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PsyMar
Hooked Joined: 17 Nov 2006
Posts: 32
Location: The Triad, North Carolina, US Posted: Fri Nov 17, 2006 5:49 pm    Post subject: Assassin 25 I found this one pretty easy -- solved it in under half an hour, without any hints. I did use a program to highlight possibilities (e.g. all cells with a 1 in them, all cells with a 2 in them, etc) and do the arithmetic for me, but that's all (I still had to tell it what rows/columns/nonets I wanted innies/outies of, and for more complex things I used a calculator and a couple of times the killer configuration tables on the site). Walkthrough coming up (I'm solving it again to write one, I didn't keep track of my steps the first time through)   PsyMar
Hooked Joined: 17 Nov 2006
Posts: 32
Location: The Triad, North Carolina, US Posted: Fri Nov 17, 2006 6:35 pm    Post subject: walkthrough Walkthrough follows (edit: no longer tinytext): 4/2 in C1 = {13} (naked pair) 4/2 in C4 = {13} (naked pair) 23/3 in N4 = {689} (naked triple) Innies of C6789 = R48C6 = 17/2 = {89} (naked pair) Innies of C6 = R15679C6 = 15/5 = {12345} (naked quintuple!) -> 13/2 in C6 = {67} (naked pair) Innies of N2 = R1C46 = 9/2 without 1367 = {45} (naked pair) -> 19/3 in C5 = {289} (naked triple) 5/2 in C5 -> {14} (naked pair) Innies of C1234 = R48C4 = 9 = {27|45} but {45} would contradict R1C4 so R48C4 = {27} (naked pair) 6/2 in R9 = {15|24} -> 3 in C6 locked in cage 20/5 -> R45C7 <> 3 21/3 in R1 = {579|489} -> 9 not in rest of R1 9/3 in R1 = {135|234} -> 3 not in rest of R1 Innies of R1 = R1C159 = 15 with no 3459 = {168|267} -> R1C19 = {16|67} 15/2 in R9 = [69|78|96] Outies of R123 = R4C1289 = 10 = {1234} (naked quad) -> R4C4 = 7 -> R8C4 = 2, R4C3 = 5 -> R4C5 = 6 21/4 in N8 = {2379} -> R8C6 = 9 -> 26/4 in N5 =  -> R4C7 = 9 20/5 in C67 = 9/1+11/4 = {12359} -> R7C6 = 5 -> R1C6 = 4 -> R1C4 = 5, 6/2 in R9 = , 5/2 in C5 =  Innies of N9 = R7C8 = 3 -> 4/2 in C1 =  -> 21/4 in N8 =  Innies of C789 = R5C7 = 1 9/3 in R1 = {135} -> R1C5 = 2 (hidden single) R1C23 = {13} (naked pair) 21/3 in R1 =  Innies of N6 = R4C89 = 5 =  -> R4C12 =  -> R1C23 =  28/5 in C34 = 5+{49}+2|3|7+6|8 -> R5C3+R7C4 = 10 =  -> 15/2 in R9 =  -> 20/5 in C67 =  -> 16/3 in R67 =  -> R7C3 = 7 -> R7C7 = 2 -> 10/2 in C9 =  15/4 in R34C12 = 4+1+10/2 -> R3C12 = {28} (naked pair) -> R23C3 = {46} (naked pair) -> R1C1 = 7, 26/5 in N7 =  -> R1C9 = 6, R3C12 =  -> R2C12 = , R5C2 = 8, 19/3 in C5 =  14/4 in R34C89 -> R3C89 = 9 = {45} (naked pair) -> lots of naked singles R6C147 = naked triple {469} 15/3 in R67C78 =  -> naked singles solve it 731524896 594386712 826197354 415768923 982453167 673912485 167845239 358279641 249631578 Thanks to SudokuEd and Andrew for the proofreads. Hopefully fixed now.Last edited by PsyMar on Sun Nov 26, 2006 6:38 pm; edited 3 times in total   sudokuEd
Grandmaster Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia Posted: Sat Nov 18, 2006 1:15 pm    Post subject: Great walk-through PsyMar. Ruud must be going soft on us this week after last week's Tetris. Thought I'd make a V2 - the same solution, just a few cages changed (the new ones have diagonal cells in n2 - the 26(5)n12 crosses from r1c4 to r2c5; the 27(6)n23 crosses from r1c5 to r2c4). I found this one a lot harder. But still has a logical solve path. Hope you like to do harder Killers PsyMar Here is the text code for Assassin 25V2. Would someone mind setting a picture version please? Thanks. 3x3::k:7936:6657:6657:6657:6917:6917:6917:6917:4872:7936:7936:7936:6917:6657:3342:4872:4872:4872:3858:3858:7936:6917:6657:3342:4872:3609:3609:3858:3858:7197:6686:6686:6686:5153:3609:3609:5924:5924:7197:7197:6686:5153:5153:4651:4651:5924:4142:4142:7197:1329:5153:3891:3891:4651:1078:4142:6712:7197:1329:5153:6972:3891:2622:1078:6712:6712:5442:5442:5442:6972:6972:2622:6712:6712:3914:3914:5442:1613:1613:6972:6972:Last edited by sudokuEd on Mon Nov 20, 2006 2:44 am; edited 1 time in total   frank
Regular Joined: 07 Oct 2006
Posts: 10 Posted: Sat Nov 18, 2006 4:23 pm    Post subject:  Last edited by frank on Tue Nov 21, 2006 10:55 pm; edited 1 time in total   PsyMar
Hooked Joined: 17 Nov 2006
Posts: 32
Location: The Triad, North Carolina, US Posted: Sat Nov 18, 2006 5:03 pm    Post subject: re: Assassin 25v2 sudokuEd wrote: Great walk-through PsyMar.

Thanks! It's actually my first.

 Quote: Ruud must be going soft on us this week after last week's Tetris.

Probably -- I tried that one and couldn't get much of anywhere. As it is I had a bit of trouble with today's One-Trick Pony.

 Quote: Hope you like to do harder Killers PsyMar I enjoy trying them, at least. I'll see just how much harder this is...

Edit: It's quite a bit harder. Took several hours. Can't do any more today as I have too much homework, but here's a walkthrough for v2...

1. 4/2 in C1 = {13} (naked pair)
2. 5/2 in C5 = {14|23}
3. 6/2 in R9 = {15|24}
4. 10/2 in C9 <> 5
5. 13/2 in C6 = {49|58|67}
6. 15/2 in R9 = {69|78}
7. 23/3 in N4 = {689} (naked triple)
8. 14/4 in R34C89 <> 9 (min with 9 is 1+2+3+9 = 15)
9. 26/4 in N5 <> 1 (max with 1 is 1+7+8+9 = 25)
10. 19/5 in N3 = {1...} -> not elsewhere in N3
11. 31/5 in N1 = {9...} -> not elsewhere in N1
12. Innies of C6789 = R1C678+R48C6 = 38
13. R1C5678+R23C4 = 27, R1C5+R23C4 >= 6 -> 21 >= R1C678
14. combine 12 and 13: R1C678+R48C6+21 >= 38+R1C678 -> R48C6 >= 17 -> R48C6 = 17 = {89} (naked pair)
15. Combine 12 and 14: R1C678 = 38-17 = 21/3 split cage; R1C5+R23C4 = 6/3 split cage = {123} (naked triple)
16. 13/2 in C6 = {67} (naked pair)
17. Outies/Innies of N2 -> R1C23 = R1C6 -> R1C23 = {13|23} (cannot be {14} because 4 is either in R1 or cage 26/5)
18. 3 locked in R1C23 -> 3 not in rest of R1 or N1
19. R1C235 = naked triple {123}
20. split cage 21/3 in R1C678 = [489|498|579|597] -> no 9 in rest of R1 or N3
21. 3 of N2 locked in C4 -> not in rest of C4
22. 9 of N2 locked in C5 -> not in rest of C5
23. Outies of N3 = 9 -> R4C89 = {13|14|23}
24. Outies of N1 = R1C4+R23C5+R4C12 = 27, R1C4+R23C5 = {(4|5)89} = 21|22 -> R4C12 = 27-21|27-22 = 5|6 -> R4C1 <> 7 && R4C2 <> 5|7
25. Innies of N7 = 15 -> R7C2+R9C3 = {69|78}
26. Innies of N9 = 8 -> R7C8+R9C2 = [71|62|35] -> R9C6 <> 2
27. Innies of N58 = R5679C46 = 38; R5679C6 = 10|11 -> R5679C4 = 27|28 = {9...} <> {(1|2)...} -> 9 not in rest of C4
28. C5: 1 and 2 must be in R167C5 -> not in rest of column
29. 2 and 3 of C6 locked in cage 20/5 in C67 -> R45C7 <> 2|3
30. R567C6 = {123|234|235} = 6|9|10 -> R45C7 = 10|11|14 = {19|46|56|59|68} <> 7
31. If R6C2 = 1 then R6C3+R7C2 = 15 -> R6C3 = 7 and R7C2 = 8 -> R9C3 = 7, CON
32. Outies of R123 = R4C1289 = 10 (why didn't I see this earlier?) = {1234} (naked quad)
33. Quints for 28/5 in C34 -> R5C3 = {123}
34. R4C3+R6C23 = {57...} -> R6C4 <> 5|7
35. R238C4 = hidden triplet {123}
36. Innies of C1234 = R1C234+R4C4 = 16, R1C234 = {135|138|234|238} = 9|12|13 -> R4C4 = 7 (First number placed finally!) -> R4C3 = 5
37. R1C234 = 9 = {135|234} -> 8 of N2 locked in C5 -> not in rest of C5 -> R4C5 = 6
38. R23C5 = hidden pair on {89} in N2
39. R79C4+R8C6 = slightly hidden triple {689} in N8
40. R1C4 = hidden single on 5 in C4 -> R1C6 = 4 -> R1C23 = {13} naked pair, R1C78 = {89} naked pair
41. R14C7 = naked pair {89}
42. Outies of N1 = R4C12 = 5 = [23|41] -> R14C2 = naked pair {13} -> R1C5 = 2
43. R4C6+R56C4 = naked triple {489}
44. 26/4 in N5 =  -> R4C7 = 9, R8C6 = 9 -> R1C78 = 
45. 6/2 in R9 = {15} naked pair
46. 7 of N4 locked in R6C23 -> R6C789+R7C2 != 7
47. 5/2 in C5 = 
48. 20/5 in C67: R5C7 = 1, R7C6 = 5 -> 6/2 in R9 =  -> R8C4 = 2
49. innies of N9 = R7C8 = 3 -> 4/2 in C1 =  -> 21/4 in N8 = 
50. R4C2 = hidden single on 1 -> R1C23 = 
51. 1 of N3 locked in R2 -> R23C4 = 
52. Outies of N1 = R4C1 = 4 -> R4C89 = 
53. 15/3 in R67C78 =  -> LOTS of naked singles
54. 14/4 -> R3C8 = 5
55. 28/5 in C34 =  -> five more naked singles
56. 15/2 in R9 =  -> naked singles solve it
731524896
594386712
826197354
415768923
982453167
673912485
167845239
358279641
249631578

Whoo! What a rush. My head is spinning. But it's done.    Andrew
Grandmaster Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta Posted: Wed Mar 17, 2010 5:20 am    Post subject: PsyMar was so quick to post his walkthrough for A25 that I never got to post my one. I only had a quick look at Ed's A25V2 at the time but now I'm working through my backlog of unfinished puzzles I've had another go at it. Once I'd found the 45s in steps 6 and 8, particularly step 8, I found this puzzle straightforward. I'll rate A25V2 at Hard 1.0, rather than 1.0, because the innies-outies for step 8 were fairly well hidden. Here is my walkthrough for A25V2. Prelims a) R23C6 = {49/58/67}, no 1,2,3 b) R67C5 = {14/23} c) R78C1 = {13} d) R78C9 = {19/28/37/46}, no 5 e) R9C23 = {69/78} f) R9C67 = {15/24} g) 23(3) cage in N6 = {689} h) 14(4) cage at R3C8 = {1238/1247/1256/1346/2345}, no 9 i) 26(4) cage in N4 = {2789/3689/4589/4679/5678}, no 1 Steps resulting from Prelims 1a. Naked pair {13} in R78C1, locked for C1 and N7 1b. Naked triple {689} in 23(3) cage, locked for N4 1c. 31(5) cage in N1 must contain 9, locked for N1 1d. 19(5) cage in N3 must contain 1, locked for N3 2. 45 rule on N7 2 innies R7C2 + R9C3 = 15 = {69/78} 3. 45 rule on R123 4 outies R4C1289= 10 = {1234}, locked for R4 4. 45 rule on N1 2 outies R4C12 = 2 innies R1C23 + 1 4a. Max R4C12 = 7 -> max R1C23 = 6, no 6,7,8,9 in R1C23 5. 45 rule on R1234 2 innies R4C37 = 1 outie R5C5 + 9 5a. Max R4C37 = 16 -> max R5C5 = 7 6. 45 rule on C1234 2 outies R23C5 = 4 innies R2348C4 + 4 6a. Min R2348C4 = 11 (cannot be 10 because R4C4 doesnt contain any of 1,2,3,4) -> min R23C5 = 15, no 1,2,3,4,5 in R23C5 6b. Max R23C5 = 17 -> max R2348C4 = 13, no 8,9, 1 locked for C4 6c. R4C4 = {567} -> no 5,6,7 in R238C4 7. 45 rule on N3 2 innies R1C78 = 2 outies R4C89 + 12 7a. Min R4C89 = 3 -> min R1C78 = 15, no 2,3,4,5 8. 45 rule on C6789 2 innies R48C6 = 3 outies R1C5 + R23C4 + 11 8a. Min R1C5 + R23C4 = 6 -> min R48C6 = 17 -> R48C6 = 17 = {89}, locked for C6, R1C5 + R23C6 = 6 = {123}, locked for N2 and 27(6) cage at R1C5, clean-up: no 4,5 in R23C6 9. Naked pair {67} in R23C6, locked for C6 and N2 9a. Naked pair {89} in R23C5, locked for C5 and N2 9b. Naked pair {45} in R1C46, locked for R1 9c. Naked triple {123} in R1C235, locked for R1 10. R1C5 + R23C4 = 6 (step 8a) -> R1C678 = 21 = {489/579} (cannot be {678} because R1C6 only contains 4,5), no 6, 9 locked for R1 and N3 11. R23C5 = {89} = 17 -> R1C234 = 9 = {135/234}, 3 locked for R1 and N1 11a. 3 in N2 only in R23C4, locked for C4 11b. R23C5 = R2348C4 + 4 (step 6) -> R2348C4 = 13 = {1237} (cannot be {1345} which clashes with R1C4) -> R238C4 = {123}, 2 locked for C4, R4C4 = 7, R4C3 = 5, R4C5 = 6, clean-up: no 8 in R9C3 12. R1C4 = 5 (hidden single in C4), R1C6 = 4, R1C23 (step 11) = {13}, locked for R1 and N1 -> R1C5 = 2, clean-up: no 7 in R1C78 (step 10), no 3 in R67C5, no 2 in R9C7 12a. Naked pair {89} in R1C78, locked for R1 and N3 13. Naked pair {13} in R23C4, locked for C4 -> R8C4 = 2, clean-up: no 8 in R7C9, no 4 in R9C7 13a. Naked pair {15} in R9C67, locked for R9 14. Naked pair {14} in R67C5, locked for C5 14a. 26(4) cage in N5 = {5678} (only remaining combination) -> R5C5 = 5, R4C6 = 8, R4C7 = 9, R1C78 = , R8C6 = 9, clean-up: no 1 in R7C9, no 6 in R9C3 15. Naked pair {49} in R56C4, locked for C4, N5 and 28(5) cage at R4C3 -> R6C5 = 1, R7C5 = 4, clean-up: no 6 in R8C9 15a. R4C3 = 5, R56C4 = {49} = 13 -> R5C3 + R7C4 = 10 = , R9C4 = 6, R9C3 = 9 16. R56C6 =  = 5, R4C7 = 9 -> R5C7 + R7C6 = 6 = , R9C67 =  17. R4C1 = 4 17a. Naked pair {23} in R4C89, locked for R4, N6 and 14(4) cage at R3C8 -> R4C2 = 1, R1C23 = , R6C23 = , R7C2 = 6, R7C3 = 7, clean-up: no 3,4 in R8C9 18. R4C89 = {23} = 5 -> R3C89 = 9 = {45}, locked for R3 and N3 19. R4C12 =  = 5 -> R3C12 = 10 = {28}, locked for R3 and N1 20. 45 rule on N9 1 remaining innie R7C8 = 3, R7C7 = 2, R7C9 = 9, R8C9 = 1 21. R7C8 = 3 -> R6C78 = 12 =  and the rest is naked singles.   Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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