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This is in some way similar to the DJAPE 20050927 puzzle I just posted earlier. Very easy to start, and then gradually becomes more and more difficult, requiring more and more "inter-cage manipulation" steps to dissolve. Even takes a relatively tricky maneuveur with just a few cells remaining. A carefully written complete walkthrough to be submitted soon.

1:
N6 16(2) = {79} (R4,N6)
C2 4(2) = {13}
Innie N6: R5C7 = [6]
Innies N7: R7C23 = 3(2) = [12]
--> C2 4(2) = [31]

2:
N45 38(8) = {12345689}, <> [7]
Hidden single N5: R5C6 = [7]
Hidden single N4: N4 8(2) = [17]
--> N7 22(3) = {589} (C1,N7)
--> N7 9(2) = [63], N7 11(2) = {47} (R9)

3:
Innie-outies R9: R9C1 = R8C8 + 2
--> R8C8 = [7], R9C1 = [9]
--> N6 16(2) = [79], R9 10(2) = {28} (R9)
--> R9 11(2) = {56} (R9,N8)
Innies N78: R789C6 = 11(3) = [812]
--> N7 22(3) = [589], N78 14(2) = [239], C5 11(2) = [74], R8 6(2) = [15], R9 10(2) = [28]
--> R8C9 = [2]

4:
N6 11(2) = {38}
Innie-outies R56789: R5C2 = R4C59 + 5
--> R5C2 = [9], R4C59 = 4 = [13]
--> N6 11(2) = [38], N9 11(3) = [731]
Hidden single N5: R5C5 = [3]
--> N2 12(2) = [48/75]
[2] of N4 locked in R4C12 (R4)

*5:
N2 12(2) + N25 11(3) + R9 11(2) = 34(7)
Min R2349C4 = 1+4+5+6 = 16
--> Max R239C5 = 34-16 = 18
--> R3C5 = [2] (or min R239C5 = 5+6+8 = 19 > 18)
--> Min R239C5 = 2+5+6 = 13
--> Max R2349C4 = 34-13 = 21
--> R3C4 = [1] (or min R2349C4 = 4+5+6+7 = 22 > 21)
--> N25 11(3) = [128]

6:
Hidden single N4: R6C3 = [8]
--> [6] of N4 locked in R4C13 (R4)
C67 16(4) <> [9] (or min sum = 1+3+4+9 = 17 > 16)
C67 16(4): R34C6 <> [3/4] (or max sum = 2+3+4+6 = 15 < 16)
--> R34C6 = [65], R23C7 = 5 = [14/23]

7:
N1 8(2) = [35/62]
--> [8] of N1 locked in R23C2
--> N1 17(3) = {278/458} (not {368} since no {36} in R23C2)
N1 17(3): R2C1 = [2/4], R23C2 = {78/58}
[1] of N1 locked in R12C3
Innies N1: R12C3+R3C13 = 20(4)
--> R123C3 <> [4] (or max total = 1+3+4+9 = 17 < 20)
--> [4] of N1 locked in R23C1

*8:
Min R23C1 = 2+4 = 6 (step 7)
--> Min R23C17 = 6+5 = 11 (step 6)
R3C17 = 3+4 = 7
--> Min R2C17 = 11-7 = 4
--> R2C1 = [4] (or R2C17 = 2+1 = 3 < 4)
--> N2 12(2) = [75], R3C1 = [3]
--> C67 16(4) = [1645]

Naked singles from here.

Steps 5 and 8 (marked with *) are obviously my pick of critical breakthroughs.

In fact I find them quite similar to the "horizontal/vertical analysis" technique I picked up when learning to solve Kakuro puzzles from the experts in these forums.

Also, in hindsight this puzzle is a touch easier than the 20050927 killer, which doesn't offer these Kakuro-like techniques.

I've seen Kakuro and understand that it based on horizontal and vertical totals but that's the limit of my knowledge; I've never tried a Kakuro puzzle. Congratulations on spotting that the three cages in C45 can be combined as a group with a known total, even though they aren't a cage! Step 5 then did nice Kakuro analysis on the two columns; I hope I'm not wrong in considering it as min-max analysis.

Simon's nice 45 in his step 4, using placements and a naked triple in zero cells, shortened the solving path compared with the way I got that result with a series of simple steps.

While Simon and I also used different ways to get the final breakthrough, it was interesting to note that there was some sort of similarity. Simon's step used the fact that R2C17 both contained 2 while my way used the fact that R23C1 both contained 4.

I'll rate my walkthrough for A231 at 1.5 because I used a short forcing chain and an unusual blocker for my final step.

I agree with that even though I've given A231 a higher rating than I did for the 20050927 killer. A231 was fairly short but had two tricky steps. The 20050927 killer was a lot longer and, although I didn't use any technically difficult steps apart from a Killer ALS block near the end, I felt that some steps were hard to find.

Prelims

a) R1C12 = {17/26/35}, no 4,8,9
b) R2C45 = {39/48/57}, no 1,2,6
c) R4C78 = {79}
d) R45C9 = {29/38/47/56}, no 1
e) R56C1 = {17/26/35}, no 4,8,9
f) R67C2 = {13}
g) R78C5 = {29/38/47/56}, no 1
h) R8C23 = {18/27/36/45}, no 9
i) R8C67 = {15/24}
j) R9C23 = {29/38/47/56}, no 1
k) R9C45 = {29/38/47/56}, no 1
l) R9C67 = {19/28/37/46}, no 5
m) 11(3) cage at R3C4 = {128/137/146/236/245}, no 9
n) 22(3) cage at R7C1 = {589/679}
o) 11(3) cage at R8C8 = {128/137/146/236/245}, no 9
p) 12(4) cage at R5C8 = {1236/1245}, no 7,8,9
q) 38(8) cage at R5C5 = {12345689}, no 7

1. 45 rule on N6 1 innie R5C7 = 6, clean-up: no 5 in R45C9, no 2 in R6C1, no 4 in R9C6

2. Naked pair {79} in R4C78, locked for R4 and N6, clean-up: no 2,4 in R45C9

3. Naked pair {38} in R45C9, locked for C9 and N6

4. 45 rule on N7 2 innies R7C23 = 3 = [12], R6C2 = 3, clean-up: no 5,7 in R1C1, no 5 in R56C1, no 7,8 in R8C2, no 6,7,8 in R8C3, no 9 in R8C5, no 9 in R9C2, no 8,9 in R9C3

5. R9C23 = [47/74/83] (cannot be {56} which clashes with 22(3) cage at R7C1), no 5,6

6. R6C1 = 7 (hidden single in R6), R5C1 = 1, clean-up: no 7 in R1C2, no 6 in 22(3) cage at R7C1

7. Naked triple {589} in 22(3) cage at R7C1, locked for C1 and N7, clean-up: no 4 in R8C23, no 3 in R9C3

8. R8C23 = [63], clean-up: no 2 in R1C1, no 5,8 in R7C5

9. Naked pair {47} in R9C23, locked for R9, clean-up: no 3,6 in R9C6, no 3 in R9C7

10. 14(3) cage at R7C3 contains 2 = {239/248/257}, no 1,6, no 9 in R7C4
10a. 1 in N8 only in R89C6, locked for C6

11. 45 rule on N78 3 remaining innies R789C6 = 11
11a. Min R78C6 = 4 -> no 8,9 in R9C6, clean-up: no 1,2 in R9C7
11b. R789C6 = 11 contains 1 = {128/146} (cannot be {137} because 3,7 only in R7C6) -> R7C6 = {68}, no 5 in R8C6, clean-up: no 1 in R8C7

12. 1 in 12(4) cage at R5C8 only in R6C789, locked for R6

13. R4C5= 1 (only remaining place in 38(8) cage at R4C5)

14. 3 in 38(8) cage at R4C5 only in R5C45, locked for R5 and N5 -> R45C9 = [38]
14a. 9 in R6 only in R6C3456, locked for 38(8) cage at R4C5, no 9 in R5C345
14b. 2 in 38(8) cage at R4C5 only in R5C56 + R6C456, locked for N5

15. R5C6 = 7, R5C2 = 9 (hidden singles in R5)

16. 45 rule on R9 1 innie R9C1 = 1 outie R8C8 + 2 -> R8C8 = 7, R9C1 = 9, R4C78 = [79], R9C7 = 8, R9C6 = 2
16a. R8C8 = 7 -> R9C89 = 4 = [31]
16b. Naked pair {56} in R9C45, locked for N8 -> R7C6 = 8, R8C5 = 4, R7C5 = 7, R78C1 = [58], R78C4 = [39], R8C6 = 1, R8C7 = 5, R8C9 = 2
16c. Clean-up: no 5,8 in R2C4, no 3,9 in R2C5

17. R5C5 = 3 (hidden single in R5)

18. 17(3) cage at R2C1 = {278/458} (cannot be {368} because 3,6 only in R2C1, cannot be {467} which clashes with R9C2), no 3,6, 8 locked for C2 and N1
18a. R2C1 = {24} -> no 2,4 in R23C2
18b. 5 of {458} must be in R3C2 (R2C12 cannot be [45] which clashes with R2C45), no 5 in R2C2
18c. Killer pair 2,5 in R1C2 and 17(3) cage, locked for N1
18d. Killer pair 7,8 in R2C2 and R2C45, locked for R2
18e. 5 in N1 only in R13C2, locked for C2

19. 11(3) cage at R3C4 = {128/146/245}, no 7
19a. 8 of {128} must be in R4C4 -> no 8 in R3C45
19b. 6 of {146} must be in R3C5 -> no 6 in R34C4
19c. {245} can only be [425] (cannot be [254,524] which clash with R2C45, combo blocker) -> no 2,5 in R3C4, no 5 in R3C5

20. 16(4) cage at R2C7 = {1249/1456/2356}
20a. 9 of {1249} must be in R3C6 -> no 9 in R23C7, no 3,4 in R3C6
20b. 4,9 of {1249}must be in R34C6, 5,6 of {1456/2356} must be in R34C6 -> no 3,4 in R3C6

21. 11(3) cage at R3C4 (step 19) = {128/146/245}
21a. Consider permutations for R2C45 = [48/75]
R2C45 = [48] => 11(3) cage = {128}
or R2C45 = [75] => R9C5 = [56] => 11(3) cage = {128}
-> 11(3) cage = {128} -> R3C45 = [12], R4C4 = 8

22. R6C3 = 8 (hidden single in R6)
22a. 6 in N4 only in R4C13, locked for R4

23. 16(4) cage at R2C7 (step 20) = {1456/2356} (cannot be {1249} because 1,2 only in R2C7) -> R3C6 = 6, R4C6 = 5
23a. 1,2 only in R2C7 -> R2C7 = {12}
23b. Naked pair {34} in R3C17, locked for R3
23c. R5C3 = 5 (hidden single in N4)

24. 45 rule on N1 4 innies R12C3 + R3C13 = 20 contains 1,9 = {1379/1469}
24a. R3C1 = {34} -> no 4 in R12C3
24b. 4 in N1 only in R23C1, locked for C1

25. R23C7 = [14] (cannot be [23] which is blocked by R23C1 = [24/43], not sure what sort of blocker this is), R6C7 = 2

and the rest is naked singles.

I think I miss a point there, but how did you conclude "no 3 in R3C7" in 20b.?

Not that it affects any subsequent step though.

Thanks for the walkthrough. However I just have some reservations about steps 19c and 25. For some reasons these "combo blockers" feels very much like forcing chains or contradiction chains to me, because they essentially first assume 1 cell to be of a certain value, which forces other cells of the same cage to be of some other values, then those cells together would clash with all possibilities of another cage.

Take 19c. as an example: "11(3) cage at R3C4 cannot be [254,524] which clash with R2C45, combo blocker", could be interpreted this way:
R3C4 = [2] --> R3C5 = [5] & R4C4 = [4] --> no {45} for 12(2) cage at R2C45
R3C4 = [5] --> R3C5 = [2] & R4C4 = [4] --> no {45} for 12(2) cage at R2C45
But that 11(3) cage can be [425] without causing any problem to the 12(2) cage.

To me, a legitimate "combo blocker" should be more along the line of: "a certain combination for a cage, no matter in what permutation, would clash with all possibilities of another cage". So then the resulting elimination would be the whole combination from the cage, instead of some particular permutation(s) of that combination in the cage.

So personally I think "whole combination blocker" is more acceptable than "permutation blocker" if we plan to make our steps smell less like chains.

With R5C5 = [3] in step 17, R4C6 cannot be [3] anyway in step 20b.

3 in R4C6 was eliminated by step 14 3 in 38(8) cage at R4C5 only in R5C45, locked for R5 and N5