Re: [math-fun] Finding equidistant points among the vertices of a cube
On 2018-08-09 13:48, Dan Asimov wrote:
If this is in OEIS, I didn't succeed in finding it.
What about the actual squared distances between the 4K points? (Ought to be a simple f(K), presuming there is just one such.) --rwg -------------
Another math-fun member has kindly pointed out to me that an Hadamard matrix in dimension 4K implies the maximal number (n+1) of equidistant points among the vertices of the (4K-1)-dimensional cube...
...and that since the Hadamard conjecture is that an Hadamard matrix exists in every dimension of form 4K, it would if true mean that n+1 equidistant points exist in each dimension of form 4K-1.
—Dan
----- The relationship is that if there is an n x n Hadamard matrix, then there exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube.
These n vectors are equal length, so their endpoints are equidistant.
I ask about a related but not identical situation: when there exist n+1 equidistant points among the 2^n points of {-1, 1}^n.
And in general, *how large* is the largest subset of {-1, 1}^n consisting of equidistant points?
—Dan
Fred Lunnon wrote: ----- See https://en.wikipedia.org/wiki/Hadamard_matrix in particular, Sylvester construction.
----- On 8/9/18, Dan Asimov <dasimov@earthlink.net> wrote: Suppose the metric space Q_n is the vertices of the n-dimensional cube
Q_n = {-1, 1}^n
with the induced metric from Euclidean space, and the questions is:
Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible?
(And in general, what is the size S(n) of the largest equidistant subset of Q_n?)
If this is in OEIS, I didn't succeed in finding it. ----- -----
there are many seqs in the OEIS related to the Hadamard max det problem. Check the Index to the OEIS See for example this page: https://oeis.org/wiki/Index_to_OEIS:_Section_Ha Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Sat, Aug 11, 2018 at 7:53 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2018-08-09 13:48, Dan Asimov wrote:
If this is in OEIS, I didn't succeed in finding it.
What about the actual squared distances between the 4K points? (Ought to be a simple f(K), presuming there is just one such.) --rwg -------------
Another math-fun member has kindly pointed out to me that an Hadamard matrix in dimension 4K implies the maximal number (n+1) of equidistant points among the vertices of the (4K-1)-dimensional cube...
...and that since the Hadamard conjecture is that an Hadamard matrix exists in every dimension of form 4K, it would if true mean that n+1 equidistant points exist in each dimension of form 4K-1.
—Dan
----- The relationship is that if there is an n x n Hadamard matrix, then there exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube.
These n vectors are equal length, so their endpoints are equidistant.
I ask about a related but not identical situation: when there exist n+1 equidistant points among the 2^n points of {-1, 1}^n.
And in general, *how large* is the largest subset of {-1, 1}^n consisting of equidistant points?
—Dan
Fred Lunnon wrote: ----- See https://en.wikipedia.org/wiki/Hadamard_matrix in particular, Sylvester construction.
----- On 8/9/18, Dan Asimov <dasimov@earthlink.net> wrote: Suppose the metric space Q_n is the vertices of the n-dimensional cube
Q_n = {-1, 1}^n
with the induced metric from Euclidean space, and the questions is:
Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible?
(And in general, what is the size S(n) of the largest equidistant subset of Q_n?)
If this is in OEIS, I didn't succeed in finding it. ----- -----
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f(K) = 2^2 * (4 K)/2 = 8 K ?! WFL On 8/12/18, Bill Gosper <billgosper@gmail.com> wrote:
On 2018-08-09 13:48, Dan Asimov wrote:
If this is in OEIS, I didn't succeed in finding it.
What about the actual squared distances between the 4K points? (Ought to be a simple f(K), presuming there is just one such.) --rwg -------------
Another math-fun member has kindly pointed out to me that an Hadamard matrix in dimension 4K implies the maximal number (n+1) of equidistant points among the vertices of the (4K-1)-dimensional cube...
...and that since the Hadamard conjecture is that an Hadamard matrix exists in every dimension of form 4K, it would if true mean that n+1 equidistant points exist in each dimension of form 4K-1.
—Dan
----- The relationship is that if there is an n x n Hadamard matrix, then there exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube.
These n vectors are equal length, so their endpoints are equidistant.
I ask about a related but not identical situation: when there exist n+1 equidistant points among the 2^n points of {-1, 1}^n.
And in general, *how large* is the largest subset of {-1, 1}^n consisting of equidistant points?
—Dan
Fred Lunnon wrote: ----- See https://en.wikipedia.org/wiki/Hadamard_matrix in particular, Sylvester construction.
----- On 8/9/18, Dan Asimov <dasimov@earthlink.net> wrote: Suppose the metric space Q_n is the vertices of the n-dimensional cube
Q_n = {-1, 1}^n
with the induced metric from Euclidean space, and the questions is:
Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible?
(And in general, what is the size S(n) of the largest equidistant subset of Q_n?)
If this is in OEIS, I didn't succeed in finding it. ----- -----
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participants (3)
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Bill Gosper -
Fred Lunnon -
Neil Sloane