Computational Geometry Some useful algorithms/formulas for solving geometric problems

Overview Representation Basic geometric problems Other issues

Representation Points Segments Lines Polygon (x, y) Cartesian coordinates (r, Θ) Polar coordinates Segments Position of the end points Lines 2-tuple (m, c) using y = mx + c 3-tuple (a, b, c) using ax + by = c Two points P0, P1 on the line using P(t) = (1-t)P0 + t P1 Polygon Ordered list of points

Basic geometric problems Area of polygon CCW Line segment intersection Distance between a point and a line Closest point Angle between two lines Point in a polygon Convex hull

Area of polygon Given a simple polygon, find its area. Area = 1/2 * [(x1.y2 + x2.y3 +…+ xn.y1) - (y1.x2 + y2.x3 + … + yn, x1)] (x2, y2) … (x1, y1) (xn, yn)

CCW Basic geometric primitive Returns 1 if the points are ccw Returns -1 if the points are cw Returns 0 if the points are collinear Area given in the previous slide is positive when the points are listed in counter-clockwise direction and negative otherwise 3 2 1 ccw(p1, p2, p3) returns 1

Line segment intersection (I) Given two line segments l1 and l2, determine if they intersect ? isIntersect = ccw(l1.p1, l1.p2, l2.p1) * ccw(l1.p1, l1.p2, l2.p2) <= 0 && ccw(l2.p1, l2.p2, l1.p1) * ccw(l2.p1, l2.p2, l1.p2) <= 0 l1.p2 l2.p1 l2.p2 l1.p1

Distance between a point and a line Given a point and a line, determine the shortest distance between them. Select two points A, B on the line ½ * AB * h = area of triangle ABP P h B A

Closest point (I) Given a line L and a point P, find the point on L closest to P Recall that if u is a unit vector and v is any vector. The projection of v on u is given by (u.v)u u.v is the number k which minimized |v – ku|

Closest point (II) The theorem in the previous slide can only be applied to lines which pass through the origin This is sufficient to compute the general case as we can apply a translation if the line does not pass through the origin

Angle between two lines Represent lines as vectors u (a, b) and v (c, d) Definition of dot product: u.v = |u||v|cos Θ = ac + bd |u| = sqrt(u.u) Θ = cos-1(u.v / |u||v|) u Θ v

Point in a polygon (I) Given a point and a closed polygon, determine whether the point lies inside the polygon. Most algorithms extends a ray from the given point and considers the interaction of the polygon with the ray Such algorithms needs to handle the following boundary cases:

Point in a polygon (II) A standard way to resolve this issue is to adopt the following set of rules Edge crossing rules an upward edge includes its starting endpoint and excludes its endpoint a downward edge excludes its starting endpoint and includes its endpoint horizontal edges are excluded the edge-ray intersection point must be strictly right of the point P

Point in a polygon (III) One of the simplest algorithm is to compute the winding number of the point

Point in a polygon (IV) Winding number Point is outside iff WN = 0 number of times the polygon winds around the point Point is outside iff WN = 0 Upward edges => WN++ Downward edges => WN--

Point in a polygon (V) //      Input:   P = a point, //               V[] = vertex points of a polygon with V[n]=V[0] //      Return: wn = the winding number (=0 only if P is outside V[]) int wn_PointInPoly(Point P, Point* V, int n) {     int wn = 0;    // the winding number counter     // loop through all edges of the polygon     for (int i=0; i<n; i++) {   // edge from V[i] to V[i+1]         if (V[i].y <= P.y) {         // start y <= P.y             if (V[i+1].y > P.y)      // an upward crossing                 if (ccw( V[i], V[i+1], P) > 0)  // P left of edge                     ++wn;            // have a valid up intersect         }         else { // V[i].y > P.y             if (V[i+1].y <= P.y)     // a downward crossing                 if (ccw( V[i], V[i+1], P) < 0)  // P right of edge                     --wn;            // have a valid down intersect         }     }     return wn; }

Convex hull (I) Given a set of points, determine the smallest convex set containing all the points A set S is convex if whenever two points P and Q are inside S, then the whole line segment PQ is also in S

Convex hull (II) One simple algorithm is Graham Scan, often cited as the first computational geometry algorithm Pseudocode Select the lowest leftmost point P[0] in S Sort the rest of the points in S according to the angle made with P[0], break ties base on distance to P[0] Let P[0..n-1] be the sorted array of points Create an empty stack ST ST.push(P[n-1]) //P[n-1] must be on the hull ST.push(P[0]) //P[0] must be on the hull For i from 2 to n – 1 Let PT1 denote the topmost point on ST Let PT2 denote the second topmost point on ST while (ccw(PT2, PT1, P[i]) <= 0) ST.pop() ST.push(P[i])

Other issues Case analysis Use of floating point numbers Overflow

References http://softsurfer.com/, Dan Sunday Algorithms, Robert Sedgewick Programming Challenges, Skiena and Revilla