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A Sudoku puzzle with order $n$ has $n^2 \times n^2$ squares and $n^2$ regions.

For example, a Sudoku puzzle of order $n = 3$ has $9 \times 9$ squares and $9$ regions.

A minimal Sudoku is one with the fewest number of starting clues.

For 1x1 Sudoku of order n=1, the minimum number of starting clues is 1.

For 4x4 Sudoku of order n=2, the minimum number of starting clues is 4.

For 9x9 Sudoku of order n=3, the minimum number of starting clues is 17.

What is the minimum number of starting clues for 16x16 Sudoku of order n=4?

Furthermore, if possible, what is the expression for the relationship between the minimum number of starting clues and order n?

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2 Answers 2

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Given that the solution of the minimal $9 \times 9$ sudoku took the equivalent of 7.1 million core hours on a supercomputer over 11 calendar months, I suspect that the minimum number for a $16 \times 16$ sudoku will not be solved in the next several decades, even allowing for Moore's Law. Forget $25 \times 25$.

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  • $\begingroup$ That does not mean we can't try to solve it mathematically instead! Humanity shall prevail. $\endgroup$ Commented Mar 4, 2017 at 13:21
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    $\begingroup$ @CarolineRudolph: I admire your confidence in "humanity" but as a former professor in both mathematics and computer science (and a former Sudoku addict), I must say that the "rock star" problem of minimal number of constraints just smacks of computation, not math. Related problems (magic squares) have resisted math attacks for millennia. Incidentally, I highly recommend the book "Taking Sudoku seriously" by Jason Rosenhouse and Laura Taalman. Also search YouTube for my TEDx (StanleyPark) talk on sudoku. $\endgroup$ Commented Mar 4, 2017 at 18:41
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Wouldn't a $1\times 1$ sudoku need $0$ clues to be solved uniquely, so the pattern would actually be, $0, 4, 17$?

If the pattern were quadratic (which we have no reason to assume with this few data points), the pattern would be explained with the following:

$$t(n) = 4.5n^2 − 9.5n + 5$$

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