Lecture 5: Computational Geometry

Transcription

1 December 9, 2015

2 What is computational geometry? Study of algorithms for geometric problem solving. Typical problems Given a description of a set of geometric objects, e.g., set of points/segments/vertices of a polygon in a certain order. Answer a query about this set, e.g.: 1 do some segments intersect? 2 what is the convex hull of the set of points?... In this lecture, we assume objects represented by a set/sequence of n points p 0, p 1,..., p n 1 where each p i = (x i, y i ) R 2

3 Properties of line segments ASSUMPTION: p1 = (x 1, y 1 ) and p 2 = (x 2, y 2 ) are distinct points A convex combination of p 1 and p 2 is any point p 3 = (x 3, y 3 ) such that { x3 = α x 1 + (1 α) x 2 y 3 = α y 1 + (1 α) y 2 where 0 α 1. Intuition: p 3 is any point that is on the line passing through p 1 and p 2, which is on or between p 1 and p 2. The line segment p 1 p 2 delimited by points p1 = (x 1, y 1 ) and p 2 = (x 2, y 2 ) is the set of convex combinations between p 1 and p 2 The directed segment (or vector) p #» 1 p 2 imposes an ordering on its endpoints: p 1 is its origin, and p 2 its destination

4 Cross product Assumption: We identify p i = (x i, y i ) with the vector op #» i where o = (0, 0) is the origin of the system of coordinates. ( ) x1 x p 1 p 2 = det 2 = x y 1 y 1 y 2 x 2 y 1 = p 2 p 1 2 Geometric interpretation: (a) p 1 p 2 is the signed area of the parallelogram. (b) The lightly shaded region contains vectors that are clockwise from p. The darkly shaded region contains vectors that are counterclockwise from p.

5 Vectors and their relative positions Given three distinct points p i = (x i, y i ), 0 i 2 Determine if p #» 0 p 1 is closer to p #» 0 p 2 in clockwise or counterclockwise direction w.r.t. their common endpoint p 0. p 2 p 2 p 0 p 1 p 1 (0,0)

6 Vectors and their relative positions Given three distinct points p i = (x i, y i ), 0 i 2 Determine if p #» 0 p 1 is closer to p #» 0 p 2 in clockwise or counterclockwise direction w.r.t. their common endpoint p 0. p 2 p 2 p 0 p 1 p 1 (0,0) SOLUTION: shift vectors p #» 0 p 1 and p #» 0 p 2 to have origin in (0, 0) p 1 = (x 1 x 0, y 1 y 0 ), p 2 = (x 2 x 0, y 2 y 0 ) compute p 1 #» p 2 ; if positive, then p 0 p 1 is clockwise from p #» 0 p 2 ; if negative, it is counterclockwise.

7 Cross product Using cross product to determine how consecutive line segments p #» 0 p 1 and p #» 0 p 2 turn at point p 1 :

8 Cross product Using cross product to determine how consecutive line segments p #» 0 p 1 and p #» 0 p 2 turn at point p 1 : Check turn of #» p 0 p 2 w.r.t. #» p 0 p 1 : if counterclockwise, the points make a left turn. if clockwise, the points make a right turn. (p 1 p 0 ) (p 2 p 0 ) = (x 1 x 0 ) (y 2 y 0 ) (x 2 x 0 ) (y 1 y 0 ) If positive: #» p 0 p 1 is clockwise from #» p 0 p 2 If negative: #» p 0 p 1 is counterclockwise from #» p 0 p 2

9 Segment intersection test ASSUMPTION: p i = (x i, y i ) are four distinct points, 1 i 4. Question: Do segments p 1 p 2 and p 3 p 4 intersect or not? REMARK: p 1 p 2 and p 3 p 4 intersect if either (or both) of the following conditions hold: 1 p 1 and p 2 are on different sides of the line p 3 p 4 ; and p 3 and p 3 are on different sides of the line p 1 p 2, 2 an endpoint of one segment lies on the other segment (this condition comes from the boundary case).

10 Segment intersection test Pseudocode /* check if p 1 p 2 p 3 p 4 */ SegmentsIntersect(p 1, p 2, p 3, p 4 ) d 1 = Direction(p 3, p 4, p 1 ) d 2 = Direction(p 3, p 4, p 2 ) d 3 = Direction(p 1, p 2, p 3 ) d 4 = Direction(p 1, p 2, p 4 ) if ((d 1 < 0 d 2 > 0) (d 1 > 0 d 2 < 0)) ((d 3 < 0 d 4 > 0) (d 3 > 0 d 4 < 0)) return TRUE elseif d 1 == 0 OnSegment(p 3, p 4, p 1 ) return TRUE elseif d 2 == 0 OnSegment(p 3, p 4, p 2 ) return TRUE elseif d 3 == 0 OnSegment(p 1, p 2, p 3 ) return TRUE elseif d 4 == 0 OnSegment(p 1, p 2, p 4 ) return TRUE else return FALSE

11 Segment intersection test Auxiliary methods Direction and OnSegment Direction(p i, p j, p k ) return (p k p i ) (p j p i ) /* check if p k is between p i and p j */ OnSegment(p i, p j, p k ) if min(x i, x j ) x k max(x i, x j ) and min(y i, y j ) y k max(y i, y j ) return TRUE else return FALSE

12 Determine if any pair of segments in a set intersects Given a set S = {s 1,..., s n } of line segments Determine if s i s j for some 1 i j n.

13 Determine if any pair of segments in a set intersects Given a set S = {s 1,..., s n } of line segments Determine if s i s j for some 1 i j n. We can do this in O(n log n) time with the sweeping technique:

14 Determine if any pair of segments in a set intersects Given a set S = {s 1,..., s n } of line segments Determine if s i s j for some 1 i j n. We can do this in O(n log n) time with the sweeping technique: An imaginary vertical sweep line passes through the given set of geometric objects, usually from left to right. We will assume that the sweeping line moves across the x-dimension

15 Determine if any pair of segments in a set intersects Given a set S = {s 1,..., s n } of line segments Determine if s i s j for some 1 i j n. We can do this in O(n log n) time with the sweeping technique: An imaginary vertical sweep line passes through the given set of geometric objects, usually from left to right. We will assume that the sweeping line moves across the x-dimension Simplifying assumptions 1 No input segment is vertical 2 No three input segment intersect at a single point

16 Auxiliary notions Ordering segments ASSUMPTIONS: s 1, s 2 S are two line segments; sw x is the vertical sweep line with x-coordinate x Example s 1, s 2 are comparable at x if sw x intersects both s 1 and s 2 s 1 x s 2 if s 1, s 2 are x-comparable, and the intersection point s 1 sw x is higher than s 2 sw x In the figure below, we have a r c, a t b, b t c, b t c, and b u c. Segment d is not comparable with any other segment. Remark: x is a total preorder relation: reflexive, transitive, but neither symmetric nor antisymmetric.

17 Detecting segment intersections When line segments e and f intersect, they reverse their orders: we have e v f and f w e. Simplifying assumption 2 implies vertical sweep line sw x for which the intersections with segments e and f are consecutive w.r.t. total preorder x. Any sweep line that passes through the shaded region in figure above (such as z) has e and f consecutive in its total preorder.

18 Moving the sweep line The sweep line moves from left to right, through the sequence of endpoints sorted in increasing order of the x-coordinate. The sweeping algorithm maintains two data structures: Sweep line status: the relationships among the objects that the sweep line intersects. Event-point schedule: a sequence of points (the event points) ordered from left to right according to their x-coordinates.

19 Moving the sweep line The sweep line moves from left to right, through the sequence of endpoints sorted in increasing order of the x-coordinate. The sweeping algorithm maintains two data structures: Sweep line status: the relationships among the objects that the sweep line intersects. Event-point schedule: a sequence of points (the event points) ordered from left to right according to their x-coordinates. Whenever the sweep line reaches the x-coordinate of an event point: the sweep halst, processes the event point, and then resumes Changes to the sweep-line status occur only at event points.

20 The sweeping algorithm for segment intersections Auxiliary data structures THE SWEEP LINE STATUS: container for a total preorder T = x between line segments from S Requirements: to perform efficiently the following operations: 1 insert(t, s) : insert segment s into T 2 delete(t, s): delete segment s fro T 3 above(t, s): return the segment immediately above segment s in T. 4 below(t, s): return the segment immediately below segment s in T. REMARK: all these operations can be performed in O(log n) time using red-black trees.

21 The sweeping algorithm for segment intersections Pseudocode AnySegmentsIntersect(S) 1. T = 2. sort the endpoints of the segments in S from left to right, breaking ties by putting left endpoints before right endpoints and breaking further ties by putting points with lower y-coordinates first 3. for each point p in the sorted list of endpoints 4. if p is the left endpoint of a segment s 5. insert(t, s) 6. if (above(t, s) exists and intersects s) or (below(t, s) exists and intersects s) 7. return TRUE 8. if p is the right endpoint of a segment s 9. if both above(t, s) and below(t, s) exist and above(t, s) intersects below(t, s) 10. return TRUE 11. delete(t,s) 12. return FALSE

22 The sweeping algorithm for segment intersection Every dashed line is the sweep line at an event point. The ordering of segment names below each sweep line corresponds to the total preorder T at the end of the for loop processing the corresponding event point. The rightmost sweep line occurs when processing the right endpoint of segment c.

23 Convex hull of a set of points ASSUMPTION: Q is a finite set of n points. The convex hull CH(Q) of Q is the smallest convex polygon P with vertices in Q, such that each point in Q is either on the boundary of P or in its interior. Intuition: each point of Q is a nail stuck in a board convex hull = the shape formed by a tight rubber band that surrounds all the nails. EXAMPLE:

24 Convex hull Graham s scan Computes CH(P) in O(n log n), where n = Q with a technique named rotational sweep: vertices are processed in the order of the polar angles they form with a reference vertex. MAIN IDEA: Maintain a stack S of candidate points for the vertices of P in counterclockwise order. each point of Q is pushed onto S one time. the points in already S, which are not in CH(Q), are popped from S. Related operations: push(p, S), pop(s), and top(s) return, but do not pop, the point on top of S nexttotop(s): return the point one entry below the top of S without changing S

25 Convex hull Graham s scan algorithm: pseudocode GrahamScan(Q) 1 let p 0 be the point in Q with the minimum y-coordinate, or the leftmost such point in case of a tie 2 let p 1, p 2,..., p m be the remaining points in Q, sorted by polar angle in counterclockwise order around p 0 (if more than one point has the same angle, remove all but the one that is farthest from p 0 ) 3 let S be an empty stack 4 push(p 0, S) 5 push(p 1, S) 6 push(p 2, S) 7 for i = 3 to m 8 while the angle formed by nexttotop(s), top(s), and p i makes a nonleft turn 9 pop(s) 10 push(p i, S) 11 return S

26 Graham s scan algorithm: pseudocode Snapshots of algorithm execution

27 References Chapters 33: Computational Geometry from the book Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest. Introduction to Algorithms. McGraw Hill, 2000.