The evidence that the Hadamard conjecture is true is overwhelming - look at how fast the number of H matrices or order n grows: see A007299 in the OEIS - and also there are a huge number of constructions. On Sun, Oct 7, 2012 at 12:31 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Right. (Warren also mentioned this in e-mail to me.)
I don't know how likely it is that the Hadamard conjecture is true, however.
--Dan
On 2012-10-07, at 9:04 AM, Veit Elser wrote:
The largest subset that forms a simplex cannot be larger than n+1, and by the Hadamard conjecture these exist for all n=3 mod 4. Here's the construction:
Take a Hadamard matrix of order n+1 and flip the sign of each row so that the first element equals 1. Delete the first element in each row and replace -1 with 0. The n+1 rows are then your elements X(n). The Hamming distance between any pair of elements is (n+1)/2; this is the square of the euclidean distance.
-Veit
On Oct 6, 2012, at 6:30 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Consider the corners of the (say) unit n-cube: {0,1}^n, in R^n.
QUESTION: What is the size |X(n)| of a largest subset X(n) of the 2^n corners, such that all nonzero distances in X(n) have the same value?
The distinct distances are {sqrt(k) | 0 <= k <= n}.
For a fixed k > 0, the number of distances = sqrt(k) is 2^(n-1) * C(n,k), where C(n,k) := n!/(k! (n-k)!).
MORE GENERAL QUESTION: What is the size |X(n,k)| of a largest subset X(k,n) having all its nonzero distances = sqrt(k) ?
(These questions can be rephrased to ask what is the size of the largest clique in the constant-distance graphs for {0,1}^n.)
F'rinstance, in {0,1}^4, the graph of all distances = sqrt(2) has two components. Each is isomorphic to what you get if you connect all pairs of vertices of an octagon (or 3-cube), except antipodes, with an edge. So each component of the graph has 8 vertices and 24 edges. The largest clique has only 4 vertices.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I have now retired from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com