1. Digitised cages have one cell value equal to the last digit of the cage total -> the other two cell values must total 10 = {19/28/37/46}, digitised cages can only contain 5 if the cage total is 15
1a. Digitised cage totals
11 = [137/146]
12 = [246]
13 = [139]
14 = [149/248]
15 = [159/258/357]
16 = [169/268]
17 = [179]
18 = [468]
19 = [379/469]
2. Digitised cages from R1C1 to R6C6 each have cage totals greater than the one to the left, the cage totals can only range between 11 and 19, the cage at R7C7 is greater than the one at R6C6 but isn’t digitised so can have a total greater than 19 -> cage at R1C1 = 11/12/13/14 … cage at R6C6 = 16/17/18/19
2a. No 5 in cage at R1C1
2b. No 5 in cage at R6C6
3. The cages at R7C8 and R7C9 have the same total -> cage at R7C7 must have an odd total = 17/19/21 (cannot total 23 which contains consecutive values) -> cages at R7C8 and R7C9 must total 12/13/14
3a. Non-consecutive cages totalling 12/13/14 must contain 1 or 2, which must be in R7C89 = {12}, locked for R7 and N9
3b. Cage at R7C8 is digitised and totals 12/13/14 -> R8C8 = {34}
3c. Non-digitised cage at R7C9 totals 12/13/14 -> R8C9 = {345}
3d. Non-digitised cage at R7C7 -> R7C7 = {345}
3e. Naked triple {345} in R7C7 + R8C89, locked for N9
3f. Non-consecutive fully ordered cages cannot contain both of 8,9 -> 8,9 in R9C789, locked for R9 and N9
4. From step 2, the maximum total for cage at R6C6 is 19 -> max total for cage at R3C3 is 16
4a. The first cells of the cages at R1C1, R2C2 and R3C3 are all in N1 and must be {123} (because 18 and 19 are the only digitised cages which can start with 4)
4b. The only digitised cage totalling 11…16 and starting with 3 is 15(3) = [357]
4c. Digitised cage at R3C3 cannot total 15 with 3 in R3C3, because cages at R1C1 and R2C2 would both require 4 in N1 -> cage at R2C2 must total 15 = [357] -> R2C2 = 3, R3C2 = 5, R4C2 = 7
4d. Digitised cage at R1C1 must total 11/12/14 -> R1C1 = {12}, R2C1 = 4, R3C1 = {6789}
4e. Digitised cage at R3C3 must total 16 -> R3C3 = {12}, R4C3 = 6, R5C3 = {89}
4f. Naked pair {12} in R1C1 + R3C3, locked for N1
5. Digitised cage at R3C3 totals 16 -> digitised cages at R4C4, R5C5 and R6C6 must total 17,18,19
5a. Digitised cage at R4C4 totals 17 = [179]
5b. Digitised cage at R5C5 totals 18 = [468]
5c. Digitised cage at R6C6 totals 19 = [379]
5d. Naked triple {258} is digitised cage at R4C5 -> R4C56 = [25], R5C6 = 8
5e. 4 in R4 only in R4C789, locked for N6
6. Digitised cage at R3C3 totals 16, R5C3 = 9 -> R3C3 = 1
6a. Naked pair {78} in R12C3, locked for C3 and N3
6b. Digitised cage at R1C1 totals less than 15, R1C1 = 2, R2C1 = 4 -> cage = [246] (only remaining permutation) -> R3C1 = 6, R1C2 = 9
7. 6 in R5 only in non-digitised cage at R5C7 = [136] (only remaining permutation) -> R5C12 = [52]
8. Digitised cage at R6C6 totals 19, non-digitised cage at R7C7 must be odd and greater so must total 21 = [579], R9C89 = [68]
8a. Non-digitised cage at R7C7 totals 21 -> equal cages at R7C8 and R7C9 must total 12
8b. Digitised cage at R7C8 = [246], R78C9 = [13]
9. Non-digitised cage at R6C7 = [257] (hidden triple in R6)
10. Non-digitised cage at R8C5 must contain 1 (R8C5 cannot be 5 because no 8,9 in R9C45) -> R8C5 = 1
11. 7,9 in C5 only in digitised cage at R1C5 = [379], R9C5 = 5 -> R9C4 = {23}, R12C3 = [78]
12. Digitised cage at R1C8 = [169] (only remaining permutation) -> R1C8 = 1, R2C78 = [69]
and the rest is naked singles.
Login
Register
FAQ
Search
