# What is the most frequent number of edges of Voronoi cells of a large set of random points?

Consider a large set of points with coordinates that are uniformly distributed within a unit-length segment. Consider a Voronoi diagram built on these points. If we consider only non-infinite cells, what would be (if any) the typical (most frequent) number of edges (that is, neighbors) for those cells? Is there a limit for this number when number of points goes to infinity? Does it have anything common with `kissing number`? If so, doest it generalize to higher dimensions, that is, 6 for 2D, 13 for 3D etc?

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## 2 Answers

I think the book Spatial tessellations : concepts and applications of Voronoi diagrams by OKABE, BOOTS,and SUGIHARA discusses this.

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All the links to content on the page you provided are dead. – mbaitoff Jan 31 '11 at 11:55
@mbaitoff: yes, it seems so, sorry. But it's easy fixable: Just  ua instead of okabe in the host part. For instance, the contents are at ua.t.u-tokyo.ac.jp/okabelab/Voronoi/contents.html – lhf Jan 31 '11 at 12:17
There are contents headlines only, no book contents. – mbaitoff Jan 31 '11 at 12:49
The book is on the net, thanks Buddha. Now starting reading. – mbaitoff Jan 31 '11 at 12:58
Well, seems that it's not easy to find an answer in 683-page book. If you know the answer, would you please post it instead? – mbaitoff Jan 31 '11 at 14:55
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I don't have access to the book that @lhf referenced, but here is a nice topological argument for the expected number of edges of a Voronoi cell in $2$D. Unfortunately, it does not generalize to more than two dimensions.

Consider the dual of the Voronoi diagram, which is the Delaunay triangulation of the given points. This is a planar connected graph, so its Euler characteristic is $\chi = 2$. The Euler characteristic is also given by $\chi = V - E + F$, where $V$, $E$, and $F$ are the number of vertices, edges, and faces in the Delaunay triangulation respectively. Now every face has $3$ edges, while all non-boundary edges are adjacent to $2$ faces. Under reasonable conditions*, the proportion of boundary edges tends to zero, so let us ignore them. This means that $3F \approx 2E$, and plugging this into $V - E + F = 2$ gives $V \approx \frac13E + 2$, where by "$\approx$" I mean the ratio tends to $1$ as $V \to \infty$. So there are about three times as many edges as vertices, and since each edge is incident on $2$ vertices, the average degree of the vertices approaches $6$. As the degree of a vertex in the Delaunay triangulation is precisely the number of edges of the corresponding Voronoi cell, this agrees with your intuition for the $2$D case. (Although, strictly speaking, the expected number of edges is not the same as the most frequent number of edges).

In three dimensions, the Euler characteristic is still $2$, but is now given by $V - E + F - C$, where $C$ is the number of tetrahedral cells in the triangulation. A similar argument as above gives $4C \approx 2F$, but we have no control over $E$. Indeed, there are triangulations on the same point set with the same boundary (the convex hull) but which have different numbers of edges. In the $2$D case, every such triangulation had exactly the same number of edges! So in $3$D one will have to think about the geometry of the Voronoi diagram and/or Delaunay triangulation, and cannot get a result purely via its topology.

* I believe it is sufficient for the points to be drawn from a uniform distribution on a strictly convex area, but I don't know the details.

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Can we rely on the fact that ration of the perimeter of the figure to the area of the figure goes to zero, and assume this fact is true for ratio number of boundary vertices to the number of internal vertices? – mbaitoff Feb 8 '12 at 12:27
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