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Looking at the diagrams on Wikipedia here and here:

  1. In 2D: Clearly every triangle in the Delaunay mesh (black lines) does not have to contain a node of the Voronoi diagram (red points) (it can contain any number 0, 1, 2,...).
  2. In 2D: Is every cell of the Voronoi diagram (red lines) guaranteed to contain exactly one node of a Delaunay triangle?
  3. What about the 3D case involving tetrahedra (same question as #2)?

Thanks!

smörkex
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2 Answers2

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  1. A Voronoi vertex is the circumcenter of a Delaunay triangle. It is inside the triangle iff the triangle is acute.

  2. Each Voronoi cell contains exactly one vertex of the Delaunay triangulation. This is by definition. And holds for all dimensions.

lhf
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Let $L$ be a point set. Then for each $l\in L$ there is a Voronoi cel $V_l$, which consists of all points of space, which are closer to $l$ than to any different $l'\in L$. Thus any Voronoi cell always contains exactly 1 point out of $L$.

If $L$ happens to be a lattice, then all Voronoi cells will be simply translated copies of oneanother. The polytopal complex of all those Voronoi cells is being called the Voronoi complex. The dual polytopal complex to that then is the Delaunay complex. Thus the vertex set of that Delaunay complex is nothing but $L$. The polytopal cells of that complex then are the Delaunay cells.

But even for lattices the set of Delaunay cells does not need to consist of copies of a single shape only, there might be various. And especially those don't need to be simplices.

E.g. for the square lattice the Voronoi cell consists of a square, centered around every lattice point. And the single type of Delaunay cells here happens to be just such a square, the Vertices of which are closest lattice points. (In fact a cirquit of 4 lattice points.) Thus in this case the Voronoi cell and the Delaunay cell are simply shifted copies of each other.

Still, even for non-lattice point sets $L$, any vertex of a Voronoi cell uniquely corresponds to a specific Delaunay cell. But that Voronoi cell vertex needs not lie within the corresponding Delaunay cell, as already @lhf pointed out. This even can be spotted in the pictures you provided.

--- rk

Dr. Richard Klitzing
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