It's a killer sudoku where the cages aren't given. There isn't any constraint on where the numbers are located in the cells, like with the Sunday telegraph puzzles though. the Killer itself isn't hard, especially considering the normal puzzles here, but the searching for where the cages are, should still make it a bit challenging.
In this Killer Sudoku the cages are hidden. Each cell with a clue number is part of a separate cage. Each cage consists of orthogonally connected cells. The number indicates the sum of the digits in those cells. No digit can repeat within a cage. Cages consist of at least 2 cells. No two cages can overlap.
Since cages are being built up, I’ve omitted clean-ups until they are needed.
1. 5 at R8C8 must be part of a 5(2) cage with the other cell at R8C9 or R9C8, since the rules say that cages cannot be diagonally connected
1a. R9C9 must be part of a 4 (or more) cell cage, either totalling 14 including R7C8 or totalling 16 including R8C7
1b. Cannot have 14(4) cage + 5(2) cage in the same nonet because minimum for 6 cells in the same nonet is 21 -> 16(4) cage at R8C7 + R9C789 and 5(2) cage at R8C89
1c. 5(2) cage = {14/23}, 16(4) cage = {1456/2356}
2. R7C789 = {789} (hidden triple for N9), locked for R7
2a. R7C7 = 7 (because part of 8(2) cage), 1 in either R6C7 or R7C6
3. R7C89 = {89} cannot both be in a 14(2 or more) cage, only other cage total which can connect to R7C9 is 18 at R5C7 -> 18(5) cage at R5C7 = {12348} -> R7C9 = 8, R5C789 + R6C9 (because R6C8 required for 14(2or more) cage at R7C8) = {1234}, locked for N6
4. R7C8 = 9, only way to form cage totalling 14 is R67C8 = 14(2) cage -> R6C8 = 5
5. No 1 in R6C7 -> R7C6 = 1 (step 2a), 8(2) cage R7C67
6. Cage totals 10 at R5C6 and R6C5 -> R6C7 must be part of a 10(3) cage = {136} but no 1 in R56C6 -> 10(3) cage R6C567 = [136]
6a. Naked triple {789} in R4C789, locked for R4
6b. 1,3 in N6 only in R5C789, locked for R5
6c. 6 in N9 only in R9C789, locked for R9
7. 7(2 or more) cage at R3C8 must include R3C9 -> 27 total cage at R2C8 cannot reach any of R4C789
7a. R4C789 = {789} = 24 cannot all be in 21 total cage at R3C7 so there must be 35(5) cage = {56789} in R4C56789, 5,6 locked for R4 and N5
8. R5C6 can only be part of a 10(2) cage R5C56 = {28}, locked for R5 and N5 -> R4C4 = 4
8a. Naked triple {134} in R5C789, locked for R5 and N6 -> R6C9 = 2
8b. Naked pair {79} in R56C4, locked for C4
9. 7(2 or more) cage at R3C8 cannot be {124} which clashes with R58C9, ALS block, so must be 7(2) cage in R3C89 = {16/34}/[25], no 7,8,9
10. R56C4 = {79} cannot be in the same cage because there is no orthogonally adjacent 1 to form a 17(3) cage -> R6C4 must be in the 19 cage at R5C3; this must be a 19(3) cage because 4+5+6+7 > 19 -> 19(3) cage in R5C3 + R6C34 = {469} (only remaining combination, cannot be {568} because R6C4 only contains 7,9, cannot be {478} because 4,8 only in R6C3) -> R6C3 = 4, R5C3 = 6, R6C4 = 9, R5C4 = 7
11. 8(2) cage in R7C34 = [26/35/53] (cannot be 8(3) cage because no available 1), no 2 in R7C4
12. 45 rule on N78 (using R7C6 = 1) outies in N4 total 20 -> R6C12 = {78} and 5 in one of R5C12 must be part of 28(5 of more cage) at R7C2 with either R7C12 = 8 = {26} or R7C12 + R8C1 = 8 = {34}1 (note that 5 cannot be in R7C12 because it’s required for one of R5C12)
13. 17(3 or more) cage at R5C4 contains R45C4 = [47] = 11 so needs 6 more. These cannot be in R4C123 (which would block access to the 9 in one of R5C12) -> 17(3 or more cage at R5C4) must contains R3C4 and possibly further cells
14. 5(2) cage at R3C3 must be in R34C3 = {23}, locked for C3 -> R7C3 = 5, R7C4 = 3 (step 11)
14a. R7C12 = {26} (step 12, because no 3 available for R7C12), locked for R7 and N7, R7C5 = 4
15. 9 in one of R5C12 must be part of 15(3 or more) cage at R3C2 (cannot be part of 20(5 or more) cage at R2C2 because cannot then form a 15 cage at R3C2; the other one of R5C12 is part of the 28(5) cage at R7C2 and max R34C2 = 12) -> R5C2 = 9, either R34C2 = [42/51] or R3C2 + R4C12 = {123} with 1 in R4C12, no 1,6,7,8 in R3C2
15a. R5C1 = 5 is part of 28(5) cage at R7C2
16. R23C6 must be part of 21(3 or more) cage at R3C7
17. R1C7 must be part of 8(2 or more cage at R2C7) -> 27(4) cage in R12C89 = {3789/4689/5679}, no 1,2
17a. 27(4) cage at R1C8 + 7(2) cage at R3C8 = 34 -> R123C7 = 11 -> R12C7 cannot be {35} because R123C7 cannot be {35}3 -> must be 8(3) cage in R1C6 + R12C7 -> R1C6 = {245}, R12C7 = {12/13/15}, 1 locked for C7 and N3
18. R3C89 (step 9) = [25/34/43], no 6
18a. Killer pair 2,3 in R3C3 and R3C89, locked for R3
19. 45 rule on N3 1 innie R3C7 = 1 outie R1C6 + 3 -> R1C6 = {25}, R3C7 = {58}
19a. 8(3) cage in R1C6 + R12C7 = {125} (only remaining combination), no 3 in R12C7
19b. R3C89 (step 18) = {34} (only remaining combination, cannot be [25] which clashes with 8(3) cage), locked for R3 and N3 -> R3C2 = 5, R34C3 = [23], R3C7 = 8, R4C7 = 9, R4C9 = 7, R4C8 = 8
19c. R12C7 = {12} (hidden pair in N3), locked for C7 and 8(3) cage -> R1C6 = 5, R4C56 = [56]
20. Naked pair {67} in R12C8, locked for C8 and N3 -> R12C9 = [95]
20a. R9C9 = 6 (hidden single in C9)
21. R2C5 must be in 11(2 or more) cage at R3C5 -> must be 21(3) cage in R23C6 + R3C7
21a. R3C7 = 8 -> R23C6 = [49]
22. R3C2 = 5, R5C2 = 9 -> 15(3) cage in R345C2 = [519], R4C1 = 2, R7C12 = [62]
23. 17(3 or more) cage at R5C4 contains R45C4 = [47] = 11 so needs 6 more (step 13) -> R3C4 = 6 (cannot be R123C4 or R23C4 cannot total 6) -> 17(3) cage in R345C4
23a. R3C5 = 7, R3C1 = 1
24. R3C5 = 7 -> cannot be 11(2) cage in R23C5 -> must be 11(3) cage in R2C45 + R3C5 = {137} -> R2C45 = [13], R12C7 = [12]
24a. Naked pair {28} in R1C45, locked for R1 -> R1C3 = 7, R12C8 = [67]
24b. R2C2 = 6 (hidden single in N1)
25. R1C45 = {28} must be in 30(5 or more) cage at R2C3 -> R2C3 = 9, R2C1 = 8, R6C12 = [78]
25a. R1C45 = {28} = 10, R2C3 = 9, R1C3 must be in 30 cage -> R1C2 = 4 makes cage total of 30, 30(5) cage in R1C2345 + R2C3
25b. R1C1 = 3, 20(5) cage in R1C1234 + R2C2
26. R89C5 = [69] (hidden pair in C5), R89C1 = [94]
27. R89C1 = [94] = 13, R89C2 = {37} = 10 must be in 31(5 or more cage at R8C2), requires further 8 -> R9C3 = 8
28. R8C3 = 1 -> R8C89 = [23], R3C89 = [34], R5C789 = [341], R89C2 = [73], R9C78 = [51], R8C7 = 4, R9C46 = [27], R1C45 = [82], R5C56 = [82], R8C46 = [58]
28a. Remaining cages 18(3) cage in R7C5 + R8C56, 24(5) cage in R8C34 + R9C456
Is it possible to work out the cage pattern first, then solve the puzzle? Probably not.
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