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.

- 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 |

'---------------------'

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!