- Code: Select all
*--------------------------------------------------------------------------------*
| 56 7 56 | 2 8 1 | 4 9 3 |
| 1 4 3 | 5 6 9 | 2 8 7 |
| 2 8 9 | 4 3 7 | 16 16 5 |
|--------------------------+--------------------------+--------------------------|
|c569 a136 7 | 16 29 26 | 8 b13456 146 |
| 4 d169-3 26 | 169 5 8 | 136 7 126 |
| 8 16 256 | 3 7 4 | 9 1256 126 |
|--------------------------+--------------------------+--------------------------|
| 3 569 4 | 69 1 256 | 7 26 8 |
| 69 2 8 | 7 49 3 | 5 146 1469 |
| 7 56 1 | 8 249 256 | 36 2346 2469 |
*--------------------------------------------------------------------------------*
Here is a single move that will solve the puzzle and only involves 4 cells. The logic is a bit more involved than we normally get in these sort of enquiries, so I'll talk you through it.
The cells are labelled a-b-c-d. Now suppose cell a was
not 3. Now there are only two 3's in Row 4, so cell b would have to be 3. In particular it would not be 5.
Now there are also only two 5's in Row 4, so cell c would have to be 5. In particular it would not be 9. There are only two 9's in Box 4, cells c and d, so cell d would have to be 9.
So let's see where we have gotten with all this. We have shown that if cell a is not 3, cell d must be 9.
In particular it is not 3.
On the other hand, if cell a
is 3 then cell d is obviously not 3, since it's in the same column or box.
So, if cell a is 3 or is not 3, cell d is not 3. Thus, since cell a can only be 3 or not 3 we can conclude that cell d can't be 3.
In fact, this one elimination solves the whole puzzle, via a cascade of singles.
Leren
PS I just noticed that you have a 9 in r4c4, but not in r4c2 or r9c2.
There is an X Wing of 9's in r57/c24 which makes all three eliminations, so I'm assuming you used this move but just missed the elimination r4c4. In any event it doesn't affect the logic of the move I described above.
Leren