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 Assassin 65 Goto page Previous  1, 2
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CathyW
Master

Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

 Posted: Thu Aug 30, 2007 12:17 pm    Post subject: Oh dear! Sorry about that! It is a really tough puzzle. Chin up though. Since you can manage the puzzles posted by members of the DJApe forum, you are surely as expert a solver as anyone else who attempts Ruud's Assassins, especially the variations.
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Thu Aug 30, 2007 2:03 pm    Post subject: Hi all This is how i finished the puzzle from Howards move 26. End of WT 65V2 Just being thorough: Prelim: 11(3) at R1C1 = {128/137/146/236/245}: no 9 21(3) at R1C8 = {489/579/678}: no 1,2,3 9(3) at R2C8 = {126/135/234}: no 7,8,9 22(3) at R3C1 = {589/679}: no 1,2,3,4; 9 locked for C1 20(3) at R8C3 = {389/479/569/678}: no 1,2 26(4) at R4C7 = {2789/3689/4589/4679/5678}: no 1 Clean up: 17(3) at R1C3 = {269/278/368/458/467}: no 1 6(2) at R3C6 = [15/42/51]: R3C6: no 2 12(3) at R6C1 = [192]/{17}[4]/{37}[2]/{26}[4]/{35}[4]: no 8 Let's begin. 27. 45 on N69: 2 outies: R3C9 + R8C6 = 13 -->> R3C9 = {46} 27a. Killer Triple {456} in 21(3) at R1C8 + 9(3) at R2C8 + R3C9 -->> locked for N3 27b. Clean up: 12(3) at R3C9 = {147/156/246/345}: no 8,9 28. 45 on R12: 3 innies: R2C258 = 9 = {126/135/234}: no 7,8,9 This move opens the whole puzzle up. 29. 9 in N3 locked within R12 29a. R2C46 = [69/78](R2C6 = R8C4 through R8C6) 29b. 17(3) at R1C3: {269} blocked, because then R2C4 = 6 -->> R2C6 = 9, which leave no room for 9 in N3 29c. 17(3) at R1C3 = {278/368/458/467}: no 9 29d. 9 in N1 locked for R3 29e. R2C6 = 9(hidden); R8C46 = [97]; R2C4 = 6; R7C1 = 4; R3c9 = 6 30. 2 in N5 locked for R4 30a. 12(3) at R3C9 = 6{15}(last combo): R4C89 = {15} -->> locked for R4 and N6 30b. R34C6 = [42]; R9C456 = [241] 31. Naked Pair {79} in R45C5 locked for C5 and N5 31a. R4C4 = 8; R1C46 = [58]; R1C5 = 1; R35C4 = [71]; R45C5 = [79] 32. 45 on N1: 1 innie and 1 outie: R4C2 + 4 = R3C1 -->> R3C1 = 8; R4C2 = 4(only possible combo) 32a. R4C1 = 9(hidden); R5C1 = 5 33. 17(3) at R1C3 = 6{47} (last combo) -->> R12C3 = {47}; locked for C3 and N1 33a. 11(3) at R1C1 = {236} (last combo) -->> locked for N1 34. 45 on R12: 3 innies: R2C258 = 9 = {135}({234} blocked by R2C2): no 2,4; locked for R2 34a. R2C1 = 2; R23C5 = [32]; R1C7 = 2(hidden); R2C7 = 7; R12C3 = [74]; R2c9 = 8 35. Naked Pair {15} in R24C8 -->> locked for C8 35a. R3C8 = 3 36. 26(4) at R4C7 = {3689/4679}: {2789} blocked by R4C7: no 2; 6 and 9 locked for N6 36a. R6C7 = 9(hidden) 37. 18(3) at R8C6 = 7{38/56} -->> R89C7 = {38/56}: no 1,4 37a. R5C7 = 4(hidden) 37b. 26(4) at R4C7 = 49{67} -->> R4C7 = 6; R5C8 = 7 37c. R89C7 = {38} -->> locked for C7 and N9 37d. R6C8 = 8(hidden) 38. 12(3) at R5C9 = {237} (last combo): locked for C9 -->> R7C9 = 7 39. 2 in C8 locked within 19(4) cage at R6C8 -->> 19(4) = 8{245}: no 1,6,9 -->> R7C78 = [52]; R8C8 = 4 And the rest is al naked singles(and 1 hidden single). greetings ParaLast edited by Para on Thu Aug 30, 2007 2:13 pm; edited 2 times in total
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Thu Aug 30, 2007 2:09 pm    Post subject:

 Howard S wrote: Cathy It is rubbish I will delete from 26 down This is beginning to get to me!!! Not sure I have earned my upgrade to Rookie Howard

I messed up the first time i posted a Walk-through here as well. Happens to all of us. We can't be perfect. Just see it as an initiation ritual .

And for that status upgrade, it just has to with how much you talk around here. So this grandmaster status of mine just means i don't know when to shut up.

greetings

Para
CathyW
Master

Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

 Posted: Thu Aug 30, 2007 3:16 pm    Post subject: Well done Para - you've made it look easy (maybe JSudoku was doing it the hard way!). Proved that Howard was correct so it wasn't rubbish at all - just needed explaining!
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Thu Aug 30, 2007 3:32 pm    Post subject:

 CathyW wrote: Well done Para - you've made it look easy (maybe JSudoku was doing it the hard way!). Proved that Howard was correct so it wasn't rubbish at all - just needed explaining!

In the Walk-throughs it always looks easier .

Step 29 isn't really a solver-step, more human creativity.

Didn't see Howards steps after step 26, so didn't know that.

greetings

Para

ps. That V3 is really deceiving, you think you're doing pretty well, placing digits, making eliminations and then all of the sudden you hit a dead end and there's seems to be no way out.
Howard S
Regular

Joined: 20 Aug 2007
Posts: 10
Location: Amsterdam

 Posted: Fri Aug 31, 2007 6:47 am    Post subject: Many thanks for all your help and encouragement. I have just been through Para's walk thru and one of my attempts was very close - seem to have my 15's mixed up in R23. Learnt a lot Cheers Howard
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

 Posted: Sat Sep 01, 2007 5:21 am    Post subject: I only finished Assassin 65 on Wednesday, not because it was particularly difficult but because of a mistake that I made. Then I worked through Cathy's and Para's walkthroughs this evening and decided my one is sufficiently different to post, even though it's more than a week since this puzzle was first posted. Cathy's Grouped X-Wing was a neat one. Like Para I also enjoyed step 26. My step 26 also used LoL but was otherwise completely different. Then I reached an impossible position because I’d missed out one permutation [167] in step 29b, which happened to be part of the solution; I’d worked out those permutations “in my head” rather than using Ruud’s excellent combination calculator as I usually do. That had given me 5 locked in 3 innies of N8 which was wrong. Fortunately, after finding my error, I then found step 30 which had the 5 locked in 5 innies and I was able to use most of my remaining moves. I'll rate this Assassin as 1.25. It's probably also 1.25 for the ways that Cathy and Para solved it. Here is my walkthrough for Assassin 65 1. R34C6 = {13}, locked for C6 2. R67C4 = {13}, locked for C4 2a. Naked pair {13} in R4C6 + R6C4, locked for N5 3. 8(3) cage at R1C7 = 1{25/34}, 1 locked in R12C7 for C7 and N3 4. 9(2) cage in N3 = {234} (only remaining combination), locked for N3 4a. Naked pair {15} in R12C7, locked for C7 and N3 -> R2C6 = 2 4b. 2 in N3 locked in R3C78, locked for R3 5. 24(3) cage in N3 = {789}, locked for N3 -> R3C9 = 6 5a. R4C89 = 11 = {29/38/47}, no 1,5 6. R345C1 = {289/379/469/478/568}, no 1 7. 21(3) cage at R8C3 = {489/579/678} 8. R9C456 = {289/379/469/478/568}, no 1 9. 45 rule on C789 1 remaining outie R8C6 = 4 -> R89C7 = 9 = {27/36}, no 8,9 10. CPE 8,9 in C7 locked in R4567C7 -> no 8,9 in R6C8 11. 45 rule on N5 4 innies R4C56 + R6C45 – 4 = 2 outies R3C4 + R7C6 11a. R4C6 + R6C4 = 4 (step 2a) -> R46C5 = R3C4 + R7C6 11b. 41(7) cage at R3C4 + R4C6 + R6C4 effectively form 45(9) cage -> R46C5 = R3C4 + R7C6 becomes LoL -> no 2 in R46C5 12. 45 rule on C123 2 outies R28C4 = 12 = [48]/{57}, no 6,9, no 8 in R2C4 13. 45 rule on N7 1 outie R8C4 – 4 = 1 innie R7C1 -> R7C1 = {134} 14. 45 rule on N7 3 innies R7C1 + R89C3 = 17 = {179/359/368/458/467} 14a. 4 of {458/467} must be in R7C1 -> no 4 in R9C3 14b. 21(3) cage at R8C3 (step 7) = {579/678} = 7{59/68}, CPE no 7 in R8C12 15. 45 rule on C9 3 outies R149C8 = 24 = {789}, locked for C8, clean-up: no 7,8,9 in R4C9 (step 5a) 15a. Min R9C8 = 7 -> max R89C9 = 5, R8C9 = {123}, R9C9 = {1234} 16. 21(4) cage at R6C8 = {1569/2469/2568/3459/3468/3567} (cannot be {1389/1479/1578/2379/2478} because 7,8,9 only in R7C7) -> R7C7 = {789} 17. 5 in C9 locked in R567C9 = 5{29/38/47}, no 1 17a. 1 in C9 locked in R89C9, locked for N9 17b. 12(3) cage in N9 = 1{29/38/47} 18. 45 rule on N9 1 innie R7C9 – 3 = 1 remaining outie R6C8 -> no 2,3 in R7C9, no 3 in R6C8 19. 45 rule on C89 2 outies R37C7 – 9 = 1 innie R5C8, max R37C7 = 13 -> max R5C8 = 4 19a. 5,6 in C8 locked in R678C8, 21(4) cage at R6C8 (step 16) = 56{19/28} (cannot be {3567} which clashes with R89C7), no 3,4,7, clean-up: no 7 in R7C9 (step 18) 19b. 4 in N9 locked in R79C9, locked for C9, clean-up: no 7 in R4C8 (step 5a) 20. 45 rule on R89 3 innies R8C258 = 10 = {127/136/235}, no 8,9 20a. 6 in {136} must be in R8C8 -> no 6 in R8C25 21. 9 in R8 locked in R13C8, locked for N7 22. 21(3) cage at R8C3 (step 14b) = {579/678} 22a. 9 of {579} only in R8C3 -> no 5 in R8C3 23. 45 rule on N6 5 innies R4C89 + R5C9 + R6C89 = 24, R4C89 = 11 (step 5a) -> R5C9 + R6C89 = 13 = {139/157/238/256} 23a. 1 of {157} and 6 of {256} only in R6C8 -> no 5 in R6C8, clean-up: no 8 in R7C9 (step 18) 23b. 5 in N6 locked in R56C9, locked for C9, clean-up: no 2 in R6C8 (step 18) 24. R567C9 (step 17) = 5{29/47} (cannot be {358} because R7C9 only contains 4,9), no 3,8 24a. 9 of {259} must be in R7C9 -> no 9 in R56C9 24b. 8 in C9 locked in R12C9, locked for N3 24c. R5C9 + R6C89 (step 23) = {157/256} 25. 4 in N6 locked in 21(4) cage at R4C7 = 4{179/269/368} (cannot be {2478} which clashes with R56C9) 26. 45 rule on N2 3 remaining innies R2C4 + R3C46 – 11 = 1 outie R4C5, min R4C5 = 4 -> min R2C4 + R3C46 = 15, max R2C4 + R3C6 = 10 -> min R3C4 = 5, clean-up: no 4 in R46C5 (LoL, step 11b) 26a. Min R4C5 = 5 -> min R2C4 + R3C46 = 16, max R2C4 + R3C6 = 10 -> no 5 in R3C4 26b. Max R2C4 + R3C46 = 19 -> max R4C5 = 8 26c. R4C5 = {5678} -> R2C4 + R3C46 = 16 or 17 or 18 or 19 = [493/583/781/593/791/783/793], no 7 in R3C4 27. 45 rule on R12 2 innies R2C258 = 13 = {139/148/346} (cannot be {157} which clashes with R2C7), no 5,7 28. 45 rule on C5 3 innies R159C5 = 11 = {128/137/146/236/245}, no 9 28a. 1 of {128/137} must be in R1C5 -> no 7,8 in R1C5 28b. 7 of {137} must be in R5C5 -> no 7 in R9C5 29. 45 rule on N8 3 remaining innies R7C46 + R8C4 – 5 = 1 outie R6C5, max R6C5 = 9 -> max R7C46 + R8C4 = 14, min R78C4 = 6 -> max R7C6 = 8 29a. Min R7C46 + R8C4 = 12 -> min R6C5 = 7 29b. R6C5 = {789} -> R7C46 + R8C4 = 12,13,14 = [165/157/175/158/167/185/365] 30. 45 rule on N8 5 remaining innies R7C456 + R8C45 = 22 = {12568/13567} (cannot be {12379} because R7C6 + R8C4 require two of 5,6,7,8 in the combination) = 156{28/37}, no 9, 5,6 locked for N8 30a. Taking the remaining combinations with the permutations in step 29b and avoiding repetition, R678C5 = [782/863/917/935/953/962/971], no 2 in R7C5 31. R9C456 = {289/379} = 9{28/37}, 9 locked for R9, clean-up: no 2 in R89C9 (step 17b) 32. Killer pair 7,8 in R9C456 and R9C8, locked for R9, clean-up: no 2 in R8C7 (step 9) 33. 21(3) cage at R8C3 (step 14b) = {579/678}, R9C3 = {56} -> no 5,6 in R8C34, 7 locked in R8C34 for R8, clean-up: no 1 in R7C1 (step 13), no 1 in R7C5 (step 30a), no 2 in R9C7 (step 9) 34. Naked pair {36} in R89C7, locked for C7 and N9 -> R8C9 = 1, R9C9 = 4, R9C8 = 7, R1C8 = 9, R4C8 = 8, R4C9 = 3, R7C9 = 9, R7C7 = 8, R6C8 = 6 (step 20a), R34C6 = [31], R67C4 = [31], R5C8 = 1 (hidden single in C8), clean-up: no 7 in R56C9 (step 24), no 7 in R6C5 (step 30a), no 7 in R7C5 (step 30a), no 3 in R9C5 (step 31) 35. Naked triple {235} in R8C258, locked for R8 -> R89C7 = [63] 35a. Naked triple {289} in R9C456, locked for R9 and N8 -> R8C4 = 7 -> R7C1 = 3 (step 13), clean-up: no 7 in R4C5 (LoL, step 11b) 35b. Naked pair {56} in R7C56, locked for R7 and N8 -> R78C8 = [25], R8C25 = [23], R3C7 = 2 (hidden single in R3) 36. R7C1 = 3 -> R6C12 = 10 = {19}/[28], no 4,5,7, no 8 in R6C1 36a. Killer pair 8,9 in R6C12 and R6C5, locked for R6 37. R159C5 (step 28) = {128/245} (cannot be {146} because only 2,8 in R9C5), no 6,7 38. Naked pair {56} in R47C5, locked for C5, clean-up: no 4 in R15C5 (step 37) -> R1C5 = 1, R12C7 = [51] [This naked pair has been there since the clean-up in step 35a but because I did step 37 first it was more effective.] 39. R3C5 = 7 (hidden single in C5), R24C5 = 10 = [46] (only remaining permutation), R2C4 = 5, R23C8 = [34], R7C56 = [56], R1C46 = [68] , R9C6 = 9, R12C9 = [78], R3C4 = 9, clean-up: R6C12 = [28] (step 36), R6C5 = 9 (LoL, step 11b) and the rest is naked singles and a couple of cage sums
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Sat Sep 01, 2007 10:41 pm    Post subject:

 Para wrote: Step 29 isn't really a solver-step, more human creativity.

Nice move, Para!

I agree with you here that it's extremely unlikely that an automated solver would (or could) find such a step. However, it's not inconceivable that a solver with a very good AIC/Nice Loop implementation could find the alternative move presented below that would serve as a replacement for your steps 29 - 29c:

 Code: .-----------------------.-----------.-----------------------------------.-----------.-----------------------. | 12345678    12345678  | 23456789  | 2568        134         2568      | 123789    | 456789      456789    | |           .-----------:           '-----------.-----------.-----------'           :-----------.           | | 12345678  | 123456    | 23456789    68        | 1234      | 79          123789    | 123456    | 456789    | :-----------:           '-----------.-----------:           :-----------.-----------'           :-----------: | 56789     | 123456789   123456789 | 156789    | 123479    | 145       | 123456      123456    | 46        | |           |           .-----------:           |           |           :-----------.-----------'           | | 56789     | 123456789 | 13456789  | 156789    | 1279      | 125       | 23456789  | 1234567     1234567   | |           :-----------'           |           '-----------'-----------:           '-----------.-----------: | 56789     | 123456789   123456789 | 156789      79          13568     | 23456789    23456789  | 123456789 | :-----------'-----------.           :-----------.-----------.           |           .-----------:           | | 123567      1235679   | 12356789  | 4         | 568       | 13568     | 2356789   | 12356789  | 12356789  | |           .-----------'-----------:           |           |           :-----------'           |           | | 24        | 12456789    12456789  | 3         | 568       | 568       | 12456789    12456789  | 12456789  | :-----------:           .-----------'-----------:           :-----------'-----------.           :-----------: | 12345678  | 12345678  | 345678      79        | 568       | 79          12345678  | 12345678  | 12345678  | |           '-----------:           .-----------'-----------'-----------.           :-----------'           | | 35678       356789    | 356789    | 12          124         124       | 356789    | 356789      356789    | '-----------------------'-----------'-----------------------------------'-----------'-----------------------'

29. Complex discontinuous Nice Loop (2 weak links at discontinuity) removes 9 from r12c3, as follows ("=>" implies strong link, "," implies direct link):
29a. if r12c3 = 9 -> r3c123 <> 9
29b. => r3c45 = 9 -> r2c6 <> 9
29c. => r2c6 = 7 -> r8c6 <> 7
29d. => r8c6 = 9 -> r8c4 <> 9
29e. => r8c4 = 7, r2c4 = 8 -> r12c3 <> 9 (contradiction)
29f. Conclusion: no 9 in r12c3

Note: Nice Loop notation avoided for clarity and due to difficulty representing direct link between 7 in r8c4 and 8 in r2c4.
_________________
Cheers,
Mike
Glyn
Major Major Major

Joined: 16 Jan 2007
Posts: 92
Location: London

Posted: Wed Sep 12, 2007 12:06 am    Post subject:

I did have a brief look at whether an automated solver could replicate Step 29 or similar. The problem is that most are geared up to follow a specific hierarchy of solving methods, and for programs with a killer ancestry this usually means slogging through all the cage combos first.

This is the reason why Jsudoku plods away for some time seemingly getting nowhere, then suddenly an avalanche of solves comes into play if a chink in the armour appears.

As an experiment I copied the candidate matrix into Jsudoku and gave it limited room for manoeuvre by only setting up three cages:- the 17(3) cage at r1c3 and hidden cages r28c4 (sum 15) and r28c6 (sum 16). It couldn't ignore a couple of simple moves, a naked pair on r9 and locked 2s in r4 for n5, but with nothing else left it had to go for chains using the direct links through the outties of r123.

The result before it stalls again is below, Jsudoku uses several chains and some cage combos which I list below with some annotation.

1) Grouped XY-Chain on 9 with 2 cells 1 links (9)R3C123=(9)R12C3-(8=6)R2C4-(7=9)R8C4 -> common buddies of R3C123 & R8C4 <> 9
(ie r3c4<>9)

2a)Complex XY-Chain on 6|7 with 2 links (6=7)Split cage 15/2 in R28C4-(7)R8C6=(7)R2C6 ->
-> R2C34 = {(6|~7)..}
(in essence as either r2c4=6 or r2c6=7).
Cage 17(3) (r12c3+r2c4) must be either
{29}|{38}|{47} with 6 or [72]|{36}|{45} with 8

2b)Complex XY-Chain on 8|9 with 2 links (8=9)Split cage 15/2 in R28C4-(9)R8C6=(9)R2C6 ->
-> R2C34 = {(8|~9)..}
(in essence as either r2c4=8 or r2c6=9)
This reduces the possibilities for cage 17(3) (r12c3+r2c4) to either
[92]|{38}|{47} with 6 or [72]|{36}|{45} with 8.

Conclusion of 2a) and 2b) r1c3<>2 and r2c3<>9.

3) Complex XY-Chain on 9 with 3 links (9)R3C123=(9)R1C3-(8=9)Split cage 15/2 in R28C4-(9)R8C6=(9)R2C6 -> common buddies of R3C123 & R2C6 <> 9.
(ie r3c5<>9).

The hidden single r2c6=9 in n2 then leads to a number of placements and other simple moves up to the point shown.

 Code: +------------------------------+------------------+-----------------------------+ | 12345678  12345678    3478   |  258  134   258  |  123789   456789    456789  | |  1234578   12345      3478   |   6   1234   9   |   12378    12345     4578   | |   5689    12345689  12345689 | 1578 12347  145  |  123456   123456      46    | +------------------------------+------------------+-----------------------------+ |   56789   13456789  13456789 | 1578  1279  125  |  3456789  134567    134567  | |   56789  123456789 123456789 | 1578   79  13568 | 23456789 23456789 123456789 | |  123567   1235679   12356789 |   4   568  13568 |  2356789 12356789  12356789 | +------------------------------+------------------+-----------------------------+ |    24     12456789  12456789 |   3   568   568  | 12456789 12456789  12456789 | |  1234568  1234568    34568   |   9   568    7   |  1234568  1234568  1234568  | |   35678    356789    356789  |  12   124   124  |  356789   356789    356789  | +------------------------------+------------------+-----------------------------+

I guess that a program could be instructed to exhaustively search for the most potent move rather than following the standard solving order, whereas it is fairly obvious to a human solver that something interesting could result from the interaction in the four innies of n258 once we had tired of pecking away at the combos. The program only took the bait when I gave it no other options.
_________________
I have 81 brain cells left, I think.
Caida
Hooked

Joined: 03 Nov 2007
Posts: 38

 Posted: Tue Nov 06, 2007 5:13 am    Post subject: Assassin 65v3 Hello, Edited to say:I have redone this puzzle with fresh eyes and managed to solve it (after looking for a hint in Afmob's walkthrough below!) I haven't found a walkthrough for Assasin 65v3. I've been pounding away at it for quite sometime and I think I'm going in circles. I've resorted to trial and error strategies and have wound up with conflicts at the end. Here is my walkthrough - I know the answer is wrong (not any more!!) - but hopefully the formatting is an improvement over my last attempt. Assassin 65V3 Preliminaries: a. 12(4)n14 = {1245/1236} (no 7..9) -> r1c2 no 1,2 (CPE) b. 9(3)n23 = {126/135/234} (no 7..9) c. 12(2)n25 = {39/48/57} (no 1,2,6) d. 22(3)n36 ={589/679} (no 1..4) -> r56c9 no 9 (CPE) e. 10(2)n58 = {19/28/37/46} (no 5) f. 11(3)n58 = {128/137/146/236/245} (no 9) g. 21(3)n89 = {489/579/678} (no 1..3) 1. Outies c123: r28c4 = 8(2) = {17/26/34} (no 4,8,9) 2. Outies c789: r28c6 = 11(2) = {29/38/56}/[47] 2a. -> r8c2 no 1; r8c6 no 4 3. Outies c1: r169 = 19(3) = {289/379/469/568} (no 1) 4. Outie and Innie n3: r3c9 - r2c6 = 7 4a. since max r3c9 = 9 the max r2c6 = 2 4b. -> r2c6 = 2; r3c9 = 9 4c. -> cleanup: 9(3)n23: r12c7 no 5 4d. -> h11(2)c6 (step 2): r8c6 = 9 4e. -> cleanup: 21(3)n89: r89c7 no 6 4f. -> cleanup: h8(2)c4 (step 1): r8c4 no 6 4g. -> cleanup: 12(2)n25 no 3 4h. -> cleanup: 10(2)n58: r6c4 no 1 5. Innies c5: r159c5 = 19(3) ={289/379/469/478/568} (no 1) 6. Outie and Innie n7: r8c4 - r7c1 = 2 6a. -> min r7c1 = 1 -> min r8c4 = 3 (r8c4 no 1,2) 6b. -> max r8c4 = 7 -> max r7c1 = 5 (r7c1 no 6,7,8,9) 6c. -> r6c1 no 2, 4 6d. -> cleanup: h8(2) (step 1): r2c4 no 6,7 6e. -> cleanup: 10(2)n58 no 3,7 (blocked by h8(2)c4 7. 17(3)n12: since max r2c4 = 5 the min r12c3 = 12 (no 1,2) 7a. 17(3)n47: since max r7c1 = 5 the min r6c12 = 12 (no 1,2) 8. Outie and Innie n6: r7c9 – r6c8 = 5 8a. -> min r6c8 = 1 -> min r7c9 = 6 (r7c9 no 1..5) 8b. -> max r7c9 = 8 -> max r6c8 = 3 (r6c8 no 4..9) 9. 14(3)n69: since min r7c9 = 6 the max r56c9 = 8 (no 8) 10. 40(7)n258 = {14/23}{56789} -> 9 locked in n5 in 40(7)n258 10a. cleanup: 10(2)n58: r7c4 no 1 11. 16(4)n69 = {1249/1348/1357/2356) -> contains either 4 or 5 Note: combos {1258/1456/2347} blocked by r89c7 11a. -> killer pair {45} locked in n9 for 16(4)n79 and 21(3)n89 11b. -> 12(3)n9 no 4,5 11c. -> 12(3)n9 = {129/138} no 6,7 -> 1 locked in 12(3) for n9 Note: combo {237} blocked by 16(4) and 21(3) 11d. -> 8 locked for n9 in 16(4), 21(3), and 12(3) -> r7c9 no 8 11e. cleanup: Innies & Outies n6 (step 8): r6c3 no 3 11f. cleanup: 12(3)n9 r9c8 no 2 (needs a 9) 11g. if r9c8 = 1 this would mean r89c9 = {38} and 16(4)n79 = [2]{356} not possible 11h. -> 1 locked for n9 and c9 in r89c9 12. 23(4)n6 = {1589/1679/2489/3479} Note: other combinations ({2579/2678/3569/3578/4568}) blocked by r4c89 12a. 7 locked for n6 in 23(4) and r4c89 -> r56c9 no 7 13. 14(3)n69 = {35}[6]/{34}[7] -> 3 locked for n6 and c9 in r56c9 Note: combo {25}[7] blocked by step 6 13a. -> r56c9 no 2,6 13b. cleanup: 12(3)n9: r9c8 no 8 14. Innies c7: r34567 = 26(5) = {12689/23489} Note: combo {24569} blocked by r12c7 (and r89c7) and combo {23579} blocked by 23(4)n6 14a. -> 8 locked for c7 in r34567 14b. -> r89c7 = {57} -> locked for c7 and n9 14c. -> single: r7c9 = 6 14d. -> r6c8 = 1 (step 8) 15. 14(3)n69 = {35}6 15a. -> r4c89 = [67] 15b. 4 locked in 14(3)n3 in r12c9 for n3 and c9 15c. -> r12c7 = {16} -> locked for c7 15d. cleanup: 14(3)n3 = {248} (only possible combination) 15e. single: r3c7 = 3 16. 40(7)n357 must contain 6 -> locked in r3c4 and n5 16a. -> r7c4 no 6 (CPE) 16b. cleanup: 10(2)n57: r7c4 no 4 16c. 10(2)n57 = {28} – locked for c4 16d. 40(7)n357 must contain 8 -> locked in r7c6 and n5 16e. -> r4c6 no 8 (CPE) 17. 16(3)n2 = {349/358/367/457} (no 1 – blocked by r1c7 and r3c6) 18. Innies r12: r2c259 = 17(3) = [197/395]/{46}[7] Note: combo [485] blocked by r2c9 18a. r2c2 no 5,8 18b. r2c5 no 1,3,5,7,8 19. Innies r89: r8c259 = 16(3) = {17}[8]/{268} -> 8 locked for r8 Note: combos {358/367/457} blocked by r8c47 19a. r8c259 no 3,4,5 19b. 4 locked in r7c78 and in r9c456 20. 12(3)n8 = {147/246/345} no 8 20a. r9c5 no 6 21. 16(4)n69 = 1{249/348} 21a. -> r7c78 no 2,8 22. pair {28} in r18c8 -> locked for c8 23. 11(3)n58: r6c5 no 5 (no 4 in r78c5) 24. Innies n8: This is where I borrowed from Afmob’s walkthrough - his step #15 (but written out in super long form so I can see it) - Thanks Afmob!!! 24a. r7c456+r8c45 = 24(5) 24b. r28c4 = 8(2) 24c. r7c456+r8c45-r28c4 = 16 24d. r7c456+r8c5-r2c4 = 16 24e. r2c4 = 1/3/5 -> r7c456+r8c5 = 17/19/21 (4) 24f. since r7c4 must be 2 or 8 then r8c5 <> 2 or 8 as this would mean that r7c56 = 7/9/11(2) and this isn’t possible with any remaining combination (two odd numbers add together to get an even number) 24g. r8c5 no 2,8 -> 8 locked for n8 and r7 in r7c456 25. 18(3)n7: r8c2 no 1 (b/c no 8 in r7c23) 25a. cleanup: h16(3) (step 19) r8c5 no 7 25b. killer pair {12}: h16(3)r8 = {268/178}; r8c9 = {1/2} -> {12} locked for r8 in c2589 26. 11(3)n58 = {28}1/{37}1/[416]/{23}6 (no 5) 26a. -> r6c5 no 6 26b. 6 in n5 locked in 40(7)n358 26c. ->r3c4 no 6 27. 14(3)n78: min r8c34 = 7 27a. -> max r9c3 = 7 (no 8,9) 27b. -> 8, 9 locked in r9 in 15(3)n7 and 12(3)n9 (each must have only 1 of these) 27c. 15(3)n7 = [519/618]/4{29}/5{28}/4{38} (no 7) 27d. -> r8c1 no 3,6; r9c12 no 5,6 27e. -> 3 locked in r8 in c45 28. 18(3)n7 = {279/378/567} -> 7 locked for n7 Note: combo {189} blocked by 15(3) (step 27b.); combo {369} blocked by r7c78 29. 14(3)n78 = {356} (only possible combination) 29a. r9c3 = 5,6 (no 3 b/c of step 27e.) 29b. hidden single: r8c1 = 4 29c. -> r9c12 = {29/38} no 1 29d. hidden single: r7c1 = 1 29e. -> r8c4 = 3 (step 6); r2c4 = 5 (h8(2)c4) 29f. -> {56} locked for n7 and c3 in r89c3 29g. hidden single: r7c6 = 5 29h. -> r4c5 = 5 29i. -> 12(2)n35 = [84] The rest is singles and cage sums - finally!!Last edited by Caida on Sun Dec 02, 2007 1:08 am; edited 1 time in total
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Sat Nov 10, 2007 4:23 pm    Post subject: After the Brick Wall I decided that I wanted to solve another tough Assassin and since this hasn't been solved yet I tried this one and succeeded. The main cracker is step 8h). Edit: Removed some steps, added step 8b) which is important for the following step 8h) and corrected some typos. Edit 2: Made some corrections suggested by Mike, thanks! A65 V3 Walkthrough: 1. C123 a) Outies C1 = 19(3) -> no 1 b) 12(4) = 12{36/45} -> R1C2 <> 2 c) Outies C123 = 8(2) = {17/26/35} 2. C789 a) Outies C789 = 11(2) = [29/38/47/56/65] b) 22(3) = 9{58/67} -> R56C9 <> 9 3. N7 a) Outies = 19(2+1): R6C12 <> 1,2 since R8C4 <= 7 b) Innies = 12(3): R7C1 <> 6,7,8,9 because R89C3 @ 14(3) >= 7 since R8C7 <= 7 c) Innies = 12(3): R8C4 <> 1,2 because R89C3 must be less than 12 d) Innies = 12(3): R7C1 <> 2 because R89C3 = 7/8/9/11 4. C456 a) Innies C5 = 19(3) -> no 1 b) Outies C123 = 8(2): R2C4 <> 6,7 c) 17(3): R12C3 <> 1,2 since R2C4 <= 5 5. N369 ! a) ! Outies N69 = 18(1+1) -> R3C9 = 9, R8C6 = 9 b) Innies+Outies N3: -7 = R2C6 - R3C9 -> R2C6 = 2 c) 9(3) = 2{16/34} d) Innies N6 = 9(3) -> no 7,8,9 6. C456 a) 21(3) = 9{48/57} b) 12(2) = {48/57} c) Outies C123 = 8(2) = [17/35/53] d) Killer pair (37) of Outies C123 blocks {37} of 10(2) e) 40(7) = 56789{14/23} -> 9 only possible @ N5 -> R4C5+R6C4 <> 9 f) 10(2) = {28/46} 7. N69 a) Innies+Outies N6: 5 = R7C9 - R6C8 -> R7C9 = (678), R6C8 = (123) b) 23(4) = 9{158/167/248/347} -> Other combos are blocked by Killer pairs (56,57,68) of 22(3) c) 16(4) = {1249/1267/1348/1357/2356} -> other combos blocked by Killer pairs (45,47,58) of 21(3) d) 16(4) <> {1267} since R7C9 would be 8 -> no combo for 21(3) e) Killer pair (45) locked in 16(4) + 21(3) for N9 f) 12(3) = {129/138/237} -> no 6 8. C789 ! a) ! Consider both possibilities of 21(3) -> 12(3) <> 7: - i) R89C7 = {57} -> 12(3) <> 7 - ii) R89C7 = {48} -> 16(4) = 35{17/26} together with R7C9 = (67) builds naked pair for N9 -> 12(3) <> 7 b) Outies C9 = 17(3): R9C8 <> 1 since R14C8 <= 15 c) 12(3) = 1{29/38} -> 1 locked for C9+N9 d) 14(3) @ N6 must have 6,7 xor 8 and R7C9 = (678) -> R56C9 <> 6 e) Innies N6 = 3{15/24} -> 3 locked for N6 f) 12(3): R9C8 <> 2 because R89C9 <> 9 g) Innies C7 = 26(5) = 29{168/348/357} because 2,9 locked in Innies and {24569} blocked by Killer pair (45) of 21(3) h) ! Combo analysis of Innies C7 and R456C7 @ 23(4) -> Innies C7 <> 5,7: - 23(4) = 9{158/167/248} and Innies = 29{168/348/357}, now consider possibilites for R456C7 - 3 candidates of Innies must be in R456C9 therefore 3 of those 5 candidates must be in 23(4) - Analysis for Innies C7 = {23579}: - i) 23(4) = {1589} has only 5,9 in common - ii) 23(4) = {1679} has only 7,9 in common - iii) 23(4) = {2489} has only 2,9 in common -> But it must have at least 3 candidates in common -> Innies C7 = 289{16/34} -> 8 locked for C7 9. N69 a) 21(3) = {579} -> {57} locked for N9 b) 16(4) = 14{29/38} -> R6C8 = 1 c) Hidden Single: R7C9 = 6 @ N9 d) 14(3) = 6{35}; {35} locked for C9+N6 e) 22(3) = {679} -> R4C8 = 6, R4C9 = 7 f) 12(3): R9C8 <> 8 since R89C9 <> 3 10. N3 a) 4 locked in C9 @ 14(3) -> locked for N3 and 14(3) = 4{28/37} b) 9 = {126} -> {16} locked for N3 c) 14(3) = {248} because (37) only possible @ R1C8; {248} locked and 2 locked for R1 d) 15(3) = {357} -> R3C7 = 3 11. N25 a) 6 locked in 40(7) between C4 and N5 -> R6C4 <> 6 b) 10(2) = {28} locked for C4 c) 8 locked in 40(7) between C6 and N5 -> R4C6 <> 8 d) 12(2) = [75/84] e) 16(3) <> 1 because R1C7 = (16) blocks {169} and R3C6 = (78) blocks {178} 12. R12 a) Innies = 17(3) must have 5 xor 7 since R2C8 = (57) -> R2C25 <> 5,7 b) Innies = 17(3) = {179/359/458/467}: R2C5 <> 1,3 because R2C28 <= 13 c) Innies = 17(3) <> 8 because {458} blocked by R2C9 = (48) 13. R789 a) Innies R89 = 16(3) = {178/268/358/367} -> no 4 since {457} blocked by R8C7 = (57) b) 4 locked in 16(4) in N9 for R7 c) 4 locked in 12(3) for R9 -> 12(3) = 4{17/26/35} d) 16(4): R7C78 <> 2 because R8C8 <> 4,9 e) 18(3): R8C2 <> 5 because R7C23 <> 6 f) 11(3): R6C5 <> 5 because R78C5 <> 4 14. N6 a) 2 locked in 23(4) @ C7 -> R5C8 <> 2 15. N258 ! a) ! Innies+Outies N25: 16 = R7C456+R8C5 - R2C4; R2C4 = (135) -> R7C456+R8C5 = 17/19/21 -> R8C5 <> 2,8 because together with R7C4 = (28) R7C56 would be 7/9/11(2) but that's not possible because only remaining candidates in R7C56 are (1357) -> sum would be even but it must be odd b) 8 locked in R7C456 for R7 16. C789 a) 8 locked in 23(4) @ C7 -> R5C8 <> 8 b) Hidden pair (28) in R18C8 @ C8 -> no other candidates 17. R89 a) Innies = 16(3) = 8{17/26/35} because R8C8 = (28) -> 8 locked for R8 b) Innies = 16(3): R8C5 <> 3 because it's the only place where 5 is possible c) Innies = 16(3): R8C2 <> 6 because R8C5 <> 2,8 d) 18(3) = {189/279/378} e) 18(3): R8C2 <> 1,3 because R7C23 <> 8 f) Innies = 16(3) = 8{17/26}: R8C5 <> 7 because R8C28 <> 1 g) 11(3) <> 5 because R8C5 = (16) h) Killer pair (12) of Innies + R8C9 locked for R8 i) 14(3): R9C3 <> 8,9 because R8C34 >= 7 j) 15(3) must have 8 xor 9 since 12(3) @ R9 has one of (89) and it's nowhere else possible in R9 -> 15(3) <> 7 k) 18(3) <> 1 because 15(3) @ N7 must have one of (89) l) 18(3) = 7{29/38} -> 7 locked for N7 18. N5 a) 11(3): R6C5 <> 6 because R78C5 <> 4 and R8C5 <> 2,3 b) 6 locked in 40(7) -> R3C4 <> 6 19. N7 a) 15(3): R8C1 <> 3 because R9C12 <> 4 b) 15(3) must have 4,5 xor 6 and R8C1 = (456) -> R9C12 <> 5,6 c) 3 locked in 14(3) for R8 -> R9C3 <> 3; 14(3) = 3{47/56} d) 14(3) = {356} because R9C3 = (56) e) Hidden Single: R8C1 = 4 f) 15(3) = 4{29/38} g) Innies = 12(3) = {156} -> R7C1 = 1 h) 14(3) = {356} -> R8C4 = 3; {56} locked for C3 i) 17(3) = {179} -> {79} locked for R6+N4 20. C456 a) Outies C123 = 8(2) = [53] -> R2C4 = 5 b) 16(3) = 3{49/67} -> 3 locked for R1 c) 17(3) = 5{39/48}; R2C3 <> 9 d) 8 locked in R3C56 for R3 21. R89 a) Hidden Single: R7C3 = 7 @ C3 b) 18(3): R7C2 <> 2 c) Innies = 16(3) = {268} -> R8C5 = 6 d) 11(6) = {236} -> R7C5 = 2, R6C5 = 3 22. C456 a) 15(3) = {159} -> R2C5 = 9, R3C5 = 1, R4C5 = 5 b) 16(3) = {367} -> R1C5 = 7, R1C6 = 3, R1C5 = 6 c) R3C4 = 4, R3C3 = 2, R3C6 = 8, R4C6 = 4, R5C5 = 8, R6C6 = 6, R7C6 = 5 d) R1C7 = 1, R2C7 = 6, R2C8 = 7, R3C8 = 5, R3C2 = 6, R3C1 = 7 23. N17 a) 12(4) = {1236} -> {13} locked for C2 b) 18(3) = {279} -> R7C2 = 9, R8C2 = 2 24. Rest is singles. Rating: 1.75 it wasn't as difficult as A74 Brick Wall but a bit more complicated than M1, it's solvable without guesses or T&E. So no contradiction moves from me this time. Last edited by Afmob on Fri Nov 16, 2007 7:57 am; edited 1 time in total
gary w

Joined: 07 Sep 2007
Posts: 84
Location: south wales

 Posted: Mon Nov 12, 2007 12:19 am    Post subject: assassin 65 v3 Being reminded of this one by Afmob's post I had a go myself and,after a tough battle..rather more than 5 hours....managed to crack it without any t&e needed.The beginning of my solution is outlined- 1. Outies on N69=18 r3c9=9 and r8c6=9 thus r2c6=2 r12c7={16} {34}->1/3 2. r56c9+r6c8=9 3. Fairly straightforward combo work -> N9 1 is in 12(3) cage and because of outies c9 1 is not at r9c8 so at r89c9 4. I-O N69 -> r7c9=r6c8+5 readily -> r7c9=6/7 and r6c8=1/2 5. r37c7+r5c8+3 thus r5c8<>1 because would -> r37c7={13} blocked by 9(3) cage N3 (see 1.) 6. If 1 is in r456c7->r12c7={34} -> r89c7={57} -> r7c9=6 ->r6c8=1 (from 2.) Contradiction.So in N6 r6c8=1.r7c9=6.r56c9=[35}.r4c89=67. r12c7={16} 7. In c7 3 at r3/7 -> r37c7<>5/7 -> r5c8=5/7 not possible.So in N3 5/7 are in c8 and conflicting combos mean they are not in r1c8.So r23c8={57}.r3c7=3.r89c7={57} 8.N369 now well under way. 9.Outies N14=6={33}/{15} and r28c4=8. Now did combo work on N12 and also worked on 1s to show that the 40(7)cage must contain a 1 (and so a 4 and doesn't contain a 2 or 3). Good luck! Regards Gary
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