Assassin 46 Lite
Assassin 46 Lite
Being accused several times of posting V2's which are way too difficult, this time I have a V2 for Assassin 46 which is slightly easier than V1. It does not require detection of conflicting combinations.
3x3::k307330734869:4869:4869:4104384035954366:4366:4104:4104:4626:4626:46263844:4366:4888:4888:4888:46261308:8734:8734:8734:4385:4385:48882084:4390:4390:873418332347:46533374:8734:8734:87342099:5685:4653:4653:465356905685:5685:5685:5183:51833641:569026192118:5183340156903917:3917
Still a very enjoyable puzzle... I think.
Ruud
3x3::k307330734869:4869:4869:4104384035954366:4366:4104:4104:4626:4626:46263844:4366:4888:4888:4888:46261308:8734:8734:8734:4385:4385:48882084:4390:4390:873418332347:46533374:8734:8734:87342099:5685:4653:4653:465356905685:5685:5685:5183:51833641:569026192118:5183340156903917:3917
Still a very enjoyable puzzle... I think.
Ruud
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
One should never underestimate any of Ruud's killers, particularly when he posts comments on them with the rolling eyes emoticon!
Ruud didn't just post a puzzle for us to solve. He was also giving us a test to see what we made of it, and we've all been sleeping...
All this stuff about V2's being way too difficult may well be true, but I strongly suspect that it's only half the story. Therefore, this post will also only cover half the story. The rest will come in a second post, so stay tuned!
In the meantime, here's the walkthrough:
1. 19/3 at R1C6: no 1
2. 5/2 at R4C2 = {14|23}
3. 17/2 at R4C7 = {89} -> no 8,9 elsewhere in R4 or N6
4. 8/2 at R5C1: no 4,8,9
5. 17/2 at R5C3 = {89} -> no 8,9 elsewhere in R5
6. 7/2 at R5C6: no 7 (8,9 already eliminated)
7. 9/2 at R5C8 = {27|36|45} (no 1)
8. 13/2 at R6C2: no 1,2,3
9. 8/2 at R6C7: no 4 (8,9 already eliminated)
10. 22/3 at R7C5 = {(58|67)9} -> no 9 elsewhere in C5 or N8
11. 10/3 at R7C6: no 8,9
12. 20/3 at R8C1: no 1,2
13. 8/3 at R8C8 = {1(25|34)} -> no 1 elsewhere in N9
14. Innies N5: R5C46 = 11/2 = [83|92] -> R5C7 = {45}
15. Innie R5: R5C5 = 4 -> R5C7 = 5 -> R5C6 = 2 -> R5C4 = 9 (step 14) -> R5C3 = 8
16. 1 in R5 locked in 8/2 at R5C1 = {17} -> no 1,7 elsewhere in R5 or N4
17. 9/2 at R5C8 = {36} -> no 3,6 elsewhere in N6
18. 13/2 at R6C2 = {49} (only combo possible) -> no 4,9 elsewhere in R6 or N4
19. 8/2 at R6C7 = {17} (only combo possible) -> no 1,7 elsewhere in R6 or N6
20. Naked Single (NS) at R6C9 = 2 -> R4C9 = 4
21. 2 in C5 locked in 15/3 at R1C5 -> {2(58|67}) (no 1,3) -> no 2 elsewhere in N2
22. Hidden Single (HS) in C5 at R4C5 = 1
23. HS in C5 at R6C5 = 3
24. Outie R123: R4C1 = 6 -> R6C1 = 5
25a. Outies N1: R123C4 = 8/3 = {134} (2 unavailable) -> no 1,3,4 elsewhere in C4 or N2
25b. 14/3 at R2C3 = {(19|37)4} -> no 2,5,6 in R2C3
Edit: Should have noticed here that 4 is locked in R23C4, allowing elimination of 4 in R2C3 and R1C4, making step 42 unnecessary.
26a. Outies N9: R789C6 = 8/3 = {134} (2 unavailable)
26b. 10/3 at R7C6 = {136} (only permutation possible) -> R8C7 = 6, R78C6 = {13}, R9C6 = 4 (step 26a)
26c. R9C78 = {29|38} (no 5,7)
27a. 7 in N9 locked in R7 -> no 7 elsewhere in R7
27b. Split 20/3 cage at R7C789 = {(49|58)7} (no 3)
28. Innies R7: R7C456 = 12/3 = {1(29|56)} ({138} unavailable, since {13} only in R7C6)
-> R7C6 = 1, no 8 in R7C45, R8C6 = 3
29a. 3 in R7 locked in N7 -> no 3 elsewhere in N7
29b. Split 13/3 cage at R7C123 = {(28|46)3} (no 9)
30. 20/3 at R8C1 = {(49|58)7} -> no 7 elsewhere in N7
31. 13/3 at R9C2 = {(17|26)5} (no 8,9) (3,4 unavailable) -> no 5 elsewhere in R9
32a. 14/3 at R7C4: no 9 in R8C3 (due to {13} unavailable in R78C4)
32b. 9 in N7 now locked in 20/3 at R8C1 = {479} -> no 4 elsewhere in N7
32c. 4 in R7 now locked in R7C78 -> no 4 in R8C8
32d. 8/3 at R8C8 = {125} (4 no longer available) -> no 2,5 elsewhere in N9
32e. NS at R9C9 = 1 -> R8C9 = 5, R8C8 = 2
32f. NS at R8C3 = 1
32g. Split 11/2 cage at R9C78 = {38} (no 9) -> no 8 elsewhere in R9 or N9
32h. Split 20/3 cage at R7C789 = {479} -> no 9 in R7C5
33. Innie N8: R9C4 = 2
34. R9C23 = {56} -> no 6 in R7C23
35. <step deleted>
36. Innies R9: R9C15 = 16/2 = {79}
37. Naked Pair (NP) on {23} in C3 at R47C3 -> no 2,3 elsewhere in C3
38a. Split 12/3 cage at R3C123: no 8 (due to {138} being unavailable in R3C3)
38b. 12/3 cage at R1C2: no 8 (due to {138} being unavailable in R1C3)
38c. 8 in N1 now locked in 15/3 cage at R1C1 = {(16|25|34)8} (no 7,9)
39. 7 in C3 locked in N1 -> no 7 elsewhere in N1
40. 8 in C9 locked in N3 -> no 8 elsewhere in N3
41. 19/3 at R1C6: no 5 (would imply {568}, impossible since R1C7 has none of these digits as candidates)
42. Innie/outie difference (i/o diff.) N1: R2C3 = R1C4 + 6 -> R1C4 = {13}, R2C3 = {79}
43. I/o diff. N3: R1C6 = R2C7 + 5 -> R2C7 = {1234}
44. 12/3 at R1C2: no 1 in R1C2 (would force 3 into R1C4, requiring 8 in R1C3 - unavailable)
45a. Innies R3: R3C456 = 18/3 -> no 2 in R3C5 (since max. sum of R3C46 = 13)
45b. Similarly, no 5 in R3C6 (since max. sum of R3C45 = 12)
46a. Innies R1: R1C159 = 14/3 -> no 8 in R1C15 (due to {1245} unavailable in R1C9)
46b. 8 in N1 now locked in R2C12 -> no 8 elsewhere in R2
47. 15/3 cage at R1C5: no 5 in R3C5 (due to {3489} unavailable in R12C5)
48. Innies N1: R1C23+R2C3 = 18/3 -> no 9 in R1C2 (since sum of R12C3 >= 11)
49. 12/3 at R1C2: no 3 in R1C2 (would imply {138}, impossible since R1C3 has none of these digits as
candidates)
50. Split 12/3 cage at R3C123: no 9 in R3C12 (would require either of {12} in R3C3 - unavailable)
51. 9 in C1 now locked in R89C1 -> no 9 in R8C2
52. HS in C2 at R6C2 = 9 -> R6C3 = 4
53. Split 12/3 cage at R3C123: no 5 in R3C2 (would force R3C3 to {79} - cage sum already reached)
54. 16/3 cage at R1C9: no 9 in R2C8 (sum of R12C9 >= 9 -> R2C8 <= 7)
55. 15/3 at R1C1: min. sum of R2C12 = 11 (due to R1C1 <= 4) -> no 1,2 in R2C12
56a. Split 15/3 at R3C789: no 9 in R3C78 (due to {1245} unavailable in R3C9)
56b. Split 15/3 at R3C789: no 1 in R3C8 (since none of {5689} in R3C7)
56c. Split 15/3 at R3C789: no 3 in R3C8 (since {45} unavailable in both of R3C79)
57. I/o diff. C9: R357C9 - R2C8 = 17 -> no 7 in R2C8 (since. max. sum of R357C9 = 23 -> R2C8 <= 6)
--- Now for the hard-to-find move that cracks the puzzle... ---
58. 16/3 at R1C9 = {169|178|349|358|367} -> must have one of {13}
Innies N3: R1C78+R2C7 = 14/3 = {149|167|239|347} (5,8 unavailable) -> must have one of {13}
-> 16/3 at R1C9+R2C89 and 14/3 at R1C78+R2C7 form killer pair on {13} -> no 1,3 elsewhere in N3
(i.e., in R3C79).
59a. Split 15/3 at R3C789 must contain a digit under 5 -> R3C7 = 2
59b. Split cage R3C89 = 13/2 = {58|67} -> no 9 in R3C9
60. HS in R2 at R2C5 = 2
61. Split 12/3 cage at R3C123 = {(17|35)4} -> no 4 elsewhere in R3 or N1, no 9 in R3C3
62. HS in R3 at R3C6 = 9 -> R2C67 = [53|71]
63. HS in C4 at R2C4 = 4
64a. R2C7 now {13} -> R1C6 = {68} (step 43)
64b. 19/3 at R1C6: no 3 in R1C78 (since {79} both unavailable in R1C6)
65. R3C45 = [18|36] (innies, see step 45a)
66. NP on {68} in N2 at R1C6 and R3C5 -> no 6 in R1C5
67. 15/3 at R1C1 = {(16|25)8} (no 3) -> R2C1 = 8
68. HS in C2 at R7C2 = 8
69. 3 in N1 locked in split 12/3 cage at R3C123 = {345} -> R3C12 = {34}, R3C3 = 5
--- and the puzzle now ends with an avalanche of naked singles... ---
Ruud didn't just post a puzzle for us to solve. He was also giving us a test to see what we made of it, and we've all been sleeping...
All this stuff about V2's being way too difficult may well be true, but I strongly suspect that it's only half the story. Therefore, this post will also only cover half the story. The rest will come in a second post, so stay tuned!
In the meantime, here's the walkthrough:
1. 19/3 at R1C6: no 1
2. 5/2 at R4C2 = {14|23}
3. 17/2 at R4C7 = {89} -> no 8,9 elsewhere in R4 or N6
4. 8/2 at R5C1: no 4,8,9
5. 17/2 at R5C3 = {89} -> no 8,9 elsewhere in R5
6. 7/2 at R5C6: no 7 (8,9 already eliminated)
7. 9/2 at R5C8 = {27|36|45} (no 1)
8. 13/2 at R6C2: no 1,2,3
9. 8/2 at R6C7: no 4 (8,9 already eliminated)
10. 22/3 at R7C5 = {(58|67)9} -> no 9 elsewhere in C5 or N8
11. 10/3 at R7C6: no 8,9
12. 20/3 at R8C1: no 1,2
13. 8/3 at R8C8 = {1(25|34)} -> no 1 elsewhere in N9
14. Innies N5: R5C46 = 11/2 = [83|92] -> R5C7 = {45}
15. Innie R5: R5C5 = 4 -> R5C7 = 5 -> R5C6 = 2 -> R5C4 = 9 (step 14) -> R5C3 = 8
16. 1 in R5 locked in 8/2 at R5C1 = {17} -> no 1,7 elsewhere in R5 or N4
17. 9/2 at R5C8 = {36} -> no 3,6 elsewhere in N6
18. 13/2 at R6C2 = {49} (only combo possible) -> no 4,9 elsewhere in R6 or N4
19. 8/2 at R6C7 = {17} (only combo possible) -> no 1,7 elsewhere in R6 or N6
20. Naked Single (NS) at R6C9 = 2 -> R4C9 = 4
21. 2 in C5 locked in 15/3 at R1C5 -> {2(58|67}) (no 1,3) -> no 2 elsewhere in N2
22. Hidden Single (HS) in C5 at R4C5 = 1
23. HS in C5 at R6C5 = 3
24. Outie R123: R4C1 = 6 -> R6C1 = 5
25a. Outies N1: R123C4 = 8/3 = {134} (2 unavailable) -> no 1,3,4 elsewhere in C4 or N2
25b. 14/3 at R2C3 = {(19|37)4} -> no 2,5,6 in R2C3
Edit: Should have noticed here that 4 is locked in R23C4, allowing elimination of 4 in R2C3 and R1C4, making step 42 unnecessary.
26a. Outies N9: R789C6 = 8/3 = {134} (2 unavailable)
26b. 10/3 at R7C6 = {136} (only permutation possible) -> R8C7 = 6, R78C6 = {13}, R9C6 = 4 (step 26a)
26c. R9C78 = {29|38} (no 5,7)
27a. 7 in N9 locked in R7 -> no 7 elsewhere in R7
27b. Split 20/3 cage at R7C789 = {(49|58)7} (no 3)
28. Innies R7: R7C456 = 12/3 = {1(29|56)} ({138} unavailable, since {13} only in R7C6)
-> R7C6 = 1, no 8 in R7C45, R8C6 = 3
29a. 3 in R7 locked in N7 -> no 3 elsewhere in N7
29b. Split 13/3 cage at R7C123 = {(28|46)3} (no 9)
30. 20/3 at R8C1 = {(49|58)7} -> no 7 elsewhere in N7
31. 13/3 at R9C2 = {(17|26)5} (no 8,9) (3,4 unavailable) -> no 5 elsewhere in R9
32a. 14/3 at R7C4: no 9 in R8C3 (due to {13} unavailable in R78C4)
32b. 9 in N7 now locked in 20/3 at R8C1 = {479} -> no 4 elsewhere in N7
32c. 4 in R7 now locked in R7C78 -> no 4 in R8C8
32d. 8/3 at R8C8 = {125} (4 no longer available) -> no 2,5 elsewhere in N9
32e. NS at R9C9 = 1 -> R8C9 = 5, R8C8 = 2
32f. NS at R8C3 = 1
32g. Split 11/2 cage at R9C78 = {38} (no 9) -> no 8 elsewhere in R9 or N9
32h. Split 20/3 cage at R7C789 = {479} -> no 9 in R7C5
33. Innie N8: R9C4 = 2
34. R9C23 = {56} -> no 6 in R7C23
35. <step deleted>
36. Innies R9: R9C15 = 16/2 = {79}
37. Naked Pair (NP) on {23} in C3 at R47C3 -> no 2,3 elsewhere in C3
38a. Split 12/3 cage at R3C123: no 8 (due to {138} being unavailable in R3C3)
38b. 12/3 cage at R1C2: no 8 (due to {138} being unavailable in R1C3)
38c. 8 in N1 now locked in 15/3 cage at R1C1 = {(16|25|34)8} (no 7,9)
39. 7 in C3 locked in N1 -> no 7 elsewhere in N1
40. 8 in C9 locked in N3 -> no 8 elsewhere in N3
41. 19/3 at R1C6: no 5 (would imply {568}, impossible since R1C7 has none of these digits as candidates)
42. Innie/outie difference (i/o diff.) N1: R2C3 = R1C4 + 6 -> R1C4 = {13}, R2C3 = {79}
43. I/o diff. N3: R1C6 = R2C7 + 5 -> R2C7 = {1234}
44. 12/3 at R1C2: no 1 in R1C2 (would force 3 into R1C4, requiring 8 in R1C3 - unavailable)
45a. Innies R3: R3C456 = 18/3 -> no 2 in R3C5 (since max. sum of R3C46 = 13)
45b. Similarly, no 5 in R3C6 (since max. sum of R3C45 = 12)
46a. Innies R1: R1C159 = 14/3 -> no 8 in R1C15 (due to {1245} unavailable in R1C9)
46b. 8 in N1 now locked in R2C12 -> no 8 elsewhere in R2
47. 15/3 cage at R1C5: no 5 in R3C5 (due to {3489} unavailable in R12C5)
48. Innies N1: R1C23+R2C3 = 18/3 -> no 9 in R1C2 (since sum of R12C3 >= 11)
49. 12/3 at R1C2: no 3 in R1C2 (would imply {138}, impossible since R1C3 has none of these digits as
candidates)
50. Split 12/3 cage at R3C123: no 9 in R3C12 (would require either of {12} in R3C3 - unavailable)
51. 9 in C1 now locked in R89C1 -> no 9 in R8C2
52. HS in C2 at R6C2 = 9 -> R6C3 = 4
53. Split 12/3 cage at R3C123: no 5 in R3C2 (would force R3C3 to {79} - cage sum already reached)
54. 16/3 cage at R1C9: no 9 in R2C8 (sum of R12C9 >= 9 -> R2C8 <= 7)
55. 15/3 at R1C1: min. sum of R2C12 = 11 (due to R1C1 <= 4) -> no 1,2 in R2C12
56a. Split 15/3 at R3C789: no 9 in R3C78 (due to {1245} unavailable in R3C9)
56b. Split 15/3 at R3C789: no 1 in R3C8 (since none of {5689} in R3C7)
56c. Split 15/3 at R3C789: no 3 in R3C8 (since {45} unavailable in both of R3C79)
57. I/o diff. C9: R357C9 - R2C8 = 17 -> no 7 in R2C8 (since. max. sum of R357C9 = 23 -> R2C8 <= 6)
--- Now for the hard-to-find move that cracks the puzzle... ---
58. 16/3 at R1C9 = {169|178|349|358|367} -> must have one of {13}
Innies N3: R1C78+R2C7 = 14/3 = {149|167|239|347} (5,8 unavailable) -> must have one of {13}
-> 16/3 at R1C9+R2C89 and 14/3 at R1C78+R2C7 form killer pair on {13} -> no 1,3 elsewhere in N3
(i.e., in R3C79).
59a. Split 15/3 at R3C789 must contain a digit under 5 -> R3C7 = 2
59b. Split cage R3C89 = 13/2 = {58|67} -> no 9 in R3C9
60. HS in R2 at R2C5 = 2
61. Split 12/3 cage at R3C123 = {(17|35)4} -> no 4 elsewhere in R3 or N1, no 9 in R3C3
62. HS in R3 at R3C6 = 9 -> R2C67 = [53|71]
63. HS in C4 at R2C4 = 4
64a. R2C7 now {13} -> R1C6 = {68} (step 43)
64b. 19/3 at R1C6: no 3 in R1C78 (since {79} both unavailable in R1C6)
65. R3C45 = [18|36] (innies, see step 45a)
66. NP on {68} in N2 at R1C6 and R3C5 -> no 6 in R1C5
67. 15/3 at R1C1 = {(16|25)8} (no 3) -> R2C1 = 8
68. HS in C2 at R7C2 = 8
69. 3 in N1 locked in split 12/3 cage at R3C123 = {345} -> R3C12 = {34}, R3C3 = 5
--- and the puzzle now ends with an avalanche of naked singles... ---
Last edited by mhparker on Fri Apr 20, 2007 8:21 pm, edited 1 time in total.
Cheers,
Mike
Mike
Re: Assassin 46 Lite
Generally you open with the message that your V2 was too difficult to put one the site. We never complain (we wouldn't dare to ). We are usually just stunned for a moment. But we never refused a challenge, don't think we ever will. (maybe once with Ed's V2 on the bullseye) (ps. anyone feel like having a crack at that one again?)Ruud wrote:Being accused several times of posting V2's which are way too difficult
Ruud wrote:The Easter Bunny came by and hid the solutions to these variations of Assassin 45.
If you can help me recover these, I would be both pleased and impressed.
Ruud wrote:Here is a V2 that will take a team effort to crack. All killer solvers on my PC (including the latest JSudoku) give up and need to use backtracking to solve this one.
Just messing around.Ruud wrote:Even though Assassin 39 is not yet available, I want to post these 2 killers using the same cage pattern.
They are ... erm ... too difficult?
General conclusions: There's no problem with "too" difficult V2's. It is generally a lot of fun to see what we can come up with to crack those real toughies (like 7 outie 45-moves or massive hypotheticals that use 20 steps to come to a final conclusion).
Question to everyone: Anyone feel like tackling "the one that got away" (Ed's V2 on the bullseye)?
greetings
Para
I said above that I wanted to make a second post on this puzzle.
Before I start, I want to say that my walkthrough for this Lite version deliberately avoided using conflicting combinations, because Ruud assured us that the puzzle can be completed without them. Furthermore, although there's nothing spectacular in it (it's all text book stuff, as would be done by an automated solver), it's main purpose is to be used as a basis for what I want to say right now, which is what makes this particular killer one of the most interesting I have ever seen.
Let's start with the puzzle state immediately after step 57 in my walkthrough, which is as far as SumoCue v1.30 gets with this puzzle:
From this position, there is namely a spectacular sequence of moves based on Unique Rectangles (UR) and deadly patterns, which are not mentioned in most Killer solving guides.
By the way, for anyone not familiar with UR-based techniques, you can find more information here:
http://www.sudocue.net/guide.php#UR
However, please bear in mind that all discussions relating to UR on the web are based on regular (vanilla) Sudoku. For Killer Sudoku, there is an important extra requirement, namely:
All (pre-defined) cages (if any) overlapping with the cells of the deadly pattern must contain exactly two UR cells of opposite polarity
Otherwise, you can't (in general) exchange the two digits of the deadly pattern to get an alternate solution, because then the cage constraints (such as the cage sums) will not be met.
UR Type 1 ("Unique Corner")
Refer now to the above candidate grid, in particular the four cells R47C23. All apart from one of the cells contains only the candidates {23}. Only R7C2 has a single extra candidate, 8. In order to avoid the so-called deadly rectangle (where the two participating digits can be freely interchanged in the solution to provide an equally valid alternate solution), R7C2 must take the value of the extra candidate, so we can directly place an 8 there, which creates a hidden single in C1 at R2C1 = 8. After making both of these moves, we end up with the following grid:
From here, we can independently make a second UR move, as follows:
UR Type 4 ("Unique Pair")
Referring to the above candidate grid, in particular the four cells R89C15, we can see that the digit 9 has only two candidate positions in R8, and thus must be located in either of R8C15, the two top corners of the UR. Therefore, neither of these corners may contain the digit 7, because this would form the deadly rectangle on {79} in R89C15. Therefore, we can eliminate candidate 7 from R8C15, giving the following grid:
As if that's not enough, we can perform yet another UR move here!:
"Killer" UR
This is a variant that is particular to Killer Sudoku, and which I've never seen documented anywhere.
Take a look at the 12/3 cage at R1C2 in the above grid. Note that, because of the candidates remaining in R1C23, if R1C2 were 5, this would force R1C3 to 6 (i.e., R1C234 = [561]). Similarly, if R1C2 were 6, this would force R1C3 to 5 (i.e., R1C234 = [651]). Because of the additional links on 5 and 6 between R1C2 and R9C2 and between R9C2 and R9C3, it can be seen that placing either of {56} in R1C2 would force a deadly rectangle on {56} in R19C23. Therefore we can eliminate both 5 and 6 from R1C2. Thereafter, we can remove candidate 6 from R1C3, because no possible permutations now use it (obsolete candidate).
Additionally, this opens up the following moves:
Unique "Swordfish"
Last but not least, take a look at the 6 cells R3C12, R4C23 and R7C13. These form a swordfish pattern, which, if every cell were to contain the same two candidates ({23} in this case), would be just as deadly as the deadly rectangle. This means that R3C12 cannot be {23}, ruling out the combination {237} for the split 12/3 cage at R3C123 (because neither of the digits {23} are available in the remaining cell R3C3). Since the only other combination for these 3 cells is {345}, this means that we can immediately eliminate the candidate 2 from R3C12 and place a "5" in R3C3.
After applying these UR-based techniques (the last one being optional), the puzzle becomes easily solvable via singles.
The sheer number and type of UR-based moves makes this puzzle for me a case study in uniqueness tests as applied to Killer Sudoku.
Before I start, I want to say that my walkthrough for this Lite version deliberately avoided using conflicting combinations, because Ruud assured us that the puzzle can be completed without them. Furthermore, although there's nothing spectacular in it (it's all text book stuff, as would be done by an automated solver), it's main purpose is to be used as a basis for what I want to say right now, which is what makes this particular killer one of the most interesting I have ever seen.
Let's start with the puzzle state immediately after step 57 in my walkthrough, which is as far as SumoCue v1.30 gets with this puzzle:
Code: Select all
.-------.-----------------------.-------.-----------------------.-------.
| 1234 | 2456 5679 13 | 2567 | 6789 23479 34679 | 36789 |
| '-------.---------------: :---------------.-------' |
| 348 34568 | 79 134 | 2567 | 5679 1234 | 13456 3679 |
:---------------'-------. | | .-------'---------------:
| 1234 1234 579 | 134 | 678 | 6789 | 1237 567 36789 |
| .---------------+-------'-------'-------+---------------. |
| 6 | 23 23 | 57 1 57 | 89 89 | 4 |
:-------'-------.-------'-------. .-------'-------.-------'-------:
| 17 17 | 8 9 | 4 | 2 5 | 36 36 |
:-------.-------'-------.-------' '-------.-------'-------.-------:
| 5 | 9 4 | 68 3 68 | 17 17 | 2 |
| '---------------+-------.-------.-------+---------------' |
| 238 238 23 | 56 | 56 | 1 | 479 479 79 |
:---------------.-------' | | '-------.---------------:
| 479 47 | 1 78 | 789 | 3 6 | 2 5 |
| .-------'---------------: :---------------'-------. |
| 79 | 56 56 2 | 79 | 4 38 38 | 1 |
'-------'-----------------------'-------'-----------------------'-------'
By the way, for anyone not familiar with UR-based techniques, you can find more information here:
http://www.sudocue.net/guide.php#UR
However, please bear in mind that all discussions relating to UR on the web are based on regular (vanilla) Sudoku. For Killer Sudoku, there is an important extra requirement, namely:
All (pre-defined) cages (if any) overlapping with the cells of the deadly pattern must contain exactly two UR cells of opposite polarity
Otherwise, you can't (in general) exchange the two digits of the deadly pattern to get an alternate solution, because then the cage constraints (such as the cage sums) will not be met.
UR Type 1 ("Unique Corner")
Refer now to the above candidate grid, in particular the four cells R47C23. All apart from one of the cells contains only the candidates {23}. Only R7C2 has a single extra candidate, 8. In order to avoid the so-called deadly rectangle (where the two participating digits can be freely interchanged in the solution to provide an equally valid alternate solution), R7C2 must take the value of the extra candidate, so we can directly place an 8 there, which creates a hidden single in C1 at R2C1 = 8. After making both of these moves, we end up with the following grid:
Code: Select all
.-------.-----------------------.-------.-----------------------.-------.
| 1234 | 2456 5679 13 | 2567 | 6789 23479 34679 | 36789 |
| '-------.---------------: :---------------.-------' |
| 8 3456 | 79 134 | 2567 | 5679 1234 | 13456 3679 |
:---------------'-------. | | .-------'---------------:
| 1234 1234 579 | 134 | 678 | 6789 | 1237 567 36789 |
| .---------------+-------'-------'-------+---------------. |
| 6 | 23 23 | 57 1 57 | 89 89 | 4 |
:-------'-------.-------'-------. .-------'-------.-------'-------:
| 17 17 | 8 9 | 4 | 2 5 | 36 36 |
:-------.-------'-------.-------' '-------.-------'-------.-------:
| 5 | 9 4 | 68 3 68 | 17 17 | 2 |
| '---------------+-------.-------.-------+---------------' |
| 23 8 23 | 56 | 56 | 1 | 479 479 79 |
:---------------.-------' | | '-------.---------------:
| 479 47 | 1 78 | 789 | 3 6 | 2 5 |
| .-------'---------------: :---------------'-------. |
| 79 | 56 56 2 | 79 | 4 38 38 | 1 |
'-------'-----------------------'-------'-----------------------'-------'
UR Type 4 ("Unique Pair")
Referring to the above candidate grid, in particular the four cells R89C15, we can see that the digit 9 has only two candidate positions in R8, and thus must be located in either of R8C15, the two top corners of the UR. Therefore, neither of these corners may contain the digit 7, because this would form the deadly rectangle on {79} in R89C15. Therefore, we can eliminate candidate 7 from R8C15, giving the following grid:
Code: Select all
.-------.-----------------------.-------.-----------------------.-------.
| 1234 | 2456 5679 13 | 2567 | 6789 23479 34679 | 36789 |
| '-------.---------------: :---------------.-------' |
| 8 3456 | 79 134 | 2567 | 5679 1234 | 13456 3679 |
:---------------'-------. | | .-------'---------------:
| 1234 1234 579 | 134 | 678 | 6789 | 1237 567 36789 |
| .---------------+-------'-------'-------+---------------. |
| 6 | 23 23 | 57 1 57 | 89 89 | 4 |
:-------'-------.-------'-------. .-------'-------.-------'-------:
| 17 17 | 8 9 | 4 | 2 5 | 36 36 |
:-------.-------'-------.-------' '-------.-------'-------.-------:
| 5 | 9 4 | 68 3 68 | 17 17 | 2 |
| '---------------+-------.-------.-------+---------------' |
| 23 8 23 | 56 | 56 | 1 | 479 479 79 |
:---------------.-------' | | '-------.---------------:
| 49 47 | 1 78 | 89 | 3 6 | 2 5 |
| .-------'---------------: :---------------'-------. |
| 79 | 56 56 2 | 79 | 4 38 38 | 1 |
'-------'-----------------------'-------'-----------------------'-------'
"Killer" UR
This is a variant that is particular to Killer Sudoku, and which I've never seen documented anywhere.
Take a look at the 12/3 cage at R1C2 in the above grid. Note that, because of the candidates remaining in R1C23, if R1C2 were 5, this would force R1C3 to 6 (i.e., R1C234 = [561]). Similarly, if R1C2 were 6, this would force R1C3 to 5 (i.e., R1C234 = [651]). Because of the additional links on 5 and 6 between R1C2 and R9C2 and between R9C2 and R9C3, it can be seen that placing either of {56} in R1C2 would force a deadly rectangle on {56} in R19C23. Therefore we can eliminate both 5 and 6 from R1C2. Thereafter, we can remove candidate 6 from R1C3, because no possible permutations now use it (obsolete candidate).
Additionally, this opens up the following moves:
- 1. Hidden single in C3 at R9C3 = 6 -> R9C2 = 5
2. Hidden single in C2 at R2C2 = 6 -> R1C1 = 1 (last digit in cage)
3. Naked single at R1C4 = 3
4. R23C4 = {14} -> R2C3 = 9
Code: Select all
.-------.-----------------------.-------.-----------------------.-------.
| 1 | 24 57 3 | 2567 | 6789 2479 4679 | 6789 |
| '-------.---------------: :---------------.-------' |
| 8 6 | 9 14 | 257 | 57 1234 | 1345 37 |
:---------------'-------. | | .-------'---------------:
| 234 234 57 | 14 | 678 | 6789 | 1237 567 36789 |
| .---------------+-------'-------'-------+---------------. |
| 6 | 23 23 | 57 1 57 | 89 89 | 4 |
:-------'-------.-------'-------. .-------'-------.-------'-------:
| 7 17 | 8 9 | 4 | 2 5 | 36 36 |
:-------.-------'-------.-------' '-------.-------'-------.-------:
| 5 | 9 4 | 68 3 68 | 17 17 | 2 |
| '---------------+-------.-------.-------+---------------' |
| 23 8 23 | 56 | 56 | 1 | 479 479 79 |
:---------------.-------' | | '-------.---------------:
| 49 47 | 1 78 | 89 | 3 6 | 2 5 |
| .-------'---------------: :---------------'-------. |
| 79 | 5 6 2 | 79 | 4 38 38 | 1 |
'-------'-----------------------'-------'-----------------------'-------'
Last but not least, take a look at the 6 cells R3C12, R4C23 and R7C13. These form a swordfish pattern, which, if every cell were to contain the same two candidates ({23} in this case), would be just as deadly as the deadly rectangle. This means that R3C12 cannot be {23}, ruling out the combination {237} for the split 12/3 cage at R3C123 (because neither of the digits {23} are available in the remaining cell R3C3). Since the only other combination for these 3 cells is {345}, this means that we can immediately eliminate the candidate 2 from R3C12 and place a "5" in R3C3.
After applying these UR-based techniques (the last one being optional), the puzzle becomes easily solvable via singles.
The sheer number and type of UR-based moves makes this puzzle for me a case study in uniqueness tests as applied to Killer Sudoku.
Last edited by mhparker on Sun Apr 22, 2007 9:30 pm, edited 2 times in total.
Cheers,
Mike
Mike
Wow - very informative Mike. Amazing to find so many of these moves in the one puzzle. (And yes - I was snoozing in LoL land )
Another example of the Killer UR elimination: which absolutely busts a very difficult puzzle wide open is in my walk-through for Assassin 44V2 step 40a : a move which Para saw as a short-cut to solve that puzzle. Richard and Para also saw the Type 1 UR in Assassin 42V2 step 50.
These techniques are clearly very powerful. Thanks again for the tut.
Ed
Another example of the Killer UR elimination: which absolutely busts a very difficult puzzle wide open is in my walk-through for Assassin 44V2 step 40a : a move which Para saw as a short-cut to solve that puzzle. Richard and Para also saw the Type 1 UR in Assassin 42V2 step 50.
These techniques are clearly very powerful. Thanks again for the tut.
Ed
Thanks for the links, Ed.
In addition to the UR's discussed above and in the other walktroughs, we need to keep a lookout for the Type 2's ("Unique Side"), where the same extra candidate appears on two corners, allowing for elimination of that candidate digit in all common peers.
Type 3's ("Unique Subset") are also conceivably useful, especially in the simplest case where only two extra candidate digits are involved, allowing them to form a killer pair with another peer cell, cage or split cage. Triples and beyond are probably not worth bothering about at first, unless we trip over them in a real example.
The swordfish pattern (abb. US?) is something that's probably quite rare, so I suspect we don't need to look out for it. But I was fascinated to see it for real, so couldn't resist mentioning it! Nevertheless, I should have referred to it as the deadly swordfish instead - sounds much better!
In addition to the UR's discussed above and in the other walktroughs, we need to keep a lookout for the Type 2's ("Unique Side"), where the same extra candidate appears on two corners, allowing for elimination of that candidate digit in all common peers.
Type 3's ("Unique Subset") are also conceivably useful, especially in the simplest case where only two extra candidate digits are involved, allowing them to form a killer pair with another peer cell, cage or split cage. Triples and beyond are probably not worth bothering about at first, unless we trip over them in a real example.
The swordfish pattern (abb. US?) is something that's probably quite rare, so I suspect we don't need to look out for it. But I was fascinated to see it for real, so couldn't resist mentioning it! Nevertheless, I should have referred to it as the deadly swordfish instead - sounds much better!
Cheers,
Mike
Mike
Swordfish has been used on this forum before. I used it in step 15 of my Assassin 41 walkthrough. At the time I didn't know it was called a Swordfish, I just called it a 3 row/column X-wing, but have since edited the walkthrough to give the proper name.
As I wrote during the solution of Assassin 42V2, I'm not keen on UR because it depends of the assumption that a puzzle has a unique solution. I said
Obviously I'm in a minority on this as others are obviously keen on UR. I certainly don't want to stop others using it but I doubt that I will myself. I prefer to show by other techniques that there is only one solution.
As I wrote during the solution of Assassin 42V2, I'm not keen on UR because it depends of the assumption that a puzzle has a unique solution. I said
The tag solution then went forward without using that UR move and successfully solved the puzzle.Andrew wrote:I suppose it's a matter of personal preference whether one uses a UR step. It's something I've never used myself. I don't doubt that Ruud has checked that this puzzle only has one solution. Therefore I have made that move in my diagram to be consistent with the thread. However it would be nice if subsequent moves can be used to show that one can logicially reach the solution without using a UR move.
Obviously I'm in a minority on this as others are obviously keen on UR. I certainly don't want to stop others using it but I doubt that I will myself. I prefer to show by other techniques that there is only one solution.
Hi Andrew,
Just a couple of quick points:
The same goes for cages in Killers where the cages spanning two or more boxes may contain duplicate digits. It's normally implicitly assumed that the cage digits must all be distinct (i.e., conform to the Killer Convention), even if the source of the puzzle doesn't explicitly mention the fact. I see no real difference between depending on this assumption and depending on the fact that a puzzle has a unique solution.
Lastly, your embedded quote suggests (as many people do), that you prefer to use logic rather than UR moves, although I would argue here that UR moves are also very logical and elegant.
Just a couple of quick points:
I meant the Swordfish-like deadly pattern, of course, not to be confused with the Swordfish solving technique. In the case of the deadly pattern, three identical pairs of digits are distributed over distributed across three rows, three columns, three boxes and three cages, such that each row, column, box and cage contains 2 cells belonging to the pattern. For an example, see the leftmost image at the very bottom of Ruud's Sudoku Solving Guide.Swordfish has been used on this forum before. I used it in step 15 of my Assassin 41 walkthrough. At the time I didn't know it was called a Swordfish, I just called it a 3 row/column X-wing, but have since edited the walkthrough to give the proper name.
It seems to be a very safe assumption, though. Indeed, in my archive of several hundred Sudokus (jigsaws, killers and vanilla), there's not a single one that I've seen yet that didn't have a unique solution. I dare say that there are some that exist, but even in the unlikely case that I would come across one, I'd just get annoyed with the puzzle maker for designing what I would then see as an invalid Sudoku, rather than start questioning the usage of UR-based techniques in general.I'm not keen on UR because it depends of the assumption that a puzzle has a unique solution.
The same goes for cages in Killers where the cages spanning two or more boxes may contain duplicate digits. It's normally implicitly assumed that the cage digits must all be distinct (i.e., conform to the Killer Convention), even if the source of the puzzle doesn't explicitly mention the fact. I see no real difference between depending on this assumption and depending on the fact that a puzzle has a unique solution.
Lastly, your embedded quote suggests (as many people do), that you prefer to use logic rather than UR moves, although I would argue here that UR moves are also very logical and elegant.
Cheers,
Mike
Mike
Like PsyMar I've only come across one sudoku that had multiple solutions. It was a vanilla sudoku in a newspaper, only the 3rd one that I did, some months before I came across any sudoku websites. It had 4 solutions which they eventually admitted after initially only giving one solution.
Here is my walkthrough for Assassin 46 Light. Fairly similar to the route that Mike took. The main differences are that I used the innie/outie differences for N1, N3, N7 and N9 earlier and more often and I wasn't quite as rigorous in responding to Ruud's "It does not require detection of conflicting combinations." as you will see in the brief discussion after step 30.
1. R4C23 = {14/23}
2. R4C78 = {89}, locked for R4 and N6
3. R5C12 = {17/26/35}, no 4,8,9
4. R5C34 = {89}, locked for R4
5. R5C67 = {16/25/34}, no 7
6. R5C89 = {27/36/45}, no 1
7. R6C23 = {49/58/67}, no 1,2,3
8. R6C78 = {17/26/35}, no 4
9. R1C678 = {289/379/469/478/568}, no 1
10. 20(3) cage in N7 = {389/479/569/578}, no 1,2
11. R789C5 = 9{58/67}, 9 locked for C5 and N8
12. 10(3) cage in N89 = {127/136/145/235}, no 8,9
13. 8(3) cage in N9 = 1{25/34}, 1 locked for N9
14. 45 rule on R5 1 innie R5C5 = 4, clean-up: no 3 in R5C67, no 5 in R5C89
15. 45 rule on N5 2 innies R5C46 = 11 = [92] (only remaining combination) -> R5C3 = 8, R5C7 = 5, clean-up: no 3,6 in R5C12, no 7 in R5C89, no 5 in R6C23, no 3 in R6C78
15a. R5C89 = {36}, locked for N6, clean-up: no 2 in R6C78
15b. R5C12 = {17}, locked for N4, clean-up: no 4 in R4C23, no 6 in R6C23
15c. R6C23 = {49}, locked for N4 and R6
15d. R4C23 = {23}, locked for R4 and N4
15e. R46C1 = {56}, locked for C1
15f. R6C78 = {17}, locked for R6 and N6
16. R46C9 = [42] (naked singles)
16a. 8(3) cage in N9 = 1{25/34} (step 13), 2,4 only in R8C8 -> no 1,3,5 in R8C8
16b. 1 in N9 locked in R89C9, locked for C9
17. 45 rule on R123 2 outies R4C19 = 10 -> R4C1 = 6 -> R6C1 = 5
18. 45 rule on N1 3 outies R123C4 = 8 = 1{25/34}, 1 locked for C4 and N2
19. 45 rule on N3 3 outies R123C6 = 22 = 9{58/67}
20. 45 rule on N7 3 outies R789C4 = 15
21. 45 rule on N9 3 outies R789C6 = 8 = {134}, locked for C6 and N8
22. 1,4 in C4 locked in R123C4 = {134}, locked for C4 and N2, no 2,5
23. R4C5 = 1, R6C5 = 3 (hidden singles in C5)
23a. 45 rule on C1234 2 innies R46C4 = 13 = [58/76]
23b. 45 rule on C6789 2 innies R46C6 = 13 = [58/76]
24. 45 rule on N1 1 innie R2C3 – 6 = 1 outie R1C4 -> R1C4 = {13}, R2C3 = {79}
25. 45 rule on N3 1 outie R1C6 – 5 = 1 innie R2C7 -> R1C6 = {6789}, R2C7 = {1234}
26. 45 rule on N7 1 outie R9C4 – 1 = 1 innie R8C3 -> R8C3 = {14567}
27. 45 rule on N9 1 innie R8C7 – 2 = 1 outie R9C6 -> R8C7 = {36}, R9C4 = {14}
28. 3 in N8 locked in R78C6 -> no 3 in R8C7
28a. 10(3) cage in N89 = {13}6 -> R78C6 = {13}, R8C7 = 6, R9C6 = 4 (step 27), clean-up: no 7 in R9C4
29. R9C6 = 4 -> R9C78 = 11 = {29/38}, no 5,7
30. 7 in N9 locked in R7C789, locked for R7, R7C789 = 7{49/58}, no 3
At this stage one could use 45 rule on R9 3 innies R9C159 = 17 = {179/368} (cannot be {359} which clashes with R9C78) … but Ruud said “It does not require detection of conflicting combinations.” (I hope that steps 42, 46b and 56a are not considered to be conflicting combinations; I didn’t feel that they were although maybe 56a is) so
31. 45 rule on R89 3 outies R7C456 = 12 = {129/156} (cannot be {138} because 1,3 only in R7C6), no 8 = [291]/{56}1 -> R7C6 = 1, R8C3 = 3
32. 3 in R7 locked in R7C123, locked for N7, R7C123 = 3{28/46}, no 9
33. 20(3) cage in N7 = {479/578} = 7{49/58}, 7 locked for N7, clean-up: no 8 in R9C4
33a. 5 only in R8C2 -> no 8 in R8C2
34. R9C234 = {256} (only remaining combination), locked for R9, clean-up: no 9 in R9C78
35. R9C78 = {38}, locked for R9 and N9 -> R9C9 = 1, R8C9 = 5, R8C8 = 2, clean-up: no 6 in R9C4
35a. No 8 in R8C1 (step 33)
36. R7C789 = {479}, locked for R7 -> no 6 in R7C123 (step 32), no 2 in R7C4 (step 31)
37. R8C3 = 1 (hidden single in R8) -> R9C4 = 2 (step 26)
38. Naked pair {23} in R47C3, locked for C3
39. 8 in C9 locked in R123C9, locked for N3
40. 7 in C3 locked in R123C3, locked for N1
41. R1C678 = {289/379/469/478} (cannot be {568} because no 5,6,8 in R1C7) [2/3/4], no 5
42. 45 rule on N1 3 innies R1C23 + R2C3 = 18 and remembering R2C3 – 6 = R1C4 (step 24)
42a. R1C234 = {129/156/237/345} (cannot be {138} because no 1,3,8 in R1C3, cannot be {147} because R1C23 + R2C3 would then be {477}, cannot be {246} because no 2,4,6 in R1C4), no 1,3,8,9 in R1C2
43. 16(3) cage in N3 = {169/178/349/358/367} (cannot be {457} because 4,5 only in R2C8), no 9 in R2C8
44. 45 rule on R12 3 outies R3C456 = 18 = {189/369/378/459/468} (cannot be {279} because no 2,7,9 in R3C4, cannot be {567} because no 5,6,7 in R3C4), no 2, no 5 in R3C6
45. R3C789 = {159/168/258/267/357} [1/2/3], no 9 in R3C7, no 1,3,9 in R3C8
46. R1C6 – 5 = R2C7 (step 25)
46a. R1C678 = {289/379/469/478} (step 41)
46b. R1C678 cannot be [946] because R1C78 + R2C7 would then be [464] -> no 6 in R1C8
47. R1C678 has 2/3/4 in R1C78 (step 41), R3C789 has 1/2/3 (step 45), R2C7 = {1234} -> 16(3) cage in N3 (step 43) = {169/178/358/367} (cannot be {349}), no 4
48. 4 in N3 locked in R1C78 + R2C7
48a. 45 rule on N3 3 innies R1C78 + R2C7 = 14 = 4{19/37}, no 2 -> no 7 in R1C6 (step 25)
48b. R1C678 = {379/469/478}
49. R3C7 = 2 (hidden single in N3) -> R3C89 = 13 = [58]/{67}, no 3,9
50. 1 in N3 locked in R2C78, locked for R2
51. R3C123 = {147/345} = 4{17/35} (cannot be {138} because no 1,3,8 in R3C3), no 8,9, 4 locked for R3 and N1
52. R2C4 = 4 (hidden single in N2), clean-up: no 9 in R1C6 (step 25)
53. R3C6 = 9 (hidden single in R3)
53a. Naked pair {68} in R16C6, locked for C6
53b. R3C456 (step 44) = [189/369], no 5,7 in R3C5
54. R123C5 = 2{58/67}, 2,5,7 only in R12C5 -> no 6,8 in R12C5
55. 4 in N3 locked in R1C78 = 4{69/78}, no 3
56. R1C78 must contain 7/9 (step 55), R3C89 must contain 7/8 (step 49) -> 16(3) cage in N3 must contain 7/8/9 = {169/358/367} (cannot be {178})
56a. No 7 in R2C8 because R12C9 = {36} would clash with R5C9
57. 8 in N1 locked in 15(3) cage = {168/258}, no 3,9
57a. 5,6 only in R2C2 -> no 2,8 in R2C2
57b. 8 in N1 locked in R12C1, locked for C1
58. R7C2 = 8 (hidden single in R7)
59. Naked pair {56} in R29C2, locked for C2 -> R1C2 = 2, R2C1 = 8, R1C1 = 1, R2C2 = 6 (step 57), R9C23 = [56]
60. Naked pair {34} in R3C12, locked for R3 -> R3C3 = 5 (step 51), R3C4 = 1, R1C4 = 3, R12C3 = [79] (cage sums)
60. R2C5 = 2, R2C6 = 7 (hidden singles in N2)
61. R1C78 = {49} -> R1C6 = 6
and the rest is naked singles
Here is my walkthrough for Assassin 46 Light. Fairly similar to the route that Mike took. The main differences are that I used the innie/outie differences for N1, N3, N7 and N9 earlier and more often and I wasn't quite as rigorous in responding to Ruud's "It does not require detection of conflicting combinations." as you will see in the brief discussion after step 30.
1. R4C23 = {14/23}
2. R4C78 = {89}, locked for R4 and N6
3. R5C12 = {17/26/35}, no 4,8,9
4. R5C34 = {89}, locked for R4
5. R5C67 = {16/25/34}, no 7
6. R5C89 = {27/36/45}, no 1
7. R6C23 = {49/58/67}, no 1,2,3
8. R6C78 = {17/26/35}, no 4
9. R1C678 = {289/379/469/478/568}, no 1
10. 20(3) cage in N7 = {389/479/569/578}, no 1,2
11. R789C5 = 9{58/67}, 9 locked for C5 and N8
12. 10(3) cage in N89 = {127/136/145/235}, no 8,9
13. 8(3) cage in N9 = 1{25/34}, 1 locked for N9
14. 45 rule on R5 1 innie R5C5 = 4, clean-up: no 3 in R5C67, no 5 in R5C89
15. 45 rule on N5 2 innies R5C46 = 11 = [92] (only remaining combination) -> R5C3 = 8, R5C7 = 5, clean-up: no 3,6 in R5C12, no 7 in R5C89, no 5 in R6C23, no 3 in R6C78
15a. R5C89 = {36}, locked for N6, clean-up: no 2 in R6C78
15b. R5C12 = {17}, locked for N4, clean-up: no 4 in R4C23, no 6 in R6C23
15c. R6C23 = {49}, locked for N4 and R6
15d. R4C23 = {23}, locked for R4 and N4
15e. R46C1 = {56}, locked for C1
15f. R6C78 = {17}, locked for R6 and N6
16. R46C9 = [42] (naked singles)
16a. 8(3) cage in N9 = 1{25/34} (step 13), 2,4 only in R8C8 -> no 1,3,5 in R8C8
16b. 1 in N9 locked in R89C9, locked for C9
17. 45 rule on R123 2 outies R4C19 = 10 -> R4C1 = 6 -> R6C1 = 5
18. 45 rule on N1 3 outies R123C4 = 8 = 1{25/34}, 1 locked for C4 and N2
19. 45 rule on N3 3 outies R123C6 = 22 = 9{58/67}
20. 45 rule on N7 3 outies R789C4 = 15
21. 45 rule on N9 3 outies R789C6 = 8 = {134}, locked for C6 and N8
22. 1,4 in C4 locked in R123C4 = {134}, locked for C4 and N2, no 2,5
23. R4C5 = 1, R6C5 = 3 (hidden singles in C5)
23a. 45 rule on C1234 2 innies R46C4 = 13 = [58/76]
23b. 45 rule on C6789 2 innies R46C6 = 13 = [58/76]
24. 45 rule on N1 1 innie R2C3 – 6 = 1 outie R1C4 -> R1C4 = {13}, R2C3 = {79}
25. 45 rule on N3 1 outie R1C6 – 5 = 1 innie R2C7 -> R1C6 = {6789}, R2C7 = {1234}
26. 45 rule on N7 1 outie R9C4 – 1 = 1 innie R8C3 -> R8C3 = {14567}
27. 45 rule on N9 1 innie R8C7 – 2 = 1 outie R9C6 -> R8C7 = {36}, R9C4 = {14}
28. 3 in N8 locked in R78C6 -> no 3 in R8C7
28a. 10(3) cage in N89 = {13}6 -> R78C6 = {13}, R8C7 = 6, R9C6 = 4 (step 27), clean-up: no 7 in R9C4
29. R9C6 = 4 -> R9C78 = 11 = {29/38}, no 5,7
30. 7 in N9 locked in R7C789, locked for R7, R7C789 = 7{49/58}, no 3
At this stage one could use 45 rule on R9 3 innies R9C159 = 17 = {179/368} (cannot be {359} which clashes with R9C78) … but Ruud said “It does not require detection of conflicting combinations.” (I hope that steps 42, 46b and 56a are not considered to be conflicting combinations; I didn’t feel that they were although maybe 56a is) so
31. 45 rule on R89 3 outies R7C456 = 12 = {129/156} (cannot be {138} because 1,3 only in R7C6), no 8 = [291]/{56}1 -> R7C6 = 1, R8C3 = 3
32. 3 in R7 locked in R7C123, locked for N7, R7C123 = 3{28/46}, no 9
33. 20(3) cage in N7 = {479/578} = 7{49/58}, 7 locked for N7, clean-up: no 8 in R9C4
33a. 5 only in R8C2 -> no 8 in R8C2
34. R9C234 = {256} (only remaining combination), locked for R9, clean-up: no 9 in R9C78
35. R9C78 = {38}, locked for R9 and N9 -> R9C9 = 1, R8C9 = 5, R8C8 = 2, clean-up: no 6 in R9C4
35a. No 8 in R8C1 (step 33)
36. R7C789 = {479}, locked for R7 -> no 6 in R7C123 (step 32), no 2 in R7C4 (step 31)
37. R8C3 = 1 (hidden single in R8) -> R9C4 = 2 (step 26)
38. Naked pair {23} in R47C3, locked for C3
39. 8 in C9 locked in R123C9, locked for N3
40. 7 in C3 locked in R123C3, locked for N1
41. R1C678 = {289/379/469/478} (cannot be {568} because no 5,6,8 in R1C7) [2/3/4], no 5
42. 45 rule on N1 3 innies R1C23 + R2C3 = 18 and remembering R2C3 – 6 = R1C4 (step 24)
42a. R1C234 = {129/156/237/345} (cannot be {138} because no 1,3,8 in R1C3, cannot be {147} because R1C23 + R2C3 would then be {477}, cannot be {246} because no 2,4,6 in R1C4), no 1,3,8,9 in R1C2
43. 16(3) cage in N3 = {169/178/349/358/367} (cannot be {457} because 4,5 only in R2C8), no 9 in R2C8
44. 45 rule on R12 3 outies R3C456 = 18 = {189/369/378/459/468} (cannot be {279} because no 2,7,9 in R3C4, cannot be {567} because no 5,6,7 in R3C4), no 2, no 5 in R3C6
45. R3C789 = {159/168/258/267/357} [1/2/3], no 9 in R3C7, no 1,3,9 in R3C8
46. R1C6 – 5 = R2C7 (step 25)
46a. R1C678 = {289/379/469/478} (step 41)
46b. R1C678 cannot be [946] because R1C78 + R2C7 would then be [464] -> no 6 in R1C8
47. R1C678 has 2/3/4 in R1C78 (step 41), R3C789 has 1/2/3 (step 45), R2C7 = {1234} -> 16(3) cage in N3 (step 43) = {169/178/358/367} (cannot be {349}), no 4
48. 4 in N3 locked in R1C78 + R2C7
48a. 45 rule on N3 3 innies R1C78 + R2C7 = 14 = 4{19/37}, no 2 -> no 7 in R1C6 (step 25)
48b. R1C678 = {379/469/478}
49. R3C7 = 2 (hidden single in N3) -> R3C89 = 13 = [58]/{67}, no 3,9
50. 1 in N3 locked in R2C78, locked for R2
51. R3C123 = {147/345} = 4{17/35} (cannot be {138} because no 1,3,8 in R3C3), no 8,9, 4 locked for R3 and N1
52. R2C4 = 4 (hidden single in N2), clean-up: no 9 in R1C6 (step 25)
53. R3C6 = 9 (hidden single in R3)
53a. Naked pair {68} in R16C6, locked for C6
53b. R3C456 (step 44) = [189/369], no 5,7 in R3C5
54. R123C5 = 2{58/67}, 2,5,7 only in R12C5 -> no 6,8 in R12C5
55. 4 in N3 locked in R1C78 = 4{69/78}, no 3
56. R1C78 must contain 7/9 (step 55), R3C89 must contain 7/8 (step 49) -> 16(3) cage in N3 must contain 7/8/9 = {169/358/367} (cannot be {178})
56a. No 7 in R2C8 because R12C9 = {36} would clash with R5C9
57. 8 in N1 locked in 15(3) cage = {168/258}, no 3,9
57a. 5,6 only in R2C2 -> no 2,8 in R2C2
57b. 8 in N1 locked in R12C1, locked for C1
58. R7C2 = 8 (hidden single in R7)
59. Naked pair {56} in R29C2, locked for C2 -> R1C2 = 2, R2C1 = 8, R1C1 = 1, R2C2 = 6 (step 57), R9C23 = [56]
60. Naked pair {34} in R3C12, locked for R3 -> R3C3 = 5 (step 51), R3C4 = 1, R1C4 = 3, R12C3 = [79] (cage sums)
60. R2C5 = 2, R2C6 = 7 (hidden singles in N2)
61. R1C78 = {49} -> R1C6 = 6
and the rest is naked singles