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1. Ar1c1 has a 3(2) = {12}: both locked for n1
1a. and 7(2): which can only be {34} ({16/25} would clash with 3(2)): both locked for n1
1b. Ar1c1 from (1234)

2. Ar2c2 also has a 7(2), which can't be {34} or it would clash with step 1a -> must be from {16/25} with the overlap cell at r2c2 = (12)
2a. 3(2) in Ar1c1 must be in c2 or r2 -> no 1,2 in r1c1

3. Ar2c7 and r7c2 both have a 16(2) = {79} and 4(2) = {13} -> naked quad 1,3,7,9: all locked for respective nonets

4. naked quad 1,3,7,9 in r2c8, r3c7, r7c3 and r8c2: all locked for D/

5. "45" on n3: 1 overlap cell r2c8 + 5 = 2 innies r1c7 + r3c9 -> no 5 in either innies (IOU)

6. 5 in n3 only in Ar1c8 = 14(2) & 6(2) -> must have {15/59}. 1 & 9 are only available in r2c8 -> r2c8 = (19)
6a. no 5 in r1c9 since cages cannot be diagonal

7. same "45" as step 6 on n7 with all the same (mirror image) eliminations -> r8c2 = (19)
7a. no 5 in r9c1

8. naked pair 1,9 in r8c2 and r2c8: both locked for D/ and no 1,9 in common peers in r2c2 and r8c8 (CPE)
8a. naked pair 3,7 in r7c3 + r3c7-> no 3,7 in r3c3 or r7c7 (CPE)

9. r2c2 = 2, placed for D\
9a. Ar2c2 must have 2 in r2c2 in the 7(2) cage since a 15(2) can't have a 2-> 5 locked in r2c3 or r3c2 for n1
9b. 15(2) in Ar2c2 must involve r3c3: no 7 in r3c3 -> no 8 in r2c3 or r3c2

10. note: Ar7c7 has a 6(2) = {15/24} = [1/2,1/4..]
10a. "45" on n9: 1 overlap cell at r8c8 - 3 = 2 innies r7c9 + r9c7
10b. min. 2 innies = {13} = 4 (can't be {12} because this would clash with 6(2) in Ar7c7) -> min. r8c8 = 7
10c. r8c8 = (78) -> 2 innies = 4,5 = {13/23}({14} blocked since it would clash with 6(2) in Ar7c7)
10c. 3 must be in 2 innies, locked for n9
10d. r7c9 & r9c7 = (123)

11. Ar7c7 cannot have 6 in 12(2) or 6(2) and cannot have 9 since no 3 is available
11a. 6 and 9 in Ar8c8 must form 13(2) = {67} and 17(2) ={89}: all locked for n9

12. Ar6c6: no 5 possible in 10(2) or 15(2) -> no 5

13. Ar3c3: 4(2) = {13} cannot involve r3c3 -> must have r4c4 -> r4c4 = (13)

14. r5c5 = 5 (hsingle D\)
14a. placed for D/

15. "45" on n5: 1 overlap cell r5c5 + 5 = 2 innies r4c4 + r6c6
15a. -> 2 innies = 10 = [19/37]

16. r8c8 + r9c9 cannot be [89] since no 8 would be available in r8c9 or r9c8 for 17(2) also cannot be [79], blocked by r6c6 -> no 9 in r9c9

17. Generalized X-wing on 9 in n9 and D/ -> 9 must be in r29c8, locked for c8; and in r8c29, locked for r8

18. 9 in n3 only in r2: locked for r2

19. 9 in n7 only in c2: locked for c2

20. In Ar2c2, r3c3 must be in 15(2) (from step 9a), no 9 in r2c3 or r3c2 -> no 6 in r3c3

21. r9c9 = 6 (hsingle D\)
21a. -> 13(2) in Ar8c8 must be in r89c9 or r9c89 = {67} -> no 7 in r8c8 (CPE)
21b. r8c8 = 8, placed for D\
21c. r3c3 = 9, placed for D\
21c. r6c6 = 7 -> r4c4 = 3 (step 15a), placed for D\
21d. r1c1 = 4 placed for d\
21e. r7c7 = 1

22. deleted

23. naked pair {28} in r1c9 + r9c1: both locked for D/
23a. naked pair {46} in r4c6 + r6c4: both locked for n5
23b. naked pair {23} in n9: 2 locked for n9

24. "45" on n7: 1 overlap cell at r8c2 + 5 = 2 innies r7c1 + r9c3 -> 2 innies = 6,14 = [24/68]= [2/8..]
24a. r7c1 = (26), r9c3 = (48)

25. Killer pair 2,8 in innies n7 (step 24) and r9c1: 2 locked for n7; 8 locked for r9; 2 locked for c1

26. 2 in r8 only in n8: locked for n8
26a. 1 in r9 only in n8: locked for n8
26b. 4 in n7 only in r9: locked for r9

27. Ar8c4 can only have 1 in 9(2) cage but no 8 is available -> no 1 in r9c45
27a. r9c6 = 1 (hsingle n8)

28. r9c45 must have 7 or 9 for r9 since the only other place is r9c8 (Hidden killer pair)
28a. Ar8c4 sums to 20 -> cannot have both 7 & 9 (= 16)since no 1 is available to get the final 2 cells to sum to 4
28b. -> no 7 in r8c45

29. no 7 in r8c3 since it can only form a 16(2) in r8 in Ar7c2: but {79} clashes with r8c9
29a. r8c9 = 7 (hsingle r8)
29b. r9c8 = 9

Now getting it to singles ASAP.

30. 7 locked in r9 in r9c45 in Ar8c4 but r9c4 can't be 7. Like this.
30a. Ar8c4 = 11(2) and 9(2) but r9c45 can't sum to 11 or 9 -> the two cages must be vertical
30b. with 7 in r9c4 -> r9c5 = 3 -> r89c5 must be 9(2) = [63] and r89c4 = 11(2) = [47]
30c. but r8c45 = [46] clashes with r8c17 = (456)
30d. -> no 7 in r9c4

31. r9c4 = 5, r9c5 = 7 (hsingle r9)

32. r89c4 cannot be 9(2) since this would force r89c5 = 11(2) but both would need a 4
32a. r89c4 = 11(2) -> r8c4 = 6
32b. r89c5 = 9(2) ->r8c5 = 2

33. r6c4 = 4 -> r4c6 = 6

34. Ar4c5 with 6 in r4c6 must have the 6 with 14(2) -> 8 locked for n5

35. Ar5c4 with 4 in r6c4 must belong to 13(2) since no 3 is available for 7(2) -> 7(2) and must have 5 -> 2 locked
35a. r5c4 = 2
35b. r6c45 must be 13(2) -> r6c5 = 9

rest is singles without considering the diagonals.