In column 1 there is a naked pair on 67.
We have 2 cells and two candidates. "That uses up" all the 6s and 7s in that column.
Thus the 7s are not possible in r7c1 and r8c1.
- Code: Select all
.---------------------.
| 2 7 68 |
| 9 4 68 |
| 5 3 1 |
:---------------------+
|#67 8 5 |
|#67 1 279 |
| 3 29 4 |
:---------------------+
| 148-7 6 279 |
| 48-7 29 3 |
| 18 5 29 |
'---------------------'
In the middle box there is a naked pair on 67.
There can be no other 6 or 7 in that box, thus r5c3 cannot be 7.
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.---------------------.
| 2 7 68 |
| 9 4 68 |
| 5 3 1 |
:---------------------+
|#67 8 5 |
|#67 1 29-7 |
| 3 29 4 |
:---------------------+
| 148 6 279 |
| 48 29 3 |
| 18 5 29 |
'---------------------'
It doesn't matter which order you do them in or whether you do them all "at once".
You could have looked at the last elimination differently, still valid, but it shows a slightly different type of elimination.
In the middle box only cells r5c3 and r6c2 have the digits 29.
That is a hidden pair and since we have 2 cells and 2 candidates there can be no other candidates in those 2 cells and we eliminate the 7 from r5c3
- Code: Select all
.---------------------.
| 2 7 68 |
| 9 4 68 |
| 5 3 1 |
:---------------------+
| 67 8 5 |
| 67 1 #29-7 |
| 3 #29 4 |
:---------------------+
| 148 6 279 |
| 48 29 3 |
| 18 5 29 |
'---------------------'
Edited to correct typos, thanks clivem!