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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!

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  • $\begingroup$ 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
    Commented Jun 24, 2018 at 23:41
  • $\begingroup$ @kodlu Yes, your understanding is correct. $\endgroup$
    – Zhi-Wei Sun
    Commented Jun 25, 2018 at 16:47

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