3
$\begingroup$

In 1967 H. J. Ryser conjectured that every Latin square of odd order has a Latin transversal. Similar to Latin squares, we may consider Latin cubes.

QUESTION: Let $n$ be any positive integer. Does every $n\times n\times n$ Latin cube contain a Latin transversal?

Let $N$ be any positive integer. In 2008, I proved that for the $N\times N\times N$ Latin cube over $\mathbb Z/N\mathbb Z$ formed by the Cayley addition table, each $n\times n\times n$ subcube with $n\le N$ contains a Latin transversal (cf. my paper An additive theorem and restricted sumsets). Motivated by this, in the 2008 paper I conjectured that my above question has a positive answer.

Any comments are welcome!

$\endgroup$
  • $\begingroup$ Is an $n\times n\times n$ subcube essentially $J_1\times J_2 \times J_3$ where $J_i\subset \mathbb{Z}/N\mathbb{Z}$ for $i=1,2,3,$ and $|J_i|=n$? $\endgroup$ – kodlu Jun 24 '18 at 23:41
  • $\begingroup$ @kodlu Yes, your understanding is correct. $\endgroup$ – Zhi-Wei Sun Jun 25 '18 at 16:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.