f(n) >= ( Volume of hypercube with side 1 ) / ( Volume of simplex with side 2^(1/2) ) = 1^n / ( 2^(n/2) . (n+1)^(1/2) / 2^(n/2) n! ) = n! / (n+1)^(1/2) -> oo as n -> oo ; therefore f(n) -> oo as n -> oo . As I tried to point out before (but also garbled) the limit is actually independent of the side lengths. For n = 3 , the volume ratio yields f(3) >= 3 > 2 ; for n = 4 , f(4) > 24/sqrt(5) > 10 . WFL On 2/28/19, Dan Asimov <dasimov@earthlink.net> wrote:
My usual "Arggh!" I should not post that late at night! Let me restate the problem this way:
Let the 2^n n-cube vertices be given by
C(n) = {0, 1}^n
with distances determined in R^n, and let the (n+1) n-simplex vertices be given by a orthonormal set
S(n) = {e_j in R^(n+1) | <e_j, e_k> = delta(j,k)}.
Then for n >= 2, define
f(n) = the smallest number of isometric copies of S(n) that cover C(n).
I.e., if ∆_1, ..., ∆_r are subsets of the n-cube C(n) each congruent to S(n) whose union is C(n):
C(n) = ∆_1 u ∆_2 u ... u ∆_r
then f(n) = the least such r.
Questions: What is f(n) and what is it asymptotically as n —> oo ???
—Dan
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