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 Post subject: Assassin 373
PostPosted: Mon Apr 01, 2019 8:00 am 
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Posts: 1040
Location: Sydney, Australia
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NOTE: Disjoint 37(8) at r5c4 with r9c78

Assassin 373
Was stuck for an eternity before finding a nice short chain to get things moving. My optimised solution gets to that step much quicker than my first solve so misses out about 30 steps! Felt hard. SudokuSolver score 1.65 JSudoku uses 6 chains.
code:
3x3::k:3328:4609:4609:3586:3586:3331:3331:3331:4356:3328:4609:4609:3586:4357:3846:4103:4103:4356:3080:3080:4357:4357:4357:3846:3846:4103:4356:3593:3593:5386:3083:3083:3083:4108:4108:4108:3593:5386:5386:9485:9485:2830:4367:1808:1808:4881:4881:4881:4881:9485:2830:4367:5650:4627:788:788:4373:9485:9485:9485:4367:5650:4627:5654:5654:4373:3863:3863:4888:5650:5650:4627:5654:5654:3863:3863:4888:4888:9485:9485:4627:
solution:
Code:
+-------+-------+-------+
| 8 6 4 | 3 9 7 | 1 5 2 |
| 5 7 1 | 2 6 4 | 3 9 8 |
| 9 3 2 | 8 1 5 | 6 4 7 |
+-------+-------+-------+
| 2 8 7 | 6 5 1 | 4 3 9 |
| 4 9 5 | 7 2 3 | 8 6 1 |
| 3 1 6 | 9 4 8 | 2 7 5 |
+-------+-------+-------+
| 1 2 9 | 5 3 6 | 7 8 4 |
| 6 4 8 | 1 7 9 | 5 2 3 |
| 7 5 3 | 4 8 2 | 9 1 6 |
+-------+-------+-------+

Cheers
Ed


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 Post subject: Re: Assassin 373
PostPosted: Sat Apr 06, 2019 7:48 pm 
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Thanks Ed! Tricky little puzzle this one. Took a while! Had to some combination analysis which is more yours and Andrew's forte than mine.
I too used one little contradiction chain (Step 8) which I'm pretty sure is different than what you used. Maybe between us we could do away with both :)
Edited twice. Missed a logical step, and corrections from Andrew.
Assassin 373 WT:
1. 3(2)n7 = {12}
17(2)n7 = {89}
Innies n7 -> r9c3 = 3
-> 17(4)n7 = {4567}

2. Outies n4 -> r6c4 = 9
(89) already in c3 -> 21(3)n4 cannot be {489}
-> 21(3)n4 = {678} or {579} with 8 or 9 in r5c2 and 7 in r45c3
IOD r4 -> r4c3 = r5c1 + 3
-> Since r4c3 from (567) -> r5c1 from (234)
-> The other of (89) in n4 in r4c12

3. Innies n78 = r7c456 = +14(3) which can only be {3(47|56)}
-> 9 in 37(8) only in r9c78
Innies n69 -> r9c78 = +10(2) = {19}
-> 37(8) in n5 = r5c45,r6c5 = {2(56|47)}

4. 9 in n8 only in r8c56
-> 17(2)n7 = [98]
-> r7c789 = {8(56|47)}
-> 8 in n8 in r9c56 (15(4)n8 already contains a 3 so cannot contain an 8)

5. 1 in n8 only in r8c45
Either 15(4)n8 = [1932] or 19(3)n8 = [9{28}]
-> 2 in n8 in r9

6. 1 in n5 only in r4
-> 1 in n4 in r6
-> 1 in n6 in r5
Since 9 already in r6 and in n9 -> 1 cannot be in r5c7 (17(3) cannot be [1{79}])
-> 7(2)n6 = {16}
-> Innies c9 = r45c9 = +10(2) = [91] or [46]

7! Since (19) already in r6 and in n9 -> neither 22(4)r6c8 nor 18(4)r6c9 can contain either 1 or 9.
(23) in n9 both in r8
Since 22(4) cannot contain a 9 -> 22(4) cannot contain both (23)
-> r8c9 from (23)
Since 22(4) contains one of (23) and not a 9 -> 22(4) must contain an 8.
-> 8 in r67c8

8! r89 in n9 contains (19) in r9c78, (23) in r8, and either (47) or (56)
-> one of (4567) in r9c9 and one of (7654) (= 11 - r9c9) in r8c78

Min r789c9 = {245} = +11
-> Max r6c9 = 7
-> 8 in c9 only in r123c9 or r7c9

Can 8 go in r7c9?

That would put 8 in r6c8, and ...
... 18(4)r6c9 = [2835] or [3825], and ...
... (since r9c9 = 5) -> 6 in r8c78 and r7c78 = {47} ...
... which leaves no solution for 22(4)r6c8

-> 8 in c9 in r123c9

9. -> HS 9 in c9 -> r45c9 = [91]
-> r5c8 = 6
Also HS 9 in n4 -> r5c2 = 9
-> r45c3 = {57}
-> 14(3)n4 = {248}
-> r6c123 = {136}

10. Innies r4 = r4c123 = +17(3) = [{28}7] or [{48}5]
-> 16(3)n6 = [{34}9]
-> r4c123 = [{28}7]
-> r5c123 = [495]
Also 12(3)n5 = {156}
-> HS 3 in n5 -> 11(2)n5 = [38]
-> HS 8 in r5 -> r5c7 = 8
-> r6c789 = {257}
-> r5c45 = {27}
-> r6c5 = 4
-> r7c456 = {356}
-> r7c789 = <487>

11. HS 8 in n8 -> r9c5 = 8
IOD c6789 (given r9c78 = +10) -> r4c6 + r7c6 = r9c5 - 1 = 8 - 1 = +7
Since r4c6 from (156) and r7c6 from (56) -> r47c6 = [16]
-> r4c45 = {56} and r7c45 = {35} -> 5 in n2/c6 in r123c6

12. 9 in n1 in c1 and since 4 already in c1 -> 12(2)n1 = [93]
-> HS 3 in n4 -> r6c1 = 3
4 in n7 in r89c2 -> at least one of (67) in c1 in r89c1 -> 13(2)n1 = {58}
-> r4c12 = [28]
-> 3(2)n7 = [12]
Also r89c1 = {67}
-> r89c2 = {45}
Innies n1 -> r3c3 = 2
-> 18(4)n1 = {1467} with 4 in r12c3 and 7 in r12c2

13! 3 in n2 in r12c45
But since 5 already in n2 in c6 and (48) already in c5 -> no solution for 17(4)n2 if r2c5 = 3
-> 3 in n2 in 14(3)
-> 14(3)n2 = {239} or {347}
-> HS 8 in n2 -> r3c4 = 8
-> HS 6 in n2 -> r23c5 = {16}
-> 12(3)n5 = [651]
-> r7c456 = [536]
Also r8c4 = 1

14! r3c6789 = {457(1|6))
Since 17(3)n3 is missing two values that sum to 9 -> r3c89 cannot be {45}
Since 6 already in n2 -> r3c67 cannot be {45}
-> One of (45) in r3c67 and the other in r3c89.
Since 3 already in c6 15(3)r2c6 cannot be [3{57}]
Also 15(3)r2c6 cannot contain both 4 and 7
-> 7 in r3 in r3c89
-> r3c7 from (16)

15. -> IOD n3 -> r1c6 from (27)
Either way 9 in n3 not in r1c78
-> 9 in n3 in r2c78

-> HS 9 in n2 -> r1c5 = 9
-> r12c4 = {23}
-> r123c6 = [7{45}]
-> r3c7 = 6
-> r23c5 = [61]
Also 9 in n3 in 16(3) in r2c78
-> HS 7 in n3 -> r3c9 = 7
-> r12c9 = {28}
-> r1c78 = {15}
-> 16(3)n3 = [{39}4]
-> r4c78 = [43]
-> r2c78 = [39]
-> r12c4 = [32]
Also 13(2)n1 = [85]
-> r12c9 = [28]
Also NS r6c9 = 5
-> r67c7 = [27]
-> 22(4)r6c8 = [7852]
etc.


Last edited by wellbeback on Fri Apr 19, 2019 9:31 pm, edited 1 time in total.

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 Post subject: Re: Assassin 373
PostPosted: Wed Apr 10, 2019 7:25 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
One of the rare puzzles to get four ! steps from wellbeback. I used a very similiar path to wellbeback for the early steps, diverged while I reduced other areas, then converged again to do the same one key elimination from both our chains (my step 23d.). Suggests it may have a very narrow solving path. Will be interesting to see if Andrew gets that as his key elimination also. I didn't use the last of wellbeback's ! steps. Must have found something else instead but can't think what.

WT for a373:
Preliminaries from SudokuSolver
Cage 3(2) n7 - cells ={12}
Cage 17(2) n7 - cells ={89}
Cage 7(2) n6 - cells do not use 789
Cage 12(2) n1 - cells do not use 126
Cage 13(2) n1 - cells do not use 123
Cage 11(2) n5 - cells do not use 1
Cage 21(3) n4 - cells do not use 123
Cage 19(3) n8 - cells do not use 1
Cage 37(8) n589 - cells ={12345679}



1. "45" on n23: 1 outie r3c3 = 2
1a. "45" on n1234: 1 outie r6c4 = 9
1b. "45" on n7: 1 innie r9c3 = 3

2. 3(2)n7 = {12} only: both locked for r7 and n7

3. "45" on n78: 3 innies r7c456 = 14 = {347/356}(no 9) = 4 or 6
3a. must have 3: 3 locked for r7 and n8 and for 37(8) in n5

4. "45" on n69: 2 innies r9c78 = 10
4a. but {46} blocked by r7c456 (step 3.)
4b. = {19} only: both locked for r9 and n9, and 1 for 37(8) in n5
[Andrew noticed that 9 which must be in 37(8) only in r9c78 = {19} so don't have to do the {46} clash in 4a].

5. hidden single 9 in r7 -> r78c3 = [98]
5a. max. r67c7 = {78} = 15 -> min. r5c7 = 2
[Alt way Andrew noticed: {179} blocked by r9c7 = (19) -> no 1 both r56c7)]

6. 15(4)r9c3: [3]{246} blocked by 4/6 in r7c456
6a. = [3]{129/147/156}(no 8)
6b. 2 in {129} must be in r9c4 -> no 2 in r8c45
[Alt way Andrew noticed: 15(4)r9c3 must contain 1 for n8]

7. 19(3)n8: must have 8 for n8 and is only in r9: 8 locked for r9
7a. = {289/478/568}
7b. 9 in {289} must be in r8c6 -> no 2 in r8c6

8. 2 in n8 only in r9: locked for r9

Took a long time to see this.
9. 22(4)r6c8 = {1678/2578/3478/3568/4567}: note- can't have both 2 & 3
9a. -> hidden killer 2,3 in r8
9b. r8c9 = (23)
9c. 22(4) must have one of 2 or 3 = {2578/3478/3568}(no 1)
9d. must have 8: 8 locked in r67c8: locked for c8
9e. no 2,3 in r6c8 since 22(4) can't have both 2,3

10. 21(3)n4: {489} blocked by 8,9 only in r5c2
10a. = {579/678}(no 4)
10b. r5c2 = (89)
10c. must have 7: locked for c3 and n4

11. "45" on r4: 1 innie r4c3 - 3 = 1 outie r5c1
11a. =[52/63/74]; r5c1 = (234)

12. 1 in r5 only in 7(2)n6 = {16} only: both locked for n6: 6 for r5

13. "45" on c9: 2 innies r45c9 = 10 = [91/46]

14. 16(3)n6 must have 4,9 for r4c9 = {259/349/457}(no 8) = 5 or 9, 4 or 5

15. "45" on n5: 3 innies r5c45+r6c5 = 13 and must have 2 for 37(8) = {247/256}
15a. 2 locked for n5

16. 1 in n5 only in 12(3) = {138/147/156}
16a. 1 locked for r4

17. "45" on r4: 3 innies r4c123 = 17
17a. but {359} blocked by 5 or 9 in 16(3)n6
17b. but {368} as {38}[6] only, blocked by 14(3)r4c1 can't be {38}[3]
17c. but {458} blocked by 16(3)n6 = 4 or 5
17d. but {467} as {46}[7] blocked by 14(3)r4c1 can't be {46}[4]
17e. = {269/278}(no 3,4,5)
17f. must have 2: locked for r4 and n4
17g. can't have both 6 & 7 -> no 6 in r4c12

18. 3 in r4 in 16(3) = {349} or in 12(3)n5 -> 4 in 12(3) must also have 3 or there would be no 3 for r4
18a. -> 12(3)n5: {147} blocked (Locking-out cages)
18b. = {138/156}(no 4,7)
[Andrew noticed a simpler way, 16(3) = {349/457}: 4 locked for r4]

19. 4 & 7 in n5 only in 37(8) or r56c6: r7c6 sees all those cells -> no 4,7 in r7c6 (Common Peer Elimination CPE)

20. "45" on c6789: (remembering r9c78 = 10), 2 innies r47c6 + 1 = 1 outie r9c5
20a. max. r9c5 = 8 -> max. r47c6 = 7
20b. -> r4c6 = 1
20c. -> r7c6 + 2 = r9c5 = [35/57/68]

21. 2 & 9 in c6 in r89c6 or must both be in r123c6
21a. "45" on n3: 3 outies r123c6 = 16
21b. ->{268/349} blocked
21c. but {367} blocked by 11(2) r56c6 = {38/47/56}
21d. h16(3)n2 = {259/358/457}(no 6)
21e. must have 5: locked for n2 and c6
21f. no 6 in r6c6
21g. no 7 in r9c5 (step 20c)

A difficult one
22. "45" on r6789: 4 outies r5c4567 = 20
22a. but {3458} blocked by r5c1 = (34)
22b. = {2378/2459}
22c. 3 in r4 in 12(3) = {38}[1] or 3 in r4c789
22d. = 8in n5 or 3 in n6
22e. -> {2378} as {27}[83] blocked from r5c4567, can only be {27}[38]
22f. -> no 8 in r5c6, no 3 in r6c6

The key chain
23. 8 in r9c5 -> 6 in r7c6 (step 20c) -> h13(3) in 37(8) in n5 = {247} -> r6c6 = 8 -> 8 which must be in 22(4)r6c8 in r7c8
or, r9c5 = 5
ie, r7c8 = 8 or r9c5 = 5
23a. 18(4)r6c9 = {2358/2367/2457/3456}
23b. but {2358} must have 2,3 in r68c9 and [85] in r97c9
23c. but this is blocked by r7c8 = 8 or r9c5 = 5!
23d. ->18(4)r6c9 = {2367/2457/3456}(no 8)

cracked.
24. 8 in c9 only in 17(3) = {278/368/458}(no 1,9)
24a. 8 locked for n3

25. r45c9 = [91] (hidden singles c9), r5c8 = 6
25a. r4c78 = 7 = {34} only: both locked for n6 and 3 for r4
25b. r4c45 = 11 = {56} only: both locked for n5 and 6 for r4
25c. r4c3 = 7, r5c23 = [95]
25d. r4c12 = {28} = 10 -> r5c1 = 4; 8 locked for n4

26. r5c45 = {27}: both locked for n5 and r5, r5c67 = [38], r67c6 = [86] -> r9c5 = 8 (step 20c), r6c5 = 4

27. 13(2)n1: {67} blocked by r89c1 = two of {567}
27a. = {58} only: both locked for c1 and n1, r4c12 = [28], r7c12 = [12]

28. 12(2)n1 = [93] only permutation

29. "45" on n3: 1 outie r1c6 - 1 = 1 innie r3c7
29a. no 4,9 in r1c6, no 5,7 in r3c7

30. no 1,3 in r1c6 -> no 9 in r1c78

31. hidden single 9 in r1 -> r1c5 = 9
31a. -> r12c4 = 5 = {14/23}(no 6,7,8)

32. hidden single 9 in n8 -> r89c6 = [92]

33. r123c6 = {457} only: 4,7 locked for n2
33a. -> r12c4 = {23} only: both locked for c4 and 3 for n2
33b. 4 in n2 must be in 15(3) = {45}[6] only
33c. r1c6 = 7
33d. -> r1c78 = 6 = {15/24}(no 3)

34. 16(3)n3:{457} blocked by r1c78
34a. = {259/349}(no 1,7)

35. r3c9 = 7 (hidden single r3)

36. 18(4)r6c9 = [5436] only permutation

on from there
Cheers
Ed


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 Post subject: Re: Assassin 373
PostPosted: Fri Apr 19, 2019 4:05 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
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It was certainly a tricky Assassin; quite an easy start but then it needed quite a lot of thought to eliminate one of the two alternative combinations in many parts of the puzzle.

Thanks to Ed for pointing out that my step 4d, now deleted, was flawed and for other corrections. Fortunately removing step 4d wasn't too critical for the rest of my solving path.
Here is my walkthrough for Assassin 373:
Prelims

a) R12C1 = {49/58/67}, no 1,2,3
b) R3C12 = {39/48/57}, no 1,2,6
c) R56C6 = {29/38/47/56}, no 1
d) R5C89 = {16/25/34}, no 7,8,9
e) R7C12 = {12}
f) R78C3 = {89}
g) 21(3) cage at R4C3 = {489/579/678}, no 1,2,3
h) 19(3) cage at R8C6 = {289/379/469/478/568}, no 1
i) 37(8) disjoint cage at R5C4 = {12345679}, no 8

Steps Resulting From Prelims and Initial Placements
1a. Naked pair {12} in R7C12, locked for R7 and N7
1b. Naked pair {89} in R78C3, locked for C3 and N7
1c. 45 rule on N1 1 innie R3C3 = 2
1d. 45 rule on N4 1 outie R6C4 = 9, clean-up: no 2 in R56C6
1e. 45 rule on N7 1 innie R9C3 = 3
1f. R6C4 = 9 -> R6C123 = 10 = {127/136/145/235}, no 8

2. 45 rule on N78 3 innies R7C456 = 14 = {347/356}, no 9, 3 locked for R7, N8 and 37(8) disjoint cage at R5C4
2a. 45 rule on N69 2 innies R9C78 = 10, must contain 9 for 37(8) disjoint cage = {19}, locked for R9 and N7, 1 also locked for 37(8) cage
2b. 2 in 37(8) cage only in R5C45 + R6C5, locked for N5
2c. R7C3 = 9 (hidden single in R7) -> R8C3 = 8
2d. 1 in N5 only in 12(3) cage at R4C4, locked for R4
2e. 19(3) cage at R8C6 = {289/478/568} (cannot be {469} which clashes with R7C456), 8 locked for R9
2f. 9 of {289} must be in R8C6 -> no 2 in R8C6

3a. 21(3) cage at R4C3 = {579/678} (cannot be {489} because 8,9 only in R5C2), no 4
3b. 8,9 of {579/678} only in R5C2 -> R5C2 = {89}
3c. 21(3) cage = {579/678}, 7 locked for C3 and N4
3d. 45 rule on R4 1 innie R4C3 = 1 outie R5C1 + 3, R4C3 = {567} -> R5C1 = {234}
3e. Hidden killer pair 8,9 in 14(3) cage at R4C1 and R5C2 for N4, R5C2 = {89} -> 14(3) cage must contain one of 8,9 = {239/248}, no 5,6, 2 locked for N4

4. 45 rule on C6789 using R9C78 = 10, 1 outie R9C5 = 2 innies R47C6 + 1
4a. Min R47C6 = 4 -> min R9C5 = 5
4b. Max R9C5 = 8 -> max R47C6 = 7, no 7 in R7C6
4c. Max R47C6 = 7, min R7C6 = 3 -> max R4C6 = 4
4d. Deleted

5a. 17(3) cage at R5C7 = {269/278/359/368/458/467} (cannot be {179} which clashes with R9C7), no 1
5b. 1 in R5 only in R5C89 = {16}, locked for N6, 6 also locked for R5 clean-up: no 5 in R6C6
5c. 45 rule on C9 2 innies R45C9 = 10 = [46/91]
5d. 16(3) cage at R4C7 = {259/349/457} (cannot be {358} because R4C9 only contains 4,9), no 8
5e. Consider placement for 2 in N4
2 in R4C12 => 16(3) cage = {349/457}
or 2 in R5C1 => R4C3 = 5 (step 3d) => 16(3) cage = {349}
-> 16(3) cage = {349/457}, no 2, 4 locked for R4 and N6
5f. 1 in N5 only in 12(3) cage at R4C4 = {138/156}, no 7
5g. 2 in R4 only in R4C12, locked for N4, clean-up: no 5 in R4C3 (step 3d)
[After this step I missed Ed’s 4,7 in N5 only in R5C456 + R6C56, CPE no 4,7 which would have led to an early placement R4C6 = 1, as in his walkthrough.]

6. R9C3 = 3 -> R8C45 + R9C4 = 12 and must contain 1 for N8 = {129/147/156}
6a. 2 of {129} must be in R9C4 -> no 2 in R8C45
6b. 2 in N8 only in R9C46, locked for R9

7. 1 in N1 only in 18(4) cage at R1C2 = {1359/1368/1458/1467}
7a. Consider combinations for R6C123 = 10 (step 1f) = {136/145}
R6C123 = {136} => 4 in C3 only in R12C3 => 18(4) cage = {1458/1467}
or R6C123 = {145} => R45C3 = [67] => R126C3 = {145} => 18(4) cage = {1359/1458/1467} (cannot be {1368} which requires 1,6 in R12C3)
-> 18(4) cage = {1359/1458/1467}
7b. R3C12 = {39/48} (cannot be {57} which clashes with 18(4) cage), no 5,7
7c. R12C1 = {58/67} (cannot be {49} which clashes with R3C12), no 4,9

8. Hidden killer pair 2,3 in 22(4) cage at R6C8 (in R8C78) and R8C9 for R8, 22(4) cannot contain both of 2,3 (because it doesn’t contain 9) -> R8C9 = {23}, R8C78 must contain one of 2,3
8a. 22(4) cage can only contain one of 2,3 in R8C78 -> no 2,3 in R6C8
8b. 22(4) contains one of 2,3 = {2578/3478/3568}, 8 locked for C8
8c. 18(4) cage at R6C9 contains at least one of 2,3 = {2358/2367/2457/3456}
8d. 8 of {2358} must be in R7C9 -> no 8 in R6C9

9. 45 rule on N3 1 outie R1C6 = 1 innie R3C7 + 1, no 1,3 in R1C6, no 9 in R3C7
9a. 45 rule on N3 3 innies R1C78 + R3C7 = 12 = {138/147/156/237/246/345} (cannot be {129} = {29}1 because 13(3) cage at R1C6 cannot be 2{29}), no 9

10. 45 rule on N12 3 innies R123C6 = 16
10a. Hidden killer pair 2,9 in R123C6 and R89C6 for C6, R89C6 must contain both or neither of 2,9 -> R123C6 must contain both or neither = {178/259/358/457} (cannot be {169/268/349} which only contain one of 2,9, cannot be {367} which clashes with R56C6), no 6, clean-up: no 5 in R3C7 (step 9)
[Before I spotted the hidden killer pair, I’d eliminated {169/349} which clash with R4567C6 and R567C6 respectively, killer ALS blocks.]
10b. Consider combinations for R123C6
R123C6 = {178} => R89C6 = [92] (hidden pair in C6)
or R123C6 = {259/358/457}, 5 locked for C6
-> no 5 in R89C6
10c. 19(3) cage at R8C6 (step 2e) = {289/478/568}
10d. 6 of {568} must be in R8C6 -> no 6 in R9C6
10e. 17(4) cage at R2C5 contains 2 = {1268/2348/2456} (cannot be {1259/2357} which clash with R123C6), no 7,9
10f. 14(3) cage at R1C4 = {167/239/347} (cannot be {149/158/248/356} which clash with 17(4) cage, cannot be {257} which clashes with R123C6), no 5,8

11a. Hidden killer pair 1,9 in 17(3) cage at R1C9 and R45C9 for C9, R45C9 contains both or neither of 1,9 -> 17(3) cage must contain both or neither of 1,9 = {179/278/368/458/467} (cannot be {269/359} which contain 9 but not 1)
11b. 9 in N3 only in 17(3) cage at R1C9 and 16(3) cage at R2C7, 17(3) cage only contain 9 in {179}, 16(3) cage cannot contain both of 7,9 so cannot contain 7 = {169/259/268/349/358}, no 7

[I had to think hard to find this forcing chain.]
12. R9C5 = R47C6 + 1 (step 4), R123C6 (step 10a) = {178/259/358/457}
12a. Consider combinations for R7C456 (step 2) = {347/356}
R7C456 = {347} => R5C45 + R6C5 = {256} => R56C6 = {38/47} => R123C6 (step 10a) = {259/358/457} (cannot be {178} which clashes with R56C6)
or R7C456 = {356}, locked for N8, R9C5 = {78} => R47C6 = 6,7 = [15/16] (cannot be [34] because no 4 in R7C6 in this path)
-> R123C6 = {259/358/457}, no 1, 5 locked for C6 and N2, clean-up: no 6 in R6C6
[Amended because step 4d deleted.]
12b. R4C6 = 1 (hidden single in C6)
[Extra sub-step, now that I know about Ed’s CPE.
4 in N5 only in R5C456 + R6C56, CPE no 4 in R7C6.]

12c. R9C5 = {58} -> R47C6 = 4,7 -> R7C6 = {36}
12d. 6 in C6 only in R78C6, locked for N8
12e. 15(4) cage at R8C4 contains 1 for N8 = {1239/1347}, no 5
12f. 17(4) cage at R2C5 (step 10e) = {1268/2348}, 8 locked for N2, clean-up: no 3 in R123C6, no 7 in R3C7 (step 9)
12g. 15(3) cage at R2C6 cannot be {249} = [294] which clashes with R3C12 -> no 4 in R3C7, clean-up: no 5 in R1C6 (step 9)

13. 45 rule on R6789 4 outies R5C4567 = 20, must contain 2 for R5 = {2378/2459}
13a. {2378} = {27}[38] (cannot be {27}[83] because 16(3) cage at R4C7 (step 5e) must contain both or neither of 3,9), {2459} = {25}[49] -> R5C45 {25/27}, R5C6 = {34}, R5C7 = {89}
13b. R5C6 = {34} -> R6C6 = {78}
13c. Hidden killer pair 4,6 in R6C123 and R6C5 for R6, R6C123 (step 1f) contains one of 4,6 -> R6C5 = {46}
13d. 4 in 37(8) disjoint cage only in R6C5 + R7C45, CPE no 4 in R8C5
13e. 17(3) cage at R5C7 (step 5a) = {269/278/359/368/458} (cannot be {467} because R5C7 only contains 8,9)
13f. R5C7 = {89} -> no 8 in R67C7
13g. 2 of {278} must be in R6C7 -> no 7 in R6C7
13h. Consider placement for 8 in N6
R5C7 = 8 => 17(3) cage = {278/368/458}
or R6C8 = 8 => R7C9 = 8 (hidden single in N9) => R689C9 = 10 = {23}5 => 17(3) cage = {269}
-> 17(3) cage = {269/278/368/458}
13i. 17(3) cage {458} = [854] -> no 5 in R7C7
13j. Consider combinations for R7C456 (step 2) = {347/356}
R7C456 = {347}, locked for R7 => R7C789 = [685] (cannot be [658] which clashes with 22(4) cage at R6C9 = [2835/3825]) => R5C7 = 8 (hidden single in N6)
or R7C456 = {356}, locked for 37(8) disjoint cage => R6C5 = 4, R5C6 = 3 => R5C4567 = {27}[38]
-> R5C7 = 8, clean-up: no 9 in R1C6 (step 9)
[Cracked. The rest is fairly straightforward.]

14a. R5C2 = 9 -> R45C3 = 12 = [75]
14b. R6C123 (step 1f) = {136} (only remaining combination), locked for R6, 3 also locked for N4
14c. R6C5 = 4, R5C6 = 3 -> R6C6 = 8
14d. 16(3) cage at R4C7 = {349} (hidden triple in R4)
14e. R7C6 = 6, R9C5 = 8 (hidden single in N8) -> R7C45 = {35} (hidden pair in N8), 5 locked for R7
14f. R3C4 = 8 (hidden single in N2) -> R3C12 = [93]
14g. R5C1 = 4, R467C1 = [231] (hidden triple in C1) -> R4C2 = 8
14h. 8 in N1 only in R12C1 = {58}, 5 locked for C1 and N1
14i. R89C2 = {45} (hidden pair in N7), 4 locked for C2
14j. R3C34 = [28] = 10 -> R23C5 = 7 = {16}, locked for C5 and N2
14k. R4C45 = [65], R7C45 = [53]
14l. Naked pair {16} in R3C57, locked for R3
14m. R3C7 = {16} -> R1C6 = {27} (step 9)
14n. R7C8 = 8 (hidden single in C8)

15. 8 in C9 only in 17(3) cage at R1C9 = {278} (only possible combination, cannot be {368} because R3C9 only contains 4,5,7, cannot be {458} which clashes with R3C8) -> R3C9 = 7, R12C9 = {28}, 2 locked for C9 and N3
15a. R45C9 = [91] (hidden pair in C9) -> R5C8 = 6
15b. 9 in N3 only in 16(3) cage at R2C7 = {349} (only possible combination, cannot be {169} which clashes with R3C7) -> R3C8 = 4, R2C78 = {39}, locked for R2, 3 also locked for N3
15c. R123C6 (step 12a) = {457} (only remaining combination) = [745], R3C7 = 6 (step 9)

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A373 at a full 1.5. I used several forcing chains, some weren't easy ones to find.


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