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I can put the 6 combinations of ABC words in a 3x3 square

ABC = 3 letters = 6 combinations in a 3x3 square

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For ABCD = 4 letters = 24 combinations in three 4x4 squares a solution was found by George Sicherman

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Now can you find solutions for

ABCDE = 5 letters = 120 combinations in twelve 5x5 squares and

ABCDEF = 6 letters = 720 combinations in sixty 6x6 squares

If not, which is the most quantity of squares with different words you can find?

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    $\begingroup$ Are you aware of the answers to the questions you pose? Specifically, about whether solutions for ABCDE / ABCDEF exist, and/or what the optimal solutions are? If you know optimal solutions which don't use all of the combinations, are you aware of a proof of opimality? $\endgroup$
    – bobble
    Commented Apr 21, 2023 at 1:15
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    $\begingroup$ I don´t know the answers. I think could exist perfect solutions for cases 5 and 6 as there are perfect solutions for cases 3 and 4 $\endgroup$ Commented Apr 21, 2023 at 1:25

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Here's an explicit solution with $12$ squares for $n=5$, obtained via integer linear programming:

\begin{matrix}ABDEC&ACDEB&ABECD&AEBDC&ACBDE&ADBEC&ABCED&AEBCD&ADEBC&ABEDC&AEDCB&ADBCE&\\BAECD&BAEDC&EADBC&DACBE&DACEB&DAECB&CAEDB&CADBE&EADCB&EABCD&CABDE&CADEB&\\CDABE&DBACE&DCAEB&EBACD&CEABD&CBADE&DCABE&ECADB&CBAED&CDAEB&ECABD&BEADC&\\DECAB&CEBAD&BDCAE&CDEAB&EBDAC&BECAD&BEDAC&DBEAC&DCBAE&BCDAE&BDEAC&EBCAD&\\ECBDA&EDCBA&CEBDA&BCDEA&BDECA&ECDBA&EDBCA&BDCEA&BECDA&DECBA&DBCEA&DCEBA&\\\end{matrix}

I forced $A$ along the diagonals to heuristically reduce the search space.

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Based off your example for $n=4$, there should be an easy generalization for any $n\ge 3$, but I'll examplify using $n=5$ here.

Instead of letting the alphabet be $\{\mathrm{A,B,C,D,E}\}$, we instead use $\{0,1,2,3,4\}$ as they represent the 5 congruence classes of addition mod 5. Fix any two distinct permutations starting with $0$, say $(0,4,2,3,1)$ and $(0,2,4,3,1)$. For the addition table mod 5 using the two permutations as axes: \begin{array}{c|ccccc}+&0&4&2&3&1\\\hline0&0&4&2&3&1\\2&2&1&4&0&3\\4&4&3&1&2&0\\3&3&2&0&1&4\\ 1&1&0&3&4&2\\\end{array} Now notice that each row or column is the corresponding initial permutation but added some constant to all terms. But each permutation $\sigma$ of $\{0,1,2,3,4\}$ is uniquely formed by (a) taking a unique corresponding permutation starting with 0, and (b) adding a fixed value from $\{0,1,2,3,4\}$ and then mod 5 to all terms of the starting permutation. For example, \begin{equation}(2,3,1,0,4)\equiv(0,1,4,3,2)+2\pmod{5}\end{equation} Thus if we take all $(n-1)!$ many permutations starting with 0, group them into pairs, and form corresponding addition tables mod $n$, then we necessarily get each permutation on $\{0,\ldots,n-1\}$ precisely once. In particular, the optimal number of squares is equal to $\frac{(n-1)!}{2}$ for all $n\ge 3$.

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