Line Segment Intersection Computational Geometry, WS 2007/08 Lecture 3 – Supplementary Prof. Dr....

Line Segment Intersection

Computational Geometry, WS 2007/08Lecture 3 – Supplementary

Prof. Dr. Thomas Ottmann

Algorithmen & Datenstrukturen, Institut für InformatikFakultät für Angewandte WissenschaftenAlbert-Ludwigs-Universität Freiburg

Computational Geometry, WS 2007/08Prof. Dr. Thomas Ottmann 2

Line Segment Intersection: Additions

• Implementation of the event queue by a heap:

The complexity of heap construction

• Iso-oriented line segments• Intersection counting• Line segment intersection by Divide & Conquer

• Polygon intersection


Heaps

Definition. A sequence F = (k1, k2, . . . , kn) of keys is a (Max-) Heap,

iff for all i {1, 2, …, n/2} : ki ≥ k2i und ki ≥ k2i+1.

This requirement for an array can be interpreted as a condition on a

complete binary tree:

• Level i has 2i nodes (except, eventually, the last level)• Node i has node 2i as ist left and node 2i + 1 as its right successor

andnode i/2 as ist predecessor (if they exist).

1 2 3 4 5 6 7

7 6 5 2 3 4 1

7

6 5

2 3 4 1

1

2 3

54 6 7


Heap Construction

Level, #nodes, #comp./exchanges

k levels, total # of nodes = 2k+1 -1


Heap Construction

Heap construction by iterative trickling down:0 20

1 21

2 2² k = 3 2k

nodes

Total number of nodes: n = 2k+1 - 1

Total number of key comparisons and exchanges:

.

)()1(*2

)22(*2

2*2)(2*2

1

1

0 1

nkn

k

jik

k

k

i

k

j

jki


Complexity of Heap Construction

Total complexity = 20 2k + 21 2(k-1) + 22 2(k-2) + … + 2(k-1) 2

= 2 S(k),

where S(k) = 20 k + 21(k-1) + 22(k-2) + … + 2(k-1)1

= 2 (n – k – 1) = O(n)


Intersections of iso-oriented Line Segments

• Report all pairs of intersecting segments• Count the number of intersecting segments

...........

......


Iso-oriented Line Segments

Report all pairs of intersecting segments:

A sweep-line algorithm can solve this problem in optimal time

O(n log n + k) and space O(n), k = # of intersections!


Iso-oriented Line Segments

Count all pairs of intersecting segments:

A sweep-line algorithm can solve this problem in optimal time

O(n log n) and space O(n)!


A Divide and Conquer Algorithm

Report all pairs of intersecting segments

...........

......


How to divide the input set

A separate representation of segments allows to divide theinput set.

A

BC

D

E

A.

B.

C.

D.

E.

.A.D

.B.C

.E


ReportCuts(S)

Input: Set S of vertical segments and endpoints of horizontal segments.

Output: All pairs of intersecting vertical segments and those horizontal segments which are represented by at least one endpoint in S.

1. Divide

if |S| > 1

then divide S by a vertical line G into two sets S1 (to the left of G) and S2 (to the right of G) of equal size

else S contains no intersections


ReportCuts

A

B

C

D

E

AD

BC

ES

S1 S2

1. Divide

2. Conquer

ReportCuts(S1); ReportCuts(S2)


ReportCuts

3. Merge: ???

Different possibilities for intersections of a horizontal segment in S1

Case 1: both endpoints in S1

h

S1 S2


ReportCuts

Case 2: only one endpoint of h in S1

2 a) right endpoint in S1

h

S1 S2


ReportCuts

2 b) left endpoint of h in S1

h

S1

right endpointin S2

h

S1

right endpoint not in S2

S2

S2


ReportCuts(S)

3. Merge:

Report intersections between vertical segments in S2 and those horizontal segments in S1, for which the

left endpoint is in S1 and the right endpoint is neither in S1 nor in S2

Analogously for S1

S1 S2


Implemention

Set S

L(S): y-coordinats of all left endpoints in S which have no right partner in S

R(S): y-coordinats of all right endpoints in S which have no left

partner in S

V(S): y-intervals of vertical segments in S


Base Cases

S contains only one element s

Case 1: s = (x,y) is left endpoint

L(S) = {y} R(S) = V(S) =

Case 2: s = (x,y) is right endpoint

L (S)= R(S) = {y} V(S) =

Case 3: s = (x,y1,y2) is vertical segment

L(S) = R(S) = V(S) = {[y1,y2]}


Merge Step

Assume L(Si), R(Si), V(Si) i=1,2 are known and S = S1 S2

L(S) =

R(S) =

V(S) =

L,R: sorted according to increasing y-coordinatslinked lists

V: sorted according to increasing lower endpoints linked list


Reporting of Intersections

V(S2)h3

h2

h1

L(S1)


Runtime

Input (vertical segments, left/right endpoints of horizontal segments)

has to be sorted initially and is stored in an array.

Divide-and-Conquer:

T(n) = 2T(n/2) + an + output-size

T(1) = O(1)

O(n log n + k) k = #intersections


Polygon Intersection by Line Sweep

A

B

CD

E

F

G



A

B

CD

E

F

G

Event-queue Q

Sweep direction



A

B

CD

E

F

G

Event-queue Q

•A•D

Status-structure T:


•G•EAD

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

Q1

ReportedIntersections1. Q1 (E--A)

Status-structure T:


GAED

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD

Q2

Q3

ReportedIntersections1. Q1 (E--A)2. Q2 (G--A)3. Q3 (E--D)

Q3

Status-structure T:


GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD


Q3

GAED

Status-structure T:


GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD


Q3

GAED (E• = •F)

Status-structure T:


GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD


Q3

GAED

Status-structure T:


GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD

ReportedIntersections1. Q1 (E--A)2. Q2 (G--A)3. Q3 (E--D)4. Q4 (G--B)

Q3

GAED

Status-structure T: AGBCDF

Q4

Q4


ABGCDF

AGBCDF

GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

Sweep direction

•A•D

•G•EAD

ReportedIntersections1. Q1 (E--A)2. Q2 (G--A)3. Q3 (E--D)4. Q4 (G--B)5. Q5 (G--C)

Q3

GAED

Status-structure T:

Q5

Q4Q5


ABGCDF

AGBCDF

GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD

ReportedIntersections1. Q1 (E--A)2. Q2 (G--A)3. Q3 (E--D)4. Q4 (G--B)5. Q5 (G--C)

Q3

GAED

Status-structure T:

Q4Q5

A•B•GCDF


ABGCDF

AGBCDF

GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD

ReportedIntersections1. Q1 (E--A)2. Q2 (G--A)3. Q3 (E--D)4. Q4 (G--B)5. Q5 (G--C)6. Q6 (G--D)

Q3

GAED

Status-structure T:

Q4Q5

A•B•GCDF

CGDF

Q6

Q6


CDGF

CGDF

ABGCDF

AGBCDF

GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD


Q3

GAED

Status-structure T:

Q4Q5

A•B•GCDF

Q6


CDGF

CGDF

ABGCDF

AGBCDF

GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD


Q3

GAED

Status-structure T:

Q4Q5

A•B•GCDF

Q6

CDG•F•


CDGF

CGDF

ABGCDF

AGBCDF

GA•B•CDF

GAD•F

GADE

Q2

Q1


A

B

CD

E

F

G

Event-queue Q

•A•D

•G•EAD


Q3

GAED

Status-structure T:

Q4Q5

A•B•GCDF

Q6

CDG•F•

C•D•

Line Segment Intersection Computational Geometry, WS 2007/08 Lecture 3 – Supplementary Prof. Dr....

Documents

Transcript of Line Segment Intersection Computational Geometry, WS 2007/08 Lecture 3 – Supplementary Prof. Dr....