SudoCue Users

The Easter Bunny came by and hid the solutions to these variations of Assassin 45.

If you can help me recover these, I would be both pleased and impressed.

Assassin 45 - Variation 2



3x3::k4097:4354:4354:6916:4101:410143604097:4354:6916:6916:6916:41014360:4882:40972068:691636074122:4882:4882591861764122:41222852:5918:59186176:617620912605591861764404:440451753375:90182355440451759018:9018:9018:4165186351753649:9018:4165:4165:3901

Assassin 45 - Variation 3



3x3::k3585:4610:4610:58923077:435920484610:5892:5892:589243593602386058923095:4359:41223602540677124122:41223364:5406:54067712:771218352861:4911:5406771233802861:5687:4911:4911:69702099:41572111:56876970:6970:6970:5957:41572111:56872113:6970:5957:5957:4157

TIA,
Ruud

Thanks for the challenge Ruud. V2 looks like a real toughie. Found lots of little ones that were very tasty and interesting. But help needed now.

Assassin 45V2
1. 17(2)n3 = {89}: locked for n3, c9

2. 9(3)n6 = {126/135/234}(no 789}

3. "45" c9: r456c8 = 20 = h20(3) = {479/569/578}(no 123)
3a. {389} blocked since {89} must be in r89c7(from step 1): not possible since r89c7 is a 16(3) cage

4. 8(2)n6 = [53/62/71]

5. 10(3)n3 = {127/136/145/235}(no 8,9)

6. h20(3)n6 = [8/9], not both ->
6a. 15(3)n9 must have [8/9] = {159/168/249/258/348}(no 7)

7. 7(2)n9 = {16/25/34}(no 789}

8. r89c8 must have [8/9] (only other place in n9 besides 15(3))
8a. 16(3)n9 = {169/178/259/268/349/358}

9. 15(3)n9 must have only 1 of 1/2/3 (step 6a)
9a. -> 10(3)n3 must have 2 of 1/2/3
9b. -> 10(3)n3 = {127/136/235}(no 4) = [1/5...]
9c. {159} blocked from 15(3)n9 = {168/249/258/348}

10. 35(5)n8 = {56789}:locked n8
10a. max. r9c6 = 4 -> min. r89c7 = 12 -> 3 minimum in each cell
10b. no 1 or 2 r89c7

11. "45" r12: r3c258 = 22 = h22(3)r3 = {89}[5]/{679}(no 1234, no 5 r3c25)
11a. = 9{58/67}: 9 locked r3
11b. 10(3)n3: no 5,6,7 r12c8

12. 11(2)n4 (no 1)

13. "45"n5: r5c37 = 11 = h11(2)(no 1)

14. "45" r1234: r4c456 = 11 = h11(3) (no 9)

15. "45" r6789: r6c456 = 19 = h19(3) (no 1)

16. 9(3)n5 = {126/135/234}(no 789)

17. "45" r5: r5c456 = 15 = h15(3)r5
17a. {159} with 1 only in r5c5 (5 in r5c5 means 9(3) = {135}: 2 1's n5)
17b. {168} with 1 only in r5c5 (6 in r5c5 means 9(3) = {126}: 2 1's n5)
17c. {249} with 2 only in r5c5 (4 in r5c5 means 9(3) = {234}: 2 2's n5)
17d. {258}: blocked by 11(2) & h11(2)r5
17e. {267}: Blocked by 11(2) & h11(2)r5
17f. {348} with 3 only in r5c5 (4 in r5c5 means 9(3) = {234}: 2 3's n5)
17g. {357}: Blocked by 11(2) & h11(2)r5
17h. {456}

18. In summary: h15(3)r5 = {159/168/249/348/456}(no 7)
18a. r5c46: no 1,2,3

19.20(3)n7 = {389/479/569/578}(no 1,2)

20. 5(2)n7 = {14/23}

21. 9(2)n1 = {18/27/36/45}

22. 16(3)n1 = {169/178/259/268/349/358/457} ({367} blocked by 20(3)n7)

23. 19(3)n1: no 1

24. 10(3)n4 = {127/136/145/235}

25. "45"c1: r456c2 = 9 = h9(3) = {126/135/234}(no 7,8,9)
25a. no 234 r5c1
25b. 1 in h9(3) only in r6c2 -> no 5 or 6 r6c2

26."45" c34:r28c4 = 15 = h15(2) = {69/78}

27. 23(4) n5 has to fit in with this h15(2) & 9(3)n5
27a. {1589}: 8/9 must be in r5c3
27b. {1679}: 7 must be in r5c3
27c. {2489}: 8/9 must be in r5c3
27d. {2579}: blocked: 7/9 must be in r5c3 but {25}n5 clash with 9(3)
27e. {2678}: 6 must be in r5c3
27f. {3479}: 7/9 must be in r5c3
27g. {3569}: 3/6 must be in r5c3: (9(3) = [3/6])
27h. {3578}: 3/5/7 in r5c3
27i. {4568}: 6/8 must be in r5c3

28. Conclusion: no 2 or 4 r5c3
28a. no 9 or 7 r5c7 (h11(2)r5)

29. "45" c67: r28c6 = 11 (no 1)
29a. r2c6 = 2..6

30. "45" r89: r7c258 = 22 = h22(3)r7 = {589/679}(no 1..4)
30a. 9 locked r7

31. "45"n9: r9c6 + 7 = r7c79 = 8..11
31a. max. r7c7 = 6 -> min r7c9 = 2
[marks pic edit: forgot step 28, 28a]

Thanks, Ed,

Taking off from where your walkthrough left off, here's my suggestion for the next few steps, leading up to the first digit placement:

32. Innie/outie difference, c1: r345c1 - r6c2 = 21 -> no {1234} in r34c1, no 4 in r6c2

33. Innie/outie difference, r123: r3c1349 - r4c7 = 9 -> no 1 in r4c7
(reason: from step 32, r3c1 >= 5 -> r3c1349 >= 11)

34. Only 2 possible locations for digit 7 in n9:
(a) in r7c9 -> r789c7 = 16 (in total) -> r7c7 = {1234} (since r89c7 >= 12) -> no 1 in r6c7
(b) within 16/3 cage -> 16/3 = {178} -> r7c7 + r7c9 = 8 -> no 1 in r6c7
(reason: would imply r7c6 + r7c7 = 8, thus - whatever value r7c7 would take - r7c6 would clash with r7c9)
Summary: no 1 in r6c7

35. 1 in n6 now locked in c9 -> no 1 in r3c9; 7/2 cage at r89c9 = {25|34}

36. No 9 now possible in 15/3 cage in n9
({159} excluded due to 10/3(n3) cage, {249} excluded due to 7/2(n9) cage)
-> 8 in n9 locked in 15/3 cage (since we know it must contain one of {89}) -> 9 in n9 locked in r89c7 -> hidden single(n9) r7c9 = 7

Great work Mike. Love those sort of sneeky contradictions.

Here's some more: a bit greedy, but have V3 yet! That's me for the day.

37. r6c89 = [91]/{46}

38. 9 in c8 now in n6: no 9 r4c7
38a. h20(3) = {479/569}
38b. no 7 r4c8

39. 16(3)n3 = {169/259/349}
39a. r4c8 = 9
39b. r56c8 = [74/56]
39c. no 2 r5c9

40. r6c89 = {46}: locked for n6, r6
40a. no 7, 5 r5c3 (h11(2)r5)

41. "45" n9: r9c6 = r7c7 = {1234}
41a. 9(3)n6 = {135/234} = 3{15/24}

42. h22(3)r7 = {589}:Locked for r7

43. 6 r7 only in n7: locked for n7
43a. 20(3) = {389/479/578}

44. h19(3)r6 = {289/379} = 9{28/37}(no 5)
44a. 9 locked n5, r6

45. 2 6's n7
45a. 6 r7c1 -> r6c12 = {13} (cage combo) -> hidden triple {235} r6c1257
45b 6 r7c3 -> r6c3 = 3/5 (cage combinations 13(3)) -> naked triple {235} r6c357
45c. hidden triple {235} in r6c12357 for r6
45d. and no 2 r6c123

46. 19(3)n1 = {478/568} = 8{47/56}(no 2,3)
46a. 8 locked c1
46b. no 3 r5c2
46c. no 1 r12c1

47. h9(3) n4 = [621/423] ({135} blocked by 1 and 3 only in r6c2
47a. r5c2 = 2, r5c1 = 9
47b. r4c2 = {46}

48. h11(2)r5 = {38}/[65] = [3/5..]
48a. [53] blocked from 8(2)n6
48b. r5c89 = [71]

49. r6c8 = 4 (h20(3)n6), r6c9 = 6

50. 10(3)n3 = 3{16/25}
50a. 3 locked n3, c3

51. 1 in n5 locked in h11(3) = 1{28/37/46}(no 5)
51a. 1 locked r4

52. 8(3)n1 = 1{25/34}
52a. 1 locked r3
52b. no 5 r3c34

53. 5 in n5 only in r5 in h15(3)n5 = {456}: locked r5, n5

54. combinations for 19/3 in r34 -> no 6 in r34c1
55. combinations for h22/3 in r3c258 -> no 6 in r3c25
56. r4c2 = hidden single 6
57. outies of c1 = r6c2 = 1
58. 19/3 in c12 = {568} -> 58 pair in c1, elim from rest of c1
59. 9/2 in n1 = {27|36}
60. r4c3 = hidden single 4
61. 10/3 in c12 = [316|712]
62. 4 of c1 locked in 5/2 in n7 -> 5/2 in n7 = [14], elim from rest of n7
63. combinations for 8/3 in r34 = {134} -> r3c34 = {13} pair -> elim from rest of r3
64. 7 of r4 locked in n5 -> elim from rest of n5
65. r6c46 = naked pair (89) -> elim from rest of n5/r6
66. innies of r6789 = r6c456 = 19/3 -> r6c5 = 2
67. 20/3 in n7 = {389|578} = {8...} -> elim 8 from rest of n7 and c2
68. combinations for 14/3 in r89 = {239|257|347} -> no 1 in r9c4 (cannot be 149 as only 1 and only 4 both in r9c4)
69. r4c456 = naked triple {137} -> elim from rest of r4
70. combinations for 16/3 in c89 = {259} -> {25} pair in r34c9 -> elim from rest of c9
71. r89c9 = naked pair {34} -> elim from rest of n9
72. r9c1469 = naked quad {1234} -> elim from rest of r9
73. 2 of r9 locked in n8 -> elim from rest of n8

Edit: Here's the rest.
73. combinations for 14/3 in r34 = {248|257} -> no 6
74. r3c8 = hidden single 6
75. r12c8 = {13} pair -> elim from rest of c8/n3
76. r79c8 = naked pair {58} -> elim from rest of n9 -> r8c8 = 2 -> r7c7 = 1
77. outies of n9 = r9c6 = 1 -> r89c1 = [14] -> r89c9 = [43] -> r9c4 = 2
78. combination for 9/3 in c67 = [531] -> r34c9 = [52] && r7c4 = 4 -> r34c1 = [85] && r4c7 = 8 -> 23/4 in c34 = [1859] && 24/4 in c67 = [7638] && 9/3 in c5 = [342] -> r3c34 = [13]
79. innies of c1234 = r28c4 = 15/2 = {78} -> r1c4 = 6
80. innies of c6789 = r28c6 = 11/2 = [29] -> r3c67 = [42] && r89c7 = [69] -> r1c6 = 5
81. r9c5 = hidden single 6
82. r7c2 = hidden single 9 -> r3c2 = 7 -> r3c5 = 9
83. combinations: 9/2 in n1 = [36] -> naked singles and last-digit-in-cage moves solve the puzzle

By the way, here's my walkthrough for V3. Haven't proof read it yet, so let me know of any significant mistakes...


Walkthrough - Assassin 45 V3

1. Preliminaries:

a) 8/2 at R1C1: no 4,8,9
b) 13/3 at R1C9: no 1,2,3
c) 9/3 at R4C5: no 7,8,9
d) 30/4 at R4C6 = {6789} -> no 6,7,8,9 in R5C45
e) 13/2 at R5C1: no 1,2,3
f) 7/2 at R5C8: no 7,8,9
g) 11/3 at R6C1: no 9
h) 19/3 at R6C3: no 1
i) 8/3 at R6C7 = {1(25|34)}
j) 22/3 at R7C2 = {(67|58)9} -> no 9 elsewhere in C2,N7
k) 8/2 at R8C1: no 4,8
l) 8/3 at R8C3 = {1(25|34)} -> no 1 in R9C1 -> no 7 in R8C1
m) 23/3 at R8C7 = {689} -> no 6,8,9 in R9C89 -> no 2 in R8C9
n) 8/2 at R8C9: no 4,8,9

2. Outies N5: R5C37 = 15/2 = {69|78} -> 13/2 at R5C1 = [94]|{58}

3. Innies C34: R28C4 = 9/2 -> no 9

4. Innies C67: R28C6 = 5/2 = {14|23}

5a. Outies C1: R456C2 = 9/3 -> no 7,8 -> 13/2 at R5C1 = [94]|[85]
5b. R46C2 = {(1|2)3} -> no 3 elsewhere in C2 or N4, no 1 in R34C1

6a. Innies R5: Split cage R5C456 = 10/3 -> no 8,9 in R5C6, no 4,5 in R5C45
6b. 1 locked in R5C45 -> no 1 elsewhere in R5 or N5 -> 7/2 at R5C8 = {25|34}

7. Outies R89: R7C258 = 22/3 = {(67|58)9} -> no 9 elsewhere in R7 -> no 2 in R6C3

8a. Naked quad on {6789} in C6 at R4569C6 -> no 6,7,8,9 in R13C6
8b. 7 in C6 now locked in N5 -> no 7 in R46C4 or R5C7 -> no 8 in R5C3 (step 2)

9. Naked triple on {689} in C7 at R589C7 -> no 6,8,9 in R1234C7

10a. 21/4 at R4C4 cannot contain both of {67} due to R5C6
10b. 9/3 at R4C5 must contain 2 of {123} -> forms naked killer triple on {123} with R5C4 -> no 2,3 in R46C4
10c. 21/4 at R4C4 must contain at least one of {45} in C4 -> no 4,5 in R28C4 (step 3)

11. Innies N2: R13C46 = 22/4 -> min. R13C4 = 13 -> no 1,2,3,4,5 -> no 9 in R34C3

12. Innies N9: R789C7+R7C9 = 21/4 -> can only contain 2 of {6789} (in R89C7) -> no 6,7,8 in R7C9

13. Outies N1: R13C4(min. 13)+R4C123(min. 3) = 24/5 -> max. R4C1 = 8 -> no 9 in R4C1

14. Innie/outie difference N3: R3C79 - R1C6 = 3 -> max. R3C79 = 8 -> no 8,9 in R3C9

--- Now comes the complicated bit that most automated solvers can't do... ---

15a. 7 in C7 locked in either of 12/3 cages at R1C6 or R3C6
15b. Only one of these 2 12/3 cages can contain a 7, since 7 not available in R13C6 = {(14|23)7}
15c. {689} not available -> the other 12/3 cage must be {345}
15d. The only other location for the second 5 in C67 is within 8/3 cage at R6C7 = {125} (no 3,4)

16a. N3: 17/3 and 13/2 cages must each contain one of {89} -> 13/2 <> {67} -> no 6,7 in R12C9
16b. 6 in N3 locked in R123C8+R3C9 -> no 6 in R4C8

17. 17/3 at R1C8: {359} and {458} both blocked due to 13/2 at R1C9 -> no 5 in R123C8

18a. Hidden 15/4 cage(N3) at R123C7+R3C9 also cannot contain a 5, as both {1257} and {1356} blocked by 8/3 at R6C7
-> 5 in N3 locked in 13/2 at R1C9 = {58} -> no 5,8 elsewhere in C9,N3 -> no 3 in R89C9
18b. Hidden 15/4 cage(N3) = {(17|26)34} -> 3,4 locked -> no 3,4 in 17/3 cage at R1C8

--- end of complicated bit ---

19a. 9 in N3 locked in 17/3 in C8 -> no 9 elsewhere in C8
19b. 9 in N6 locked in R46C9 -> no 9 in R5C7 -> no 6 in R5C3 (step 2)
19c. 9 in N9 locked in R89C7 -> no 9 in R9C6 -> 9 in N8 locked in C5 -> no 9 elsewhere in C5
19d. 9 in N2 locked in R13C4 -> no 9 elsewhere in C4, no 7,8 in R4C1 (step 13)

20. 6 in R5 locked in R5C67 -> no 6 in R46C6

21a. {34} in C6 locked in R1238
21b. R28C6 must contain exactly 1 of {34} (step 4) -> R13C6 must contain exactly 1 of {34}
-> 23/5 at R1C5 cannot contain both of {34} -> must contain 5 in C5 (since 9 not available)
-> no 5 elsewhere in C5 or N2

22a. Naked quad on {1234} in in R1238C6 -> no 1,2,3,4 elsewhere in C6 -> R7C6 = 5
22b. Naked pair on {12} in R67C7 -> no {12} elsewhere in C7

23a. Hidden single in C7 at R4C7 = 5
23b. 7/2 at R5C8 = {34} -> no 3,4 elsewhere in R5 or N6 -> R5C12 = [85], R5C67 = [76] -> R5C3 = 9

24. Hidden single in C1 at R3C1 = 9 -> no 6 in R4C1

25. Hidden single in C4,N55 at R6C4 = 5 -> R45C4 = [61], R5C5 = 2

26. Hidden single in C4 at R1C4 = 9

27. Hidden single in C6 at R9C6 = 6

28. Hidden single in C8 at R2C8 = 9

29a. 22/3 at R789C2 = {679} -> no 6,7 elsewhere in C2 or N7
29b. 14/3 at R123C2 = {248} -> no 2,4,8 elsewhere in C2 or N1
29c. 8/2 at R8C1 = {35} -> no 3,5 elsewhere in C1 or N7
29d. 8/2 at R1C1 = {17} -> no 1,7 elsewhere in C1 or N1

30. Hidden single in C1 at R6C1 = 6

31. Hidden single in C3 at R7C3 = 8

... and so on (no twist in the tail in this one)