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 Post subject: Assassin 396
PostPosted: Mon Jun 01, 2020 5:57 am 
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Posts: 852
Location: Sydney, Australia
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Assassin 396

This is a really fun puzzle, then has a mighty sting in the tail. Perfect Assassin really....Come a long way since Ruud got this series started 2nd June, 2006. SudokuSolver gives it a 1.75, JSudoku uses 7 complex intersections (not default solvers)

triple click code:
3x3::k:2816:4097:4097:1794:3331:3331:5380:5380:5893:2816:6662:1794:1794:5380:5380:5380:5893:5893:6662:6662:6662:4103:1288:4361:4361:5898:5893:6662:6662:6667:4103:1288:7948:7948:5898:5893:3853:3853:6667:6667:6667:2574:7948:5898:5898:2837:2837:3853:0000:6667:2574:7948:7948:7948:3601:5394:5394:0000:6667:0000:0000:0000:3348:3601:3601:5394:0000:0000:3091:3091:3091:3348:2576:2576:5394:0000:0000:0000:4623:4623:4623:
Solution:
Code:
+-------+-------+-------+
| 3 9 7 | 4 8 5 | 1 2 6 |
| 8 5 2 | 1 6 3 | 9 4 7 |
| 1 6 4 | 7 2 9 | 8 3 5 |
+-------+-------+-------+
| 2 8 6 | 9 3 7 | 4 5 1 |
| 9 1 3 | 5 4 2 | 6 7 8 |
| 4 7 5 | 6 1 8 | 3 9 2 |
+-------+-------+-------+
| 5 3 1 | 8 7 4 | 2 6 9 |
| 7 2 8 | 3 9 6 | 5 1 4 |
| 6 4 9 | 2 5 1 | 7 8 3 |
+-------+-------+-------+


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 Post subject: Re: Assassin 396
PostPosted: Fri Jun 05, 2020 12:40 am 
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Posts: 208
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Thanks Ed! No shortcuts on this one! :) A few easy placements to start followed by a tough middle and a tougher end!
I would expect to see some very different solving paths after the start.
Unusually for me I have a complete WT - which means there is more likely to be some errors/typos.
Some corrections & clarifications thanks to Ed and Andrew.
Assassin 396 WT:
1. 16(2)n1 = {79}
-> 13(2)n2 = {58}
-> 17(2)r3 = [98]
-> 16(2)c4 = [79]

2. 7(3)r1c4 = {124}
-> Whichever of (124) is in r2c3 goes in n2 in r3c5
-> r2c56 = {36}
-> Cells in 21(5)r1c7 in n3 are [{12}9] or [{14}7]

3. Remaining outie n36 -> r4c6 = 7
-> (79) both in n8 in r789c5

4. -> 26(6)r2c2 has no (79) -> must be {124568}
-> 3 in n1 in r1c1
-> 11(2)n1 = [38]
-> r4c2 = 8
Also (HS) -> r1c9 = 6

5. Remaining IOD n6 -> r3c8 = r4c9 + 2
Since r3c8 is max 5 -> r4c9 is max 3
-> Whatever goes in r4c9 must go in n3 in r1c78 (I.e., from (12))

6! Whichever of (124) is in r2c3 also goes in r4c1 and r3c5
-> r4c15 = +5(2)
It also goes in n3 in r1c78
-> r4c19 have the same values as in r1c78 i.e., either {12} or [41]
But since r4c15 = +5(2) -> r4c19 cannot also be +5(2)
-> r4c19 = {12}
-> r1c78 = {12} and r1c4 = 4 and r2c7 = 9

7. 7 in n6 only in r5c89
-> 7 in n4 in r6c123
-> 7 in 15(3)n4 or 11(2)n4

Innies n14 = r245c3 = +11(3) - > No 9
-> 9 also in 15(3)n4 or 11(2)n4

-> One of:
(A) 11(2)n4 = {29} -> 15(3)n4 = [537] -> r4c15 = [14] -> r45c3 = [64]
(B) 11(2)n7 = {47} -> 15{3}n4 = {159} -> r4c15 = [23] -> r45c3 = [63]

I.e., r4c3 = 6 and r4c5 = r5c3 (either 3 or 4)

8! IOD r123456 -> r7c5 = r6c4 + 1
-> 26(6) must contain a 1 which can only go in n5

6 already in 26(6)-> r7c5 cannot be 6 -> r6c4 cannot be 5
-> 5 in n5 in 26(6)

Since r4c5 from (34) -> [r6c4,r7c5] cannot be [34]
-> 3 must be in 26(6) either in r5c3 or in n5

-> 26(6) = {1356(29|47)} (No 8)

-> Either 10(2)n5 = {28} and r6c4 = 6 (and r7c5 = 7)
or 10(2)n5 = {46} and r6c4 = 8 (and r7c5 = 9)

9! r1c78 = {12} prevents 31(6)r4c6 = [7{124}{89}] (Simpler - r4c9 from (12) does the same thing)
-> In n6 one of (89) in r6c89 and the other in r5c89
Since one of (89) in r6c89 and one of (82) in r6c46 -> 11(2)n4 cannot be {29}!
-> 11(2)n4 = {47}
-> 15(3)n4 = {159}
-> r4c1 = 2 -> r2c34 = [21] -> 5(2)r3c5 = [23] and r5c3 = 3
Also -> r4c9 = 1 -> r3c8 = 3
Also r2c56 = [63]

10. 23(5)n3 = [6{457}1] with 7 in r2c89
-> 7 in n9 in c7r789
-> 13(2)n9 from {49} or {58}
-> r2c8 from (45)
-> r2c9 = 7
-> r5c8 = 7
Also -> r6c8 from (89)

11. Either r6c4 = 6 or 10(2)n5 = [46]
-> HS 6 in n6 -> r5c7 = 6
r4c78 = {45}
-> 23(4) = [3578] or [3479]
-> (HP) r6c79 = {23}
-> 8 in n5 either in r6c4 or r6c6
-> r6c8 = 9
-> 23(4) = [3578]
-> r4c7 = 4

12. Also -> 13(2)n9 = {49}
-> r2c8 = 4 and r3c9 = 5
-> r2c2 = 5 and r3c123 = {146}

13. Only possible solution for 18(3)n9 = [783]
-> r6c79 = [32]

14. 8 in n7 in r78c3
HP r78c2 = {23}
-> 2 in r9 in n8 in r9c46
Either r9c4 = 2 -> 10(2)n5 = [28] -> 13(2)n2 = [85]
Or r9c6 = 2 -> r5c4 = 2 -> 5 in r56c4 -> 13(2)n2 = [85]
Either way -> 13(2)n2 = [85]
(Simpler is: 8 in c5 in r1c5 or r8c5.
But the latter puts r7c5 = 7 puts r6c4 = 6 and 10(2)n5 = [28] which contradicts 13(2)n2 = {58}
-> 8 in c5 only in r1c5)


15. r8c2 is either 2 in r8c2 -> r78c1 = {57}
or r8c2 is 3 -> r78c1 = {56}
Either way 5 in n7 in r78c1
-> 5 in n8/r9 in r9c45

16! Trying 2 in r9c6 puts 10(2)n5 = [46] puts r56c4 = [28] puts r789c5 = [974] puts r9c4 = 5
This leaves no place for 6 in r9
-> HS r9c4 = 2

17. -> 10(2)n5 = [28]
-> r6c4 = 6 and r7c5 = 7
Also r9c5 = 5
-> r8c5 = 9
-> 13(2)n9 = [94]
Also r5c4 = 5
-> r56c5 = [41]
Also -> 15(3)n4 = [{19}5]

18. Also r78c4 = {38} and r789c6 = {146} with r8c6 from (16)
-> 12(3)r8 only from [156] or [651]
-> 2 in n9 in r7c78
-> r78c2 = [32]
-> r78c4 = [83] and r8c3 = 8
Also r78c1 = [57]
-> 11(2)n4 = [47]
-> 16(2)n1 = [97]
-> (HS 6 in n7) 10(2)n7 = [64]
-> r79c3 = [19]
-> r3c123 = [164]
-> r5c12 = [91]
Also -> r789c6 = [461]
-> 12(3)r8 = [651]
-> r1c78 = [12]
-> r7c78 = [26]


Last edited by wellbeback on Sun Jun 14, 2020 7:02 pm, edited 1 time in total.

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 Post subject: Re: Assassin 396
PostPosted: Wed Jun 10, 2020 8:06 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 852
Location: Sydney, Australia
Really enjoyed wellbeback's WT. So many lovely clone sequences. Love 6. Really like the simplicity of 8 as well. Think I'd call that a type of "Locking-out cages". In all, I counted (a record?) 5 !s in wellbeback's WT! (note: two ! in step 9, which deserves them)

I only managed to find one clone which was very important for my solution (step 13) [edit: check out Andrew's step 3 for a really neat way to avoid this clone]. So, as predicted, quite a different WT but as often happens, we worked in the same areas. My 10, 13, 14 & 28 are the main differences. [Edit: ahhh, so many typos! Thanks Andrew and wellbeback for checking my WT]

Assassin 396 walk-through.
32 steps:
Prelims courtesy of SudokuSolver
Cage 16(2) n1 - cells ={79}
Cage 16(2) n25 - cells ={79}
Cage 17(2) n23 - cells ={89}
Cage 5(2) n25 - cells only uses 1234
Cage 13(2) n2 - cells do not use 123
Cage 13(2) n9 - cells do not use 123
Cage 10(2) n7 - cells do not use 5
Cage 10(2) n5 - cells do not use 5
Cage 11(2) n4 - cells do not use 1
Cage 11(2) n1 - cells do not use 1
Cage 7(3) n12 - cells ={124}

note: this WT is optimised so not all clean-up done, only what is shown. Also, naked singles may be available but not placed because they are not essential at that point.
1. 17(2)r3c6 = {89}: both locked for r3
1a. r34c4 = [79]

2. 16(2)n1 = {79}: both locked for r1 and n1

3. 13(2)n2 = {58}: both locked for r1 and n2
3a. r3c67 = [98]

4. 7(3)r1c4 = {124}, r2c56 sees all those -> no 1,2,4 (Common Peer Elimination, CPE)

5. r2c56 = {36}: both locked for r2, 21(5), 3 for n2
5a. r2c56 = 9 -> r1c78 + r2c7 = 12 = {129/147}
5b. r2c7 = (79)
5c. 1 locked for n3, r1
5d. naked triple 1,2,4 in r1c478, 2 & 4 locked for r1

6. r2c56 = 9, r3c6 = 9
6a. -> from "45" on n36: 1 remaining outie r4c6 = 7
6b. 7 in n6 only in r5: locked for r5

7. 26(6)r2c2 = {124568}(no 3)
7a. r1c1 = 3 (hsingle n1), r2c1 = 8
7b. r1c9 = 6
7c. 26(6) must have 8 -> r4c2 = 8
7d. 5 & 6 in n1 only in 26(6) -> no 5,6 in r4c1

8. "45" on c123: 3 innies r245c3 = 11 (no 9)

9. hidden killer pair 7,9 in n4
9a. -> 11(2)n4 must have one of 7,9 = {29/47}(no 3,5,6)
9b. -> 15(3)n6 must have one of 7,9: but {249/267} blocked by 11(2)
9c. = {159/357}(no 2,4,6)
9d. must have 5: locked for n4

10. "45" on n12346: 1 remaining outie r4c5 + 6 = 2 innies r45c3
10a. since r4c35 see each other, the 3rd cell cannot be the Innie/outie Difference of 6 (IOU)
10b. no 6 in r5c3
10c. r4c3 = 6 (hsingle n4)

11. 6 in n5 only in 10(2) = {46} or in r6c4
11a. -> no 4 in r6c4 (Locking out cages)

12. 10(2)n5 = {28/46}(no 1,3)
12a. 5(2)r3c5 = {14}/[23]

13. "45" on n12346: 1 remaining outie r4c5 = 1 innie r5c3 = (134)
13a. -> from cage sum 26(6)r4c3, r4567c5+r5c4 = 20
13b. -> but {12458} blocked since must have 4 in r7c5 to escape clash with 10(2)n5, and also means [41] in r34c5: but two 4's in c5
13c. = {12359/13457}(no 8)
13d. must have 7 or 9 with are only in r7c5 -> r7c5 = (79)
13e. and 26(6)r4c3 = [6]{12359/13457}: note, has both 3 & 5

14. 15(3)n4 = {159}/{357}
14a. but {357} as [537] only, leaves no place for both 3 & 5 in 26(6)r4c3
14b. = {159} only: 1 & 9 locked for n4

15. 11(2)n4 = {47}: 4 locked for r6 & n4
15a. r4c1 = 2, r5c3 = 3 -> r4c5 = 3 (step 13), r3c5 = 2, r2c56 = [63], r12c4 = [41], r2c3 = 2
15b. no 6 in r5c6

16. r1c78 = {12} = 3 -> r2c7 = 9 (cage sum)

17. 23(5)r1c9 = {14567} only (no 3)
17a. -> r4c9 = 1
17b. r3c8 = 3 (hsingle n3)
17c. -> 23(4)r3c8, which must have 7 for r5 = [3]{479/578}(no 2,6), r5c89 = {78/79}
17d. r5c7 = 6 (hsingle r5)

18. 13(2)n9 = {49/58}(no 7)

19. killer pair 4,5 in r3c9 with 13(2)n9: both locked for c9
19a. r2c9 = 7
19b. r5c8 = 7 (hsingle n6)

20. r24c8 = {45}: both locked for c8

21. killer pair 8,9 in r5c9 with 13(2)n9: both locked for c9

22. 10(2)n7 = {19/46}/[73] (no 2, 7 in r9c2)
22a. 18(3)n9: must have 2 or 3 for r9c9
22b. but {369} blocked by 10(2)n7
22c. = {279/378}(no 1,4,5,6)
22d. -> r9c7 = 7, r9c89 = 11 = [92/83]
22e. no 3 in 10(2)n7

23. killer pair 8,9 in r9c8 and 13(2)n9: both locked for n9

24. hidden pair 2,3 in n7 -> r78c2 = {23}

25. 14(3)n7 must have one of 2 or 3 for r8c2
25a. but {47}[3] blocked by r6c1 = (47)
25b. = {57}[2]/{56}[3](no 1,4,9)
25c. 5 locked for c1 and n7

26. "45" on r1..6: 1 innie r6c4 + 1 = 1 outie r7c5
26a. = [67/89]

27. 26(6)r4c3 = [63]([2]{15}[9]/[5417])
27a. note: has 5 in r5c4 or 9 in r7c5
27b. 1 locked for c5

Took many permutations of this step to get it to the simplest. Hope it's correct.
28. "45" on r9: 4 innies r9c3456 = 17 and must have 5 for r9
28a. = {1259/1358/2456}
28b. 4 in {2456} must be in r9c3 -> no 4 in r9c56
28c. 5 only in n8, locked for n8
28d. the only permutation with 5 in r9c4 = [1592] but this is blocked by 26(6)r4c3 (step 27a)
28e. -> no 5 in r9c4
28f. -> r5c4 = 5 (hsingle c4)
28g. -> r567c5 = [417] (step 27)

29. 10(2)n5 = {28}: both locked for c6, 8 for n5

30. r5c12 = {19}: 9 locked for r5 and n4
30a. r5c9 = 8
30b. -> r9c8 = 8 (hsingle c8)
30c. r9c9 = 3 (cage sum)

31. r1c56 = [85], r8c5 = 9
31a. r78c9 = [94]
31b. r89c6 = {16}: both locked for n8

32. 12(3)r8c6 = {156} only
32a. -> r8c7 = 5, 1 & 6 both locked for r8

Singles now.
Cheers
Ed


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 Post subject: Re: Assassin 396
PostPosted: Wed Jun 10, 2020 11:04 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1718
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Thanks Ed for this latest Assassin. A hard one right to the end, definitely a sting in the tail; it took me a while to find an acceptable way to do that bit.

Ed mentioned clones in his and wellbeback's walkthoughs; I didn't knowingly use any but perhaps in at least one place I used an implied one.

Thanks Ed for pointing out the simpler way for step 11a; I also then spotted a simpler way for it.

Here is my walkthrough for Assassin 396:
Prelims

a) R12C1 = {29/38/47/56}, no 1
b) R1C23 = {79}
c) R1C56 = {49/58/67}, no 1,2,3
d) R34C4 = {79}
e) R34C5 = {14/23}
f) R3C67 = {89}
g) R56C6 = {19/28/37/46}, no 5
h) R6C12 = {29/38/47/56}, no 1
i) R78C9 = {49/58/67}, no 1,2,3
j) R9C12 = {19/28/37/46}, no 5
k) 7(3) cage at R1C4 = {124}

1a. Naked pair {79} in R1C23, locked for R1 and N1, clean-up: no 2,4 in R12C1, no 4,6 in R1C56
1b. Naked pair {58} in R1C56, locked for R1 and N2, clean-up: no 3,6 in R2C1
1c. R3C67 = [98] -> R34C3 = [79], clean-up: no 1 in R56C6
1d. 7(3) cage at R1C4 = {124}, CPE no 1,2,4 in R2C56
1e. Naked pair {36} in R2C56, locked for R2 and 21(5) cage at R1C7, 3 locked for N2, clean-up: no 2 in R4C5
1f. R2C56 = 9 -> R1C78 + R2C7 = 12 = {129/147} -> R2C7 = {79}, R1C78 = {12/14}, 1 locked for R1 and N3
1g. Naked triple {124} in R1C478, 2,4 locked for R1
1h. 7(3) cage at R1C4 = {124}, 1 locked for R2

2a. 45 rule on N36 using R1C78 + R2C7 = 12, 1 remaining outie R4C6 = 7, clean-up: no 3 in R56C6
2b. 7 in N6 only R5C89, 7 locked for R5
2c. 1 in C6 only in R789C6, locked for N8
2d. 45 rule on N14 3 innies R245C3 = 11 = {128/146/236/245}, no 9

3a. 45 rule on R123456 1 outie R7C5 = 1 innie R6C4 + 1, no 8 in R7C5
3b. 26(6) cage at R4C3 = {123569/123578/124568/134567} (cannot be {123479} because 7,9 only in R7C5
3c. 7,9 of {123569/123578/134567} must be in R7C5 -> no 3 in R7C5, clean-up: no 2 in R6C4
3d. R6C4 + R7C5 = [12/45/56/67/89] (cannot be [34] which clashes with R34C5 = [41]), no 3 in R6C4, no 4 in R7C5
3e. R6C4 + R7C5 = [45/56/67/89] (cannot be [12] which clashes with R34C5 = [23] because 3 in N5 must be in R4C5 when 26(6) cage = {124568}), no 1 in R6C4, no 2 in R7C5
3f. R6C4 + R7C5 = [56/67/89] (cannot be [45] because R56C6 = {28} and 26(6) cage = {124568} would require 2,4,8 in R45C3), no 4 in R6C4, no 5 in R7C5
3g. R6C4 + R7C5 = [67/89] (cannot be [56] because R56C6 = {46}, only other place for 6 in N5 and R7C5 clash with R2C56), no 5 in R6C4, no 6 in R7C5
[Some of those can be considered to be equivalent to forcing chains.]
3h. 26(6) cage at R4C3 = {123569/123578/134567}
3i. 5 in N5 only in R5C45 + R6C5, locked for 26(6) cage, no 5 in R45C3
3j. Killer pair 6,8 in R56C6 and R6C4, locked for N5

4a. 45 rule on N3 using R1C78 + R2C7 = 12, 1 innie R3C8 = 1 outie R4C9 + 2 -> R3C8 = {3456}, R4C9 = {1234}
4b. 23(5) cage at R1C9 = {12569/13469/14567/23459/23567} (cannot be {12479} because R1C9 must contain one of 3,6)
4c. 23(5) cage = {12569/14567/23459/23567} (cannot be {13469} which clashes with R3C8 + R4C9 = [31]), 5 locked for N3, clean-up: no 3 in R4C9

5a. 45 rule on N124 using R2C56 = 9, 2 innies R45C3 = 1 outie R4C5 + 6, IOU no 6 in R5C3
5b. R245C3 (step 2d) = {128/146/236}
5c. 6 of {146/236} must be in R4C3 -> no 3,4 in R4C3

[I ought to have spotted this obvious step earlier but hadn’t checked and found that 26(6) cage without 7 or 9 has one combination; I ought to have realised that since it’s missing 7 and 9 it must also be missing 3 to make a total of 19 missing. I’d also used combinations in step 3 for the other 26(6) cage.]
6a. 26(6) cage at R2C2 = {124568}, no 3
6b. R1C1 = 3 (hidden single in N1) -> R2C1 = 8, R1C9 = 6, clean-up: no 3,8 in R6C2, no 7 in R78C9, no 2,7 in R9C2, no 4 in R4C9 (step 4a)
6c. 26(6) cage = {124568} -> R4C2 = 8, clean-up: no 2 in R9C1
6d. 5,6 in N1 only in R2C2 + R3C123, locked for 26(6) cage, no 5,6 in R4C1
6e. 26(6) cage at R4C3 (step 3h) = {123569/134567} -> R4C3 = 6, clean-up: no 5 in R6C12
6f. 5 in N4 only in 15(3) cage at R5C1 = {159/357}, no 2,4
6g. 7 of {357} must be in R6C3 -> no 3 in R6C3
6h. 45 rule on N4 2 remaining innies R4C1 + R5C3 = 5 = [14/23/41], no 2 in R5C3
6i. 18(3) cage at R9C7 = {279/369/378/567} (cannot be {189/459/468} which clash with R78C9), no 1,4
6j. R9C12 = {19/46} (cannot be [73] which clashes with 18(3) cage), no 3,7
6k. 18(3) cage = {279/378/567} (cannot be {369} which clash with R9C12), 7 locked for R9 and N9
6l. 14(3) cage at R7C1 = {239/257/347/356} (cannot be {149/167} which clash with R9C12), no 1
6m. 3 of {239/347/356} must be in R8C2 -> no 4,6,9 in R8C2
6n. 45 rule on N9 4 innies R78C78 = 14 = {1238/1256/1346} (cannot be {2345} which clashes with R78C9), no 9

7a. 5 in R4 only in R4C78, locked for N6
7b. 7 in R5 only in 23(4) cage at R3C8 = [35]{78}/{34}{79}, 3 locked for C8
7c. 45 rule on R1234 2 outies R5C89 = 1 remaining innie R4C7 + 11
7d. R5C89 = {78/79} = 15/16 -> R4C7 = {45}

[Something else I ought to have spotted earlier.]
8a. 3 in N4 only in R5C23, locked for R5
8b. 3 in N5 only in R46C5, locked for C5 -> R2C56 = [63]

9a. 15(3) cage at R5C1 (step 6f) = {159/357}, 14(3) cage at R7C1 (step 6l) = {239/257/347/356}
9b. Consider combinations for R4C1 + R5C3 (step 6h) = [14/23/41]
R4C1 + R5C3 = {14} => 15(3) cage = [537] => 14(3) cage = {27}5
or R4C1 + R5C3 = [23] => 15(3) cage = {159} => R6C12 = {47}=> 14(3) cage = {257/356} (cannot be {29}3 which clashes with R4C1, cannot be {47}3 which clashes with R6C1)
-> 14(3) cage = {257/356}, no 4,9, 5 locked for N7

10a. 1 in C9 only in R46C9, locked for N6
10b. 31(6) cage at R4C6 = {145678/234679/235678} (cannot be {124789} which clashes with R5C89, cannot be {135679} which clashes with 23(4) cage at R3C8, step 7b)
10c. 31(6) cage = {234679/235678} (cannot be {145678} = [75{46}81] which clashes with R6C12 + R6C46, killer ALS block), no 1, 3 locked for R6 and N6
10d. Naked pair {45} in R4C78, 4 locked for R4 and N6
10e. R4C9 = 1 (hidden single in N6) -> R3C8 = 3 (step 4a), R4C1 = 2, R4C5 = 3 -> R3C5 = 2, R12C4 = [41], R2C3 = 2, R5C3 = 3 (step 6h), clean-up: no 9 in R6C12
10f. Naked pair {47} in R6C12, 4 locked for R6, 7 locked for N4, clean-up: no 6 in R5C6
10g. R1C78 = {12}, R2C56 = [63] -> R2C7 = 9 (cage sum)
10h. R9C7 = 7 (hidden single in C7) -> R9C89 = 11 = {29}/[65/83], no 5 in R9C8, no 8 in R9C9
10i. 14(3) cage at R7C1 (step 9b) = {257/356} -> R8C2 = {23}, R78C1 = {56/57}, 5 locked for C1
10j. R78C2 = {23} (hidden pair in C2)

11a. Consider combinations for 21(4) cage at R7C2 = 2{478}/3{189}
21(4) cage = 2{478}, 7 locked for C3 => R1C3 = 9 => R6C3 = {15}
or 21(4) cage = 3{189}, 1,9 locked for C3 => R6C3 = 5
-> R6C3 = {15}
[Ed pointed out 21(4) cage at R7C2 = 2{478}/3{189} -> killer pair 7,9 in R1C3 and R789C3, locked for C3.
I also realised that R6C35 = {15} (hidden pair for R6) was simpler.
So I’d seen this step as a forcing chain when both killer pair and hidden pair were simpler.]

11b. 9 in N4 only in R5C12, locked for R5
11c. R5C89 = {78}, 8 locked for R5 and N6, R3C8 = 3 -> R4C8 = 5 (cage total), clean-up: no 2 in R6C6
11d. R4C7 = 4, R5C7 = 6 (hidden single in R5)
11e. 5 in N3 only in R23C9, locked for C9, clean-up: no 8 in R78C9, no 6 in R9C8 (step 10h)
11f. Naked pair {49} in R78C9, locked for C9 and N9 -> R23C9 = [75], R2C28 = [54], clean-up: no 2 in R9C89 (step 10h)
11g. R6C3 = 5 (hidden single in N4) -> R6C5 = 1
11h. R9C89 = [83]
11i. 2,5 in R9 only in R9C456, locked for N8
11j. 12(3) cage at R8C6 = {156/246}, no 8, 6 locked for R8
11k. 12(3) cage = {156/246} = [156/651/426], no 1 in R8C7 = {25}, no 2 in R8C8
11l. R59C4 = {25} (hidden pair in C4)

[I’ve now reached Ed’s “sting in the tail”, how to eliminate 12(3) cage at R8C6 = [156]. I refuse to use Unique Rectangle, R8C6 cannot be 1 => R1C78 and R7C78 both {12} whose order cannot be resolved, as that relies on the solution being unique but walkthroughs are meant to reach a unique solution.]

[Or maybe that’s not initially the right approach.]
12a. R56C6 = [28/46], R7C5 = R6C4 + 1 (step 3a), 14(3) cage at R7C1 (step 9b) = {257/356}
12b. Consider permutations for 12(3) cage at R8C6 (step 11k) = [156/651/426]
12(3) cage = [156] => 14(3) cage = [572], 6 in N7 only in R9C12 = {46}, R9C35 = [19] (hidden pair in R9) => R7C5 = 7, R6C4 = 6 => R56C6 = [28]
or 12(3) cage = [426/651] => R8C6 = {46} => R56C6 = [28] (cannot be [46] which clashes with R8C6)
-> R56C6 = [28], R5C45 = [54], R6C4 = 6, R7C5 = 7, R1C56 = [85], R89C5 = [95], R78C9 = [94]
12c. 12(3) cage = [156/651] -> R8C7 = 5, 1 locked for R8
12d. R8C1 = 7 -> R7C1 + R8C2 = 7 = [52], R6C1 = 4
12e. 6 in N7 only in R9C12 = [64]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A396 at 1.5. I used IOD analysis and later a few forcing chains, the last of which feels like a full 1.5.


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