Re: [math-fun] What is the ratio, Kenneth?
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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Turns out Dan is considering just vertices, not polytope interiors --- Tom seems to have managed to work that out before I did! In this case, the limit is even easier: the number r of simplexes (simplices?) required to cover the hypercube is bounded below by r >= 2^n / (n+1) -> oo as n -> oo . For n = 4 we can fill up two opposite cubes of the tesseract with r = 2+2 = 4 simplexes, and there is surely no way this can be improved. But as n increases, interaction with Hadamard matrices probably further reduces the minimum below r = (2^n)/4 , at least whenever n+1 = 2^k ... WFL On 3/1/19, Fred Lunnon <fred.lunnon@gmail.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Newly Created Water Phase Helps Solve Planetary Mystery | | | | | | | | | | | Newly Created Water Phase Helps Solve Planetary Mystery | | |
https://str.llnl.gov/2018-12/millot On Saturday, March 2, 2019, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
Newly Created Water Phase Helps Solve Planetary Mystery
| | | | | |
|
| | | | Newly Created Water Phase Helps Solve Planetary Mystery
|
|
|
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Dan Asimov -
Eugene Salamin -
Fred Lunnon -
James Propp