# Hexagonal Grid from Cube

It's fairly well-known that an intersection with the integer cube grid (i.e., the vertices of the grid are in $$\mathbb{Z}^3$$) with the plane $$u-v+w=0$$ creates a hexagonal grid (usually people use $$u+v+w=0$$, but I need it this way for other reasons).

it is also well known that the grid vertices of a hexagonal grid lie on the Eisenstein integers $$a+b\omega$$ where $$a,b \in \mathbb{Z}^2$$ and $$\omega = \frac{-1 + i\sqrt{3}}{2}$$ (notice it's a complex number). Is there a canonical way to parameterize the plane $$u-v+w=0$$ s.t. every $$u,v,w \in \mathbb{Z}^2$$ would produce $$a,b \in \mathbb{Z}$$? I didn't find anything concrete---people just either use either set of coordinates.

Given $$\,u,v,w \in \mathbb{Z}\,$$ with $$\, u-v+w=0, \,$$ then $$\,v + w\,\omega\,$$ is one candidate for the corresponding Eisenstein integer. Here $$\,(1,1,0) \,$$ maps to $$\,1\,$$ and $$\,(-1,0,1)\,$$ maps to $$\,\omega.\,$$ These two vectors form a linear basis for the plane and moreover they have the same magnitude and have angular difference of $$\,2\pi/3.\,$$ There are other choices which have the same properties. For example, swap $$\,u\,$$ and $$\,w.\,$$ Another way is to negate $$\,u,v,w.\,$$
• Interesting. So map to $(a,b)$ with non-orthogonal basis, rather than find one explicitly. – Amir Vaxman Sep 28 '18 at 13:40