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.


  • 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.

  • $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.