# Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?

Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?

This one has $1728$ such submatrices, which is as low as I've gotten: $$\begin{bmatrix} 6 & 7 & 8 & 9 & 1 & 4 & 2 & 3 & 5 \\ 5 & 6 & 1 & 7 & 2 & 8 & 3 & 4 & 9 \\ 9 & 1 & 6 & 2 & 4 & 3 & 7 & 5 & 8 \\ 4 & 5 & 3 & 6 & 8 & 7 & 1 & 9 & 2 \\ 1 & 2 & 4 & 8 & 3 & 5 & 9 & 6 & 7 \\ 2 & 3 & 7 & 4 & 9 & 6 & 5 & 8 & 1 \\ 8 & 9 & 2 & 3 & 5 & 1 & 6 & 7 & 4 \\ 7 & 8 & 5 & 1 & 6 & 9 & 4 & 2 & 3 \\ 3 & 4 & 9 & 5 & 7 & 2 & 8 & 1 & 6 \\ \end{bmatrix}$$

It doesn't seem likely that random Latin squares will help much; they average in the thousands of such submatrices. The one above is the best random Latin square I've found so far (although, I haven't busted a gut doing this; it seems like it won't work anyway).

The groups of order $9$ have lots ($C_9$ has $5832$ and $C_3 \times C_3$ has $19440$).

This question was motivated by answering this math.SE question which asks if any $9 \times 9$ Latin square can have its rows and columns permuted to give a sudoku square.

One way to find an explicit counterexample would be to find a $9 \times 9$ Latin square with no $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$. But this attempt didn't work since I couldn't find one. Hence my question.

$$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 4 & 3 & 6 & 7 & 8 & 9 & 5 \\ 3 & 4 & 2 & 1 & 8 & 9 & 5 & 6 & 7 \\ 4 & 3 & 1 & 2 & 7 & 8 & 9 & 5 & 6 \\ 5 & 6 & 9 & 8 & 2 & 3 & 4 & 7 & 1 \\ 6 & 7 & 5 & 9 & 1 & 2 & 3 & 4 & 8 \\ 7 & 8 & 6 & 5 & 9 & 1 & 2 & 3 & 4 \\ 8 & 9 & 7 & 6 & 4 & 5 & 1 & 2 & 3 \\ 9 & 5 & 8 & 7 & 3 & 4 & 6 & 1 & 2 \\ \end{bmatrix}$$

I discovered it by modifying a Latin Square solver to add the restriction that no 3x3 submatrix can contain 9 different values. It found this in about 1 hour.

• Fascinating. Nice find! (Actually, I'm a bit surprised one even exists: I was obviously looking in the wrong spot.) Jun 9 '15 at 0:22
• We cited your answer in our recent paper Covers and partial transversals of Latin squares. Jun 20 '18 at 14:12
• Thanks, Rebecca! :) Jun 21 '18 at 14:39

We might consider the general concept of a saturated sub-matrix, or $SSM$, being an $M \times K$ sub-matrix within a Latin square of of rank $N = M \times K$, that contains all $N$ values.

We are looking for Latin squares that are completely void of $SSM$s, so they might well be called poly-unsaturated!

For example $6 = 3 \times 2$. Of the 9408 reduced forms, just 6 are poly-unsaturated. Here is one of them:

$$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \\ 3 & 1 & 5 & 6 & 2 & 4 \\ 4 & 5 & 6 & 2 & 1 & 3 \\ 5 & 6 & 1 & 3 & 4 & 2 \\ 6 & 4 & 2 & 5 & 3 & 1 \end{bmatrix}$$

The counter is still running for $9 = 3 \times 3$, but these are definitely scarce objects.

I have also confirmed existence for $8 = 4 \times 2$. One is tempted to conjecture that for all $M,K \gt 1$, whenever $M \neq N$ they will be found to exist.

For $M = K$, well we now know that $N=9$ has poly-unsaturated forms, and $N=4$ does not. I don't know enough about cyclic groups to predict the result for $N=16$.