SudokuSolver Forum

1. In D\(r1c1): (r1c1, r4c4, r6c6, r9c9) = {6789}
But 6 cannot be in r4c4 or r6c6. Must be in r1c1 or r9c9.
-> either 15/2@r1c1 = [69] or 15/2@r8c9 = [96]
-> either way - 15/2@r1c8 and 15/2@r9c1 both = {78}
-> Both 15/2@r1c1 and 15/2@r8c9 = {69}
-> (r4c4,r6c6) = {78} and r6c4 = 9
-> 13/2@r7c3 = [49]
-> 15/2@r8c9 = [69]
-> 15/2@r1c1 = [69]
-> HS 9 in n3 -> r1c7 = 9
Also HS 9 in n8 r7c5 = 9
-> HS 9 in n2 -> r3c6 = 9

2. 6 in n3 must be in r2c78
-> HS 6 in n2 r3c5 = 6

3. Neither 7 or 8 can go in r2c3 since that would leave no place for that number in n2
-> (78) in c3 in n4 in r456c3
-> r456c3 = [7{18}] or [8{27}]
-> HS 6 in c3 -> r9c3 = 6
-> 6 in n8 in r7c4 or r7c6
-> HS 6 in n5 -> r4c6 = 6
-> HS 6 in n8 -> r7c4 = 6
-> HS 6 in n3 -> r2c7 = 6.

4. Following on...
HP (78) in n1 -> r3c12 = {78}
-> (78) in n2 in r2c456
Also r123c3 in n1 from (135) or (235). 5/2@r1c2 means it cannot be the latter.
-> r123c3 = {135}
-> 9/2@r5c3 = {27}
-> r4c3 = 8
-> r3c12 = [87]
-> r9c12 = [78]
-> r1c89 = [78]
-> r56c3 = [72]
Also r4c4 = 7, r6c6 = 8

5. Whichever two numbers go in 5/2@r3c7 - at least one of them must go in r1 in n2 since (78) already in r2 in n2.
-> Whichever of [14] or {23} goes in 5/2@r3c7 - the other pair must go in the 5/2@r1c2
-> Whichever of [14] or {23} goes in 5/2@r3c7 - they must also go in r2c23 in n1
-> NS 5 in n1 -> r3c3 = 5
Also r12c2 = {42} and r12c3 = {13}
-> HS 2 in n7 -> r7c1 = 2

6. (78) in n8 can only go in r8c456
-> HP (78) in n9 -> r7c89 = [87]
-> HP (78) in n6 -> r56c7 = [87]

7.(There must be a simpler way to do this next step)
Setup...
In D\(r1c2) there is no place for a 9 -> must include a 5 in r4c5 or r5c6
In D/(r1c8) there is no place for a 8 -> must include a 4 in r5c4 or r4c5
In D/(r2c9) there is no place for a 7 -> Unknown cells are {12345}.
-> Whatever goes in r3c7 must be in r5c6 or r6c5.
2 in n5 can only go in r5c56
Whatever goes in r3c4 - the only place for it in n1 is r1c3 - must be from (13)

Given the above...
->(r2c1,r4c5,r5c6) either [245] or [452]

If the former... i.e. if 5/2@r1c2 = [23]
-> 5/2@r3c7=[14]
also -> r5c5 = 2 and r3c4 = 3
-> r6c5 = 3
which leaves no place for 1 in D/(r2c9)

-> 5/2@r1c2 = [41]
-> r2c23 = [23]
-> HS 4 in D\(r1c1) -> r8c8 = 4.
Also r3c4 = 1, r5c4 = 4, r4c5 = 5, r5c6 = 2

After this it is more or less just filling in the blanks

Prelims

a) R12C1 = {69/78}
b) R1C23 = {14/23}
c) R1C89 = {69/78}
d) R3C78 = {14/23}
e) 24(3) disjoint cage at R4C4 = {789}
f) R56C3 = {18/27/36/45}, no 9
g) R78C3 = {49/58/67}, no 1,2,3
h) R89C9 = {69/78}
i) R9C12 = {69/78}

1. R78C3 = {49/58} (cannot be {67} which clashes with R9C12), no 6,7
1a. Killer pair 8,9 in R78C3 and R9C12, locked for N7
1b. R56C5 = {18/27/36} (cannot be {45} which clashes with R78C3), no 4,5

2. Naked triple {789} in 24(3) disjoint cage at R4C4, locked for N5

3. Naked quad {6789} in R1C1 + R4C4 + R6C6 + R9C9, locked for D\ R1C1
3a. 6 on D\ R1C1 only in R1C1 + R9C9, CPE 6 in R1C9 + R9C1, clean-up: no 9 in R1C8, no 9 in R9C2

4. Naked triple {789} in R1C9 + R6C4 + R9C1, locked for D/ R1C9, clean-up: no 4,5 in R8C3

5. 15(2) cages at R1C1 and R8C9 must have the same combination, as must 15(2) cages at R1C8 and R8C1, because the 15(2) cages around the outside of the grid “see” adjacent 15(2) cages
5a. 6 on D\ R1C1 only in R1C1 + R9C9 -> R12C1 = {69}, locked for C1 and N1 and R89C1 = {69}, locked for C9 and N9, clean-up: no 6 in R1C8, no 6 in R9C2
5b. Naked pair {78} in R1C89, locked for R1 and N3
5c. Naked pair {78} in R9C12, locked for R9 and N7 -> R8C3 = 9, placed for D/ R2C9, R7C3 = 4, placed for D/ R1C9, R8C9 = 6, placed for D\ R1C2, R9C9 = 9, placed for D\ R1C1, R1C1 = 6, R2C1 = 9, placed for D\ R2C1, clean-up: no 1 in R1C2, no 1 in R3C8
5d. Naked pair {78} in R1C9 + R9C1, locked for D/ R1C9 -> R6C4 = 9

6. R1C7 = 9 (hidden single in N3)
6a. 5 in R1 only in R1C456, locked for N2
6b. 6 in N3 only in R2C78, locked for R2
6c. R7C5 = 9 (hidden single in N8)
6d. R3C6 = 9, then R3C5 = 6 (hidden singles in N2)
6e. R28C5 = {78} (hidden pair on C5)
6f. 6 on D/ R1C9 only in R2C8 + R4C6, CPE no 6 in R4C8

7. 6 in R7 only in R7C246, CPE no 6 in R5C4 using D/ R1C8 and D\ R2C1
7a. R4C6 = 6 (hidden single in N5), placed for D/ R1C9
7b. R2C7 = 6 (hidden single in N3), placed for D/ R1C8, clean-up: no 3 in R5C3
7c. R9C3 = 6 (hidden single in N7), clean-up: no 3 in R6C3
7d. R7C4 = 6 (hidden single in R7)
7e. R5C28 = {69} (hidden pair in R5)

8. 5 on D/ R1C9 only in R2C8 + R5C5 + R8C2, CPE no 5 in R2C2+ R8C8 using D\ R1C1

9. 7,8 in R2 only in R2C3456, CPE no 7,8 in R3C4 using D\ R1C2
9a. R3C12 = {78} (hidden pair in R3), locked for N1
9b. 4 in N1 only in R12C2, locked for C2

10. 7,8 in C3 only in R456C3, locked for N4
10a. Hidden killer pair 7,8 in R4C3 and R56C3 for C3, R56C3 contains one of 7,8 -> R4C3 = {78}
10b. Naked pair {78} in R4C34, locked for R4
10c. Naked pair {78} in R3C2 + R4C3, locked for D\ R2C1
10d. R7C89 = {78} (hidden pair in N9)
10e. R56C7 = {78} (hidden pair in C7), locked for N6
10f. Naked pair {78} in R6C67, locked for R6, clean-up: no 1,2 in R5C3
10g. 3 in C3 only in R123C3, locked for N1, clean-up: no 2 in R1C3

[At this stage all the short diagonals must contain all of 1,2,3,4,5]

11. 4 on D/ R1C8 only in R4C5 + R5C4, locked for N5
11a. 4 in R1 only in R1C2 + R1C456, CPE no 4 in R3C4 using D\ R1C2
11b. 4 in R3 only in R3C89, locked for N3
11c. 4 on D/ R2C9 only in R3C8 + R4C7, CPE no 4 in R46C8

12. R1C23 = [23/41], R2C2 = {124} -> R12C2 must contain 2, locked for C2 and N1
12a. Naked triple {135} in R123C3, locked for C3 and N1 -> R6C3 = 2, placed for D/ R1C8, R5C3 = 7, R4C3 = 8, placed for D\ R2C1, R4C4 = 7, R6C6 = 8, R3C12 = [87], R5C7 = 8, R6C7 = 7, placed for D\ R1C2, R7C89 = [87], R1C89 = [78], R9C12 = [78]
12b. R7C1 = 2 (hidden single in N7)

13. R3C4 = {123}, R3C78 = [14]/{23} -> R3C47 must contain 1, locked for R3

14. 2 in N5 only in R5C56, locked for N6
14a. 2 on D\ R1C2 only in R1C2 + R3C4 + R5C6, CPE no 2 in R1C6
14b. 4 on D\ R1C2 only in R1C2 + R4C5, CPE no 4 in R1C5
14c. 2 on D\ R1C1 only in R2C2 + R5C5 + R8C8, CPE no 2 in R2C8 using D/ R1C9
14d. 2 on D\ R2C1 only in R8C7 + R9C8, locked for N9

[I’ll do this next step in two parts.]
15. Consider placements for 4 on D\ R1C1
R2C2 = 4 => R1C23 = [23], 2 placed for D\ R1C2, R1C456 = {145}, locked for N2, R3C4 = 3 => R3C78 = [14]
or R8C8 = 4 => R2C2 = 2, R1C23 = [41], R3C78 = {23}
-> R1C23 and R3C78 must have different combinations
15a. R1C23 = [23] or R3C78 = {23}, locked for R3 -> R3C3 = 5, placed for D/ R1C1
15b. 5 on D\ R1C2 only in R4C5 + R5C6, locked for N5

16. R1C23 = [23] or R3C78 = {23} (step 15a) -> 1 in R2C3 + R3C4, locked for D/ R1C2
16a. 1 in R2C3 + R3C4, CPE no 1 in R2C46

17. Consider placements for [23] in N1
R1C23 = [23], 2 placed for D\ R1C2 => R1C456 = {145} => R3C4 = 3
or R2C23 = [23]
-> 3 in R2C3 + R3C4, locked for D\ R1C2
17a. R2C3 + R3C4 = {13} (hidden pair on D\ R1C2)
17b. R2C3 + R3C4 = {13}, CPE no 3 in R2C46

18. Hidden killer pair 2,4 in R3C78 and R3C9 for R3, R3C78 contains one of 2,4 -> R3C9 = {24}
18a. 2 in R3 only in R3C789, locked for N3

[From here my steps may seem to be slow but I’m trying to make progress with forcing chains while avoiding using contradiction moves. It would have been quicker if I’d seen key steps 21 and 22 sooner.]

19. Consider placements for 4 on D\ R1C2
R1C2 = 4, R1C3 = 1, R2C3 = 3, R3C4 = 1
or R4C5 = 4, R35C4 = {13}
-> 1 in R35C4, locked for C4
19a. R1C2 = 4, R4C5 = 5, R5C6 = 2, R56C5 = {13}
or R4C5 = 4, R1C2 = 2, R1C3 = 3, R2C3 = 1, R3C4 = 3, 3 in N5 only in R56C5
-> 3 in R56C5, locked for C5 and N5
19b. 3 on D/ R1C8 only in R7C2 + R8C1, locked for N7

20. Consider placements for 4 in C2
R1C2 = 4, R1C3 = 1, R2C3 = 3, R3C4 = 1, R5C4 = 4
or R2C2 = 4
-> no 4 in R2C4

21. Consider placements for 2 in C2
R1C2 = 2, R1C3 = 3 => R3C78 must have a different combination (step 15) = [14]
or R2C2 = 2 => R3C7 = 2 (hidden single in D/ R1C9
-> R3C7 = {12}, R3C8 = {34}
21a. 3 on D/ R1C9 only in R2C8 + R5C5, CPE no 3 in R8C8 using D\ R1C1

22. Consider placements for 5 in R2
R2C8 = 5
or R2C9 = 5, R5C6 = 2, naked pair {13} in R5C5 + R7C7, locked for D\ R1C1 => R8C8 = 4 => R3C8 = 3
-> no 3 in R2C8
[Cracked.]

23. 3 in N3 only in R2C9 + R3C8, locked for D/ R2C9 -> R6C5 = 1, placed for D\ R2C1 and D/ R2C9, R5C4 = 4, placed for D\ R2C1, R4C5 = 5, R1C5 = 2, R1C2 = 4, R1C3 = 1, R2C3 = 3, placed for D\ R1C2, R3C4 = 1, R2C9 = 5, R2C8 = 1, placed for D/ R1C9, R8C2 = 5

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

I'll rate my walkthrough at least 1.5. I used short forcing chains for some of the later steps.