HATMAN said that this puzzle is almost Paper Solvable so I’ll try to do it that way, or at least start that way and try not to go further than insertion solving.
1. 17(2) cage at R2C6, 28(4) cage at R3C6 and 16(2) cage at R3C8 must contain all 8s for C67 and all 9s for C678 -> 28(5) cage at R6C6 cannot contain 9 so must contain 8 -> R6C8 = 8
1a. Similarly 17(2) cage at R2C6, 28(4) cage at R3C6 and 16(2) cage at R3C8 must contain all 9s for R234
2. 9 in N9 only in 18(3) cage at R7C8, 8 in N9 only in 18(3) cage or R9C9 -> 18(3) cage = {189} or R9C89 = [18] (locking cages), no other 1 in N9, no other 8,9 in C9
3. 17(2) cage at R2C6, 28(4) cage at R3C6 don’t contain 1, 28(5) cage at R6C6 without 9 cannot contain 1 -> R1C7 = 1 (hidden single in C7), R1C6 = 5
4. 16(4) cage at R1C8 without 1 must contain both of 2,3 -> R3C9 = 4, R4C9 = 1
4a. 16(4) cage = {2356} (only remaining combination), locked for N3
5. 18(5) cage at R2C4 must contain 1 -> R23C2 and 18(5) cage must contain 1 for R23, R23C2 contains 1 for N1 -> 18(5) cage must contain 1 for N2
45 rule on C67 2 innies R57C6 = 1 remaining outie R8C8
Max R8C8 = 6 (because 7 clashes with R34C8) -> max R57C6 = 6 contains 1,2,3,4 with at least one of 1,2
-> 7 in C6 must be in R34C6 or in R6C6
[I don’t seem to be making any more progress with this approach so I’ve now changed to elimination solving, continuing from the steps above.]
Prelims (after the steps above)
a) R12C1 = {78}/[96]
b) R23C2 = {13}
c) R2C67 = {89}
d) R34C8 = {79}
e) R45C3 = {28/37/46}, no 1,5,9 (because 1,9 already eliminated from R4C3)
f) R5C12 = {49/58/67}, no 1,2,3
g) R67C1 = {17/26/35}, no 4,8,9
h) R7C56 = [47/56/74/83/92], no 1, no 2,3,6 in R6C5
i) R89C5 = {39/48/57}, no 1,2,6
j) R8C78 = {37/46}, no 2,5
k) R9C34 = {29/38/47/56}, no 1
l) R9C89 = [17/45]{27/36}, no 5 in R9C8
m) 9(3) cage at R8C6 = {126/135/234}, no 7
n) 28(4) cage at R3C6 = {4789/5689}, no 2,3
6. Naked pair {13} in R23C2, locked for C2 and N1
7. Naked pair {89} in R2C67, locked for R2, clean-up: no 7 in R1C1
8. Naked pair {79} in R34C8, locked for C8, clean-up: no 3 in R8C7, no 2 in R9C9
9. Naked quad {2356} in 16(4) cage at R1C8, 5 locked for R2
10. 9 in N1 only in R1C123, locked for R1
10a. 7 in N3 only in R3C78, locked for R3
10b. 9 in N6 only in R4C78, locked for R4
11. 45 rule on N7 2 innies R7C1 + R9C3 = 12 = [39/57/75] -> R6C1= {135}, R9C4 = {246}
12. 45 rule on N9 2 innies R79C7 = 8 = {26/35}, no 4,7
12a. Killer pair 3,6 in R79C7 and R8C78, locked for N9
13. 28(4) cage at R3C6 = {4789/5689}
13a. 5 of {5689} must be in R4C7 -> no 6 in R4C7
14. 9 in N9 only in 18(3) cage at R7C8 = {189/279/459}
14a. 2 of {279} must be in R7C8 -> no 2 in R78C9
14b. 4 of {459} must be in R7C8 -> no 5 in R7C8
15. 45 rule on C67 2 innies R57C6 = 1 remaining outie R8C8
15a. R8C8 = {346} -> R57C6 = [12/13/24/42], R5C6 = {124}, R7C6 = {234}, R7C5 = {789}
16. 9(3) cage at R8C6 = {126/135/234}
16a. 6 of {126} must be in R89C6 (R89C6 cannot be {12} which clashes with R57C6), no 6 in R9C7, clean-up: no 2 in R7C7 (step 12)
17. 28(5) cage at R5C7 = {25678/34678}
17a. Hidden killer pair 2,3 in 28(5) cage and R9C7 for C7, 28(5) cage contains one of {23} -> R9C7 = {23}, clean-up: no 3 in R7C7 (step 12)
17b. 2,3 of 28(5) cage must be in C7 -> no 2,3 in R6C6
17c. 3 in N9 only in R8C8 + R9C7, CPE no 3 in R8C6
18. 3 in C6 only in R79C6, locked for N8, clean-up: no 9 in R89C5
18a. 9 in R9 only in R9C123, locked for N7
19. 9(3) cage at R8C6 = {126/234}, CPE no 2 in R9C4, clean-up: no 9 in R9C3, no 3 in R7C1 (step 11), no 5 in R6C1
19a. Killer pair 4,6 in R9C4 and 9(3) cage, locked for N8, clean-up: no 7 in R7C5, no 8 in R89C5
20. Naked pair {57} in R7C1 + R9C3, locked for N7
21. Naked pair {57} in R89C5, locked for C5 and N8
22. Naked pair {57} in R9C35, locked for R9 -> R9C9 = 8, R9C8 = 1
23. 9(3) cage at R8C6 (step 19) = {126/234}
23a. 1 of {126} must be in R8C6 -> no 6 in R8C6
23b. 6 in N8 only in R9C46, locked for R9
24. 28(4) cage at R3C6 = {4789/5689} must contain one of 4,6,7 in C6 (both of 8,9 in C6 or C7 would clash with R2C6 or R2C7)
24a. Killer triple 4,6,7 in 28(4) cage at R3C6, R6C6 and 9(3) cage at R8C6, locked for C6
25. R57C6 (step 15a) = [12/13] -> R5C6 = 1
25a. R57C6 = [12/13] = 3,4 -> R8C8 = {34} (step 15), clean-up: no 4 in R8C7
[I’ve now reached the same stage that Ed did with the neat step
45 rule on C67 3(2+1) innies R57C6 + R8C7 = 10 cannot contain more than one of 4,5 because R1C67, 28(4) cage at R3C6 and 28(5) cage at R5C7 each contain one of 4,5 for C67.]
26. 9(3) cage at R8C6 (step 19) = {234} (only remaining combination), 4 locked for C6 and N8 -> R9C4 = 6, R9C3 = 5, R7C1 = 7, R6C1 = 1, R89C5 = [57], R2C1 = 6, R1C1 = 9
26a. R9C2 = 9 (hidden single in R9)
27. R5C12 = {58} (only remaining combination, cannot be {67} because 6,7 only in R5C2), locked for R5 and N4, clean-up: no 2 in R45C3
27a. R6C3 = 9 (hidden single in N4)
28. R3C5 = 1 (hidden single in C5), R23C2 = [13]
29. Naked pair {28} in R3C34, locked for R3 -> R3C1 = 5
30. R3C6 = 6 (hidden single in R3), R6C6 = 7, R4C6 = 8, R2C67 = [98]
31. R3C6 = 6 -> 28(4) cage at R3C6 = {5689} (only remaining combination) -> R4C7 = 5, R3C7 = 9, R7C7 = 6, R8C7 = 7, R8C8 = 3, R9C7 = 2, R89C6 = [43], R7C6 = 2, R7C5 = 9, R7C89 = [45], R8C9 = 9, R7C234 = [831], R8C1234 = [2618], R9C1 = 4
32. R34C1 = [53], R4C2 = 7 (cage sum)
33. R3C34 [82] = 10, R4C4 = 4 -> R2C4 = 3 (cage sum)
and the rest is naked singles.
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