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frank
Regular

Joined: 07 Oct 2006
Posts: 10
Location: Victoria, B.C., Canada

 Posted: Thu Oct 11, 2007 11:19 pm    Post subject: Assassin 72 Very nice . The weekly Assassins are posted at 3 pm Thursday afternoon here on the west coast of Canada, so I am usually the first one to concede defeat . It is nice to claim victory for a change.
Andrew
Grandmaster

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

 Posted: Fri Oct 12, 2007 7:20 am    Post subject: Congratulations Frank for solving it so quickly! You finished it before I started. Fairly straightforward but I did need some hidden killers, expressed directly or indirectly so I can't rate it lower than 1.0. Here is my walkthrough for A72. There could be quicker routes near the end but it was getting late so I may have missed some. Edit. I've now checked through my walkthrough (it was too late to do that on Thursday evening, well actually the "wee sma' hours" of Friday morning), tidying up the later steps and correcting an error near the end. Hope I've got it right now. 1. R12C1 = {49/58/67}, no 1,2,3 2. R12C5 = {29/38/47/56}, no 1 3. R12C9 = {59/68} 4. R89C1 = {59/68} 5. R89C5 = {{29/38/47/56}, no 1 6. R89C9 = {14/23} 7. 9(3) cage at R2C3 = {126/135/234}, no 7,8,9 8. 10(3) cage at R3C1 = {127/136/145/235}, no 8,9 9. R5C123 = {128/137/146/236/245}, no 9 10. R5C789 = {127/136/145/235}, no 8,9 11. 19(3) cage at R6C1 = {289/379/469/478/568}, no 1 12. 10(3) cage at R6C8 = {127/136/145/235}, no 8,9 13. 10(3) cage at R7C6 = {127/136/145/235}, no 8,9 14. 32(5) cage at R2C8 = {26789/35789/45689} = 89{267/357/456}, no 1 15. 32(5) cage at R6C6 = {26789/35789/45689} = 89{267/357/456}, no 1 16. 39(7) cage at R3C5 = {1356789/2346789} = 36789{15/24} "It starts with a little present" 17. 45 rule on R5 and C5, R5C5 counts toward both row and column, total 82 -> R5C5 = 8, clean-up: no 3 in R12C5, no 3 in R89C5 "There are in fact two more little presents" 18. 45 rule on C1234 1 innie R5C4 = 9 19. 45 rule on C6789 1 innie R5C6 = 7 20. 9 in C5 locked in R12C5 or R89C5 -> one of these must be {29} -> 2 in C5 locked in R12C5 or R89C5 21. 45 rule on R1234 2 innies R34C5 = 6 = {15} (only remaining combination) 21a. Naked pair {15} in R34C5, locked for C5, clean-up: no 4 in 39(7) cage (step 16), no 6 in R12C5, no 6 in R89C5 22. 45 rule on R1 3 innies R1C159 = 16 = {259/268/457} 22a. 2 of {259} must be in R1C5 -> no 9 in R1C5, clean-up: no 2 in R2C5 23. 45 rule on R1 3 outies R2C159 = 22 = {589/679} = 9{58/67}, no 4, 9 locked for R2, clean-up: no 9 in R1C1, no 7 in R1C5 24. R1C159 (step 22) = {259/268/457} 24a. 4 of {457} must be in R1C5 -> no 4 in R1C1, clean-up: no 9 in R2C1 25. Killer pair 5,6 in R12C1 and R89C1, locked for C1 [I missed the clash between R12C1 and R89C1. However that allowed the interesting step 36.] 26. 1 in C1 locked in R345C1 26a. CPE no 1 in R4C2 27. 45 rule on R9 3 innies R9C159 = {169/178/259/349/367/457} (cannot be {268} because 6,8 only in R9C1, cannot be {358} because no 3,5,8 in R9C5) 27a. 2 of {259} must be in R9C9 -> no 2 in R9C5, clean-up: no 9 in R8C5 28. 9 in N8 locked in R9C56, locked for R9, clean-up: no 5 in R8C1 29. R9C159 (step 27) = {169/178/259/367/457} (cannot be {349} because no 3,4,9 in R9C1) 29a. 4 of {457} must be in R9C9 -> no 4 in R9C5, clean-up: no 7 in R8C5 30. R12C9 contains 8/9 30a. 32(5) cage at R2C8 (step 14) = 89{267/357/456} so must contain 8/9 in N3 -> R4C7 = {89} 30b. Killer pair 8,9 in R12C9 and 32(5) cage, locked for N3 31. Hidden killer pair 8,9 in R12C9 and R4C9 -> R4C9 = {89} 32. Killer pair 8,9 in R4C79, locked for R4 and N6 33. 32(5) cage at R6C6 (step 15) = 89{267/357/456}, 8,9 locked for N9 34. 10(3) cage at R3C1 (step 8) = {127/136/145/235} 34a. No 4 in R3C1 because R4C12 = [15] clashes with R4C5 35. 4 in C1 locked in R4567C1 35a. CPE no 4 in R6C2 36. R12C1 contains 7/8, R89C1 contains 8/9 -> R67C1 cannot contain more than one of 7,8,9 -> max R67C1 = 13 -> min R6C2 = 6 36a. Max R6C2 = 9 -> min R67C1 = 10 -> R67C1 must contain one of 7,8,9 36b. Killer triple 7,8,9 in R12C1, R67C1 and R89C1, locked for C1 37. 10(3) cage at R3C1 (step 8) = {127/136/145/235} 37a. 5,6,7 only in R4C2 -> R4C2 = {567} 38. 45 rule on C9 4 outies R4C8 + R5C78 + R6C8 = 11 = {1235}, locked for N6 38a. 7 in N6 locked in R6C79, locked for R6 39. R5C789 (step 10) = {136/145} (cannot be {127/235} because R5C9 only contains 4,6), no 2, 1 locked for R5 and N6 39a. 2 in N6 locked in R46C8, locked for C8 40. 2 in R5 locked in R5C123, locked for N4 40a. R5C123 (step 9) = {236/245} 41. 7 in R4 only in R4C23 41a. CPE no 7 in R23C2 42. 19(3) cage at R6C1 (step 11) = {289/379/469/478} 42a. 7 of {379} must be in R7C1 -> no 3 in R7C1 42b. 2 of {289} and 7 of {478} must be in R7C1 -> no 8 in R7C1 43. 10(3) cage at R6C8 (step 12) = {127/145} (cannot be {136} because R6C89 = [36] clashes with R6C5, cannot be {235} because R6C9 only contains 4,6,7) -> R7C9 = 1, R6C89 = [27/54], clean-up: no 4 in R89C9 44. Naked pair {23} in R89C9, locked for C9 and N9 45. 10(3) cage at R7C6 (step 13) = {127/136/145/235} 45a. Min R8C7 = 4 -> max R78C6 = 6, no 6 45b. 1 of {145} must be in R8C6 45c. 5 of {235} must be in R8C7 45d. -> no 4,5 in R8C6 46. R9C159 (step 29) = {259/367}, no 8, clean-up: no 6 in R8C1 47. R9C678 = {367/457} (cannot be {169/178/259/268/349/358} because 1,2,3,8,9 only in R9C6), no 1,2,8,9 in R9C6, 7 locked in R9C78 for R9 and N9 47a. 3 of {367} must be in R9C6 -> no 6 in R9C6 48. R9C5 = 9 (naked single), R8C5 = 2, R12C5 = [47], R89C9 = [32], R8C6 = 1, clean-up: no 6 in R1C1 [Could now have fixed R12C19 using step 22. This also applies after step 50.] 49. 1 in R9 locked in R9C23 49a. R9C234 = {148} (only remaining combination), locked for R9, clean-up: no 5 in R9C678 (step 47) -> R9C6 = 3, R67C6 = [36], clean-up: no 6 in R8C7 (step 13) 50. Naked pair {67} in R9C78, locked for R9 and N9 -> R9C1 = 5, R8C1 = 9, clean-up: no 8 in R12C1 -> R12C1 = [76], clean-up: no 8 in R1C9 51. 7 in N8 locked in R78C4, locked for 18(3) cage -> no 7 in R8C3 51a. 18(3) cage at R7C4 = {567} (only remaining combination), no 4,8 -> R8C3 = 6 52. Naked pair {57} in R78C4, locked for C4 and N8 -> R7C6 = 4, R7C1 = 2, R8C7 = 5, R78C4 = [57], R9C4 = 8 52a. Naked pair {14} in R9C23, locked for N7 -> R8C2 = 8, R8C8 = 4 53. R6C1 = 8 (hidden single in C1), R6C2 = 9 54. 4 in C1 locked in R45C1, locked for N4 55. 23(5) cage at R6C3, R7C23 = {37}, R8C2 = 8 -> R6C34 = 5 = [14] I seem to have deleted too much when editing this walkthrough so the rest is straightforward but not quite down to naked singles.Last edited by Andrew on Tue Oct 23, 2007 4:36 am; edited 6 times in total
Afmob
Expert

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

 Posted: Fri Oct 12, 2007 1:54 pm    Post subject: I think the problem with this assassin was not to solve it but to make the solving path not too long so I tried to keep it as small as possible. Assassin 72 Walkthrough: 1. Opening the present aka position of cells of 39(7) a) Innies C1234 = R5C4 = 9 b) Innies C6789 = R5C6 = 7 c) Innies R5 = R5C5 = 8 d) Innies R1234 = 6(2) = {15/24} e) Innies R6789 = 9(2) = {36/45} f) Killer pair (45) of Innies R1234 = 6(2) blocks {45} of Innies R6789 = 9(2) g) Innies R6789 = {36} locked for C5 h) Both 11(2) in C5 = {29/47} locked for C5 2. R1 a) Outies = 22(3) = 9{58/67} -> 9 locked for R2 b) 11(2) = [29/47] 3. C1 a) Killer pair (58) of 14(2) blocks {58} of 13(2) b) 13(2) = [49/67/76] c) Killer pair (69) in 13(2) + 14(2) locked 4. C89 a) Outies C9 = 11(4) = {1235} locked for N6 b) 10(3) @ R5 = 1{36/45} since R5C9 = (46) -> 1 locked for R5+N6 c) 2 locked in R46C8 for C8 d) 17(3): R3C9 <> 1,2 because R4C89 <= 14 e) 1,2 locked in R789C9 for N9 5. C1 a) 11(3) = 2{36/45} -> 2 locked for N4 6. R1 a) Outies = 22(3) = {679} locked for R2 because R2C15 <> 5,8 b) 14(2) = [59/86] c) Innies = 16(3) = {268/457}: R1C1 <> 4 since Innies must have 2 xor 4 and R1C5 = (24) d) 13(2) = {67} locked for C1+N1 7. C1 a) 14(2) = {59} locked for C1+N7 b) 8 locked in 19(3) = 8{29/47/56} -> no 3; R6C2 <> 8 8. C9 a) 10(3) @ R6: R7C9 <> 5,6,7 because R6C89 >= 6 b) 5 locked in R13C9 for N3 9. R3 a) 7 locked in R3C789 for N3 10. R67 a) 9 locked in R7C78 for N9 b) 9 locked in 32(5) in R7C78 -> R6C7 <> 9 c) 9 locked in R6C23 for N4 d) 10(3) = 1{27/36/45} since R6C9 = (467) -> R7C9 = 1 11. R89 a) 5(2) = {23} locked for C9+N9 b) Innies R9 = 16(3) = 9{25/34}; R9C5 <> 2 because R9C19 <= 12 c) 11(2): [29/74] d) Hidden Single: R8C1 = 9 @ R8 e) R9C1 = 5 f) Innies R9 = 16(3) = {259} locked for R9 g) R9C9 = 2, R8C9 = 3, R9C5 = 9, R8C5 = 2 i) R1C5 = 4, R2C5 = 7, R2C1 = 6, R1C1 = 7, R2C9 = 9, R1C9 = 5 j) Hidden Single: R4C7 = 9 @ N6, R7C8 = 9 @ N9 12. R89 a) 13(3) = 4{18/36} -> 4 locked for R9 b) 10(3) = 1{36/45} -> R8C6 = 1 c) 16(3) = {367} locked for R9 -> R9C6 = 3; {67} locked for N9 d) R7C5 = 6, R6C5 = 3 e) 10(3) = {145}; R7C6+R8C7 = {45} -> R7C7 <> 4,5 f) R7C7 = 8 g) Hidden Single: R6C1 = 8 @ C1, R4C9 = 8 @ R4 h) Naked pair (45) locked in R8C78 for R8 i) 18(3) = 6[48/57] -> R8C4 = 6 13. C4 a) Hidden Single: R8C4 = 7 b) R8C2 = 8, R7C4 = 5, R7C6 = 4, R8C7 = 5, R8C8 = 4 c) R9C4 = 8, R7C1 = 2, R6C2 = 9 14. R6 a) 32(5) = {45689} -> R6C7 = 6, R6C6 = 5 15. Rest is clean-up and singles. Rating: 0.75 - 1.0 but more 1.0 than 0.75.Last edited by Afmob on Fri Oct 19, 2007 4:45 am; edited 2 times in total
CathyW
Master

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

 Posted: Fri Oct 12, 2007 8:39 pm    Post subject: I have 19 steps (excluding the prelims) in this WT before it gets to singles! Prelims: a) 13(2) r12c1 = {49/58/67} b) 11(2) r12c5, r89c5 = {29/38/47/56} c) 14(2) r12c9, r89c1 = {59/68} -> blocks {58} for 13(2) r12c1 d) 9(3) @ r2c3 = {126/135/234} e) 32(5) @ r2c8, r6c6 = {26789/35789/45689} must have 8,9 f) 10(3) @ r3c1, r5c7, r6c8, r7c6 = {127/136/145/235} g) 39(7) @ r3c5 = {1356789/2346789} h) 11(3) r5c123: no 9 i) 19(3) @ r6c1: no 1 j) 5(2) r89c9 = {14/23} 1. Overlap r5, c5 -> r5c5 = 8 -> r12c5, r89c5 <> 3 2. Innies c1234 -> r5c4 = 9 3. Innies c6789 -> r5c6 = 7 4. Innies r1234 -> r34c5 = 6 -> r34c5 = {15/24} 5. Innies r6789 -> r67c5 = 9 -> r67c5 = {36} – only possible combo. a) r12c5, r89c5 <> 5 b) 11(2) r12c5, r89c5 = {29/47} c) r34c5 = {15} 6. Outies c9 -> r456c8 + r5c7 = 11 = {1235} n/e N6 a) 10(3) r5c789 = {13}6/{15}4 -> r5c123, r46c8 <> 1 b) 11(3) r5c123 = {236/245} -> r5c78 <> 2 c) 2 locked r46c8 n/e c8 d) 10(3) @ r6c8: r6c8 = (235), r6c9 = (467) -> r7c9 = 1 (only option) e) 5(2) r89c9 = {23} n/e N9/c9 f) r6c89 = [27/54] ([36]) blocked by r6c5 7. Outies c1 -> r456c2 + r5c3 = 22 a) max from r4c2 + r5c23 = 5+6+7 = 18 -> r6c2 min 4 b) r4c2 <> 1 (can’t make 21 from other cells) 8. 5 locked r123c9 n/e N3 9. Innies r9: r9c159 = 16 = [592/943/673] a) r8c1 <> 6 b) r8c5 <> 9 10. Outies r9: r8c159 = 14 = [923/842/572] 11. Innies r1: r1c159 = 16 = [925/628/475/745] a) r2c5 <> 2 b) r2c9 <> 5,8 12. Outies r1: r2c159 = 22 = {679} – only possible combo, n/e r2 a) r1c1 <> 9 b) r1c5 <> 7 c) split 16(3) r1c159 = [628/745] d) 13(2) r12c1 = {67} n/e N1/c1 e) 14(2) r89c1 = {59} n/e N7/c1 13. 8 locked r67c1 within 19(3) -> 19(3) = 8{29/47} -> r67c1 <> 3, r6c2 = (79) 14. Split 16(3) r9c159 = [592/943] -> r8c5 <> 4 15. 9 locked r9c56 in N7 n/e r9 a) r9c1 = 5, r8c1 = 9 b) r9c5 = 9, r8c5 = 2 c) r9c9 = 2, r8c9 = 3 d) r1c5 = 4, r2c5 = 7 e) r2c1 = 6, r1c1 = 7 f) r2c9 = 9, r1c9 = 5 16. HS r4c7 = 9, r7c8 = 9 17. 1 locked r6c34 within 23(5) -> r8c2 <> 1 18. 18(3) @ r7c4 <> 1 a) HS r8c6 = 1 -> r7c6 = (345), r8c7 = (456) b) 1 locked r9c23 within 13(3) -> 13(3) = {148} -> 16(3) r9c678 = 3{67} -> {67} n/e N9 c) r7c5 = 6, r6c5 = 3 d) r7c6, r8c7 = {45} -> pointing cells r7c7, r8c4 <> 4,5 -> r7c7 = 8 19. 18(3) @ r7c4 = {468/567} -> r8c3 = 6 a) NP r8c78 = {45} n/e r8 b) r7c4 = (45) -> NP r7c46: 4,5 n/e r7/N8 c) NS: r9c4 = 8, r8c4 = 7, r8c2 = 8, r7c1 = 2 d) HS: r6c1 = 8 e) r7c23 = {37} -> r6c34 = {14} n/e r6 Singles and simple cage combos from here. Enjoyed this one. I'd agree with the 1.0 rating. Would be good to see more puzzles which make use of overlaps.Last edited by CathyW on Fri Oct 26, 2007 4:18 pm; edited 3 times in total
gary w

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

 Posted: Fri Oct 12, 2007 11:15 pm    Post subject: assassin 72 A nice easy one after last week's horror! Early steps after which it is pretty straightforward... 1. r5c5=8 overlap of c5 and r5 2. 45 rule c1234 -> r5c4=9 c6789 -> r5c6=7 r1234-> r34c5=6 r6789->r67c5=9 (must be3/6 as 4/5 blocks the 6/2 bit)..and r34c5={15} 3.O 0n r1 -> r2c159=22 so contains a 9 and also r2c5=7/9 4. O on c9 -> r5c7+ r456c8=11 {1235} 5. Thus r5c9= 4/6 6.So cannot be a 1 in the 11/2 cage N4 (remaining 10/2 then blocked ) 7. Thus in r5 1 at c7/8 8. Thus r7c9 is a HS=1 9. r89c9={23} 10.9 cannot be in r9c6 now (blocked by 123 in N9) 11. So 11/2 cage N8 = {29} 12.Thus r2c5=7 (from 3. above) r1c5=4r2c9=9 r1c9=5 and r12c1=76 Mop up now If there aren't any really fiendish variants I might get a chance to do something else this weekend! Regards GaryLast edited by gary w on Wed Oct 24, 2007 6:57 pm; edited 1 time in total
Andrew
Grandmaster

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

Posted: Fri Oct 12, 2007 11:43 pm    Post subject:

 Afmob wrote: I think the problem with this assassin was not to solve it but to make the solving path not too long so I tried to keep it as small as possible.

My solving path would probably have been a bit shorter if I'd seen 4 outies on C9 = 11 = {1235} earlier. I suppose I could have then put it in earlier but I didn't because it wasn't something that I thought I ought to have seen earlier. Anyway it was getting late and I didn't want to have to restart and check all the later moves. Also I wanted to get my walkthrough posted on Thursday evening (well actually it wasn't, it was after midnight).

Last edited by Andrew on Thu Oct 18, 2007 11:29 pm; edited 1 time in total
Para
Yokozuna

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

 Posted: Tue Oct 16, 2007 4:16 pm    Post subject: Hi all This was probably the simplest V2 to create because the choice was too obvious. It solves very nicely and it opens up some interesting moves from the the original pattern that weren't needed to solve the puzzle before. It rates about a V1.75 the way i solved it. Old School Assassin 72V2 PS: 3x3::k:3328:4353:4353:4353:2820:3077:3077:3077:3592:3328:6154:2315:2315:2820:3854:3854:8208:3592:2578:6154:6154:2315:5910:3854:8208:8208:4378:2578:2578:6154:6154:5910:8208:8208:4378:4378:5156:5156:5156:5156:5910:4393:4393:4393:4393:4909:4909:5935:5935:5910:8242:8242:2612:2612:4909:5935:5935:4665:5910:2619:8242:8242:2612:3647:5935:4665:4665:2883:2619:2619:8242:1351:3647:3401:3401:3401:2883:4173:4173:4173:1351: Enjoy. greetings Para
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

 Posted: Wed Oct 17, 2007 2:47 pm    Post subject: Hi folks, Is anyone else doing Para's "Old School" A72V2? This is the way I did it, which is probably not the way Para intended it to be done. In particular, there's one high-level (but simple and very powerful) move (step 22) that effectively killed the puzzle off without having to resort to a lot of low-level combination crunching (which I personally prefer to leave to the automated solvers, which are better at it). Would be interesting to see how others solve it. The way I did it would probably correspond to a rating of around 1.5. Many thanks to Para for creating this variant. Fortunately, I didn't do the original this time, so it was like a brand new puzzle for me. Maybe this is a small advantage because I didn't have to suffer the negative psychological impact of seeing my V1 solving path brutally shut off? On the other hand, I didn't have the V1 as "training" and no doubt overlooked several moves you guys made in the original. So I guess it's a case of swings and roundabouts. Assassin 72 V2 Walkthrough Prelims: 13(2)n56 = {49/58/67}: no 1..3 11(2)n2 and 11(2)n8 = {29/38/47/56}: no 1 14(2)n3 and 14(2)n7 = {59/68} 9(3)n12 = {126/135/234}: no 7..9 32(5)n356 and 32(5)n569 = {(267/357/456)89}: no 1 10(3)n14, 10(3)n69 and 10(3)n89 = {127/136/145/235}: no 8,9 19(3)n47 = {289/379/469/478/568}: no 1 5(2)n9 = {14/23} 1. Innie r5: r5c5 = 8 1a. cleanup: no 3 in r1289c5 1b. 11(2)n2 and 11(2)n8 = {29/47/56} 2. Innies r1234: r34c5 = 6(2) = {15}, locked for c5 (Note: {24} blocked by 2 11(2) cages in c5 (step 1b)) 2a. cleanup: no 6 in r1289c5 3. Hidden pair {36} in c5 at r67c5 3a. -> r67c5 = {36} 4. 8 of 32(5)n356 and 32(5)n569 now locked in c78 4a. -> 32(5)n356 and 32(5)n569 form generalized X-Wing on 8 in c78 4b. -> no 8 elsewhere in c78 5. 8 in n9 locked in 32(5)n569 5a. -> no 8 in r6c7 6. 8 in r6 locked in n4 -> not elsewhere in n4 7. {58} combo for 13(2)n1 blocked by 14(2)n7 7a. -> no 5,8 in r12c1 8. Outies r1: r2c159 = 22(3) = {679}, locked for r2 (Note: {589} blocked because {58} only in r2c9) 8a. cleanup: no 9 in r1c1; no 7,9 in r1c5; no 6,9 in r1c9 9. Innies r1: r1c159 = 16(3) = [628/745] 9a. -> no 4 in r1c1 9b. cleanup: no 9 in r2c1 10. Naked pair {67} at r12c1 -> no 6,7 elsewhere in c1 and n1 10a. cleanup: no 8 in r89c1 11. Naked pair {59} at r89c1 -> no 5,9 elsewhere in c1 and n7 12. 12(3)n23 = {129/138/147/345} (no 6) (Note: {156/237/246} blocked by r1 innies (step 9)) 13. 17(3)n12 = {179/269/359/368} (no 4) (Note: {467} unplaceable, and {278/458} blocked by r1 innies (step 9)) 13a. if {179} then 7 must go in r1c4; if {269/368} then 6 must go in r1c4 13b. -> no 1,2,8 in r1c4 14. 8 in c4 locked in n8 -> not elsewhere in n8 15. 8 in r9 locked in 13(3)n78 = {148/238} (no 5,6,7,9) = {(2/4)..}, {(3/4)..} 16. Innies r9: r9c159 = [574/592] (Note: [943] blocked by 13(3)n78 (step 15)) 16a. -> r8c1 = 9 16b. cleanup: no 7,9 in r8c5; no 2,4 in r8c9 17. 13(3)n78 and r9c9 form killer pair on {24} in r9 -> no 2,4 elsewhere in r9 18. 8 in c1 locked in 19(3)n47 = {289/478} (no 3,5,6) (Note: {568} unplaceable) 18a. must have 1 of {79}, only available in r6c2 18b. -> r6c2 = {79} 19. 10(3)n14 can only contain 2 of {1..4}, which must go in r34c1 19a. -> no 1..4 in r4c2 19b. [415] permutation blocked by r4c5 19c. -> no 4 in r3c1 20. I/O diff. c1: r46c2 = r5c1 + 11 20a. r5c1 = 1..4 -> r46c2 = 12..15 = [57/59/67/69] 20b. -> no 7 in r4c2 21. I/O diff. c9: r46c8 = r5c9 + 1 21a. -> no 1 in r46c8 (IOU) 21b. max. r46c8 = 10 21c. -> no 9 in r4c8 --- this next move cracks it --- 22. no 1 in r8c9. Here's how. 22a. 1 in r8c9 -> no 1 in r8c3 22b. 1 in r8c9 -> no 1 in r6c9 -> 1 in r6c34 -> no 1 in r7c23+r8c2 22c. 1 in r8c9 -> 4 in r9c9 -> 13(3)n78 = {238} -> no 1 in r9c23 22d. but this leaves nowhere to place the 1 in n7 22e. -> no 1 in r8c9 --- just mop-up now --- 23. r89c9 = [32] 23a. -> r9c5 = 9 (step 16) 23b. -> r8c5 = 2 24. r12c5 = [47] 24a. -> r12c1 = [76] 24b. -> r12c9 = [59] 25. 16(3)n89 = {367} (only remaining combo) 25a. -> r9c6 = 3 25b. r9c78 = {67}, locked for n9 26. r67c5 = [36] 27. Hidden single (HS) in r1 at r1c4 = 6 27a. cleanup: no 1 in r1c23 28. 18(3)n78 = {468/567} (no 1) 28a. -> r8c3 = 6 29. 10(3)n89 = {145} (no 7) (only remaining combo) 30. Innies n8: r78c6+r9c4 = 13(3) = {148} (no 5) (only remaining combo), locked for n8 30a. -> r9c4 = 8 31. r8c7 = 5 (step 29) 31a. -> r78c4 = [57] 32. HS in n9 at r7c9 = 1 32a. -> r6c89 = [27/54] 33. r78c6 = [41] 34. HS in n9 at r8c8 = 4 34a. -> r8c2 = 8 34b. cleanup: no 3 in r1c3 35. Naked single (NS) at r7c1 = 2 35a. -> r6c12 = [89] 36. split 28(4) at r6c67+r7c78 = {5689} (only remaining combo) 36a. -> r6c67 = [56] Now just naked singles and cage sums to end._________________Cheers, MikeLast edited by mhparker on Thu Oct 18, 2007 6:48 am; edited 1 time in total
gary w

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

 Posted: Wed Oct 17, 2007 6:47 pm    Post subject: 72 v2 Mike, Very much enjoyed your wt.My path was completely different.Didn't keep a wt but it was based on; the outies of c9=18 and for c1=31.This much restricts the combos possible especially in N6.Thsi together with the x-wing on the 8s as you use to great advantage) enabled me to place 2 3/5 in r46c8.Then the available combos for the 32/5 cage N3 together with combos in 12/3 cage N23 (eventually) enabled me to eliminate both 2 and 5 from r4c8.For me this was a "messy" route..I do much prefer solutions where some symmetry or patterns are apparent.For example..my solution to your concentric squares killer which I was quite pleased with..excuse the lack of humility..these darned things are so hard that when you see a neat solution it's difficult not to feel rather pleased with yourself Just a question though,Mike.Step 22 was great but was it a hypothetical? Regards GaryLast edited by gary w on Wed Oct 24, 2007 6:58 pm; edited 1 time in total
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Wed Oct 17, 2007 9:00 pm    Post subject:

Hi Gary,

 gary w wrote: Just a question though,Mike.Step 22 was great but was it a hypothetical?

Thanks for raising this question. Don't know whether this is the answer you're looking for, but here's my response:

 Select text in box (e.g., by triple-clicking it) to see what I wrote: Possibly, but arguably no more so than a complex XY-Loop, aka. complex Nice Loop. If there were only 2 places (i.e., cells or cell groups) for the 1 in N7, with 2 inference chains emanating from the starting cell (r8c9), then there would have been a grouped strong link between the 2 cell groups in N7, allowing the logic to be re-expressed as a loop. The loop would have been a discontinuous complex Nice Loop, with two strong links at the discontinuity (r8c9), and would have shown that "if r8c9 is not 3, then it must be 3" (with the conclusion r8c9 = 3). In this case (step 22 of my A72V2 WT), there are however three places for the 1 in N7, and three chains emanating from r8c9 instead of two, as in the theoretical example above. As such, it is no longer possible to create a simple loop. However, the chains themselves are bona fide bidirectional inference chains, just as in the former case. So it's arguably just an extension of the 2-way (loop) scenario to the 3-way case.

I hope this has answered your question and at least been partially comprehensible!
_________________
Cheers,
Mike
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Thu Oct 18, 2007 10:42 pm    Post subject:

Hi folks,

Since no-one else has submitted a full WT for the Assassin 72 V2 , I decided to compare notes with JSudoku instead, in an attempt to see how much effort my high-level breakthrough move (step 22) saved compared to the traditional low(er)-level approach. The difference was huge, illustrating once again just how powerful some of these complex steps can be. In case some of you also might find the comparison interesting, I've decided to publish my results here.

Here's the grid state after my step 21, which I took as a common starting position:

 Code: .-----------.-----------------------------------.-----------.-----------------------------------.-----------. | 67        | 123589      123589      35679     | 24        | 12345789    1234579     1234579   | 58        | |           :-----------.-----------------------:           :-----------------------.-----------:           | | 67        | 123458    | 12345       12345     | 79        | 123458      12345     | 23458     | 69        | :-----------:           '-----------.           :-----------:           .-----------'           :-----------: | 123       | 1234589     1234589   | 123456    | 15        | 123456789 | 23456789    23456789  | 123456789 | |           '-----------.           '-----------:           :-----------'           .-----------'           | | 1234        56        | 12345679    12345679  | 15        | 2345679     23456789  | 234567      123456789 | :-----------------------'-----------------------:           :-----------------------'-----------------------: | 1234        12345679    12345679    12345679  | 8         | 12345679    12345679    12345679    12345679  | :-----------------------.-----------------------:           :-----------------------.-----------------------: | 248         79        | 123456789   12345679  | 36        | 2345679     2345679   | 234567      1234567   | |           .-----------'           .-----------:           :-----------.           '-----------.           | | 248       | 1234678     1234678   | 123456789 | 36        | 123457    | 23456789    23456789  | 1234567   | :-----------:           .-----------'           :-----------:           '-----------.           :-----------: | 9         | 1234678   | 1234678     12345678  | 24        | 1234567     1234567   | 2345678   | 13        | |           :-----------'-----------------------:           :-----------------------'-----------:           | | 5         | 12348       12348       12348     | 79        | 13679       13679       13679     | 24        | '-----------'-----------------------------------'-----------'-----------------------------------'-----------'

From this position, recall that I used the following single high-level move to crack the puzzle:

 mhparker (in WT above) wrote: 22. no 1 in r8c9. Here's how. 22a. 1 in r8c9 -> no 1 in r8c3 22b. 1 in r8c9 -> no 1 in r6c9 -> 1 in r6c34 -> no 1 in r7c23+r8c2 22c. 1 in r8c9 -> 4 in r9c9 -> 13(3)n78 = {238} -> no 1 in r9c23 22d. but this leaves nowhere to place the 1 in n7 22e. -> no 1 in r8c9

From the same position, JSudoku required (with a bit of help from me) the following 24 (!) lower-level moves to crack the puzzle:

 JSudoku (after heavy optimization and help from me, esp. with final step) wrote: 22. 18(3)n78 = {189/279/369/378/459/468/567} 22a. 9 only in r7c4 22b. -> no 1,2 in r7c4 22c. r78c4 cannot contain both of {79} due to r9c5 22d. -> no 2 in r8c3 23. 2 in r9 locked in 13(3)n78 or 5(2)n9 23a. if 13(3)n78 contains a 2, it must also contain a 3 ({238}) 23b. if 5(2)n9 contains a 2, it must also contain a 3 ([32]) 23c. -> 3 locked in r9c234 or r8c9 23d. -> no 3 in r9c78 24. 4 in r9 locked in 13(3)n78 or 5(2)n9 24a. if 13(3)n78 contains a 4, it must also contain a 1 ({148}) 24b. if 5(2)n9 contains a 4, it must also contain a 1 ([14]) 24c. -> 1 locked in r9c234 or r8c9 24d. -> no 1 in r9c78 25. Naked triple on {679} in r9 at r9c578 -> no 6,7,9 in r9c6 26. 6 in r9 locked in n9 -> not elsewhere in n9 27. Grouped XY-Chain eliminates 3 from r7c4. Here's how. 27a. Either r9c6 = 3, or... 27b. ...r9c6 <> 3 => r9c6 = 1 (strong link, bivalue cell) 27c. -> r9c78 <> {7..} (weak link, combinations 16(3)) 27d. => r9c5 = 7 (strong link, r9) 27e. -> r9c5 <> 9 (weak link, r9c5) 27f. => r7c4 = 9 (strong link, n8) 27g. Either way, no 3 in r7c4 28. I/O difference n8: r7c5 + r9c46 = r8c37 + 6 28a. no 1,3 in r9c4. Here's how. 28b. min. r8c37 = 3 -> min. r7c5 + r9c46 = 9 28c. if r9c4 = 1/3, then r7c5 + r9c46 would be forced to {136} = 10 28d. -> r8c37 would have to sum to 4 28e. but this is impossible, since {13} in r8c37 blocked by r8c9 28f. -> no 1,3 in r9c4 29. (from steps 23c and 24c) {13} now locked in r9c23 or r8c9 29a. -> no 1,3 in r8c23 30. no 3 in r7c5. Here's how. 30a. min. r8c37 now 5 ([41]) 30b. -> (from step 28) min. r7c5 + r9c46 = 11 30c. -> only possible permutation for r7c5 + r9c46 with a 3 in r7c5 is [381] = 12 30d. -> r8c37 = 6 = [42] (only possible permutation) 30e. but [42] for r8c37 is blocked by r8c5 30f. -> no 3 in r7c5 31. r67c5 = [36] 32. 10(3)n69 = {127/136/145/235} = {(1/3)..} 32a. {13} only available in c9 32b. -> 10(3)n69 and r8c9 form killer pair on {13} -> no 1,3 elsewhere in c9 33. 3 in c9 locked in n9 -> not elsewhere in n9 34. no 2 in r9c4. Here's how. 34a. I/O diff. n8: r9c46 = r8c37 34b. only possible permutation with 2 in r9c4 is r8c37+r9c46 = [4123] 34c. but this is blocked by r8c5 34d. -> no 2 in r9c4 35. CPE: r8c7 sees all 2's in n8 -> no 2 in r8c7 36. 10(3)n89 cannot have both of {27} within n8 due to 11(2)n8 36a. -> no 7 in r78c6 36b. 10(3)n89 = {127/145/235} = {(2/4)..} 37. 10(3)n89 (step 36b) and r8c5 form killer pair on {24} 37a. -> no 2,4 in r8c4 38. r78c4 and r9c5 form hidden killer pair on {79} within n8 38a. -> r78c4 = {(7/9)..} 38b. -> {468} combo blocked for 18(3)n78 = {189/369/378/459/567} 38c. -> possible permuations for 18(3)n78 = [981/963/783/945/567/765] 38d. -> r7c4 = {579}(no 4,8), r8c3 = {468}(no 7), r8c4 = {1357}(no 8) 39. Hidden single in c4 at r9c4 = 8 40. CPE: r8c7 sees all 4's in n8 -> no 4 in r8c7 41. I/O diff. n8: r8c37 = r9c6 + 8 = 9 or 11 41a. -> r8c37+r9c6 = [451/653/811] (Note: [473] blocked by 13(3)n78) 41b. -> no 7 in r8c7 42. 7 now unavailable for 10(3)n89 (step 36b) = {145/235} 42a. 5 locked -> no 5 in r8c4 42b. -> no 4 in r8c3 (step 38c) 43. 2 in n8 locked in 10(3)n89 or split 14(3) at r8c159 (i.e., r9 outies) 43a. if 10(3)n89 contains a 2, it must also contain a 3 ({23}[5]) 43b. if r8c159 contains a 2, it must also contain a 3 ([923]) 43c. -> 3 locked in r78c6 or r8c9 43d. -> no 3 in r8c4 44. (from step 38c) 18(3)n78 = [981/567] (only remaining permutations) 44a. -> no 7 in r7c4 45. I/O diff. n8: r78c4 + r9c6 = r8c7 + 10 = 11 or 15 45a. -> r78c4 + r9c6 = [573] (only possible permutation) 45b. -> r8c7 = 5, r8c3 = 6 (18(3) cage sum)

Note that this is after a lot of analysis and simplification by me, with step 45 coming completely from me to shorten the solving path, otherwise JSudoku would have taken even longer to crack it! The fact that JSudoku didn't see step 45 is due to a known deficiency in innie/outie difference handling that I've already mentioned on this forum. Believe it or not, I've cut out the non-essential steps! I also took step 30 a bit further than JSudoku took it (where JSudoku for some reason failed to show that r7c5 cannot be a 3).

Note also that although I used the term "lower-level" for the JSudoku moves above, JSudoku still needed an AIC (step 27). It's also interesting to note that JSudoku made extensive use of Locked Cages moves (steps 23, 24 and 43).

BTW, for those who haven't yet tried interpreting JSudoku logs and converting them into standard walkthrough form, optimizing out the unnecessary steps, be warned that it is a very time-consuming process. The above analysis must have taken me at least 3 hours (!). However, I usually find the results very instructive, as is the case here.

I suppose that (to be fair) I should really do the same with Sudoku Solver. Trouble is, I've now run out of energy!
_________________
Cheers,
Mike
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Tue Oct 30, 2007 9:52 am    Post subject:

 about "old school" A72V2 Para wrote: It solves very nicely
Not by me though. Have given up. But thanks for the challenge Para. Wow. What a brute - but not for Mike! You made it look real easy Mike - nice work. I really like all the lead up steps before step 22 too. Took me forever just to get to there - generalized X-wing was about step 38, innies r1234 step 39 !! Time to just do V1s for a while - too much pacing . !

I really enjoyed the JSudoku log Mike. Really easy to follow and many, many clever things. Thanks for doing that.

Great work Para!
Ed
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