Here is my walkthrough & the solution:
1. Innies of R5 -> R5C5 = 6
2. Innies of R234 -> R4C5 = 9
3. Cage 20/3 in C5 -> R6C5 = 5
4. Innies of C5 -> R19C5 = 15 = {78}
5. Innies of C1234 -> R19C4 = 12
6. Innies of C6789 -> R19C6 = 6
7. Cage 18/3 in R9C456 = {9(18|27)} -> R9C4 = 9, R9C6 = {12}, R1C4 = 3
8. Cage 5/2 in R23C5 = {14} (NP @ N2, C5)
9. Cage 5/2 in R78C5 = {23} (NP @ N8)
10. R9C56 = [81], R1C56 = [75]
11. Cage 12/3 in R1C123 = {129|246} = {2..} -> no 2 elsewhere in N1, R1
12. 8 of R1 locked in Cage 18/3 = {8(19|46)} -> no 8 elsewhere in N3
13. Since R2C5 = {14} -> Cage 11/3 in R2C234 <> {146} = {128|236|245} = {2..} -> R2C4 = 2
14. Innies of C1 -> R159C1 = 6 = {123} (NT @ C1)
15. Cage 8/3 in R8C234 = {1(25|34)} -> no 1 elsewhere in N7, R8 -> R8C23 = {1(2|3)}, R8C4 = {45}
16. R8C23&R9C1 = NT {123} @ N7
17. R8C235 = NT {123} @ R8
18a. Cage 13/2 in R5C89 = {49|58}
18b. -> Cage 9/2 in R5C34 <> {45} = {18}|[27], Cage 12/2 in R5C67 <> {48} = [39|75]
19. 7 of R5 locked in R5C46 -> not elsewhere in N5
20a. Cage 16/3 in N125 = {169|178} = {1..}
20b. Since R3C4 <> {17} -> R3C3&R4C4 <> 8
20c. Since R4C4 <> {6789} -> R4C4 = 1, R3C34 = [78|96]
21. Cage 9/2 in R5C34 -> R5C3 = {12}
22. R5C3 & Cage 5/2 in R5C12 = complex naked pair on {12} -> not elsewhere in N4
23. R5C7 & Cage 13/2 in R5C89 = complex naked pair on {59} -> not elsewhere in N6
24a. Innies of R2 -> R2C159 = 14 = {149|158|167|347}
24b. Since each in R2C59 <= 6 -> R2C1 = {789}
25. Cage 18/3 in R234C1 -> R34C1 = {456}
26a. Outies of N1 -> R3C4+R4C123 = 25
26b. R3C4 = {68} -> R4C123 = 17|19 with 5 locked in R4C123 = {458|568} = {58(4|6)}
26c. Since R4C1 = {456} (step 25) -> 8 locked in R4C23 -> Cage 16/3 in N14 = {358}
26d. -> R4C23 = {58} (NP @ R4, N4), R3C2 = 3, R4C1 = {46}
27a. 7 of R4 locked in R4C78 -> Cage 14/3 in N36 = {167|257|347}
27b. Since R4C78 <> {57} -> R3C8 <> 2
28. Max R34C6=[94]=13 -> Min R3C7=17-13=4. Cannot repeat 4 in cage 17/3 -> Min R3C7=5
29. R3C9 = 2 (HS @ R3), R24C9 = [16]|{34}
30. 14/3 in R2C159 = {149|347} = {4(7|9)..} -> no 4 elsewhere in R2, R2C1 = {79}
31. R2C1&R3C4 = NP {79} @ N1
32. Cage 12/3 in R1 = {246} -> R159C1 = [213]
33. R5C234 = [427], R8C234 = [215], R78C5 = [23], R1C23 = [64]
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Solution
Code: Select all
+-------+-------+-------+
| 2 6 4 | 3 7 5 | 1 9 8 |
| 7 1 8 | 2 4 9 | 6 5 3 |
| 5 3 9 | 6 1 8 | 7 4 2 |
+-------+-------+-------+
| 6 8 5 | 1 9 2 | 3 7 4 |
| 1 4 2 | 7 6 3 | 9 8 5 |
| 9 7 3 | 8 5 4 | 2 1 6 |
+-------+-------+-------+
| 8 9 7 | 4 2 6 | 5 3 1 |
| 4 2 1 | 5 3 7 | 8 6 9 |
| 3 5 6 | 9 8 1 | 4 2 7 |
+-------+-------+-------+
However there was a nice X Wing on 5 in there
I'll list that alternating path with the X-Wing because I think it's interesting:
24a. Innies of R2 -> R2C159 = 14 = {149|158|167|347}
24b. Did not remember how I eliminated {167} here. Assume this is the case
24c. R2C159 = 14 = {149|158|347} = {(4|5)..}
24d. Since each in R2C59 <= 6 -> R2C1 = {789}
24e. Since R2C159 = {(4|5)..} -> Cage 11/3 in R2 <> {245} = {128|236}
25. Cage 18/3 in R234C1 -> R34C1 = {456}
26. 5 of R2 locked in R2C789 -> not elsewhere in N3
27a. 5 of C2 locked in R3479C2, 5 of C3 locked in R479C3
27b. This forms a grouped X Wing on 5 with Cage 16/3 & N7
27c. -> no 5 elsewhere in N7, Cage 16/3 must include a 5
27d. -> Cage 16/3 in N14 = {5(38|47)} (no {169})
27e. Alternatively, there is also a grouped X Wing on 5 with R34 * Cage 18/3 in C1 & Cage 16/3 in N14
27f. -> both cages must include a 5
28. 5 of C1 locked in R34C1 -> Cage 18/3 in C1 = {5(49|67)} (no 8)
29. Cage 21/3 in C1 = {8(49|67)}
30. R2C1&R3C4 = NP {79} @ N1
...