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Since "difficult" is undefined, let me define it arbitrarily: A Sudoku is difficult if it can't be solved by only considering singles (naked or hidden), the most basic solving strategy.

How many leads can be given such that a Sudoku is still difficult? (Note: I thought I ran into one candidate, but https://www.sudoku-solutions.com/ proved otherwise. You can verify your attempt there, use "Preferences".)

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  • $\begingroup$ You might want to add that the sudoku has to have a unique solution. Otherwise I can make one which has 77 clues and can't be solved by singles because it doesn't have a unique solution. $\endgroup$
    – Wheat Wizard
    Oct 1, 2021 at 18:24
  • $\begingroup$ It's also a good question to ask this about doubles, triples, etc. as well. I have ran into needing Y-wings and/or uniqueness very late in puzzles quite a few times. $\endgroup$
    – Max Xiong
    Oct 1, 2021 at 22:06
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    $\begingroup$ @WheatWizard: For me the name "Sudoku" implies it has a solution and it is unique, but that of course is just my naming convention. $\endgroup$
    – Hauke Reddmann
    Oct 2, 2021 at 8:36

1 Answer 1

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Have a look at the thread below which catalogs the most clueful puzzles discovered at each Sudoku Explainer difficulty rating:

http://forum.enjoysudoku.com/maximum-number-of-clues-for-a-given-se-rating-t6210-120.html

SE assigns a rating of 1.2 to puzzles that can be solved only with singles, so the max clue puzzles for anything beyond 1.2 would qualify. The thread lists two such "hard" puzzles with 70 clues:

4.2 | 70 | 96173254884395672172581463919467385268752.3.4352..8.67419385276238.67..55762...83
5.6 | 70 | 12345678945678912378913256423.9.58.66.52.893.8973612453625174985.86943.29..82365.

For example, the second one is this:

clueful sudoku not singles

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  • $\begingroup$ Looking at the Sudoku, this was exactly what I searched. My gut feeling was "What? Unique?" And I would solve it by testing (probably in my head). $\endgroup$
    – Hauke Reddmann
    Jan 26, 2022 at 20:38
  • $\begingroup$ Just curious, how to solve the second one? (Beyond the obvious one of simply trying one value and complete the board, I mean, it's doable in this case, but what general technique is needed?) $\endgroup$
    – justhalf
    Jan 27, 2022 at 6:50
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    $\begingroup$ @justhalf If r5c2 was 1, then r8c2 is 7 and r8c8 is 1. Now you can't put a 1 into the center right box. $\endgroup$
    – Kruga
    Jan 27, 2022 at 9:19

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