given eight vertices, how to verify they form a cube?

Question

I am given eight coordinate points in xyz plane, how can I verify that they form a cube?

Thanks,

Answer 1

In computer graphics one can find some amazing, non-trivial and very efficient tests (see chapter $16$ of the Real-Time Rendering book, for what I mean by this).

I do not claim to have such a solution.

What you could do is to test for certain properties of a cube.

(Source) (Large Version)

• only certain distances should show up if you calculate the symmetric matrix of distances $$ d_{ij} = \lVert u_i - u_j \rVert $$ where $u_i = (x_i, y_i, z_i)^T$, like base length $a$, face diagonal length $\sqrt{2} \, a$, volume diagonal length $\sqrt{3} \, a$. For faster calculation, one might stick with the squared distances $d_{ij}^2 = \lVert u_i - u_j \rVert^2$ and check for the occurrence of $0$, $a^2$, $2 a^2$ and $3 a^2$ entries only, in the proper amounts (see below).
• the smallest non-zero distance should give the base length $a$
• there should be $8$ distinct vertices $u_i$ (only the diagonal entries $d_{ii}$ should be zero)
• of the $(8\cdot 8 - 8)/2 = 28$ distinct segments $(u_i, u_j)$ with $i < j$, $s_{ij} = u_i - u_j$ there should be
  • $12$ distinct segments of base length $a$
  • $6\cdot 2 = 12$ distinct face diagonals
  • $4$ distinct volume diagonals
• the $i$-th row of $d_{ij}$ should contain
  • one zero at $d_{ii}$
  • three base length entries
  • three face diagonal entries
  • one volume diagonal entry
• the same counts should occur for the $i$-th column
• relative angles between segments
  • line segments of base length $a$ should have relative angles of $0^\circ$, $\pm 90^\circ$, $180^\circ$ only, or $\cos \varphi = (s_{ij} \cdot s_{kl})/ a^2 \in \{ 0, \pm 1 \}$
  • face diagonals have relative angles of $0^\circ$, $\pm 90^\circ$, $180^\circ$, $\pm 60^\circ$
  • volume diagonals have relative angles $\cos \varphi = \pm (1/3)$

The general idea is to come up with a list of properties that fully characterize a cube, i.e. if all are fulfilled it can only be a cube. If at least one property is not fulfilled, it can not be a cube.

For efficiency one should sort the list of properties such that the most discriminating ones are tested first.

Answer 2

You can check with vectors. Suppose you are at a corner of the cube and you have these 3 vectors popping out of the corner corresponding to the 3 edges:

$\vec a=\begin{pmatrix}x_1\\y_1\\ z_1\end{pmatrix}, \vec b=\begin{pmatrix}x_2\\y_2\\ z_2\end{pmatrix},\vec c=\begin{pmatrix}x_3\\y_3\\ z_3\end{pmatrix}$

You want the vectors to have the same length so : $|\vec a|=|\vec b|=|\vec c|$

$|\vec a|= \sqrt{x^2+y^2+z^2} $

So, check if they have the same length.

Afterwards you need to show that the vectorial product $ \vec a \times \vec b $ is parallel to vector $\vec c$. You can calculate the vectorial product using determinants. To check if $ \vec a \times \vec b $ is parallel to $\vec c$ check if their coordinates are proportional.This would mean that $ \space \space \vec c$ is perpendicular to the plane set by vectors $\vec a, \vec b $.

Finally you need to show that the angle betwenn $\vec a$ and $ \vec b$ is 90 degrees.

$ |\vec a \times \vec b|= |\vec a| \cdot |\vec b| \cdot sin \theta$ , $\theta$ being the angle between vectors $ \vec a $ and $ \vec b$. So, $ sin\theta = \frac{|\vec a \times \vec b|}{|\vec a| \cdot |\vec b|}$ Use your solution from the previous step to find $|\vec a| \cdot |\vec b|$. You should have $sin \theta = 1$