Instructor: Scot Drysdale | June 28, 1996 |

The intersection of two convex sets is a convex set. Select two points in the intersection. Both are in the first set, so the segment joining them is in the first set by the definitions of intersection and convexity. Similarly the segment joining the two points is in the second set. Therefore the segment joining them is in the intersection by definition of intersection. All the regions in a Voronoi diagram are convex, because they are intersections of n-1 half-planes (which are all convex).

You can't triangulate k collinear points, you would get a degenerate triangle as the Delaunay Triangulation. However, if you then add a site off of the line its region will be adjacent to the Voronoi region of every site on the line, so will be connected to every point in the Delaunay Triangulation.

(Polyginal - parabola or Lou's polygon. The shape of the boundary of the Voronoi region of the point off of the line in the example just given. )

Vertices on the convex hull have infinite regions. This time, the points on the edges of the convex hull between the extremal points are also included. Every point on the convex hull, including points between the extremal points, has an infinite region.

**Voronoi diagrams
**

**Knuth's Post-Office Problem**(or Nearest-Search Problem or Helicopter Rescue Problem)

**Closest Pair Problem**

**All-Nearest-Neighbor Problems**

**Euclidean Minimum Spanning Tree**

**Largest Empty Circle Problem**(or Toxic Waste Dump Location Problem)

**Fixed-Radius Near Neighbor Problem**

**All-Points-k- Nearest-Neighbor Problem**

**Enumerating Interpoint Distances In Increasing Order**

Page 173 for proof of size of diagram which refers you back to page 119

At most there are 3n - 6 ( really 3n - 3 - # of hull edges) Voronoi edges q(n)

2n - 4 Voronoi vertices

For an obtuse triangle the circumcenter is outside the region

For an acute triangle the circumcenter is inside the region

For a right triangle the circumcenter is on the edge

Euler's Formula

V + F - E = 2

Voronoi Diagrams in three-dimensions are used in Crystallography.

If Crystals start growing at

If crystals grow at some constant rate, but start at

{x| dc(p,x) + w(p) = dc(q,x) + w(q)}

This can be arranged:

d(p,x) - d(q,x) = C + w(q) - w(p)

which shows that the shape of the "bisector" curves is a hyperbola.

What if crystals start at the

Multiplicutively weighted Voronoi Diagram. See slide C 1, C2, C3.

sep(p,q) = { x | dc(p,x)/ wp = dc(q,x)/ wq}

In this case the separator is a circle, which encloses the lighter-weight point but is not centered at it.

- Johnson & Mehl (1939) - All sites with same growth rate, simultaneous or sequential
nucleation. These correspond to standard and additively weighted Voronoi diagrams,
respectively.

- Tamemura & Hasegawa (1980) - Territories of animals of equal strength. Similar models
to JM.

- Bülow-Olsen, Sakevile-Hamilton, & Hutchings (1984) - Colonies of clover. Uses same
growth rate assumption, so same models.

These models (standard and additively weighted) are O(n log n) algorithms to compute them. (Shamos, Guibas & Stolfi, Fortune,...

- Frost & Thompson (1987, 1988) - Many different growth models, including multiplicatively
weighted. Use standard multiplicatively weighted diagram, but note it is not an
accurate model.

Multiplicatively weighted diagram studied by Aurenhammer & Edelsbrunner. O(n2) size and time, because regions can be disconnected.

The Multiplicative model is *not*
an accurate model of crystal growth. Distances measured in* straight lines*
through *other*
crystals.

Crystals cannot grow through areas that have already crystallized.

Should grow "around" slower growing crystals.

Treat other crystals as *obstacles*
.

Regions in multiplicatively weighted diagrams can be disconnected but clearly crystals
can not grow this way. Shaudt and Drysdale studied this. The separators are sometimes
part of circles, but can be much more complicated curves (e.g. logarithmic spirals).

What would it mean to be a Voronoi diagrams for line segments? or for polygons? Examples:
territorial waters for islands (each island gets the part of the ocean closer to
it than to other islands, within the 200-mile limit), sprinklers watering plants,
parking in Princeton

Robots - segments are walls and obstacles -- compute shortest distances to see if
robot can fit through and throw out any edges where the robot won't fit. Then analyze
the graph theory problem and see if you can get from start to finish. This can be
done in n log n time.

2. The Delaunay Triangulation has many useful properties. Here is another property which it may or may not have: is it the minimum weight triangulation? That is, if you add up the lengths of all edges in the Delaunay triangulation the sum of these lengths be the smallest of all possible ways to triangulate the points? (Try figuring this out for 4 points first.)

http://www.supercomp.org/sc96/education --- To apply for funding to attend Supercomputing 96 in Pittsburg in Nov.

Mailto: dimacs-www@dimacs.rutgers.edu

Last modified: October 3, 1996