btree applications

CSE 2331

Search Tree Applications

7.1

CSE 2331

Line Segment Intersection: Detection

7.2

CSE 2331

Horizontal-Vertical Line Segment Intersection

Given:

• set LH of horizontal line segments;

• set LV of vertical line segments.

Does any line segment in LV intersect a line segment in LH?

7.3

CSE 2331


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7.4

CSE 2331


Brute force algorithm:

1 foreach vertical line segment σV ∈ LV do

2 foreach horizontal line segment σH ∈ LH do

3 if σH ∩ σV 6= ∅ then

4 q ← σH ∩ σV ;

5 return (q);

6 end

7 end

8 end

9 return (“No intersection”);

7.5

CSE 2331

Sort Horizontal Lines

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7.6

CSE 2331

Sort Horizontal Lines

1 LV ← vertical line segments;

2 LH ← horizontal line segments sorted by increasing y;


4 σH ← first line segment in LH with σH .y ≥ σV .miny;

5 while (σH 6= NIL) and (σH .y ≤ σv.maxy) do


7 q ← σH ∩ σV ;

8 return (q);

9 end

/* Get next horizontal segment in LH */

10 σH ← σH .Next;

11 end

12 end


7.7

CSE 2331

Line Sweep Algorithm

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7.8

CSE 2331


1 LV ← vertical line segments sorted by increasing x;

2 LH ← horizontal line segments sorted by increasing y;


4 σH ← first line segment in LH with σH .y ≥ σV .miny;

5 while (σH 6= NIL) and (σH .y ≤ σv.maxy) do


7 q ← σH ∩ σV ;

8 return (q);

9 end

/* Get next horizontal segment in LH */

10 σH ← σH .Next;

11 end

12 end


7.9

CSE 2331


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7.10

CSE 2331

Tree Minimum Greater Than or Equal to

Find minimum key greater than or equal to 12.

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7.11

CSE 2331

Tree Minimum Greater Than or Equal to

function TreeMinGE(T , K)

/* Return node with min key greater than or equal to K */

1 u← NIL;

2 v ← T.root;

3 while (v 6= NIL) do4 if (K ≤ v.key) then5 u← v;

6 v ← v.left;

7 else8 v ← v.right;

9 end

10 end

11 return (u);

7.12

CSE 2331


1 T.Init();

2 L← endpoints of all line segments sorted by x-coordinate;

3 foreach point p ∈ L do4 if (p is the left endpoint of a horizontal segment σH) then5 Create new tree node u with

6 u.key = σH .y and u.data.lineseg = σH ;

7 TreeInsert (T , u);

8 else if (p is the lower endpoint of a vertical segment σV ) then9 u← TreeMinGE (T , p.y);

10 if (u 6= NIL) and (u.y ≤ σV .maxy) then

11 return (u.data.lineseg ∩ σV );

12 else if (p is the right endpoint of a horizontal segment σH) then13 u← tree node representing σH ;

14 TreeDelete (T ,u);

15 end

16 end


7.13

CSE 2331

Line Segment Intersection: Reporting

7.14

CSE 2331

Report All HV Line Segment Intersections

Given:



Report all the intersections of line segments in LH and LV .

7.15

CSE 2331


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7.16

CSE 2331


Brute force algorithm:

1 foreach horizontal line segment σH ∈ LH do



4 q ← σH ∩ σV ;

5 Report intersection point q;

6 end

7 end

8 end

7.17

CSE 2331

Running Times of Reporting

Let I = number of intersections.

Desired running time is Θ(f(n) + I) where f(n) is small

(e.g., f(n) ∈ Θ(n log2(n)).)

7.18

CSE 2331

Report: Line Sweep Algorithm

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7.19

CSE 2331


1 T.Init();




7 u.data.lineseg← σH ;


9 else if (p is the lower endpoint of a vertical segment σV ) then10 Report line segments in T intersecting σV ;



14 end

15 end

7.20

CSE 2331

Report in Range

Report all nodes of the following tree with keys in range [8, 25]:(Range [8, 25] includes 8 and 25.)

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7.21

CSE 2331

Report in Range: Running Time

procedure TreeRangeReport(x, kmin, kmax)

1 if (x 6= NIL) then

2 if (kmin ≤ x.key) then

3 TreeRangeReport(x.left, kmin, kmax);

4 end

5 if (kmin ≤ x.key ≤ kmax) then print x.key;

6 if (x.key ≤ kmax) then

7 TreeRangeReport(x.right, kmin, kmax);

8 end

9 end

h = tree heightI = number of nodes reported.

Running time:

7.22

CSE 2331


1 T.Init();






9 else if (p is the lower endpoint of a vertical segment σV ) then10 TreeRangeReport (T.root, σV .miny, σV .maxy);



14 end

15 end

7.23

CSE 2331

Line Segment Intersection: Counting

7.24

CSE 2331

Count HV Line Segment Intersections

Given:



For each σV ∈ LV , count number of intersections with lines in LH .

7.25

CSE 2331


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7.26

CSE 2331


1 count← 0;

2 T.Init();


4 foreach point p ∈ L do5 if (p is the left endpoint of a horizontal segment σH) then6 Create new tree node u;

7 u.key← σH .y;



10 else if (p is the lower endpoint of a vertical segment σV ) then11 count← count+ CountNumIntersections (T , σV );



15 end

16 end

7.27

CSE 2331

Binary Search Tree: Counting

For each node u in the binary search tree, add a field u.size which

stores the number of nodes in the subtree rooted at u.

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7.28

CSE 2331

Count Nodes Greater Than or Equal to

function TreeCountGE(x, K)

/* Return number of nodes with keys greater than or equalto K in subtree rooted at x */

1 count← 0;

2 v ← x;

3 while (v 6= NIL) do4 if (v.key ≥ K) then5 count← count + 1;

6 if (v.right 6= NIL) then count← count + v.right.size;

7 v ← v.left;

8 else9 v ← v.right;

10 end

11 end

12 return (count);

7.29

CSE 2331

Count Number of Nodes in Range

function TreeRangeCount(x, kmin, kmax)

/* Return number of nodes with keys in range [kmin, kmax] in

subtree rooted at x */

1 v ← x;

2 while (v 6= NIL) and (v.key 6∈ [kmin, kmax]) do

3 if (v.key ≤ kmin) then v ← v.right;

4 else if (v.key ≥ kmax) then v ← v.left;

5 end

6 if (v 6= NIL) then

7 countL ← TreeCountGE(v.left, kmin);

8 countR ← TreeCountLE(v.right, kmax);

9 return (1 + countL + countR);

10 else return (0);

7.30

CSE 2331


1 count← 0;

2 T.Init();


4 foreach point p ∈ L do5 if (p is the left endpoint of a horizontal segment σH) then6 Create new tree node u;

7 u.key← σH .y;



10 else if (p is the lower endpoint of a vertical segment σV ) then11 count← count+ TreeRangeCount (T , σV .miny, σV .maxy);



15 end

16 end

7.31

CSE 2331

Point Location

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7.32

CSE 2331

Point Location

Given:

• set Srect of non-intersecting, axis-parallel rectangles;

• set Spoint of points.

Report the rectangle containing each point.

7.33

CSE 2331

Data Structures: Applications

7.34

CSE 2331

Data Structures

• Stack;

• Heap (HeapSort, Prim’s MST algorithm, Dijkstra’s shortest

path algorithm);

• Red-black tree (Line segment intersection);

• Hash table (Duplicate elements, counting number of elements);

• Union-find structure (Connected components).

7.35

CSE 2331

Exercise: Deep Thought - Highest Priority

You are in charge of the supercomputer Deep Thought.

• Deep Thought has an “Input List” of programs which it needs

to process;

• Every minute new programs are being added to the Input List

by scientists from across the world;

• Each program has a priority number from 1 to 2128 − 1;

• Each time Deep Thought completes a program, it selects the

next program with highest priority from the Input List and

works on that program.

Give a data structure for quickly adding new programs to the list

and selecting the next program for Deep Thought to process.

7.36

CSE 2331

Exercise: Deep Thought - Count

Competitive scientists always want to know how many other programs in

the Input List have priority less than their program, and how many have

priority greater than their program.

• Every minute new programs are being added to the Input List by

scientists from across the world;

• Each program has a priority number from 1 to 2128 − 1;

• Each time Deep Thought completes a program, it selects the next

program with highest priority from the Input List and works on that

program.

Give a data structure which allows a scientist to quickly query how many

programs in the Input List have higher priority and how many programs

have lower priority.

7.37

CSE 2331

Exercise: Deep Thought - Wait

Because Deep Thought is such a valuable resource, some scientists

want to hog Deep Thought for their research. For this reason, you

instituted a rule that after a scientist has submitted a program, she

must wait 24 hours until she can submit another program.

• Each scientist has a 128 bit password which she must submit

with her program;

• Each program has a submission time which is the time the

program was added to the Input List.

Give a data structure for quickly determining if the scientist who

submitted a program has waited 24 hours from the last time she

submitted a program.

7.38

CSE 2331

Exercise: Deep Thought - Median Priority

Some selfish scientists have been assigning high priority to their programs

to get them processed first. To “punish” such selfishness, Deep Thought

no longer selects the program with highest priority from the Input List.

Instead, Deep Thought selects the program with median priority.

• Every minute new programs are being added to the Input List by

scientists from across the world;

• Each time Deep Thought completes a program, it selects the next

program with median priority from the Input List and works on that

program.

• Let n be the number of programs in the Input List. The median

priority is an integer I such that ⌊n/2⌋ elements have priority less

than or equal to I and ⌈n/2⌉ elements have priority greater than or

equal to I.

Give a data structure for quickly adding new programs to the list and

selecting the next program for Deep Thought to process.

7.39

btree applications

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