A Voronoi diagram is a kind of tesselation that divided the medium into polygons in 2D and polyhedrons in 3D. In two dimensions, any Voronoi diagram has vertexes(V), edges(E) and regions(F) that equal the number of sites. Eulerβs formula: V - E + F = 2 demonstrates the relationship between these variables. Furthermore, the relation between the vertexes and edges is obtained as: π = 3π£ β 6 with some assumptions. Also in three dimensions, any Voronoi diagram has vertexes, edges, regions and faces. I want to know in three dimensions, is there any relation between these variables like two dimensions? I am new to computational geometry and happy with any kind of regard.
Relation between variables (vertexes, edges, regions and faces) in three dimensional Voronoi diagram
1 Answer
Euler's formula $V-E+F=2$ is not specific to Voronoi diagrams. Rather, it counts those quantities for a planar graph, or for a polyhedron in $\mathbb{R}^3$ (whose $1$-skeleton is a planar graph). See David Eppstein's Twenty Proofs of Euler's Formula.
Euler's formula for polytopes in $\mathbb{R}^4$ (and Voronoi diagrams in $\mathbb{R}^3$) is $$V-E+F-C=0 \;,$$ where $C$ is the number of three-dimensional cells. So the hypercube becomes $8 - 16 + 32 - 24 = 0$. The constants $=2$ and $=0$ in 3D and 4D are the Euler characteristics of the respective complexes.
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$\begingroup$ Thank you for your regard. Also, I want to know, based on some assumptions and Voronoi diagram, is there any relation like "π = 3π£ β 6" between some variables for three dimensional? or any relationship that determines the number of vertexes? $\endgroup$– ALINCommented Jul 1, 2020 at 17:33
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$\begingroup$ @ALIN: A "generic" Voronoi diagram has vertices of degree-$3$. That's from where $E=3V-6$ derives. Generic for 3D Voronoi diagrams would mean that no $5$ points are co-spherical, so vertices have degree-$4$. $\endgroup$ Commented Jul 1, 2020 at 17:44