I have the following scheme:


Please look at the lower left subgrid. The 8 may go in R7C1 or R9C1.

If I put the 8 in R7C1 I'd have a 4 in R9C7 and a 5 in R8C7

pass 1_1

If I put the 8 in R9C1 I'd have the following situation:

pass 2_1 | pass 2_2 | pass 2_3

In this last scheme it doesn't matter where I put 4, 5 and 6. It will be correct anyway.

So, based on these considerations, may I put a 4 in R9C7 and a 5 in R8C7 and be sure they will be correct?


If you put the $8$ into R9C1 you have multiple solutions, so you might as well go ahead and assume it is R7C1 and see if this gives a unique solution or even a solution at all.

If not, write to the publisher of the Sudoku and ask for your money back!

  • $\begingroup$ mmm no, in the second case R7C6=8, because 4 and 6 would go in R8C5 and R9C5 :) $\endgroup$
    – Massimo
    Sep 26 '19 at 13:14
  • $\begingroup$ sorry, getting confused in my old age! @Massimo $\endgroup$
    – JMP
    Sep 26 '19 at 13:28

What you are describing is called a deadly pattern. Essentially, if some situation would result in multiple solutions, then you can deduce that the situation is invalid (assuming the sudoku has a unique solution). This leads to advanced solving strategies where you can spot potential deadly patterns and rule out possible cell values that would lead to those patterns.

In your case, you have found that putting an 8 in (9,1) causes a deadly pattern, which means that you can rule out that possibility, which forces 8 to be in (7,1) instead.

But there is a more conventional way of proceeding:

  1. In the middle sector, the 1 can only be in (6,4) or (6,5). This eliminates 1 as a possibility in (6,7).
  2. Now look at column 7. The 1 can only be in (2,7) or (3,7). This eliminates 1 as a possibility in (3,8) and (3,9).
  3. In the middle right sector, a 6 can only be in (4,7) or (6, 7). This eliminates 6 as a possibility in (1,7) and (3,7). At this point, here is the situation:

enter image description here

  1. You can use a technique known as XY-Wing to proceed. Notice that the cell marked with green can only be 27. It is connected to the two red cells that can only be 67 and 26 respectively. If the green cell were 2, the red 26 cell must be 6. If the green cell were 7, the red 67 cell must be 6. In other words the green cell forces one of the two red cells to be 6. Since one of the two red cells must be 6, any cell connected to both red cells (marked in purple) cannot be 6. This causes cell (1,9) to be 9 and cell (3,6) to be 4. From there the rest is solved easily.
  • $\begingroup$ I have to admit I hadn't noticed the 1s in the middle sector... It was surely a more conventional way :) Thanks for your very helpful answer! $\endgroup$
    – Massimo
    Sep 26 '19 at 20:49

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