1. Boaz BenMoshe Matthew Katz Joseph Mitchell
SoCG 2001   Farthest Neighbors and Center Points in the Presence of Rectangular Obstacles 
Boaz BenMoshe
Matthew Katz
Joseph Mitchell
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2. Model and Problems § Input: {O,P}
O: a set of m disjoint Obstacles (axis-aligned rectangles) in the plane.
P: a set of n points in the free space F=R2 - O.
§        Distances:
The distance between two points d(a,b) is the shortest L1 obstacle-avoiding path between a and b.
§        Motivation:
VLSI systems design.
Plotter movement.
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3. Problems:
Preprocess for farthest neighbor queries (FVD representation).
Farthest (L1) Voronoi Diagram:
(n*m) complexity
Farthest Neighbor Query:
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4. Problems: Center point (set):
a point c that minimizes max{d(p,c) | pP}.
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5. Problems: Approximations:
given e>0, filter out a large portion of the points.
only O(1/e) of the points remain.
approximate the diameter and radius, using the remaining points.
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6. Previous Related Results
|P| = n , |O| = m.
Nearest neighbor queries: O((m+n) log(m+n)) preprocessing, O(log(m+n)) query [CKV], [M]
All farthest neighbors: O(m2 + n2 log m) [AC]
Center Point: O(m2 * n) (center set) [CSK]
No previous results for the approximation problems.
Our results:
Farthest neighbor queries: O(m*n log(m+n)) preprocessing, O(log(m+n)) query
All farthest neighbors: O(m*n log(m+n))
Center Point: O(m*n log(m+n))
Near linear-time approximations.
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7. Preliminaries
2.1  Claim: 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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8. Preliminaries
2.2  Claim: 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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9. Preliminaries
2.3  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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10. 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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11. Preliminaries 2.4 Claim: The farthest neighbor from p is
max (p.left, p.right, p.above , p.below).
2.5 Claim: 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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12. 3.1 Theorem d(p2,q)  d(p1,q) + dx – dy
Given 3 points p1, p2, q such that:
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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13. Proving Theorem 3.1 Proof (Sketch only):
if dy = 0, d(p2,q)  d(p1,q) + dx
 if there are no obstacles in between p2 and p1 – easy.
else the obstacles can only affect d(p2,q) by making it longer.
if dx = 0, d(p2,q)  d(p1,q) - dy
 if there are no obstacles in between p2 and p1 - easy.
else the obstacles can only affect d(p2,q) by making it longer.
Let p’=(p1.x,p2.y) : d(p2,q)  d(p’,q) + dx  d(p1,q) +dx - dy.
d(p2,q)  d(p1,q) + dx - dy.
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14. Constructing the Filter
For each point in the input scene check whether it is a dominant point
Defining: X-, X+, Y-, Y+.
Focus on X-.
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15. Filtering example
In the “average “ case the size of the remaining set of points is O(n).
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16. Constructing a DS for farthest neighbor queries (on the left)
Events:
Obstacles
Points of X-
A compact representation of all the sweep-lines is kept (DS(X-)).
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17. Updating the sweep-line
The sweep line is the upper envelope of the distance functions from the points.
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18. Computing the Query Given query point q
The farthest neighbor on the left is p3 (using DS(X-)).
d(p3,q)=d(p3,q’)+d(q’,q)
The farthest neighbor from below is one of the two extreme lower points.
The farthest neighbor of q is the maximum over the 4 directional maximums.
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19. Center Point
We may encounter local minima while approaching the radius.
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20. Computing the bisector
bs(X-,X+): all points b s.t. b.left = b.right
Resp. bs(Y-,Y+)
Computing.
Complexity.
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21. Computing the center point
Possible positions:
On the FVD edges.
bisectors:
bs(X-,X+), bs(Y-,Y+)
Obstacle edges.
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22. Examples for Bisectors, Center Points
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23. Complexity of the Center Point algorithm
Filtering: O(n log(n))
Computing DS(X-) : O(n m log(n+m)
Computing the bs(X-, X+) : O(n m log(n+m))
Searching for Center Point : O(n m log(n+m))
Total complexity: O(n m log(n+m))
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24. Approximations Given e > 0, find: Approximate Diameter D’:
D(1 - e)  D’  D
Approximate Radius R’:
R  R’  R(1+ e)
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25. Constructing the approximation filter
Given e > 0 ,
Divide the Y-range into O(1/e) horizontal slabs of height, e*D/2  h  e*D
In each slab mark the leftmost and the rightmost points.
Do the same with the vertical slabs.
Remove all unmarked points.
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26. Conclusion Future research: Other types of obstacles.
Improving the FVD representation / Proving a lower bound.
Higher dimensions – a natural generalization exists.
More info: (implementations, papers etc.)
Questions ??
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