ONE DIMENSIONAL RANGE SEARCH - Daniel G. Graham PhD · 2019. 2. 13. · 1d range search: BST...

O N E D I M E N S I O N A L R A N G E S E A R C H

1d range search

Extension of ordered Binary Search Tree.

・Insert key-value pair.

・Search for key k.

・Delete key k.

・Range search: find all keys between k1 and k2.

・Range count: number of keys between k1 and k2.

Application. Database queries.

Geometric interpretation.

・Keys are point on a line.

・Find/count points in a given 1d interval.

insert B B

insert D B D

insert A A B D

insert I A B D I

insert H A B D H I

insert F A B D F H I

insert P A B D F H I P

search G to K H I

count G to K 2

1d range search: elementary implementations

Unordered list. Fast insert, slow range search.

Ordered array. Slow insert, binary search for k1 and k2 to do range search.

data structure insert range count range search

unordered list 1 N N

ordered array N log N R + log N

goal log N log N R + log N

order of growth of running time for 1d range search

N = number of keys

R = number of keys that match

L E T ’ S F O C U S O N R A N G E C O U N T I N G・Range count: number of keys between k1 and k2.

1d range count: BST implementation

Calculate the rank each node in the BST

・The Rank of a Node: is number of keys that are less than it

・Question: How do we determine the number of keys between k1 and k2

・Step1: Figure how many items less than k1

・Step2: Subtract the rank(k2) from rank(k1)

H O W W O U L D W E D O S E A R C H

・Range search: find all keys between k1 and k2.

1d range search. Find all keys between lo and hi.

・Recursively find all keys in left subtree (if any could fall in range).

・Check key in current node.

・Recursively find all keys in right subtree (if any could fall in range).

Proposition. Running time proportional to R + log N.Pf. Nodes examined = search path to lo + search path to hi + matches.

1d range search: BST implementation

black keys arein the range

red keys are used in comparesbut are not in the range

searching in the range [F..T]

Range search in a BST

・Recursively find all keys in right subtree (if any could fall in rang

Start at root S

Check E 1) E is not the range 2) is less than smallest key F 3)Only need to recurse right

Start at root S

Check R 1) R is range [F … T] 2) Recurse left

Start at root S

Check H 1) H is range [F … T] 2) Recurse left (null) 3) Recurse right

Same process for M, L, P

Start at root S

Check X 1) X is not range [F … T] 2) end recursion

L E T S E X T E N D T H I S I D E A T O T H E 2 D C A S E

Orthogonal line segment intersection

Given N horizontal and vertical line segments, find all intersections.

Quadratic algorithm. Check all pairs of line segments for intersection.

Nondegeneracy assumption. All x- and y-coordinates are distinct.

・Remove all the lines that touch with intersecting.

Sweep vertical line from left to right.

・x-coordinates define events.

・h-segment (left endpoint): insert y-coordinate into BST.

Orthogonal line segment intersection: sweep-line algorithm

y-coordinates

・h-segment (right endpoint): remove y-coordinate from BST.

y-coordinates

・h-segment (right endpoint): remove y-coordinate from BST.

・v-segment: range search for interval of y-endpoints.

1d rangesearch

y-coordinates

Orthogonal line segment intersection: sweep-line analysis

Proposition. The sweep-line algorithm takes time proportional to N log N + R

to find all R intersections among N orthogonal line segments.

・Put x-coordinates on a PQ (or sort).

・Insert y-coordinates into BST.

・Delete y-coordinates from BST.

・Range searches in BST.

Bottom line. Sweep line reduces 2d orthogonal line segment intersection

search to 1d range search.

N log N

N log N + R

L E T S E X T E N D T H E A L G O R I T H M T O 2 D S PA C E

2-d orthogonal range search

Extension to 2d keys.

・Insert a 2d key. (Points)

・Delete a 2d key.

・Search for a 2d key.

・Range search: find all keys that lie in a 2d range.

・Range count: number of keys that lie in a 2d range.

Geometric interpretation.

・Keys are point in the plane.

・Find/count points in a given rectangle

2d orthogonal range search: grid implementation

Grid implementation.

・Divide space into M-by-M grid of squares.

・Create list of points contained in each square.

・Use 2d array to directly index relevant square.

・Insert: add (x, y) to list for corresponding square.

・Range search: examine only squares that intersect 2d range query.

RT = Right Top LB = Left Bottom

・M-by-M grid

Cell ID 1 2 MxMPoint ID 5 6 7

2d orthogonal range search: grid implementation analysis

Choose grid square size to tune performance.

・Too small: wastes space.

– (Searching square that don’t contain anything)

・Too large: too many points per square.

– (Can’t distinguish between points)

H O W E V E R , P O I N T S A R E A LW AY S E V E N I N G D I S T R I B U T E D

Grid implementation. Fast, simple solution for evenly-distributed points.

Problem. Clustering a well-known phenomenon in geometric data.

Ex. USA map data.

Clustering

half the squares are empty half the points arein 10% of the squares

13,000 points, 1000 grid squares

Grid implementation. Fast, simple solution for evenly-distributed points.

Problem. Clustering a well-known phenomenon in geometric data.

・Need data structure that adapts gracefully to data.

Clustering

N E E D PA R T I T I O N I N G A P P R O A C H T H AT F I T S T H E D I S T R I B U T I O N O F T H E D ATA

Use a tree to represent a recursive subdivision of 2d space.

Grid. Divide space uniformly into squares.

2d tree. Recursively divide space into two halfplanes.

Space-partitioning trees

Grid 2d tree

C O N S T R U C T I N G A 2 D T R E E

Recursively partition plane into two halfplanes.

2d tree construction

Alternative between And at each level in the tree

Data structure. BST, but alternate using x- and y-coordinates as key.

・Search gives rectangle containing point.

・Insert further subdivides the plane.

2d tree implementation

even levelsof the tree

pointsleft of p

pointsright of p

pointsbelow q

pointsabove q

odd levelsof the tree

S O W H AT A B O U T R A N G E S E A R C H

Goal. Find all points in a query axis-aligned rectangle.

・Check if point in node lies in given rectangle.

・Recursively search left (if any points could fall in rectangle).

・Recursively search right (if any points could fall in rectangle).

2d tree demo: range search

・Recursively search left/bottom (if any could fall in rectangle).

・Recursively search right/top (if any could fall in rectangle).

search root nodecheck if query rectangle contains point 1

query rectangle to left of splitting linesearch only in left subtree

search left subtreecheck if query rectangle contains point 3

query rectangle intersects splitting linesearch bottom and top subtrees

(search hit)

query rectangle intersects splitting linesearch bottom and top subtrees

search bottom subtreestop since empty

search top subtreestop since empty

return from function call

search top subtreecheck if query rectangle contains point 6

search left subtreestop since empty

N E A R E S T N E I G H B O R

Goal. Find closest point to query point.

2d tree demo: nearest neighbor

query point

・Check distance from point in node to query point.

・Recursively search left/bottom (if it could contain a closer point).

・Recursively search right/top (if it could contain a closer point).

search root nodecompute distance from query point to 1

(update champion nearest neighbor)

query point is to the left of splitting linesearch left subtree first

search left subtreecompute distance from query point to 3

(update champion)

query point is above splitting linesearch top subtree first

search top subtreecompute distance from query point to 6

query point is to left of splitting linesearch left subtree first

search left subtreereturn since empty

search right subtreeprune since nearest neighbor

can't be here

return from function callsearch bottom subtree next

search bottom subtreecompute distance from query point to 4

query point is to left of splitting linesearch left subtree first

search left subtreecompute distance from query point to 5

(update champion)

query point is above splitting linesearch top subtree first

search top subtreereturn since empty

search bottom subtreereturn since empty

return from function callsearch right subtree next

can't be here(drawing not quite to scale)

Distancelarger

return from function callsearch right subtree next

can't be here

nearest neighbor = 5

Flocking birds

Q. What "natural algorithm" do starlings, migrating geese, starlings,

cranes, bait balls of fish, and flashing fireflies use to flock?

http://www.youtube.com/watch?v=XH-groCeKbE

Flocking boids [Craig Reynolds, 1986]

Boids. Three simple rules lead to complex emergent flocking behavior:

・Collision avoidance: point away from k nearest boids.

・Flock centering: point towards the center of mass of k nearest boids.

・Velocity matching: update velocity to the average of k nearest boids.

https://www.youtube.com/watch?v=SJyRkeq4Mgw

K - D T R E E K D I M E N S I O N S

Kd tree

Kd tree. Recursively partition k-dimensional space into 2 halfspaces.

・Just cycle through each dimension at each level of there tree

Implementation. BST, but cycle through dimensions ala 2d trees.

Efficient, simple data structure for processing k-dimensional data.

・Widely used.

・Adapts well to high-dimensional and clustered data.

level ≡ i (mod k)

points

whose ithcoordinate

is less than p’s

points

whose ithcoordinate

is greater than p’s

Jon Bentley

I N T E R VA L S E A R C H T R E E S

1d interval search. Data structure to hold set of (overlapping) intervals.

・Insert an interval ( lo, hi ).

・Search for an interval ( lo, hi ).

・Delete an interval ( lo, hi ).

・Interval intersection query: given an interval ( lo, hi ), find all intervals

(or one interval) in data structure that intersects ( lo, hi ). Q. Which intervals intersect ( 9, 16 ) ? A. ( 7, 10 ) and ( 15, 18 ).

1d interval search

(7, 10)

(5, 8)

(4, 8) (15, 18)

(17, 19)

(21, 24)

Create BST, where each node stores an interval ( lo, hi ).

・Use left endpoint as BST key.

・Store max endpoint in subtree rooted at node.

Interval search trees

binary search tree (left endpoint is key)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

max endpoint insubtree rooted at node

To insert an interval ( lo, hi ) :

・Insert into BST, using lo as the key.

・Update max in each node on search path.

Interval search tree demo: insertion

insert interval (16, 22)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

compare 16 and 17 (go left)

(16, 22)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

(16, 22)

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

(16, 22)

compare 16 and 5 (go right)

24(5, 8)

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

(16, 22)

24(5, 8)

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

(16, 22)

compare 16 and 15 (go right)

(15, 18)

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

(15, 18)

(16, 22)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

(16, 22) no more tree (insert here)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

no more tree (insert here)

(16, 22)(16, 22)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

update max in each node on search path

(16, 22)(16, 22)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

update max in each node on search path

22(16, 22)(16, 22)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

I N T E R VA L S E A R C H

Interval search tree demo: intersection

To search for any one interval that intersects query interval ( lo, hi ) :

・If interval in node intersects query interval, return it.

・Else if left subtree is null, go right.

・Else if max endpoint in left subtree is less than lo, go right.

・Else go left.

interval intersectionsearch for (23, 25)

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

compare (23, 25) to (17, 19) (no intersection)

(23, 25) (17, 19)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

compare 22 to 23 (no intersection in left, go right)

(23, 25) (17, 19)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(23, 25)

(17, 19)

・Else go left.

(17, 19)

(5, 8)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(23, 25)

compare (23, 25) to (21, 24) (intersection!)

(21, 24)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(12, 14)

(17, 19)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(12, 14)

compare 22 to 12 (go left)

(17, 19)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(12, 14)

(17, 19)

・Else go left.

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(12, 14)

(5, 8)

・Else go left.

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(12, 14)

compare 8 to 12 (go right)

(5, 8)

・Else go left.

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(12, 14)

(5, 8)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

22(16, 22)

(12, 14) compare (12, 14) to (15, 18) (no intersection)

(15, 18)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

22(16, 22)

(12, 14) compare 10 to 12 (go right)

(15, 18)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

22(16, 22) (12, 14)

(15, 18)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22 (12, 14)(16, 22)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22 (12, 14)

(16, 22)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22 (12, 14)

left subtree is null (go right)

(16, 22)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

node is null (no intersection)

(16, 22)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(21, 23)

(17, 19)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(21, 23)

compare 22 to 21 (go left)

(17, 19)

・Else go left.

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(21, 23)

(17, 19)

・Else go left.

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(21, 23)

(5, 8)

・Else go left.

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(21, 23)

compare 8 to 21 (go right)

(5, 8)

・Else go left.

(17, 19)

(21, 24)

(4, 8) (15, 18)

(7, 10)

22(16, 22)

(21, 23)

(5, 8)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

22(16, 22)

(21, 23) compare (21, 23) to (15, 18) (no intersection)

(15, 18)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

22(16, 22)

(21, 23) compare 10 to 21 (go right)

(15, 18)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8)

(7, 10)

22(16, 22) (21, 23)

(15, 18)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22 (21, 23)(16, 22)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22 (21, 23)

(16, 22)

・Else go left.

(17, 19)

(5, 8) (21, 24)

(4, 8) (15, 18)

(7, 10)

22 (21, 23)

(16, 22)

Orthogonal rectangle intersection

Goal. Find all intersections among a set of N orthogonal rectangles.

Quadratic algorithm. Check all pairs of rectangles for intersection.

Non-degeneracy assumption. All x- and y-coordinates are distinct.

I N T E R S E C T I O N R E C TA N G L E S

Microprocessors and geometry

Early 1970s. microprocessor design became a geometric problem.

・Very Large Scale Integration (VLSI).

・Computer-Aided Design (CAD).

Design-rule checking.

・Certain wires cannot intersect.

・Certain spacing needed between different types of wires.

・Debugging = orthogonal rectangle intersection search.

・x-coordinates of left and right endpoints define events.

・Maintain set of rectangles that intersect the sweep line in an interval

search tree (using y-intervals of rectangle).

・Left endpoint: interval search for y-interval of rectangle; insert y-interval.

・Right endpoint: remove y-interval.

Orthogonal rectangle intersection: sweep-line algorithm

y-coordinates

Orthogonal rectangle intersection: sweep-line analysis

Proposition. Sweep line algorithm takes time proportional to N log N + R log Nto find R intersections among a set of N rectangles.

・Put x-coordinates on a PQ (or sort).

・Insert y-intervals into ST.

・Delete y-intervals from ST.

・Interval searches for y-intervals.

Bottom line. Sweep line reduces 2d orthogonal rectangle intersection

search to 1d interval search.

N log N

N log N + R log N

problem example solution

1d range search BST

2d orthogonal linesegment intersection

sweep line reduces to 1d range search

kd range search kd tree

1d interval search interval search tree

2d orthogonalrectangle intersection

sweep line reduces to 1d interval search

Geometric applications of BSTs

http://algs4.cs.princeton.edu

Algorithms ROBERT SEDGEWICK | KEVIN WAYNE

ROBERT SEDGEWICK | KEVIN WAYNE

F O U R T H E D I T I O N

Algorithms

GEOMETRIC APPLICATIONS OF

‣ 1d range search‣ line segment intersection‣ kd trees‣ interval search trees‣ rectangle intersection

ONE DIMENSIONAL RANGE SEARCH - Daniel G. Graham PhD · 2019. 2. 13. · 1d range search: BST...

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