There are a few patterns, not so rare though, all yielding the same set of eliminations, totally useless for solving the puzzle:
- Code: Select all
+---------------------+-----------------------+-------------------+
| 6 1 257 | 28* 3 259-8* | 89 4 257 |
| 24 78 257 | 14-28 6 12459-8* | 189 3 257 |
| 234 38 9 | 7 58* 1245-8* | 18 6 25 |
+---------------------+-----------------------+-------------------+
| 8 9 1 | 5 7 34* | 34 2 6 |
| 5 2 6 | 48-3 1 348* | 7 9 34 |
| 7 34 34 | 9 2 6 | 5 8 1 |
+---------------------+-----------------------+-------------------+
| 23 5 8 | 23 4 7 | 6 1 9 |
| 19 6 234 | 123 59 25-13 | 234 7 8 |
| 19 347 2347 | 6 89 18* | 234 5 34 |
+---------------------+-----------------------+-------------------+
Look at the 8 cells b2p13689, r469c6; they contain 7 digits (1234589) => one digit must there twice; only 8 can be twice (in b2p18 and r59c6), all other digits are there.
=> - 28r2c4, -3r5c4, -13r8c6, -8r123c6
(cannibalistic 8123c6 eliminations: this group is in sight of both 8b2p18, 8r59c6, one of which is true)
Same eliminations with:
- Doubly linked ALS's: (124589)b2p13689, (1348)r459c6, RC 1,4
- Sue de Coq (extended): (124589)r123c6, (258)b2p18, (1348)r459c6, RC (258) resp. (14), or RC (25) resp. (148)
- loop (1=8)r9c6 - r5c6 = (8-4)r5c4 = (4-1)r2c4 = (1)r8c4
(RC = Restricted Commons)
Anyhow, puzzle solved in one step by one AIC:
(8=2)r1c4 - (2=3)r7c4 -r7c1 = r3c1 - (3=8)r3c2 =>-8r3c5; ste