COSC 6114
Prof. Andy Mirzaian
References:
• [M. de Berge et al] chapter 10
Data Structures:
• Interval Trees
• Priority Search Trees
• Segment Trees
Applications:
•
•
•
•
•
Windowing Queries
Vehicle navigation systems
Geographic Information Systems
Flight simulation in computer graphics
CAD/CAM of printed circuit design
Windowing
PROBLEM 1: Preprocess a set S of non-crossing line-segments in the
plane for efficient query processing of the following type:
Query: given an axis-parallel rectangular query window W,
report all segments in S that intersect W.
Windowing
PROBLEM 2: Preprocess a set S of horizontal or vertical line-segments
in the plane for efficient query processing of the following type:
Query: given an axis-parallel rectangular query window W,
report all segments in S that intersect W.
INTERVAL TREES
PROBLEM 2: Preprocess a set S of horizontal or vertical line-segments
in the plane for efficient query processing of the following type:
Query: given an axis-parallel rectangular query window W,
report all segments in S that intersect W.
W
INTERVAL TREES
SUB-PROBLEM 1.1 & 2.1:
Let S be a set of n line-segments in the plane. Given an axis-parallel query
window W, the segments of S that have at least one end-point inside W
can be reported in O(K + log n) time with a data structure that uses
O(n log n) space and O(n log n) preprocessing time, where K is the
number of reported segments.
Method:
Use 2D Range Tree on segment end-points and fractional cascading.
INTERVAL TREES
Now consider horizontal (similarly, vertical) segments in S that intersect W,
but their end-points are outside W.
They must all cross the left edge of W.
W
SUB-PROBLEM 2.2:
Preprocess a set SH of horizontal line-segments in the plane, so that the subset of
SH that intersects a query vertical line can be reported efficiently.
Method: Use Interval Trees.
INTERVAL TREES
Imed
Imed
Ileft
Ileft
Iright
Iright
xmed
Associated structure for Imed :
L left = list of segments in Imed sorted by their left end-points,
L right = list of segments in Imed sorted by their right end-points.
L left = 3,4,5
L left =1,2
L right = 5,3,4
L left =6,7
L right = 1,2
2
1
L right = 7,6
5
3
4
6
7
INTERVAL TREES
THEOREM: Interval Tree for a set of n horizontal intervals:
• O(n)
storage space
• O(n log n)
construction time
• O(K + log n)
query time
[report all K data intervals that contain
a query x-coordinate.]
INTERVAL TREES
SUB-PROBLEM 2.3:
Now instead of the query being on a vertical line, suppose it is on
a vertical line-segment.
The primary structure of Interval Trees is still valid.
Modify the associated secondary structure.
(qx, q’y)
q
(qx, qy)
SOLUTION:
L left = Range Tree on left end-points of Imed ,
L right = Range Tree on right end-points of Imed .
(-: qx]  [qy : q’y]
q
xmed
INTERVAL TREES
THEOREM: Interval Tree for a set of n horizontal intervals:
• O(n log n)
storage space
• O(n log n)
construction time
• O(K + log2 n)
query time
[report all K data intervals that intersect
a query vertical line-segment.]
COROLLARY: Let S be a set of n horizontal or vertical line-segments in
the plane. We can preprocess S for axis-parallel rectangular query window
intersection with the following complexities:
• O(n log n)
storage space
• O(n log n)
construction time
• O(K + log2 n)
query time
[report all K data intervals that intersect
the query window.]
PRIORITY SEARCH TREES
Improving the previous solution:
the associated structure can be implemented by Priority Search Trees,
instead of Range Trees.
P = {p1, p2, … , pn }  2.
A Priority Search Tree (PST) T on P is:
• a binary tree, one point per node,
• heap-ordered by x-coordinates,
• (almost) symmetrically ordered by y-coordinates.
PRIORITY SEARCH TREES
pmin
pmin  point in P with minimum x-coordinate.
ymin  min y-coordinate of points in P
ymax  max y-coordinate of points in P
P’

ymed 
Pbelow 
Pabove 
ymin , ymax
P – {pmin}
y-median of points in P’
{ p P’ | py  ymed }
{ p P’ | py > ymed }
PST
on
PST
on
Pbelow
Pabove
p7
p1
p3
y7 y7
y6 y3
p6
p3
pmin p1
p6
y6 y6
y4 y3
p2
p5
y4 y2
y5 y5
p4
y4 y4
p7
ymed
p2
p5
p4
PRIORITY SEARCH TREES
Priority Search Tree T on n points in the plane requires:
• O(n)
storage space
• O(n log n) construction time:
 either recursively, or
 pre-sort P on y-axis, then construct T
in O(n) time bottom-up. (How?)
Priority Search Trees can replace the secondary structures (range trees)
in Interval Trees.
• simpler (no fractional cascading)
• linear space for secondary structure.
How to use PST to search for a query range R = (-: qx]  [qy : q’y] ?
ALGORITHM QueryPST (v, R)
if v = nil or pmin x(v) > qx or ymin(v) > q’y or ymax(v) < qy
then return
if pmin x (v)  qx and qy  ymin(v)  ymax(v)  q’y
then Report.In.Subtree (v, qx)
else do
if pmin x (v)  R then report pmin x(v)
QueryPST (lc(v), R)
QueryPST (rc(v), R)
end else
end
PROCEDURE Report.In.Subtree (v, qx)
if v=nil then return
if pmin x (v)  qx then do
report pmin x(v)
Report.In.Subtree (lc(v), qx)
Report.In.Subtree (rc(v), qx)
end if
end
q’y
ymax
pmin qx T v
R
ymin
qy
Truncated Pre-Order on
the Heap: O(1 + Kv) time.
v
PST
LEMMA: Report.In.Subtree(v, qx) takes O(1 + Kv) time to report all points in the
subtree rooted at v whose x-cooridnate is  qx , where Kv is the number
of reported points.
THEOREM: Priority Search Tree for a set P of n points in the plane has complexities:
• O(n)
• O(n log n)
• O(K + log n)
Storage space
Construction time
Query time
[report all K points of P in a query range
R = (-: qx]  [qy : q’y] .]
PST
qy
q’y
SEGMENT TREES
Back to Problem 1: Arbitrarily oriented line segments.
Solution 1: Bounding box method.
W
Bad worst-case.
Many false hits.
W
SEGMENT TREES
Back to Problem 1: Arbitrarily oriented line segments.
Solution 2: Use Segment Trees.
a) Segments with end-points in W can be reported
using range trees (as before).
b) Segments that intersect the boundary of W can
be reported by Segment Trees.
SUB-PROBLEM 1.1: Preprocess a set S of n non-crossing line-segments
in the plane into a data structure to report those segments in
S that intersect a given vertical query segment
q = qx  [qy : q’y] efficiently.
SEGMENT TREES
Elementary x-intervals of S
p1 p2
p3
...
pm
(-: p1), [p1 : p1], (p1 : p2), [p2 : p2], … , (pm-1 : pm), [pm : pm], (pm : +).
Build a balanced search tree with each leaf corresponding (left-to-right)
to an elementary interval (in increasing x-order).
Leaf v:
Int(v) = set of intervals (in S) that contain the elementary interval
corresponding to v.
IDEA 1: Store Int(v) with each leaf v.
Storage O(n2), because intervals in S that span many elementary intervals will be
stored in many leaves.
SEGMENT TREES
IDEA 2:
 internal node v:
Int(v) = union of elementary intervals corresponding to the leaf-descendents of v.
Store an interval [x:x’] of S at a node v iff Int(v)  [x:x’] but Int(parent(v))  [x:x’].
Each interval of S is stored in at most 2 nodes per level (i.e., O(log n) nodes).
Thus, storage space reduces to O(n log n).
What should the
associated structure be?
s2,s5
s1
s5
s1
s3
s1
s3,s4
s3
s2,s5
s1
s2
s5
s4
s3
s4
SEGMENT TREES
v1 S(v1) = {s3}
v2
v3
S(v2) = {s1 , s2}
S(v3) = {s5 , s7}
s7
s3
s6
s2
s5
s4
s1
SEGMENT TREES
Associated structure
is a balanced search tree based on the vertical ordering
of segments S(v) that cross the slab Int(v)  (- : +).
s6
s5
s4
s6
s5
s4
s3
s2
s3
s2
s1
s1
SEGMENT TREES
THEOREM:
Segment Tree for a set S of n non-crossing line-segments in the plane:
• O(n log n)
• O(n log n)
• O(K + log2 n)
Storage space
Construction time
Query time
[report all K segments of S that intersect a
vertical query line-segment.
COROLLARY:
Segment Trees can be used to solve Problem 1 with the above
complexities. That is, the above complexities applies if the query is
with respect to an axis-parallel rectangular window.
Exercises
1.
Let P be a set of n point sin the plane, sorted on y-coordinate. Show that, because S is sorted,
a Priority Search Tree of the points in P can be constructed in O(n) time.
2.
Windowing queries in sets of non-crossing segments are performed using range query on the
set of end-points and intersection queries with the four boundary edges of the query window.
Explain how to avoid reporting segments more than once. To this end, make a list of all possible
ways in which an arbitrary oriented segment can intersect a query window.
3.
Segment Trees can be used for multi-level data structures.
(a) Let R be a set of n axis-parallel rectangles in the plane. Design a data structure for R such
that the rectangles in R that contain a query point q can be reported efficiently. Analyze the
amount of storage and the query time of your data structure. [Hint: Use a segment tree on
the x-intervals of the rectangles, and store canonical subsets of the nodes in this segment
tree in an appropriate associated structure.]
(b) Generalize this data structure to d-dimensional space. Here we are given a set of axisparallel hyper-rectangles, i.e., polytopes of the form [x1 : x’1]  [x2 : x’2]  …  [xd : x’d] , and
we want to report the hyper-rectangles that contain a query point.
4.
Let I be a set of intervals on the real line. We want to store these intervals such that we can
efficiently determine
(a) those intervals that are completely contained in a given interval [x : x’]. Describe a data
structure that uses O(n log n) storage and solves such queries in O(K + log n) time,
where K is the output size. [Hint: Use a range tree.]
(b) The same question as in part (a), except that we want to report the intervals that contain
a query interval [x : x’].
5.
Consider the following alternative approach to solve the 2-dimensional range searching
problem: We construct a balanced search tree on the x-coordinate of the points. For a
node v in the tree, let P(v) be the set of points stored in the subtree rooted at v. For each
node v we store two associated priority search trees of P(v), a tree T left allowing for range
queries that are unbounded to the left, and a tree T right for range queries that are unbounded
to the right.
A query with a range [x : x’]  [y : y’] is performed as follows. We search for the node vsplit
where the search paths toward x and x’ split in the tree. Now we perform a query with range
[x : +  )  [y : y’] on T right(lc(v)) and a query with range (- : x’]  [y : y’] on T left(rc(v)).
This gives all the answers (there is no need to search further down in the tree!).
(a) Work out the details of this approach.
(b) Prove that the data structure correctly solves range queries.
(c) What are the bounds for preprocessing time, storage space, and query time of this
structure? Prove your answers.
END