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 Post subject: Assassin 369
PostPosted: Fri Feb 01, 2019 6:39 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
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a369.JPG
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Note: disjoint 9(2) in n69. Also, its an x-killer so 1-9 cannot repeat on either diagonal.

Assassin 369
This is what I aim for every Assassin! Wish they were all this quality. SudokuSolver gives it 1.75 but don't let that scare you. I used a few tricky to see steps but nothing heavy. It just resists and resists and resists till the end. It takes me about 30 steps to crack it. This is one to savour.
code:
3x3:d:k:0000:0000:0000:0000:0000:0000:4354:4354:4354:5891:0000:2564:0000:4102:4102:3847:3847:3847:5891:0000:2564:0000:8968:4102:3849:4362:4362:5891:5891:5131:5131:8968:4102:3849:3849:4362:4620:5891:5131:8968:8968:8968:9997:2305:4362:4620:1550:1550:4623:8968:3856:9997:9997:9997:4620:4620:1550:4623:8968:3856:9997:3857:4370:5395:5395:5395:4623:4623:9997:9997:3857:4370:4608:4608:4608:2565:2565:9997:2305:3857:3857:
solution:
Code:
+-------+-------+-------+
| 5 2 6 | 8 3 4 | 9 7 1 |
| 3 8 9 | 2 1 7 | 4 5 6 |
| 4 7 1 | 6 9 5 | 8 2 3 |
+-------+-------+-------+
| 2 5 4 | 9 8 3 | 6 1 7 |
| 6 9 7 | 1 4 2 | 3 8 5 |
| 8 1 3 | 7 5 6 | 2 9 4 |
+-------+-------+-------+
| 1 3 2 | 5 6 9 | 7 4 8 |
| 7 6 8 | 4 2 1 | 5 3 9 |
| 9 4 5 | 3 7 8 | 1 6 2 |
+-------+-------+-------+

Cheers
Ed


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 Post subject: Re: Assassin 369
PostPosted: Fri Feb 08, 2019 6:22 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
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Thanks Ed for this Assassin. It certainly resisted until the end, in fact the end of my solving path was possibly the hardest part to work out.

Thanks Ed for your comments, particularly pointing out what I missed near the start.
Here is my walkthrough for Assassin 369:
Prelims

a) R23C3 = {19/28/37/46}, no 5
b) Disjoint cage R5C8 + R9C7 = {18/27/36/45}, no 9
c) R67C6 = {69/78}
d) R78C9 = {89}
e) R9C45 = {19/28/37/46}, no 5
f) 20(3) cage at R4C3 = {389/479/569/578}, no 1,2
g) 6(3) cage at R6C2 = {123}
h) 21(3) cage at R8C1 = {489/579/678}, no 1,2,3
i) 39(8) cage at R5C7 = {12345789}, no 6

Steps Resulting From Prelims
1a. Naked pair {89} in R78C9, locked for C9 and N9, clean-up: no 1 in R5C8
1b. Naked triple {123} in 6(3) cage at R6C2, CPE no 3 in R45C3
[I overlooked that this CPE also eliminated 1,2,3 from R6C4 using D/. This would have simplified my solving path from step 7d onward.]

2. 45 rule on N7 3 innies R7C123 = 6 = {123}, locked for R7 and N7
2a. Max R7C12 = 5 -> min R56C1 = 13, no 1,2,3 in R56C1

3. 21(3) cage at R8C1 = {579/678} (cannot be {489} which clashes with R8C9), no 4, 7 locked for R8 and N7
3a. Killer pair 8,9 in 21(3) cage and R8C9, locked for R8
3c. 4 in N7 only in 18(3) cage at R9C1, locked for R9, clean-up: no 5 in R5C8, no 6 in R9C45

4. 45 rule on C789 2 outies R89C6 = 9 = [18/27/45] -> R8C6 = {124}, R9C6 = {578}
4a. 9 of 39(8) cage at R5C7 only in R5C7 + R6C78, locked for N6

5. 45 rule on N9 3 innies R789C7 = 13 = {157/247/256/346}
5a. 6 of {346} must be in R9C7 -> no 3 in R9C7, clean-up: no 6 in R5C8

6. 4 in N7 only in 18(3) cage at R9C1 = {459/468}, R89C6 (step 4) = [18/27/45]
6a. Consider combinations for 18(3) cage
18(3) cage = {459}, locked for R9 => R89C6 = [18/27], 2 or 8
or 18(3) cage = {468}, locked for R9
-> R9C45 = {19/37} (cannot be {28}, no 2,8
6b. 2 in R9 only in R9C789, locked for N9
6c. R789C7 (step 5) = {157/247/346} (cannot be {256} because 2,6 only in R9C7)
6d. {157} can only be [751] (cannot be [517/715] which clash with R89C6), no 5 in R7C7, no 1 in R8C7, no 5,7 in R9C7, clean-up: no 2,4 in R5C8
6e. R89C6 = [18/27] (cannot be [45] which clashes with R78C7), no 4,5
6f. Killer pair 1,7 in R89C6 and R9C45, locked for N8, clean-up: no 8 in R6C6
6g. 7 in N8 only in R9C456, locked for R9
6h. 7 in R7 only in R7C78, CPE no 7 in R6C8
6i. R67C6 = {69} (cannot be [78] which clashes with R9C6), locked for C6
6j. 16(4) cage at R2C5 = {1258/1267/1348/1357/1456/2347/2356} (cannot be {1249} which clashed with R8C6), no 9

7. Consider permutations for R789C7 (step 6c) = [436/742/751]
R789C7 = [436] => 18(3) cage at R9C1 (step 6) = {459}, locked for R9 => R9C45 = {37} => R89C6 = [18]
or R789C7 = [742/751] => R89C6 = [18] => R9C45 = {37}
-> R89C6 = [18], R9C45 = {37}, 3 locked for R9 and N8
7a. 18(3) cage at R9C1 = {459} (only remaining combination), 5,9 locked for R9 and N7
7b. Naked triple {678} in 21(3) cage at R8C1, 6,8 locked for R8 -> R78C9 = [89]
7c. 39(8) cage at R5C7 = {12345789}, CPE no 2,3,4,5,7 in R4C7
7d. 2 in N8 only in R8C45 -> 18(4) cage at R6C4 = {2349/2457} (cannot be {1269/1278/2367} because R8C45 only contain 2,4,5, cannot be {2358} because 3,8 only in R6C4), no 1,6,8
7e. 3,7 of {2349/2457} can only be in R6C4 -> R6C4 = {37}
[If I’d spotted the elimination of 1,2,3 from R6C4 in step 1b, then step 7d would have just been 2 in 18(4) cage at R6C4 = {2457} (only possible combination), and step 7e would then give R6C4 = 7, placed for D/, which would have simplified later steps.]
7f. 18(4) cage at R6C4 = {2349/2457}, 4 locked for N8
7g. 3 in R8 only in R8C78, CPE no 3 in R6C8
7h. Naked pair {37} in R69C4, locked for C4
7i. 16(4) cage at R2C5 (step 6j) = {1357/2347/2356} (cannot be {1258/1267/1348/1456} because R234C6 need to contain three of 2,3,4,5,7), no 8, CPE no 3 in R1C6
7j. 1,6 of {1357/2356} must be in R234C6 -> no 5 in R2C5
7k. Consider combinations for 16(4) cage = {1357/2347/2356}
16(4) cage = {1357/2356} => 5 in R234C6
or 16(4) cage = {2347}, CPE no 2,3,4,7 in R1C6 => R1C6 = 5
-> 5 in R1234C6, locked for C6

8. 45 rule on N58 2 innies R4C46 = 1 outie R3C5 + 3
8a. Min R4C46 = 6 -> min R3C5 = 3
8b. 35(7) cage must be missing two digits which total 10 so must contain three of the pairs 1,9, 2,8, 3,7 and 4,6
8c. Consider placements for R6C4 = {37}
R6C4 = 3 => R9C45 = [73] => no 3 in 35(7) cage
or R6C4 = 7 => R9C45 = [37] => no 7 in 35(7) cage
-> 35(7) cage = {124589} (only remaining combination, since it doesn’t contain both of 3,7), no 3,7
[Alternatively R69C4 = {37}, R9C45 = {37} -> R6C4 = R9C5
35(7) cage = {124589} (only remaining combination, other combinations clash with R6C4 + R9C5), no 3,7]
8d. R4C6 + R6C4 = {37} (hidden pair in N5), locked for D/
8e. 3 of 6(3) cage at R6C2 = {123} only in R6C12, locked for R6 and N4 -> R6C4 = 7, R4C6 = 3, R9C45 = [37]
[Now I’m back to where I should have been if I’d spotted the elimination of 1,2,3 from R6C4 in step 1b, but I wouldn’t have found some interesting steps including 8c.]
8f. 39(8) cage at R5C7 = {12345789}, 3,7 locked for C7
8g. R6C4 = 7 -> R7C4 + R8C45 = 11 = {245}, 5 locked for N8
8h. Naked pair {69} in R7C56, 6 locked for R7
8i. 5 in C6 only in R123C6, locked for N2
8j. R1C5 = 3 (hidden single in C5)
8k. 8,9 in C5 only in R34567C5, locked for 35(7) cage, no 8,9 in R5C4

9. 1,6 in N6 only in R4C789 + R5C9
9a. 45 rule on N3 4 outies R4C789 + R5C9 = 19 = {1468/1567}, no 2,3
9b. 8 of {1468} must be in R4C78, which cannot also contain 4 (because no 3 in R3C7), no 4 in R4C8
9c. R4C78 cannot total 10 -> no 5 in R3C7

10. 15(4) cage at R7C8 = {1257/1356/2346} (cannot be {1347} which clashes with R7C7)
10a. 3 of {2346} must be in R8C8 -> no 4 in R8C8

11. Hidden killer pair for 3,7 in R5C7 and R78C7 for C7, R789C7 (step 6c) must contain one of 3,7 in R78C7 -> R5C7 = {37}
11a. 39(8) cage at R5C7 = {12345789}, 2,9 locked for R6 -> R6C6 = 6, placed for D\, R7C56 = [69], clean-up: no 4 in R2C3
11b. Naked pair {13} in R6C23, 1 locked for R6, N4 and 6(3) cage at R6C2 -> R7C3 = 2, placed for D/, clean-up: no 8 in R23C3
11c. Naked pair {13} in R67C2, locked for C2
11d. R7C12 = {13} -> R56C1 = 14 = [68/95]
11e. 20(3) cage at R4C3 = {479/578} (cannot be {569} which clashes with R56C1), no 6, 7 locked for C3 and N4, clean-up: no 3 in R23C3
11f. R8C1 = 7 (hidden single in N7)
11g. 2 in N4 only in R4C12 + R5C2, locked for 23(5) cage at R2C1, no 2 in R23C1
11h. 23(5) cage = {23459/23468} (cannot be {12389} which clashes with R7C1, cannot be {12569} which clashes with R56C1), no 1, 3 locked for C1 -> R7C12 = [13], R6C23 = [13]
[Note. 23(5) cage and R56C1 form a combined 37(7) cage = {2345689}.]
11i. 16(4) cage at R2C5 (step 7h) = {1357/2347}, 7 locked for C6

12. 20(3) cage at R4C3 (step 11e) = {479/578}
12a. 9 of {479} must be in R4C4 (cannot be R45C3 + R4C4 = {79}4 which clashes with R23C3 using D\) -> no 4 in R4C4, no 9 in R45C3
12b. 45 rule on N5 1 remaining outie R3C5 = 1 innie R4C4 -> no 4 in R3C5, no 5 in R4C4
12c. R4C4 = {89} -> no 8 in R45C3
12d. 8,9 in N4 only in R45C12 + R6C1, CPE no 8,9 in R23C1
12e. Killer triple 4,5,9 in R23C3, R45C3 and R9C3, locked for C3
12f. 45 rule on N4 3(2+1) outies R23C1 + R4C4 = 16
12g. R4C4 = {89} -> R23C1 = 7,8 = {34/35}, no 6

13. R8C8 = 3 (hidden single on D\) -> R57C7 = [37] (hidden pair in C7), 7 placed for D\, clean-up: no 6 in R9C7
13a. R9C8 = 6 (hidden single in N9)
13b. 7 in N6 only in R45C89, CPE no 7 in R3C8

14. R789C7 (step 6c) = [742/751] -> R89C7 = [42/51] -> R7C8 + R89C7 = [451/542]
14a. 15(3) cage at R3C7 = {168}/[465] (cannot be {159} = [915] which clashes with R7C8 + R89C7), no 9 in R3C7, no 7 in R4C8, 6 locked for C7
14b. 45 rule on N3 3 innies R3C789 = 13 = {139/157/238/256/346} (cannot be {148} which clashes with R3C35, ALS block, cannot be {247} = [427] because no 3 in R45C9)
14c. 1 of {139/157} must be in R3C7 -> no 1 in R3C89
14d. 8 of {238} must be in R3C7 -> no 8 in R3C8
14e. R3C789 = {346} can only be [643] -> no 4 in R3C79
14f. 6 of {256/346} must be in R3C7 -> no 6 in R3C9
14e. 15(3) cage = {168} (only remaining combination), no 5 in R4C8

15. R89C7 (step 14) = [42/51]
15a. 17(3) cage at R1C7 = {179/269/458/467} (cannot be {278} because no 2,7,8 in R1C9)
15b. 17(3) cage = {179/269/458} (cannot be {467} = [476] because R1C9 + R3C7 + R8C2 = [618] and R13C7 = [41] clashes with R89C7)
15c. 1 of {179} must be in R1C9 -> no 1 in R1C78
15d. 15(3) cage at R2C7 = {249/258/348/456} (cannot be {159/168} which clash with 17(3) cage, cannot be {267/357} because 3,6,7 only in R2C9), no 1,7
15e. 17(3) cage = {179/269} (cannot be {458} which clashes with 15(3) cage), no 4,5,8
15f. 17(3) cage can only be [926/971] (cannot be [296] because then R3C7 = 1, hidden single in N3, and R13C7 = [21] clash with R9C7) -> R1C7 = 9, R6C8 = 9 (hidden single in N6)
15g. 1 in N3 only in R1C9 + R3C9, locked for D/

16. 17(4) cage at R3C8 = {1457/2357/2456} (cannot be {1367} because R3C8 only contains 2,4,5)
16a. Consider permutations for R5C8 + R9C7 = [72/81]
R5C8 + R9C7 = [72] => R9C9 = 1 => 17(4) cage = {2357/2456}
or R5C8 + R9C7 = [81] => R4C8 = 1 => 17(4) cage = {2357/2456}
-> 17(4) cage = {2357/2456}, no 1, 2 locked for R3 and N3 -> R1C8 = 7, R1C9 = 1 (cage sum)
16b. R45C8 = [18], R4C7 = 6, R3C7 = 8, placed for D/, R9C9 = 2, placed for D\, R7C8 = 4 (cage sum), R2C8 = 5, placed for D/
16c. R3C3 = 1 (hidden single on D\) -> R2C3 = 9
16d. R3C5 = 9 -> R5C5 = 4, placed for both diagonals

and the rest is naked singles, without using the diagonals.

Rating Comment:
I'll rate my walkthrough for A369 at 1.5. There were enough forcing chains for that rating, plus the final stages probably also qualify for that rating.


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 Post subject: Re: Assassin 369
PostPosted: Sun Feb 10, 2019 6:53 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Thanks Andrew for your WT. As usual, we saw the key steps quite differently especially my 7 & 10. I worked in the same areas as the first two thirds then I veered off from my step 23 and everything was working toward 29 which was my final cracker. A very challenging puzzle!

WT:
Preliminaries courtesy of SudokuSolver
Cage 17(2) n9 - cells ={89}
Cage 15(2) n58 - cells only uses 6789
Cage 9(2) n69 - cells do not use 9
Cage 10(2) n1 - cells do not use 5
Cage 10(2) n8 - cells do not use 5
Cage 6(3) n47 - cells ={123}
Cage 21(3) n7 - cells do not use 123
Cage 20(3) n45 - cells do not use 12
Cage 39(8) n689 - cells ={12345789}

Note: no routine clean-up done unless stated.
1. 17(2)n9 = {89}: both locked for c9 and n9

2. "45" on n7: 3 innies r7c123 = 6 = {123}: all locked for n7 and r7

3. 21(3)n7: {489} blocked by r8c9
3a. = {579/678}(no 4)
3b. must have 7: locked for r8 and n7

4. 4 in n7 only in r9: locked for r9
4a. 18(3) = {459/468} = 5 or 8
4b. no 6 in 10(2)n8

5. "45" on c789: 2 outies r89c6 = 9
5a. = {18}[27/45](no 3,9, no 5 in r8c6, no 2 in r9c6)

6. "45" on n9: 3 innies r789c7 = 13 = {157/247/256/346}:
6a. note: can't have both 1 and 4 (no eliminations yet)

7. "45" on r9: 4 innies r9c6789 = 17
7a. but {1358} blocked by 18(3)n7 = 5 or 8 (step 4a)
7b. but {2357} blocked by r78c7 (which sees all of the h17(4)) can't be [41] (step 6a)
7c. = {1268/1367}(no 5)
7d. must have 1 & 6: both locked for r9, and no 1 in r8c7 since it sees all those cells (Common Peer Elimination CPE)
7e. no 4 in r8c6 (h9(2)r89c6)

8. 18(3)n7 = {459} only: 5,9 locked for n7 and 9 for r9
8a. 21(3)n7 = {678} only: 6,8 locked for r8
8b. r78c9 = [89]

9. h9(2)r89c6: [27] blocked by 10(2)n8 = {28/37} = 2 or 7
9a. -> r89c6 = [18]
9b. 10(2)n8 = {37}: both locked for r9 and n8
9c. 15(2)r6c6 = {69} only: both locked for c6
9d. r9c789 = naked triple {126}: 2,6 locked for n9

10. 6(3)r6c2 = {123}: r6c4 sees all those cells -> no 1,2,3 in r6c4 (CPE)

11. 2 in n8 only in r8c45 in 18(4)
11a. = {2457} only valid combination
11b. -> r6c4 = 7 (Placed for D/), naked triple {245} in n8: 4,5 locked for n8
11c. r9c45 = [37]

12. 35(7)r3c5 = {1245689} only (no 3)
12a. -> hidden single 3 in n5: r4c6 = 3, placed for D/
12b. -> hsingle 3 in n2 -> r1c5 = 3

13. 3 must be in 6(3)r6c2, only in r6c23: 3 locked for n4 and r6

14. h13(3)r789c7 can't have both 3 & 7 since can't be {337}
14a. & 39(8)r5c7 must have both 3 & 7
14b. -> r5c7 = (37) (hidden killer pair)
14c. 3 and 7 both locked for c7

15. 39(8)r5c7 must have both 2,9: only in r6c789: both locked for r6 and n6
15a. r67c6 = [69]: 6 placed for D\
15b. naked pair {13} in r6c23: 1 locked for r6, n4 and no 1 in r7c3
15c. r7c3 = 2 (Placed for d/)
15d. no 8 in 10(2)n1
15e. r7c5 = 6

16. Naked quad {2457} in r1235c6: r3c5 sees all of those -> no 2,4,5 in r3c5 (CPE)

17. "45" on n5: 1 outie r3c5 = 1 innie r4c4 = (89)

18. "45" on n7: 2 remaining outies r56c1 = 14 = [68/95] = 5/6

19. 20(3)r4c4: {56}[9] blocked by r56c1 (step 18)
19a. = {479/578}(no 6)
19b. must have 7: 7 locked for n4 and c3
19c. no 3 in 10(2)n1

20. hsingle 7 in n7 -> r8c1 = 7

21. r23c1 see all 2 & 6 in n4 -> no 2,6 (CPE)

22. "45" on n47: 3 outies r23c1 + r4c4 = 16
22a. r4c4 = (89) -> r23c1 = 7/8 = {34/35}(no 1,8,9) = 4/5
22b. must have 3 -> 3 locked for c1 and n1
22c. r7c12 = [13], r6c23 = [13]

23. r56c1 = 14 = [68/95] (step 18)
23a. but r569c1 = [954] blocked by r23c1 = {345}
23b. -> r56c1 = [68]

24. 16(4)r2c5: must have two of {2457} for r23c6 = {1357/2347/2356}(no 8,9)

25. 8 & 9 in c5 only in 35(7)r3c5 -> no 8,9 in r5c4

26. hidden single 3 on d\ -> r8c8 = 3
26a. 15(4)n9 must have two of 1,2,6 for r9c89 = {1356/2346}(no 7)
26b. must have 6 -> r9c8 = 6

27. hidden single 7 in n9 -> r7c7 = 7: placed for D\
27a. r5c7 = 3

28. "45" on r6789: 1 remaining outie r5c8 - 3 = 1 innie r6c5
28a. = [74/85]: note, 8 in r5c8 must have 5 in r6c5

29. 8 in r5 only in r5c58 -> with step 28a. 5 blocked from r5c5 since it would leave no 8 for r5

30. 5 on d\ only in n1: 5 locked for n1
30a. r23c1 = {34}: 4 locked for n1, c1 and no 4 in r45c2

31. 4 in n4 only in 20(3)r4c3 = {47}[9] only, 4 locked for c3, r4c4 = 9, placed for d\

cracked. Naked singles to the end- don't forget the diagonals!
Cheers
Ed


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 Post subject: Re: Assassin 369
PostPosted: Sun Feb 10, 2019 11:18 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
And here's mine! I think we all started in the same areas with only differences in specific details, but I veered off in my Step 11 which opened it up considerably for me. Wonderful puzzle Ed!
Assassin 369 WT:
1. Innies n7 r7c123 = +6(3) = {123}
17(2)n9 = {89}
-> 21(3)n7 cannot be {489} - Must be {7(59|68)} (7 locked in r8/n7)
-> 18(3)n7 = {4(68|59)} (4 locked in r9/n7)
-> Whichever of (89) is in r7c9 is in r8 in n7, and in r9 in n8

2. Outies c789 = r89c7 = +9(2)
Since 6 not in 39(8) -> r89c6 not {36}

3! Innies n9 = r789c7 = +13(3)
-> Possibilities for r789c7 are {256}, {346}, {157}, {247}
-> One of (45) in r789c7
Consider r9c7. If it was from (123) it could also be in r8c6.
But if it was 4 or greater it could not be repeated in r8c6 since that would leave no place for it in r7
-> Since one of (45) in r789c7 -> r89c6 cannot be {45}
-> r89c6 from {18} or [27]
-> 10(2)n8 cannot be {28}

4. Since one of (89) in r9 in n8 but with 8 not in r9c45 and 9 not in r9c6 -> One of:
a) 18(3)n7 = {468} -> 10(2)n8 = {19} -> r89c6 = [27]
b) 18(3)n8 = {459} -> r89c6 = [18] -> 10(2)n8 = {37}
But the former puts 6 in n9 in r78c8 which leaves no place for 7 in n9
(D'oh - put the wrong line here at first).
-> 18(3)n8 = {459} -> r89c6 = [18] -> 10(2)n8 = {37}
Also -> 21(3)n7 = {678}
Also r9c789 = {126}
Also 17(2)n9 = [89]
Also 7 in r7c78
Also 3 in r8c78
Also one each of (45) in r7c78 and r8c78.
Also 15(2)r6c6 = {69}

5! r8c45 = {24} or {25}
r6c4 sees all of 6(3)r6c2 -> r6c4 is Min 4.
r7c4 sees all of r7c123 = {123} -> r7c4 is Min 4.
-> Only possibility for 18(4)r6c4 = [7{245}]
-> 10(2)n8 = [37]
-> 7 not in 35(7)r3c5
-> 35(7)r3c5 = {1245689}
-> HS 3 in n5 -> r4c6 = 3
-> HS 3 in c5 -> r1c5 = 3

6! 3 in r4c6 -> 3 in r7c12
-> 3 in r6c23
Whichever of (123) is in r7c2 goes in n4 in r6c3
-> Whichever of (123) is in r7c1 goes in n4 in r6c2
-> 3 in one of r67c2
-> 3 in n1 in one of r23c13
If 3 in 10(2)n1 -> 10(2) = {37}
-> Not both of (37) in D\ in n1
-> At least one of (37) in D\ in n9
-> Whether 7 goes in r7c7 or r7c8 -> 3 cannot go in r8c7
-> r8c8 = 3
-> r5c8 not 3 -> r9c7 not 6
-> 6 in r9c89
-> r7c7 = 7
-> r7c8,r8c7 = {45}

7. -> HS 3 in 39(8) -> r5c7 = 3
-> 9 in 39(8) in r6c78
-> 15(2)r6c6 = [69]
-> r7c5 = 6
Also r9c8 = 6
-> r9c79 = {12}

8. (37) on D\ in n9 -> 10(2)n1 cannot be {37}
-> 3 in n1 in r23c1
-> r7c2 = 3
-> r6c3 = 3

9. 2 in 39(8) only in r6c789
-> r7c3 = 2
-> r7c1 = 1
-> r6c2 = 1

10. -> 1 not in 23(5)r2c1
Also r56c1 = +14(2)
-> Combined cage 23(5)r2c1 + H+14(2) = {2345689}
-> 7 in n4 in r45c3
-> 7 in n1 in r13c2
-> 21(3)n7 = [7{68}]

11! Whatever goes in r3c5 must go in n5 in r4c4 and -> in n8 in r789c6 (= [918])
-> r3c5 = r4c4 from (89) (Cannot be 1 since 20(3) cannot contain a 1).
a) r4c4 = 8 -> r45c3 = {57}
b) r4c4 = 9 -> r45c3 = {47}
-> Whichever of (45) is in r45c3 is also in 23(5) in r23c1 ...
and also in D\ in r5c5 (!)
and -> in r4c3
-> r5c3 = 7!

12. -> Disjoint 9(2)r5c8 can only be [81]
-> r8c7 = 5
-> 15(4)n9 = [4362]
-> r6c789 = [{29}4]
-> r4c3 = r5c5 cannot be 5
-> r4c3 = r5c5 = 4
-> r4c4 = r3c5 = 9
Also r23c1 = {34}
-> 10(2)n1 = [91]
Also r9c2 = 4

13. Remaining cells in D\ -> r1c1,r2c2 = {58}
-> HS 8 in c3 -> r8c23 = [68]
Also HS 5 in c3 -> 18(3)n7 = [945]
-> NS r1c3 = 6
Also r56c1 = [68]
-> r1c1,r2c2 = [58]
-> NP r13c2 = {27}
-> 23(5)r2c1 = [{34}259]
-> HS 8 in r4 -> r4c5 = 8
-> r4c789 = [6{17}]
-> r5c9 = 5
Since 2 already on D/ -> 15(3)r3c7 = [861]
-> r4c9 = 7
-> 17(3)r3c8 = [2375]
-> r6c78 = [29]
-> Remaining cell on D/ -> r1c9 = 1, r2c8 = 5
-> 17(3)n3 = [971]
-> 15(3)n3 = [456]
Also r13c2 = [27]
Also r23c1 = [34]
Also r1c46 = [84]
Also r3c46 = [65]
-> r2c456 = [217]
-> r5c46 = [12]
-> r6c5 = 5
Also 18(4)r6c4 = [7542]


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