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It is known that the minimum number of clues a sudoku must have to have a unique solution is 17. But on every website I've seen for them, I haven't found any that are rotationally symmetrical.

I once saw a puzzle in a book which only had 19 clues and was rotationally symmetrical, but I don't remember whether the solver I ran it through said that there was a unique solution or not. My question is, what is the minimum number of already-filled-in squares that a rotationally symmetrical Sudoku must have in order to have a unique solution?

Note: by rotationally symmetrical, I'm referring to it in the standard sense that if a clue appears in one position, a clue will also appear in the position opposite it on the board.

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  • $\begingroup$ rotationally symmetric how, just the center square stops it from being symmetric $\endgroup$
    – ratchet freak
    Commented May 16, 2014 at 14:11
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    $\begingroup$ The center square is symmetrical with itself. $\endgroup$
    – user88
    Commented May 16, 2014 at 14:12
  • $\begingroup$ then it isn't a valid sudoku, all 9 digits must occur exactly once, any symmetry (besides 360°) would require a number to occur twice $\endgroup$
    – ratchet freak
    Commented May 16, 2014 at 14:13
  • $\begingroup$ By rotationally symmetrical, I mean the positions of the clues can be rotated 180 degrees and will overlap itself perfectly. $\endgroup$
    – user88
    Commented May 16, 2014 at 14:15
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    $\begingroup$ @ratchetfreak why do you think the center can't be symmetric? Like Joe said, it is trivially symmetric. $\endgroup$
    – Kevin
    Commented May 16, 2014 at 14:59

1 Answer 1

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This book appears to have a puzzle with only 18 clues that is rotationally symmetrical.

7 2 . | . . . | . . .
. 5 . | . . 9 | . . .
. . . | . 3 8 | . . .
------+-------+-------
. . . | 4 . . | 5 . .
. . 3 | . . . | 9 . .
. . 1 | . . 3 | . . .
------+-------+-------
. . . | 2 5 . | . . .
. . . | 6 . . | . 3 .
. . . | . . . | . 1 9

But I still don't know if a rotationally symmetrical 17 exists.

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    $\begingroup$ I am totally bluffed, because these numbers are as well the first 18 digits of PI $\endgroup$
    – Wa Kai
    Commented Sep 14, 2015 at 15:26
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    $\begingroup$ @WaKai, At first blush that seems interesting, but the specific digits are arbitrary, that is, if we changed the two 1s to 3s, and the four 3s to 1s, it would still work, without being the first 18 digits of PI. $\endgroup$
    – agc
    Commented Jul 11, 2019 at 5:19