Assassin 44

[Para]

Hi

Ever heard of quadruple innies and outies? If you want to practice this technique, use this week's Assassin.

I missed the use of quadruple innies and outies in this puzzle. This puzzle was fairly easy. Just the end was interesting. It needed some thought. First time i implemented a xyz-wing in a Killer Sudoku. But i can be by-passed. Just looks fun.

1. R12C5 = {16/25/34}: no 7,8,9
2. R1C678 = {389/479/569/578}: no 1,2
3. R2C123 = {489/579/678}: no 1,2,3
4. R23C4 = {17/26/35}: no 4,8,9
5. 12(4) in R3C1 = {1236/1245}: no 7,8,9; 1,2 locked in 12(4) -->> R1C2: no 1,2
6. R3C56 = {39/48/57}: no 1,2,6
7. R7C45 = {49/58/67}: no 1,2,3
8. R78C6 = {15/24}: no 3,6,7,8,9
9. R89C5 = {59/68}: no 1,2,3,4,7
10. R9C234 = {289/379/469/478/568}: no 1
11. 45 on R1: 2 innies: R1C59 = 7 = {16/25/34}: no 7,8,9
12. 45 on N1: 2 outies: R1C4 + R4C2 = 6 = {15/24/33}: no 6,7,8,9
13. 45 on R9: 2 innies: R9C15 = 6 = [15]
13a. R8C5 = 9
13b. R7C45 = {67}: locked for R7 and N8
13c. R78C6 = {24}: locked for C6 and N8
13d. Naked Pair: R9C46 = {38}; locked for R9 and N8
13e. R8C4 = 1
13f. Clean up: R12C5: no 2; R3C5: no 8; R3C6: no 3, 7; R23C4: no 7
14. 45 on C123 : 2 outies : R19C4 = 10 = [28]
14a. R9C6 = 3; R4C2 = 4 (step 12)
14b. 12(4) in R3C1 = {125}4: R3C123 = {125} -->> locked for R3 and N1
14c. 8(2) in R23C4 = [53]; 12(2) in R3C56 = [48]
14d. R12C5 = {16} -->> locked for C5 and N2
14e. R7C45 = [67]
14f. Naked Triple {679} in R3C789: locked for N3
14g. Naked pair {79} in R12C6-->> locked for C6
15. 45 on N9: 1 outie: R6C8 = 4
16. 20(3) in R1C6 = {389/578}: no {479} can’t have both {79} -->> R1C7: no 4; 8 locked in R1C78 for R1 and N3
17. 18(4) in R1C1 = {349/367}2 -->> 3 locked in 18(4) for R1
17a. 20(3) = [7]{58}; R2C6 = 9
17b. R1C78 = {58}: locked for N3
17c. 18(4) in R1C1 = {349}2 : R1C123 = {349} locked for R1 and N1
17d. R1C9 = 1; R12C5 = [61]
18. 17(4) in R1C9 = 1{349}: no 1{367} because it needs 2 of {234} in R2C89
18a. R2C89 = [34]; R3C8 = 9; R2C7 = 2
19. 24(4) in R2C6 = 92{67}: needs one of {67} in R3C7
19a. Naked Pair {67}in R34C7: locked for C7
20. R5C4 = 4(hidden, that must have been there for ages)
21. 19(3) in R9C2 = {{29}/[74]}8: no 6
21a. 6 locked in N7 for R8
21b. 13(3) in R8C7 = {238/247}-->> 2 locked for R8 and N9; 13(3): no 5
21c. R78C6 = [24]
21d. 13(3) in R8C7 = {238} -->> locked for R8 and N9
21e. 20(4) in R9C6 = 3[4]{67}; locked for R9; (R9C7 = 4)
21f. Naked Triple {159} in R7C789 -->> locked for R7
22. 22(4) in R7C2 = [8]{67}1 -->> locked for N7 (R7C2 = 8)
22a. R8C3 = 5
22b. Naked Pair {67} in R28C2 -->> locked for C2
22c. Naked Pair {67} in R39C9 -->> locked for C9
23. 15(4) in R6C3 = [63]51 (R67C3 = [63])
23a. R7C1 = 4; R1C3 = 4(hidden)
23b. 16(4) in R4C1 = {237}4 -->> R456C1 = {237} locked for C1 and N4
23c. R1C12 = [93]; R3C1 = 5; R8C12 = [67]; R2C123 = [867]
24. 13(3) in R5C4 = 4[36/81]: no 2,5
25. 21(4) in R3C9 = {2379/2568}: no {3567} because only room for 1 of {67} -->> 2 locked in R456C9 for C9 and N6
25a. R8C8 = 2(hidden)
26. 16(3) in R6C4 = [781/925]: {358} clashes with R5C5 -->> no 3
26a.16(3) in R4C4 = [781/925]: {358} clashes with R5C5, {169} clashes with R5C6, {268} doesn’t have 7 or 9 in R4C4, {367} clashes with R4C7 -->> no 3,6
26b. R5C5 = 3, R5C6 = 6(both hidden in N5)
27. 20(4) in R4C8 = {1379/1568}
27a. Only place for 6 is R4C8: no 5, 8
27b. Only place for 3 is R6C7: no 9
28. 16(3) in R4C4 = [925]: [781] clashes with R4C78
28a. R6C456 = [781]
28b. Naked Triple {358} in R168C7 -->> locked for C7
29. XYZ-wing in R5C27 + R6C2 -->> R5C3: no 9
29a. Naked Pair {18} in R45C3 -->> locked for C3 and N4
29b. R3C23 = [12]; R9C23 = [29]
30. R4C79 needs at least one of {68} because of R4C1
30a. 20(4) in R4C8 = {1379}: {1568} clashes with step 30 -->> locked for N6
30b. R4C7 = 6; R4C9 = 8; R3C79 = [76]; R45C3 = [18]; R4C8 = 7; R456C1 = [372]
30c. R56C9 = [25]; R56C2 = [59]; R5C78 = [91]; R6C7 = 3; R7C789 = [159]
30d. R8C79 = [83]; R9C89 = [67]; R1C78 = [58]

And we are done.

greetings

Para

[sudokuEd]

Thanks for the hint Para. Was totally stuck at this spot - but now realize how to unlock it. Very clever. Now just have to work out what's next! So nice to have a Killer that's hard at the end. Thanks Ruud.

Marks pic

Code: Select all

.-----------------------.-----.-----------------.-----.
| 9     3     4     2   | 6   | 7     58    58  | 1   |
:-----------------.-----:     :-----------.-----'     |
| 8     6     7   | 5   | 1   | 9     2   | 3     4   |
:-----------------:     :-----'-----.     |     .-----:
| 5     12    12  | 3   | 4     8   | 67  | 9   | 67  |
:-----.     .-----+-----'-----------:     :-----:     |
| 37  | 4   | 18  | 9     2     5   | 67  | 167 | 38  |
|     :-----'     :-----------------+-----'     |     |
| 27  | 159   189 | 4     3     6   | 19    178 | 259 |
|     |     .-----+-----------------:     .-----:     |
| 23  | 59  | 6   | 7     8     1   | 35  | 4   | 259 |
|     :-----:     :-----------.-----+-----'     '-----:
| 4   | 8   | 3   | 6     7   | 2   | 19    15    59  |
:-----'     |     '-----.-----:     :-----------------:
| 6     7   | 5     1   | 9   | 4   | 38    2     38  |
|     .-----'-----------:     :-----'-----------------:
| 1   | 29    29    8   | 5   | 3     4     67    67  |
'-----'-----------------'-----'-----------------------'

[Andrew]

Nice walkthrough Para. Your steps 30 and 30a were really neat!

Ever heard of quadruple innies and outies? If you want to practice this technique, use this week's Assassin.

The only place I could see to use quadruple innies and outies was in N4 where there are 4 innies after R12C6 have been fixed as a pair. However at that stage it was simpler to use 45 rule on N36 to fix R6C8. Para came from the other side and used 45 rule on N9 to fix F6C8.

Was there somewhere else to use quadruple innies and outies?

BTW at Para's step 29 there are actually two different XYZ-wings! XYZ-wing in R45C3 + R5C7: no 1 in R5C2 gives the same effect after the eliminations in N4, N1 and N7.

As Para said in his message, the XYZ-wing is fun but can be bypassed. That's true. One can go directly from step 28 to step 30.

[Andrew]

Here is my walkthrough. It followed a fairly similar path to Para's one but I think it is different enough to be posted. I've added some comments after working through Para's walkthrough.

Thanks Para for your comments on my walkthrough which have been added in red. There is also a minor correction to step 27 which I found while working through the comments.

Clean-up is used in various steps, using the combinations in steps 1 to 6 for further eliminations from these two cell cages. In some of the later steps, clean-up is followed by further moves and sometimes more clean-up.

1. R12C5 = {16/25/34}, no 7,8,9

2. R23C4 = {17/26/35}, no 4,8,9

3. R3C56 = {39/48/57}, no 1,2,6

4. R78C6 = {15/24}

5. R89C5 = {59/68}

6. R7C45 = {49/67} (cannot be {58} which would clash with R89C5)

7. Killer pair 6/9 in R7C45 and R89C5 for N8

8. R1C678 = {389/479/569/578}, no 1,2

9. R2C123 = {489/579/678}, no 1,2,3

10. R9C234 = {289/379/469/478/568}, no 1

11. 12(4) cage in N14 = 12{36/45}, no 7,8,9
[I missed no 1,2 in R1C2 due to R4C2 “pointing” into N1]

12. 45 rule on N1 2 outies R1C4 + R4C2 = 6 = {15/24/33} (double possible), no 6,7,8,9, R1C4 = {12345}, R4C2 = {12345}

13. 45 rule on N9 2 outies R6C8 + R9C6 = 7 = {25/34}/[61], no 7,8,9, no 1 in R6C8

14. 45 rule on R1 2 innies R1C59 = 7 = {16/25/34}, no 7,8,9

15. 45 rule on R9 2 innies R9C15 = 6 -> R9C1 = 1, R9C5 = 5, R8C9 = 9, clean-up: no 2 in R12C5, no 2,5 in R1C9, no 3,7 in R3C6, no 4 in R7C45, no 1 in R78C6
15a. R7C45 = {67}, locked for R7 and N8
15b. R78C6 = {24}, locked for C6 and N8, clean-up: no 8 in R3C5
15c. R9C6 = 3, R9C4 = 8, R8C4 = 1 (naked singles), clean-up: no 7 in R23C4, no 5 in R4C2
15d. 8 in C5 locked in R456C5, locked for N5
[Clean-up should also have included no 2,6 in R6C8 (step 13) in step 15 and then R6C8 = 4 (step 13) in step 15c. R6C8 was fixed in step 31.]

16. 45 rule on C123 1 remaining outie R1C4 = 2, clean-up: no 6 in R23C4, R4C2 = 4 (step 12)
16a. R23C4 = {35}, locked for C4 and N2, clean-up: no 4 in R12C5, no 3,4 in R1C9, no 7,9 in R3C56
16b. R3C56 = [48] (naked singles)
16c. R12C5 = {16}, locked for C5 and N2
16d. R7C45 = [67] (naked singles)
16e. R12C6 = {79}, locked for C6
16f. R1C59 = {16}, locked for R1

17. 12(4) cage in N14 = 12{36/45} (step 11), R4C2 = 4 -> R3C123 = {125}, locked for R3 and N1

18. R23C4 = [53] (naked singles)

19. R2C123 = {489/678} = 8{49/67}, 8 locked for R2 and N1

20. R3C789 = {679}, locked for N3

21. R1C9 = 1, R12C5 = [61] (naked singles)

22. R2C789 = {234}, locked for R2 and N3

23. R1C78 = {58} -> R1C6 = 7, R2C6 = 9

24. 17(4) cage in N3 = {1349} (cannot be {1367} because 6,7 in same cell) -> R3C8 = 9, R2C89 = {34}, locked for R2 -> R2C7 = 2

25. 24(4) cage in N236, R2C6 = 9, R2C7 = 2 -> R34C7 = 13 = {67}, locked for C7

26. R9C234 = {29}8/[748] [7/9], no 6, no 7 in R9C3

27. 6 in R9 locked in R9C89 corrected from R9C789, locked for N9
27a. 20(4) cage in R9C6789 = 36{29/47}, no 4,9 in R9C89

28. 22(4) cage in N7 = 1{489/678} (cannot be {1579} which would clash with R9C23) = 18{49/67} [7/9], no 2,3,5, 8 locked for N7
[Should be 22(4) cage in N7 = {1678} (cannot be {1489/1579} which would clash with R9C23. Very strange that I saw one clash but not the other one!]

29. Killer pair 7/9 in 22(4) cage and R9C23 for N7

30. 45 rule on N7 1 outie R6C3 – 2 = 1 innie R7C1, R7C1 = {345} -> R6C3 = {567}

31. 45 rule on N36 1 innie R6C8 = 4
31a. R2C89 = [34]

32. R5C4 = 4 (hidden single in C4)
32a. R5C56 = [36/81], no 2,5

33. R8C789 = {238/247} = 2{38/47}, 2 locked for R8 and N9, no 5

34. R78C6 = [24] (naked singles)

35. R9C89 = {67}, locked for R9 and N9 -> R9C7 = 4
35a. R39C9 = {67}, locked for C9

36. R9C23 = {29}, locked for N7 -> R7C2 = 8, R8C12 = {67}, locked for N7

37. R8C789 = {238}, locked for R8 and N9 -> R8C3 = 5

38. 15(4) cage in N47 R8C34 = [51] -> R67C3 = 9 = [63] (only remaining combination) -> R7C1 = 4
[Alternatively step 30 could have been used to fix R67C3 and R7C1 but the cage combination is the more obvious way.]

39. R1C3 = 4 (hidden single in R1)

40. R6C456 = [781/925], no 3 in R6C5
40a. R4C456 = [736/781/925]
[See comment after 46a]

41. R4567C1 = {2347} (only remaining combination) -> R456C1 = {237}, locked for C1 and N4

42. R1C12 = [93] , R8C12 = [67], R2C123 = [867], R3C1 = 5 (naked singles)

43. 21(4) cage in N36 = {2379/2568} (cannot be {3567} because 6,7 only in R3C9) = 2{379/568}, 2 locked for C9 and N6

44. R8C8 = 2 (hidden single in N9)

45. 20(4) cage in N6 = {1379/1568} = 1{379/568}

46. If R5C456 = [481] => R6C456 = [925] => R4C456 = [736] clashes with R4C7 = 67} -> R5C456 cannot be [481]
46a. R5C456 = [436]
[At step 40a I missed the fact that [736] clashes with R4C7. If I’d eliminated [736] there it would have fixed R5C456 and I wouldn’t have needed the chain in step 46.]

47. 20(4) cage in N6, 3 only in R6C7 -> no 9 in R6C7, 6 only in R4C8 -> no 5,8 in R4C8

48. If R4C456 = [781] => R4C7 = 6 clashes with R4C8 -> R4C456 cannot be [781]
48a. R4C456 = [925], R6C456 = [781]

49. 20(4) cage in N6, R6C7 = {35} -> no 5 in R5C78

50. R48C9 = {38}, locked for C9

51. Naked triple {358} in R168C7, locked for C7

52. If 20(4) cage in N6 = {1568}, R6C7 => 5, R4C8 => 6, R4C7 => 7, R4C1 => 3, R4C9 => 8 => no 3 in N6 (This can alternatively be seen as two 8s in N6) -> 20(4) cage in N6 cannot be {1568}
[Para’s step was more elegant. The reason I did it this way is that I was looking for interactions between R456C1 and the 20(4) cage to make further eliminations in the 20(4) cage.]

53. 20(4) cage in N6 = {1379}, locked for N6 -> R6C7 = 3, R5C7 = 9, R456C1 = [372], R456C9 = [825], R45C8 = [71]

and the rest is naked singles, naked pairs and simple elimination

[Para]

Another way to finish the puzzle. Almost the same. Think it looks more Killer-like.

a. 21(4) in R3C9 = [73]{29}/[68]{25} -->> R34C9 = [73/68]
b. 45-test on N3: innie = outie: R3C9 = R4C7 -->> R4C79 = [73/68]
c. R4C79 -->> no [73]: clashes with R4C1
d. R4C79 = [68]
And that leads to the same finish.

greetings

Para

[Para]

Hi all

Here is a V2 for this weeks assassin. It grinds to a halt a bit earlier. But after struggling through the middle, you can finish it in the same way as the original puzzle.

Assassin 44V2



PS:
3x3::k:4608:4608:4608:4608:4100:5125:5125:5125:4360:5385:5385:53854100:41004360:43603090206030944360:5402:41235917:4126:4126:41265154:5402:4123:5917:591733675154:5154:5402:4123:59174144:4144:4144:5154:4916:5402:4123:568733851595:4916:4916:4916:5687:56873906159533975687:4937:4937:49375197:5197:5197:5197:

greetings

Para

[sudokuEd]

It grinds to a halt a bit earlier. But after struggling through the middle

Sure does grind. Masterful V2 Para. Really enjoyed it. Way, way harder than the original. Takes a Richard-style innie move to finally pick the lock. Did I miss any shortcuts?

Assassin 44V2 Walkthrough - Please let me know if anything can be improved. [Thanks Andrew and Para for a couple of corrections. Also, Para has shown a huge shortcut from a uniqueness move at step 40. It busts the puzzle!]

1."45" n9: r6c8 + r9c6 = 7 (no 789)

2. "45" r9: r9c15 = 6 = h6(2) = {15/24}

3. "45" n8: r9c46 = 11 = h11(2) = [92/83/74/{56}]
3a. r9c4 = {5..9}
3b. r9c6 = {2..6}
3c. no 6 r6c8 (step 1)

4. 6(2)n8 = {15/24}

5. 13(2)n8 = {49/58/67}

6. 15(3)n8 = {159/249/267/348} = [2/5/8..]. Other combo's blocked: Here's how.
6a. {168} blocked by 13(2) & 6(2) (2 4s n8)
6b. {258} blocked by 6(2)n8
6c. {357} blocked by 13(2) & 6(2) (2 4s n8)
6d. {456} by 6(2)n8

7. {58} cannot be in 13(2)n8. Here's how.
7a. 15(3)n8 = [2/5/8..] (step 6)
7b. 13(2) = {58} -> 6(2) = {24} = {258}: clash with 15(3)

8. 13(2)n8 = {49/67} = [4/7..]
8a. ->[74] cannot be in h11(2)n8
8b. no 3 r6c8 (step 1)

9. {56} blocked from h11(2)n8 by 13(2) & 6(2)n8 (2 4s n8)
9a. no 1, 2 r6c8 (step 1)

10. "45" n147: r19c4 = 10 = h10(2)c4 = [19/28]

11. "45" n1: r1c4 + r4c2 = 6 = [15/24]
11a. r1c4 = {12}
11b. r4c2 = {45}

12. 12(4)n1 = {1245}(no 3,6)(since r4c2 = {45})
12a. 1 and 2 locked in r3c123 for n1, r3
12b. no 4,5 r12c2 (4,5 locked in 12(4)n1)

13. "45" n2: r1c46 = 9 = h9(2) = [18/27]

14. 18(4)n1 must have 1/2 not both (r1c4)
14a. any combo with 1 cannot have 8 (because r1c46 = [18])
14b. any combo with 2 cannot have 7 (because r1c46 = [27])
14c. {1269/1278/1368/1458/2367/2457/3456} all blocked

15. 18(4)n1 = {1359/1467/2349/2358} = [4/5..]
15a. Killer pair 4/5 in n1 between 18(4) and 12(4): locked for n1

16. 21(3)n1 = {678}:locked for n1 and r2

17. 18(4)n1 = {1359/2349} = 39{15/24}
17a. 3, 9 locked for r1

18. 20(3)n2 = {578} only
18a. 5,7,8 locked for r1
18b. 5 locked for n3

19. 18(4)n1 = {2349}
19a. r1c4 = 2
19b. 3,4,9 locked for r1, n1

20. r4c2 = 4 (step 11)
20a. 5 locked in r3c123 for r3


21. r1c6 = 7 (h9(2)n2)
21a. r1c78 = {58}:locked for n3

22. r9c4 = 8 (h10(2)c4)
22a. r9c23 = 11(2) = {29/56}/[74] (no 1,3 & no 7 r9c3) [edit]

23. r9c6 = 3 (h11(2)n8)
23a. r9c789 = 17(3) = {179/269/467}(no 5) = [1/6,2/7..]

24. r6c8 = 4 (step 1)
24a. r7c789 = 15(3) = {159/258/357}(no 6) ({168/267} blocked by 17(3)n9 step 23a)
24b. 15(3)n9 = 5{19/28/37}
24c. 5 locked for n9, r7
24d. no 1 r8c6

25. 13(3)n9 = {139/148/238/346}(no 7) ({247} blocked by 17(3)n9)

26. r23c4 = 8 = [53]

27. r3c56 = 12 = {48}: locked for n2, r3

28. r2c56 = {19}:locked for n2, r2

29. r1c59 = [61]
29a. no 7 r7c4

30. 17(4)n3 now 16(3) = {349} ({367} blocked since 6 and 7 in same cell)
30a. r2c89 = [34], r3c8 = 9

31. r2c7 = 2

32. r34c7 = 13 = {67}: locked for c7

33. 1 in n6 only in 20(4) = 1{289/379/568}

34. 6 in n8 only in c4: locked c4

Now: time to dig very carefully
35. deleted:easier way

36. 16(3)r4&6 {268/358} combo's blocked by r46c4
36a. 16(3)r4 = {169/178/259}(no 3) ({367} blocked by r4c7)

37. 13(3)n5 must have 4 for n5 = 4{18/27/36}(no 5,9)

38. 16(3)r6 = {178/259/367} ({169} blocked by 16(3)r4 step 36a)

39. 16(3)r4 = {178/259}(no 6) ({169} blocked by 16(3)r6 step 38)

40. 13(3)n5 = 4{18/36}(no 2,7) ({247} blocked by 16(3)r4 step 39)
Para's shortcut:
(40aWhen 13(3) in N5 = {148} -->> 13(3) = [1]{48}
But then we would have R3C56 = {48} and R5C56 = {48}.
This would mean a (non-)unique rectangle (, both sets of cells could either
be [48] or [84] without messing with any sudoku rules). As we know this
puzzle is unique, this situation can't be correct.
40b.Thus 13(3) can't be [1]{48}.
40c.Therefore R4C5 = 4)

41. 9 in n5 only in {259} combo in 1 of 16(3)s
41a. -> 9 in n5 can only be in r46c4
41b. 9 locked c4

42. r2c56 = [19] (hsingle 9 c6)
42a. no 5 r9c1 (h6(2)r9)

43. no 7 r8c4. Here's how.
43a. 15(3)n8 = {159/249/267}
43b. {267} must have 6 in r8c4 and is only combo with 7
43c. no 7 r8c4
43d. 7 in n8 in c5 only: 7 locked for c5

44. r46c4 = {79} (hidden pair n5):locked for c4

45. r78c5 = {79} (hidden pair n8)

46. no {249} combo in 15(3)n8
46a. 15(3)n8 = {249} = [492] only
46b. -> r9c1 = 4 (h6(2)r9)
46c. -> 4 in n9 must be in r7: not possible
46d. no {249} combo

47. 15(3)n8 = {159/267}(no 4)
47a. no 2 r9c1 (h6(2)r9)

Must now be some XYZ wing or ALS at this spot but just can't see it.

Unlocking time - (Steps 48 - 50: dedicated to rcbroughton - thnks Richard!).
48. r9c1 = {14} and r3c1 = {125}
48a. ->16(4)n4 {1249/1258/1348/1456} blocked
48b. 16(4)n4 = {1267/1357/2347/2356}(no 8,9)

49. "45"c1: 5 innies = 29 and must have 1/4
49a. = {14789/15689/24689/34589} ({34679} clash with 16(4)n4)

50. no 4 r1c1 because of 5 innies:step 49a. Here's how
50a. {14789}: 1 must be in r3c1 -> 4 must be in r9c1
50b. {24689/34589}: 4 must be in r9c1
50c. no 4 r1c1

51. r1c3 = 4 (hsingle)

52. 11(2)n7 = {29/56} (no 7)

53. 7 in r9 only in 17(3)n9 = 7{19/46}(no 2)
53a. 7 locked for n9

54. 15(3)n9 = 5{19/28}(no 3)

55. 3 for n9 only in 13(3) = 3{19/28/46}
55a. 3 locked r8

56. "45"n7: r6c3 - 2 = r7c1
56a. -> r7c1 no 2,
56b. r6c3 no 1,2,7

57. 14(3)n4 = {158/167/239/257/356}

58. 22(4)n7 must have 1/4.
58a. {1579/3469} blocked by 11(2)n9 (step 52)
58b. ={1489/1678/2479/3478/4567}
58c. only combo with 3 is {3478} -> r7c2 = 3, r9c15 = [42], r8c12 = {78} and r8c45 = {67} (2 7's r8)
58d. {3478} blocked
58e. no 3 r7c2

59. 3 in n7 only in r7c13

60. no 3 r7c1. Here's how.
60a. 3 in r7c1 -> rest of 16(4) in n4 = {157/256} = 5{17/26}
60b. but must also have 5 in r6c3 (step 56)
60c. no 3 r7c1

61. r7c3 = 3 (hsingle)
61a. no 5 r6c3 (step 56)

62. r68c3 = 11 = [92/65]
62. r7c1 = {47} (step 56)

63. 16(4)n4 must have 4/7 = {1267/1357/2347} = 7{126/135/234}
63a. 7 locked c1

64. Killer pair 2/5 in r8c3 & 11(2)n7: locked for n7

65. Killer pair 4/7 in r7c1 & 13(2)n8: locked for r7
65a. no 2 r8c6

66. 13(3)n9: {346} combo forces 1 in both r8c4 and r7c6: 2 1s n8
66a. {346} combo blocked
66b. 13(3) = 3{19/28} (no 4,6)

67. 6 in n9 in 17(3) = {467}:locked for r9, n9
67a. r9c7 = 4
67b. r9c15 = [15] (h6(2)r9)
67c. r8c45 = [19], 11(2)r9 = {29}:locked n7 [edit]

68. 13(3)n9 = {238}:locked n9,r8

69. 22(4)n7 now 21(3) = {678}

the rest unfolds from here and then links back into the final steps for the original. Very unusual to have the same hard ending in both variations of this puzzle.

[Andrew]

I had another look at Assassin 44V2 this week after the discussion about variants in the Assassin 47 thread.

A really challenging puzzle. It’s amazing how small changes to cages can make such a huge difference! The change to N2 wasn’t too much of a problem; maybe that was done to maintain symmetry as well as to take away an easy move. It was the change to N8 that really made this a V2!

Fortunately the changes provided some new pairs of innies to give something to work with and plenty of interactions between them. A V2 where things were taken away and nothing extra given in return would be even harder.

Ed did some useful combination eliminations for the 15(3) cage in N8 as early as step 6. I assume this cage was looked at early because it is one of the differences from the original puzzle, trying to get back to that one as soon as possible. That's an approach that hadn't occurred to me until yesterday although it now seems such an obvious thing to do. It was a pity that Ed's step 6 didn't also eliminate any candidates although it did allow him to remove a pair of candidates from the 13(2) cage in N8 in his next step.


Here is my walkthrough. I've included Para's very interesting shortcut as a comment. This shortcut is also in Ed's walkthrough.

1. R23C4 = {17/26/35}, no 4,8,9

2. R3C56 = {39/48/57}, no 1,2,6

3. R7C45 = {49/58/67}

4. R78C6 = {15/24}

5. R1C678 = {389/479/569/578}, no 1,2

6. R2C123 = {489/579/678}, no 1,2,3

7. R9C234 = {289/379/469/478/568}, no 1

8. 12(4) cage in N14 = 12{36/45}, no 7,8,9, no 1,2 in R1C2

9. 45 rule on N1 2 outies R1C4 + R4C2 = 6 = {15/24/33} (double possible), no 6,7,8,9, R1C4 = {12345}, R4C2 = {12345}

10. 45 rule on N9 2 outies R6C8 + R9C6 = 7 = {16/25/34}, no 7,8,9

11. 45 rule on R1 2 innies R1C59 = 7 = {16/25/34}, no 7,8,9

12. 45 rule on R9 2 innies R9C15 = 6 = {15/24}

13. 45 rule on C123 2 outies R19C4 = 10 = [19/28/37/46], clean-up: no 1 in R4C2

14. 1 in 12(4) cage locked in R3C123, locked for R3 and N1, clean-up: no 7 in R2C4

15. 45 rule on N8 2 innies R9C46 = 11 = [65/74/83/92], clean-up: no 1,6 in R6C8

16. 45 rule on N2 2 innies R1C46 = 9 = [18/27/36/45]

17. 45 rule on C789 2 outies R19C6 = 10 = [64/73/82], clean-up: no 4 in R1C4, no 2 in R4C2, no 2 in R6C8, no 6 in R9C4

18. 2 in 12(4) cage locked in R3C123, locked for R3 and N1, clean-up: no 6 in R2C4

19. 45 rule on N7 2 outies R6C3 + R9C4 – 10 = 1 innie R7C1, max R6C3 + R9C4 = 18 (doubles possible) -> max R7C1 = 8, min R6C3 + R9C4 = 11 -> min R6C3 = 2

20. R7C45 = {49/58/67} [7/8/9], R9C4 = {789} -> 15(3) cage must contain 7/8/9
20a. Valid combinations for 15(3) cage {159/168/249/258/267/348/357} [1/2/3, 4/5/6, 7/8/9]
20b. 15(3) cage [1/2/3], R78C6 = {15/24} [1/2] -> R9C6 = {23}, clean-up: no 6 in R1C6, no 3 in R1C4, no 3 in R4C2, no 3 in R6C8, no 7 in R9C4

21. R4C2 = {45} -> 12(4) cage in N14 = {1245}, no 3,6

22. 3 in N1 locked in R1C123, locked for R1, clean-up: no 4 in R1C59

23. R1C678 = {479/578} = 7{49/58} (cannot be {569} because no 5,6,9 in R1C6) [4/5,8/9], no 6, 7 locked for R1

24. 7 in N1 locked in R2C123, locked for R2, R2C123 = 7{59/68}, no 4

25. 8/9 in R1C678 -> 8/9 in R1C123
25a. R1C1234 = 3{159/168/249/258}

26. 45 rule on N3 2 outies R1C6 + R4C7 – 7 = 1 innie R3C9, min R3C9 = 3 -> min R1C6 + R4C7 = 10 -> min R4C7 = 2

27. R3C56 = {39/48/57} [7/8/9], R1C6 = {78} and 8/9 in 16(3) cage
[Any 16(3) cage must have at least one of 7/8/9 and, in this case, the 7 has already been eliminated] -> no 7 in R3C4, clean-up: no 1 in R2C4
27a. Valid combinations for 16(3) cage = {169/268/358} (cannot be {259} which clashes with R23C4, cannot be {349} because no 3,4,9 in R1C5) [1/2, 5/6, 8/9], no 4

28. R3C56 = {48} (only 4s in N2), locked for R3 and N2

29. R1C6 = 7 (naked single), clean-up: R1C4 = 2, no 5 in R1C59, no 6 in R3C4, R4C2 = 4, R9C4 = 8, R9C6 = 3, R6C8 = 4, no 9 in R1C7, no 5 in R7C45
29a. R9C4 = 8 -> R9C23 = 11, no 7 in R9C3

30. R3C123 = {125}, locked for R3 and N1 -> no 9 in R2C123 (step 24)
30a. R2C123 = {678}, locked for R2 and N1
30b. R1C123 = {349}, locked for R1

31. R23C4 = [53]

32. R2C56 = {19}, locked for R2 and N2

33. R1C59 = [61] (naked singles), clean-up: no 7 in R7C4

34. 17(4) cage in N3 = {1349} (only valid combination) -> R3C8 = 9, R2C89 = [34]

35. R2C7 = 2 (naked single), R34C7 = 13 = {67}, locked for C7

36. R9C6789 = 3{179/269/467} [1/6, 4/9], no 5

37. 15(3) cage in N8 = {159/249/267}, 6 only in R8C4 -> no 7 in R8C4
37a. 7 in N8 locked in R78C5, locked for C5
37b. 6 in N8 locked in R78C4, locked for C4

38. R6C3 + R9C4 – 10 = R7C1 (step 19), R9C4 = 8 -> R6C3 – 2 = R7C1, no 2 in R6C3, no 2,8 in R7C1

39. R4C456 = [196/718/781/916/925/952] (cannot be [736] which clashes with R4C7) [7/9], no 3 in R4C5, no 9 in R4C6

40. 4 in N5 locked in R5C456 = 4{18/27/36}, no 5,9
40a. R6C456 = [196/718/736/781/916/925/952] [7/9], no 9 in R6C6
40b. Killer pair 7/9 in R4C456 and R6C456 (steps 39 and 40a) for N5, no 7 in R5C4
-> no 2 in R5C56 (step 40)

[Para’s shortcut would work at this stage. Surprisingly it is also after Ed's step 40 even though he and I have used somewhat different steps. Here it is, taken from Ed’s walkthrough and slightly edited.
40a. When 13(3) in N5 = {148} it can be either [1]{48} or [4]{18]
If 13(3) = [1]{48} then we would have R3C56 = {48} and R5C56 = {48}.
This would mean a (non-)unique rectangle (both sets of cells could either
be [48] or [84] without messing with any sudoku rules). As we know this
puzzle is unique, this situation can't be correct.
40b.Thus 13(3) can't be [1]{48}.
40c.Therefore R4C5 = 4]

41. R2C6 = 9 (hidden single in C6), R2C5 = 1, clean-up: no 5 in R9C1
41a. R4C456 = [196/781/925/952], no 8 in R4C6
41b. R6C456 = [196/736/781/925/952], no 8 in R6C6

42. 3 in N5 can only be in [436] in R5C456 or in [736] in R6C456 which each require 6 -> no [196] in R4C456 or R6C456 -> no 1 in R46C4, no 9 in R46C5
42a. R46C4 = {79}, locked for C4, clean-up: no 4 in R7C5
42b. R4C456 = [781/925/952], no 6 in R4C6
42c. R5C456 = [148/184/436/481]
42d. R6C456 = [736/781/925/952]

43. Hidden pair {79} for N8 in R78C5
43a. 15(3) cage in N8 = {159/249/267}, no 4 in R9C5, clean-up: no 2 in R9C1

44. 22(4) cage in N7 must have R9C1 = {14}, valid combinations are {1489/1678/2479/3478} (cannot be {1579/3469} which clash with R9C23, cannot be {4567} -> R9C23 = {29} clashes with R9C15 = [42]) [1/2/3], no 5
44a. {1489} cannot have 1 in R9C1 because R9C23 = {56} would then clash with R9C15 = [15] -> no 4 in R8C1

45. 20(4) cage in N6 = 1{289/379/568}
45a. 21(4) cage in R3456C9 = {2379/2568/3567}

“Takes a Richard-style innie move to finally pick the lock.”

That was the hint that I needed to get me going after I seemed to be grinding to a halt. Thanks Ed. Typically I used different innies than he did and had to work a lot harder.

Some of the following steps may seem a bit long winded. I’ve kept the different ways that I’ve worked on the 3 innies in N7 as separate groups of steps for clarity.

46. 45 rule on N7 3 innies R7C13 + R8C3 = 12 = {129/138/147/156/237/246/345}
46a. R7C13 + R8C3 = {129} => R9C1 = 4 => R9C23 = {56}
46b. R7C13 + R8C3 = {138} => R9C15 = [42] => R9C23 = {56}
46c. R7C13 + R8C3 = {147} clashes with R9C1 (it also clashes with all remaining combinations in the 22(4) cage in step 44)
46d. R7C13 + R8C3 = {156} => R9C15 = [42] -> no remaining combinations for R9C23
46e. R7C13 + R8C3 = {237} => R9C23 = {56} => R9C15 = [42]
46f. R7C13 + R8C3 = {246} clashes with all combinations in R9C23
46g. R7C13 + R8C3 = {345} => R9C23 = {29}, R9C1 = 1

47. Summary of step 46
47a. R7C13 + R8C3 = {129/138/237/345}, no 6, no 8 in R6C3
47b. R9C23 = {29/56}, no 4,7
47c. R9C123 = 1{29}/4{56}

48. 7 in R9 locked in R9C89, locked for N9
48a. R9C6789 = 37{19/46}, no 2

49. 19(4) cage in N69 = 4{159/168/258}, no 3

50. 3 in N9 locked in R8C789, locked for R8, R8C789 = 3{19/28/46}, no 5, no 1 in R8C7, no 6 in R8C9

51. 5 in N9 locked in R7C789, locked for R7, 19(4) cage = 45{19/28}, no 6, clean-up: no 7 in R6C3, no 1 in R8C6

Now to consider each of the remaining combinations for R7C13 + R8C3 in a bit more detail; this is something that I would never have attempted before the tag solution for Assassin 42V2.

52. R7C13 + R8C3 = {129} => R9C15 = [42] => R78C6 = [15]
52a. R7C13 + R8C3 = {129} must have 1 in R7C1 which clashes with R7C6 -> R7C13 + R8C3 cannot be {129}, no 9 in R78C3

53. R7C13 + R8C3 = {138} => R9C15 = [42] => R78C6 = [15] -> no 1 in R7C13, no 8 in R8C3, clean-up: no 3 in R6C3

54. R7C13 + R8C3 = {237} => R9C23 = {56} => R9C15 = [42] => R78C6 = [15] => R7C789 = {258} -> no 2 in R7C3

55. R7C13 + R8C3 = {345} -> no 4 in R8C3

56. R7C13 + R8C3 = {345} => R7C13 = {34} (it would actually be [43], see step 60) -> no 4 in R7C6 (the other combinations for the 3 innies had R7C6 = 1), clean-up: no 2 in R8C6

57. Each of the combinations of the 3 innies is associated with a different combination for the 22(4) cage. R7C13 + R8C3 = {129} was eliminated in step 52. This eliminates {3478} from the 22(4) cage.

58. R7C13 + R8C3 = {138/237/345}
58a. R7C13 + R8C3 = {138} => R8C3 = 1
58b. R7C13 + R8C3 = {237} => R8C3 = 2
58c. R7C13 + R8C3 = {345} => R8C3 = 5
58d. No 7 in R8C3

59. R7C13 + R8C3 = {138/237/345}
59a. R7C13 + R8C3 = {138} => R7C13 = [38], R9C15 = [42] => R78C6 = [15] => R7C789 = {258} clashes with R7C3 -> R7C13 + R8C3 cannot be {138}
59b. R7C13 + R8C3 = {237} => R8C3 = 2, R7C13 = {37}, R9C23 = {56} => R9C15 = [42], R78C6 = [15] => R7C789 = {258} => R7C45 = [49] => R7C2 = 6 clashes with R9C23 -> R7C13 + R8C3 cannot be {237}

60. R7C13 + R8C3 = {345} (only remaining combination), R8C3 = 5 -> no 3 in R7C1 (step 38), R7C13 = [43], R6C3 = 6 (step 38), R7C45 = [67], R78C6 = [24], R8C45 = [19], R9C1 = 1, R9C5 = 5, R5C4 = 4, R3C56 = [48]
60a. Clean-up: no 6 in R9C2, R9C23 = {29}, locked for R9 and N7 -> R9C7 = 4, R7C2 = 8, R8C12 = {67}, locked for R8

61. R1C3 = 4 (hidden single in R1)

62. Naked pair {67} in R28C2, locked for C2

63. Naked pair {67} in R39C9, locked for C9

64. 1 in N4 locked in 23(4) cage = {1589} (only remaining combination), no 2,3,7, locked for N4
[This locks 5 in R56C2 for C2 and 8 in R45C3 which is an alternative way to achieve the same result as step 65.]

65. R456C1 = {237}, locked for C1 -> R8C12 = [67], R2C123 = [867], R1C12 = [93], R3C1 = 5 (naked singles)

66. R5C6 = 6 (hidden single in C6) -> R5C5 = 3

And now we return to moves from the solution to the original Assassin 44. We were actually back to the original puzzle after R8C3 = 5 in step 60 with the rest of step 60 and steps 61-65 getting back to the original solution path.

67. 21(4) cage in N36 = {2379/2568} (cannot be {3567} because 6,7 only in R3C9) = 2{379/568}, 2 locked for C9 and N6

68. R8C8 = 2 (hidden single in N9)

69. 20(4) cage in N6 = {1379/1568} = 1{379/568}
69a. 3 only in R6C7 -> no 9 in R6C7
69b. 6 only in R4C8 -> no 5,8 in R4C8

70. If R4C456 = [781] => R4C7 = 6 clashes with R4C8 -> R4C456 cannot be [781]
70a. R4C456 = [925], R6C456 = [781]

71. 20(4) cage in N6, R6C7 = {35} -> no 5 in R5C78

72. Naked pair {38} in R48C9, locked for C9

73. Naked triple {358} in R168C7, locked for C7
[Alternatively hidden pair {19} in R57C7.]

74. If 20(4) cage in N6 = {1568}, R6C7 => 5, R4C8 => 6, R4C7 => 7, R4C1 => 3, R4C9 => 8 => no 3 in N6 (or if one prefers, two 8s in N6) -> 20(4) cage in N6 cannot be {1568}

75. 20(4) cage in N6 = {1379}, no 5,6,8, locked for N6

and the rest is naked singles

Alternatively, instead of step 74, there’s Para’s neat alternative finish from earlier in this thread

“Another way to finish the puzzle. Almost the same. Think it looks more Killer-like.

a. 21(4) in R3C9 = [73]{29}/[68]{25} -->> R34C9 = [73/68]
b. 45-test on N3: innie = outie: R3C9 = R4C7 -->> R4C79 = [73/68]
c. R4C79 -->> no [73]: clashes with R4C1
d. R4C79 = [68]
And that leads to the same finish.”

[Para]

A really challenging puzzle. It’s amazing how small changes to cages can make such a huge difference! The change to N2 wasn’t too much of a problem; maybe that was done to maintain symmetry as well as to take away an easy move. It was the change to N8 that really made this a V2!

Hi

Yes N2 was to keep symmetry and it didn't mess much with the solving progress so that was ok. I played around with the 14(2) cage in N8 a little because that was the big opener of the puzzle. I also tried combining the 13(2) cage with the 14(2) (and also kept it symmetric in N2 by combining the 7(2) and 12(2) cage) but that one was far easier than this one. Needed only one additional move compared to the V1.

You could also combine the 14(2) cage in N8 with the 7(2) cage in N2 forming a 21(4) toroidal cage. That gives a good puzzle as well, somewhere in between the V1 and the V2 for difficulty.

In the end i liked the solving process and the look of this V2 the most.

greetings

Para

[Andrew]

You could also combine the 14(2) cage in N8 with the 7(2) cage in N2 forming a 21(4) toroidal cage. That gives a good puzzle as well, somewhere in between the V1 and the V2 for difficulty.

Looks like you've just posted a new puzzle! Ed ought to enjoy it; it doesn't even need a new diagram. He posted the Blackhole Remote Killer. I haven't yet attempted that, or any other remote killer, so I'll take your word that the toroidal one is "somewhere in between the V1 and the V2 for difficulty". Maybe I'll try the toroidal before I attempt the Blackhole.

In the end i liked the solving process and the look of this V2 the most.

Thanks again. It was a good, very challenging puzzle.

[Andrew]

You could also combine the 14(2) cage in N8 with the 7(2) cage in N2 forming a 21(4) toroidal cage. That gives a good puzzle as well, somewhere in between the V1 and the V2 for difficulty.

I'm confused. I've had another look at the Assassin 44 diagram. The 7(2) cage in N2 is R12C5 and the 14(2) cage in N8 is R89C5. Therefore the 21(4) toroidal cage would be R1289C5. This is already linked by C5 so I don't see how combining those two cages makes any difference. Is there something special about a toroidal cage that I'm missing? I've never seen one before although I have seen them mentioned on a different forum.

[Para]

Hi

21(4) has more combinations than 14(2) and 7(2) apart. It also changes the 45-tests that uses this cage. Just try it , you'll see.

greetings

Para

[sudokuEd]

somewhere in between the V1 and the V2

Think this is the pic for Para's Assassin 44V1.5 It has a toroidal/remote 21(4) cage r1289c5. .

3x3::k:4608:4608:4608:4608:5380:5125:5125:5125:4360:5385:5385:53855380:6158:6158:4360:4360309020603094:6158:4360:5402:41235917:4126:4126:4126:6158:5154:5402:4123:5917:591733675154:5154:5402:4123:59174144:4144:4144:5154:4916:5402:4123:568733851595:4916:4916:4916:5687:56873887:538033973397:5687:4937:4937:4937:5380:5197:5197:5197:5197:

[Andrew]

21(4) has more combinations than 14(2) and 7(2) apart. It also changes the 45-tests that uses this cage. Just try it , you'll see.

Thanks for explaining that. I obviously hadn't thought hard enough about it. Still it was worth asking the question on the forum so that everyone got to read your answer.

Thanks for the diagram Ed. Guess I'd better have a go at it soon.

[Andrew]

I did have a look at it, fairly soon after I posted my previous message, but didn't get too far. It's now over a month since then and it's not near the top of my backlog pile so maybe someone else would like to have a go and post a walkthrough.

Combining those two 2-cell cages into one 4-cell "toroidal" cage makes a big difference because it takes away a lot of 45s.

somewhere in between the V1 and the V2

My current feeling is that "toroidal" V1.5 is harder than V2. Both challenging puzzles by Para!