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614 579 382
732 684 915
958 213 674
846 137 259
123 956 748
597 842 136
389 725 461
271 468 593
465 391 824

Prelims courtesy of SudokuSolver
Preliminaries
Cage 5(2) n1 - cells only uses 1234
Cage 14(2) n6 - cells only uses 5689
Cage 13(4) n9 - cells do not use 89
Cage 39(6) n12 - cells ={456789}

1. "45" on n6: 3 innies r4c7+r6c78 = 6 = {123} only: all locked for n6

2. "45" on n69: 3 innies r467c7 = 7 = {12}[4] only
2a. r7c7 = 4: placed for D\
2b. 1 & 2 locked in r46c7 for c7 and n6
2c. -> r6c8 = 3
2d. r7c7 = 4 -> r78c6 = 13 = {58/67}(no 1,2,3,9)
2e. no 1 in r2c3

3. "45" on n3: 1 innie r2c7 - 7 = 1 outie r4c7 = [81/92]

4. 17(4)r2c6 must have 8 or 9 for r2c7 = {1259/1268/1349/1358/2348}
4a. Can't have more than two of 1,2,3 -> r3c4 must have one of 1,2,3 for n2 (Hidden killer triple)
4b. r3c4 = (123)

5. "45" on r12: Two outies r34c1 - 3 = 3 innies r2c678
5a. Min. r2c678 = {128} = 11 -> min. r34c1 = 14 (no 1,2,3,4)
[note: {128} clashes with 5(2)n1 which must have 1 or 2 but this is an extra dimension that is not necessary to this optimised solution]

6. "45" on n78: 2 remaining outies r56c1 = 6 = {15/24}(no 3,6,7,8,9) = one of 1 or 2
6a. 30(5)r4c2 = can't have more than one of 1,2,3
6b. 30(7)r4c3 = {1234569/1234578}, ie, must have each of 1,2,3, two of which must be in n45
6c. -> the remaining two out of six of 1,2,3 for n45 must be in r4c456+r5c6 in 32(7)r3c2 (Hidden killer sextuple)
6d. -> 1,2,3 taken with r3c4 for 32(7)r3c2 -> no 1,2,3 in r3c23 (Killer triple)
[note: wellbeback noticed that a powerful follow on step would have made step 7d. redundant
6e. -> Since 31/5@r1c1 contains at most one of (123) and r1c3 is 4 or more -> two of (123) in n1 must be in r2c23 -> 5/2@r2c2 = {23}.]

7. r4c1 must repeat in n1 in one of r1c3 or r3c23 since no common digits with r2c23
7a. "45" on n1: 1 outie r4c1 + 9 = 3 innies r1c3 + r3c23
7b. since r4c1 repeats in one of those 3 innies -> the other two innies make up the Innie Outie Difference (IOD) of 9 -> must be {45} only
7c. 4 and 5 locked for n1
7d. 5(2)n1 = {23} only: both locked for n1 and r2

8. deleted

9. 2 & 3 in n2 only in r3: both locked for r3

10. 20(5)r2c8 = {12458/12467}(no 9)
10a. must have 2 -> r4c7 = 2
10b. -> r2c7 = 9 (IODn3=+7)
10c. r6c7 = 1
10d. 20(5)r2c8 must have 1 & 4 which are only in n3 -> both locked for n3
10e. no 5 in r5c1 (h6(2)r56c1)

11. 3 in c1 only in r789c1: locked for n7

12. 5 in c1 only in r6789c1: r7c23 see all those -> no 5 in r7c23 (Common Peer Elimination CPE)

13. r56c1 = 6 -> r7c123 = 20 = {389/569/578}(no 1,2)
13a. Must have 3 or 5 which are only in r7c1 -> r7c1 = (35)

14. 1 in c3 only in r89c3: 1 locked for n7 and for 25(6)r7c4 -> no 1 in r7c45+r8c4

15. 1 in r1 only in r1c12 -> no 1 in r2c1

16. 1 in D/ only in r2c8 or r4c6: r2c6 sees both of those -> no 1 in r2c6 (CPE)
16a. Hidden single 1 in r2 -> r2c8 = 1: Placed for D/
16b. -> 4 in n3 only in r3: locked for r3
16c. -> Hidden single 4 in n1 -> r1c3 = 4
16d. -> Hidden single 4 in n2 -> r2c6 = 4

17. r2c67 = [49] = 13 -> r3c56 = 4 = {13} only: both locked for n2
17a. r3c4 = 2

18. "45" on r12: 2 outies r34c1 = 17 = {89} only: both locked for c1 and 31(5)r1c1 -> no 8,9 in r1c2
18b. Naked triple {167} in r1c12+r2c1: locked for n1
18c. Naked triple {589} in r3c123: all locked for r3
18d. & 5 locked for 32(7)r3c2 -> no 5 in r4c456+r5c6
18e. Naked triple {467} in r3c789: all locked for n3

19. Hidden single 1 in r7 -> r7c9 = 1
19a. 13(4)n9 must have 1 & 3 for n9 = {1237} only: locked for n9 and 3 for c9

20. Hidden single 2 in r7 -> r7c5 = 2
20a. Hidden single 2 in n5 -> r6c6 = 2: Placed for D\
20b. no 4 in r5c1 (h6(2)r56c1)
20c. r2c23 = [32]: 3 placed for D\ -> r9c89 = [27], 7 placed for D\, r8c9 = 3

21. Hidden single 2 in c9 -> r1c9 = 2: Placed for D/
21a. Hidden single 2 in n7 -> r8c1 = 2, r56c1 = [15] (h6(2)), r1c1 = 6, Placed for D\, r2c1 = 7, r1c2 = 1

22. r7c1 = 3 -> r7c23 = 17 = {89} only (h20(3)r7c123): both locked for r7 and n7
22a. r9c1 = 4, Placed for D/, r8c2 = 7 (cage sum), Placed for D/

23. Hidden single 9 in n9 -> r8c8 = 9: Placed for D\

24. Naked pair {58} in r3c3 + r5c5, 8 locked for D\ and no 8 in r7c3 since it sees both 5 & 8 through D/

Easy from there.

Prelims

a) R2C23 = {14/23}
b) R56C9 = {59/68}
c) 13(4) cage at R7C9 = {1237/1246/1345}, no 8,9
d) 39(6) cage at R1C3 = {456789}, no 1,2,3

1. 45 rule on N6 3 innies R4C7 + R6C78 = 6 = {123}, locked for N6
1a. Max R6C8 = 3 -> min R7C8 + R8C78 + R9C7 = 28, no 1,2,3 in R7C8 + R8C78 + R9C7

2. 45 rule on N3 1 innie R2C7 = 1 outie R4C7 + 7, R2C7 = {89}, R4C7 = {12}
2a. Min R2C7 = 8 -> max R2C6 + R3C56 = 9, no 7,8,9 in R2C6 + R3C56
2b. 3 in N6 only in R6C78, locked for R6

3. Hidden killer triple 1,2,3 in 17(4) cage at R2C6 and R3C4 for N2, 17(4) cage cannot contain all of 1,2,3 -> R3C4 = {123}
3a. 7,8,9 in N2 only in R1C456 + R2C45, locked for 39(6) cage at R1C3, no 7,8,9 in R1C3

4. 45 rule on N9 1 innie R7C7 = 1 outie R6C8 + 1, R6C8 = {123} -> R7C7 = {234}
4a. 31(5) cage at R6C8 cannot be {34789}, which clashes with R6C8 + R7C7 = [34], no 4 in R7C8 + R8C78 + R9C7

5. 45 rule on N69 3 innies R467C7 = 7 = {124} -> R7C7 = 4, placed for D\, R46C7 = {12}, locked for C7 and N6 -> R6C8 = 3, clean-up: no 1 in R2C3
5a. R7C7 = 4 -> R78C6 = 13 = {58/67}
5b. 3 in C7 only in R13C7, locked for N3
5c. Naked pair {12} in R46C7, CPE no 1,2 in R4C3
5d. 30(7) cage at R4C3 must contain 3 in R4C3 + R5C345, CPE no 3 in R5C2

6. 1,2,3 in N9 only in 13(4) cage at R7C9 = {1237}, 7 locked for N9

7. 45 rule on N78 2 remaining outies R56C1 = 6 = {15/24}
7a. 45 rule on N78 3 remaining innies R7C123 = 20 = {389/569/578}, no 1,2
7b. R7C123 + R7C8 must contain 8, locked for R7, clean-up: no 5 in R8C6 (step 5a)

8. 45 rule on N8 3 remaining innies R7C45 + R8C4 = 13 = {139/238/247} (cannot be {148} because 4,8 only in R8C4, cannot be {157/256} which clash with R78C6, cannot be {346} which clashes with R7C123 + R7C6), no 5,6 in R7C45 + R8C4
8a. 4,8 of {238/247} must be in R8C4 -> no 2,7 in R8C4
8b. 45 rule on N8 3 remaining outies R8C3 + R9C23 = 12 = {138/147/156/246} (cannot be {129/237/345} which clash with R7C45 + R8C4 (other part of 25(6) cage), no 9 in R8C3 + R9C23
8c. 13(3) cage at R8C1 must have the same combination as R7C45 + R8C4, because of interactions with R7C123 and R8C3 + R9C23 -> 13(3) cage = {139/238/247}, no 5,6

9. R8C3 + R9C23 (step 8b) = {138/147/156/246}, R7C45 + R8C4 (step 8) = {139/238/247}
9a. Consider combinations for R7C123 (step 7a) = {389/569/578}
R7C123 = {389/578}, 8 locked for N7
or R7C123 = {569}, locked for R7 => R7C6 = 7 => R7C45 + R8C4 = {139/238}, 3 locked for 25(6) cage at R7C4
-> R8C3 + R9C23 = {147/156/246}, no 3,8
[Note. I can see that 26(5) cage at R5C1 must contain 5 because {24}{389} clashes with 13(3) cage at R8C1 = {247}. However this doesn’t help at this stage.]

10. 17(4) cage at R2C6 = {1259/1268/1349/1358/2348} (other combinations don’t contain one of 8,9 for R2C7)
10a. 17(4) cage = {1259/1268/1349/2348} (cannot be {1358} because R3C4 = 2, 2 in N5 only in R5C5 + R6C56, R6C7 = 1, R4C7 = 2 clashes with R24C7 = [81/92], step 2)

11. Consider combinations for R2C23 = [14]/{23}
R2C23 = [14] => 4 in 39(6) cage at R1C3 must be in R1C456 and 1 in N2 in R3C56 => 18(4) cage at R1C7 must contain 1 but not 4
or R2C23 = {23} => 2,3 in N2 only in R3C456 => 2,3 in N3 only in 18(4) cage at R1C7
-> 18(4) cage = {1269/1278/1359/1368/2349/2358/2367} (cannot be {1458/1467/2457/3456}
11a. 18(4) cage = {1269/1278/1359/2358/2367} (cannot be {1368/2349} because R2C7 + R3C789 + R4C7 = {2457}1/{1567}2 don’t total 20), no 4
11b. 1 of {1269/1278/1359} must be in R1C89 -> no 1 in R2C9
11c. 3 of {1359/2358} must be in R1C7 -> no 5 in R1C7

12. Deleted; when I checked my walkthrough, I found that I’d carelessly omitted a combination in an earlier step.

13. 17(4) cage at R2C6 (step 10a) = {1259/1268/1349/2348}
13a. Consider combinations for R2C23 = [14]/{23}
R2C23 = [14] => 4 in 39(6) cage at R1C3 must be in R1C456, locked for N2 => 17(4) cage = {1259/1268} => R3C4 = 3 (hidden single in N2) => R1C7 = 3 (hidden single in N3)
or R2C23 = {23} => 2,3 in N2 only in R3C456 => R1C7 = 3 (hidden single in N3)
-> R1C7 = 3
13b. 7 in C7 only in R35C7, CPE no 7 in R5C5 using D/

14. R1C7 = 3 -> 18(4) cage at R1C7 (step 11a) = {1359/2358/2367}
14a. 18(4) cage = {1359/2358}, killer pair 8,9 in 18(4) cage and R2C7, locked for N3
or 18(4) cage = {2367} => 1 in N3 only in 20(5) cage at R2C8 => R4C7 = 2, R2C7 = 9 (step 2)
-> no 9 in R2C8 + R3C789
14b. 9 in R3 only in R3C123, locked for N1

15. Consider combinations for 17(4) cage at R2C6 (step 10a) = {1259/1268/1349/2348/2348}
15a. 17(4) cage = {1259/1268} => R3C4 = 3
or 17(4) cage {1349} => 4 in 39(6) cage at R1C3 only in R1C3 => R2C23 = {23}, locked for R2 and N1
-> no 3 in R2C6 + R3C123
26b. 3 in 32(7) cage at R3C2 only in R3C4 + R4C456 + R5C6, CPE no 3 in R5C4

[With hindsight I ought to have seen this next step earlier. After it the puzzle is cracked.]
16. 18(4) cage at R1C7 (step 14) = {1359/2358/2367}
16a. Hidden killer pair 1,2 in 31(5) cage at R1C1 and 18(4) cage for R1, 18(4) cage contains one of 1,2, 31(5) cage cannot contain both of 1,2 -> 31(5) cage contains one of 1,2 in R1C12, 18(4) cage contains one of 1,2 in R1C89, no 1,2 in R234C1, no 2 in R2C9
16b. Killer pair 1,2 in 31(5) cage and R2C23, locked for N1
16c. 31(5) cage at R1C1 contains one of 1,2 = {16789/25789}, no 3,4

17. Deleted

18. 3 in N1 only in R2C23 = {23}, locked for R2 and N1
18a. 2 in R1 only in R1C89, locked for N3
18b. 18(4) cage at R1C7 (step 12a) = {2358/2367}, no 1,9

19. 1 in N3 only in R2C8 + R3C89, locked for 20(5) cage at R2C8 -> R4C7 = 2, R2C7 = 9 (step 2), R6C7 = 1, clean-up: 5 in R5C1 (step 7)
19a. 9 in N9 only in R78C8, locked for C8

20. R2C8 = 9 -> 17(4) cage at R2C6 (step 10a) = {1259/1349}, no 6, 1 locked for N2
20a. 6 in N2 only in R1C456 + R2C45, locked for 39(6) cage at R1C3, no 6 in R1C3

21. 31(5) cage at R1C1 (step 16c) = {16789} (only remaining combination), no 5, 9 locked for C1
21a. 3 in C1 only in R789C1, locked for N7

22. 1 in C3 only in R89C3, locked for N7 and 25(6) cage at R7C4, no 1 in R7C45 + R8C4
22a. R7C45 + R8C4 (step 8) = {238/247}, no 9, 2 locked for R7, N8 and 25(6) cage at R7C4, no 2 in R8C3 + R9C23
22b. 4,8 of {238/247} must be in R8C4 -> R8C4 = {48}

23. R7C9 = 1 (hidden single in R7)
23a. 1 on D/ only in R2C8 + R4C6, CPE no 1 in R2C6
23b. 1 in N2 only in R3C56, locked for R3
23c. R2C8 = 1 (hidden single in N3), placed for D/

24. 4 in N3 only in R3C89, locked for R3
24a. R1C3 = 4 (hidden single in N1)
24b. R2C6 = 4 (hidden single in N2) -> 17(4) cage at R2C6 (step 20) = {1349} (only remaining combination)

25. R3C4 = 2 (hidden single in N2), R7C5 = 2 (hidden single in R7)
25a. R6C6 = 2 (hidden single in N5), placed for D\ -> R2C2 = 3, placed for D\, R2C3 = 2, R9C9 = 7, placed for D\, R9C8 = 2, R8C9 = 3, clean-up: no 4 in R5C1 (step 7)

26. R1C9 = 2 (hidden single in C9), placed for D/
26a. R8C1 = 2 (hidden single in N7), R5C1 = 1, R6C1 = 5 (step 7), clean-up: no 9 in R5C9
26b. R7C123 (step 7a) = {389} (only remaining combination) -> R7C1 = 3, R7C23 = {89}, locked for R7 and N7, R7C4 = 7, R9C1 = 4, R8C2 = 7, both placed for D/, clean-up: no 6 in R78C6 (step 5a)
26c. R78C6 = [58], R8C4 = 4, R7C8 = 6, R8C7 = 5, R8C8 = 9, placed for D\, R9C7 = 8, R3C7 = 6, placed for D/, R5C7 = 7

27. Naked pair {89} in R6C4 + R7C3, locked for D/ -> R4C6 = 3, R5C5 = 5, placed for D\, clean-up: no 9 in R6C9
27a. R3C3 = 8, placed for D\, R7C3 = 9, placed for D/
27b. Naked pair {67} in R46C3, locked for N4 -> R5C3 = 3

28. Naked pair {68} in R56C9, locked for C9 and N6 -> R2C9 = 5

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

I'll rate my walkthrough for A279 at 1.75. This is partly based on my deleted step 12. I would probably have gone for a lower rating if I'd spotted step 16 earlier.

1. Innies n6 -> r4c7+r6c78 = +6 = {123}
Innies n69 => r467c7 = +7 = {124}
-> r6c8 = 3, r7c7 = 4, r46c7 = {12}
Innies n36 -> r26c7 = +10 -> r26c7 = [91] or [82]

2. 3 in n9 must go in 13/4@r7c9 which cannot contain a 4.
-> 13/4@r7c9 = {1237} with 3 in c9
-> 31/5@r6c8(n9) = {5689}

3. 39/6@r1c3 = {456789}
-> (123) in n2 all in r2c6+r3c456
Whatever is in r6c7 (1 or 2) must go in 32/7@r3c2(n5)
-> it must go in 17/4@r2c6(n2)
and since r26c7 = +10 -> Other two values in 17/4@r2c6(n2) = +7
-> 17/4@r2c7 from {19(25|34)} or {28(16|34)}
-> one of (456) in 17/4@r2c6(n2)
-> r1c3 from (456)

4. Outies r1 -> r2c1459+r34c1 = +43
Since r2c7 from (89) -> Max r2c1459 = +27 (9765)
-> Min r34c1 = +16 - i.e. is {79} or {89}
-> Whatever goes in r4c1 (from 789) must go in n1 in r3c23
-> Innies n1 (r1c3 + Other value in r3c23) = +9
-> since r1c3 from (456) -> Other value in r3c23 from (543)

5. 32/7@r3c2 and 30/7@r4c3 both contain (12)
-> Whatever is in r4c7 (1 or 2) must be in 30/7@r4c3 in r5c345 or r6c56
All those locations are buddies with r5c6
-> Only location for that value (1 or 2) in the 32/7@r3c2 is r3c4
I.e., r3c4 from (12) and is the same value as in r4c7

6. Conclusions from Step 5
-> 3 in 17/4@r2c6(n2)
-> 4 in 17/4@r2c6(n2)
-> r1c3 = 4
-> 5/2@r2c2 = {23}
-> (23) in r3 in n2 (r3c456)
-> (23) in r1 in n3 (r1c789)
-> 1 in 20/5@r2c8(n3)
-> r4c7 = 2 -> r3c4 = 2
-> r6c7 = 1
-> r2c7 = 9
-> 17/4@r2c6 = {1349} with 9 in r2c7
Also 1 in r1c12
-> 31/5@r1c1 = {16789} with 1 in r1c12

7. Outies r1 -> r2c1459+r34c1 = +43
Since 9 in r2c7 -> Max r2c1459 = +26
-> Min r34c1 = +17
-> r34c1 = {89} and r1c12+r2c1 = {167}
-> (67) are also in 39/6@r1c3(n2)
-> (67) in r3 in n3 in r3c789
-> 20/5@r2c8 = {12467} with 2 in r4c7
-> 18/4@r1c7 = {2358} with r1c7 = 3 and 2 in r1c89

8. Outies r789 = +9
Since r6c8 = 3 -> r56c1 = {15} or {24}

9. Breakthrough!!
30/7@r4c3 contains exactly one of (67)
32/7@r3c2 contains either both or neither of (67)

Either: a) Putting 6 or 7 in r6c4 would put that same value in n4 in r45c3
which would put the other of (67) in 32/7@r3c2(n5)

or b) Putting some other value in r6c4 puts at least one of (67) in 32/7@r3c2(n5)

Either way - at least one of (67) in 32/7@r3c2(n5)
-> (Since (67) are in r3c789) -> both of (67) in 32/7@r3c2(n5)

10. Conclusions from Step 9
-> One of (67) in 30/7@r4c3(n4) (I.e., in r45c3)
-> One of (67) in r6 in n6
-> Only possibility is 6 in r6c9
-> 14/2@r5c9 = [86]
-> 25/4@r4c8 = {4579} with 9 in r4c9
Also -> 6 in r45c3
Also -> r1c8 = 8
-> r12c9 = [25]
-> r9c8 = 2
-> r789c9 = {137}
-> r3c9 = 4
-> HS (4 in n1) -> r1c3 = 4
-> 17/4@r2c6 = [49{13}]
-> r2c8 = 1 and r3c78 = {67}

11. More conclusions
Also -> r3c4 = 2
-> 2 in n8 can only go in r78c5
-> HS 2 in c6 -> r6c6 = 2
-> (D\) 5/2@r2c2 = [32]
-> 3 in n4 can only go in r45c3
-> r45c3 = {36}
-> 30/7@r4c3 must contain a 4 which can only go in n5
-> 4 not in 32/7@r3c2
-> 9 not in 32/7@r3c2
-> r3c1 = 9, r3c23 = {58}
-> r4c1 = 8
Also 32/7@r3c2(n5) = {1376}
-> 30/7@r4c3(n5) = {2459} (with 2 in r6c6)
-> r6c4 = 8

12. r7c123 = +20 -> No (12)
-> HS 2 in r7 -> r7c5 = 2
-> 2 in n7 in r8c12. Must be r8c1 since 2 already in D/ in r1c9.
-> 1 in n7 in 25/6@r7c4(n7)
-> 1 in n8 in 19/4@r8c5
-> r789c9 = [137]
Also r78c4 = +11 must be [74] or {56}
-> r78c6 (= +13) must be {58}
-> r78c4 = [74]
-> 13/3@r8c1 = [{27}4]
-> HS 3 in c1 -> r7c1 = 3
-> r7c23 = {89} - must be [89] since 8 already in D/ in r6c4
-> r78c6 = [58]
Also -> r56c1 = [15]
-> r1c2 = 1, r12c1 = {67} - must be [67] since 7 already in D\ in r9c9.
-> r8c12 = [27]
-> r6c3 = 7
-> r456c2 = [429]
etc.