p-Piercing Problems Liao Chung-Shou

“On Piercing Sets of Axis-Parallel Rectangles and Rings”, Michael Segal, International Journal of Computational Geometry and Applications, vol 9(3), (1999), “Rectilinear and Polygonal p-Piercing and p-Center Problems”, Micha Sharir, Emo Welzl, Proceeding 12 th ACM Symposium on Computational Geometry, (1996), “Obnoxious Facility Location: Complete Service with Minimal Harm”, Michael Segal,…etc, International Journal of Computational Geometry and Applications, vol 10(6), (2000),

Topic 1.Introduction 2.The latest results 3.Axis-transformation way 4.Location domain way 5.My conception 6.Future work and open problems

Introduction Given a collection of axis-parallel rectangles in the plane. Determine whether there exists a set of p points whose union intersects all the given rectangles. (for fixed p) related problems : p-center, piercing rings, piercing squares,… p = 3 * **

The latest results 1999 p = 2,3 => O(n) time p = 4 => O(nlogn) time p = 5 => O(nlogn) time p  6 => O(n logn) time p p = 2,3 => O(n) time p = 4 => O(nlog n) time p = 5 => O(nlog n) time p  6 => O(n log n) time p

Axis-transformation way ( c x, c y, d x, d y ) centroid ( c x, c y ) * * dxdx dydy cxcx dxdx cycy dydy A B C D x y *A C B * D * * A *B * C * D

query point p ( P x, 0, P y, 0 )p * * * * * Cone p = 1 x-cone cover all y-cone cover all * * * * * * *

p = 2C1 C2 ** * * * * X ** * * * * C3 C4 Y first, second, (C1  C3)  (C2  C4) cover all rectangles. or (C1  C4)  (C2  C3) cover all rectangles.

p = 3 first, second, C1, C2, C3 cover all in X-axis. C4, C5, C6 cover all in Y-axis. assume C1, C3, and C4, C6 are constrained cones (since there exists at least one constrained-cones pair) 4 possibilities => C1  C4, C1  C6, C3  C4, C3  C6. the rest of rectangles => 2-piercing

p = 4 first, second, similar to p = 3 C1, C2, C3, C4 cover all in X-axis. C5, C6, C7, C8 cover all in Y-axis. assume C1, C4, and C5, C8 are constrained cones find a pair Ci  Cj such that the rest of rectangles are 3-piercable, where i  {1,2,3,4} and j  {5,6,7,8}. (a) i  {1,4} and j  {5,8} (b) i  {2,3} and j  {6,7} imply i'  {1,4} and j'  {5,8} (c) each constrained cone maps to an unconstrained cone (Worst)

insert or delete 6 combinations =>independent each need O(n) steps each step need O(logn) updates => O(nlogn) time C1 C4 * ** * * * * * * * * * C3 before C3 after

Location domain way if an axis-parallel line traverses all rectangles else * T B L R * * * Observation

p = 2 p = 3 2-piercable => diagonal pair 3-piercable => at least one vertex in corner the rest of rectangles => similar to 2-piercing way 4 possibilities

p = 4 try each intersection part => O(n) time the rest of rectangles => 3-piercable => O(log n) time 3 => O(nlog n) time 3

p = 5 case1. there exists some point in corner 4 possibilities the rest of rectangles => 4-piercable case2. there exist 5 points in boundary (not in any corner)

case3. there exist 4 points in boundary (not in any corner), and 1 point inside. case2,3  O(logn) time However, it can’t work as p = 6.

My conception (as p = 4) Step1. pre-check whether there exists one point in corner. 4 possibilities, and the rest of rectangles => 3-piercable. Step2. there exists 4 points in boundary (not in any corner). (1) the intersection of outside rectangles in each side call the line segments T0, B0, L0, R0 T T0

(2) the intersection of the rectangles that intersect only one boundary. the set of the rectangles that intersect the boundary T called Tr, similar to Br, Lr, Rr. e.g. Tr \ (Br  Lr  Rr) call the line segments T1,B1,L1,R1 T T1 consider the intersection of T0 and T1 if there are at least two segment, not 4-piercable. else …………….

consider no rectangles that intersect three or four of T,B,L,R. (3) the intersection of the rectangles that intersect exact two boundaries (a) the boundaries are adjacent e.g. For Tr,Lr,Rr T LR B call the line segments T2a,B2a,L2a,R2a consider the intersection of T0 and T2a T2a

Note in (a) the boundaries are adjacent T B LR T2a consider the intersection of T0 and T2a in above case => O(n) time

(b) the boundaries are opposite e.g. For Tr,Br T Bcall the line segments T2b,B2b,L2b,R2b consider the intersection of T0 and T2b T2b

We have to estimate the time complexity only in Step2. (1). Scan from L to R or from B to T => O(n) time (2)and(3). Similar to (1) => O(n) time Now, we calculate the combination of the line segments we get  except (3)(a)note, the number of other line segments is fixed.  except (3)(a)note, the time complexity of combination is O(1) goal : O(n)  O(1) = O(n) time and extend to p = 5.

Future work p = 2~5, special case p  6, how ? In graph theory, transformation to intersection graph

The properties of the intersection graphs The clique cover problem in this graphs

Open problems 1.By my concept, p = 4,5 => O(n) time ? 2.As p  6, can we reduce the complexity ?