Joined: Wed Apr 16, 2008 1:16 am Posts: 1044 Location: Sydney, Australia
In July Ed wrote:
Would you guys mind putting your numerical estimation and numerical rating of a puzzle into hide or collapse tags? I think it would really help my enjoyment of our forum if I don't see any numerical ratings...This is really important to me.
Indeed it did! But, time for me to have a long break from the Assassin forum. I have not coped well with the non-covered ratings comments on this A194 thread. Which means that all your hiding them till now hasn't helped me calm down. I don't know why I can't just accept the ratings are a difference of interpretation. I can't help getting upset when your ratings don't correlate with my experience of a (harder-than-that-to-me) puzzle. For some reason I expect you guys to anticipate precisely how hard I will find a puzzle! Ridicuolous! Thanks for helping me out the last year.
I won't be able to give you this weeks Assassin after all. No joy. Sorry.
I have not coped well with the non-covered ratings comments on this A194 thread.
I am sorry to hear that, especially as you are possibly referring to my rating comment that I have since deleted.
Ed wrote:
I can't help getting upset when your ratings don't correlate with my experience of a (harder-than-that-to-me) puzzle.
On reflection, maybe the rating I gave should have been one notch (0.25) higher, due to my step 14, and the fact that Andrew and I used a couple of bifurcative steps. But even before this adjustment, I was putting this puzzle on a par with the Assassin 60 RP-Lite and Maverick 1, which both put up quite a big fight. With the adjustment, I'm putting the puzzle on the same level as the Assassin 62 V2, which I eventually managed to solve using several complicated AICs. Anything higher than that would correspond to the Assassin 48 Hevvie and Assassin 74 "Brick Wall", which were both even tougher than this puzzle, requiring ugly tryfurcation and/or very complex permutation analysis. In short, I hope you weren't falsely interpreting my comment as implying that I found this puzzle in any way straightforward.
Ed wrote:
For some reason I expect you guys to anticipate precisely how hard I will find a puzzle! Ridicuolous!
This is only possible for puzzles that don't have a narrow solving path. Otherwise the perceived difficulty is often much higher than the quoted (i.e., nominal) rating because, if one overlooks one or more important moves on the "intended" path, it's sometimes very difficult to find any alternative route.
Ed wrote:
I won't be able to give you this weeks Assassin after all. No joy. Sorry.
Sad. Ed
That's no problem, Ed. I can cope with the loss of a puzzle, as long as I know that you're still around to inspire me.
I'm sure I'm also speaking for the others when I say that I hope you can change your mind and stay with us. In any case, thanks for telling us how you feel.
I was also very sorry that Ed has decided to take a break from Assasins although I'm sure he'll still remain active on the Other Variants forum. Hope to see you back here soon, Ed!
I got distracted by another hard puzzle so didn't get to work through the whole of the "tag" or Afmob's walkthrough until the last couple of days.
I found it interesting that Afmob, Mike and I found three very different ways to make the important placement for R4C6. This also applies for some of the later steps. There were clearly several narrow points in solving paths including placing R4C6, the 16(3) cage in N4 and near the end outies from C123 (or from N14) but apart from that there was scope for three (or maybe more) different solving paths.
After seeing that Mike had used a 45 on N89 to start his remaining steps for the "tag", I took that as a "hint" to make my own attempt to finish the puzzle.
Here is my complete alternative walkthrough:
No Prelims
1. 45 rule on N7 2 innies R79C3 = 6 = {15/24}
2. 45 rule on N9 2 innies R79C7 = 14 = {59/68} 2a. Min R7C7 = 5 -> max R5C8 + R6C7 = 7, no 7,8,9 in R5C8 + R6C7
3. 45 rule on R123 2 outies R4C19 = 14 = {59/68} 3a. Min R4C1 = 5 -> max R3C12 = 9, no 9 in R3C12
4. 45 rule on R1234 2 outies R5C37 = 1 innie R4C5 + 14 4a. Max R5C37 = 17 -> max R4C5 = 3 4b. Min R5C37 = 15, no 1,2,3,4,5 in R5C37
5. 45 rule on R1234 4 outies R5C3467 = 27 = {3789/4689/5679}, no 1,2, 9 locked for R5
6. 45 rule on N4 2(1+1) outies R4C4 + R7C3 = 1 innie R4C1 + 5, IOU no 5 in R7C3, clean-up: no 1 in R9C3 (step 1)
7. 45 rule on R789 2 innies R7C37 = 1 outie R6C5 + 9 7a. Max R7C37 = 13 -> max R6C5 = 4 7b. Min R7C37 = 10 -> min R7C7 = 6, clean-up: no 9 in R9C7 (step 2) 7c. Min R7C7 = 6 -> max R5C8 + R6C7 = 6, no 6 in R5C8 + R6C7
8. 45 rule on N1 3(2+1) outies R12C4 + R4C1 = 14 8a. Min R4C1 = 5 -> max R12C4 = 9, no 9 in R12C4
9. 45 rule on N3 3(2+1) outies R12C6 + R4C9 = 20 9a. Min R12C6 = 11, no 1 in R12C6
10. 45 rule on N5689 2 innies R4C49 = 1 outie R9C3 + 13 10a. Min R9C3 = 2 -> min R4C49 = 15, no 1,2,3,4,5 in R4C49, clean-up: no 9 in R4C1 (step 3) 10b. Max R4C49 = 17 -> max R9C3 = 4, clean-up: no 1 in R7C3 (step 1)
11. Naked pair {24} in R79C3, locked for C3 and N7 11a. Max R5C2 + R7C3 = 12 -> no 1 in R6C3
12. 25(4) cage at R4C6 = {1789/2689/3589/3679/4579/4678} contains one of 1,2,3,4 in R4C678 12a. Hidden killer quad 1,2,3,4 in R4C235 and R4C678 for R4 -> R4C235 must contain three of {1234}, no 5,6,7,8,9 in R4C23
13. 45 rule on R789 4 innies R7C3467 = 23 14a. Max R7C3 = 4 -> min R7C467 = 19, no 1 in R7C46
14. 45 rule on N5 3 innies R4C46 + R6C5 = 14 14a. Min R4C4 + R6C5 = 7 -> max R4C6 = 7
15. 45 rule on N6 4 innies R4C789 + R5C7 = 1 outie R7C7 + 21 15a. R7C7 = {689} -> R4C789 + R5C7 = 27,29,30 = {3789/5679/5789/6789} (cannot be {4689} because R79C7 = [68] when R4C789 + R5C7 = 27 and R4C89 cannot be {68} which clashes with R4C19, CCC), no 1,2,4, 7,9 locked for N6, also 7 locked for 25(4) cage at R4C6, no 7 in R4C6
16. 25(4) cage at R4C6 must contain 7 = {1789/3679/4579/4678}, no 2 16a. 4 of {4579} must be in R4C6 -> no 5 in R4C6 16b. 4 of {4678} must be in R4C6, 6 of {3679} must be R4C78 + R5C7 (R4C78 + R5C7 cannot be {379} => R4C69 = [68] clashes with R4C19, CCC) -> no 6 in R4C6
17. Naked quad {1234} in R4C2356, locked for R4
18. R4C789 + R5C7 (step 15a) = {5679/5789/6789} 18a. 12(3) cage at R5C9 = {138/246/345} (cannot be {156} which clashes with R4C789 + R5C7) 18b. 12(3) cage at R5C8 = {129/138/246} (cannot be {345} because R7C7 only contains 6,8,9, cannot be {156} which clashes with R4C789 + R5C7 which must be {5679} when R7C7 = 6), no 5 in R5C8 + R6C7
19. R7C7 = {689} -> R4C789 + R5C7 (step 15a) = {5679/5789/6789} 19a. For R7C7 = {68} => R4C789 + R5C7 = {5679/5789} => R4C19 = {68} => R7C7 = R4C9 because R4C789 + R5C7 only contains one of 6,8 19b. For R7C7 = 9 => R4C789 + R5C7 = {6789} => R4C1 = 5 (hidden single in R4) => R4C9 = 9 (step 3) 19c. From steps 19a and 19b, R4C9 = R7C7 for all three values in R7C7 19d. 45 rule on N6 2(1+1) outies R4C6 + R7C7 = 1 innie R4C9, R4C9 = R7C7 (from above) -> R4C6 = 4 19e. R4C46 + R6C5 = 14 (step 14), R4C6 = 4 -> R4C4 + R6C5 = 10, no 6 in R4C4
20. 13(3) cage in N5 = {139/157/238/256} 20a. 1,2 of {139/238} must be in R4C5 -> no 3 in R4C5 20b. 3 in R4 only in R4C23, locked for N4
21. R5C3467 (step 5) = {3789/5679}, 7 locked for R5
22. 14(3) cage at R5C2 = {149/248/257} (cannot be {158/167} because R7C3 only contains 2,4), no 6 22a. 7,9 only in R6C3, 8 of {248} must be in R6C3 -> R6C3 = {789}, no 8 in R5C2
23. 18(3) cage in N5 = {189/369/378/567} (cannot be {279} which clashes with 13(3) cage), no 2 23a. 2 in N5 only in R46C5, locked for C5
24. R7C3467 = 23 (step 13) 24a. 14(3) cage at R6C5 = {149/158/167/239/248/257} (cannot be {347} because R7C3467 cannot be [4478]) 24b. 3 of {239} must be in R6C5 (R7C46 cannot be {39} because R7C37 would both then be even so R7C3467 would be even), no 3 in R7C46
26. 12(3) at R5C9 (step 18a) = {138/246/345} 26a. 8 of {138} must be in R6C89 (R6C89 cannot be {13} => R6C5 = 2, R4C4 = 8 (step 19e) => R5C7 = 8 (step 25a) -> no 8 in R5C9
27. 16(3) cage in N4 = {169/259/457} (cannot be {178} which clashes with 14(3) cage at R5C2, cannot be {268} which clashes with 14(3) cage at R5C2 = [482] because 4 in N4 must be in either 16(3) cage or R5C2), no 8 27a. 4 in N4 only in 16(3) cage and R5C2 27b. 16(3) cage = {169/259} => R5C2 = 4 27c. 16(3) cage = {457} => 14(3) cage at R5C2 (step 22) = {149/248} => R7C3 = 4 27d. From steps 27b and 27c R5C2 = 4 or R7C3 = 4 -> 14(3) cage at R5C2 = {149/248}, no 5,7 [Alternatively for steps 27a to 27d 14(3) cage at R5C2 = {149/248} (cannot be {257} because then cannot place 4 in N4 because 16(3) cage requires {39}/{57} to contain 4), no 5,7]
28. 45 rule on C89 3 innies R245C8 = 1 outie R8C7 + 14 28a. Max R245C8 = 21 -> max R8C7 = 7 28b. But R8C7 cannot be 7, here’s how R8C7 = 7 => R4C8 = 7 (hidden single in N6) => max R245C8 = [974] = 20 -> max R8C7 = 6 28c. Min R245C8 = 15, max R45C8 = 13 -> min R2C8 = 2
29. 8 in R5 only in R5C34567 29a. R5C3467 (step 21) = {3789/5679} 29b. R5C3467 = {3789} => 3 locked for R5 R5C3467 = {5679} => R5C5 = 8 -> no 3 in R5C5
30. 18(3) cage in N5 (step 23) = {189/369/378/567} 30a. 1 of {189} must be in R6C46 (R6C46 cannot be {89} which clashes with R6C3) -> no 1 in R5C5 30b. 8 of {189/378} must be in R5C5 -> no 8 in R6C46
31. Killer pair 7,9 in 16(3) cage in N4 and 18(3) cage in N5, locked for R6 -> R6C3 = 8, clean-up: no 6 in R4C9 (step 3), no 1 in R5C2 (step 27d), no 6 in R7C7 (step 19d), no 8 in R9C7 (step 2) [With hindsight this killer pair was available immediately after the first part of step 27 but at the time it seemed natural, the way I work, to continue analysing the cages in N4.]
32. 12(3) cage at R5C8 (step 18b) = {129/138}, no 4, 1 locked for N6
33. Killer pair 5,6 in R4C1 and 16(3) cage, locked for N4
35. 17(3) cage at R1C3 = {179/269/359/368/467} (cannot be {278/458} because 2,4,8 only in R1C4) 35a. 7 of {179} must be in R1C4 (R12C3 cannot be {79} which clashes with R5C3) -> no 1 in R1C4 35b. 2,4,8 of {269/368/467} must be in R1C4 -> no 6 in R1C4 35c. R12C4 + R4C1 = 14 (step 8), min R1C4 + R4C1 = 7 -> max R2C4 = 7
36. 14(3) cage at R3C1 = {158/167/257/356} (cannot be {248/347} because R4C1 only contains 5,6), no 4
37. 18(3) cage at R3C8 = {189/279/369/378/459/468} (cannot be {567} because R3C9 only contains 8,9) 37a. 15(3) cage in N3 must contain at least one of 1,2,3,4 37b. Hidden killer quad for 1,2,3,4 in R123C7 + R2C8, 15(3) cage and 18(3) cage at R3C8 for N3, 15(3) cage and 18(3) cage must contain at least two of 1,2,3,4 -> R123C7 + R2C8 cannot contain more than two of 1,2,3,4 37c. Hidden killer quad for 1,2,3,4 in R123C7, R6C7 and R8C7 for C7, R123C7 cannot contain more than two of 1,2,3,4 -> R6C7 and R8C7 must each contain one of 1,2,3,4 and R123C7 must contain two of R123C7, no 5,6 in R8C7 37d. R123C7 contains two of 1,2,3,4 and R123C7 + R2C8 cannot contain more than two of 1,2,3,4 -> no 2,3,4 in R2C8 37e. 15(3) cage in N3 can only contain one of 1,2,3,4
At this stage I merged my steps in with the “tag” already started by Mike and Ed and added appropriate steps from my partial walkthrough to their steps.
Then Mike finished off the “tag”, continuing after “tag” step 40 (my original step 37) having decided to omit “tag” step 41 (my original step 38).
If I’d spotted the 45 he used for his next step I might have been able to finish the puzzle so I used that knowledge as a "hint".
39. 45 rule on N89 4(1+2+1) outies R5C8 + R6C5 + R6C7 + R9C3 = 9 = [1134/2214/3312] (cannot be [2232] because R5C8+R6C7 must contain 1, other permutations for R5C8 + R6C5 + R6C7 excluded because only 2,4 in R9C3), no 2 in R6C7, 1 locked for R6 39a. From the above permutations R5C8 = R6C5
40. R46C5 = {12} (hidden pair in N5), locked for C5, clean-up: no 7 in R4C4 (step 19e), no 3 in R5C8 (step 39a)
41. 7 in R4 only in R4C78, locked for N6 41a. Naked pair {89} in R4C49, locked for R4 41b. Naked pair {89} in R57C7, locked for C7
42. R5C3 = 7 (hidden single in R5) 42a. 16(3) cage in N4 (step 27) = {169/259}, no 4, 9 locked for R6 42b. 1 of {169} must be in R5C1 -> no 6 in R5C1 42c. 17(3) cage at R1C3 (step 35) = {179/269/359/368} (cannot be {467} because 4,7 only in R1C4), no 4
43. R5C2 = 4 (hidden single in N4), R7C3 = 2, R9C3 = 4 43a. Min R6C5 + R7C6 = 6 -> no 9 in R7C4 [If one wants to omit my step 38, then that elimination can be made now using R7C3467 = 23 (step 13), R7C3 = 2 -> R7C467 = 21 = {489/579/678}, 4 of {489} must be in R7C4, 8,9 of {579/678} must be in R7C7 -> no 8,9 in R7C4.]
44. 21(5) cage at R8C5 = {12459/12468/13458/13467/23457} 44a. R9C7 = {56} -> no 5,6 in R8C5 + R9C46 44b. 9 of {12459} must be in R8C5 -> no 9 in R9C46
45. 4 in N1 only in 15(3) cage = {249/348/456}, no 1,7
46. R245C8 = R8C7 + 14 (step 28) 46a. R8C7 = {1234} -> min R245C8 = 15 46b. Max R45C8 = 9 -> min R2C8 = 6 46c. R245C8 cannot be {67}2 (because R5C8 is only place for 1 in N6 when R8C8 = 1) -> no 6,7 in R2C8
47. 18(3) cage at R2C6 = {189/279/369/378/459/468} (cannot be {567} because R2C8 only contains 8,9) 47a. 4 of {459} must be in R3C7 -> no 5 in R3C7
48. 1 in N2 only in R2C4 + R3C46, CPE no 1 in R3C3 [Just spotted this, it’s been there since step 35a.]
49. 14(3) cage at R1C6 = {149/158/167/239/248/257/347/356} 49a. 8,9 of {239/248} must be in R1C6, 2 of {257} must be in R12C7 (R12C7 cannot be {57} which clashes with R49C7, ALS block), no 2 in R1C6
50. 7 in N5 only 18(3) cage (step 30) = {378/567} 50a. Cannot be {378}, here’s how 18(3) cage = {378} => R5C5 = 8, R6C46 = {37}, R6C7 =1, R6C5 = 2, R4C4 = 8 (step 19e) clashes with R5C5 [Alternatively R4C4 + R6C5 = [82/91] -> R4C4 + R6C57 = [821/913] 7 in N5 only 18(3) cage (step 30) = {567} (cannot be {378} = 8{37} which clashes with R4C4 + R6C57)] 50b. 18(3) cage = {567}, locked for N5 50c. 3 in N5 only in R5C46, locked for R5
51. 45 rule on C123 3 outies R124C4 = 17 = {179/269/278/359/368/458} (cannot be {467} because R4C4 only contains 8,9) 51a. R4C4 = {89} -> no 8 in R1C4 [Another step I ought to have spotted earlier although, if I had, I might not have found the interesting step 50.]
52. 17(3) cage at R1C3 (step 42c) = {179/269/359}, 9 locked for C1 and N1, clean-up: no 2 in 15(3) cage in N1 (step 45)
53. 13(3) cage at R2C2 = {157/238/256/346} (cannot be {148/247} because R3C3 only contains 3,5,6) 53a. 3 of {238} must be in R3C3, 4 of {346} must be in R2C4 -> no 3 in R2C4
54. Hidden killer pair 2,7 in R2C2 and R3C12 for N1, R2C2 can only contain one of 2,7 -> R3C12 must contain at least one of 2,7 54a. 14(3) cage at R3C1 (step 36) = {167/257} (cannot be {158/356} which don’t contain 2 or 7), no 3,8, 7 locked for R3 and N1 54b. R4C1 = {56} -> no 5,6 in R3C12
55. 18(3) cage at R3C8 (step 37) = {189/369/459/468}, no 2
56. 15(3) cage in N1 (step 45) = {348/456}, R3C3 = {356} -> 15(3) cage + R3C3 must contain 3, locked for N1 56a. 17(3) cage at R1C3 (step 52) = {179/269/359} 56b. 2,3,7 only in R1C4 -> R1C4 = {237}
57. 13(3) cage at R2C2 (step 53) = {157/238/256/346} 57a. 7 of {157} must be in R2C4 -> no 1 in R2C4
58. 1 in N2 only in R3C46, locked for R3 -> R3C12 = {27}, locked for R3 and N1, R4C1 = 5 (step 54a), R4C9 = 9 (step 3), R4C4 = 8, R5C7 = 8, R7C7 = 9, R9C7 = 5 (step 2), clean-up: no 2 in 16(3) cage in N4 (step 42a)
60. Naked pair {69} in R6C12, locked for R6 60a. Naked pair {57} in R6C46, locked for R6 and N5 -> R5C5 = 6, R5C9 = 5
61. 45 rule on R789 2 remaining innies R7C46 = 12 = [48/57/75], no 6 [At this stage there’s the interesting R7C46 = [48] (cannot be {57} because R67C46 would form UR) but I don’t use URs; anyway it looks like this puzzle is almost finished without using it. It is, of course, still possible that a later placement of a 5 or 7 in C4 or C6 will happen. Still that would also be applicable for an UR in a plain vanilla sudoku, which is why I’m assuming that this would be an UR.]
62. 6 in N8 only in R8C46, locked for R8 62a. 6 in C3 only in R123C3, locked for N1, clean-up: no 5 in 15(3) cage in N1 (step 56) 62b. Naked triple {348} in 15(3) cage, locked for N1
63. 13(3) cage at R2C2 (step 53) = {157/256} 63a. 2,7 only in R2C4 -> R2C4 = {27} 63b. 13(3) cage at R2C2 = {157/256}, 5 locked for N1
64. 45 rule on N1 2 remaining outies R12C4 = 9 = {27}, locked for C4 and N2 -> R6C46 = [57], R7C4 = 4, R7C6 = 8 (step 61)
65. 45 rule on N3 2 remaining outies R12C6 = 11 = {56}, locked for C6 and N2, CPE no 6 in R2C7
68. 14(3) cage at R1C6 (step 49) = {257/356} (cannot be {347} because R1C6 only contains 5,6) -> R1C6 = 5, R2C6 = 6, R12C7 = [27/63/72], no 4, no 3 in R1C7
69. 15(3) cage in N3 can only contain one of 1,2,3,4 (step 37e) = {168) (only remaining combination, cannot be {249/348} which contain two of 1,2,3,4, cannot be {267} which clashes with 14(3) cage at R1C6), locked for N3 -> R2C8 = 9, R3C7 = 3 (step 47)
70. Naked pair {27} in R12C7, locked for C7 -> R4C78 = [67], R8C7 = 4
71. 15(4) cage in N9 = {1347} (only remaining combination) -> R7C9 = 7, R78C8 = {13}, locked for C8 and N9 71a. Naked pair {68} in R9C89, locked for R9 and N9 -> R8C9 = 2
Here are my remaining steps, from the marks pic in my earlier post (if you prefer you can ignore my original "tag" step 41, as Mike did, and start with R7C4 = {2456789}), numbering them from 41 onward as Mike did and amending the step references to be consistent (I hope) with the "tag" steps.
41. 45 rule on N89 4(1+2+1) outies R5C8 + R6C5 + R6C7 + R9C3 = 9 = [1134/2214/3312] (cannot be [2232] because R5C8+R6C7 must contain 1, other permutations for R5C8 + R6C5 + R6C7 excluded because only 2,4 in R9C3), no 2 in R6C7, 1 locked for R6 41a. From the above permutations R5C8 = R6C5
42. R46C5 = {12} (hidden pair in N5), locked for C5, clean-up: no 7 in R4C4 (step 15), no 3 in R5C8 (step 41a)
43. 7 in R4 only in R4C78, locked for N6 43a. Naked pair {89} in R4C49, locked for R4 43b. Naked pair {89} in R57C7, locked for C7
44. R5C3 = 7 (hidden single in R5) 44a. 16(3) cage in N4 (step 30) = {169/259}, no 4, 9 locked for R6 44b. 1 of {169} must be in R5C1 -> no 6 in R5C1 44c. 17(3) cage at R1C3 (step 38) = {179/269/359/368} (cannot be {467} because 4,7 only in R1C4), no 4
45. R5C2 = 4 (hidden single in N4), R7C3 = 2, R9C3 = 4 45a. Min R6C5 + R7C6 = 6 -> no 9 in R7C4 [If one wants to omit my original "tag" step 41, then that elimination can be made now using R7C3467 = 23 (step 25), R7C3 = 2 -> R7C467 = 21 = {489/579/678}, 4 of {489} must be in R7C4, 8,9 of {579/678} must be in R7C7 -> no 8,9 in R7C4.]
46. 21(5) cage at R8C5 = {12459/12468/13458/13467/23457} 46a. R9C7 = {56} -> no 5,6 in R8C5 + R9C46 46b. 9 of {12459} must be in R8C5 -> no 9 in R9C46
47. 4 in N1 only in 15(3) cage = {249/348/456}, no 1,7
48. R245C8 = R8C7 + 14 (step 31) 48a. R8C7 = {1234} -> min R245C8 = 15 48b. Max R45C8 = 9 -> min R2C8 = 6 48c. R245C8 cannot be {67}2 (because R5C8 is only place for 1 in N6 when R8C8 = 1) -> no 6,7 in R2C8
49. 18(3) cage at R2C6 = {189/279/369/378/459/468} (cannot be {567} because R2C8 only contains 8,9) 49a. 4 of {459} must be in R3C7 -> no 5 in R3C7
50. 1 in N2 only in R2C4 + R3C46, CPE no 1 in R3C3 [Just spotted this, it’s been there since step 38a.]
51. 14(3) cage at R1C6 = {149/158/167/239/248/257/347/356} 51a. 8,9 of {239/248} must be in R1C6, 2 of {257} must be in R12C7 (R12C7 cannot be {57} which clashes with R49C7, ALS block), no 2 in R1C6
52. 7 in N5 only 18(3) cage (step 33) = {378/567} 52a. Cannot be {378}, here’s how 18(3) cage = {378} => R5C5 = 8, R6C46 = {37}, R6C7 =1, R6C5 = 2, R4C4 = 8 (step 15) clashes with R5C5 [Alternatively R4C4 + R6C5 = [82/91] -> R4C4 + R6C57 = [821/913] 7 in N5 only 18(3) cage (step 33) = {567} (cannot be {378} = 8{37} which clashes with R4C4 + R6C57)] 52b. 18(3) cage = {567}, locked for N5 52c. 3 in N5 only in R5C46, locked for R5
53. 45 rule on C123 3 outies R124C4 = 17 = {179/269/278/359/368/458} (cannot be {467} because R4C4 only contains 8,9) 53a. R4C4 = {89} -> no 8 in R1C4 [Another step I ought to have spotted earlier although, if I had, I might not have found the interesting step 52.]
54. 17(3) cage at R1C3 (step 44c) = {179/269/359}, 9 locked for C1 and N1, clean-up: no 2 in 15(3) cage in N1 (step 47)
55. 13(3) cage at R2C2 = {157/238/256/346} (cannot be {148/247} because R3C3 only contains 3,5,6) 55a. 3 of {238} must be in R3C3, 4 of {346} must be in R2C4 -> no 3 in R2C4
56. Hidden killer pair 2,7 in R2C2 and R3C12 for N1, R2C2 can only contain one of 2,7 -> R3C12 must contain at least one of 2,7 56a. 14(3) cage at R3C1 (step 39) = {167/257} (cannot be {158/356} which don’t contain 2 or 7), no 3,8, 7 locked for R3 and N1 56b. R4C1 = {56} -> no 5,6 in R3C12
57. 18(3) cage at R3C8 (step 40) = {189/369/459/468}, no 2
58. 15(3) cage in N1 (step 47) = {348/456}, R3C3 = {356} -> 15(3) cage + R3C3 must contain 3, locked for N1 58a. 17(3) cage at R1C3 (step 54) = {179/269/359} 58b. 2,3,7 only in R1C4 -> R1C4 = {237}
59. 13(3) cage at R2C2 (step 55) = {157/238/256/346} 59a. 7 of {157} must be in R2C4 -> no 1 in R2C4
60. 1 in N2 only in R3C46, locked for R3 -> R3C12 = {27}, locked for R3 and N1, R4C1 = 5 (step 56a), R4C9 = 9 (step 1), R4C4 = 8, R5C7 = 8, R7C7 = 9, R9C7 = 5 (step 3), clean-up: no 2 in 16(3) cage in N4 (step 44a)
62. Naked pair {69} in R6C12, locked for R6 62a. Naked pair {57} in R6C46, locked for R6 and N5 -> R5C5 = 6, R5C9 = 5
63. 45 rule on R789 2 remaining innies R7C46 = 12 = [48/57/75], no 6 [At this stage there’s the interesting R7C46 = [48] (cannot be {57} because R67C46 would form UR) but I don’t use URs; anyway it looks like this puzzle is almost finished without using it. It is, of course, still possible that a later placement of a 5 or 7 in C4 or C6 will happen. Still that would also be applicable for an UR in a plain vanilla sudoku, which is why I’m assuming that this would be an UR.]
64. 6 in N8 only in R8C46, locked for R8 64a. 6 in C3 only in R123C3, locked for N1, clean-up: no 5 in 15(3) cage in N1 (step 58) 64b. Naked triple {348} in 15(3) cage, locked for N1
65. 13(3) cage at R2C2 (step 55) = {157/256} 65a. 2,7 only in R2C4 -> R2C4 = {27} 65b. 13(3) cage at R2C2 = {157/256}, 5 locked for N1
66. 45 rule on N1 2 remaining outies R12C4 = 9 = {27}, locked for C4 and N2 -> R6C46 = [57], R7C4 = 4, R7C6 = 8 (step 63)
67. 45 rule on N3 2 remaining outies R12C6 = 11 = {56}, locked for C6 and N2, CPE no 6 in R2C7
70. 14(3) cage at R1C6 (step 51) = {257/356} (cannot be {347} because R1C6 only contains 5,6) -> R1C6 = 5, R2C6 = 6, R12C7 = [27/63/72], no 4, no 3 in R1C7
71. 15(3) cage in N3 can only contain one of 1,2,3,4 (step 40e) = {168) (only remaining combination, cannot be {249/348} which contain two of 1,2,3,4, cannot be {267} which clashes with 14(3) cage at R1C6), locked for N3 -> R2C8 = 9, R3C7 = 3 (step 49)
72. Naked pair {27} in R12C7, locked for C7 -> R4C78 = [67], R8C7 = 4
73. 15(4) cage in N9 = {1347} (only remaining combination) -> R7C9 = 7, R78C8 = {13}, locked for C8 and N9 73a. Naked pair {68} in R9C89, locked for R9 and N9 -> R8C9 = 2
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