3x3::k:4608:4608:1794:1794:5892:1797:1797:2055:2055:4617:4608:2571:2571:5892:3342:3342:2055:6673:4617:4617:4617:4373:5892:2839:6673:6673:6673:3099:5404:4373:4373:5892:2839:2839:5154:3107:3099:5404:5926:5926:5926:5926:5926:5154:3107:3099:5404:3119:3119:5425:4658:4658:5154:3107:6198:6198:6198:3119:5425:4658:4924:4924:4924:6198:2624:1601:1601:5425:3908:3908:3910:4924:2624:2624:2634:2634:5425:2381:2381:3910:3910:

Preliminaries courtesy of SudokuSolver
Cage 6(2) n78 - cells only uses 1245
Cage 15(2) n89 - cells only uses 6789
Cage 7(2) n12 - cells do not use 789
Cage 7(2) n23 - cells do not use 789
Cage 13(2) n23 - cells do not use 123
Cage 9(2) n89 - cells do not use 9
Cage 10(2) n78 - cells do not use 5
Cage 10(2) n12 - cells do not use 5
Cage 8(3) n3 - cells do not use 6789
Cage 21(3) n4 - cells do not use 123
Cage 10(3) n7 - cells do not use 89
Cage 20(3) n6 - cells do not use 12
Cage 11(3) n256 - cells do not use 9
Cage 26(4) n3 - cells do not use 1

No clean-up done unless stated. This is a highly optimised solution.
1. "45" on c5: 1 innie r5c5 = 1

2. "45" on n7: 2 outies r89c4 = 5 = {14}/[23]
2a. 2 innies r89c3 = 11 = [29/47/56]

3. hidden triple 789 in r1 -> r1c125 = {789}

4. 6 in r1 only in one of two 7(2) cages -> 1 locked for r1 (Locking cages)

5. 8(3)n3 = {125/134}
5a. -> r2c8 = 1 (the tag WT did this differently, very nicely done)
5b. no 9 in r2c34

6. 18(3)n1 must have two of 789 for r1c12 = {279/378}
6a. -> r2c2 = (23)
6b. must have 7: locked for r1 and n1
6c. no 3 in r2c4

7. "45" on n1: 2 outies r12c4 = 8 = [17]/{26} = 2 or 7
7a. r1c3 = (156)
7b. r2c3 = (348)

8. from steps 2. & 7,: r1289c4 = [1723]/{26}[14]
8a. must have both 1 & 2 (killer pair): locked for c4
8b. note: must have 1 in r1c4 or 4 in r9c4

9. "45" on n3: 2 outies r12c6 = 9
9a. but {27} blocked by h8(2)r12c4
9b. and {36} as [36] only, forces r12c46 = [1376], but this is blocked by 7 also in r2c7 ie, two 7s in r2
9c. -> r12c6 = [18]/{45} = 5 or 8
9d. r1c7 = (236)
9e. r2c7 = (589)

10. 3 in r1 only in n3: locked for n3

11. "45" on n9: 2 outies r89c6 = 13
11a. but {58} blocked by r12c6
11b. = [94]/{67}
11c. r8c7 = (689)
11d. r9c7 = (235)

This took a long time to see
12. combined cage r1289c6 = 22(4)
12a. but [1894] blocks all 1 in c4 (step 8b)
12b. = [18]/{67}/{45}{67}
12c. -> r89c6 = {67}: both locked for c6 and n8
12c. no 6 in r8c7
12d. no 5 in r9c7

13. "45" on c6789: 2 innies r5c67 = 7 = {25/34}(no 6789)

14. "45" on c89: 2 outies r37c7 = 10 = {28/46}/[73/91](no 5, no 79 in r7c7)

15. "45" on n3: 2 innies r12c7 = 11 = [29/38/65]
15a. "45" on n9: 2 innies r89c7 = 11 = [83/92]
15b. note: no eliminations yet

16. hidden killer triple 147 in c7: r37c7 can only have one of them -> r456c7 must have two of 147
16a. and "45" on n6: those 3 innies = 13
16b. = {148/157/247}(no 369)
16c. -> no 4 in r5c6 (h7(2)r5c67)

17. 9 in c6 only in 18(3)r6c6 = {189/279/459}(no 3)

18. 11(3)r3c6 = {128/137/245}
18a. 3 in {137} can only be in r4c6 -> no 3 in r3c6

19. 3 in c6 only in n5: locked for n5

Now to the other side
20. "45" on n7: 2 innies r89c3 = 11 = [29/47/56] = 2 or 4 or 6
20a. "45" on c1234: 2 innies r5c34 = 15 = {69/78}

21. "45" on n4: 3 innies r456c3 = 12
21a. the only combinations with 6 possible are {156}: but this is blocked by r1c3 = (156)
21b. and {246}: blocked by r89c3 (step 20)
21c. -> no 6 in r456c3
21d. no 9 in r5c4 (h15(2))

22. 12(3)r6c3 cannot have 9 since 12 are only in r6c3
22a. -> no 9
22a. min. r6c34 = [14] = 5 -> max. r7c7 = 7 (no 8)

23. 9 in c4 is only in 17(3)r3c4
23a. = {179/269/359}(no 48)
23b. no 9 in r4c3

24. 8 in c4 only in r56c4: locked for n5
24a. 8 in r6c4 -> r6c3 + r7c4 = 4 = [13] only
24b. 8 in r5c4 -> r5c3 = 7
24c. -> h12(3)r456c3 must have 1 or 7
24d. but {147} blocked by 21(3)n4 = 4 or 7
24e. = {129/138/237}(no 4,5)
24f. has one of 789 which must go in r5c3 -> no 78 in r46c3

25. 17(3)r3c4 = {179/269/359}
25a. 3 in {359} must go in r4c3 -> no 3 in r3c4

26. 3 in c4 only in n8: locked for n8

27. 21(4)r6c5 = {2469/2478/2568}
27a. must have 6 or 7 which are only in r6c5 -> r6c5 = (67)
27b. must have 2: locked for n8 & c5

28. r89c4 = {14} only: both locked for c4 and n8
28a. -> 21(4)r6c5 = [6]{258}: 5 & 8 locked for c5 and n8
28a. -> r5c34 = {78} only: both locked for r5

29. r1c5 = 9

30. r1c12 = {78} = 15 -> r2c2 = 3
30a. 8 locked for n1

Much easier now

Prelims

a) R1C34 = {16/25/34}, no 7,8,9
b) R1C67 = {16/25/34}, no 7,8,9
c) R2C34 = {19/28/37/46}, no 5
d) R2C67 = {49/58/67}, no 1,2,3
e) R8C34 = {15/24}
f) R8C67 = {69/78}
g) R9C34 = {19/28/37/46}, no 5
h) R9C67 = {18/27/36/45}, no 9
i) 8(3) cage at R1C8 = {125/134}
j) 11(3) cage at R3C6 = {128/137/146/236/245}, no 9
k) 21(3) cage at R4C2 = {489/579/678}, no 1,2,3
l) 20(3) cage at R4C8 = {389/479/569/578}, no 1,2
m) 10(3) cage at R8C2 = {127/136/145/235}, no 8,9
n) 26(4) at R2C9 = {2789/3689/4589/4679/5678}, no 1

1a. 45 rule on C5 1 innie R5C5 = 1
1b. 45 rule on C1234 2 innies R5C34 = 15 = {69/78}
1c. 45 rule on C6789 2 innies R5C67 = 7 = {25/34}

2a. Hidden triple {789} in R1C125
2b. Killer triple 7,8,9 in R1C2 and 21(3) cage at R4C2, locked for C2
2c. 6 in R1 only in R1C34 = {16} or R1C67 = {16} (locking cages), 1 locked for R1
2d. 8(3) cage at R1C8 = {125/134} -> R2C8 = 1, clean-up: no 6 in R1C6, no 9 in R2C34
2e. 18(3) cage at R1C1 = {279/378}, 7 locked for R1 and N1, clean-up: no 3 in R2C4
2f. 18(3) cage = {279/378} -> R2C2 = {23}
2g. 45 rule on R9 1 innie R9C5 = 2 outies R8C28 + 1
2h. Min R8C28 = 3 -> min R9C5 = 4
2i. Max R8C28 = 8 -> no 8,9 in R8C8

3a. 45 rule on N1 2 outies R12C4 = 8 = [17]/{26}, clean-up: no 2,3,4 in R1C3, no 2,6 in R2C3
3b. 45 rule on N3 2 outies R12C6 = 9 = [18/27/36]/{45}, no 9, clean-up: no 4 in R2C7
3c. 45 rule on N7 2 outies R89C4 = 5 = [14/23/41] -> R9C3 = {679}, clean-up: no 1 in R8C3
3d. 45 rule on N9 2 outies R89C6 = 13 = {67}/[85/94] -> R9C7 = {2345}
3e. 45 rule on C12 2 outies R37C3 = 13 = {49/58}/[67]
3f. 45 rule on C89 2 outies R37C7 = 10 = [28]/{37/46}/[91], no 5, no 9 in R7C7
3g. Killer pair 1,2 in R12C4 and R89C4, locked for C4
3h. Max R67C4 = 11 -> no 9 in R67C4
3i. Min R67C4 = 8 (cannot be {34} which clashes with R89C4) -> max R6C3 = 4

4a. 45 rule on N4 3 innies R456C3 = 12 = {129/138/147/237} (cannot be {156} which clashes with R1C3, cannot be {246} which clashes with R89C3, cannot be {345} because no 3,4,5 in R5C3), no 6, clean-up: no 9 in R5C4 (step 1b)
4b. R5C3 = {789} -> no 7,8,9 in R4C3
4c. Killer triple 7,8,9 in 21(3) cage at R4C2 and R5C3, locked for N4
4d. 9 in C4 only in 17(3) cage at R3C4 = {179/269/359}, no 4,8
4e. 12(3) cage at R4C1 = {156/246/345}
4f. Killer triple 4,5,6 in 12(3) cage and 21(3) cage, locked for N4
4g. Max R4C3 = 3 -> min R34C4 = 14, no 3
[I spent time looking at other interactions in C34 before finding a relatively simple forcing chain …]
4h. R5C34 = 15 (step 1b), R12C4 (step 3a) = 8 = [17]/{26}, R89C4 (step 3c) = 5 = [14/23/41]
4i. Consider placement for 8 in C4
R5C4 = 8 => R5C3 = 7 => R46C3 = {23} => R1C3 = 1 (hidden single in C3) => R12C4 = [62]
or 8 in 12(3) cage at R6C3 = 1{38}, 3 locked for C4
-> R89C4 = {14}, locked for C4 and N8, R12C4 = {26}, locked for C4 and N2, clean-up: no 6 in R1C3, no 3 in R1C6 (step 3b), no 4,5 in R1C7, no 3 in R2C3, no 7 in R2C6 (step 3b), no 6,7 in R2C7, no 9 in R5C3 (step 1b), no 4 in R8C4, no 9 in R8C6 (step 3d), no 6 in R8C7, no 7 in R9C4, no 5 in R9C7
4j. Naked pair {78} in R5C34, locked for R5
4k. Killer pair 4,8 in R2C3 and R2C67, locked for R2
4l. 21(3) cage at R4C2 = {489/579} (cannot be {678} which clashes with R5C3), no 6, 9 locked for C2
4m. 6 in N4 only in 12(3) cage at R4C1, locked for N1
4n. 3 in C3 only in R46C3, locked for N4
4o. 6 in N1 only in R3C23, locked for R3
4p. R37C3 (step 3e) = {49}/[67] (cannot be {58} which clashes with R12C3), no 5,8
4q. R89C6 (step 3d) = {67} (cannot be [85] which clashes with R12C6), locked for C6 and N8, clean-up: no 7 in R8C7, no 4 in R9C7
4r. Hidden killer triple 4,6,7 in 23(4) cage at R1C5 and R6C5 for C5, 23(4) cage cannot contain all of 4,6,7 -> R6C5 = {467}
4s. 12(3) cage at R6C3 = 1{38}/[273], no 5
4t. R34C4 = {59} (hidden pair in C4) -> R4C3 = 3 (cage sum)

[I was then doing the more routine steps after the key step 4i before I spotted this final breakthrough.]
5a. 18(3) cage at R1C1 (step 2e) = {279/378}
5b. R2C34 = [46] (cannot be [82] which clashes with 18(3) cage), R1C4 = 2 -> R1C3 = 5, R8C3 = 2 -> R8C4 = 4, R9C4 = 1 -> R9C3 = 9, R367C3 = [617] -> R5C34 = [87]
5c. Naked pair {34} in R1C89, locked for R1 and N3 -> R1C67 = [16], R2C6 = 8 (step 3b) -> R2C7 = 5, R1C5 = 9, R34C4 = [59]
5d. 21(3) cage at R4C2 = {579} (last remaining combination), 5,7 locked for C2, 5 locked for N4, R1C12 = [78] -> R2C2 = 3 (cage sum)
5e. R12C5 = [97] = 16 -> R34C5 = 7 = [34]
5f. R3C6 = 4 -> R4C67 = 7 = [52], R9C7 = 3 -> R9C6 = 6, R8C6 = 7 -> R8C7 = 8
5g. R5C67 (step 1c) = 7 = [34]
5h. R37C7 (step 3f) = 10 = [91]
5i. R8C58 = [56], R6C7 = 7
5j. R4C8 = 8 -> R56C8 = 12 = [93]

and the rest is naked singles.

Preliminaries courtesy of SudokuSolver
Cage 6(2) n78 - cells only uses 1245
Cage 15(2) n89 - cells only uses 6789
Cage 7(2) n12 - cells do not use 789
Cage 7(2) n23 - cells do not use 789
Cage 13(2) n23 - cells do not use 123
Cage 9(2) n89 - cells do not use 9
Cage 10(2) n78 - cells do not use 5
Cage 10(2) n12 - cells do not use 5
Cage 8(3) n3 - cells do not use 6789
Cage 21(3) n4 - cells do not use 123
Cage 10(3) n7 - cells do not use 89
Cage 20(3) n6 - cells do not use 12
Cage 11(3) n256 - cells do not use 9
Cage 26(4) n3 - cells do not use 1

No clean-up done unless stated. This is a highly optimised solution.
1. "45" on c5: 1 innie r5c5 = 1

2. "45" on n7: 2 outies r89c4 = 5 = {14}/[23]
2a. 2 innies r89c3 = 11 = [29/47/56]

3. hidden triple 789 in r1 -> r1c125 = {789}

4. 6 in r1 only in one of two 7(2) cages -> 1 locked for r1 (Locking cages)

5. 8(3)n3 = {125/134}
5a. -> r2c8 = 1 (the tag WT did this differently, very nicely done)
5b. no 9 in r2c34

6. 18(3)n1 must have two of 789 for r1c12 = {279/378}
6a. -> r2c2 = (23)
6b. must have 7: locked for r1 and n1
6c. note: must have 8 in r1 or 2 in r2c2
6d. no 3 in r2c4

7. "45" on n1: 2 outies r12c4 = 8 = [17]/{26} = 2 or 7
7a. r1c3 = (156)
7b. r2c3 = (348)

8. from steps 2. & 7,: r1289c4 = [1723]/{26}[14]
8a. must have both 1 & 2 (killer pair): locked for c4

Now to Andrew's step 5a&b which effectively cracks the puzzle
9. 10(2)r2c3: [82] blocked by 18(3)n1 (step 6c)
9a. = [37/46]
9b. no 6 in r1c4 (h8(2)r12c4)
9c. no 1 in r1c3

10. "45" on c12: 2 outies c37c3 = 13 = {49/58}/[67](no 1,2,3, no 6 in r7c3)

11. "45" on c1234: 2 innies r5c34 = 15 = {69/78}

12. "45" on n4: 3 innies r456c3 = 12 and must have 1 for c3
12a. but {147} blocked by 21(3)n4 = 4 or 7
12b. and {156} blocked by r1c3 = (56)
12c. = {129/138}(no 4,5,6,7)
12d. 8,9 must be in r5c3 -> no 8,9 in r46c3
12e. 1 locked for n4
12f. no 8,9 in r5c4 (h15(2))

13. 12(3)r6c3 can't have 9 since 1&2 are only in r6c3

14. 9 in c4 only in 17(3)r4c3 = {179/269/359}(no 4,8)

15. 8 in c4 only in 12(3)r6c3
15. = [1]{38} only: 3 locked for c4

Much easier from here