Aniedunai wrote:Yes in c3 thats where the two possible places the 4 can go along with the other two 4 on c7. The number 7 is already done I had just forgotten to put it before.

- Code: Select all
`. . 7 | 4 . . | . . .`

1 . . | . . . | . . 7

. 2 . | 9 . 7 | . 8 .

------+-------+------

. 1 2 | 3 7 4 | . 9 8

7 4 9 | 8 6 5 | . 3 .

3 . 8 | 1 2 9 | 7 . .

------+-------+------

4 . . | . . . | 8 7 .

2 7 6 | 5 . . | 3 . .

. . 1 | 7 4 3 | . 5 .

Still looking at your initial board, the 89 pair at r9c12 implies a 26 pair at r9c79.

Now looking at the givens 2 and 6 at boxes 5,7, we see that at row 7 the only places for 2 and 6 are at r7c46 (as a 26 pair).

That also implies that r2c4 is 2 or 6. These three (26) pair cells (r2c4,r7c46), called my attention

due to a potential UR (unique rectangle), so we must have (8)r2c6.

Now r8c6 has to be 1 and now there is only one place for 8 and 9 at box 8, etc

EDIT: re-phrased the text for clarity (my thanks to Maxito_Bahiense)