Computational Geometry: Line Arrangements
Computational Geometry: Line Arrangements Luis Pe˜ naranda IMPA
January 21-25, 2013
Luis Pe˜ naranda | IMPA | January 21-25, 2013
1 / 33
Computational Geometry: Line Arrangements
Course Outline
Lectures 1. Line Segment Arrangements in the Plane 2. Robustness in Geometric Computations 3. Arrangements in CGAL
Luis Pe˜ naranda | IMPA | January 21-25, 2013
2 / 33
Computational Geometry: Line Arrangements
Bibliography Bibliography I
M. de Berg, M. van Kreveld, M. Overmars, O. Cheong. Computational Geometry: Algorithms and Applications. 3rd edition, Springer, 2008.
I
F. Preparata and M. Shamos. Computational Geometry: an Introduction. Springer, 1985.
I
H. Edelsbrunner. Algorithms in Combinatorial Geometry. Springer, 1987.
I
J. Goodman and J. O’Rourke, eds. Handbook of Discrete and Computational Geometry. 2nd edition, Chapman & Hall, 2004.
I
J.-D. Boissonnat and M. Teillaud, eds. Effective Computational Geometry for Curves and Surfaces. Springer, 2007.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
3 / 33
Computational Geometry: Line Arrangements
Today’s Lecture
Line Segment Arrangements in the Plane I
Line Segment Intersection
I
The DCEL
I
Overlays
I
Application: Boolean Operations
Luis Pe˜ naranda | IMPA | January 21-25, 2013
4 / 33
Computational Geometry: Line Arrangements
Introduction maps information I
cities
I
rivers
I
railroads
I
regions
I
...
Luis Pe˜ naranda | IMPA | January 21-25, 2013
5 / 33
Computational Geometry: Line Arrangements
Introduction maps information I
cities
I
rivers
I
railroads
I
regions
I
...
the data I
are related
I
are stored independently
I
provide information when overlapped
Luis Pe˜ naranda | IMPA | January 21-25, 2013
5 / 33
Computational Geometry: Line Arrangements
First Approach segment intersection I
a map is a set of n segments
I
find all their intersections
Luis Pe˜ naranda | IMPA | January 21-25, 2013
6 / 33
Computational Geometry: Line Arrangements
First Approach segment intersection I
a map is a set of n segments
I
find all their intersections
in other words I
compute the arrangement of the n segments
Luis Pe˜ naranda | IMPA | January 21-25, 2013
6 / 33
Computational Geometry: Line Arrangements
First Approach segment intersection I
a map is a set of n segments
I
find all their intersections
in other words I
compute the arrangement of the n segments
complexity I
brute force O(n2 )
I
in some cases, Ω(n2 )
I
can we do better?
Luis Pe˜ naranda | IMPA | January 21-25, 2013
6 / 33
Computational Geometry: Line Arrangements
Definitions 1 y-interval
Luis Pe˜ naranda | IMPA | January 21-25, 2013
7 / 33
Computational Geometry: Line Arrangements
Definitions 1 y-interval
sweep line an horizontal line that sweeps the plane, from top to bottom
Luis Pe˜ naranda | IMPA | January 21-25, 2013
7 / 33
Computational Geometry: Line Arrangements
More Definitions
status segments intersected by the sweep line at a given moment
Luis Pe˜ naranda | IMPA | January 21-25, 2013
8 / 33
Computational Geometry: Line Arrangements
More Definitions
status segments intersected by the sweep line at a given moment
event points the points on which an update of the status is required
Luis Pe˜ naranda | IMPA | January 21-25, 2013
8 / 33
Computational Geometry: Line Arrangements
Algorithm Outline Sweep-line Algorithm I
order segments intersecting the sweep from left to right
Luis Pe˜ naranda | IMPA | January 21-25, 2013
9 / 33
Computational Geometry: Line Arrangements
Algorithm Outline Sweep-line Algorithm I
order segments intersecting the sweep from left to right
I
test segments only when they are adjacent
Luis Pe˜ naranda | IMPA | January 21-25, 2013
9 / 33
Computational Geometry: Line Arrangements
Algorithm Outline Sweep-line Algorithm I
order segments intersecting the sweep from left to right
I
test segments only when they are adjacent
I
update the status at each move of the sweep line
Luis Pe˜ naranda | IMPA | January 21-25, 2013
9 / 33
Computational Geometry: Line Arrangements
Algorithm Outline Sweep-line Algorithm I
order segments intersecting the sweep from left to right
I
test segments only when they are adjacent
I
update the status at each move of the sweep line
Invariant All intersection points above the sweep line were computed correctly. Luis Pe˜ naranda | IMPA | January 21-25, 2013
9 / 33
Computational Geometry: Line Arrangements
Correctness Lemma 1 Let si and sj be two non-horizontal segments whose interiors intersect in a single point p, and assume there is no third segment passing through p. Then there is an event point above p where si and sj become adjacent and are tested for intersection.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
10 / 33
Computational Geometry: Line Arrangements
Correctness Lemma 1 Let si and sj be two non-horizontal segments whose interiors intersect in a single point p, and assume there is no third segment passing through p. Then there is an event point above p where si and sj become adjacent and are tested for intersection.
Proof
Luis Pe˜ naranda | IMPA | January 21-25, 2013
10 / 33
Computational Geometry: Line Arrangements
Data structures event queue Q I
operations: remove next and return; insert
I
implementation: balanced BST with event points, ordered w.r.t. ≺
I
all operations in O(log m)
I
sorting of event points
I
why not a heap?
Luis Pe˜ naranda | IMPA | January 21-25, 2013
symbolic perturbation
11 / 33
Computational Geometry: Line Arrangements
Data structures event queue Q I
operations: remove next and return; insert
I
implementation: balanced BST with event points, ordered w.r.t. ≺
I
all operations in O(log m)
I
sorting of event points
I
why not a heap?
symbolic perturbation
status structure T I
operations: insert; delete; find neighbor
I
implementation: balanced BST (because it must always be sorted), leaves are segments intersecting sweep line
I
all operations in O(log n)
Luis Pe˜ naranda | IMPA | January 21-25, 2013
11 / 33
Computational Geometry: Line Arrangements
Bentley-Ottman Algorithm #1
FindIntersections(S) Input: a set S of line segments in the plane Output: the set of intersection points among the segments in S, with for each intersection point the segments that contain it I
initialize an empty event queue Q and insert the segment points into it;
I
initialize an empty status structure T ;
I
while Q is not empty do determine the next event point p; HandleEventPoint(p);
Luis Pe˜ naranda | IMPA | January 21-25, 2013
12 / 33
Computational Geometry: Line Arrangements
Bentley-Ottman Algorithm #2 HandleEventPoint(p) I
U(p) := {segments whose upper endpoint is p} (these segments are stored with the event point p);
I
Search in T the set S(p) of all segments that contain p (they are adjacent in T ).
I
Let L(p) ⊂ S(p) be the set of segments whose lower endpoint is p;
I
C(p) := {segments that contain p in their interior};
I
if L(p) ∪ U(p) ∪ C(p) contains more than one segment then report p as an intersection, and L(p), U(p) and C(p);
I
Delete the segments in L(p) ∪ C(p) from T ;
I
Insert the segments in U(p) ∪ C(p) into T (the order of the segments in T should correspond to the order in which they are intersected by a sweep line just below p);
Luis Pe˜ naranda | IMPA | January 21-25, 2013
13 / 33
Computational Geometry: Line Arrangements
Bentley-Ottman Algorithm #3 HandleEventPoint(p) (continued) I
if U(p) ∪ C(p) = ∅ then Let sl and sr be the neighbors of p in T ; FindNewEvent(sl ,sr ,p); else Let s 0 be the leftmost segment of U(p) ∪ C(p) in T ; Let sl be the left neighbor of s 0 in T ; FindNewEvent(sl ,s 0 ,p); Let s 00 be the rightmost segment of U(p) ∪ C(p) in T ; Let sr be the right neighbor of s 00 in T ; FindNewEvent(s 00 ,sr ,p);
Luis Pe˜ naranda | IMPA | January 21-25, 2013
14 / 33
Computational Geometry: Line Arrangements
Bentley-Ottman Algorithm #4
FindNewEvent(sl ,sr ,p) I
if sl and sr intersect below the sweep line or (on it and to the right of the current event point p) and the intersection is not yet present as an event in Q then Insert the intersection point as an event into Q;
Luis Pe˜ naranda | IMPA | January 21-25, 2013
15 / 33
Computational Geometry: Line Arrangements
Performance (exercise) Lemma 2 FindIntersections computes all intersection points and the segments that contain it correctly.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
16 / 33
Computational Geometry: Line Arrangements
Performance (exercise) Lemma 2 FindIntersections computes all intersection points and the segments that contain it correctly.
Lemma 3 The running time of FindIntersections for a set S of n line segments in the plane is O(n log n + I log n), where I is the number of intersection points of segments in S.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
16 / 33
Computational Geometry: Line Arrangements
Performance (exercise) Lemma 2 FindIntersections computes all intersection points and the segments that contain it correctly.
Lemma 3 The running time of FindIntersections for a set S of n line segments in the plane is O(n log n + I log n), where I is the number of intersection points of segments in S.
Lemma 4 FindIntersections runs in O(n) space.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
16 / 33
Computational Geometry: Line Arrangements
DCEL: the Doubly-Connected Edge List Why? I
maps contain subdivisions in regions
I
DCEL is a representation of that
I
it stores incidence relations. . . and whatever you want
Luis Pe˜ naranda | IMPA | January 21-25, 2013
17 / 33
Computational Geometry: Line Arrangements
DCEL: the Doubly-Connected Edge List Why? I
maps contain subdivisions in regions
I
DCEL is a representation of that
I
it stores incidence relations. . . and whatever you want
DCEL permits I
point-location queries
I
visit edges around a vertex
Luis Pe˜ naranda | IMPA | January 21-25, 2013
17 / 33
Computational Geometry: Line Arrangements
DCEL: the Doubly-Connected Edge List Why? I
maps contain subdivisions in regions
I
DCEL is a representation of that
I
it stores incidence relations. . . and whatever you want
DCEL permits I
point-location queries
I
visit edges around a vertex
Complexity of a subdivision Sum of the number of vertices, edges and faces.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
17 / 33
Computational Geometry: Line Arrangements
DCEL Definitions
Half-edges I
twin half-edges
I
origin and destination
Luis Pe˜ naranda | IMPA | January 21-25, 2013
18 / 33
Computational Geometry: Line Arrangements
DCEL Definitions
Half-edges I
twin half-edges
I
origin and destination
Implementation I
vertex table
I
face table
I
half-edge table
Luis Pe˜ naranda | IMPA | January 21-25, 2013
18 / 33
Computational Geometry: Line Arrangements
Stored Data Vertex records I
coordinates of the vertex
I
pointer to an incident edge
Luis Pe˜ naranda | IMPA | January 21-25, 2013
19 / 33
Computational Geometry: Line Arrangements
Stored Data Vertex records I
coordinates of the vertex
I
pointer to an incident edge
Face records I
outer component
I
inner components
Luis Pe˜ naranda | IMPA | January 21-25, 2013
19 / 33
Computational Geometry: Line Arrangements
Stored Data Vertex records I
coordinates of the vertex
I
pointer to an incident edge
Face records I
outer component
I
inner components
Half-edge records I
origin
I
twin
I
incident face
I
previous and next
Luis Pe˜ naranda | IMPA | January 21-25, 2013
19 / 33
Computational Geometry: Line Arrangements
DCEL Example A simple arrangement
Luis Pe˜ naranda | IMPA | January 21-25, 2013
20 / 33
Computational Geometry: Line Arrangements
General Version of the DCEL
Variants I
enough to walk around
I
associate data with vertex, faces and half-edges
I
store less information
I
higher dimensions
Luis Pe˜ naranda | IMPA | January 21-25, 2013
21 / 33
Computational Geometry: Line Arrangements
Overlay of Subdivisions
Luis Pe˜ naranda | IMPA | January 21-25, 2013
22 / 33
Computational Geometry: Line Arrangements
Definition of Overlay Formal The overlay of two subdivisions S1 and S2 is the subdivision O(S1 , S2 ) such that there is a face f in O(S1 , S2 ) if and only if there are faces f1 in S1 and f2 in S2 such that f is a maximal connected subset of f1 ∩ f2 .
Luis Pe˜ naranda | IMPA | January 21-25, 2013
23 / 33
Computational Geometry: Line Arrangements
Definition of Overlay Formal The overlay of two subdivisions S1 and S2 is the subdivision O(S1 , S2 ) such that there is a face f in O(S1 , S2 ) if and only if there are faces f1 in S1 and f2 in S2 such that f is a maximal connected subset of f1 ∩ f2 .
Informal The overlay O(S1 , S2 ) is the subdivision of the plane induced by the edges from S1 and S2 .
Luis Pe˜ naranda | IMPA | January 21-25, 2013
23 / 33
Computational Geometry: Line Arrangements
Definition of Overlay Formal The overlay of two subdivisions S1 and S2 is the subdivision O(S1 , S2 ) such that there is a face f in O(S1 , S2 ) if and only if there are faces f1 in S1 and f2 in S2 such that f is a maximal connected subset of f1 ∩ f2 .
Informal The overlay O(S1 , S2 ) is the subdivision of the plane induced by the edges from S1 and S2 .
I
each face in O(S1 , S2 ) must be labelled with the faces which contain it
I
we can do it with the DCEL’s of S1 and S2
Luis Pe˜ naranda | IMPA | January 21-25, 2013
23 / 33
Computational Geometry: Line Arrangements
How?
Sweep first
Luis Pe˜ naranda | IMPA | January 21-25, 2013
24 / 33
Computational Geometry: Line Arrangements
And Handle intersections
Luis Pe˜ naranda | IMPA | January 21-25, 2013
25 / 33
Computational Geometry: Line Arrangements
And Handle intersections
What is left?
Luis Pe˜ naranda | IMPA | January 21-25, 2013
25 / 33
Computational Geometry: Line Arrangements
Overlay Faces Graph G I
one node for each boundary cycle (inner or outer)
I
one node for the imaginary outer boundary
I
one arc between two cycles iff one of them is the boundary of a hole and the other has a half-edgde immediatly to the left of the leftmost vertex of that hole cycle
I
no half edge to the left the unbounded face
Luis Pe˜ naranda | IMPA | January 21-25, 2013
link the node representing the cycle with
26 / 33
Computational Geometry: Line Arrangements
Overlay Faces Graph G I
one node for each boundary cycle (inner or outer)
I
one node for the imaginary outer boundary
I
one arc between two cycles iff one of them is the boundary of a hole and the other has a half-edgde immediatly to the left of the leftmost vertex of that hole cycle
I
no half edge to the left the unbounded face
link the node representing the cycle with
Lemma Each connected component of the graph G corresponds exactly to the set of cycles incident to one face.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
26 / 33
Computational Geometry: Line Arrangements
Example of Graph
Luis Pe˜ naranda | IMPA | January 21-25, 2013
27 / 33
Computational Geometry: Line Arrangements
Overlay Algorithm #1 MapOverlay(S1 ,S2 ) Input: subdivisions S1 and S2 , stored in DCEL’s Output: their overlay, stored in a DCEL D I
copy DCEL’s from S1 and S2 to a new DCEL D;
I
compute all intersections between edges from S1 and S2 with the plane sweep algorithm. In addition to the actions on T and Q, do the following: I I
Update D if the event involves edges of both S1 and S2 ; store the half-edge immediately to the left of the event point at the vertex in D representing it;
(now D is the DCEL for O(S1 , S2 ), except that the information about the faces has not been computed yet) I
determine the boundary cycles in O(S1 , S2 ) by traversing D;
Luis Pe˜ naranda | IMPA | January 21-25, 2013
28 / 33
Computational Geometry: Line Arrangements
Overlay Algorithm #2 MapOverlay(S1 ,S2 ) (continued) I
construct the graph G whose nodes correspond to boundary cycles and whose arcs connect each hole cycle to the cycle to the left of its leftmost vertex; and compute its connected components;
I
for each connected component in G do let C be the unique outer boundary cycle in the component; let f denote the face bounded by the cycle; create a face record for f; OuterComponent(f) := some half-edge of C; InnerComponents(f) := {pointers to one half-edge in each hole cycle in the component}; let the IncidentFace() pointers of all half-edges in the cycles point to the face record of f;
I
label each face of O(S1 , S2 ) with the names of the faces of S1 and S2 containing it;
Luis Pe˜ naranda | IMPA | January 21-25, 2013
29 / 33
Computational Geometry: Line Arrangements
Complexity of the overlay algorithm
Theorem Let S1 be a planar subdivision of complexity n1 , let S2 be a subdivision of complexity n2 , and let n = n1 + n2 . The overlay of S1 and S2 can be constructed in O(n log n + k log n) time, where k is the complexity of the overlay.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
30 / 33
Computational Geometry: Line Arrangements
Complexity of the overlay algorithm
Theorem Let S1 be a planar subdivision of complexity n1 , let S2 be a subdivision of complexity n2 , and let n = n1 + n2 . The overlay of S1 and S2 can be constructed in O(n log n + k log n) time, where k is the complexity of the overlay.
Proof Exercise
Luis Pe˜ naranda | IMPA | January 21-25, 2013
30 / 33
Computational Geometry: Line Arrangements
Application: Boolean Operations
Luis Pe˜ naranda | IMPA | January 21-25, 2013
31 / 33
Computational Geometry: Line Arrangements
Boolean Operations
I
a very important application
I
computed as overlays of polygonal regions (can contain holes)
How to compute them I
use the overlay algorithm, but. . .
I
intersection: extract only faces labeled with P1 and P2 ,
I
union: extract P1 or P2 , and
I
difference: extract faces labeled with P1 and not with P2
Luis Pe˜ naranda | IMPA | January 21-25, 2013
32 / 33
Computational Geometry: Line Arrangements
Polygon Operations
Corollary 7 Let P1 be a polygon with n1 vertices and P2 a polygon with n2 vertices, and let n = n1 + n2 . Then P1 ∩ P2 , P1 ∪ P2 and P1 \ P2 can each be computed in O(n log n + k log n) time, where k is the complexity of the output.
Luis Pe˜ naranda | IMPA | January 21-25, 2013
33 / 33
