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PostPosted: Sun Jul 06, 2008 5:42 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Old SSv3.2.1 scores:
Killer rating table      
Rounded Score from SSv3.2.1
! = 0.10+ change from previous Score
pg# on this thread - PART B
(E) = Easy (H) = Hard
======================================================================
|A ## Rate Score|A ## Rate Score|A ## Rate Score|
|----------------------+----------------------+----------------------|
|A.63 1.00 1.05|A.64v2 1.75 !2.10|A.66 1.00 0.95|
|A.63v1.5 1.25 0.95|A.65 1.25 1.05| |
|A.63v2 H1.50 1.20|A.65v2 H1.50 2.65| |
|A.64 1.25 1.35|A.65v3 1.50 1.85| |
|====================================================================|
page #4
Old scores SSv3.3.0:
Rounded Score from SSv3.3.0 
! = 0.10 change from previous Score
pg# on this thread - PART B
(E) = Easy (H) = Hard
======================================================================
|A ## Rate Score|A ## Rate Score|A ## Rate Score|
|----------------------+----------------------+----------------------|
|A.63 1.00 1.05|A.64v2 1.75 2.15|A.66 1.00 !1.25|
|A.63v1.5 1.25 1.00|A.65 1.25 1.05| |
|A.63v2 H1.50 !1.35|A.65v2 H1.50 !3.50| |
|A.64 1.25 !1.25|A.65v3 1.50 !1.95| |
|====================================================================|
page #4
Killer rating table
SudokuSolver Target range v3.6.3
Rating.....Score
0.50 = 0.85
0.75 = 0.90-0.95
1.00 = 1.00-1.20
1.25 = 1.25-1.45
1.50 = 1.50-1.70 (E) = Easy (H) = Hard
===========================================================================================
|A ## by Rate Score|A ## by Rate Score|A ## by Rate Score|
|-----------------------------+-----------------------------+-----------------------------|
|A.63 Ruud 1.00 1.10|A.64v2 Ruud 1.75 1.85|A.66 Ruud 1.00 1.10|
|A63v1.5 mhp 1.25 1.15|A.65 Ruud 1.25 1.25| |
|A.63v2 mhp H1.50 1.35|A.65v2 Ruud H1.50 2.75| |
|A.64 Ruud 1.25 1.35|A.65v3 Ruud 1.50 1.75| |
|=========================================================================================|
page #4


Assassin 63 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5376:3329:3329:3329:5124:3333:3333:3333:5128:5376:6410:6410:5124:5124:5124:4111:4111:5128:5376:5376:6410:4373:4373:4373:4111:5128:5128:3355:3355:6410:6430:2079:4896:4111:2594:2594:2084:2084:6430:6430:2079:4896:4896:3627:3627:2093:2093:5935:6430:2079:4896:6195:2100:2100:4150:4150:5935:3897:3897:3897:6195:5693:5693:4150:5935:5935:4930:4930:4930:6195:6195:5693:4150:3145:3145:3145:4930:4173:4173:4173:5693:
Solution:
+-------+-------+-------+
| 9 7 1 | 5 6 3 | 2 8 4 |
| 5 8 2 | 1 9 4 | 7 3 6 |
| 3 4 6 | 8 2 7 | 5 9 1 |
+-------+-------+-------+
| 8 5 9 | 6 4 2 | 1 7 3 |
| 6 2 3 | 7 1 8 | 4 5 9 |
| 7 1 4 | 9 3 5 | 8 6 2 |
+-------+-------+-------+
| 1 9 8 | 4 5 6 | 3 2 7 |
| 2 6 5 | 3 7 1 | 9 4 8 |
| 4 3 7 | 2 8 9 | 6 1 5 |
+-------+-------+-------+
Quote:
goooders: isnt this very easy?
Andrew: This was a smooth puzzle except right at the end..about 0.75 which is probably about right
herschko: found this one relatively easy.... I just wouldn't call it an "assassin."
Andrew, after doing v1.5: in Mike's definition of ratings, that average Assassins now have a rating of 1.25, I would put A63 as 1.0.
Walkthrough by Andrew including (Hidden &) Killer pair:
Ruud wrote:
Although this Assassin cannot be solved by SumoCue, it only requires a single advanced move. Can you find it?
Maybe that was step 18?

This was a smooth puzzle except right at the end where I had to work harder to finish it off. Maybe I missed something there. Here is my walkthrough.

Thanks to Mike and Para for their comments which have been added below. The comment after step 33d and the typo corrections in steps 48 and 51 were made earlier.

1. R4C12 = {49/58/67}, no 1,2,3

2. R4C89 = {19/28/37/46}, no 5

3. R5C12 = {17/26/35}, no 4,8,9

4. R5C89 = {59/68}

5. R6C12 = {17/26/35}, no 4,8,9

6. R6C89 = {17/26/35}, no 4,8,9

7. R456C5 = 1{25/34}, 1 locked for C5 and N5

8. 45 rule on N1 2 outies R1C4 + R4C3 = 14 = {59/68/77}
8a. Min R1C4 = 5 -> max R1C23 = 8, no 8,9

9. 45 rule on N2 2 innies R1C46 = 8 = [53/62/71], clean-up: no 5,6 in R4C3 (step 8)

10. 45 rule on N3 2 outies R1C6 + R4C7 = 4 = {13/22}

11. 45 rule on N7 2 outies R6C3 + R9C4 = 6 = {15/24/33}

12. 45 rule on N8 2 innies R9C46 = 11 = [29/38/47/56], clean-up: no 5 in R6C3 (step 11)

13. 45 rule on N9 2 outies R6C7 + R9C6 = 17 = {89}, clean-up: no 4,5 in R9C4 (step 12), no 1,2 in R6C3 (step 11)
13a. Min R9C6 = 8 -> max R9C78 = 8, no 8,9
[Mike pointed out that after step 13 there is the CPE no 8,9 in R6C6. This then sets up a hidden pair {89} in R6C47 -> R6C4 = {89}.]

14. Killer pair 8,9 in R5C89 and R6C7, locked for N6, clean-up: no 1,2 in R4C89

15. 23(4) cage at R6C3 with R6C3 = {34} = {2489/3479/3569/3578/4568}, no 1

16. 45 rule on N5 2 outies R5C37 = 7 = {16/25/34}, no 7,8,9

17. R5C34567 must contain 4 and 8/9
17a. 45 rule on R5 5 innies R5C34567 = 23 = {13478/23459/23468} (cannot be {12479} which clashes with R5C37, cannot be {13469} which clashes with R5C89) = 34{178/259/268}, 3 locked for R5, clean-up: no 5 in R5C12
[Para pointed out that I missed R5C37 cannot be {16} because there are no R5C34567 combinations with both 1 and 6. Alternatively, and possibly more obvious, R5C37 cannot be {16} because this would clash with R5C12.]

18. Hidden killer pair 8,9 for N4 in R4C12 and R4C3 -> R4C12 = {49/58}, R4C3 = {89}, 8,9 locked for R4, clean-up: no 7 in R1C4 (step 8), no 1 in R1C6 (step 9), no 3 in R4C7 (step 10)

19. 7 in N4 must be in R5C12 or R6C12, one of which must be {17} -> no 1 in R5C3, clean-up: no 6 in R5C7 (step 16)

20. 45 rule on C123 2 outies R19C4 – 4 = 1 innie R5C3, min R19C4 = 7 -> min R5C3 = 3, clean-up: no 5 in R5C7 (step 16)
[Para pointed out that if I had done step 20 before step 19, then there would have been a hidden killer double instead of hidden killer singles. However this is the order in which I spotted the moves. The fact that we see different moves, or see the same moves in different orders, is one of the things that makes this forum interesting and makes it worthwhile working through other people’s walkthroughs.]

21. R5C12 must contain 1/2 -> R6C12 must contain 1/2 -> R6C12 = {17/26}, 6 locked in R56C12 for N4, clean-up: no 1 in R5C7 (step 16)

22. 3 in N4 locked in R56C3, locked for C3

23. 25(4) cage at R4C4 with R5C3 = {345} = {3589/3679/4579/4678}, no 2
23a. 19(4) cage at R4C6 = {2359/2368/2458/2467/3457}

24. 45 rule on R1 3 innies R1C159 = 19 = {289/379/469/478/568}, no 1

25. 45 rule on N6 3 innies R456C7 = 13 = [139/148/238] -> no 2 in R5C7, clean-up: no 5 in R5C3 (step 16)

26. Naked pair {34} in R5C37, locked for R5

27. Naked pair {34} in R56C3, locked for C3 and N4, clean-up: no 9 in R4C12

28. Naked pair {58} in R4C12, locked for R4 -> R4C3 = 9 -> R1C4 = 5 (step 8), R1C6 = 3 (step 9), R4C7 = 1 (step 10), clean-up: no 7 in R6C89

29. 7 in N6 locked in R4C89 = {37}, locked for R4 and N6 -> R5C7 = 4, R5C3 = 3, R6C3 = 4, R9C4 = 2 (step 11), R9C6 = 9 (step 12), R6C7 = 8 (step 13), clean-up: no 6 in R5C89, no 5 in R6C89
[Para “As i always check hidden singles. There is a hidden single 3 in R6C5 at this point. Doesn't matter much as it becomes naked in 2 steps.” Maybe I forgot to eliminate 3 from R6C4. It’s so easy to overlook those round-the-corner eliminations. In this puzzle I always remembered the ones for the L-shaped 4-cell cages but often forgot them for the sideways T-shaped cages, both when originally solving this puzzle and later when checking walkthroughs.]

30. Naked pair {26} in R6C89, locked for R6

31. Naked pair {17} in R6C12, locked for R6 and N4 -> R6C456 = [935]

32. R5C5 = 1 (hidden single in R5), R4C5 = 4 (cage sum), R4C46 = [62], R5C4 = 7 (cage sum), R5C6 = 8

33a. R1C4 = 5 -> R1C23 = 8 = {17/26}, no 4
33b. R1C6 = 3 -> R1C78 = 10 = [28/64/91]
33c. R9C4 = 2 -> R9C23 = 10 = [37/46]
33d. R9C6 = 9 -> R9C78 = 7 = [34/61]
[I missed R9C78 cannot be [34] which clashes with R9C23, as in goooders message and Para's walkthrough, but fortunately it didn’t matter as I fixed those cells with the next step.]

34. 5 in R9 locked in R9C159
34a. 45 rule on R9 3 innies R9C159 = 17 = {458} (only remaining combination), locked for R9 -> R9C23 = [37], R9C78 = [61], clean-up: no 1 in R1C2 (step 33a), no 4 in R1C8 (step 33b) -> R1C78 = [28], R1C23 = [71], R6C12 = [71]

35. 16(4) cage at R2C7 = {1357/1456} = 15{37/46}, no 9, 5 locked for N3
35a. 20(4) cage in N3 = 19{37/46}
[Para pointed out that {1456} is blocked because 4,6 only in R2C8. Oops! :oops: I should have spotted that. It was getting very late when I finished this puzzle and first posted my walkthrough.]

36. 9 in C7 locked in R78C7, locked for N9
[Alternatively grouped X-Wing for 9 in 20(4) cage in N3 and R5C89, locked for C89, but I spotted the locked 9 in C7 first.]

37. 24(4) cage at R6C7 = {2589/3489} = 89{25/34}, no 7
37a. R78C7 = {359} -> R8C8 = {24}

38. 7 in C7 locked in R23C7, locked for N3
38a. 16(4) cage at R2C7 (step 35) = {1357} (only remaining combination), 3 locked for N3

39. 23(4) cage at R6C3 = {2489/4568} = 48{29/56}, 8 locked for N7
39a. 16(4) cage in N7 = 14{29/56}

40. 25(4) cage at R2C2 = {2689} (only remaining combination), 2,6,8 locked for N1

41. 5 in C3 locked in R78C3, locked for N7 -> R9C1 = 4, R1C159 = [964]

42. Naked pair {35} in R23C1, locked for C1 and N1 -> R3C2 = 4, R4C12 = [85]

43. 23(4) cage at R6C3 (step 39) = {4568} (only remaining combination), locked for N7

44. Naked pair {12} in R78C1, locked for C1 and N7 -> R5C12 = [62], R7C2 = 9

45. Naked triple {578} in R789C5, locked for C5 and N8

46. Naked pair {35} in R2C18, locked for R2 -> R2C7 = 7

47. R3C6 = 7 (hidden single in R3)

48. R8C7 = 9 (hidden single in C7)

49. R7C456 = [186/384/456] (cannot be {357} which clashes with R7C7), no 7 in R7C5, no 1 in R7C6

50. R8C5 = 7 (hidden single in C5)

51. 19(4) cage at R8C5 = [1765/3718/3745], no 4 in R8C4

52. 22(4) cage in N9 = {2578/3478} = 78{25/34}
52a. For {3478} 4 must be in R7C8 and 7 in R7C9 -> no 3 in R7C89

53. 20(4) cage at R1C5 = {1469} (only remaining combination because [6824] clashes with R2C23) -> R2C5 = 9, R2C46 = {14}, locked for R2 and N2

and the rest is naked singles, hidden singles and cage sums
Walkthrough by Para including Killer Triple:
Hi all

Here's the Walk-through for the original as well.
Ruud wrote:
Although this Assassin cannot be solved by SumoCue, it only requires a single advanced move. Can you find it?
They don't really fall in the really advanced moves category, but by giving SumoCue step 6c, 9a and 9b, SumoCue will solve the puzzle.

Walk-through Assassin 63

1. R4C12 = {49/58/67}: no 1,2,3

2. R4C89 = {19/28/37/46}: no 5

3. R5C12, R6C12 and R6C89 = {17/26/35}: no 4,8,9

4. R5C89 = {59/68}: no 1,2,3,4,7

5. 8(3) at R4C5 = {125/134}: no 6,7,8,9; 1 locked for C5 and N5

6. 45 on N9: 2 outies: R6C7 + R9C6 = 17 = {89}
6a. Killer Pair {89} in R5C89 + R6C7 -->> locked for N6
6b. Clean up: R4C89 = {37/46}: no 1,2
6c. R4C12 = {49/58}: {67} blocked by R4C89: no 6,7
6d. Killer Triple {567} in R4C89, R5C89 and R6C89 -->> locked for N6

7. 45 on R789: 2 outies: R6C37 = 12 = [39/48] -->> R6C3 = {34}

8. 45 on N7: 2 outies: R6C3 + R9C4 = 6 = [42/33] -->> R9C4 = {23}

9. 45 on N5: 2 outies: R5C37 = [34/43/52/61] -->> R5C3 = {3456}
9a. R5C12 = {1|5|6..}; R5C89 = {5|6..} -->> R5C37: [61] blocked: R5C3: no 6; R5C7: no 1
9b. Killer Triple {345} in R4C12, R5C3 and R6C3 -->> locked for N4
9c. Naked Quad {1267} in R5C12 + R6C12 -->> locked for N4

10. 45 on N1: 2 outies: R1C4 + R4C3 = 14 = [59/68] -->> R1C4 = {56}

11. 45 on R123: 2 outies: R4C37 = [82/91] -->> R4C7 = {12}

12. 45 on N3: 2 outies: R1C6 + R4C7 = 4 = [31/22] -->> R1C6 = {23}

13. 45 on N6: 3 innies: R456C7 = [139/148/238] -->> R5C7: no 2
13a. Clean up: R5C37 = {34} -->> locked for R5
13b. Killer Pair {34} in R4C89 + R5C7 -->> locked for N6
13c. Clean up: R6C89: no 5
13d. Naked Quad {1267} in R6C1289 -->> locked for R6

14. 5 in N6 locked within 14(2) cage at R5C89 -->> R5C89 = {59} -->> locked for R5 and N6
14a. R6C7 = 8; R9C6 = 9; R6C3 = 4; R9C4 = 2; R5C37 = [34]
14b. Clean up: R4C12 = {58} -->> locked for R4
14c. R4C3 = 9; R1C4 = 5; R4C7 = 1; R1C6 = 3; R6C4 = 9; R6C6 = 5; R6C5 = 3
14d. R5C5 = 1(hidden); R4C5 = 4; R4C6 = 2(hidden); R5C6 = 8
14e. Clean up: R4c89 = {37} -->> locked for R4 and N6
14f. R45C4 = [67]
14g. R5C12 = {26} -->> locked for N4

15. 12(3) at R9C2 = [37/46]2: R9C2 = {34]; R9C3 = {67}

16. 16(3) at R9C6 = 9[61]: [934] blocked by R9C2 -->>R9C78 = [61]
16a. R9C23 = [37]

17. 13(3) at R1C6 = 3[28]: 3{46} blocked by R1C7 -->> R1C78 = [28]

18. 13(3) at R1C2 = [71]3 -->> R1C23 = [71]
18a. R6C12 = [71]

19. 25(4) at R2C2 = 9{268}(last combo) -->> R2C23 + R3C3 = {268} -->> locked for N1

20. 5 in C3 locked within 23(4) cage at R7C3 -->> 23(4) = 4{568} -->> R7C3 + R8C23 = {568} -->> locked for N7; R8C2 = {68}
20a. R9C1 = 4; R1C1 = 9; R1C5 = 6; R1C9 = 4; R3C2 = 4(hidden); R7C2 = 9(hidden)
20b. R4C1 = 8(hidden); R4C2 = 5; R5C1 = 6(hidden); R5C2 = 2

21. 16(4) at R2C7 = 1{357}(last combo) -->> R2C78 + R3C7 = {357} -->> locked for N3
21a. R8C7 = 9(hidden)
21b. Naked Triple {578} in R789C5 -->> locked for C5 and N8

22. 17(3) at R3C4 = [197/827] -->> R3C6 = 7

23. 24(4) at R6C7 = 89[52/34] -->> R7C7 = {35}; R8C8 = {24}
23a. R2C7 = 7(hidden)

24. 20(4) at R1C5 = 6{149/248} -->> R2C456 = {149}: {248} clashes with R2C23: no 2,8 -->> R2C5 = 9; R2C46 = {14} -->> locked for R2 and N2
And this leaves you with all naked singles.

greetings

Para


Last edited by Ed on Sun Dec 28, 2008 9:03 am, edited 3 times in total.

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PostPosted: Sun Jul 06, 2008 5:45 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 63v1.5 by mhparker (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4608:6145:6145:6145:4356:2309:2309:2309:5896:4608:3338:3338:4356:4356:4356:4879:4879:5896:4608:4608:3338:4117:4117:4117:4879:5896:5896:3611:3611:3338:4126:4639:5152:4879:2850:2850:2340:2340:4126:4126:4639:5152:5152:3627:3627:2605:2605:5935:4126:4639:5152:7219:1844:1844:3894:3894:5935:3897:3897:3897:7219:4157:4157:3894:5935:5935:4674:4674:4674:7219:7219:4157:3894:5961:5961:5961:4674:2381:2381:2381:4157:
Solution:
+-------+-------+-------+
| 2 9 8 | 7 4 5 | 3 1 6 |
| 6 4 1 | 8 3 2 | 9 5 7 |
| 3 7 5 | 1 6 9 | 4 8 2 |
+-------+-------+-------+
| 8 6 3 | 5 7 4 | 1 2 9 |
| 4 5 2 | 3 9 1 | 7 6 8 |
| 9 1 7 | 6 2 8 | 5 3 4 |
+-------+-------+-------+
| 1 2 9 | 4 5 6 | 8 7 3 |
| 5 3 4 | 2 8 7 | 6 9 1 |
| 7 8 6 | 9 1 3 | 2 4 5 |
+-------+-------+-------+
Quote:
Para: 1.5 seems overrated. Didn't have any more trouble with this one than with the V1. Nothing special needed...and one Killer Triple
Andrew: Para's step 21 was a powerful move...would probably be rated as 1.0 for Para's solving path...but 1.25 for the way that I solved it.
Walkthrough by Para including Killer Triple:
Hi

V1.5 seems overrated. Didn't have any more trouble with this one than with the V1. Nothing special needed: Only 45-tests, naked pairs/simple combination work and one Killer Triple needed.

Walk-through Assassin 63 V1.5(?)

1. 24(3) at R1C2 = {789} -->> locked for R1

2. 9(3) at R1C6 = {126/135/234}: no 7,8,9

3. 13(4) at R2C2 = {1237/1246/1345}: no 8,9

4. R4C12 and R5C89 = {59/68}: no 1,2,3,4,7

5. R4C89 = {29/38/47/56}: no 1

6. R5C12 = {18/27/36/45}: no 9

7. R6C12 = {19/28/37/46}: no 5

8. R6C89 = {16/25/34}: no 7,8,9

9. 28(4) at R6C7 = {4789/5689}: no 1,2,3

10. 23(3) at R9C2 = {689} -->> locked for R9

11. 9(3) at R9C6 = {135/234}: no 7; 3 locked for R9

12. 45 on N7: 2 outies: R6C3 + R9C4 = 16 = [79/88] -->> R6C3 = {78}; R9C4 = {89}

13. 45 on R789: 2 outies: R6C37 = 12 = [84/75] -->> R6C7 = {45}
13a. Killer Triple {456} in R5C89 + R6C7 + R6C89 -->> locked for N6
13b. Clean up: R4C89: no 7

14. 45 on R123: 2 outies: R4C37 = 4 = {13} -->> locked for R4
14a. R5C7 = 7(hidden)
14b. 45 on N5: 1 outie: R5C3 = 2

The Big Clean up:
15. R4C89 = {29} -->> locked for R4 and N6
15a. R4C12 = {68} -->> locked for R4 and N4
15b. R6C3 = 7; R9C4 = 9(step 12); R6C7 = 5(step 13)
15c. R5C89 = {68} -->> locked for R5 and N6
15d. R6C89 = {34} -->> locked for R6 and N6
15e. R4C7 = 1; R4C3 = 3
15f. R6C12 = {19} -->> locked for R6 and N4
15g. R5C12 = {45} -->> locked for R5

The free outies
16. 45 on N1: 1 outie: R1C4 = 7
16a. 45 on N3: 1 outie: R1C6 = 5
16b. 45 on N9: 1 outie: R9C6 = 3
16c. R4C6 = 4; R4C4 = 5; R4C5 = 7

17. R1C23 = {89} -->> locked for N1
17a. R9C23 = {68} -->> locked for N7
17b. R4C1 = 8(hidden); R4C2 = 6; R9C23 = [86]; R1C23 = [98]; R4C12 = [91]

18. 9(3) at R1C6 = 5[31](last combo) -->> R1C78 = [31]

19. 20(4) at R4C6 = 47{18}(last combo) -->> R56C6 = [18]
19a. R56C4 = [36]; R56C5 = [92]
19b. Clean up: R9C78 = {24} -->> locked for R9 and N9

20. 28(4) at R6C7 = 5{689} (last combo) -->> R7C7 + R8C78 = {689} -->> locked for N9

21. 45 on C89: 3 innies: R289C8 = 18 = [594]: ([792] blocked by R4C8; {68}[4] blocked by R5C8)
21a. R9C7 = 2; R4C89 = [29]; R6C89 = [34]; R7C8 = 7
21b. R7C3 = 9(hidden); R8C6 = 7(hidden); R9C1 = 7(hidden)

22. 13(4) at R2C2 = 3{145}: (3{127} blocked by R2C2(only place for {27}): no 2,7 -->> R2C2 = 4; R2C3 = 1; R3C3 = 5
22a. R5C12 = [45]; R8C3 = 4; R8C2 = 3; R7C2 = 2; R3C2 = 7
And the rest is all naked singles.

greetings

Para
Walkthrough by Andrew:
I've now worked through both of Para's walkthroughs, for Ruud's original puzzle and Mike's V1.5.

For the original puzzle we have different first breakthrough moves but otherwise had fairly similar solving paths.

Similarly for V1.5 we had different first breakthrough moves. Para had a killer triple which I didn't spot but a few moves later I had a hidden killer triple and we both reached the same position. Then Para's step 21 was a powerful move that really broke it open for him. I've commented more on that below.

Ed has asked me to give my estimated ratings whenever I post walkthroughs and all other regulars are encouraged to do the same.

On the basis, in Mike's definition of ratings, that average Assassins now have a rating of 1.25, I would put A63 as 1.0. For Mike's V1.5 it would probably be rated as 1.0 for Para's solving path but 1.25 for the way that I solved it.

Here is my walkthrough for A63V1.5. Thanks Para for the correction and the comments that have been added in the walkthrough.

1. R4C12 = {59/68}

2. R4C89 = {29/38/47} (cannot be {56} which clashes with R4C12), no 1,5,6

3. R5C12 = {18/27/36/45}, no 9

4. R5C89 = {59/68}

5. R6C12 = {19/28/37/46}, no 5

6. R6C89 = {16/25/34}, no 7,8,9

7. R1C234 = {789}, locked for R1

8. R1C678 = {126/135/234}

9. R9C234 = {689}, locked for R9

10. R9C678 = {135/234} = 3{15/24}, no 7, 3 locked for R9
10a. R9C159 = 7{15/24}

11. 13(4) cage at R2C2 = {1237/1246/1345}, no 8,9

12. 28(4) cage at R6C7 = {4789/5689}, no 1,2,3

13. 45 rule on N1 2 outies R1C4 + R4C3 = 10 = [73/82/91]

14. 45 rule on N2 2 innies R1C46 = 12 = [75/84/93]
14a. R1C678 (step 8) = {135/234} (cannot be {126} because only 3,4,5 in R1C6) = 3{15/24}, no 6, 3 locked for R1
14b. R1C159 = 6{15/24}

15. 45 rule on N3 2 outies R1C6 + R4C7 = 6 = [33/42/51]

16. 45 rule on R123 2 outies R4C37 = 4 = {13}, locked for R4, clean-up: no 8 in R1C4 (step 13), no 4 in R1C6 (step 14 or step 15), no 8 in R4C89
16a. 8 in R1 locked in R1C23, locked for N1

17. 45 rule on N7 2 outies R6C3 + R9C4 = 16 = [79/88]
17a. 6 in R9 locked in R9C23, locked for N7

18. 45 rule on N8 2 innies R9C46 = 12 = [84/93]

19. 45 rule on N9 2 outies R6C7 + R9C6 = 8 = [44/53]
19a. 28(4) cage at R6C7 (step 12) = {4789/5689} = 89{47/56}, R6C7 = {45} -> no 4,5 in R7C7 + R8C78
[At this stage I missed Para’s first breakthrough, killer triple 4,5,6 in R5C89, R6C7 and R6C89 for N6. My first breakthrough is in step 23.]
19b. 8,9 locked in R7C7 + R8C78 for N9

20. 45 rule on N5 2 outies R5C37 = 9 = {18/27/36/45}, no 9

21. 45 rule on N4 3 innies R456C3 = 12 = [138/147/318/327], R5C3 = {1234}, clean-up: R5C7 = {5678} (step 20)

22. 45 rule on N6 3 innies R456C7 = 13 = [175/184/364], no 5 in R5C7, clean-up: no 4 in R5C3 (step 20)
22a. R456C3 (step 21) = [138/318/327]
[Para pointed out that I missed that 3 is locked for C3 and N4.]

23. 45 rule on R6789 3 innies R6C456 = 16 = {169/259/268/349/358/367} (cannot be {178} which clashes with R6C3, cannot be {457} which clashes with R6C7) [7/8/9]
23a. Hidden killer triple 7,8,9 in R6C12, R6C3 and R6C456 -> R6C12 = {19/28/37}, no 4,6
23b. Killer triple 1,2,3 in R4C3, R5C3 and R6C12, locked for R5C12, clean-up: no 6,7,8 in R5C12
23c. Naked pair {45} in R5C12, locked for R5 and N4, clean-up: no 9 in R4C12, no 9 in R5C89
23d. Naked pair {68} in R5C89, locked for R5 and N6 -> R5C7 = 7, R5C3 = 2 (step 20), clean-up: no 4 in R4C89, no 8 in R6C12, no 1 in R6C89
[Para suggested that it’s easier to go directly from step 23a to step 23c. Agreed but step 23b is what I spotted.]

24. R4C7 = 1 (hidden single in N6), R4C3 = 3, R6C3 = 7 (step 22a), R6C7 = 5 (step 22), R1C4 = 7 (step 13), R1C6 = 5 (step 14 or 15), R9C4 = 9 (step 17), R9C6 = 3 (step 18 or 19), clean-up: no 2 in R6C89
24a. Naked pair {89} in R1C23, locked for N1
24b. R1C6 = 5 -> R1C78 = 4 = [31]
24c. Naked pair {68} in R9C23, locked for N7
24d. R9C6 = 3 -> R9C78 = {24} (only remaining combination), locked for R9 and N9

25. R4C5 = 7 (hidden single in R4), R56C5 = 11 = [38/92]

26. R4C1 = 8 (hidden single in C1), R4C2 = 6, R9C23 = [86], R1C23 = [98], R6C12 = [91]

27. Naked triple {139) in R5C456, locked for N5

28. R6C456 (step 23) = {268} (only remaining combination), locked for N5 -> R4C46 = [54]

29. R4C6 + R5C7 = [47] -> R56C6 = 9 = [18], R6C45 = [62], R5C45 = [39]

30. 28(4) cage at R6C7 (step 12) = {5689} (only remaining combination) -> 6 locked for N9
[At this stage Para had his second breakthrough 45 rule on C89 3 innies R289C8 = 18 = [594]: ([792] blocked by R4C8; {68}[4] blocked by R5C8). I’d stopped looking for any more 45s, possibly because the original Assassin 63 came out with straightforward combination work so that’s what I used but it’s quite a bit longer for this puzzle.]

31. 23(4) cage at R6C3 = {2579/3479} = 79{25/34}, no 1
31a. 2,3 only in R8C2 -> R8C2 = {23}
31b. 1 in C3 locked in R23C3, locked for N1
31c. 15(4) cage in N7 = {1257/1347} = 17{25/34}

32. 13(4) cage at R2C2 = {1345} (only remaining combination, cannot be {1237} because 2,7 only in R2), 4,5 locked for N1

33. Naked pair {45} in R25C2, locked for C2

34. R3C456 = {169/268/349} [4/6,…]
34a. 1 of {169} must be in R3C4 -> no 1 in R3C5
34b. 3 of {349} must be in R3C5 -> no 4 in R3C5

35. Killer pair 4,6 in R1C5 and R3C456, locked for N2

36. 17(4) cage in N2 = {1268/1349/2348}
36a. 2 of {1268/2348} must be in R2C6 -> no 2 in R2C4

37. R7C456 = {168/258/267/456}
37a. 4 of {456} must be in R7C4 -> no 4 in R7C5

38. 18(4) cage in N8 = {1278/1467/2457} (cannot be {1458} because R8C6 only contains 2,6,7) = 7{128/146/245} -> R8C6 = 7
38a. 1 of {1278/1467} must be in R9C5 -> no 1 in R8C45
38b. 5 of {2457} must be in R9C5 -> no 5 in R8C5
38c. 2 of {1278} must be in R8C4 -> no 8 in R8C4

39. R7C456 (step 37) = {168/258/456}
39a. 2 of {258} must be in R7C6 -> no 2 in R7C4
39b. 6 of {168/456} must be in R7C6 -> no 6 in R7C5

40. 19(4) cage at R2C7 = {1279/1459/1468} (cannot be {1567} because 5,7 only in R2C8)
40a. 5,7 only in R2C8 -> no 9 in R2C8
40b. {1279} must have 7 in R2C8 -> no 2 in R2C8
40c. {1468} must have 4 in R23C7 because {68} there would clash with R78C7 -> no 4 in R2C8

41. R2C78 + R3C7 contain 2/4 -> 23(4) cage in N3 must contain 2/4 = {2579/2678/4568}
41a. {2579} must have 2 in R1C9 and then 9 in R3C8 to avoid clash with R4C9 -> no 9 in R23C9
41b. {2678} cannot have {268} in R123C9 which would clash with R5C9 -> no 7 in R3C8
41c. {2678} cannot have {678} in R123C9 which would clash with R5C9 -> no 2 in R3C8
41d. {4568} cannot have {468/568} in R123C9 which would clash with R5C9 -> no 4,5 in R3C8

42. Naked triple 6,8,9 in R358C8, locked for C8 -> R4C89 = [29], R9C78 = [24], R6C89 = [34]
[Alternatively after step 41a, R4C9 = 9 (hidden single in C9). Just spotted that while checking this walkthrough. Anyway it was good to continue with the eliminations from R3C8.]

43. 23(4) cage in N3 (step 41) = {2579/2678} = 27{59/68}, 7 locked R23C9 for C9 and N3 -> R2C8 = 5, R7C8 = 7
43a. 23(4) cage = {2678} (only remaining combination), 6,8 locked for N3

and the rest is naked singles and a cage sum


Last edited by Ed on Sun Jul 06, 2008 10:34 pm, edited 1 time in total.

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PostPosted: Sun Jul 06, 2008 5:46 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 63v2 by mhparker (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4608:5121:5121:5121:6148:2821:2821:2821:6664:4608:3594:3594:6148:6148:6148:5391:5391:6664:4608:4608:3594:3605:3605:3605:5391:6664:6664:2587:2587:3594:3870:3359:6688:5391:802:802:2084:2084:3870:3870:3359:6688:6688:2091:2091:3373:3373:6447:3870:3359:6688:5939:2868:2868:5686:5686:6447:3641:3641:3641:5939:6205:6205:5686:6447:6447:4418:4418:4418:5939:5939:6205:5686:3401:3401:3401:4418:3149:3149:3149:6205:
Solution:
+-------+-------+-------+
| 5 9 8 | 3 2 4 | 1 6 7 |
| 6 1 7 | 8 9 5 | 4 3 2 |
| 3 4 2 | 1 6 7 | 5 9 8 |
+-------+-------+-------+
| 7 3 4 | 5 8 6 | 9 2 1 |
| 2 6 1 | 7 4 9 | 8 5 3 |
| 8 5 9 | 2 1 3 | 6 7 4 |
+-------+-------+-------+
| 1 8 6 | 9 3 2 | 7 4 5 |
| 9 7 3 | 4 5 1 | 2 8 6 |
| 4 2 5 | 6 7 8 | 3 1 9 |
+-------+-------+-------+
Quote:
mhparker, lead-in: (Estimated rating: 1.75)
Andrew in June 2011: congratulations to nouggie for your first posted walkthrough. :applause:... this was the only walkthrough posted at the time (where was everybody? ;) ) ... I'll rate my walkthrough at Hard 1.5
Stream of consciousness, first effort ever Walkthrough by Nouggie:
Hi all.

This is my first effort to write a walk through. Some of my notation may not be correct or standard. Also, I suspect that there are unneeded entries, as this was a "stream of consciousness" solution. Suggestions welcome.

Here goes for V2....

a. Automatic cleanup after placing each number and locked numbers.
b. Bold print indicates a "celled" number within the cage.

1. R4C89={12}-->R5C89={35}-->R6C89={47}-->R456C7={689}

2. 45 Rule N5 outies sum to 9={36/18}; conflict with R5C89-->R5C7=8 and R5C3=1

3. From Step 2, R5C12={26} and R5C456={479}

4. R4C12={37}-->R6C12={58}-->R46C3={49}; conflict w/ R6C89-->R6C3=9, R6C7=6, R4C7=9, and R4C3=4

5. 45 Rule: R1C4=3, R1C6=4, R9C6=8, and R9C4=6

6. From Step 5, R9C78={13}, R9C23={25}, R1C3=8, and R1C2=9

7. 23(4) Cage N69={6278/6458}–>R8C8=8 and R78C7={27/45}

8. R5C4 no 9 because R4C4 cannot have a {123}; R5C5 no 9 because R4C5 cannot have a {13}–>R5C6=9

9. 25(4) cage N47={9367–>no 7 in R9C1; 22(4) cage N7={1489}

10. 14(4) cage N14={4127/{4235}({4136} isn't possible because of block by 25(4)cage N47)-->18(4) cage N1= {3456/1467-->18(4) cage N1 must have a {46}-->14(4) cage N14 no 6-->R8C2 no 6-->R48C2={37}; R3C2 no 3, R7C3 no 9, and R1C1 no 34

11. 26(4) cage N56 no 1-->R6C5=1-->R4C5={58} and R6C46={23};conflict w/ R1C4-->R6C4=2, R6C6=3, R4C6=6, and R4C45={58}

12. R1C7={125}-->R1C8={256}

13. R3C456={158/167/257}; If R3C456={257}-->R3C3=3-->R2C23={25}; therefore R3C456<>{257} and R3C3<>3 because UR for 25 at R2C23 and R9C23; R3C46 locked for 1-->R3C12 no 1

14. R7C456={149/239/257/347}; If R7C456={149}-->R7C1 and R7C2=8; therefore R7C456<>{149}; 24(4) cage N9 no 1 because R8C8=8-->R7C12 locked for 1-->killer pair 49 at R89C1-->killer pair 18 at R7C12

15. Hidden single R3C2=4-->R123C1 locked for 6-->R5C1=2 and R5C2=6

16. 26(4) cage N3 no 1-->hidden single R4C9=1-->R4C8=2; R1C7={12}; R1C8={56}

17. 24(4) cage N2={2589/2679} locked for 2 and 9

18. ??45 Rule for C9, R23C7 must sum to 8, 9, or 10-->R2C8={34}-->R23C7 must sum to 8 or 9-->R19C7 must sum to 5 or 4-->R9C7=3 and R9C8=1

19. 45 Rule for C8, R37C9 must sum to 13, 15, or 16-->R3C8={679} and R7C8={4679}

20. 45 Rule for R1C159=14={167/257}

21. 24(4) cage N2={2589/2679}; 14(3) cage N2={167/158}-->R3C7 no 1; 26(4) cage N3={2789/3689/4589}; 14(3) cage N8={239/257/347}; 17(4) cage N8={1457/1349/1259}

22. R1C9={2567}-->R3C8={79}-->45 Rule R7C8={469}

23. Eliminate 6 from R1C9 as follows:

23a. If R1C9=6-->R7C8=6-->R7C3<>6 locked pair

23b. If R1C9=6-->R1C8=5-->R1C7=2-->R1C5=7-->R2C4=9 and R5C5=4-->R5C4=7-->R4C4=5-->R7C4=4-->R7C5=3 and R7C6=7-->R7C3=6

23c. Steps 23a and 23b conflict therefore R1C9<>6-->R78C9 locked for 6-->hidden single R1C8=6-->R1C1=5-->18(4) cage N1={3456}

24. 24(4) cage N2={2589}-->14(3) cage N2={167}

25. Singles, cage set reductions, and cage sums to end
2011 Walkthrough by Andrew:
Thanks Mike for a challenging variant.

Mike wrote:
outies on the corner nonets give you defined numbers straightaway then...middle third falls...after that it just seems to fall out
All of the above three assertions except one could be also be claimed to apply to the following puzzle. Guess which?
Thanks for the easy question. Now to try the puzzle. ;)

Belated congratulations to nouggie for your first posted walkthrough. :applause:

It's surprising that this was the only walkthrough posted at the time (where was everybody? ;) ) so here, at last, is the second one.

Here is my walkthrough for A63 V2.

Prelims

a) R4C12 = {19/28/37/46}, no 5
b) R4C89 = {12}
c) R5C12 = {17/26/35}, no 4,8,9
d) R5C89 = {17/26/35}, no 4,8,9
e) R6C12 = {49/58/67}, no 1,2,3
f) R6C89 = {29/38/47/56}, no 1
g) 20(3) cage at R1C2 = {389/479/569/578}, no 1,2
h) 11(3) cage at R1C6 = {128/137/146/236/245}, no 9
i) 14(4) cage at R2C2 = {1238/1247/1256/1346/2345}, no 9
j) 26(4) cage in N3 = {2789/3689/4589/4679/5678}, no 1
k) 26(4) cage at R4C6 = {2789/3689/4589/4679/5678}, no 1

1. Naked pair {12} in R4C89, locked for R4 and N6, clean-up: no 8,9 in R4C12, no 6,7 in R5C89, no 9 in R6C89

2. Naked pair {35} in R5C89, locked for R5 and N6, clean-up: no 6,8 in R6C89

3. Naked pair {47} in R6C89, locked for R6 and N6, clean-up: no 6,9 in R6C12

4. Naked pair {58} in R6C12, locked for R6 and N4

5. Naked triple {689} in R456C7, locked for C7

6. Killer pair 6,7 in R4C12 and R5C12, locked for N4

7. Killer pair 3,4 in R4C12 and R4C3, locked for R4 and N4

8. 9 in N4 only in R56C3, locked for C3, CPE no 9 in R6C4

9. 45 rule on N5 2 outies R5C37 = 9 = [18], clean-up: no 7 in R5C12
9a. Naked pair {26} in R5C12, locked for R5 and N4 -> R6C3 = 9, R46C7 = [96], clean-up: no 4 in R4C12
9b. Naked pair {37} in R4C12, locked for R4 -> R4C3 = 4
9c. R6C5 = 1 (hidden single in R6), R45C5 = 12 = [57/84], no 6 in R4C5, no 9 in R5C5

10. 45 rule on N1 1 remaining outie R1C4 = 3
10a. 20(3) cage at R1C2 = {389} (only remaining combination) -> R1C23 = [98]
10b. R6C4 = 2, R5C3 = 1 -> R45C4 = 12 = [57/84], no 6 in R4C4, no 9 in R5C4
10c. R6C6 = 3, R45C6 = [69] (hidden pair in N5)

11. 45 rule on N3 1 remaining outie R1C6 = 4, R1C78 = 7 = [16/25/52], no 7, no 1 in R1C8

12. 45 rule on N7 1 remaining outie R9C4 = 6, R9C23 = 7 = [25/43/52], no 1,7,8, no 3 in R9C2

13. 45 rule on N9 1 remaining outie R9C6 = 8, R9C78 = 4 = {13}, locked for R9 and N9, clean-up: no 4 in R9C2
13a. Naked pair {25} in R9C23, locked for R9 and N7
13b. R4C9 = 1 (hidden single in C9), R4C8 = 2, clean-up: no 5 in R1C7 (step 11)

14. 14(4) cage at R2C2 contains 4 = {1247/1346/2345}
14a. 1 of {1247/1346} must be in R2C2 -> no 6,7 in R2C2
14b. 3 of {2345} must be in R23C3 (R23C3 cannot be {25} which clashes with R9C3) -> no 3 in R2C2
[With hindsight I could have eliminated {1346} = 1{36}4 which clashes with R78C3, ALS block. However step 15b makes this elimination.]

15. 25(4) cage at R6C3 contains 9 = {3679} (only remaining combination because R78C3 must contain two of 3,6,7), 3,6,7 locked for N7
15a. 6 of {3679} must be in R78C3 (R78C3 cannot be {37} which clashes with 14(4) cage at R2C2) -> no 6 in R8C2
15b. 6 in N7 only in R78C3, locked for C3
15c. Naked pair {37} in R48C2, locked for C2

16. 23(4) cage at R6C7 contains 6 = {2678/4568} -> R8C8 = 8
16a. R78C7 = {27/45}
[I’ve put this step first to simplify step 17.]

17. 21(4) cage at R2C7 contains 9 = {1479/2379/3459} (cannot be {1569/2469} which clash with 11(3) cage at R1C6), no 6
17a. 1 of {1479} must be in R23C7 (R23C7 cannot be {47} which clashes with R78C7) -> no 1 in R2C8

18. 1 in N3 only in R123C7, locked for C7 -> R9C78 = [31]
18a. 21(4) cage at R2C7 (step 17) = {1479/2379/3459}
18b. 3 of {3459} must be in R2C8 -> no 5 in R2C8

19. 14(3) cage in N2 = {158/167/257}, no 9
19a. 9 in N2 only in R2C45, locked for N2

20. 14(3) cage in N8 = {239/257/347} (cannot be {149} which clashes with R7C12, ALS block), no 1
20a. 3 of {239/347} must be in R7C5 -> no 4,9 in R7C5
20b. 1 in N8 only in R8C46, locked for R8

21. Naked pair {49} in R89C1, locked for C1 and N7

22. R3C2 = 4 (hidden single in N1)
22a. R5C2 = 6 (hidden single in C2), R5C1 = 2

23. 45 rule on C89 2 remaining innies R12C8 = 9 = [54/63] -> R2C8 = {34}
23a. 21(4) cage at R2C7 (step 17) = {1479/2379/3459}
23b. 5 of {3459} must be in R3C7 -> no 5 in R2C7
23c. Killer pair 3,5 in R12C8 and R5C8, locked for C8

24. 17(4) cage in N8 must contain 1 = {1259/1349/1457}
24a. 3 of {1349} must be in R8C5, 9 of {1259} must be in R9C5 -> no 9 in R8C5

25. 26(4) cage in N3 must contain 8,9 = {2789/3689/4589}
25a. 5,6 on {3689/4589} must be in R1C9 -> no 5,6 in R2C9 + R3C89
25b. 6 in N3 only in R1C89, locked for R1

26. 14(3) cage in N2 (step 19) = {158/167} (cannot be {257} which clashes with R1C5), no 2, 1 locked for R3 and N2
26a. 6 of {167} must be in R3C5 -> no 7 in R3C5

27. 24(4) cage in N2 must contain 9 = {2589/2679}
27a. 6,8,9 only in R2C45 -> R2C4 = {89}, R2C5 = {689}

28. 18(4) cage in N1 contains 4,6 = {1467/3456}
28a. 5 of {3456} must be in R1C1 -> no 5 in R23C1

29. 45 rule on C1234 4 innies R2378C4 = 22 must contain 1,9 = {1489/1579}
29a. 1 of {1579} must be in R8C4 (R78C4 cannot be {57} which clashes with 14(3) cage in N8 (step 20) = {257}, CCC; note that 14(3) cage = {347} must have 4 in R7C4 so is not relevant to this clash) -> no 5,7 in R8C4

30. 14(3) cage in N8 (step 20) = {239/257/347}
30a. Consider combinations for 17(4) cage in N8 (step 24) = {1259/1349/1457}
17(4) cage = {1259/1457}, 5 locked for N8
17(4) cage = {1349} => R8C56 = [31], R8C4 + R9C5 = {49} => R2378C4 (step 29) = {1489} (cannot be {1579} because no 1,5,7 in R28C4) => no 5,7 in R7C4
30b. -> 14(3) cage in N8 = {239/347}, no 5 -> R7C5 = 3, R7C6 = {27}, R7C4 = {49}
30c. 5 in N8 only in R8C56, locked for R8, clean-up: no 4 in R7C7 (step 16a)

31. R2378C4 (step 29) = {1489} (cannot be {1579} because 5,7 only in R3C4), locked for C4 -> R45C4 = [57], R45C5 = [84]

32. Killer pair 7,9 in 14(3) cage and R9C5, locked for N8

33. 14(3) cage in N2 (step 26) = {158/167}
33a. R3C5 = {56} -> no 5 in R3C6

34. 14(3) cage in N8 (step 30b) = {239/347}, 17(4) cage in N8 (step 33) = {1259/1457}, R78C7 (step 16a) = {27/45}
34a. Consider combinations for 14(3) cage
14(3) cage = {239} => R9C5 = 7, R8C456 = [451] => R78C7 = [72] => R7C3 = 6
14(3) cage = {347} = [437] => R7C3 = 6
34b. -> R7C3 = 6, R8C23 = {37}, locked for R8, clean-up: no 2 in R7C7 (step 16a)

35. Naked quad {1245} in R8C4567, locked for R8 -> R89C1 = [94], R8C9 = 6
35a. 24(4) cage in N9 must contain 6,9 = {2679/4569}
35b. 2,5 only in R7C9 -> R7C9 = {25}
35c. Naked triple {257} in R7C679, locked for R7

36. Naked triple {479} in R367C8, locked for C8 -> R2C8 = 3, R5C89 = [53], R1C8 = 6, R1C7 = 1 (step 11)

37. 18(4) cage in N1 (step 28) = {1467/3456}
37a. R1C1 = {57} -> no 7 in R23C1

38. 14(4) cage at R2C2 (step 14) = {1247/2345}
38a. 3 of {2345} must be in R3C3 -> no 5 in R3C3

39. 24(4) cage in N9 (step 35a) = {2679/4569} = [9267/4569]
39a. R6C89 = [74] (cannot be [47] which clashes with 24(4) cage in N9, combo blocker)

and the rest is naked singles.


Rating comment. I'll rate my walkthrough at Hard 1.5 because of the analysis in some of my later steps.


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PostPosted: Sun Jul 06, 2008 5:53 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 64 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:3328:3585:3585:3585:6916:4613:4613:4613:1544:3328:2058:2058:6916:6916:1806:3599:3599:1544:7698:7698:2058:6916:11542:1806:3599:5145:5145:2843:7698:7698:11542:11542:11542:5145:5145:4899:2843:3877:3877:3877:11542:2857:2857:2857:4899:2843:3630:3630:11542:11542:11542:3379:3379:4899:3630:3630:5176:2361:11542:5947:5948:3379:3379:2111:5176:5176:2361:5947:5947:5948:5948:2631:2111:3401:3401:3401:5947:3661:3661:3661:2631:
Solution:
+-------+-------+-------+
| 4 6 3 | 5 8 2 | 7 9 1 |
| 9 1 2 | 7 3 6 | 8 4 5 |
| 8 7 5 | 9 4 1 | 2 3 6 |
+-------+-------+-------+
| 1 9 6 | 8 5 3 | 4 7 2 |
| 3 5 8 | 2 7 4 | 1 6 9 |
| 7 2 4 | 1 6 9 | 3 5 8 |
+-------+-------+-------+
| 5 3 7 | 6 2 8 | 9 1 4 |
| 2 4 9 | 3 1 5 | 6 8 7 |
| 6 8 1 | 4 9 7 | 5 2 3 |
+-------+-------+-------+
Quote:
Para: Was having some fun with this one. Also used a technique we don't get to use much in Assassins (LoL). Guess this assassin rates somewhere between 1.00 and 1.25
Andrew: I agree with Para that it's rated between 1.0 and 1.25
CathyW: Hmm - more a 1.5 in my book - quite a challenge
Andrew: Lots of killers
Walkthrough by Para:
Hi all

That was fun. Just being a bit creative on this one. Didn't really take the more conventional opening. Was having some fun with this one. Also used a technique we don't get to use much in Assassins. Guess this assassin rates somewhere between 1.00 and 1.25.

Walk-Through Assassin 64

1. R12C1 = {49/58/67}: no 1,2,3

2. R12C9 = {15/24}: no 3,6,7,8,9

3. 27(4) at R1C5 = {3789/4689/5679}: no 1,2; 9 locked for N2

4. 8(3) at R2C2 = {125/134}: no 6,7,8,9; 1 locked for N1

5. R23C6 = {16/25/34}: no 7,8

6. 30(4) at R3C1 = {6789}

7. 19(3) at R4C9 = {289/379/469/478/568}: no 1

8. 11(3) at R4C1 and R5C6 = {128/137/146/236/245}: no 9

9. 14(4) at R6C2 = {1238/1247/1256/1346/2345}: no 9

10. 13(4) at R6C7 = {1237/1246/1345}: no 8,9

11. 20(3) at R7C3 = {389/479/569/578}: no 1,2

12. R78C4 = {18/27/36/45}: no 9

13. 23(3) at R7C7 = {689} -->> locked for N9

14. R89C1 = {17/26/35}: no 4,8,9
14a. 9 in C1 locked for N1

15. R89C9 = {37} -->> locked for C9 and N9

16. 45 on C9: 2 innies: R37C9 = 10 = [64/82/91] -->> R3C9 = {689}; R7C9 = {124

17. 45 on R89: 4 outies: R7C3467 = 30 = {6789} -->> locked for R7
17a. Clean up: R8C4 = {123}

18. 14(3) at R9C6 = [7]{25}/[8]{15/24}/[9]{14} -->> R9C6 = {789}; R9C78 = {14/15/24/25}= {1|2..},{4|5..}
18a. R7C89 = {14/15/24/25} = {1|2..},{4|5..}

19. 45 on C1: 2 innies: R37C1 = 13 = [94/85] -->> R3C1 = {89}; R3C7 = {45}
19a. Killer Pair {45} in R7C1 + R7C89 -->> locked for R7
19b. R12C1 = {4|5|6..}; R7C1 = {45} -->> 11(3) in R4C1 = {128/137/236} = {6|7|8..}: {146/245} blocked by R12C1 + R7C1 -->> R456C1: no 4,5

20. 45 on R12: 4 outies: R3C3467 = 17 = {1259/1349/1457/2348/2357/2456} = {6|7|8|9..}: {1268/1367} blocked by R3C129
20a. Killer Quad {6789} in R3C129 + R3C3467 -->> locked for R3

21. LOL on N5: R37C5 = R5C46 = {12345} -->> R5C46: no 6,7,8,9

22. 15(3) at R5C2 = {159/168/249/258/267/348/357/456} -->> {6|7|8|9..} in R5C23(no room for these in R5C4)
22a. Killer Quad {6789} in R4C23 + R456C1 + R5C23 -->> locked for N4
22b. 14(3) at R6C2 = {2345}(last combo): no 1 -->> R6C23 = {2|3..},{4|5..}
22c. 11(3) at R4C1 = {128/137}: {236} blocked by R6C23 : no 6; 1 locked for C1 and N4 R456C1 = {2|3..}
22d. Killer Pair {23} in R456C1 + R6C23 -->> locked for N4
22e. Clean up: R89C1 = {26/35}: no 7 = {2|3..},{5|6..}

23. Killer Pair {23} in R7C2 + R89C1 -->> locked for N7
23a. 20(3) at R7C3 = {479/578} = {4|5..}: {569} blocked by R89C1: no 6; 7 locked for N7
23b. Killer Pair {45} in R7C1 + 20(3) cage at R7C3 -->> locked for N7
23c. R89C1 = {26} -->> locked for C1 and N7
23d. R7C2 = 3
23e. Clean up: R456C1 = {137} -->> locked for C1 and N4
23f. R3C2 = 7(hidden in 30(4) cage at R3C1)
23g. 6 in 30(4) cage locked for R4 and N4

24. 13(3) at R9C2 = {19}[3]/[18}[4] : R9C4 = {34}; 1 locked for R9
24a. 1 in N9 locked for R7 and 13(4) cage at R6C7
24b. R7C5 = 2; R8C1 = 2(hidden); R9C1 = 6
24c. 2 in N4 locked for R6
24d. 2 in N5 locked for R5

25. 13(4) at R6C7 = {1345}(last combo): no 6,7
25a. 3 in 13(4) locked for R6 and N6

26. 45 on N1: 1 innie and 1 outie: R1C4 + 3 = R3C1 -->> R1C4 = {56}
26a. R23C6 = {4|5|6..} -->> 27(4) at R1C5 = {3789}: ({4689/5679} blocked by R1C4 + R23C6) -->> {3789} locked for N2
26b. Clean up: R23C6: ={16/25}: no 4 = {5|6..}
26c. Killer Pair {56} in R1C4 + R23C6 -->> locked for N2
26d. Clean up: R5C46: no 3,5(LOL N5)

Now for the Finish
27. 14(3) at R1C2 = {356}(last combo with R1C4 = {56}) -->> locked for R1; R1C3 = 3(only place in 14(3) cage)
27a. 8(3) at R2C2 = {125}(last combo) -->> locked for N1
27b. R1C24 = [65]; R3C1 = 8(step 26); R4C23 = [96]; R7C1 = 5(hidden)

28. R23C6 = {16}(last combo) -->> locked for N2 and C6
28a. R3C5 = 4; R1C6 = 2; R5C6 = 4; R5C4 = 2(LOL N5)
28b. R9C4 = 4

29. R9C23 = {18}(last combo within 13(3) at R9C2) -->> locked for R9 and N7
29a. R8C2 = 4; R6C23 = [24]

30. 14(3) at R9C6 = [7]{25}(last combo) -->> R9C6 = 7; R9C78 = {25} -->> locked for R9
30a. R89C9 = [73]; R78C3 = [79]; R9C5 = 9; R7C6 = 8; R78C4 = [63]
30b. R3C4 = 9; R3C9 = 6; R8C56 = [15]; R23C6 = [61]; R46C6 = [39]; R7C7 = 9

31. 19(3) at R4C9 = [298](last combo)
31a. R2C9 = 5(hidden)

32. 45 on N3: 1 innie: R3C8 = 3

And the rest is all naked and hidden singles

greetings

Para
Walkthrough by CathyW with xy-loop:
Hmm - more a 1.5 in my book - quite a challenge. Contrary to Ruud's comment on the puzzle page, I thought the 45(9) cage was quite helpful for some later eliminations. Key move at step 29 - is there a particular name for this?

Edit: Answering my own question! Having gone through this using JSudoku, my step 29 is apparently an xy-loop.

Further edit - see note after step 17.

Here's my WT - a bit longer than Para's!

Prelims:

a) 8(3) N1 = {125/134} 1 not elsewhere in N1
b) 30(4) r3c12+r4c23 = {6789}
c) 13(2) N1 = {49/58/67}
d) 27(4) N2 = {3789/4689/5679} 9 not elsewhere in N2
e) 7(2) N2 = {16/25/34}
f) 6(2) N3 = {15/24}
g) 13(4) r6c78+r7c89 = {1237/1246/1345} must have 1
h) 23(3) N9 = {689} not elsewhere in N9 -> 10(2) N9 = {37} not elsewhere in N9/c9
-> r7c89+r9c78 = {1245}
-> 14(3) r9c678 = {149/248/158/257} -> r9c6 = (789)
i) 8(2) N7 = {17/26/35}
j) 11(3) r456c1 and r5c678: no 9
k) 20(3) N7: no 1 or 2

1. Innies c1: r37c1 = 13 = [67/76/85/94]: r7c1 = (4567) -> 9 locked r123c1, not elsewhere in N1

2. Innies c9: r37c9 = 10 = [91/82/64] -> r7c9 <> 5
-> 19(3) r456c9 = {289/469/568}

3. Innies r1: r1c159 = 13 = [931/832/841/751/742/634/652/571/562/472/481] -> r1c9 <> 5 -> r2c9 <> 1
Edit: Andrew noted I could also have had r1c5 <> 9 here

4. Innies r5: r5c159 = 19 (no 1) = {289/379/469/478/568}

5. Innies r9: r9c159 = 18 = [693/783/297/387/567/657]
-> r9c1 <> 1 -> r8c1 <> 7
-> r9c5 = (5689)

6. Innies N2: r1c46 + r3c5 = 11 = {128/137/146/236/245}

7. Innies N8: r9c46 + r7c5 = 13
r9c6 = (789) -> r9c4 + r7c5 = 4 or 5 or 6 = {13} or {14/23} or {15/24}
-> r9c4 + r7c5 = (1…5)

8. Outies N1: r1c4 + r4c23 = 20 = {677/686/695/785/794/893}
-> r1c4 = (34567)

9. Outies N3: r1c6 + r4c78 = 13

10. Outies N7: r9c4 + r6c23 = 10 = [127/136/145/217/226/235/316/325/334/415/424/514/523]
({118} blocked by r7c1 min 4) -> r6c23 <> 8

11. Outies N9: r6c78 + r9c6 = 15 = {17/26/35}7; {16/25/34}8; {15/24}9

12. Innies c1234: r2346c4 = 25 = {1789/2689/3589/3679/4579/4678}

13. Innies c6789: r4678c6 = 25 (same options as above!)
(Andrew noted that actually {1789} is not possible for this split cage)
14. Outies r12: r3c3467 = 17 -> r3c12589 = 28
Min from r3c129 is 21 {678} -> r3c58 is max 7 -> r3c58 = (1…6)

15. Outies r89: r7c3467 = 30 = {6789}
-> r8c4 = (123)
-> r7c1 = (45) -> r3c1 = (89)
-> pointing cells: r56c2 <> 6,7

16. Outies c123: r159c4 = 11
r1c4 min 3, r9c4 min 1 -> r5c4 max 7
r1c4 max 7, r9c4 max 5 -> r5c4 min 1
-> r5c4 = (1…7)

17. Outies c789: r159c6 = 13
r9c6 min 7 -> r15c6 max 6 -> r15c6 = (1…5) -> r1c78 = (4…9)
-> 27(4) N2 must have both 89 = {3789/4689}
-> split 11(3) N2 must have one of 3 or 4 {137/146/245} ({236} blocked by 27(4))
-> 7(2) N2 <> {34}

Edit: Evidently when I did this, I neglected to remove the 5s from the 27(4) cage which affects step 19 and leads to a naked quad in c1.
Sorry - I don't have the inclination to rework the rest of this WT!


18. Outies – Innies / Law of Leftovers N5: r5c46 = r37c5
-> r3c5, r5c4 = (1…6); r5c6, r7c5 = (1…5)

19. 9 locked to r3456c9 -> r4c78 <> 9
> 9 locked to r456c9 -> r3c9 <> 9 -> r7c9 <> 1
-> 6(2) r12c9 = [15] -> r3c6 <> 2, r1c1 <> 8
-> r1c15 = [48/57/75/93] (r1c1 <> 6, r1c5 <> 4,6,9) -> r2c1 <> 7
-> r3c3 <> 2
-> 14(3) N3 = {239/248/347} no 6
-> Options for split 13(3) r1c6+r4c78 = {157/247/337/355/445}

20. 1 locked to r23c6/r3c5 in N2 -> r46c6 <> 1

21. 2 locked to r12c6/r3c5 in N2 -> r46c6 <> 2

22. 1 locked to r6c78+r7c8 of 13(4) -> r45c8 <> 1

23. 2 and 4 locked to r4567c9 -> r6c78 <> 2, 4

24. 6 and 8 locked to r3456c9 -> r4c78 <> 6, 8

25. 3 locked to r7c25
a) r7c2 = 3 -> r5c2 <> 3
b) r7c5 = 3 -> one of r5c46 = 3 -> r5c2 <> 3
Either case, r5c2 <> 3

26. 14(4) r6c23+r7c12: Max from r6c2+r7c12 = 12 {345} -> r6c3 <> 1

27. 8 locked to cage 30(4) r3c12+r4c23 and r23456c1
If one of r456c1 = 8 -> r3c2 = 8 -> r1c23 <> 8
If one of r4c23 = 8 -> r2c1 = 8 -> r1c23 <> 8
If one of r2c1,r3c12 = 8 -> r1c23 <> 8
All options r1c23 <> 8 -> 14(3) r1c234 = {257/347/356}

28. Killer pair {12} in c6: 7(2) r23c6 = {16}/[25]; split 13(3) r159c6 = {157/247/148/238/139}
-> r8c6 <> 1,2

29. a) r7c1 = 4 -> r3c1 = 9
-> r7c9 = 2 -> r3c9 = 8 -> r2c1 = 8
b) r7c1 = 5 -> r3c1 = 8
-> r3c9 = 6 -> r7c9 = 4
Either case r3c247 <> 8; r7c258 <> 4
-> 8 locked to r23c1 -> r456c1 <> 8

30. Split 11(3) N2 = {137/146/245} Combo analysis: r3c5 = (1245); r1c4 <> 3
-> r5c4 <> 6

31. Combo analysis of split 13(3) r159c6: r5c6 <> 5

32. 13(4) r6c78+r7c89: r7c89 can’t be [12] (forces {45} to r9c78 not possible for 14(3))
-> combination {1237} not possible -> r6c78 <> 7
-> options now: {1246/1345} -> r7c9 = 4
-> r3c9 = 6 -> r3c1 = 8 -> r7c1 = 5
-> 19(3) r456c9 = {289} 2,8 not elsewhere in N6
-> r3c2 = 7 -> r4c23 = {69} not elsewhere in N4/r4
-> 13(2) N1 = {49} not elsewhere in N1/c1
-> 8(3) N1 = {125} -> r3c3 = 5 -> r3c6 = 1 -> r2c6 = 6

Fairly straightforward singles and cage combinations from here
Walkthrough by Andrew:
Only finished it yesterday and then worked through Para's and Cathy's walkthroughs today before deciding to post it. I agree with Para that it's rated between 1.0 and 1.25

I seem to have done more with the combinations for sub-cages, which is unusual for me, but they were helpful this time. I would have finished earlier but I found I'd deleted one combination too many from a sub-cage, which happened to be part of the solution. :oops: Still that's better than deleting a different combination and having a flawed solution path. :wink: Lots of killers, which I enjoyed.

Here is my walkthrough for A64.

1. R12C1 = {49/58/67}, no ,1,2,3

2. R12C9 = {15/24}

3. R23C6 = {16/25/34}, no 7,8,9

4. R78C4 = {18/27/36/45}, no 9

5. R89C1 = {17/26/35}, no 4,8,9

6. R89C9 = {19/28/37/46}, no 5

7. 8(3) cage in N1 = 1{25/34}, 1 locked for N1

8. R456C1 = {128/137/146/236/245}, no 9

9. R456C9 = {289/379/469/478/568}, no 1

10. R5C678 = {128/137/146/236/245}, no 9

11. 20(3) cage in N7 = {389/479/569/578}, no 1,2

12. 23(3) cage in N9 = {689}, locked for N9, clean-up: no 1,2,4 in R89C9
12a. Naked pair {37} in R89C9, locked for C9 and N9

13. 27(4) cage in N2 = {3789/4689/5679} = 9{378/468/567}, no 1,2, 9 locked for N2

14. 30(4) cage at R3C1 = {6789}

15. 14(4) cage at R6C2 = {1238/1247/1256/1346/2345}, no 9

16. 13(4) cage at R6C7 = {1237/1246/1345} = 1{237/246/345}, no 8,9

17. 45 rule on R5 3 innies R5C159 = 19 = {289/379/469/478/568}, no 1

18. 45 rule on N5 2 outies R37C5 = 2 innies R5C46, specifically that R37C5 and R5C46 must contain the same two numbers, in other words Law of Leftovers.

19. 45 rule on C1 2 innies R37C1 = 13 = {67}/[85/94], R7C1 = {4567}

20. 9 in C1 locked in R123C1, locked for N1
20a. 9 in C1 locked in R123C1 -> 4 in C1 must be in R12C1 or R7C1 (R12C3 and R37C1 are both 13(2) cages), locked for C1
20b. R456C1 (step 8) = {128/137/236}, no 5

21. 45 rule on C9 2 innies R37C9 = 10 = [64/82/91]

22. R9C678 = {149/158/248/257} -> R9C6 = {789}
22a. R9C78 = {14/15/24/25} -> R7C89 = {14/15/24/25} (cannot be {12/45} which clash with R9C78)
22b. 13(4) cage at R6C7 = {1246/1345} (cannot be {1237} because R7C89 cannot be {12} from step 22a), no 7
22c. {1246} = {16/26} in R6C78, {14/24} in R7C89
22d. {1345} = {34/35} in R6C78, {14}/[51] in R7C89
22e. R6C78 = {16/26/34/35}, R7C89 = {14/24}/[51] -> R9C78 = {15/24/25}
22f. R456C9 (step 9) = {289/469/568}
22g. R6C78 = {16/34/35} (cannot be {26} which clashes with R456C9), no 2

23. 45 rule on C789 3 outies R159C6 = 13 = {139/148/157/238/247} (cannot be {256/346} because only 7,8,9 in R9C6) -> R15C6 = {13/14/15/23/24}, no 6,7,8
23a. If R15C6 = {23/24} -> R23C6 = {16}
23b. Killer single 1 in R15C6 and R23C6, locked for C6
[Alternatively 45 rule on C6 4 innies R4678C6 = 25 but cannot be {1789} which clashes with R9C6 -> no 1]

24. 45 rule on C123 3 outies R159C4 = 11 = {128/137/146/236/245}, no 9

25. 45 rule on R89 4 outies R7C3467 = 30 = {6789}, locked for R7, clean-up: no 6,7 in R3C1 (step 19), R8C4 = {123}

26. Killer pair 4,5 in R7C1 and R7C89 (step 22e), locked for R7

27. 45 rule on N1 2 innies R3C12 – 10 = 1 outie R1C4, min R3C12 = 14, max R3C12 = 17 -> R1C4 = {4567}

28. 45 rule on N7 2 innies R7C12 – 4 = 1 outie R9C4
28a. R7C12 = [42/43/51/52/53] (cannot be [41] which clash with R7C89) -> R9C4 = {234}

29. 45 rule on N8 3 innies R7C5 + R9C46 = 13 = {139/148/247} (cannot be {238} which clashes with R78C4)
29a. 1 of {139} in R7C5 -> no 3 in R7C5
29b. 2 of {247} in R7C5 -> no 2 in R9C4

30. R7C2 = 3 (hidden single in R7), clean-up: no 5 in R89C1
30a. 14(4) cage at R6C2 = {1346/2345} (cannot be {1238 because R7C1 only contains 4,5} = 34{16/25}, no 7,8

31. 3 in C1 locked in R456C1 (step 20b) = {137/236}, no 8, 3 locked for N4
31a. Killer pair 1,2 in R6C23 (step 30a) and R456C1, locked for N4
31b. Min R5C23 = 9 -> max R5C4 = 6, clean-up: no 7,8 in R3C5 (step 18)

32. Hidden killer pair 1,2 in R89C1 and R9C23 -> R9C23 must contain 1/2
32a. R9C234 = {139/148/238/247} (cannot be {157/256} because R9C4 only contains 3,4), no 5,6
32b. 20(3) cage in N7 = {479/569/578}
32c. Killer pair 4,5 in R7C1 and 20(3) cage, locked for N7
32d. Killer pair 6,7 in R89C1 and 20(3) cage, locked for N7
32e. R9C234 = {139/148/238}

32. 8 in C1 locked in R123C1, locked for N1
32a. Killer pair 6,7 in R456C1 (step 31) and R89C1, locked for C1
32b. Killer pair 4,5 in R12C1 and 8(3) cage, locked for N1

33. R1C234 = {257/347/356}
33a. 4,5 only in R1C4 -> R1C4 = {45}
33b. 3 of {356} in R1C3 -> no 6 in R1C3

34. R3C12 – 10 = R1C4 (step 27)
34a. R1C4 = {45} -> R3C12 = 14,15 = [86/87/96], R4C23 = {69/78/79}
34b. 45 rule on R123 5 innies R3C12589 = 28, min R3C1259 = 21 -> max R3C8 = 7

35. 45 rule on N2 3 innies R1C46 + R3C5 = 11, max R1C46 = 9 -> min R3C5 = 2
35a. R1C46 + R3C5 = {146/245} (cannot be {236} because only 4,5 in R1C4) = 4{16/25}, no 3, 4 locked for N2, clean-up: no 3 in R23C6, no 3 in R5C46 (step 18)
35b. Killer pair 5,6 in R1C46 + R3C5 and R23C6, locked for N2
35c. 1 in N2 locked in R123C6, locked for C6

36. R5C678 = {128/146/236/245} (cannot be {137} because R5C6 only contains 2,4,5), no 7

37. 7 in N6 locked in R4C78, locked for R4 and 20(4) cage at R3C8
[This makes R3C2 a hidden single in the 30(4) cage but step 37a does more!]
37a. R4C23 (step 34a) = {69} (only remaining combination), locked for R4, N4 and 30(4) cage at R3C1 -> R3C12 = [87], R7C1 = 5 (step 19), clean-up: no 2 in R456C1 (step 31), no 2 in R7C9 (step 21)

38. Naked triple {137} in R456C1, locked for C1 and N4

39. Naked pair {24} in R6C23, locked for R6 and N4
39a. Naked pair {58} in R5C23, locked for R5 -> R5C4 = 2, R5C6 = 4 -> R37C5 = [42] (step 18), R9C46 = [47] (step 29), R1C4 = 5, R89C9 = [73], clean-up: no 2 in R23C6, no 1 in R2C9

40. R7C3 = 7 (hidden single in C3), R1C6 = 2 (hidden single in N2), R1C23 = [63], clean-up: no 4 in R2C9

41. Naked pair {26} in R89C1, locked for N7
41a. 20(3) cage in N7 = {479} (only remaining combination) -> R8C23 = {49}, locked for R8 and N7

41. Naked pair {14} in R7C89, locked for 13(4) cage at R6C7
41a. 13(4) cage at R6C7 = {1345} (last remaining combination) -> R6C78 = {35}, locked for R6 and N6

42. 9 in N6 locked in R456C9, locked for C9 -> R3C9 = 6, R5C9 = 9, R6C9 = 8, R4C9 = 2, R7C9 = 4 (step 21), R7C8 = 1, R12C9 = [15], R23C6 = [61]

43. Naked pair {68} in R8C78, locked for R8 and N9 -> R7C7 = 9, R89C1 = [26]

44. R5C78 = [16]

45. R1C6 = 2 -> R1C78 = 16 = [79]

46. R3C9 = 6, R4C78 = {47} -> R3C8 = 3 (cage sum)

and the rest is naked singles


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PostPosted: Sun Jul 06, 2008 5:57 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 64v2 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:2816:4353:4353:4353:5124:3845:3845:3845:3080:2816:4106:4106:5124:5124:2062:4367:4367:3080:3602:3602:4106:5124:11542:2062:4367:4121:4121:5147:3602:3602:11542:11542:11542:4121:4121:3107:5147:2341:2341:2341:11542:5673:5673:5673:3107:5147:5934:5934:11542:11542:11542:3891:3891:3107:5934:5934:2872:1593:11542:6459:4412:3891:3891:2367:2872:2872:1593:6459:6459:4412:4412:2887:2367:6217:6217:6217:6459:2637:2637:2637:2887:
Solution:
+-------+-------+-------+
| 2 1 7 | 9 6 3 | 4 8 5 |
| 9 5 3 | 4 8 1 | 2 6 7 |
| 4 6 8 | 2 5 7 | 9 3 1 |
+-------+-------+-------+
| 8 3 1 | 6 4 9 | 5 7 2 |
| 7 4 2 | 3 1 5 | 8 9 6 |
| 5 9 6 | 8 7 2 | 3 1 4 |
+-------+-------+-------+
| 1 7 4 | 5 3 8 | 6 2 9 |
| 3 2 5 | 1 9 6 | 7 4 8 |
| 6 8 9 | 7 2 4 | 1 5 3 |
+-------+-------+-------+
Quote:
CathyW: According to Ed's rating list from Sudoku Solver... the V2 has a marginally lower rating than the V1 - 1.28 compared to 1.29 presumably indicating that it shouldn't be any more difficult. But I beg to differ - completely stumped after 26 steps.
mhparker: Just shows that there's some more tweaking for Ed to do before we can replace the manual ratings with the automatic ones.
mhparker: it was considerably harder than the V1, because of the very narrow solving paths however it was no monster. I would give it a clear rating of 1.75
Tag solution: by CathyW, mhparker, Howard S, rcbroughton, Para, Glyn & goooders
Andrew (in 2014): When this puzzle was originally posted, I wrote the Consolidated Walkthrough but hadn't taken part in the "tag", so I've now tried this puzzle.
It's hard to know what rating to give my walkthrough. All I'll say is that it was harder than A65 V2 and A65 V3, both of which I've solved recently; also that my hardest step for this puzzle was more difficult than any steps I used for those two puzzles.
A 2021 Revisit thread here
Consolidated Walkthrough by Andrew:
Assassin 64V2 was solved as a "tag" solution by Cathy, Mike, Howard, Richard, Para, Glyn & goooders. The original moves are earlier in this thread.

I had hoped to take part in the later stages, after I had solved Assassin 64 but the late spurt beat me to it. That was why I volunteered to write this after Cathy asked if anybody was inclined to do it.


Consolidated Walkthrough for Assassin 64V2

I’ve tried to keep fairly closely to the moves as posted in the “tag” solution in this thread. Some editing has been done to maintain a consistent format. I've retained the original step numbers; un-numbered steps have been added to existing steps. Alternatives, given as [Alternative …] after the original step, were posted in the “tag” solution. A few “withdrawn” moves have been retained for completeness.

Prelims

a) 24(3) r9c234 = {789) not elsewhere in r9 -> r8c1 <> 1,2; r8c9 <> 2,3,4
b) 22(3) r5c678 = {589/679} 9 not elsewhere in r5
c) 11(2) n1 and n9 – no 1
d) 8(2) n2 = {17/26/35}
e) 12(2) n3 = {39/48/57}
f) 20(3) r456c1 = {389/479/469/578} – no 1,2
g) 9(2) n7 = [36/45/54/63/72/81]
h) 6(2) n8 = {15/24}
i) 14(4) at r3c1 = {1238/1247/1256/1346/2345} – no 9
j) 9(3) r5c234 = {126/135/234}
k) 11(3) n7 = {128/137/146/236/245} – no 9

1. Innies c1: r37c1 = 5 = {14/23} -> max 4 in r7c1 -> r6c23, r7c2 <> 1

2. Innies c9: r37c9 = 10 = {19/28/37/46} – no 5

3. Outies r12: r3c3467 = 26 -> r3c3467 <> 1 -> r2c6 <> 7

4. Outies r89: r7c3467 = 23

5. Outies n1: r1c4 + r4c23 = 13

6. Outies n3: r1c6 + r4c78 = 15

7. Outies n7: r6c23 + r9c4 = 22 = {499/589/679/688/778} -> r6c23 <> 2,3

8. Outies n9: r6c78 + r9c6 = 8 = {116/125/134/224/233} -> r6c78 <> 7,8,9, r9c6 <> 6

9. Innies r1: r1c159 = 13 -> r1c5 <> 9 (no 1 in r1c19)

10. Innies r5: r5c159 = 14

11. Innies r9: r9c159 = 11 = {146/236/245} -> 10(3) r9 = {235/145/136}

12. LOL n5: r37c5 = r5c46

13. r9c5 max 6 -> r7c6, r8c56 <> 1

14. 25(4) n8 = {1789/2689/3589/3679/4579/4678} Must have at least two of 7,8,9
-> Killer triple with r9c4 -> r7c5 <> 7,8,9

15. Innies n2: r1c46 + r3c5 = 17

16. Innies n8: r7c5 + r9c46 = 14 = {167/347/158/248/239} (options {257/149} excluded as would eliminate all options for 6(2) n8)

17. Outies c123: r159c4 = 19 -> r15c4 <> 1
Max from r59c4 = 6+9 = 15 -> r1c4 <> 2,3
Options for split 19(3): [469/568/649/658/739/748/829/847/928/937]

18. Outies c789: r159c6 = 12
Min from r59c6 = 5+1 = 6 -> r1c6 max 6

19. No 1 in r5c46 -> r37c5 <> 1 (step 12)

20. Innies c1234: r2346c4 = 20

21. Innies c6789: r4678c6 = 25

22. 6(2) r78c4 = {15/24} -> r23c4 of 20(4) and r46c4 of 45(9) can’t have both 12/14/25/45

23. Innies n1: r1c23 + r3c12 = 18

24. Innies n3: r1c78 + r3c89 = 16

25. Innies n7: r7c12 + r9c23 = 25 -> r7c1 max 4 -> r7c2 min 5

26. Innies n9: r7c89 + r9c78 = 17

27. 25(4) n8 (step 14): {1789} combo blocked by r9c4
27a. -> no 1 in r9c5

28. Outies n6: r37c89 + r5c6 = 20(5)
28a. from step 2: r37c9 = 10(2) -> r37c8 + r5c6 = 10(3) -> no 8,9 in r5c6
28b. Combos for split 10(3) r5c6 + r37c8 = {127/136/145/235}; r5c6 = {567} -> r37c8 = {1234}

29. Split 12(3) r159c6 (step 18): {237/156} both blocked by 8(2) n2; 8,9 no longer available
29a. -> r159c6 = {147/246/345}
29b. -> must have 2 of {1234}, only available in r19c6
29c. -> no 5,6 in r19c6
29d. 4 locked in r19c6 for c6

30. 9 of 22(3) locked in r5c78 for n6
30a. {89} only available in r5c78
30b. -> no 5 in r5c78

31. LoL n5 (step 12): no 8,9 in r3c5

32. 10(3) r9 (step 11) must contain 1 of {56}, only available within n9
32a. -> 11(2) n9 <> {56}

33. r7c12 = 8, 9 or 10 (because r7c12 + r9c23 = 25)
-> r6c23 = 13, 14, or 15
-> r45c23 = 10, 11 or 12
-> r4c23 <> 7,8

34. Combination options for split 17(3) n2: {179/269/278/359/368/458/467}
Analysis: r1c4 <> 4,5, r3c5 <> 2,3,4

35. Outies n4: r37c2 + r5c4 = 16 (from step 1, r37c1 = 5)

36. r1c6 + r4c78 = 15 (step 6) = {168/258/267/348/357/447/456}
Analysis: r4c78 <> 1,2
-> CPE 1,2 in n6 locked to r45c9, r6c789 -> r7c9 <> 1,2 -> r3c9 <> 8,9 (step 2)

37. 15(3) r1c678: r1c6 is max 4 -> r1c78 min 11 -> r1c78 <> 1

38. Split 19(3) r159c4 (step 17) = {289/379/469/478/568}
Analysis: r5c4 <> 6
-> (from step 35) max from r5c4+r7c2 = 5+9 = 14 -> r3c2 <> 1

39. CPE 6 locked to r12346c4 -> r3c5 <> 6
-> split 17(3) n2 must now have one of 5,7. If combo includes 7 it must go in r3c5 -> r1c4 <> 7

40. Split 17(4) r7c89+r9c78 = {1259/1268/1349/1358/1367/1457/2348/2357/2456}
Analysis: r7c9 <> 3 -> r3c9 <> 7 (step 2)

41. 9 locked to r3c347 within split 26(4) -> 26(4) r3c3467 = {9...}
Options: {2789/3689/4589/4679}
If {4589} r3c6 = 5 -> r3c347 <> 5

42. No 4 in r6c23. Here's how:
42a. If r9c4=7, then r6c23 = 15
42b. If r9c4=8, then r6c23 = 14
42c. If r9c4=9, then r6c23 = 13 -> r7c12 = 10 = [19/46] = [4/9, ..]
(not [28/37] because 9 in r9c4 forces {78} into r9c23)
-> {49} combo in r6c23 blocked
42d. -> no 4 in r6c23

43. Outies n3: r1c6 + r4c78 = 15 (step 6)
43a. r4c78 max is 13 to fit with the 16(4) cage
43b. -> r1c6 <> 1

44. Outies n9: r6c78 + r9c6 = 8 (step 8)
44a. Options for 12(3) r456c9 are {138/147/156/246} ({237} is blocked by 11(2) n9, {345} is blocked 12(2) n3)
44b. Only possibility for r6c78 + r9c6 with 6 is {16}1 but {16} would block all possibilities in 12(3)n6
44c. -> r6c78 <> 6

45. 15(4) at r6c7 - only combinations with a 4 also require an 8, 7 or 6 which only occur at r7c9
45a. -> r7c9 <> 4
45b. -> r3c9 <> 6 (step 2)
[Alternatively. Innies/outie n9: r9c6 + 7 = r7c89 -> r7c89 = 8/9/10/11 -> r7c9 <> 4, r3c9 <> 6 (step 2).]

46. Split 25(4) in r4678c6 must have 8 and 9. Options: {1789/2689/3589}

47. Innies r123: r3c12589 = 19. Max from r1589 = 2+3+4+7 = 16 -> r3c2 <> 2

48. r3c12589 (step 47) = 19(5) = {12349}/{12358}/{12367}/{13457}/{13456}
48a. {12349} - no 9 so not possible
48b. {12358} - r3c5 = 5, r3c189={123} -> r3c2=8
48c. {12367} - r3c5 = 7, r3c189={123} -> r3c2=6
48d. {13457} - r3c5 = 5/7, r3c189={134} -> r3c2=7/5
48e. {13456} - r3c5 = 5, r3c189={134} -> r3c2=6
48f. -> r3c2 <> 3,4
[Two alternatives were also posted.
Alternative 1. r3c12589 (step 47) = 19(5) = {12358/12367/12457/13456} -> needs 3 of {1234} which need to go in r3c189, so nowhere else in r3c12589 -> r3c2 = {5678}.
Alternative 2. If r3c2 = 3/4 it forms quadruplet r3c1289={1234}, requires r3c5 = 9 -> r3c2 <> 3,4.]

49. No 5 in 20(4) n2. Here's how:
49a) If r3c5 = 7, r1c46 + r3c5 = 17 (step 15) -> r1c46 = 10 = [64/82]
Taken together they block {17} and {26} combos for 8(2) cage. Only remaining alternative is {35}
49b) Alternative is r3c5=5
49c) -> no 5 in 20(4) cage -> r1c5, r2c45 <> 5

50. CPE r3c5 = 5 or r23c6 = {35} (step 49) -> r46c6 <> 5

51. If 12(3) n6 = {147} => 12(2) n3 = {39} together block all combos for 11(2) n9 ->
r456c9 <> 7
[The poster of step 51 hinted that r3c2 <> 5 from a forcing chain through n6 and back into r3. This wasn’t used in the “tag” solution. You can work it out for your own amusement; the first part of the chain was used in step 54.]

52. Outies n9: r6c78 + r9c6 = 8 (step 8). = {152}/{134}/{224}/{233} (we already removed {116})
52a. {224} - means r7c89 must total 9= [18/36] and r9c78 totals 8 ={53}
52aa. r7c89 = [18] is possible
52ab. r7c89 = [36] and r9c78 = {53} contradiction - so not possible
52b. {233} - means r7c89 must total 10 = [19/46] and r9c78 totals 7 = {25} ({16} would clash with r7c89)
52ba. r7c89 = [19] would force 12(3) n6 = 1{47}/{56} and r3c9 = 1 (step 2) contradiction - so not possible
52bb. r7c89 = [46] would force r9c78 = {25} but that would eliminate all candidates in r9c9
52c. Remaining options for r6c78 + r9c6 are {152}/{134}/{224}

53. 9(2) n7 cannot be {45} - blocks 11(2) n1 and 20(3) n4 - 11(2) would be forced to {29} or {38} and 20(3) to {389} – contradiction

54) Forcing Chain
If r3c5 = 7 => r5c678 = 7{69} => r7c9 = 6 (hidden single in c9) => 25(4) n8 must contain 6 whilst blocked for 7 in c56. Only combo is {2689} forcing r78c4 = {15}
Alternative r3c5=5
CPE r3c5 =5 or r78c4 = {15} -> r46c4, r789c5 <> 5
[Alternatively. If r3c5 = 7 => r5c678 = 7{69} => r7c5 <> 6 (LoL n5, step 12) …]

At this stage an implication chain and its results were expressed verbally. The first parts of this were re-phrased after steps 55 to 65 had been posted and are included as step 66.

55. 5 in r9 locked in r9c78, not elsewhere in n9
55a. 10(3) r9c678 = {145/235}, no 6

56. No 6 in r7c5. Here's how.
56a. Outies c123 (step 17): r159c4 = 19(3)
56b. Innies n8 (step 16): r7c5 + r9c46 = 14(3), r7c5 can only be 6 if r9c4 = 7
56c. Only options with 7 in r9c4 are [847/937]
56d. -> r7c5 = {34} (LoL n5, step 12)
56e. -> Innies n8 = {347}
56f. -> {167} blocked for n8 innies (step 16)
56g. -> No 6 in r7c5

57. LoL, n5 (step 12): no 6 in r5c6

58. 22(3)n56 = [5]{89}/[7]{69}
58a. -> no 7 in r5c78

59. 7 in n6 locked in r4c78, not elsewhere in r4
59a. 16(4) at r3c8 = {(126/135/234)7}, no 8

60. 6 in n8 locked in 25(4) = {2689/3679/4678}, no 5

61. 5 in n8 locked in r78c4
61a. -> 6(2) n8 = {15}, locked for c4 and n8

62. Both of {16} now unavailable to c789 outies
62a. -> r159c6 = {345} (only remaining combo, see step 29a)
62b. -> r5c6 = 5
62c. r19c6 = {34}, locked for c6

63. r3c5 = 5 (hidden single in n2)

64. From 62b 22(3) r5c678 = 5{89}, {89} locked for r5/n6

65. 12(3) n6 - {138} not now possible because no 8, so no 3 either (combo {345} was eliminated in step 44a)
65a. -> 12(3) n6 = 6{15}/{24} - 6 locked for n6 and c9
65b. -> no 4 at r3c9 (step 2)

This is the first part of the implication chain that was expressed verbally after step 54.
66. 1 in c1 locked to split 5(2) r37c1 and r9c1
66a. If r9c1 = 1 -> r8c1 = 8 -> r9c23 = {79}
-> split 5(2) r37c1 = {23}
-> 11(3) n7 = {245}
-> r7c1 = 3, r7c2 = 6 -> r6c23 = {59}
BUT 20(3) r456c1 requires at least one of 8,9 CONFLICT -> r89c1 <> [81]
66b. Split 5(2) r37c1 = {14} -> 9(2) r89c1 = {36}/[72]
If 9(2) = [72] no options for 11(3) n7 ({146} clashes with r7c1)
-> 9(2) = {36} not elsewhere in n7
-> 11(2) r12c1 = {29} not elsewhere in n1
-> 20(3) r456c1 = {578} not elsewhere in n4

Now follow through newly revealed naked single and naked pairs and the rest is straightforward.



Mike has posted an alternative solution starting from the position after step 27. He has also posted a message showing how SudokuSolver and JSudoku came up with different ways to solve this puzzle
Alternative ending to tag solution by mhparker:
Marks pic after original step 27:

Code:
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 23456789  | 123456789   123456789   456789    | 12345678  | 123456      123456789   123456789 | 345789    |
|           :-----------------------.-----------'           :-----------.-----------------------:           |
| 23456789  | 123456789   123456789 | 123456789   123456789 | 12356     | 123456789   123456789 | 345789    |
:-----------'-----------.           |           .-----------:           |           .-----------'-----------:
| 1234        12345678  | 23456789  | 23456789  | 23456789  | 23567     | 23456789  | 123456789   12346789  |
:-----------.           '-----------+-----------'           '-----------+-----------'           .-----------:
| 3456789   | 12345678    12345678  | 123456789   123456789   123456789 | 123456789   123456789 | 123456789 |
|           :-----------------------'-----------.           .-----------'-----------------------:           |
| 345678    | 123456      123456      23456     | 12345678  | 56789       56789       56789     | 12345678  |
|           :-----------------------.-----------'           '-----------.-----------------------:           |
| 3456789   | 456789      456789    | 123456789   123456789   123456789 | 123456      123456    | 123456789 |
:-----------'           .-----------+-----------.           .-----------+-----------.           '-----------:
| 1234        56789     | 12345678  | 1245      | 23456     | 23456789  | 123456789 | 123456789   12346789  |
:-----------.-----------'           |           :-----------'           |           '-----------.-----------:
| 345678    | 12345678    12345678  | 1245      | 23456789    23456789  | 123456789   123456789 | 56789     |
|           :-----------------------'-----------:           .-----------'-----------------------:           |
| 123456    | 789         789         789       | 23456     | 12345       123456      123456    | 23456     |
'-----------'-----------------------------------'-----------'-----------------------------------'-----------'


28. 9(2)n7 cannot be {45}. Here's how:
28a. Either 11(2)n1 = {(4/5)..}, or
28b. 11(2)n1 = {(2/3)..} -> h5(2) at r37c1 = {14}
28c. either way, {45} blocked for 9(2)n7

29. 9(2)n7 = {36}/[72]/[81] = {(1/3/7)..}
29a. -> {137} combo blocked for 11(3)n7
29b. -> 11(3)n7 = {128/146/236/245} (no 7)

30. 11(3)n7 cannot be {146}. Here's how:
30a. 1 in c1 locked in r379
30b. thus, either r9c1 = 1, or
30c. r37c1 = {14}
30d. either way, {146} is blocked for 11(3)n7

31. 11(3)n7 now = {128/236/245}
31a. 2 locked for n7
31a. cleanup: no 7 in r8c1, no 3 in r3c1

32. 9(2)n7 = {36}/[81] = {(1/3)..},{(3/8)...}
32a. -> {38} combo blocked for 11(2)n1
32b. -> {389} combo blocked for 20(3)n4 = {479/569/578} (no 3) = {(4/6/8)..}, {(5/9)..}, {(7/9)..}

33. 2 in c1 locked in n1 -> not elsewhere in n1

34. 3 in c1 locked in n7 -> not elsewhere in n7

35. 11(3)n7 now = {128/245} (no 6)

36. r7c12 cannot contain both of {45}. Here's how:
36a. r7c12 = [45] -> 23(4)n47 = {68}[45]
36b. but this is blocked by 20(3)n4 (step 32b)

37. only other place for {45} in n7 is 11(3)n7
37a. -> 11(3)n7 = {(4/5)..}
37b. -> 11(3)n7 = {245}, locked for n7
37c. cleanup: no 1 in r3c1

38. r7c2 cannot contain a 6. Here's how:
38a. r7c2 = 6 -> 23(4)n47 = {79}[16]/{59}[36]
38b. but both of these are blocked by 20(3)n4 (step 32b)
38c. -> no 6 in r7c2

39. 6 in n7 locked in 9(2)n7
39a. -> 9(2)n7 = {36}, locked for c1
39b. cleanup: no 5 in r12c1

40. HS in c1 at r7c1 = 1
40a. -> r3c1 = 4
40b. cleanup: no 7 in r12c1

41. 11(2)n1 = {29}
41a. 9 locked for c1 and n1

42. 20(3)n4) = {578}, locked for n4

43. Split 22(3) at r6c23+r7c2 = {679} (only possible combo)
43a. -> r7c2 = 7; r6c23 = {69}, locked for r6 and n4

44. HS in r7 at r7c2 = 7

..
How Solvers tackled A64v2 by mhparker:
I wanted to make a post on how automated solvers tackled this problem:

SudokuSolver (SS)

The key to the solution path used by SS were the conflicting partial combinations between 20(3)n4 and 23(4)n47, including several I did not list above. Although the two cages do not completely "see" each other (r7c2 is not a peer of r456c1), the interaction between them was greatly enhanced by the fact that there was no overlap between the candidates of r7c1 and r7c2. Thus, for any digit pair in r7c12, there was only one possible permutation.

However, SS did not use my step 30 above, and hence did not deduce 9(2)n7 = {36} by locking the 6 of n7 within these two cells. Instead it locked the 8 into the h25(4) n7 innies, thus eliminating [81] as an option for 9(2)n7, leaving {36} as the only remaining combo.


JSudoku (JS)

As mentioned by Glyn above, JSudoku (version 0.6b2 in my case) loops on this puzzle. So (as Ruud said), it does indeed require a "large number of steps" to solve this puzzle. An infinite number in fact!

But before it looped, it noticed the following move (which I've simplified quite a bit), which could have been used to break through the deadlock around our original step 53, where we could not find a way to progress without using hypotheticals (see marks pic below):

Code:
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 23456789  | 123456789   123456789   689       | 1234678   | 234         23456789    23456789  | 345789    |
|           :-----------------------.-----------'           :-----------.-----------------------:           |
| 23456789  | 123456789   123456789 | 12346789    12346789  | 12356     | 123456789   123456789 | 345789    |
:-----------'-----------.           |           .-----------:           |           .-----------'-----------:
| 1234        5678      | 2346789   | 2346789   | 57        | 23567     | 2346789   | 1234        1234      |
:-----------.           '-----------+-----------'           '-----------+-----------'           .-----------:
| 3456789   | 123456      123456    | 123456789   123456789   1236789   | 345678      345678    | 1234568   |
|           :-----------------------'-----------.           .-----------'-----------------------:           |
| 345678    | 123456      123456      2345      | 12345678  | 567         6789        6789      | 1234568   |
|           :-----------------------.-----------'           '-----------.-----------------------:           |
| 3456789   | 56789       56789     | 123456789   123456789   1236789   | 12345       12345     | 1234568   |
:-----------'           .-----------+-----------.           .-----------+-----------.           '-----------:
| 1234        56789     | 12345678  | 1245      | 23456     | 2356789   | 123456789 | 1234        6789      |
:-----------.-----------'           |           :-----------'           |           '-----------.-----------:
| 3678      | 12345678    12345678  | 1245      | 23456789    2356789   | 123456789   123456789 | 789       |
|           :-----------------------'-----------:           .-----------'-----------------------:           |
| 1236      | 789         789         789       | 23456     | 1234        123456      123456    | 234       |
'-----------'-----------------------------------'-----------'-----------------------------------'-----------'


The key move (and its follow-up) JS found here can be simplified down as follows:

54. outies c123 = r159c4 = 19(3) cannot contain both of {67} ({667} not possible)
54a. -> split 20(4) at r2346c4 must contain at least 1 of {67}
54b. -> innies n2 (r1c46+r3c5) cannot contain both of {67}
54c. -> {467} blocked
54d. -> no 6 in r1c4
54e. r159c4 = [829/847/928/937]
54f. -> no 5 in r5c4

55. LoL(n5): r3c5 = r5c6, r5c4 = r7c5
55a. -> no 6 in r5c6; no 5,6 in r7c5
55b. cleanup: no 7 in r5c78

now continue as for step 59 in original tag WT


This begs the question as to why JS did not see the key conflicting partial combinations that SS did. Maybe it would have if it had not looped, but I doubt it. I suspect that the reason is that JS only detects conflicting partial combinations occuring within a single house. For example, if 20(3)n4 = {389/479/569/578} = {(5/9}..}, {(7/9}..}, {(8/9}..}, then JS would recognize that 23(4)n47 cannot have both of {59}, both of {79} or both of {89} within n4, but would ignore the fact that r7c1 (which is also a peer of all cells of 20(3)n4) also needs to be considered.

In that sense, it appears that SudokuSolver handles conflicting partial combinations better than JSudoku does. Congratulations, Richard, you seem to have gone one up there!
Walkthrough by Andrew:
This puzzle was originally solved as a “tag” for which I wrote a consolidated walkthrough but didn’t take part in the “tag”, so I’ve now tried it.

Prelims

a) R12C1 = {29/38/47/56}, no 1
b) R12C9 = {39/48/57}, no 1,2,6
c) R23C6 = {17/26/35}, no 4,8,9
d) R78C4 = {15/24}
e) R89C1 = {18/27/36/45}, no 9
f) R89C9 = {29/38/47/56}, no 1
g) 20(3) cage at R4C1 = {389/479/569/578}, no 1,2
h) 9(3) cage at R5C2 = {126/135/234}, no 7,8,9
i) 22(3) cage at R5C6 = {589/679}
j) 11(3) cage at R7C3 = {128/137/146/236/245}, no 9
k) 24(3) cage at R9C2 = {789}
l) 10(3) cage at R9C6 = {127/136/145/235}, no 8,9
m) 14(4) cage at R3C1 = {1238/1247/1256/1346/2345}, no 9
n) and, of course, 45(9) cage at R3C5 = {123456789}

Steps resulting from Prelims
1a. 22(3) cage at R5C6 = {589/679}, 9 locked for R5
1b. 24(3) cage at R9C2 = {789}, locked for R9, clean-up: no 1,2 in R8C1, no 2,3,4 in R8C9

2. 45 rule on C1 2 innies R37C1 = 5 = {14/23}

3. 45 rule on C9 2 innies R37C9 = 10 = {19/28/37/46}, no 5

4. 45 rule on C123 3 outies R159C4 = 19 = {289/379/469/478/568}, no 1
4a. 2,3 of {289/379} must be in R5C4 -> no 2,3 in R1C4

5. 45 rule on C789 3 outies R159C6 = 12 = {129/138/147/246/345} (cannot be {156/237} which clash with R23C6)
5a. R5C6 = {56789} -> no 5,6,7,8,9 in R19C6
5b. Max R1C6 = 4 -> min R1C78 = 11, no 1 in R1C78

6. 25(4) cage at R7C6 = {2689/3589/3679/4678} (cannot be {1789} which clashes with R9C4, cannot be {4579} which clashes with R78C4), no 1
6a. Killer triple 7,8,9 in 25(4) cage and R9C4, locked for N8

7. 45 rule on R12 4 outies R3C3467 = 26 = {2789/3689/4589/4679/5678}, no 1, clean-up: no 7 in R2C6

8. 9 in N7 only in R7C2 + R9C23
8a. 45 rule on N7 4 innies R7C12 + R9C23 = 25 = {1789/2689/3589/3679/4579}
8b. R7C1 = {1234} -> no 1,2,3,4 in R7C2

9. 45 rule on N3 2 innies R3C89 = 1 outie R1C6 + 1
9a. Max R1C6 = 4 -> max R3C89 = 5, no 5,6,7,8,9, clean-up: no 1,2,3,4 in R7C9 (step 3)
9b. Min R3C89 = 3 -> min R1C6 = 2
9c. 15(4) cage at R6C7 contains one of 6,7,8,9, R7C9 = {6789} -> no 6,7,8,9 in R6C78 + R7C8
9d. 1 in C9 only in R3456C9, CPE no 1 in R4C78

10. 45 rule on R1 2 outies R2C19 = 1 innie R1C5 + 10
10a. Max R2C19 = 17 -> max R1C5 = 7

11. 45 rule on N2 3 innies R1C46 + R3C5 = 17 = {269/278/359/368/458/467} (cannot be {179} because R1C6 only contains 2,3,4), no 1
11a. R1C6 = {234} -> no 2,3,4 in R1C4 + R3C5
11b. 1 in N5 only in 45(9) cage at R4C4 -> no 1 in R7C5
[Alternatively Law of Leftovers (LoL) for N5, two outies R37C5 must exactly equal two innies R5C46, no 1 in R5C46 -> no 1 in R7C5.]

12. 1 in C1 only in R37C1 = {14} (step 2) or R89C1 = [81] -> R89C1 = {36}/[72/81] (cannot be {45}, locking-out cages), no 4,5 in R89C1
12a. 11(3) cage at R7C3 = {128/146/236/245} (cannot be {137} which clashes with R89C1), no 7
12b. R7C12 + R9C23 (step 8a) = {1789/3589/3679/4579} (cannot be {2689} which clashes with R89C1), no 2, clean-up: no 3 in R3C1 (step 2)

13. 45 rule on N1 2 innies R3C12 = 1 outie R1C4 + 1
13a. Min R1C4 = 5 -> min R3C12 = 6, max R3C1 = 4 -> min R3C2 = 2

14. 45 rule on N8 3 innies R7C5 + R9C46 = 14
14a. LoL for N5, two outies R37C5 must exactly equal two innies R5C46
14b. R159C4 (step 4) = {289/379/469/478/568}
14c. 2,3,4 of {289/379/469/478} must be in R5C4 and also in R7C5 using the LoL, 8 of {568} must be in R9C4 -> R15C4 = {56} -> R7C5 = 5 using the LoL (because max R7C5 + R9C4 = 13) -> no 6 in R7C5

15. R7C12 + R9C23 (step 12b) = {1789/3589/3679/4579}
15a. Consider combinations for R37C1 (step 2) = {14}/[23]
R37C1 = {14} => R7C12 + R9C23 = {1789/4579}
or R37C1 = [23] => R89C1 = [81] => R7C12 + R9C23 = {3679}
-> R7C12 + R9C23 = {1789/3679/4579}, 7 locked for N7, clean-up: no 2 in R9C1
[Alternatively R7C12 + R9C23 = {1789/3679/4579} (cannot be {3589} which clashes with R3789C1 = [2381]) …]
15b. 2 in C1 only in R123C1, locked for N1

16. Killer pair 1,3 in R37C1 and R89C1, locked for C1, clean-up: no 8 in R12C1
16a. 3 in C1 only in R789C1, locked for N7
16b. 2 in N7 only in 11(3) cage at R7C3 = {128/245}, no 6

17. 45 rule on N7 3(2+1) outies R6C23 + R9C4 = 22 -> min R6C23 = 13, no 1,2,3

18. 45 rule on R9 3 innies R9C159 = 11 = {146/236} (cannot be {245} because R9C1 only contains 1,3,6), 6 locked for R9
18a. 10(3) cage at R9C6 = {145/235}, 5 locked for R9 and N9, clean-up: no 6 in R89C9
18b. R9C159 = {146/236}
18c. 4 of {146} must be in R9C9 -> no 4 in R9C5

19. R7C5 + R9C46 (step 14) = 14 = {158/239/248/347} (cannot be {149/257} which clash with R78C4)
19a. 1 in N8 only in R78C4 = {15} or R7C5 + R9C46 = {158}, 5 locked for N8 (locking cages)
[With hindsight 6 in N8 only in 25(4) cage at R7C6 (step 6) = {2689/3679/4678}, no 5
I also noticed that I’d missed 5 in N8 only in R78C4 + R8C5, CPE no 5 in R46C4. I got this elimination, and more, using LoL in step 23.]

20. 45 rule on N6 5(2+2+1) outies R3C89 + R5C6 + R7C89 = 20, R37C9 = 10 (step 3) -> R37C8 + R5C6 = 10
20a. Min R37C8 = 3 -> max R5C6 = 7
20b. 22(3) cage at R5C6 = {589/679}, 9 locked for N6
20c. 5 of {589} must be in R5C6 -> no 5 in R5C78
20d. R159C6 (step 5) = {147/246/345}, 4 locked for C6
20e. LoL for N5, no 8,9 in R5C6 -> no 8,9 in R3C5
20f. R1C46 + R3C5 = 17 (step 11), max R1C6 + R3C5 = 11 -> min R1C4 = 6

21. R159C4 (step 4) = {289/379/469/478/568}
21a. 4,5 of {469/568} must be in R5C4 -> no 6 in R5C4

22. 12(3) cage at R4C9 = {138/147/156/246} (cannot be {237} which clashes with R89C9, cannot be {345} which clashes with R12C9)
22a. 5 in C9 only in R12C9 = {57} or 12(3) cage at {156} -> 12(3) cage = {138/156/246} (cannot be {147}, locking-out cages), no 7

23. LoL for N5 two outies R37C5 must exactly equal two innies R5C46 -> R3C5 = R5C6 and R5C4 = R7C5 (5 in R7C5 would make R3C5 = {67} -> R3C5 = R5C6)
23a. 5 in N8 only in R78C4 + R7C5, R7C5 = R5C4 -> 5 in R578C4, locked for C4

24. R9C159 (step 18b) = {146/236}
24a. 6 in N8 only in 25(4) cage at R7C6 = {2689/3679/4678}
24b. Consider combinations for R7C12 + R9C23 (step 15a) = {1789/3679/4579}
R7C12 + R9C23 = {1789} => R89C1 = {36} => R9C159 = {236}, locked for R9 => R9C6 = {14}, killer pair 1,4 in R78C4 and R9C6, locked for N8 => 25(4) cage at R7C6 = {2689/3679}
or R7C12 + R9C23 = {3679/4579} => R9C23 = {79} => R9C4 = 8 => 25(4) cage at R7C6 = {3679}
-> 25(4) cage at R7C6 = {2689/3679}, no 4, 9 locked for N8

25. 9 in R9 only in R9C23, locked for N7
25a. R6C23 + R9C4 = 22 (step 17), max R9C4 = 8 -> min R7C23 = 14, no 4 in R7C23

26. R3C3467 (step 7) = {2789/3689/4589/4679} (cannot be {5678} which clashes with R3C5)
26a. Killer quad 1,2,3,4 in R3C1, R3C3467 and R3C89, locked for R3
26b. 14(4) cage at R3C1 = {1238/1247/1256/1346/2345}
26c. 7,8 of {1238/1247} must be in R3C2 -> no 7,8 in R4C23

27. R7C12 + R9C23 (step 15a) = {1789/3679/4579}
27a. R6C23 + R9C4 = 22 (step 17), R9C4 = {78} -> R6C23 = 14,15 -> R7C12 = 8,9 = {17/18/36/45} -> R6C23 = {59/68/69} (cannot be {78} which clashes with R7C12 = {17}), no 7 in R6C23
27b. 7 in N4 only in 20(3) cage at R4C1, locked for C1, clean-up: no 4 in R12C1
27c. 20(3) cage = {479/578}, no 6
27d. R6C23 = {68/69} (cannot be {59} which clashes with 20(3) cage), no 5, 6 locked for R6, N4 and 23(4) cage at R6C2, no 6 in R7C2
27e. R7C12 + R9C23 = {1789/4579}, no 3, clean-up: no 2 in R3C1 (step 2)
[Cracked. The rest is fairly straightforward.]

28. Naked pair {14} in R37C1, locked for C1, clean-up: no 8 in R8C1
28a. 2 in C1 only in R12C1 = {29}, locked for C1 and N1
28b. Naked triple {578} in 20(3) cage at R4C1, locked for N4
28c. Naked pair {69} in R6C23, locked for R6

29. R6C23 + R9C4 = 22 (step 17), R6C23 = {69} = 15 -> R9C4 = 7
29a. Naked pair {89} in R9C23, locked for N7
29b. 11(3) cage at R7C3 = {245} (only remaining combination), locked for N7 -> R7C12 = [17], R3C1 = 4, clean-up: no 5 in R8C4
29c. R9C159 (step 18b) = {236} (only remaining combination), locked for R9, clean-up: no 7 in R8C9

30. R7C5 + R9C46 (step 19) = {347} (only remaining combination) -> R9C6 = 4, R7C5 = 3, clean-up: no 2 in R78C4
30a. Naked pair {15} in R9C78, locked for N9
30b. R5C4 = 3 (hidden single in N5) -> R5C24 = 6 = {24}, locked for R5 and N4
30c. Naked pair {13} in R4C23, locked for R4
30d. R3C1 = 4, R4C23 = {13} -> R3C2 = 6 (cage sum), R6C23 = [96], R9C23 = [89], clean-up: no 2 in R2C6

31. LoL for N5 no 6 in R3C5 -> no 6 in R5C6
31a. R159C6 = 12 (step 5), R9C6 = 4 -> R15C6 = 8 = [35], clean-up: no 9 in R2C9
31b. R159C4 = 19 (step 4), R59C4 = [37] -> R1C4 = 9 (cage sum), R12C1 = [29], clean-up: no 3 in R2C9
31c. LoL for N5 R5C6 = 5 -> R3C5 = 5
31d. R3C7 = 9 (hidden single in R3)
31e. R5C6 = 5 -> R5C78 = 17 = [89]
31f. R1C4 = 9 -> R1C23 = 8 = [17], R4C23 = [31], R2C2 = 5
31g. R1C6 = 3 -> R1C78 = 12 = [48], R12C9 = [57], R1C5 = 6, R2C6 = 1 -> R3C6 = 7
31h. R3C7 = 9 -> R2C78 = 8 = {26}, locked for R2 and N3

32. R5C15 = [71], R5C9 = 6 -> R46C9 = 6 = {24}, locked for C9 and N6, R9C9 = 3 -> R8C9 = 8

33. R6C78 = {13} (hidden pair in N6), R7C9 = 9 -> R7C8 = 2 (cage sum)

and the rest is naked singles.


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PostPosted: Sun Jul 06, 2008 6:26 am 
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A64 thread Bifurcation/chain/T&E discussion
Post 1: goooders on A64v2:
maybe there is a different way to start (A64v2)

column1 1 can only go in the 5(row3 and7) or the 9 in the bottom nonet

1 will not go in the 9 because that requires542 for the 11(neither 641 nor 623 works) but if that is so columns 1 and 2 of row 7 are 63 which works with neither 68 nor 59 required by the bottom left outies of 22 (note 59 cant work with the 20 in middle left nonet)

so that means rows 3 and 7 of column 1 are 14

that means the 9 in bottom left nonet can only be 72 or 63

but 72 will not work because the 11 cant be made at all

so the 9 in the bottom left nonet is 36

that makes the 11 cage 542 and columns 1 and 2 of row 7 are 17

that makes the other part of the 23 cage 69 ( given the 22 outie of the bottom left nonet)

so in column 1 the 11 is 29 and the 20 is 875

the 14 cage in top left nonet has a 4 in it which means the 9 cage in middle left nonet must be 423 and therefore the 14 cage above is 4631`

that means column 4 row 5 is 3

that means column 4 row 1 is 9 ( outies of leftside 3 nonets are 19)

easy now because columns 4 8 and 9 of row 3 have to be 135

the only other info you need is that row 1 column 6 which looks like it is either 3 or 5 or 7 can only be 3 or 5 because the outies of the 3 rightside nonets are12

should also say 16 cage in top left nonet has to be 358 because of either 3 or 5 in row 1 column 6 and row 3 columns 8 and 9 being two of 135
Post 2: Glyn:
The chain supplied by Gooders :applause: tore the problem right open and would have done so even right at the start, none of the programs is geared up to start a puzzle that way. Either human inspiration or a radically different type of computer solver would be required to investigate the hypotheticals.
Post 3: mhparker:
This (unexpected) post is not going to be easy to write. But what I have to say has to be said. Sometimes remaining silent is not the most diplomatic approach.
Glyn wrote:
The chain supplied by Gooders =D> tore the problem right open and would have done so even right at the start, none of the programs is geared up to start a puzzle that way. Either human inspiration or a radically different type of computer solver would be required to investigate the hypotheticals.
They (i.e., the "computer solvers") can already do it: it's called bifurcation!

You know that of course, but what I'm trying to say is that if you allow a computer-based solver to simply save the grid state, try out the elimination or (as in this case) cage combination, continue with the puzzle using basic techniques until (potentially) some sort of conflict is found, restoring the old grid state (if so) and carrying out the original hypothesis, then even nightmarish puzzles like the A60RP and A50V2 are a breeze! For many people, myself included, using T&E based on simple techniques is not as satisfying as trying to solve a puzzle using advanced techniques without T&E.

A good example is TJK18. I originally spent days on this, unsuccessfully. There were 2 possibilities for everything (i.e., one solution, and one near-solution). If T&E were acceptable, I could have easily finished on one evening by doing (part of) the puzzle twice and publishing the walkthrough as such!

Going back to the implication chain in question, I (and probably others on this forum) find conflicts like the one Goooders did all the time, but unless they can be encapsulated in simpler, less T&E-based terms, I don't publish them, otherwise I'd be (metaphorically-speaking) torn apart limb-from-limb by the other forum members! For a newbie it may be forgivable, but for any one of the regulars here, their reputation as a solver would quickly be in tatters!

In that sense, one can indeed argue that newbies make for more effective (but not necessarily better) solvers than the experts, because (in general) the more experience one gets, the less willing one is to sacrifice all those great solving techniques one has picked up over the years and simply use a try-it-and-see approach instead. This certainly applies to me, where there are several moves I made some time back (e.g. step 26 in the A48Hevvie consolidated WT) that I would not make now, even if it means not being able to complete the puzzle.

Sure, we can start using extensive T&E in our WTs (although I won't), and I'm sure we'll often reach our "destination" faster if we do. But the end result is hardly likely to impress the Ruuds, Jean-Christophes, Myth Jellies or Andrew Stuarts of this world, is it?

As Ruud put it on the "Assassin 60 - the rejected pattern" thread:
Ruud wrote:
There is no value in trying to solve [the puzzle] with guesses.
Incidentally, Ruud, when posting a puzzle, often refers to us as the "advanced solvers on the forum". If we were to routinely use T&E to solve the puzzles he sets us, I don't think it would be long before this accolade is withdrawn...
Post 4: mhparker:
Hi all,

Just want to clarify a few points:

Although I made it pretty clear yesterday how much I object to T&E, people may be asking themselves what the point is of trying to struggle through a puzzle using primitive logic, when a newbie like Goooders can quickly come up with a few T&E spoiler steps that crack the puzzle very quickly, by-passing and negating all those more complex moves the rest of us had been making up to that point.

How can this be? Isn't something wrong here? Is advanced logic failing us? Should we (heaven forbid!) acknowledge that T&E was better here?

Yes and no. Yes, something is wrong, but no, advanced logic is not failing us. The first point is that the rest of us were mainly concentrating on the wrong part of the grid (n258), where we should have been looking where Goooders was looking (c1/n7). The second point is that the fact that Goooders' T&E steps led to a quick contradiction indicates that there is a lot of "contention" in this part of the grid. In such cases, it is often possible to find advanced logical steps to replace the simpler T&E ones, with similar effect.

And so it is here. Indeed SudokuSolver did precisely this. It was using advanced non-T&E logic, in the same area that Goooders was looking at, to bust open the puzzle very quickly, thus accounting for its low rating.

To give Goooders the credit for a good basic idea, and to illustrate that the power of Goooders' steps can be achieved without T&E, I am including an alternative path, starting from our original step 27. The ideas are taken from Goooders, SudokuSolver (mainly) and some of my own (in particular, step 30). It should just be taken as an example of how we could/should have done it.
Post 5: CathyW:
Appreciate the clarification Mike. I was still mulling over your previous post. It depends on how you define T&E, and where you draw the line for T&E. For myself, Goooders contradiction chain was perfectly logical but it is a bifurcation with one route leading to conflict and one route leading to solution.

Your post today has convinced me that the A64 V2 is indeed solvable without hypotheticals. I wouldn't like to say that one way is definitely better than the other but as a still (I hope!) improving solver I would also prefer not to use bifurcation or contradiction chains if possible. Having said that, I'm not going to rule it out from future puzzles if I can't find another way forward.

I'd be interested to read other members' views.
Post 6: mhparker:
CathyW wrote:
but it is a bifurcation with one route leading to conflict and one route leading to solution....I wouldn't like to say that one way is definitely better than the other but ... I would also prefer not to use bifurcation or contradiction chains if possible.
Hi, Cathy. Sounds like you may be misunderstanding the term bifurcation. It simply means forking the state of the puzzle, not simply going in one of two directions (possibly in turn). For example, all AICs are also based on a 2-state starting premise (i.e. premise is true or false). And yet AICs are not regarded as bifurcation. What is the crucial difference? I'll answer that below.

As a second example, consider Goooders' contradiction "chain":
goooders wrote:
1 will not go in the 9 because that requires542 for the 11(neither 641 nor 623 works) but if that is so columns 1 and 2 of row 7 are 63 which works with neither 68 nor 59 required by the bottom left outies of 22 (note 59 cant work with the 20 in middle left nonet)
Let me simply ask the questions:

Why doesn't {146} work for 11(3)n7?

Similarly why doesn't {236} work?

"But", I hear you say, "that's obvious! {146} can't work because we've already used the 1 in the 9(2)n7 cage. And {236} can't work, because a 1 in 9(2)n7 will force h5(2) at r37c1 to be {23}, thus blocking {236} for 11(3)n7".

If so, that would be correct. But let's think it through in a bit more detail first. The problem is, in order to know that the 1 is not available for the 11(3) cage, the hypothesis 9(2) = [81] must already have been committed in order that the candidates 1 and 8 are eliminated in the corresponding peer cells. Furthermore, in order to know that r37c1 is now {23}, the combinations must be re-evaluated. Not only that, but conflicting combination analysis needs to be performed to detect that {23} in r37c1 blocks the {236} in 11(3)n7. In an automated solver, all this updated information, needed for later parts of the "chain" has to be stored somewhere. Of course, at this stage we don't even know whether the original hypothesis is correct, so we can't "burn our bridges", as it were, in case we need to go back.

In computer terms, this means we need to bifurcate. In other words, the current grid state must be saved first. Then the initial hypothesis is committed. After that, the solver is simply allowed to run on, using every solving technique (naked/hidden singles, combinational analysis, conflicting combinations, etc., etc.) it has at its disposal until (if at all) a contradiction is reached. If this happens, the effects of applying the initial hypothesis (now known to be false) have to be reversed. This is done by restoring the original saved grid state, a process also known as backtracking. The word bifurcation implies (in computer terms) "forking", because when the state of the grid is saved, we initially have two identical copies of it (i.e., the working copy and the saved one).

This is the computer equivalent of making a photocopy of a puzzle being done on paper, making a guess and carrying on accordingly, throwing the working copy in the bin if a contradiction is reached, and returning to the saved photocopy.

Such bifurcation is almost universally regarded as trial-and-error. Of course one can try to distinguish between "limited T&E" (I use the term hypothetical for that) and full-blown T&E, but it's T&E all the same.

This may seem obvious. But let's go back to the first example, namely AICs. Why isn't this bifurcation? Why isn't it T&E?

The crucial difference is that AICs, Nice Loops, X-Cycles, and so on, work purely by analysis of the strong and weak links currently present in the grid. No hypothesis or other intermediate results have to be committed, and no other solving technique (like conflicting combinations, etc.) needs to be invoked. No commitments implies that no copy of the grid needs to be made, therefore neither bifurcation nor backtracking is involved. They are therefore pure solving techniques and not T&E, as in the first case.
Post 7: goooders:
i sort of agree with mhparker and sort of disagree

i agree theres no point in doing a puzzle based upon a guess also theres no point in trial and error however you then run into the issue of what is trial and error my own view is that if i can work through a combination in my head which leads to a contradiction that is fine but if i were to resort to pen and paper i have crossed a line(i fully accept this is "how long is a piece of string" but i feel comfortable with it)

i suspect i maybe do assassins slightly differently from some others my own rule is not to fill in all the numbers and then cut them back bit by bit however if there is 23 cage i will fill in a 689

i rely on combinations and patterns so for example in the 64v2 i noticed straightaway that in the bottom left nonet the 9 cage would not support a 54 because of the other numbers in column 1 nor then a 72 so given the 789 and the 22 outie i thought it was worth having a look at

in pure mathematics all proofs derive essentially from either induction or contradiction(for example that the square root of 2 is irrational) thus if there are only two paths and one is wrong then a fortiori the other is right

newbie or otherwise of the 64 assassins to date i thought 3 or 4 were very hard the last being 55

it is perhaps worth saying that the very fast solvers in the ordinary sudoku world champioships are all "pattern recognisers" rather than number eliminators(vis a vis devising a quick strategy to a solution)

finally finally some illustrious names are called in aid i wondered if anyone had come across "udosuk " whom i havent seen on this website but who demonstrated some brilliant logic in solving some of nate dorwards 10 puzzles
Post 8: Para:
Hi

I guess for me it all has to do with the satisfaction of solving a puzzle. Implication chains are a sound form of logic which can solve any puzzle, but it's not really the most beautiful logic. I rather take a few hour more on a puzzle to find a nicer way to the solution. I guess it is the same as i feel about uniqueness moves. I rather solve a puzzle without uniqueness moves, so i can also prove the puzzle only has one solution.
goooders wrote:
it is perhaps worth saying that the very fast solvers in the ordinary sudoku world champioships are all "pattern recognisers" rather than number eliminators(vis a vis devising a quick strategy to a solution)
The big difference in this, is that on sudoku and logic puzzle championships it is all about speed, solving as many puzzle as possible in a certain time limit, or all puzzles as fast as possible. This of course gives a whole different attitude to solving. The solving techniques i use when speed solving are a lot different then when i do it for fun. In such championships use anything that will bring me closer to a solution. I sometimes solve these assassins on speed as well. Then i tend to use a lot of implication chains and uniqueness eliminations and don't work the same way as i normally do. This will always work faster.
When i solve puzzles for fun, i just want to reach a nice path to a solution. I have a lot of logic puzzles laying that i haven't solved yet, but if i wanted to could easily solve using a T&E/implication chain approach. But i choose to try and solve it using nicer logic.

So i guess it all comes down to a choice you make when solving logic puzzles. And i hope we can keep it at that, because these discussions have been going on everywhere and the arguments and discussions are always the same. Beauty vs speed, logic?, mathematicians use it to solve problems as well. Seen it, heard it and it all still keeps coming down to the choice you make as solver.
Post 9: CathyW:
Well said Para. I expect discussions on what constitutes acceptable T&E techniques will continue from time to time. I certainly find it more satisfying to solve a puzzle without T&E but I still have a lot to learn on some of the advanced techniques that Mike mentioned, especially as they might apply in killers. So, for now at least, I choose not to rule out hypotheticals and UR moves.
Post 10: Andrew:
CathyW wrote:
but it is a bifurcation with one route leading to conflict and one route leading to solution....
mhparker wrote:
Sounds like you may be misunderstanding the term bifurcation. It simply means forking the state of the puzzle, not simply going in one of two directions (possibly in turn). For example, all AICs are also based on a 2-state starting premise (i.e. premise is true or false). And yet AICs are not regarded as bifurcation.
and later in the same message
mhparker wrote:
This may seem obvious. But let's go back to the first example, namely AICs. Why isn't this bifurcation? Why isn't it T&E?

The crucial difference is that AICs, Nice Loops, X-Cycles, and so on, work purely by analysis of the strong and weak links currently present in the grid. No hypothesis or other intermediate results have to be committed, and no other solving technique (like conflicting combinations, etc.) needs to be invoked. No commitments implies that no copy of the grid needs to be made, therefore neither bifurcation nor backtracking is involved. They are therefore pure solving techniques and not T&E, as in the first case.
I hadn't planned to get involved in this discussion about bifurcation, T&E, etc. but I can't completely agree with what Mike has said.

In its original sense bifurcation means forking in two directions. The people who have invented advanced techniques in the last couple of years didn't like that. They have therefore taken the view that techniques that start with forking but are thereafter logical are not considered to be bifurcation even though they really are in the pure sense of that word.

To take an example, simple colouring (why I'll admit I don't yet understand well enough to use except in the very simplest case) has two chains from one cell to another cell. If one chain has an even number of links and the other an odd number of links, then a candidate in the destination cell can be eliminated. The simplest case, which can occur in jigsaw sudokus, has a two link chain and a single link chain. This is pure logic but still starts with bifurcation.

The same applies to the more advanced techniques that Mike has listed. They are not T&E but they are still bifurcation in the pure sense of that word. To repeat myself, they are not considered to be bifurcation because the people who have invented advanced techniques in the last couple of years didn't like that and consider it to be a dirty word.

I don't think that Cathy was misunderstanding the term bifurcation; she was just using it in a different sense than Mike does, in fact in the same way that I do.
Post 11: Glyn:
Hi everyone

Now that the thorny subject of bifurcation and T&E has come up there are always likely to be differences of opinion, however our goal must be to extend the number of solving methods available as we have a whole load of unsolvables which it would be nice to crack. It is quite possible that they may remain unsolvable without some sort form of chaining and maybe this chaining will be required to violate standard rules of AICs. (IMHO the word maybe can be deleted here).

I know that Mike is not entirely against AIC which formalises the processes involved. In vanilla Sudoku these chains can not only be long, but also be comprise of multithreads, groups, fish structures, ALS etc. They can usually be laid out fairly clearly on an unmodified grid. They do however require strong links for their construction and such relationships are sadly lacking in Killers unless one is prepared to bite the bullet and allow the links to modify the cage values to a certain extent. Perhaps we could call them strong active links. The chains would preferably be short as any of the active links would start to make the grid unrecognisable and carry the strong odour of T&E. JC has already described killer variants of some chain structures and hinted at ALS extensions which I think may be have some similarities to what we might require.

I hope that mentioning AIC hasn't put people off, I think we would really only use it to encapsulate the tricky moves and a description of what was going on would be necessary.
Post 12: Andrew:
I've got no problems with advanced techniques being used so long as they are clearly presented and the outcomes, such as fixed cells, candidate eliminations or combo eliminations, are stated.
Post 13: mhparker:
Andrew wrote:
In its original sense bifurcation means forking in two directions. The people who have invented advanced techniques in the last couple of years didn't like that. They have therefore taken the view that techniques that start with forking but are thereafter logical are not considered to be bifurcation even though they really are in the pure sense of that word.
Andrew wrote:
I've had Andrew Stuart's book since it first came out but, unlike Cathy and Mike who have well thumbed copies, I've only managed quick glances so far.
Yes, bifurcation does simply mean forking in its original sense. However, in Sudoku circles it's invariably reserved for a specific type of forking, namely forking of the entire grid state. For an example, see chapter 1 of Andrew Stuart's book, where he associates bifurcation with Ariadne's Thread. Another distinguishing feature of bifurcation in this sense is that the starting choice (e.g., cell candidate, cage combination, ...) is essentially arbitrary.

With the regular chain-based techniques in Sudoku, the situation is very different for a number of reasons:
  1. The starting choice is not arbitrary. For example, for AICs, the starting choice is the candidate premise at one end of a strong link being false.
  2. The grid is not modified. Note: This is important! Even the slightest modification to the grid would in general require saving the entire grid state at each branching point in the chain! I use AICs and Nice Loops quite a lot, so if anybody sees me using any strong links that weren't there at the start of the chain, I've made a mistake and would like to know about it!
  3. The chain is built up "merely" by following weak and strong links, not by potentially using every solving technique under the sun. For automated solvers, this means using dedicated chain processing rather than simply allowing the solver to run further.
  4. Although regular chains do often involve finding contradictions, one is looking out for particular termination conditions here, not simply responding to any unexpected error condition.
Glyn wrote:
...unless one is prepared to bite the bullet and allow the links to modify the cage values to a certain extent. Perhaps we could call them strong active links. The chains would preferably be short as any of the active links would start to make the grid unrecognisable and carry the strong odour of T&E.
I use the term dynamic links for these, because the strong links arise dynamically as a result of side effects of earlier links in the chain. But the difference between these dynamic links and the "active" links you describe is merely one of terminology. The concept is identical.

All chains containing dynamic ("active") links are hypotheticals in my book. Hypotheticals are generally regarded as limited T&E, or even "guessing" (as Ruud has just expressed it on the A65 thread). However, whilst this is often true, to uniformly apply this equation would belittle the tremendous aptitude humans have for recognizing promising patterns and cutting off unpromising avenues of exploration in order to dramatically reduce the amount of searching to be performed. The problem is, how can hypotheticals be formalized? I agree with Glyn that a lot of research work still needs to be done here (and that it's best to do all this on another thread).
Post 14: Ruud:
Interesting discussion developing here. Being a moderator on the Eureka forum, I never imagined that people could also discuss this subject in a civilized manner.

In my solving guide, I've written a section about Sudoku Controversies. It badly needs an update.

If you want to see bifurcation in action, take a puzzle that has no more logical solving steps and make an extra copy. Now pick a cell with 2 candidates or a cage with 2 combinations or a row, column or box where a certain number can only go in 2 places. Select one of the choices in the first copy and the other in the second. Now continue to solve both puzzles.

If you find the solution in one of the copies and you accept it, you have used a Guess.

If you find a contradiction in one of the copies and you reject it, you have used Trial & Error. Some might say that this is the same as a bad guess. I would not entirely disagree with this opinion. There is a special case to consider here, where the initial choice is part of the contradiction. When a path A => NOT A exists, there must also be a reverse path. Advanced solvers really appreciate reversible chains.

If you find a common truth in both copies, i.e. a candidate that is eliminated in both puzzles, or a placement made in both, you have used a Forcing Net. In my SudoCue program, the Tabling Verity technique works this way. Again, reversibility comes to the rescue here. When A => B and NOT A => B then there must be a path NOT B => NOT A => B which confirms B.

We do not need to limit ourselves to bifurcation. Why not trifurcation or quadruple ramification (who invented these words?). Well, the problem is that these will never lead to a reversibile proof without using branching in either direction.

This brings me to my current position on the acceptability of solving techniques:

A solving technique is acceptable when it can be rewritten as a (reversible) sequence of implications where each node in this sequence is a coherent group of candidates.

Because every sequence of implications is reversible, the use of this term is superfluous.

In regular Sudoku, a group of candidates is coherent when they all belong to the same row, column or box. In Killer Sudoku, candidates in a single (split) cage are also a coherent group.

This leaves room for more than just singles. Any pattern that can connect two coherent groups could be placed in the chain, as long as that pattern creates a reversible link between these nodes. Almost Locked Sets, Almost UR's and Almost Fish can be placed between the nodes.

Finally, this is just a personal opinion and it is certainly not carved in stone. I'm learning new things about Sudoku everyday. I'm not even sure which techniques would qualify or fail this test. I'd really appreciate your input on this so we may refine this definition in a way that we can all agree with.
Post 15: mhparker:
Hi Ruud,

Thanks for the suggestions. Here are a few quick points to get the ball rolling. Maybe other forum members can add a few more.
Ruud wrote:
If you want to see bifurcation in action, take a puzzle that has no more logical solving steps and make an extra copy. Now pick a cell with 2 candidates or a cage with 2 combinations or a row, column or box where a certain number can only go in 2 places.
Actually (just in case anyone's getting the wrong impression), as far as the first two solving techniques you mentioned are concerned (Guess and Trial & Error), one can pick a cell with more than two candidates (or a cage with more than two combinations), and it's still only bifurcation (not trifurcation, etc.). As such, only one copy of the grid state needs to be made, and the two branches are essentially cell = chosen candidate (or cage = chosen combination) and cell <> chosen candidate (or cage <> chosen combination).

For the Tabling Verity technique, one clearly needs as many copies of the grid (including the original) as the number of candidates in the chosen cell, combinations in the chosen cage, or candidate positions in the chosen unit, respectively, in order to be able to compare the results. So this is real N-furcation. I presume this is what Andrew had in mind in his previous post.
Ruud wrote:
If you find the solution in one of the copies and you accept it, you have used a Guess.
Care must be taken here. Strictly speaking, one has found a solution, not necessarily the solution. So, by completing the grid after making a guess, one is also making the implicit assumption that the puzzle only has one solution.
Ruud wrote:
If you find a common truth in both copies, i.e. a candidate that is eliminated in both puzzles, or a placement made in both, you have used a Forcing Net.
Just for info: We also recently had a slightly more complicated case (Cathy's step 22 in her A61X WT), where trifurcation was involved that, whilst not yielding a common placement, nevertheless locked a digit into one of two positions within a unit, thus allowing that candidate digit to be eliminated elsewhere in that unit. A placement common to all branches can in this sense just be seen as being the special case where a particular digit is locked into a single candidate position within a particular unit.
Ruud wrote:
A solving technique is acceptable when it can be rewritten as a (reversible) sequence of implications where each node in this sequence is a coherent group of candidates. Because every sequence of implications is reversible, the use of this term is superfluous.
True, it should in theory be superfluous. However, this requirement is clearly intended to filter out chains with dynamic links (see my previous post), which are what I usually mean when I refer to the term hypothetical. For example, if a side effect of a link in the chain changes the grid state from G to G', then a subsequent link may not be reversible, because the underlying grid state in the forward direction (G') differs from that in the reverse direction (G).


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PostPosted: Sun Jul 06, 2008 6:30 am 
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Location: Sydney, Australia
Assassin 65 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:3584:3584:4098:3843:3843:3843:2054:6151:6151:3584:4874:4098:4098:4365:2054:2054:2320:6151:4882:4874:4874:10517:4365:1047:2320:2320:4378:4882:4874:4381:10517:4365:1047:5409:4378:4378:4882:4381:4381:10517:10517:10517:5409:5409:4140:3373:3373:4381:1072:4401:10517:5409:5428:4140:3373:3383:3383:1072:4401:10517:5428:5428:4140:3903:3383:5441:5441:4401:3396:3396:5428:3143:3903:3903:5441:4939:4939:4939:3396:3143:3143:
Solution:
+-------+-------+-------+
| 4 3 2 | 6 1 8 | 5 9 7 |
| 7 6 9 | 5 4 2 | 1 3 8 |
| 8 1 5 | 9 7 3 | 2 4 6 |
+-------+-------+-------+
| 5 7 4 | 2 6 1 | 9 8 3 |
| 6 9 3 | 4 8 5 | 7 1 2 |
| 2 8 1 | 3 9 7 | 4 6 5 |
+-------+-------+-------+
| 3 4 7 | 1 5 6 | 8 2 9 |
| 9 2 8 | 7 3 4 | 6 5 1 |
| 1 5 6 | 8 2 9 | 3 7 4 |
+-------+-------+-------+
Quote:
CathyW: Excellent puzzle - kept me occupied most of this afternoon... Was stuck for quite a while
Andrew: not ..... particularly difficult..I rate this Assassin as 1.25..also 1.25 for the ways that Cathy and Para solved it.
Walkthrough by CathyW:
Excellent puzzle - kept me occupied most of this afternoon despite a few early placements (it's OK I wasn't skiving, had a day off work!) Was stuck for quite a while before my step 20. Looking forward to the variations.

Prelims:

a) 41(7) @ r3c4 – no 1,3
b) 4(2) r67c4 and r34c6 = {13} not elsewhere in c4, c6
-> NP {13} r6c4 + r4c6 not elsewhere in N5
c) 24(3) N3 = {789} not elsewhere in N3
d) 8(3) r12c7+r2c6 = {125/134} must have 1 in r12c7 not elsewhere in c7/N3
-> 9(3) N3 = {234} not elsewhere in N3
-> r12c7 = {15} not elsewhere in c7 -> r2c6 = 2 -> r3c9 = 6
-> r4c89 = {29/38/47}
-> 2 locked to r3c78 not elsewhere in r3
-> 2 locked to r45c4,r5c5 of 41(7) -> r46c5 <> 2
e) 19(3) r345c1 and r9c456 – no 1
f) 21(3) r89c3 + r8c4 – no 1,2,3


1. Outies c789: r28c6 = 6 -> r8c6 = 4 -> r89c7 = {27/36}

2. Outies c123: r28c4 = 12 = [48]/{57}

3. Innies r12: r2c258 = 13 = {19/46}3 / {18/36}4

4. Innies r89: r8c258 = 10 = {127/136/235}

5. O-I N9: r7c9 – r6c8 = 3 -> r7c9 = (45789), r6c8 = (12456)

6. Outies c1: r169c2 = 16

7. Outies c9: r149c8 = 24 = {789} not elsewhere in c8
-> r4c9 <> 7,8,9
-> 12(3) r9c8+r89c9 = [714], 8{13}, 9{12} -> r89c9 = (1234) ({723} would block both options for r89c7) -> 1 locked to r89c9, not elsewhere in c9/N9

8. Innies c5: r159c5 = 11 = {128/137/146/236/245}
If {128/137/146} r1c5 = 1 -> r1c5 <> 7, 8
If {137} r5c5 = 7 -> r9c5 <> 7

9. O-I N1: r2c4 + r4c2 - r3c1 = 4 -> r4c2 – r3c1 = 0 or -1 or -3

10. O-I c12: r37c3 – r5c2 = 3

11. O-I c89: r37c7 – r5c8 = 9
Max from r37c7 = 13 -> r5c8 is max 4 -> 6 locked to r678c8 of 21(4) -> r7c7 <> 6
r5c8 = (1234) -> r37c7 = 10, 11, 12, 13, r3c7 = (234) -> r7c7 is min 7

12. 21(3) r89c3 + r8c4 = [948]/{579/678}

13. 16(3) r12c3 + r2c4 = {39/57}4 / [29]/{38/47}5 / {18/36/45}7

14. Outies N14: r2c4+r7c1 = 8 = [44/53/71] -> r7c1 = (134)

15. 9 locked to r8c13, not elsewhere in N7 -> r8c3 <> 5.

16. 8 locked to r8c134 -> r9c3 <> 8

17. Killer Triple: Split 10(3) in r8 c258 = {127/136/235} (each option has 2 of 1,2,3), r8c9 = (123) -> r8c17 <> 1,2,3 -> r9c7 <> 6,7

18. 15(3) r1c456 = {168/348/357/456} ({159} blocked by r1c7)

15(3) contains one of 5,8 -> 17(3) r234c5 <> [854]

19. 21(4) r678c8+r7c7 must have 5 and 6 within r678c8
-> 21(4) = 56+{19/28/37} -> r67c8 <> 4 -> r7c9 <> 7 (step 5)
-> 4 locked to r79c9 not elsewhere in c9 -> r4c8 <> 7

20. 9 locked to r4569c4 -> r7c6 <> 9
9 also locked to r569c6
(Grouped x-wing) -> r7c5 <> 9, r9c8 <> 9 -> r89c9 <> 2
-> 2 locked to r456c9 not elsewhere in N6 -> r7c9 <> 5
-> 5 locked to r56c9 -> r6c8 <> 5 -> r7c9 <> 8

21. 16(3) r567c9 = {57}4 / {25}9 ({39}4 blocked by r89c9) -> r56c9 <> 3, 8, 9
-> 8 locked to r12c9 -> r1c8 <> 8

22. Looking at 3s to eliminate 3 in common peer cell r7c8
a) if 3 in r23c8 -> r7c8 <> 3
b) if 3 in r3c7 -> 3 locked to r12c5 -> r7c4 = 3 -> r7c8 <> 3
Either option r7c8 <> 3

23. Looking at 3s to eliminate 3 in common peer cell r8c8
a) if 3 in r23c8 -> r8c8 <> 3
b) if 3 in r3c7 -> r3c6 = 1 -> r4c6 = 3 -> r4c9 = 2 -> 16(3) = {57}4 -> r89c9 = {13} -> r8c8 <> 3
Either option r8c8 <> 3

24. Revisiting 21(4) options now: 56{19/28} -> r7c7 <> 7

25. Split (10)3 r8c258 can’t now be {136} – blocked by r8c9
Remaining options {127/235}

26. Looking at 6s to eliminate 6 in common peer cell r6c5
a) r6c8 = 6 -> r6c5 <> 6
b) r7c8 = 6 -> 6 locked to 41(7) within N5 -> r6c5 <> 6
Either option r6c5 <> 6

27. 17(3) r678c5 = {179/269/278/359/368/458/467}
If {359} r6c5 = 9; if {458} r6c5 = 4 -> r6c5 <> 5

28. 19(3) r9c456 must have 9. Options: {289/379} -> r9c456 <> 5,6
-> 5,6 locked to r9c123 not elsewhere in N7
-> r8c1 of 15(3) is min 7 -> r9c12 <> 7,8
-> HS r8c7 = 6 -> r9c7 = 3 -> r8c9 = 1 -> r9c9 = 4 -> r9c8 = 7
-> lots more singles and reasonably straightforward with cage combos and singles from here
Walkthrough by Para:
Hi all

Here's my walk-through. Was a fun puzzle. Check out step 26. That was a fun one and it also broke the puzzle for me. Hope i can find some time for Ruud's variations.

Walk-Through Assassin 65

1. 8(3) at R1C7 = {125/134}: no 6,7,8,9

2. 24(3) at R1C8 = {789} -->> locked for N3

3. 19(3) at R3C1 and R9C4 = {289/379/469/478/568}: no 1

4. 41(7) at R3C4 = {2456789}: no 1,3

5. R34C6 = {13} -->> locked for C6

6. R67C4 = {13} -->> locked for C4
6a. Naked Pair {13} in R4C6 + R6C4 -->> locked for N5

7. 21(3) at R8C3 = {489/579/678}: no 1,2,3

8. 45 on C123: 2 outies: R28C4 = 12 = {48/57}: no 2,6,9

9. 45 on C789: 2 outies: R28C6 = 6 = {24} -->> locked for C6

10. 45 on N3: 1 innie and 1 outie: R2C6 + 4 = R3C9 -->> R2C6 = 2; R3C9 = 6(only possible combo)
10a. R8C6 = 4
10b. Clean up: R2C4: no 8(step 8)

11. 8(3) at R1C7 = 2{15} -->> R12C7 = {15} -->> locked for C7 and N3
11a. 17(3) at R3C9 = 6{29/38/47}-->> R4C89 = {29/38/47}: no 1,5
11b. 13(4) at R8C6 = 4{27/36} -->> R89C7 = {27/36}: no 8,9

12. 45 on N7: 1 innie and 1 outie: R7C1 + 4 = R8C4: R8C4 = {578} -->> R7C1 = {134}

13. 45 on C9: 3 outies: R149C8 = {789} -->> locked for C8
13a. Clean up: R4C9: no 7,8,9
13b. 12(3) at R8C9 = {129/138/147}: {237} blocked by R89C7; {345} blocked by R9C8 -->> R89C9 = {12/13}/[14]: no 5,7,8,9; 1 locked for C9 and N9

14. 45 on N9: 1 innie and 1 outie: R6C8 + 3 = R7C9: Min R6C8 = 1 -->> Min R7C9 = 4, R7C9 = {45789} -->> R6C8 = {12456}

15. 21(4) at R6C8 = {1569/2469/2568/3459/3468/3567}: other combos blocked because {789} only in R7C7 -->> 21(4) needs one of {789} and can only go in R7C7 -->> R7C7 = {789}

16. 5 in C9 locked within 16(3) cage at R5C9 -->> 16(3) = {259/358/457}

17. 45 on N9: 3 outies: R56C9 + R6C8 = {139/148/157/238/247/256/346} = {39}[1]/{48}[1]/{57}[1]/{38}[2]/{47}[2]/{27}[4]/{25}[6]/{34}[6] -->> R6C8 = {1246}: no 5
17a. Clean up: R7C9: no 8(step 14)
17b. R56C9 = {57/38/47/25}: {39/48/27/34} blocked by 16(3) at R5C9: no 9 -->> R6C8 = {126}(step 17): no 4
17b. Clean up: R7C9: no 7

18. 21(4) at R6C8 = {1569/2568/3567}: {2469/3459/3468} blocked by R23C8(can’t have 2 of {234} in R678C8): no 4; 6 locked within R678C8 -->> locked for C8; 5 locked within R78C8 -->> locked for C8 and N9
18a. Clean up: R6C8: no 2

19. 16(3) at R5C9 = {25}[9]/{57}[4]; R56C9 = {25/57}: no 3,4,8

20. 4 in N9 locked for C9
20a. 8 in C9 locked for N3
20b. Clean up: R4C8: no 7

21. 45 on R89: 3 innies: R8C258 = 10 = {127/136/235}: no 8,9
21a. 9 in R8 locked for N7

22. 45 on N7: 3 innies: R7C1 + R89C3 = 17 = [197/395]/[3]{68}/[4]{58}/[4]{67} -->> R9C3: no 4
22a. 21(3) at R8C3 = {579/678} -->> 7 locked within cage : R8C12: no 7
22b. R7C1 + R89C3 = [4]{58} blocked by 21(3) cage at R8C3 -->> R8C3: no 5

23. 45 on N8: 2 innies and 2 outies: R6C45 + 1 = R7C6 + R8C4: Min R7C6 + R8C4 = 11 -->> Min R6C45 = 10; Max R6C45 = 12 -->> Max R7C6 + R8C4 = 13: R6C45 = [19/37/38/39]; R7C6 + R8C4 = [65]/{57}/{58}/[67] = {5|6..}-->> R6C5 = {789}; R7C6: no 9

24. 19(3) at R9C4 = {289/379}: {568} blocked by R7C6 + R8C4: no 5,6; 9 locked for R9 and N8
24a. Killer Pair {78} in 19(3) at R9C4 + R9C8 -->> locked for R9
24b. Clean up: R89C9: no 2(13b); R8C3: no 6; R8C4: no 5(step 22a)
24c. Killer Pair {78} in R8C4 + 19(3) cage at R9C4 -->> locked for N8

25. R7C6 + R8C4 = [57/58/67](step 23)
25a. R6C45 = [38/39](step 23 + 24c) -->> R6C4 = 3; R6C5 = {89}
25b. R34C6 = [31]; R7C4 = 1; R2C8 = 3(hidden)
25c. Naked Pair {24} in R3C78 -->> locked for R3

26. LOL on N5: R3C4 + R7C6 = R46C5 -->> R3C4 = {89}; R4C5 = {56}(this works because the 41(7) cage contains exactly the same digits as the remaining 7 empty cells in N5)
26a. 17(3) at R2C5 = [485/476]: {179} blocked by R4C5 -->> R2C5 = 4; R3C5 = {78}

27. 6 in N2 locked within 15(3) at R1C4 -->> 15(3) = {168} -->> R1C5 = 1; R1C46 = {68} -->> locked for R1 and N2
27a. R12C7 = [51]; R3C45 = [97]; R2C4 = 5; R4C5 = 6
27b. R2C9 = 8(hidden); R7C6 = 6(hidden); R1C46 = [68]; R6C5 = 9(hidden)
27c. R9C6 = 9(hidden); R8C4 = 7(step 25)

28. R9C45 = {28} (step 24) -->> locked for R9 and N8
28a. R9C8 = 7; R1C8 = 9; R4C8 = 8; R14C9 = [73]; R89C9 = [14]; R7C79 = [89]
28b. R6C8 = 6(step 14)

29. 15(3) at R8C1 = [9]{15}/[8]{16} -->> R8C1 = {89}; R9C12 = {156}
29a. Naked Triple {156} in R9C123 -->> locked for R9 and N7
29b. R89C7 = [63]; R7C1 = 3(step 12)(missed this earlier); R8C2 = 2
29c. R78C5 = [53]; R78C8 = [25]; R3C78 = [24]; R5C8 = 1

30. 14(3) at R1C1 = {239/347}: no 6; R1C2 = 3(only place in cage)
30a. R5C3 = 3(hidden)

31. 13(3) at R6C1 = 3[28] -->> R6C12 = [28]
And the rest is all naked singles

greetings

Para
Walkthrough by Andrew:
I only finished Assassin 65 on Wednesday, not because it was particularly difficult but because of a mistake that I made. Then I worked through Cathy's and Para's walkthroughs this evening and decided my one is sufficiently different to post, even though it's more than a week since this puzzle was first posted.

Cathy's Grouped X-Wing was a neat one.

Like Para I also enjoyed step 26. My step 26 also used LoL but was otherwise completely different.

Then I reached an impossible position because I’d missed out one permutation [167] in step 29b, which happened to be part of the solution; I’d worked out those permutations “in my head” rather than using Ruud’s excellent combination calculator as I usually do. That had given me 5 locked in 3 innies of N8 which was wrong. Fortunately, after finding my error, I then found step 30 which had the 5 locked in 5 innies and I was able to use most of my remaining moves.

I'll rate this Assassin as 1.25. It's probably also 1.25 for the ways that Cathy and Para solved it.


Here is my walkthrough for Assassin 65

1. R34C6 = {13}, locked for C6

2. R67C4 = {13}, locked for C4
2a. Naked pair {13} in R4C6 + R6C4, locked for N5

3. 8(3) cage at R1C7 = 1{25/34}, 1 locked in R12C7 for C7 and N3

4. 9(2) cage in N3 = {234} (only remaining combination), locked for N3
4a. Naked pair {15} in R12C7, locked for C7 and N3 -> R2C6 = 2
4b. 2 in N3 locked in R3C78, locked for R3

5. 24(3) cage in N3 = {789}, locked for N3 -> R3C9 = 6
5a. R4C89 = 11 = {29/38/47}, no 1,5

6. R345C1 = {289/379/469/478/568}, no 1

7. 21(3) cage at R8C3 = {489/579/678}

8. R9C456 = {289/379/469/478/568}, no 1

9. 45 rule on C789 1 remaining outie R8C6 = 4 -> R89C7 = 9 = {27/36}, no 8,9

10. CPE 8,9 in C7 locked in R4567C7 -> no 8,9 in R6C8

11. 45 rule on N5 4 innies R4C56 + R6C45 – 4 = 2 outies R3C4 + R7C6
11a. R4C6 + R6C4 = 4 (step 2a) -> R46C5 = R3C4 + R7C6
11b. 41(7) cage at R3C4 + R4C6 + R6C4 effectively form 45(9) cage -> R46C5 = R3C4 + R7C6 becomes LoL -> no 2 in R46C5

12. 45 rule on C123 2 outies R28C4 = 12 = [48]/{57}, no 6,9, no 8 in R2C4

13. 45 rule on N7 1 outie R8C4 – 4 = 1 innie R7C1 -> R7C1 = {134}

14. 45 rule on N7 3 innies R7C1 + R89C3 = 17 = {179/359/368/458/467}
14a. 4 of {458/467} must be in R7C1 -> no 4 in R9C3
14b. 21(3) cage at R8C3 (step 7) = {579/678} = 7{59/68}, CPE no 7 in R8C12

15. 45 rule on C9 3 outies R149C8 = 24 = {789}, locked for C8, clean-up: no 7,8,9 in R4C9 (step 5a)
15a. Min R9C8 = 7 -> max R89C9 = 5, R8C9 = {123}, R9C9 = {1234}

16. 21(4) cage at R6C8 = {1569/2469/2568/3459/3468/3567} (cannot be {1389/1479/1578/2379/2478} because 7,8,9 only in R7C7) -> R7C7 = {789}

17. 5 in C9 locked in R567C9 = 5{29/38/47}, no 1
17a. 1 in C9 locked in R89C9, locked for N9
17b. 12(3) cage in N9 = 1{29/38/47}

18. 45 rule on N9 1 innie R7C9 – 3 = 1 remaining outie R6C8 -> no 2,3 in R7C9, no 3 in R6C8

19. 45 rule on C89 2 outies R37C7 – 9 = 1 innie R5C8, max R37C7 = 13 -> max R5C8 = 4
19a. 5,6 in C8 locked in R678C8, 21(4) cage at R6C8 (step 16) = 56{19/28} (cannot be {3567} which clashes with R89C7), no 3,4,7, clean-up: no 7 in R7C9 (step 18)
19b. 4 in N9 locked in R79C9, locked for C9, clean-up: no 7 in R4C8 (step 5a)

20. 45 rule on R89 3 innies R8C258 = 10 = {127/136/235}, no 8,9
20a. 6 in {136} must be in R8C8 -> no 6 in R8C25

21. 9 in R8 locked in R13C8, locked for N7

22. 21(3) cage at R8C3 (step 14b) = {579/678}
22a. 9 of {579} only in R8C3 -> no 5 in R8C3

23. 45 rule on N6 5 innies R4C89 + R5C9 + R6C89 = 24, R4C89 = 11 (step 5a) -> R5C9 + R6C89 = 13 = {139/157/238/256}
23a. 1 of {157} and 6 of {256} only in R6C8 -> no 5 in R6C8, clean-up: no 8 in R7C9 (step 18)
23b. 5 in N6 locked in R56C9, locked for C9, clean-up: no 2 in R6C8 (step 18)

24. R567C9 (step 17) = 5{29/47} (cannot be {358} because R7C9 only contains 4,9), no 3,8
24a. 9 of {259} must be in R7C9 -> no 9 in R56C9
24b. 8 in C9 locked in R12C9, locked for N3
24c. R5C9 + R6C89 (step 23) = {157/256}

25. 4 in N6 locked in 21(4) cage at R4C7 = 4{179/269/368} (cannot be {2478} which clashes with R56C9)

26. 45 rule on N2 3 remaining innies R2C4 + R3C46 – 11 = 1 outie R4C5, min R4C5 = 4 -> min R2C4 + R3C46 = 15, max R2C4 + R3C6 = 10 -> min R3C4 = 5, clean-up: no 4 in R46C5 (LoL, step 11b)
26a. Min R4C5 = 5 -> min R2C4 + R3C46 = 16, max R2C4 + R3C6 = 10 -> no 5 in R3C4
26b. Max R2C4 + R3C46 = 19 -> max R4C5 = 8
26c. R4C5 = {5678} -> R2C4 + R3C46 = 16 or 17 or 18 or 19 = [493/583/781/593/791/783/793], no 7 in R3C4

27. 45 rule on R12 2 innies R2C258 = 13 = {139/148/346} (cannot be {157} which clashes with R2C7), no 5,7

28. 45 rule on C5 3 innies R159C5 = 11 = {128/137/146/236/245}, no 9
28a. 1 of {128/137} must be in R1C5 -> no 7,8 in R1C5
28b. 7 of {137} must be in R5C5 -> no 7 in R9C5

29. 45 rule on N8 3 remaining innies R7C46 + R8C4 – 5 = 1 outie R6C5, max R6C5 = 9 -> max R7C46 + R8C4 = 14, min R78C4 = 6 -> max R7C6 = 8
29a. Min R7C46 + R8C4 = 12 -> min R6C5 = 7
29b. R6C5 = {789} -> R7C46 + R8C4 = 12,13,14 = [165/157/175/158/167/185/365]

30. 45 rule on N8 5 remaining innies R7C456 + R8C45 = 22 = {12568/13567} (cannot be {12379} because R7C6 + R8C4 require two of 5,6,7,8 in the combination) = 156{28/37}, no 9, 5,6 locked for N8
30a. Taking the remaining combinations with the permutations in step 29b and avoiding repetition, R678C5 = [782/863/917/935/953/962/971], no 2 in R7C5

31. R9C456 = {289/379} = 9{28/37}, 9 locked for R9, clean-up: no 2 in R89C9 (step 17b)

32. Killer pair 7,8 in R9C456 and R9C8, locked for R9, clean-up: no 2 in R8C7 (step 9)

33. 21(3) cage at R8C3 (step 14b) = {579/678}, R9C3 = {56} -> no 5,6 in R8C34, 7 locked in R8C34 for R8, clean-up: no 1 in R7C1 (step 13), no 1 in R7C5 (step 30a), no 2 in R9C7 (step 9)

34. Naked pair {36} in R89C7, locked for C7 and N9 -> R8C9 = 1, R9C9 = 4, R9C8 = 7, R1C8 = 9, R4C8 = 8, R4C9 = 3, R7C9 = 9, R7C7 = 8, R6C8 = 6 (step 20a), R34C6 = [31], R67C4 = [31], R5C8 = 1 (hidden single in C8), clean-up: no 7 in R56C9 (step 24), no 7 in R6C5 (step 30a), no 7 in R7C5 (step 30a), no 3 in R9C5 (step 31)

35. Naked triple {235} in R8C258, locked for R8 -> R89C7 = [63]
35a. Naked triple {289} in R9C456, locked for R9 and N8 -> R8C4 = 7 -> R7C1 = 3 (step 13), clean-up: no 7 in R4C5 (LoL, step 11b)
35b. Naked pair {56} in R7C56, locked for R7 and N8 -> R78C8 = [25], R8C25 = [23], R3C7 = 2 (hidden single in R3)

36. R7C1 = 3 -> R6C12 = 10 = {19}/[28], no 4,5,7, no 8 in R6C1
36a. Killer pair 8,9 in R6C12 and R6C5, locked for R6

37. R159C5 (step 28) = {128/245} (cannot be {146} because only 2,8 in R9C5), no 6,7

38. Naked pair {56} in R47C5, locked for C5, clean-up: no 4 in R15C5 (step 37) -> R1C5 = 1, R12C7 = [51]
[This naked pair has been there since the clean-up in step 35a but because I did step 37 first it was more effective.]

39. R3C5 = 7 (hidden single in C5), R24C5 = 10 = [46] (only remaining permutation), R2C4 = 5, R23C8 = [34], R7C56 = [56], R1C46 = [68] , R9C6 = 9, R12C9 = [78], R3C4 = 9, clean-up: R6C12 = [28] (step 36), R6C5 = 9 (LoL, step 11b)

and the rest is naked singles and a couple of cage sums


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PostPosted: Sun Jul 06, 2008 6:31 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 65 v2 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:2816:2816:4354:3587:3587:3587:4614:5383:5383:2816:4874:4354:4354:3085:4614:4614:2320:5383:5650:4874:4874:10005:3085:1559:2320:2320:3098:5650:4874:4893:10005:3085:1559:6689:3098:3098:5650:4893:4893:10005:10005:10005:6689:6689:3116:3117:3117:4893:1840:4913:10005:6689:4916:3116:3117:3127:3127:1840:4913:10005:4916:4916:3116:4671:3127:5185:5185:4913:4676:4676:4916:4167:4671:4671:5185:1867:1867:1867:4676:4167:4167:
Solution:
+-------+-------+-------+
| 6 3 7 | 5 1 8 | 2 9 4 |
| 2 1 4 | 6 3 9 | 7 5 8 |
| 8 5 9 | 7 2 4 | 1 3 6 |
+-------+-------+-------+
| 9 4 3 | 8 7 2 | 6 1 5 |
| 5 6 8 | 1 9 3 | 4 7 2 |
| 1 7 2 | 4 6 5 | 9 8 3 |
+-------+-------+-------+
| 4 9 1 | 3 8 6 | 5 2 7 |
| 3 2 6 | 9 5 7 | 8 4 1 |
| 7 8 5 | 2 4 1 | 3 6 9 |
+-------+-------+-------+
Quote:
Ruud, lead-in: Tough, but no hypotheticals (guesses) needed.
CathyW: Hmmph! Confused. Can't say that the V2 is a fun puzzle.
Howard S: Learnt a lot
Tag solution: Howard S and Para - moral support from CathyW and mhparker
Para: Step 29 isn't really a solver-step, more human creativity
Andrew (in 2014): In some ways my solving path was the other way round from the Tag solution. I used an unusual, possibly creative, step for my first key move but a more regular one for my final breakthrough. Rating Hard 1.5
Tag Walkthrough:
Howard S:
I have taken into account your comments and had another unsuccessful attempt at V2. Any help with the below would be much appreciated. Hopefully no more typo's

1 Cage 7/2 N8 = {124}
2 Outies C789=> R2C6+R8C6=16={97}
3 Outies C123=> R2C4+R8C4=15={87I96}
4 Outies N14=>R2C4+R7C1=10=[91I82I73I64]
5 39/7 N258=98763{42I51}
6 97 Locked in R345C45
7 97 cannot be both in C4=> R5C5=9I7
8 => 19/3 C5 <>3
9 => 3 in R7 locked in N8
10 => R7C1<>3 => R2C4 <>7 => R8C4<>8
11 Cage 6/2 C6 = {42I51)
12 Cage 7/2 C4 =[16I25I43]
13 Combinations on R2C46 & R8C46 => 9 locked in R8C46
14 => R78C5= 2 of 5678
15 => Cage 19/3 C5 = {8{74I65}}
16 Rule 45=> R159C5=14=>R1C5=1,3,4,5,6
17 Cage 14/3 R1 <>97=>{851I842I653}
18 Cage 12/3 C5 = {921I741I732I651}
19 => R1C5 <>56 (as Cage 19/3 C5 = 865 with all except 651)
20 => 3 Locked in C6 R567 => R345C4 <>3
21 => Cage 7/2 C4 = [43] => Cage 19/3 C5 = {865} => R8C46={97}
22 =>R2C4<>9 R7C1<>1
23 => Cage 12/3 C5 <> {651} =>97 locked in C5R2345
24 => Cage 39/7 = 9876531
25 => 3 locked in C5 N2
26 97 Locked in C4 R3458[/quote]

Para:
Hi all

This is how i finished the puzzle from Howards move 26.

End of WT 65V2

Just being thorough:

Prelim:
11(3) at R1C1 = {128/137/146/236/245}: no 9
21(3) at R1C8 = {489/579/678}: no 1,2,3
9(3) at R2C8 = {126/135/234}: no 7,8,9
22(3) at R3C1 = {589/679}: no 1,2,3,4; 9 locked for C1
20(3) at R8C3 = {389/479/569/678}: no 1,2
26(4) at R4C7 = {2789/3689/4589/4679/5678}: no 1

Clean up:

17(3) at R1C3 = {269/278/368/458/467}: no 1
6(2) at R3C6 = [15/42/51]: R3C6: no 2
12(3) at R6C1 = [192]/{17}[4]/{37}[2]/{26}[4]/{35}[4]: no 8

Let's begin.

27. 45 on N69: 2 outies: R3C9 + R8C6 = 13 -->> R3C9 = {46}
27a. Killer Triple {456} in 21(3) at R1C8 + 9(3) at R2C8 + R3C9 -->> locked for N3
27b. Clean up: 12(3) at R3C9 = {147/156/246/345}: no 8,9

28. 45 on R12: 3 innies: R2C258 = 9 = {126/135/234}: no 7,8,9

This move opens the whole puzzle up.
29. 9 in N3 locked within R12
29a. R2C46 = [69/78](R2C6 = R8C4 through R8C6)
29b. 17(3) at R1C3: {269} blocked, because then R2C4 = 6 -->> R2C6 = 9, which leave no room for 9 in N3
29c. 17(3) at R1C3 = {278/368/458/467}: no 9
29d. 9 in N1 locked for R3
29e. R2C6 = 9(hidden); R8C46 = [97]; R2C4 = 6; R7C1 = 4; R3c9 = 6

30. 2 in N5 locked for R4
30a. 12(3) at R3C9 = 6{15}(last combo): R4C89 = {15} -->> locked for R4 and N6
30b. R34C6 = [42]; R9C456 = [241]

31. Naked Pair {79} in R45C5 locked for C5 and N5
31a. R4C4 = 8; R1C46 = [58]; R1C5 = 1; R35C4 = [71]; R45C5 = [79]

32. 45 on N1: 1 innie and 1 outie: R4C2 + 4 = R3C1 -->> R3C1 = 8; R4C2 = 4(only possible combo)
32a. R4C1 = 9(hidden); R5C1 = 5

33. 17(3) at R1C3 = 6{47} (last combo) -->> R12C3 = {47}; locked for C3 and N1
33a. 11(3) at R1C1 = {236} (last combo) -->> locked for N1

34. 45 on R12: 3 innies: R2C258 = 9 = {135}({234} blocked by R2C2): no 2,4; locked for R2
34a. R2C1 = 2; R23C5 = [32]; R1C7 = 2(hidden); R2C7 = 7; R12C3 = [74]; R2c9 = 8

35. Naked Pair {15} in R24C8 -->> locked for C8
35a. R3C8 = 3

36. 26(4) at R4C7 = {3689/4679}: {2789} blocked by R4C7: no 2; 6 and 9 locked for N6
36a. R6C7 = 9(hidden)

37. 18(3) at R8C6 = 7{38/56} -->> R89C7 = {38/56}: no 1,4
37a. R5C7 = 4(hidden)
37b. 26(4) at R4C7 = 49{67} -->> R4C7 = 6; R5C8 = 7
37c. R89C7 = {38} -->> locked for C7 and N9
37d. R6C8 = 8(hidden)

38. 12(3) at R5C9 = {237} (last combo): locked for C9 -->> R7C9 = 7

39. 2 in C8 locked within 19(4) cage at R6C8 -->> 19(4) = 8{245}: no 1,6,9 -->> R7C78 = [52]; R8C8 = 4

And the rest is al naked singles(and 1 hidden single).

greetings

Para
Alternative to tag solutions' step 29 by mhparker using discontinuous Nice Loop:
Para wrote:
Step 29 isn't really a solver-step, more human creativity.
Nice move, Para!

I agree with you here that it's extremely unlikely that an automated solver would (or could) find such a step. However, it's not inconceivable that a solver with a very good AIC/Nice Loop implementation could find the alternative move presented below that would serve as a replacement for your steps 29 - 29c:

Code:
.-----------------------.-----------.-----------------------------------.-----------.-----------------------.
| 12345678    12345678  | 23456789  | 2568        134         2568      | 123789    | 456789      456789    |
|           .-----------:           '-----------.-----------.-----------'           :-----------.           |
| 12345678  | 123456    | 23456789    68        | 1234      | 79          123789    | 123456    | 456789    |
:-----------:           '-----------.-----------:           :-----------.-----------'           :-----------:
| 56789     | 123456789   123456789 | 156789    | 123479    | 145       | 123456      123456    | 46        |
|           |           .-----------:           |           |           :-----------.-----------'           |
| 56789     | 123456789 | 13456789  | 156789    | 1279      | 125       | 23456789  | 1234567     1234567   |
|           :-----------'           |           '-----------'-----------:           '-----------.-----------:
| 56789     | 123456789   123456789 | 156789      79          13568     | 23456789    23456789  | 123456789 |
:-----------'-----------.           :-----------.-----------.           |           .-----------:           |
| 123567      1235679   | 12356789  | 4         | 568       | 13568     | 2356789   | 12356789  | 12356789  |
|           .-----------'-----------:           |           |           :-----------'           |           |
| 24        | 12456789    12456789  | 3         | 568       | 568       | 12456789    12456789  | 12456789  |
:-----------:           .-----------'-----------:           :-----------'-----------.           :-----------:
| 12345678  | 12345678  | 345678      79        | 568       | 79          12345678  | 12345678  | 12345678  |
|           '-----------:           .-----------'-----------'-----------.           :-----------'           |
| 35678       356789    | 356789    | 12          124         124       | 356789    | 356789      356789    |
'-----------------------'-----------'-----------------------------------'-----------'-----------------------'


29. Complex discontinuous Nice Loop (2 weak links at discontinuity) removes 9 from r12c3, as follows ("=>" implies strong link, "," implies direct link):
29a. if r12c3 = 9 -> r3c123 <> 9
29b. => r3c45 = 9 -> r2c6 <> 9
29c. => r2c6 = 7 -> r8c6 <> 7
29d. => r8c6 = 9 -> r8c4 <> 9
29e. => r8c4 = 7, r2c4 = 8 -> r12c3 <> 9 (contradiction)
29f. Conclusion: no 9 in r12c3

Note: Nice Loop notation avoided for clarity and due to difficulty representing direct link between 7 in r8c4 and 8 in r2c4
Walkthrough by Andrew:
Prelims

a) R34C6 = {15/24}
b) R67C4 = {16/25/34}, no 7,8,9
c) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
d) 21(3) cage at R1C8 = {489/579/678}, no 1,2,3
e) 9(3) cage at R2C8 = {126/135/234}, no 7,8,9
f) 22(3) cage at R3C1 = {589/679}
g) 19(3) cage at R6C5 = {289/379/469/478/568}, no 1
h) 20(3) cage at R8C3 = {389/479/569/578}, no 1,2
i) 7(3) cage at R9C4 = {124}
j) 26(4) cage at R4C7 = {2789/3689/4589/4679/5678}, no 1

Steps resulting from Prelims
1a. 22(3) cage at R3C1 = {589/679}, 9 locked for C1
1b. 7(3) cage at R9C4 = {124}, locked for R9 and N8, clean-up: no 3,5,6 in R6C4
1c. Min R9C89 = 8 -> max R8C9 = 8

2. 45 rule on C123 2 outies R28C4 = 15 = {69/78}

3. 45 rule on C789 2 outies R28C6 = 16 = {79}, locked for C6
3a. 17(3) cage at R1C3 = {179/269/278/359/368/458/467}
3b. 1 of {179} must be in R2C3 (R2C34 cannot be {79} which clashes with R2C6) -> no 1 in R1C3
3c. 19(3) cage at R6C5 cannot be 3{79} which clashes with R8C6 -> no 3 in R6C5

4. 45 rule on N3 1 outie R2C6 = 1 innie R3C9 + 3, R2C6 = {79} -> R3C9 = {46}
4a. 12(3) cage at R3C9 = {147/156/246/345} (cannot be {129/138/237} because R3C9 only contains 4,6), no 8,9

5. 45 rule on N7 1 outie R8C4 = 1 innie R7C1 + 5, R8C4 = {6789} -> R7C1 = {1234}

6. 45 rule on R12 3 innies R2C258 = 9 = {126/135/234}, no 7,8,9

7. 45 rule on R89 3 innies R8C258 = 11 = {128/137/146/236/245}, no 9
7a. 5,8 of {128/245} must be in R8C5 -> no 5,8 in R8C28

8. R34C6 = {15/24}, R9C6 = {124} -> combined half cage R349C6 = {15}2/{15}4/{24}1, 1 locked for C6

9. 45 rule on N3 3 innies R12C7 + R3C9 = 15 = {168/249/267/348} (cannot be {159/258/357} because R3C9 only contains 4,6, cannot be {456} which clashes with 21(3) cage at R1C8), no 5
9a. R3C9 = {46} -> no 4,6 in R12C7

10. 45 rule on C89 1 innie R5C8 = 2 outies R37C7 + 1
10a. Min R37C7 = 3 -> min R5C8 = 4
10b. Max R37C3 = 8, no 8,9 in R7C7

11. 45 rule on N1 2(1+1) outies R2C4 + R4C2 = 1 innie R3C1 + 2
11a. Max R2C4 + R4C2 = 11, min R2C4 = 6 -> max R4C2 = 5

12. R28C4 = 15 (step 2), R28C6 = 16 (step 3), R2C46 cannot have the same value -> R8C6 cannot be 1 more than R8C4 -> R8C46 cannot be [67] or [89] -> R78C5 cannot be [98] -> no 2 in R6C5

13. 45 rule on C5 3 innies R159C5 = 14 = {149/158/239/248/257/347} (cannot be {356} because R9C5 only contains 1,2,4, cannot be {167} which clashes with 19(3) cage at R6C5), no 6

14. 39(7) cage at R3C4 = {1356789/2346789}
14a. Hidden killer pair 7,9 in R28C4 and 39(7) cage at R3C4 for C4, R28C4 contains one of 7,9 -> R345C4 contains one of 7,9 -> R5C5 = {79} (only other place for one of 7,9 in 39(7) cage)
14b. Killer pair 7,9 in R28C4 and R345C4, locked for C4
14c. R159C5 (step 13) = {149/239/257/347} (cannot be {158/248} because R5C5 only contains 7,9)
14d. R5C5 = {79} -> no 7,9 in R1C5
14e. 2 of {239/257} must be in R9C5 -> no 2 in R1C5

15. 19(3) cage at R6C5 = {469/478/568} (cannot be {379} which clashes with R5C5), no 3
15a. 3 in N8 only in R7C46, locked for R7, clean-up: no 8 in R8C4 (step 5), no 7 in R2C4 (step 2)
15b. 19(3) cage = {478/568} (cannot be {469} = 4{69} which clashes with R8C68, ALS block), no 9, 8 locked for C5
15c. 4 of {478} must be in R6C5 -> no 7 in R6C5
15d. 9 in N8 only in R8C46, locked for R8
15e. R159C5 (step 14c) = {149/239/347} (cannot be {257} which clashes with 19(3) cage), no 5
15f. R8C258 (step 7) = {128/137/146/236/245}
15g. 6,7 of {137/146/236} must be in R8C5 -> no 6,7 in R8C28

16. 45 rule on N7 3 innies R7C1 + R89C3 = 15 = {159/249/258/348/456} (cannot be {168/267} because 20(3) cage at R8C3 cannot be {68}6/{67}7, cannot be {357} because R7C1 only contains 1,2,4), no 7
16a. 18(3) cage at R8C1= {189/279/369/378/468/567} (cannot be {459} which clashes with R7C1 + R89C3)
16b. 9 of {369} must be in R9C2, 3 of {378} must be in R89C1 (R89C1 cannot be {78} which clashes with 22(3) cage at R3C1), no 3 in R9C2

17. 3 in N8 only in R7C46, CPE no 3 in R345C4
17a. 39(7) cage at R3C4 = {1356789/2346789}, 3 locked for C6

18. 14(3) cage at R1C4 = {158/248/356}
18a. R1C5 = {134} -> no 1,3,4 in R1C46

19. R7C4 = 3 (hidden single in C4) -> R6C4 = 4, clean-up: no 2 in R3C6
19a. 3 in C6 only in R56C6, locked for N5

20. 39(7) cage at R3C4 = {1356789} (only remaining combination), no 2, 1 locked for C4 -> R9C4 = 2

21. 19(3) cage at R6C5 (step 15b) = {568} (only remaining combination), locked for C5

22. R8C46 = {79} (hidden pair in N8), locked for R8, clean-up: no 9 in R2C4 (step 2), no 1 in R7C1 (step 5)
22a. Killer pair 6,8 in 14(3) cage at R1C4 and R2C4, locked for N2
22b. 6,8 in N2 only in R1C46 + R2C4, CPE no 6,8 in R1C3

23. 17(3) cage at R1C3 = {269/278/368/458/467} (cannot be {179/359} because R2C4 only contains 6,8), no 1
23a. 3 of {368} must be in R1C3 -> no 3 in R2C3
23b. {368} must be [386] (cannot be [368] which clashes with 14(3) cage at R1C4) -> no 6 in R2C3

24. 2 in C5 only in 12(3) cage at R2C5 = {129/237}, no 4
[With hindsight, one can now jump forward to step 31 for a shorter solving path, although some of the intermediate steps may still be needed.]

25. R7C1 + R89C3 (step 16) = {249/258/348/456}
25a. 18(3) cage at R8C1 (step 16a) = {189/279/369/378/567} (cannot be {468} which clashes with R7C1 + R89C3), no 4
25b. 12(3) cage at R7C2 = {129/138/147/237} (cannot be {246} which clashes with R7C1, cannot be {345} which clashes with R7C1 + R89C3, cannot be {156} which clashes with R7C56, ALS block), no 6

26. 12(3) cage at R6C1 = {129/147/237/246/345} (cannot be {138/156} because R7C1 only contains 2,4), no 8
26a. 11(3) cage at R1C1 = {128/137/146/236/245}
26b. 5 of {245} must be in R12C1 (R12C1 cannot be {24} which clashes with R7C1), no 5 in R1C2

27. R2C4 = {68} -> R12C3 must total 9 or 11
27a. 45 rule on C123 4 innies R1289C3 = 22 = {2389/2479/2569/2578/3478/3568/4567} (cannot be {3469} because no 6 in R12C3)
27b. {29} of {2569} and {27} of {2578} must be in R12C3, {56} of {3568} must be in R89C3 (R12C3 cannot be {56} because 17(3) cage at R1C3 cannot be {56}6), {47} of {4567} must be in R12C3 -> no 5 in R12C3
[With hindsight a simpler way is R1289C3 = 22, R12C3 = 9,11 -> R89C3 = 11,13
R12C3 cannot be {45} = 9 which clashes with R89C3 = 13 = [49/58/85] -> no 5 in R12C3.]


28. 2 in N5 only in R4C56, locked for R4
28a. 12(3) cage at R3C9 (step 4a) = {147/156/345}
28b. R3C9 = {46} -> no 4,6 in R4C89
28c. 26(4) cage at R4C7 = {2789/3689/4589/4679} (cannot be {5678} which clashes with 12(3) cage), 9 locked for N6

29. 12(3) cage at R5C9 = {129/138/147/156/237/345} (cannot be {246} which clashes with R3C9)
29a. 8 of {138} must be in R56C9 (R56C9 cannot be {13} which clashes with 12(3) cage at R3C9 ), no 8 in R7C9
29b. 6 of {156} must be in R56C9 (R56C9 cannot be {15} which clashes with 12(3) cage at R3C9), no 6 in R7C9
29c. 1 of {147} must be in R7C9 (12(3) cage at R5C9 cannot be {14}7/{17}4 which clash with 12(3) cage at R3C9), 5 of {345} must be in R7C9 (12(3) cage at R5C9 cannot be {35}4 which clash with 12(3) cage at R3C9) -> no 4 in R7C9

30. 45 rule on N9 2(1+1) outies R6C8 + R8C6 = 1 innie R7C9 + 8
30a. R8C6 = {79} -> R6C8 must either be 1 more or 1 less than R7C9, R7C9 = {12579} -> R6C8 = {12368}, no 5,7

[At last some more placements …]
31. R2C6 = R3C9 + 3 (step 4) -> R2C6 + R3C9 = [74/96]
31a. R28C4 = 15 (step 2), R28C6 = 16 (step 3), R8C46 = {79} = 16 -> R2C46 = 15
31b. 45 rule on N2 4 remaining innies R2C5 + R3C456 = 16 = {1357/2347} (cannot be {1249} which clashes with R2C6 + R3C9, IOD clash), 7 locked for R3 and N2 -> R2C6 = 9, R2C4 = 6, R8C46 = [97], R3C9 = 6, R7C1 = 4 (step 5)
[I’d seen the 4 innies a long time ago but only just spotted the IOD clash.
The puzzle is cracked, the rest is fairly straightforward.]

31c. 39(7) cage at R3C4 must contain 9 -> R5C5 = 9
31d. 12(3) cage at R2C5 (step 24b) = {237} (only remaining combination), locked for C5
31e. 9 in N6 only in 26(4) cage at R4C7, locked for C7

32. R3C9 = 6 -> R4C89 = 6 = {15}, locked for R4 and N6 -> R4C6 = 2, R3C6 = 4, R9C56 = [41], R4C45 = [87], 14(3) cage at R1C4 = [518], R35C4 = [71]
32a. R2C6 = 9 -> R12C7 = 9 = {27}, locked for C7 and N3
32b. Naked pair {49} in R1C89, locked for R1 and N3, R2C9 = 8 (cage sum)

33. 6 in R1 only in 11(3) cage at R1C1 = {236} (only remaining combination), locked for N1 -> R12C3 = [74], R12C7 = [27], R2C1 = 2 (hidden single in N1), R23C5 = [32]

34. 19(4) cage at R2C2 = {1459} (only remaining combination) -> R4C2 = 4
34a. R3C1 = 8 (hidden single in N1) -> R45C1 = 14 = [95]
34b. R8C6 = 7 -> R89C7 = 11 = {38/56}
34c. R56C7 = [49] (hidden pair in C7) = 13 -> R4C7 + R5C8 = 13 = [67] -> R89C7 = {38}, locked for C7 and N9
34d. Naked pair {15} in R24C8, locked for C8 -> R3C8 = 3

35. R4C3 = 3
35a. R8C4 = 9 -> R89C3 = 11 = {56}, locked for C3 and N7
35b. R4C3 = 3 -> 19(4) cage at R4C3 = {2368} (only remaining combination) -> R5C2 = 6, R5C6 = 3, R56C9 = [23], R56C3 = [82]
35c. R56C9 = [23] -> R7C9 = 7 (cage sum)

36. R6C8 = 8 -> 19(4) cage at R6C8 = {1468/2458} -> R8C8 = 4, R1C89 = [94], R9C8 = 6 -> R7C78 = [52]

and the rest is naked singles.

Easier than expected from the SSscore but harder than the V3 as there aren't any immediate placements. I'll rate my walkthrough at Hard 1.5. Step 12 is a rarely used one, which I can only remember using one, or maybe twice, before.


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PostPosted: Sun Jul 06, 2008 6:34 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 65 v3 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:4352:4352:4354:4099:4099:4099:2310:3591:3591:4352:3082:4354:4354:3853:2310:2310:3856:3591:3858:3082:3082:10261:3853:3095:3856:3856:5658:3858:3082:4637:10261:3853:3095:5921:5658:5658:3858:4637:4637:10261:10261:10261:5921:5921:3628:4397:4397:4637:2608:2865:10261:5921:4148:3628:4397:4663:4663:2608:2865:10261:4148:4148:3628:3903:4663:3649:3649:2865:5444:5444:4148:3143:3903:3903:3649:3147:3147:3147:5444:3143:3143:
Solution:
+-------+-------+-------+
| 5 4 9 | 6 7 3 | 1 2 8 |
| 8 1 3 | 5 9 2 | 6 7 4 |
| 7 6 2 | 4 1 8 | 3 5 9 |
+-------+-------+-------+
| 2 3 8 | 1 5 4 | 9 6 7 |
| 6 5 1 | 9 8 7 | 2 4 3 |
| 9 7 4 | 2 3 6 | 8 1 5 |
+-------+-------+-------+
| 1 9 7 | 8 2 5 | 4 3 6 |
| 4 2 5 | 3 6 9 | 7 8 1 |
| 3 8 6 | 7 4 1 | 5 9 2 |
+-------+-------+-------+
Quote:
Ruud, lead-in: hot hot hot, probably needs some creative moves (guesses)
Para: That V3 is really deceiving, you think you're doing pretty well, placing digits, making eliminations and then all of the sudden you hit a dead end and there's seems to be no way out.
Caida: I've been pounding away at it for quite sometime and I think I'm going in circles. I've resorted to trial and error
Afmob: Rating: 1.75 it wasn't as difficult as A74 Brick Wall but a bit more complicated than M1, it's solvable without guesses or T&E
gary w: after a tough battle..rather more than 5 hours....managed to crack it

Andrew (in 2014): I've only just tried this puzzle. Not as hard as I'd expected from Ruud's and Para's comments. Strange that this puzzle was the V3; maybe Ruud's solver couldn't find the immediate placements from the starting grid (the V2 doesn't have any). Rating 1.5
A 2021 forum Revisit here
No contradiction moves Walkthrough by Afmob:
After the Brick Wall I decided that I wanted to solve another tough Assassin and since this hasn't been solved yet I tried this one and succeeded. The main cracker is step 8h).

Edit: Removed some steps, added step 8b) which is important for the following step 8h) and corrected some typos.
Edit 2: Made some corrections suggested by Mike, thanks!

A65 V3 Walkthrough:

1. C123
a) Outies C1 = 19(3) -> no 1
b) 12(4) = 12{36/45} -> R1C2 <> 2
c) Outies C123 = 8(2) = {17/26/35}

2. C789
a) Outies C789 = 11(2) = [29/38/47/56/65]
b) 22(3) = 9{58/67} -> R56C9 <> 9

3. N7
a) Outies = 19(2+1): R6C12 <> 1,2 since R8C4 <= 7
b) Innies = 12(3): R7C1 <> 6,7,8,9 because R89C3 @ 14(3) >= 7 since R8C7 <= 7
c) Innies = 12(3): R8C4 <> 1,2 because R89C3 must be less than 12
d) Innies = 12(3): R7C1 <> 2 because R89C3 = 7/8/9/11

4. C456
a) Innies C5 = 19(3) -> no 1
b) Outies C123 = 8(2): R2C4 <> 6,7
c) 17(3): R12C3 <> 1,2 since R2C4 <= 5

5. N369 !
a) ! Outies N69 = 18(1+1) -> R3C9 = 9, R8C6 = 9
b) Innies+Outies N3: -7 = R2C6 - R3C9 -> R2C6 = 2
c) 9(3) = 2{16/34}
d) Innies N6 = 9(3) -> no 7,8,9

6. C456
a) 21(3) = 9{48/57}
b) 12(2) = {48/57}
c) Outies C123 = 8(2) = [17/35/53]
d) Killer pair (37) of Outies C123 blocks {37} of 10(2)
e) 40(7) = 56789{14/23} -> 9 only possible @ N5 -> R4C5+R6C4 <> 9
f) 10(2) = {28/46}

7. N69
a) Innies+Outies N6: 5 = R7C9 - R6C8 -> R7C9 = (678), R6C8 = (123)
b) 23(4) = 9{158/167/248/347} -> Other combos are blocked by Killer pairs (56,57,68) of 22(3)
c) 16(4) = {1249/1267/1348/1357/2356} -> other combos blocked by Killer pairs (45,47,58) of 21(3)
d) 16(4) <> {1267} since R7C9 would be 8 -> no combo for 21(3)
e) Killer pair (45) locked in 16(4) + 21(3) for N9
f) 12(3) = {129/138/237} -> no 6

8. C789 !
a) ! Consider both possibilities of 21(3) -> 12(3) <> 7:
- i) R89C7 = {57} -> 12(3) <> 7
- ii) R89C7 = {48} -> 16(4) = 35{17/26} together with R7C9 = (67) builds naked pair for N9 -> 12(3) <> 7
b) Outies C9 = 17(3): R9C8 <> 1 since R14C8 <= 15
c) 12(3) = 1{29/38} -> 1 locked for C9+N9
d) 14(3) @ N6 must have 6,7 xor 8 and R7C9 = (678) -> R56C9 <> 6
e) Innies N6 = 3{15/24} -> 3 locked for N6
f) 12(3): R9C8 <> 2 because R89C9 <> 9
g) Innies C7 = 26(5) = 29{168/348/357} because 2,9 locked in Innies and {24569}
blocked by Killer pair (45) of 21(3)
h) ! Combo analysis of Innies C7 and R456C7 @ 23(4) -> Innies C7 <> 5,7:
- 23(4) = 9{158/167/248} and Innies = 29{168/348/357}, now consider possibilites for R456C7
- 3 candidates of Innies must be in R456C9 therefore 3 of those 5 candidates must be in 23(4)
- Analysis for Innies C7 = {23579}:
- i) 23(4) = {1589} has only 5,9 in common
- ii) 23(4) = {1679} has only 7,9 in common
- iii) 23(4) = {2489} has only 2,9 in common
-> But it must have at least 3 candidates in common
-> Innies C7 = 289{16/34} -> 8 locked for C7

9. N69
a) 21(3) = {579} -> {57} locked for N9
b) 16(4) = 14{29/38} -> R6C8 = 1
c) Hidden Single: R7C9 = 6 @ N9
d) 14(3) = 6{35}; {35} locked for C9+N6
e) 22(3) = {679} -> R4C8 = 6, R4C9 = 7
f) 12(3): R9C8 <> 8 since R89C9 <> 3

10. N3
a) 4 locked in C9 @ 14(3) -> locked for N3 and 14(3) = 4{28/37}
b) 9 = {126} -> {16} locked for N3
c) 14(3) = {248} because (37) only possible @ R1C8; {248} locked and 2 locked for R1
d) 15(3) = {357} -> R3C7 = 3

11. N25
a) 6 locked in 40(7) between C4 and N5 -> R6C4 <> 6
b) 10(2) = {28} locked for C4
c) 8 locked in 40(7) between C6 and N5 -> R4C6 <> 8
d) 12(2) = [75/84]
e) 16(3) <> 1 because R1C7 = (16) blocks {169} and R3C6 = (78) blocks {178}

12. R12
a) Innies = 17(3) must have 5 xor 7 since R2C8 = (57) -> R2C25 <> 5,7
b) Innies = 17(3) = {179/359/458/467}: R2C5 <> 1,3 because R2C28 <= 13
c) Innies = 17(3) <> 8 because {458} blocked by R2C9 = (48)

13. R789
a) Innies R89 = 16(3) = {178/268/358/367} -> no 4 since {457} blocked by R8C7 = (57)
b) 4 locked in 16(4) in N9 for R7
c) 4 locked in 12(3) for R9 -> 12(3) = 4{17/26/35}
d) 16(4): R7C78 <> 2 because R8C8 <> 4,9
e) 18(3): R8C2 <> 5 because R7C23 <> 6
f) 11(3): R6C5 <> 5 because R78C5 <> 4

14. N6
a) 2 locked in 23(4) @ C7 -> R5C8 <> 2

15. N258 !
a) ! Innies+Outies N25: 16 = R7C456+R8C5 - R2C4; R2C4 = (135)
-> R7C456+R8C5 = 17/19/21
-> R8C5 <> 2,8 because together with R7C4 = (28) R7C56 would be 7/9/11(2) but
that's not possible because only remaining candidates in R7C56 are (1357)
-> sum would be even but it must be odd
b) 8 locked in R7C456 for R7

16. C789
a) 8 locked in 23(4) @ C7 -> R5C8 <> 8
b) Hidden pair (28) in R18C8 @ C8 -> no other candidates

17. R89
a) Innies = 16(3) = 8{17/26/35} because R8C8 = (28) -> 8 locked for R8
b) Innies = 16(3): R8C5 <> 3 because it's the only place where 5 is possible
c) Innies = 16(3): R8C2 <> 6 because R8C5 <> 2,8
d) 18(3) = {189/279/378}
e) 18(3): R8C2 <> 1,3 because R7C23 <> 8
f) Innies = 16(3) = 8{17/26}: R8C5 <> 7 because R8C28 <> 1
g) 11(3) <> 5 because R8C5 = (16)
h) Killer pair (12) of Innies + R8C9 locked for R8
i) 14(3): R9C3 <> 8,9 because R8C34 >= 7
j) 15(3) must have 8 xor 9 since 12(3) @ R9 has one of (89) and it's nowhere else possible in R9
-> 15(3) <> 7
k) 18(3) <> 1 because 15(3) @ N7 must have one of (89)
l) 18(3) = 7{29/38} -> 7 locked for N7

18. N5
a) 11(3): R6C5 <> 6 because R78C5 <> 4 and R8C5 <> 2,3
b) 6 locked in 40(7) -> R3C4 <> 6

19. N7
a) 15(3): R8C1 <> 3 because R9C12 <> 4
b) 15(3) must have 4,5 xor 6 and R8C1 = (456) -> R9C12 <> 5,6
c) 3 locked in 14(3) for R8 -> R9C3 <> 3; 14(3) = 3{47/56}
d) 14(3) = {356} because R9C3 = (56)
e) Hidden Single: R8C1 = 4
f) 15(3) = 4{29/38}
g) Innies = 12(3) = {156} -> R7C1 = 1
h) 14(3) = {356} -> R8C4 = 3; {56} locked for C3
i) 17(3) = {179} -> {79} locked for R6+N4

20. C456
a) Outies C123 = 8(2) = [53] -> R2C4 = 5
b) 16(3) = 3{49/67} -> 3 locked for R1
c) 17(3) = 5{39/48}; R2C3 <> 9
d) 8 locked in R3C56 for R3

21. R89
a) Hidden Single: R7C3 = 7 @ C3
b) 18(3): R7C2 <> 2
c) Innies = 16(3) = {268} -> R8C5 = 6
d) 11(6) = {236} -> R7C5 = 2, R6C5 = 3

22. C456
a) 15(3) = {159} -> R2C5 = 9, R3C5 = 1, R4C5 = 5
b) 16(3) = {367} -> R1C5 = 7, R1C6 = 3, R1C4 = 6
c) R3C4 = 4, R3C3 = 2, R3C6 = 8, R4C6 = 4, R5C5 = 8, R6C6 = 6, R7C6 = 5
d) R1C7 = 1, R2C7 = 6, R2C8 = 7, R3C8 = 5, R3C2 = 6, R3C1 = 7

23. N17
a) 12(4) = {1236} -> {13} locked for C2
b) 18(3) = {279} -> R7C2 = 9, R8C2 = 2

24. Rest is singles.

Rating: 1.75 it wasn't as difficult as A74 Brick Wall but a bit more complicated than M1, it's solvable without guesses or T&E. So no contradiction moves from me this time. :lol:
Fresh-eyes Walkthrough by Caida:
Hello,

Edited to say:I have redone this puzzle with fresh eyes and managed to solve it (after looking for a hint in Afmob's walkthrough below!)


I haven't found a walkthrough for Assasin 65v3. I've been pounding away at it for quite sometime and I think I'm going in circles. I've resorted to trial and error strategies and have wound up with conflicts at the end.

Here is my walkthrough - I know the answer is wrong (not any more!!) - but hopefully the formatting is an improvement over my last attempt.


Assassin 65V3

Preliminaries:
a. 12(4)n14 = {1245/1236} (no 7..9) -> r1c2 no 1,2 (CPE)
b. 9(3)n23 = {126/135/234} (no 7..9)
c. 12(2)n25 = {39/48/57} (no 1,2,6)
d. 22(3)n36 ={589/679} (no 1..4) -> r56c9 no 9 (CPE)
e. 10(2)n58 = {19/28/37/46} (no 5)
f. 11(3)n58 = {128/137/146/236/245} (no 9)
g. 21(3)n89 = {489/579/678} (no 1..3)

1. Outies c123: r28c4 = 8(2) = {17/26/35} (no 4,8,9)

2. Outies c789: r28c6 = 11(2) = {29/38/56}/[47]
2a. -> r8c2 no 1; r8c6 no 4

3. Outies c1: r169c2 = 19(3) = {289/379/469/478/568} (no 1)

4. Outie and Innie n3: r3c9 - r2c6 = 7
4a. since max r3c9 = 9 the max r2c6 = 2
4b. -> r2c6 = 2; r3c9 = 9
4c. -> cleanup: 9(3)n23: r12c7 no 5
4d. -> h11(2)c6 (step 2): r8c6 = 9
4e. -> cleanup: 21(3)n89: r89c7 no 6
4f. -> cleanup: h8(2)c4 (step 1): r8c4 no 6
4g. -> cleanup: 12(2)n25 no 3
4h. -> cleanup: 10(2)n58: r6c4 no 1

5. Innies c5: r159c5 = 19(3) ={289/379/469/478/568} (no 1)

6. Outie and Innie n7: r8c4 - r7c1 = 2
6a. -> min r7c1 = 1 -> min r8c4 = 3 (r8c4 no 1,2)
6b. -> max r8c4 = 7 -> max r7c1 = 5 (r7c1 no 6,7,8,9)
6c. -> r7c1 no 2, 4
6d. -> cleanup: h8(2) (step 1): r2c4 no 6,7
6e. -> cleanup: 10(2)n58 no 3,7 (blocked by h8(2)c4

7. 17(3)n12: since max r2c4 = 5 the min r12c3 = 12 (no 1,2)
7a. 17(3)n47: since max r7c1 = 5 the min r6c12 = 12 (no 1,2)

8. Outie and Innie n6: r7c9 – r6c8 = 5
8a. -> min r6c8 = 1 -> min r7c9 = 6 (r7c9 no 1..5)
8b. -> max r7c9 = 8 -> max r6c8 = 3 (r6c8 no 4..9)

9. 14(3)n69: since min r7c9 = 6 the max r56c9 = 8 (no 8)

10. 40(7)n258 = {14/23}{56789} -> 9 locked in n5 in 40(7)n258
10a. cleanup: 10(2)n58: r7c4 no 1

11. 16(4)n69 = {1249/1348/1357/2356) -> contains either 4 or 5
Note: combos {1258/1456/2347} blocked by r89c7 {1267} blocked by r7c9 and r89c7
11a. -> killer pair {45} locked in n9 for 16(4)n79 and 21(3)n89
11b. -> 12(3)n9 no 4,5
11c. -> 12(3)n9 = {129/138} no 6,7 -> 1 locked in 12(3) for n9
Note: combo {237} blocked by 16(4) and 21(3)
11d. -> 8 locked for n9 in 16(4), 21(3), and 12(3) -> r7c9 no 8
11e. cleanup: Innies & Outies n6 (step 8): r6c3 no 3
11f. cleanup: 12(3)n9 r9c8 no 2 (needs a 9)
11g. if r9c8 = 1 this would mean r89c9 = {38} and 16(4)n79 = [2]{356} not possible
11h. -> 1 locked for n9 and c9 in r89c9

12. 23(4)n6 = {1589/1679/2489/3479}
Note: other combinations ({2579/2678/3569/3578/4568}) blocked by r4c89
12a. 7 locked for n6 in 23(4) and r4c89 -> r56c9 no 7

13. 14(3)n69 = {35}[6]/{34}[7] -> 3 locked for n6 and c9 in r56c9
Note: combo {25}[7] blocked by step 6
13a. -> r56c9 no 2,6
13b. cleanup: 12(3)n9: r9c8 no 8

14. Innies c7: r34567c7 = 26(5) = {12689/23489}
Note: combo {24569} blocked by r12c7 (and r89c7) and combo {23579} blocked by 23(4)n6
14a. -> 8 locked for c7 in r34567
14b. -> r89c7 = {57} -> locked for c7 and n9
14c. -> single: r7c9 = 6
14d. -> r6c8 = 1 (step 8)

15. 14(3)n69 = {35}6
15a. -> r4c89 = [67]
15b. 4 locked in 14(3)n3 in r12c9 for n3 and c9
15c. -> r12c7 = {16} -> locked for c7
15d. cleanup: 14(3)n3 = {248} (only possible combination)
15e. single: r3c7 = 3

16. 40(7)n357 must contain 6 -> locked in r3c4 and n5
16a. -> r6c4 no 6 (CPE)
16b. cleanup: 10(2)n57: r7c4 no 4
16c. 10(2)n57 = {28} – locked for c4
16d. 40(7)n357 must contain 8 -> locked in r7c6 and n5
16e. -> r4c6 no 8 (CPE)

17. 16(3)n2 = {349/358/367/457} (no 1 – blocked by r1c7 and r3c6)

18. Innies r12: r2c258 = 17(3) = [197/395]/{46}[7]
Note: combo [485] blocked by r2c9
18a. r2c2 no 5
18b. r2c5 no 1,3,5,7,8

19. Innies r89: r8c258 = 16(3) = {17}[8]/{268} -> 8 locked for r8
Note: combos {358/367/457} blocked by r8c47
19a. r8c258 no 3,4,5
19b. 4 locked in r7c78 and in r9c456
20. 12(3)n8 = {147/246/345} no 8
20a. r9c5 no 6

21. 16(4)n69 = 1{249/348}
21a. -> r7c78 no 2,8

22. pair {28} in r18c8 -> locked for c8

23. 11(3)n58: r6c5 no 5 (no 4 in r78c5)

24. Innies n8: This is where I borrowed from Afmob’s walkthrough - his step #15 (but written out in super long form so I can see it) - Thanks Afmob!!!
24a. r7c456+r8c45 = 24(5)
24b. r28c4 = 8(2)
24c. r7c456+r8c45-r28c4 = 16
24d. r7c456+r8c5-r2c4 = 16
24e. r2c4 = 1/3/5 -> r7c456+r8c5 = 17/19/21 (4)
24f. since r7c4 must be 2 or 8 then r8c5 <> 2 or 8 as this would mean that r7c56 = 7/9/11(2) and this isn’t possible with any remaining combination (two odd numbers add together to get an even number)
24g. r8c5 no 2,8 -> 8 locked for n8 and r7 in r7c456

25. 18(3)n7: r8c2 no 1 (b/c no 8 in r7c23)
25a. cleanup: h16(3) (step 19) r8c5 no 7
25b. killer pair {12}: h16(3)r8 = {268/178}; r8c9 = {1/2} -> {12} locked for r8 in c2589

26. 11(3)n58 = {28}1/{37}1/[416]/{23}6 (no 5)
26a. -> r6c5 no 6
26b. 6 in n5 locked in 40(7)n358
26c. ->r3c4 no 6

27. 14(3)n78: min r8c34 = 7
27a. -> max r9c3 = 7 (no 8,9)
27b. -> 8, 9 locked in r9 in 15(3)n7 and 12(3)n9 (each must have only 1 of these)
27c. 15(3)n7 = [519/618]/4{29}/5{28}/4{38} (no 7)
27d. -> r8c1 no 3; r9c12 no 5,6
27e. -> 3 locked in r8 in c34

28. 18(3)n7 = {279/378/567} -> 7 locked for n7
Note: combo {189} blocked by 15(3) (step 27b.); combo {369} blocked by r7c78

29. 14(3)n78 = {356} (only possible combination)
29a. r9c3 = 5,6 (no 3 b/c of step 27e.)
29b. hidden single: r8c1 = 4
29c. -> r9c12 = {29/38} no 1
29d. hidden single: r7c1 = 1
29e. -> r8c4 = 3 (step 6); r2c4 = 5 (h8(2)c4)
29f. -> {56} locked for n7 and c3 in r89c3
29g. hidden single: r7c6 = 5
29h. -> r4c5 = 5
29i. -> 12(2)n35 = [84]

The rest is singles and cage sums - finally!!
(Archive Note) Typos and a few other things corrected.
Beginning solution outline by gary w without any t&e:
Being reminded of this one by Afmob's post I had a go myself and,after a tough battle..rather more than 5 hours....managed to crack it without any t&e needed.The beginning of my solution is outlined-



1. Outies on N69=18 r3c9=9 and r8c6=9 thus r2c6=2 r12c7={16} {34}->1/3
2. r56c9+r6c8=9
3. Fairly straightforward combo work -> N9 1 is in 12(3) cage and because of outies c9 1 is not at r9c8 so at r89c9
4. I-O N69 -> r7c9=r6c8+5 readily -> r7c9=6/7 and r6c8=1/2
5. r37c7=r5c8+3 thus r5c8<>1 because would -> r37c7={13} blocked by 9(3) cage N3 (see 1.)
6. If 1 is in r456c7->r12c7={34} -> r89c7={57} -> r7c9=6 ->r6c8=1 (from 2.) Contradiction.So in N6 r6c8=1.r7c9=6.r56c9=[35}.r4c89=67.
r12c7={16}
7. In c7 3 at r3/7 -> r37c7<>5/7 -> r5c8=5/7 not possible.So in N3 5/7 are in c8 and conflicting combos mean they are not in r1c8.So r23c8={57}.r3c7=3.r89c7={57}
8.N369 now well under way.
9.Outies N14=6={33}/{15} and r28c4=8.

Now did combo work on N12 and also worked on 1s to show that the 40(7)cage must contain a 1 (and so a 4 and doesn't contain a 2 or 3).



Good luck!

Regards

Gary
Walkthrough by Andrew:
Here is my walkthrough for Assassin 65 V3
Thanks Ed for your comments.

Prelims

a) R34C6 = {39/48/57}, no 1,2,6
b) R67C4 = {19/28/37/46}, no 5
c) 9(3) cage at R1C7 = {126/135/234}, no 7,8,9
d) 22(3) cage at R3C9 = {589/679}
e) 11(3) cage at R6C5 = {128/137/146/236/245}, no 9
f) 21(3) cage at R8C6 = {489/579/678}, no 1,2,3
g) 12(4) cage at R2C2 = {1236/1245}, no 7,8,9

1. 45 rule on N69 2(1+1) outies R3C9 + R8C6 = 18 -> R3C9 = 9, R8C6 = 9, clean-up: no 3 in R34C6, no 1 in R6C4
1a. R8C6 = 9 -> R89C7 = 12 = {48/57}, no 6
1b. 45 rule on N3 1 outie R2C6 = 2 -> R12C7 = 7 = {16/34}, no 5
1c. 45 rule on N9 3 remaining outies R5C9 + R6C89 = 9 = {126/135/234}, no 7,8,9
1d.45 rule on N9 1 innie R7C9 = 1 remaining outie R6C8 + 5 -> R7C9 = {678}, R6C8 = {123}
1e. 40(7) cage at R3C4 must contain 9 in R4C4 + R5C45, locked for N5, clean-up: no 1 in R7C4

2. 45 rule on C5 3 innies R159C5 = 19 = {289/379/469/478/568}, no 1

3. 45 rule on C123 2 outies R28C4 = 8 = {17/35}/[62], no 4,8,9, no 6 in R8C4

4. 45 rule on C1 3 outies R169C2 = 19 = {289/379/469/478/568}, no 1

5. 45 rule on R12 3 innies R2C258 = 17 = {179/359/368/458/467}
5a. 1 of {179} must be in R2C2 -> no 1 in R2C58

6. 45 rule on C9 3 outies R149C8 = 17 = {179/269/278/359/368/458/467}
6a. 9 of{179} must be in R9C8 -> no 1 in R9C8

7. 45 rule on N7 1 outie R8C4 = 1 innie R7C1 + 2 -> R8C4 = {357}, R7C1 = {135}, clean-up: no 6,7 in R2C4 (step 3)
7a. Max R7C1 = 5 -> min R6C12 = 12, no 1,2 in R6C12
7b. R67C4 = {28/46} (cannot be {37} which clashes with R28C4), no 3,7 in R67C4
7c. 17(3) cage at R1C3 = {179/359/368/458} (cannot be {269/278/467} because R2C4 only contains 1,3,5), no 2
7d. 1 of {179} must be in R2C4 -> no 1 in R12C3

8. 45 rule on C89 2 outies R37C7 = 1 innie R5C8 + 3
8a. Min R37C7 = 5 (cannot be {13} which clashes with R12C7) -> min R5C8 = 2

9. 45 rule on N9 4 innies R7C789 + R8C8 = 21 = {1389/1569/2379/2469/3468/3567} (cannot be {1479/1578/2478/2568/3459} which clash with R89C7)
9a. R7C789 + R8C8 = {2469/3468/3567} (cannot be {1389/1569/2379} because 16(4) cage at R6C4 cannot be 3{139}/1{159}/2{239}), no 1, 6 locked for N9
[Alternatively these combinations are blocked by R7C9 = R6C8 + 5, step 1d.]
9b. 1 in N9 only in R89C9, locked for C9
9c. 12(3) cage at R8C9 contains 1 = {129/138} (cannot be {147} which clashes with R89C7), no 4,5,7
9d. 9 of {129} must be in R9C8 -> no 2 in R9C8

10. 14(3) cage at R5C9 = {248/347/356} (cannot be {257} because R56C9 + R6C8 = 9 (step 1c) cannot be {25}2)
10a. 14(3) cage = {347/356} (cannot be {248} = {24}8 which clashes with 12(3) cage at R8C9 = {13}8, overlap clash), no 2,8, 3 locked for C9 and N6
[Ed wrote: I think its just a normal block since the 12(3) = 2/8, it’s just the 2 is only in C9.]
10b. 6 of {356} must be in R7C9 -> no 6 in R56C9
10c. 12(3) cage at R8C9 (step 9c) = {129/138}
10d. 3,9 only in R9C8 -> R9C8 = {39}

11. R149C8 (step 6) = {179/269/359/368} (cannot be {278/458/467} because R9C8 only contains 3,9), no 4
11a. 5,7 of {179/359} must be in R4C8 -> no 5,7 in R1C8

12. 14(3) cage at R1C8 = {158/248/347} (cannot be {167} which clashes with R7C9, cannot be {257} which clashes with 14(3) cage at R5C9, cannot be {356} which clashes with R12C7), no 6
12a. Killer pair 1,4 in R12C7 and 14(3) cage at R1C8, locked for N3
12b. Killer pair 4,5 in 14(3) cage at R1C8 and 14(3) cage at R5C9, locked for C9
12c. R3C9 = 9 -> R4C89 = [58]/{67}, no 8 in R4C8

13. 14(3) cage at R1C8 (step 12) = {158/248/347}, R4C89 = [58]/{67} (step 12c)
13a. Consider combinations for 14(3) cage at R5C9 (step 10a) = {347/356}
14(3) cage at R5C9 = {347}, 4 locked for C9 => 14(3) cage at R1C8 = {158}
or 14(3) cage at R5C9 = {356} = {35}6, 5 locked for N6 => R4C89 = [67] => 14(3) cage at R1C8 = {158/248}
-> 14(3) cage at R1C8 = {158/248}, no 3,7, 8 locked for N3

14. R47C9 = {67} (hidden pair in C9), clean-up: no 5 in R4C8 (step 12c)
14a. Naked pair {67} in R4C89, locked for R4 and N6, clean-up: no 5 in R3C6

15. 7 in N3 only in 15(3) cage at R2C8 = {267/357}
15a. 2 of {267} must be in R3C8 (R23C8 cannot be {67} which clashes with R4C8), no 2 in R3C7

16. 2 in C7 only in R4567C7, CPE no 2 in R6C8 -> R6C8 = 1, R7C9 = 6 (step 1d), R4C89 = [67], clean-up: no 4 in R6C4
16a. R7C9 = 6 -> R56C9 = 8 = {35}, locked for C9 and N6
16b. 1 in C7 only in R12C7 (step 1b) = {16}, 6 locked for C7
16c. Naked triple {248} in 14(3) cage at R1C8, 2 locked for R1 and N3

17. R37C7 = R5C8 + 3 (step 8)
17 a. 3 in C7 only in R37C7 -> whichever value is in R5C8 must be in one of R37C7, no 5,7 in R5C8 -> no 5,7 in R37C7 -> R3C7 = 3, R7C7 = {2489}
17b. R89C7 = {57} (hidden pair in C7), locked for N9
[Cracked. The rest is a bit easier.]

18. R2C258 (step 5) = {179/359/467} (cannot be {368} because R2C8 only contains 5,7, cannot be {458} which clashes with R2C9), no 8
18a. R2C8 = {57} -> no 5,7 in R2C25
18b. 9 of {359} must be in R2C5 -> no 3 in R2C5

19. 45 rule on R89 3 innies R8C258 = 16 = {178/268} (cannot be {457} which clashes with R8C7, cannot be {358/367} which clash with R8C47, ALS block), no 3,4,5, 8 locked for R8
19a. Naked pair {28} in R18C8, locked for C8
19b. Naked triple {128} in R8C89 + R9C9, locked for N9
19c. Killer pair 1,2 in R8C258 and R8C9, locked for R8

20. 4 in N9 only in R7C78, locked for R7, clean-up: no 6 in R6C4
20a. 4 in R8 only in R8C13, locked for N7
20b. 4 in N8 only in 12(3) cage at R9C4 = {147/246/345}, no 8
20c. Naked pair {28} in R67C4, locked for C4

21. 14(3) cage at R8C3 = {347/356} (cannot be {149/158/239/248} because 1,2,8,9 only in R9C3, cannot be {167} which clashes with R8C258, cannot be {257} which clashes with R8C7), no 1,2,8,9
21a. 14(3) cage = {347/356}, CPE no 3 in R8C1
21b. 3 in R8 only in R8C34, locked for 14(3) cage, no 3 in R9C3

22. 18(3) cage at R7C2 = {189/279/378} (cannot be {369} which clashes with R7C78, ALS block, cannot be {567} which clashes with R9C3), no 5,6

[The final breakthrough.]
23. 45 rule on C123 4 innies R1289C3 = 23 = {3479/3569/3578/4568}
23a. 7 of {3479} must be in R9C3 -> 17(3) cage at R1C3 (step 7c) = {359/368/458} (cannot be {179} because R12C3 cannot contain both of 7,9), no 1 in R2C4, clean-up: no 7 in R8C4 (step 3)
23b. Naked pair {35} in R28C4, locked for C4
23c. 12(3) cage at R9C4 (step 20b) = {147/246} (cannot be {345} which clashes with R8C4), no 3,5

24. 40(7) cage at R3C4 must contain 5,8 in R5C56 + R67C6, CPE no 5,8 in R4C6 -> R4C6 = 4 -> R3C6 = 8
24a. 40(7) cage must contain 8 -> R5C5 = 8, R67C4 = [28]
24b. 40(7) cage at R3C4 = {1456789} (only remaining combination), no 3 -> R3C4 = 4

25. 3 in N5 only in R46C5, locked for C5
25a. 11(3) cage at R6C3 = {137/236} -> R6C5 = 3, R78C5 = {17}/[26], no 5, no 2 in R8C5
25b. R56C9 = [35]
25c. R8C4 = 3 (hidden single in N8) -> R2C4 = 5, R23C8 = [75]
25d. R7C6 = 5 (hidden single in N8) -> R4C5 = 5 (hidden single in N5)

26. 45 rule on N14 1 remaining outie R7C1 = 1 -> R6C12 = 16 = {79}, locked for R6 and N4 -> R6C6 = 6

27. R2C4 = 5 -> R12C3 = 12 = {39} (cannot be {48} which clashes with R6C3), locked for C3 and N1

28. 12(4) cage at R2C2 = {1236} (only remaining combination) -> R4C2 = 3

29. 18(3) cage at R7C2 (step 22) = {279} (only remaining combination) -> R7C2 = 9, R6C12 = [97], R8C2 = 2, R7C3 = 7, R7C5 = 2 -> R8C6 = 6 (cage sum)

and the rest is naked singles.

Rating 1.5. I used some combination analysis and one short forcing chain.


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PostPosted: Sun Jul 06, 2008 6:37 am 
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Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Assassin 66 by Ruud (Aug 07)
Puzzle pic:
Image
Code: Select, Copy & Paste into solver:
3x3::k:5120:5120:5120:2307:2307:4357:2310:2310:2310:3337:3337:4619:2307:4357:4357:3855:5648:5648:7186:3337:4619:4619:3350:3855:3855:5648:6682:7186:3356:3356:3356:3350:4640:4640:4640:6682:7186:1317:1317:4647:4647:4647:1834:1834:6682:7186:5678:5678:5678:817:2610:2610:2610:6682:7186:4151:4408:4408:817:3899:3899:4413:6682:4151:4151:4408:3906:3906:4676:3899:4413:4413:2632:2632:2632:3906:4676:4676:2894:2894:2894:
Solution:
+-------+-------+-------+
| 4 7 9 | 2 6 8 | 3 5 1 |
| 2 3 6 | 1 5 4 | 8 7 9 |
| 1 8 5 | 7 9 3 | 4 6 2 |
+-------+-------+-------+
| 3 5 2 | 6 4 1 | 9 8 7 |
| 9 4 1 | 3 8 7 | 5 2 6 |
| 8 6 7 | 9 2 5 | 1 4 3 |
+-------+-------+-------+
| 7 9 4 | 5 1 6 | 2 3 8 |
| 6 1 8 | 4 3 2 | 7 9 5 |
| 5 2 3 | 8 7 9 | 6 1 4 |
+-------+-------+-------+
Quote:
Howard S: One of the easier ones - I am torn between 0.75 and 1.0
Andrew: I think I'll rate this as 1.0. Step 32 isn't a difficult one but I found it was hard to spot
CathyW: 0.5 because I only used Innies, Outies, Outies-Innies, cage combinations, killer combos and combination analysis and because it only took me just over an hour including writing down my steps as I did them
azpaull: this was the first Assassin in several weeks that I have been able to do on my own....I'll give it a 0.75....let me thank you all for your walk-throughs and your commentary
gary w: both a66 and a67 took me about 1.5-2 hrs to do.A "deadly" killer in The Times normally takes me about 0.5 hrs so i'ld rate both of these as significantly harder
Walkthrough by Andrew:
mhparker wrote:
That was quick! BTW, if it's just a matter of checking for typos, I'm sure no-one would have minded if you would have posted your WT "as is", especially if accompanied by an appropriate remark to that effect.
If it had been just a case of possible typos I would have done that. In the later stages of solving I had spotted several hidden singles and wondered how long they had been there so I wanted to check. As it happened they hadn't been there long but while checking I found one or two others.

I also had to decide whether to move my key step 32 to the earliest position that I might have been able to use it but have decided not to, see comment in the walkthrough.
CathyW wrote:
Super quick - though you do have a time advantage!
Yes I do have a time advantage but only if I start a puzzle on the day that it is posted and manage to solve it the same day. Otherwise people in Europe, for example, have a time advantage over me.

I think I'll rate this as 1.0. Step 32 isn't a difficult one but I found it was hard to spot; I'd got to a stage where I was looking around for the next step and a relationship that I hadn't yet used; there's also step 31 which was easier to see.

I'll be interested to see Cathy's walkthrough and see why she thinks the rating is as low as 0.5.


Here is my walkthrough

1. R34C5 = {49/58/67}, no 1,2,3

2. R5C23 = {14/23}

3. R5C78 = {16/25} (cannot be {34} which clashes with R5C23)
3a. Killer pair 1,2 in R5C23 and R5C78, locked for R5

4. R67C5 = {12}, locked for C5

5. R1C123 = {389/479/569/578}, no 1,2

6. 9(3) cage in N2 = {126/135/234}, no 7,8,9
6a. 1,2 only in R12C4 -> no 6 in R12C4

7. R1C789 = {126/135/234}, no 7,8,9

8. 22(3) cage in N3 = 9{58/67}, 9 locked for N3

9. R6C234 = 9{58/67}, 9 locked for R6

10. R6C678 = {136/145/235} (cannot be {127} which clashes with R6C5), no 7,8,9
10a. Killer pair 5,6 in R6C234 and R6C678, locked for R6

11. R9C123 = {127/136/145/235}, no 8,9

12. R9C789 = {128/137/146/236/245}, no 9

13. 45 rule on R9 3 innies R9C456 = 24 = {789}, locked for R9 and N8
13a. 45 rule on R9 1 innie R9C4 – 6 = 1 outie R8C6 -> R8C6 = {123}
13b. R9C123 (step 11) = {136/145/235}
13c. R9C789 (step 12) = {146/236/245}

14. 45 rule on R1 1 innie R1C6 – 7 = 1 outie R2C4 -> R1C6 = {89}, R2C4 = {12}
14a. 17(3) cage in N2 = {179/269/278/359/368/458}
14b. R1C6 = {89} -> no 8,9 in R2C56
14c. 7 on {179/278} must be in R2C5 -> no 7 in R2C6
14d. R1C123 (step 5) = {479/569/578} (cannot be {389} which clashes with R1C6), no 3

15. 45 rule on N1 1 outie R3C4 – 6 = 1 innie R3C1 -> R3C1 = {123}, R3C4 = {789}

16. 45 rule on N2 3 innies R3C456 = 19 = {379/469/478/568} (cannot be {289} which clashes with 17(3) cage), no 1,2

17. 45 rule on N3 1 outie R3C6 – 1 = 1 innie R3C9 -> no 1 in R3C9

18. 45 rule on N7 3 innies R7C13 + R8C3 = 19 = {289/379/469/478} (cannot be {568} which clashes with R9C789), no 1,5
18a. 45 rule on N7 1 innie R7C1 – 2 = 1 outie R7C4 -> no 2,9 in R7C1, no 3 in R7C4

19. 45 rule on N8 3 innies R7C456 = 12 = {156/246} (cannot be {345} because R7C5 only contains 1,2) = 6{15/24}, no 3, 6 locked for R7 and N8, clean-up: no 4 in R7C4 (step 18a)
19a. R7C5 = {12} -> no 1,2 in R7C46, clean-up: no 3,4 in R7C1 (step 18a)
19b. R8C456 = 3{15/24}, 3 locked for R8

20. 45 rule on N9 1 innie R7C9 – 2 = 1 outie R7C6 -> R7C9 = {78}, no 4 in R7C6

21. Naked pair {56} in R7C46, locked for R7 and N8 -> R7C5 = 1 (step 19), R6C5 = 2
21a. 4 in N8 locked in R8C45, locked for R8
21b. 15(3) cage = 4{29/38), no 7 in R9C4
21c. 18(3) cage = 7{29/38}
[With hindsight it would have been useful to spot step 32 now. It might have made the solution a bit shorter. However since I don’t consider it to have been an “I ought to have seen it now” move, I haven’t moved it to step 22.]

22. Naked pair {78} in R7C19, locked for R7

23. 45 rule on R6 2 remaining innies R6C19 = 11 = {38/47}, no 1

24. 45 rule on R5 2 innies R5C19 = 15 = {69/78}
24a. R5C456 = {369/378/459} (cannot be {468/567} which clash with R5C19)

25. 45 rule on R123 3 innies R3C159 = 12 = {129/138/147/156/237/246/345}
25a. 8 of {138} must be in R3C5 -> no 8 in R3C9, clean-up: no 9 in R3C6 (step 17)

26. 45 rule on N3 3 innies R2C7 + R3C79 = 14 = {158/167/248/347} (cannot be {257/356} which clash with 22(3) cage)
26a. 5 of {158} must be in R3C9 -> no 5 in R23C7

27. 17(3) cage at R7C3 = {269/359/368/458/467} (cannot be {278} because R7C4 only contains 5,6)
27a. R7C4 = {56} -> no 6 in R8C3

28. 15(3) cage at R7C6 = {159/258/267/357/456} (cannot be {249/348} because R7C6 only contains 5,6, cannot be {168} because 1,8 only in R8C7)
28a. 2 of {258/267} must be in R7C7 -> no 2 in R8C7
28b. 9 of {159} must be in R7C7 -> no 9 in R8C7

29. 45 rule on N4 4 outies R37C1 + R46C4 = 23, max R37C1 = 11 -> min R46C4 = 12 -> no 1 in R4C4
29a. 1 in N5 locked in R46C6, locked for C6
29b. 1 in N2 locked in 9(3) cage = 1{26/35}, no 4
29c. 17(3) cage in N2 (step 14a) = {269/278/359/458} (cannot be {368} which clashes with 9(3) cage
29d. 2 of {269} must be in R2C6 -> no 6 in R2C6

30. 45 rule on N6 4 outies R46C6 + R37C9 = 16
30a. Min R46C6 = 4 -> max R37C9 = 12, min R7C9 = 7 -> max R3C9 = 5, clean-up: no 7,8 in R3C6 (step 17)
30b. Min R37C9 = 9 -> max R46C6 = 7, no 7,8,9

31. Hidden killer triple 7,8,9 in R159C6 -> R5C6 = {789}
31a. R5C456 (step 24a) = {369/378/459}
31b. 9 on {369/459}must be in R5C6 -> no 9 in R5C45

32. 45 rule on C123 4 outies R3467C4 = 27 = {5679} (cannot be {3789/4689} which clash with R9C4), locked for C4 -> R9C4 = 8
32a. R9C56 = {79} -> R8C6 = 2

33. Naked pair {34} in R58C4, locked for C4
33a. R12C4 = {12} -> R1C5 = 6

34. 45 rule on C789 4 outies R3467C6 = 15 = {1356} (only remaining combination), locked for C6 -> R2C6 = 4

35. 17(3) cage in N2 = [854] (only remaining permutation), R3C6 = 3, R3C9 = 2 (step 17), R3C1= 1, R3C4 = 7 (step 15), R3C5 = 9, R9C56 = [79], R5C6 = 7, R4C5 = 4, R5C45 = [38], R8C45 = [43], clean-up: no 2 in R5C23
[I should also have included clean-up: no 8 in R3C8 (step 8) but this got fixed in step 40.]
35a. R23C3 = 11 = [38/65]

36. Naked pair {14} in R5C23, locked for R5 and N4, clean-up: no 6 in R5C78

37. Naked pair {25} in R5C78, locked for N6

38. R1C789 (step 7) = {135} (only remaining permutation), locked for R1 and N3 -> R12C4 = [21]
[R1C4 had been a hidden single since step 35; missed that one!]

39. Naked triple {479} in R1C123, locked for N1

40. R3C7 = 4 (hidden single in R3), R2C7 = 8, R3C8 = 6

41. R34567C9 = {23489/23678} (cannot be {12689} because 1 would have to be in R4C9 when the only remaining 6,9 are both in R5C9) = 238{49/67}, no 1, 3,8 locked for C9
41a. 9 of {23489} must be in R5C9 -> no 9 in R4C9
41b. 6 of {23678} must be in R5C9 -> no 6 in R4C9

42. Killer pair 7,9 in R2C9 and R34567C9, locked for C9

43. R4C234 = {256} (only remaining combination, cannot be {238} because R4C4 only contains 5,6,9), locked for R4 -> R4C6 = 1, R6C4 = 9 (hidden single in C4)
43a. R4C78 = 17 = [98] (only remaining permutation), R5C9 = 6, R5C1 = 9, R467C9 = {378} (step 41) -> R7C9 = 8, R46C9 = {37}, locked for C9 and N6 -> R2C89 = [79], R6C78 = [14] -> R6C6 = 5, R7C46 [56], R4C4 = 6, R7C1 = 7, R46C1 = [38], R46C9 = [73], R1C1 = 4

44. R9C9 = 4 (hidden single in C9)
44a. R9C78 (step 13c) = {25}/[61], no 3
44b. R9C123 (step 13b) = 3{16/25}, 3 locked for N7

45. Killer pair 1,5 in R8C9 and R9C78, locked for N9 -> R8C78 = [79], R8C3 = 8
[R8C7 had been a hidden single for at least a couple of steps.]

46. R9C7 = 6 (hidden single in C7), R9C8 = 1

47. Naked triple {235} in R9C123, locked for N7

and the rest is naked singles and cage sums
Walkthrough by CathyW:
And here's my WT in 22 steps (though it's not to the point of all singles)! 0.5 because I only used Innies, Outies, Outies-Innies, cage combinations, killer combos and combination analysis and because it only took me just over an hour including writing down my steps as I did them, though it's taken me as long again to type it up and check through!

Prelims

a) 20(3) N1 – no 1 or 2
b) 9(3) N2 & N3 = {126/135/234}
c) 22(3) N3 = {589/679} 9 n/e N3
d) 13(2) r34c5 = {49/58/67}
e) 5(2) r5c23 = {14/23}
f) 7(2) r5c78 = {16/25} ({34} blocked by 5(2)) -> 5(2) and 7(2) form KP 1,2 n/e r5
g) 22(3) r6c234 = {589/679} 9 n/e r6
h) 3(2) r67c5 = {12} n/e c5 -> 10(3) r6c678 = {136/145/235} ({127} blocked by r6c5}
-> 10(3) and r6c5 form KP 1,2 -> r6c19 <> 1,2


1. Innies r5: r5c19 = 15 = {69/78}
-> 18(3) r5c456 = {369/378/459} ({468/567} blocked by split 15(2))

2. Innies r9: r9c456 = 24 = {789} n/e r9/N8
-> r8c6 = (123)
-> 10(3) r9c123 = {136/145/235}, 11(3) r9c789 = {146/236/245}

3. O-I N1: r3c4 – r3c1 = 6 -> r3c4 = (789), r3c1 = (123)

4. O-I N3: r3c6 – r3c9 = 1 -> r3c6 = (2…9), r3c9 = (1…8)

5. Innies N2: r3c456 = 19. Min from r3c45 = 7+4 = 11 -> r3c6 max 8 (i.e. <> 9) -> r3c9 <> 8

6. Innies r4: r4c159 = 14

7. Innies r6: r6c159 = 13

8. O-I N7: r7c1 – r7c4 = 2 -> r7c1 = (3…8), r7c4 = (1…6)

9. O-I N9: r7c9 – r7c6 = 2 -> r7c9 = (3…8), r7c6 = (1…6)

10. Innies r123: r3c159 = 12 = {129/138/147/156/237/246/345}

11. Innies r789: r7c159 = 16 = {178/268} 8 locked to r7c19 n/e r7
-> r7c19 <> 3,4,5
clean up: r7c46 <> 1,2,3 (from steps 8 and 9)

12. Innies N8: r7c456 = 12 = {156/246} -> 6 locked to r7c46 n/e r7/N8
-> r7c19 = {78} -> r7c5 = 1, r6c5 = 2
-> r7c46 = {56} -> r7c2378 = {2349}
-> r8c456 = {234} -> r8c123789 = {156789}

13. 10(3) r6c678 = {136/145} Killer combo with 22(3) -> r6c19 <> 5,6 -> r6c19 = {38/47}
-> r5c19 <> {78} since r7c19 = {78} (would block both options for r6c19)
-> r5c19 = {69} -> 7(2) = {25} n/e N6, 5(2) = {14} n/e N4 -> r6c9 <> 7, 18(3) = {378} n/e N5 -> r3c5 <> 5,6
Options for 10(3): If {145}, r6c6 = 5 -> r6c6 <> 4

14. 17(3) r7c34+r8c3: Max from r7c34 = 9+6 = 15 -> r8c3 <> 1
Min from r7c4+r8c3 = 5+6 = 11 -> r7c3 <> 9
-> 17(3) = {269/359/368/458/467}
Combo analysis: r8c3 <> 5,6

15. 18(3) N8 = {279/378} -> r9c4 <> 7 -> 15(3) N8 = {834/924}

16. 15(3) r7c67 + r8c7 = [591/528/537/546/645/627] -> r8c7 <> 9

17. Outies c123: r3467c4 = 27 = {4689/5679}
Combo analysis: r3c4 <> 9 -> r3c1 <> 3
-> split 27(4) = [8496]/7{569}
How did I manage to miss the conflict with r9c4 here?! Thanks to Andrew for pointing it out.

18. 28(5) r34567c1 = {13789/14689/23689/24589/24679/25678} ({15679} not possible)
Combo analysis: r4c1 <> 2,6,7,8

19. Split 19(3) r3c456 = [793/784/892/874/847]
-> r3c6 <> 5,6,8 -> r3c9 <> 4,5,7

20. Split 14(3) r4c159 = [347/356/563/941] -> r4c9 = (1367)

21. 26(5) r34567c9 = {13679/14678/23489/23678} ({12689} not possible)
Combo analysis: r3c9 <> 3 -> r3c6 <> 4
-> 9(3) r1c789 = {135/234} ({126} blocked by r3c9)
-> 3 locked to r1c789 n/e r1/N3 -> 20(3) r1c123 = {479/469/578}

22. 15(3) r2c7 + r3c67 = {258/267/348} since {456} no longer possible and {357} would be blocked by combo with 22(3). If {267}, r3c6 = 2 (r23c7 can't be {26} blocked by combo with r3c9 and 9(3)) -> r3c6 <> 7 -> r3c9 <> 6.
-> NP {12} r3c19
-> r3c6 = 3 -> r23c7 = {48} n/e N3/c7
-> r3c9 = 2, r3c1 = 1
-> r3c4 = 7, r3c5 = 9, r4c5 = 4

Fairly straightforward from here with cage combos and singles


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