Thanks Afmob for a challenging puzzle. It took me quite a long time and I found it harder than the estimated rating.
I'll rate A129 at least 1.25 the way I solved it, mainly because of step 20.
Here is my walkthrough.
Prelims
a) 8(2) diagonal cage at R5C4 = {17/26/35}, no 4,8,9 b) 12(2) diagonal cage at R5C6 = {39/48/57}, no 1,2,6 c) R89C5 = {79}, locked for C5 and N8 d) R1C123 = {489/579/678}, no 1,2,3 e) R123C5 = {126/135/234}, no 7,8,9 f) R1C789 = {127/136/145/235}, no 8,9 g) R4C456 = {389/479/569/578}, no 1,2 h) 21(3) diagonal cage at R5C2 = {489/579/678}, no 1,2,3 i) 13(4) diagonal cage at R5C1 = {1237/1246/1345}, no 8,9, CPE no 1 in R789C1 j) 14(4) cage at R8C3 = {1238/1247/1256/1346/2345}, no 9
1. 45 rule on N1 3 innies R23C1 + R3C3 = 7 = {124}, locked for N1 1a. R1C123 = {579/678}, 7 locked for R1 and N1 1b. R1C789 = {136/145/235} 1c. Killer pair 5,6 in R1C123 and R1C789, locked for R1
2. 45 rule on N1 1 outie R4C1 = 1 innie R3C3 + 5, R4C1 = {679} 2a. 3 in C1 locked in R56789C1, CPE no 3 in R78C2
4. 45 rule on R123 4 outies R4C1379 = 13 = {1237/1246} (cannot be {1345} because no 1,3,4,5 in R4C1), no 5,8,9, 1,2 locked for R4, clean-up: no 4 in R3C3 (step 2) 4a. R4C1 = {67} -> no 6,7 in R4C37 4b. 4 in N1 locked in R23C1, locked for C1 4c. 1,2 in C2 locked in R789C2, locked for N7 4d. Min R89C3 = 7 -> max R89C4 = 7, no 8 in R89C4 4e. Min R789C1 = 14 -> max R9C2 = 8
5. 45 rule on R1234 2 innies R4C28 = 12 = {39/48/57}, no 6 5a. R4C456 = {389/569/578} (cannot be {479} which clashes with R4C28), no 4
6. Hidden killer pair 1,2 in R23C1 and R56C1 for C1 -> R56C1 must contain one of 1,2 6a. R56C1 contains one of 1,2 -> R78C2 cannot contain more than one of 1,2 6b. Hidden killer pair 1,2 in R78C2 and R9C2 for C2 -> R9C2 = {12}, R78C2 must contain one of 1,2 6c. 13(4) diagonal cage at R5C1 must contain both of 1,2 = {1237/1246}, no 5 6d. 3 of {1237} must be in R56C1 -> no 7 in R56C1 (because R56C1 must contain one of 1,2) 6e. 4 of {1246} must be in R78C2 -> no 6 in R78C2 (because R78C2 must contain one of 1,2)
7. Max R34C3 = 6 -> min R123C4 = 21 -> R123C4 = {579/678/589/679/689/789} (cannot be {489} when [24] in R34C3), no 1,2,3,4 7a. Killer pair 8,9 in R1C123 and R1C4, locked for R1
8. 45 rule on C1234 2 innies R47C4 = 8 = [35/53/62/71], no 8,9 in R4C4, no 4,6,8 in R7C4
9. 45 rule on C6789 2 innies R47C6 = 13 = [58/76/85/94], no 3,6 in R4C6, no 1,2,3 in R7C6
10. 45 rule on C5 2 outies R7C46 = 1 innie R4C5 + 1, min R7C46 = 5 -> no 3 in R4C5
11. Hidden killer triple 1,2,3 in R123C5 and R567C5 for C5, R123C5 must contain two of 1,2,3 -> R567C5 must contain one of 1,2,3 11a. 21(5) cage at R5C5 = {12468/13458} 11b. R567C5 contains one of {123} -> R7C4 = {123}, clean-up: no 3 in R4C4 (step 8) 11c. R4C456 (step 4a) = {569/578}, 5 locked for R4 and N5, clean-up: no 7 in R4C28 (step 5), no 3 in R6C3, no 7 in R6C7
12. Hidden killer triple 1,2,3 in R123C5 and R123C6 for N2, R123C5 must contain two of 1,2,3 -> R123C6 must contain one of 1,2,3 12a. 22(5) cage at R1C6 must contain two of 1,2,3 -> R4C7 = {123}, 22(5) cage at R1C6 cannot be {12379}
13. 45 rule on C12 3 outies R127C3 = 1 innie R4C2 + 16 13a. Max R127C3 = 24 -> max R4C2 = 8, clean-up: no 3 in R4C8 (step 5) 13b. Min R4C8 = 4 -> max R5C7 + R6C6 = 9, no 9 in R5C7 + R6C6
14. 17(3) diagonal cage at R4C2 = {278/359/368/458/467} (cannot be {179/269} because R4C2 only contains 3,4,8), no 1 14a. 5 of {359} must be in R5C3 -> no 9 in R5C3 14b. 5 of {458} must be in R5C3, 4 of {467} must be in R4C2 -> no 4 in R5C3
15. 9 in N4 locked in R56C2, locked for C2 and 21(3) diagonal cage at R5C2, no 9 in R7C3 15a. 21(3) diagonal cage at R5C2 = {489/579}, no 6 15b. 6 in C2 locked in R123C2, locked for N1 15c. 9 in N7 locked in R789C1, locked for C1
16. R1C123 (step 1a) = {579/678} 16a. 9 of {579} must be in R1C3 -> no 5 in R1C3 16b. 17(3) cage in N1 = {359/368} 16c. 9 of {359} must be in R2C3 -> no 5 in R2C3
17. 13(3) diagonal cage at R4C8 = {139/148/238/247/346} (cannot be {157/256} because R4C8 only contains 4,8,9), no 5
18. 45 rule on C123 5 innies R34689C3 = 1 outie R6C4 + 12 18a. Min R34689C3 = 15 -> min R6C4 = 3
19. 22(4) cage in N7 = {1579/2389/2569} 19a. Hidden killer pair 5,8 in R1C1 and R789C1 for C1 -> R1C1 = {58} 19b. R1C123 (step 1a) = {579/678} 19c. R1C1 = {58} -> no 5,8 in R1C23
20. R789C1 “see” all cells of 13(4) cage at R5C1, 13(4) cage (step 6c) = {1237/1246} 20a. 3 in C1 locked in R56C1 + R789C1 -> if R789C1 contains 7 it must also contain 3 20b. 22(4) cage in N7 = {2389/2569} (cannot be {1579}), no 1,7 20c. R9C2 = 2 20d. 1 in C2 locked in R78C2, locked for 13(4) cage at R5C1, no 1 in R56C1 [Steps 20a and 20b can either be considered as a fairly difficult CPE or as an implied chain.]
21. R4C1 = 7 (hidden single in C1), R23C1 = {14} (hidden pair in C1), locked for N1 -> R3C3 = 2, clean-up: no 8 in R4C56 (step 11c), no 1,6 in R5C4 21a. Naked pair {56} in R4C45, locked for R4 and N5 -> R4C6 = 9, clean-up: no 3 in R4C2 (step 5), no 3 in R6C7 21b. Naked pair {48} in R4C28, locked for R4, clean-up: no 9 in R3C7 (step 3) 21c. 2 in R4 locked in R4C79, locked for N6 21d. 21(3) diagonal cage at R5C2 (step 15a) = {489/579} 21e. 7 of {579} must be in R7C3 -> no 5 in R7C3
22. 3 in C2 locked in R23C2, locked for N1 22a. 17(3) cage in N1 (step 16b) = {359/368} 22b. R2C3 = {89} -> no 8 in R23C2 22c. 8 in C2 locked in R456C2, locked for N4 22d. 4 in N4 locked locked in R456C2, locked for C1
23. Naked pair {17} in R78C2, locked for C2 and N7 -> R1C2 = 6, R1C13 = [87] (step 1a), R1C4 = 9, R2C3 = 9, clean-up: no 3 in R789C1 (step 20b) 23a. Naked triple {569} in R789C1, locked for C1 and N7 23b. Naked pair {23} in R56C1, locked for N4 -> R4C3 = 1, clean-up: no 7 in R5C4 23c. Naked pair {35} in R23C2, locked for C2 23d. Naked pair {23} in R5C14, locked for R5, clean-up: no 9 in R6C7 23e. Naked pair {23} in R4C79, locked for N6
24. 27(5) cage at R1C4 = {12789} (only remaining combination) -> R23C4 = {78}, locked for C4 and N2
25. 17(3) diagonal cage at R4C2 (step 14) = {368/458} -> R4C2 = 8, R4C8 = 4, clean-up: no 8 in R5C6
26. 3 in N7 locked R89C3, locked for 14(4) cage at R8C3, no 3 in R89C4 26a. 14(4) cage at R8C3 = {1238/1346/2345} 26b. Killer triple 1,2,3 in R5C4, R7C4 and R89C4, locked for C4 -> R6C4 = 4, R5C3 = 5 (step 25), R6C3 = 6, R5C4 = 2, R5C6 = 7, R6C7 = 5, R56C1 = [32], R56C2 = [49], R7C3 = 8 26c. 14(4) cage at R8C3 = {1346} (only remaining combination) -> R89C4 = {16}, locked for C4 and N8 -> R4C45 = [56], R7C4 = 3 26d. Naked pair {18} in R56C5, locked for C5 and N5 -> R6C6 = 3, R5C7 = 6 (step 17)
Alright, here is V2 or maybe more appropiate V1.5 depending on your solving path. I hope you have as much fun solving this Killer as I did since it offers some really nice and elegant (chainless) moves.
Hint: Remember: Even your odds! Though in this case it might help to even the evens.
As suggested by Andrew, here is how I cracked A129 using only (Naked and Hidden) Killer pairs.
A129 Walkthrough snippet: 1. R1234 a) Innies N1 = 7(3) = {124} locked for N1 b) 21(3) = 7{59/68} -> 7 locked for R1+N1 c) 12(3) = {129/147/246} because R23C1 = (124) -> R4C1 = (679) d) Outies R123 = 13(4) = 12{37/46} because R4C1 >= 6 -> 1,2 locked for R4; R4C379 <> 6,7 e) Innies R1234 = 12(2) <> 6 f) 12(3) = 4{17/26} -> 4 locked for C1+N1 g) Innies+Outies N3: -5 = R4C9 - R3C7 -> R3C7 = (6789)
2. C123! a) 1,2 locked in R789C2 for N7 b) R78C2 <> 3 since it sees all 3 of C1 c) 13(4): R78C2 <> 6 because {1246} blocked by R23C1 = (124) d) Hidden Killer pair (12) in R56C1 @ 13(4) for C1 since 12(3) <> 9 e) ! Hidden Killer pair (12) in R9C2 for N8 since R78C2 @ 13(4) cannot be {12} -> R9C2 = (12) -> 22(4) = {1579/1678/2389/2569/2578} f) From step 2e und 2f -> 13(4) must have 1 and 2 -> 13(4) = 12{37/46} g) ! 13(4): R78C2 <> 2 because (27) is a Killer pair of 22(4) and 13(4) = {16}{24} blocked by Killer pair (16) of 12(3) h) Hidden Single: R9C2 = 2 @ N7
All 5 cells of 26/5 @ r5c5 can see 15/2 @ r8c5={69|78} Also 26/5 @ r5c5 must include at least 2 of {6789} => 2 cells of 26/5 must be {69|78}=15 => the 3 remaining cells must be from {12345}=26-15=11={245} Now r7c4 is from {245} & r7c6 is from {689} => r567c5 must have 2 cells from {245} & 1 cell from {679} => at least one of r56c5 must be from {245} => 11/3 @ r4c4 can't be {245} => r4c6 can't be 5
Innies @ c6789: r47c6=14={68} (NP @ c6, PP @ r56c5+r7c3) => r7c6 & 15/2 @ r8c5 form KNP {68} @ n8 Innies @ c5: r4567c5=17 from {1234579} must have 1|3 => r4c5 must be from {13} => r4c5+r5c4={13} (NP @ n5) Innies @ c1234: r47c4=6={24} (NP @ c4, PP @ r56c5+r7c7) => r56c5={57|59} (5 @ c5,n5 locked) => r56c5 & 15/2 @ r8c5 form KNP {79} @ c5 => r7c45={24} (NP @ r7,n8)
HP @ n5: /46={68} (NP @ d/) 12/2 @ r5c6=[48|75|93] Innie-outies @ c12: r4c2=r127c3-18 can't exceed 24-18=6 Innies @ r1234: r4c28=12=[39|48|57] Innie-outies @ c12: r127c3=r4c2+18 => r127c3 from 21..23 must be from {456789} 15/3 @ r4c2=[348|456|528|546] => r5c3 must be from {245}
Critical Step: Innie-outies @ r1234: r5c3+r6c4=r4c8+3 But r5c3 can't be 3 => r4c8 & r6c4 can't be equal => r4c68+r6c4 can't be [688] => r4c8 can't be 8 => Innies @ r1234: r4c28=[39|57] => 15/3 @ r4c2=[348|528|546] blocking 12/2 @ r5c6 to be [48] => 12/2 @ r5c6=[75|93]
HP @ n5: \46={24} (NP @ d\) Innie-outies @ n3: r3c7=r4c9+4 => r3c7 from {579}, r4c9 from {135} => r4c259={135} (NT @ r4) => /357={579} (NT @ d/, PT @ \37) Innie-outies @ n1: r4c1=r3c3+4 => r3c3=3, r4c1=7
I used an odd-even move instead of a (very) small chain. I am not sure about the rating for this particular move (step 5b), so I'd like to hear your thoughts about it.
A129 V2 Walkthrough: 1. C456 ! a) ! 26(5) = 245{69/78} since other combos blocked by Killer pairs (68,79,89) of 15(2) b) Innies C1234 = 6(2) = [15/24/42] c) Innies C6789 = 14(2) = [59/68/86] d) 26(5) must have 2,4,5 -> at least one of them must be @ R56C5 e) ! 11(3) = {128/146/236} <> 5,7 because R4C6 = (568) and {245} blocked by Killer triple (245) of R56C5 f) Innies C6789 = 14(2) = {68} locked for C6; CPE: R7C3+R56C5 <> 6,8 g) Killer pair (79) locked in 15(2) + 26(5) for C5 h) Hidden Killer pair (13) in 13(3) + R4C5 for C5 since 13(3) <> 9 -> R4C5 = (13) and 13(3) = {148/238/346} <> 5 i) 11(3): R4C4 <> 1 because (24) only possible there j) Innies C1234 = {24} locked for C4; CPE: R7C7+R56C5 <> 2,4
2. C456 a) 26(5) must have 2,4 -> R7C45 = {24} locked for R7+N8 b) Hidden pair (68) in R4C6+R6C4 @ N5 locked for D/; R6C4 = (68); CPE: R4C2 <> 6,8 c) 2 locked in R4C4+R6C6 @ N5 for D\; CPE: R4C8 <> 2 d) Naked pair (13) locked in R4C5+R5C4 for N5 e) 26(5) must have 5 -> 5 locked for N5 f) 12(2): R6C7 = (358) g) 15(3) <> 9 since R6C4 >= 6
4. C123 + D/ a) Innies+Outies C12: 18 = R127C3 - R4C2; R4C2 >= 3 -> R127C3 <> 1,2,3 and R4C2 <> 7 b) Naked triple (579) locked in R3C7+R5C5+R7C3 for D/; CPE: R3C3+R7C7 <> 5,7,9 c) Innies+Outies N1: 4 = R4C1 - R3C3 -> R4C1 <> 9
5. R1234 ! a) Innies R1234 = 12(2): R4C8 <> 5 b) ! 27(5) @ R1C6 must have an even number of even candidates and R123C6 cannot have 2 and 4 since it's a Killer pair of 13(3) @ N2 -> R4C7 must have even candidates -> R4C7 = (2468) c) Outies R123 = 22(4): R4C3 <> 1 since R4C179 <= 20 d) 17(3) @ R4C8: R5C7 <> 9 because R4C8+R6C6 >= 9
6. C789 a) 16(4) @ R8C6: R89C7 <> 9 since R89C6 <> 2,4 b) 9 locked in R13C7 @ C7 for N3 c) 17(3) @ N3 = 8{36/45} because R4C9 = (135) -> 8 locked for C9+N3; R23C9 <> 5 d) Hidden Single: R4C5 = 1 @ R4 e) R5C4 = 3, R6C3 = 1 f) 8 locked in Innies N3 = 8{49/67}
Joined: Wed Apr 23, 2008 5:29 am Posts: 302 Location: Sydney, Australia
Afmob wrote:
I used an odd-even move instead of a (very) small chain. I am not sure about the rating for this particular move (step 5b), so I'd like to hear your thoughts about it.
I love the logic of the technique itself, it's very nice and elegant. If you've seen me working with BossaNova or KenKen puzzles before you'd know I'm a big digger of parity (odd-even) tricks.
However, the application of your move for this particular puzzle which requires some cage blocking action from an adjacent cage reduces the elegancy a little. As a matter of fact I can't see your move being any more elegant (i.e. non-chain-like) than my critical move in my walkthrough. Also you didn't specifically eliminate a minor (obvious) combo in that move. Just for the sake of it this is how I'd write out for that step:
Hidden Text:
27/5 @ r1c6 with r123c6 from {1234579} & r3c7 from {579}: 27/5 @ r1c6 can't be {13579}=25, so must have at least two even digits. (It can't have only one even digit since the sum would then be even not odd.) r123c6 can't have both {24} since 13/3 @ r1c5 must have at least one of {24}. Thus r123c6+r3c7 can have at most one even digit. Hence r4c7 must be even.
Also, have to say your step 7b also doesn't feel any more elegant than my critical step (i.e. I can't see it as any less "chainy" than any of my moves).
Perhaps I can rewrite my critical step using a bit of odd-even analysis:
Hidden Text:
Critical Step: Innie-outies @ r1234: r5c3+r6c4=r4c8+3 But r5c3 can't be 3 => r4c8 & r6c4 can't be equal i.e. can't be [88] Also, r4c6 from {68} blocks r4c8+r6c4 to be [86] Thus r4c8 & r6c4 can't be of the same parity Hence r5c3=r4c8+3-r6c4 must be even, i.e. can't be 5 15/3 @ r4c2 with r5c3 from {24} & r6c4 from {68} This cage must have at least one odd digit => r4c2 must be from {35} ...
Another variant from my backlog of unfinished puzzles. Thanks Afmob for a challenging variant.
When I first tried this puzzle I didn't get very far. It was only when I came back to it recently that I found the next step, the first one in Afmob's optimised walkthrough. The solving path seems narrow at first but the posted walkthroughs show that there are then several ways to make the key breakthrough and fix R4C28.
Rating Comment:
I'll rate my walkthrough for A129V2 at Easy 1.5. I used a couple of contradiction moves, steps 19b and 26a, each of which made an important elimination from R4C2.
Afmob wrote:
I used an odd-even move instead of a (very) small chain. I am not sure about the rating for this particular move (step 5b), so I'd like to hear your thoughts about it.
It's hard to know how to rate this move, which I also used. I don't think it was the hardest move in either of our walkthroughs; I felt that the ratings were determined by other steps in each case.
Here is my walkthrough for A129 V2.
Prelims
a) 4(2) diagonal cage at R5C4 = {13}, CPE no 1,3 in R5C123 + R6C456 b) 12(2) diagonal cage at R5C6 = {39/48/57}, no 1,2,6 c) R89C5 = {69/78} d) 20(3) cage in N1 = {389/479/569/578}, no 1,2 e) 10(3) cage in N3 = {127/136/145/235}, no 8,9 f) R4C456 = {128/137/146/236/245}, no 9
1. 45 rule on N1 3 innies R23C1 + R3C3 = 9 = {126/135/234}, no 7,8,9 1a. 45 rule on N1 1 outie R4C1 = 1 innie R3C3 + 4, no 6 in R3C3, no 1,2,3,4 in R4C1
2. 45 rule on N3 3 innies R23C9 + R3C7 = 21 = {489/579/678}, no 1,2,3 2a. 45 rule on N3 1 innie R3C7 = 1 outie R4C9 + 4, no 4 in R3C7, no 6,7,8,9 in R4C9
3. 45 rule on R1234 2 innies R4C28 = 12 = {39/48/57}, no 1,2,6
6. R4C456 = {128/146/236/245} (cannot be {137} which clashes with R5C4), no 7 6a. R4C6 = {568} -> no 5,6,8 in R4C45, clean-up: no 1 in R7C4 (step 4)
7. 45 rule on C12 3 outies R127C3 = 1 innie R4C2 + 18 7a. Max R127C3 = 24 -> no 7,8,9 in R4C2, clean-up: no 3,4,5 in R4C8 (step 3) 7b. Min R4C2 = 3 -> min R127C3 = 21, no 1,2,3
8. 45 rule on C9 3 innies R156C9 = 1 outie R9C8 + 4 8a. Min R156C9 = 6 -> min R9C8 = 2
9. 45 rule on C123 5 innies R34689C3 = 1 outie R6C4 + 12 9a. Min R34689C3 = 15 -> no 2 in R6C4 9b. Min R4C2 + R6C4 = 7 -> max R5C3 = 8
[This is how far I got when this variant originally appeared. Back then I didn’t spot the next step, even though I did something similar in C12 for the original Assassin.]
10. 26(5) cage at R5C5 = {24569/24578} (all other combinations clash with R89C5 because all cells of the 26(5) cage “see” R89C5), no 1,3 10a. 8 of {24578} must be in R7C6 -> no 8 in R567C5 10b. Combined cage 26(5) at R5C5 + R89C5 = {2456789}, 7 locked for C5
11. R123C5 = {139/148/238/346} (cannot be {256} which clashes with 26(5) cage at R5C5, ALS block), no 5
12. 5 in C5 only in R567C5, locked for 26(5) cage at R5C5, no 5 in R7C4, clean-up: no 1 in R4C4 (step 4)
13. Naked pair {24} in R49C4, locked for C4, CPE no 2,4 in R7C7 using D\ 13a. Min R4C2 + R6C4 = 8 -> max R5C3 = 7 [Having looked at the posted walkthroughs by udosuk and Afmob, I realise that I missed CPEs for R56C5 in steps 13 and 15.]
14. R4C456 (step 6) = {128/146/236} (cannot be {245} which clashes with 26(5) cage at R5C5, ALS block because 26(5) must have at least one of 2,4,5 in R56C5), no 5, clean-up: no 9 in R7C6 (step 5) 14a. 1,3 only in R4C5 -> R4C5 = {13} 14b. Naked pair {13} in R4C5 + R5C4, locked for N5, clean-up: no 9 in R6C7
15. Naked pair {68} in R47C6, locked for C6, CPE no 6,8 in R7C3 using D/, clean-up: no 4 in R6C7 15a. Killer pair 6,8 in R7C6 and R89C5, locked for N8
16. 26(5) cage at R5C5 (step 10) = {24569/24578} 16a. 6 of {24569} must be in R7C6 -> no 6 in R56C5
17. Killer pair 7,9 in 26(5) cage at R5C5 and R89C5, locked for C5 [Alternatively 9 in combined cage 26(5) at R5C5 + R89C5 only in R56789C5, locked for C5.]
18. R4C6 + R6C4 = {68} (hidden pair in N5), locked for D/, clean-up: no 2,4 in R4C9 (step 2a)
19. 15(3) cage at R4C2 = {258/348/456} (cannot be {267} because 2,7 only in R5C3, cannot be {357} because R6C4 only contains 6,8), no 7 19a. R6C4 = {68} -> no 6 in R5C3 19b. 2 of {258} must be in R5C3, 4 of {348/456} must be in R5C3 (cannot be [456] => R4C6 = 8 clashes with R4C28 = [48]) -> no 4 in R4C2, no 5 in R5C3, clean-up: no 8 in R4C8 (step 3)
20. Naked triple {135} in R4C259, locked for R4, clean-up: no 1 in R3C3 (step 1a)
21. R23C1 + R3C3 (step 1) = {135/234} (cannot be {126} because R234C1 cannot be {16}6), no 6 in R23C1, 3 locked for N1 21a. Killer pair 4,5 in 20(3) cage and R23C1 + R3C3, locked for N1
22. 17(3) cage at R4C8 = {179/269/278/359/467} (cannot be {368/458} because R4C8 only contains 7,9) 22a. 1 of {179} must be in R5C7, 9 of {269} must be in R4C8 -> no 9 in R5C7 22b. 6,8 of {269/278} must be in R5C7 -> no 2 in R5C7 22c. 3,6 of {359/467} must be in R5C7 -> no 4,5 in R5C7 22d. 1,6,8 of {179/278/467} must be in R5C7 -> no 7 in R5C7
23. 27(5) cage at R1C6 must contain two even numbers (cannot be {24678} because 6,8 only in R4C7) 23a. R123C6 cannot contain both of 2,4 which would clash with R123C5, R3C7 is odd -> R4C7 must be even -> R4C7 = {2468}
24. 16(3) cage at R5C2 = {169/178/259/349/358/367/457} (cannot be {268} because no 2,6,8 in R7C3) 24a. 1,3 of {169/367} must be in R6C2 -> no 6 in R6C2 24b. 1,3 of {178/358} must be in R6C2 -> no 8 in R6C2
25. R1C123 = {169/178/268} 25a. 45 rule on R1 3 innies R1C456 = 15 = {159/249/258/357/456} (cannot be {168/267} which clash with R1C123, cannot be {348} which clashes with R123C5, ALS block even with overlap at R1C5) 25b. R1C789 = {149/239/248/356} (cannot be {158/347} which clash with 10(3) cage in N3, cannot be {167} which clashes with R1C123, cannot be {257} which clashes with R1C456), no 7 25c. Hidden killer pair 3,4 in R1C456 and R1C789 for R1, R1C789 contains one of 3,4 -> R1C456 must contain one of 3,4 25d. R1C456 = {249/357/456} (cannot be {159/258} which don’t contain 3 or 4), no 1,8 25e. 3 of {357} must be in R1C5 -> no 3 in R1C46 25f. 9 of {249} must be in R1C4 -> no 9 in R1C6
[At this stage I had a long look at the interactions between the 22(4) cage at R5C1, 21(4) cage at R7C1 and 16(3) cage at R5C2 (and also the equivalent cages in N69), each of which “see” most, but not all, of the cells of the adjacent cage but I couldn’t find any way to use them. Then I found the contradiction move in my next step, which cracked this puzzle.]
26. R127C3 = R4C2 + 18 (step 7) 26a. R4C2 cannot be 5, here’s how R4C2 = 5 => R127C3 = 23 = {689} => R12C3 = {68} => 20(3) cage in N3 = {569/578} => R23C2 = {57/59} which clash with R4C2 26b. -> R4C2 = 3, R4C5 = 1, R4C9 = 5, R3C7 = 9 (step 2a), placed for D/, R4C8 = 9 (step 3), R5C4 = 3, R6C3 = 1, clean-up: no 7 in R5C6 26c. 3 in C5 only in R123C5, locked for N2
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