Orthogonal range searching

1
Orthogonal range searching
2
The problem (1-D)
Given a set of points S on the line, preprocess
them to build structure that allows efficient
queries of the from Given an interval Ix1,x2
find all points in S that are in the interval.
2
4
5
7
8
12
15
19
3
1-D solution

• Build a balanced tree using the points
  co-ordinates as the keys.

2
4
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query O(log nk)
space O(n)
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12
2
5
8
15
7
19
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12
8
2
4
5
4
The problem (2-D)
Given a set of points S on the plane, preprocess
them to build structure that allows efficient
queries of the from Given an rectangle
Rx1,x2y1,y2 find all points in S that are
in the rectangle.
P3
P8
P5
P2
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P4
P7
P1
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Range Trees (2-D)
Maintain the points in a balanced search tree
ordered by x-coor. In each internal node maintain
the points in its subtree in a balanced search
tree ordered by y
P3
P3
P2
P2
P4
P4
P1
P1
P4
P3
P2
P1
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Range Trees (Contd.)
P3
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P5
P2
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P4
P7
P1
P3
P2
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P1
P4
P3
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P1
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Range Trees (Contd.)
P3
P8
P5
P2
P6
P4
P7
P1
P3
P2
P4
P1
P4
P3
P2
P1
P8
P7
P6
P5
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Query processing

• Search by the first dimension gives us O(logn)
  trees which together contain the output.
• We search each of these trees to get the answer

9
Analysis (2-D)

• Space O(nlog n)
• Query O(log2 nk)
• Preprocessing O(nlog n)

10
Further facts

• Generalizes to d-dimensions
• Query time can be reduced using a technique
  called fractional cascading (we may talk about it
  later on)

11
The dynamic case

• Suppose the set S is not fixed
• We want to be able to insert and delete points,
  and make queries intermixed with insertions and
  deletions.

12

• Easy in 1-D, just use a red-black tree or some
  other balance search tree

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P3
P8
P5
P2
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P4
P7
P1
P3
P2
P4
P1
P4
P3
P2
P1
P8
P7
P6
P5
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P3
P8
P5
P2
P6
P4
P7
P1
P3
P2
P4
P1
P4
P2
P1
P8
P7
P6
P5
15

• May need to rebalance the primary tree by doing a
  rotation

y
x
ltgt
A
x
C
y
B
B
C
A
? We have to rebuild the secondary data
structures at x and y
16
Dynamic range trees (analysis)

• So we use BB(a) trees
• Then the amortized cost of rebalancing the
  primary tree is O(log n)
• and we get

Query O(log2 nk)
Insert, delete O(log2 n) (amortized)
17
Priority search trees

• Suppose the query is unbouded from below, can we
  take advantage of this ?

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P11
P3
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P1
18
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
A
P1
A
P3
P12
We put the lowest point at the root and the
x-median (A) of the points other than the root
P4
P6
P7
P2
P5
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P9
P10
P11
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
A
B
A
P4
P7
B
How do we answer a query ?
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P6
P5
P2
P11
P10
P9
P3
P12
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P10
P11
P12
P3
P9
P8
P5
P2
P6
P4
P7
P1
P1
A
P4
P7
B
P8
P6
P5
P2
P11
P10
P9
P3
P12
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P10
You stop the search in the when you hit a point
that is not in the answer
P11
P12
P3
P9
P8
P5
P2
P6
P4
P7
P1
P1
A
P4
P4
P7
B
P8
P6
P6
P5
P2
P2
P11
P10
P9
P3
P12
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Priority search trees (analysis)

• query O(log(n) k)
• space O(n)
• preprocessing O(nlogn)
• How do we make them dynamic ?

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Dynamic priority search trees
P10
P11
P3
P12
First put the points at the leaves of a red-black
tree
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P2
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P4
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P1
P2
P3
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P1
P4
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P11
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P6
P7
P9
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Top-down At each internal node put the smallest
point in its subtree not already assigned to an
internal node
P1
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P2
How does the insertion go ?
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P2
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P1
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
P7
P2
P4
P6
P3
P11
P6
P5
P10
P2
P3
P12
P1
P4
P10
P11
P5
P6
P7
P8
P9
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
P7
P2
P4
P6
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
P7
P2
P4
P6
P3
P11
P6
P5
P10
N
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
P7
P2
P4
P6
P3
N
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
P7
P2
N
P4
P6
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
N
P7
P2
P4
P6
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
P1
N
N
P7
P2
P4
P6
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P2
P4
P6
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P2
P4
P6
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
We may also need to rebalance the tree Is this a
problem ?
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
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P10
P11
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P8
P9
P3
N
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M
P10
P11
P3
P12
P9
P8
P5
P2
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P4
P7
P1
N
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P5
P6
P7
P8
P9
P3
N
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M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
P4
P10
P11
P6
P7
P8
P9
P3
N
M
P5
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M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
M
P4
P10
P11
P6
P7
P8
P9
P3
N
M
P5
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M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
M
P4
P10
P11
P6
P7
P8
P9
P3
N
M
P5
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M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
M
P4
P10
P11
P6
P7
P8
P9
P3
N
M
P5
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M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
N
P1
N
P7
P1
P4
P6
P2
P3
P11
P6
P5
P10
P12
P1
P2
M
P4
P10
P11
P6
P7
P8
P9
P3
N
M
P5
42
P
M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
P1
N
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P
M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
N
P1
N
P7
P1
P4
P2
P6
P3
P11
P6
P5
P10
P12
P2
P1
P
P10
P11
P8
P6
P7
P9
P3
N
M
P
P4
P5
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Rotation, rotations
X
X
ltgt
Y
Z
C
A
B
C
B
A
Z is bubbled down into B or C Y is bubbled up
from A or B
45
P
M
P10
P11
P3
P12
P9
P8
P5
P2
P6
P4
P7
N
P1
N
P7
P1
P4
P2
P6
P3
P11
P6
P5
P10
P12
P2
P1
P4
M
P10
P11
P8
P6
P7
P9
P3
N
M
P
P4
P5