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Afmob
 Post subject:
PostPosted: Sun Mar 14, 2010 6:01 pm 
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thanks for this challenging variant!

I hope there is a better way to solve this Killer. A lot of times Grouped Killer subsets were almost available but one candidate was missing, so maybe I missed something.

A188 V2 Walkthrough:

1. N6789
a) 23(3) = {689} locked for C8
b) Outies N6789 = 9(3) <> 7,8,9
c) 9(2) @ N5: R5C7 <> 1,2
d) 25(6) must have 1 and 2
e) 1,2 locked in 25(6) + 7(3) for R67 (Caged X-Wing)
f) 11(2): R8C1 <> 9
g) Innies R89 = 12(2) <> 1,2,6; R8C9 <> 3
h) 11(2): R7C1 <> 5,9

2. R6789 !
a) Innies+Outies R6789: 2 = R5C89 - R6C5: R6C5 <> 3,4 and R5C9 <> 6,7,8,9 since R5C8 >= 6
b) Innies N7 = 12(3) <> {345} since it's a Killer triple of 11(2)
c) ! Killer quad (6789) locked in 11(2) + 15(2) + Innies N7 for N7
d) 7(2) <> 1
e) Innies+Outies R89: -1 = R7C1 - R8C9: R7C1+R8C9 <> [45] since it's a Killer pair of 7(2)
-> R7C1 <> 4, R8C9 <> 5
f) 1,6 locked in Outies R9 = 26(5) = 16{289/379/478} <> 5
g) 5 locked in R8C123 @ R8 for N7
h) 1 locked in Innies N7 = 12(3) <> 6

3. C159+R12
a) Innies C5 = 7(2) <> 7,8,9
b) 8(2): R1C9 <> 2
c) Innies C9 = 11(3) <> 9; R59C9 <> 5 since R1C9 <> 2,4
d) Innies R12 = 9(2) <> 9; R2C1 <> 6,7,8

4. R456 !
a) Innies R5 = 21(5) <> {23457} since R5C8 = (689)
b) Innies R5 = 21(5) = 1{2369/2378/2459/3458} since other combos blocked
by Killer pairs (67,68) of 15(2) -> 1 locked for R5
c) 9(2) <> 8
d) Innies+Outies R6789: 2 = R5C89 - R6C5: R5C9 <> 4,5 since R5C89 = 6[4/5] blocked by
Killer triples (467,567) of 9(2) + 15(2)
e) ! Consider placement of 9 in 22(3) -> R4C7 <> 9:
- i) R3C3 = 9 -> 9 locked in R12C7 @ N3 for C7
- ii) 9 locked in R4C34 for R4
f) 15(3) <> 1 because {168} blocked by R56C8 = (689)

5. N56 !
a) ! Innies+Outies N6789: -7 = R6C4 - R567C7: R5C7 <> 4 since R67C7 = (124) and R567C7 > 7
b) Innies N6 = 30(6): R6C9 <> 6,8,9 since R56C8 = (689) and R5C7 <> 1,2,4
c) 9 locked in 23(3) @ N6 for 23(3)

6. R789 !
a) ! Hidden Killer pair (59) locked in 25(6) + R7C9 for R7 since 25(6) can only have one of them
-> R6C4 <> 5 and R7C9 = (59), 25(6) = 126{349/358/457}
b) ! Hidden Killer pair (37) locked in R7C1 + 25(6) for R7 since 25(6) can only have one of them
-> R6C4 <> 3 and R7C1 = (37)
c) 6 locked in 15(2) @ N7 = {69} locked for R9+N7
d) Innies N7 = 12(3) = 1{38/47}
e) 7(2) = {25} locked for R8 since (34) is a Killer pair of Innies N7
f) 11(2): R8C1 <> 3,7
g) Innies R89 = 12(2) = {48} locked for R8

7. C789
a) 21(4) = {1389/1479/1578/3459} <> 2 since R7C9 = (59) and R8C9 = (48)
b) 15(3) <> {357} since (37) is a Killer pair of 21(4)
c) Killer triple (689) locked in 15(3) + R56C8 for N6
d) 9(2): R5C6 = (246)

8. N5+C5
a) Outies N6789 = 9(3) = {126} locked for N5; 1 also locked for R6
b) 17(3) = 5{39/48} -> 5 locked for C5+N5
c) Innies C5 = 7(2) = {16} locked for C5 since (34) is a Killer pair of 12(2)
d) 9(2) @ C5 = {27} -> R8C5 = 7, R9C5 = 2
e) 25(6) must have 2 -> R6C4 = 2
f) R5C6 = 6 -> R5C7 = 3

9. R789
a) Innies N78 = 13(2) = [94] -> R8C6 = 9, R9C6 = 4
b) Cage sum: R9C7 = 5
c) R7C9 = 9
d) 21(4) = {1479} -> R5C9 = 1, R8C9 = 4, R6C9 = 7
e) R8C1 = 8 -> R7C1 = 3

10. C789
a) Innies C9 = 11(3) = {236} -> R9C9 = 3, R1C9 = 6, R4C9 = 2
b) Cage sum: R1C8 = 2
c) R8C8 = 1
d) 10(2) = {37} -> R2C7 = 7, R2C8 = 3,
e) R6C7 = 4, R4C8 = 5 -> R4C7 = 8, R3C8 = 4
f) 24(6) = 1234{59/68} -> R4C6 = 3, R3C6 = 2; R3C4 = (58); 1 locked for R3
g) 16(3) = {178} -> R1C7 = 1, R2C6 = 8, R1C6 = 7

11. Rest is singles.

Rating:
1.5. I used a forcing chain.


Last edited by Afmob on Sat Feb 11, 2012 8:22 am, edited 2 times in total.

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Ed
 Post subject: Re:
PostPosted: Sun Apr 18, 2010 9:07 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Afmob wrote:
A lot of times Grouped Killer subsets were almost available but one candidate was missing.
This is what got me hooked on this puzzle many weeks ago but couldn't do anything with it. Have now (step 14)! :) While I'm happy with the resulting walkthrough, it was a terrible time to get something decent. Kept making errors but fortunately, kept finding simpler ways. Finally found my steps 10 & 11 which gave the same elimination as Afmob's last ! (step 6b) so it seems that was the crucial elimination for this puzzle. I got it very differently since I missed Afmob's step 5b.

I like Afmob's use of "Caged X-Wing" so have used it too.

Finally, thanks for a very interesting puzzle Mike!

A188 V2 Walkthrough:
Note: this is an optimized solution so many obvious eliminations are left out though I do try and do clean-up along the way.

prelims
i. 13(2)n1: no 1,2,3
ii. 12(2)n2: no 1,2,6
iii. 8(2)n3: no 4,8,9
iv. 14(4)n1: no 9
v. 10(2)n3: no 5
vi. 13(2)n3: no 1,2,3
vii. 22(3)n1: no 1,2,3,4
viii. 15(2)n4 = {69/78}
ix. 9(2)n5: no 9
x. 23(3)n6 = {689}
xi. 7(3)n5 = {124}
xii. 11(2)n7: no 1
xiii. 7(2)n7: no 7,8,9
xiv. 9(2)n8: no 9
xv. 15(2)n7 = {69/78}
xvi. 9(2)n8: no 9

1. 25(6)n5 must have 1 & 2. 7(3)n5 must have 1 & 2 -> 1 & 2 locked for r67 (Caged X-Wing)
1a. no 9 in r8c1

2. "45" on r89: 2 innies r8c19 = 12 (no 1,2,6), no 3 in r8c9
2a. no 5,9 in r7c1

3. "45" on n7: 3 innies r7c23 + r9c3 = 12
3a. 9 in n7 in h12(3) = {129} or in 15(2) = {69} = [1/6..]-> {16} blocked from 7(2) and {156/246} from h12(3) (Locking-out cages)

4. h12(3)n7 must have 1 for n7 = {129/138/147}(no 5,6)

5. 5 in n7 only in r8: 5 locked for r8
5a. no 4 in r9c5
5b. no 7 in r8c1 (h12(2)r8c19)
5c. no 4 in r7c1

6. "45" on n6789: 3 outies r5c6 + r6c46 = 9 = {126/135/234}(no 789)
6a. no 1,2 in r5c7

Time to start going big.
7. "45" on n6789: 1 outie r6c4 + 7 = 3 innies r567c7 ie, r567c7 must be more than 7
7a. -> no 4 in r5c7 since it forces r567c7 = {124} = 7
7b. no 5 in r5c6

8. 15(2)n4 = {69/78}, ie both 6,9 or it's 8 -> 8 locked for r5 in 15(2) or r5c8
8a. no 1 in r5c6

9. "45" on n78: 1 outie r6c4 + 11 = 2 innies r89c6
9a. min. 2 innies = 12 (no 1,2)

Crucial observation. No eliminations yet.
10. r6c4 must be cloned in one cell of 9(2)n8 since Innie Outie Difference (IOD) n78 is greater than 9 (step 9) (Large IOD ie r6c4 cannot equal either one of r89c6) and r6c4 sees all other cells in n8 apart from r89c5. See pic below to illustrate.

Image

11. no 3 in r6c4. Like this.
11a. 25(6)n5 = {123469/123478/123568/124567}
11b. all combos with each of 1,2 & 4 must have 1/2/4 in r6c4 so it will not clash with r7c6
11c. the only combo without each of 1,2,4 is {123568} but 3 cannot be in r6c4 since the 6 in this combo would have to be in n8 which would clash with {36} in 9(2)(from step 10. Clone CCC) -> no 3 in r6c4

12. h9(3)n5 = {126/135/234}
12a. 3 in {234} must be in r5c6 -> no 4 in r5c6
12b. no 5 in r5c7

13. 9(2)n5 = [27]/{36} = [6/7..]
13a. Killer pair 6,7 with 15(2)n4: both locked for r5

14. 25(6)n5 = {123469/123478/123568/124567}
14a. all combinations except one form a caged x-wing on 6 with 23(3)n6.
14b. the one exception is {123478}: with {378} in r7 which forces r7c1 = 6
14c. -> 6 locked for r67 in 25(6)n5, r67c8 and r7c1

15. 17(3)n4 = {359/458}(no 7)
15a. must have 5 -> 5 locked for n4 and r6

16. "45" on r6789: 2 outies r5c89 - 2 = 1 innie r6c5
16a. -> max. 2 outies = 11 -> max. r5c9 = 3
16b. min. 2 outies = 9 -> min. r6c5 = 7

17. r5c5 = 5 (hsingle r5)
17a. no 4 in 9(2)n8
17b. no 7 in 12(2)n2
17b. r46c5 = 12 = [39/48]

18. naked quad {3489} in r1246c5: all locked for c5
18a. -> 9(2)n8 = {27}: both locked for n8, c5
18a. no 2,7 in r9c7

19. r6c4 = 2 (step 10)

On from there. Much easier now.

cheers
Ed
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Andrew
 Post subject: Re: Assassin 188
PostPosted: Fri Feb 10, 2012 6:00 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
After A123 V2, which I think is definitely the hardest V2 on this forum, I found A188 V2 to be one of the hardest of the other V2s.

When I first worked on this puzzle almost two years ago, I'd as far as step 28 and then got stuck. When I looked at it again this year I still took quite a long time to find a way forward; then I found steps 31, 37 and 38.

Afmob and Ed used very different ways to solve this puzzle. My solving path was more like Ed's way, although quite a lot longer; of course that might be partly because I saw things that weren't needed for the best solving path, as often happens in my walkthroughs. I really liked Afmob's steps 6a and 6b and Ed's step 14!

Ed wrote Finally found my steps 10 & 11 which gave the same elimination as Afmob's last ! (step 6b) so it seems that was the crucial elimination for this puzzle.:
I'm inclined to broaden Ed's statement. I think that eliminating 3,5 and 6 from R6C4 are the crucial eliminations for this puzzle but there's no particular order in which they can be done.

As will be seen in my walkthrough, I discussed what I'd done with Ed at the time I was stuck and he pointed out alternative, and in two cases definitely better, ways to do three of my steps. Thanks Ed for those comments and for not giving away anything about how you solved it; I've had to work things out for myself after coming back to this puzzle.

Rating Comment:
I'll rate my walkthrough for A188 V2 at 1.75, mainly because of the difficulty of finding some of the steps; I'd probably have gone a bit lower if I'd found the steps more easily.

Here is my walkthrough for A188 V2:
Prelims

a) R1C34 = {49/58/67}, no 1,2,3
b) R12C5 = {39/48/57}, no 1,2,6
c) R1C89 = {17/26/35}, no 4,8,9
d) R2C78 = {19/28/37/46}, no 5
e) R23C9 = {49/58/67}, no 1,2,3
f) R5C34 = {69/78}
g) R5C67 = {18/27/36/45}, no 9
h) R78C1 = {29/38/47/56}, no 1
i) R8C23 = {16/25/34}, no 7,8,9
j) R89C5 = {18/27/36/45}, no 9
k) R9C12 = {69/78}
l) R9C67 = {18/27/36/45}, no 9
m) 22(3) cage at R3C3 = {589/679}
n) 23(3) cage at R5C8 = {689}
o) 7(3) cage at R6C6 = {124}
p) 14(4) cage at R2C1 = {1238/1247/1256/1346/2345}, no 9

1. Naked triple {689} in 23(3) cage at R5C8, locked for C8, clean-up: no 2 in R1C9, no 1,2,4 in R2C7

2. 45 rule on R12 2 innies R2C19 = 9, no 6,7,8 in R2C1, no 9 in R2C9, clean-up: no 4 in R3C9

3. 45 rule on R89 2 innies R8C19 = 12 = {39/48/57}, no 1,2,6, clean-up: no 5,9 in R7C1

4. R5C34 + R5C8 must contain 8 (think this might be an ALS), locked for R5, clean-up: no 1 in R5C67
[Alternatively combined cage R5C348 = 15(2) + {689} = 21,23,24 = {678/689/789}, 8 locked for R5 ...]

5. 45 rule on N14 4 outies R1245C4 = 26 = {2789/3689/4589/4679/5678}, no 1
5a. 45 rule on N14 4 innies R1345C3 = 1 outie R2C4 + 24, max R1345C3 = 30 -> max R2C4 = 6

6. 45 rule on N69 4 innies R5679C7 = 1 outie R8C6 + 5
6a. Min R5679C7 = 10 -> min R8C6 = 5

7. 45 rule on N78 2 innies R89C6 = 1 outie R6C4 + 11
7a. Min R89C6 = 12, no 1,2, clean-up: no 7,8 in R9C7
7b. Max R89C6 = 17 -> max R6C4 = 6

8. 45 rule on N23 2 innies R12C4 = 1 outie R4C6 + 7
8a. Max R12C4 = 15 -> max R4C6 = 8

9. 45 rule on C1 3 innies R169C1 = 20 = {389/479/569/578}, no 1,2

10. 45 rule on C9 3 innies R149C9 = 11 = {128/137/146/236/245}, no 9
10a. 5 of {245} must be in R1C9 -> no 5 in R49C9

11. 45 rule on C5 2 innies R37C5 = 7 = {16/25/34}, no 7,8,9

12. Hidden killer triple 5,6,7 in R5C34, R5C67 and R5C12589 for R5, R5C34 contains one of 6,7, R5C67 contains one of 5,6,7 -> R5C12589 must contain one of 5,6,7
12a. 45 rule on R6789 2 outies R5C89 = 1 innie R6C5 + 2
12b. Min R5C89 = 7 -> min R6C5 = 5
12c. Max R5C89 = 11 -> max R5C9 = 4 (R5C89 cannot be [65] because R5C12589 only contains one of 5,6,7)

13. 15(3) cage at R4C7 = {159/249/258/267/348/357/456} (cannot be {168} which clashes with 23(3) cage at R5C8, ALS block)
13a. 1 of {159} must be in R4C9 -> no 1 in R4C78

14. 45 rule on N1234 3 outies R4C46 + R5C4 = 19 = {289/379/469/478/568}, no 1
14a. 3,4 of {379/478} must be in R4C6 -> no 7 in R4C6

15. 45 rule on N6789 3 outies R5C6 + R6C46 = 9 = {126/135/234}, no 7, clean-up: no 2 in R5C7

16. 24(6) cage at R3C4 = {123459/123468/123567}, 1 locked for R3

17. 45 rule on N6789 1 innie R5C7 = 2 outies R6C46
17a. R5C7 cannot be 4, here’s how
R5C7 = 4 => R6C46 = 4 = [31] => cannot place 4 in 7(3) cage at R6C6
17b. -> no 4 in R5C7, clean-up: no 5 in R5C6
[Ed’s alternative way was
45 rule on N6789 3 innies R567C7 = 1 outie R6C4 + 7
R5C7 cannot be 4 because that would make R567C7 = 4{12} = 7 and R567C7 must be greater than 7; this would also have eliminated 1,2 from R5C7 if they hadn’t already been eliminated.]

18. 45 rule on N1234 2 outies R4C46 = 1 innie R5C3 + 4
18a. R5C3 less than 9 => max R4C46 = 12 => no 8 in R4C6
or R5C3 = 9 => 9 in 22(3) cage at R3C3 must be in R4C4 => R4C46 = 13 = [94]
-> no 8 in R4C6

19. Deleted, this step was flawed because I’d overlooked one valid permutation.

[At this stage I initially found
20. 25(6) cage at R6C4 = {123469/123478/123568/124567}
20a. R6C4 = {124} => naked triple {124}, locked for R6
R6C4 = {356} => 1,2 locked in R7 => R7C7 = 4 => R6C67 = {12}, locked for R6
20b. -> 1,2 locked in R6C467, locked for R6
but later I found the more useful and technically simpler …]

20. 25(6) cage at R6C4 = {123469/123478/123568/124567}
20a. Caged X-Wing for 1,2 in 25(6) cage at R6C4 and 7(3) cage at R6C6 for R67, no other 1,2 in R67, clean-up: no 9 in R8C1, clean-up: no 3 in R8C9 (step 3)

[I’ve slightly re-worked this step to make it simpler and less “chainy” …]
21. R78C1 = {38/47/56}, R9C12 = {69/78} -> combined cage R78C1 + R9C12 = {38}{69}/{47}{69}/{56}{78}, 6 locked for N7, clean-up: no 1 in R8C23
21a. 1 in N7 only in R7C23 + R9C3
21b. 45 rule on N7 3 innies R7C23 + R9C3 = 12 = {129/138/147}, no 5

22. 5 in N7 only in R8C123, locked for R8, clean-up: no 7 in R8C1 (step 3), no 4 in R7C1, no 4 in R9C5

23. 15(3) cage at R4C7 (step 13) = {159/249/258/267/348/357/456}
23a. 15(3) cage contains one of 6,8,9 or 15(3) cage = {357} => R5C7 = 6
23b. Killer triple 6,8,9 in 15(3) cage + R5C7 and R56C8, locked for N6

24. Deleted because step 19 deleted.

25. R4C46 + R5C4 (step 14) = {289/379/469/478/568}
25a. 5 of {568} must be in R4C4 (R45C4 cannot be [68/86] because R345C3 cannot be {79}7/{59}9) -> no 5 in R4C6

26. 45 rule on N3 6 outies R12C6 + R3C456 + R4C6 form a 26(6) cage, because these cells all “see” each other = {123479/123569/123578/124568/134567}]
[I originally saw this as
45 rule on N3 2 outies R12C6 = 2 innies R3C78 + 2
As a result R12C6 + R3C456 + R4C6 form a 26(6) cage, because these cells all “see” each other
When I mentioned this to Ed, he pointed out that 6 outies for N3 is simpler.]
26a. R12C5 = {39/48} (cannot be {57} which clashes with R12C6 + R3C456 + R4C6 because no 5,7 in R4C6)
26b. R37C5 (step 11) = {16/25} (cannot be {34} which clashes with R12C5

27. 45 rule on N7 6 outies R6C4 + R7C456 + R89C4 form a 25(6) cage, because these cells all “see” each other = {123469/123478/123568/124567}
[In similar manner to step 26, I’d originally seen this as
45 rule on N7 2 innies R7C23 = 2 outies R89C4
As a result R6C4 + R7C456 + R89C4 form a 25(6) cage, because these cells all “see” each other = {123469/123478/123568/124567}]

28. R89C6 = R6C4 + 11 (step 7) -> the number in R6C4 must also be in R89C5 (because the innie-outie difference is greater than 9)
28a. R6C4 = 1 => R89C6 = 12 cannot contain 6
or R6C4 = 2 => 2 must be in R89C5 = {27} => R89C6 = 13 cannot be {67}
or R6C4 = 3 => 3 must be in R89C5 = {36} => no 6 in R89C6
or R6C4 = 4 => R89C6 = 15, max R9C6 = 8 => no 6 in R8C6
or R6C4 = {56} => R89C6 = 16,17 cannot contain 6
-> no 6 in R8C6
[There is a similar step based on R12C4 = R4C6 + 7 (step 8) which eliminates 3 from R2C4. I haven’t included it because it doesn’t contribute much at this stage.]

29. 45 rule on C6789 2 outies R3C45 = 1 innie R7C6 + 1
29a. Min R3C45 = 3 -> min R7C6 = 2

30. Innies for N7 contain 1, outies for N7 contain 1 (step 27), 25(6) cage at R6C4 contains 1 -> 12(3) cage at R8C4 must contain 1, CPE no 1 in R9C5, clean-up: no 8 in R8C5

31. The number in R6C4 must also be in R89C5 (step 28)
31a. 25(6) cage at R6C4 = {123469/123478/123568/124567}
31b. 25(6) cage = {123469/123478/124567} must have one of 1,2,4 in R6C4 (because 1,2,4 all in R7C23456 clashes with R7C7)
31c. 25(6) cage = {123568} must have one of 1,2,5,6,8 in R6C4 (R6C4 cannot be 3 because no 6 in R7C23 so 3{12568} would clash with R89C5 = {36}), no 3 in R6C4
[I had to remind myself that 25(6) cage at R6C4 and 25(6) outies for N7 (step 28) do not necessarily have the same combination, so the method in step 31c cannot be used to eliminate 6 from R6C4.

I also noticed that if innies for N7 are {129} then 12(3) cage at R8C4 must also be {129} because 9 in R9 is only in R9C1234. However the converse doesn’t necessarily apply; 12(3) cage at R8C4 can be {129} without innies for N7 having the same combination, the innies have a different combination when 9 is in R8C4.]

32. R5C6 + R6C46 (step 15) = {126/135/234}
32a. 3 of {234} must be in R5C6 -> no 4 in R5C6, clean-up: no 5 in R5C7

33. Killer quad 6,7,8,9 in R5C34, R5C67 and R5C8, locked for R5
33a. Killer pair 6,7 in R5C34 and R5C67, locked for R5

34. 45 rule on R1234 2 outies R5C12 = 1 innie R4C5 + 2
34a. Max R5C12 = 9 -> max R4C5 = 7

35. 17(3) cage at R4C5 = {179/278/359/458/467} (cannot be {269} which clashes with R37C5, cannot be {368} which clashes with R12C5)
35a. 1,2 of {179/278} must be in R5C5 -> no 1,2 in R4C5
35b. 8,9 of {359/458} must be in R6C5 -> no 5 in R6C5

36. R5C89 = R6C5 + 2 (step 12a)
36a. Min R5C89 = 9 -> min R6C5 = 7
36b. Max R5C89 = 11 -> no 4 in R5C9

37. 17(3) cage at R6C1 = {359/368/458/467}
37a. R5C12 = R4C5 + 2 (step 34)
R4C5 = {34567} -> R5C12 = 5,6,7,8,9 = {14/15/24/25/35/45} (cannot be {23} which clashes with R5C67, cannot be {34} which clashes with 17(3) cage)
37b. 17(3) cage = {359/368/467} (cannot be {458} which clashes with R5C12)
37c. 17(3) cage = {359}, locked for R6 and N4 => R5C5 = 5 (hidden single in R5) => 17(3) cage at R4C5 (step 35) = [458] => R6C8 = 6
or 17(3) cage = {368/467}, 6 locked for R6
-> no 6 in R6C4

[I quickly spotted a contradiction chain to eliminate 5 from R6C4
R6C4 = 5 =>25(6) cage at R6C4 = {123568} (other combinations require one of 1,2,4 in R6C4, steps 30b and 30c), 6,8 locked for R7 => R7C8 = 9, R56C8 = [86] => 17(3) cage at R6C1 clashes with R6C48 = [56]
but decided to try to find another way, possibly using a forcing chain rather than a contradiction. At that stage I went to bed and read a book for a while. Next morning, just as I was about to get up, the forcing chain came into my head …]

38. Consider the combinations for 17(3) cage at R6C1 (step 37b) = {359/368/467}
17(3) cage = {359}, 5 locked for R5
or 17(3) cage = {368/467}, 6 locked for R5 => R7C8 = 6 (hidden single in C8) => 25(6) cage at R6C4 (step 31a) = {123478} (only combination which doesn’t contain 6) => R6C4 must contain one of 1,2,4
-> no 5 in R6C4
[Now the puzzle is cracked; the rest is straightforward.]

39. Naked triple {124} in R6C467, locked for R6
39a. 17(3) cage at R6C1 (step 37b) = {359/368}, no 7, 3 locked for R6 and N4

40. R5C6 + R6C46 (step 15) = {126/234}
40a. 3,6 only in R5C6 -> R5C6 = {36}, clean-up: no 7 in R5C7
40b. R5C6 + R6C46 = {126/234}, 2 locked for R6 and N5

41. Naked pair {36} in R5C67, locked for R5, clean-up: no 9 in R5C34
41a. R5C8 = 9 (hidden single in R5)

42. 24(6) cage at R3C4 = {123459/123468/123567}, 2 locked for R3

43. 15(3) cage at R4C7 (step 13) = {258/267/348/357/456}, no 1

44. 1 in R4 only in R4C12, locked for N4
44a. R5C12 = {24/25/45} = 6,7,9 -> R4C5 (step 36a) = {457}, no 3,6

45. R4C46 + R5C4 (step 14) = {379/478/568} (cannot be {469} because R5C4 only contains 7,8)
45a. 5 of {568} must be in R4C4 -> no 6 in R4C4
45b. 6 in N5 only in R45C6, locked for C6, clean-up: no 3 in R9C7

46. 17(3) cage at R4C5 (step 35) = {179/458}
46a. 8,9 only in R6C5 -> R6C5 = {89}

47. 17(3) cage at R6C1 (step 39a) = {359/368}
47a. Killer pair 8,9 in 17(3) cage and R6C5, locked for R6 -> R6C8 = 6, R7C8 = 8, R5C67 = [63]
47b. 17(3) cage at R6C1 = {359} (only remaining combination), locked for R6 and N4 -> R6C5 = 8, R5C34 = [87], R6C9 = 7
47c. 17(3) cage at R4C5 (step 46) = {458} (only remaining combination) -> R45C5 = {45}, locked for C5 and N5 -> R4C46 = [93]
47d. Naked pair {39} in R12C5, locked for C5 and N2
47e. R89C5 = {27} (only remaining combination), locked for C5 and N8
47f. Naked pair {24} in R5C12, locked for R5 and N4 -> R45C5 = [45], R5C9 = 1, R6C7 = 4
47g. Clean-up: no 4,6 in R1C3, no 5 in R1C4, no 6 in R23C9, no 3 in R8C1 (I’ll skip the other eliminations for R8C19), no 5 in R9C6, no 2,6 in R9C7

48. R8C19 (step 3) = {48} (only remaining combination) -> R8C9 = 4, R8C1 = 8, R7C1 = 3, R8C6 = 9, R7C9 = 9 (cage sum), clean-up: no 7 in R9C12

49. Naked pair {58} in R23C9, locked for C9 and N3 -> R4C9 = 2, R4C78 = [85]

[Clean-ups omitted in the remaining steps]
50. R1C89 = {26} (only remaining combination) -> R1C8 = 2, R1C9 = 6, R9C9 = 3

51. Naked pair {17} in R89C8, locked for C8 and N9 -> R3C8 = 4, R2C8 = 3, R2C7 = 7, R7C7 = 2, R6C46 = [21], R8C7 = 6, R9C7 = 5, R9C6 = 4, R7C6 = 5

52. Naked pair {16} in R7C45, locked for R7 and N8 -> R8C4 = 3, R9C4 = 8, R9C3 = 1 (hidden single in N7), R1C4 = 4, R1C3 = 9

53. R1C7 = 1 -> R12C6 = 15 = [78], R23C9 = [58], R2C4 = 6, R3C45 = [51]

54. Naked pair {67} in R34C3, locked for C3 -> R7C3 = 4

and the rest is naked singles.
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