introduction kd-trees range searching and kd-treeslaber/range_kd_trees.pdf · 2012-07-17 ·...

IntroductionKd-trees

Range searching and kd-trees

Computational Geometry

Lecture 7: Range searching and kd-trees

Computational Geometry Lecture 7: Range searching and kd-trees


Database queries1D range trees

Databases

Databases store records or objects

Personnel database: Each employee has a name, id code, dateof birth, function, salary, start date of employment, . . .

Fields are textual or numerical




Database queries

A database query may ask forall employees with agebetween a1 and a2, and salarybetween s1 and s2

date of birth

salary

19,500,000 19,559,999

G. Ometerborn: Aug 16, 1954salary: $3,500




Database queries

When we see numerical fields of objects as coordinates, adatabase stores a point set in higher dimensions

Exact match query: Asks for the objects whose coordinatesmatch query coordinates exactly

Partial match query: Same but not all coordinates arespecified

Range query: Asks for the objects whose coordinates lie in aspecified query range (interval)




Database queries

Example of a 3-dimensional(orthogonal) range query:children in [2 , 4], salary in[3000 , 4000], date of birth in[19,500,000 , 19,559,999]

19,500,000 19,559,999

3,000

4,000

2

4




Data structures

Idea of data structures

Representation of structure, for convenience (like DCEL)

Preprocessing of data, to be able to solve futurequestions really fast (sub-linear time)

A (search) data structure has a storage requirement, a querytime, and a construction time (and an update time)




1D range query problem

1D range query problem: Preprocess a set of n points onthe real line such that the ones inside a 1D query range(interval) can be reported fast

The points p1, . . . ,pn are known beforehand, the query [x,x′]only later

A solution to a query problem is a data structure description,a query algorithm, and a construction algorithm

Question: What are the most important factors for theefficiency of a solution?




Balanced binary search trees

A balanced binary search tree with the points in the leaves

3 10 19 23 30 37 59 62 70 80

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The search path for 25

3 10 19 23 30 37 59 62 70 80

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30 59 70

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The search paths for 25 and for 90

3 10 19 23 30 37 59 62 70 80

893 19

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30 59 70

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49




Example 1D range query

A 1-dimensional range query with [25, 90]

3 10 19 23 30 37 59 62 70 80

893 19

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30 59 70

62

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37

49




Node types for a query

Three types of nodes for a given query:

White nodes: never visited by the query

Grey nodes: visited by the query, unclear if they lead tooutput

Black nodes: visited by the query, whole subtree isoutput

Question: What query time do we hope for?




Node types for a query

The query algorithm comes down to what we do at each typeof node

Grey nodes: use query range to decide how to proceed: tonot visit a subtree (pruning), to report a complete subtree, orjust continue

Black nodes: traverse and enumerate all points in the leaves






3 10 19 23 30 37 59 62 70 80

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split node




1D range query algorithm

Algorithm 1DRangeQuery(T, [x : x′])1. νsplit ←FindSplitNode(T,x,x′)2. if νsplit is a leaf3. then Check if the point in νsplit must be reported.4. else ν ← lc(νsplit)5. while ν is not a leaf6. do if x≤ xν

7. then ReportSubtree(rc(ν))8. ν ← lc(ν)9. else ν ← rc(ν)10. Check if the point stored in ν must be reported.11. ν ← rc(νsplit)12. Similarly, follow the path to x′, and . . .




Query time analysis

The efficiency analysis is based on counting the numbers ofnodes visited for each type

White nodes: never visited by the query; no time spent

Grey nodes: visited by the query, unclear if they lead tooutput; time determines dependency on n

Black nodes: visited by the query, whole subtree isoutput; time determines dependency on k, the output size




Query time analysis

Grey nodes: they occur on only two paths in the tree, andsince the tree is balanced, its depth is O(logn)

Black nodes: a (sub)tree with m leaves has m−1 internalnodes; traversal visits O(m) nodes and finds m points for theoutput

The time spent at each node is O(1) ⇒ O(logn+ k) querytime




Storage requirement and preprocessing

A (balanced) binary search tree storing n points uses O(n)storage

A balanced binary search tree storing n points can be built inO(n) time after sorting, so in O(n logn) time overall(or by repeated insertion in O(n logn) time)




Result

Theorem: A set of n points on the real line can bepreprocessed in O(n logn) time into a data structure of O(n)size so that any 1D range query can be answered inO(logn+ k) time, where k is the number of answers reported




Example 1D range counting query

A 1-dimensional range tree for range counting queries

3 10 19 23 30 37 59 62 70 80

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1 1 1 1 1 1 1 1 1 1 1 1

112 22222

3 34 4

7 7

14




Example 1D range counting query

A 1-dimensional range counting query with [25, 90]

3 10 19 23 30 37 59 62 70 80

893 19

10

30 59 70

62

93

89

8023

49

93 97

37

49

1 1 1 1 1 1 1 1 1 1 1 1

112 22222

3 34 4

7 7

14




Result

Theorem: A set of n points on the real line can bepreprocessed in O(n logn) time into a data structure of O(n)size so that any 1D range counting query can be answered inO(logn) time

Note: The number of points does not influence the outputsize so it should not show up in the query time



Kd-treesQuerying in kd-treesKd-tree query time analysisHigher-dimensional kd-trees

Range queries in 2D




Range queries in 2D

Question: Why can’t we simply use a balanced binary tree inx-coordinate?

Or, use one tree on x-coordinate and one on y-coordinate, andquery the one where we think querying is more efficient?




Kd-trees

Kd-trees, the idea: Split the point set alternatingly byx-coordinate and by y-coordinate

split by x-coordinate: split by a vertical line that has half thepoints left and half right

split by y-coordinate: split by a horizontal line that has halfthe points below and half above




Kd-trees

Kd-trees, the idea: Split the point set alternatingly byx-coordinate and by y-coordinate

split by x-coordinate: split by a vertical line that has half thepoints left or on, and half right

split by y-coordinate: split by a horizontal line that has halfthe points below or on, and half above




Kd-trees

p4

p1

p5

p3

p2

p7

p9

p10

p6

p8

ℓ1

ℓ2

ℓ3

ℓ4

ℓ5

ℓ6

ℓ7

ℓ8

ℓ9

p1 p2

ℓ8 p3 p4

ℓ4 ℓ5

p5

p6 p7

p8 p9 p10

ℓ7

ℓ3

ℓ1

ℓ6

ℓ9

ℓ2




Kd-tree construction

Algorithm BuildKdTree(P,depth)

1. if P contains only one point2. then return a leaf storing this point3. else if depth is even4. then Split P with a vertical line ℓ through the

median x-coordinate into P1 (left of oron ℓ) and P2 (right of ℓ)

5. else Split P with a horizontal line ℓ throughthe median y-coordinate into P1 (belowor on ℓ) and P2 (above ℓ)

6. νleft ← BuildKdTree(P1,depth+1)

7. νright ← BuildKdTree(P2,depth+1)

8. Create a node ν storing ℓ, make νleft the leftchild of ν , and make νright the right child of ν .

9. return ν




Kd-tree construction

The median of a set of n values can be computed in O(n)time (randomized: easy; worst case: much harder)

Let T(n) be the time needed to build a kd-tree on n points

T(1) = O(1)

T(n) = 2 ·T(n/2)+O(n)

A kd-tree can be built in O(n logn) time

Question: What is the storage requirement?




Kd-tree regions of nodes

ℓ1

ℓ2

ℓ3region(ν)

ν

ℓ1

ℓ2

ℓ3




Kd-tree regions of nodes

How do we know region(ν) when we are at a node ν?

Option 1: store it explicitly with every node

Option 2: compute it on-the-fly, when going fromthe root to ν

Question: What are reasons to choose one or the otheroption?




Kd-tree querying

p1 p2

p2

p1

p3

p3 p4

p4

p5

p5

p6p6

p7

p7 p8

p8

p9

p9

p10

p10

p11

p11

p12

p12 p13

p13




Kd-tree querying

Algorithm SearchKdTree(ν ,R)

Input. The root of (a subtree of) a kd-tree, and a range R

Output. All points at leaves below ν that lie in the range.1. if ν is a leaf2. then Report the point stored at ν if it lies in R

3. else if region(lc(ν)) is fully contained in R

4. then ReportSubtree(lc(ν))5. else if region(lc(ν)) intersects R

6. then SearchKdTree(lc(ν),R)

7. if region(rc(ν)) is fully contained in R

8. then ReportSubtree(rc(ν))9. else if region(rc(ν)) intersects R

10. then SearchKdTree(rc(ν),R)




Kd-tree querying

Question: How about a range counting query?How should the code be adapted?




Kd-tree query time analysis

To analyze the query time of kd-trees, we use the concept ofwhite, grey, and black nodes

White nodes: never visited by the query; no time spent

Grey nodes: visited by the query, unclear if they lead tooutput; time determines dependency on n

Black nodes: visited by the query, whole subtree isoutput; time determines dependency on k, the output size





p1 p2

p2

p1

p3

p3 p4

p4

p5

p5

p6p6

p7

p7 p8

p8

p9

p9

p10

p10

p11

p11

p12

p12 p13

p13





White, grey, and black nodes with respect to region(ν):

White node ν: R does not intersect region(ν)

Grey node ν: R intersects region(ν), but region(ν) 6⊆ R

Black node ν: region(ν)⊆ R





Question: How many grey and how many black leaves?Computational Geometry Lecture 7: Range searching and kd-trees




Question: How many grey and how many black nodes?Computational Geometry Lecture 7: Range searching and kd-trees




Grey node ν : R intersects region(ν), but region(ν) 6⊆ R

It implies that the boundaries of R and region(ν) intersect

Advice: If you don’t know what to do, simplify until you do

Instead of taking the boundary of R, let’s analyze the numberof grey nodes if the query is with a vertical line ℓ





Question: How many grey and how many black leaves?Computational Geometry Lecture 7: Range searching and kd-trees




We observe: At every vertical split, ℓ is only to one side, whileat every horizontal split ℓ is to both sides

Let G(n) be the number of grey nodes in a kd-tree with n

points (leaves). Then G(1) = 1 and:

If a subtree has n leaves: G(n) = 1+G(n/2) at even depth

If a subtree has n leaves: G(n) = 1+2 ·G(n/2) at odd depth

If we use two levels at once, we get:

G(n) = 2+2 ·G(n/4) or G(n) = 3+2 ·G(n/4)





x

y y

y

x x

n leaves n leaves





G(1) = 1

G(n) = 2 ·G(n/4)+O(1)

Question: What does this recurrence solve to?





The grey subtree has unary and binary nodes





The depth is logn, so the binary depth is 12· logn

Important: The logarithm is base-2

Counting only binary nodes, there are

212·logn = 2logn1/2

= n1/2 =√

n

Every unary grey node has a unique binary parent (except theroot), so there are at most twice as many unary nodes asbinary nodes, plus 1





The number of grey nodes if the query were a vertical lineis O(

√n)

The same is true if the query were a horizontal line

How about a query rectangle?





The number of grey nodes for a query rectangle is at mostthe number of grey nodes for two vertical and two horizontallines, so it is at most 4 ·O(

√n) = O(

√n) !

For black nodes, reporting a whole subtree with k leaves,takes O(k) time (there are k−1 internal black nodes)




Result

Theorem: A set of n points in the plane can be preprocessedin O(n logn) time into a data structure of O(n) size so thatany 2D range query can be answered in O(

√n+ k) time,

where k is the number of answers reported

For range counting queries, we need O(√

n) time




Efficiency

n logn√

n

4 2 216 4 464 6 8

256 8 161024 10 324096 12 64

1.000.000 20 1000




Higher dimensions

A 3-dimensional kd-tree alternates splits on x-, y-, andz-coordinate

A 3D range query is performed with a box




Higher dimensions

The construction of a 3D kd-tree is a trivial adaptation of the2D version

The 3D range query algorithm is exactly the same as the 2Dversion

The 3D kd-tree still requires O(n) storage if it stores n points




Higher dimensions

How does the query time analysis change?

Intersection of B and region(ν) depends on intersection offacets of B ⇒ analyze by axes-parallel planes (B has no moregrey nodes than six planes)




Higher dimensions

m leaves

x

y

z

y

z zz





Let G3(n) be the number of grey nodes for a query with anaxes-parallel plane in a 3D kd-tree

G3(1) = 1

G3(n) = 4 ·G3(n/8)+O(1)

Question: What does this recurrence solve to?

Question: How many leaves does a perfectly balanced binarysearch tree with depth 2

3logn have?




Result

Theorem: A set of n points in d-space can be preprocessed inO(n logn) time into a data structure of O(n) size so that anyd-dimensional range query can be answered in O(n1−1/d + k)time, where k is the number of answers reported


Introduction2D Range trees

Degenerate cases

Range trees

Computational Geometry

Lecture 8: Range trees

Computational Geometry Lecture 8: Range trees


Degenerate casesRange queries

Database queries

A database query may ask forall employees with agebetween a1 and a2, and salarybetween s1 and s2

date of birth

salary

19,500,000 19,559,999

G. Ometerborn: Aug 16, 1954salary: $3,500




Result

Theorem: A set of n points on the real line can bepreprocessed in O(n logn) time into a data structure of O(n)size so that any 1D range [counting] query can be answered inO(logn [+k]) time




Result

Theorem: A set of n points in the plane can be preprocessedin O(n logn) time into a data structure of O(n) size so thatany 2D range query can be answered in O(

√n+ k) time,

where k is the number of answers reported

For range counting queries, we need O(√

n) time




Faster queries

Can we achieve O(logn [+k]) query time?




Faster queries

If the corners of the query rectangle fall in specific cells of thegrid, the answer is fixed (even for lower left and upper rightcorner)




Faster queries

Build a tree so that the leaves correspond to the differentpossible query rectangle types (corners in same cells of grid),and with each leaf, store all answers (points) [or: the count]

Build a tree on the different x-coordinates (to search with leftside of R), in each of the leaves, build a tree on the differentx-coordinates (to search with the right side of R), in each ofthe leaves, . . .




Faster queries

n

n

n

n




Faster queries

Question: What are the storage requirements of thisstructure, and what is the query time?




Faster queries

Recall the 1D range tree and range query:

Two search paths (grey nodes)

Subtrees in between have answers exclusively (black)






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split node




Examining 1D range queries

Observation: Ignoring the search path leaves, all answers arejointly represented by the highest nodes strictly between thetwo search paths

Question: How many highest nodes between the searchpaths can there be?




Examining 1D range queries

For any 1D range query, we can identify O(logn) nodes thattogether represent all answers to a 1D range query




Toward 2D range queries

For any 2d range query, we can identify O(logn) nodes thattogether represent all points that have a correct firstcoordinate





(3, 8)(1, 5) (4, 2) (5, 9) (6, 7) (8, 1)(7, 3) (9, 4)





(3, 8)(1, 5) (4, 2) (5, 9) (6, 7) (8, 1)(7, 3) (9, 4)

data structurefor searching ony-coordinate





(3, 8)(1, 5) (4, 2) (5, 9) (6, 7) (8, 1)(7, 3) (9, 4)

(3, 8)

(1, 5)

(4, 2)

(5, 9)

(6, 7)

(8, 1)

(7, 3)

(9, 4)



Degenerate cases

ConstructionQueryingHigher dimensionsFractional cascading

2D range trees

Every internal node stores a whole tree in an associated

structure, on y-coordinate

Question: How much storage does this take?



Degenerate cases


Storage of 2D range trees

To analyze storage, two arguments can be used:

By level: On each level, any point is stored exactly once.So all associated trees on one level together have O(n)size

By point: For any point, it is stored in the associatedstructures of its search path. So it is stored in O(logn) ofthem



Degenerate cases


Construction algorithm

Algorithm Build2DRangeTree(P)

1. Construct the associated structure: Build a binary searchtree Tassoc on the set Py of y-coordinates in P

2. if P contains only one point3. then Create a leaf ν storing this point, and make

Tassoc the associated structure of ν .4. else Split P into Pleft and Pright, the subsets ≤ and >

the median x-coordinate xmid

5. νleft ← Build2DRangeTree(Pleft)

6. νright ← Build2DRangeTree(Pright)

7. Create a node ν storing xmid, make νleft the leftchild of ν , make νright the right child of ν , andmake Tassoc the associated structure of ν

8. return ν



Degenerate cases


Efficiency of construction

The construction algorithm takes O(n log2 n) time

T(1) = O(1)

T(n) = 2 ·T(n/2)+O(n logn)

which solves to O(n log2 n) time



Degenerate cases



Suppose we pre-sort P on y-coordinate, and whenever we splitP into Pleft and Pright, we keep the y-order in both subsets

For a sorted set, the associated structure can be built in lineartime



Degenerate cases



The adapted construction algorithm takes O(n logn) time

T(1) = O(1)

T(n) = 2 ·T(n/2)+O(n)

which solves to O(n logn) time



Degenerate cases


2D range queries

How are queries performed and why are they correct?

Are we sure that each answer is found?

Are we sure that the same point is found only once?



Degenerate cases


2D range queries

ν

µ µ′ p

p

p

p



Degenerate cases


Query algorithm

Algorithm 2DRangeQuery(T, [x : x′]× [y : y′])1. νsplit ←FindSplitNode(T,x,x′)2. if νsplit is a leaf3. then report the point stored at νsplit, if an answer4. else ν ← lc(νsplit)5. while ν is not a leaf6. do if x≤ xν

7. then 1DRangeQ(Tassoc(rc(ν)), [y : y′])8. ν ← lc(ν)9. else ν ← rc(ν)10. Check if the point stored at ν must be reported.11. Similarly, follow the path from rc(νsplit) to x′ . . .



Degenerate cases


2D range query time

Question: How much time does a 2D range query take?

Subquestions: In how many associated structures do wesearch? How much time does each such search take?



Degenerate cases


2D range queries

ν

µ µ′



Degenerate cases


2D range query efficiency

We search in O(logn) associated structures to perform a 1Drange query; at most two per level of the main tree

The query time is O(logn)×O(logm+ k′), or

∑ν

O(lognν + kν)

where ∑kν = k the number of points reported



Degenerate cases



Use the concept of grey and black nodes again:



Degenerate cases



The number of grey nodes is O(log2 n)

The number of black nodes is O(k) if k points are reported

The query time is O(log2 n+ k), where k is the size of theoutput



Degenerate cases


Result

Theorem: A set of n points in the plane can be preprocessedin O(n logn) time into a data structure of O(n logn) size sothat any 2D range query can be answered in O(log2 n+ k)time, where k is the number of answers reported

Recall that a kd-tree has O(n) size and answers queries inO(√

n+ k) time



Degenerate cases


Efficiency

n logn log2 n√

n

16 4 16 464 6 36 8

256 8 64 161024 10 100 324096 12 144 64

16384 14 196 12865536 16 256 256

1M 20 400 1K16M 24 576 4K



Degenerate cases



Question: How about range counting queries?



Degenerate cases


Higher dimensional range trees

A d-dimensional range tree hasa main tree which is aone-dimensional balancedbinary search tree on the firstcoordinate, where every nodehas a pointer to an associatedstructure that is a(d−1)-dimensional range treeon the other coordinates



Degenerate cases


Storage

The size Sd(n) of a d-dimensional range tree satisfies:

S1(n) = O(n) for all n

Sd(1) = O(1) for all d

Sd(n)≤ 2 ·Sd(n/2)+Sd−1(n) for d ≥ 2

This solves to Sd(n) = O(n logd−1 n)



Degenerate cases


Query time

The number of grey nodes Gd(n) satisfies:

G1(n) = O(logn) for all n

Gd(1) = O(1) for all d

Gd(n)≤ 2 · logn+2 · logn ·Gd−1(n) for d ≥ 2

This solves to Gd(n) = O(logd n)



Degenerate cases


Result

Theorem: A set of n points in d-dimensional space can bepreprocessed in O(n logd−1 n) time into a data structure ofO(n logd−1 n) size so that any d-dimensional range query canbe answered in O(logd n+ k) time, where k is the number ofanswers reported

Recall that a kd-tree has O(n) size and answers queries inO(n1−1/d + k) time



Degenerate cases


Comparison for d = 4

n logn log4 n n3/4

1024 10 10,000 18165,536 16 65,536 4096

1M 20 160,000 32,7681G 30 810,000 5,931,6411T 40 2,560,000 1G



Degenerate cases


Improving the query time

We can improve the query time of a 2D range tree fromO(log2 n) to O(logn) by a technique called fractionalcascading

This automatically lowers the query time in d dimensions toO(logd−1 n) time



Degenerate cases



The idea illustrated best by a different query problem:

Suppose that we have a collection of sets S1, . . . ,Sm, where|S1|= n and where Si+1 ⊆ Si

We want a data structure that can report for a querynumber x, the smallest value ≥ x in all sets S1, . . . ,Sm



Degenerate cases



1 2 3 5 8 13 21 34 55

1 3 5 8 13 21 34 55

1 3 13 34 55

3 34 55

21

S1

S2

S3

S4



Degenerate cases



Suppose that we have a collection of sets S1, . . . ,Sm, where|S1|= n and where Si+1 ⊆ Si

We want a data structure that can report for a querynumber x, the smallest value ≥ x in all sets S1, . . . ,Sm

This query problem can be solved in O(logn+m) time insteadof O(m · logn) time



Degenerate cases



Can we do something similar for m 1-dimensional rangequeries on m sets S1, . . . ,Sm?

We hope to get a query time of O(logn+m+ k) with k thetotal number of points reported



Degenerate cases



1 2 3 5 8 13 21 34 55

1 3 5 8 13 21 34 55

1 3 13 34 55

3 34 55

21

S1

S2

S3

S4



Degenerate cases



1 2 3 5 8 13 21 34 55

1 3 5 8 13 21 34 55

1 3 13 34 55

3 34 55

21

[6,35]

S1

S2

S3

S4



Degenerate cases


Fractional cascading

Now we do “the same” on the associated structures of a2-dimensional range tree

Note that in every associated structure, we search with thesame values y and y′

Replace all associated structures except for the one ofthe root by a linked list

For every list element (and leaf of the associatedstructure of the root), store two pointers to theappropriate list elements in the lists of the left child andof the right child



Degenerate cases





Degenerate cases



(2, 19)

(5, 80)

(7, 10)

(8, 37)

(12, 3)

(15, 99)

(17, 62) (21, 49)

(33, 30)

(41, 95)

(52, 23)

(58, 59)

(67, 89)

(93, 70)

2

5 7 8 12 15

17

21 33 41 52

58

67

17

8

155

7 12 21 41

33

52

58 67

932



Degenerate cases



3 99

10 19 37 80

30 4962

3 10 19 23 30 37 49 59 62 70 80 89 95 99

89705923 9530 49803 9962371910

3 9962 89705923 9530 49

897023 9510 3719 80

70892395371019

59

4980 3 3099



Degenerate cases



(2, 19)

(5, 80)

(7, 10)

(8, 37)

(12, 3)

(15, 99)

(17, 62) (21, 49)

(33, 30)

(41, 95)

(52, 23)

(58, 59)

(67, 89)

(93, 70)

2

5 7 8 12 15

17

21 33 41 52

58

67

17

8

155

7 12 21 41

33

52

58 67

932

[4, 58]× [19, 65]



Degenerate cases



3 99

10 19 37 80

30 4962

3 10 19 23 30 37 49 59 62 70 80 89 95 99

89705923 9530 49803 9962371910

3 9962 89705923 9530 49

897023 9510 3719 80

70892395371019

59

4980 3 3099



Degenerate cases



Instead of doing a 1D range query on the associated structureof some node ν , we find the leaf where the search to y wouldend in O(1) time via the direct pointer in the associatedstructure in the parent of ν

The number of grey nodes reduces to O(logn)



Degenerate cases


Result

Theorem: A set of n points in d-dimensional space can bepreprocessed in O(n logd−1 n) time into a data structure ofO(n logd−1 n) size so that any d-dimensional range query canbe answered in O(logd−1 n+ k) time, where k is the number ofanswers reported

Recall that a kd-tree has O(n) size and answers queries inO(n1−1/d + k) time



Degenerate cases

Degenerate cases

Both for kd-trees and for rangetrees we have to take care ofmultiple points with the samex- or y-coordinate



Degenerate cases

Degenerate cases

Treat a point p = (px,py) with two reals as coordinates as apoint with two composite numbers as coordinates

A composite number is a pair of reals, denoted (a|b)

We let (a|b)< (c|d) iff a < c or ( a = c and b < d ); thisdefines a total order on composite numbers



Degenerate cases

Degenerate cases

The point p = (px,py) becomes ((px|py) , (py|px)). Then notwo points have the same first or second coordinate

The median x-coordinate or y-coordinate is a compositenumber

The query range [x : x′]× [y : y′] becomes

[(x|−∞) : (x′|+∞)]× [(y|−∞) : (y′|+∞)]

We have (px,py) ∈ [x : x′]× [y : y′] iff

((px|py) , (py|px)) ∈ [(x|−∞) : (x′|+∞)]× [(y|−∞) : (y′|+∞)]


introduction kd-trees range searching and kd-treeslaber/range_kd_trees.pdf · 2012-07-17 ·...

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