Assassin 20

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Nasenbaer
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Assassin 20

Post by Nasenbaer »

That was another nice killer, thanks Ruud. Luckily I had some spare time this morning, so I was able to solve it.

A walkthrough is already written, but I won't post it until sunday, so eveybody gets a chance to solve it. :D

If you have any objections (don't post a walkthrough, post it earlier/later, I'm stuck give me a hint) please inform me.

Have fun solving it!

Peter
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Post by Marlie »

I can't paste it in sudocue, i have the latest version but notthing appears.
How did you do that?
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Post by Nasenbaer »

Marlie,

you are using the wrong program, you have to use SumoCue for the killer sudokus.

Peter
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Post by Marlie »

Ow... i am ashamed :oops:
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Post by sudokuEd »

Hi Marlie,
I had the same problem http://www.sudocue.net/forum/viewtopic.php?t=183

You need to download suMocue - not the suDocue version - the green one not the grey/black(?) one. The suD one is for a variant of Killer apparently :o

I see Peter has just beaten me - first with a walkthrough and now again :evil:
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Post by Marlie »

Thank you both.
But it is nice to know that persons in Germany and Australia can help me within a few minutes. :D
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Post by Nasenbaer »

SudokuEd convinced me that with tiny font there are no spoilers, so here it is:
Walkthrough Assassin 20
EDIT: Thanks to sudokuEd and Andrew for the additional input and corrections!

1. r1 : 4(2) = {13} -> 1,3 locked in r1
2. -> (step 1) r1 : 8(2) = {26} -> 2,6 locked in r1
3. "45" on n1 (2 outies) : r1c4 + r4c2 = 4 -> r4c2 = {13}
4. "45" on n3 (2 outies) : r1c6 + r4c8 = 7 -> r4c8 = {15}
5. "45" on r1234 (2 innies) : r4c19 = 13 = {49}/{58}/{67} -> no 1,2,3
6. "45" on r12345 (3 innies) : r5c456 = 12
7. c7 : 15(2) = {69}/{78} -> no 1,2,3,4,5
8. r9 : 15(2) = {69}/{78} -> no 1,2,3,4,5
9. n9 : 5(2) = {14}/{23} -> no 5,6,7,8,9
10. n7 : 16(2) = {79} -> 7,9 locked in c2 and in n7
11. -> n7 : 8(2) = {26}/{35}
12. "45" on r9 -> r9c37 = 10
12a. -> no 5,6 in r9c3 -> r9c3 = {23} -> no 2,3 in r8c3 -> r8c3 = {56}
12b. -> no 6,9 in r9c7 -> r9c7 = {78} -> no 6,9 in r8c7 -> r8c7 = {78} -> 7,8 locked in c7 and n9
13. -> r9 : 15(2) = {69} -> 6,9 locked in r9 and n9
14. n7 : 7(2) = {34} ({16} not possible because of locked 6 in step 13, {25} not possible because 8(2) in n7 has to have either a 2 or a 5 (step 11)) -> 3,4 locked in r9 and n7
15. -> r9c3 = 2 -> r8c3 = 6
16. -> r9c7 = 8 -> r8c7 = 7
17. -> r8c2 = 9 -> r7c2 = 7
18. r9 : 13(3) = {157} -> 1,5,7 locked in n8
19. "45" on c5 : r89c5 = 8 = {35} (only possible combination left from step 18) -> r9c5 = 5 -> r8c5 = 3
20. -> no 3 in r8c8 -> no 2 in r7c8
21. c1 : look at 11(3) : r78c1 = {18}/{15} = 1{5/8} (leftover from n7)
21a. -> {15} not possible because it would require 5 in r6c1, but already used in 11(3)
21b. -> r78c1 = {18} -> r6c1 = 2
21c. -> no 8 in r4c1 -> no 5 in r4c9 (step 5)
22. -> r7c3 = 5
23. -> r8c9 = 5 -> r67c9 = {14}/{23} (or should it read {[2]3} because 2 is not possible in r6c9 -> no 3 in r7c9 ?)
24. r8 : 15(3) = {348} (only possible combination left)
24a. -> r8c1 = 1 -> r7c1 = 8
24b. -> r8c8 = 2 -> r7c8 = 3
25. -> r67c9 = {14} -> 1,4 locked in c9 (who cares now about my rambling in step 23? :-) )
26. -> no 4 in r4c9 -> no 9 in r4c1 (step 5)
27. r7c7 = {14} (leftover from n9) -> 6(2) in c7 has either 1 or 4 -> 1,4 locked in c7
28. 25(4) in n478 : possible combinations: {3589}/{4579} -> no 1,2,6 -> r7c4 = 9
28a. -> r6c23 = 11 = {38}/{4[7]}
29. "45" on n4 -> r4c23 = 10 -> r4c3 = {79}
30. c3 : 10(2) = {19}/{37}
30a. -> has to have 1 or 3 -> because of r1c3 = {13} 1,3 is locked for c3 -> no 3 in r6c3 -> no 8 in r6c2
30b. -> has to have 7 or 9 -> because of r4c3 = {79} 7,9 is locked for c3 -> no 7 in r6c3 -> r6c3 = 8 -> r6c2 = 3 -> r5c3 = 4
31. -> r9c2 = 4 -> r9c1 = 3
32. -> r4c2 = 1 -> r4c3 = 9 -> r5c3 = 4 -> no 4 or 9 in r4c19 (step 5)
33. -> r1c4 = 3 (step 3) -> r1c3 = 1
34. n1 : 10(2) = {37} -> 3,7 locked in c3 and n1
35. r4c8 = 5 -> r1c6 = 2 (step 4) -> r1c7 = 6
36. no 5 in r4c1 -> no 8 in r4c9 (step 5)
37. c6 : 6(2) = {15} (no 2 -> no 4)
38. r7c6 = 6 -> r7c5 = 2
39. 9(3) in n14 : {126} only possible combination left -> r3c1 = 6 -> r3c2 = 2
39a. no 2 in r3c7 -> no 4 in r2c7
40. r4c1 = 7 -> r4c9 = 6 ->r9c9 = 9 -> r9c8 = 6
41. r5c2 = 6 -> r5c1 = 5
42. r5c6 = 1 -> r6c6 = 5
43. r9c6 = 7 -> r9c4 = 1
44. c4 : 9(2) = {27} (only possible combination left) -> r6c4 = 7 -> r5c4 = 2
45. r5c5 = 9 (from step 6) -> r6c5 = 6 (because of 17(3))
46. r6c7 = 9 (because 1,4 locked in c7 from step 27) -> r6c8 = {14}
47. r4c6 = 3
48. 24 in n245 : {4569} only possible combination left -> r4c4 = 4 -> r3c4 = 5 -> r2c4 = 6
49. r8c4 = 8 -> r8c6 = 4
50. r4c5 = 8 -> r4c7 = 2 -> no 2 in r2c7 -> no 4 in r3c7
51. -> r3c7 = 1 -> r2c7 = 5
52. r7c7 = 4 -> r7c9 = 1 -> r6c9 = 4 -> r6c8 = 1
53. r5c7 = 3 -> r5c89 = {78}
54. c2 : r12c2 = {58} -> r1c2 = 5 -> r2c2 = 8
55. c6 : r23c6 = {89} -> r2c6 = 9 -> r3c6 = 8
56. 17(3) in n36 : {359} only possible combination left -> r3c8 = 9 -> r3c9 = 3
57. r3c3 = 7 -> r2c3 = 3
58. r1c2 = 4 -> r1c1 = 9
59. r3c5 = 4 -> r2c5 = 1 -> r1c5 = 7
60. r1c8 = 4 -> r1c9 = 8
61. r5c9 = 7 -> r5c8 = 8
62. r2c8 = 7 -> r2c9 = 2


Well, if there is still an empty cell then you missed a step. :-)

Please feel free to comment my steps. I know there are very obvious steps included, but it helps me not to overlook anything.

Have fun!

Peter
Last edited by Nasenbaer on Sat Oct 21, 2006 8:00 am, edited 2 times in total.
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Post by Ruud »

sudokuEd wrote:The suD one is for a variant of Killer apparently :o
Indeed. It is a Zero-Killer where all cages have size 1. A very limited Killer variant.
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Post by sudokuEd »

Ruud wrote:It is a Zero-Killer where all cages have size 1. A very limited Killer variant.
Limited indeed - but not in popularity!

Now, Peter obviously found Assassin 20 way too easy (BTW -great walk-through Peter), so hope to give him a bit more challenge with V2. This puzzle has the same solution, just a few cage changes. But can still be solved logically.

The text code is below. Would someone set it for me? Thanks. You can also print a big version using SumoCue and Paint.

Assassin 20V2
3x3::k:6656:6656:1026:1026:5124:2053:2053:5383:5383:6656:6656:2571:6156:5124:5646:1551:5383:5383:2322:2322:2571:6156:5124:5646:1551:4377:4377:5659:2322:6156:6156:5124:5646:5646:4377:6179:5659:5659:5659:2343:4392:1577:6179:6179:6179:2861:6446:6446:2343:4392:1577:5171:5171:2613:2861:4151:6446:6446:4392:5171:5171:1341:2613:2861:4151:2113:7234:7234:7234:3909:1341:2613:1864:1864:2113:7234:7234:7234:3909:3919:3919:
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Post by Nasenbaer »

sudokuEd,

do you want to torture me??? :wink:

I just had a quick look at at V2. The combination of those two cages in n8 makes a lot of difference. My walkthrough stops with step 11 in V2, and now I have to look for other clues. A LOT more difficult.

But that has to wait until sunday. First I have to solve the new clueless. :D

Peter
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Post by Ruud »

Image
:mrgreen:
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Post by Nasenbaer »

Wow. Version 2 is MUCH harder. It took me ages to place the first number, even with parts of the solution still in my memory. Heavy use of innies and outies is required, and always have a look which combinations are possible. :D

SudokuEd, since you gave this new version to us, I leave the honor of creating the walkthrough for you... :twisted: :wink:

Peter
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Post by sudokuEd »

Nasenbaer wrote:do you want to torture me???
Glad you survived the torture chamber :twisted:

And thankyou Ruud for setting V2.

In this tip-toe-through, I've included Peter's first 10 steps and then let you find your own way back into Peter's steps after the initial placements.

1. r1 : 4(2) = {13} -> 1,3 locked in r1
2. -> (step 1) r1 : 8(2) = {26} -> 2,6 locked in r1
3. "45" on n1 (2 outies) : r1c4 + r4c2 = 4 -> r4c2 = {13}
4. "45" on n3 (2 outies) : r1c6 + r4c8 = 7 -> r4c8 = {15}
5. "45" on r1234 (2 innies) : r4c19 = 13 = {49}/{58}/{67} -> no 1,2,3
6. "45" on r12345 (3 innies) : r5c456 = 12
7. c7 : 15(2) = {69}/{78} -> no 1,2,3,4,5
8. r9 : 15(2) = {69}/{78} -> no 1,2,3,4,5
9. n9 : 5(2) = {14}/{23} -> no 5,6,7,8,9
10. n7 : 16(2) = {79} -> 7,9 locked in c2 and in n7

11. -> n7 : 8(2) = {26}/{35} = [2/5, 3/6 ...]-> no 1
12. 7(2) cage in N7 = {16/34} (not {25} step 11)-> no 2 or 5 -> 7(2) cage = [3/6...]
13. Killer pair on 3 and 6 in N7 in 8(2) and 7(2) cages -> no 3 or 6 elsewhere in N7
14. "45" on N7 -> 1 outie + 3 = 1 innie -> min. r7c3 = 4 -> no 1 or 2. r7c3 = {458} -> r6c1 = {125}
15. 11(3) cage in N47 now {128/245} = 2{18/45} -> 2 in c1 locked in -> no 4 or 6 in r3c2
16. In N9 the two 15(2) cages take {69/78} between them -> no 6,7,8 and 9 elsewhere in N9
17. "45" on N9 -> i innie = 1 outie -> no 6 or 7 in r6c9
18. 10(3) cage in N69 now {145/235} = 5{14/23} -> no 5 elsewhere c9
19. (from step 16) 6 in r7 now locked in N8 -> no 6 in 28(6) cage in N8 -> 28(6) = {123589/124579/134578} = 15{2389/2479/3478} -> no 1 or 5 in r7c456

20. "45" on r89 -> 4 innies = 17. Min. r8c2 = 7 -> max. r8c189 = 10 -> max 7 in any one cell -> no 8 in r8c1.
21. 8 in N7 now locked in r7 -> no 8 elsewhere in r7
22. "45" on N8 -> r7c456 = 17 -> h17(3) cage in N8 must have 6 (step 19) = 6{29/47} -> no 3
23. 3 in N8 locked in 28(6) cage = 1358{29/47}
24. 3 in r7 only in N9 -> no 3 elsewhere in N9 -> no 2 in r7c8
25. from step 22, 8 in 11(3) cage in N47 can only go in r7c1 -> no 1 in r7c1 (since 1 is only in {128} combination in 11(3) cage with 8 now locked in r7c1)
26. 1 in r7 now locked in N9 -> no 1 elsewhere in N9 -> no 4 in r7c8. 5(2) cage in N9 = [32/14]

27. "45" on r89 -> 4 innies = 17 -> 4 cells (r8c1289) = must sum from six candidates (124579). These six sum to 28 -> excess two candidates = 28-17= 11 [Edit:typo] which can only be {29/47} -> must have 1 and 5 -> r8c1 = 1, r8c9 = 5, r67c1 = 10 = [28]
28. "45" on N7 -> r7c3 = 5, r9c12 = 7 = {34} -> 3 and 4 locked for N7 and r9
29. 8(2) cage in N7 = {26} -> 2 and 6 locked for c3
30. r23c3 = 10 = {19/37}(no 4 or 8) = [1/3..] -> Killer pair on 1 and 3 with r1c3 -> 1 and 3 locked for N1 and c3
31. 9(3) cage in N14 = {126/234} (not {135} since 1 and 3 are only in r4c2) ->r3c12 = [62/42] -> r3c2 = 2, r3c1 = {46}
32. r67c9 =5 = {14}/[32]
33. "45" on N9 -> 2 innies = 5 -> r7c79 = {14}/[32]
34. r6c23 + r7c4 = 20 = [389/4{79}] ->r6c2 = {34}, r7c3 = {79}
35. {34} naked pair r69c -> 3 and 4 locked for c2 -> r4c2 = 1, r3c1 = 6
36. "45" on c12 -> 1 outie - 1 = 1 innie -> r5c3 = 4, r6c2 = 3
36. "45" on c1234 -> r89c4 = 9 -> r8c4 = {2478}, r9c4 = {1257}
37. "45" on c5 -> r89c5 = 8 -> r8c5 = {37}, r9c5 = {15}
38. "45" on c6789 -> r89c7 = 11 -> r8c6 = {2349}, r9c6 = {2789}

and the rest goes from there - feeling too tired to go any further - so back to Peter's every cell

walk-through
Edited to add in a few more steps at the end. Too tired to do any more
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Post by Andrew »

I've just looked through Peter's walkthrough for Ruud's original version of this puzzle. Nice walkthough.

A lot of it is the same as the way that I solved it but there were some differences that may be of interest so here is my walkthrough. I think I missed seeing a few of Peter's steps and it's possible that he may also have missed some of mine.

Step 1
R1C34 = {13}, 45 rule on N1 2 outies = 4 -> R4C2 = {13}, R1C67 = {26}

Step 2
R9C89 = {69/78}, R89C7 = {69/78}, no other 6, 7, 8 or 9 in N9, R78C2 = {79}, R89C3 = {26/35}, 45 rule on R9 2 innies R9C37 = 10 -> R9C3 = {23}, R9C7 = {78}, R8C3 = {56}, R8C7 = {78}, R9C89 = {69}, R9C12 = {25/34}, killer pair 2/3 in R9C123, no other 2/3 in R9 or N7, 1 in R9 locked in N8 -> 13(3) cage = 1{48/57}

Step 3
45 rule on N7 1 innie – 1 outie = 3, 8 in N7 is in R78C1 or in R7C3
8 cannot be in R7C3 as follows. If R7C3 = 8, R6C1 = 5, R78C1 = {24} which is not possible as these clash with {25/34} in R9C12
-> 8 in R78C1, other two cells of R678C1 = {12}, R7C3 = {45}, R78C1 = {18} (hidden single 1 in N7), R6C1 = 2, R7C3 = 5, R89C3 = [62], R9C12 = {34}, R9C456 = {157}, R89C7 = [78], R78C2 = [79]

Step 4
15(3) cage in N8 must be {348} -> R78C1 = [81], R78C8 = [32], R8C9 = 5 (hidden single in N9), R7C79 = {14}, R6C9 = {14}, R7C456 = {269}

Step 5
R8C456 = {348} from step 4 and R9C456 = {157} from step 3
45 rule on C56789 2 outies R89C4 = 9
45 rule on C12345 2 outies R89C6 = 11 -> R89C5 = 8
R89C6 = 11 -> [47], R89C5 = 8 -> [35] -> R89C4 = [81], R1C34 = [13], R4C2 = 1, R56C6 = {15}, R4C6 = 3 (hidden single in N5), R56C4 = [27]
45 rule on N3 1 innie – 1 outie = 1 R1C7 = {26} (step 1) -> R4C8 = {15} but R4C2 = 1 -> R4C8 = 5, R1C7 = 6, R1C6 = 2, R7C5 = 2, R56C5 = {69}, R4C45 = [48]

Step 6
R23C3 = {37}, R3C12 = [62], 26(4) cage in N1 = {4589}, 7 in N4 locked in R45C1

Step 7
45 rule on C4 1 innie = 1 outie R4C3 = R7C4, R2347C4 = {4569}, R7C4 = R4C3 = 9 (4 and 5 in R7, 6 in C3), R34C4 = [65], R7C6 = 6, R6C23 = [38], R9C12 = [34], R12C2 = {58}, R12C1 = {49}
22(4) cage in N4 = {4567} -> R5C3 = 4, R5C2 = 6, R45C1 = [75], R56C6 = [15]
45 rule on R1234 2 innies R4C19 = 13 -> R4C9 = 6, R4C7 = 2, R23C6 = {89}

Step 8
R56C5 = [96], R9C89 = [69], R123C5 = {147}, R23C7 = [51], R5C7 = 3 (hidden single in C7), R5C89 = {78}, R7C7 = 4, R7C9 = 1, R6C9 = 4, R6C8 = 1 (hidden single in N5), R6C7 = 9

Step 10
R12C2 = [58], R23C6 = [98], R3C89 = [93], 21(4) cage in N3 = {2478} and carry on … with simple elimination

Andrew
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Post by Oscar »

Direct solving only specific tip would be: Establish the link between couples of A5(3) and 13(3) at N8, which becomes clear when you see the V2 version.
I would rate it 1.0
Nothing can both be and not be
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