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This is a Sudoku puzzle created by the puzzle maker Shye. It has a very clever and ingenious trick to solve it.

  • Normal Sudoku rules apply in this puzzle.
  • This is an incredibly difficult puzzle to solve and has a very clever and ingenious trick in it.
  • I request solvers to show their work at different stages of the puzzle (as they see fit).

EDIT 1 - Partial Answers are encouraged even if you have done only the Naked Singles.

This puzzle is not my own. Proper attribution will be provided once the puzzle is solved. This is because the page where I saw this puzzle also has its solution in it. For some partial attribution, I have been told that this puzzle was created by the puzzle maker Shye.

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  • 1
    $\begingroup$ Other than perhaps being a difficult sudoku, is there actually any "trick" involved? Because I plugged this into a sudoku solver and it quickly spit out the solution, leading me to believe there's nothing especially interesting about the puzzle. $\endgroup$
    – SQLnoob
    Jun 15, 2021 at 13:28
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    $\begingroup$ @SQLnoob does your solver use only human-like deduction? Because solving a sudoku automatically using brute-force and backtracking is a trivial task for a computer, but that's not what a human would do. I tried a little bit to solve this sudoku but I couldn't get much further apart from writing down ~10 numbers using some basic deduction rules. $\endgroup$
    – melfnt
    Jun 15, 2021 at 14:09
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    $\begingroup$ Having watched the solve of this on Cracking The Cryptic youtube channel there is a very nice trick in it and they titled their video "Classic Sudoku: A new incredible trick" - given how much sudoku they do I assume therefore that this trick is a bit out of the ordinary. $\endgroup$
    – Chris
    Jun 15, 2021 at 16:53
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    $\begingroup$ @KabirKanhaArora: Not sure what I said that spoiled anything? The link to logic masters germany is just the same puzzle as far as I can see and saying that it was solved on Cracking The Cryptic channel doesn't spoil anything. Sure, people can go there and look at a full solve but we're not children. If somebody wants to look at the solution they can. If they don't want spoilers they just don't go there... Unless you think me saying "its a nice trick" is a spoiler in which case I would argue that it is no more of a spoiler than the original question... $\endgroup$
    – Chris
    Jun 16, 2021 at 8:14
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    $\begingroup$ @SQLnoob "you'd solve it the same way you solve every other normal sudoku" did you mean by only using some of the tecniques listed here? I tried and I failed, hence there must be some "trick". Also, human-like deduction is not slightly-refined brute force, they are two different algorithms. Anyway I won't comment further, not here nor on the answer below since it's clear that neither of us is going to change their mind. Sorry $\endgroup$
    – melfnt
    Jun 16, 2021 at 12:54

2 Answers 2

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


         a   b   c       d   e   f       g   h   i

New entries are marked by *



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


         a   b   c       d   e   f       g   h   i

Multi-digit entries are understood to show all that is not ruled out yet.

    1    .   1   2   |   .   7   .   |   3   5 *89
    2    3   .   .   |   2  *69  1   |   4 *89   7
    3    4   .   .   |   5   3   .   |   2   1   6
         ------------+---------------+------------
    4    2   .   .   |   .   .   .   |   .   7   .
    5    1   6   .   |   .   8   .   |   .   4   2
    6    .   4   8   |   .   .   .   |   6   3   .
         ------------+---------------+------------
    7    .   3   1   |   8   .   .   |   .   .   4
    8    .   2   4   |   1   5   .   |   .   .   3
    9    .   .   .   |   .   4   3   |   1   2   .


         a   b   c       d   e   f       g   h   i

if e2 = 6 must have a1 = 6 if e2 = 9 cannot have a1 = 9 therefore never a1 = 9, leaving 68 

    1   *68  1   2   |   .   7   .   |   3   5  89
    2    3   .   .   |   2  69   1   |   4  89   7
    3    4   .   .   |   5   3   .   |   2   1   6
         ------------+---------------+------------
    4    2   .   .   |   .   .   .   |   .   7   .
    5    1   6   .   |   .   8   .   |   .   4   2
    6    .   4   8   |   .   .   .   |   6   3   .
         ------------+---------------+------------
    7    .   3   1   |   8   .   .   |   .   .   4
    8    .   2   4   |   1   5   .   |   .   .   3
    9    .   .   .   |   .   4   3   |   1   2   .


         a   b   c       d   e   f       g   h   i

Very clever and ingenious trick / shameless cheat:

If a1 = 8 then b9 = 8, h2 = 8, g8 = 8, i4 = 8 the last forces e4 = 1 and we see that d1,f1,d4,f4 must be 2x4 + 2x6 If we are allowed to use the assumption that the puzzle has a unique solution then this is a contradiction Therefore a1 = 6



    1   *6   1   2   |  *4   7  *89  |   3   5  89
    2    3   .   .   |   2  *6   1   |   4  89   7
    3    4   .   .   |   5   3  *89  |   2   1   6
         ------------+---------------+------------
    4    2   .   .   |  *6   .  *4   |   .   7   .
    5    1   6   .   |   .   8   .   |   .   4   2
    6    .   4   8   |   .   .   .   |   6   3   .
         ------------+---------------+------------
    7    .   3   1   |   8   .   .   |   .   .   4
    8    .   2   4   |   1   5   .   |   .   .   3
    9    .   .  *6   |   .   4   3   |   1   2   .


         a   b   c       d   e   f       g   h   i

If f5 = 7 must have a6 = 7 and d9 = 7 leaving no room for 7 in bottom left box Therefore f5 != 7 leaving 5
The rest is routine

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


         a   b   c       d   e   f       g   h   i

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I must have missed the trick somehow. You can fill in a few 1/2/3/4s straight away. After that there are only two spaces left in the top right, which are 8/9 in some order. I tried the one that seemed to give most information, hoping to get a contradiction and eliminate it. After that I just followed my nose - you can get the rest of the 8s, then the 6s and a couple of 5s, then that gives you the 7 in the bottom right which starts another chain of deductions. Somewhat to my surprise, I didn't get a contradiction but a solution.

enter image description here

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  • 1
    $\begingroup$ Having watched the video of the solution another commenter posted, I can confirm there's no trick. It's just a sudoku. $\endgroup$
    – SQLnoob
    Jun 16, 2021 at 12:38
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    $\begingroup$ Trying one possibility and hoping to find a contradiction is a form of bifurcation which is strongly avoided in a proper logical-deduction & grid-deduction setting. I'm pretty sure finding a smart argument i.e. why the other possibility is wrong without bifurcation, is what this question intended to. $\endgroup$
    – athin
    Jun 16, 2021 at 12:52
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    $\begingroup$ @melfnt the problem is that the logical deduction steps used in sudoku all amount to the same thing as following your nose - "if this cell was X, then this other cell would have to be Y, but it can't be Y so therefore this cell isn't X" etc. all sudoku "tricks" are just variants of the same exact thing. $\endgroup$
    – SQLnoob
    Jun 16, 2021 at 12:52
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    $\begingroup$ @SQLnoob True that logical deduction steps are somewhat finding an contradiction and such, but to some extent. I love to link this blogpost by Thomas Snyder (a.k.a Dr Sudoku): gmpuzzles.com/blog/2013/03/… to define the "limit" of trial-and-error. In short, here is his quote: "When I can solve a puzzle in ink, without erasures, with all deductions either positive or negative coming from visualization in my head and not making scratch-work on the paper, the puzzle is solvable by logic." $\endgroup$
    – athin
    Jun 16, 2021 at 12:57
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    $\begingroup$ @athin that sounds like a limit of his working memory, not logic. he even says "If you’ve never reasoned out before why a 7 cannot go into R8C2, this may feel like guesswork. The Y-wing step, while mathematically solid in a sudoku language of strong and weak links, feels like guesswork until you get used to seeing it." that's what all sudoku "techniques" are, they're just variants of "if this cell is an X then this other cell can't be a Y..." etc. there's nothing fundamentally different between "guessing" that a cell is a 7 and seeing what happens, and "using a Y-wing" or whatever. $\endgroup$
    – SQLnoob
    Jun 16, 2021 at 13:06

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