P-Piercing Problems Liao Chung-Shou 2001. 8.29. “On Piercing Sets of Axis-Parallel Rectangles and...
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Transcript of P-Piercing Problems Liao Chung-Shou 2001. 8.29. “On Piercing Sets of Axis-Parallel Rectangles and...
p-Piercing Problems
Liao Chung-Shou
2001. 8.29
“On Piercing Sets of Axis-Parallel Rectangles and Rings”,Michael Segal, International Journal of Computational Geometry and Applications, vol 9(3), (1999), 219-233.
“Rectilinear and Polygonal p-Piercing and p-Center Problems”,Micha Sharir, Emo Welzl, Proceeding 12th ACM Symposium onComputational Geometry, (1996), 122-132.
“Obnoxious Facility Location: Complete Service with Minimal Harm”,Michael Segal,…etc, International Journal of Computational Geometry and Applications, vol 10(6), (2000), 219-233.
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-4
1996
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-4
3
4
5
Axis-transformation way
( cx , cy , dx , dy )
centroid( cx , cy )** dx
dy
cx
dx
cy
dy
AB
C
D
x
y
*AC B
* D*
*A * B*C*
D
query point p
( Px , 0 , Py , 0 )
p*
*** *
Cone
p = 1
x-cone cover ally-cone cover all
*
*
*
** *
*
p = 2
C1 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 => C1C4, C1C6, C3C4, C3C6.
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 = 4try each intersection part
=> O(n) time
the rest of rectangles => 3-piercable
=> O(log n) time3
=> O(nlog n) time3
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 workas 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
TT0
(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
L R
B
call the line segmentsT2a,B2a,L2a,R2a
consider the intersection of T0 and T2a
T2a T2a
Note in (a) the boundaries are adjacent
T
B
L R
T2a
consider the intersection of T0 and T2a in above case=> O(n) time
(b) the boundaries are opposite
e.g. For Tr,BrT
B
call the line segments
T2b,B2b,L2b,R2b
consider the intersection of T0 and T2b
T2bT2b
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 ?