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So I recently learnt that supposedly for any mesh, and pretty much any scheme, the valence of regular vertices must be 6. It seems to be related to the Euler-Poincare formula but I have not been able to find the justification.

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  • $\begingroup$ This does not really apply to a tetradedral outcropping but it does apply most quatrilateral turned to triangles meshes where the repetition of triangulation is uniform $\endgroup$ – joojaa Sep 24 '20 at 6:13
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If require that all faces have the same number of sides $s$ and require that all vertices also have a certain valency $t$. We see that the following relation between edges, and faces hold for a regular mesh:

$$s\cdot f = 2e,$$ $$t\cdot v = 2e.$$ Substitution in the Euler-Poincare formula yields:

$$\left(\frac{1}{s} + \frac{1}{t} - \frac{1}{2}\right)e = 1 - g$$

If we then take for instance a regular plane which can be said to have the topology of a torus with genus $g = 1$. We then set the valency of faces $s = 3$ then

$$\left(\frac{1}{3} + \frac{1}{t} - \frac{1}{2}\right)e = 0,$$ $$\left(-\frac{1}{6} + \frac{1}{t}\right)e = 0.$$

The solution for $t$ is $6$, which says the valency of a vertex in a regular triangulation is 6.

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