3
$\begingroup$

Let $Q_{4n-1}$ be a unit hypercube of dimension $4n-1$. Has the following statement been proven?

There are $4n$ vertices in $Q_{4n-1}$ such that the distance between each pair of them is $2\sqrt{2n}$.

In other words, such vertices form a complete graph of equal sides (=$2\sqrt{2n}$).

$\endgroup$

1 Answer 1

10
$\begingroup$

You probably mean that the hypercube is $Q_{4n-1}=\{-1,+1\}^{4n-1}$. If $u,v\in Q_{4n-1}$ and $\|u-v\|=2\sqrt{2n}$, then $8n-2-2(u,v)=(u-v)^2=8n$, $(u,v)=-1$. Add $(4n)$-th coordinate 1 to $u$, $v$. We get two vectors $U,V\in Q_{4n}$ such that $(U,V)=(u,v)+1=0$. So your question reduces to the famous Hadamard conjecture about Hadamard matrices of order $4k$. Actually two questions are equivalent: if Hadamard's matrix exists, we may suppose without loss of generality that the last column consists only on 1's, and removing the last columns we get $4n$ rows of length $4n-1$ satisfying your property.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.