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JFFK 7 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=558 |
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Author: | manu [ Mon Jun 29, 2009 7:36 pm ] |
Post subject: | JFFK 7 |
My first attempt to solve this puzzle forced me to use some hard moves, but working again on my WT, I realized it could be cracked without too much effort. My rating should not exceed 1.5 and I guess there are many ways to tackle this killer, so I hope you will enjoy it. JFFK 7 3x3::k:2048:5633:5633:3075:2052:1285:1285:2567:2567:2048:5633:3075:3075:2052:6158:3087:3087:2321:3858:2835:2835:2835:6158:6158:4632:4632:2321:3858:3858:1565:1565:11551:6158:11551:4632:3875:3876:3876:3876:11551:11551:11551:11551:11551:3875:2093:5166:4143:4143:11551:5170:11551:3892:3875:2093:5166:4143:5689:5170:5170:3388:3892:3892:3391:5166:5166:5689:5689:5689:3388:3388:3143:3391:2889:2889:2379:2379:2893:2893:2893:3143: Solution : Hidden Text: SSscore : 2.10 (estimated much lower) Enjoy ! |
Author: | Ed [ Sun Jul 05, 2009 9:32 am ] |
Post subject: | Re: JFFK 7 |
Completely stuck with no clue how to go any further. Tag anyone? JFFK 7 Prelims i. 8(2)n1: no 4,8,9 ii. 22(3)n1: no 1,2,3,4 iii. 8(2)N2: NO 4,8,9 iv. 5(2)n2 = {14/23} v. 10(2)n3: no 5 vi. 12(2)n3: no 1,2,6 vii. 9(2)n3: no 9 viii. 11(3)n1: no 9 ix. 6(2)n4 = {15/24} x. 8(2)n4: no 4,8,9 xi. 20(3)n5: no 1,2 xii. 13(2)n7: no 1,2,3 xiii. 12(2)n9: no 1,2,6 xiv. 11(2)n7: no 1 xv. 9(2)n8: no 9 xvi. 11(3)n8: no 9 1. "45" n8: 1 outie r6c6 - 6 = 1 innie r9c6 1a. r6c6 = (789), r9c6 = (123) 2. "45" n89: 2 outies r6c68 + r9c6 = 12 2a. r6c8 = (345) 3. "45" r9: 2 innies r9c19 = 14 = {59}/[68](no 3,4,7; no 8 in r9c1) 3a. r8c1 = (478), r8c9 = (347) 4. "45" c1: 3 innies r345c1 = 16 4a. hidden killer pair 8,9 in c1 in h16(3) & 13(2)n7 since a h16(3) cannot have both of 8 + 9 = 17 4b. 13(2)n7 = [49/85] 4c. h16(3) must have both 4 & 9 or neither (only other place for 4 & 9 in c1 is 13(2)n7) or 8 = {178/268/349/358} 4d. {178} must have 7 in r5c1 since a 15(3) cage cannot be {17}[7] nor {78}[0] -> no 7 in r34c1 5. h14(2)r9 = {59}: both locked for r9 5a. no 2 or 6 in 11(2)n7 5b. no 4 in r8c9 5c. no 4 in 9(2)n8 6. 11(2)n7 = {38/47} = [4/8..] 6a. Killer Pair 4,8 in 11(2) and r9c1: both locked for n7 6b. {1478/2468/3458} all blocked from 20(4)n4 7. "45" n3: 2 outies r1c6 + r4c8 = 9 7a. r4c8 = (5..8) 8. 22(3)n1 = {589/679}: 9 locked for n1 9. "45" n1: 4 innies r2c3 + r3c123 = 15 and must have 4 for n1 9a. = 4{128/137/236}(no 5) 10. "45" n1: 1 outie r3c4 + 4 = 2 innies r2c3 + r3c1 10a. -> no 4 in r2c3 (IOU) 10b. 4 in n1 only in r3: locked for r3 10c. no 5 in r2c9 11. "45" r12: 2 innies r2c69 = 13 (no 1,2,3; no 4,8 in r2c6) 11a. no 6,7,8 in r3c9 12. "45" c1: 1 outie r4c2 + 1 = r5c1 12a. no 9 in r4c2, no 1 in r5c1 13. "45" n12: 1 outie r4c6 + 5 = 2 innies r1c6 + r3c1 13a. max. 2 innies = [48] = 12 -> max. r4c6 = 7 Marks: "Paste Into" JFFK7 in Sudoku Solver: Ed |
Author: | Afmob [ Sun Jul 05, 2009 11:56 am ] |
Post subject: | |
It's been a while since we did a tag, let the fun begin! 14. Innies R1234 = 20(3) <> 1,2 15. Innies R6789 = 10(3) <> 8,9 16. Innies C9 = 9(2) <> 9; R7C9 <> 4 16a. R1C8 <> 1 17. 15(3) @ R4C9 <> 5{37/19} since they are blocked by R8C9 = (37) and R9C9 = (59) 17a. 15(3) = {168/249/267/456} <> 3 since 8{25/34} blocked by Killer quads (2348,2358) of 9(2) @ C9 + Innies C9 = 9(2) 17b. R6C5 <> 3 (CPE @ N6) 17c. Innies R6789 = 10(3): R6C7 <> 6 because 3 only possible there Marks: |
Author: | Ed [ Mon Jul 06, 2009 11:16 am ] |
Post subject: | Re: JFFK 7 |
Thanks Afmob! There are so many multiples of 5 through r5 and n56. Can anyone make anything of that? The only definite things I can really see about that is step 20 but no eliminations. 15(3) cages at r5c1 and r4c9 45(9) cage at n5 h20(3)r4c579 h10(3)r6c579 6 outies r5 = 30 1 innie r5c9 + 15 = 4 outies r46c57 4 outies n39: r19c6 + r46c8 = 15 Anyhow, a couple of actual eliminations but not much. 18. "45" r123: 4 outies r4c1268 = 19 18a. {1279/1459/1468/2359/2458/3457} all blocked by 6(2)n4 [edit] 18b. h19(4) = {1369/1378/1567/2368/2467} [edit] 18c. 9 in {1369} must be in r4c1 but none of (136) will fit into the 15(3)n1: can't have [096/393] and no 5 in r3c1 for [591] -> no 9 in r4c1 18d. min. r4c68 = [15] = 6 -> max. r4c12 = 13 -> min. r3c1 = 2 19. 9 in r4 only in h20(3)r4 (step 14) = 9{38/47/56} 19a. CPE -> no 9 in r5c78 20. "45" n56: after taking out h12(2)r6c68 -> r4c468 + r6c4 = 18 20a. note: this is an implied h18(4) cage since it cannot have repeats because of the 45(9)n56. But can't see any way to use this. 20b. The same candidates in the 15(3)r5c1 must be in 3 of the 4 cells of the 45(9) that are in r46c57 with the 4th cell being equal to r5c9 and must be one of r46c5. Can't take that any further. marks: |
Author: | Afmob [ Mon Jul 06, 2009 5:32 pm ] |
Post subject: | |
Here some further small moves: (I hope that step 22a is still acceptable as combo analysis) 21. 11(3) @ R9 <> 7 since {37} is a Killer pair 11(2) @ R9 22. 15(3) @ R6C8: R7C8 <> 1 because R6C8+R7C9 <= 13 22a. Outies C9 = 16(3): R1C8 <> 4 because [439] places (39) in 15(3) @ R6C8 and [457] forces 15(3) @ R6C8 = [573] which is blocked by R8C9 = (37) 22b. 10(2) @ N3: R1C9 <> 6 22c. Innies C9 = 9(2): R7C9 <> 3 23. Innies+Outies N7: -1 = R6C2 - R7C13 -> R6C2 <> 1 Marks: |
Author: | Ed [ Tue Jul 07, 2009 9:53 am ] |
Post subject: | Re: JFFK 7 |
Another 3-ways type step in 26. This is about as desperate as I'm prepared to get. If we don't find anything for a couple of days, can you give us a hint manu? 24. h16(3)r345c1 = {178/268/349/358} 24a. {349} must have 9 in r5c1 -> no 4 in r5c1 24b. -> no 3 in r4c2 (i/o c1 = -1) 25. 20(4)n4 = {1289/1379/1469/1568/2369/2567} ([8]{237}/[4]{367} blocked by 11(2)n7; [4]{259} blocked by r9c1) 25a. = only one of 5/9. This led to finding the next step. 26. no 9 in r6c2. Like this. 9 in r6c2 -> no 5 in 20(4)n4 (step 25a) 26a. 9 in r6c2 ->2 innies = 10 (step 23)-> neither can be 5 26b. -> only 5 remaining in n7 is in r9c1 -> 9 in c1 is in r5c1 but this clashes with 9 in r6c2 26c. max. r6c2 = 8 -> max. r7c13 = 9 (no 9) marks: Ed |
Author: | Joe Casey [ Tue Jul 07, 2009 3:35 pm ] |
Post subject: | Re: JFFK 7 |
Here's a hint - I think (at your own risk): You're down to 2 possibilities for R9C19; Hidden Text: cheers |
Author: | Afmob [ Tue Jul 07, 2009 5:11 pm ] |
Post subject: | |
Another little combo analysis step: 27. Outies R123 = 19(4) = {1378/1567/2368/2467} since other combos blocked by 6(2) @ R4 27a. R4C12 @ 15(3) <> 4 since 15(3) can neither be 9{24} nor 5{46} and 15(3) cannot have 4 and 7 27b. IOD C1: -1 = R4C2 - R5C1: R5C1 <> 5 |
Author: | Ed [ Wed Jul 08, 2009 7:48 am ] |
Post subject: | Re: JFFK 7 |
Thanks Afmob for showing what may be a fault line for this puzzle. Some implied cage blocks combined with combo analysis coming up. Starting to get complicated. If there is an easier way or it can be clearer, please share! 28. 15(3)r4c9 = {168/249/267/456} = [2/6,4/6,..] 28a. 45(9) = all of 2,4,6 -> there is no room elsewhere in n56 for both 4 & 6 nor both 2 & 6 (no elims yet) 29. {2467} in h19(4)r4 (step 27) blocked by step 28a. & 15(3)n1. Like this. 29a. {2467} must have r4c68 = [47] ([46] blocked by step 28a) 29b. this leaves r4c12 as {26}: but no 7 available in r3c1 in 15(3) 29c. no 4 in r4c6 30. {2368} in h19(4)r4 (step 27) blocked by step 28a & 15(3)n1 30a. {2368} must have 6 or 8 in r4c12 for the 15(3)n1 to reach the cage sum 30ai. = {26}[38]: blocked by no 7 in r3c1 30aii. = {28}[36]: blocked by no 5 in r3c1 30aiii. = {36}[28]: blocked by the 15(3)n1 cannot be [6]{36} 30aiv. = {38}[26]: blocked by step 28a 30b. h19(4)r4 = {1378/1567}(no 2) 30c. no 3 in r5c1 (i/o C1 = -1) 30d. 1 & 7 locked for r4 31. 6(2)n4 = {24}: 4 locked for r4 31a. deleted 31b. {258} combo blocked from 15(3)n1 since h19(4)r4 cannot have both 5 & 8 (step 30) marks: |
Author: | Ed [ Thu Jul 09, 2009 6:19 am ] |
Post subject: | Re: JFFK 7 |
Afmob has verified the previous few steps [edit: and thanks to Andrew for some corrections]. So, continuing on. 32. r4c4 = (24) -> 45(9)n56 must have at least one of 2 or 4 in n6 -> {249} combo blocked from 15(3)r4c9 32a. 15(3)r4c9 = {168/267/456}(no 9) Looks like it might be cracked now so will miss lots of clean-up. 33. r9c9 = 9 (hsingle c1) 33a. r8c9 = 3 33b. r89c1 = [85] 34. r5c1 = 9 (hsingle c1) 34a. r5c23 = 6 = {15} ({24} blocked by r4c3): both locked for r5 & n4 35. r4c7 = 9 (hsingle n6) 36. no 3 in the two 8(2) cages in c1 36a. 3 only in 15(3)n1 = [438] (only valid permutation) (must have been a hidden single 4 for c1 here too) 37. r4c6 = 1 (hsingle r4) 37a. r4c8 = 7 (hsingle r4) 38. 5 in n1 only in 22(3) = {589} 38a. r1c3 = 8, r12c2 = {59}: both locked for c2 39. r8c3 = 9 (hsingle n7) 39a. r5c23 = [15] 40. 11(2)n7 = {47}: both locked for n7 & r9 41. 20(4)n4 = {236}[9] -> r7c2 = 3 41a. r68c2 = {26}: locked for c2 42. r3c2 = 7 -> r3c34 = 4 = {13}: both locked for r3 43. 8(2)n1 = {26}: both locked for n1 & c1 43a. r67c1 = [71] 44. "45" n3: 1 remaining outie r1c6 = 2, r1c7 = 3 44a. r9c6 = 3 44b. r9c78 = 8 = {26}: both locked for r9 & n9 45. "45" n9: 1 remaining outie r6c8 = 3 46. "45" n7: 4 outies r6c1234 = 23 = [7]{26}[8] (only permutation) 46a. 2 & 6 both locked for n4 & r6 46b. r4c34 = [42] 47. 8 in 45(9)n5 only in n6: locked for n6 47a. 5 in 45(9) only in c5: locked for c5 47b. r12c5 = 8 = {17} only: both locked for c5 and n2 48. 15(3)r4c9 = {456}: all locked for c9 & n6 48a. h9(2)r17c9 = [18] 48b. r7c8 = 4 (cage sum) 48c. r1c8 = 9 48d. r12c2 = [59] 48e. r12c5 = 8 = [71] 49. r23c3 = [31] 49a. r12c4 = 9 = [45] 49b. r6c6 = 9 50. r7c56 = 11 = [65] Rest are singles. Turns out that steps 28a, 29 & 30 were the keys ones. Very impressed that manu did this one on his own. Quite satisfied with that solution. Thanks again manu and Afmob. Cheers Ed |
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