Figure 93: Delaunay triangulations and convex hull.
3D Convex Hulls and Delaunay Triangulations There is a fascinating relationship between delaunay triangulations and the convex hull of a particular set of 3D points.
At first, Delaunay triangulations and convex hulls appear to be quite different structures, one is based on metric properties (distances) and the other on affine properties (collinearity, coplanarity). Today we show that it is possible to con vert the problem of computing a Delaunay triangulation in dimension d to that of computing a convex hull in dimension d + 1. Thus, there is a remarkable relationship between these two structures.
We will demonstrate the connection in dimension 2 (by computing a convex hull in dimension 3). Some of this may be hard to visualize. The connection between the two structures is the paraboloid z = x2 + y2. Observe that this equation defines a surface whose vertical cross sections (constant x or constant y) are parabolas, and whose horizontal cross sections (constant z) are circles. For each point in the plane, (x, y), the vertical projection of this point onto this paraboloid is (x,y, x2 +y2) in 3 space.
Given a set of points S in the plane, let S0 denote the projection of every point in S onto this paraboloid. Consider the lower convex hull of S0. This is the portion of the convex hull of S0 which is visible to a viewer standing at z = -1. We claim that if we take the lower convex hull of S0, and project it back onto the plane, then we get the Delaunay triangulation of S. In particular, let (p, q, r) be elements of S, and let p0 , q0 , r0 denote the projections of these points onto the paraboloid. Then p0 q0 r0 define a face of the lower convex hull of S0 if and only if pqr is a triangle of the Delaunay triangulation of S. The process is illustrated in the following figure.
Figure 93: Delaunay triangulations and convex hull.
The question is, why does this work? To see why, we need to establish the connection between the triangles of the Delaunay triangulation and the faces of the convex hull of transformed points. In particular, recall that
Delaunay condition: Three points p, q, r, in S form a Delaunay triangle if and only if the circumcircle of these points contains no other point of S.
Convex hull condition: Three points p0, q0, r0 in S0 form a face of the convex hull of S0 if and only if the plane passing through p0, q0, and r0 has all the points of S0 lying to one side.
Clearly, the connection we need to establish is between the emptiness of circumcircles in the plane and the emptiness of halfspaces in 3 space. We will prove the following claim.
Lemma: Consider 4 distinct points p, q, r, s in the plane, and let p0, q0, r0 s0 be their respective projections onto the paraboloid, z = x2 + y2 . The point s lies within the circumcircle of p, q, r if and only if s0 lies on the lower side of the plane passing through p0, q0, r0.
To prove the lemma, first consider an arbitrary (nonvertical) plane in 3 space, which we assume is tangent to the paraboloid above some point (a, b) in the plane. To determine the equation of this tangent plane, we take derivatives of the equation z = x2 + y2 with respect to x and y, giving:
dz/dx = 2x
dz/dy = 2y
At the point (a, b, a2 +b2 ) these evaluate to 2a and 2b. It follows that the plane passing through these point has the form
To solve for k we know that the plane passes through
so we solve giving
Implying that k = - (a2 + b2 ). Thus the plane equation is:
If we shift the plane upwards by some positive amount r2 we get the plane
This is just a circle. Thus, we have shown that the intersection of a plane with the paraboloid produces a space curve (which turns out to be an ellipse), which when projected back onto the (x, y) coordinate plane is a circle centered at (a, b).
Thus, we conclude that the intersection of an arbitrary lower halfspace with the paraboloid, when projected onto the (x, y) plane is the interior of a circle. Going back to the lemma, when we project the points p; q; r onto the paraboloid, the projected points p0, q0 and r0 define a plane. Since p0, q0, and r0, lie at the intersection of the plane and paraboloid, the original points p, q, r lie on the projected circle. Thus this circle is the (unique) circumcircle passing through these p, q, and r. Thus, the point s lies within this circumcircle, if and only if its projection s0 onto the paraboloid lies within the lower halfspace of the plane passing through
An applet that show simultaneously the delaunay triangulation and the 3D convex hull can be found here It may be necessary to try a collection of patterns or sets of points (click on the left coordinate system near the bottom of the page).
