Here is a V2 that will take a team effort to crack. All killer solvers on my PC (including the latest JSudoku) give up and need to use backtracking to solve this one.
I think maybe Ruud (:twisted:) has found a new hobby. While creating his assassins, he dumps every assassin he figures is too difficult to put on the site on the forum and trying to give us nightmares in the process.
Not sure that we need a V2 for this one...but, of course....
PS4 gives Assassin 42(original) a rating of 33862. A 30000+, and with no zero's, usually means hard . Para's X-K has a similiar rating. Looks like my weekend's just got filled up.
Will have to abandon my normal contrarian support and hope Australia knocks over The Netherlands real quick.
Assassin 42V2
Last edited by sudokuEd on Fri Mar 16, 2007 8:54 pm, edited 1 time in total.
Here is the walk-through for Assassin 42 V1, not this monster above.
Was a nice puzzle, but didn't really get stuck anywhere.
Walk-Through Assassin 42
1. R12C1 = {16/25/34}
2. 35(5) in R1C2 = {56789} -->> R2C2: no 5,6,7,8,9; R2C1: no 5,6
2a. Clean up : R1C1: no {12}
3. R12C9 = {89} -->> locked for N3 and C9
4. R234C5 = {124} -->> locked for C5
5. 14(4) in R3C4 = {1238/1247/1256/1346/2345} : no 9
6. 29(4) in R3C6 = {5789}
7. R56C1 = {29/38/47/56}: no 1
8. R56C2 = {18/27/36/45} : no 9
9. 21(3) in R5C7 = {489/579/678} : no 1,2,3
10. R56C8 = {19/28/37/46} : no 5
11. R56C9 = {13} -->> locked for C9 and N6
11a. Clean up : R56C8 : no 7, 9
12. 45 on N1: 2 outies : R2C4 + R4C1 = 17 = {89} -->> R4C4 : no 8
12a. 5,6,7 locked in 35(5) in R1C2 for N1: no 5,6 and 7 anywhere else in N1
12b. R12C1 = {34} -->> locked for C1 and N1
12c. Clean up: R56C1: no 7,8
12d. 7 locked in C1 for N7
12e. Naked pair {89} in R2C49 -->> locked for R2
13. 3 in R4 locked in 14(4) in R3C4 -->> R3C4 : no 3
13a. 14(4) = {1238/1346/2345} : no 7
13b. 3 in R3 locked for N3
14. 29(4) in R3C6 can’t have both {89} in R4C678 because of R4C1 -->> R3C6 = {89}
14a. Naked Pair {89} in R2C4 + R3C6 -->> locked for N2
14b. Killer Pair {89} in R4C1 + R4C678 -->> locked for R4
14c. Killer Pair {89} in R3C12 + R3C6 -->> locked for R3
14d. {57} locked in ‘29(4) in R3C6’ in R4C567 -->> locked for R4: no 5,7 anywhere else in R4
15. 14(3) in R1C4 = {167/257/356}: {347} blocked by R1C1 -->> no 4
15a. Killer Triple {567} in R1C23 + R1C456 -->> locked for R1
16. 45 on N3: 2 outies: R2C6 + R4C9 = 9 = [36/54/72] : R2C6 = {357}
17. 45 on N7: 2 outies: R79C4 = 7 = {16/25/34} : no 7,8,9
18. 45 on N9: 2 outies: R79C6 = 7 = {16/25/34} : no 7,8,9
19. 45 on R89: 3 outies: R7C159 = 20 = {389/479/569/578}: no 1,2
19a. {389} not possible: only option for 3 is R7C5: no place for 8 or 9 in R7C9 -->> R7C5: no 3
20. 45 on C6789: 3 innies: R158C6 = 10 = {127/136/145/235} : no 8,9
21. 45 on C1: 3 outies: R239C2 = 14 = [185/194/284/293] -->> R3C2 = {89}; R9C2 = {345}
21a. Clean up: R3C1: no 8,9
22. 45 on C9: 3 outies: R239C8 = 9 = {126/135/234} : no 7,8,9
22a. R23C8 can’t be {12} or {24}: clashes with R1C78 -->> R9C8: no 6
22b. 45 on C9: 2 innies + 1 outie: R34C9 – R9C8 = 8 -->> min. R9C8 = 1; min R34C9 = 9 -->> R3C9: no 2
23. R23C8: no 2; R23C8 see all 2’s in N6
23a. 2 in N3 locked in 20(5) in R1C7. 20(5) = {12467/23456}: 4 and 6 locked in 20(5) for N3: R23C8: no 4, 6
23b. When 20(5) = {12467}, 7 in R2C6 -->> R23C7: no 7
23c. Hidden single 7 in R3C9
23d. Clean up : R9C8: no 2,4 (step 22b)
24. 16(4) in R7C9 = {1456/2356} -->> 5 locked in 16(4) for N9; 6 locked in16(4) for N9 and C9
24a. Clean up: R9C8: no 5 (step 22b)
24b. Naked triple {135} in R239C8 -->> locked for C8
25. 6 in C8 locked for N6 -->> 6 locked in 10(2) cage in R5C8 -->> R56C8 = {46} -->> locked for C8 and N6
25a. R1C8 = 2; R4C9 = 2; R9C8 = 1 (step 22b); R2C8 = 5; R3C8 = 3; R2C6 = 7 (step 16)
25b. R2C3 = 6; R3C3 = 5
25c. Naked Pair {14} in R12C7 -->> locked for C7; R3C7 = 6
26. 14{3} in R1C4 = {356} -->> locked for R1 and N2
26a. R12C1 = [43]; R12C7 = [14]
27. 7 locked in R4 for N6
27a. 21 in R5C7 = {489} -->> R6C6 = 4; R56C7 = {89} -->> locked for C7 and N6
27b.R4C5 = 1; R23C5 = [24]; R3C4 = 1; R2C2 = 1; R3C1 = 2
27c. R56C8 = [46]; R4C78 = [57]; R56C1 = [65](only option left for 11(2))
27d. R4C23 = {34} -->> locked for R4 and N4; R4C4 = 6
28. 12(3) in R7C6 = [129/138] -->> R7C6 = 1; R7C7: no 7
28a. R9C6 = 6 (step 18); Hidden single 6 in R1C5
28b. R8C1 = 1(hidden)
29. 18(4) in R5C4 = {2358}: no 7, 9 -->> {2358} locked for N5
29a. R4C6 = 9; R6C4 = 7; R3C6 = 8; R2C4 = 9; R2C9 = 8; R1C9 = 9; R4C1 = 8; R3C2 = 9
29b. R56C2 = [72]; R1C23 = [87]
29c. R9C2 = 4 (step 21); R4C23 = [34]; R9C9 = 5
This move is purely informative. No need to result to uniqueness based techniques but just something I noticed. Ed said he liked to see some other techniques in killer walk-throughs so here you go.
There is now a BUG-Lite move in R56C3579. If R5C5 = {38} there would be 2 ways to fill in cells R56C3579, so R5C5 can’t be 3 or 8 so must be 5.
Now to continue with the rest of the walk-through.
30. 18(4) in R8C2 = {2358} : no 6, 9
30a. R8C2 = 5; R7C2 = 6; R7C9 = 4; R8C9 = 6
30b. 13(3) in R7C2 = [625], so R7C34 = [25]; R7C78 = [38]; R8C8 = 9
30c. R1C46 = [35]; R9C4 = 2; R9C7 = 7; R9C1 = 9; R8C7 = 2; R8C6 = 3
30d. R89C3 = [83]; R8C45 = [47]; R79C5 = [98]; R7C1 = 7
30e. R6C5 = 3; R5C456 = [852]; R56C7 = [98]; R56C3 = [19]; R56C9 = [31]
Voilà, c’est fini.
Para
Last edited by Para on Thu Mar 22, 2007 11:24 pm, edited 3 times in total.
For quite a long way my solving path was fairly similar except for a couple of steps where I'd missed Para's extensions where cells "see" other cells. I must train myself to look out for those moves. I ought to have seen the "crossover" one at the end of Para's step 12 because I've used those before in Killer-Xs. The other type, used in Para's step 2, where a cell "points" into a different nonet are harder to spot.
In the later stages my path was quite a lot different so here is my walkthrough.
Clean-up is used in various steps, using the combinations in steps 1 and 6 for further eliminations from these two cell cages; it is also used for the two cell sub-cages that are produced by applying the 45 rule.
1. R12C1 = {16/25/34}, no 7,8,9
2. R12C9 = {89}, locked for C9 and N3
3. R56C1 = {29/38/47/56}, no 1
4. R56C2 = {18/27/36/45}, no 9
5. R56C8 = {19/28/37/46}, no 5
6. R56C9 = {13}, locked for C9 and N6, clean-up: no 7,9 in R56C8
7. R234C5 = {124}, locked for C5
8. 21(3) cage in N56 = {489/579/678}, no 1,2,3
9. 14(4) cage in N254 = {1238/1247/1256/1346/2345}, no 9
10. 29(4) cage in N256 = {5789}
11. 35(5) cage in N12 = {56789}
12. 20(5) cage in N23 = {12359/12368/12458/12467/13457/23456}, must contain three of 1,2,3,4
13. 31(5) cage in N8 = {16789/25789/34789/35689/45679}, 9 locked for N8
14. 16(4) cage in N9, R789C9 = {24567}, only valid combinations {1267/1456/2347/2356} -> R9C8 = {13}
32. 21(3) cage in N56 (step 8) = {489} -> R6C6 = 4, R56C7 = {89}, locked for C7 and N6, clean-up: no 5 in R5C2, no 3 in R79C6, no 2 in R56C8 = [46] -> R4C9 = 2, R4C5 = 1; clean-up: no 5 in R5C1, no 3 in R5C2, no 5 in R6C2
33. R23C5 = {24}, locked for N2
34. R2C4 + R3C6 = {89}, locked for N2
35. 3 in N2 locked in R1C456, locked for R1 -> R1C1 = 4, R2C1 = 3
35a. R1C456 = {356} (only valid combination), locked for R1 and N2 -> R2C6 = 7, R1C8 = 2, R1C7 = 1, R3C4 = 1, R3C1 = 2, R23C5 = [24], R2C8 = 5, R3C8 = 3, R3C7 = 6, R2C7 = 4, R3C9 = 7, R9C8 = 1, R4C78 = [57] (naked singles), R2C2 = 1, R3C3 = 5, R2C3 = 6 (hidden singles in N1), clean-up: no 8 in R56C2, no 6 in R7C6, no 9 in R56C1 = [65], clean-up: no 3 in R6C2
35b. R4C23 = {34}, locked for R4 and N4 -> R4C4 = 6
35c. R56C2 = {27}, locked for C2 and N4
I've started on Assassin 42V2 but haven't got too far into it.
Ed has suggested that I post the easier moves first to get it started and I'm expecting him to take it up from here.
It's some time since I've been involved in a tag solution so I hope I'll be able to make more contributions later.
Here are the starting moves, in tiny text while Assassin 42 is still live, because V2 uses the same cage pattern and therefore a lot of the moves also apply to V1 but using different numbers.
Edits in red. Thanks Ed and Para.
1. R12C1 = {69/78}
2. R12C9 = {17/26/35}, no 4,8,9
3. R56C1 = {49/58} (cannot be {67} which would clash with R12C1)
3a. Killer pair 8/9 in R1256C1 for C1
4. R56C2 = {69/78}
4a. Killer pair 8/9 in R56C12 for N4
5. R56C8 = {18/27/36/45}, no 9
6. R56C9 = {15/24}
7. 20(3) cage in R234C5 = {389/479/569/578}, no 1,2
8. 7(3) cage in N45 = {124}
9. 24(3) cage in R7C234 = {789}, locked for R7
10. 13(4) cage in N14 = {1237/1246/1345}, no 8,9
11. 26(4) cage in N36, no 1
12. 14(4) cage in N7, no 9
13. 45 rule on N7 2 outies R79C4 = 16 = {79}, locked for C4 and N8
14. 8 in R7 locked in R7C23, locked for N7
15. 45 rule on N9 2 outies R79C6 = 7 = {16/25/34}
15a. 8 in N8 locked in 22(5) cage
20. 45 rule on C9 2 innies R34C9 – 9 = 1 outie R9C8, max R34C9 = 17 -> max R9C8 = 8
Time to start thinking harder! Probably need to work the combinations more. It looks like there are still too many candidates to be able to do any useful contradiction moves yet.
I also noticed the following which I haven't included above because they don't (yet) lead directly to any eliminations
Valid combinations for 23(4) cage in N78 are {1679/2579/3479/3569}
Thanks for getting us started Andrew. Had to use my favourite combining steps and then doing hypotheticals to make any impression. Probably too many steps for a tag solution.
[edits:thanks Andrew]
21. 7(3)n4 = {124} -> no 4 r6c1
21a. no 9 r5c1
22. 8(2)c9 = {17/26/35}
22a. 6(2)c9 = {15/24}
22b. -> 2 locked for c9 in these 2 cages
23. 26(4) n3 = 9{278/368/458/467} ({5678} blocked by 8(2)n3)
24. 1 locked for r6 in r6c349. Here's how.
24a. r6c34 = {24} -> r6c9 = 1 ([15] in r56c9 blocked by 1 in r5c3)
24b. only other options for r6c34 = {12/14} include 1
24c. -> 1 locked for r6
24d. no 8 r5c8
25. 13(4)n1 = 1{237/246/345}(no 8,9)
25a. 15(2)n1 must have [8/9], not both
25b. -> 24(5)n1 must have [8/9] for n1 -> {24567} combo not possible
25c. 24(5)n1 cannot have both 8 and 9 because of 15(2) -> {12489} combo not possible
25d. 24(5)n1 cannot have both 7 and 9 because of 15(2) -> {12579/13479} blocked
26. 9 in c2 locked in 15(2) or r7c2
27. Now need some hypotheticals to make progress
27a. "45" n4 -> r6c4 + 10 = r4c123 = 11, 12 or 14 and must have 3
27b. r6c4 = 1 -> r4c123 = 11 = {137} ({236} blocked by 2 in r56c3)
27c. r6c4 = 2 -> r4c123 = 12 = {237} ({345} blocked by 4 in r56c3)
27d. r6c4 = 4 -> r4c123 = 14 = {347}
27e. r6c4 = 4 -> r4c123 = 14 = {356}
28. However, r4c123 = 14 = {356} is blocked. Here's how.
28a. 20(4)n2 now = {1478/1568/2378/2468/2567/3458/3467}
28b. "45" n1 -> 2 outies = 7.
28c. Since r6c4 = 4 (step 27e) in this hypothetical -> r4c1 != 3
28c. -> r4c23 = 3{5/6}
28d. the only combo's in 20(4) that allow 3{5/6} are {3458/3467} with {48/47} in r34c4
28e. but this means 2 4's in c4
28f. -> r4c123 cannot be {356}
29. r4c123 = {137/237/347} = 37{1/2/4}(no 5 or 6)
29a. no 12 r2c4
30. 7 Locked in r4c23 for n4, r4 and must be in 20(4)n2 = 7{148/238/256/346}
30a. no 6 r2c6 (step 18)
31. 13(2)n4 = {58}(hidden 5 n4): Locked n4, c1
32. 15(2)n1 and n4 = {69}:locked for c12, n14
33. 14(4)n7 = {1247/2345} = 24{17/35}: 2 and 4 locked n7
33a. 5 only in r9c2 -> no 3 r9c2
34. 23(4)n7 = {1679/3569} = 69{17/35}
35. 13(4)n1 = {1237/1345} = 13{27/45}
36. 24(5) n1 must have 8 for n1
36a. 13(4) must have both 1 and 3, only 1 of which can 'hide' in r4c1 -> {13578} blocked from 24(5) (note: 3 can't hide in r2c4 when 1 in r4c1 since 2 outies n1 = 7) 36aa. 24(5)n1 now = {12678/14568/23478/23568}
36b. 24(5), only combo's with 5 also have 6 {14568/23568} which is only in r2c4 -> no 5 r2c4
36c. no 2 r4c1
Last edited by sudokuEd on Fri Mar 23, 2007 9:59 am, edited 4 times in total.
39. Now try looking at those hypotheticals starting from R2C4 and the effect on the 20(4) cage in N254 which contains 7{148/238/256/346} (step 30)
39a. R2C4 = 3 -> R6C4 = 4, R4C1 = 4, R4C23 = {37}, 20(4) = {2378} ({3467} clashes with R6C4)
39b. R2C4 = 4 and R6C4 = 1 -> R4C1 = 3, R4C23 = {17} but 20(4) = {1478} clashes with R2C4 so cannot have R2C4 = 4 and R6C4 = 1
39c. R2C4 = 4 and R6C4 = 2 -> R4C1 = 3, R4C23 = {27}, 20(4) = 7{238/256}
39d. R2C4 = 6 -> R6C4 = 1, R4C1 = 1, R4C23 = {37}, 20(4) = {2378} ({3467} clashed with R2C4)
40. To summarise steps 38 and 39
40a. There is a one-one relationship between R2C4 and R6C4, R2C4 = 3,4,6 -> R6C4 = 4,2,1
40b. 20(4) cage in N254 = 7{238/256} = 27{38/56}, no 1,4
41. To continue further and look at the effect of these hypotheticals on the 24(5) cage in N12
41a. R2C4 = 3, R6C4 = 4, R56C3 = {12}, 24(5) cage = {23478} -> R1C2 = 2, R123C3 = {478}
41b. R2C4 = 4, R6C4 = 2, R56C3 = {14}, 24(5) cage = {23478} -> R1C2 = {2378}, R123C3 = {2378} In the next message Ed points out why there cannot be 8 in R1C2. Nice one Ed
41c. R2C4 = 6, R6C4 = 1, R56C3 = {24}, 24(5) cage = {12678} -> R1C2 = 2, R123C3 = {178}
41d. R2C4 = 6, R6C4 = 1, R56C3 = {24}, 24(5) cage = {14568} -> R1C2 = 4, R123C3 = {158}
41e. In summary, combining all these hypotheticals, R1C2 = {23478}, no 1,5, R123C3 = {1234578} See comment about step 41b. R1C2 = {2347}, no 1,5,8
Last edited by Andrew on Thu Mar 22, 2007 10:56 pm, edited 3 times in total.
Look good - only just missed the first placement. I see my hypothetical disease is contagious. Great work Andrew.
We are probably going to have to move to other areas now - one of Para's far-flung relationship between two digits in a cage combination, or one of Richard's wild "45"s.
44. 9(2) in R5C8 can't be {45} because of 6(2) in R5C9.
45. 45-test on C6789: R158C6 = 20: no options with 4 in R15C6
Two more steps
46. 1 locked in R12356C3.
46a. 24(5) = {12678/14568}: 1 locked in R123C3.
46b. 24(5) = {23478}: R2C4 = 3 -->> R4C1 = 4 -->> R56C3 = {12}
46c. 24(5) = {23478}: R2C4 = 4 -->> R4C1 = 3 -->> R4C3 = {27} -->> one of {27} in R123C3 : Killer Pair {27} in R1234C7 -->> R56C3 = {14}
46d. R89C3: no 1
46e. 23(4)in R8C2: only 1 in R8C2. So R8C2: no 7.
47. Building on 46-->> R1C2: no 3
47a. 24(5) = {12678/14568}: R1C2: no 3
47b. {23478}: R2C4 = 3: R1C2: no 3
47c. 24(5) = {23478}: R2C4 = 4 -->> R4C1 = 3 -->> R4C3 = {27} -->> one of {27} in R123C3: can't have both {27} in R123C3 so R1C2 = {27}: R1C2: no 3
greetings
Para
Last edited by Para on Tue Apr 03, 2007 2:31 pm, edited 2 times in total.
It was a good puzzle, but is definitely solvable. I don't regularly check this forum, so if there are specific questions email it to me - heartdr@northwestern.edu. I won't post the answer unless Ruud asks, and the only hint I will give is to use the left and right columns and work inwards, then use the 4 box totals in the lower left and right corners.
Good luck all.
just trying to catch up and see a couple of things. Hope I've picked up the marks pick correctly !! [Small edit to step 49 to clarify - thanks Andrew, Para)
48. 13(3)n2 - no 6,8 in r1c5
48a. only options with 8 are {148} -r1c45 must be {14} or {238} r1c5 must be 2
48b. only options with 6 is {256} - r1c5 must be 2
49. 45 on n5 - 5 innies total 23. no placement with 4 or 8 at r6c6
49a. {12479} - 7 must be at r6c6(thanks Andrew
49b. {14567} - 7 must be at r6c6
49c. {23459} - 3 must be at r6c6
49d {23468} - ditto
49e. {13469} - ditto
49f. {12569} - no 4 or 8
49g. {12578} - 7 must be at r6c6
49h {13568} - 3 must be at r6c6
49i. {12389} - ditto
50. 45 on n6. r4c789 minus r6c6 is 12 -> no 5 at r6c6 because:
50a. when r6c6 is 5, r4c789=17(3)={269} or {458}
50b. {269} would eliminate all possibilities for 6(2) and 9(2) n6 since -> 6(2)={15}, 9(2)=no options)
50c. {458} would eliminate all possibilies from 6(2)n6
That brings up the half century, which is generally more than England's batsmen can do.
Richard
Last edited by rcbroughton on Sat Mar 24, 2007 9:59 am, edited 2 times in total.
54. R6C6 = 7 -->> R4C789 = {289} (part of step 52b), R56C9 = {15}, R56C8 = {36}, R56C7 = {47} (4 hadn't been eliminated at step 52b) but 18(3) cage cannot be {477} -> R4C789 cannot be {289}
Then an extension to one of Richard’s moves
55. “49. 45 on n5 - 5 innies total 23. no placement with 4 or 8 at r6c6
49a. {12479} – 7 must be at r6c6
49b. {14567} - 7 must be at r6c6”
In both cases from step 52b as amended by step 54, R6C6 = 7 -->> R4C789 = {469} which clashes with {1249/1456} in R4C456 -> 5 innies in N5 cannot be {12479/14567} “49c. {23459} - 3 must be at r6c6
49e. {13469} – ditto
49i. {12389} – ditto”
In each of these cases from step 52d, R6C6 = 3 -->> R4C789 = {159/249} which clashes with the 9 in the N5 innies -> 5 innies in N5 cannot be {12389/13469/23459}
55a. The remaining combinations for 5 innies in N5 are {12569/12578/13568/23468}
Last edited by Andrew on Thu Mar 22, 2007 10:57 pm, edited 2 times in total.
3331
3840
6145:6145:5133:6150:6150:6672
3877
5671:5671:5671:4650
1580
6199:6199:6199:5690
2619
