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1. Innies - Outies n1 -> r2c4 = r3c1+8
-> r2c4 = 9
-> 6/2@r3c1 = [15]

Since 12/2@r1c1 from [93] or {48}
-> r23c3 = {25}


2. Innies c9 r145 = +13
Since r1c9 <= 6 -> r45c9 >= +7
-> r3c78 <= +13

Innies n23 (and given r2c4=9) -> r3c478 = +19
-> r3c4 >= 6

-> 18/5@r3c4 does not contain a 5. Must be {12348}.
-> r3c4 = 8
-> (Innies n2) r23c6 = +9
-> r2c78 = +15 - Must be {78}

Also r3c78 = +11 and r4c89+r5c9 = +10.

(Step 3 redone.)
3. Innies - Outies r1234 -> r4c567 = r5c29+20
-> r4c567 = {689} and r5c29 = {12} or r4c567 = {789} and r5c29 = {13}


4. Following on...
r5c29 from {12} or {13}
-> 4 in r4c234

In n3 3 locked in 9/3@r1c7
-> Max r1c9 = 5
-> Min r45c9 = +8
-> Max r4c8 = 2

Also in n3 6 locked in r2c9 or r3c789
-> 6 not in hidden group +10 at r4c89+r5c9.
Also 4,5 cannot be in that hidden group.

-> Hidden group +10 r4c89+r5c9 = {127} with 7 in r4c9
-> r3c78 can only be {56}
-> 10/2@r2c9 = [19]
-> r4c8 = 1, r5c9 = 2

Also 22/3@r4c6 = [{89}5]
-> r4c5 = 6
-> r5c2 = 1

Easier from here:

He made excellent use of min-max steps to break into this puzzle.

Corrections and clarifications courtesy of Andrew
Thanks for that credit. It was wellbeback who found the innies-outies in the new, much simpler version of step 3, after I'd asked for clarification of the original step 3.

My key step was in a very different area.

Prelims

a) R12C1 = {39/48/57}, no 1,2,6
b) R23C9 = {19/28/37/46}, no 5
c) R34C1 = {15/24}
d) R67C9 = {18/27/36/45}, no 9
e) R78C1 = {29/38/47/56}, no 1
f) R89C9 = {49/58/67}, no 1,2,3
g) 9(3) cage at R1C7 = {126/135/234}, no 7,8,9
h) 22(3) cage at R4C6 = {589/679}
i) 18(5) cage at R3C4 = {12348/12357/12456}, no 9

Steps resulting from Prelims
1a. 22(3) cage at R4C6 = {589/679}, CPE no 9 in R4C89
1b. 19(5) cage at R1C4 must contain 1, locked for N2

2. 45 rule on N1 1 outie R2C4 = 1 innie R3C1 + 8 -> R2C4 = 9, R3C1 = 1, R4C1 = 5, clean-up: no 3,7 in R1C1, no 7 in R2C1, no 6 in R78C1
2a. R2C4 = 9 -> R23C3 = 7 = {25} (only remaining combination, cannot be {34} which clashes with R12C1), locked for C3 and N1
2b. 9 in R1 only in R1C123, locked for N1
2c. 22(3) cage at R4C6 = {589/679}
2d. 5 of {589} must be in R5C7 -> no 8 in R5C7

3. R78C1 = {29/47} (cannot be {38} which clashes with R12C1), no 3,8
3a. Killer pair 4,9 in R12C1 and R78C1, locked for C1
3b. 6 in C1 only in R569C1, CPE no 6 in R7C23

4. 45 rule on N9 1 outie R8C6 = 1 innie R7C9 + 1, no 1 in R8C6

5. 45 rule on R789 4 innies R7C2369 = 13 = {1237/1246/1345}, no 8,9, 1 locked for R7, clean-up: no 1 in R6C9, no 9 in R8C6 (step 4)

6. 18(5) cage at R3C4 = {12348/12357/12456}
6a. 5 in {12357/12456} must be in R3C4 -> no 6,7 in R3C4

7. 45 rule on R1234 3 outies R5C279 = 1 innie R4C5 + 2
7a. Min R5C279 = 8 -> min R4C5 = 6
7b. Max R5C279 = 11, no 9 in R5C79
7c. 22(3) cage at R4C6 = {589/679}, 9 locked for R4
7d. Max R5C279 = 10, min R5C7 = 5 -> max R5C29 = 5, no 5,6,7,8
[As wellbeback pointed out to me in a PM, while discussing his step 3 (since re-written), R5C7 cannot be 5 when R4C5 is 8 because of 22(3) cage at R4C6, limiting R5C29 to a maximum of 4 so R5C29 = {12/13}, 1 locked for R5.
He also pointed out that there’s a better way to do this step
45 rule on R1234 3 innies R4C567 = 2 outies R5C29 + 20 -> R4C567 = 23,24, R5C29 = 3,4 …
which is now his re-written step 3.]

8. 45 rule on N3 4 innies R23C78 = 26 = {2789/4589/4679/5678} (cannot be {3689} which clashes with 9(3) cage at R1C7), no 1,3

9. 24(4) cage at R2C6 = {3678/4578}, no 2, CPE no 7,8 in R2C5

10. 45 rule on N5 4 innies R46C46 = 16
10a. Min R4C6 = 6 -> max R4C4 + R6C46 = 10, no 8,9 in R4C4 + R6C46

11. 45 rule on C9 3 innies R145C9 = 13 = {148/157/238/247/256/346}
11a. 7,8 of {148/157/238/247} must be in R4C9, 2 of {256} must be in R5C9 -> no 1,2 in R4C9

12. 6 in C1 only in R569C1
12a. 45 rule on C1 3 innies R569C1 = 16 = {268/367}
12b. Killer CPE 2,7 in R569C1 and R78C1, no 2,7 in R7C23
12c. Max R56C1 = {68} (cannot be {78} because of combinations in R569C1) = 14, max R7C23 = {45} = 9 -> min R6C2 = 3

13. 17(3) cage at R9C1 = {269/359/368/458/467} (cannot be {179/278} which clash with R78C1), no 1

14. 1 in N7 only in R78C23
14a. 45 rule on N7 4 innies R78C23 = 17 = {1259/1358/1367/1457} (cannot be {1349} which clashes with R78C1, cannot be {1268} because no 2,6,8 in R7C23)
14b. 2 of {1259} must be in R8C2 -> no 9 in R8C2
14c. 17(3) cage at R9C1 (step 13) = {269/368/458/467} (cannot be {359} which clashes with R78C23)

15. R23C78 = 26 (step 8), max R2C78 = 15 -> min R3C78 = 11, min R45C9 = [31] = 4 -> max R4C8 = 6

16. R569C1 (step 12a) = {268/367}
16a. 26(5) cage at R5C1 = {14678/24578/34568} (cannot be {12689} because no 2,6,8 in R7C23, cannot be {23489/23579/24569} which clash with R78C1 because 9 only in R6C2, cannot be {13679} which clashes with R569C1, CCC, because 9 only in R6C2, cannot be {13589} because R569C1 cannot contain both of 3,8, cannot be {14579} because 1,4,5,9 only in R6C2 + R7C23, cannot be {23678} because R7C23 only contain 1,3,4,5), no 9, 8 locked for N4
16b. 5 in {34568} must be in R7C2 -> no 3 in R7C2
16c. 26(5) cage = {14678/24578/34568}, CPE no 4 in R89C2

17. 9 in N4 only in R56C3, locked for C3
17a. 15(3) cage at R5C3 = {159/249}, no 3,6,7
17b. 2,5 only in R6C4 -> R6C4 = {25}

18. 18(5) cage at R3C4 = {12348/12357/12456}
18a. 5,8 only in R3C4 -> R3C4 = {58}

19. 45 rule on N2 3 remaining innies R2C6 + R3C46 = 17 = {368/458} (cannot be {467} because R2C6 only contains 5,8), no 7, 8 locked for N2

20. R7C2369 (step 5) = {1246/1345} (cannot be {1237} because 26(5) cage at R5C1 cannot contain both of 1,3), no 7, 4 locked for R4, clean-up: no 2 in R6C9, no 7 in R8C1, no 8 in R8C6 (step 4)
20a. 17(3) cage at R9C1 (step 14c) = {269/368/458/467}
20b. 9 of {269} must be in R9C2 -> no 2 in R9C2
20c. 4 of {467} must be in R9C3 -> no 7 in R9C3

21. 45 rule on N23 3 outies R4C89 + R5C9 = 1 innie R3C4 + 2
21a. R2C4 = {58} -> R4C89 + R5C9 = 7,10 = {124/127/136}, no 8, 1 locked for N6
21b. 4 of {124} must be in R4C9 -> no 4 in R4C8 + R5C9
21c. R145C9 (step 11) = {157/247/346} (cannot be {256} because R4C89 + R5C9 cannot contain both of 2,6)
21d. 6 of {346} must be in R4C9 (R45C9 cannot contain both of 3,4), 7 of {157/247} must be in R4C9 -> R4C9 = {67}
21e. 4,5 of R145C9 only in R1C9 -> R1C9 = {45}
21f. R67C9 = [36/63/72/81] (cannot be {45} which clashes with R1C9), no 4,5, clean-up: no 5,6 in R8C4 (step 4)

22. Naked quad {6789} in R4C5679, locked for R4
22a. 18(5) cage at R3C4 = {12348} (only remaining combination) -> R3C4 = 8, clean-up: no 2 in R2C9

23. 24(4) cage at R2C6 (step 9) = {3678/4578} -> R2C78 = {78}, locked for R2 and N3, clean-up: no 4 in R1C1, no 3 in R2C9, no 2,3 in R3C9
23a. R23C78 (step 8) = 26, R2C78 = {78} = 15 -> R3C78 = {29/56}, no 4
23b. Killer pair 6,9 in R23C9 and R3C78, locked for N3
23c. Killer pair 2,5 in R3C3 and R3C78, locked for R3

24. 9(3) cage at R1C7 = {135/234}, 3 locked for R1
24a. R1C9 = {45} -> no 4,5 in R1C78
24b. Hidden killer pair 1,2 in 19(5) cage at R1C4 and 9(3) cage for R1, 9(3) cage contains one of 1,2 -> 19(5) cage must contain one of 1,2 in R1
24c. 19(5) cage contains both of 1,2 -> R2C5 = {12}

[First time through I’d forgotten that this was a Killer-X. It was only a few steps later, when I’d reduced both R78C1 and R78C7 to {29} that I realised that the diagonals were needed to give a unique solution. I’ve re-worked from step 26a, which has the first placement on a diagonal.]
25. R56C1 + R6C2 = {678} (hidden triple in N4) = 21 -> R7C23 = 5 = {14}, locked for R7 and N7, clean-up: no 8 in R6C9, no 7 in R7C1, no 2 in R8C6 (step 4)
25a. Naked pair {29} in R78C1, locked for C1 and N1 -> R1C1 = 8, placed for D\, R2C1 = 4, clean-up: no 6 in R3C9, no 5 in R8C9
25b. Naked pair {67} in R56C1, locked for C1 and N4 -> R6C2 = 8, R9C1 = 3, placed for D/, R9C23 = 14 = [68], R8C3 = 7, R8C2, = 5, placed for D/, R1C9 = 4, placed for D/, R3C9 = 9, R2C9 = 1, R2C2 = 3, placed for D\, R13C2 = [97], R1C3 = 6, clean-up: no 2 in R3C78 (step 24), no 6 in R7C9 (step 4), no 3 in R6C9
25c. R8C23 = [57] = 12 -> R78C4 = 7 = [34/61]

26. R3C7 = 6, placed for D/, R3C8 = 5, R3C3 = 2, placed for D\, R4C9 = 7, R6C9 = 6, R7C9 = 3, R5C9 = 2, R4C8 = 1 (cage sum), R8C6 = 4 (step 4)
26a. R8C6 = 4 -> R78C7 = 11 = {29} -> R7C7 = 9, placed for D\, R8C7 = 2
26b. R9C9 = 5, placed for D\

27. R2C78 = [78] = 15, R3C6 = 3 -> R2C6 = 6 (cage sum)

and the rest is naked singles, without using the diagonals.

I'll rate my walkthrough for Pinata #30 at 1.5, based on the heavy analysis in step 16a.