In my solution to the 24 Nov 2006 Nightmare, in two (consecutive!) steps I found patterns in which the surplus candidates did not quite form a locked set, but instead formed an Almost Locked Set (ALS) with other cells, and these could be used with other ALS's in an ALS XZ rule elimination.
Code: Select all
.------------------.----------------------.------------------.
| 148 9 478 | 15 6 1245 | 1257 3 245 |
| 6 *25 47 | 1359 *25+49 12345| 1257 125 8 |
| 14 *25 3 | 7 *25+4 8 | 9 1256 2456 |
:------------------+----------------------+------------------:
| 89 7 2 | 16 3 16 | 4 589 59 |
| 5 3 48 | 2 A48 9 | 6 7 1 |
| 489 6 1 | 58 7 45 | 3 289 29 |
:------------------+----------------------+------------------:
| 3 1 5 | 4 A29 7 | 8 B269 B269 |
| 2 4 69 | 3-5689 A589 3-56 |B15 B159 7 |
| 7 8 69 | 569 1 256 | 25 4 3 |
'------------------'----------------------'------------------'
In this position, the potential non-unique rectangle in r23c25 (marked with "*") contains surplus candidates 4 and 9, in column 5 only. These candidates form an ALS with r578c5 (marked with "A"), containing digits {2,4,5,8,9}. The other ALS, marked with "B", is in r7c89 and r8c78, containing digits {1,2,5,6,9}. The "restricted common" digit is X=2, and the eliminated digit is Z=5, in r8c46.
After these eliminations:
Code: Select all
.------------------.---------------------.------------------.
| 148 9 478 |A15 6 -1245 | 1257 3 245 |
| 6 25 47 | 1359 2459 -12345| 1257 125 8 |
| 14 25 3 | 7 245 8 | 9 1256 2456 |
:------------------+---------------------+------------------:
| 89 7 2 | 16 3 B16 | 4 589 59 |
| 5 3 48 | 2 48 9 | 6 7 1 |
| 489 6 1 |A58 7 45 | 3 289 29 |
:------------------+---------------------+------------------:
| 3 1 5 | 4 29 7 | 8 269 269 |
| 2 4 *69 |*69+38 589 B36 | 15 159 7 |
| 7 8 *69 |*69+5 1 256 | 25 4 3 |
'------------------'---------------------'------------------'
Here the potential non-unique rectangle, again marked with "*", lies in r89c34, with surplus candidates 3,5,8 in column 4 only. These combine with r16c4 (marked with "A") to form an ALS containing digits {1,3,5,8}. The other ALS, marked with "B", lies in r48c6 and contains digits {1,3,6}. The "restricted common" digit in this case is X=3, one of the surplus candidates. The eliminated digit is Z=1 in r12c6.
The steps preceding and following these steps were not entirely straightforward, as some grouped chains were involved, but for brevity I won't discuss these (unless someone requests).