Cartesian coordinates for vertices of a regular 16-simplex?

I am using amoeba to solve an optimization problem and want to distribute the initial perturbations uniformly about the initial estimate. With only Excel at my disposal, I cannot figure out how to compute the Cartesian coordinates for the vertices of a regular 16-simplex.

There are published values up to a 10-simplex available on the 'net, but I've not been able to locate any solutions for higher dimensions. Perhaps someone out there has the tools to compute these coordinates (e.g. Mathematica, Maple, or Matlab), or can point me as a procedure to accomplish this in Excel.

Solution 1:

Hint:

You could start to define the vertices $q_i^{(n)}$ and midpoints $m^{(n)}$ of a regular n-simplex (edge length equalling 1) recursively, e.g. $$ q_0^{(1)} = (0),\\ q_1^{(1)} = (g_1), \quad\text{with}\quad g_1 = 1,\\ m^{(1)} = (h_1), \quad\text{with}\quad h_1 = \frac{1}{2},\\ q_i^{(n+1)} = (q_i^{(n)},0), \quad\text{with}\quad 0 \le i \le n,\\ q_{n+1}^{(n+1)} = (m^{(n)},g_{n+1})\\ m^{(n+1)} = (m^{(n)},h_{n+1}). $$

If one then imposes the conditions $(*)$: $$ \|q_{n+1}^{(n+1)} - q_{n}^{(n+1)} \|^2 = 1\\ \|q_{n+1}^{(n+1)} - m^{(n+1)}\|^2 = \|q_{n}^{(n+1)} - m^{(n+1)}\|^2. $$ This gives $$ \|q_{n+1}^{(n+1)} - q_{i}^{(n+1)} \|^2 = 1$$ and then $$ \|q_{j}^{(n+1)} - q_{i}^{(n+1)} \|^2 = 1.$$

It follows from the conditions $(*)$, that we have $(2*)$: $$ (g_{n+1}-h_{n+1})^2 = (1-g_{n+1}^2) + h_{n+1}^2,\\ (g_{n+1}-h_{n+1})^2 = (g_n-h_n)^2 + h_{n+1}^2. $$

From which one can deduce $$ g_{n+2}^2 = \frac{4g_{n+1}^2-1}{4g_{n+1}^2}. $$

Which can be used to get by induction $$g_n^2 = \frac{n+1}{2n}.$$ and hence $$h_n^2 = \frac{1}{2n(n+1)}.$$

I hope there is no typo and the details can be filled by the reader.

The definition of $q_i^{(n)}$ and the determinant formula for the simplex volume, give additionally for this simplex $$ Vol(S_n) = \frac{(n+1)^{1/2}}{n!} \left(\frac{1}{2}\right)^{n/2} $$