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Given a $9 \times 9$ solved Sudoku game with $3 \times 3$ regions, is it possible that one (or more) of the regions are invalid if all rows and columns are valid (i.e. have a unique sequence of $1-9$)?

dragonfly
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1 Answers1

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Yes, it can happen that all $3 \times 3$ regions are invalid:

\begin{array}{|ccc|ccc|ccc|} \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 \\ 3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 \\ \hline 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 \\ 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 \\ 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5 \\ \hline 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5 & 6 \\ 8 & 9 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 9 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \end{array}

user133281
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  • Thanks! Given this special case where all are invalid, is there any other case? – dragonfly Jun 21 '17 at 12:12
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    @dragonfly Swap rows 2 and 4, and rows 3 and 7. The three top regions are now valid, while the rest are not. – jvdhooft Jun 21 '17 at 12:15
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    Permutating the numbers $1$ to $9$ in this example arbitarily gives other solutions, but there will be tons of further solutions. An interesting question would be : What is the probability that a random $9\times 9$-Latin square has the desired property ? And further : What is the probability that exactly $k$ regions are invalid ($k\in [0,9])$ ? – Peter Jun 21 '17 at 12:15
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    Concerning other cases: if two regions in the same row are valid, then all three are, and dually for columns. On the other hand, any pattern of valid and invalid regions that obeys this rule is possible to realize. This, and much more, can be found at https://mathoverflow.net/q/129143 . – Emil Jeřábek Jun 21 '17 at 16:38