Assassin 61

[CathyW]

Tougher than it looked from the preliminaries but got there eventually - hopefully with no errors this time (or at least only minor typos)! First time I've knowingly used grouped x-wings to make eliminations.

Preliminaries:

a) 30(4) r12c34 = {6789}
b) 17(2) r89c1 = {89} not elsewhere in c1/N7 -> 10(2) r78c3 = {37/46}
c) 4(2) r89c9 = {13} not elsewhere in c9/N9
d) 8(3) r8c4+r9c34 = {125/134} Must have 1 -> r9c56 <> 1
e) 12(4) r12c67 = {1236/1245} Must have 1 and 2
f) 20(3) N3, no 1,2
g) 11(3) N1, no 9
h) 15(2) N9 = {69/78}
i) 18(5) at r3c6 = {12348/12357/12456} Must have 1 and 2

1. Outies c1234: r6c5 = 5

2. Innies c5: r45c5 = 5 = {14/23} -> r3c6+r4c67 = 13 (must have 1 or 2, can’t have both 3,4) = {148/157/238/256}

3. Innies N3: r12c7+r3c9 = 8 = {125/134} Must have 1, not elsewhere in N3
-> 1 locked to r12c7, not elsewhere in c7; r12c6 <> 1
No 3 in r3c9 -> r12c7 <> 4

4. Innies N9: r7c9+r9c7 = 11 = {29/47/56}

5. Outies c12: r35c3 = 10 = {19/28/37/46}

6. Outies c89: r35c7 = 10 = {28/37/46}

7. Outies – Innies r12: r3c5 – r2c28 = 1
-> r2c28 max 8, min 3 -> r3c5 = (46789), r2c2 = (1…6), r2c8 = (2…7)
Options: [412]/[614]/6{23}/[715]/7{24}/[816]/8{25/34}/[917]/9{26/35}

8. Innies r123: r3c1469 = 11 = {1235} not elsewhere in r3
-> r5c3 <> 7,8,9; r5c7 <> 7,8
-> split 8(3) N3 = {125}
-> 20(3) N3 = {389/479} Must have 9; If {389}, r1c8 = 3 -> r1c3 <> 8
-> 17(3) N3 = {368/467} Must have 6
-> 3 locked to r12c8, not elsewhere in c8 -> 3 locked to r456c7
-> r3c5 <> 4 (step 7)

9. Innies N7: r7c1+r9c3 = 6 = {15/24}

10. Grouped x-wing (1): 8(3) and 4(2) both in r89, 1 not elsewhere in r89

11. 12(3) N7 = {156/246/237} {147} blocked by 10(2) and {345} blocked by r7c1+r9c3 -> r7c2 <> 5

12. 15(3) N9 = {249/258/456} {267} blocked by 15(2)

13. 17(4) r6c345+r7c4 = 5{129/138/147/237/246} Must have at least one of 1,2

14. Innies N1: r12c3+r3c1 = 18 = {189/279/369/378/567}

15. Outies r6789: r5c1678 = 14 (no 9) = {1238/1247/1256/1346/2345}

16. 22(4) r349c9+r4c8 = 2{479/569/578}, 5{179/269/278/467}

17. O-I N8: r7c46 – r9c37 = 4
-> r7c46 min 7, max 17; r9c37 min 3, max 13

18. Grouped x-wing (5): in c34 5 locked within 20(4) and 8(3)
-> 8(3) = {125} -> r7c1 <> 2
-> r9c56 <>2, r9c6 <> 5 (cells see all of 8(3))

19. Grouped x-wing (9): in r12 both 30(4) and 20(3) must have 9
-> r12c5 <> 9

20. 16(3) N1, min from r3c23 = {46} = 10 but can’t repeat so r2c2 <> 6, r3c23 not {46}
Options: {169/178/268/349/367/457}
If {349/457} r2c2 = 3/5 -> r2c2 <> 4

21. Pointing cells: 6 locked to r4567c9 -> r5c78, r6c8 <> 6 -> r3c7 <> 4

22. Pointing cells: 1 locked to r45c5+r34c6 of 18(5) -> r56c6 <> 1

23. r3c9 = (25)
a) r3c9 = 2 -> r12c7 = {15} -> r12c6 = {24}
-> in N2 5 locked to r3c46 -> r3c1 <> 5
b) r3c9 = 5 -> r3c1 <> 5
Either case r3c1 <> 5
Further, if r3c9 = 5, r12c7 = {12}, then 5 locked to r12c6, so in either case 12(4) = {1245}
-> 4 locked to r12c6 not elsewhere in N2/c6
-> 18(3) N2 options: {189/279/369/378} -> r3c5 <> 6

24. 16(3) r8c6+r9c67 = {259/268/349/358/367/457}
If {259} r9c6 = 9; if {349} r9c7 = 4 -> r9c7 <> 9 -> r7c9 <> 2

25. r3c9 = (25)
a) r3c9 = 2 -> r4c8 <> 2
b) r3c9 = 5 -> r12c7 = {12} -> 2 locked to r789c8 -> r4c8 <> 2
Either case, r4c8 <> 2

26. 30(4) N12 = {6789}
a) if 9 within r12c3 -> r6c3 <> 9
b) if 9 within r12c4 -> r6c6 = 9 -> r6c3 <> 9
Either case r6c3 <> 9

27. r3c9 = (25)
a) r3c9 = 2 -> r12c7 = {15} -> r4c7 <> 5
b) r3c9 = 5 -> r3c4 <> 5 -> r89c4 must have 5 -> r9c3 <> 5 -> r4c3 = 5 -> r4c7 <> 5
Either case r4c7 <> 5

28. 18(5) = {12348/12357/12456}
Because r45c5 must be {14} or {23}, if 18(5) = {12357}, r45c5 = {23}, r3c6 = 5, r4c6 = 1, r4c7 = 7 -> r4c6 <> 7

29. 18(3) N2 = {189/279/369/378} Must have one of 7,9
-> 17(3) N8 not {179} -> r7c5 <> 1
-> 17(3) N8 = {269/278/368/467}

30. Split 18(3) N1 = {189/279/369/378} Must have one of 3,9
-> 16(3) N1 not {349}
-> 16(3) N1 = {169/178/268/367/457}

31. 16(3) N1 must have at least one of 6,7 within r3c23
17(3) N3 must have at least one of 6,7
If r2c8 = 6, r3c78 = [74]; else 6 in r3c78
-> r3c5 <> 7
-> 18(3) N2 = {18/27/36}9/{37}8

32. 1 locked to r45c5+r34c6 of 18(5)
a) if r3c6 = 1 -> 1 locked to r789c4 -> r456c4 <> 1
b) if r45c5 or r4c6 = 1 -> r456c4 <> 1
Either case r456c4 <> 1
-> 1 locked to r45c5+r4c6 of 18(5) -> r3c6 <> 1

33. In r3, 1 in r3c1 or r3c4
a) r3c1 = 1 -> r7c1 <> 1
b) r3c4 = 1 -> r7c6 = 1 -> r7c1 <> 1
Either case r7c1 <> 1
-> r9c3 <> 5
-> 5 locked to r89c4, not elsewhere in c4/N8
-> HS r4c3 = 5
-> r345c4 = 15 = {168/249/267/348} -> r45c4 <> 2,3

34. 9 locked tor123c3 -> r3c2 <> 9

35. Pointing pair: 5 locked to r5c89 -> r7c9 <> 5 -> r9c7 <> 6

36. 16(3) r8c6+r9c67 = {259/268/349/358/367}
If {367} r9c7 = 7 -> r89c6 <> 7

37. 7 locked to r567c6 -> 24(4) = {7…}-> r6c7 <> 7
24(4) = {2679/3678} Must also have 6

38. 1 locked to r789c4 -> r3c4 <> 1
-> r3c1 = 1 -> r12c3 = {89} not elsewhere in c3/N1 -> r12c4 = {67} not elsewhere in c4/N2
-> 11(3) N1 = {236/245}
-> 16(3) N1 = {367/457}, 7 not elsewhere in r3
-> HS r3c5 = 9 -> r12c5 = {18} not elsewhere in c5
-> r45c5 = {23} not elsewhere in c5/N5/18(5) -> r3c6 = 5
-> r3c9 = 2, r3c4 = 3
-> r45c4 = {48} not elsewhere in c4/N5 -> r6c4 = 9 -> r789c4 = {125} not elsewhere in N8
-> 8 locked to r3c78 -> 17(3) = 3{68} -> 20(3) = {479}

… Straightforward cage combinations and singles from here

2 3 8 6 1 4 5 7 9
6 5 9 7 8 2 1 3 4
1 7 4 3 9 5 8 6 2
3 8 5 4 2 1 7 9 6
4 9 6 8 3 7 2 1 5
7 1 2 9 5 6 3 4 8
5 4 3 1 6 8 9 2 7
9 2 7 5 4 3 6 8 1
8 6 1 2 7 9 4 5 3

[Andrew]

I liked Cathy's grouped x-wings, especially the one she used to fix the 8(3) cell as {125}, the pointing cells moves and the chains. Maybe the most impressive step was 27b. When I was adding comments while going through her walkthrough I typed "Wow! That’s not an obvious chain."

Put aside your Deathly Hallows, for they will not help you beat this Assassin.

I found this one fairly straightforward and didn't need any wizard moves, unlike most recent Assassins which have required lots of them. This was the first Assassin that I've solved in one session for many weeks.

Thanks to Para for his comments.

Here is my walkthrough.

1. R78C3 = {19/28/37/46}, no 5

2. R78C7 = {69/78}

3. R89C1 = {89}, locked for C1 and N7, clean-up: no 1,2 in R78C3

4. R89C9 = {13}, locked for C9 and N9

5. 11(3) cage in N1 = {128/137/146/236/245}, no 9

6. 20(3) cage in N3 = {389/479/569/578}, no 1,2

7. 8(3) cage at R8C4 = 1{25/34}, no 1 in R9C56

8. 30(4) cage at R1C3 = {6789}

9. 12(4) cage at R1C6 = 12{36/45}, no 7,8,9

10. 18(5) cage at R3C6 = {12348/12357/12456} = 12{348/357/456}, no 9

11. 15(3) cage in N9 = {249/258/456} (cannot be {267} which clashes with R78C7), no 7

12. 12(3) cage in N7 = {156/237/246} (cannot be {147/345} which clash with R78C3)

13. Killer pair 6,7 in 12(3) cage (step 12) and R78C3, locked for N7

14. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = {15/24}, no 3

15. 45 rule on N9 2 innies R7C9 + R9C7 = 11 = {29/47/56}, no 8

16. 45 rule on N1 3 innies R12C3 + R3C1 = 18, min R12C3 = 13 -> max R3C1 = 5
[Para commented. Actually minimum R12C3 = 14 already as {67} is blocked by R78C3. Doesn't matter much in the solving path though.]

17. 45 rule on N3 3 innies R12C7 + R3C9 = 8 = 1{25/34}, no 6,7,8,9, 1 locked in R12C7 for C7, N3 and 12(4) cage -> no 1 in R12C6
17a. {134} must have 4 in R3C9 -> no 4 in R12C7

18. 45 rule on R123 4 innies R3C1469 = 11 = {1235}, locked for R3

19. R12C7 + R3C9 (step 16) = {125} (only remaining combination), locked for N3

20. 45 rule on C12 2 outies R35C3 = 10 = {46}/[73/82/91] -> R5C3 = {12346}

21. 45 rule on C1234 1 outie R6C5 = 5
[I ought to have seen this much earlier! If I’d been looking at the original puzzle grid rather than my elimination grid I probably would have done so.]

22. 45 rule on C6789 3 innies R34C6 + R4C7 = 13 -> R34C5 = 5 = {14/23}
[Alternatively 45 rule on C5 2 innies R34C5 = 5 -> R34C6 + R4C7 = 13. In retrospect this is more obvious but after step 21 I was looking for something similar.]

23. 45 rule on C89 2 outies R35C7 = 10 = {46}/[73/82] -> R5C7 = {2346}, no 9 in R3C7

24. 45 rule on R6789 4 outies R5C1678 = 14 = {1238/1247/1256/1346/2345}, no 9

25. 45 rule on R1234 5 outies R5C23459 = 31 = {16789/25789/34789/35689/45679}, R5C5 = {1234} -> no 1,2 in R5C2349, clean-up: no 8,9 in R3C3
[Alternatively, using the result from step 24, 45 rule on R5 5 innies R5C23459 = 31]
25a. Cannot be {25789} because R5C3 only contains 3,4,6 -> R5C23459 = {16789/34789/35689/45679}, no 2, clean-up: no 3 in R4C5
25b. 2 in R5 locked in R5C1678 = {1238/1247/1256/2345}

26. Naked quad {3467} in R3578C3, locked for C3, clean-up: no 8,9 in R12C4, no 2 in R7C1 (step 14)

27. Naked pair {89} in R12C3, locked for C3 and N1
[Para commented. This of course sets R3C1 to 1 from innies N1 (step 16), which you eventually get to through combination analysis of 2 cages. But it doesn't matter much really as the puzzle is now broken down to the easier bits.

Agreed. Interesting that I missed two things about step 16. The combinations are still needed for the other eliminations although those steps would have been simpler if I'd fixed R3C1 first.]

28. Naked pair {67} in R12C4, locked for C4 and N2
[With hindsight, this reduced the 12(4) cage at R1C6 to one combination. I did that in step 33 so missing that here didn’t affect the solving path.]

29. 16(3) cage in N1 = {367/457} = 7{36/45}, no 1,2, 7 locked for N1
29a. No 3,5 in R3C23 -> R2C2 = {35}, R3C23 = {47/67}
29b. 7 locked in R3C23, locked for R3

30. 11(3) cage in N1 = {236/245} (cannot be {146} which clashes with R3C23), no 1, 2 locked for N1

31. R3C1 = 1 (hidden single in N1), clean-up: no 5 in R9C3 (step 14)

32. Naked pair {12} in R69C3, locked for C3 -> R4C3 = 5

33. 3 in R3 locked in R3C46, locked for N2

34. Hidden triple {189} in R123C5, locked for C5, clean-up: no 4 in R45C5 (step 22) -> R45C5 = [23] -> R3C6 = 5, R3C9 = 2, R3C4 = 3

35. Naked pair {24} in R12C6, locked for C6

36. Naked pair {15} in R12C7, locked for C7

37. Naked triple {467} in R789C5, locked for N8

38. R3C6 + R45C5 = [523] = 10 -> R4C67 = 8 = [17] (only remaining permutation), clean-up: no 8 in R78C7
[At this stage I saw grouped X-wing (9) with 20(3) cage in N3 and 22(4) cage at R3C9, 9 locked for C89, but then decided to use the naked pair in the next step.]

39. Naked pair {69} in R78C7, locked for C7 and N9

40. Naked pair {24} in R59C7, locked for C7 -> R3C7 = 8, R6C7 = 3, R3C5 = 9
[Alternatively R6C7 was hidden single in C7.]

41. R9C7 = {24} -> total of R89C6 must be even = {39} -> R9C7 = 4, R5C7 = 2
[I’ll admit that 16(3) cage at R8C6 = {39}4 (only remaining combination) is more direct but I like using the property of even/odd numbers when I can.]

42. Naked pair {39} in R89C6, locked for C6 and N8 -> R7C6 = 8

43. R7C9 = 7 (hidden single in N9), clean-up: no 3 in R8C3

44. R3C4 + R4C3 = [35] -> R45C4 = {48} (only remaining combination), locked for C4 -> R6C4 = 9

45. R3C7 = 8 -> R23C8 = 9 = [36] (only remaining permutation), R2C2 = 5, R12C7 = [51], R12C5 = [18], R12C3 = [89], R12C9 = [94], R1C8 = 7, R12C6 = [42], R2C1 = 6, R12C4 = [67]

46. R7C1 = 5 (hidden single in C1) -> R9C3 = 1 (step 14), R6C3 = 2, R7C4 = 1, R89C9 = [13], R89C6 = [39], R89C1 = [98], R78C7 = [96], R789C8 = [285], R89C4 = [52], clean-up: no 4 in R7C3

47. R6C2 = 1 (hidden single in C2)
[There is also a hidden single in R8C2 but I didn’t notice it originally. It’s possible that it wasn’t a hidden single when I first solved the puzzle. I had to edit some steps because I hadn’t made all the eliminations after fixing a couple of earlier cells.]

48. Naked pair {47} in R56C1, locked for C1 and N4 -> R4C1 = 3, R5C3 = 6

and the rest is naked singles

[mhparker]

Hi guys,

Running out of suitable puzzles? Here's one to do on the subway home.

Note: It's a Killer-X, where no repeated digits are allowed on either diagonal. (Est. difficulty: V1.75 level)


Assassin 61X



3x3:d:k:4608:4608:4610:46105381:5381:4615:4615:46084610:46105381:53814615:53943338:4885539930885394:5394:4885:4885:5399:5399:53993354:5412:5394:5394:6183:6183:6183:6698:66985412:5412363136346698:6698:5412:5431363136343901:6698:831:5431416241643901831:5431:4162:41624164:4164:3901

Happy journey!

[Para]

(Est. difficulty: V1.75 level)

When did the number behind the V become an estimation of difficulty instead of just a second harder version?

Para

[Para]

(Est. difficulty: V1.75 level)

When did the number behind the V become an estimation of difficulty instead of just a second harder version? I mean V2 was just second version right?

Para

[mhparker]

Hi Para,

When did the number behind the V become an estimation of difficulty instead of just a second harder version? I mean V2 was just second version right?

Firstly, congratulations on your new Grandmaster status, even though you reached it by posting more or less the same message twice!

It was Ruud, wasn't it, who first released a "V1.5" instead of calling it a "V2" here.

The obvious implication at the time was that the puzzle wasn't quite up to the difficulty level of a traditional V2.

This was backed up recently by Andrew on the Assassin 58 thread:

It took me several hours to solve this one so I feel it was hard enough for a V1. In my opinion V1 Assassins shouldn't become so hard that they drive away solvers like myself. Real evil is fine for V2 and V2+. V1.5 should be somewhere between those levels.

I intend to publish an estimated difficulty level with all of my own puzzles in the future. I wanted to make a post on this, but haven't got round to it yet. The idea was to use a numeric rating scheme based around the traditional "V1" and "V2" difficulty levels, which would have the numeric ratings 1.0 and 2.0, respectively. I would then just use the numeric rating "1.75", instead of the term "V1.75" (which is maybe confusing).

Looks like I'd better hurry up and make that other post!

[mhparker]

Hi folks,

As mentioned above, here's my suggestion for an informal numeric rating scheme. I'll be using this for any future puzzles of mine, unless we agree to use something different.

All this shouldn't become "cast in stone", but it's useful to have some way (as a puzzle setter) of communicating the estimated difficulty of a puzzle.

Any feedback and suggestions welcome.

0.5: Typical newspaper "Deadly". Intended to be done using only limited pencilmarks.

0.75: Easy Assassin, like some of the very early ones, such as A1. Rarely seen now: all recent Assassins would have at least a 1.0 rating on this scale.

1.0: "Average" V1 Assassin (looking back over a longer period of time). Something like A57, perhaps.

1.25: Harder Assassin. Actually, most recent Assassins seem to have become more difficult than they traditionally used to be. So a rating of "1.25" would be considered the norm now. Typical example: A59.

1.5: Hard Assassin, having a significantly longer and/or narrower solution path, and/or requiring more advanced techniques. The A60 was definitely one of these.

1.75: Very hard Assassin, but still not hard enough to require a team effort to solve. Does not require any hypotheticals. The A60RP-Lite could maybe deserve such a rating.

2.0: Traditional "V2" standard, typically requiring a team effort and maybe (but not necessarily) involving limited use of hypotheticals. Example: A55V2.

2.5: Requires a team effort and several short to medium length hypotheticals. The TJK18 and A48-Hevvie would probably fall into this category.

3.0: "Ruudiculous", requiring a team effort and massive hypotheticals to solve, if it can be solved at all. The A50V2 and (possibly) A60RP could be considered examples of this.

4.0: Puzzles with unique solutions, but which can only realistically be solved by computer programs using backtracking (brute force).

P.S. The puzzle I posted above could possibly be rated 1.5 instead of 1.75 on this scale. Difficult to tell. Better to just try it and let me know!

[Para]

Wasn't Ruud working on some difficulty estimation for Killers as well? I remember him mentioning it once.

Para

ps. What about adding a 4.0 for puzzles that have no way of solving without T&E for the puzzle i posted in the 60RP thread.

[Andrew]

Since both hypotheticals and T&E have been mentioned in the above discussion, plus there are also contradiction moves, I was wondering what the difference is between these three categories.

Let me have a try at answering that myself.

A contradiction move is normally of the type If r.c. = . followed by a short chain to show that r.c. can't have that value. That is what I did for R3C7 + R7C3 in Assassin 60 to eliminate [22] and leave the remaining options {13} which I then solved by direct steps.

Hypotheticals are when one looks at the range of values for a cell, or possibly combos for a cage, works through a short chain and eliminates some candidates from other cells in the chain and possibly also from the original cell.

T&E is more like bifurcation where one tries the values for a cell and finds that all except one lead to an impossible position. That was what I did by accident in Assassin 24 where I started as a hypothetical and it just happened that one candidate led to an impossible position; BTW it did also work as a hypothetical for a different cell.

[Para]

Firstly, congratulations on your new Grandmaster status, even though you reached it by posting more or less the same message twice!

Doesn't it look like the colour of my grandmaster status is a faded red from expert, as to say i've been here too long and my colour has started to fade from overexposure?

Para

[CathyW]

Running out of suitable puzzles? Here's one to do on the subway home.

Note: It's a Killer-X, where no repeated digits are allowed on either diagonal. (Est. difficulty: V1.75 level)

I'm lucky - I can walk home in 5 minutes! I'm not exactly running out of puzzles (a small pile waiting!) but being a true addict can't help but be tempted by new ones.

Not sure about the estimated difficulty level on this one. I found it easier than the A61 (though just as many steps). More often than not, I find the extra constraint of diagonals more of a help than a hindrance. Will post WT tomorrow once I've checked it again.

PS I've never managed a Times Deadly without pencilmarks

[Para]

Hi all

Just finished this one. Don't know why i couldn't break it the first time. But i found the breakthrough move (a hidden killer Quad in C4) at the second look.

Walk-Through Assassin 61

1. 11(3) at R1C1 = {128/137/146/236/245}: no 9

2. 30(4) at R1C3 = {6789}

3. 12(3) at R1C6 = {1236/1245}: no 7,8,9

4. 20(3) at R1C8 = {389/479/569/578}: no 1,2

5. 18(5) at R3C6 = {12348/12357/12456}: no 9

6. R78C3 = {19/28/37/46}: no 5

7. R78C7 = {69/78}: no 1,2,3,4,5

8. R89C1 = {89} -->> locked for C1 and N7
8a. Clean up: R78C3: no 1,2

9. 8(3) at R8C4 = {125/134}: no 6,7,8,9; 1 locked in 8(3) cage: R9C56: no 1

10. R89C9 = {13} -->> locked for C9 and N9

11. 45 on C1234: 1 outie: R6C5 = 5

12. 45 on N7: 2 innies: R7C1 + R9C3 = 6 = {15/24}: no 3,6,7

13. 45 on N9: 2 innies: R7C9 + R9C7 = 11 = {29/47/56}: no 8

14. 45 on C5: 2 innies: R45C5 = 5 = {14/23}: no 6,7,8

15. 45 on R123: 4 innies: R3C1469 = 11 = {1235} -->> locked for R3

16. 45 on N3: R12C7 + R3C9 = 8 = {125}({134} blocked by R3C9) -->> locked for N3
16a. 1 in N3 locked for C7 and 12(4) cage at R1C6
16b. 3 in N3 locked for C8
16c. 17(3) at R2C8 = {368/467}: no 9; 6 locked for N3

17. 45 on C89: 2 outies: R35C7 = 10 = [46/64/73/82]: R5C7 = {2346}

18. 45 on C12: 2 outies: R35C3 = 10 = [46/64/73/82/91]: R5C3 = {12346}

19. Killer Quad {6789} in R12C3 + R35C3 + R78C3 -->> locked for C3
19a. 8 in N4 locked for C2
19b. 9 in N4 locked for C2

20. 45 on N1: 3 innies: R12C3 + R3C1 = 18 = {89}[1]/{79}[2]/{69/78}[3]: {67}[5] blocked by R78C3: R3C1: no 5

21. 16(3) at R2C2 = {169/178/268/349/367/457}: {259/358 blocked by R3C2)

22. 5 in N1 either in 11(3) at R1C1 or 16(3) at R2C2; 11(3) = {245} or 16(3) = {457}
22a. 11(3) at R1C1: {137/146} blocked(clash with 16(3) = {457}; 16(3): {268/349} blocked(clash with 11(3) = {245})
22b. 11(3) at R1C1 = {236/245}: no 1,7; 2 locked for N1

23. 7 in C1 locked for N4

24. 17(4) at R6C3 = 5{129/138/147/237/246} = {6|7|8|9..} -->> R67C4 needs one of {6789}

25. Hidden Killer Quad {6789} in R12C4 + R67C4 + R45C4 -->> R45C4 needs one of {6789}
25a. 20(4) at R3C4 needs one of {6789} in R45C4 and no room for {6789} in any of ther other cells of that cage -->> 20(4) contains max. one of {6789} -->> 20(4) = {2459/3458}: no 1,6,7

26. CPE: R45C5: no 1, see all 1’s in N2
26a. R45C5 = {23}(step 14) -->> locked for C5, N5 and 18(5) cage at R3C6

27. 18(3) at R1C5 = {189/468}: no 7; 8 locked for N2 and C5
27a. 7 in N2 locked for C4 and 30(4) cage at R1C3
27b. 7 in N5 locked for C6

28. 18(5) in R3C6 = {12348/12357}: ({12456} blocked by R45C5): no 6; 1 locked within R34C6 -->> locked for C6

29. 45 on N1: R12C3 + R3C1 = 18 = {69}[3]/{89}[1] -->> 9 locked in R12C3 for N1 and 30(4) cage at R1C3
29a. Naked Pair {67} in R12C4 -->> locked for N2, C4 and 20(4) at R1C3
29b. Naked Pair {89} in R12C3 -->> locked for C3
29c. R12C3 = {89} -->> R3C1 = 1(step 29)
29d. R3C6 = 5; R3C9 = 2; R3C4 = 3; R4C6 = 1(hidden)

30. 18(5) at R3C6 = 15{237} -->> R4C7 = 7
30a. 20(4) at R3C4 = 3{458}: no 2,9 -->> R4C3 = 5; R45C4 = {48} -->> locked for N5 and C4
30b. R6C4 = 9

31. R78C7 = {69} -->> locked for C7 and N9
31a. Clean up: R35C7 = [82](step 17); R7C9 = 7 and R9C7 = 4(step 13); R8C3: no 3
31b. R6C7 = 3; R45C5 = [23]; R5C8 = 1(hidden); R5C9 = 5(hidden)
31c. R5C2 = 9(hidden); R7C3 = 3(hidden); R8C4 = 7; R9C5 = 7(hidden)
31d. R5C4 = 8(hidden); R4C4 = 4

32. 17(3) at R2C8 = 8{36} -->> R23C8 = [36]
And the rest is all singles

greetings

Para

[mhparker]

What about adding a 4.0 for puzzles that have no way of solving without T&E for the puzzle i posted in the 60RP thread.

Done. Rating table in my post above updated.

PS I've never managed a Times Deadly without pencilmarks

Also done. BTW, I was going to change that earlier, but thought no-one would notice if I didn't! How wrong I was!

Not sure about the estimated difficulty level on this one. I found it easier than the A61 (though just as many steps).

That was quick! Was hoping to keep you (all) busy for a bit longer than that! Look forward to reading the WT.

Note: I could also have posted a real V2, requiring a team effort (rating 2.0 or above on my scale). But the problem at the moment is that we have two team puzzles (TJK32 and A60RP) still in the "to do" queue, so I don't want to open up a new one. I guess Ruud probably feels the same. Therefore, it's important to make some sort of response to these puzzles, whether it be to do them, reject them, or simply take them as far as we can and "officially" stop. That way, Ruud and I (and anyone else wanting to contribute a puzzle) will know that we're ready for more. It's a sort of deadlock situation. In order to get more, we need to somehow clear the queue. But how does one clear the queue when both puzzles appear to be pretty "ruudiculous"? Anyone got any suggestions here?

My personal opinion is that the A60RP should be just "signed off" with a marks pic of the final ("brick wall") state, unless someone has some unexpected inspiration. Ruud's post was not wasted here, because I managed to successfully create the RP-Lite from his cage pattern.

For the TJK32, the same (first part) may end up applying. But the only way to really find out is to start the tag.

[mhparker]

Forgot to respond to this one:

Wasn't Ruud working on some difficulty estimation for Killers as well? I remember him mentioning it once.

True, but I suspect it's extremely difficult (if not impossible) to come up with a computer-generated rating that accurately reflects how easy a puzzle is for a human to solve. Such an approach would probably work well up to a point (i.e., up to a certain difficulty level), but rapidly break down when the program approaches its (solving) limits. This "breakdown point" would no doubt occur too early to reliably rate Assassins, which would often (or even usually) end up being "off the scale".

Also, computer programs don't tend to combine logical steps into compound moves like humans do, and don't usually contain the kind of Artificial Intelligence (AI) needed to distinguish the hypotheticals that humans can relatively easily detect from brute force T&E. Furthermore, computer-generated ratings tend to ignore an important difficulty factor: namely how narrow the solution path is.

In the end, I expect we would be no better off with computer-generated ratings than with a relative, subjective rating scheme as proposed above, where puzzles are assigned to relatively small number of difficulty "bands".

[rcbroughton]

My personal opinion is that the A60RP should be just "signed off" with a marks pic of the final ("brick wall") state, unless someone has some unexpected inspiration. Ruud's post was not wasted here, because I managed to successfully create the RP-Lite from his cage pattern.

I've been banging my head on that brick wall on and off for a few days, but really can't find any way into it. Seems a shame to plant a white flag on it though . . .

For the TJK32, the same (first part) may end up applying. But the only way to really find out is to start the tag.

This one is more solvable - but so far I still need one great big Hypothetical (aka WAG) to move past a small brick wall in the middle game.