Thanks Ruud for another great Killer-X. I know that others solved this one before I did but, since they haven't posted their walkthroughs, here is my one for SKX3.Ruud wrote:Here is my third Killer-X. A combination of Killer and regular Sudoku techniques is required, unless you manage to find a shortcut that I have missed. As usual, the forum welcomes your walkthrough.
I look forward to the next SKX after Ed's current tag killer-X PANIV has been solved. I know from Ed's comment of SKX1 that he is looking forward to it too.
As with SKX1, many thanks to Ed for reviewing my walkthrough. I've included his comments below.
Clean-up is used in various steps, using the combinations in steps 1 to 12 for further eliminations from these two cell cages. In some of the later steps, clean-up is followed by further moves and sometimes more clean-up.
1. R1C23 = {13}, locked for R1 and N1
2. R1C78 = {24}, locked for R1 and N3
3. R23C5 = {12}, locked for C5 and N2
4. R5C67 = {13}, locked for R5
5. R5C34 = {48/57}, no 2,6,9
6. R9C23 = {29/38/47/56}, no 1
7. R9C78 = {49/58/67}, no 1,2,3
8. R78C5 = {59/68}, no 3,4,7
9. R34C4 = {49/58/67}, no 1,2,3
10. R34C6 = {17/26/35}, no 4,8,9, no 6,7 in R4C6
11. R67C4 = {17/26/35}, no 4,8,9
12. R67C6 = {16/25/34}, no 7,8,9
13. 28(4) cage in N1 = {4789/5689} = 89{47/56}, no 2, 8,9 locked for N1
14. 28(4) cage in N47 = {4789/5689} = 89{47/56}, no 1,2,3
15. 13(4) cage in N4 = {1237/1246/1345}, no 8,9, 1 locked in R46C1, locked for C1 and N4
16. 14(4) cage in N9 = {1238/1247/1256/1346/2345}, no 9, must contain at least two of 1,2,3
17. 1 in N7 locked in R8C23, locked for R8
18. 42(7) cage in N123 = {3456789}, no 1,2
18a. 3,4 locked in R2C3467, locked for R2
19. R2C5 = 2 (hidden single in R2), R3C5 = 1
20. Only valid combinations for 20(4) cage in N3 are 1{379/568} -> R2C7 + R3C78 = {379/568}
21. If {1379} combination in 20(4) cage in N3, 3 must be in R3C9 -> no 7,9 in R3C9
22. 2 in R3 locked in R3C23, no 2 in R4C23
23. 2 in N4 locked in 13(4) cage = {1237/1246} = 12{37/46}, no 5
23a. The {1237} would have {13} in R46C1 -> no 7 in R46C1
24. 37(7) cage in N789 must contain 4,8,9 [Note that the two excluded numbers form a 8(2) pair {17/26/35} so that if one of these numbers is locked in the 37(7) cage in a later step, the other number will also be locked in the 37(7) cage.]
-> no {89} in R8C5 [Thanks Ed, missed that.]
25. 14(2) cage in N8 = {59/68} -> the part of the 37(7) which is in N8 must contain 8 or 9 and R8C37 must contain 8 or 9
[If Ed’s addition to step 24 had been used, this would be 12(2) = [95/86] (see note on step 24). The conclusion about R8C37 = [8/9] is correct.]
26. Only remaining {34} in C5 are in R4569C5, they cannot both be in the 17(3) cage in R456C9 -> R9C5 = {34} with 3 or 4 in R456C5
27. R456C5 = 3{59/68} or {467} ({458} would clash with R78C5, if R456C5 = {467}, then R78C5 = {59}), killer triple 5/6/9 in R45678C5, no 5,6,9 in R1C5 = {78} [Alternatively I could have used 45 rule on C5 but I spotted the more interesting and complicated way first]
28. 45 rule on R12 2 remaining outies R3C19 = 12, no 6, no 5,8 in R3C1
29. 45 rule on C789 3 innies R258C7 = 18, R5C7 = {13} (step 3), R258C7 = {189/369/378}, no 2,3,4,5 in R28C7
30. 3 in 42(7) cage locked in R2C46, locked for N2, clean-up: no 5 in R4C6
31. 45 rule on C123 3 innies R258C3 = 12, min R25C3 = 9 -> R8C3 = {123}, min R28C3 = 5 -> R5C3 = {457}, clean-up: R5C4 = {578}
32. R8C3 = {123} -> R8C7 = {89} (step 25)
33. 4 in 37(7) cage locked in N8, locked for N8, clean-up: no 3 in R6C6
34. 45 rule on N8 2 outies R8C37 – 6 = 2 innies R7C46, max R8C37 = 12 -> max R7C46 = 6, no 6,7 in R7C46, clean-up: no 1,2 in R6C4, no 1 in R6C6
35. 1 in N5 locked in C6, locked for C6, clean-up: no 6 in R6C6, 1 in N8 locked in C4
36. 7 in N8 locked in 37(7) cage -> 37(7) cage must contain 1 = 14789{26/35}
37. 45 rule on C6 5 innies R12589C6 = 30 and must contain 8,9 which aren’t in R3467C6 = {15789/34689}(cannot be {24789/25689} which don’t have {13}) = 89{157/346} -> no 2 in R89C6, for the {34689} combination the 3 must be in R5C6 -> no 3 in R289C6
38. R4567C6 must contain 123{4/5}, R1289C6 must contain 4/5 (step 37), killer pair 4/5 -> no 5 in R3C6, clean-up: no 3 in R4C6
Ed commented “An easier way to show that no 5 in R3C6. R34C6 = [53] blocks 7(2) in C6 -> R34C6 cannot be [53]”. I’ll accept that as an alternative way. Whether it’s easier depends on what one happens to see.
39. 2 in N5 locked in C6 -> no 2 in R7C6, clean-up: no 5 in R6C6
40. R2C4 = 3 (hidden single in N2), clean-up: no 5 in R67C4
41. R456C6 must contain 12{3/4}, R456C5 must contain 3/4 (step 27), no 4 in R4C4, clean-up: no 9 in R3C4
42. 9 in N2 locked in 42(7) cage -> no 9 in R2C7
43. R2C7 + R3C78 = {379/568} (step 20) -> no 7 in R3C78
44. 45 rule on N36 4 innies R25C7 + R6C78 = 16, min R25C7 = 7 -> max R6C78 = 9, no 9 in R6C78
45. 45 rule on N14 4 innies R25C3 + R6C23 = 25, min R25C3 = 9 -> max R6C23 = 16 -> R6C23 cannot contain 8 and 9, 20(4) cage in N14 does not contain 1 so R4C23 cannot contain 8 and 9, R46C23 must contain 8 and 9 -> R4C23 must contain 8 or 9, R6C23 must contain 8 or 9, R7C23 must contain 8 or 9 [Sorry that’s rather complicated, hope people understand it.]
46. 20(4) cage in N14 must contain 2 and 8 or 9 = 2{369/378/459/468} [3/4, 5/6/7]
47. 45 rule on R89 3 outies R7C159 = 18, R7C23 must contain 8 or 9 (step 45) so R7C15 cannot be {89} -> no 1 in R7C9
48. 14(4) cage in N9, min R7C9 + R8C78 = {234} = 9 -> max R9C9 = 5
49. 45 rule on N3 1 innie R2C7 + 5 = 2 outies R4C78, min R2C7 = 6 -> min R4C78 = 11, no 1
50. 45 rule on N9 1 innie R8C7 – 2 = 2 outies R6C78, R8C7 = {89} -> R6C78 = 6 or 7, no 7,8
51. 45 rule on N7 1 innie R8C3 + 13 = 2 outies R6C23 -> min R6C23 = 14, no 4
52. 28(4) cage in N47 = 89{47/56} (step 14), the 8 and 9 must be in different rows (step 45) so cannot have {47} in R7C23 -> no 7 in R7C23
53. 45 rule on N1 1 innie R2C3 + 7 = 2 outies R4C23, R2C3 = {4567} -> R4C23 must total 11 to 14, R4C23 must contain 8 or 9 (step 45) -> no 7 in R4C23
54. 45 rule on C1 3 outies R258C2 = 15, min R25C2 = 7, max R8C2 = 8, no 9
55. 45 rule on R1234 3 innies R4C159 = 12, min R4C15 = 4, max R4C9 = 8, no 9
56. 28(4) cage in N1 = 89{47/56} (step 13), 4 only in R3C1 -> no 7 in R3C1, no 5 in R3C9 (step 28)
57. 20(4) cage in N3 = 1{379/568} (step 20), for the {1568} combination the 8 must be in R3C9 -> no 8 in R1C9 + R2C89
58. In R9 the 11(2) and 13(2) cages form a 24(4) cage which must contain 8 and/or 9
59. Valid combinations for R9C14569, which must contain {34} in R9C5 and must contain 1 are {12369/12378/12459/12468/13458/13467}
60. 37(7) cage must have 5/6 in N8, R78C5 must have 5/6 -> 5 locked for N8, R7C6 = 3, clean-up: R6C6 = 4, 4 locked for D\, R5C67 = [13], R4C6 = 2, 2 locked for D/, clean-up: R3C6 = 6, no 7 in R4C4
61. R9C5 = 4, clean-up: no 7 in R9C23, no 9 in R9C78, R456C5 = 3{59/68}, no 7 -> R1C5 = 7, R3C4 = 4 (hidden singles in N2), clean-up: R4C4 = 9, 9 locked for D\, R3C1 = 9 (naked single in N1)
62. 9 locked in R12C6 for N2, locked for C6, 9 in N8 locked in R78C5 -> R78C5 = {59}, 5 locked for C5 and N8 -> no 3 in 37(7) cage (step 24), R456C5 = {368}, locked for N5, R6C4 = 7, locked for D/, clean-up: R7C4 = 1, R5C4 = 5, R5C3 = 7, R1C4 = 8 (naked single), R12C6 = {59}, locked for the 42(7) cage, R2C7 = 6, R2C3 = 4, clean-up: no 7 in R9C8
63. 37(7) cage = {1246789}, R89C4 = {26}, R89C6 = {78}, R8C3 = 1, R8C7 = 9, R1C23 = [13], clean-up: no 8 in R9C2, R78C5 = [95]
64. R9C9 = 1 -> R2C8 = 1 -> R6C7 = 1 -> R4C1 = 1 (all hidden singles)
65. No 9 in R7C23 -> 8 locked in R7C23 (step 45), no other 8 in R7, N7 and R6C23, clean-up: no 3 in R9C2, 9 locked in R6C23 (step 45), locked for R6
66. 28(4) cage in N1 = {5689} (subtraction combo) -> R1C1 = 6, locked for D\, R2C12 = {58}, locked for R2 and N1
67. R3C23 = [72], 2 locked for D\, clean-up: no 9 in R9C2
68. R5C5 = 8 (naked single), locked for both diagonals, R2C2 = 5, R7C7 = 7, R8C8 = 3, (naked singles on D\), R2C1 = 8, R46C5 = {36}, clean-up: no 6 in R9C8, R9C78 = {58}, locked for R9 and N9, more clean-up: no 6 in R9C23 = [29]
69. R2C6 = 9, R1C6 = 5, R1C9 = 9, R2C9 = 7, R3C9 = 3
70. R3C78 = [58], 5 locked for D/, no 5,8 in R4C78 -> R4C78 = [47]
71. R7C3 = 6, R8C2 = 4, R9C1 = 3 (naked singles on D/)
72. R8C1 = 7, R89C6 = [87]
and the rest is simple elimination
