Klee’s Measure Problem
Computational Geometry, WS 2007/08
Group Work
Prof. Dr. Thomas Ottmann
Khaireel A. Mohamed
Algorithmen & Datenstrukturen, Institut für Informatik
Fakultät für Angewandte Wissenschaften
Albert-Ludwigs-Universität Freiburg
Overview
• The measure problem
• 1D analysis
• 2D analysis
• Naïve solution
• Implicit area computation by segment tree
• Improved solution
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
2
The Measure Problem
• Victor Klee, 1977
• Given a set of n horizontal line-segments, compute the length of
their union on the real line.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
3
1D Problem (Original)
• Sorting and partitioning
– Total partitions (space):
– Time:
• Sweep direction: Left to right
– Event Q:
– Status-measure T:
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
4
1D Problem (Original)
• Time complexities
– Time per event:
– Total sweep time:
– Total runtime:
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
5
The Measure Problem in 2D
• Jon Bentley, 1977
• Given a set of n rectangles on the 2D plane, compute the area of
their union.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
6
Representing a Rectangle
• Assume that all rectangles in the set are non-trivial
• Sufficient to store bottom-left and top-right coordinates
(xhi, yhi)
yhi
r
ylo
(xlo, ylo)
xlo
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
xhi
7
2D Problem – Sweepline Approach
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
epi
epi+1
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2D Problem – Sweepline Approach
• Sorting and partitioning
– Total partitions (space):
– Time:
• Sweep direction: Left to right
– Event Q:
epi
– Status-structure T:
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
9
2D Sweepline – Naïve Solution
epi
epi+1
• Let m(i) be the 1D measure of the active vertical segments at epi.
• Then the measure for the 2D slab between epi and epi+1:
• In general, for k 2D slabs, the measure for the entire set of
rectangles:
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
10
Analyses – 2D Naïve
• Maintaining the status-structure T, per event point epi
– Time (per insertion):
– Time (per deletion):
– Time to compute m(i):
• Total insertions (into T) = Total deletions (from T) =
• Overall performance of the naïve algorithm
– Time complexity:
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
11
Analyses – Naïve Worst Case
• Worst case runtime:
• Can this be avoided?
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
12
Improving the Naïve Approach
• Main reason for inefficiency:
• Overcoming the inefficiency:
– Limit the amount of work done at epi by
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
13
Observation for Improvement
• Consider the slab between epi and epi+1 when a new rectangle q
becomes active.
epi
epi+1
epi
q
epi+1
q
epi
epi+1
q
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
epi
epi+1
q
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Decomposing the Vertical Sweep
Idea:
• All segments appearing on the vertical sweepline can be composed
of a collection of consecutive fragments [vj, vj+1].
• A single offline structure is sufficient to represent the vertical sweeps
to collectively compute the measure m(i) at each epi.
• Insertions and deletions of segments affect the structure only by
changing the augmented values in the internal nodes.
• At the start of the event point epi, the measure for the previous
segment m(i-1) can be read off the root immediately.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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The Segment Tree as Status-Structure
• The ordered (vertical) segments are entered as leaves in a balanced
binary search tree .
• Each vj (except v1 and vm) will occur twice, representing the closed
non-overlapping segment [vj, vj+1].
9
vm
8
9
C
8
8
7
6
6
6
5
5
4
B
E
A
3
5
D
2
3
3
1
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
v1
1
0
0
1
2
3
4
5
6
7
8
9 10
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Keeping Track of Segments
• The segment tree  represents all O(n) partitions on the vertical axis.
• When a rectangle becomes active (or inactive), we mark the internal
nodes in  accordingly; i.e. we should be able to know
– if the subtree under a node x in  fully or partially covers a set of
intervals – and thus, the measure at node x, and
– the number of segments that fully covers the range of x.
x’
vk
x
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
vk
x
vj
vj
vj
vj
vi
vi
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Implicit Handling
• It is NOT necessary to mark all the individual fragments from which
a segment q is composed. Why?
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Classifying Nodes based on a Segment q
• Given a segment q = [vi, vj], then a node in  is said to be
– q-full if it covers a segment fully contained in q
– q-partial if it is not q-full but has a child that is either q-partial or q-full
• Consider the search paths for ‘[vi’ and ‘vj]’; let the node where the
two paths diverge be t.
– All nodes encountered before t must have been
vm

v j]
t
q
[vi
v1
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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1-Umbrella of the Segment q
v j]
hi-tip
hi-line
t
q
lo-line
lo-tip
[vi
• The 1-umbrella of a segment q is the subtree of  rooted at t
consisting of
– all the q-partial nodes on the hi-line from t to hi-tip,
– all the q-partial nodes on the lo-line from t to lo-tip, and
– all the q-full nodes which are directly connected to the hi-line or lo-line.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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1-Umbrella of the Segment q
Lemma:
• The 1-umbrella of the segment q can be built in O(log n) steps.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Marking Nodes on the 1-Umbrella
• We maintain the following information in  for any sets of segments,
where augment in the node x
– x.val : The measure of the fragments in the subtree of x which are
covered by 1-umbrellas through it or below it
– x.cnt : A count of the number of times in which x becomes a q-full node
of some umbrella.
vm
t
vk
vj
vj
vi
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Inserting a Segment q into 
• Input:
– The root of the segment tree 
– The range q = [vi, vj]
• Output:
– The measure of the vertical line-segments at root.val
• Idea:
– Percolating measure differences between successive levels in 
•
•
•
•
Determine the 1-umbrella for q.
Handle measure differences from lo-tip to t.
Handle measure differences from hi-tip to t.
Combine measure differences and percolate to root.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Example – Activate qA
[8,9]
[6,9]
[5,9]
v=0
c=0
[5,6]
v=0
c=0
[1,5]
v=0
c=0
9
8
[6,8]
[1,9]
v=0
c=0
9
C
8
8
7
6
6
B
A
5
6
E
4
5
qA
3
[3,5]
v=0
c=0
5
3
1
[1,3]
3
0
1
D
2
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Example – Activate qB
[8,9]
[6,9]
v=0
c=0
[5,9]
v=1
c=0
[1,9]
v=3
c=0
[1,5]
v=2
c=0
[6,8]
v=0
c=0
9
9
8
8
8
7
6
6
qB
C
B
A
5
[5,6]
v=1
c=1
6
[3,5]
v=2
c=1
5
3
1
[1,3]
3
0
E
4
5
1
3
D
2
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Example – Activate qC
[8,9]
[6,9]
v=2
c=0
[5,9]
v=3
c=0
[1,9]
v=5
c=0
[1,5]
v=2
c=0
[6,8]
v=2
c=1
9
9
8
8
8
7
6
6
C
B
A
5
[5,6]
v=1
c=2
6
[3,5]
v=2
c=1
5
3
1
[1,3]
3
0
E
4
5
1
qC
3
D
2
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Special Cases
• When t equals lo-tip, then t also equals hi-tip. Why?
• The percolated measure difference have to be ‘killed’
if x.parent.cnt ≥ 1. Why?
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
27
Inserting a Segment
Lemma:
• Inserting a segment and maintaining the information in  requires
O(log n) time.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Deleting a Segment
Lemma:
• Deleting a segment and maintaining the information in  requires
O(log n) time.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
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Example – Deactivate qA
[8,9]
[6,9]
v=2
c=0
[5,9]
v=3
c=0
[1,9]
v=8
c=1
[1,5]
v=2
c=0
[6,8]
v=2
c=1
9
9
8
8
8
7
6
6
C
B
A
5
[5,6]
v=1
c=2
6
[3,5]
v=2
c=1
5
3
1
[1,3]
3
0
E
4
5
1
qA
3
D
2
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
30
Example – Deactivate qC
[8,9]
[6,9]
v=2
c=0
[5,9]
v=3
c=0
[1,9]
v=8
c=1
[1,5]
v=0
c=0
[6,8]
v=2
c=1
9
9
8
8
8
7
6
6
C
B
A
5
[5,6]
v=1
c=1
6
[3,5]
v=0
c=0
5
3
1
[1,3]
3
0
E
4
5
1
qC
3
D
2
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
31
Example – Activate qD, qE
[8,9]
[6,9]
v=2
c=0
[5,9]
v=3
c=0
[1,9]
v=3
c=0
[1,5]
v=0
c=0
[6,8]
v=2
c=1
9
9
8
8
8
7
6
[5,6]
v=1
c=1
6
[3,5]
v=0
c=0
5
[1,3]
v=0
c=0
3
qE
C
B
6
E
A
5
4
5
3
3
1
D
2
1
qD
0
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
32
Example – Deactivate qD
[8,9]
[6,9]
v=2
c=0
[5,9]
v=3
c=0
[1,9]
v=5
c=0
[1,5]
v=2
c=0
[6,8]
v=2
c=2
9
9
8
8
8
7
6
6
[5,6]
v=1
c=1
6
[3,5]
v=0
c=0
5
[1,3]
v=2
c=1
3
C
B
E
A
5
4
5
3
3
1
D
2
1
qD
0
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
33
Example – Deactivate qE, qB
[8,9]
[6,9]
v=2
c=0
[5,9]
v=3
c=0
[1,9]
v=3
c=0
[1,5]
v=0
c=0
[6,8]
v=2
c=2
9
9
8
8
8
7
6
qE
C
B
6
A
5
[5,6]
v=1
c=1
6
[3,5]
v=0
c=0
5
3
1
[1,3]
v=2
c=0
3
0
E
4
5
1
3
D
2
0
1
2
3
4
5
6
7
8
9 10
Sweep direction
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
34
The Improved Solution
Theorem:
• The measure problem for a set of n rectangles in the plane can be
solved on O(n log n) time.
Computational Geometry, WS 2007/08
Prof. Dr. Thomas Ottmann
35