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I think I am missing something in the instance below, since I could solve it only with a long-winded chain of deductions starting with the assumption that (coordinates are (row,column)):

8 is in (8,7),

and continuing as follows:

=> 8 is in (7,6) => 9 is in (1,4) => 9 is in (5,5) => 9 is in (4,1) => 9 is in (7,3) => 3 is in (9,2) => 3 is in (8,7)

a contradiction.

I just started learning Sudoku and it never happened before to have to do this, so maybe I am missing something, or I am not aware of a simple(r) technique applicable in this instance.

EDIT:

  • Coordinates are (row,column), starting with the upper left square which is (1,1);

  • Candidates for a number in a box are filled in only if there are at most two of them. With the exception of 3 and 8 in the bottom-center box, that are filled in even if there are 3 candidates.

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  • $\begingroup$ Which site is this Sudoku from? $\endgroup$ – JMP Sep 23 at 20:58
  • $\begingroup$ It is from an app aptly called "Sudoku" $\endgroup$ – user445082 Sep 23 at 21:00
  • $\begingroup$ You may want to double check your candidate numbers, they are not complete or consistent. For instance, the bottom-left of the center box could be a "9" but this is not listed. $\endgroup$ – Mikey T.K. Sep 23 at 21:10
  • $\begingroup$ @Mikey You are right, I should have specified the criteria I used to fill in the candidates: they are there only if there are at most 2 candidates of a given number in a box. In the center box, as you notice, there are three cells 9 can be in, so I didn't fill it in. The exception is the bottom-center box, where 3 and 8 are filled in, even if there are more than 2 squares they can be in. $\endgroup$ – user445082 Sep 23 at 21:15
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You can deduce:

Cell (1,6) must be an 8.

The reason is that the rows and columns of (1,6) contain 2345679 leaving only 1 and 8 as possibilities. But if you look at the upper left sector, the 1 must be in one of the top two cells, because the other 2 cells cannot be 1. Therefore, since 1 must be in either (1,1) or (1,2), (1,6) can not be a 1. This leaves 8 as the only possibility.

From, there, cell (7,6) must be a 3. I didn't keep going after that.

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A different approach:

8 in (7,6) => 9 in (7,1) and 3 in (7,3), which disagrees with (4,1)=9.

The logic bit:

Row 7 should read (389)4(389) | (56)(56)(389) | 721, but (2,1), (2,3) and (9,6) mean that it becomes (89)4(39) | (56)(56)(38) | 721. So, 8 in (7,6) makes (7,1)=9 and (7,3)=3.

The (56) is forced by the 5 and 6 in row 8 and column 6, and the 3 filled cells on row 9 in block {3,2}.

To summarize, if (7,6)=8, then (4,1)=9, which removes all candidates from (7,1).

You can remove some of your original logic by noticing a contradiction on row 7:

9 in (4,1) means (7,1) is 8, but so is (7,6).

Another way:

Let (7,1)=9. Then {(4,1),(4,2)}={2,7}, so (4,7)=9 -> (6,4)=9 -> (1,5)=9 -> (1,6)=8 -> (7,6)=3, which means (7,3) is stranded.

Finally, but a bit sneaky:

The 5,6 in block {3,2} is forced, and 5 in block {1,2} is in the middle row, so the 6 must be on top in that block (assuming that the Sudoku has a unique solution), which makes (2,2)=6.

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  • $\begingroup$ Actually, if 8 in (7,6) then 9 in (7,3) follows. Thank you for your answer, I know it's hard to follow, that's why I'm asking if there was some other easier fact I was missing. $\endgroup$ – user445082 Sep 23 at 21:51
  • $\begingroup$ Wait, I think your first 2 spoilers are assuming my candidates are row/column-wise, but actually they are box-wise (that's why you find contradictions). The third spoiler is interesting, but I can't understand why, if 5 is in the middle row of {1,2}, then I can conclude that 6 is in its upper row. $\endgroup$ – user445082 Sep 23 at 22:11

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