Decidoku, candidates 0-9. Non-consecutive (NC) including 9 and 0 NC.
I’ve numbered the rows and columns 0 to 9, with the decidoku windows (decets) DR0C0, etc. Hidden decets, as in wellbeback’s diagram will be HR0C4, etc.
Decidoku has hidden cages, as in previous decidoku puzzle.
Prelims
a) R0C12 = {09/18/27/36} (cannot be {45} because of NC)
b) R0C78 = {79}
c) R12C0 = {79}
d) R12C9 = {09/18/27/36} (cannot be {45} because of NC)
e) R78C0 = {09/18/27/36} (cannot be {45} because of NC)
f) R78C9 = {79}
g) R9C12 = {79}
h) R9C78 = {09/18/27/36} (cannot be {45} because of NC)
i) 6(3) cage at R4C0 = {015/024} (cannot be {123} because of NC)
j) 6(3) cage at R5C7 = {015/024} (cannot be {123} because of NC)
k) 9(3) cage at R0C4 cannot contain 9
l) 9(3) cage at R7C5 cannot contain 9
Steps resulting from Prelims
1a. Naked pair {79} in R0C78, locked for R0 and D0C6, clean-up: no 0,2 in R0C12 + R12C9
1b. Naked pair {79} in R12C0, locked for C0 and D0C0, clean-up: no 0,2 in R78C0
1c. Naked pair {79} in R78C9, locked for C9 and D6C9, clean-up: no 0,2 in R9C78
1d. Naked pair {79} in R9C12, locked for R9 and D0C6
1e. Naked pair {79} in R0C78 -> no 8 in R0C69 + R1C78 (NC)
1f. Naked pair {79} in R12C0 -> no 8 in R03C0 + R12C1 (NC)
1g. Naked pair {79} in R78C9 -> no 8 in R69C9 + R78C8 (NC)
1h. Naked pair {79} in R9C12 -> no 8 in R8C12 + R9C03 (NC)
1i. 6(3) cage at R4C0 = {015/024}, 0 locked for R4 and DR1C3
1j. 6(3) cage at R5C7 = {015/024}, 0 locked for R5 and DR5C6
[NC effects on the placements in the 6(3) cages have been omitted because of the next step.]
[As HATMAN said the hidden windows helped a lot. I agree, without them this step wouldn’t get me into the puzzle quickly.]
2. Hidden decet DR0C5 contains R0123C5 + R4C012345
This hidden window contains 10(3) cage at R0C5, 6(3) cage at R4C0 and R3C5 + R4C345
2a. 45 rule on hidden window 4 remaining cells R3C5 + R4C345 = 29 = {5789} -> 6(3) cage = {024}, locked for R4 and DR1C3, 10(3) cage = {136}, locked for C5 and DR0C5
2b. Similarly for one of the hidden decets at R5C4 containing R5C456789 + R6789C4 which contains 6(3) cage at R5C7, 10(3) cage at R7C4 and R5C456 + R6C4
45 rule on hidden window 4 remaining cells R5C456 + R6C4 = 29 = {5789} -> 6(3) cage = {024}, locked for R5 and DR5C6, 10(4) cage = {136}, locked for C4 and DR6C1
2c. 9(3) cage at R0C4 = {027/045}, 0 locked for C4
2d. 4,5 of {045} must be in R02C4 because of NC -> no 4,5 in R1C4
2e. 9(3) cage at R7C5 = {027/045}, 0 locked for C5
2f. 4,5 of {045} must be in R79C5 because of NC -> no 4,5 in R8C5
2g. 8 in R0 only in R0C123, locked for DR0C0
2h. 8 in R9 only in R9C678, locked for DR6C9
2i. R3C5 + R4C345 = {5789} -> 5 must be in R4C45 and 8 in R3C5 + R4C3 because of NC, no 5 in R3C5 + R4C3, no 8 in R4C45, 5 locked for R4
2j. R5C456 + R6C4 = {5789} -> 5 must be in R5C45 and 8 in R5C6 + R6C4 because of NC, no 5 in R5C6 + R6C4, no 8 in R5C45, 5 locked for R5
2k. Grouped X-Wing for 5 in R4C45 and R5C45, no other 5 in C45
3. 15(3) cage at R4C7 = {168}, locked for R4 and DR4C5
3a. 15(3) cage at R5C0 = {168}, locked for R5 and DR0C4
3b. R4C6 = 3 (hidden single in R4) -> no 2,4 in R3C6 (NC)
3c. R5C3 = 3 (hidden single in R5) -> no 2,4 in R6C3 (NC)
3d. 9(3) cage at R0C4 = {027}, locked for C4
3e. 9(3) cage at R7C5 = {027}, locked for C5
3f. R3C4 = 4 (hidden single in C4) -> no 5 in R3C3 + R4C4 (NC)
3g. R6C4 = 8 (hidden single in C4) -> no 9 in R5C4 + R6C5, no 7,9 in R6C3 (NC)
3h. R4C45 = [95], R5C45 = [59]
3i. R4C3 = 7 -> no 6,8 in R3C3 (NC)
3j. R3C5 = 8 -> no 7,9 in R3C6 (NC)
3k. R5C6 = 7 -> no 6 in R6C6 (NC)
3m. R6C5 = 4 -> no 5 in R6C6 (NC)
3n. R0C4 = {02} -> no 1 in R0C35 (NC)
3o. R9C5 = {02} -> no 1 in R9C46 (NC)
3p. 7 in R3 only in R3C78, locked for DR05
3q. 7 in R6 only in R6C12, locked for DR6C1
[Looks like there’s nothing more to get from the hidden windows.]
4. R0C12 = {18} (cannot be {36} which clashes with R0C5), locked for R0 and DR0C0
4a. R9C78 = {18} (cannot be {36} which clashes with R9C4), locked for R9 and DR6C9
4b. 1,8 in C3 only in R123C3, locked for DR1C3
4c. 8 in R12C3 -> no 9 in R12C3 (NC)
4d. 1,8 in C6 only in R678C6, locked for DR5C6
4e. 8 in R78C6 -> no 9 in R78C6 (NC)
[Unless I’m missing something it looks like it’s now time to nibble away using interactions and NC.]
5. Consider placements for R0C5 = {36}
R0C5 = 3 => no 2 in R0C46 (NC)
or R0C5 = 6 => R12C5 = {13} => no 2 in R12C4 (NC) => R0C4 = 2
-> R0C45 = [03/26], no 2 in R0C6
5a. Similarly R9C45 = [03/26], no 2 in R9C3 (NC)
6. Taking steps 3p and 3q further
6a. 7 in R3 only in R3C78 -> R4C78 must contain 1 (cannot both be {68} because of NC) -> no 1 in R4C9
6b. 1 in C9 only in R123C9, locked for DR0C6
6c. Killer pair 6,8 in R12C9 and R4C9, locked for C9
6d. 7 in R6 only in R6C12 -> R5C12 must contain 1 (cannot both be {68} because of NC) -> no 1 in R5C0
6e. 1 in C0 only in R678C0, locked for DR6C0
6f. Killer pair 6,8 in R5C0 and R78C0, locked for C0
6g. 4 in R4 must be in R4C12 (R4C12 cannot be {02} which clashes with 1 in R5C12), no 4 in R4C0
6h. 4 in R5 must be in R5C78 (R5C78 cannot be {02} which clashes with 1 in R4C78), no 4 in R5C9
6i. 6 in R3 only in R3C12, locked for DR1C1
6j. 6 in R3 only in R3C12 -> no 5 in R3C12 (NC)
6k. 6 in R6 only in R6C78, locked for DR5C6
6l. 6 in R6 only in R6C78 -> no 5 in R6C78 (NC)
7. Consider placements for 8 in C3
R1C3 = 8 => no 7 in R1C4 (NC) => R2C4 = 7 (hidden single in C4) => R12C0 = [79]
or R2C3 = 8 => no 9 in R2C2 + R3C3 (NC)
-> no 9 in R2C2
7a. 9 in DR1C3 only in R3C123, locked for R3
7b. R2C2 = {35} -> no 4 in R1C2 + R2C1 (NC)
7c. Similarly no 9 in R7C7, 9 in DR5C6 only in R6C678, locked for R6
7d. R7C7 = {35} -> no 4 in R7C8 + R8C7 (NC)
8. Consider placements for R3C3 = {19}
R3C3 = 1 => R3C12 = {69} (hidden pair in DR1C1) => R2C2 = 3 (hidden single in DR1C1)
or R3C3 = 9 => R3C12 = {36} => R2C2 = 5 (hidden single in DR1C1) => no 6 in R3C2 => R3C12 = [63]
-> 3 in R23C2, locked for C2 and DR1C1
8a. Similarly 3 in R67C7, locked for C2 and DR5C6
[I wasn’t happy with this step, which cracks this puzzle. It only stops just short of being a contradiction, but I couldn’t see any other forcing chain to achieve this result.
There is a long forcing chain based on the placement of 1 in R4C78 but it seems to depend on symmetry and, while the puzzle clearly has symmetry, it’s not easy to prove.]
9. Consider combinations for R12C9 = {18/36}
R12C9 = {18}, locked for DR0C6
or R12C9 = {36} => R4C9 = 8, R3C9 = 1 (hidden single in C9) => no 0,2 in R3C8 (NC), R3C8 = {57} => no 6 in R4C8 => R4C8 = 1 => R9C8 = 8
-> no 8 in R2C8
9a. 8 in DR0C6 only in R12C9 = {18}, locked for C9
9b. R3C3 = 1 (hidden single in R3), naked pair {58} in R12C3, locked for C3 and DR1C3, R2C2 = 3
9c. R2C2 = 3 -> no 2 in R1C2 + R2C1 (NC)
9d. R6C3 = 0 -> no 9 in R7C3 (NC wrap)
9e. R8C3 = 9 (hidden single in C3) -> R78C9 = [97]
9f. R7C5 = 7 (hidden single in C5) -> no 6 in R7C4, no 8 in R7C6 (both NC)
9g. R8C6 = 8 (hidden single in C6)
9h. Similarly considering combinations for R78C0 = {36}/[81] and using NC where necessary -> no 8 in R7C1 -> R78C1 = [81], R6C6 = 1, R7C67 = [53], R3C6 = 0, R1C6 = 9, R12C0 = [79], R2C4 = 7, R12C3 = [85], R12C9 = [18], also no 6 in R2C6, no 2 in R7C8 + R8C7 (NC)
10. R2C5 = 1 -> no 2 in R2C6 (NC)
10a. R2C67 = [42]
10b. R0C6 = 6 -> no 7 in R0C7 (NC)
10c. R0C5 = 3 -> no 2 in R0C4 (NC)
10d. R0C78 = [97], R3C8 = 5
10e. R3C7 = 7 -> no 6,8 in R4C7 (NC)
10f. R4C7 = 1 -> no 0 in R5C7 (NC)
10g. R9C78 = [81] -> no 0,2 in R8C8 + R9C9 (NC)
10h. Similarly R7C234 = [241], R9C34 = [63], R9C12 = [79], R6C12 = [57], R5C2 = 1, R0C12 = [18], no 0,2 in R0C0 + R1C1, no 0 in R4C2 (NC)
and the rest is naked singles, without using non-consecutive or hidden groups.