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Preliminaries courtesy of SudokuSolver (Solver menu-Options-Scoring-OK-F10)
Cage 4(2) n58 - cells ={13}
Cage 6(2) n2 - cells only uses 1245
Cage 12(2) n8 - cells do not use 126
Cage 9(2) n8 - cells do not use 9
Cage 9(2) n25 - cells do not use 9
Cage 10(2) n2 - cells do not use 5
Cage 10(2) n7 - cells do not use 5
Cage 11(2) n3 - cells do not use 1
Cage 6(3) n9 - cells ={123}
Cage 9(3) n125 - cells do not use 789
Cage 19(3) n5 - cells do not use 1
Cage 19(3) n1 - cells do not use 1
Cage 33(5) n69 - cells do not use 12

1. 4(2)r6c5 = {13} only: both locked for c5
1a. no 5 in r2c4
1b. no 6,8 in r8c6
1c. no 6,8 in 9(2)r3c5

2. "45" on n1: 1 outie r4c1 - 4 = 2 innies r13c3
2a. min. r13c3 = {12} = 3 -> min. r4c1 = 7 (no 1,2,3,4,5,6)
2b. max. r13c3 = 5 (no 5,6,7,8,9)

3. Whatever is in r4c1 can only repeat in n1 in r1c2 (no common digits with r13c3) -> r1c2 = r4c1 = (789) (Clone)

4. Since r4c1 = r1c2 (which is in the 19(3)r1c1) -> r124c1 = h19(3) cage
4a. "45" on c1 (including that h19(3)r124c1): 1 outie r5c2 - 6 = 1 innie r3c1
4b. r5c2 = (789)
4c. r3c1 = (123)

Learnt this next move from manu.
5. "45" on c12: 1 outie r2c3 + 9 = 5 innies r46789c2
5a. min. any four of those innies = {1234} = 10 which is greater than the IOD of +9 -> r2c3 cannot equal any one of those five innies
5b. -> r2c3 = r5c2 = (789) (Clone)

6. Since r5c2 = r2c3 -> r5c2 cannot repeat in c1 in r12c1 (same nonet as r2c3)
6a. -> r5c2 must repeat in c1 in r89c1 (Clone)
6b. -> 10(2)n7 = {19/28/37}(no 4,6)

7. r78c2 'sees' all r9 except r9c89 -> must repeat there (Clone)
7a. r9c89 = {123} -> r78c2 = {123}

8. r6c2 'sees' all r9 except r9c189: however, r9c89 are already taken by r78c2
8a. -> r6c2 = r9c1 (Clone)
8b. we already know that r5c2 repeats in r89c1 -> r56c2 = h10(2) = [73/82/91]
8c. since r6c2 = r9c1 -> r9c1 = (123)
8d. -> r5c2 = r8c1 = (789)

9. Naked triple {123} in r678c2: all locked for c2 and 45(9)r6c2
9a. no 1,2,3 in r9c234567
9c. Naked triple {123} in n7 in r78c2 + r9c1: all locked for n7

This is the step that took me the longest to find: [edit: Andrew's step 10 is probably a simpler way]
10. "45" on c5: 2 outies r2c4 + r8c6 + 17 = 3 innies r159c5
10a. Max r159c5 = 24 -> max. 2 outies = 7 (no 7 in r8c6)
10b. no 2 in r8c5

11. 2 in n8 only in c6: 2 locked for c6
11a. no 8 in 10(2)n2

12. 8 in n2 only in r1: 8 locked for r1 and for 45(9)r1c3
12a. no 8 in r234c8
12b. no 3 in r2c9
12c. no 8 in r4c1 (r1c2=r4c1)

13. 19(3)r1c1: must have 7 or 9 for r1c2 -> {568} blocked
13a. no other way for a 5 to be in a 19(3) -> no 5 in r12c1

14. 5 in c1 only in 22(4)r5c1: but {1579} is blocked by r4c1 = (79)
14a. = {2569/2578/3568/4567}(no 1)

15. 5 in n1 is only in r23c2: 5 locked for c2
15a. ->30(5)r2c1 must have 5 = {15789/25689/35679/45678}[Andrew noticed that {45678} blocked because r3c1 = (123). Ooops, I really should have seen that]

16. 1 in c1 is only in 10(2)r8c1 = [91] or in r3c1 in 30(5)r2c2 -> 9 in r4c1 must also have 1 in r3c1 or there would be no 1 for c1 (Locking-out cages)
16a. ->{25689} blocked from 30(5)r2c2
16b. 30(5)r2c2 = {15789/35679/45678}(no 2)
16c. no 8 in r5c2 (r5c2-6=r3c1)
16d. no 2 in r6c2 (h10(2)r56c2)
16e. no 2 in r9c1 (r6c2 = r9c1)
16f. no 8 in r8c1

17. Naked pair {79} in r4c1 + r5c2: both locked for n4
17a. Naked pair {79} in r1c2 + r5c2: both locked for c2
17b. Naked pair {79} in r4c1 + r8c1: both locked for c1
17c. Naked pair {13} in r3c1 + r9c1: 3 locked for c1

18. 17(3)n4 = {368/458}(no 1,2)
18a. must have 8: 8 locked for n4

19. 1 in n4 only r6: 1 locked for r6
19a. r67c5 = [31]
19b. r6c2 = 1
19c. r5c2 = 9 (h10(2)r56c2)
19d. r9c1 = 1 (r6c2=r9c1)

On from there. Still have to keep thinking for a long time yet but much easier.

Ed found more 'clones' than I did and made excellent use of them. I really ought to have spotted the one in his step 3. The next one in step 5 was a very clever one; I'd seen the 45 but not how to use it.

After locking 2 in R78C6 for N8 I spent a long time working on these 2s before I moved on and realised that the first three columns are a more important area of this puzzle. Then I found 5(4+1) outies for C1, an alternative way to some of the 'clones' and, I think, the way used by SudokuSolver.

Prelims

a) R12C9 = {29/38/47/56}, no 1
b) R2C45 = {15/24}
c) R23C6 = {19/28/37/46}, no 5
d) R34C5 = {18/27/36/45}, no 9
e) R67C5 = {13}
f) R78C4 = {39/48/57}, no 1,2,6
g) R89C1 = {19/28/37/46}, no 5
h) R8C56 = {18/27/36/45}, no 9
i) 19(3) cage at R1C1 = {289/379/469/478/568}, no 1
j) 9(3) cage at R3C3 = {126/135/234}, no 7,8,9
k) 19(3) cage at R5C4 = {289/379/469/478/568}, no 1
l) 6(3) cage at R8C9 = {123}
m) 33(5) cage at R6C9 = {36789/45789}, no 1,2
n) and, of course, both 45(9) cages = {123456789}

Steps resulting from Prelims
1a. Naked pair {13} in R67C5, locked for C5, clean-up: no 5 in R2C4, no 6,8 in R34C5, no 6,8 in R8C6
1b. Naked triple {123} in 6(3) cage at R8C9, locked for N9
1c. 1 in R1 only in R1C34678, locked for 45(9) cage at R1C3, no 1 in R234C8
1d. 5 in R9 only in R9C234567, locked for 45(9) cage at R6C2, no 5 in R678C2

2. 45 rule on C9 2 innies R67C9 = 2 outies R59C8 + 12
2a. Max R67C9 = 17 -> max R59C8 = 5 -> R5C8 = {1234}
2b. Min R59C8 = 3 -> min R67C9 = 15, no 3,4,5 in R67C9
2c. 33(5) cage at R6C9 = {45789} (only remaining combination), no 6, 4,5 locked for N9
2d. R8C56 = [28/63/72/81] (cannot be {45} which clashes with 33(5) cage, ALS block), no 4,5 in R8C56
2e. 6 in N9 only in R79C7, locked for C7
2f. 6 in N9 only in R79C7, CPE no 6 in R7C2

3. 45 rule on N1 1 outie R4C1 = 2 innies R13C3 + 4
3a. Max R13C3 = 5 = {12/13/14/23}
3b. Min R13C3 = 3 -> min R4C1 = 7

4. 45 rule on R1234 3 outies R5C389 = 10 = {127/136/145/235}, no 8,9

5. 45 rule on N47 5(1+4) outies R6C4 + R9C4567 = 1 innie R4C1 + 29
5a. Min R4C1 = 7 -> min R6C4 + R9C4567 = 36
5b. Max R9C4567 = 30 -> min R6C4 = 6
5c. Max R6C4 = 9 -> min R9C4567 = 27, no 1,2 in R9C456
5d. R9C4567 = 27 … 30 must contain 9, locked for R9 and 45(9) cage at R6C2, no 9 in R678C2, clean-up: no 1 in R8C1

6. R34C2 = [27/54/72] (cannot be [45] which clashes with R2C45 = [42]), no 4 in R3C2, no 5 in R4C2

7. Max R2C5 + R34C5 = 14 must contain 2 (because no 1,3 in R234C5), locked for C5, clean-up: no 7 in R8C6
7a. 2 in N8 only in R78C6, locked for C6, clean-up: no 8 in R23C6
7b. 8 in N2 only in R1C456, locked for R1 and 45(9) cage at R1C3, no 8 in R234C8, clean-up: no 3 in R2C9
7c. 8 in C8 only in R678C8, CPE no 8 in R6C9
7d. Naked quint {45789} in 33(5) cage at R6C9, 8 locked for N9

8. 19(3) cage at R1C1 = {289/379/469/478/568}
8a. 8 of {289/568} must be in R2C1 -> no 2,5 in R2C1

9. 45 rule on N478 2(1+1) outies R6C4 + R9C7 = 3(1+2) innies R4C1 + R7C56 + 5
9a. Max R6C4 + R9C7 = 18 -> max R4C1 + R7C56 = 13, min R4C1 = 7 -> max R7C56 = 6, no 6,7,8,9 in R7C6
9b. 6 in N8 only in R8C5 + R9C456, CPE no 6 in R8C2

10. 45 rule on N236 3(2+1) innies R3C34 + R6C9 = 2(1+1) outies R1C3 + R4C6 + 14
10a. Max R3C34 + R6C9 = 22 -> max R1C3 + R4C6 = 8, no 8,9 in R4C6

11. R9C1 “sees” all the cells of 45(9) cage at R6C2 except for R6C2 -> R6C2 = R9C1
11a. Law of Leftovers (LoL) on R9, three outies R678C2 must exactly equal three innies R9C189, R6C2 = R9C1 -> R78C2 = R9C89 -> R78C2 = {123}
[There are similar steps for R1 and 45(9) cage at R1C3; I’ll use them when I need them.]

12. Naked triple {123} in R8C269, locked for R8, clean-up: no 9 in R7C4, no 7,8 in R9C1, no 7,8 in R6C2 (step 11, or from 7,8 locked in R9)

13. 1,2 in 45(9) cage at R6C2 only in R6789C2 + R9C3, R6C2 = R9C1 (step 11) -> 1,2 in R789C2 + R9C13, locked for N7

14. Consider the placements for 2 in N8
R7C6 = 2 => R7C5 + R8C6 = {13}, locked for N8
or R8C6 = 2, R8C5 = 7, R34C3 = [54], R2C45 = [42] => R78C4 = {39} = [39]
-> 3 in R7C45 + R8C6, locked for N8
14a. 1,2,3 in 45(9) cage at R6C2 only in R678C2 + R9C23, R6C2 = R9C1 (step 11) -> 1,2,3 in R78C2 + R9C123, locked for N7

[Taking the forcing chain in step 14 further]
15. R7C6 = 2 => R7C5 + R8C6 = {13}, locked for N8, 9 in N8 only in R9C456, locked for R9, R9C7 = {67} => 15(3) cage at R6C6 = [429/627] (cannot be [726] which clashes with 33(5) cage at R6C9 = 9{4578})
or R8C6 = 2, R8C5 = 7, R34C3 = [54], R2C45 = [42] => R78C4 = [39], R7C5 = 1, R7C6 = {45}, R9C7 = 9 (hidden single in R9), R7C7 = 6 (hidden single in N8) => R67C6 = 9 = [54]
-> R6C6 = {456}, R7C6 = {24}
15a. 15(3) cage at R6C6 = [429/456/546/627]
15b. R23C6 = {19/37} (cannot be {46} which clashes with 15(3) cage), no 4,6 in R23C6

16. Consider combinations for R23C6 = {19/37}
R23C6 = {19}, locked for N2 => R2C45 = {24}
or R23C6 = {37}, locked for N2, no 2 in R4C5 => 2 in C5 only in R23C5
-> 2 in R2C45 + R3C5, locked for N2

17. Consider placements for 2 in N8
17a. R7C6 = 2 => killer pair 1,3 in R23C6 and R8C6, locked for C6
or R8C6 = 2, R7C5 = 1 (hidden single in N8), R6C5 = 3
-> no 3 in R45C6

18. 45 rule on N4 3 innies R4C1 + R6C23 = 1 outie R7C1 + 6
18a. Max R4C1 + R6C23 = 15 but cannot be [717] -> no 7,8,9 in R6C3

19. 45 rule on C1 5(4+1) outies R1235C2 + R2C3 = 36
19a. Max R1235C2 = 30 -> min R2C3 = 6
19b. Max R123C2 + R2C3 = 30 -> min R5C2 = 6
19c. Min R1235C2 = 27, no 1,2 in R123C2
19d. Min R1235C2 = 27, must contain 9, locked for C2
19e. Min R123C2 + R2C3 = 27, must contain 9, locked for N1

20. 19(3) cage at R1C1 = {289/379/469/478/568}
20a. 9 of {379} must be in R1C2 -> no 3 in R1C2

21. 30(5) cage at R2C2 = {15789/24789/25689/35679/45678} (cannot be {34689} which clashes with 19(3) cage at R1C1 because the 30(5) cage “sees” all cells of the 19(3) cage except that R4C1 and R1C2 don’t “see” each other)
21a. 5 in N1 only in 19(3) cage at R1C1 = {568} or 30(5) cage -> 30(5) cage = {15789/25689/35679/45678} (cannot be {24789} which clashes with 19(3) cage = {568}, locking-out cages), 5 locked for N1
[With hindsight 19(3) cage = {289/379/469/478} (cannot be {568} which clashes with 30(5) cage), no 5 is simpler.]
21b. 19(3) cage at R1C1 = {289/379/469/478}
21c. 9 of {469} must be in R1C2 -> no 6 in R1C2

22. R1C9 “sees” all the cells of 45(9) cage at R1C3 except for R4C8 -> R1C9 = R4C8
22a. Law of Leftovers (LoL) on R1, three outies R234C8 must exactly equal three innies R1C129, R6C2 R1C9 = R4C8 -> R78C2 = R1C12 -> R23C8, no 5 in R1C12 -> no 5 in R23C8

[I ought to have spotted this step a lot earlier; I was also slow in spotting step 19, but that was less obvious.]
23. 45 rule on C1 2 outies R15C2 = 2 innies R34C1 + 6
23a. Min R34C1 = 8 -> min R15C2 = 14, no 4 in R1C2
23b. Max R15C2 = 17 -> max R34C1 = 11, min R4C1 = 7 -> max R3C1 = 4

24. 30(5) cage at R2C2 (step 21a) = {15789/25689/35679/45678}, 5 locked for C2
24a. R3C1 = {1234} -> no 3,4 in R23C2

25. R1235C2 + R2C3 = 36 (step 19)
25a. R23C2 contain 5 (step 24)
25b. Max R1235C2 = {5789} = 29 -> min R2C3 = 7
25c. Max R123C2 + R2C3 = {5789} = 29 -> min R5C2 = 7

26. 17(3) cage at R4C2 = {269/359/368/458/467} (cannot be {179/278} which clash with R4C1 + R5C2, ALS block), no 1
26a. 9 of {269} must be in R4C3 -> no 2 in R4C3
26b. Killer triple 7,8,9 in R4C1, 17(3) cage and R5C2, locked for N4

27. 1 in C2 only in R6789C2, locked for 45(9) cage at R6C2, no 1 in R9C3

28. R4C1 = R13C3 + 4 (step 3)
28a. 30(5) cage at R2C2 (step 21a) = {15789/25689/35679/45678}
28b. 9 of {15789} must be in R4C1 (because R13C3 = {23} = 5 when 1 in R3C1), 5,6 of {25689/35679/45678} must be in R23C2 -> no 9 in R23C2

29. 30(5) cage at R2C2 (step 21a) = {15789/25689/35679/45678}, 19(3) cage at R1C1 (step 21b) = {289/379/469/478}
29a. Consider placements for 6 in N1
6 in 19(3) cage = {469}, locked for C1, no 6 in R9C1 => no 6 in R6C2 (step 11)
or 6 in 30(5) cage = {25689/35679/45678}, 6 locked for C2, no 6 in R6C2 => no 6 in R9C1 (step 11)
-> no 6 in R6C2, no 6 in R9C1, clean-up: no 4 in R8C1

30. 30(5) cage at R2C2 (step 21a) = {15789/25689/35679} (cannot be {45678} which clashes with 19(3) cage at R1C1 because R2C3 and R4C1 both “see” R12C1), no 4
30a. 30(5) cage = {15789/25689/35679}, CPE no 9 in R4C3
30b. 9 in N4 only in R4C1 + R5C2, CPE no 9 in R7C1

31. 17(3) cage at R4C2 (step 26) = {368/458/467}, no 2

32. 5 in C1 only in 22(4) cage at R5C1 = {1579/2569/2578/3568/4567}
32a. 9 in C1 only in R4C1 or in R89C1 = [91] -> 22(4) cage = {2569/2578/3568/4567} (cannot be {1579}, locking-out cages), no 1

33. 1 in N4 only in R6C23, locked for R6 -> R6C5 = 3, R7C5 = 1, clean-up: no 8 in R8C5, no 3 in R9C1 (step 11), no 7 in R8C1
[What a long time until the first placements!]
33a. 1 in N5 only in R4C46, locked for R4

34. R4C1 + R6C23 = R7C1 + 6 (step 18), IOU R6C23 cannot total 6 and contains 1 -> no 5 in R6C3

35. 30(5) cage at R2C2 (step 30a) = {15789/25689/35679}, 22(4) cage at R5C1 (step 32a) = {2569/2578/3568/4567}
35a. Consider placements for 9 in N4
R4C1 = 9, R3C1 = 1 (hidden single in C1), R2C23 + R3C2 = {578}, locked for N1, R1C2 = 9, R12C1 = 10 = {46}, locked for C1 => 22(4) cage = {2578/3568}
or R5C2 = 9 => 22(4) cage = {2569}
-> 22(4) cage = {2569/2578/3568}, no 4

36. Consider combinations for 22(4) cage at R5C1 (step 35a) = {2569/2578/3568}
22(4) cage = {2569/2578}, 2 locked for C1
or 22(4) cage = {3568}, 3,5,6 locked for C1 => R1C2 = 9 (hidden single in C2), R12C1 = 10 = [28]
-> 2 in R156C1, locked for C1, clean-up: no 2 in R6C2 (step 11), no 8 in R8C1
36a. Looking at the combinations slightly differently
22(4) cage = {2569/3568}, 6 locked for C1
or 22(4) cage = {2578}, R4C1 = 9 (hidden single in N4), R8C1 = 6
-> 6 in R568C1, locked for C1, clean-up: no 6 in R23C8 (step 22a)

37. R23C2 = {56} (hidden pair in N1), locked for C2
37a. 30(5) cage at R2C2 (step 30a) = {25689/35679} (cannot be {15789} which doesn’t contain 6), no 1
[Cracked, at last …]

38. R9C1 = 1 (hidden single in C1), R8C1 = 9, clean-up: no 3 in R7C4
38a. R8C6 = 3 (hidden single in N8), R8C5 = 6, R78C2 = [32], R8C9 = 1, clean-up: no 7 in R23C6
38b. Naked pair {19} in R23C6, locked for C6 and N2, clean-up: no 5 in R2C5
38c. Naked pair {24} in R2C45, locked for R2 and N2, clean-up: no 7,9 in R1C9, no 7 in R4C5, no 7,9 in R4C8 (step 22)
38d. R6C2 = 1 (hidden single in C2)

39. R7C6 = 2 (hidden single in N8), R6C6 + R7C7 = 13 = [49/67]
39a. R9C7 = 6 (hidden single in N9)

40. R5C389 (step 4) = {127/136/145/235}
40a. 1 of {145} must be in R5C8 -> no 4 in R5C8

41. 30(5) cage at R2C2 (step 30a) = {25689/35679} -> R2C3 = 9, R23C6 = [19], R1C2 = 7, R2C1 = 8 (hidden single in N1), R1C1 = 4 (cage sum), R4C1 = 7, R23C2 = {56} = 11, R3C1 = 3 (cage sum); clean-up: no 2,3 in R1C9, no 7 in R2C9, no 2,3,4 in R4C8 (step 22)
41a. R23C8 = [74] (hidden pair in 45(9) cage at R1C3, or using step 22a)
41b. Naked pair {56} in R12C9, locked for C9 and N3 -> R2C7 = 3

42. R5C2 = 9 (hidden single in N4)
42a. 19(3) cage at R5C4 = {478/568}, 8 locked for R5 and N5
42b. Killer triple 4,5,6 in R4C6, 19(3) cage and R6C6, locked for N5 -> R4C5 = 2, R3C5 = 7, R2C45 = [24], R4C4 = 1, R3C3 = 2, R3C4 = 6 (cage sum), R3C9 = 8, R3C7 = 1, R23C2 = [65], R12C9 = [65], R4C8 = 6 (hidden single in 45(9) cage at R1C3, or using step 22)

43. R23C7 = [31] = 4 -> R4C67 = 14 = [59], R67C9 = [79], R78C8 = {58}, locked for C8 and N9, R6C8 = 2, R7C7 = 7, R6C6 = 6 (cage sum)

44. R7C34 = [49] = 13 -> R78C3 = 12 = [57]

and the rest is naked singles.

I'll rate my walkthrough for Pinata #42 at least 1.75. Ed's clever use of 'clones' would probably put the optimum solving path somewhere in the 1.5 range.