# Why is the valence of regular vertices 6?

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.

• 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 – joojaa Sep 24 '20 at 6:13

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.