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Assassin 62

 
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Para
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PostPosted: Fri Aug 03, 2007 10:51 am    Post subject: Assassin 62 Reply with quote

Hi all

This one was very basic. Nothing special to solve it. It was a relaxing puzzle.

Walk-through Assassin 62

1. R23C6 = {29/38/47/56}: no 1

2. 26(4) at R2C8 = {2789/3689/4589/4679/5678}: no 1

3. 6(3) at R3C9 = {123} -->> locked for C9

4. 8(3) at R5C4 = {125/134}: no 6,7,8,9; 1 locked in 8(3) cage -->> R6C56: no 1

5. 20(3) at R5C7 = {389/479/569/578}: no 1,2

6. R78C4 = {89} -->> locked for C4 and N8

7. 45 on R89: 2 innies and 1 outie: R8C24 = R7C6 + 2: Min R8C24 = 9 -->> Min R7C6 = 7 -->> R7C6 = 7; R8C24 = [18]
7a. R7C4 = 9
7b. Clean up: R23C6: no 4

8. 15(3) at R8C5 = {456} -->> locked for N8

9. 45 on N9: 2 outies: R6C9 + R8C6 = 9 = [63/72] -->> R6C9 = {67}

10. 18(4) at R7C6 = 7{236/245} ({128} blocked because {18} only in R9C7, {146} blocked by R8C6): no 1,8,9

11. 7 and 9 in N9 locked within 24(4) cage at R8C8 -->> 24(4) = {2679/3579}: no 1,4,8
11a. R9C4 = 1(hidden)
11b. R6C3 = 1(only place within 8(3) cage(step 4))

12. 1 and 8 within N9 locked in 19(4) cage at R6C9 -->> 19(4) = {1378/1468}: no 2,5; 8 locked for R7
12a. 19(4) needs one of {67} which has to go in R6C9 -->> R7C789: no 6
12b. 5 and 6 in R7 locked for N7

13. 45 on N7: 3 innies: R7C123 = 14 = {356}(5 and 6 locked in these cells) -->> locked for N7 and R7
13a. R7C5 = 2; R8C6 = 3; R6C9 = 6(step 9); R9C8 = 3(hidden)
13b. Clean up: R23C6: no 8; R89C7 = {26}(step 10) -->> locked for C7 and N9
13c. R89C3 = [48](last possible combination within 13(3) at R8C3)

14. 45 on C12: 3 outies: R357C3 = 19 = {379} -->> R7C3 = 3; R35C3 = {79} -->> locked for R3

15. 45 on N1: 3 outies: R23C4 + R4C1 = 10 = {26}[2]/{35}[2]/{25}[3]/{34}[3]/{24}[4]/{23}[5] -->> R23C4 = {23456}; R4C1 = {2345}

16. 18(4) at R1C3 = {3456}(last possible combination): no 2 -->> R12C3 = {56} -->> locked for C3 and N1; R23C4 = {34} -->> locked for C4 and N2
16a. R4C3 = 2; R4C1 = 3(step 15); R4C9 = 1; R5C9 = 2(hidden); R56C4 = [52]
16b. R3C9 = 3; R23C4 = [34]; R1C2 = 3(hidden)

17. 17(3) at R4C2 = {467}: ({458} blocked by R5C3) -->> R5C3 = 7; R45C2 = {46} -->> locked for C2 and N4
17a. R3C3 = 9; R7C12 = [65]; R6C2 = 9; R56C1 = [85]

18. 14(3) at R1C4 = {167}: ({158} blocked by R1C4) -->> locked for N2
18a. R5C6 = 1(hidden); R3C1 = 1(hidden); R3C2 = 8

And the rest is all naked singles

greetings

Para


Last edited by Para on Fri Aug 10, 2007 9:19 am; edited 2 times in total
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CathyW
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PostPosted: Fri Aug 03, 2007 2:13 pm    Post subject: Reply with quote

Yup! Probably a 0.5 on Mike's scale but still a good challenge during lunch break.

I didn't keep a complete WT this time but early placements made with outies - innies of r89. Thereafter some combo crunching did the job.
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Ruud
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PostPosted: Fri Aug 03, 2007 6:56 pm    Post subject: Reply with quote

Here is A62V2 (?)



3x3::k:7168:7168:3074:4099:4099:4869:4869:1799:1799:7168:7168:3074:3074:4099:2830:4869:5136:1799:4626:4626:4626:3074:4886:2830:5136:5136:4122:4626:5660:4886:4886:4886:5920:5920:5136:4122:2084:5660:5660:4647:5920:5920:3626:3626:4122:2084:4910:4647:4647:3889:3889:3889:3626:6197:2084:4910:4910:2873:3889:4667:6197:6197:6197:2367:4910:5697:2873:4419:4667:4667:4934:4934:2367:2367:5697:5697:4419:4419:4667:4934:4934:

If you cannot find any shortcuts, this killer will keep you busy for quite a while.

Ruud
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Last edited by Ruud on Mon Aug 06, 2007 9:23 pm; edited 1 time in total
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Para
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PostPosted: Fri Aug 03, 2007 9:00 pm    Post subject: Reply with quote

CathyW wrote:
Yup! Probably a 0.5 on Mike's scale but still a good challenge during lunch break.


Don't know if i will manage this one without pencil marks. Might be ok try it. Let's say 0.75(it is harder than Assassin 1, which falls in the no PM category). Somewhere between 0.5-0.75 probably.

Para
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CathyW
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PostPosted: Fri Aug 03, 2007 9:06 pm    Post subject: Reply with quote

Just checked Mike's rating list in the other thread. He amended the bit about pencil marks after I said I'd never managed a Times Deadly without them though I think it's supposed to say "requiring only limited pencil marks"! The A62 did need more than limited marks for the combo crunching so 0.75 is a fairer assessment.

Had a quick look at the V2(?) using JSudoku. Don't think I'll get very far on my own so I'll leave it for the masochists while I'm on holiday. Laughing
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PostPosted: Fri Aug 03, 2007 10:11 pm    Post subject: Reply with quote

CathyW wrote:
Just checked Mike's rating list in the other thread...though I think it's supposed to say "requiring only limited pencil marks"!

Oops! Thanks for pointing that out. I've corrected the table now.

CathyW wrote:
The A62 did need more than limited marks for the combo crunching so 0.75 is a fairer assessment.

Yes, I'll settle for 0.75.

CathyW wrote:
Had a quick look at the V2(?) using JSudoku. Don't think I'll get very far on my own so I'll leave it for the masochists while I'm on holiday. Laughing

I have also been busy looking into variants (but with a slightly modified cage pattern). Looks like Ruud's leap-frogged me, because all of mine that I had labelled 'promising' were at least one notch easier than Ruud's V2.

Maybe I'll post one of them sometime later as a "coming home" present for when you get back! Smile
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Para
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PostPosted: Fri Aug 03, 2007 11:56 pm    Post subject: Reply with quote

mhparker wrote:

I have also been busy looking into variants (but with a slightly modified cage pattern). Looks like Ruud's leap-frogged me, because all of mine that I had labelled 'promising' were at least one notch easier than Ruud's V2.


I also have a changed cage pattern/same solution V2. But Ruud already beat me to it. But i guess mine is more a V1.5. We'll see if we need more puzzles. But i think with all these monsters lurking around, it would be nice to have a properly solvable Version too some times as well.

Para
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Andrew
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PostPosted: Sat Aug 04, 2007 2:27 am    Post subject: Reply with quote

Para wrote:
But i think with all these monsters lurking around, it would be nice to have a properly solvable Version too some times as well.

Para

That's what the V1s are most of the time. Ruud has the balance right with more than half of the Assassins around the V1 level; 1 or 2 each month usually up to the V1.5 level.

I haven't yet started on Assassin 62 but from the length of Para's tiny text walkthrough and comments in this thread I'm expecting it to be an easier one. I'm still working on A61X, think I'm nearly finished now, plus I've still got to find time to look at the walkthroughs for A60. So many puzzles and walkthroughs but keep them coming!

Finally, so we don't forget where it is posted, may I suggest that Mike post his ratings in the Rating Assassins thread and then Ruud make that thread a Sticky.
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PostPosted: Sat Aug 04, 2007 6:39 pm    Post subject: Reply with quote

Andrew wrote:
Para wrote:
But i think with all these monsters lurking around, it would be nice to have a properly solvable Version too some times as well.

Para

That's what the V1s are most of the time. Ruud has the balance right with more than half of the Assassins around the V1 level; 1 or 2 each month usually up to the V1.5 level.


What i meant by this, is that i rather have V1.5-1.75 second versions than V2.0 and above. Especially as/when there are still V2.0 and above rated puzzles being tagged.
I guess i just feel that one big tagpuzzle every 2-3 weeks is enough.

Para
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mhparker
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PostPosted: Sun Aug 05, 2007 9:10 am    Post subject: Reply with quote

Hi all,

Ruud wrote:
If you cannot find any shortcuts, this killer will keep you busy for quite a while.

Indeed, looks like we may be in for a rough ride here. So I suggest we do this as a team (tag) solution. Unless anyone has any major objections, I would also suggest that we use normal text for convenience.

If we can get the tag solution off the ground soon, maybe there's time for a V1.5 for those who don't want to participate in the tag. But I'd prefer such a puzzle to be posted after the V2 tag solution is well underway, otherwise it would probably serve as too much of a distraction.

Agreed?

P.S. If anyone wants to get the tag started, feel free. Don't feel obliged to discover a particular number of complex moves alone first. Even listing the prologue und more basic starting moves would be a big help in terms of getting this one in motion!
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PostPosted: Mon Aug 06, 2007 11:38 pm    Post subject: Reply with quote

Para wrote:
This one was very basic. Nothing special to solve it. It was a relaxing puzzle.

Somewhat surprising considering the complicated cage pattern with so many 4-cell cages. Still we know by now that the choice of cage totals tends to contribute much more than the cage pattern to the level of difficulty. I haven't (yet?) looked at V2 but that could well be a monster since nothing has been posted so far.

Cathy wrote:
I didn't keep a complete WT this time but early placements made with outies - innies of r89. Thereafter some combo crunching did the job.

It was interesting that there were two ways to look at outies - innies of r89 depending whether one included r78c4. It just happened that Para did it one way and I used the other way; both lead to the same early placements.

Para's steps 14, 15 and 16 made his more direct than my solving path. I've made a comment after my step 18.

Here is my walkthrough for A62.

1. R23C6 = {29/38/47/56}, no 1

2. R78C4 = {89}, locked for C4 and N8

3. R345C9 = {123}, locked for C9
3a. Min R12C9 = 9 -> max R1C8 = 6

4. R567C1 = {289/379/469/478}, no 1

5. 8(3) cage at R5C4 = 1{25/34}, no 1 in R6C56

6. 20(3) cage in N6 = {389/479/569/578}, no 1,2

7. 26(4) cage at R2C8 = {2789/3689/4589/4679/5678}, no 1

8. 45 rule on N1 3 outies R23C4 + R4C1 = 10, min R23C4 = 3 -> max R4C1 = 7

9. 45 rule on R89 2 outies R7C46 – 15 = 1 innie R8C2 -> R7C46 = [97] (only possible permutation greater than 15), R8C2 = 1, R8C4 = 8, clean-up: no 4 in R23C6

10. 15(3) cage in N8 = {456} (only remaining combination), locked for N8

11. 18(4) cage at R7C6 = 7{236/245} (cannot be {1278} because 1,8 only in R9C7, cannot be {1467} because only 2,3 in R8C6) = 27{36/45}, no 1,8,9, no 2 in R8C8

12. 45 rule on N9 2 remaining outies R6C9 + R8C6 = 9 = [63/72]

13. 18(4) cage at R6C2 max R7C23 + R8C2 = 15 -> min R6C2 = 3

14. 24(4) cage in N9 must contain 7,9 = 79{26/35}, no 1,4,8
14a. 2 only in R9C8 -> no 6 in R9C8

15. 1,8 in N9 locked in R7C789, locked for R7
15a. 19(4) cage at R6C9 = 18{37/46}, no 2,5
15b. 6 in {1468} must be in R6C9 -> no 6 in R7C789
15c. 5,6 in R7 locked in R7C123, locked for N7
15d. 18(4) cage at R6C2 max R7C23 + R8C2 (step 13) = 12 -> min R6C2 = 6

16. R9C4 = 1 (hidden single in R9)
16a. R89C3 = 12 = {39}/[48], no 2,7, no 4 in R9C3

17. R6C3 = 1 (only remaining cell for 1 in 8(3) cage, step 5)
17a. R56C4 = {25/34}

18. 45 rule on C89 3 outies R357C7 = 20 = {389/479/569/578}, no 1,2
[I didn’t do the corresponding 3 outies for C12 at this stage because it didn’t do anything useful and had forgotten about it by the time that R7C3 had been reduced to 3 candidates, at which stage Para’s step 14 showed that it was very useful.]

19. R7C8 = 1 (hidden single in R7)

20. R23C4 + R4C1 = 10 (step 8)
20a. Min R4C1 + either R2C4 or R3C4 = 4 -> max R2C4 or R3C4 = 6
20b. Min R23C4 = 5 -> max R4C1 = 5

21. 7 in N7 locked in 18(3) cage = 7{29/38}, no 4

22. R89C123 = 18 + 1 + 12 (step 16a) = 31 -> R7C123 = 14 and must contain 5,6 = {356}, locked for R7 and N7 -> R7C5 = 2, R8C6 = 3, clean-up: no 8 in R23C6, no 9 in R89C3 (step 16a) = [48]

23. Naked pair {48} in R7C79, locked for N9 -> R6C9 = 6 (step 15a)

24. R9C8 = 3 (hidden single in N9)
24a. 24(4) cage in N9 = {3579} (step 14), locked for N9

25. Naked pair {26} in R78C6, locked for C7

26. 18(4) cage at R6C2 = {1359/1368} = 13{59/68}, no 7
26a. 3 locked in R7C23, locked for R7
26b. Max R7C1 = 6 -> min R56C1 = 13, no 2,3

Ruud wrote:
... but there are some nice quadruple outies in here.

Now for a couple of quadruple outies, one of which was very useful.

27. 45 rule on N3 4 outies R1C6 + R4C89 + R5C9 = 16, max R4C89 + R5C9 = 14 -> min R1C6 = 2

28. 45 rule on N7 4 outies R5C1 + R6C12 + R9C4 = 23, R9C4 = 1 -> R5C1 + R6C12 = 22 = 9{58/67}, 9 locked for N4
28a. R567C1 = {568} (only remaining combination), locked for C1
28b. R5C1 + R6C12 = {589} (only remaining combination)
28c. Naked pair {58} in R56C1, locked for N4 -> R7C1 = 6, R6C2 = 9
28d. 9 in N7 locked in R89C1, locked for C1

29. 17(3) cage in N4 = {467} (only remaining combination), locked for N4
29a. 4 in N4 locked in R45C2, locked for C2

30. Naked pair {23} in R4C13, locked for R4 -> R4C9 = 1
30a. R5C9 = 2 (hidden single in N6) -> R3C9 = 3, clean-up: no 5 in R6C4 (step 17a)

31. R5C6 = 1 (hidden single in C6)

32. 1 in N3 locked in R12C7, 14(3) cage at R1C6 = 1{49/58/67}, no 2 in R1C6

33. 2 in C6 locked in R23C6 = {29}, locked for C6 and N2
33a. No 9 in R1C6 -> no 4 in R12C7 (step 32)

34. R6C4 = 2 (hidden single in C4) -> R5C4 = 5 (step 17a), R56C1 = [85]

35. R7C5 = 2 -> R6C567 = 15 = {348} (only remaining combination), locked for R6 -> R6C8 = 7
35a. R5C78 = 13 = {49} (only remaining combination), locked for R5 and N6

36. R6C7 = 3 (hidden single in N6) -> R6C56 = {48}, locked for N5 -> R4C6 = 6, R4C45 = [79], R5C5 = 3, R4C2 = 4, R4C7 = 8 (cage sum), R4C8 = 5, R7C79 = [48], R5C78 = [94], R8C8 = 9

37. Naked pair {57} in R89C9, locked for C9
37a. R12C9 = {49} -> R1C8 = 2

38. Naked triple {346} in R123C4, locked for N2

39. R4C45 = [79] -> R3C5 + R4C3 = 7 = [52] (only remaining permutation) -> R4C1 = 3

40. R8C5 = 6 -> R9C56 = [45], R6C56 = [84], R89C7 = [26], R8C1 = 7, R9C12 = [92], R89C9 = [57], R1C6 = 8, R3C7 = 7

41. Naked pair {68} in R3C28, locked for R3 -> R3C3 = 9, R3C4 = 4, R23C6 = [92], R3C1 = 1, R1C12 = [42], R12C9 = [94], R3C2 = 8 (cage sum), R23C8 = [86]

41. R12C5 = {17} -> R1C4 = 6 (cage sum), R2C4 = 3

42. R1C2 = 3 (hidden single in N1), R2C2 = 7 (cage sum)

and the rest is naked singles


Last edited by Andrew on Sat Aug 18, 2007 5:01 am; edited 1 time in total
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mhparker
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PostPosted: Thu Aug 09, 2007 12:15 pm    Post subject: Reply with quote

Para wrote:
This one was very basic. Nothing special to solve it. It was a relaxing puzzle.

Andrew wrote:
Somewhat surprising considering the complicated cage pattern with so many 4-cell cages.

Indeed, Ruud seems to be going through a phase (i.e., last two or three weeks) of using very complicated patterns at the moment. I've noticed this when trying to create V2's (as yet unpublished). Normally, one would expect (harder) variants to be created by combining smaller cages. However, recently I've found myself doing the exact opposite, namely splitting more complex cages. Even then, it's still very easy to come up with ruudiculous puzzles...

Andrew wrote:
Still we know by now that the choice of cage totals tends to contribute much more than the cage pattern to the level of difficulty.

Wasn't the chink in the armor this time your step 9 (Para's step 7)? Without that, the puzzle would have been dramatically more difficult.

Andrew wrote:
I haven't (yet?) looked at V2 but that could well be a monster since nothing has been posted so far.

CathyW wrote:
Had a quick look at the V2(?) using JSudoku. Don't think I'll get very far on my own so I'll leave it for the masochists while I'm on holiday.

Ruud wrote:
If you cannot find any shortcuts, this killer will keep you busy for quite a while.

True, I am in fact a masochist - how could you tell? Smile . Been looking at this puzzle intensively for the past couple of days. Got up to well over 50 moves, some of them consisting of heavy duty ammunition, but apart from managing 3 placements, couldn't make any serious impact on the brute.

Indeed, not so much of an Assassin, this one. More like a Terminator!

So, with no further ado, here's my vote: Give up
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PostPosted: Thu Aug 09, 2007 1:04 pm    Post subject: Reply with quote

I wrote:
So, with no further ado, here's my vote: Give up

Hang on, maybe I gave up too early. Looks like I may have found something! Stay tuned!
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PostPosted: Thu Dec 06, 2007 10:53 pm    Post subject: Here's one I made earlier Reply with quote

"Here's one I made earlier..."

Just noticed the Magic Roundabout topic on the DJApe Killer forum, so thought I'd "join in" by starting this thread with a now-famous quote from another cult British children's TV program. (Hint: do the words "sticky-back plastic" mean anything to you? If not, see here).

The walkthrough below really is one I made earlier. More than 3 months ago to be exact. It's very long, even in fairly optimized form, and uses several AICs, and some other creative moves. I note that SudokuSolver (SS) can now complete this puzzle, without using any chains. However, some of the "45" moves it uses are quite involved and very difficult to "spot" (read: work out using pen and paper), even if they're not classified as "extreme" or "insane". I personally prefer to use inference chains when the going gets tough. However, the SS log also makes interesting reading, and I may follow up this WT by elaborating on one or two of the steps that SS makes.

For now, here's my WT as I left it on August 20, in unmodified form:


Assassin 62V2 Walkthrough

I wrote:
----
Note: New shorthand notation (example): "CPE(R7): no 3 in R6C8"
=> "Common Peer Elimination(CPE): R6C8 sees all 3's in R7 -> no 3 in R6C8".
----

1. 28/4 at R1C1 = {(47/56)89}: no 1,2,3
1a. {89} locked for N1

2. 12/4 at R1C3 = {12(36/45)}: no 7,8,9

3. 19/3 at R1C6: no 1

4. 7/3 at R1C8 = {124}, locked for N3

5. 11/2 at R2C6: no 1

6. 22/3 at R4C2 = {(58/67)9}: no 1,2,3,4
6a. 9 locked for N4

7. 8/3 at R5C1 = {1(25/34)}: no 6,7,8,9
7a. 1 locked for C1

8. 11/2 at R7C4: no 1

9. 22/3 at R8C3 = {(58/67)9}: no 1,2,3,4
9a. no 9 in R9C12

10. 9/3 at R8C1 = {126/234}: no 5,7,8,9
10a. (Note: {135} blocked by 8/3 at R5C1 (step 7))
10b. 2 locked for N7
10c. 1 only in R9C2 -> no 6 in R9C2
10d. no 3 in R9C2 (requires {24} in R89C1 - blocked by 8/3 at R5C1)
10e. no 4 in R9C2 (requires {23} in R89C1 - blocked by 8/3 at R5C1)
10f. Summary: 9/3 at R8C1 = {26}[1]/{34}[2]

11. 9 in C1 locked in R12C1 -> not elsewhere in N1

12. Outies C12: R357C3 = 14/3 = {149/158/167/239/248/257/347/356}
12a. min. R5C3 = 5
12b. -> max. R37C3 = 9
12c. -> no 9 in R7C3

13. Outies C89: R357C7 = 10/3 = {127/136/145/235} (no 8,9)
13a. {12} only in R57C7
13b. -> no 7 in R57C7

14. Outies R6789: R5C1478 = 14/4 = {1238/1247/1256/1346/2345} (no 9)
14a. Cleanup: no 1,2 in R6C4

15. CPE(C1): no 7 in R3C23

16. CPE(N1): no 1,2,3 in R3C4

17. {123} in 12/4 at R1C3 (step 2) only in R1C3+R2C34
17a. -> no 6 in R1C3+R2C34

18. 12/4 at R1C3 and 7/3 at R1C8 form grouped X-Wing on {12} in R12
18a. -> no 1,2 elsewhere in R12
18b. Cleanup: no 9 in R3C6
18c. CPE(N2): no 1 in R4C4

19. 20/4 at R2C8: {124} only available in R4C8
19a. -> {1289},{1469},{1478},{2459},{2468} all blocked
19b. -> 20/4 at R2C8 = {1379/1568/2369/2378/2567/3458/3467} = {(1/2/4)..}
19c. -> R4C8 = {124}

20. from steps 7 and 10f: 8/3 at R5C1 and 9/3 at R8C1 together lock {234} in C1
20a. -> no 2,3,4 in R1234C1

21. Outies N7: R5C1+R6C12+R9C4 = 13/4
21a. min. R5C1+R6C12 = 6
21b. -> max. R9C4 = 7 (no 8,9)
21c. min. R9C4 = 5
21d. -> max. R5C1+R6C12 = 8
21e. -> R5C1+R6C12 = {123/124/125/134} (no 6,7,8)
21f. 1 locked in R5C1+R6C12 for N4

22. 9 in 22/3 at R8C3 locked in R89C3 for C3 and N7
22a. {89} in 22/3 only in R89C3
22b. -> no 5 in R89C3

23. 18/4 at R3C1 = {1368/1458/1467/2358/2367/2457/3456}
23a. (Note: {1278} blocked by 12/4 at R1C3)
23b. -> must contain 2 of {1234}
23c. {1234} only in R3C23
23d. -> R3C23 = {1234} (no 5,6)
23e. Cleanup: no 1 in R7C3 (step 12)

24. 1 in C3 locked in N1 -> not elsewhere in N1

25. Innies N1: R123C3+R3C12 = 17/5, w/ {123} locked = {123(47/56)}
25a. 6 only in R3C1
25b. -> no 5 in R3C1

26. Outies N1: R23C4+R4C1 = 13/3
26a. min. R4C1 = 5
26b. -> max. R23C4 = 8
26c. -> max. R2C4 = 3 (no 4,5)

27. R3C23 cannot contain {34}. Here's how:
27a. R3C123 = [6]{34} blocked by 28/4 at R1C1 (step 1), and...
27b. ...[7]{34} would require 4 in R4C1 for 18/4 cage sum - unavailable
27c. -> R3C23 must contain exactly 1 of {12} (step 23)
27d. -> R12C3 must contain exactly 1 of {12} (only other place in N1)
27e. -> other of {12} for 12/4 at R1C3 (step 2) must go in R2C4
27f. -> no 3 in R2C4

28. Only other place for {12} in R3/N2 is R3C56
28a. -> either R3C5 = {12} -> R2C4+R3C5 = {12} -> R1C6+R3C4 = 15 (innies N2) = [96], or...
28b. ...R3C6 = 2 -> R2C6 = 9
28c. -> 9 locked in R12C6 for C6 and N2
28d. from steps 28a/28b: either way, no 6 in 11/2 at R2C6
28e. -> R23C6 <> {56}

29. 9 in R3 locked in N3 -> not elsewhere in N3

30. 19/3 at R1C6 = {379/478/568}
30a. (Note: {469} now blocked because {49} only available in R1C6)
30b. 9 only in R1C6
30c. -> no 3,7 in R1C6

31. I/O difference R12: R2C68 = R3C4 + 8 = 12, 13 or 14
31a. if 12 (R3C4 = 4): R2C68 = [93] ([75] blocked because it would require 4 in R3C6)
31b. if 13 (R3C4 = 5): R2C68 = [76/85]
31c. if 14 (R3C4 = 6): R2C68 = [86/95]
31d. Summary: R2C6 = {789} (no 3,4), R2C8 = {356} (no 7,8)
31e. Cleanup: no 7,8 in R3C6

32. Innies C1234: R1C4+R4C34 = 13/3
32a. Min. R1C4+R4C3 = 5
32b. -> Max. R4C4 = 8 (no 9)

33. Hidden killer pair on {78} in C1 as follows:
33a. {78} in C1 locked in R1234C1
33b. R34C1 cannot contain both of {78} (step 23a)
33c. 9 in C1 already locked in R12C1 (step 11)
33d. -> R12C1 = {79/89} (no 5,6), 18/4 at R3C1 = {(7/8)..}

34. I/O difference R89: R7C46 = R8C2 + 6
34a. R7C46 cannot sum to 11, otherwise R7C6 would clash w/ R8C4
34b. -> no 5 in R8C2

35. 5 in N7 locked in R7 -> not elsewhere in R7

36. no 7 in R4C1. Here's how:
36a. 1 in R3 locked in N12 innies (= 23/5) within R3C35
36b. -> if R3C3 <> 1, then R3C5 must be 1 -> max. R3C5+R1C6 = 10 -> min. R3C123 = 13
36c. -> if R4C1 = 7, then R3C123 <> [6]{23}, as this only sums to 11
36d. furthermore, if R4C1 = 7, then R3C123 cannot be [6]{14} as this blocked by 28/4 at R1C1 (step 1)
36e. but these are the only 2 options available with 7 in R4C1
36f. -> no 7 in R4C1

37. 7 in C1 locked in N1 -> not elsewhere in N1

38. from step 6: if 22/3 at R4C2 <> {679}, then it must be {589} -> R4C1 = 6
38a. -> 6 in N4 locked in R4C1 or 22/3 at R4C2
38b. -> no 6 elsewhere in N4 (R46C3)

39. 18/4 at R3C1 = {1368/1467/2367/2457}
39a. {2457} blocked by grouped AIC. Here's how ('=>' = strong link, '->' = weak link):
39b. if R34C1 <> {6..} => R89C1 = {6..} -> R789C3 <> {6..} => R5C3 = 6, R3C3 <> {24} (step 12)
39c. In other words, if R34C1 doesn't contain a 6, R3C3 cannot contain a 2 or 4
39d. -> {2457} combo blocked
39e. -> 18/4 at R3C1 = {1368/1467/2367}
39f. -> no 5 in R4C1

40. 5 in C1 now locked in 8/3 at R5C1 = {125} (no 3,4)
40a. 2 locked in R56C1 for C1 and N4
40b. Cleanup: no 8 in R4C4 (step 32)

41. Hidden single (HS) in N7 at R9C2 = 2
41a. Cleanup: R89C1 = {34}, locked for N7

42. Recall step 27c
42a. {12} now not available in R3C2
42b. -> R3C3 = {12} (no 3,4)

43. R4C1 and 22/3 at R4C2 (step 9) form killer pair on {68} within N4
43a. -> no 8 in R46C3

44. Outies C12 revisited: R357C3 = 14/3 = {158/167/257}
44a. -> R57C3 can only contain at most 1 of {689}
44b. Only other places for {689} in C3 are R89C3
44c. -> R57C3 and R89C3 form hidden killer triple on {689} in C3
44d. -> R57C3 must contain exactly 1 of {689} -> R357C3 = {1(58/67)} (no 2) and...
44e. ...R89C3 must contain exactly 2 of {689} -> no 7 in R89C3
44f. Cleanup: no 6 in R9C4

45. Naked single (NS) at R3C3 = 1
45a. Note: R57C3 now = {58/67} (step 44)

46. HS in N2 at R2C4 = 1

47. 1 in C2 locked in 19/4 at R6C2
47a. 7 in N7 locked in 19/4 at R6C2
47b. -> 19/4 at R6C2 = {17..} = {17(38/56)} (no 4) = {(3/5)..}

48. 4 in C2 locked in N1 -> not elsewhere in N1

49. 12/4 at R1C3: 4 now only available in R3C4
49a. -> no 5 in R3C4 (step 2)

50. Split 17/3 at R3C12+R4C1 = [638/746]
50a. -> R3C12 = [63/74] = {(4/6)..}
50b. -> R3C12 and R3C4 form killer pair on {46} in R3
50c. -> no 4,6 elsewhere in R3
50d. Cleanup: no 7 in R2C6

51. Innies N2: R1C6+R3C45 = 17/3 = {269/368/458/467}
51a. R3C45 cannot be [67] due to R3C1
51b. -> if {467}: no 4 in R1C6
51c. if {458}: 4 must go in R3C4
51d. Conclusion: no 4 in R1C6

52. 4 now unavailable to 19/3 at R1C6 = {379/568} = {(6/7)..} (no eliminations yet)

53. Another grouped AIC:
53a. Either R3C2 = 3, or...
53b. R3C2 <> 3 => R3C2 = 4
53c. -> R3C4 <> 4 => R3C4 = 6
53d. -> R3C1 <> 6 => R3C1 = 7
53e. -> R3C789 <> 7 => R12C7 = {7..} = {37} (step 52)
53f. Thus, either or both of R3C2 and R12C7 contain(s) a 3
53g. -> no 3 in R3C789
53h. Cleanup: no 5,6 in R57C7 (step 13)

54. Hidden killer pair on {36} in N3, as follows:
54a. Only places for {36} in N3 are R2C8 and within 19/3 (R12C7)
54b. 19/3 cannot contain both of {36} (step 52)
54c. -> R2C8 and R12C7 must each have one of {36}
54d. -> R2C8 = {36} (no 5); no 6 in R1C6

55. {467} combo now blocked for N2 innies (step 51), as none of these digits present in R1C6
55a. -> R1C6+R3C45 = {269/368/458} (no 7)

56. 7 in N2 now locked in 16/3 at R1C4 = {367/457} (no 8)

57. Another hidden killer pair, this time in C4:
57a. Only places for {89} in C4 are within 18/3 (R56C4) and within 11/2 (R78C4)
57b. Neither can contain both of {89} (due to no 1 in R6C3), so both must contain exactly one of {89}
57c. -> 11/2 at R7C4 = {29/38} (no 4,5,6,7); {567} combo blocked for 18/3 at R5C4

58. Back to grouped AIC's:
58a. Either R7C1 = 5, or...
58b. R7C1 <> 5 => R7C1 = 1
58c. -> R56C1 <> 1 => R6C2 = 1 (strong link N4)
58d. -> R6C2 <> 3 => R3C2 = 3 (strong link C2)
58e. -> R12C3 <> 3 => R12C3 = {25} (ALS node)
58f. Thus, either R7C1 = 5 or R12C3 = {25}
58g. -> no 5 in R7C3 (common peer)
58h. Cleanup: no 8 in R5C3 (step 45a)

59. 8 in C3 locked in N7 -> not elsewhere in N7

60. Outies C1: R123C2 = 15/3, w/ 4 locked = {348/456} = {(3/5)..}
60a. {35} not available in R7C3+R8C2
60b. -> R123C2 and R67C2 (step 47b) form killer pair on {35} within C2
60c. -> no 5 in R45C2

61. I/O difference R89: R8C24 = R7C6 + 5. Analysis follows:
61a. if R8C2 = 1, R7C6+R8C4 = [48]
61b. if R8C2 = 6, R7C6+R8C4 = [32/43]
61c. if R8C2 = 7, R7C6+R8C4 = [42]
61d. Summary: R7C6 = {34} (no 1,2,6,7,8}; R8C4 = {238} (no 9)
61e. Cleanup: no 2 in R7C4

62. 17/3 at R8C5 = {179/269/368/458/467} = {(3/4/9)..}
62a. (Note: {278} and {359} blocked by 11/2 at R7C4 (step 57c))
62b. 11/2 at R7C4 = {(3/9)..} (step 57c)
62c. -> 17/3 at R8C5, 11/2 at R7C4 and R7C6 form killer triple on {349} in N8
62d. -> no 3,4,9 elsewhere in N8 (R7C5+R8C6)

63. 24/4 at R6C9:
63a. Max. R7C7 = 4 -> Min. R6C9+R7C89 = 20
63b. -> no 1,2 in R6C9+R7C89

64. CPE(R7): no 4 in R89C7

65. CPE(C7): no 4 in R6C9

66. Outies N9: R6C9+R78C6 = 16
66a. R7C6 = {34} -> R6C9+R8C6 = 12 or 13
66b. -> no 3,9 in R6C9; no 1,2 in R8C6

67. 17/3 at R8C5 must contain exactly one digit in range {1..4} (step 62)
67a. 11/2 at R7C4 must contain exactly one digit in range {1..4} (step 57c)
67b. Only other places for remaining 2 of {1..4} in N8 are R7C56
67c. -> 17/3 at R8C5, 11/2 at R7C4 and R7C56 form hidden killer quad on {1234} in N8
67d. -> no 6,7,8 in R7C5

68. I/O difference R789: R6C29 = R7C15 + 4
68a. R7C15 = [12/51/52] -> sum to 3, 6 or 7
68b. -> R6C29 = 7, 10 or 11. Analysis follows:
68c. if 7, R6C29 = [16]
68d. if 10, R6C29 = [37]
68e. if 11, R6C29 = [38/56]
68f. -> R6C9 = {678} (no 5)

69. 24/4 at R6C9 (revisited): 5 no longer available
69a. -> valid combos are {1689/2679/3678} (no 4)
69b. (Note: {3489} blocked by R7C6)
69c. -> 24/4 at R6C9 must have exactly one of {123}, which must be in R7C7
69d. -> no 3 in R7C89
69e. 6 locked in R6C9+R7C89
69f. -> no 6 in R89C9

70. HS in R7 at R7C6 = 4

71. 4 in C7 locked in N6 -> not elsewhere in N6

72. Split 14/3 at R8C67+R9C7 = {158/167/257/356} (no 9)
72a. (Note: {239} combo blocked, as none of these digits present in R8C6)

73. 9 in C7 locked in N6 -> not elsewhere in N6

74. Innies N8: R7C5+R8C6+R9C4 = 13/3, w/ {349} unavailable
74a. -> valid combos are {157/256} (no 8)
74b. 5 locked in R8C6+R9C4 for N8
74c. -> no 5 in R9C7 (CPE)

75. Permutations 20/4 at R2C8:
75a. 20/4 cannot have both of {67} due to 19/3 at R1C6 (step 52)
75b. -> 20/4 at R2C8 = {1379/1568/2378} = {(6/7)..}
75c. -> must have exactly one of {57}, which must go in R3C7
75d. -> no 5,7 in R3C8

76. 19/3 at R1C6 (step 52) and 20/4 at R2C8 (step 75b) form killer pair on {67} within N3
76a. -> no 7 in R3C9

77. 7 in N3 locked in C7 -> not elsewhere in C7

78. {257} combo now blocked for split 14/3 at R8C67+R9C7 (step 72), as none of these digits present in R9C7
78a. -> 14/3 at R8C67+R9C7 = {158/167/356} (no 2) = {(1/3)..}
78b. {36} in R89C7 blocked by R12C7
78c. -> no 3 in R8C7

79. 10/3 at R357C7 (step 13) = [7]{12}/[541]/[5]{23} = {(1/3)..}
79a. R8C6 has neither of {13}
79b. -> R57C7 and R89C7 (step 78a) form killer pair on {13} within C7
79c. -> no 1,3 elsewhere in C7

80. HS in N3 at R2C8 = 3

81. 19/3 at R1C6 = {568} (no 7,9)
81a. 6 only available in R12C7, locked for C7

82. HS in C6/N2 at R2C6 = 9
82a. -> R3C6 = 2

83. HS in C1/N1 at R1C1 = 9

84. HS in C7/N3 at R3C7 = 7

85. NS at R3C1 = 6
85a. -> R4C1 = 8, R3C4 = 4
85b. -> R2C1 = 7, R3C2 = 3

86. Naked pair (NP) on {25} at R12C3 -> no 2,5 elsewhere in C3 and N1

87. Hidden pair (HP) on {34} in C3 at R46C3
87a. -> R46C3 = {34} (no 7)

88. HP on {37} in R1/N2 at R1C45
88a. -> R1C45 = {37} (no 5,6)
88b. -> R2C5 = 6 (cage-split)

89. HS in R1/C7/N3 at R1C7 = 6

90. 6 in C4 locked in N5 -> not elsewhere in N5

91. 3 no longer available to 19/4 at R6C2
91. -> (from step 47b) 19/4 at R6C2 = {1567} (no 8)
91a. 6 locked for N7

92. Cage-split of 22/3 at R8C3: R89C3 = {89} = 17
92a. -> R9C4 = 5

93. Revisit split 14/3 at R8C67+R9C7 (step 78a):
93a. {158} blocked because R8C6 has none of these digits
93b. {167} blocked because {67} only in R8C6
93c. -> 14/3 at R8C67+R9C7 = {356} = [653] (only permutation possible)

Now it's just naked and hidden singles to end.

_________________
Cheers,
Mike
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