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3x3::k:3072:5377:5377:5377:5377:7682:3331:3331:5636:3072:5377:3845:5377:7682:7682:3331:5636:5636:4102:3845:3845:2311:2311:7682:7682:5636:4616:4102:5129:3082:3082:3082:7435:7435:7948:4616:4102:5129:5133:5133:3086:3086:7435:7948:4616:4367:5129:5133:5133:4368:7435:7435:7948:7948:4367:4113:4113:4113:4368:7435:9490:7948:2067:4367:5652:5652:9490:4368:4368:9490:2067:2067:4367:5652:5652:9490:9490:9490:9490:2069:2069:

1. 21(6)n1 = {123456}
Of the numbers (789) in n1 at most one in 15(3)
Also one in 12(2)n1
-> r3c1 from (789)

2. Innies r123 = r3c19 = +13(2)
-> r3c9 from (456)
Innies n3 = r3c79 = +10(2) (No 5)
-> r3c179 from [746] or [964]

3! IOD n2 -> Innies n2 (=r1c45,r2c4) = r3c7 + 6.
Since whatever is in r3c7 is in innies n2 -> Others in Innies n2 = +6(2)
-> Trying r3c179 = [746] puts innies n2 = {145} which leaves no solution for 9(2)n2
-> r3c179 = [964]

4. -> 6 in Innies n2
-> (HS 6 in n1) r2c3 = 6
-> 6 in n2 in r1c45

5. Innies c89 -> r1c8 = 3
H+9(2)r3c23 and 9(2)n2 between them are {18} and {27}
-> (HS 3 in n1) -> r2c2 = 3
-> (HS 3 in n2) -> r3c6 = 3
-> (HS 5 in r3) -> r3c8 = 5

6. Innies n9 = r7c8 + r789c7 = +29(4) = {5789}
-> (HS 4 in n9) -> r8c8 = 4
-> r78c9 = {13}
-> 8(2)n9 = {26}
-> 37(7)n89 = {1345789} with (134) in n8 and 5 in n9 in r789c7.
Also 5 in n6 in r456c9

7. Remaining IOD n6 -> +r456c7 = r7c8
-> Whatever is in r7c8 goes in n6 in r45c9
Since r45c9 = +14(2) = {59} or {68} -> r7c8 from (89)
-> r456c7 = {134} or {234}

8! Remaining cells in 30(5)n2 = r1c6 + r2c56 = +21(3) = {489} or {579}
Remaining Outies n6 = r467c6 = +21(3) or +20(3) (No (12))
-> HP r89c6 = [21]
-> (24) in r7 only in r7c123
Also whichever of (13) is in r8c9 is in n8 in r9 -> is in n7 in r7c123
I.e., r7c123 = {124} or {234}
-> 16(3)r7 = [439]!

9. -> r7c1 = 2
Also (Outies n9) -> r6c1 = 1
-> 17(4)c1 = [1268]
-> 16(3)c1 = [9{34}]
-> 12(2)n1 = {57}
-> r3c23 = {18}
-> r1c23 = [24]
Also 9(2)n2 = {27}
-> 30(5)n2 = [{489}36]
-> Innies n2 = {156}

10. Also r7c8 = 8
-> 18(3)c9 = [4{68}]
-> r6c9 = 5
Also r456c7 = {134}
Also r456c8 = {279}
Also r789c7 = {579}
-> r12c7 = [82]
-> r2c8 = 1
Also r2c56 = {48}
-> r1c6 = 9
-> r12c9 = [79]
-> 12(2)n1 = [57]
-> r2c4 = 5 and r1c45 = {16}

11. Also remaining cells in 29(6)r4c7 = r467c6 = +21(3) = {678} with 8 in r46c6
-> r2c56 = [84]
-> (NS 5 in c6) -> 12(2)n5 = [75]
-> r7c6 = 7 and r46c7 = {68}
Also (HS 8 in n8) -> r8c4 = 8
-> r9c45 = {34}
-> r78c5 = [65]
-> r6c5 = 4
-> r9c45 = [43]

12. Also HS 9 in n5 -> 12(3)r4 = [219]
-> r56c4 = {23}
-> r56c3 = [87]
etc.

Prelims

a) R12C1 = {39/48/57}, no 1,2,6
b) R3C45 = {18/27/36/45}, no 9
c) R5C56 = {39/48/57}, no 1,2,6
f) R9C89 = {17/26/35}, no 4,8,9
e) 20(3) cage at R4C2 = {389/479/569/578}, no 1,2
f) 8(3) cage at R8C9 = {125/134}
[Oops! I overlooked 21(6) cage at R1C2 = {123456}. I solved the puzzle without using that.]

1a. 45 rule on C89 1 innie R1C8 = 3, clean-up: no 9 in R2C1
1b. 45 rule on N9 4 innies R7C78 + R89C7 = 29 = {5789}, 5,7 locked for N9, clean-up: no 1,3 in R9C89
1c. Naked pair {26} in R9C89, locked for R9, 2 locked for N9
1d. 8(3) cage at R8C9 = {134}, 3 locked for C9
1e. R1C8 = 3 -> R12C7 = 10 = {19/28/46}, no 5,7
1f. 45 rule on N3 2 innies R3C79 = 10 = {19/28/46}, no 5,7
1g. 3 in N6 only in R456C7, locked for 29(6) cage at R4C6

2a. 37(7) cage at R7C7 = {1345789} (cannot be {1246789/2345689} because 2,6 only in R8C4), no 2,6, 1,3,4 locked for N8
2b. 45 rule on N89 3 innies R7C468 = 1 outie R6C5 + 20
2c. Max R7C468 = 24 -> max R6C5 = 4
2d. Min R7C468 = 21, no 2
2e. 2 in N8 only in R7C5 + R8C56, locked for 17(4) cage at R6C5

3a. 45 rule on N7 2(1+1) outies R6C1 + R7C4 = 10, R7C4 = {56789} -> R6C1 = {12345}
3b. 45 rule on R123 2 innies R3C19 = 13 = [49/58/76/94], no 1,2,3, no 6,8 in R3C1, clean-up: no 8,9 in R3C7 (step 1f)

4. Hidden killer triple 1,3,4 in R7C1, 16(3) cage at R7C2 and R7C9 for R7, R7C1 cannot contain more than one of 1,3,4, R7C9 = {134} -> 16(3) cage must contain at least one of 1,3,4 = {169/178/349/358/367/457} (cannot be {259/268} which don’t contain any of 1,3,4), no 2
[I looked at using a forcing chain on the combinations but, at this stage, {349} doesn’t lead to much although the other combinations significantly reduce the candidates for R7C1 and R7C5. However {349} = {34}9 does limit 17(4) cage at R6C1 to two combinations, which may be useful later.]

5a. R3C19 (step 3b) = [49/58/76/94], R3C79 (step 1f) = [19/28]/{46} -> combined half-cage R3C179 = [419/528/746/964]
5b. R3C45 = {18/27/36} (cannot be {45} which clashes with R3C179), no 4,5

6a. 45 rule on N2 3 innies R1C45 + R2C4 = 1 outie R3C7 + 6
6b. Whichever value of {1246} is in R3C7 can only be in R1C45 + R2C4 in N2 -> the other two values in R1C45 + R2C4 must total 6 = {15/24} -> no 3,7,8,9 in R1C45 + R2C4
6c. R3C179 (step 5a) = [419/528/746/964]
6d. Consider combinations for R3C45 (step 5b) = {18/27/36}
R3C45 = {18} => R1C45 + R2C4 = {246}, with 2,4 which total 6 not repeated in R3C7 => R3C7 = 6
or R3C45 = {27} => R3C179 = [419/964] => R1C45 + R2C4 = {156}, with 1,5 which total 6 not repeated in R3C7 => R3C7 = 6
or R3C45 = {36}, locked for R3
-> 6 in R3C45 + R3C7, locked for R3, clean-up: no 7 in R3C1 (step 3b), no 4 in R3C7 (step 1f)
6e. 6 in R3C45 + R3C7, CPE no 6 in R1C6 + R2C56
6f. Consider placements for R3C7 = {126}
R3C7 = 1 => R1C45 + R2C4 = {124}
or R3C7 = 2 => R1C45 + R2C4 = {125}
or R3C7 = 6 => R1C45 + R2C4 = {156/246}, R3C45 = {18/27}, killer pair 1,2 in N2
-> no 1,2 in R13C6 + R2C45
[I’ve realised that the later part of my original step 6h was incorrect. I’ve kept step 6g and the first part of step 6h for the record, but they can be ignored and one can move directly to step 6ha.]
6g. Hidden killer pair 3,6 in R12C456 + R3C6 and R3C45 for N2, R3C45 must contain both or neither of 3,6 -> R12C456 + R3C6 must contain both or neither of 3,6
6h. If R12C456 + R3C6 contains both of 3,6 then 6 in R1C45 + R2C4, 3 in R1C6 + R2C56 + R3C6 and 6 in R3C7
6ha. 21(6) cage at R1C2 = {123456} [The Prelim which I overlooked.]
6hb. Hidden killer triple 7,8,9 in R12C1, 15(3) cage at R2C3 and R3C1 for N1, R12C1 contains one of 7,8,9, 15(3) cage cannot contain more than one of 7,8,9 -> R3C1 = 9 -> R3C9 = 4 (step 3b), R3C7 = 6 (step 1f), clean-up: no 3 in R2C1, no 3 in R3C45
6i. Naked pair {13} in R78C9, 1 locked for C9 and N9
6j. R3C1 = 9 -> R45C1 = 7 = {16/25/34}, no 7,8
6k. R3C9 = 4 -> R45C9 = 14 = {59/68}
6l. Killer quad 5,7,8,9 in R12C7 and R789C7, locked for C7, 5,7 also locked for N9 and 37(7) cage at R7C7

7a. 4 in C7 only in R456C7, locked for N6 and 29(6) cage at R4C7
7b. R456C7 = {134/234} = 8,9 -> R467C6 = 20,21 = {569/578/579/678}, no 1,2
7c. R89C6 = [21] (hidden pair in C6)
7d. R8C8 = 4
7e. 4 in N8 only in R9C45, locked for R9
7f. Naked quint {56789} in R7C45678, locked for R7, 6 locked for N8

8a. 6 in N2 only in R1C45 + R2C4, locked for 21(6) cage at R1C2
8b. R2C3 = 6 (hidden single in N1)
8c. R2C3 = 6 -> R3C23 = 9 = {18/27}, no 3,5
8d. Killer pair 7,8 in R12C1 and R3C23, locked for N1
8e. R2C2 = 3 (hidden single in N1)
8f. R3C68 = [35] (hidden pair in R3), clean-up: no 9 in R5C5

9a. 16(3) cage at R7C2 (step 4) = {349} (only remaining combination, cannot be {169/178/358/367/457} because 5,6,7,8,9 only in R7C4) = [439]
9b. R7C19 = [21], R8C9 = 3, R7C8 = 8, R6C4 = 8, clean-up: no 1 in R3C5, no 5 in R45C1 (step 6j)
9c. 8 in C7 only in R12C7 (step 1e) = {28}, locked for N3, 2 locked for C7
9d. 8 in N6 only in R45C9 (step 6k) = {68}, 6 locked for C9 and N6 -> R9C89 = [62]
9e. Naked pair {79} in R12C9, locked for C9 and N3 -> R2C8 = 1, R6C9 = 5
9f. Naked triple {134} in R6C157, locked for R6
9g. R7C4 = 9 -> R6C1 = 1 (step 3a), clean-up: no 6 in R45C1 (step 6j)
9h. Naked pair {34} in R45C1, 4 locked for C1 and N4, clean-up: no 8 in R12C1
9i. Naked pair {57} in R12C1, locked for C1 and N1, clean-up: no 2 in R3C23 (step 8c)
9j. Naked pair {18} in R3C23, locked for R3, 1 locked for N1
9k. R1C23 = [24] -> R2C4 = 5
9l. 6 in N4 only in 20(3) cage at R4C2 = {569}, locked for C2, 5,9 locked for N4

10a. Naked pair {34} in R69C5, locked for C5, clean-up: no 8,9 in R5C6
10b. R2C56 = [84] (hidden pair in R2), R5C5 = 7 -> R5C6 = 5
10c. R78C5 = [65], R8C6 = 2 -> R6C5 = 4 (cage sum)
10d. R4C5 = 9 (hidden single in C5) -> R4C34 = 3 = [21]

and the rest is naked singles.

I'll rate my walkthrough for A402 at 1.5

Preliminaries courtesy of SudokuSolver
Cage 8(2) n9 - cells do not use 489
Cage 12(2) n5 - cells do not use 126
Cage 12(2) n1 - cells do not use 126
Cage 9(2) n2 - cells do not use 9
Cage 8(3) n9 - cells do not use 6789
Cage 20(3) n4 - cells do not use 12
Cage 21(6) n12 - cells ={123456}

1. "45" on c89: 1 innie r1c8 = 3
1a. -> r12c7 = 10 = {19/28/46}(no 5,7)

2. 21(6)r1c2 = {123456}. Must have 3 which is only in r2: locked for r2

3. 12(2)n1 = {48/57}(no 9)
3a. = 4 or 5
3b. -> 21(6), which must have both 4 & 5 must have at least one in n2
3c. -> {45} blocked from 9(2)n2 [Another way to do this, much funner, at 6b)
3d. = {18/27/36}(no 4,5)

4. "45" on r123: 2 innies r3c19 = 13 = {49/58/67}(no 1,2,3)

5. "45" on n3: 2 innies r3c79 = 10 = [19/28]{46}(no 5, no 789 in r3c7, no 7 in r3c9)

Key step
6. 9(2)r3c4 sees all the 30(5)r1c6 -> forms a combined 39(7) with no repeats
6a. -> must be missing a 6(2) = missing {15/24}.
6b. ie, can't have both 1 and 4 (nor both 4 & 5, so {45} blocked from 9(2)r3c4 if it wasn't already blocked)
6c. no eliminations yet

7. combined half cage r3c179 = h13(2) and h10(2) can't be [746] since this forces combined 39(7) to have {18} in r3c45 and 4 in r3c7 which is impossible (step 6a)
7a. -> no 7 in r3c1, no 4 in r3c7 and no 6 in r3c9

8. 30(5)r1c6 must have 9 for n2
8a. but {25689} blocked by 9(2)n2
8b. = {15789/24789/34689/35679}
8c. must have 4 or 5 in n2, but not both

9. hidden killer pair 4,5 in n2 -> 21(6)r1c3 must have exactly one of 4,5 in n2 and one of 4,5 in n1
9a. -> killer pair 4,5 with 12(2)n1: both locked for n1
9b. -> r3c19 = [94] only permutation (h13(2))
9c. -> r3c7 = 6 (h10(2))
9d. -> 30(5)r1c6 = {3489/3579}[6](no 1,2)
9e. must have 3 -> r3c6 = 3
9f. -> r2c2 = 3 (hsingle n1)

10. 15(3)n1 = {168/267}
10a. -> r2c3 = 6

11. hsingle 5 in r3 -> r3c8 = 5

12. 8(3)n9 must have 1: locked for n9
12a. -> 8(2)n9 = {26} only: both locked for r9 and n9
12b. -> 8(3)n9 = {134} only
12c. -> r8c8 = 4, r78c9 = {13}: both locked for c9, 3 for n9

13. 3 & 4 in c7 only in 29(6)r4c6: 4 locked for 29(6)
13a. -> = 34{1579/1678/2569/2578}
13b. note: can't have both 1 & 2
13c. hidden killer pair 1,2 in c7 -> r12c7 has one of 1,2 and 29(6) has one of 1,2 for c7
13d. -> 1,2 locked for 29(6) (note: SudokuSolver can't do this type of clean-up step in killer pairs)

14. hidden singles 1,2 in c6 -> r89c6 = [21]
(note: from a v2 I was working on, this placement of the 2 is the key to cracking this puzzle)

The final key step
15. r7c7 sees all of n8 apart from r8c5 so they must be the same = {5789}
15a. "45" on r89: 4 outies r67c1 + r7c79 - 4 = 1 remaining innie r8c5.
15a. since r7c7 = r8c5 -> r67c1 + r7c9 = 4 = [121] only permutation
15b. r8c9 = 3

16. r67c1 = [12] = 3 -> r89c1 = 14 = [68] only

17. "45" on n7: 1 remaining outie r7c4 = 9
17a. -> r7c23 = 7 = [43] only

18. 5 & 9 in n9 only in r789c7: both locked for c7 and 5 for 37(7)

19. 12(2)n1 = {57} only: both locked for c1 and n3

pretty easy from here. Sorry, a bit lazy on my sabbatical WT.