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PostPosted: Fri Jun 17, 2011 10:46 pm 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
SOK (Special Occasion Killer) 2 by Nasenbaer (September 2008) here
Puzzle Diagram:
Image
Attention: To unite the continents symbolized by the rings there is a remote cage in r3c258 r7c37 !

For better visibility here's a picture without colors:

Image
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3::k:5376:5376:5376:2563:2563:2563:3590:3590:3590:4617:4617:5643:4620:4620:4620:4879:4879:4879:4617:7443:5643:6165:7443:4620:4879:7443:4634:4617:5643:5643:6165:6165:6165:4634:4634:4634:2852:2852:2342:2342:2856:2601:2601:4139:4139:2852:6446:6446:4912:2856:5682:5682:5682:4139:3638:6446:7443:4912:3130:5435:7443:5682:2878:3638:6446:4912:4912:3130:5435:5435:5435:2878:3638:3638:1866:1866:3130:2637:2637:3663:3663:
Solution:
+-------+-------+-------+
| 9 5 7 | 1 3 6 | 4 8 2 |
| 4 1 8 | 7 5 2 | 9 3 6 |
| 6 3 2 | 9 8 4 | 1 5 7 |
+-------+-------+-------+
| 7 9 3 | 8 6 1 | 5 2 4 |
| 8 2 5 | 4 9 7 | 3 6 1 |
| 1 6 4 | 5 2 3 | 8 7 9 |
+-------+-------+-------+
| 5 8 6 | 2 1 9 | 7 4 3 |
| 2 7 9 | 3 4 5 | 6 1 8 |
| 3 4 1 | 6 7 8 | 2 9 5 |
+-------+-------+-------+
Quote:
Nasenbaer: Due to my computer meltdown I couldn't realize this Special Occasion Killer earlier. It should have been posted during the Olympic Games 2008. So now have a look at the Olympic flag and remember all the glorious moments.
Rating SSolver: 1.25.
Too hard??? Here's a hint: ***SPOILER*** Each ring was designed as a single cage, so no repetitions in the two cages it has now.
Have fun!

Børge: Hopefully there are no IOC members among the visitors here. The IOC is extremely protective of their symbols. Here a screen shot from en.wikipedia.org/wiki/Olympic_rings

Image

This is the reason for my 8 ring sudoku. If you remove the 3 bottom rings and apply the correct colours, you have the 5 OL rings.

Afmob: Thanks for this fun killer! Nice cage pattern, too.
SudokuSolver's rating is so high because even though it spots step 2a, it didn't apply step 2c afterwards but decided to continue with some more difficult moves.
Rating: 1.0.

Andrew: Thanks Nasenbaer for a fun puzzle with an interesting cage pattern.
I'll also rate SOK 2 at 1.0.
I was intrigued by Nasenbaer's hidden hint. I set up a diagram with the appropriate 4-cell cages combined into rings and quickly realised that this appeared to make an extremely difficult puzzle because it took away most of the moves that Afmob and I used so I'm not planning to try to solve it. Just out of curiosity does anyone know what the SS score is with the five rings as 8-cell cages? Edit. I've now been told that combining those cages wouldn't give a unique solution so obviously it can't then have a score.

Walkthrough by Afmob:
Thanks for this fun killer! Nice cage pattern, too.

SudokuSolver's rating is so high because even though it spots step 2a, it didn't apply step 2c afterwards but decided to continue with some more difficult moves.

SOK 2 Walkthrough:

1. R1234
a) Innies N2 = 17(2) = {89} locked for R3+N2
b) Innies N3 = 12(2) = {57} locked for R3+N3
c) 14(3) = {149/239/248} <> 6
d) 21(3) = 7{59/68} because {489} blocked by Killer pair (89) of 14(3)
-> 7 locked for R1+N1
e) Innies R234 = 16(3) must have one of (123) since R3C5 = (89) -> R3C2 = (123)
f) Innies R234 = 16(3) = {178/259/358}
g) Outies R234 = 13(2) = {49/67} since (58) is a Killer pair of Innies R234

2. C123 !
a) ! Innies+Outies C1: -2 = R259C2 - R1C1 -> R1C1 = (89), R259C2 = 12{3/4} -> 1,2 locked for C2
b) R3C2 = 3
c) Innies+Outies C1: -2 = R259C2 - R1C1: R259C2 = {124} locked for C2 -> R1C1 = 9
d) 21(3) = {579} locked for R1+N1
e) 25(4) must have one of (1234) -> R6C3 = (1234)

3. R1234
a) 10(3) = {136} locked for R1+N2
b) Naked triple (248) locked in R1C789 for N3
c) Innies R234 = 13(2) = {58} -> R3C5 = 8, R3C8 = 5
d) R3C4 = 9, R3C9 = 7
e) 18(4) @ N3 = 7{128/146/236/245} <> 9
f) 9 locked in R4C23 @ R4 for N4
g) Outies N1 = 19(3) <> 1
h) 18(4) @ N1 must have one of (357) -> R4C1 = (357)
i) 18(4) @ N1 = 1{278/368/458/467} because (357) only possible @ R4C1 -> 1 locked for N1
j) Outies N1 = 19(3) = {379/478/568} <> 2 because R4C1 = (357); R4C23 <> 5
k) 22(4) can only have two of (2468) and R23C2 = (2468) -> R4C23 <> 4,6,8

4. R456
a) Outies N1 = 19(3) = {379} locked for R4+N4
b) 6,8 locked in 25(4) @ C2 = 68{29/47} <> 1,5
c) Hidden Single: R1C2 = 5 @ C2, R1C3 = 7
d) 11(3) = 1{28/46} because R6C3 = (24) blocks {245} -> 1 locked for N4
e) Hidden Single: R5C3 = 5 @ N4
f) Cage sum: R5C4 = 4
g) 22(4) @ R2C3 = 3{289/469/478} because {2479} blocked by R6C3 = (24) -> R4C3 = 3
h) R4C1 = 7, R4C2 = 9
i) 25(4) = {4678} -> R6C3 = 4
j) 22(4) @ R2C3 = {2389} -> R3C3 = 2, R2C3 = 8

5. R789
a) 7(2) = {16} locked for R9
b) 14(2) = {59} -> R9C8 = 9, R9C9 = 5
c) 11(2) @ N9 = {38} locked for C9+N9
d) 10(2) = [37/82]
e) Hidden quad (2345) in 14(4) for N7
f) Hidden Single: R9C6 = 8 @ R9
g) Cage sum: R9C7 = 2
h) 12(3) <> 9 because R9C5 = (347)

6. N589
a) 9 locked in 11(2) @ C5 = {29} locked for C5+N5
b) 24(4) = {1689} -> R4C4 = 8, {16} locked for R4+N5
c) 10(2) = {37} locked for R5
d) Hidden Single: R9C5 = 7 @ R9
e) 12(3) = {147} locked for C5+N8
f) R9C4 = 6, R9C3 = 1
g) 19(4) = {2359} -> R8C3 = 9, {235} locked for C4 and 2 also locked for N8
h) R7C3 = 6
i) Outie R234 = R7C7 = 7
j) 21(4) = 59{16/34} -> R7C6 = 9, R8C6 = 5
k) 21(4) = {1569} -> 1,6 locked for R8+N9

7. Rest is singles.

Rating: 1.0.
Walkthrough by Andrew:
Thanks Nasenbaer for a fun puzzle with an interesting cage pattern.

I'll also rate SOK 2 at 1.0.

Here is my walkthrough. I was a bit slow finding the important breakthrough step 7. I've kept steps 5 and 6 because they may have simplified the later steps.

Prelims

a) R5C34 = {18/27/36/45}, no 9
b) R56C5 = {29/38/47/56}, no 1
c) R5C67 = {19/28/37/46}, no 5
d) R78C9 = {29/38/47/56}, no 1
e) R9C34 = {16/25/34}, no 7,8 9
f) R9C67 = {19/28/37/46}, no 5
g) R9C89 = {59/68}
h) R1C123 = {489/579/678}, no 1,2,3
i) R1C456 = {127/136/145/235}, no 8,9
j) 11(3) cage in N4 = {128/137/146/236/245}, no 9
k) 14(4) cage in N7 = {1238/1247/1256/1346/2345}, no 9

1. 45 rule on N2 2 innies R3C45 = 17 = {89}, locked for R3 and N2

2. 45 rule on N3 2 innies R3C89 = 12 = {57}, locked for R3 and N3

3. 45 rule on R1234 3 innies R3C258 = 16 = {178/259/358} (cannot be {169/268/349/367} because 1,2,3,4,6 in only R3C2, cannot be {457} because 5,7 only in R3C8), no 4,6
3a. R3C258 = 16 -> R7C37 = 13 = {49/67} (cannot be {58} which clashes with R3C258)

4. R1C789 = {149/239/248}, no 6
4a. R1C456 = {136/145/235} (cannot be {127} which clashes with R1C789), no 7
4b. 7 in R1 locked in R1C123, locked for N1
4c. R1C123 = {579/678}, no 4

5. 45 rule on C12 2 innies R34C2 = 1 outie R16C3 + 1
5a. Min R16C3 = 6 -> min R34C2 = 7, no 1,2,3 in R4C2
5b. Max R34C2 = 12 -> max R16C3 = 11, no 7,8,9 in R6C3
5c. Max R6C3 = 6 -> min R678C2 = 19, no 1

6. 45 rule on R9 3 innies R9C125 = 14 = {149/167/239/248/257/347} (cannot be {158/356} which clash with R9C89)
6a. Hidden killer triple 7,8,9 in R9C125, R9C67 and R9C89 -> R9C67 must contain one of 7,8,9 = {19/28/37}, no 4,6
[There is also Hidden killer triple 1,2,3 in R9C125, R9C34 and R9C67 -> R9C125 must contain one of 1,2,3 but I didn’t use this for my solution path.]

7. 45 rule on C1 1 innie R1C1 = 3 outies R259C2 + 2
7a. Min R259C2 = 6 -> min R1C1 = 8
7b. Max R1C1 = 9 -> max R259C2 = 7
7c. R259C2 = {123/124}, 1,2 locked for C2 -> R3C2 = 3, R259C2 = {124}, locked for C2, R1C1 = 9 (step 7), clean-up: no 6,8 in R1C23 (step 4c), no 1,3 in R1C789 (step 4)

8. R3C2 = 3 -> R3C58 = [85] (step 3), R3C4 = 9, R3C9 = 7, clean-up: no 3 in R56C5, no 4 in R78C9, no 9 in R9C9

9. Naked triple {248} in R1C789, locked for R1 and N3, clean-up: no 5 in R1C456 (step 4a)
9a. Naked pair {57} in R1C23, locked for N1
9b. Naked triple {136} in R1C456, locked for N2

10. 45 rule on N1 3 outies R4C123 = 19 = {289/379/469/478/568}, no 1
10a. R3C9 = 7 -> R4C789 = 11 = {128/137/146/236/245}, no 9

11. 45 rule on C5 3 innies R124C5 = 14 = {167/347/356} (cannot be {257} because R1C5 only contains 1,3,6), no 2
11a. 5 of {356} must be in R2C5 -> no 5 in R4C5

12. 18(4) cage at R2C1 cannot be {246}6, {248}4 or {268}2 -> must contain 1 in R2C12 + R3C1, locked for N1
12a. R2C12 + R3C1 contains 1 plus two even numbers -> R4C1 must be odd = {357}

13. R4C123 (step 10) = {379} (cannot be {289/469} because R4C1 only contains 3,5,7, cannot be {478/568} because R23C3 + R4C23 cannot be {2468} which only totals 20), locked for R4 and N4, clean-up: no 2,6 in R5C4

14. 6,8 in C2 locked in R678C2 -> 25(4) cage at R6C2 = {2689/4678}, no 1,5
14a. 2,4 only in R6C3 -> R6C3 = {24}

15. R1C2 = 5 (hidden single in C2), R1C3 = 7, clean-up: no 6 in R7C7 (step 3a)

16. 22(4) cage at R2C3 = {2389/3469} (cannot be {2479} which clashes with R6C3) -> R4C23 = [93], R4C1 = 7, clean-up: no 4 in R9C4
16a. R678C2 = {678} -> R6C3 = 4 (step 14), R23C3 = [82], R3C6 = 4, clean-up: no 1,5,7 in R5C4, no 7 in R5C5, no 6 in R5C7, no 9 in R7C7 (step 3a), no 3,5 in R9C4
16b. 6 in N1 locked in R23C1, locked for C1

17. 11(3) cage in N4 = {128} (only remaining combination), locked for N4 -> R6C2 = 6, R5C3 = 5, R5C4 = 4, clean-up: no 6 in R5C6, no 7 in R6C5, no 2 in R9C4
17a. 8 in N4 locked in R56C1, locked for C1
17b. Naked triple {169} in R789C3, locked for N7

18. Naked pair {16} in R9C34, locked for R9, clean-up: no 9 in R9C67, no 8 in R9C89
18a. R9C89 = [95], clean-up: no 2,6 in R78C9
18b. Naked pair {38} in R78C9, locked for C9 and N9, clean-up: no 2,7 in R9C6

19. R9C6 = 8 (hidden single in R9), R9C7 = 2, R9C12 = [34], R9C5 = 7, R2C2 = 1, R23C1 = [46], R5C2 = 2, R2C5 = 5, R3C7 = 1, clean-up: no 6 in R5C5, no 9 in R5C6, no 8 in R5C7, no 9 in R6C5
19a. R56C5 = [92] , clean-up: no 1 in R5C6

20. R124C5 (step 11) = {356} (only remaining combination) -> R14C5 = [36]
20a. Naked pair {37} in R5C67, locked for R5
20b. Naked pair {14} in R78C5, locked for N8 -> R9C4 = 6, R9C3 = 1, R1C46 = [16]

21. 1 in N9 locked in R78C8, locked for C8
21a. 6 in N6 locked in 16(3) cage = {169} (only remaining combination) -> R5C89 = [61], R6C9 = 9, R56C1 = [81], R2C789 = [936]

22. R8C7 = 6 (hidden single in C7), R78C3 = [69], R7C7 = 7 (step 3a), R5C67 = [73], R2C46 = [72], R78C2 = [87], R78C9 = [38]

23. R4C6 = 1, R7C6 = 9 (hidden singles in C6)

24. R4C789 (step 10a) = {245} (only remaining combination), locked for R4 and N6 -> R4C4 = 8

25. R7C6 = 9, R8C7 = 6 -> R8C68 = 6 = [51]

and the rest is naked singles

I was intrigued by Nasenbaer's hidden hint. I set up a diagram with the appropriate 4-cell cages combined into rings and quickly realised that this appeared to make an extremely difficult puzzle because it took away most of the moves that Afmob and I used so I'm not planning to try to solve it. Just out of curiosity does anyone know what the SS score is with the five rings as 8-cell cages? Edit. I've now been told that combining those cages wouldn't give a unique solution so obviously it can't then have a score.


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PostPosted: Fri Jun 17, 2011 10:52 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Old SSv3.2.1 scores:
Score = SudokuSolver v3.2.1 Score, rounded to nearest 0.05
E = Easy
H = Hard
In these tables, Rating is the lowest of the ratings given by Afmob,
Andrew and Mike, including estimates for puzzles by Afmob and Mike

+--------------------------+--------------------------+--------------------------+
| Puzzle Rating Score | Puzzle Rating Score | Puzzle Rating Score |
+--------------------------+--------------------------+--------------------------+
| A120 1.25 1.25 | HS3x 1.50 2.00 | A123 H1.25 2.50 |
| A120V2 H1.50 | A122 1.00 1.65 | A123V0.9 1.25 1.55 |
| A121 1.00 1.20 | A122V1.5 1.25 1.85 | A123V2 2.00 2.70 |
+--------------------------+--------------------------+--------------------------+

Puzzle rating table, with links to archive entries; each of these has a link to the puzzle thread.
Assassin 123 "Roulette" and Assassin 123 "Roulette" V0.9 are in the same archive entry.

Abbreviations used in Rating Table on this page:
Est = Estimated rating by puzzle maker
E = Easy
H = Hard
N = Nasenbaer
Score = SudokuSolver v3.3.0 score, rounded to nearest 0.05
! indicates that the Score has changed at least 0.10 from the SS v3.2.1 score
R indicates lowest score out of 4 rotations of puzzle, for some harder puzzles
** in the Afmob column indicates that these puzzles were made by him,
for these ones the estimate is his rating.
+------------------------+-----------+-------+-------+-------+-----------------------+-------+
| Puzzle | Made By | Est | Afmob | Andrew| Other Raters | Score |
+------------------------+-----------+-------+-------+-------+-----------------------+-------+
| Ass. 120 "Twin Towers" | Ed | 1.25 | E1.25 | 1.25 | | 1.35 |
| Assassin 120V2 | Ed | | H1.50 | 1.75 |(Ed)1.75 | 3.70 |
| Assassin 121 | Nasenbaer | 1.00 | 1.00 | H1.00 | | 1.20 |
| Assassin 121V2 | Nasenbaer | 1.25 | 1.25 | 1.25 | | 1.30 |
| Human Solvable 3X | HATMAN | H0.75 | 1.50 | 1.50 |(N)1.50(Ed)1.50 |!R3.65 |
| Assassin 122 | Afmob | 1.00 | ** | 1.25 |(Ed)H1.0 | 1.85 |
| Assassin 122V1.5 | Afmob | H1.25 | ** | 1.25 | | 1.90 |
| Ass. 123 "Roulette" | udosuk | | H1.25 | |(Ed)E1.50 |!R3.25 |
| Assassin A123V0.9 | udosuk | | | 1.25 | | 1.55 |
| Assassin 123V2 | udosuk | | 2.00 | 2.50 | |!R3.85 |
+------------------------+-----------+-------+-------+-------+-----------------------+-------+


Some of the selected quotes in the puzzle entries have been edited to remove "spoilers"; the full rating comments are included with the walkthroughs. In some cases the puzzle makers gave hints; these are included in tiny text in the selected quotes.


Last edited by Andrew on Thu Jan 26, 2012 4:38 am, edited 4 times in total.

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PostPosted: Thu Jul 07, 2011 3:07 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 120 Twin Towers by Ed (September 2008) here
Puzzle Diagram:
Image
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3::k:5376:5376:5376:4355:4355:4355:4614:4614:4614:2057:5376:3595:3595:5122:5122:5122:4614:2577:2057:3603:3603:3603:5122:5143:5143:5143:2577:3585:3603:5405:3614:3614:3614:5153:5143:4643:3585:3585:5405:3879:3879:3879:5153:4643:4643:3585:4654:5405:4144:4144:4144:5153:5428:4643:4415:4654:4654:4654:4154:5428:5428:5428:1086:4415:4424:4154:4154:4154:3332:3332:5198:1086:4424:4424:4424:4939:4939:4939:5198:5198:5198:
Solution:
+-------+-------+-------+
| 6 3 4 | 7 8 2 | 1 5 9 |
| 7 8 9 | 5 4 1 | 6 3 2 |
| 1 2 5 | 6 9 3 | 4 7 8 |
+-------+-------+-------+
| 2 1 8 | 4 3 7 | 9 6 5 |
| 3 4 6 | 9 1 5 | 8 2 7 |
| 5 9 7 | 8 2 6 | 3 1 4 |
+-------+-------+-------+
| 8 6 2 | 1 5 4 | 7 9 3 |
| 9 7 3 | 2 6 8 | 5 4 1 |
| 4 5 1 | 3 7 9 | 2 8 6 |
+-------+-------+-------+
Quote:
Ed (after posting the V2 first): :bigoops: Sorry guys. Goofed badly. Did a double-check after seeing Afmob needed forcing chains and found a really silly mistake. Back to plan A! I'll make the first one the V2.

Ed (on the V1): This time the Twins are not quite identical.
Estimated rating of 1.25. SS(V3.2.1)score = 1.27. JSudoku = "15 guesses".
[edit: forgot to say welcome to herschko!]
Enjoy this time!

Andrew: Thanks Ed. Great timing. I'd reached the stage with what is now the V2 ...
The SS score looks spot on. I'll rate A120 at 1.25.

Afmob: It seems that the most important move is ...
BTW, welcome herschko!
Rating: Easy 1.25.

Walkthrough by Andrew:
Thanks Ed. Great timing. I'd reached the stage with what is now the V2 where after prelims and 18 steps I was looking for contradiction moves to make progress. I don't like using those moves or forcing chains, which I rarely find, so early in a puzzle.

The SS score looks spot on. I'll rate A120 at 1.25 because I used combined cages which were partly formed from hidden cages and then for the final breakthrough I used 4 innies.

Here is my walkthrough. Thanks Afmob for pointing out that step 18 was incorrect because of an elimination that I'd made earlier. I've re-worked step 18 and edited the remaining steps. I've also corrected typos in steps 18c and 25.

Prelims

a) R23C1 = {17/26/35}, no 4,8,9
b) R2C34 = {59/68}
c) R23C9 = {19/28/37/46}, no 5
d) R78C1 = {89}, locked for C1 and N7
e) R78C9 = {13}, locked for C9 and N9, clean-up: no 7,9 in R23C1
f) R8C67 = {49/58/67}, no 1,2,3
g) R456C3 = {489/579/678}, no 1,2,3
h) R456C7 = {389/479/569/578}, no 1,2
i) R9C456 = {289/379/469/478/568}, no 1
j) 14(4) cage at R3C2 = {1238/1247/1256/1346/2345}, no 9
k) 14(4) cage at R4C1 = {1238/1247/1256/1346/2345}, no 9

1. 45 rule on R1 2 outies R2C28 = 11 = {29/38/47/56}, no 1

2. 45 rule on R123 2 outies R4C28 = 7 = {16/25/34}, no 7,8,9

3. 45 rule on R9 2 outies R8C28 = 11 = [29/38]/{47/56}, no 1 in R8C2, no 2 in R8C8

4. 45 rule on R789 2 outies R6C28 = 10 = {19/28/37/46}, no 5

5. 45 rule on N4 2 innies R46C2 = 10 = [19/28/37/46/64], no 5 in R4C2, no 1,2,3 in R6C2, clean-up: no 2 in R4C8 (step 2), no 7,8,9 in R6C8 (step 4)
5a. 1 in N6 locked in R456C8, locked for C8

6. 45 rule on R12 3 outies R3C159 = 18 = {189/279/369/378/459/468/567}
6a. 9 of {189} must be in R3C5 -> no 1 in R3C5
6b. 2 of {279} must be in R3C9 -> no 2 in R3C15, clean-up: no 6 in R2C1
6c. 9 of {369} must be in R3C5, 6 of {468} must be in R3C1 -> no 6 in R3C5

7. 45 rule on N9 3 innies R7C78 + R8C7 = 21 = {489/579/678}, no 2
7a. 2 in N9 locked in R9C789, locked for R9

8. 1 in R9 locked in R9C123, locked for N7
8a. 17(4) cage in N7 = {1367/1457}, no 2, 7 locked for N7, clean-up: no 9 in R8C8 (step 3)

9. R9C456 = {379/469/478/568}
9a. Hidden killer pair 8,9 in R9C456 and R9C789 for R9 -> R9C789 must have one of 8,9
9b. 20(4) cage at N9 = {2459/2468} (cannot be {2567} which doesn’t contain 8 or 9), no 7, 4 locked for N9, clean-up: no 4 in R8C2 (step 3), no 9 in R8C6
9c. 8,9 must be in R9C789 -> no 8 in R8C8, clean-up: no 3 in R8C2 (step 3)

10. 45 rule on R89 3 outies R7C159 = 16 = {169/178/349/358} (cannot be {259/268/457} because R7C9 only contains 1,3, cannot be {367} because R7C1 only contains 8,9)
10a. 4,5,6,7 only in R7C5 -> R7C5 = {4567}

11. Combined cage R8C2678 using R8C28 (step 3) = 24 = {4569/4578}, 4,5 locked for R8
11a. 4,9 of {4569} must be in R8C67, 5,8 of {4578} must be in R8C67 -> no 6,7 in R8C67
11b. Killer pair 8,9 in R8C1 and R8C67, locked for R8
11c. 7 in N9 locked in R7C78, locked for R7

12. Combined cage R2C2348 using R2C28 (step 1) = 25 = {2689/3589/4579/4678} (cannot be {3679} which clashes with R2C34)
12a. 5 or 6 must be in R2C34 -> no 5,6 in R2C28

13. 2 in R8 locked in R8C345, 16(4) cage at R7C5 = {2347/2356} (cannot be {1267} because 6,7 of {1267} must both be in N8 clashing with R9C456), no 1, 2,3 locked in R8C345, locked for R8 -> R78C9 = [31]
13a. 4,5 only in R7C5 -> R7C5 = {45}

14. 2 in N7 locked in R7C23 + R8C3
14a. 45 rule of N7 3 innies R7C23 + R8C3 = 11 = {236} (only remaining combination, cannot be {245} which clashes with R7C5) -> R8C3 = 3, R7C23 = {26} locked for R7, N7 and 18(4) cage at R6C2, clean-up: no 4 in R4C2 (step 5), no 3 in R4C8 (step 2), no 4 in R6C8 (step 4), no 8 in R7C78 + R8C7 (step 7), no 5 in R8C6, no 5 in R8C8 (step 3)

15. R7C23 = {26} = 8 -> R6C2 + R7C4 = 10 = [91], R4C2 = 1 (step 5), R4C8 = 6 (step 2), R6C8 = 1 (step 4), R8C8 = 4, R8C2 = 7 (step 3), R8C6 = 8, R78C1 = [89], R8C7 = 5, R7C5 = 5 (step 13), R7C6 = 4 (hidden single in R7), clean-up: no 4,5 in R456C3 (prelim g), no 6 in R9C456 (step 9)
15a. Naked triple {678} in R456C3, locked for C3 and N4 -> R7C23 = [62], clean-up: no 6,8 in R2C4
15b. Naked pair {59} in R2C34, locked for R2, clean-up: no 3 in R3C1
15c. Naked triple {379} in R9C456, locked for R9

16. 45 rule on N1 2 remaining outies R23C4 = 11 = [56/92]

17. 1 in R5 locked in R5C56 -> R5C456 = {159/168}, no 2,3,4,7

18. R456C7 = {389} (only remaining combination, cannot be {479} which clashes with R7C7), locked for C7 and N6 -> R7C78 = [79]
18a. 4 in N6 locked in R456C9, locked for C9, clean-up: no 6 in R23C9
18b. Naked pair {28} in R23C9, locked for C9 and N3 -> R9C9 = 6, R9C78 = [28]
18c. R1C9 = 9 (hidden single in C9), R5C8 = 2 (hidden single in C8)

19. 14(4) cage at R3C2 = {1256/1346} (cannot be {1238} because 3,8 only in R3C2) -> R3C4 = 6, R3C23 = [25/34], R8C45 = [26], clean-up: no 2 in R2C1

20. R1C1 = 6 (hidden single in C1), 8 in N1 locked in R12C2 -> 21(4) cage = {2568/3468}, no 1
20a. Naked pair {45} in R13C3, locked for C3 and N1 -> R29C3 = [91], R2C4 = 5, clean-up: no 3 in R2C1
20b. Naked triple {238} in R123C2, locked for C2

21. R5C456 (step 17) = {159/168}
21a. R5C4 = {89} -> R5C5 = 1, R5C6 = {56}

22. R6C456 = {268/358/367/457}
22a. 5,6 only in R6C6 -> R6C6 = {56}
22b. Naked pair {56} in R56C6, locked for N5

23. Deleted

24. Deleted
24a. Deleted
24b. R2C7 = 6 (hidden single in C7)

25. 18(4) cage in N3 = {1359} (only remaining combination) -> R1C7 = 1, R12C8 = [53], R3C78 = [47], R13C3 = [45], R23C1 = [71], R3C6 = 3 (cage sum), R123C2 = [382], R23C9 = [28], R2C56 = [41], R3C5 = 9

26. R4C456 = {239/248/347}
26a. 7 of {347} must be in R4C6 -> no 7 in R4C45
26b. 4 of {248} must be in R4C4 -> no 8 in R4C4

27. R6C456 = {268/367/457} (cannot be {358} which clashes with R6C7)
27a. 2 of {268} must be in R6C5 -> no 8 in R6C5

28. 45 rule on R4 4 remaining innies R4C1379 = 24 = {2589/3579/4578} (cannot be {3489} which clashes with R4C456)
28a. 9 of {3579} must be in R4C7 -> no 3 in R4C7
28b. 7 of {3579/4578} must be in R4C3 -> no 7 in R4C9

29. 45 rule on R6 4 remaining innies R6C1379 = 19 = {2467/3457} (cannot be {2368} because R6C9 only contains 4,5,7, cannot be {2458} because 2,4,5 only in R6C19), no 8, 4,7 locked for R6
29a. R6C7 = 3, R6C139 = {457} -> R6C3 = 7, R6C19 = {45}, locked for R6

and the rest is naked singles

With hindsight, I've realised that I didn't need to use combined cages in steps 11 and 12 although they seemed obvious and attractive at the time; I like using combined cages, particularly pairs of 2-cell cages. 6,7 can be eliminated from R8C67 in step 11 because of the clash with R8C28, leading to killer pair 4,5 in R8C28 and R8C67. Step 12 was just a simple clash with R2C34 that I ought to have spotted at step 1 :oops: . I also missed that in step 27 {457} clashes with R6C9. That additional elimination would have made the final steps simpler.
Walkthrough by Afmob:
It seems that the most important move is step 1e which Andrew also found quite early but afterwards I took a different path to finish this Killer.

BTW, welcome herschko!

A120 Walkthrough:

1. R789 !
a) 17(2) = {89} locked for C1+N7
b) 4(2) = {13} locked for C9+N9
c) Outies R9 = 11(2) <> 1; R8C8 <> 2
d) 20(4) = 2{459/468/567} -> 2 locked for R9+N9
e) ! Hidden Killer pair (89) in 20(4) for R9 since 19(3) cannot be {289}
-> 20(4) = 24{59/68} -> 4 locked for N9; R8C8 <> 8,9
f) Outies R9 = 11(2) = [56/65/74]
g) 17(4) = 17{36/45} -> 1,7 locked for N7
h) 13(2) <> {67} because it's a Killer pair of Outies R9
i) Killer pair (89) locked in R8C1 + 13(2) for R8
j) Killer pair (45) locked in Outies R9 + 13(2) for R8
k) 13(2): R8C6 <> 9

2. R789
a) Outies R89 = 16(3) = {169/178/349/358} <> 2 because R7C1 = (89)
and R7C9 = (13); R7C5 = (4567)
b) 2 locked in 16(4) @ R8 = 2{167/347/356}
c) Innies N9 = 21(3) = 7{59/68} -> 7 locked for R7+21(4)
d) Innies N9 = 21(3): R7C78 <> 8 because 6,7 only possible there

3. R456+C1
a) Innies N6 = 7(2) <> 7,8,9
b) Outies R789 = 10(2) <> 5; R6C2 <> 1,2,3
c) Innies N4 = 10(2): R4C2 <> 5,7,8
d) Innies N6 = 7(2): R4C8 <> 2
e) 14(4) <> 7 because (47) is a Killer pair of 21(3)
f) Innies+Outies C1: -6 = R5C2 - R19C1 -> R5C2 <> 8 because R19C1 <= 13
g) Innies N4 = 10(2) <> {46} since it's a Killer pair of 14(4)
h) Outies R789 = 10(2) <> 4,6
i) Innies N6 = 7(2): R4C8 <> 1,3

4. R789
a) 18(4): R7C4 <> 8,9 because R6C2+R7C23 >= 12
b) Outies N9 = 13(2+1): R7C6 <> 9 because R8C6 >= 4
c) 9 locked in 19(3) @ N8 for R9 -> 19(3) = 9{37/46}
d) 20(4) = {2468} locked for N9
e) 13(2): R8C6 <> 5
f) 5 locked in R7C456 @ N8 for R7
g) Hidden Single: R8C7 = 5 @ N8 -> R8C6 = 8
h) Outies R9 = 11(2) = {47} -> R8C2 = 7, R8C8 = 4
i) 16(4) = {2356} -> R7C5 = 5; 3 locked for R8
j) R8C9 = 1, R7C9 = 3
k) 18(4) = {1269} -> R6C2 = 9, R7C4 = 1; {26} locked for R7+N7

5. R456
a) Innie N4 = R4C2 = 1
b) Outie R789 = R6C8 = 1
c) Innie N6 = R4C8 = 6
d) 21(3) = {678} locked for C3+N4
e) 20(3) = {389} locked for C7+N6 since {479} blocked by R7C7 = (79)
f) 18(4) @ N6 = {2457} -> 4 locked for C9

6. R123
a) 10(2) = {28} locked for C9+N3
b) Hidden Single: R1C9 = 9 @ C9
c) 18(4) = {1359} -> R1C7 = 1, {35} locked for C8
d) R3C8 = 7, R3C7 = 4 -> R3C6 = 3
e) 14(2) = {59} locked for R2
f) Outies R1 = 11(2) = {38} -> R2C8 = 3, R2C2 = 8

7. R456
a) 1 locked in 15(3) @ N5 = 1{59/68}
b) 16(3) <> 4 because {457} blocked R6C9 = (457)
c) 4 locked in 14(3) @ N5 for R4 -> 14(3) = 4{28/37}
d) 9 locked in 15(3) @ N5 = {159} locked for R5+N5

8. Rest is singles.

Rating: Easy 1.25. I used a Hidden Killer pair.


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PostPosted: Thu Jul 07, 2011 5:34 am 
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Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 120 V2 by Ed (September 2008) here
Puzzle Diagram:
Image
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3::k:4096:4096:4096:5123:5123:5123:3846:3846:3846:3337:4096:2571:2571:6402:6402:6402:3846:1553:3337:5139:5139:5139:6402:5399:5399:5399:1553:5633:5139:2845:4638:4638:4638:2849:5399:6179:5633:5633:2845:4135:4135:4135:2849:6179:6179:5633:6190:2845:2864:2864:2864:2849:4660:6179:1855:6190:6190:6190:5178:4660:4660:4660:3134:1855:5704:5178:5178:5178:2564:2564:6222:3134:5704:5704:5704:2379:2379:2379:6222:6222:6222:
Solution:
+-------+-------+-------+
| 2 9 4 | 7 8 5 | 3 6 1 |
| 6 1 8 | 2 9 3 | 7 5 4 |
| 7 3 5 | 4 6 1 | 8 9 2 |
+-------+-------+-------+
| 1 8 2 | 6 5 7 | 4 3 9 |
| 5 7 6 | 3 4 9 | 2 1 8 |
| 9 4 3 | 8 1 2 | 5 7 6 |
+-------+-------+-------+
| 4 6 9 | 5 3 8 | 1 2 7 |
| 3 2 1 | 9 7 4 | 6 8 5 |
| 8 5 7 | 1 2 6 | 9 4 3 |
+-------+-------+-------+
Quote:
Ed: Change of plan. :D I enjoyed this "V2" so much it's now "V1"! [edit: back to being the V2. A120 follows]. I have the old one ready to go, so if anyone is desperate for an easier one (though no pushover) I can post that also [Gone: see below]. But only if asked for.
The cage design was inspired by a solving strategy idea, which didn't work for this one. Inadvertently ended up with the Twin Towers! Still have vivid memories of that terrible day 7 years ago.
I'll give it an estimated rating of "old-style" 1.50 (ie Weekly Assassin level 1.5..and yes, still bothered by the "new" 1.5). You don't want to know the SSscore. It's so far wrong it's not worth publishing. :oops: Instead, JSudoku, after "recursively solve" says, "17 guesses". I normally look for "16 guesses" to (often) get an SSscore of around 1.25. Hence, didn't throw the "17 guesses" one out. Boy, am I glad. Gremlins fixed.

herschko: This was a hard one for me.

Afmob: What a challenging Killer!
It took me quite some time to find the breakthrough move ...
Rating: (Hard) 1.5.

Andrew (in 2010): I'll rate my walkthrough for A120 V2 at 1.75 because I found this puzzle almost as hard as A108 V2 which I finished recently.
herschko's key step:
This was a hard one for me. I did have to guess twice. The most important guess was
R1C4:R1C6 is either 479 or 578.
Walkthrough by Afmob:
What a challenging Killer!

It took me quite some time to find the breakthrough move (step 8c,d) but this was only because I was looking for a different breakthrough path (combo elimination via Killer subsets) but this led to nothing. After I changed my approach and looked for different moves to progress (e.g. using forcing chains) the first placement came quicky and then it was over.

A120 V2 Walkthrough:

1. R123
a) Outies R1 = 6(2) = {15/24}
b) Innies N3 = 24(3) = {789} locked for N3
c) 21(4) = {1389/1479/1578/2379/2478} <> 6 because R3C78 = (789); R3C6+R4C8 = (12345)
d) 15(4) = 36{15/24} -> 3,6 locked for R1
e) 20(3) = 7{49/58} -> 7 locked for R1+N2
f) 16(4) = 12{49/58} -> 1,2 locked for N1
g) 10(2): R2C3 <> 3 and R2C4 <> 8,9
h) Innies+Outies N2: R2C7 = R23C4+R3C6 -> R3C4 <> 8,9; R23C4+R3C6 cannot be {123} because R3C7 >= 7
i) Hidden Killer triple (123) in 25(4) for N2 since R23C4+R3C6 <> {123} -> 25(4) <> 4
j) 25(4) can only have one of (789) @ N2 since 20(3) must have two of them -> 25(4) <> 1
k) 3 locked in R3C23 @ 20(4) @ N1 for R3; R4C2 <> 3

2. R456
a) Outies R123 = 11(2) = [65/74/83/92]
b) Innies N6 = 10(2) = [28/37/46]
c) Outies R123 = 11(2) <> 6
d) Innies N4 = 12(2) = [75/84/93]
e) 21(4) <> 5 because (15) only possible @ R3C6

3. R123
a) R3C5 <> 8,9 since it sees all 8,9 of N3
b) 25(4) = 9{268/358/367} -> 9 locked for R2
c) 10(2) <> 1
d) 1 locked in R3C46 @ N2 for R3
e) 20(4) = 3{179/269/278/458/467} (from step 1k)
f) Outies R12 = 15(3) = {258/267/456} <> 9 because {249} blocked by Killer triple (124) of
20(4) and R3C6 = (124)
g) 13(2) <> 4
h) 6(2): R2C9 <> 5
i) 3 locked in Innies N1 = 16(3) = 3{49/58/67}: R3C23 <> 4,8 because R2C3 <> 3,5,9

4. N6
a) 9 locked in 24(4) = 9{168/258/267/348/357/456}
b) 24(4) @ N6 <> 9{267/348} because (267,348) are Killer triples of Innies N6
c) 11(3) <> {236} since it's a Killer triple of Innies N6

5. R456 !
a) ! 21(4): R3C6 <> 4 because (478) is a Killer triple of Outies R12
b) 21(4) must have 3 xor 4 -> R4C8 <> 2
c) Outies R123 = 11(2) <> 9
d) Innies N4 = 12(2) <> 3
e) Innies N6 = 10(2) <> 8
f) 11(3) @ N4, 11(3) @ N5 <> 5 because {245} blocked by R6C2 = (45)
g) 11(3) @ N5 <> 4 since (46) is a Killer pair of Outies R789
h) Innies N46 = 12(2) + 10(2) = 22(4) = [7456/8347]
-> CPE: R4C13+R6C79 <> 4, R4C79+R6C13 <> 7
i) 4 locked in R6C123 @ R6 for N4
j) 18(3) <> 2 because (27) is a Killer pair of 11(3) @ N5

6. C789 !
a) 6(2) + 12(2) = 18(4) = {1359/1458/2349/2457}
b) ! 24(4) @ N6 <> 2,4 because combined cage 18(4) has two of (2589, 4569) @ C9
c) 2 locked in 11(3) @ N6 = 2{18/45}; 2 locked for C7
d) 24(4) @ N6: R5C8 <> 3,7 because (59) is a Killer pair of combined cage 18(4)
e) 24(4) @ N6: R456C9 <> 5 because (35) is a Killer pair of combined cage 18(4)

7. R456 !
a) 9 locked in Innies R6789 = 23(4) = 9{158/167/248/257/347/356}
b) Innies R6789 = 23(4) <> 7 because {1679} blocked by R6C8 = (67), {2579} blocked by
Killer pair (57) of Outies R789 and {3479} unplaceable since R6C7 = (1258)
c) ! Consider placement of 7 in R6 -> R4C456+R6C9 <> 3
- i) 7 in 11(3) @ N5 = {137} locked for R6+N5
- ii) 7 in Innies N6 = [37] -> R4C8 = 3
d) 24(4) @ N6: R5C9 <> 3 because 7 only possible there
e) 3 locked in R4C89 @ N6 for R4

8. R456 !
a) Hidden Killer pair (69) in Innies R1234 because 18(3) cannot be {369}
-> 2 locked in Innies R1234 = 16(4) = 2{149/167/356} <> 8
b) Innies R1234 = 16(4): R4C7 <> 5 because 3 must be in R4C9 @ 24(4) = {3579} -> R4C7 <> 5
c) ! Innies N46 = 22(4) = [7456/8347] -> Either R6C2 = 5 xor R4C8 = 3
d) ! Innies R1234 = 16(4) = 12{49/67} because [5623] leaves neither 5 in R6C2 nor 3 in R4C8
-> 1 locked for R4
e) Hidden Single: R4C8 = 3 @ N6
f) Innies N46 = 21(4) = [8347] -> R4C2 = 8, R6C2 = 4, R6C8 = 7
g) 24(4) @ N6 = {1689} locked for N6
h) Naked triple (245) locked in R456C7 for C7
i) 21(4) = 39{18/27} -> 9 locked for R3+N3

9. R123
a) 9 locked in R2C56 @ R2 for N2
b) 20(3) = {578} locked for R1+N2
c) 16(4) = {1249} -> 4 locked for R1+N1
d) 20(4) = 38{27/45}
e) Hidden Single: R3C6 = 1 @ N2
f) 21(4) = {1389} -> 8 locked for R3+N3
g) R2C7 = 7
h) 25(4) = {3679} -> R3C5 = 6; 3 locked for R2
i) 13(2) = [67/85]
j) Hidden Single: R2C8 = 5 @ R2

10. R789
a) Innies N9 = 9(3) = {126} -> R7C8 = 2; {16} locked for C7+N9
b) 24(4) @ N9 = {3489} locked for N9 because R89C8+R9C7 = (3489); 3 also locked for R9
c) 9(3) = {126} locked for R9+N8
d) 18(4) = 27{18/36} -> R7C6 = (38)
e) 24(4) @ N7 = 49{38/56} because R7C9 = (57) blocks 4{578} -> 9 locked for R7
f) 4,7 locked in Outies R89 = 14(3) = {347} -> R7C9 = 7, {34} locked for R7
g) R8C9 = 5, R7C6 = 8 -> R7C7 = 1, R8C7 = 6 -> R8C6 = 4
h) 20(4) = {1379} since R7C5+R8C45 = (379) -> R7C5 = 3, R8C3 = 1; {79} locked for R8

11. N145
a) 11(3) @ N4 = {236} locked for C3+N4
b) R2C3 = 8 -> R2C4 = 2, R3C4 = 4, R2C1 = 6 -> R3C1 = 7
c) 11(3) @ N5 = {128} because {236} blocked by R6C3 = (236) -> R6C6 = 2; {18} locked for R6+N5

12. Rest is singles.

Rating: (Hard) 1.5. I used lots of Killer triples and a small forcing chain.
Ed's Analysis of Afmob's Key Steps:
Afmob's WT for A120 V2 has a number of really neat moves, all of which I either missed or never got near (eg 3a :oops: ; 3f and 5h). But two more of the stand-outs deserve some pics. The first of these is what Afmob refers to as a "forcing chain", a name which has a very negative connotation for me. But Afmob's one is actually very elegant!

End of step 7b. here
Code:
.-------------------------------.-------------------------------.-------------------------------.
| 124589    124589    124589    | 45789     45789     45789     | 13456     123456    123456    |
| 5678      1245      4678      | 2346      235689    235689    | 789       1245      124       |
| 5678      35679     35679     | 12456     256       12        | 789       789       245       |
:-------------------------------+-------------------------------+-------------------------------:
| 12356789  78        123678    | 13456789  13456789  13456789  | 12458     34        13689     |
| 12356789  12356789  123678    | 123456789 123456789 123456789 | 12458     15689     136789    |
| 12345689  45        123468    | 123678    123678    123678    | 1258      67        13689     |
:-------------------------------+-------------------------------+-------------------------------:
| 123456    123456789 123456789 | 123456789 123456789 123456789 | 13456789  123456789 345789    |
| 123456    123456789 123456789 | 123456789 123456789 12346789  | 1346789   123456789 345789    |
| 123456789 123456789 123456789 | 123456    123456    123456    | 13456789  123456789 123456789 |
'-------------------------------.-------------------------------.-------------------------------'

Afmob wrote:
7c) ! Consider placement of 7 in R6 -> R4C456+R6C9 <> 3
- i) 7 in 11(3) @ N5 = {137} locked for R6+N5
- ii) 7 in Innies N6 = [37] -> R4C8 = 3

Image


Afmob tells me this "forcing chain" is called "Locked Cages" on sudopedia. I personally prefer SudokuSolver's "Dependant Cages" as a name. It is really neat!

That move is essential to get to Afmob's even cooler breakthrough. But first, a couple more steps in-between.
7d) 24(4) @ N6: R5C9 <> 3 because 7 only possible there
e) 3 locked in R4C89 @ N6 for R4

8. R456 !
a) Hidden Killer pair (69) in Innies R1234 because 18(3) cannot be {369}
-> 2 locked in Innies R1234 = 16(4) = 2{149/167/356} <> 8
b) Innies R1234 = 16(4): R4C7 <> 5 because 3 must be in R4C9 @ 24(4) = {3579} -> R4C7 <> 5

Now ready for the puzzle breaker move.
Image
Afmob wrote:
8c) ! Innies N46 = 22(4) = [7456/8347] -> Either R6C2 = 5 xor R4C8 = 3
d) ! Innies R1234 = 16(4) = 12{49/67} because [5623] leaves neither 5 in R6C2 nor 3 in R4C8
A really hidden hidden-cage blocker!

ALT way to see this move
I think I'd have a better chance of being able to find that elimination this way:

From step 8c, r6c2 & r4c8 cannot BOTH be 4, so anything that forces both to 4 cannot be true.
(i) If h16(4)r4 = {2356} -> r4c8 = 4.
(ii) {2356} must have 5 in r4c1. This forces r6c2 = 4
(iii) -> {2356} blocked since it forces both r4c8 & r6c2 to 4.

Well done Afmob!

Personally, I have no hesitation in giving this puzzle a rating of 1.75. Because of the number of key solving techniques needed (none of which are, in themselves, too complicated), no way would I consider it as the Weekly Assassin (and very glad I changed it to the V2!). It reminds me a lot of the A60 RP-Lite.
Walkthrough by Andrew in 2010:
Almost two years ago I wrote:
Thanks Ed. Great timing. I'd reached the stage with what is now the V2 where after prelims and 18 steps I was looking for contradiction moves to make progress. I don't like using those moves or forcing chains, which I rarely find, so early in a puzzle.
That was when Ed posted what is now the V1.

At the time Ed and I discussed my first 18 steps and discovered that I'd missed a CPE which I've now added to step 6; that led to some changes to step 13. Then after two or three more steps I moved on to other puzzles and didn't come back to A120 V2 until this week. It took me some time to find the first key breakthrough in step 29a (Afmob's step 6b) but after that I was making progress although I still needed a contradiction move for my final breakthrough.

Afmob wrote:
What a challenging Killer!
It certainly was!

Rating Comment. I'll rate my walkthrough for A120 V2 at 1.75 because I found this puzzle almost as hard as A108 V2 which I finished recently. For A120 V2 I used combination and permutation analysis, a "ruudiculous" ALS block using a combined cage for the first key breakthrough and a fairly short contradiction move for my final breakthrough.

Here is my walkthrough for A120 V2.

Prelims

a) R23C1 = {49/58/67}, no 1,2,3
b) R2C34 = {19/28/37/46}, no 5
c) R23C9 = {15/24}
d) R78C1 = {16/25/34}, no 7,8,9
e) R78C9 = {39/48/57}, no 1,2,6
f) R8C67 = {19/28/37/46}, no 5
g) R1C456 = {389/479/569/578}, no 1,2
h) R456C3 = {128/137/146/236/245}, no 9
i) R456C7 = {128/137/146/236/245}, no 9
j) R6C456 = {128/137/146/236/245}, no 9
k) R9C456 = {126/135/234}, no 7,8,9

1. 45 rule on R1 2 outies R2C28 = 6 = {15/24}

2. 45 rule on R123 2 outies R4C28 = 11 = {29/38/47/56}, no 1

3. 45 rule on R789 2 outies R6C28 = 11 = {29/38/47/56}, no 1

4. 45 rule on R9 2 outies R8C28 = 10 = {19/28/37/46}, no 5

5. 45 rule on N4 2 innies R46C2 = 12 = {39/48/57}, no 2,6
5a. 45 rule on N6 2 innies R46C8 = 10 = {28/37/46}, no 5,9

6. 45 rule on N3 3 innies R2C7 + R3C78 = 24 = {789}, locked for N3, CPE no 7,8,9 in R3C5
6a. 15(4) cage in N3 = {1356/2346}, 3,6 locked for R1
[I’ve added the CPE to step 6 after Ed pointed it out in discussions about my first 18 moves, at the time I originally got stuck.]

7. R1C456 = {479/578}, 7 locked for R1 and N2, clean-up: no 3 in R2C3
7a. 3 in R2 locked in R2C456, locked for N2

8. 21(4) cage at R3C6 contains two of 7,8,9 in R3C78 = {1389/1479/2379/2478} (cannot be {1578} because 1,5 only in R3C6, other combinations only have one of 7,8,9), no 5,6, clean-up: no 5 in R4C2 (step 2), no 7 in R6C2 (step 5), no 4 in R6C8 (step 5a)
8a. R3C78 = {789} -> no 7,8,9 in R3C6 + R4C8, clean-up: no 3,4 in R4C2 (step 2), no 8,9 in R6C2 (step 5), no 2,3 in R6C8 (step 5a)

9. 45 rule on N9 3 innies R7C78 + R8C7 = 9 = {126/135/234}, no 7,8,9, clean-up: no 1,2,3 in R8C6

10. 45 rule on N9 3 outies R6C8 + R78C6 = 19, max R6C8 + R8C6 = 17 -> min R7C6 = 2
10a. R6C8 + R8C6 cannot total 13 (because R7C6 “sees” both of R6C8 + R8C6 and there’s no 4,5,9 in R6C8 and no 5 in R8C6) -> no 6 in R7C6

11. 45 rule on N89 2 outies R6C8 + R8C3 = 1 innie R7C4 + 3
11a. Min R6C8 + R8C3 = 7 -> min R7C4 = 4
11b. Max R7C4 = 9 -> max R6C8 + R8C3 = 12, max R8C3 = 6

12. 45 rule on N1 3 outies R23C4 + R4C2 = 14
12a. R4C2 = {789} -> R23C4 = 5,6,7, no 8,9, clean-up: no 1,2 in R2C3

13. 25(4) cage at R2C5 can only contain 7 if it also contains 3 (because R2C34 = [73] is the only other way to place 3 in R2) -> 25(4) cage = {2689/3589/3679} (cannot be {1789/4579/4678} which contain 7 but not 3), no 1,4, 9 locked in R2C567, locked for R2, clean-up: no 1 in R2C4, no 4 in R3C1
13b. 1 in N2 only in R3C46, locked for R3, clean-up: no 5 in R2C9
13c. 5 of {3589} must be in R3C5 -> no 5 in R2C56
[Before I added the CPE to step 6, I used hidden killer quint 1,2,3,4,5 in R2C156, R2C2, R2C34, R2C8 and R2C9 for R2 -> R2C156 can only contain one of 1,2,3,4,5 to make step 13c work. This step is simpler after the CPE. Also I think that 9 locked ... in step 13 and the whole of step 13b require that there to be no 9 in R3C5.]

14. 16(4) cage in N1 = {1249/1258}, 2 locked for N1

15. 45 rule on N2 3 innies R23C4 + R3C6 = 1 outie R2C7
15a. R2C7 = {789} -> R23C4 + R3C6 (must contain 1, step 13b) = 7,8,9 = {124/125/134/126} (cannot be {135} because R23C4 must total 5,6,7 (step 12a) and 3 only in R2C4)
15b. 4 of {124/134} must be R23C4 because R23C4 = 5,6,7 (step 12a) -> no 4 in R3C6
15c. 2 of {126} must be in R3C6 (R23C4 cannot be {26} because R23C4 = 5,6,7) -> {126} must be [612], no 6 in R3C4

16. 25(4) cage at R2C5 (step 13a) = {2689/3589/3679}
16a. R23C4 + R3C6 (step 15a) = {124/134/126} (cannot be {125} which clashes with {2689/3589} and for {3679} R2C7 = 7 => R23C4 + R3C6 = {124} using step 15) -> no 5 in R3C4
[Ed suggested that simpler is R2C7 = 8 => R23C4 + R3C6 (step 15) = 8. R2C7 = 8 => no 8 in R2C3, no 2 in R2C4 => min R2C4 = 3 => max R3C46 = 5, no 5.
Alternatively R2C7 = 8 => R23C4 + R3C6 (step 15) = 8. R2C7 = 8 => no 8 in R2C3, no 2 in R2C4 => R23C4 + R3C6 cannot be {125}, no 5.]

17. 45 rule on R12 3 outies R3C159 = 15 = {258/267/456} (cannot be {249} which clashes with R3C46, ALS block), no 9, clean-up: no 4 in R2C1

18. 21(4) cage at R3C6 (step 8) = {1389/1479/2379/2478}
18a. R3C6 = {12} -> no 2 in R4C8, clean-up: no 9 in R4C2 (step 2), no 3 in R6C2 (step 5), no 8 in R6C8 (step 5a)
18b. Max R6C2 = 5 -> min R7C234 = 19, no 1
18c. R6C8 + R78C6 = 19 (step 10), max R6C8 = 7 -> min R78C6 = 12, no 2 in R7C6

19. R456C3 = {128/137/146/236} (cannot be {245} which clashes with R6C2), no 5

20. R6C456 = {128/137/236} (cannot be {146} which clashes with R6C28, cannot be {245} which clashes with R6C2), no 4,5
20a. R4C456 = {189/369/459/567} (cannot be {279} which clashes with R6C456, cannot be {378} which clashes with R4C2, cannot be {468} which clashes with R4C28), no 2

21. 3 in R3 only in R3C23 -> 20(4) cage at R3C2 = {1379/2378/3458/3467} (cannot be {2369} because R4C2 only contains 7,8)
21a. 4 of {3458/3467} must be in R3C4 -> no 4 in R3C23

22. 3 in R3 only in R3C23
22a. 45 rule on N1 3 innies R2C3 + R3C23 = 16 = {349/358/367}
22b. 8 of {358} must be in R2C3 -> no 8 in R3C23

23. 45 rule on C89 4 innies R3467C8 = 2 outies R19C7 + 9
23a. R46C8 = 10 (step 5a) -> R19C7 = R37C8 + 1
23b. Min R37C8 = 8 -> min R19C7 = 9, no 1,2 in R9C7
23c. Min R37C8 = 8 = [71] => R78C7 = {26/35} (step 9) -> min R19C7 = 9 but cannot be [63] which clashes with R78C7 -> no 3 in R9C7
23d. Max R19C7 = [69] = 15 -> max R37C8 = 14 = [95] (because 9 in R9C7 forces 9 in R3C8)
Max R19C7 without using 9 = [68] = 14 -> max R37C8 = 13 = [85] (9 in N3 must be in R23C7 and R27C8 cannot be [76] which clashes with R6C8)
-> no 6 in R7C8

24. R4C28 (step 2) = [74/83], R46C2 (step 5) = [75/84] -> R4C28 + R6C2 = [745/834], CPE no 4 in R4C13 + R6C79

25. R4C28 (step 2) = [74/83], R46C8 (step 5a) = [37/46] -> R4C28 + R6C8 = [746/837], CPE no 7 in R4C79 + R6C13

26. 9 in R6 only in R6C19
26a. 45 rule on R6789 4 innies R6C1379 = 23 = {1589/2489/3479/3569} (cannot be {1679} which clashes with R6C8, cannot be {2579} which clashes with R6C28)
26b. 4 of {2489/3479} must be in R6C13
{1589/3569}, 5 locked for R6 -> R6C2 = 4
-> 4 locked in R6C123, locked for N4

27. 9 in N6 only in 24(4) cage = {1689/2589/3579/4569} (cannot be {2679} which clashes with R6C8, cannot be {3489} which clashes with R4C8)
27a. R23C9 = [15]/{24}, R78C9 = {39/48/57} -> combined cage R2378C9 = [15]{39}/[15]{48}/{24}{39}/{24}{57}
27b. 24(4) cage = {1689/3579} (cannot be {2589/4569} which clash with R2378C9, ALS block), no 2,4
27c. Killer pair 6,7 in 24(4) cage and R6C8, locked for N6

28. 2 in N6 only in R456C7, locked for C7, clean-up: no 8 in R8C6
28a. R456C7 = {128/245}, no 3

29. R7C78 + R8C7 (step 9) = {126/135/234} -> R7C78 = {12/15/23/24/26/35} (cannot be {13/16/34} because no 2,5 in R8C7)
29a. 18(4) cage at R6C8 = {1269/1278/2367/2457/3456} (cannot be {1359/1458/2349/2358} because R6C8 only contains 6,7, cannot be {1368/1467} which don’t contain any combinations for R7C78)
29b. R7C78 = {12/23/24/26/35} (cannot be {15} because no combinations for 18(4) cage contain both of 1,5)
29c. 2 of {12/24} must be in R7C8 -> no 1,4 in R7C8

30. R7C78 + R8C7 (step 9) = {126/135/234}
30a. 5 of {135} must be in R7C8 (R78C7 cannot be [51] which clashes with R456C7), no 5 in R7C7
30b. R78C7 = {13/16/34}
30c. Killer pair 1,4 in R456C7 and R78C7, locked for C7

31. R78C7 (step 30b) = {13/16/34}
31a. 24(4) cage in N6 (step 27b) = {1689/3579} cannot be {3579}, here’s how
{3579} => R78C9 = {48} (cannot be {39/57} which clash with {3579}, ALS block) => R78C7 = {13/16} => R456C7 (step 28a) = {245} clashes with {3579}
[Almost a complete loop around N69. ;) ]
31b. -> 24(4) cage = {1689}, locked for N6 -> R456C7 = {245}, locked for C7 and N6, R46C8 = [37], R4C2 = 8 (step 2), R6C2 = 4 (step 3), clean-up: no 2 in R2C8 (step 1), no 3,7 in R8C2 (step 4), no 6 in R8C6, no 2,6 in R8C8 (step 4)
31c. 7 in N3 only in R23C7, locked for C7

32. 15(4) cage in N3 (step 6a) = {1356/2346}
32a. 4 of {2346} must be in R2C8 -> no 4 in R1C89

33. 18(4) cage at R6C8 (step 29a) = {1278/2367} (cannot be {2457} because R7C7 only contains 1,3,6) -> R7C8 = 2, clean-up: no 5 in R8C1
33a. R7C67 = [36/81], no 4,5,9 in R7C6, no 3 in R7C7

34. R7C78 + R8C7 (step 9) = {126} (only remaining combination) -> R78C7 = {16}, locked for C7 and N9 -> R1C7 = 3, clean-up: no 9 in R8C2 (step 4), no 7 in R8C6

35. R23C4 + R4C2 = 14 (step 12), R4C2 = 8 -> R23C4 = 6 = {24}, locked for C4 and N2 -> R3C6 = 1, clean-up: no 9 in R1C456 (step 7), no 4,7 in R2C3

36. Naked triple {578} in R1C456, locked for R1 and N2 -> R3C5 = 6, clean-up: no 7 in R2C1

37. 25(4) cage at R2C5 (step 13) = {3679} (only remaining combination) -> R2C7 = 7
37a. Naked pair {89} in R3C78, locked for R3, clean-up: no 5 in R2C1

38. 16(4) cage in N1 (step 14) = {1249} (only remaining combination), no 5, clean-up: no 1 in R2C8 (step 1)

39. R2C8 = 5 (hidden single in R2), R2C2 = 1 (step 1), clean-up: no 9 in R8C8 (step 4)
39a. Naked pair {24} in R23C9, locked for C9, clean-up: no 8 in R78C9

40. Naked triple {489} in R389C8, locked for C8
40a. Naked triple {489} in R8C8 + R9C78, locked for N9, clean-up: no 3 in R78C9

41. R9C9 = 3 (hidden single in N9)

42. R9C456 = {126} (only remaining combination), locked for R9 and N8

43. R6C456 (step 20) = {128/236}, 2 locked for R6 and N5 -> R6C7 = 5

44. 24(4) cage at R6C2 = 4{569} (only remaining combination, cannot be 4{389} which clashes with R7C6, cannot be 4{578} which clashes with R7C9), 5,6,9 locked for R7, 6 also locked for N7 -> R78C9 = [75], R7C7 = 1, R8C7 = 6, R8C6 = 4, R7C6 = 8 (step 33a), R7C5 = 3, R7C1 = 4, R8C1 = 3, R8C2 = 2, R8C3 = 1, R8C8 = 8

45. R7C4 = 5 (hidden single in N8)

46. R456C3 (step 19) = {236} (only remaining combination), locked for C3 and N4 -> R2C3 = 8, R2C4 = 2, R2C1 = 6, R3C1 = 7

47. 7 in R4 only in R4C456, locked for N5
47a. R4C456 (step 20a) = {567} (only remaining combination), locked for R4 and N5 -> R4C3 = 2

48. R6C456 (step 43) = {128} (only remaining combination) -> R6C6 = 2, R6C45 = {18}, locked for R6 and N5 -> R6C1 = 9

and the rest is naked singles.


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PostPosted: Fri Jul 08, 2011 4:49 am 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 121 by Nasenbaer (September 2008) here
Puzzle Diagram:
Image
Attention: This is a Killer X (1-9 cannot repeat on the diagonals)
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3:d:k:4096:4096:3586:3586:5636:1285:1285:5127:5127:4096:3082:3082:5636:5636:5636:2575:2575:5127:3602:3082:4628:4628:7958:4631:4631:2575:4378:3602:3612:4628:4628:7958:4631:4631:2594:4378:3602:3612:3878:7958:7958:7958:2858:2594:4378:3885:3885:3878:3120:7958:1330:2858:3636:3636:3885:5431:5431:3120:7958:1330:4412:4412:3636:2879:2879:5431:5442:5442:5442:4412:3910:3910:2879:2633:2633:2633:5442:4429:4429:4429:3910:
Solution:
+-------+-------+-------+
| 5 3 6 | 8 2 4 | 1 9 7 |
| 8 7 1 | 5 9 6 | 3 2 4 |
| 9 4 2 | 7 3 1 | 8 5 6 |
+-------+-------+-------+
| 1 9 3 | 6 8 5 | 4 7 2 |
| 4 5 8 | 2 1 7 | 6 3 9 |
| 6 2 7 | 9 4 3 | 5 8 1 |
+-------+-------+-------+
| 7 8 4 | 3 6 2 | 9 1 5 |
| 2 6 9 | 1 5 8 | 7 4 3 |
| 3 1 5 | 4 7 9 | 2 6 8 |
+-------+-------+-------+
Quote:
Nasenbaer: This week a little bit earlier than usual so you guys in Australia have something for breakfast) ;)
I almost changed my plan for this weeks Assassin because of a new one that just popped up (rating: 1.5). It would be really great for a v2 but time is running short and I'm still working on a v1 so it will have to wait. :( But don't be disappointed, this newborn killer is already reciting Arnold Schwarzenegger: "I'll be back!" :twisted:
But first you'll have to work on this one. Nothing fancy (except that it is a Killer X), just good old basic stuff. :whistle:
Rating SSolver(3.2.1): 1.20. My personal rating: 1.0
V2 is already waiting to be posted.
Have fun!

Andrew: Thanks Nasenbaer for a fun puzzle! I look forward to V2.
I'll rate A121 at Hard 1.0.

Afmob: Thanks Nasenbaer for this fun killer!
Rating: 1.0.
Nasenbaer: Wonderful! Two different walkthroughs in such a short time! Bravo! :applause: And we can always rely on Afmob to produce a brilliant move (step 1j) that shortens the solving path. :cheesey:

Walkthrough by Andrew:
Thanks Nasenbaer for a fun puzzle! I look forward to V2.

I'll rate A121 at Hard 1.0 because I used a Hidden Killer Pair as well as fairly routine combination analysis.

Here is my walkthrough. I've given eliminations on the diagonals; it's so easy for those of us doing manual eliminations to overlook them.

Probably not the most direct solving path. I was trying to get it finished before people in Europe. ;) Thanks Nasenbaer for the feedback. Now I know some of the things that I missed, one of which would have made my walkthrough quite a lot shorter.

Prelims

a) R1C34 = {59/68}
b) R1C67 = {14/23}
c) R45C2 = {59/68}
d) R45C8 = {19/28/37/46}, no 5
e) R56C3 = {78} (cannot be {69} which clashes with R45C2), locked for C3 and N4, clean-up: no 6 in R1C4, no 6 in R45C2
f) R56C7 = {29/38/47/56}, no 1
g) R67C4 = {39/48/57}, no 1,2,6
h) R67C6 = {14/23}
i) 20(3) cage in N3 = {389/479/569/578}, no 1,2
j) 10(3) cage in N3 = {127/136/145/235}, no 8,9
k) 21(3) cage in N7 = {489/579/678}, no 1,2,3
l) 11(3) cage in N7 = {128/137/146/236/245}, no 9
m) R9C234 = {127/136/145/235}, no 8,9
n) 31(7) cage at R3C5 = {1234579/1234678}

1. Naked pair {59} in R45C2, locked for C2 and N4

2. 45 rule on R89 2 innies R8C37 = 16 = [97], clean-up: no 5 in R1C4, no 4 in R67C7
2a. R8C3 = 9 -> R7C23 = 12 = [75/84]
2b. R8C7 = 7 -> R7C78 = 10 = {19/28/46}, no 3,5
2c. 11(3) cage in N7 = {128/137/146/236} (cannot be {245} which clashes with R7C3), no 5

3. 45 rule on C1234 3 innies R258C4 = 8 = {125/134}, 1 locked for C4

4. 45 rule on C6789 3 innies R258C6 = 21 = {489/579/678}, no 1,2,3
4a. 5 of {579} must be in R8C6 -> no 5 in R25C6

5. 45 rule on C9 3 outies R168C8 = 21 = {489/579/678}, no 1,2,3
5a. 5 of {579} must be in R8C8 -> no 5 in R16C8

6. 45 rule on C89 4 innies R2379C8 = 14 = {1238/1256/1346/2345} (cannot be {1247} which clashes with R168C8), no 7,9, clean-up: no 1 in R7C7 (step 2b)

7. 45 rule on N1 3 innies R1C3 + R3C13 = 17 = {269/359/368/458/467} (cannot be {179/278} because R1C3 only contains 5,6), no 1
7a. 7,8,9 only in R3C1 -> R3C1 = {789}
7b. R1C3 = {56} -> no 5,6 in R3C3

8. 45 rule on N7 1 innie R7C1 = 1 outie R9C4 + 3, no 1,2,3,4 in R7C1, no 6,7 in R9C4

9. 45 rule on N47 2 outies R3C1 + R9C4 = 1 innie R4C3 + 10, max R3C1 + R9C4 = 14 -> max R4C3 = 4

10. 45 rule on N9 1 outie R9C6 = 1 innie R7C9 + 4, no 1,2,3,4 in R9C6, no 6,8,9 in R7C9

11. 45 rule on R12 2 outies R3C28 = 9 = [36/45/63/72/81], no 1,2 in R3C2, no 4 in R3C8

12. 10(3) cage in N3 = {136/145/235}
12a. Hidden killer pair 1,2 in 10(3) cage and R1C7 + R3C79 for N3 -> R1C7 + R3C79 must contain one of 1,2
12b. 45 rule on N3 3 innies R1C7 + R3C79 = 15 = {168/249/258/267} (cannot be {159} which clashes with 10(3) cage, cannot be {348/357/456} which don’t contain 1 or 2), no 3, clean-up: no 2 in R1C6
12c. 1 of {168} must be in R1C7 -> no 1 in R3C79
12d. 20(3) cage in N3 = {389/479/578} (cannot be {569} which clashes with 10(3) cage), no 6

13. 45 rule on C123 4 outies R1349C4 = 25 = {2689/3679/4678} (cannot be {3589/4579} which clash with R258C4), no 5, clean-up: no 8 in R7C1 (step 8)
13a. R9C4 = {234} -> no 2,3,4 in R34C4

14. 6 in C4 locked in R34C4, locked in 18(4) cage at R3C3 = {1269/1368/1467/2367}
14a. 1 of {1467} must be in R4C3 -> no 4 in R4C3

15. R9C234 = {127/136/145/235}
15a. 4 of {145} must be in R9C4 -> no 4 in R9C23

16. 45 rule on C789 4 outies R1349C6 = {1459/1468/1567/2359/2368} (cannot be {1279/1369/1378/2458/2467/3457} which clash with R67C6)
16a. 3 of {2359/2368} must be in R1C6 -> no 3 in R34C6

17. 45 rule on N69 2 outies R3C9 + R9C6 = 1 innie R4C7 + 11, max R3C9 + R9C6 = 18 -> no 8,9 in R4C7
[Nasenbaer pointed out that I missed Min R4C7 = 1 -> min R3C9 + R9C6 = 12, no 2 in R3C9.]

18. 15(3) cage at R6C1 = {267/456} (cannot be {357} because 5,7 only in R7C1), no 1,3
18a. 5,7 only in R7C1 -> R7C1 = {57}
[Here I missed clean-up: no 3 in R9C4 (step 8) which proves to be an important elimination.]
18b. 6 locked in R6C12, locked for R6 and N4, clean-up: no 5 in R5C7
[Nasenbaer pointed out that I missed R7C23 (step 2a) = [84] (cannot be [75] which clashes with R7C1. Oops! I must have been getting tired. That would have made the later stages much quicker because it leads directly to R7C78 = [91] …]

19. 45 rule on N4 2 outies R37C1 = 1 innie R4C3 + 13
19a. R4C3 = {123} -> R37C1 = 14,15,16 = [95/87/97], no 7 in R3C1

20. R345C1 = {149/239} (cannot be {248} which clashes with 15(3) cage at R6C1) -> R3C1 = 9
20a. Killer pair 2,4 in R45C1 and R6C12, locked for N4

21. R1C3 + R3C13 (step 7) = {269/359}, no 4

22. R3C1 + R9C4 = R4C3 + 10 (step 9), R3C1 = 9 -> R9C4 = R4C3 + 1, no 3 in R9C4

23. R9C234 = {127/145/235} (cannot be {136} because R9C4 only contains 2,4), no 6
23a. R9C4 = {24} -> no 2,4 in R9C23

24. 2,6 in N7 locked in 11(3) cage = {236}, locked for N7
24a. 1 in N7 locked in R9C23, locked for R9

25. R7C2 = 8 (hidden single in N7), R7C3 = 4 (step 2a), locked for D/, clean-up: no 8 in R6C4, no 1 in R6C6, no 2,6 in R7C78 (step 2b), no 8 in R9C6 (step 10)
25a. R7C78 = [91], 9 locked for D\, clean-up: no 9 in R45C8, no 2 in R56C7, no 3 in R6C4, no 4 in R6C6, no 5 in R9C6 (step 10)
25b. Naked pair {23} in R3C3 and R6C6, locked for D\
25c. Naked pair {23} in R67C6, locked for C6, clean-up: no 2 in R1C7
25d. Naked pair {14} in R1C67, locked for R1

26. R7C5 = 6 (hidden single in R7), clean-up: no 5,9 in 31(7) cage at R3C5 (prelim n), no 2 in R7C9 (step 10)
26a. R7C6 = 2 (hidden single in R7), R6C6 = 3, R3C3 = 2 (hidden single on D\), clean-up: no 8 in R5C7
[I should also have given R9C4 = 4 here (I must have been careless with candidate eliminations) but since it’s done in step 27 it’s not worth changing.]
26b. R3C5 = 3 (hidden single in 31(7) cage)

27. 18(4) cage at R3C3 (step 14) = {2367} (only remaining combination, cannot be {1368} because 1,3 only in R4C3) -> R4C3 = 3, R34C4 = {67}, locked for C4, R9C4 = 4 (step 22), clean-up: no 2 in R45C1 (step 20), no 7 in R5C8, no 5 in R67C4
27a. R67C4 = [93], 9 locked for D/, R1C4 = 8, R1C3 = 6, R7C9 = 5, R9C6 = 9 (step 10), R7C1 = 7, R9C23 = [15], R1C1 = 5, R2C3 = 1
27a. Naked pair {14} in R45C1, locked for C1 and N4
27b. Naked pair {26} in R6C12, locked for R6

28. 10(3) cage in N3 (step 12) = {235} (only remaining combination) -> R3C8 = 5, R2C78 = {23}, locked for R2 and N3, R1C89 = [97], 7 locked for D/, R1C25 = [32]

29. Naked triple {236} in R2C8 + R8C2 + R9C1, locked for D/, R3C7 = 8, locked for D/, R5C5 = 1, R4C6 = 5
29a. 6 on D/ locked in R8C2 + R9C1, locked for N7
29b. R23C9 = [46], R2C2 = 7, locked for D\, R4C4 = 6, locked for D\
29c. R6C7 = 5, R5C7 = 6

and the rest is naked singles.
Walkthrough by Afmob:
Thanks Nasenbaer for this fun killer!

A121 Walkthrough:

1. C123 !
a) 15(2) = {78} locked for C3+N4 since (69) is a Killer pair of 14(2)
b) 14(2) @ N4 = {59} locked for C2+N4
c) 21(3) = 9{48/57} because (78) only possible @ R7C2 -> 9 locked for C3+N7; R7C2 = (78)
d) 14(2) @ N1 = [59/68]
e) Innies N1 = 17(3) = {269/359/368/458/467} <> 1 because R1C3 = (56); R3C1 = (789), R3C3 = (234)
f) 15(3) must have one of (578) -> R7C1 = (578)
g) Innies+Outies N7: -3 = R9C4 - R7C1 -> R9C4 = (245)
h) 10(3) <> 6 because R9C4 = (245)
i) 6 locked in 11(3) @ N7 = 6{14/23}
j) ! Innies+Outies C12: -8 = R2C3 - R79C2: R9C2 <> 7 and R2C3 <> 5,6 because
R2C3 <> 7, R7C2 = (78) and R9C2 = (1234)

2. R789
a) Hidden pair (78) locked in R7C12 @ N7 for R7 -> R7C1 <> 5
b) Innies R89 = 16(2) = {79} -> R8C3 = 9, R8C7 = 7
c) Innies+Outies N7: -3 = R9C4 - R7C1 -> R9C4 <> 2
d) 5 locked in R79C3 @ N7 for C3
e) R1C3 = 6 -> R1C4 = 8
f) 12(2) = [39/75/93]

3. C456
a) Innies C1234 = 8(3) = 1{25/34} -> 1 locked for C4
b) Killer pair (45) locked in Innies C1234 + R9C4 for C4
c) 12(2) = {39} locked for C4
d) Innies C1234 = 8(3) = {125} locked for C4
e) R9C4 = 4
f) 10(3) = {145} -> R9C2 = 1, R9C3 = 5
g) R7C3 = 4 -> R7C2 = 8
h) R7C1 = 7 -> R6C12 = 8(2) = {26} locked for R6+N4

4. R123
a) 18(4) @ N1 = {2367} -> R4C3 = 3, R3C3 = 2
b) 12(3) = {147} -> R2C3 = 1, {47} locked for N1
c) 16(3) = {358} -> R1C2 = 3, R1C1 = 5, R2C1 = 8
d) 5(2) = {14} locked for R1
e) 20(3) = {479} because R1C89 = (79) -> R2C9 = 4, {79} locked for R1+N3
f) R1C7 = 1, R1C6 = 4, R2C2 = 7

5. N569
a) 17(3) @ R7C7 = {179} -> R7C7 = 9, R7C8 = 1
b) 5(2) = {23} -> R6C6 = 3, R7C6 = 2
c) R6C4 = 9, R4C4 = 6
d) R9C9 = 8, R8C8 = 4 -> R8C9 = 3
e) 17(3) @ R9 = {269} -> R9C6 = 9, {26} locked for R9+N9
f) R9C1 = 3
g) 14(3) = {158} -> R7C9 = 5, R6C8 = 8, R6C9 = 1
h) 10(2) = {37} -> R4C8 = 7, R5C8 = 3
i) R3C9 = 6, R3C8 = 5, R3C7 = 8, R2C8 = 2,

6. Rest is singles without considering diagonals.

Rating: 1.0.


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PostPosted: Fri Jul 08, 2011 4:59 am 
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Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 121 V2 by Nasenbaer (September 2008) here
Puzzle Diagram:
Image
Attention: This is a Killer X (1-9 cannot repeat on the diagonals)
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3:d:k:4096:4096:3586:3586:5636:1285:1285:5127:5127:4096:3082:3082:5636:5636:5636:2575:2575:5127:3602:3082:2324:2324:7958:2327:2327:2575:4378:3602:5916:5916:5916:7958:4896:4896:4896:4378:3602:5916:3878:7958:7958:7958:4138:4896:4378:3885:3885:3878:3120:7958:4138:4138:3636:3636:3885:5431:5431:3120:7958:4138:4412:4412:3636:2879:2879:5431:5442:5442:5442:4412:3910:3910:2879:2633:2633:2633:5442:4429:4429:4429:3910:
Solution:
+-------+-------+-------+
| 5 3 6 | 8 2 4 | 1 9 7 |
| 8 7 1 | 5 9 6 | 3 2 4 |
| 9 4 2 | 7 3 1 | 8 5 6 |
+-------+-------+-------+
| 1 9 3 | 6 8 5 | 4 7 2 |
| 4 5 8 | 2 1 7 | 6 3 9 |
| 6 2 7 | 9 4 3 | 5 8 1 |
+-------+-------+-------+
| 7 8 4 | 3 6 2 | 9 1 5 |
| 2 6 9 | 1 5 8 | 7 4 3 |
| 3 1 5 | 4 7 9 | 2 6 8 |
+-------+-------+-------+
Quote:
Nasenbaer: This version has the same solution but a slightly altered cage pattern which removes the easy start from v1. It's only slightly harder than v1. Maybe I should have called it v1.5. Oh well.
Rating SSolver(3.2.1): 1.31. My personal rating: 1.25
(I would have liked to join the cages 15(2) @ r5c3 and 12(2) @ r6c4 because of the symmetry but that would produce a rating of 2.64 with extended T&E) :(
Have fun!

Afmob: Like Nasenbaer said A121 V2 is more a V1.5 than a V2 but fun nonetheless. Thanks Nasenbaer!
I checked SudokuSolver's and JSudoku's solving path and it seems quite narrow though my ...

Andrew: Thanks Nasenbaer. Another enjoyable variant. As you said the cage changes took away the early entry ...
My impression after going through Afmob's walkthrough and mine is that it can't be too narrow ... I'll assume the comment meant that Afmob's solving path was very similar to that used by Sudoku Solver and JSudoku. Mine looks very different.
I'll also rate A121 V2 at 1.25.

Walkthrough by Afmob:
Like Nasenbaer said A121 V2 is more a V1.5 than a V2 but fun nonetheless. Thanks Nasenbaer!

I checked SudokuSolver's and JSudoku's solving path and it seems quite narrow though my Killer quint could have been replaced by some easier moves but since I like such moves I didn't change my walkthrough.

A121 V2 Walkthrough:

1. R123
a) Innies R12 = 13(4) = 1{237/246/345} <> 8,9; 1 locked for R2
b) Outies R12 = 9(2): R3C2 <> 1,9
c) Innies+Outies N1: 6 = R13C4 - R3C1 -> R1C4 <> 6 (IOU @ R3)

2. R789
a) Innies R89 = 16(2) = {79} locked for R8
b) Innies+Outies N7: -3 = R9C4 - R7C1 -> R9C4 <> 7 and R7C1 <> 1,2,3
c) Innies+Outies N9: 4 = R9C6 - R7C9 -> R9C6 = (56789) and R7C9 = (12345)

3. C1234 !
a) Innies C1234 = 8(3) = 1{25/34} -> 1 locked for C4
b) Killer pair (79) locked in 15(2) + R8C3 for C3
c) 14(2) = [59/68]
d) 9(2): R3C3 <> 8 and R3C4 <> 2
e) Innies N1 = 17(3) = {269/359/368/458/467} <> 1 because R1C3 = (56); R3C1 = (789), R3C3 = (234)
f) 9(2): R3C4 = (567)
g) Outies C1 = 11(3) <> 9
h) 9 locked in R123C1 @ N1 for C1
i) 9 locked in 21(3) @ N7 = 9{48/57}
j) Innies+Outies N7: -3 = R9C4 - R7C1 -> R9C4 <> 6 and R7C1 <> 4
k) ! Killer quint (12345) locked in R2589 + 12(2) for C4

4. C123 !
a) 9(2) <> 4
b) Innies N1 = 17(3) <> 7
c) Outies N4 = 22(2+1): R7C1 <> 8 because R3C1 = (89) and R4C4 >= 6
d) Innies+Outies N7: -3 = R9C4 - R7C1 -> R9C4 <> 5
e) 10(3): R9C23 <> 4 since R9C4 <> 1,5
f) ! Innies N7 = 13(3) = 5{17/26} -> 5 locked for N7
g) 21(3) = {489} -> R8C3 = 9; {48} locked for R7+N7
h) 15(2) = {78} locked for C3+N4
i) R7C3 = 4, R7C2 = 8, R8C7 = 7

5. N89
a) 17(3) @ R7C7 = {179} -> 1,7 locked for R7+N9
b) Innies+Outies N9: 4 = R9C6 - R7C9 -> R9C6 <> 5,8
c) 12(2): R6C4 <> 3,8
d) 1,8 locked in 21(4) @ N8 = 18{39/57}

6. C1234
a) Killer pair (35) locked in Innies C1234 + 12(2) for C4
b) 10(3) = 1{27/45} -> 1 locked for R9+N7
c) 11(3) = {236} locked for N7
d) 12(3) = 4{17/26/35} because {156/237} blocked by R1C3 = (56) and R3C3 = (23)
-> 4 locked for C2+N1
e) 14(3) <> 3,6 because R3C1 = (89) and {239} blocked by R89C1 = (236)
f) 15(3) = 7{26/35} since {456} blocked by Killer pair (45) of 14(3) -> R7C1 = 7
g) 10(3) = {145} -> R9C4 = 4; 5 locked for R9
h) Innies C1234 = {125} locked for C4
i) R7C4 = 3 -> R6C4 = 9
j) R1C4 = 8 -> R1C3 = 6
k) 21(4) = {1578} -> R9C5 = 7; 5,8 locked for R8+N8

7. C6789
a) 15(3) = {348} -> R9C9 = 8; {34} locked for R8+N9
b) 17(3) @ R9 = {269} -> R9C6 = 9; 2,6 locked for R9+N9
c) R9C1 = 3, R7C9 = 5, R1C9 = 7
d) Outies C9 = 21(3) = {489} -> R8C8 = 4, R6C8 = 8, R1C8 = 9
e) Cage sums: R2C9 = 4, R6C9 = 1
f) Innies C6789 = 21(3) = {678} -> R8C6 = 8, {67} locked for C6
g) 16(4) = {2356} -> R7C6 = 2; 6 locked for C7+N6
h) 17(3) @ N3 = {269} -> R3C9 = 6; {29} locked for N6
i) 10(3) = {235} locked for N3
j) R1C7 = 1, R3C7 = 8 -> R3C6 = 1, R4C6 = 5, R2C8 = 2

8. Rest is singles.

Rating: 1.25. I used a Killer quint.
Walkthrough by Andrew:
Thanks Nasenbaer. Another enjoyable variant. As you said the cage changes took away the early entry but I still had a naked pair in step 1 and a killer pair in step 2.

Afmob wrote:
I checked SudokuSolver's and JSudoku's solving path and it seems quite narrow .....
My impression after going through Afmob's walkthrough and mine is that it can't be too narrow although we both used the critical area in C4 but even then used different ways. I'll assume the comment meant that Afmob's solving path was very similar to that used by Sudoku Solver and JSudoku. Mine looks very different.

I'll also rate A121 V2 at 1.25. It wasn't much harder than A121.

Here is my walkthrough. I've given eliminations on the diagonals; it's so easy for those of us doing manual eliminations to overlook them.

At the end I've also given some other moves that I was looking at before I found a more powerful step 18. From step 22a onward there may have been a more direct path; I was concentrating on finding hidden singles on the diagonals so that when I reached naked singles there were no more eliminations on the diagonals.

Prelims

a) R1C34 = {59/68}
b) R1C67 = {14/23}
c) R3C34 = {18/27/36/45}, no 9
d) R3C67 = {18/27/36/45}, no 9
e) R56C3 = {69/78}
f) R67C4 = {39/48/57}, no 1,2,6
g) 20(3) cage in N3 = {389/479/569/578}, no 1,2
h) 10(3) cage in N3 = {127/136/145/235}, no 8,9
i) 21(3) cage in N7 = {489/579/678}, no 1,2,3
j) 11(3) cage in N7 = {128/137/146/236/245}, no 9
k) R9C234 = {127/136/145/235}, no 8,9
l) 31(7) cage at R3C5 = {1234579/1234678}

1. 45 rule on R89 2 innies R8C37 = 16 = {79}, locked for R8

2. Killer pair 7,9 in R56C3 and R8C3, locked for C3, clean-up: no 5 in R1C4, no 2 in R3C4
2a. 21(3) cage in N7 = {489/579/678}
2b. 5 of {579} must be in R7C3 -> no 5 in R7C2

3. 45 rule on R12 2 outies R3C28 = 9 = {27/36/45}/[81], no 1,9 in R3C2

4. 45 rule on C1234 3 innies R258C4 = 8 = {125/134}, 1 locked for C4, clean-up: no 8 in R3C3

5. 45 rule on C6789 3 innies R258C6 = 21 = {489/579/678}
5a. 5 of {579} must be in R8C6 -> no 5 in R25C6

6. 45 rule on C9 3 outies R168C8 = 21 = {489/579/678}
6a. 5 of {579} must be in R8C8 -> no 5 in R16C8

7. 45 rule on C1 3 outies R168C2 = 11 = {128/137/146/236/245}, no 9

8. 45 rule on N7 1 innie R7C1 = 1 outie R9C4 + 3, no 1,2,3,4 in R7C1, no 7 in R9C4

9. 45 rule on N47 3 outies R3C1 + R49C4 = 19
9a. Max R9C4 = 6 -> min R3C1 + R4C4 = 13, no 1,2,3

10. 45 rule on N9 1 outie R9C6 = 1 innie R7C9 + 4, no 6,7,8,9 in R7C9, no 1,2,3,4 in R9C6

11. 45 rule on N4 3 outies R37C1 + R4C4 = 22, max R37C1 = 17 -> min R4C4 = 5

12. 6 in C4 locked in R1349C4
12a. 45 rule on C123 4 outies R1349C4 = 25 = {2689/3679/4678}, no 5, clean-up no 4 in R3C3, no 8 in R7C1 (step 8)

13. R9C234 = {127/136/145/235}
13a. 4 of {145} must be in R9C4 -> no 4 in R9C23

14. 45 rule on N7 3 innies R7C1 + R9C23 = 13 = {139/157/256}
14a. 6 of {256} must be in R7C1 (R9C23 cannot be {26} because R9C234 cannot be {26}2), no 6 in R9C23

15. 21(3) cage in N7 = {489/678} (cannot be {579} which clashes with R7C1 + R9C23), no 5, 8 locked in R7C23, locked for R7 and N7, clean-up: no 4 in R6C4
15a. R8C3 = {79} -> no 7,9 in R7C2
15b. 4 of {489} must be in R7C3 (R78C3 cannot be {89} which clashes with R56C3), no 4 in R7C2
15c. 6 of {678} must be in R7C2 (R78C3 cannot be {67} which clashes with R56C3), no 6 in R7C3
15d. -> 21(3) cage = [849/687]
15e. Killer pair 8,9 in R56C3 and R78C3, locked for C3, clean-up: no 6 in R1C4

16. 17(3) cage in N9 = {179/269/359} (cannot be {467} which clashes with R7C23), no 4, 9 locked for N9
16a. 9 in R9 locked in R9C56, locked for N8, clean-up: no 3 in R6C4

17. 45 rule on N1 3 innies R1C3 + R3C13 = 17 = {269/359/368} (cannot be {179/278} because R1C3 only contains 5,6, cannot be {458/467} because 4,7,8 only in R3C1), no 1,4,7, clean-up: no 8 in R3C4
17a. 8,9 only in R3C1 -> R3C1 = {89}
17b. 2,3 only in R3C3 -> R3C3 = {23}, clean-up: no 3,4 in R3C4

This was where I had my interesting original step 18, see comment at the end.

18. R1349C4 (step 12a) = {2689/3679/4678}
18a. 2,3,4 only in R9C4 -> R9C4 = {234}, clean-up: no 9 in R7C1 (step 8)

19. R8C3 = 9 (hidden single in N7), R8C7 = 7, R7C23 = [84] (step 15d), 4 locked for D/, clean-up: no 2,5 in R3C6, no 1 in R3C8 (step 3), no 6 in R56C3, no 8 in R6C4, no 8 in R9C6 (step 10)
19a. Naked pair {78} in R56C3, locked for N4
19b. R8C7 = 7 -> R7C78 = {19} (step 16), locked for R7 and N9, clean-up: no 5 in R9C6 (step 10)

20. R1349C4 (step 12a) = {2689/4678} (cannot be {3679} which clashes with R56C4), no 3, clean-up: no 6 in R7C1 (step 8)
[Alternatively Killer pair 3,5 in R258C4 and R67C4, locked for C4 gives the same eliminations.]
20a. 6 in R7 locked in R7C56, locked for N8, clean-up: no 2 in R7C9 (step 10)

21. R7C56 = {26} (hidden pair in R7), locked for N8 -> R9C4 = 4, R7C1 = 7 (step 8), clean-up: no 9 in R14C4 (step 20), no 3 in R258C4 (step 4), no 5 in R6C4, no 2,3 in R9C23 (step 13)
21a. R1C4 = 8, R1C3 = 6, clean-up: no 1 in R3C7
21b. R67C4 = [93] (hidden pair in C4), 9 locked for D/, R7C9 = 5, R9C6 = 9 (step 10), clean-up: no 4,5 in R258C6 (step 5)
21c. R8C6 = 8, R9C5 = 7 (hidden single in N8)
21d. Naked pair {15} in R9C23, locked for N7

22. Naked pair {67} in R25C6, locked for C6 -> R7C56 = [62], R5C6 = 7, R2C6 = 6, R4C4 = 6, locked for D\, R3C4 = 7, R3C3 = 2, locked for D\, R3C1 = 9 (step 17), R56C3 = [87], clean-up: no 3 in R1C7, no 3 in R3C7, no 5 in 31(7) cage at R3C5 (prelim l)
[While checking through this walkthrough, I realised that I could have started this step with R5C6 = 7 (hidden single in 31(7) cage at R3C5).]
22a. R2C2 = 7, R7C7 = 9, R9C9 = 8 (hidden singles in D\), R7C8 = 1, R8C8 = 4, locked for D\, R8C9 = 3 (cage sum), R1C89 = [97], R2C9 = 4, clean-up: no 1 in R1C6
22b. R3C7 = 8 (hidden single in D/), R3C6 = 1, R3C9 = 6
[It would have been more fun to use 1 on D/ locked in R4C6 + R5C5, locked for N5 first but the hidden single on D/ is more logical.]
22c. Naked pair {34} in R1C6 + R3C5, locked for N2
22d. R2C5 = 9 (hidden single in N2)

23. R5C5 = 1 (hidden single on D/), locked for D\, R5C4 = 2, R2C4 = 5, R8C45 = [15], R1C5 = 2, R1C7 = 1, R1C6 = 4, R3C5 = 3, R3C28 = [45]

24. R9C1 = 3 (hidden single in R9), locked for D/ -> R2C78 = [32], 2 locked for D/

25. R7C9 = 5 -> R6C89 = 9 = [81]
25a. R3C1 = 9 -> R45C1 = 5 = [14]

and the rest is naked singles

My original step 18 was

18. 8 in R7 locked in R7C23 = [84/68] (step 15d)
18a. 45 rule on R89 4 outies R7C2378 = 22 = {1489/1678/3478/3568} (cannot be {2389/2578} which don’t contain 4 or 6), no 2
18b. R7C2378 must contain 2 odd numbers -> R7C78 = {13579}

I'd also got the following in my notes for possible later use

15(3) cage in N9 must have at least one odd number
45 rule on N9 3 innies R7C9 + R9C78 = 13 must have at least one odd number
Hidden killer quint for odd numbers in N9, 17(3) cage in N9 contains three odd numbers -> 15(3) cage in N9 and R7C9 + R9C78 must each contain only one odd number

A pity that I wasn't able to use these steps. :(


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PostPosted: Mon Jul 11, 2011 4:58 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Human Solvable 3X by HATMAN (September 2008) here
Puzzle Diagram:
Image
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3:d:k:2315:18:3330:3330:19:3329:3329:20:2313:2315:2315:21:22:3089:23:2313:2313:24:3331:25:2575:26:3089:27:2064:28:3336:3331:29:30:2575:31:2064:32:33:3336:34:35:36:37:38:39:40:41:42:3332:43:44:3853:45:3598:46:47:3335:3332:48:3853:49:50:51:3598:52:3335:53:2314:2314:54:55:56:57:2316:2316:2314:58:3333:3333:59:3334:3334:60:2316:
Nasenbaer: BTW, the code string doesn't work with SudoCue. Here's one that does:
3x3:d:k:2304:1:3330:3330:4:3333:3333:7:2312:2304:2304:11:12:3085:14:2312:2312:17:3346:19:2580:21:3085:23:2072:25:3354:3346:28:29:2580:31:2072:33:34:3354:36:37:38:39:40:41:42:43:44:3373:46:47:3888:49:3634:51:52:3381:3373:55:3888:57:58:59:3634:61:3381:63:2368:2368:66:67:68:69:2374:2374:2368:73:3402:3402:76:3405:3405:79:2374:
Solution:
+-------+-------+-------+
| 1 8 4 | 9 3 6 | 7 5 2 |
| 2 6 5 | 8 7 1 | 3 4 9 |
| 9 3 7 | 2 5 4 | 1 6 8 |
+-------+-------+-------+
| 4 9 2 | 3 1 7 | 6 8 5 |
| 6 7 3 | 5 8 2 | 4 9 1 |
| 5 1 8 | 6 4 9 | 2 3 7 |
+-------+-------+-------+
| 8 4 9 | 1 2 3 | 5 7 6 |
| 7 5 1 | 4 6 8 | 9 2 3 |
| 3 2 6 | 7 9 5 | 8 1 4 |
+-------+-------+-------+
Quote:
HATMAN: Quite interesting JSudoku takes about 23 fishes (afmob points out that my version is not the latest, so) Sudoku Solver gives 2.01 with T&E which is very high, I'd put my solution below 1.0.

Nasenbaer: That was a very interesting puzzle, HATMAN. Thanks a lot! ;clapclap;

HATMAN: sketch of solution:
How did you find the difficulty? It was very hard to create. I had the basic idea but more than ten attempts before I could get one that I could solve without the computer finding it easy.
I'll create some less taxing ones for now. Hopefully someone will do it more neatly.

Afmob: Thanks for this interesting Killer!
Though all walkthroughs posted so far are quite short the moves themselves are advanced ... Rating: 1.5.

udosuk: This rating thing is a joke. :rambo: From now on (after this post) I've decided I'm not talking about any numerical rating anymore.
The "experts" might rate my step 4 as 1.5 or 15.0 or whatever but for me the only "rating" is this: EASY
It is EASIER than most recent assassins which were rated 1.25. So a rating of 1.5 looks ridiculous to me, no matter what the "experts" claim. End of argument. :evil:
Lastly, here is another curious observation about the puzzle:

Andrew: For that reason I'll rate my solution as 1.5. However that doesn't mean that HS3 should be rated as 1.5.

Andrew: udosuk was the only one of us to spot that :applause: although with hindsight it's not that difficult.
was the most difficult move, even though it may have been obvious to intuitive solvers.

HATMAN: Andrew. Thank you for your classification as solvers as “intuitive” as opposed to “detailed”. In many ways that is the core of the problem, when I occasionally solve an assassin it is because I have grasped some odd relationship (particularly multiple innies/outies – which is one of my few strengths) , hence intuitively solved it.

There was a lot more discussion about ratings, and some discussion about grammar/style. If you are interested, please read the full thread here

Walkthrough by Nasenbaer:
That was a very interesting puzzle, HATMAN. Thanks a lot! ;clapclap;

My walkthrough isn't very nice because I used a chain.

0. Preliminaries
0a. 9(3) @ r1c1, r1c9, r9c1, r9c9 = {126|135|234} -> no 7,8,9
0b. 13(2) @ r1c3, r1c6, r3c1, r3c9, r6c1, r6c9, r9c3, r9c6 = {49|58|67} -> no 1,2,3
0c. 12(2) @ r2c5 = {39|48|57} -> no 1,2,6
0d. 10(2) @ r3c3 = {19|28|37|46} -> no 5
0e. 8(2) @r3c7 = {17|26|35} -> no 4,8,9
0f. 15(2) @r6c4 = {69|78} -> no 1,2,3,4,5
0g. 14(2) @r6c6 = {59|68} -> no 1,2,3,4,7

1. hidden killer pair {89} in 15(2) and r5c5 for D/ -> r5c5 = {89}

2. 7 locked in 10(2) for D\ -> 10(2) = {37} -> 3,7 locked for D\
2a. no 3,7 in r4c3 and r3c4
2b. no 4,5 in r2c1 and r8c9

3. 4 locked in r1c1 r2c2 r8c8 r9c9 for D\ -> either 9(3) @ r1c1 or 9(3) @r9c9 is {234}
3a. 45 on D\ (1 innie r5c5, 2 outies r2c1 r8c9) : r5c5 = r2c1 + r8c9 + 3 -> r2c1 + r8c9 = 5 or 6 -> r2c1 r8c9 = {23|33} (step 3)
3b. -> 3 locked in r2c1 r8c9 -> no 3 in r2c9 and r8c1

4. 45 on D/ (1 innie r5c5, 2 outies r2c7 r8c3) : r5c5 = r2c7 + r8c3 + 4 -> r2c7 + r8c3 = 4 or 5 -> no 5,6 in r2c7 and r8c3
4a. no 4 in r1c9 (blocked by r2c1)
4b. no 4 in r9c1 (blocked by r8c9)
4c. 4 locked in r8c2 r2c8 for D/ -> no 4 in r2c2 and r8c8
4d. -> either 9(3) @ r9c1 or 9(3) @r1c9 is {234} -> no 4 in r8c3 r2c7 (step 4)

5. 8,9 locked r5c5 r6c46 for n5
5a. 7 locked in r4c46 r6c4 for n5

6. 9(3) @r1c1 : no 2 in r1c1
6a. 9(3) @r9c9 : no 2 in r9c9
6b. 2 locked in r2c2 r8c8 for D\ -> no 2 in r8c2 r2c8

7. 45 on D/ (5 innies) : r9c1 r8c2 r5c5 r2c8 r1c9 = 22(5) = 4{1269|1359|1368|2358}
7a. 3 in r8c2 (only possible for {23458}) places 2 in r1c9 -> blocked by r8c9 -> no 3 in r8c2
7b. 3 in r2c8 (only possible for {23458}) places 2 in r9c1 -> blocked by r2c1 -> no 3 in r2c8

8. no 4 in r1c1
8a. r1c1 = 4 -> r2c23 = [32], r3c3 = 7, r4c4 = 3, r2c7 = 1 -> r8c2 = 4, r9c1 = 2, r8c3 = 3, r1c9 = 3, r2c8 = 5 -> can't place anything in 8(2) -> no 4 in r1c1
8b. -> r9c9 = 4, r8c89 = [23], r8c3 = 1 -> no 4 in r8c2, no 5,6 in r9c1
8c. -> r2c8 = 4, r2c7 = 3, r1c9 = 2, r9c1 = 3, r8c2 = 5, r2c1 = 2 -> no 5 in r1c1

9. 1,6 locked in r1c1 r2c2 for D\ and n1
9a. no 7 in r4c1 and r1c4, 14(2) in D\ = {59} -> r5c5 = 8
9b. 15(2) in D/ = {69}, 8(2) in D/ = {17}
9c. r34 : 7 locked in 10(2) and 8(2) for r34
9d. r67 : 9 locked in 15(2) and 14(2) for r67

10. cleanup: no 6 in r4c1, no 9 in r1c6, no 8 in r6c1, no 4 in r67c1, no 8 in r9c4, no 9 in r9c3467
10a. 13(2) @ r3c9 = {58} -> 5,8 locked for c9
10b. 13(2) @ r6c9 = {67} -> 6,7 locked for c9
10c. r67 : 6 locked in 15(2) and 13(2) @ r6c9 for r67
10d. -> r67c1 = [58], r6c6 = 9, r7c7 = 5, r6c49 = [67], r7c39 = [96], r9c34 = [67], r9c67 = [58], r9c2 = 2, r4c4 = 3, r3c3 = 7, r3c7 = 1, r4c6 = 7, r25c9 = [91]
10e. 13(2) @r3c1 = {49} -> 4,9 locked for c1
10f. r58c1 = [67], r7c2 = 4, r1c1 = 1, r2c2 = 6, r8c7 = 9, r79c8 = [71], r9c5 = 9
10g. r1c67 = [67], r8c5 = 6, r23c5 = [75], r34c9 = [85], r13c8 = [56], r1c34 = [49], r34c1 = [94], ...

the rest is singles

Hopefully someone can show me a solving path without a chain...

Edit: Typo corrected in 8b. Thanks, Andrew.
HATMAN's sketch of solution:
Nasenbaer
I used a chain myself, sketch of solution:

The puzzle was intended to revolve around r2c17r8c39
D/ complex HP 89 -> r5c5 = 8|9
Innies minus outies on D\ -> r5c5 - r2c1r8c9 = 3
Innies minus outies on D/ -> r5c5 - r2c7r8c3 = 4
if r5c5 = 9 r2c1r8c9 = [33]
-> r2c7r8c3 = {14} but which ever one is 4 must be part of 423
but r1c9r2c8r8c2r9c1<>3 -> r5c5 = 8
do naked and hidden pairs
r1c9r2c8r8c2r9c1 = 2... -> r2c7r8c3<>[22] -> {13}
r1c1r2c2r8c8r9c9 = {1246} -> r2c1r8c9<>{14} -> {23}
do naked and hidden pairs
r2c17r8c39 = [3132]|[2313]
but r3c3r4c4 = {37} and r3c7r4c6 = {17} -> r2c17<>[13]
r2c17r8c39 = [2313]
simple from here
How did you find the difficulty? It was very hard to create. I had the basic idea but more than ten attempts before I could get one that I could solve without the computer finding it easy.
I'll create some less taxing ones for now.
Hopefully someone will do it more neatly.
Walkthrough by Afmob:
Thanks for this interesting Killer!

Though all walkthroughs posted so far are quite short the moves themselves are advanced and therefore this Killer which probably needs chains should have a high rating (definitely not below 1.0). The most difficult move is step 2b.

HS 3 Walkthrough:

1. D\/
a) Hidden Killer pair in R5C5 for D/ since 15(2) can only have one of (89) -> R5C5 = (89)
b) 7 in D\ locked in 10(2) = {37} locked for D\, CPE: R3C4+R4C3 <> 3,7
c) Both 9(3) @ D\: R2C1+R8C9 <> 4,5 because 3 only possible there
d) Innies+Outies D\: -3 = R2C1+R8C9 - R5C5; R5C5 = (89)
-> R2C1+R8C9 = 5/6(1+1) = 3{2/3} -> CPE: R2C9+R8C1 <> 3
e) Innies+Outies D/: -4 = R2C7+R8C3 - R5C5 -> R2C7+R8C3 <> 5,6
f) Both 9(3) @ D/: R1C9+R9C1 <> 4 because {234} blocked by R2C1+R8C9 = (23)
g) 4 in D/ locked in R2C8+R8C2 -> CPE: R2C2+R8C8 <> 4
h) Both 9(3) @ D\: R1C1+R9C9 <> 2 because 4 only possible there

2. D\/ !
a) 4 in D/ must be in one of 9(3) -> one of 9(3) must be {234}
b) ! Innies+Outies D\: -3 = R2C1+R8C9 - R5C5 -> R5C5 <> 9 because
R2C1+R8C9 = {33} sees all 3 of both 9(3) @ D/
-> R2C1+R8C9 = {23} -> CPE: R2C9+R8C1 <> 2
c) R5C5 = 8
d) 14(2) = {59} locked for D\; CPE: R6C7+R6C7 <> 5,9
e) 15(2) = {69} locked for D/; CPE: R6C3+R7C4 <> 6,9

3. D\/
a) 7 in D\ locked in 8(2) @ D/ = {17} locked for D/; CPE: R3C6+R4C7 <> 1,7
b) Both 9(3) @ D\ = 2{16/34} -> 2 locked for R28 + N19
c) Grouped X-Wing (7) locked in 10(2) + 8(2) for R34
d) Both 13(2) @ R34 <> 6
e) Grouped X-Wing (9) locked in 15(2) + 14(2) for R67
f) Both 13(2) @ R67 <> 4
g) 12(2) <> 4 and R2C5 <> 5
h) Outies D/ = 4(1+1) = {13} -> CPE: R2C3+R8C7 <> 1,3

4. D\/ !
a) ! Consider candidates of R3C3 -> R2C1 <> 3
- i) R3C3 = 3
- ii) R3C3 = 7 -> R3C7 = 1 -> R2C7 = 3
b) R2C1 = 2
c) 9(3) @ N1 = {126} -> {16} locked for N1+D\
d) R8C8 = 2, R8C9 = 3, R9C9 = 4, R8C3 = 1
e) 9(3) @ N7 = {135} -> R8C2 = 5, R9C1 = 3

5. R6789+C9
a) 13(2) @ R3C9 = {58} locked for C9
b) 13(2) @ R6C9 = {67} locked for C9
c) Grouped X-Wing (6) in 15(2) + 13(2) @ R6C9 locked for R67
d) 13(2) @ R6C1 = {58} -> R7C1 = 8, R6C1 = 5
e) 13(2) @ R9C3 = {67} locked for R9
f) R6C6 = 9, R6C4 = 6, R9C4 = 7, R4C4 = 3, R3C3 = 7, R3C7 = 1, R2C7 = 3, R2C9 = 9

6. R123
a) R2C5 = 7 -> R3C5 = 5
b) 13(2) @ R1C3: R1C3 <> 8

7. Rest is singles without considering diagonals.

Rating: 1.5. I used combo analysis and a chain to crack this Killer.
Nasenbaer's comments on the earlier posts:
HATMAN wrote:
How did you find the difficulty?

The path of solution was narrow and visible so there wasn't any time lost looking for new openings. But the actual moves to crack it were not easy, so I would go with Afmob's rating of 1.5.
Quote:
It was very hard to create. I had the basic idea but more than ten attempts before I could get one that I could solve without the computer finding it easy.

Yes, I believe that. My own experience shows me that this process can be really frustrating.
Quote:
I'll create some less taxing ones for now.

Hopefully not!
Quote:
Hopefully someone will do it more neatly.

Afmob's walkthrough is already neater than mine. But still some tough moves...
Andrew's general comment on the rating:
Afmob wrote:
... and therefore this Killer which probably needs chains should have a high rating (definitely not below 1.0).

Rating: 1.5. I used combo analysis and a chain to crack this Killer.
I haven't (yet?) tried this puzzle so just a comment on Afmob's rating statement. Mike has given the general view that puzzles that require chains should be rated at least 1.5. Also I've given the general view, in another thread, that any puzzle requiring elimination solving should be rated at least 1.0. That IMHO applies even when the SS score is just below 1.0.

My impression from reading the threads, although I haven't so far attempted the puzzles, is that all of HATMAN's Human Solvables are significantly harder than 1.0.
Walkthrough by udosuk:
This rating thing is a joke. :rambo: From now on (after this post) I've decided I'm not talking about any numerical rating anymore.

Anyway, here is my walkthrough:

0. Prelim
All 9/3={126|135|234}
All 13/2={49|58|67}
12/2 @ r2c5={39|48|57}
10/2 @ r3c3={19|28|37|46}
8/2 @ r3c7={17|26|35}
15/2 @ r6c4={69|78}
14/2 @ r6c6={59|68}

1.
/567 form KHP {89} @ d/
=> r5c5 from {89}
=> 7 @ d\ locked @ 10/2 @ r3c3={37} (NP @ d\)
=> \1289 has no 3
=> 9/3 @ r1c1,r8c8 can't be [243|342|153|351]
=> r2c1,r8c9 can't have {45}

2.
Innie-outies @ d\: r5c5=r2c1+r8c9+3
=> r2c1+r8c9=5|6 from {1236}={23|33}
But 4 @ d/ locked @ /1289
=> one of 9/3 @ r1c9,r8c2 must be {234}
=> one of the 6 cells r1c9+r2c78+r8c23+r9c1 must be 3
=> r2c1+r8c9, seeing all these 6 cells, can't be {33}

3.
Now r2c1+r8c9={23}
=> r5c5=2+3+3=8 (NE @ d\/)
=> 15/2 @ r6c4 from {679}={69} (NP @ d/)
=> 14/2 @ r6c6 from {569}={59} (NP @ d\)
=> 7 @ d/ locked @ 8/2 @ r3c7={17} (NP @ d/)
Generalised x-wing: 7 @ r34,n5 locked @ \34+/34
Generalised x-wing: 9 @ r67,n5 locked @ \67+/67

4.
Outies @ d/: r2c7+r8c3=4={13|22}
But r2c1+r8c9={23} => r2c7+r8c3 can't be {22}, must be {13}
Now r3c37=[37|71]
=> r2c7+r8c3 can't be [13], must be [31]
=> r2c1+r8c9=[23]
=> 9/3 @ r1c9=[234], 9/3 @ r8c2=[513], 9/3 @ r8c8=[234]

5.
13/2 @ r3c9 from {5689}={58} (NP @ c9)
=> 13/2 @ r6c9={67} (NP @ c9)
Generalised x-wing: 6 @ r67 locked @ /67+r67c9
=> 13/2 @ r6c1 from {4578}=[58]
=> 14/2 @ \67=[95]
=> 15/2 @ /67=[69]
=> 13/2 @ r6c9=[76]
=> 13/2 @ r9c3=[67]
=> 13/2 @ r9c6 from {589}=[58]
=> 10/2 @ r3c3=[73]
=> 8/2 @ r3c7=[17]
=> r25c9=[91]
=> 13/2 @ r1c6 from {4678}=[67]
=> 9/3 @ r1c1=[126]
=> 12/2 @ r2c5=[75]
=> 13/2 @ r3c9=[85]
=> r13c8=[56]
=> 13/2 @ r1c3 from {489}=[49]
=> 13/2 @ r3c1=[94]

All naked singles from here.

The "experts" might rate my step 4 as 1.5 or 15.0 or whatever but for me the only "rating" is this:

EASY

It is EASIER than most recent assassins which were rated 1.25. So a rating of 1.5 looks ridiculous to me, no matter what the "experts" claim. End of argument. :evil:

Lastly, here is another curious observation about the puzzle:

The 12/2 @ r2c5 could be changed to 9/2, 10/2 & 11/2, each with a different unique solution. The key thing is without that cage the puzzle has 4 solutions, each with a different value @ r3c5, a pretty interesting property.

(Edited: minor change in step 4 thanks to Andrew's suggestion on page 2)
HATMAN's general observations:
Gentlemen (and Ladies observing)

I am very pleased with the argument about ratings. A(n)* Human Solvable is meant to be difficult for computers, but intuitive for human beings; therefore it should also cause difficulty for a formal rating system.

In this one you were meant to instantly see that the kernel of the problem was r2c17r8c39, but then to wonder how to attack it.

Given our history it is not surprising that Matt's solution was a much more acceptable (in a rigorous sense) version of mine - note he used X-wings which he knows are the only fishes I can do easily.

* depends on which version of English you use and your accent
Walkthrough by Andrew:
An interesting start to Ed's message, replying to HATMAN's comment about whether Human Solvable is preceded by a or an. I think I always use a, rather than an, before words starting with h. From my many years living in Britain, I get the impression that "an hotel" or "an hospital" would only be used by the "upper classes" or by those who consider themselves to be linguistically correct.

I was surprised that Ed says that "a", "an" and "the" are on the way out. I haven't come across that on BBC internet radio or in our weekly overseas edition of a British newspaper and I haven't noticed it in Canadian useage. What annoys me in north American useage are people who omit "and" when saying the year, for example "two thousand eight" and those who omit "on" before days or dates, for example "he went to Toronto Friday". The politest word I can think of for this is laziness.

Returning to topic, since I made some comments earlier in this thread I felt I ought to have a go at HS3. At first I found it hard, then I realised that it might involve different thought processes than are normally used for Assassins and other forum puzzles. Then I discovered that either/or moves were very helpful; in fact after finishing the puzzle and going through the posted walkthroughs I found that I'd used them more than was necessary. Still by missing more direct paths I think I found some interesting moves. I suspect I found HS3 harder than it really was because I have a methodical approach to solving rather than an intuitive one.

I went as far as I could without using a chain but eventually found that I couldn't go any further without using a short one. For that reason I'll rate my solution as 1.5.

However that doesn't mean that HS3 should be rated as 1.5. The key point about HATMAN's outline is the use of the hidden split cage R2C17 + R8C39 which are outies of D/. Although I often looked at the values in those cells while I was looking for the next step, it never occurred to me to treat them as a cage. It's hard to give a rating for HATMAN's breakthrough move which is based on not having 7 in both of R4C46, particularly after the discussions that ensued from the breakthrough moves that Afmob and I used for A113. However it isn't necessary to rate that move. udosuk's step 4 is simpler leading to a more direct solution for HS3. My personal feeling is that udosuk's solution should be rated 1.25, possible Easy 1.25. Whether it's easier than recent 1.25 Assassins depends to some extent if one is experienced with zero killers or is fairly new to them as I am. I found it harder than those killers but I can understand udosuk finding HS3 easier.

Anyway, having mentioned my walkthrough I'd better post it. It's fairly long because I went as far as I could before trying a chain. I've given eliminations on the diagonals; it's so easy for those of us doing manual eliminations to overlook them.

Prelims

a) R1C34 = {49/58/67}, no 1,2,3
b) R1C67 = {49/58/67}, no 1,2,3
c) R23C5 = {39/48/57}, no 1,2,6
d) R34C1 = {49/58/67}, no 1,2,3
e) R3C3 + R4C4 = {19/28/37/46}, no 5
f) R3C7 + R4C6 = {17/26/35}, no 4,8,9
g) R34C9 = {49/58/67}, no 1,2,3
h) R67C1 = {49/58/67}, no 1,2,3
i) R6C4 + R7C3 = {69/78}
j) R6C6 + R7C7 = {59/68}
k) R67C9 = {49/58/67}, no 1,2,3
l) R9C34 = {49/58/67}, no 1,2,3
m) R9C67 = {49/58/67}, no 1,2,3
n) 9(3) cage in N1 = {126/135/234}, no 7,8,9
o) 9(3) cage in N3 = {126/135/234}, no 7,8,9
p) 9(3) cage in N7 = {126/135/234}, no 7,8,9
q) 9(3) cage in N9 = {126/135/234}, no 7,8,9

1. Hidden killer pair 8,9 in R5C5 and R6C4 + R7C3 for D/ -> R5C5 = {89}

2. 7 in D\ locked in R3C3 + R4C4 -> R3C3 + R4C4 = {37}, locked for D\, CPE no 3,7 in R3C4 + R4C3

3. 9(3) cage in N1 = {126/135/234}
3a. 3 of {135/234} must be in R2C1 -> no 4,5 in R2C1
3b. 6 of {126} must be in R1C1 + R2C2 (R1C1 + R2C2 cannot be {12} because R8C8 + R9C9 cannot be {45/46}) -> no 6 in R2C1

4. 9(3) cage in N9 = {126/135/234}
4a. 3 of {135/234} must be in R8C9 -> no 4,5 in R8C9
4b. 6 of {126} must be in R8C8 + R9C9 (R8C8 + R9C9 cannot be {12} because R1C1 + R2C2 cannot be {45/46}) -> no 6 in R8C9

5. 4 in D\ locked in R1C1 + R2C2 + R8C8 + R9C9 -> either R1C1 + R2C2 or R8C8 + R9C9 must be {24} with the other pair {15/16} -> R2C1 = {23}, R8C9 = {23}

6. 9(3) cage in N3 = {126/135/234}
6a. 4 of {234} must be in R2C78 (R2C78 cannot be {23} which clashes with R2C1), no 4 in R1C9

7. 9(3) cage in N7 = {126/135/234}
7a. 4 of {234} must be in R8C23 (R8C23 cannot be {23} which clashes with R8C9), no 4 in R9C1

8. 4 in D/ locked in R2C8 + R8C2, CPE no 4 in R2C2 + R8C8
8a. 4 in D\ locked in R1C1 + R9C9 -> no 2 in R1C1 + R9C9 (step 5)

9. 4 in D/ locked in R2C8 + R8C2 -> either R1C9 + R2C8 or R8C2 + R9C1 must be {24/34} -> other pair cannot be {23} -> no 4 in R2C7 + R8C3

10. 4 in D\ locked in R1C1 + R9C9
10a. Killer triple 4,5,6 in R1C1 + R1C34 + R1C67 for R1 or in R9C9 + R34C9 + R67C9 for C9 -> no 5,6 in R1C9
10b. Killer triple 4,5,6 in R1C1 + R34C1 + R67C1 for C1 or in R9C9 + R9C34 + R9C67 for R9 -> no 5,6 in R9C1

11. Killer triple 1,2,3 in R1C9, R3C7 + R4C6 and R9C1, locked for D/
11a. R2C8 = {456} -> no 5,6 in R2C7 (step 6)
11b. R8C2 = {456} -> no 5,6 in R8C3 (step 7)

12. Hidden killer pair for 7 on both diagonals
12a. 7 in D/ locked in R3C7 + R4C6 + R6C4 + R7C3, 7 in D\ locked in R3C3 + R4C4
12b. R3C3 + R3C7 + R7C3 are common peers which must contain one 7 -> the other
7 must be locked in R4C4 + R4C6 + R6C4 for N5, no 7 in R4C5 + R5C46 + R6C5

13. 15(2) cage at R6C4 and 14(2) cage at R6C6 must have one of 8,9 in R6C46 (8,9 cannot both be in R7C37 because they are common peers of R5C5) -> killer pair 8,9 in R5C5 and R6C46, locked for N5

14. 4 in D\ locked in R1C1 + R9C9 -> either R2C12 = [32] or R8C89 = [23] -> either R2C7 or R8C3 = 1, CPE no 1 in R2C3 + R8C7
14a. 45 rule on D/ 1 innie R5C5 = 2 outies R2C7 + R8C3 + 4
14b. Min R5C5 = 8 -> min R2C7 + R8C3 = 4 = {13} (only remaining combination because it must contain 1), no 2, CPE no 3 in R2C3 + R8C7
14c. R2C7 + R8C3 = 4 -> R5C5 = 8, locked for D/ and D\, clean-up: no 4 in R23C5, no 7 in R6C4 + R7C3, no 6 in R6C6 + R7C7

15. Naked pair {69} in R6C4 + R7C3, locked for D/, CPE no 6,9 in R6C3 + R7C4, clean-up: no 2 in R3C7 + R4C6
15a. Naked pair {45} in R2C8 + R8C2, locked for D/, clean-up: no 3 in R3C7 + R4C6
15b. Naked pair {17} in R3C7 + R4C6, locked for D/, CPE no 1,7 in R3C6 + R4C7
15c. Naked pair {23} in R29C1, locked for C1
15d. Naked pair {23} in R18C9, locked for C9

16. Naked pair {59} in R6C6 + R7C7, locked for D\, CPE no 5,9 in R6C7 + R7C6

17. X-Wing for 9 in R6C4 + R7C3 and R6C6 + R7C7 for R67, no other 9 in R67, clean-up: no 4 in R67C19
17a. X-Wing for 7 in R3C3 + R4C4 and R3C7 + R4C6 for R34, no other 7 in R34, clean-up: no 5 in R2C5, no 6 in R34C19

18. 9(3) cage in N1 = {126/234}, 2 locked in R2C12, locked for R2 and N1
18a. 2 in C3 locked in R456C3, locked for N4
18b. 9(3) cage in N9 = {126/234}, 2 locked in R8C89, locked for R8 and N9
18b. 2 in C7 locked in R456C7, locked for N6

19. 9(3) cage in N3 = {135/234}, 3 locked for N3
19a. 9(3) cage in N7 = {135/234}, 3 locked for N7

20. Naked pair {23} in R1C9 + R9C1, naked pair {23} in R29C1 -> R1C9 = R2C1
20a. 3 in 9(3) cage in N3 locked in R1C9 + R2C7 -> 3 locked in R2C17, locked for R2, clean-up: no 9 in R3C5

21. Naked pair {23} in R1C9 + R9C1, naked pair {23} in R18C9 -> R9C1 = R8C9
21a. 3 in 9(3) cage in N7 locked in R8C3 + R9C1 -> 3 locked in R8C39, locked for R8

22. R6C4 + R7C3 = {69}, R6C6 + R7C7 = {59} -> one of R67C1 and R67C9 must be {58} and the other {67} -> X-Wings for 5 and 6 for R67, no other 5,6 in R67
22a. One of R67C1 and R67C9 must be {58} -> one of R34C1 and R34C9 must be {49} and because 4 in D\ locked in R1C1 + R9C9, the other of R34C1 and R34C9 must be {58}
22b. Killer single 4 in R1C1 + R34C1, locked for C1
22c. Killer single 4 in R34C9 + R9C9, locked for C9
22d. Killer pair 5,8 in R34C1 and R67C1, locked for C1
22e. Killer pair 5,8 in R34C9 and R67C9, locked for C9
[Steps 22d and 22e are because of step 22a.]

23. R2C7 cannot be 1
23a. R2C7 = 1 => R3C7 = 7, R1C9 = 3 (step 19) => R9C1 = 2 => R2C1 = 3, R3C3 = 7 clashes with R3C7
23b. R2C7 = 3, R1C9 = 2, R2C8 = 4 (step 19), both locked for D/, R9C1 = 3, R8C23 = [51], R8C9 = 3, R8C8 = 2, R9C9 = 4 (both step 18b), both locked for D\, R2C1 = 2, clean-up: no 9 in R1C6, no 9 in R34C9, no 8 in R6C1, no 9 in R9C3, no 8,9 in R9C4, no 9 in R9C67
23c. Naked pair {16} in R1C1 + R2C2, locked for N1, clean-up: no 7 in R1C4

24. Naked pair {58} in R34C9, locked for C9
24a. Naked pair {67} in R67C9, locked for C9
24b. R67C9 = {67} -> R67C1 = [58] (step 22, alternatively this comes from X-Wing for 6 in R6C4 + R7C3 and R67C9 followed by clean-up in R67C1), R6C6 + R7C7 = [95], R6C4 + R7C3 = [69], R67C9 = [76], clean-up: no 7 in R1C3, no 4 in R1C4, no 8 in R1C6, no 7 in R9C3, no 5 in R9C4, no 7,8 in R9C6
24c. R9C34 = [67], R9C67 = [58], R9C2 = 2, R8C1 = 7, R7C2 = 4, R8C7 = 9, R79C8 = [71], R9C5 = 9, R2C5 = 7, R3C5 = 5, R4C4 = 3, R3C3 = 7, R3C7 = 1, R4C6 = 7, R34C9 = [85], R25C9 = [91], R3C8 = 6, R1C78 = [75], R1C6 = 6, R1C1 = 1, R2C2 = 6, clean-up: no 8 in R1C34

and the rest is naked singles.


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PostPosted: Mon Jul 11, 2011 5:55 am 
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Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 122 by Afmob (September 2008) here
Puzzle Diagram:
Image
Note the remote cage R2378C5 = 13(4)
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3:d:k:5120:5120:5120:4355:4355:4355:7942:7942:7942:5120:4106:5131:5131:3341:5134:5134:2320:7942:5120:3859:4106:5131:3341:5134:2320:2329:7942:5659:3859:3859:4106:6175:2320:2329:2329:5155:5659:1829:1829:6175:6175:6175:3882:3882:5155:5659:2094:2094:5424:6175:4658:3123:3123:5155:6966:2094:5424:4409:3341:2875:4658:3123:4670:6966:5424:4409:4409:3341:2875:2875:4658:4670:6966:6966:6966:3915:3915:3915:4670:4670:4670:
Solution:
+-------+-------+-------+
| 1 2 9 | 4 6 7 | 8 5 3 |
| 3 4 8 | 9 1 5 | 7 2 6 |
| 5 6 7 | 3 2 8 | 1 4 9 |
+-------+-------+-------+
| 9 8 1 | 5 4 6 | 2 3 7 |
| 7 3 4 | 2 8 1 | 6 9 5 |
| 6 5 2 | 7 9 3 | 4 1 8 |
+-------+-------+-------+
| 8 1 5 | 6 3 2 | 9 7 4 |
| 2 9 3 | 8 7 4 | 5 6 1 |
| 4 7 6 | 1 5 9 | 3 8 2 |
+-------+-------+-------+
Quote:
Afmob: Apart from the remote cage that I had to put in for reasons of difficulty, this Killer came out just right. Like A116 this is also a concept Killer though there is no symbol used here but if you want you can look for hidden signs. :lol:
And don't listen to the software rating (JSudoku has problems with A122, too)! This one can be solved in a matter of seconds (though minutes is more realistic).
SS Score: 1.63. Estimated rating: 1.0.
A tougher version with perfect symmetry will be posted on Sunday.

udosuk: It seems these days the common goal is to attack the weakness of JSudoku... :ugeek:
Next week you'll get an assassin with a similar cage layout, but I'm sure it's just a pure coincidence! :alien:

Andrew: When I coloured the cages, the pattern looked like a flower.
I assume you mean that the breakthrough can be achieved in that time. Congratulations to everyone who does that! Unfortunately I didn't. It took me about 2 hours until I eventually spotted the breakthrough and then about another hour to complete my walkthrough. Still I'm not a quick solver.
I'm surprised at the SS rating. I'll rate A122 at 1.25 the way I solved it. If one gets the breakthrough move early then it's probably Hard 1.0.

Joe Casey: Thank you Afmob! A new first for me

Ed: Thanks for these very enjoyable puzzles Afmob. Feel very happy to have found the break-through tricks relatively easily for both these, though not as elegantly as Andrew for V1.5.
To celebrate, here's a full walkthrough for the V1 using Andrew's V1.5 beginning. 8-) I see now that udosuk started the V1 this way too :applause: . Sounds like Joe-Casey got the step 1 part. Great! Any chance of a walkthrough next time? :D
I'll rate the V1 as a Hard 1.0 based mostly on the first 2 steps.

udosuk's critical step:
It seems these days the common goal is to attack the weakness of JSudoku... :ugeek:

No time to write a complete walkthrough, but here is the most critical step right at the start:

Overlaps @ r19c19: r19c19=10={1234}
(r19c19=r1+r9+c1+c9-(20+17+31+22+20+27+15+18)=45*4-170=10)
Innies @ d\: \159=11
Innies @ d/: /159=15
Overlaps @ d\/: r5c5=8 (NE @ d\/)
(\159+/159=r19c19+r5c5*2, r5c5=(11+15-10)/2=8)
=> \19=11-8=3={12} (NP @ d\)
=> /19=15-8=7={34} (NP @ d/)
Innie-outies @ n7: r6c6=r7c2+r8c3+3
=> Min r6c6=1+2+3=6

Afterwards the puzzle is dead easy.

Next week you'll get an assassin with a similar cage layout, but I'm sure it's just a pure coincidence! :alien:
Walkthrough by Andrew:
Afmob wrote:
... but if you want you can look for hidden signs
When I coloured the cages, the pattern looked like a flower.

Afmob wrote:
This one can be solved in a matter of seconds (though minutes is more realistic).
I assume you mean that the breakthrough can be achieved in that time. Congratulations to everyone who does that! Unfortunately I didn't. It took me about 2 hours until I eventually spotted the breakthrough and then about another hour to complete my walkthrough. Still I'm not a quick solver.

I'm surprised at the SS rating. I'll rate A122 at 1.25 the way I solved it. If one gets the breakthrough move early then it's probably Hard 1.0.

Here is my walkthrough. I've left the breakthrough move where I found it. Some of my earlier moves were interesting and might possibly be useful for Afmob's harder version.

Prelims

a) R5C23 = {16/25/34}, no 7,8,9
b) R5C78 = {69/78}
c) 20(3) cage at R2C3 = {389/479/569/578}, no 1,2
d) 20(3) cage at R2C6 = {389/479/569/578}, no 1,2
e) 9(3) diagonal cage at R2C8 = {126/135/234}, no 7,8,9
f) 9(3) cage at R3C8 = {126/135/234}, no 7,8,9
g) R456C1 = {589/679}, 9 locked for C1 and N4
h) R456C9 = {569/578} (cannot be {389/479} which clash with R5C78), 5 locked for C9 and N6
i) 8(3) cage at R6C2 = {125/134}, CPE no 1 in R45C2, clean-up: no 6 in R5C3
j) 21(3) diagonal cage at R6C4 = {489/579/678}, no 1,2,3
k) 11(3) cage at R7C6 = {128/137/146/236/245}, no 9
l) 13(4) remote cage at R2C5 = {1237/1246/1345}, no 8,9, 1 locked for C5
m) 31(5) cage in N3 = {16789/25789/34789/35689/45679}, 9 locked for N3
n) 18(5) cage in N9 = {12348/12357/12456}, 1,2 locked for N9

1. Naked quint {56789} in R5C78 and R456C9, locked for N6

2. 45 rule on C1234 3 innies R159C4 = 7 = {124}, locked for C4
2a. Max R1C4 = 4 -> min R1C56 = 13, no 1,2,3,4 (because R1C456 cannot be 4{49})
2b. Max R9C4 = 4 -> min R9C56 = 11, no 1 in R9C6

3. 45 rule on N4 2 outies R37C2 = 7 = {25/34}/[61], no 1,7,8,9 in R3C2

4. 45 rule on N6 2 outies R37C8 = 11 = [29/38/47/56/65], no 1 in R3C8, no 3,4 in R7C8

5. 45 rule on D\ 3 innies R1C1 + R5C5 + R9C9 = 11 = {128/137/146/236/245}, no 9

6. 45 rule on N2 2 outies R2C37 = 2 innies R23C5 + 12
6a. Min R23C5 = 3 -> min R2C37 = 15, no 3,4,5, no 6 in R2C7
6b. Max R2C37 = 17 -> max R23C5 = 5, no 5,6,7

7. 45 rule on N3 4 innies R23C78 = 14 = {1238/1247/1256/1346} (cannot be {2345} because R2C7 only contains 6,7,8), 1 locked in R2C8 + R3C7, locked for N3 and D/
7a. 6 of {1256/1346} must be in R2C7 -> no 6 in R2C8 + R3C78, clean-up: no 5 in R7C8 (step 4)
7b. 1 in C9 locked in R789C9, locked for N9
7c. 9(3) cage at R3C8 = {135/234}, CPE no 3 in R6C8

8. 9(3) diagonal cage at R2C8 = {126/135}, no 4
8a. 6 of {126} must be in R4C6 -> no 2 in R4C6

9. 45 rule on N8 4 outies R23C5 + R8C37 = 11
9a. Min R23C5 = 3 -> max R8C37 = 8, no 6,7,8,9 in R8C3, no 8 in R8C7

10. 45 rule on N7 1 outie R6C4 = 2 innies R7C2 + R8C3 + 3
10a. Min R7C2 + R8C3 = 3 -> min R6C4 = 6

11. 11(3) cage at R7C6 = {137/146/236/245} (cannot be {128} because no 1,2,8) in R8C7), no 8

12. Hidden killer pair 8,9 in R78C4 and R9C56 for N8, neither can have both of 8,9 -> each must contain one of 8,9
12a. R9C456 = {159/168/249/258/348} (cannot be {267/357/456} which don’t contain 8 or 9), no 7

13. 45 rule on N1 2 innies R2C3 + R3C2 = 1 outie R4C4 + 9
13a. Min R4C4 = 3 -> min R2C3 + R3C2 = 12, no 2 in R3C2, clean-up: no 5 in R7C2 (step 3)
13b. Max R2C3 + R3C2 = 15 -> max R4C4 = 6

14. Hidden killer pair 7,8 in R456C1 and R4C23 for N4 -> R4C23 must contain one of 7,8
14a. 15(3) cage at R3C2 = {168/258/267/348/357} (cannot be {456} which doesn’t contain 7 or 8)
14b. 6 of {168/267} must be in R3C2 -> no 6 in R4C23

15. 45 rule on N5 4 innies R46C46 = 21 = {1569/2568/3459/3468/3567} (other combinations don’t contain at least two of 3,5,6)
15a. 1,2 of {1569/2568} must be in R6C6, 8,9 of {3459/3468} must be in R6C4 -> no 8,9 in R6C6

16. Hidden killer quad 1,2,3,4 in R1C4 + R23C5 + R23C46 for N2 -> R23C46 must contain 3 or 4 -> one of 20(3) cage at R2C3 + 20(3) cage at R2C6 must contain {389/479} and the other must contain {569/578} -> 5 locked in R23C46, locked for N2

17. 45 rule on N3 3 outies R4C678 = 1 innie R2C7 + 4
17a. Min R2C7 = 6 -> min R4C678 = 10, max R4C78 = 7 -> no 3 in R4C6 (because R4C678 cannot be 3{34})
17b. 9(3) diagonal cage at R2C8 (step 8) = {126/135}
17c. 5 of {135} must be in R4C6 -> no 5 in R2C8 + R3C7

18. 45 rule on N1 4 innies R23C23 = 25 = {2689/3589/3679/4579/4678} (cannot be {1789} because no 7,8,9 in R3C2), no 1

Only just spotted an important step!
19. 45 rule on R1 + R9 + C1 + C9 4 innies R19C19 = 10 = {1234} (because total of the outside cages = 170 and is the only combination because R19C19 are common peers)

20. 45 rule on D/ 3 innies R1C9 + R5C5 + R9C1 = 15 = {348} (only remaining combination, cannot be {258/267/357/456} which only contain one of 2,3,4) -> R5C5 = 8, locked for D/ and D\, R1C9 + R9C1 = {34}, locked for D/, CPE no 3,4 in R1C1 + R9C9, clean-up: no 5 in R4C6 (step 8), no 7 in R5C78
20a. Naked pair {12} in R1C1 + R9C9, locked for D\
20b. R4C6 = 6, locked for D/
20c. Naked pair {69} in R5C78, locked for R5 and N6, clean-up: no 1 in R5C3
20d. Naked triple {578} in R456C9, locked for C9
20e. Naked triple {579} in 21(3) diagonal cage at R6C4, 5 locked for N7
20f. Naked pair {12} in R2C8 + R3C7, locked for N3, clean-up: no 9 in R7C8 (step 4)

21. Naked pair {57} in R5C19, locked for R5, clean-up: no 2 in R5C23
21a. Naked pair {34} in R5C23, locked for R5 and N4

22. R5C456 = {128} = 11 -> R46C5 = 13 = {49} (only remaining combination), locked for C5 and N5 -> R6C4 = 7
22a. Naked pair {59} in R7C3 + R8C2, locked for N7
22b. Naked pair {35} in R4C4 + R6C6, locked for D\

23. 13(4) remote cage at R2C5 (prelim l) = {1237} (only remaining combination), locked for C5, 7 locked in R78C5 for N8
23a. R1C5 = 6 -> R1C46 = 11 = [29/47], no 1 in R1C4, no 8 in R1C6
23b. R9C5 = 5 -> R9C46 = 10 = [19/28], no 4 in R9C4, no 2,3,4 in R9C6
23c. R1C4 = 4 (hidden single in C4), R1C6 = 7 (step 23a), R1C9 = 3, locked for D/ -> R9C1 = 4, clean-up: no 8 in R2C7 (step 7), no 3 in R3C2 (step 3), no 8 in R7C8 (step 4)

24. R6C1 = 6 (hidden single in N4), R45C1 = [97] (prelim g), R5C9 = 5, R46C9, = [78], R46C5 = [49]
24a. 8 in R4 locked in R4C23 -> 15(3) cage at R3C2 = {168/258}, no 4, clean-up: no 3 in R7C2 (step 3)
24b. 5 of {258} must be in R3C2 -> no 5 in R4C23
24c. 5 in N4 locked in R6C23, locked for R6 -> R6C6 = 3, R4C4 = 5

25. 20(3) cage at R2C3 = {389} (step 16), 3 locked in R23C4, locked for C4 and N2
25a. Naked pair {12} in R23C5, locked for C5

26. R2C7 = 7 (hidden single in N3), clean-up: no 9 in R23C6 (prelim d)
26a. Naked pair {58} in R23C6, locked for C6 and N2 -> R9C6 = 9, R9C4 = 1 (step 23b), R9C9 = 2, locked for D\, R1C1 = 1, R5C46 = [21]
26b. Naked pair {39} in R23C4, locked for C4 and 20(3) cage -> R2C3 = 8, R23C6 = [58]

27. Naked quad {2359} in R1C23 + R23C1, locked for N1 -> R3C2 = 6, R2C2 = 4, R3C3 = 7, both locked for D\, R5C23 = [34]

28. 2 in R1 locked in R1C23, locked for N1 -> R23C1 = [35], R23C4 = [93], R2C9 = 6, R3C89 = [49], R7C8 = 7 (step 4), R78C5 = [37]

29. Naked triple {124} in R7C269, locked for R7 -> R78C1 = [82]

and the rest is naked singles.
Joe Casey's note:
Thank you Afmob! A new first for me: the first time I've found for myself one of these perimeter-sum tricks and got it to work
Indeed it's such a pure example of this, it should be archived as a demo for learners.
Though fond of sashimi in real life, I've found it hard to get the hang of these various fishy techniques in samunamu. Since different techniques suit different players, 'objective' measures of difficulty are likely to seem counter-intuitive to some players. I suspect there are too many dimensions of difficulty for a single linear scale.
Ed's walkthrough:
Thanks for these very enjoyable puzzles Afmob. Feel very happy to have found the break-through tricks relatively easily for both these, though not as elegantly as Andrew for V1.5.

To celebrate, here's a full walkthrough for the V1 using Andrew's V1.5 beginning. 8-) I see now that udosuk started the V1 this way too :applause: . Sounds like Joe-Casey got the step 1 part. Great! Any chance of a walkthrough next time? :D

I'll rate the V1 as a Hard 1.0 based mostly on the first 2 steps. Also, it kept me thinking throughout and ended up with over 30 steps before getting it to singles.

The new Aweek is about to begin, so no hiding this time. Please let me know of any errors or things that are unclear.

Walkthrough for Assassin 122

Prelims
i. 20(3)n1: no 1 or 2
ii. 13(4)n2 (remote cage r2378c5): no 8 or 9
iii. 20(3)n2: no 1 or 2
iv. 2 X 9(3)n3: no 7, 8 or 9
v. 22(3)n4: no 1...4
vi. 7(2)n4: no 7, 8 or 9
vii. 8(3)n4: no 6..9
viii. 21(3)n5: no 1, 2 or 3
ix. 20(3)n6: no 1 or 2
x. 15(2)n6 = {69/78}
xi. 11(3)n8: no 9
xii. 18(5)n9: no 9

1. "45" r1 + r9 + c1 + c9 -> 4 corners = 180 - (cage sums: 170) = 10
1a. because of diagonals, all 4 corners see each other -> they cannot have repeats = h10(4) = {1234}
1b. Common Peer Elimination (CPE): no 1, 2, 3 or 4 in r5c5 because it sees all 4 corners = h10(4)

2. "45" D\ and D/ (incuding the hidden 10(4) in step 1a)
2a. 1 innie r5c5 = (90 - cage sums )/2 (/2 because r5c5 is counted twice on both diagonals) = 16/2
2b. -> r5c5 = 8 (including placed for diagonals)

3. 15(2)n6 = {69}: both locked for r5 & n6
3a. no 1 in r5c23

4. 20(3)n6 = {578}: all locked for n6 & c9

5. naked pair (57) in r5c19: both locked for r5

6. 7(2)n4 = {34}: both locked for r5 & n4

7. naked pair (12) in r5c46: both locked for n5

8. 21(3)n5 diagonal cage = {579}: locked for D/

9. 6 in D/ only in 9(3) = {126}: 1 & 2 locked for D/ and n3
9a. r4c6 = 6

10. r6c1 = 6 (h single n4)
10a. r45c1 = [97] (last permutation)
10b. r5c9 = 5

11. naked pair {34} in r1c9 + r9c1: CPE, no 3 or 4 in r1c1 or r9c9

12. naked pair {12} in r1c1 & r9c9: both locked for D\

13. 9 in c9 only in n3: locked for n3

14. 8(3)n4 = {125}: last valid combination
14a. CPE -> no 1, 2 or 5 in r4c2
14b. r4c2 = 8
14c. r46c9 = [78]

15. 15(3)n1 = [8][61/52/25], r3c2 = (256)

16. "45" n7: r6c4 - 3 = r7c2 + r8c3
16a. -> r6c4 = 7 or 9, 2 innies = 4 or 6
16b. 5 in D/ only in n7: 5 locked for n7
16c. 2 innies = 4 or 6 = [13/24]: r8c3 = (34)

17. naked pair (34) in r7c3 with both r9c1 & r5c3: 3 & 4 both locked for n7 & c3

18. 5 required in 8(3)n4 only in r6 & n4: 5 locked for r6 & n4
18a. max r4c3 = 2 -> min r3c2 = 5 (cage sum)

19. 24(5)n5 = 12489 (no 357)
19a. r46c5 = [49]
19b. r4c4 = 5 (hsingle n5) (5 placed for D\)

20. r2c2 + r3c3 = 11 = [47] (only permution) (both placed for D\)

21. r6c46 = [73]: placed for Ds

22. naked pair (59) in n7: 9 locked for n7

23. naked pair (69) in n9: both locked for n9

24. r6c4 = 7 -> 2 innies n7 = 4 (step 16) = [13]

25. naked pair (28) in n7: locked for n7 & c1

26. r9c123 = [476]

27. naked pair (25) in n4: 2 locked for n4 & r6

28. naked pair (14) in n6: 1 locked for n6
28a. r7c8 = 7 (cage sum)

29. r4c78 = (23) -> r3c8 = 4 (cage sum)

30. r4c3 = 1 -> r3c2 = 6 (cage sum)
30a. 1 innie n1 = 8

31. r123c9 = [369]

32. 2 innies c5 = 11 = [65]

33. r23c4 = [93]
33a. r1c46 = 11 = [47]

Rest is naked.


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PostPosted: Mon Jul 11, 2011 6:13 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 122 V1.5 by Afmob (September 2008) here
Puzzle Diagram:
Image
Note that there is no remote cage. :mrgreen:
Images with "udosuk style Killer Cages" by Børge:
Image     Image
Code: Select, Copy & Paste into solver:
3x3:d:k:6912:6912:6912:3843:3843:3843:6918:6918:6918:6912:2314:3083:3083:3085:4878:4878:3088:6918:6912:5395:2314:3083:3085:4878:3088:2841:6918:2587:5395:5395:2314:5919:3088:2841:2841:2851:2587:3365:3365:5919:5919:5919:4138:4138:2851:2587:3118:3118:5168:5919:4658:3635:3635:2851:5430:3118:5168:4409:2618:3387:4658:3635:6974:5430:5168:4409:4409:2618:3387:3387:4658:6974:5430:5430:5430:3915:3915:3915:6974:6974:6974:
Solution:
+-------+-------+-------+
| 7 2 4 | 8 1 6 | 5 9 3 |
| 9 1 6 | 2 5 3 | 7 4 8 |
| 5 8 3 | 4 7 9 | 1 6 2 |
+-------+-------+-------+
| 1 4 9 | 5 8 7 | 2 3 6 |
| 6 5 8 | 3 2 1 | 9 7 4 |
| 3 7 2 | 6 9 4 | 8 5 1 |
+-------+-------+-------+
| 2 3 5 | 9 4 8 | 6 1 7 |
| 4 9 7 | 1 6 2 | 3 8 5 |
| 8 6 1 | 7 3 5 | 4 2 9 |
+-------+-------+-------+
Quote:
Afmob: This Killer has an easy start but it gets tougher as you progress. Have fun with it and don't think that (only) the cracking move from V1 will solve it for you!
SS Score: 1.84. Estimated rating: 1.25 - (Hard) 1.25.

Para: Funny fact about V2. Remove the diagonals and it scores lower in SudokuSolver.

Andrew: Thanks Afmob for a nice variant. As you say there is more work after the cracking move for V1 and its logical extension which gives the first placement.
I'll rate A122 V1.5 at 1.25. Maybe that's a bit high but I'll stick with it because ...
And on Para's comment That surprises me. It must be something about the order of doing moves. It looks to me as if it would make this puzzle much harder because ...

udosuk: I suspect if you put the priority of the ... techniques (provided they're there) higher it would score much lower with diagonals.
Meanwhile, here is my walkthrough for the "non-diagonal" version. I don't think it's that much harder than the one using diagonals (see Andrew's walkthrough above) but you need to think "instinctively/intuitively" and think outside of the way a computer thinks to appreciate my tricks.

Walkthrough by Andrew:
Thanks Afmob for a nice variant. As you say there is more work after the cracking move for V1 and its logical extension which gives the first placement.

I'll rate A122 V1.5 at 1.25. Maybe that's a bit high but I'll stick with it because of the I-O work on the four corner nonets.

Here is my walkthrough. I've given eliminations on the diagonals; it's so easy for those of us doing manual eliminations to overlook them.

Prelims

a) R23C5 = {39/48/57}, no 1,2,6
b) R5C78 = {79}, locked for R5 and N6
c) R5C23 = {58} (only remaining combination), locked for R5 and N4
d) R78C5 = {19/28/37/46}, no 5
e) 9(3) diagonal cage at R2C2 = {126/135/234}, no 7,8,9
f) 19(3) cage at R2C6 = {289/379/469/478/568}, no 1
g) 21(3) cage at R3C2 = {489/579/678}, no 1,2,3
h) 11(3) cage at R3C8 = {128/137/146/236/245}, no 9
i) R456C1 = {127/136}, no 4,9, 1 locked for C1 and N4
j) 20(3) diagonal cage at R6C4 = {389/479/569/578}, no 1,2

1. 45 rule on R1 + R9 + C1 + C9 4 innies R19C19 = 27 = {3789/4689/5679} (because total of the outside cages = 153 and the combinations apply because R19C19 are common peers), no 1,2

2. 45 rule on D/ and D\, (90-86)/2 (because the centre cell is on both diagonals) -> R5C5 = 2, locked for D/ and D\, clean-up: no 4,6 in 9(3) diagonal cage at R2C2 (prelim e), no 8 in R78C5
2a. Naked triple {135} in 9(3) diagonal cage at R2C2, locked for D\

3. 45 rule on D\, 2 remaining innies R1C1 + R9C9 = 16 = {79}, locked for D\, CPE no 7,9 in R1C9 + R9C1
3a. R1C9 + R9C1 = 27-16 = 11 = {38/56}, no 4
3b. 1 on D/ locked in 12(3) diagonal cage at R2C8 = {138/147/156}, no 9

4. 21(3) cage at R3C2 = {489/579} (cannot be {678} which clashes with R456C1), no 6
4a. 5,8 only in R3C2 -> R3C2 = {58}
4b. 9 locked in R4C23, locked for R4 and N4
4c. Naked pair {58} in R35C2, locked for C2

5. 45 rule on N4 2 outies R37C2 = 11 = [56/83]
5a. 12(3) cage at R6C2 = {237/246}, 2 locked in R6C23, locked for R6 and N4, clean-up: no 7 in R46C1 (prelim i)
5b. Naked triple {136} in R456C1, locked for C1 and N4, clean-up: no 5,8 in R1C9 (step 3a)

6. 45 rule on N2 2 outies R2C37 = 13 = {49/58/67}, no 1,2,3

7. 45 rule on N6 2 outies R37C8 = 7 = {16/25/34}

8. 45 rule on N8 2 outies R8C37 = 10 = {19/28/37/46}, no 5

9. R456C9 = {128/146/245}, (cannot be {236} which clashes with R1C9), no 3
9a. 2 of {128/245} must be in R4C9 -> no 5,8 in R4C9

10. 14(3) cage at R6C7 = {158/248/356}
10a. 2 of {248} must be in R7C8 -> no 4 in R7C8, clean-up: no 3 in R3C8 (step 7)

11. 45 rule on N9 1 outie R6C6 = 2 innies R7C8 + R8C7
11a. Max R6C6 = 8 -> Max R7C8 + R8C7 = 8, no 8,9 in R8C7, clean-up: no 1,2 in R8C3 (step 8)

12. 45 rule on N1 2 innies R2C3 + R3C2 = 1 outie R4C4 + 9
12a. R2C3 + R3C2 cannot total 10 -> no 1 in R4C4
12b. 1 on D\ locked in R2C2 + R3C3, locked for N1
12c. R4C4 = {35} -> R2C3 + R3C2 = 12,14 = [48/75/68/95], no 5,8 in R2C3, clean-up: no 5,8 in R2C7 (step 6)

13. 45 rule on R1234 3 innies R4C159 = 15 = {168/267/348/456} (cannot be {258} because R4C1 only contains 1,3,6, cannot be {357} because no 3,5,7 in R4C9)
13a. 5,7,8 only in R4C5 -> R4C5 = {578}

14. 45 rule on R6789 3 innies R6C159 = 13 = {139/148/157/346}
14a. 9 of {139} must be in R6C5, 1 of {148/157} must be in R6C1 -> no 1 in R6C5
14b. 7 of {157} must be in R6C5 -> no 5 in R6C5

15. 45 rule on C1234 3 innies R159C4 = 18 = {189/369/378/459/468/567} (cannot be {279} because R5C4 doesn’t contain 2,7,9), no 2
15a. 1 of {189} must be in R5C4 -> no 1 in R19C4

16. 2 in N1 locked in 27(5) cage = {24579/24678} (cannot be {23589} which clashes with R3C2, cannot be {23679} which clashes with R1C9), no 3, 4,7 locked for N1, clean-up: no 6,9 in R2C7 (step 6)

17. Hidden pair {13} in N1 -> R4C4 = 5 (step 2a)

18. 12(3) cage at R2C3 = {129/246} (cannot be {138/147/237} because R2C3 only contains 6,9), no 3,7,8, 2 locked in R23C4, locked for C4 and N2
18a. R2C3 = {69} -> no 6,9 in R23C4

19. 19(3) cage at R2C6 = {379/469/478} (cannot be {568} because R2C7 only contains 4,7), no 5

20. 45 rule on N3 2 innies R2C7 + R3C8 = 1 outie R4C6 + 6
20a. Max R2C7 + R3C8 = 13 -> max R4C6 = 7
20b. R2C7 + R3C8 cannot total 7 -> no 1 in R4C6
20c. Min R4C6 = 3 -> min R2C7 + R3C8 = 9, no 1 in R3C8, clean-up: no 6 in R7C8 (step 7)
20d. No 5 in R4C6 -> R2C7 + R3C8 cannot total 11 -> no 4 in R3C8, clean-up: no 3 in R7C8 (step 7)

21. 12(3) diagonal cage at R2C8 (step 3b) = {138/147/156}, 1 locked in R2C8 + R3C7, locked for N3
21a. 3,6 of {138/156} must be in R4C6 -> no 3,6 in R2C8 + R3C7

22. 1 in R1 locked in R1C456, locked for N2
22a. Naked pair {24} in R23C4, locked for C4 and N2, R2C3 = 6 (step 18), R2C7 = 7 (step 6), R5C78 = [97], R3C2 = 8 (step 12c), R5C23 = [58], R7C2 = 3 (step 5), R2C2 = 1, R3C3 = 3, clean-up: no 8,9 in R2C5, no 5 in R3C5, no 8,9 in R2C6, no 6 in R3C6 (both step 19), no 7 in R4C23 (step 4), no 4 in R6C23 (step 5a), no 7 in R8C5, no 2,4 in R8C7 (step 8)
22b. R23C6 = [39], R23C5 = [57], R4C5 = 8, clean-up: no 8 in R2C8 (step 3b), no 3 in R8C5
22c. R2C8 = 4, R3C7 + R4C6 = [17] (step 3b), 4,7 locked for D/, R23C4 = [24], R2C19 = [98], R1C1 = 7, locked for D\, R9C9 = 9, clean-up: no 9 in R8C3 (step 8)
22d. Naked triple {168} in R1C456, locked for R1 -> R1C9 = 3, locked for D/, R9C1 = 8 (step 3a)

23. 9 in N7 locked in R7C3 + R8C2, locked for D/ -> R6C3 = 6, R8C2 = 9, R7C3 = 5, R56C6 = [14], R5C4 = 3, R6C5 = 9

24. 45 rule on N3 1 remaining innie R3C8 = 6, R7C8 = 1 (step 7), R8C8 = 8, R78C7 = [63]

and the rest is naked singles.

Para wrote:
Funny fact about V2. Remove the diagonals and it scores lower in SudokuSolver.
That surprises me. It must be something about the order of doing moves. It looks to me as if it would make this puzzle much harder because it takes away the second part of step 1 and also takes away steps 2 and 3.
udosuk's walkthrough (without using diagonals):
Para wrote:
Funny fact about V2. Remove the diagonals and it scores lower in SudokuSolver.

I suspect if you put the priority of the "overlaps" techniques (provided they're there) higher it would score much lower with diagonals.

Meanwhile, here is my walkthrough for the "non-diagonal" version. I don't think it's that much harder than the one using diagonals (see Andrew's walkthrough above) but you need to think "instinctively/intuitively" and think outside of the way a computer thinks to appreciate my tricks.

1.
16/2 @ r5c7={79} (NP @ r5,n6)
13/2 @ r5c2 from {4568}={58} (NP @ r5,n4)
21/3 @ r3c2={489|579|678} must have 5|8
=> r3c2 must be from {58} (only cell for 21/3)
=> r35c2={58} (NP @ c2)
Outies @ n4: r37c2=11=[56|83]
12/3 @ r6c2 can't have both of {36}
=> r6c23 can't have {36}
=> 3 @ c1,n4 locked @ 10/3 @ r4c1={136} (NT @ c1,n4)
=> 21/3 @ r3c2=[5{79}|8{49}] (9 @ r4,n4 locked @ r4c23)

2.
Outies @ n8: r8c37=10={19|28|37|46} can't have 5
Innie-outies @ n9: r7c8+r8c7=r6c6 (max value=9)
=> r7c8+r8c7 can't have 9
=> r8c3=10-r8c7 can't be 1
20/3 @ r6c4 can't have 1|2
Innie-outies @ n7: r6c234=r8c3+8 can't be 13={247}
=> r6c4 can't be 4|7, must be from {35689}
Innie-outies @ n7: r7c2+r8c3=r6c4+4=7|9|10|12|13
=> r7c2+r8c3 can't be [32|62|38|68]
=> r8c3 can't be 2|8, must be from {34679}
Innies @ n7: r78c23=24 from {3456789} must have 5|8
(subtraction combo: {34679} can't make 24/4)
=> r7c3 must be from {58} (only cell for r78c23)
=> r57c3={58} (NP @ c3)

3.
Cell cloning: r5c23 & r57c3 both are {58}
=> r5c2=r7c3
Now r357c2=[586|853]
=> r7c23=[35|68]
But r8c37=10
=> r8c23 can't sum to 10
=> r7c23=24-r8c23 can't be [68]=14, must be [35]
=> 12/3 @ r6c2: r6c23=12-3=9={27} (NP @ r6,n4)
=> r8c23=24-3-5=16={79} (NP @ r8,n7)
=> r789c1={248} (NT @ c1,n7)
=> r123c1={579} (NT @ n1)
=> 27/5 @ r1c1: r1c23=27-5-7-9=6={24} (NP @ r1,n1)

4.
Outies @ n2: r2c37=13=[67]
=> 9/3 @ r2c2=[135]
=> 12/3 @ r2c3: r23c4=12-6=6={24} (NP @ c4,n2)
=> 19/3 @ r2c6: r23c6=19-7=12=[39]
=> 12/2 @ r2c5 from {15678}=[57]
HS @ n5: r4c6=7
=> 12/3 @ r2c8: r2c8+r3c7=12-7=5 from {1245689}=[41]
Innies @ n3: r3c8=6
Outies @ n6: r7c8=1
Outies @ n8: r8c37=10=[73]
=> 17/3 @ r7c4: r78c4=17-7=10=[91]
Innies @ n5: r6c46=10 from {13468}=[64]
=> 18/3 @ r6c6: r7c7+r8c8=18-4=14=[68]
=> 13/3 @ r7c6: r78c6=13-3=10=[82]

All naked singles from here.

Maurice will be familiar with my "cell cloning" trick, which I don't think is that easy to be composed into programs. Also, I wonder how many programs can see an innie-outies move in 2 or more different ways. I know JSudoku can't. :ugeek:


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PostPosted: Thu Jul 14, 2011 4:56 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Assassin 123 "Roulette" by udosuk (October 2008) here
Puzzle Diagram:
Image
Images with "udosuk style Killer Cages" by Børge:
Image     Image     Image
Code: Select, Copy & Paste into solver:
3x3::k:7194:7194:7194:1539:1539:1539:8731:8731:8731:7194:7194:4358:4358:3335:4104:4104:8731:8731:7194:5641:11530:4358:3335:4104:11530:5387:8731:3084:5641:5641:11530:3335:11530:5387:5387:4621:3084:5390:5390:5390:11530:2319:2319:2319:4621:3084:2320:2320:11530:4369:11530:2066:2066:4621:6684:2320:11530:3604:4369:3349:11530:2066:8221:6684:6684:3604:3604:4369:3349:3349:8221:8221:6684:6684:6684:6169:6169:6169:8221:8221:8221:
Solution:
+-------+-------+-------+
| 6 5 4 | 1 3 2 | 9 7 8 |
| 9 1 2 | 6 8 7 | 4 3 5 |
| 3 8 7 | 9 4 5 | 1 6 2 |
+-------+-------+-------+
| 4 9 5 | 2 1 6 | 7 8 3 |
| 1 6 8 | 7 5 3 | 2 4 9 |
| 7 2 3 | 4 9 8 | 5 1 6 |
+-------+-------+-------+
| 8 4 9 | 5 6 1 | 3 2 7 |
| 5 7 6 | 3 2 4 | 8 9 1 |
| 2 3 1 | 8 7 9 | 6 5 4 |
+-------+-------+-------+
Quote:
udosuk: SSR: 2.51 :!: "True" SSR: 1.26.
Bear in mind all puzzles in this thread which I have posted or will post are valid ones. Use this information smartly this puzzle can be solved without any fish or chain (corresponding to the "true" SSR). :ugeek:
Good luck! V2 will be posted when (if) this thread (ever) gets to page 2, or one week later. :alien:
(Added later: A hearty note of thanks to Ed, who helped immensely on the SSR department. Good on you mate! :mrgreen:
Although personally I'm not into the "personal rating system", I don't oppose to anyone posting a (subjective) personal rating on this puzzle or his/her own walkthrough. :geek:)

Afmob: Thanks for this week's Assassin!
It took me quite some time to find the first breakthrough move (step 2d) but after that I found the final cracking move (step 3d) in no time ...
By the way, V2 shouldn't be posted later than this mid-week since in one week the next Assassin awaits.
Rating: (Hard) 1.25.

udosuk: Nice work Afmob. ;clapclap;
I can understand every move you did in step 2, but I probably will never do it that way myself, just like you'll probably never do it my way. And that's perfectly fine because everyone has a unique style of his/her own, right? ;)
Your step 3d is exactly how I picked as my "critical step". (Although I used a totally different way to get to that stage - you guys will see how later.) So congrats for spotting it! :thumbs:
I'll post v2 no earlier than Wednesday. Hopefully some others will have made more responses about v1 and v0.9 by then.

Para: This puzzle has a special property, which i can't really explain properly. Just something i noticed. Pretty sure udosuk had it in mind. (see below for full explanation)

Ed: Thanks udosuk for a really interesting puzzle! Sounds like Para has worked out why the "true" SSR is 1.25! Don't really understand it yet!
Here's another way to solve Assassin 123. I really like this way 8-) .

Andrew: Having got stuck on A123 I went back to it about 3-4 weeks later and managed to solve it then. However I've only just gone through Afmob's and Ed's posted walkthroughs.
My solving path was much more like Afmob's than Ed's but I took much longer to find the breakthrough moves so there are some heavier moves before each of them.
Ed's breakthrough step 7 was neat!
... but I'm not going to give a rating for it since it's clearly harder than either of the earlier ones.

Walkthrough by Afmob:
Thanks for this week's Assassin!

It took me quite some time to find the first breakthrough move (step 2d) but after that I found the final cracking move (step 3d) in no time which is an interesting mix of IOU and combo analysis.

By the way, V2 shouldn't be posted later than this mid-week since in one week the next Assassin awaits.

A123 Walkthrough:

1. C456 !
a) 6(3) = {123} locked for R1+N2
b) 13(3) @ C5 must have one of (123) -> R4C5 = (123)
c) 24(3) = {789} locked for R9+N8
d) 17(3) @ C5 must have one of (789) -> R6C5 = (789)
e) 17(3) @ C5 <> 1 because R78C5 <> 7,9
f) 13(3) @ C5 <> 9 because R23C5 <> 1,3
g) Innies C5 = 15(3) must have one of (456) -> R5C5 = (456)
h) 13(3) @ C5 <> {238} since R23C5 <> 2,3
i) 17(3) @ C5 <> {278} since R78C5 <> 7,8
j) ! Hidden Killer triple (456) in 13(3) + R5C5 + 17(3) for C5
-> Each of them must have exactly one of (456)
-> 13(3) = {148/157/247} <> 3,6
-> 17(3) = {269/359/368} <> 4,7
k) Outies N2 = 7(2+1) <> 6,7,8,9

2. C123 !
a) 22(3) = 9{58/67} -> CPE: R5C2 <> 9
b) Innies+Outies C12: 7 = R1469C3 - R5C2; R5C2 = (45678)
- R5C2 <> 4 since R14C3 >= 9
- R1469C3 = 12/13/14/15(4) must have two of (123) -> R69C3 = (123)
- R1C3 <> 8,9 because R4C3 >= 5 and R4C3 <> 9 because R1C3 >= 4
c) 22(3) = 9{58/67} -> 9 locked for C2
d) ! Innies+Outies C12: 7 = R1469C3 - R5C2 -> R169C3 cannot be 7(3) = {124} (IOU @ N4)
e) Innies+Outies C12: 7 = R1469C3 - R5C2: R4C3 <> 8 since {1248} not placeable
f) 22(3): R34C2 <> 5 because R4C3 <> 8,9
g) Innies N7 = 19(3) <> 1; R78C3 <> 2,3 because R7C2 <= 6
h) Hidden Killer triple (123) in R23C3 for C3
i) Innies N1 = 17(3) = {179/269/278/359/368} <> 4
j) ! Innies+Outies C12: 7 = R1469C3 - R5C2; R5C2 = (5678)
- R1469C3 must have one of (35) since else smallest combo would be {1267} = 16(4) (using step 2d)
k) Innies N1 = 17(3) <> 5 because {359} blocked by Killer pair (35) of R1469C3

3. R123 !
a) Innies N1 = 17(3) can only have one of (123) -> R3C3 <> 1,2,3
b) Innies+Outies N1: R3C23 = R23C4 -> R3C23 and R23C4 cannot be {79} since R3C23 sees R3C4
-> 17(3) <> 1
c) 17(3) = {269/359/368} <> 4,7 because {278} blocked by Killer pair (78) of 13(3) and R2C3 = (23)
d) ! Innies+Outies N1: R3C23 = R23C4 (R3C23 sees R3C4)
- R3C23 cannot be {69} since it's only possible combo for R23C4 = 15(2)
- R23C4 cannot be {68} since it's only possible combo for R3C23 = 14(2)
-> R3C23 = 8{6/7} -> 8 locked for R3+N1
-> R23C4 = 9{5/6} -> 9 locked for C4+N2
e) Hidden Single: R4C2 = 9 @ C2, R5C9 = 9 @ R5
f) Innies N3 = 11(3) <> 9
g) Outies N2 = 7(2+1): R2C7 = (234) since R2C3 = (23) and R4C5 = (12)

4. N246
a) 21(3) @ N4 = {678} locked for R5
b) 21(3) @ N6 = {678} -> 8 locked for R4+N6
c) 16(3) can only have one of (234) -> R23C6 <> 4
d) 4 locked in R23C5 @ N2 for C5
e) R5C5 = 5
f) 9(3) @ R5 = {234} locked for R5
g) R5C1 = 1
h) 9(3) @ N4 = {234} -> 4 locked for C2 and 2 locked for R6+N4
i) 18(3) = 9{27/36/45} <> 1
j) Killer triple (234) locked in 18(3) + R5C78 for N6
k) 8(3) = {125} since R6C78 = (15) -> R7C8 = 2; {15} locked for R6+N6

5. R3
a) Naked triple (678) locked in R3C238 for R3
b) R3C6 = 5, R3C4 = 9, R2C4 = 6 -> R2C3 = 2

6. C789
a) Innies N9 = 11(2) <> 1,9
b) Hidden Single: R1C7 = 9 @ C7
c) Innies+Outies C89: 14 = R469C7 - R5C8; R5C8 = (34)
-> R469C7 = 17/18(3) = 5{48/67} -> R6C7 = 5
-> R49C7 = 12/13(2) = [76/84]
d) Hidden Single: R3C7 = 1 @ C7

7. N78
a) Hidden Single: R4C5 = 1 @ N5
b) R3C5 = 4 -> R2C5 = 8
c) R6C3 = 3, R7C2 = 4
d) 9 locked in R78C3 @ C3 for N7
e) Innies N7 = 19(3) = {469} -> 6,9 locked for C3+N7
f) 17(3) = {269} -> R6C5 = 9, R7C5 = 6, R8C5 = 2
g) 13(3) = 4{18/36} -> R8C7 = (68) and R8C6 = 4

8. Rest is singles.

Rating: (Hard) 1.25. I used Killer triples and some IOD Combo analysis.
Para's Comment on Special Property:
This puzzle has a special property, which i can't really explain properly. Just something i noticed. Pretty sure udosuk had it in mind.

Because udosuk promises it is unique, you can start by placing a 5 in R5C5.
I can't really explain it properly. But all cages 180 degree rotationally symmetric are mirrors of eachother. By mirrors i mean for example, 6(3) contains the 3 lowest numbers, and 24(3) contains the 3 highest numbers. Basically the lowest of the 2 cages is exactly the same amount from the lowest possible total, as the highest cage is from the highest possible total.
Because it is unique this means, that both cages have exactly the same configuration, one from the highest and one from the lowest numbers. So if there is a 9 in the one cage there will be a 1 in the cell 180 degree rotational from it in the other cage. This also goes for 2 and 8, 3 and 7, 4 and 6, 5 and 5. So all these cells add to 10. Which gives us 40 pairs of 10 + R5C5. Which makes R5C5 equal 5.
This puzzle has to have this property, otherwise it isn't unique.

Okay this is really badly explained, and i'm sure udosuk can explain it better. But i'm also sure he had this in mind. For this whole puzzle it means you can make all the same eliminations on both sides of the board.

I hope someone understands this. Otherwise Udosuk will explain it.
Walkthrough by Ed:
Thanks udosuk for a really interesting puzzle! Sounds like Para has worked out why the "true" SSR is 1.25! Don't really understand it yet!

Here's another way to solve Assassin 123. I really like this way 8-) . Even though it is basically creative combo work (no IOD though), it is much simpler than Afmob's really hard work in his step 2. I'll rate my way as Easy 1.50. I think Afmob's step 2 feels more like a 1.75 rating step [edit: changed my mind on that. See follow-up post] but some more subjective opinions on that would be useful feedback. Either way, we are way under the SSscore.

I really hope my way is valid! [edit: Afmob has verified it :D . He also found a really important omission. Thanks Afmob]. The puzzle is cracked after step 9. I've optimized it to make steps 5-9 as simple as possible. I originally saw a version of step 7 before step 5a.

Walkthrough for Assassin 123 "28 steps"

Prelims:
i. 22(3)n1: no 1..4
ii. 6(2)n2 = {123}
iii. 21(3)n3: no 1..3
iv. 21(3)n4: no 1..3
v. 9(3)n5: no 7..9
vi. 24(3)n8 = {789}


1. 24(3)n8 = {789}: all locked for n8 & r9
1a. max. r78c5 = {56} = 11 -> min. r6c5 = 7 (can't be 6 since 6 is already in r78c5)
1b. r6c5 = (789)
1c. => 17(3)n5 can only have 1 of 7/8/9 = {269/359/368/458/467}(no 1) = [5/6, 3/4/6..]

2. 6(3)n2 = {123}: all locked for n2 & r1
2a. min. r23c5 = {45} = 9 -> max r4c5 = 3 (can't be 4 because it's already in r23c5)
2b. => 13(3)n2 cannot have more than 1 of 1/2/3 = {148/157/247}(no 3, 6, 9) = [4/5, 4/7, 7/8..] ({256/346} blocked by 17(3)n5 step 1c)
2c. r4c5 = (12)

3. "45" c5: r159c5 = 15
3a. max. r19c5 = [39] = 12 -> min. r5c5 = 4 (can't be 3 since it's already in r1c5)
3b. min. r19c5 = [17] = 8 -> max. r5c5 = 6 (can't be 7 etc...)
3c. r5c5 = (456)

4. "45" n2: 3 outies = 7
4a. -> max r2c37 = 6 (no 6..9)

5. "45" n1: 3 innies = h17(3)n1
5a. the h17(3)n1 and 17(3)n1 have the same cage sum and r3c4 sees all of h17(3) -> each candidate in r2c3 must have a different combination in both cages, otherwise r3c4 would clash with r3c23
5b. -> {179} blocked because no other combination uses 1 and the 1 must be in r2c3!
5c. no 1 in r2c3
5d. min r2c3 + r4c5 = [21] = 3 -> max r2c7 = 4 (step 4)
5e. Max. r2c3 = 5 -> min. r3c23 = 12 (no 1, 2). Missed this crucial substep to make step 7 work!

6. "45" n7: 3 innies = 19 (no 1)
6a. max. r7c2 = 6 -> min. r78c3 = 13 (no 2, 3)

7. since r4c5 = (12) and the 45(9) cage must have 1/2 -> one of r37c7 = r4c5 = (12)
7a. -> outies n2 cannot have repeats in r2c7 & r4c5
7b. min. r2c7 + r4c5 = {12} = 3 -> max. r2c3 = 4 (no 5)
7c. max. r2c3 + r3c2 = [49] = 13 -> from h17(3)n1: min. r3c3 = 5 (can't be 4 as it's already in r2c3)

8. hidden triple {123} in c3r269
8a. r2c3 = (23)
8b. r69c3 = (123)

9. Outies n2 = 7 and remembering cannot have repeats in r2c7 & r4c5
9a. = [232/241]
9b. -> r2c3 = 2, r2c7 = (34)
9c. r23c4 = 15 = {69} ({78} clashes with 13(3)n2 step 2b)
9d. 6 & 9 both locked for n2 & c4
9e. h17(3)n1: r3c23 = 15 = {78} ({69} clashes with r3c4)
9f. 7 & 8 locked for n1 & r3

10. naked pair {45} in r3c56: both locked for n2 & r3

11. innies n3 = 11 (no 9)
11a. r3c7 = 6
11b. r23c7 = 5 = [32/41] (no 3 in r3c7)
11c. r23c4 = [69]

12. 21(3)n3: r3c8 = 6 -> r4c78 = 15 = {78}
12a. both locked for r4 & n6

13. 22(3)n1 = {589/679} = 9{..}
13a. 9 locked for r4 & n4

14. 21(3)n4: no 9 available => {678}: all locked for r5
14a. 6 locked for n4

15. naked pair {59} in r4c23: 5 locked for r4 & n4 and r3c2 = 8 (cage sum)
15a. r3c3 = 7 (7 placed for 45(9) cage also)
15b. hidden triple (678) in n4: r6c1 = (78)

16. "45" n4: 2 outies r5c4 + r7c2 = 11 = [74/83]
16a. r7c2 = (34)

17. Innies n7 = 19 and cannot have 7 = [4]{69} (only valid combo)
17. r7c2 = 4 -> r5c4 = 7 (step 16)
17a. r78c3 = {69}: both locked for n7 & c3

18. 9(3)n4: r7c2 = 4 -> r6c23 = 5 = [23]

19. 12(3)n4 = {14}[7]
19a. 1 & 4 locked for c1

20. 21(3)n4 = [687]

21. 9(3)n5 = {135/234} = {4/5..]
21a. Killer pair (45) with r5c5: both locked for r5
21b. 3 locked for r5
21b. r45c1 = [41]
21c. 9(3)n5 = {234}: all locked for r5

22. r9c34 = [18]
22a. hidden pair {89} in n5 in r6c56 -> r6c6c (89)

23. r4c6 = 6 (hsingle n5)(placed for 45(9))

24. r78c3 = [96] (9 placed for 45(9))

25. r6c56 = [98] (8 place for 45(9))

26. 1 outie n8, r8c7 = 8

27. r5c59 = [59] (5 placed for 45(9))

28. 17(3)n5: r6c5 = 9 -> r78c5 = 8 = [62]

Rest unfolds easily!

Fingers crossed!
Discussion/Explanation of the Special Property and udosuk's comments on walkthroughs:
udosuk wrote:
Guys, gone through all your nice walkthroughs. Excellent work all around! :applause:

Andrew, thanks for your considerate decision to limit your combo crunching. :thumbs: I mostly have problems with deep crunching of 4- or 5-cell groups so for this puzzle (mostly 3-cell groups) there shouldn't be any issues anyway. However you don't need to change your style for me, just go ahead with your most comfortable moves for all future puzzles! ;) Note your walkthrough also demonstrates most the property as observed by Para, with most moves applied coming in "pairs".

Ed, really ingenious breakthrough move in your step 7! :applause: :applause: :applause:

This makes your walkthrough the most elegant one posted so far. The trick I used myself (as observed by Para), despite being simpler, has to make use of the uniqueness property so your approach should be the better one. Hopefully you can find a similar brilliant cracking move in v2 which I'll release some time on or after tomorrow. :mrgreen:

Para, I'll do a pictorial explanation about that property later. Everyone should be able to see it clearly. :alien:

azpaull wrote:
Wowwwww. I've been struggling with v0.9, and thought I'd drop by to see what the discussion was. I suspected exactly what you described, para (I even had also used the term "rotationally symmetric" yesterday); but, I hadn't been able to figure it out. I'm looking forward to your pictorial explanation, udo.

Thanks again, to all of you.

Andrew wrote:
Since I'm stuck again on A123 I had a peek at Para's hidden message.

I can see what he's saying but I don't know that I'm convinced by the whole argument or that R5C5 is fixed immediately. It seems to me very much like UR, which some people use but by doing so avoid solving the whole puzzle.

I hope there are reasonable ways of solving A123 without needing to use this shortcut.

There are statements that the "true SS score" for this puzzle is about 1.25. Are they on the assumption that R5C5 is fixed immediately and working from there? I don't know what the rating of Para's suggestion is but I'd be surprised if it isn't considerably higher than 1.25. Of course if, in future, it becomes an established and well known technique then it might then get a lower rating but that wouldn't be appropriate at the moment with it not being widely known or understood.

Maybe udosuk will be able to show us how "rotationally symmetric", which azpaull uses in his message, leads logically to what Para said without relying in any way on uniqueness. That would certainly help it to become an established and well known technique.

BTW I did spot one interesting feature in the cage diagram. The outer cages total 180 making R19C19 = R28C28 but this doesn't seem to lead to anything useful.

Para wrote:
Don't see it as a technique. It's just an observation, which completely depends on the cage totals. I mean it only happens if the person who made the puzzle designs it that way on purpose. This also means you have to have a certain looking solution. That is also why udosuk has a puzzle with the same solution and a different cage pattern instead of a different solution in the same cage pattern. (well this is my assumption of course).

udosuk wrote:
Okay, here is a brief graphical explanation:

Firstly, in the solution grid, label each cell value A11 to A99 as the following:

Image

Now, subtract each of these values from 10, and rotate the whole grid 180 degrees:

Image

We know the cage structure in the original grid also applies in this new grid, because if two cells belong in the same cage in the original grid, they must also belong to the same cage in the new grid.

Now what do you notice about the cage sums in the new grid?

Yes, we can work them all out algebraically. For example, the cage sum at r1c456 in the new grid = (10-A96)+(10-A95)+(10-A94) = 30-(A94+A95+A96) = 30-24 = 6.

(We know A94+A95+A96 = 24 from cage at r9c456 in the original grid.)

So, working all the new cage sums out, we obtain the following grid:

Image

Just take a look at all the cage sums, they are the exact copy of the original puzzle! :idea:

Because the original puzzle is a valid puzzle, we can immediately determine all the 81 cell values of this new grid. For example:

10-A99 = A11
10-A98 = A12
...

In particular:
10-A55 = A55
=> 2 A55 = 10
=> A55 = 5

Also this is an established technique in Vanilla Sudoku. Check out this thread. :ugeek:

Ed wrote:
OK, hmm. Thanks for all the work to make up the pics udosuk. I don't really understand it (or the link) yet... from what I can work out, as Para says and udosuk has implied, more like an interesting feature rather than a technique as such.

Don't know if this is related or not, but "45" rule on r1 + r9 + c1 + c9 = r2c2 + r2c8 + r8c2 + r8c8 = 180. Combined with the 45(9) cage that runs along the diagonals, does this imply that 1-9 cannot repeat on the diagonals? Does this make it an implied Diagonals puzzle? I can't see that it does. Don't see how this would help get r5c5 placed for example. (Note: it is "diagonal" since V2X has the same solution).

My main reason for this post is to follow up to my earlier rating of Afmob's step 2 in his Assassin 123 Walkthrough. By PM, Afmob asked, quite rightly, what I found difficult in his step 2. In actual fact, it's not as difficult as I first thought. :cheers: So, I'll withdraw that rating. Though steps 2 e & j still seem above the average Weekly Assassin and I never found them. Which is also why I called my solution (not udosuk's trick above!) an Easy 1.50. My key move is more like a one-off for this puzzle and could easily be missed. Of course, if most people found it, then it's average!! And my rating's wrong. But, it is finding this sort of trick that keeps me comin' back for more :) (though won't be trying A123 V2: lots of T&E according to the SS log). Bring it on Frank... A nice Canadian Thanksgiving puzzle perhaps...?

udosuk wrote:
Ed wrote:
OK, hmm. Thanks for all the work to make up the pics udosuk. I don't really understand it (or the link) yet... from what I can work out, as Para says and udosuk has implied, more like an interesting feature rather than a technique as such.
Hopefully others can see it better. I'm not sure if I can explain it more clearly. :bouncy:


Ed wrote:
Don't know if this is related or not, but "45" rule on r1 + r9 + c1 + c9 = r2c2 + r2c8 + r8c2 + r8c8 = 180. Combined with the 45(9) cage that runs along the diagonals, does this imply that 1-9 cannot repeat on the diagonals? Does this make it an implied Diagonals puzzle? I can't see that it does. Don't see how this would help get r5c5 placed for example. (Note: it is "diagonal" since V2X has the same solution).
With my symmetry property, r28c28 always sums to 20. And no it doesn't (directly) imply the diagonals must have no repeat. You only know r19c19 & r28c28 both consist of two pairs from {19|28|37|46|55}. Initially you can't even eliminate the possibilities of say, \19=\28. I think you're on the totally wrong track if you want to "see" the fact r5c5=5 from the X property.


Ed wrote:
My main reason for this post is to follow up to my earlier rating of Afmob's step 2 in his Assassin 123 Walkthrough. By PM, Afmob asked, quite rightly, what I found difficult in his step 2. In actual fact, it's not as difficult as I first thought. :cheers: So, I'll withdraw that rating. Though steps 2 e & j still seem above the average Weekly Assassin and I never found them.
I already mentioned, all the moves in step 2 can be understood (fairly) easily, but it's a matter of whether you actually want to apply them. I think (at least for me) elegancy is an important factor. For example, the relationship 7 = R1469C3 - R5C2 (and the subsequent crunching deductions) feels not as elegant as your establishment one of r37c7 = r4c5 (I dubbed this move "cell cloning" although you've outdone me in using a "one of two cells cloning" :thumbs:). Of course, it's all a matter of personal taste. I always think cracking moves involving fewer cells are more elegant than ones involving more cells. In this case the "cells involved" ratio is 5:3. However, if for some the process of sitting in front of the computer for a few minutes to crunch on a 5-cell group gives them the hots, I have nothing against that. :ugeek:


Ed wrote:
Which is also why I called my solution (not udosuk's trick above!) an Easy 1.50. My key move is more like a one-off for this puzzle and could easily be missed. Of course, if most people found it, then it's average!! And my rating's wrong. But, it is finding this sort of trick that keeps me comin' back for more :) (though won't be trying A123 V2: lots of T&E according to the SS log).
I decided I'm not commenting on any personal rating anymore myself, but I think you need to make up your mind here. Which do you want to choose as your measurement, easiness/difficulty of the spotting of cracking moves, or simplicity/elegancy of the logic involved? :geek:

Andrew wrote:
Thanks udosuk for a very clear explanation about how Rotational Symmetry works for this puzzle. It clearly only works for the very few cases where the cage pattern is symmetrical about the centre cell and all the cage totals are complementary except, of course, the cage containing the centre cell. I take this to mean that the totals of a complementary pair of cages, for example 17(3) cage at R2C3 and 13(3) cage at R7C6, must total 10 times the number of cells in each of those cages. This brief description appears to be a way that it can be recognised by a software solver, if the programmer considers it worth including for the very occasional cases where it appears; however it could well be a situation that will appear more in the future. Edit. However after looking at Gurth's Symmetrical Placement on the linked thread, maybe it's not as simple as I thought above, unless there is something specific for Killers that makes what I've said above correct. udosuk, please tell me whether I was right or wrong. In any case I'll leave my above statement as it is, even if it is wrong.

Unlike Ed, I've got no problem in accepting it as a technique. It's clearly logical. udosuk has told us that it's an established technique for Vanilla Sudoku. My first reaction to that was that it is an advanced technique but then I realised that most/all advanced techniques for Vanilla Sudoku start with a "what if" assumption whereas Rotational Symmetry is completely logical.

udosuk wrote:
Andrew wrote:
Thanks udosuk for a very clear explanation about how Rotational Symmetry works for this puzzle. It clearly only works for the very few cases where the cage pattern is symmetrical about the centre cell and all the cage totals are complementary except, of course, the cage containing the centre cell.
Thanks for confirming that I've explained it at least clear enough for some. :alien: You're right that this technique only works for the few puzzles that display symmetry. However if the property is there it doesn't hurt to use it. For example, for a puzzle that isn't specified as an X puzzle, if you can somehow deduce that one or both of the diagonals must have no repeat, I don't see any problem to apply diagonal techniques (e.g. CPE) on your solving path. However, most randomly generated puzzles don't have non-repeating diagonals, just as most randomly generated puzzles don't have symmetry. If one decides to only use techniques that can apply to *all* puzzles, that's his/her personal choice but I'd think such a mindset is too narrow-sighted. :ugeek:


Andrew wrote:
I take this to mean that the totals of a complementary pair of cages, for example 17(3) cage at R2C3 and 13(3) cage at R7C6, must total 10 times the number of cells in each of those cages.
Spot on! ;)


Andrew wrote:
This brief description appears to be a way that it can be recognised by a software solver, if the programmer considers it worth including for the very occasional cases where it appears; however it could well be a situation that will appear more in the future.
Yes, of course it's not difficult at all to be implemented into a program. It's up to the programmers (e.g. Richard, JC) if they want to implement it as it is rarely occured in randomly generated puzzles. As a matter of fact there are 5 main categories of symmetry (aka automorphism) in Sudoku puzzles (rotation is just one of them) and they each have quite different characteristics. There is a lot of info available in the sudoku.com forums but unless someone shows more interest I'll spare you all the (quite intimidating) details. :geek:


Andrew wrote:
Edit. However after looking at Gurth's Symmetrical Placement on the linked thread, maybe it's not as simple as I thought above, unless there is something specific for Killers that makes what I've said above correct. udosuk, please tell me whether I was right or wrong. In any case I'll leave my above statement as it is, even if it is wrong.
What you said above is perfectly correct. However, in vanilla puzzles you can freely permute the 9 digits so the "opposite cells must sum to 10" constraint is no longer necessary. For example, you can have {13|25|49|68|77} as the opposite pairs. In this case r5c5 must be the self-mapping digit, i.e. 7. But in killer puzzles I don't see any obvious pairing other than {19|28|37|46|55} (I call this the "orthodox" pairing) to easily transfer the symmetry to the cage sums.


Andrew wrote:
Unlike Ed, I've got no problem in accepting it as a technique. It's clearly logical. udosuk has told us that it's an established technique for Vanilla Sudoku. My first reaction to that was that it is an advanced technique but then I realised that most/all advanced techniques for Vanilla Sudoku start with a "what if" assumption whereas Rotational Symmetry is completely logical.
Thanks for accepting the technique! :salute: Everyone has his/her own taste and limitation of what techniques/methods he/she would choose to use. Some people simply refuse to use any techniques related to the uniqueness/validity of the puzzle and consider it a disgrace to use the extra help from this information. Some people don't use elimination solving and only solve the puzzle from a completely blank grid. Some people even refuse to use pencilmarks and only do it using pen and paper. But there are too many puzzles out there to suit everyone's own scope of acceptable puzzles so if it happens to be out of your scope just feel free to ignore it. :ok: Life is too short to waste time on every puzzle out there! ;)

Andrew wrote:
Earlier in this thread I wrote:
Thanks udosuk for a very clear explanation about how Rotational Symmetry works for this puzzle. It clearly only works for the very few cases where the cage pattern is symmetrical about the centre cell and all the cage totals are complementary except, of course, the cage containing the centre cell. ...

Unlike Ed, I've got no problem in accepting it as a technique. It's clearly logical. udosuk has told us that it's an established technique for Vanilla Sudoku. My first reaction to that was that it is an advanced technique but then I realised that most/all advanced techniques for Vanilla Sudoku start with a "what if" assumption whereas Rotational Symmetry is completely logical.

Thinking further about that I'd like to qualify my acceptance slightly.

If the uniqueness is as a result of the complete rotational symmetry then I think I would find this technique acceptable although I'm not sure whether I'd use it. My A123 walkthrough in the previous message didn't use it. However if the uniqueness is only because a software solver has checked and said that a puzzle has a unique solution, then I would consider it to be like a UR and definitely wouldn't use it.

Can udosuk, or anyone else, please tell us whether the uniqueness of A123 and the other puzzles in this thread is as a direct result of their rotational symmetry?

Para wrote:
Nah. You can create puzzles which are not unique, which have a rotational cage pattern.
Walkthrough by Andrew:
Having got stuck on A123 I went back to it about 3-4 weeks later and managed to solve it then. However I've only just gone through Afmob's and Ed's posted walkthroughs.

My solving path was much more like Afmob's than Ed's but I took much longer to find the breakthrough moves so there are some heavier moves before each of them.

Ed's breakthrough step 7 was neat! I'd quickly spotted that one of R37C7 must be the same as R4C5 but hadn't made the mental leap that R2C7 and R4C5 must therefore be different. I felt that Ed's walkthrough was easier because of that step although I've no idea what rating ought to be given to step 7. On a human level it's an easy step but only if one spots it.

I decided to post my walkthrough because I think some of the steps are interesting, for example I got more out of the IOUs in steps 6 to 9, but I'm not going to give a rating for it since it's clearly harder than either of the earlier ones.

Here is my walkthrough for A123. Steps 15 to 18 would probably have been unnecessary if I'd spotted the common cells (steps 19 and 20) after step 14. Even after that I got into some harder work until step 29 when I eventually revisited step 8, this time writing out the combinations and spotting that they provided a clash that gave the key breakthrough.

Prelims

a) R1C456 = {123}, locked for R1 and N2
b) 22(3) cage at R3C2 = {589/679}, CPE no 9 in R56C2
c) 21(3) cage at R3C8 = {489/579/678}, no 1,2,3
d) R5C234 = {489/579/678}, no 1,2,3
e) R5C678 = {126/135/234}, no 7,8,9
f) 9(3) cage at R6C2 = {126/135/234}, no 7,8,9
g) 8(3) cage at R6C7 = {125/134}, CPE no 1 in R5C8
h) R9C456 = {789}, locked for R9 and N8
i) 34(6) cage in N3 must contain 9, locked for N3
j) 26(6) cage in N7 must contain 1, locked for N7

1. Min R23C5 = {45} = 9 -> max R4C5 = 3 because R234 cannot be {45}4
1a. Max R23C5 = 12, no 9
1b. 45 rule on N2 3(2+1) outies R2C37 + R4C5 = 7, R2C37 = {12345}

2. Max R78C5 = {56} = 11 -> min R6C5 = 7 because R678C5 cannot be 6{56}
2a. Min R78C5 = 8, no 1
2b. 45 rule on N8 3(2+1) outies R6C5 + R8C37 = 23 -> min R8C37 = 14, no 1,2,3,4

3. 45 rule on C5 3 innies R159C5 = 15
3a. All combinations must contain at least one of 4,5,6 -> R5C5 = {456}
3b. Killer triple 4,5,6 in R5C234, R5C5 and R5C678, locked for R5

4. Hidden killer triple 7,8,9 in R23C5, R6C5 and R9C5 for C5 -> R23C5 must contain one of 7,8
4a. R234C5 cannot contain 6 and one of 7,8 -> no 6 in R23C5
4b. Min R23C5 containing one of 7,8 = 11 -> max R4C5 = 2

5. Hidden killer triple 1,2,3 in R1C5, R4C5 and R78C5 for C5 -> R78C5 must contain one of 2,3
5a. R678C5 cannot contain 4 and one of 2,3 -> no 4 in R78C5
5b. Max R78C5 containing one of 2,3 = 9 -> min R6C5 = 8

6. 45 rule on C89 4 outies R1469C7 = 1 innie R5C8 + 23
6a. Min R5C8 = 2 -> min R1469C7 = 25
6b. Max R14C7 = 17 -> min R69C7 = 8, no 1 in R6C7, no 1,2 in R9C7
6c. All combinations for R1469C7 totalling at least 25 need at least two of 7,8,9 -> R14C7 = {789}
6d. IOU R149C7 cannot total 23 and no 7 in R9C7 -> max R149C7 = 22 -> R6C7 must be greater than R5C8 -> max R5C8 = 4, min R6C7 = 3

[I only spotted steps 7 and 9, which are logical extensions of steps 6d and 8e, after step 14. They have been moved here for clarity.]

7. R5C8 less than R6C7 (step 6d) -> R567C8 = 6,7 = {123/124}, no 5, 1,2 locked for C8

8. 45 rule on C12 4 outies R1469C3 = 1 innie R5C2 + 7
8a. IOU R169C3 cannot total 7 and no 3 in R1C3 -> min R169C3 = 8
8b. Min R1469C3 = 13 -> min R5C2 = 6
8c. Max R5C2 = 8 -> max R1469C3 = 15, min R69C3 = 3 -> max R14C3 = 12, no 8,9 in R1C3, no 9 in R4C3
8d. Max R1469C3 = 15, all combinations for R1469C3 totalling 15 or less need at least two of 1,2,3 -> R69C3 = {123}
8e. Min R169C3 = 8 -> R5C2 must be greater than R4C3 -> no 8 in R4C3

9. R5C2 greater than R4C3 (step 8e) -> R345C2 = 23,24 = {689/789}, no 5, 8,9 locked for C2

10. 45 rule on N1 3 innies R2C3 + R3C23 = 17
10a. Max R2C3 + R3C2 = 14 -> min R3C3 = 3

11. 45 rule on N3 3 innies R2C7 + R3C78 = 11
11a. Min R2C7 + R3C8 = 5 -> max R3C7 = 6

12. 45 rule on N7 3 innies R7C23 + R8C3 = 19
12a. Max R7C2 + R8C3 = 15 -> min R7C3 = 4

13. 45 rule on N9 3 innies R7C78 + R8C7 = 13
13a. Min R7C8 + R8C7 = 6 -> max R7C7 = 7
13b. Hidden killer triple 7,8,9 in R1C7, R4C7 and R78C7 for C7 -> R78C7 must contain one of 7,8,9
13c. R7C78 + R8C7 cannot contain 6 and one of 7,8,9 -> no 6 in R78C7

14. Hidden killer triple 1,2,3 in R23C3, R6C3 and R9C3 for C3 -> R23C3 must contain one of 1,2,3
14a. R2C3 + R3C23 cannot contain 4 and one of 1,2,3 -> no 4 in R23C3

15. 6,9 in N2 locked in R23C46
15a. 45 rule on N2 4 innies R23C46 = 1 outie R4C5 + 26
15b. R4C5 = {12} -> R23C46 = 27,28 = {4689/5679/5689}
15c. R23C4 cannot be {79} (because 16(3) cage at R2C6 cannot be {556}) -> no 1 in R2C3
15d. Min R2C3 + R4C5 = 3 -> no 5 in R2C7 (step 1b)
15e. R23C4 and R23C6 cannot be {78} (which clashes with R23C5) or {79} -> R23C4 and R23C6 must each contain one of 7,8,9

16. R23C46 = {4689/5679/5689} (step 15b)
16a. R23C4 cannot total 1 more than R23C6 because that would make R2C3 equal R2C7
16b. R23C46 = {4689} can only be R23C4 = {48} and R23C6 = {69} (R23C4 cannot be {49} because 17(3) cage at R2C3 cannot be {449}, R23C6 cannot be {48} because 16(3) cage at R2C6 cannot be {448}, R23C6 cannot be {49} because R23C4 = {68} would be 1 more than R23C6) -> no 4 in R23C6
16c. R23C46 = {5679} can only be R23C4 = {69} and R23C6 = {57} (R23C4 cannot be {57} because 17(3) cage at R2C3 cannot be {557}, R23C4 cannot be {59} because it would be 1 more than R23C6 = {67}, R23C4 cannot be {67} because R2C3 doesn’t contain 4) -> no 7 in R23C4
16d. R23C46 = {5689} can be R23C4 = {59/68/69} (R23C4 cannot be {58} because no 4 in R2C3), R23C6 = {58/59/68}

17. 1,4 in N8 locked in R78C46
17a. 45 rule on N8 4 innies R78C46 = 1 outie R6C5 + 4
17b. R6C5 = {89} -> R78C46 = 12,13 = {1245/1246/1345}
17c. R78C6 cannot be {13} (because 14(3) cage at R7C4 cannot be {455}) -> no 9 in R8C7
17d. Max R6C5 + R8C7 = 17 -> min R8C3 = 6 (step 2b)
17e. R78C4 and R78C6 cannot be {12}, {13} or {23} which clashes with R78C5 -> R78C4 and R78C6 must each contain one of 1,2,3

18. R78C46 = {1245/1246/1345} (step 17b)
18a. R78C4 cannot total 1 more than R78C6 because that would make R8C3 equal R8C7
18b. R78C46 = {1246} can only be R78C4 = {14} and R78C6 = {26} (R78C4 cannot be {16} because it would be 1 more than R78C6 = {24}, R78C4 cannot be {24} because 13(3) cage at R7C6 cannot be {166} even if there was still a 6 in R8C7, R78C4 cannot be {26} because 14(3) cage at R78C4 cannot be {266}) -> no 6 in R78C4
18c. R78C46 = {1345} can only be R78C4 = {35} and R78C6 = {14} (R78C4 cannot be {14} because 13(3) cage at R7C6 cannot be {355}, R78C4 cannot be {34} which would be 1 more than R78C6 = {15}, R78C6 cannot be {34} because no 6 in R8C7 -> R78C4 cannot be {15}) -> no 3 in R78C6
18d. R78C46 = {1245} can be R78C6 = {14/15/24} (cannot be {25} because no 6 in R8C7), R78C4 = {15/24/25}

[I’ve just spotted that the total of the outer cages is 180 which means that R19C19 = R28C28 but it’s no help at this stage and might only help in the final mop-up, if at all.]

19. R2C3 + R3C23 = 17 (step 10) and 17(3) cage at R2C3 share a common cell at R2C3
19a. R23C4 can only be {48/59/68/69} (steps 16b to 16d)
19b. R2C3 + R3C23 = {278/359/368} (cannot be {269} which clashes with 17(3) cage at R2C3 = {269})
19c. 9 of {359} must be in R3C2 -> no 9 in R3C3
19d. R3C23 = {68/78}/[93/95]

20. R7C78 + R8C7 = 13 (step 13) and 13(3) cage at R7C6 share a common cell at R8C7
20a. R78C6 can only be {14/15/24/26} (steps 18b to 18d)
20b. R7C78 + R8C7 = {157/238/247} (cannot be {148} which clashes with 13(3) cage at R7C6 = {148})
20c. 1 of {157} must be in R7C8 -> no 1 in R7C7
20d. R7C78 = {23/24}/[51/71]

21. R2C3 + R3C23 = 17 (step 10), R2C7 + R3C78 = 11 (step 11) -> R3C23 cannot be 6 more than R3C78 because that would make R2C3 equal to R2C7, also R3C78 cannot be greater than 10
21a. Hidden killer triple 1,2,3 in R3C1, R3C37 and R3C9 for R3 -> R3C37 must contain at least one of 1,2,3
21b. 45 rule on R123 3 innies R3C2378 = 1 outie R4C5 + 21, R4C5 = {12} -> R3C2378 = 22,23
21c. R3C2378 = 22 = {1678/2389/2578/3469} (cannot be {1489/2479/4567} which clash with R3C23, cannot be {1579/2569/3568} because R3C23 would be 6 more than R3C78, cannot be {3478} because R2C7 + R3C78 cannot be [434])
21d. R3C2378 = 23 = {1589/2678/3569} (cannot be {1679/2489} which clash with R3C23, cannot be {2579} because R2C7 + R3C78 cannot be [227], cannot be {3479} because R3C78 would be greater than 10, cannot be {3578} because R2C7 + R3C78 cannot be [335], cannot be {4568} which doesn’t contain any of 1,2,3)
21e. R3C23 (step 19d) = {78}/[93/95] ({1678} cannot be {68}[17] because R23C23 would be 6 more than R3C78, {2678} cannot be {68}[27] because R2C7 + R3C78 cannot be [227]), no 6
21f. R3C78 = [16/18/25/26/28/36/46/64] ({1678} cannot be {68}[17] because R23C23 would be 6 more than R3C78, {2678} cannot be {68}[27] because R2C7 + R3C78 cannot be [227], {3569} cannot be [93]{56} because R3C78 would be greater than 10), no 5 in R3C7, no 7 in R3C8

22. 21(3) cage at R3C8 = {489/579/678}
22a. 5 of {579} must be in R3C8 -> no 5 in R4C8

23. R7C23 + R8C3 = 19 (step 12), R7C78 + R8C7 = 13 -> R7C23 cannot be 6 more than R7C78 because that would make R8C3 equal to R8C7, also R7C23 must be at least 10
23a. Hidden killer triple 7,8,9 in R7C1, R7C37 and R7C9 for R7 -> R7C37 must contain at least one of 7,8,9
23b. 45 rule on R789 4 innies R7C2378 = 1 outie R6C5 + 9, R6C5 = {89} -> R7C2378 = 17,18
23c. R7C2378 = 17 = {1259/1457/2348} (cannot be {1268/1349} which clash with R7C78, cannot be {1358} because R7C23 + R8C3 cannot be [388], cannot be {1367} because R7C23 would be less than 10, cannot be {2357} because R7C23 + R8C3 cannot be [577], cannot be {2456} which doesn’t contain any of 7,8,9)
23d. R7C2378 = 18 = {1278/1467/2349/2358} (cannot be {1269/1368} which clash with R7C78, cannot be {1359/1458/2457} because R7C23 would be 6 more than R7C78, cannot be {2367} because R7C23 + R8C3 cannot be [676], cannot be {3456} which doesn’t contain any of 7,8,9)
23e. R7C23 = [28/29/46/47/48/49/58/64] ({1457} cannot be {45}[71] because R7C23 would be less than 10, {2348} cannot be [38]{24} because R7C23 + R8C3 cannot be [388] , {2349} cannot be [39]{24} because R7C23 would be 6 more than R7C78), no 3 in R7C2, no 5 in R7C3
23f. R7C78 (step 20d) = {23}/[51/71] ({2348} cannot be [38]{24} because R7C23 + R8C3 cannot be [388], {2349} cannot be [39]{24} because R7C23 would be 6 more than R7C78), no 4

24. 9(3) cage at R6C2 = {126/135/234}
24a. 5 of {135} must be in R7C2 -> no 5 in R6C2

25. R2C3 + R3C23 = 17 and 17(3) cage at R2C3 share a common cell at R2C3 (step 19)
25a. R3C23 (step 21e) = {78}/[93/95]
25b. R2C3 + R3C23 (step 19b) = {278/359}
25c. R23C4 (step 19a) = {48/68/69} (cannot be {59} which clashes with R3C23 = [95]), no 5

26. R7C78 + R8C7 = 13 and 13(3) cage at R7C6 share a common cell at R8C7 (step 20a)
26a. R7C78 (step 23f) = {23}/[51/71]
26b. R7C78 + R8C7 (step 20b) = {157/238}
26c. R78C6 (step 20a) = {14/24/26} (cannot be {15} which clashes with R7C78 = [51]), no 5

27. R456C1 = {129/147/237/246/345} (cannot be {138} because R6C23 cannot be [42/62] since no 1,3 in R7C2, cannot be {156} because R4C23 + R5C2 cannot be [97]8 since no 6 in R3C2), no 8
27a. Cannot be {237} => R6C23 = [61] (R6C23 cannot be [41] because 8(3) cage cannot be [414]) => R45C3 = [54] => R45C2 = [98], R4C23 = [59] => R3C2 = 8 clashes with R5C2
27b. -> R456C1 = {129/147/246/345}

28. R456C9 = {189/369/378/468/567} (cannot be {279} because R4C78 cannot be {48/68} since no 7,9 in R3C8, cannot be {459} because R6C78 cannot be [31/32] since no 4 in R7C8 and 8(3) cage cannot be [323]), no 2
28a. Cannot be {378} => R4C78 = [94] (R4C78 cannot be [96] because 21(3) cage at R3C8 cannot be [696]) => R56C7 = [65] => R5C8 = 2 clashes with R67C8 = [12]
28b. -> R456C9 = {189/369/468/567}

29. R1469C3 = R5C2 + 7 (step 8), min R169C3 = 8 (step 8a)
29a. R5C2 = {678} -> R1469C3 = 13,14,15 = {1345/1256/1346/2345/1257/1347/1356/2346} (cannot be {1237/1246/1247} because min R169C3 = 8)
29b. R2C3 + R3C23 (step 25b) = {278} (cannot be {359} which clashes with R1469C3) -> R2C3 = 2, R3C23 = {78}, locked for R3 and N1
29c. R2C3 = 2 -> R23C4 = 15 = {69}, locked for C4 and N2 -> R3C6 = 5, R3C5 = 4, R3C8 = 6, R23C4 = [69]
29d. Max R2C3 + R4C5 = 4 -> min R2C7 = 3 (step 1b)

30. R3C8 = 6 -> R4C78 = 15 = {78}, locked for R4 and N6
30a. R4C2 = 9 (hidden single in C2, and in 22(3) cage at R3C2), clean-up: no 4,5 in R5C34 (prelim d)
30b. Naked triple {678} in R5C234, locked for R5, 6 locked in R5C23 for N4 -> R5C5 = 5, clean-up: no 1 in R5C67 (prelim e)
30c. Naked triple {234} in R5C678, locked for R5 -> R5C19 = [19], R3C1 = 3, R69C3 = [31]

31. R5C1 = 1 -> R46C1 = [47] (step 27)
31a. Naked pair {68} in R5C23, locked for R5 and N4 -> R4C3 = 5, R5C4 = 7, R6C2 = 2, R3C2 = 8 (prelim b), R3C3 = 7, R5C23 = [68], R7C2 = 4 (prelim f), R12C2 = [51], R12C1 = [69], R1C3 = 4, R89C2 = [73], R9C4 = 8
31b. Naked triple {789} in R1C789, locked for N3

32. R3C7 = 1 (hidden single in C7), R3C9 = 2, R6C4 = 4, R6C78 = [51], R6C9 = 6, R4C9 = 3, R4C456 = [216], R5C6 = 3, R7C37 = [93], R6C6 = 8, R6C5 = 9, R9C56 = [79], R2C56 = [87], R2C7 = 4 (step 1b)

and the rest is naked singles.


Assassin 123 "Roulette" V0.9 by udosuk (October 2008) here
Puzzle Diagram:
Image
Images with "udosuk style Killer Cages" by Børge:
Image     Image     Image
Code: Select, Copy & Paste into solver:
3x3::k:5121:5121:2050:1539:1539:1539:3588:5125:5125:5121:2050:4358:4358:3335:4104:4104:3588:5125:2050:5641:11530:4358:3335:4104:11530:5387:3588:3084:5641:5641:11530:3335:11530:5387:5387:4621:3084:5390:5390:5390:11530:2319:2319:2319:4621:3084:2320:2320:11530:4369:11530:2066:2066:4621:4115:2320:11530:3604:4369:3349:11530:2066:5654:2583:4115:3604:3604:4369:3349:3349:5654:2584:2583:2583:4115:6169:6169:6169:5654:2584:2584:
Solution:
+-------+-------+-------+
| 6 5 4 | 1 3 2 | 9 7 8 |
| 9 1 2 | 6 8 7 | 4 3 5 |
| 3 8 7 | 9 4 5 | 1 6 2 |
+-------+-------+-------+
| 4 9 5 | 2 1 6 | 7 8 3 |
| 1 6 8 | 7 5 3 | 2 4 9 |
| 7 2 3 | 4 9 8 | 5 1 6 |
+-------+-------+-------+
| 8 4 9 | 5 6 1 | 3 2 7 |
| 5 7 6 | 3 2 4 | 8 9 1 |
| 2 3 1 | 8 7 9 | 6 5 4 |
+-------+-------+-------+
Quote:
udosuk: However, here is a v0.9 which can be solved without fishes/chains even if one doesn't use the info above. :idea:
(Warning: potentional spoiler for the default version)
SSR: 1.56. "True" SSR: 1.14.

Andrew: I got stuck on A123 so decided to have a go at the v0.9.
Even this wasn't an easy puzzle although breaking up the six-cell cages in the four corner nonets was a great help.
I'll rate A123 v0.9 at 1.25.
Walkthrough by Andrew:
I got stuck on A123 so decided to have a go at the v0.9.

Even this wasn't an easy puzzle although breaking up the six-cell cages in the four corner nonets was a great help.

I'll rate A123 v0.9 at 1.25 because I used lots of triple - hidden triples, naked triples, killer triples and hidden killer triples - I don't think I've used all those types in a walkthrough before, probably not even as pairs let alone triples.

Here is my walkthrough for A123 v0.9. After comments in another thread, I've limited my use of strings of combinations/permutations as much as possible. However no promises that I won't revert to using them in later walkthroughs. ;)

Prelims

a) 20(3) cage in N1 = {389/479/569/578}, no 1,2
b) 8(3) diagonal cage in N1 = {125/134}, 1 locked for N1
c) R1C456 = {123}, locked for R1 and N2
d) 20(3) cage in N3 = {389/479/569/578}, no 1,2
e) 22(3) cage at R3C2 = {589/679}, CPE no 9 in R56C2
f) 21(3) cage at R3C8 = {489/579/678}, no 1,2,3
g) R5C234 = {489/579/678}, no 1,2,3
h) R5C678 = {126/135/234}, no 7,8,9
i) 9(3) cage at R6C2 = {126/135/234}, no 7,8,9
j) 8(3) cage at R6C7 = {125/134}, CPE no 1 in R5C8
k) 10(3) cage in N7 = {127/136/145/235}, no 8,9
l) R9C456 = {789}, locked for R9 and N8
m) 22(3) cage in N9 = {589/679}, 9 locked for N9
n) 10(3) cage in N9 = {127/136/145/235}, no 8

1. 8(3) diagonal cage in N1 = {125/134}
1a. R1C3 = {45} -> no 4,5 in R2C2 + R3C1

2. 22(3) cage in N9 = {589/679}
2a. R9C7 = {56} -> no 5,6 in R8C8 + R7C9

3. Min R23C5 = {45} = 9 -> max R4C5 = 3 because R234 cannot be {45}4
3a. Max R23C5 = 12, no 9
3b. 45 rule on N2 3(2+1) outies R2C37 + R4C5 = 7, max R2C37 = 6 -> R2C3 = {2345}, R2C7 = {1234}

4. Max R78C5 = {56} = 11 -> min R6C5 = 7 because R678C5 cannot be 6{56}
4a. 45 rule on N8 3(2+1) outies R6C5 + R8C37 = 23 -> min R8C37 = 14, no 1,2,3,4, no 5 in R8C3
4b. Min R78C5 = 8, no 1

5. 45 rule on N1 3 innies R2C3 + R3C23 = 17 cannot contain 5 because {359/458} clash with 20(3) cage
5a. Max R2C3 + R3C2 = 13 -> min R3C3 = 6 because R2C3 + R3C23 cannot be [494]

6. Hidden triple {123} in R3C179 => R3C79 = {123}
6a. Max R3C9 + R1C7 = 12 -> no 1 in R2C8

7. 45 rule on N3 3 innies R2C7 + R3C78 = 11, no 9

8. 45 rule on N7 3 innies R7C23 + R8C3 = 19, no 1
8a. Max R7C2 + R8C3 = 15 -> min R7C3 = 4

9. Hidden triple {123} in R269C3 -> R2C3 = {23}, R69C3 = {123}
9a. Naked triple {123} in R2C23 + R3C1, locked for N1
9b. Max R9C3 = 3 -> min R7C1 + R8C2 = 13, no 1,2,3

10. 45 rule on C5 3 innies R159C5 = 15
10a. All combinations for 15(3) must contain at least one of 4,5,6 -> R5C5 = {456}
10b. Killer triple 4,5,6 in R5C234, R5C5 and R5C678, locked for R5

11. Hidden killer triple 7,8,9 in R23C5, R6C5 and R9C5 for C5 -> R23C5 must contain one of 7,8
11a. R234C5 cannot contain 6 and one of 7,8 -> no 6 in R23C5
11b. Min R23C5 containing one of 7,8 = 11 -> max R4C5 = 2

12. R2C37 + R4C5 = 7 (step 3b), max R2C3 + R4C5 = 5 -> min R2C7 = 2
12a. R2C37 = 5,6 = [23/24/32], 2 locked for R2
12b. R2C2 = 1 (hidden single in R2)

13. Hidden killer triple 1,2,3 in R1C5, R4C5 and R78C5 for C5 -> R78C5 must contain one of 2,3
13a. R678C5 cannot contain 4 and one of 2,3 -> no 4 in R78C5
13b. Max R78C5 containing one of 2,3 = 9 -> min R6C5 = 8

14. 45 rule on N9 3 innies R7C78 + R8C7 = 13

15. Hidden triple {789} in R1C7, R4C7 and R78C7 for C7, R78C7 can only contain one of 7,8 -> R14C7 = {789}
15a. Killer triple 7,8,9 in R78C7, R7C9 and R8C8, locked for N9
15b. 10(3) cage in N9 = {136/145/235}
15c. Killer pair 5,6 in 10(3) cage and R9C7, locked for N9
15d. Naked triple {789} in R148C7, locked for C7
15e. Min R8C7 = 7 -> max R78C6 = 6, no 6

16. R6C5 + R8C37 = 23 (step 4a)
16a. R6C5 = {89} -> R8C37 = 14,15 = [68/78/87], no 9 in R8C3, 8 locked for R8

17. R2C7 + R3C78 = 11 (step 7)
17a. Max R2C7 + R3C7 = 7 -> min R3C8 = 5 because R2C7 + R3C78 cannot be [434]
17b. 4 in R3 locked in R3C456, locked for N2, clean-up: no 8 in R3C5 (step 3a)

18. Hidden triple {789} in R7C139 -> R7C13 = {789}
18a. R7C23 + R8C3 = 19 (step 8)
18b. Min R78C3 = 13 -> max R7C2 = 5 because R7C23 + R8C3 cannot be [676]
18c. 6 in R7 locked in R7C45, locked for N8, clean-up: no 2 in R7C5 (step 4b)

19. Min R7C1 + R9C3 = 8 -> no 9 in R8C2
19a. R8C8 = 9 (hidden single in R8)

20. Min R1C7 + R3C9 = 8 -> max R2C8 = 6

21. Hidden killer pair 7,8 in R1C8 and R34C8 for C8, R34C8 can only contain one of 7,8 because it must contain one of 4,5,6 -> R1C8 = {78}

22. 20(3) cage in N3 = {389/479/578} (cannot be {569} because R1C8 only contains 7,8), no 6
22a. Killer triple 7,8,9 in R1C7 and 20(3) cage, locked for N3
22b. 6 in N3 locked in R23C8, locked for C8
22c. 21(3) cage at R3C8 = {579/678} (cannot be {489} because R3C8 only contains 5,6), no 4
22d. 5 of {578} must be in R3C8 -> no 5 in R4C8

23. 6 in R1 locked in R1C12, locked for N1
23a. 20(3) cage in N1 = {569}, locked for N1 -> R1C3 = 4, R3C1 = 3 (step 1), R2C3 = 2
23b. R2C3 = 2 -> R23C4 = 15 = {69} (cannot be {78} which clashes with R23C5), locked for C4 and N2
23c. Naked pair {78} in R3C23, locked for R3
23d. Naked pair {45} in R3C56, locked for R3 and N2 -> R3C8 = 6, R23C4 = [69]
23e. Naked pair {78} in R2C56, locked for R2

24. 16(3) cage at R2C7 = {358/457} -> R3C6 = 5, R3C5 = 4

25. R3C8 = 6 -> R4C78 = {78} (step 22c), locked for R4 and N6

26. R1C7 = 9 (hidden single in C7)
26a. 20(3) cage in N3 (step 22) = {578} (only remaining combination) -> R2C9 = 5, R2C1 = 9

27. R456C9 = {369} (only remaining combination), locked for C9 and N6

28. 4 in C9 locked in R89C9 -> 10(3) cage in N9 = {145} -> R9C8 = 5, R89C9 = {14}, locked for C9 and N9
28a. R3C9 = 2, R2C8 = 3 (cage sum), R23C7 = [41], R7C78 = [32], R56C8 = [41], R6C3 = 3, R9C3 = 1, R89C9 = [14]
28b. R9C7 = 6 (I could have made this a hidden single for N9 after step 27 but preferred to do the 10(3) cage first), R7C9 = 7 (step 2), R8C7 = 8, R1C89 = [78], R4C78 = [78], R7C13 = [89], R8C2 = 7 (cage sum), R8C3 = 6, R6C5 = 9 (step 4a)

29. R9C12 = [23], R8C1 = 5 (cage sum)

and the rest is naked singles.

While working on this puzzle, I found a couple of interesting moves that I later found that I didn't need because easier moves were available. Maybe they will be useful when I try A123 again.


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