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Leonhard is designing a puzzle for Katharina. It has nine squares in a 3 × 3 grid and a number of clues. Each clue is a number 1, 2 or 3 placed in one of the squares. Katharina then has to find a solution by placing 1, 2 or 3 in each of the remaining squares so that no row or column has a repeated number. What is the smallest number of clues that Leonhard could include so that his puzzle has exactly one solution?

I tried to solved this problem and came up with the answer 4 but this isn't correct. My solution was for Leonhard to provide Katharina with clues for the top row middle, middle square, middle row left and top row left. Is this correct or can Leonhard use even less clues to provide only 1 solution for his puzzle?

Thanks.

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1 Answer 1

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The answer should be 2. Place a "1" in the center, then a "2" in one of the corners. The black numbers are given while the red numbers are deduced.

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It is also easy to see that this bound is tight because only 1 given number regardless of position will have multiple solutions.

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  • $\begingroup$ Also any two starting squares will do the job as long as they are not in the same row or same column. $\endgroup$
    – Soumik Mukherjee
    May 31 at 21:13

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