SudokuSolver Forum

For convenience I’ve numbered the rows and columns 0 to 9, rather than 1 to 10; hope this won’t be too confusing. Each row and column contains the numbers 0 to 9. There are also decets (tenets) which also each contain the numbers 0 to 9, identified as DR0C0, etc.

The grid is fully non-consecutive, horizontally and vertically, including 0 and 9.

Cages are digitised, two cells consecutive and the other non-consecutive. Black cages are ordered, red cages reverse-ordered.

1. Each cage must contain the last digit of the cage sum, with the other two digits totalling 10, so cage sums must be in the range 10 to 19.

10 = [019]
11 = [128]
12 = [129/237]
13 = [238/346]
14 = [347]
No possible permutations for 15, since [456] would have three consecutive values
16 = [367]
17 = [278/467]
18 = [189/378]
19 = [289]

2a. Second and third cells of reverse-ordered cage at R1C6 cage overlap with first and second cells of reverse-ordered cage at R2C7 can only be [764] and [643], [873] and [732] or [982] and [821]
2b. Third cell of reverse-ordered cage at R1C6 overlaps with third cell of reverse-ordered cage at R2C5 and all the cells of these cages are in the same decet so can only be [873] and [643] or [982] and [732] -> R1C6 = {89}, R2C5 = {67}, R2C6 = {34}, R2C7 = {78}, R3C7 = {23}, 3,7,8 all locked for NR0C5, R4C8 = {12}
2c. R2C5 = {67} -> ordered cage at R1C4 = [467], [278] or [378] -> R1C4 = {234}, R3C4 = {78}

3a. Reverse-ordered cage at R5C6 can only be [643/732/743/821/832/910/921] because R5C67 cannot have consecutive numbers -> R5C6 = {6789}, R5C7 = {1234}, R6C6 = {0123)
3b. Ordered cage at R6C6 can only be [189/278/289/367/378] because R6C67 cannot be consecutive numbers and no 4 in R6C6 -> no 0 in R6C6 -> no 1 in R5C7, R6C7 = {678}, R7C8 = {789}

4a. R3C7 + R4C8 = [21/32] -> ordered cage at R3C8 can only start with 3 or 4 in R3C8 because R3C78 and R34C8 cannot be consecutive -> R3C8 = {34}
4b. Naked pair 3,4 in R2C6 + R3C8, locked for DR0C5 -> R3C7 = 2, R4C8 = 1, R2C7 = 8, R1C6 = 9, R2C56 = [73], R3C8 = 4
4c. Ordered cage at R1C4 = [278] or [378] -> R1C4 = {23}, R3C4 = 8
4d. R3C8 = 4 -> ordered cage at R3C8 = [467] -> R4C79 = [67]
4e. R6C7 = 7 -> ordered cage at R6C6 = [278] -> R6C6 = 2, R7C8 = 8
4f. R6C6 = 2 -> reverse-ordered cage at R5C6 = [832] -> R5C67 = [83]
4g. R5C6 = 8 -> ordered cage at R4C5 = [278] or [378] -> R4C5 = {23}, R5C4 = 7

5. Ordered cage at R0C1 must end with 6 or 9 = [019], [129], [346] or [289] -> R0C1 = {0123}, R1C2 = {1248}, R2C1 = {69}

Now the rest of this puzzle needs the use of the 14 hidden windows or Law of Leftovers (LoL) and the non-consecutive property.

6a. LoL for R01234, 5 innies R4C56789 must exactly equal 5 outies R5C01234 (which is equivalent to there being hidden windows R45C01234 and R45C56789), no 3 in R5C01234 -> no 3 in R4C56789 -> R4C5 = 2
6b. R4C5 = 2 -> R5C0123 must contain 2 -> R1C4 = 3
6c. Remaining candidates for hidden window R45C56789 are 0,4,5,9, 9 can only be in R5C89 (because of 8 in R5C6) -> R0C4 = 9 (only available place for 9 in DR0C4, cannot be next to 8 in R3C4)

7a. R4C7 = 6 -> 6 must be in R013C5 but not next to 7 in R2C5 -> R0C5 = 6
7b. Similarly R4C8 = 1 must be in R13C5 but not next to 2 in R4C5 -> R1C5 = 1
7c. Remaining candidates in DR0C5 are 0,5, whichever is in R3C5 must be in R4C6 and cannot be 5 because of 6 in R4C7 -> R3C5 = 0, R3C6 = 5, R4C6 = 0
7d. Ordered cage at R0C1 (step 5) = [129], [346] or [289] -> no 0 in R0C1

8a. 7 in C6 cannot be in R0C6, because of 6 in R0C5, DR5C6 already contains 7 -> R9C6 = 7
8b. 6 in C6 cannot be in R0C6 or R8C6, because of 6 in R0C5 and 7 in R9C6 -> R7C6 = 6

9. LoL for R01234, 5 innies R4C56789 must exactly equal 5 outies R5C01234, R4C56789 = [20617], R5C4 = 7 -> R5C0123 must contain 0,1,2,6 -> remaining candidates in DR0C4 are 4,5 -> R4C4 = 4 (because 3 in R1C4), R2C4 = 5

10a. 9 in DR0C6 must be in R23C9 (cannot be in R2C8 because of 8 in R2C7)
10b. 9 in R5 cannot be in R5C5 because of 8 in R5C6 -> R5C8 = 9

11a. 3 in NR0C6 must be in R0C89 (cannot be in R3C9 because of 4 in R3C8) -> no 3 in R0C1
11b. Ordered cage at R0C1 (step 7f) = [129] or [289] -> R2C1 = 9, no 4 in R1C2
11c. R3C9 = 9 (from step 10a)

[Just remembered the complete non-consecutive rule]
12a. 0 cannot be in R5C9 + R6C8, because of 9 in R5C8 -> R7C7 = 0 (only remaining place for 0 in DR5C6)
12b. Remaining candidates in DR5C6 are 1,4,5 -> R6C8 = 5
12c. 1 in DR0C4 must be in R5C0123 -> R5C9 = 4, R8C6 = 1, R0C6 = 4 (remaining candidate for C6), R5C5 = 5 (remaining place for 5 in 5)

13a. Remaining candidates for C7 are 1,4,5,9 -> R1C7 = 5, R0C7 = 1
13b. R0C1 = 2 -> ordered cage at R0C1 (step 11b) = [289] -> R1C2 = 8
13c. R0C9 = 8 (only remaining place for 8 in R0)
13d. R1C8 = 7 (only remaining place in DR0C6 because 8 in R0C9)
13e. R2C9 = 6 (only remaining place in DR0C6 because 7 in R1C8)
13f. R0C8 = 3 (only remaining place in DR0C6)
13g. 2 in R1 cannot be in R1C3 because 3 in R1C4 -> R1C9 = 2 -> R2C8 = 0

14a. R23C0 = [13] (only remaining placed in DR0C0)
14b. 4 in R2 cannot be in R2C3 because 5 in R2C4 -> R2C2 = 4, R2C3 = 2
14c. 4,6 in DR0C0 must be in R1C01 -> R1C3 = 0

15a. Remaining candidates for R5 are 0,1,2,6, 6 cannot be in R5C3 because 7 in R5C4 -> R5C3 = 1
15b. R5C2 cannot be 0 or 2, because 1 in R5C3 -> R5C2 = 6 -> R5C01 = [20]

16a. Remaining candidates for C4 are 0,1,2,6, 6 cannot be in R6C4 because of 7 in R5C4 -> R67C4 only contain 0,1,2 -> R67C4 = [02] because of non-consecutive rule -> R89C4 = [61]
16b. R89C8 = [26] (remaining candidates in C8)

17a. Remaining candidates for C9 are 0,1,3,5, 3 cannot be in R6C9 because of 4 in R5C9 -> R6C9 = 1
17b. 5 in DR6C1 cannot be in R8C3 because of 6 in R8C4 so must be in R7C23
17c. 5 in C9 cannot be in R9C9 because of 6 in R9C8 -> R8C9 = 5
17d. 0 in C9 cannot be in R7C9 because of 1 in R6C9 -> R9C9 = 0, R7C9 = 3

18a. R7C1 = 1 (only remaining place in DR6C1)
18b. R3C2 = 1 (only remaining place in R3)

19a. R6C0 = 6 (only remaining place in R6)
19b. R1C01 = [46] (only remaining places in R1)
19c. R3C13 = [76] (only remaining places in R3)

20a. 7 in R7 cannot be in R7C0, because of 6 in R6C0 -> 7 must be in R7C23, 5 must be in R7C23 (step 17b)
20b. R7C05 = [94] (on remaining places in R7)
20c. 3 in C5 cannot be in R68C5 because of 4 in R7C5 -> R9C5 = 3
20d. 5,7 in R7C23 -> remaining candidates in DR6C1 must be 3,4,8,9 in R6C123 + R8C3, remaining candidates in C5 are 8,9 in R68C5 -> R8C35 must contain 9 -> R89C7 = [49]

21a. Remaining candidates in R9 are 2,4,5,8 -> 0,3,7 must be in R8C012 -> R8C1 = 3
21b. R9C1 cannot be 2,4 because of 3 in R8C1 -> R9C3 = 4, R9C2 = 2
21c. R6C1 = 4 (only remaining place in C1)
21d. 3 in R6 cannot be in R6C2 because of 4 in R6C1 -> R6C3 = 3
21e. 5,7 in R7C23 -> remaining candidates in DR6C1 are 8,9 -> R6C2 = 9, R8C3 = 8
21f. 7 in R7 cannot be in R7C3 because of 8 in R8C3 -> R7C23 = [75]
21g. R68C5 = [89] (remaining places in C5)

22a. Remaining candidates in C1 are 5,8, 8 cannot be in R4C1 because of 7 in R3C1 -> R9C1 = 8, R4C1 = 5
22b. R9C0 = 5 (remaining candidate in R9)

23a. R8C02 = [70] (remaining places in R8)
23b. Remaining candidates in R0 are 0,5,7 -> R0C0 = 0
23c. 7 cannot be in R0C2 because of 8 in R1C2 -> R0C23 = [57]
23d. Remaining candidates in R4 are 3,8,9 -> R4C0 = 8, R4C23 = [39]