SudokuSolver Forum

Prelims

a) R12C1 = {59/68}
b) R1C89 = {59/68}
c) R2C23 = {15/24}
d) R23C8 = {15/24}
e) R34C3 = {49/58/67}, no 1,2,3
f) R3C78 = {49/58/67}, no 1,2,3
g) R6C45 = {89}
h) R67C7 = {59/68}
i) R78C2 = {15/24}
j) R7C34 = {49/58/67}, no 1,2,3
k) R8C78 = {15/24}
l) R89C9 = {59/68}
m) R9C12 = {59/68}

1a. 14(2) cages at R1C1 and R89C9 must have the same combination {59} or {68} while 14(2) cages at R1C8 and R9C1 must have the other combination, because of the way these cages “see” each other. As a result they form a naked quad {5689} in R19C19, no 5,6,8,9 in R5C5 using the diagonals
1b. Similarly with the 6(2) cages which form pairs, no 1,2,4,5 in R5C5
-> R5C5 = 3
[Alternatively 3 on the diagonals only in N5 -> R5C5 = 3, only way to place 3 on both diagonals.]

[The “ring” of three 13(2) cages and one 14(2) cage on the 3,7 rows and columns is also important, but will need a bit more thought, so I’ll do the obvious routine steps next.]

2. Naked pair {89} in R6C45, locked for R6 and N5, clean-up: no 5,6 in R7C7
2a. Killer pair 8,9 in R7C7 and R89C9, locked for N9

3. R1C9 + R9C1 must contain one of 8,9 (because of step 1a) -> killer pair 8,9 in R1C9 + R9C1 and R6C4, locked for D/, clean-up: no 4,5 in R3C6, no 4,5 in R7C4
3a. R1C1 + R9C9 must contain one of 8,9 (because of step 1a) -> killer pair 8,9 in R1C1 + R9C9 and R7C7, locked for D\, clean-up: no 4,5 in R4C3

4. If the 14(2) cages are {59}, then the 6(2) cages in those nonets must be {24} and the other two 6(2) cages must be {15} -> 14(2) and 6(2) cages in the four corner nonets must be {59){24} or {68}{15}, 5 locked in 14(2) and 6(2) cages for N1, N3, N7 and N9, clean-up: no 8 in R3C6, no 8 in R4C3, no 8 in R7C4
4a. Similarly there must be {59){24} or {68}{15} in R1C1 + R2C2 + R8C8 + R9C9, 5 locked for D\
4b. And {59){24} or {68}{15} in R1C9 + R2C8 + R8C2 + R9C1, 5 locked for D/

5. R1C9 + R2C8 + R3C7 + R7C3 + R8C2 + R9C1 = {59){24}{67} or {68}{15}{47}, 7 in R3C7 + R7C3, locked for D/
5a. 7 in R3C7 + R7C3, CPE no 7 in R3C3, clean-up: no 6 in R4C3
5b. Killer pair 4,6 in R3C3 and R3C67, locked for R3, clean-up: no 2 in R2C8

[There are several more 4,6 killer pairs, but the next step is much more powerful, and this puzzle is cracked …]

6. 4 in R3 only in R3C37, CPE using D/ no 4 in R7C7, clean-up: no 9 in R7C4
6a. Naked pair {67} in R7C34, locked for R7

7. R34C3 = {49} (cannot be {67} which clashes with R7C3) -> R3C3 = 4, placed for D\, R4C3 = 9, clean-up: no 2 in R2C23, no 9 in R3C6, no 2 in R8C7
7a. Naked pair {15} in R2C23, locked for R2 and N1 -> R2C8 = 4, placed for D/, R3C8 = 2, clean-up: no 9 in R12C1, no 2 in R7C2, no 4 in R8C7
7b. Naked pair {15} in R8C78, locked for R8 and N9 -> R7C8 = 3, R8C2 = 2, placed for D/, R7C2 = 4, R7C9 = 2, clean-up: no 9 in R89C9

8. Naked pair {67} in R3C7 + R7C3, locked for D/ -> R4C6 = 1

9. Naked pair {68} in R12C1, locked for C1 and N1, clean-up: no 6,8 in R9C2

10. Naked pair {68} in R89C9, locked for C9 and N9 -> R7C7 = 9, R6C7 = 5, R8C7 = 1, R8C8 = 5, placed for D\, R2C23 = [15], R9C78 = [47], R7C1 = 1

11. R1C9 = 5 (hidden single in C9), placed for D/, R1C8 = 9, R9C1 = 9, placed for D/, R9C2 = 5, R6C45 = [89]

12. Naked pair {67} in R3C67, locked for R3 -> R3C12 = [39], R3C9 = 1, R3C45 = [58], R7C56 = [58], R1C23 = [72]
12a. R8C1 = 7, R7C3 = 6, placed for D/, R7C4 = 7, R3C67 = [67], R2C9 = 3

13. R6C6 = 7, then R4C4 = 2 (hidden singles on D\)
13a. R2C4 = 9, R2C6 = 2, R9C6 = 3, R9C3 = 8, R9C9 = 6, placed for D\, R1C1 = 8

and the rest is naked singles, without using the diagonals.

I'll rate my walkthrough for Human Solvable 15X at Easy 1.5. Several either/or steps but they aren't really chains.