A linear time algorithm for finding all farthest neighbors in a convex polygon

An efficient algorithm for the three-dimensional diameter problem

“…In the power diagram of S the distance from a point p in S to a point q is measured as power pow(p, q) = d(p, q) 2 …”

“…We can test whether it belongs to the edge E' in 0(1) time . 2 If the edge E' does not contain the point r, we pass it and take the next edge on the boundary V (p j) .…”

“…Aggarwal and Kravets [2] gave a linear-time algorithm for finding all furthest neighbors in a convex polygon. We consider a three-dimensional analog, the all-furthest neighbors (AFN) for points on a convex polytope.…”

“…fl CH(S) = { p}. Let w be the ray starting at p, perpendicular to ,B and lying in the same halfspace formed by ,B as the set S. It is easy to show that the Voronoi cell of p intersects w. (2) Using the previous case it is sufficient to find a support plane intersecting the convex hull of S only at p to determine a side of the plane a. Let f be a support face of the convex hull of S and let (p, q) and (p, r) be the consecutive edges of f. By a slight perturbation of the plane containing f (it suffices to rotate it around lines pq and pr) we can find a plane fi such that • 8 is still a support plane of the convex hull of S, • 8 intersects the convex hull of S only at the point p, and • fi is not perpendicular to the plane a.…”

“…The direction of the motion in the chain Phi can be defined using the halfplane {t I dist(t, pj) = dist(t, pi) and dist(t, pj) < dist(t, pi)} fore 54 eo and the halfplane n+ fl {tj dist(t, pi) = dist(t, Pk)} for e = el 2. The edge E' does not contain the point r if and only if the line passing through endpoints of E' separates the points r and (p + p,,)/2 in the bisector y(j, s).…”