It certainly is a hard one. However, it has rank zero logic which will result in a reduction of 53 candidates:
16 Truths = {1368R3, 1368R4, 1368R5, 1368R9}
16 Links = {136c1, 6c3, 38c4, 8c8, 136c9, 3n6, 4n5, 4n7, 5n2, 9n5, 9n6}
23 Eliminations: 36<>4, 45<>2, 47<>9, 52<>2, 95<>2, 95<>7, 96<>2, 13<>6, 14<>3, 19<>1, 19<>3, 19<>6, 21<>1, 28<>8, 29<>1, 61<>6, 63<>6, 69<>6, 71<>3, 74<>3, 81<>3, 81<>6, 89<>3
then
Naked quads of 4579 at r1c8, r1c9 and r2c8 and r2c9
5s at r3c1 and r3c3 only ones in row.
Naked triplets of 249 at r1c3, r2c1 and r2c2
Puzzle state after this:
- Code: Select all
8 16 249 | 27 12357 123 | 36 4579 4579
249 24 3 | 6 2578 28 | 1 4579 4579
156 7 56 | 4 9 13 | 2 8 36
---------------------+----------------------+---------------------
123469 5 2469 | 2389 368 7 | 68 249 12469
12369 1368 269 | 2389 4 5 | 7 29 1269
2479 2468 2479 | 1 268 2689 | 5689 3 2459
---------------------+----------------------+---------------------
2457 234 1 | 279 237 2349 | 359 6 8
247 2346 8 | 5 2367 23469 | 39 1 279
23567 9 2567 | 2378 1368 1368 | 4 257 2357
Alternatively, use could use the following multi-segment locked sets rank zero logic with same eliminations:
Base: 1368
20 Cell Truths: r12678 c2567
20 links: 136r1, 18r2, 68r6, 3r7, 36r8, 24c2, 257c5, 249c6, 59c7
After this, there is still a lot of hard work to do with complex chains.
Good luck,
Phil
PS there are voluminous threads on these rank zero techniques in the "advanced techniques" section