I think the intent was for an n-simplex to have n+1 vertices, but in R^n, not in R^(n+1). For example, a tetrahedron is a 3-simplex, has 4 vertices, and exists in R^3. So when n=3, you are covering the 2^3 = 8 vertices of a cube with two 3-simplices, i.e. with two tetrahedrons, which works nicely (and all in R^3). One other slight adjustment: For a simplex edge length of sqrt(2), you want the n-cube's vertices to be {0, 1}^n. If they're {-1, 1}^n then you need the simplex edge length to be 2*sqrt(2), at least for the n=3 case. Regarding the question, an obvious lower bound on f(n) is: ceiling((2^n)/(n+1)) So one question to ask is, what is the smallest value of n for which f(n) is greater than this (if any)? I'm also thinking one might want the simplex edge length to be a function of n, rather than a fixed sqrt(2), but I'm not certain about that. Tom Fred Lunnon writes:
(A) As defined, your simplex has only n vertices, not n+1 ?!
(B) If the missing simplex vertex was intended to be supplied by the origin, then the contents when n = 2 are respectively 4 and 1/2 ; so f(2) = 8 ?!
(C) If also the hypercube was intended to have unit edges, then the contents when n = 3 are respectively 1 and 1/6 ; so f(3) >= 6 ?!
Whatever the shapes and edge lengths, their content ratio must inevitably approach zero as n -> oo ; so f(n) -> 0 ?!
Beyond my ken, I'm afraid ... WFL
On 2/28/19, Dan Asimov <dasimov@earthlink.net> wrote:
Let C(n) denote the 2^n vertices of the n-cube [-1, 1]^n in R^n:
C(n) = {-1, 1}^n
Let S(n) denote the n+1 vertices of the n-simplex in R^(n+1):
S(n) = {e_j | 1 <= j <= n}
where { 0, i ≠ j <e_j, e_k> = delta(i, j) = { { 1, i = j
(so all interpoint distances = sqrt(2)).
Question: --------- For n >= 2, what is the smallest number f(n) of isometric copies of S(n) needed to cover C(n) ???
E.g., f(2) = 2; f(3) = 2, f(4) = ???
Question: --------- What is a nice function asymptotic to f(n) as n —> oo ???
—Dan
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