Puzzle "Extreme 71"
here Starting grid
2 . . | 1 . 9 | 4 . .
. 8 . | . . 3 | . . .
. . 7 | . 8 . | . . .
------+-------+------
9 2 . | 5 . . | . . .
6 . . | . . . | . . 1
. . . | . . 4 | . 2 5
------+-------+------
. . . | . 2 . | 8 . .
. . . | 9 . . | . 7 .
. . 8 | 4 . 1 | . . 3
Position after basic eliminations including finned x-wings (ok they aren't that basic)
as per post (Mike Barker here)
+-------------------------+------------------+---------------------+
| 2 356 356 | 1 567 9 | 4 3568 678 |
| 15 8 1569 | 267 4 3 | 1567 156 2679 |
| 1345 134569 7 | 26 8 56 | 13569 13569 269 |
+-------------------------+------------------+---------------------+
| 9 2 134 | 5 1367 67 | 367 3468 4678 |
| 6 3457 345 | 8 379 2 | 379 349 1 |
| 8 137 13 | 367 13679 4 | 3679 2 5 |
+-------------------------+------------------+---------------------+
| 13457 134569 134569 | 367 2 567 | 8 14569 469 |
| 1345 13456 2 | 9 356 8 | 156 7 46 |
| 57 569 8 | 4 567 1 | 2 569 3 |
+-------------------------+------------------+---------------------+
Solution (algebraic notation)
1. Discontinuous nice loop from 1D5 => elim D5 <6> <7>
2. Discontinuous nice loop from 9F5 => elim F5 <7>
3. AIC : strong link between 6F5 6D6 => elim <6> F4
4. Triple (137) in F2/F3/F4 => elim F5 <1> <3>+elim F7 <3> <7>
5. Pointing pairs 1F2+1F3 => elim D3 <1>. Placement D5 : 1
6. Discontinuous nice loop from 5J1 =>elim J1 <7>. Placement J1 : 7. Elim : G1 J5 <7>.
7. Intro : next move can be thought of as ALS-XY wing style. ALS-XY wing works like this : the eliminated item-to-be locks two Almost Locked Sets (ALS) one of which then generates a double lock on the other set, which is impossible, so eliminating the generator of the impossibility. Here 5A8 locks J8+H9+G9 and A2+A3. So the idea is to see whether one of those locked sets then double locks the other and if so, 5A8 can be eliminated.
In effect the first set further locks the second (although untidily in that it empties A2).
8. Discontinuous loop from 6C6 : leads quickly to elim C6 <6>. Placement C6 :5. Elims A5 C1 C2 C7 C8 G6 <5>.
9 Pointing pairs 5A2+5A3 => Placement B1 : 1. Elim B3 <5>, B7 B8 C1 G1 H1 B3 C2 <1>.
10 X-wing 5 A2+A3/E2+E3 => elim G2 G3 H2 J2 <5>
11. AIC establishing strong liink between 6D6/6D9 => elim D7 D8 <6>
12. AIC establishing strong link between 6F7/6B7 =>elim C7 H7 <6>
13. RGT from bivalue (3,7) in F4 : establishes strong link between 3H5/5H7 => elim H5 <5>=> placement J5 :5 => elim J8 <5> => triple (469) J8/H9/G9 =>elim G8 <4> <6> <9>.
14. It is fairly easy to see that F5 cannot be 9. One of the problems in sudoku theory is that certain people will not allow contradiction as a valid elimination method. When you do encounter a contradiction, especially an obvious one what do you do ?
The best approach is to learn from the situation : see if some existing method will achieve the same. Or why not...see if a new pattern can be developed.
Otherwise, just accept the situation or re-engineer to something acceptable, which is always possible, and a little educational : any elimination can be re-presented as a form of discontinous loop.
15. Discontinous nice loop from 9E7 => 7D7 => elim E7 <7>
16. RGT from bivalue (3,7) D7 : establishes strong link between 6A8/6B8. Elim all other 6s box 3, placement 9 in J8. Multiple other eliminations to solution