For $d=3$, vertex coordinates of a regular simplex have a simple expression since vertices correspond to four vertices of a cube. Is there a simple expression for higher dimensions? In particular I'm interested in $d=2^n-1$, integer $n$.
Edit: by coordinates I mean points in $\mathbb{R}^d$. Every $d$-simplex has a simple expression for coordinates in $\mathbb{R}^{d+1}$, as Mariano shows below
It is known that there is a regular simplex of side length $\sqrt{(d+1)/2}$ whose vertices are vertices of the cube $[-1,1]^d$ in $\Bbb{R}^d$ if and only if there exists a Hadamard matrix of order $d+1$; this is a square matrix of $\pm 1$-entries with pairwise orthogonal columns.
In particular, there exist Hadamard matrices of order $2^n$, one of which can be constructed using the recursive Sylvester's construction as explained on the above linked wikipedia page:
Let $H_0=[1]$ and $H_{n+1}=\left[\array{H_n & H_n \\\\ H_n & -H_n}\right].$
Note that the first column of $H_n$ consists only of ones. Delete it to obtain $2^n$ row vectors in $\Bbb{R}^{2^n-1}$. These are the coordinates of a regular simplex.
One interesting problem is to determine the $n$ such that there is a regular simplex in $\mathbb{R}^n$ with rational/integer coordinates. This is a well-known old problem which I discussed in this 1998 usenet thread, and is a nice application of the Hasse-Minowski theory of rational quadratic forms. The answer is yes iff $n+1$ is the sum of one, two, four or eight odd squares.
The $d$ points $(0,\dots,0,1,0,\dots,0)$ are the vertices of a regular $(d-1)$-simplex. If you want it to be centered at the origin, just substract their barycenter from them.
Compute the full Q matrix from the QR decomposition of a column vector of ones, and drop the first column.
eg in R+
simplex <- function(n) {
qr.Q(qr(matrix(1,nrow=n)),complete=T)[,-1]
}
n=10;[q,r] = qr(ones(n,1)); q=q(:,2:end). The 10 rows are the 10 vertices of a 9 dimensional simplex.
$\endgroup$
– Pushpendre
May 3 '18 at 3:09
Since this question is back on the front page, I wanted to mention the simplex code. For $b\ge 2$, and $n\ge1$, the code consists of $b^n$ code words, each of which is a $b$-ary string of length $s=(b^n-1)/(b-1)$. The Hamming distance between any two codewords is $b^{n-1}$. Therefore, if you write the $(b-1)$-simplex using your favorite coordinates in $\mathbf{R}^{b-1}$, then substituting these $b$ vertices for the $b$ digits in your codewords and concatenating the resulting words into vectors in $\mathbf{R}^{(b-1)s}$ gives you a nice set of coordinates for the $(b^n-1)$-simplex.
For example, taking $b=2$ and $n=2$ gives the 4 alternating vertices of the 3-cube.
The coordinates of the vertices of a regular ''n''-dimensional simplex can be obtained from these two properties,
If we initialize our first vertex to be along the first dimension, we can iteratively find the subsequent vertices by following the criteria above. Here's the code in Julia:
function regular_simplex(ndims, radius)
verts = [zeros(ndims) for i in 1:(ndims+1)];
for i in 1:(ndims+1)
if i > 1
for j in i:(ndims+1)
verts[j][i-1] = (-dot(verts[i-1], verts[j]) - 1 / ndims) / verts[i-1][i-1];
end
end
if i <= ndims
verts[i][i] = sqrt(1 - sum(verts[i].^2))
end
end
return verts .* radius;
end
