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Assassin 34

 
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Andrew
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PostPosted: Fri Jan 19, 2007 1:49 am    Post subject: Assassin 34 Reply with quote

Ruud wrote:
There were some remarks on the forum that Assassin number 33 was not as difficult as expected. Be careful what you ask for...

I'm surprised at that. I'm still working on Assassin 33 and I expect some others are too.

Also I haven't finished Assassins 26 or 31 yet but one of those did require a joint effort on the forum to solve it.
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nd
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PostPosted: Fri Jan 19, 2007 5:47 pm    Post subject: Reply with quote

Yeah this is an eeevil one. Not sure there's any way to solve it that doesn't use extensive T&E. I'm still only halfway through it.

I was pleased though to see it has a patch of "hidden subset elimination" (the technique I was using in my most recent puzzle) in column 7. (R145C7 = {56789}, and the 18(3) cage must contain 2 instances of {56789}, producing a hidden quint in C7.) Anyway, if I ever solve this monster I'll post a walkthrough.
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Nasenbaer
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PostPosted: Fri Jan 19, 2007 6:20 pm    Post subject: Reply with quote

Yes, that's a good one. After Ruud's announcement I specifically looked for something like that and was very proud when I found it. But you are right, it's a real evil assassin. There is still a lot of work ahead.

Peter
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rcbroughton
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PostPosted: Fri Jan 19, 2007 6:33 pm    Post subject: Reply with quote

Am I looking at the right one - Assassin 34. I get the following steps - not too bad I thought - certainly no T&E required [but quite a few limited what if's on permutations - I must admit Confused ]

I'll put it in small text in case it spoils the fun.

I start numbering after I've set all marks based on internal cage combos

1. Cannot have combination {56} in cage 11(2) at r1c2 - conflicts with 14(2) at r1c7

2. Cannot have combination {29} in cage 11(2) at r8c5 - conflicts with cages 11(2) at r8c5 & 3(2) at r7c6

3. Naked pair {12} at r7c6 r8c6

4. Only combinations {347} {356} {239} {248} {149} {257} {158} {167} allowed in cage 14(3) at r2c6 - cannot have 8/9 in r3c7

5. Must use 1 in cage 19(5) at r1c9 -> 1 cannot be in r3c79

6. Must use 9 in cage 22(3) at r2c4 -> 9 cannot be in r3c6

7. 45 rule on column 5. Innies r6c5 r7c5 minus outies r1c4 r1c6 equals 6
7a. Max of r6c5+r7c5 is 17 so max of r1c4 r1c6 is 11.

9. 45 rule on column 7. Innies cells r1234597 equal 27 -> {124578} {123579} {124569} {123678} {134568} {234567}
9a. No valid combination with 5/6/7/8/9 in r2c7
9b. No valid combination with 5/6/7 in r3c7
9c. No valid combination with 5/6/7/8/9 in r9c7

10. 45 rule on column 9. Innies cells r1c9 r2c9 minus outies r4c8 r5c8 equals 5
10a Max of r1c9+r2c9 is 13 -> max of r4c8+r5c8 is 8.

11. 18(3) at r3c9 only combinations with 1 is {189} - not possible for 1 to be in r4c9

12. 45 rule on column 9. Outies r2c7 r2345c8 equal 14
12a Cage 10(2) at r6c8 restricts possible values in r2345c8 {1236} & {1234} not possible
12b r2c7 cannot be 4
12c r2c8 can't now be 8/9
12d r3c8 can't now be 8/9

13. No possible combinations left with 8 in 19(5) at r1c9

14. Must use 2 in cage 19(5) at r1c9 -> cannot have 2 in r3c79

15. No possible combinations left in 14(3) at r2c6 with 8 or 9

16. Must use 3 in cage 14(3) at r2c6 -> cannot have 3 in r3c5

17. 45 rule on N1. Innies r3c1 r3c3 equal 11
17a Only possibilities are {56}/[47]/[38]/[29]
17b r3c1 can't be 1/7/8/9

18. 45 rule on N2. Innies r3c5 minus outies r3c3 r3c7 equals -8
18a Max val in r3c5 is 5 so max of r3x3+r3c7 is 13
18b No valid combination with 8
19c Remembering that r3c13 total 11 r3c1 cannot now be 3

20. 45 rule on N2. Innies r2c4 r3c4 r2c6 r3c6 r3c5 equal 28
20a r3c13 limits possibilities in r3c456 - no possible combo with 6 at r3c4
20b.this combo now means 9 cannot be in 17(4) at r1c5, r1c6 or r2c5

21. 22(3) at r2c4 can now only be {679} or [5]{89} with 5 at r4c4

22. 9 locked in column 4 of N2 - nowhere else in the column or in r3c3

23. 12(2) at r7c4 can now only be {48} {57} - no 3
23a. 11(2) at r8c5 cannot now be {47} otherwise no valid combo left in 12(2)

24. 45 rule on column 5. r1c5 r2c5 r6c5 r7c5 total 23.
24a 11(3) at r3c5 doesn't allow a combo with {15}
24b 11(2) at r8c5 doesn't allow a combo with {35}
24c so no calid combo with 1/5 at r6c5 or 5 at r7c5

25. 17(4) at r1c4 {1457} only combo with 7 - but no valid pattern with a 7 at r1c4 or r1c6

26. h11(2) found at step 17 cannot now be {29} (no 9) - so remove the 2 from r3c1

27. h28(5) found at step 20 can't now have a combo with 2&7 because of 17(4) at r1c4. So no valid pattern with 6 @ r2c4
27a. 22(3) at r2c4 now cannot have 7 at r3c3
27b h11(2) found at step 17 can now only be {56}

28. Naked pair {56} @ r3c13 - nowhere else in n1 or row 3

29. Only combination left @ 14(3) at r2c6 is {347} - no 5 or 6

30. h28(5) found at step 20 can't now have a combo with 7 @ r2c4 or r3c4 or a 4 at r2c6 or r3c6. 1 must be in r3c5

31. 17(4) at r1c4 must now be {2456} (step 30 combos have either 3/8, so 17(4) can't have both).
31a no 2/3/8 in r1c4
31b no 3/6/7/8 in r1c5 or r2c5
31c no 3/8 in r1c6

32 Only possibilty in 22(3) at r2c4 is [5]{89} with 5 in r3c3

33. Only possibility in 14(3) ar r2c6 is [4]{37} with 4 locked in r3c7

34. 19(5) at r1c9 can only be {12367} - no 5 or 9

35. 14(2) at r1c7 can now only be {59} - no 6 or 8
35a.11(2) at r1c2 can't now be {29}

36. 15(3) at r3c1 can only nw be [6]{18}/{27}/{45} - no 3 or 9

37. 11(3) at r3c5 can only now be [1]{28}/{37}/{46} - no 5

38. No combo left in 18(3) at r6c7 with a 5

39. Innies found at step 24 can only now be {2489}/{2579}
39a no 2/3/4/6 in r6c5
39b no 3/4/6 in r7c5
39c 11(3) at r3c5 must now be [1]{37}/{46} - 2 and 8 locked - no 2 or 8

40. 25(4) at r6c4 - limited permutations left - no valid pattern with 7/8 at r6c4 or 9 at r6c6

41. Innies found at step 9 can only now be {124578} -> 5 only at r1c7, no 9 no 3
41a. 18(3) at r6c7 can now only be {369} - no 1/2/7/8
41b 10(2) at r6c8 - {19} not now possible so no 1

42. 17(4) at r1c4 now must be [25]{46} - 2 at r1c5 and 5 @ r2c5
42a 11(2) at r8c5 can't now be {56} - so no 6
42b 11(3) at r3c5 can't now be {137} (3 used in 42a) - so no 3/7
42c 12(2) at r7c4 can't now be {48} (8 used in 42a) - so no 4/8
42d. innies found at step 24 now has to be [2][5]{79} - no 8

43. 18(3) at r3c9 - only combo with 2 is {279} and can't have the 2 at r4c9

44. 29(4) at r4c6 - only possibility is {5789} with r45c6 {59} and r45c7{78} - no 7/8 in r45c6

45. 21(4) at r8c8 now has 1/2 in r9c7. no 1/2 anywhere else in the cage and r9c6 cannot be 3

46 25(4) at r6c4 has r6c5 {7/9} and r7c5 {7/9} - only combos with 7&9 are {1789}/{3679}/{4579} - no 2, 7 can't be in r6c6

. . . the rest falls out fairly easily from here
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nd
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PostPosted: Sat Jan 20, 2007 3:55 am    Post subject: Reply with quote

Hm. Just been working through the walkthrough, & I must confess I don't really understand the ordering of steps or the key stage at step 20.

Re: the ordering: I don't understand why you've consistently chosen the most complicated & laborious innie-outies & combinations, & neglected to do all the basic steps first. In particular, fully half the first two dozen steps would be unnecessary if you simply applied the 45 rule to N1 (R3C13 = 11(2)), N3 (R3C79 = 12(2)), C1234 (R16C4 = 7(2)), and C5678 (R16C6 = 12(2)).

Secondly, I just don't follow what you're doing in step 20; probably if I sat down with a combination chart & did the branching myself I could verify the results, but since the point of a walkthrough is to provide an explanation that doesn't need extensive further work from the reader, then I stopped at this point.

Some remarks--

rcbroughton wrote:
9. 45 rule on column 7. Innies cells r1234597 equal 27 -> {124578} {123579} {124569} {123678} {134568} {234567}
9a. No valid combination with 5/6/7/8/9 in r2c7
9b. No valid combination with 5/6/7 in r3c7
9c. No valid combination with 5/6/7/8/9 in r9c7

I donít understand the notation (r1234597) or what youíre doing here. The deductions are correct, though. But the best (clearest, most logical) way to deal with this is via the hidden subset move I mentioned earlier.

Quote:
12. 45 rule on column 9. Outies r2c7 r2345c8 equal 14
12a Cage 10(2) at r6c8 restricts possible values in r2345c8 {1236} & {1234} not possible
12b r2c7 cannot be 4
12c r2c8 can't now be 8/9
12d r3c8 can't now be 8/9

13. No possible combinations left with 8 in 19(5) at r1c9

14. Must use 2 in cage 19(5) at r1c9 -> cannot have 2 in r3c79

Youíre doing things the exceedingly hard way in all these steps. All you need to do is: 45 rule on N3 => R3C79 = 12 = [39|48] => in conjunction with the 14(2) cage in R1C78 (={59|68}) we have a hidden pair on {89} so those candidates are eliminated in all other cells of N3. This places {127} in the 19(5) cage & excludes {89}.

Quote:
18. 45 rule on N2. Innies r3c5 minus outies r3c3 r3c7 equals -8
18a Max val in r3c5 is 5 so max of r3x3+r3c7 is 13
18b No valid combination with 8
19c Remembering that r3c13 total 11 r3c1 cannot now be 3

Again this is doing things the hard way. {38} is excluded in R3C13 by R3C79 = [39|48].

Quote:
20. 45 rule on N2. Innies r2c4 r3c4 r2c6 r3c6 r3c5 equal 28
20a r3c13 limits possibilities in r3c456 - no possible combo with 6 at r3c4

This is not at all immediately clear to me from looking at the grid. Please explain this one.

Quote:
20b.this combo now means 9 cannot be in 17(4) at r1c5, r1c6 or r2c5

Again this step isnít clear to me at all.
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nd
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PostPosted: Sat Jan 20, 2007 6:22 am    Post subject: Reply with quote

Haven't proofed this yet but here's a walkthrough.

Step 1. 12(4) cage at R4C3 = {1236|1245}. 29(4) cage at R4C6 = {5789}. R78C6 = {12}. R1C78 = {59|68}. 22(3) cage at R2C4+R3C34 = {589|679}.

Step 2. 18(3) cage at R6C7 must have 2 cells containing {56789}. (If it had only one, then the max cage-sum would be 3+4+9 = 16.) So the 18(2) cage forms a hidden quint on {56789} in conjunction with R145C7 => R239C7 = {1..4}.

Step 3. 45 rule on N3 => R3C79 = 12 = [39|48] => R23C6 = {47|56|37}. This means that R45C6 cannot be {57}. 45 rule on C6789 => R16C6 = 12 = {39|48} (again, this cannot be {57} because of R23C6). This means that R45C6 cannot be {89}. So R45C6 must contain exactly one of {57} and one of {89}. This means we have a hidden quad on {5789} within R123456C6!!! So R9C6 = {346}.

Step 4. 45 rule on N1 => R3C13 = 11 = {56} or [29] or [47] ([38] is blocked because R3C79 = [39|48], see step 3).

Step 5. 45 rule on N2 => R3C37 = R3C5 + 8 => maximum value of R3C5 = 5 (since max val of R3C37 = 13).

Step 6. 45 rule on N2 => R2C46 + R3C456 = 28(5). Obviously if the 28(5) cage has a 1 or 2 in it, then R3C5 = {12}. If it does not, the only possible combos are {34579} or {34678}, i.e. it must have both 3 and 4 in it. But this is impossible, because R3C7 = {34} "sees" all three cells R2C6 + R3C56. Therefore R3C5 = {12}. All combinations of 28(5) with 1 or 2 in them include a 9 => the 9 is locked in R23C4 within N2/C4/the 22(3) cage.

Step 7. So at this point there is only one square in N8 where the 9 can go (since 11(2) = {29} is blocked by the 3(2) cage). So R7C5 = 9 => 9 is locked in N5/the 29(4) cage in R45C6 => R16C6 = {48}, R45C6 = {(5|7)9}, R45C7 = {(5|7)8}.

Step 8. R78C4 = {48|57}, R89C5 = {38|56} ({47} is blocked by the 12(2) cage). 45 rule on N8 => R9C46 = 10 = [73|46]. Thus R78C4 + R9C4 contain a hidden {47} pair => {47} is excluded in R1..6C4 => R23C4 = {(6|8)9}, R3C3 = {57}, R3C1 = {46}.

Step 9. 45 rule on C1234 => R16C4 = 7 = {16|25}. Therefore the only spot left in C4 for the 3 is R45C4 => the 12(4) cage is {1236} with the 3 in R45C4.

Step 10. This leaves only two possible combinations for the 25(4) cage at R6C4: {1789} or {2689} => R6C6 = 8, R1C6 = 4, and the 4 is locked in C5/N5/the 11(3) cage within R45C5 => R45C5 = {4(6|7)}. This in turn forms a hidden {67} pair with the 25(4) cage => R45C6 = {59}!

Step 11. The rest is mop-up. R45C7 = {78}. R23C6 = {37} (see step 3), R3C7 = 4, R3C9 = 8, R9C6 = 6, R89C5 = {38}, R78C4 = {57}, R9C4 = 4, R23C4 = [89] (because this is the only spot for 8 in C4), R3C3 = 5, R3C1 = 6, R16C4 = [61], R45C4 = {23}, R45C3 = {16}. 45 rule on N2 => R3C5 = 1. R1C78 = {59}, R12C5 = [25], R6C5 = 7, R45C5 = {46}.

Step 12. In C9 the only spot for the 9 is R4C9 (can't be in a 11(3) cage) => R4C89 = [19], R45C6 = [59], R4C12 = {27}, R45C3 = [61], R45C4 = [32], R45C7 = [87], R45C5 = [46]. 45 rule on N4 => R6C23 = 13 = {49}, 45 rule on N6 => R6C78 = {36}, R6C19 = [52], R789C1 = {489}, R5C12 = [38], R1C23 = [38] (only place for 8 to go in N1).

Step 13. 11(3) cage in N9 must have a 1 in it (since it must have 1 and/or 2, & 2 is blocked) => R2C7 = 1, R1C1 = 1. 45 rule on C7 => R19C7 = 7 = [52], R1C8 = 9, R1C9 = 7, R89C8 = {58}. R89C58 form an x-wing on 8 => R789C1 = [849], and you carry on........
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rcbroughton
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PostPosted: Sat Jan 20, 2007 9:05 am    Post subject: Reply with quote

nd wrote:
Re: the ordering: I don't understand why you've consistently chosen the most complicated & laborious innie-outies & combinations, & neglected to do all the basic steps first. In particular, fully half the first two dozen steps would be unnecessary if you simply applied the 45 rule to N1 (R3C13 = 11(2)), N3 (R3C79 = 12(2)), C1234 (R16C4 = 7(2)), and C5678 (R16C6 = 12(2)).

true enough - I tend to work more with combinations and permutations - I sometimes find I make simple arithmetical errors trying to tally up cage totals - mental arithmetic not necessarily my strong point. (Plus I use a littel spreadsheet that shows my combinations & permutations for cage totals)

Quote:
Secondly, I just don't follow what you're doing in step 20; probably if I sat down with a combination chart & did the branching myself I could verify the results, but since the point of a walkthrough is to provide an explanation that doesn't need extensive further work from the reader, then I stopped at this point.

20. 45 rule on N2. Innies r2c4 r3c4 r2c6 r3c6 r3c5 equal 28
20a r3c13 limits possibilities in r3c456 - no possible combo with 6 at r3c4

This is not at all immediately clear to me from looking at the grid. Please explain this one.

OK - at this point we know r2c4 r3c4 r2c6 r3c6 r3c5 equal 28 and we have{6/7/8/9} {7/8/9} {3/4/5/6/7} {3/4/5/6/7} and {1/2/4/5} in those cells
The 17(4) in n2 must use either a 2 or a 7 - so we can't have any combinations with both. That leaves us with {13789}{14689} {15679} {23689} {24589} {34579} {34678}
Looking at the possibilities for placing the 6 - we have combos {14689}{15679} and {34678}
for {14689} the 1 must be in r3c5 - r23c4 can only be {89}
for {34678} the 8 must be in r23c4, the 3 in r23c6 and the 4 in r3c5 - but this leaves us with nothing for r3c7 - so combination cn't be used
for {15679} the 1 must be in r3c5, the 5 in r23c6 and the 9 in r23c4 - the 14(2) cage can only therefore be {563} - so r23c4 can only be {79}
therefore no 6 in r2c4
(a bit long-winded, but I do like playing with permutations!)
Quote:
20b.this combo now means 9 cannot be in 17(4) at r1c5, r1c6 or r2c5
Again this step isnít clear to me at all.

now we've eliminated 6 from r2c4, the 22(3) cage can only be {589}/{679} - so 9 must be in r23c4

The point of my post was more to see if I was looking at the same puzzle as everyone else. The comment had been that it was going to be very difficult and needed trial & error. Although there were a few laborious steps - I didn't see this one as particularly difficult - didn't need to use any advanced techniques like x-wing, chains, ALS, etc - and just wanted to check I wasn't being a bit slow and missing something. But I didn't just want to say it was solvable by siple logic without at least showing my working - however inelegant it might be !!
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PostPosted: Sat Jan 20, 2007 7:16 pm    Post subject: Reply with quote

Thanks--I don't think you can omit such a lengthy amount of permutation, involving extensive trial & error, in a walkthrough--otherwise the walkthrough doesn't really explain how you got your result.
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PostPosted: Sat Jan 20, 2007 9:44 pm    Post subject: Reply with quote

Point taken. By the time i've written out & crossed off the invalids i end up just jotting down the main part of the step and the end result. i'll try to be more explicit in future!
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PostPosted: Wed Jan 28, 2009 7:56 am    Post subject: Reply with quote

As I commented in this thread and the A71 thread, I didn't manage to solve three of Ruud's Assassins when they first appeared. Now having caught up with my backlog of other walkthroughs I'm having another go at them.

I originally used a contradiction move for the key breakthrough but, after I'd finished I had another look and found an alternative more direct way; I've given both in step 17.

Here is my walkthrough.

Prelims

a) R1C78 = {59/68}
b) R1C23 = {29/38/47} (cannot be {56} which clashes with R1C78), no 1,5,6
c) R67C2 = {19/28/37/46}, no 5
d) R67C8 = {19/28/37/46}, no 5
e) R78C4 = {39/48/57}, no 1,2,6
f) R78C6 = {12}, locked for C6 and N8
g) R89C5 = {38/47/56}, no 1,9
h) 22(3) cage at R2C4 = {589/679}, CPE no 9 in R3C56
i) R345C5 = {128/137/146/236/245}, no 9
j) 11(3) cage in N6 = {128/137/146/236/245}, no 9
k) R789C1 = {489/579/678}, no 1,2,3
l) R789C9 = {128/137/146/236/245}, no 9
m) 12(4) cage at R4C3 = {1236/1245}, no 7,8,9
n) 29(4) cage at R4C6 = {5789}
o) 19(5) cage in N3 = {12349/12358/12367/12457/13456}, 1 locked for N3

1. 45 rule on N1 2 innies R3C13 = 11 = [29/38/47/56/65], no 1,7,8,9 in R3C1

2. Min R23C6 = 7 -> max R3C7 = 7
2a. 45 rule on N3 2 innies R3C79 = 12 = [39/48/57/75], no 2,6, no 3,4 in R3C9

3. 45 rule on R6789 2 innies R6C19 = 7 = {16/25/34}, no 7,8,9

4. 45 rule on C1234 2 innies R16C4 = 7 = {16/25/34}, no 7,8,9

5. 45 rule on C6789 2 innies R16C6 = 12 = {39/48/57}, no 6

6. R678C7 = {189/279/369/378/459/468/567}, must contain at least two of 5,6,7,8,9
6a. Killer quint 5,6,7,8,9 in R145C7 and 18(3) cage in N69, locked for C7, clean-up: no 5,7 in R3C9 (step 2a)
6b. R678C7 can only contain two of 5,6,7 = {189/279/369/378/459/468}

7. Killer pair 8,9 in R1C78 and R3C9, locked for N3

8. 14(3) cage at R2C6 = {347/356}, no 8,9, CPE no 3 in R3C5
8a. R16C6 (step 6) = {39/48} (cannot be {57} which clashes with R23C6), no 5,7
8b. Killer quad 5,7,8,9 in R16C6, R23C6 and R45C6, locked for C6

9. Hidden killer pair 8,9 in R16C6 and R45C6 for C6 -> R45C6 must contain one of 8,9 -> R45C7 must contain one of 8,9
9a. R678C7 (step 6b) = {279/369/378/459/468} (cannot be {189} which clashes with R45C7), no 1
9b. Killer pair 8,9 in R45C7 and R678C7, locked for C7, clean-up: no 5,6 in R1C8
9c. 45 rule on N3 2 remaining innies R13C7 = 9 = [54/63]
9d. R678C7 = {279/369/378/459} (cannot be {468} which clashes with R13C7)

10. 45 rule on N2 2 outies R3C37 = 1 innie R3C5 + 8, IOU no 8 in R3C3, clean-up: no 3 in R3C1 (step 1)
10a. Max R3C37 = 13 -> max R3C5 = 5

11. 22(3) cage at R2C4 = {589/679}
11a. 5 of {589} must be in R3C34 (R3C34 cannot be {89} which clashes with R3C9), no 5 in R2C4

12. 45 rule on C89 3 innies R189C8 = 1 outie R2C7 + 21
12a. Min R189C8 = 22, no 1,2,3,4, 9 locked for C8, clean-up: no 1 in R67C8
12b. Max R189C8 = 24 -> max R2C7 = 3

13. 45 rule on C9 2 innies R12C9 = 2 outies R45C8 + 5
13a. Max R12C9 = 13 -> max R45C8 = 8, no 8
13b. Max R3C9 + R4C8 = 16 -> min R4C9 = 2

14. 45 rule on R123 3 innies R3C159 = 15 = {159/168/249/258} (cannot be {267/456} because R3C9 only contains 8,9)
14a. R3C13 = 11 (step 1) -> R3C19 cannot be 11 (Iíll call that an overlap IOU) -> no 4 in R3C5

15. 45 rule on N8 3 innies R7C5 + R9C46 = 19 = {379/469/478/568}
15a. 3 of {379} must be in R9C6 -> no 3 in R7C5 + R9C4

16. 25(4) cage at R6C4 = {1789/2689/3589/3679/4579/4678}
16a. 1 of {1789} must be in R6C4 -> no 1 in R6C5

My original step 17
17. 17(4) cage in N2 = {1268/1349/1358/2348/2456} (cannot be {1259} which clashes with R3C5, cannot be {1367/1457/2357} which clash with R23C6), no 7
17a. Cannot be {1268}, hereís how
17(4) cage = {1268} => 14(3) cage at R2C6 = {347} => R23C4 = [95] is impossible because no 8 in R3C3
17b. 17(4) cage = {1349/1358/2348/2456}
17c. Hidden killer pair 1,2 in 17(4) cage and R3C5 for N2 -> R3C5 = {12}

Alternative step 17 which avoids using a contradiction move
17. 17(4) cage in N2 = {1268/1349/1358/2348/2456} (cannot be {1259} which clashes with R3C5, cannot be {1367/1457/2357} which clash with R23C6), no 7
17a. 45 rule on N2 (from result of step 17) 5 innies R2C46 + R3C456 = 28 = {13789/15679/24679/25678} (cannot be {34579} which clashes with 14(3) cage at R2C6)
17b. 1,2 only in R3C5 -> R3C5 = {12}
17c. 17(4) cage = {1349/1358/2348/2456}

18. R3C159 (step 14) = {159/168/249/258}
18a. R3C5 = {12} -> no 2 in R3C1, clean-up: no 9 in R3C3 (step 1)
18b. 9 in 22(3) cage at R2C4 locked in R23C4, locked for C4 and N2, clean-up: no 3 in R6C6 (step 5), no 3 in R78C4

19. R7C5 = 9 (hidden single in N8), clean-up: no 3 in R1C6 (step 5), no 1 in R6C2
19a. 45 rule on N8 2 remaining innies R9C46 = 10 = [46/64/73], no 5,8 in R9C4
19b. R89C5 = {38/56} (cannot be {47} which clashes with R78C4), no 4,7

20. Naked pair {48} in R16C6, locked for C6, clean-up: no 6 in R9C4 (step 19a)

21. 9 in C6 locked in R45C6, locked for 29(4) cage at R4C6
21a. 8 in 29(4) cage at R4C6 locked in R45C7, locked for C7 and N6, clean-up: no 2 in R7C8

22. 22(3) cage at R2C4 = {589/679}
22a. 5 of {589} must be in R3C3 -> no 5 in R3C4

23. 17(4) cage (original step 17b or alternative step 17c) = {1358/2348/2456}
23a. 1 of {1358} must be in R1C45 (R1C456 cannot be {358} which clashes with R1C78), no 1 in R2C5

24. 45 rule on C5 3 remaining innies R126C5 = 14 = {167/248/257/347} (cannot be {158/356} which clash with R89C5)
24a. 1 of {167} must be in R1C5 -> no 6 in R1C5
24b. 7 of {167/257/347} must be in R6C5 -> no 3,5,6 in R6C5

25. 25(4) cage at R6C4 = {1789/2689} (cannot be {3589/3679} because 3,5,6 only in R6C4, cannot be {4579} which clashes with R45C6), no 3,4,5 -> R6C6 = 8, R1C6 = 4, clean-up: no 7 in R1C23, no 2,3 in R1C4 (step 4), no 1 in R1C45 (step 23), no 6 in R6C4 (step 4)
25a. 25(4) cage at R6C4 = {1789} (only remaining combination) -> R6C45 = [17], R1C4 = 6 (step 4), R1C7 = 5, R1C8 = 9, R3C9 = 8, R3C7 = 4 (step 2a), clean-up: no 2 in R1C23, no 3,8 in R12C5 (step 23), no 7 in R3C3 (step 1), no 6 in R6C19 (step 3), no 2,3 in R7C2, no 3 in R7C8
25b. R123C5 = [251], clean-up: no 6 in R89C5
25c. R3C37 = R3C5 + 8 (step 10), R3C5 = 1, R3C7 = 4 -> R3C3 = 5, R3C1 = 6

26. R9C6 = 6 (hidden single in C6), R9C4 = 4 (step 19a), clean-up: no 8 in R78C4
26a. Naked pair {57} in R78C4, locked for C4 -> R23C4 = [89]
26b. Naked pair {23} in R45C4, locked for N5 and 12(4) cage at R4C3
26c. 12(4) cage at R4C3 = {1236} (only remaining combination), no 4
26d. Naked pair {16} in R45C3, locked for C3 and N4, clean-up: no 4 in R7C2
26e. Naked pair {78} in R45C7, locked for C7 and N6

27. R678C7 (step 9d) = {369} (only remaining combination), locked for C7

28. 21(4) cage at R8C8 = {2568} (only remaining combination) -> R9C7 = 2, R2C7 = 1, R89C8 = {58}, locked for C8 and N9, clean-up: no 2 in R6C8

29. R4C9 = 9 (hidden single in C9), R4C8 = 1 (cage sum), R45C3 = [61], R45C5 = [46], R45C6 = [59]

30. R3C9 = 6 -> R4C12 = 9 = {27}, locked for R4 and N4 -> R45C4 = [32], R45C7 = [87], clean-up: no 8 in R7C2

31. R6C9 = 2 (hidden single in R6), R5C89 = 9 = [45], R6C1 = 5 (step 3), clean-up: no 6 in R67C8
31a. R67C8 = [37], R678C7 = [639], R78C4 = [57], R9C9 = 1

32. Naked pair {48} in R78C1, locked for C1 and N7, R7C3 = 2, R7C6 = 1, R7C2 = 6, R6C2 = 4

and the rest is naked singles
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