1. On the Geodesic Centers of Polygonal Domains
Haitao Wang
Utah State University
ESA 2016, Aarhus, Denmark

2. The center of a disk
The center s of D is the point that minimizes the maximum distance from s to all points t of D
d(s,t): the distance from s to t, which is the length of the line segment connecting s and t
All points on the boundary of D are farthest points of D to s D t s

3. The “center” of a polygon P
The center s is a point of P that minimizes the maximum distance from s to all points t of P
d(s,t): the length of the shortest path from s to t in P
shortest path distance is also called geodesic distance
s is called the geodesic center of P P s t

4. Previous work: the geodesic center of a simple polygon P
The geodesic center s is unique
Each farthest point of s must be a vertex of P
Algorithms:
O(n4 log n) time
Asano and Toussaint, 1985
O(n log n) time
Pollack, Sharir, Rote, 1989
O(n) time
Ahn et al., 2015 P s

5. Geodesic centers in a polygonal domain P (polygon with holes)
n: the total number of vertices (including those on holes)
Previous work by Bae, Korman, Okamoto, 2014
An O(n12+ԑ) time algorithm
An O(n12) upper bound on the total number of centers in P
Our new result
An O(n11 log n) time algorithm
An O(n10) upper bound on the total number of centers in P
Several other interesting observations t s

6. What makes the problem in polygonal domains difficult?
The geodesic center may not be unique

7. What makes the problem in polygonal domains difficult? (cont.)
A farthest point of a geodesic center may not be a vertex of P
A center s may have a single farthest point, even when s is in the interior of P s s
Picture from Bae et a. 2014 t t

8. Motivation of our algorithm
Picture from Bae et a. 2014

9. Motivation of our algorithm
s has three farthest points t1, t2, t3
Observation: all nine shortest paths have the same length
Form a system of eight equations, with the eight coordinates of s, t1, t2, t3, as eight variables
Compute (s, t1, t2, t3) by solving the equations

10. An exhaustive-search algorithm
Consider any combination of nine vertices of P as vij, for i=1,2,3 and j=1,2,3
Compute the shortest path map for each vij
Compute the overlap of these nine shortest path maps
For each cell C of the overlap, use the nine roots of C as uij, for i=1,2,3 and j=1,2,3
Form the equations using uij and vij as above
Solve the equations to determine a constant number of candidate centers s
Total time: O(n11 log n)
There are O(n9) combinations of nine vertices
The size of each shortest path overlap is O(n2)
The algorithm produces a set of O(n11) candidate centers in O(n11 log n) time

11. Some issues of the algorithm
What if a center s has only two farthest points t1 and t2?
all six shortest paths have the same length
form a system of five equations, with the six coordinates of s, t1, t2, as six variables
need one additional constraint to determine s
What if a center s has only one farthest points t1?
all three shortest paths have the same length
form a system of two equations, with the four coordinates of s, t1 as four variables
need two additional constraints to determine s
Question: Where are those additional constraints?

12. Notation For any point s in P
R(s): the radius of s, which is the distance from s to its farthest point
s is a center if it minimizes R(s)
A fact: each farthest point of s must be a vertex of SPM(s), the shortest path map of s, which has O(n) vertices

13. The one-farthest point case
Assume
s is a center that has only one farthest point t
both s and t are in the interior of P
there are exactly three shortest paths from s to t
What happens if we slightly move s to s’?
R(s’) < R(s) is not possible
SPM(s’) has a vertex t’ corresponding to t
the three paths through ui and vi are also shortest paths from s’ to t’
t’ is the only farthest point of s’
d(s’,t’) < d(s,t) is not possible u2 s u3 s’ u1 v3 t t’ v1 v2

14. An observation Let s be a point of P such that
t is a farthest point of s
both s and t are in the interior of P
there are exactly three shortest paths from s to t
How do we know whether there exists s’ in the vicinity of s such that R(s’) < R(s)?
Observation: The answer is yes if and only if there is a good direction to move s such that we can find a direction to move t simultaneously (with a different speed) and the lengths of all three paths strictly decrease u2 s u3 u1 v3 t v1 v2

15. New questions How do we know whether a good direction exits?
How to find all such good directions?
find the range of all good directions u2 s s s u2 u1 u3 u1 u3 u v3 v3 v3 t t t v1 v2 v1 v2 v1 v2

16. Our solution: the π–range property
There exists an open range of size π for good directions of s
The range is delimited by an open half-plane whose bounding line contains s s u v3 t v1 v2

17. Our solution: the π–range property
There exists an open range of size π for good directions of s
The range is delimited by an open half-plane whose bounding line contains s
This is true for all cases
This range can be uniquely determined by the locations of the vertices s, t, u1, u2, u3, v1, v2, v3
Recall: a direction is good to move s if we can find a direction to move t simultaneously (with a different speed) and the lengths of all three paths strictly decrease
Except for a special case u2 s u3 u1 v3 t v1 v2

18. Our solution: the π–range property
There exists an open range of size π for good directions of s
The range is delimited by an open half-plane whose bounding line contains s
This is true for all cases
This range can be uniquely determined by the locations of the vertices s, t, u1, u2, u3, v1, v2, v3
Recall: a direction is good to move s if we can find a direction to move t simultaneously (with a different speed) and the lengths of all three paths strictly decrease
Except for a special case
a1 = b1, a2 = b2, a3 = b3
no good directions for s exist u2 a1 a2 s u3 u1 a3 v3 b2 b3 t v1 v2 b1

19. How to use the π–range property?
What if a center s has only one farthest point t?
all three shortest paths have the same length
form a system of two equations, with the four coordinates of s, t as four variables
need two additional constraints to determine s
No good directions exist for s
The special case must happen
a1 = b1, a2 = b2, a3 = b3
This provides two additional (independent) constraints u2 a1 a2 s u3 u1 a3 v3 b2 b3 t v1 v2 b1

20. How to use the π–range property? (cont.)
What if a center s has only two farthest points t1 and t2?
all six shortest paths have the same length
form a system of five equations, with the six coordinates of s, t1, t2, as six variables
need one additional constraint to determine s
Observation: The π–range of t1 cannot intersect the π–range of t2
The bounding lines of the two ranges overlap
This provides an additional constraint

21. Other results by using the π–range property
There is at most one geodesic center among all points of P that have topologically equivalent shortest path maps
Each cell or edge of the shortest path map equivalence decomposition of P has at most one geodesic center
The size of the decomposition is O(n10), Chiang and Mitchell, 1999
The total number of all geodesic centers of P is O(n10)

22. The main theorem about the π–range property

23. Illustrating the π–range property

24. The π–range property when t is on an edge of P

25. s does not have a good direction because the special case happens

26. A closely related problem – geodesic diameter
Geodesic diameter: the maximum distance between any two points of P
When P is a simple polygon
O(n2) time, Chazelle 1982
O(n log n) time, Suri 1989
O(n) time, Hershberger and Suri 1997
When P is a general polygonal domain
O(n7.76) time, Bae, Korman, Okamoto, 2010

27. A pruning procedure
Our exhaustive-search algorithm produces O(n11) candidate centers
The true centers can be found by choosing those candidates whose radius are the smallest
An easy method is to compute the radius of every candidate in O(n log n) time
Total time: O(n12 log n)
Our approach: Use a pruning procedure to prune most of the candidates and only keep O(n10) candidates
Each cell of the shortest path map equivalence decomposition keeps at most one candidate
No true center is pruned
With the help of the π–range property
Total time: O(n11 log n)

28. The main theorem when t is on an edge of P

29. New questions How do we know whether a good direction exits?
How many such good directions exit?
find the range of good directions u2 s s s s u2 u1 u3 u1 u3 u v3 v3 v3 v3 t t t t v1 v1 v1 v2 v2 v2 v1 v2

30. How to use the π–range property? (cont.)
What if a center s has only two farthest points t1 and t2?
all six shortest paths have the same length
form a system of five equations, with the six coordinates of s, t1, t2, as six variables
need one additional constraint to determine s
Observation: The π–range of t1 cannot intersect the π–range of t2
The bounding lines of the two ranges overlap
This provides an additional constraint