x)
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23. Farthest point query mechanism.
Given query point q:
Project q on the obstacle (r(q)) to its left q’ (ray-shooting).
Find the persistent tree snap-shoot associated with r(q).
Binary search to find the farthest from q’ using b & u portal points.
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24. Conclusion This work: Implicit representation for the FVD.
All farthest pairs & diameter: O(N3/2 log2 N).
Current research:
Center Point O(N3/2 log2 N).
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25. Conclusion Future research:
Improving the two source points query mechanism.
Higher dimensions (a naive generalization exists).
Median.
More complex models:
weighted version.
Other type of obstacles.
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26. Fin
Questions: ???
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