circumcenter of the $n$-simplex

Question

Given $m$ points $v_i\in\mathbb{R}^n$, $m<n$. How to find the circumcenter of the simplex formed by the points?

Answer 1

In Miroslav Fiedler's lovely book Matrices and Graphs in Geometry, he shows how one can use the Cayley-Menger matrix,

$$\mathbf M=\begin{pmatrix} 0&1&1&\cdots&1\\ 1&0&d_{1,2}^2&\cdots&d_{1,n+1}^2\\ 1&d_{2,1}^2&\ddots&&\vdots\\ \vdots&\vdots&&\ddots&d_{n,n+1}^2\\ 1&d_{n+1,1}^2&\cdots&d_{n+1,n}^2&0\end{pmatrix}$$

to determine the circumsphere of an $n$-simplex determined by $n+1$ $n$-dimensional points. Here, $d_{j,k}$ signifies the distance between vertices $v_j$ and $v_k$. Using the formulae from this page, and letting $\mathbf Q=-2\mathbf M^{-1}$, the circumcenter is given by

$$\left(\frac{q_{1,2}}{q_{1,2}+\cdots+q_{1,n+2}}v_1,\cdots,\frac{q_{1,n+2}}{q_{1,2}+\cdots+q_{1,n+2}}v_{n+1}\right)$$

and the circumradius is given by $\dfrac{\sqrt{q_{11}}}{2}$. The book also discusses how to use the Cayley-Menger matrix to get the insphere. (I wrote up a Mathematica implementation of Fiedler's formulae here.)

Answer 2

Another approach is to form a tall matrix $A\in\mathbb{R}^{N\times n}$, where each row contains the displacement between unique pairs of vertices and $N=\frac{m(m-1)}{2}$ and a vector $b\in\mathbb{R}^N$ where each element is the half of the corresponding difference between the squared distance of pairs of vertices from the origin. The circumcenter $x\in\mathbb{R}^n$ is simply the least squares solution to this overdetermined system. This approach has the disadvantage of evaluating more terms (a quantity that scaled quadratically witgh the number of vertices), however, it should not present singular matrix problems.

$$ \begin{pmatrix} v_2 - v_1 \\ \vdots \\ v_{m} - v_{m-1} \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}=\frac{1}{2} \begin{pmatrix} \|v_2\|^2-\|v_1\|^2\\ \vdots\\ \|v_m\|^2-\|v_{m-1}\|^2 \end{pmatrix} $$

Answer 3

This is expansion on Mainfred Weis's answer.

The most straightforward approach is to recognize that the circumcenter, the point equidistant to all vertices, also has the property that it will lie on all (hyper)planes which bisect each combination of vertices. That is, for all unique pairs of vertices $v_i$ and $v_j$: $$\frac{(c-v_i)\cdot(v_j-v_i)}{(v_j-v_i)\cdot(v_j-v_i)}=\frac{1}{2}$$

Next, we recognize that if we have $m$ points, we only need $m-1$ hyperplanes to fully determine the system.

We can arbitrate $\vec{v}_1$ to be the "origin", and set up the following $m-1 \times n$ matrix system, where $\vec{c}'=\vec{c}-\vec{v}_1$, the relative circumcenter, and $\vec{d}_{i,j}=\vec{v}_j-\vec{v}_i$, the difference vectors:

$$\begin{pmatrix} \vec{d}_{1,2}\\ \vdots \\ \vec{d}_{1,m} \end{pmatrix} \cdot \vec{c}'=\frac{1}{2}\begin{pmatrix} \left\| \vec{d}_{1,2} \right\|^2 \\ \vdots \\ \left\| \vec{d}_{1,m} \right\|^2 \end{pmatrix}$$

When $m \leq n$

If we are so lucky to have $n + 1$ points, then we have a square matrix, and we can solve for $\vec{c}'$ directly. However, if we have $m\leq n$, then we must recognize that our relative circumcenter can be represented as a linear combination of the difference vectors, $\vec{c}'= \begin{pmatrix} \vec{d}_{1,2} & \cdots & \vec{d}_{1,m} \end{pmatrix}\cdot\begin{pmatrix} t_{1,2}\\ \vdots\\ t_{1,m} \end{pmatrix}$, and so:

$$\begin{pmatrix} \vec{d}_{1,2}\\ \vdots \\ \vec{d}_{1,m} \end{pmatrix} \cdot \begin{pmatrix} \vec{d}_{1,2} & \cdots & \vec{d}_{1,m} \end{pmatrix}\cdot\begin{pmatrix} t_{1,2}\\ \vdots\\ t_{1,m} \end{pmatrix}=\frac{1}{2}\begin{pmatrix} \left\| \vec{d}_{1,2} \right\|^2 \\ \vdots \\ \left\| \vec{d}_{1,m} \right\|^2 \end{pmatrix}$$

Allowing you to solve for $t$ and then plugging in for $\vec{c}$

A more symmetric approach

Numerically, this probably gives one of the more accurate results. The only disappointing thing about this solution so far is that we must arbitrarily choose a vertex to be our "origin".

However, if we take a page out of projective geometry, we can recognize that we can instead place all of our vertices $\vec{v}_i$ on a projective plane by appending a perpendicular element ($(x, y, z) \rightarrow (1, x, y, z)$). We can then introduce a new vertex, $\vec{v}_0 = \vec{0}$, using $\vec{v}_0$ as the "origin", disambiguating our choice.

This works because adding a new vertex can only move the circumcenter along an axis perpendicular to the existing surface spanned by our initial vertices.

This results in the following system:

$$\begin{pmatrix} 1 & \vec{v}_1\\ \vdots & \vdots \\ 1 & \vec{v}_m \end{pmatrix} \cdot \begin{pmatrix} 1 & \cdots & 1 \\ \vec{v}_1 & \cdots & \vec{v}_m \end{pmatrix}\cdot\begin{pmatrix} t_{1}\\ \vdots\\ t_{m} \end{pmatrix}=\frac{1}{2}\begin{pmatrix} \left\| \vec{v}_{1} \right\|^2 + 1 \\ \vdots \\ \left\| \vec{v}_{m} \right\|^2 + 1 \end{pmatrix}$$

The solution being: $$\begin{pmatrix} \vec{v}_1 & \cdots & \vec{v}_m \end{pmatrix}\cdot\begin{pmatrix} t_{1}\\ \vdots\\ t_{m} \end{pmatrix}$$

Answer 4

the center of the circum-sphere of a simplex can be trivially calculated as the solution of a sytem of linear of equations: calculate any spanning tree of the corners and take the intersection of the hyperplanes that contain the midpoints of the tree edges and are orthogonal to them.