Assassin 50

[Ruud]

Well, the much anticipated Assassin 50 (cheers! ) has arrived.

Thanks to all the top notch solvers for their wonderful walkthroughs. I am very glad that so many people enjoy solving these Assassins, which I still create with great pleasure.

Back to the core business...

There is a V2 for Assassin 50, which could be one of the toughest killers I made so far. You can decide whether this claim is true. Here it is:

Assassin 50 V2



3x3::k3072:4098:4098:5636256533354618:4618:5636:5636:5636:5647:56472834:4618:461811542:4119:5647:56472834:591628374119:4119:51545916:591611542115425154:5154591628634658:4658:51542605363911542:4658:6204:620441593639:5698:5698:5698:6204:6204:4167:4159:41593146:56981357:4167:4167:

cheers,
Ruud

[Glyn]

Ruud

I thought to celebrate we'd have a nice easy one V0.2 so we can have a few beers while we solve it. Maybe we will need to have a few beers when we can't.

All the best,

Glyn

[CathyW]

1. 4(2) at r34c1 = {13}, not elsewhere in c1.

2. Innies N1: r1c3 + r3c1 = 7 -> r1c3 = 46 -> r1c4 = 79 (Outies N1 = 10)

3. Innies N3: r1c7 + r3c9 = 12 -> r1c7 = 357, r3c9 = 579 -> r1c6 = 135, r4c9 = 357 (Outies N3 = 8)

4. Innies N7: r7c1 + r9c3 = 6 -> r7c1 = 45, r9c3 = 12 -> r6c1 = 89, r9c4 = 89 (Outies N7 = 17)

5. Innies N9: r7c9 + r9c7 = 17 = {89} not elsewhere in N9 -> r9c47 form Naked Pair, not elsewhere in r9 -> Pointing pair {89} at r6c1, r9c4 -> r6c4 <> 8

6. Innies r12: r2c2378 = 18 -> r3c2378 = 26, r3c2378 <> 1

7. Innies r89: r8c2378 = 22 -> r7c2378 = 15

8. Innies c12: r2378c2 = 25 -> r2378c3 = 20

9. Innies c89: r2378c8 = 17 -> r2378c7 = 19

10. 14(2) in r67c9 must be 6+8 since 5+9 would eliminate all options for 12(2) in r34c9
-> r6c9 = 6, r7c9 = 8 -> r9c7 = 9, r9c6 = 4, r9c4 = 8, r9c3 = 2, r7c1 = 4, r6c1 = 9
-> 19(3) in N7 = {568} -> r8c1 = 8
-> r9c12 = {56} not elsewhere in r9/N7 -> 20(4) in N7 = {1379}
-> r9c589 = {137} -> 11(3) in N9 = {137} -> r8c9 <> 2,4,5
-> 17(4) in N9 = {2456}

11. Innies N8: r7c456 = 18 = {279/369/567} -> r7c456 <> 1
-> 15(4) in N8 must have 1: (1{239/257/356})
-> 1 locked to r7c23 in N7 -> r8c23 <> 1

12. 13(3) in N1: r12c1 is min 7 -> r1c2 max 6, r1c2 <> 789
-> 13(3) = {157/247/256} -> r1c2 <> 3
25(4) in N1 must have 8,9: {1789/2689/3589}, 25(4) <> 4
-> 4 locked to r1c23 -> r1c589 <> 4

13. If 12(2) in r34c9 = 39 -> r89c9 = 17, r9c8 = 3
If 12(2) in r34c9 = 57 -> r89c9 = 13, r9c8 = 7
-> r9c8 <> 1 -> 1 locked to r89c9 -> r125c9 <> 1

14. Split cage 22(4) r8c2378 = {2479/3469} -> r8c78 <> 5
-> 9 locked to r8c23 -> r7c23, r8c456 <> 9
Split 18 (3) r7c456 must have 9 = {279/369} -> r7c456 <> 5
-> Split 15(4) r7c2378 = {1257/1356}
-> 15(4) in N8 = {1257/1356}

15. Split cage 25(4) r2378c2 = {1789/3589/3679} -> r23c2 <> 2

16. Split cage 20(4) r2378c3 = {1379} -> r456c3 <> 1,3,7
-> r23c3 <> 568
-> 25(4) in N1 = {1789/3589} -> r456c2 <> 8
-> 5 locked to r456c3 -> 16(4) in N4 <> 5 -> 16(4) in N4 = {1267/2347}

17. Cage 13(3) in N1 must have 2: {247/256} -> r1c2 <> 1

18. 9 locked to r7c56 -> forms pointing pair: r5c6 <> 9

19. 11(3) in r6c34+r7c4 = {128/146/245} -> r67c4 <> 3,7
-> Split 18(3) in r7c456 = {279/369} -> r7c56 <> 2,6
-> combination analysis of 11(3) -> r6c4 <> 2

20. 18(3) in r6c67+r7c6 = {189/279/378/459}: combination analysis -> r6c7 <> 5

21. 13(3) in r3c4+r4c34: Since r4c3 is min 4, r34c4 <> 9

22. 9 locked to r125c4 -> forms pointing triple: r3c5 <> 9

23. Innies N2: r1c46 + r3c456 = 21 = {12369/12378/12459/12567/13467/23457}
-> r3c5 <> 7, r3c6 <> 9

24. If r7c5 = 9 -> r4c6 = 9 -> r2c6 <> 9
If r7c6 = 9 -> r2c6 <> 9
Either case, r2c6 <> 9.

25. Killer combination in N1: If 13(3) = 247, r1c3 = 6; if 13(3) = 256, r1c3 = 4
-> r23c2 <> 6
-> split cage 25(4) r2378c2 = {1789/3589}

26. Killer combination in c9: If 12(2) = 39, r89c9 = 17; if 12(2) = 57, r89c9 = 13
-> r125c9 <> 3,7

27. Outies - Innies c9: r19c8 - r5c9 = 6 -> r19c8 = 8, 10, 11 or 15
-> If 8, r19c8 = {17/35}; if 10, r19c8 = 37; if 11, r19c8 = 38; if 15, r19c8 = 78
-> r1c8 = 13578

28. Outies - Innies c1: r19c2 - r5c1 = 4 -> r5c1 <> 2 since can't make 6 from candidates in r19c2.
If r5c1 = 6, r19c2 = 10 = 46; if r5c1 = 7, r19c2 = 11 = 56 -> r1c2 <> 2
-> at least one of r1c2, r5c1, r9c2 = 6 -> r45c2 <> 6.

29. Outies - Innies r1: r2c19 - r1c5 = 3 -> r1c5 <> 1 since can't make 4 from candidates in r2c19; r1c5 <> 2 since can't make 5 from candidates in r2c19.

30. 14(3) in N3 = {149/239/248/257} -> combination analysis: r1c8 <> 5

31. Split 25(4) in r2378c2 = {1789/3589}
If 3589, r7c2 must be 3 -> r238c2 <> 3

32. Breakthrough move!!
If r1c4 = 7 -> r1c3 = 6 -> 13(3) in N1 = 247 -> r1c1 = 2, r1c2 = 4, r2c1 = 7
If r1c4 = 9 -> r1c3 = 4 -> 13(3) in N1 = 256 -> r1c1 = 2 (Can't have both 56 in r1c12 due to {56} in r9c12).
Either case r1c1 = 2 -> r1c9 = 59 -> 14(3) in N3 = {257/149/239} -> r1c8 <> 8, r2c9 <> 5,9
-> 19(4) in N3 must have 6 and 8: {1468/2368}; 19(4) <> 5,7,9
-> 9 locked to r13c9 -> r5c9 <> 9

33. HS: r1c5 = 8 -> r2c19 = 11 (from step 29) -> r2c1 = 7, r2c9 = 4
leading to several naked and hidden singles and it is relatively straightforward from there with a few cages then having only one combination option.

Thanks to Para for pointing out an error in my original step 5 - now amended. Fortunately only one other step needed amendment as a result.
Took somewhat longer than no. 49 but got there in the end. Think I'll leave the V2 for the experts who have more free time.

Edit: Where is everybody today? I think the V2 is impossible to solve by humans without serious T&E - even JC's software cannot solve it - actually not even a single placement!!

[Glyn]

Hi Cathy

I guess that's right, I am struggling with it here. Tried making a huge implication chain round the outside, knocks out a few candidates but nothing worth reporting. Think I'll have a look at the standard first maybe I'll get some inspiration from that. The cage patterns are the same.

All the best,

Glyn

[Ruud]

Assassin 50 V0.2



3x3::k:4608:4608:4354:4354:410017973591:4608:4106:4106:4100:4100:4100:4623:46231554:4106:4106:51415655:4623:46231554:6428:5141:51415655:5655:54106428:642811542115425410:541064283631389054102093:4919:4919115425692:56923647:4919:491933945692:5692:49353647314628934935:4935:

This version 0.2 has the same difficulty rating as a recent Moderate on www.sudoku.org.uk

Ruud

[Glyn]

Thanks for the V0.2 Ruud, I did it before I had the beer though.

I thought I would point out a little discussion that's going on over at Djape looking at Killer Wings and Loops. You'll find under Other as '3 vanilla techniques: Y-Wing, XY-Wing and XYZ-Wing'.

Three of you will already know about it, as you are participating, but maybe others might like to check it out. We might need these for the V2, and it looks as though the notation is being ironed out there.

All the best,

Glyn

[Andrew]

I took a break from Assassin 50, where I've currently ground to a halt, and did V0.2

This version 0.2 has the same difficulty rating as a recent Moderate on www.sudoku.org.uk

Don't know about that. It was definitely routine so V0.2 is a fair rating. Maybe SumoCue rates it the same as that recent Moderate. However as someone who still does the killers on the other website it took me longer than any daily killer there.

50V0.2 doesn't really need a posted walkthrough. However I'm posting my one because I feel that all Assassins and other puzzles posted on this forum should have at least one posted walkthrough or, for the hardest puzzles, a tag solution.

1. R1C34 = {89}, locked for R1

2. R1C67 = {16/25/34}, no 7

3. R34C1 = {15/24}

4. R34C9 = {18/27/36/45}, no 9

5. R67C1 = {17/26/35}, no 4,8,9

6. R67C9 = {13}, locked for C9, clean-up: no 6,8 in R34C9

7. R9C34 = {39/48/57}, no 1,2,6

8. R9C67 = {29/38/47/56}, no 1

9. 18(3) cage in N1 = {279/369/378/459/468/567} (cannot be {189} because 8,9 only in R2C1), no 1
9a. 8,9 only in R2C1 -> no 2,3,4 in R2C1

10. 20(3) cage in N254 = {389/479/569/578}, no 1,2

11. 22(3) cage in N256 = 9{58/67}

12. 19(3) cage in N9 = {289/379/469/478/568}, no 1

13. 45 rule on N1 2 innies R1C3 + R3C1 = 11 -> R1C3 = 9, R3C1 = 2, R1C4 = 8, R4C1 = 4, clean-up: no 5 in R3C9, no 7 in R4C9, no 6 in R67C1, no 4 in R9C3, no 3 in R9C4

14. 45 rule on N3 2 innies R1C7 + R3C9 = 13 -> R1C7 = 6, R3C9 = 7, R1C6 = 1, R4C9 = 2, clean-up: no 5 in R9C6

15. 18(3) cage in N1 (step 9) = {378/567} (cannot be {468} because no 4,6,8 in R1C1), no 4 = 7{38/56}, 7 locked for N1
15a. 6,8 only in R2C1 -> no 5,7 in R2C1
15b. 7 locked in R1C12 for R1
15c. 16(4) cage in N1 = 14{38/56}

16. 45 rule on N7 2 innies R7C1 + R9C3 = 12 = {57}, locked for N7, clean-up: R6C1 = {13}, R9C4 = {57}
16a. Naked pair {57} in R9C34, locked for R9, clean-up: no 4,6 in R9C67

17. 45 rule on N9 2 innies R7C9 + R9C7 = 4 -> R7C9 = 1, R9C7 = 3, R6C9 = 3, R9C6 = 8, R6C1 = 1, R7C1 = 7, R9C34 = [57]

18. R1C2 = 7 (hidden single in N1)

19. 14(3) cage in N3 = {248} (cannot be {239} because no 2,3,9 in R1C9) -> R1C89 = [24], R2C9 = 8 -> R2C1 = 6, R1C1 = 5 (step 15), R1C5 = 3, R9C9 = 9
19a. R9C1 = 9 -> R8C1 + R9C2 = 5 -> R8C1 = 3, R9C2 = 2, R5C1 = 8, R9C89 = [46], R9C5 = 1, R8C9 = 9 (cage sum), R5C9 = 5

20. At last a digit in the 45 cage, R5C4 = 1 (hidden single in C4) -> R4C8 = 1 (hidden single in N6)

21. 13(4) cage in N8 = {1246} (only remaining combination), 2,4,6 locked for R8 and N8

22. Naked pair {18} in R8C23, locked for R8 and N7

23. Naked pair {57} in R8C78, locked for N9 -> R7C78 = [28]
[R7C7 was a hidden single after step 21 but I used the naked pairs which are more obvious.]

24. 25(4) cage in N4 = {3589} (only remaining combination), locked for N4, 3 locked for C2

25. R7C2 = 6 (hidden single in C2), R7C3 = 4,

26. 16(3) combination in N2 = {2347} (only remaining combination), locked for N2, 4 locked for R2 -> R2C23 = [13], R3C23 = [48], R8C23 = [81]

27. Naked pair {59} in R2C78, locked for N3 -> R3C78 = [13]

28. R56C8 = 15 = {69}, locked for C8 and N6 -> R2C78 = [95], R8C78 = [57]

29. R5C6 = 3 (hidden single in 45 cage, may have been there for some time) -> R5C2 = 9, R6C2 = 5, R4C2 = 3, R56C8 =[69]

30. 20(3) cage in N254 = {569} (only remaining combination) -> R4C3 = 6, R34C4 = {59}, locked for C4 -> R7C4 = 3, R6C34 = 11 = [74], R5C3 = 2, R6C7 = 8, R4C7 = 7, R5C57 = [74], R2C456 = [247], R8C456 = [624]

31. R4C7 = 7 -> R34C6 = 15 = [69]

and the rest is naked singles

[Glyn]

Is anybody having any luck with the V2, without extensive 'tryfurcation'? I may be able to use this method again to crack it, lots of scribbled notes to write out neatly.

[CathyW]

I haven't attempted it Glyn. Since JSudoku couldn't solve it, I certainly can't!!

It's been quiet in the Assassin forum this week - I'm surprised no-one else has posted a walkthrough to the A50. I'd be interested to see how others solved it.

[Glyn]

Specially for Cathy.

Tryfurcated Walkthrough of V2 up to the first block of placements. JSudoku can solve from here if you wish or carry on yourself, I have removed the main blocking cage combo.

But there is still much to do.



My goal was to solve the complete perimeter of the puzzles given the number of 2 cell cages, connected by known Innies/Outties for N1,N3,N7,N9.

1) 16(2) cage r1c34={79}. Not elsewhere in R1.

2) 10(2) cage r1c67={28}|{46}.

3) Innies N1=15 => r3c1=6|8.

4) 11(2) cage r34c1=[65]|[83]

5) Innies N9=5 => r7c9,r9c7<5.

6) 9(2) cage r67c9=[81]|[72]|[63]|[54]

7) Innies N3=10 r3c9=2|4|6|8.

8) 9(2) cage r34c9=[27]|[45]|[63]|[81].

9) Innies N7=15. r7c1,r9c3>6.

10) r9c3<>6 (in cage 12(2)) => r7c1<>9.

11) 10(2) cage r67c1=[28]|[37]|[46]

12) 12(2) cage r9c34=[75]|[84]|[93]

13) 11(2) cage r34c1 either contains 6 in r3 or 3 in r4. Together they block[63] combo in r67c9.
13a) r1c6<>6, r1c7<>4. (Innies of N3 and cage 10(2) in r1.)

14) a)Either r7c1 and r9c3 form a naked pair {78} in N9
or b) r9c3=9 => r1c3=7.
Common peers of r7c1,r19c3 are r78c3<>7.

15) a)Either r9c3=8 or
b) r9c3=7|9 => r7c1=8|6 forming naked pair {68} with r3c1.
Common peers or r37c1 and r9c3 are r89c1<>8.


16) Tryfurcation steps Branch 1. Assumption 10(2) cage r67c1=[46]. Must roll back any moves done here.
=========================================================

16a) r67c1=[46] => 11(2) cage r34c1=[83] ([65] is blocked).
16b) 12(2) cage r9c34=[93]. (Innies of N7)
16c) 16(2) cage r1c34=[79]. (Innies of N1)
16d) Step 16b) forces 5(2) cage r9c67={14}.

17a) 16(3) cage r8c1+r9c12 can only be {178}. ({457} is blocked on r1 and c1 see step 16a) and 16d).
17b) 8 is blocked from c1 see 16a) => r9c2=8, r89c1={17}.

18a) The 12(2) cage r1c12+r2c1 cannot contain {16},{46},{34}. (Killer pairs formed by r89c1={17},r34c1=[83],r67c1=[46]).
18b) Sole remaining combo is {129}.
18c) using 16c) the 9 cannot be in r1. => r2c1=9, r1c12={12}.

19. 10(2) cage r1=[46] ({28} is blocked by 18c).

20) Unfilled cells of r1 r1c589={358}.

21a) r1c89<>[58] (Cage 13(3) will exceed sum
21b) r1c89<>[35] (Cage 13(3) requires a second digit 5)
21c) r1c89={38}, r2c9=2.

22) Sole remaining combo 9(2) cage r34c9=[45] => r67c9=[63]|[81].

23a) Innies N9=5 => r9c7=2|4 => r9c6=1|3.
23b) r9c6<>3 blocked by conflict with 16d) => r9c6=1, r9c7=4.

24) Feeding data back into c9. r67c9=[81] (Innies N9) => r1c9=3. (using 21c) ).

25a) Remaining candidates in r89c9=6|7|9. Only possible cage sums for 16(3) cage containing 2 from this set are 367 and 169.
25b) Both combos require r9c8=1|3. (both blocked on r9 by r9c4=4 step 16b) and r9c67={14} step 16d)).

26) Conclusion r67c1<>[46].

======End of tryfurcation===========


27) Cage 11(2) r34c1=[65] (Option [83] blocked as 10(2) cage r67c1 contains 3|8).

28) Innies N1 force r1c34=[97].

29) N7 Naked pair {78} (r7c1r9c3)

29) Cage 12(2) r9c4<>3.

30) Cage 9(2) r3c9<>4.




Have fun

Glyn

[Para]

It's been quiet in the Assassin forum this week - I'm surprised no-one else has posted a walkthrough to the A50. I'd be interested to see how others solved it.

First have to completely solve it before i can post a walk-through.

Para

[Glyn]

Hope to finish V2 tomorrow. Many steps on scrap paper. No more placements yet. Sumocue can still not complete it with pattern checking!

[CathyW]

Para hasn't solved Assassin 50 V1 yet, Andrew is stuck! Better hurry up guys A51 is coming soon (I might have to wait until Monday to tackle it though).

[Para]

Hi

This is how i solved it. I spent a while looking for a nicer way (without a uniqueness move like Cathy). But can't find anything. some interesting moves in there. Especially step 33, which i hoped would break it but just stalled a bit further along again.


Walkthrough Assassin 50

1. R1C34, R67C1 and R9C67 = {49/58/67}: no 1,2,3

2. R1C67 = {17/26/35}: no 4,8,9

3. R34C1 = {13}, locked for C1

4. R34C9 = {39/48/57}: no 1,2,6

5. 11(3) in R6C3 and R8C9 = {128/137/146/236/245}: no 9

6. R67C9 = {59/68}: no 1,2,3,4,7

7. 19(3) in R8C1 = {289/379/469/478/568}: no 1

8. R9C34 = {19/28/37/46}: no 5

9. 45 on N1: 2 innies: R1C3 + R3C1 = 7 = [43/61]
9a. Clean up: R1C4 = {79}

10. In N1 no combinations with {13}, {14}, {36} and {46}
10a. 13(3) in R1C1 = {157/238/247/256}: no 9; Only place for 3 in R1C2 -->> R1C2: no 8
10b. 25(4)in R2C2 = {1789/2689/3589/4579}

11. 45 on N3: 2 innies : R1C7+ R3C9 = 12 = [39/57/75]
11a. Clean up: R1C6 = {135}; R4C9 = {357}
11b. R67C9 = {68}: {59} clashes with R34C9 -->> {68} locked for C9 in R67C9

12. In N3 no combinations with {35}, {37}, {59} and {79}
12a. 14(3) in R1C8 = {149/158/167/239/248/257}
12b. 19(4) in R2C7 = {1369/1468/1567/2368/2458/2467}

13. 45 on N7: 2 innies: R7C1 + R9C3 = 6 = [42/51]
13a. Clean up: R6C1 = {89}; R9C4 = {89}

14. In N7 no combinations with {12}, {14}, {25} and {45}
14a. 20(4) in R7C2 = {1379/1568/2369/2378/2468/3467}

15. 45 on N9: 2 innies: R7C9 + R9C7 = 17 = [89]
15a. R6C9 = 6; R9C6 = 4; R9C34 = [28]; R67C1 = [94] (step 13)

16. 19(3) in R8C1 = {568}, locked for N7.
16a. R8C1 = 8(hidden single)
16b. Naked Pair {56} locked for R9

17. 11(3) in R8C9 = {137}: needs 2 of {137} in R9C89 -->> {137} in 11(3) locked for N9

18. Killer Triple {137} in R34C9 + R89C9 -->> locked for C9

19. 15(4) in R8C4 = {1239/1257/1356}: 1 locked in 15(4) for N8
19a. 1 in R7 locked for N7
19b. 1 in C9 locked for N9

20. 45 on R89: 4 outies: R7C2378 = 15 = {1356/1725}: R7C23: no 9; 5 locked in R7C78 for R7 and N9
20a. 9 in N7 locked for R8

21. 45 on R12: 4 outies: R3C2378 = 26 = {2789/3689/4589/4679/5678}: no 1

22. 45 on C1: 2 outies – 1 innie: R19C2 – R5C1 = 4
22a. R5C1 = 2; R19C2 = [15]
22b. R5C1 = 5; R19C2 = [45/36]
22c. R5C1 = 6; R19C2 = [46]
22d. R5C1 = 7; R19C2 = [65/56]
22e. Conclusion: R1C2 = {13456}

23. 8 and 9 in N1 locked in 25(4) in R2C2 -->> 25(4) = {1789/2689/3589}: no 4
23a. 4 in N1 locked for R1

24. 13(3) in R1C1 = {157/247/256}: no 3

25. 14(3) in R1C8 needs 2 of {2459} in R12C9 -->> 14(3) = [1]{49}/[3]{29}/[7]{25}/[8]{24}: R1C8 = {1378}

26. 45 on C12 : 4 outies: R2378C3 = 20 = {1379}(9 locked and no 2 or 4, so only combination left) -->> locked for C3
26a. 25(4) in R2C2 = {1789/3589} (needs 2 of {1379} in R23C3): no 2, 6
26b. 2 in N1 locked for C1; 2 locked in 13(3) in R1C1 cage -->> 13(3) = {247/256}: no 1
26c. 5 and 8 in C3 locked for N4

27. 11(3) in R6C3 needs one of {458} in R6C3 -->> 11(3) = {45}[2]/[416]/[812]: no 3,7; R6C4: no 2
27a. Naked triple {256} in R7C478 locked for R7

28. 45 on R1: 2 outies – 1 innie: R2C19 – R1C5 = 3: min R2C19 = 6 -->> Min R1C5 = 3

29. 9 in N6 locked in 22(4) in R4C8 cage -->> 22(4) = {1489/1579/2389/2479}

30. 13(3) in R3C4 needs one of {4568} in R4C3 -->> 13(3) = {148/157/238/247/256/346}: no 9

31. 24(4) in R1C5 can’t have both {79}(clashes with R1C4) -->> 24(4) = {1689/2589/2489/3678/4569/4578} (needs one of {79})
31a. Killer Pair {79} in R1C4 + 24(4) in R1C5: locked for N2

32. 19(4) in R2C7 can’t have both {12},{24} and {29} because of 14 (3) in R1C8(step 25)
32a. 19(4) = {1369/1468/1567/2368}

Pushing it now (this is more readable i think):
33. 45 on R12: 4 outies = R3C2378 = 26 = {2789/3689/4589/4679/5678}: combining with combinations for 25(4) in R2C2 + 19(4) in R2C7.

Code: Select all

33a. 26(4) R3C23  R3C78   R2C23     R2C78
33b. 2789: [87]    --
           {79}   {28}     [81]     {36}   +
           [89]    --
33c. 3689: {39}   {68}     {58}--
           [83]   [69]     [59]     {13}   +
           [89]   {36}   {17}/[53] {19/28} +
33d. 4589: {58}--    
           [59]   {48}     [83]     {16}   +
           {89}   {45}    {17/35}    --
33e. 4679: {79}   {46}     {18}     {18}--
33f. 5678: {57}   {68}      --     
           {58}--
           {78}   {56}     {19}     {17}--
-- means contradiction
+ means proper combination

33g. Conclusions: R2C23 = [17/53/59/71/81/83]: R2C2: no 3,9; R3C23 = [59/79/83/89/97]R3C2: no 3; R2C78 = {13/16/19/28/36}: no 5, 7; R3C78 = {28/36/48}/[69]: no 5, 7

34. 45 on C89: 4 outis: R2378C7 = 19 = [1864]/{38}{26}/{28}[54]/{48}[52] -->> no 6 in R23C7; 8 locked in R23C7 for C7 and N3
34a. 8 locked in 19(4) cage in R2C7 -->> 19(4) = {1468/2368}: no 9
34b. 9 in N3 locked for C9

35. 8 and 9 in N6 locked in 22(4) cage in R4C8 -->> 22(4) = {1489/2389}: no 5,7
35a. 22(4) needs one of {24} and it has to go in R5C9 -->> R456C8: no 2,4

36. 6 in N3 locked for C8

37. Looking back at step 33:

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37a. R3C23   R3C78   R2C23   R2C78
37b. {79}    [82]    [81]     [36]
37c. [83]    [69]--
37d. [89]    [36]  {17}/[53]  [82]
37e. [59]    [84]    [83]     [16]

37d. Conclusions: R2C3: no 9, R2C7: no 2, 4; R2C8: no 1, 3, 4; R3C3: no 3; R3C7: no 2, 4; R3C8: no 3

38. 9 in N1 locked for R3
38a. Clean up: R34C9 = {57} -->> locked for C9; R1C7: no 3(step 11); R1C6: no 5
38b. Naked pair {13} in R89C9 -->> locked for N9
38c. R9C8 = 7
38d. Naked Pair {13} in R1C68 -->> locked for R1

39. Hidden singles: R7C8 = 5; R1C5 = 8

40. 45 on R1: 2 outies: R2C19 = 11 = [29/74]: R2C1: no 5, 6; R2C9: no 2

41. Building on step 37: R2C2378 = [8136]/[5382]/[8316]: no {17}[82] clashes with R2C19(step 40:needs one of {27})
41a. R2C2: no 1,7; R2C3: no 7
41b. 3 locked in R2C2378 for R2

42. 13(3) in R1C1 = [247]/{256}: [742] clashes with R1C34 -->> R1C1: no 7

43. 18(3) in R6C6 = [819/{27}[9]/[8]{37}/[549] -->> R6C6: no 1,3; R6C7 = no 5

44. 45 on C12: 4 innies: R2378C2 = 25 = [8719/8917/8539/5839] -->> R7C2: no 7; R8C2: no 3

Ok this breaks it, but there must be something nicer. But as number 51 is almost up this will do.
45. Small chain from 4’s in N1: either R1C2 or R1C3 = 4
45a. R1C2 = 4 -> R2C1 = 7 -> 25(4) in R2C2 = {3589} -> R3C1 = 1
45b. R1C3 = 4 -> R1C4 = 9 -> R1C7 = 7(hidden) -> R1C6 = 1
45c. Either way R3C456 <> 1

46. R3C1 = 1(hidden); R4C1 = 3; R2C3 = 3; R1C3 = 6 (step 9); R1C4 = 7
46a. R1C67 = [35]; R1C12 = [24]; R2C1 = 7; R1C89 = [19]; R2C9 = 4
46b. R23C7 = [83]; R2C2 = 5; R3C23 = [89]; R34C9 = [75]; R5C9 = 2; R5C1 = 6
46c. R9C12 = [56]; R8C23 = [97]; R7C23 = [31]; R8C8 = 4(hidden)

47. 15(4) in R8C4 = {1356}: no 2
47a. R7C4 = 2(hidden); R78C7 = [62]

48. 24(4) in R1C5 = 8{169} -->> {169} locked for R2 and N2
48a. R23C8 = [26]

49. 18(3) in R6C6 = [279/549/819] -->> R7C6 = 9; R6C6: no 7
49a. R7C5 = 7; R4C6 = 7(hidden); R5C2 = 7(hidden); R6C7 = 7(hidden)
49b. R6C6 = 2; R3C6 = 5; R3C45 = [42]; R45C7 = [41]; R46C2 = [21]
49c. R4C34 = [81]; R6C34 = [45]; R5C3 = 5; R6C5 = 3; R6C8 = 8
49d. R5C456 = [948]; R45C8 = [93]; R4C5 = 6; R2C456 = [691]; R9C5 = 1
49e. R8C456 = [356]; R89C9 = [13]

And we are done.
I read Cathy’s walk-through, which has a uniqueness shortcut that I tried to by-pass but it is not the easiest to by-pass. But I rather not use uniqueness moves in killer solving. Just a personal taste.

greetings

Para

[mhparker]

Thanks for the walkthrough, Para. Much appreciated.

In case anyone's interested, I found a variation on Para's step 45 (the move that finally broke the puzzle), which only involves a single loop, and which does not require using the 25/4 cage at R2C2:

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45. R1C6=1 -> R3C456<>1 -> R3C1=1 -> R1C3=6 -> R1C4=7 -> R1C67<>&#123;17&#125; -> R1C6<>1 &#40;contradiction&#41;

Conclusion&#58; R1C6<>1

This leaves a hidden single in R1 at R1C8 = 1, which is also easily enough to break the puzzle.