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Assassin 383 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1507 |
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Author: | Ed [ Sun Sep 01, 2019 8:08 am ] |
Post subject: | Assassin 383 |
Attachment: a383.JPG [ 62.82 KiB | Viewed 6691 times ] Started out as a 1.35 puzzle but had a hard time - 3 times actually, until finally finding what I missed. This is now the more challenging version below, gets a 2.00 but can make lots of progress before it gets really hard. JSudoku uses 3 easy chains which suggests it is a bit easier than I found it. code-triple click: 3x3::k:2048:2305:2305:7682:7682:7682:7682:3331:3331:2048:8964:3845:7682:5638:7682:7682:2311:2311:8964:8964:3845:3845:5638:5638:4104:4104:4104:8964:8964:8964:3853:5638:0000:0000:5643:5643:3852:3852:0000:3853:0000:0000:0000:5643:5643:2064:3852:0000:3853:0000:0000:6417:4626:5139:2064:2325:2325:5391:5391:6417:6417:4626:5139:4362:2830:2830:4372:5391:6417:6417:4626:5139:4362:4362:4372:4372:5391:6417:2313:2313:5139: solution: Code: +-------+-------+-------+ | 7 6 3 | 2 9 4 | 1 5 8 | | 1 4 9 | 3 8 5 | 6 2 7 | | 8 2 5 | 1 6 7 | 9 4 3 | +-------+-------+-------+ | 5 9 7 | 4 1 3 | 2 8 6 | | 4 8 6 | 5 2 9 | 3 7 1 | | 2 3 1 | 6 7 8 | 4 9 5 | +-------+-------+-------+ | 6 1 8 | 9 5 2 | 7 3 4 | | 9 7 4 | 8 3 1 | 5 6 2 | | 3 5 2 | 7 4 6 | 8 1 9 | +-------+-------+-------+ Ed |
Author: | Andrew [ Tue Sep 10, 2019 5:31 am ] |
Post subject: | Re: Assassin 383 |
Thanks Ed for a very challenging Assassin. As you said one can make a lot of early progress, then it gets really hard but more in the sense of searching for what to do next rather than technically harder steps. The SS score of 2.00 seems very high; maybe I found some steps which SS isn't programmed for. Some of my combination analysis was harder than I've used in recent Assassins; I've tried to explain it as clearly as possible. Ed said that JSudoku used 3 easy chains; so did I, although I won't claim that they were all easy ones. Here is my walkthrough for Assassin 383: Prelims a) R12C1 = {17/26/35}, no 4,8,9 b) R1C23 = {18/27/36/45}, no 9 c) R1C89 = {49/58/67}, no 1,2,3 d) R2C89 = {18/27/36/45}, no 9 e) R67C1 = {17/26/35}, no 4,8,9 f) R7C23 = {18/27/36/45}, no 9 g) R8C23 = {29/38/47/56}, no 1 h) R9C78 = {18/27/36/45}, no 9 1. 45 rule on C89 2 outies R39C7 = 17 = [98] -> R9C8 = 1, clean-up: no 4 in R1C89, no 8 in R2C9 1a. R3C7 = 9 -> R3C89 = 7 = {25/34}/[61], no 7,8, no 6 in R3C9 1b. 45 rule on N3 2 innies R12C7 = {16/25/34}, no 7 1c. Hidden killer pair 7,8 in R1C89 and R2C89 for N3, R1C89 contains one of 7,8 -> R2C89 must contain one of 7,8 = {27}[81], no 3,4,5,6 [Alternatively killer triple 4,5,6 in R12C7, R1C89 and R3C89 …] 1d. 45 rule on N23 1 innie R3C4 = 1 outie R4C5, no 9 in R3C4 -> no 9 in R4C5 1e. 9 in R1 only in R1C456, locked for N2 1f. 30(7) cage at R1C4 contains 9 = {1234569}, no 7,8 1g. 8 in R1 only in R1C23 = {18} or R1C89 = {58} -> R1C23 = {18/27/36} (cannot be {45}, locking-out cages), no 4,5 1h. 4 in R1 only in R1C4567, locked for 30(7) cage, clean-up: no 3 in R1C7 1i. 9 in N1 only in R2C23, CPE no 9 in R4C3 1j. 45 rule on R12 3 innies R2C235 = 21 and must contain 4 for R2 = {489}, 8 locked for R2, clean-up: no 1 in R2C9 1k. Naked pair {27} in R2C89, locked for R2 and N3, clean-up: no 1,6 in R1C1, no 6 in R1C89 1l. Naked pair {58} in R1C89, locked for R1, 5 locked for N3, clean-up: no 1 in R1C23, no 3 in R2C1 1m. 2 in 30(7) cage only in R1C456, locked for R1 and N2, clean-up: no 7 in R1C23, no 6 in R2C1, no 2 in R4C5 1n. Naked pair {36} in R1C23, locked for R1 and N1, R1C1 = 7 -> R2C1 = 1 1o. 5 in R2 only in R2C46, locked for N2, clean-up: no 5 in R4C5 1p. Naked quint {24589} in R2C23 + R3C123, CPE no 2,4,5,8 in R4C3 1q. 1 in N7 only in R7C23 = {18}, locked for R7, 8 locked for N7, clean-up: no 3 in R8C23 1r. 18(3) cage at R6C8 = {369/378/459/468/567} (cannot be {279} which clashes with R2C8), no 2 2. 45 rule on N7 2 innies R7C1 + R9C3 = 8 = {26/35} 2a. R8C23 = {29/47} (cannot be {56} which clashes with R7C1 + R9C3), no 5,6 2b. 17(3) cage at R8C1 = {269/359/467} 2c. 7 of {467} must be in R9C2 -> no 4 in R9C2 2d. 17(3) cage at R8C4 = {269/278/359/368/458/467} (cannot be {179} because no 1,7,9 in R9C3), no 1 3. 15(3) cage at R2C3 = {159/249/348/456} (cannot be {168/267/357} because 1,3,6,7 only in R3C4, cannot be {258} because 2,5 only in R3C3), no 7, clean-up: no 7 in R4C5 (step 1d) 3a. 3 of {348} must be in R3C4 -> no 8 in R3C4, clean-up: no 8 in R4C5 (step 1d) [Fairly heavy combination analysis; I’ve tried to make it as clear as possible.] 4. 35(6) cage at R2C2 = {236789/245789/345689} (cannot be {146789} = {489}{167} which clashes with R2C3), no 1 4a. 15(3) cage at R2C3 (step 3) = {159/249/348/456}, R3C4 = R4C5 (step 1d) -> split 15(3) cage R23C3 + R4C5 = 15 = {159/249/348/456} [Alternatively 45 rule on N23 3 outies R23C3 + R4C5 = 15 …] 4b. 35(6) cage at R2C2 = {245789/345689} (cannot be {236789} = 9{28}{367} which clashes with split 15(3) cage) 4c. 35(6) cage = {345689} can only be 4{58}{369} (cannot be {489}{356} which clashes with R2C3, cannot be 9{45}{368} which clashes with 15(3) cage at R2C3) 4d. 15(3) cage at R5C1= {159/168/249/258/267/348/357/456} 4e. 35(6) cage = {245789} (cannot be {345689} = 4{58}{369} which clashes with seven of the combinations of 15(3) cage at R5C1 while 4{58}{369} clashes with combined cage 15(3) at R5C1 = {258} + R6C1 = {2356}) 4f. 35(6) cage = {245789}, no 3,6 -> R4C3 = 7, clean-up: no 4 in R8C2 5. 15(3) cage at R5C1= {159/249/258/348/456} (cannot be {168} because 15(3) cage = {168} + R7C2 = {18} block all 8s in 35(6) cage at R2C2) 5a. 45 rule on R123 3 remaining outies R4C125 = 15, R4C5 = {1346} -> R4C12 = 9,11,12,14 5b. 15(3) cage = {159/348/456} (cannot be {249/258} because R4C12 cannot be {49/58} = 13), no 2 5c. 45 rule on N7 1 outie R6C1 = R9C3 5d. Consider placement for 5 in C3 R3C3 = 5 => 5 in 35(6) cage at R2C2 must be in R4C12, locked for N4 or 5 in R56C3, locked for N4 or 5 in R9C3 => R6C1 = 5 -> 15(3) cage = {348} (only remaining combination), locked for N4, clean-up: no 5 in R7C1, no 3 in R9C3 (step 2) 5e. 35(6) cage at R2C2 = {245789}, 4,8 locked for N1 -> R2C3 = 9, R3C34 = 5 = [24/51], clean-up: no 3,6 in R4C5 (step 1d) 5f. 9 in N4 only in R4C12, locked for R4 5g. R7C3 = 8 (hidden single in C3) 5h. R1C3 = 3 (hidden single in C3) -> R1C2 = 6 5i. R8C3 = 4 (hidden single in C3) -> R8C2 = 7 5j. Naked quad {1249} in R1C456 + R3C4, locked for N2 -> R2C5 = 8, R2C2 = 4 5k. Naked pair {38} in R56C2, locked for C2 and N4 -> R5C1 = 4 5l. R3C1 = 8 (hidden single in N1) 6. 17(3) cage at R8C4 (step 2d) = {269/278/359/368/458} (cannot be {467} because 4,7 only in R9C4) 6a. 17(3) cage = {269/278/359/368} (cannot be {458} = [854] which clashes with R3C34), no 4 6b. 5 of {359} must be in R9C3 -> no 5 in R89C4 7. 15(3) cage at R4C4 = {159/168/258/267/348/357/456} (cannot be {249} which clashes with R13C4, ALS block) 7a. Deleted; this step has been replaced by step 8g. [This one took me a long time to spot. I wonder whether steps 8b and 8e are “human” steps.] 8. 17(3) cage at R8C4 (step 6a) = {269/278/359/368} 8a. 45 rule on N8 3 innies R789C6 = 1 outie R9C3 + 7 8b. Consider placement for 8 in N8 R8C4 = 8 => 17(3) cage = {278/368}, no 5 or R8C6 = 8 -> R789C6 must total at least 13 (cannot be {138} because 1,8 only in R8C6) => R9C3 must be 6 -> no 5 in R9C3, clean-up: no 3 in R7C1 (step 2), no 5 in R6C1 8c. Naked pair {26} in R67C1, locked for C1 8d. Naked pair {26} in R7C1 + R9C3, 2 locked for N7 [Taking step 8b further …] 8e. R789C6 + R9C3 cannot be {238}6 which clashes with R89C4 = {29/38} -> no 8 in R8C6 8f. R8C4 = 8 (hidden single in N8) -> R9C34 = 9 = [27/63] 8g. 15(3) cage at R4C4 (step 7) = {159/267/456} (cannot be {357} which clashes with R9C4), no 3 9. 45 rule on N6789 2 innies R45C7 = 1 outie R6C1 + 3 9a. R6C1 = {26} -> R45C7 = 5,9 = {23}/[27/45] (cannot be {14} which clashes with R1C7, cannot be {36} which clashes with R2C7), no 1,6, no 5 in R4C7 10. 45 rule on N8 4 remaining innies R9C4 + R789C6 = 16 = {1267/1357/2347/2356} (cannot be {1249/1456} because R9C4 only contains 3,7), no 9 10a. 25(6) cage at R6C7 = {124567} (only remaining combination), no 3 10b. R9C4 + R789C6 = {1267/2347} (cannot be {1357} = 3{157} because R678C7 = {246} which clashes with R12C7, cannot be {2356} = 3{256} because R678C6 = {147} clashes with R1C7), no 5, 2,7 locked for N8, 2 locked for C6 and 25(6) cage at R6C7 10c. 2 in C7 only in R45C7, locked for N6 10d. R45C7 = R6C1 + 3 (step 9), R6C1 = {26} -> R45C7 = 5,9 contains 2 = {23}/[27], no 4,5 10e. R9C4 + R789C6 = {1267/2347} -> R789C6 = {126/247} 10f. 1 of {126} must be in R8C6 -> no 6 in R8C6 10g. R789C6 = {126/247} -> R678C7 = {457/156} 10h. 5 of {457} must be in R8C7, 1 of {156} must be in R6C7 -> no 5,6 in R6C7 10i. 5 in C7 only on R78C7, locked for N9 10j. R2C8 = 2 (hidden single in C8) -> R2C9 = 7 10k. 7 in N9 only in R7C78, locked for R7 10l. 2 in N9 only in 20(4) cage at R6C9 = {2369/2459/2468} (cannot be {1289} because 1,8 only in R6C9), no 1 10m. 5,9 in N8 only in 21(4) cage at R7C4 = {1569/3459} 10n. 1 of {1569} must be in R8C5 -> no 6 in R8C5 10o. 6 in R8 only in R8C789, locked for N9 11. R789C6 = {126/247} -> R678C7 = {457/156} (step 10g), 21(4) cage at R7C4 (step 10m) = {1569/3459} 11a. Consider placements for 5 in R8 R8C1 = 5 => R9C1 = 3 (hidden single in N7) or R8C5 = 5 => 21(4) cage = {3459}, 3 locked for N8 or R8C7 = 5 => R678C7 = {457} => R789C6 = {126} => R8C5 => 21(4) cage = {3459}, 3 locked for N8 -> R9C4 = 7, R9C3 = 2 (cage sum), R67C1 = [26], R3C23 = [25], R3C4 = 1 (cage sum) [Cracked, at last.] 11b. 15(3) cage at R4C4 (step 8g) = {456} (only remaining combination), locked for C4 and N5 -> R127C4 = [239], R1C7 = 1 (hidden single in R1), R2C7 = 6, R8C7 = 5 11c. Naked pair {47} in R67C7, 7 locked for C7, 4 locked for 25(6) cage at R6C7 -> R789C6 = [216], R8C5 = 3, R8C1 = 9, R9C12 = [35], R4C1 = 5, R8C89 = [62] 11d. Naked pair {34} in R37C9, locked for C9 12. R5C5 = 2 (hidden single in C5), R5C7 = 3, R56C2 = [83] 12a. R79C5 = [54], R4C5 = 1, R16C5 = [97], R5C6 = 9 12b. R5C9 = 1 (hidden single in N6), R5C8 = 7 (hidden single in R5) -> R4C89 = 14 = [86] 12b. R689C9 = [529] = 16 -> R7C9 = 4 (cage sum) and the rest is naked singles. Rating Comment: I'll rate my walkthrough for A383 at Hard 1.5; at a human level Very Hard 1.5, because of difficulty in finding the later steps but I won't rate it any higher. |
Author: | Ed [ Thu Sep 12, 2019 8:46 am ] |
Post subject: | Re: Assassin 383 |
Thanks for getting this one out Andrew! We both like a challenge! Some really nice steps in there, especially 4b and 8b. I did it a little differently (esp step 16 & 28). Thanks to wellbeback for telling me about a typo.[edit2: And Andrew for more I missed and some suggestions. Thanks!] WT for A383: Preliminaries Cage 8(2) n1 - cells do not use 489 Cage 8(2) n47 - cells do not use 489 Cage 13(2) n3 - cells do not use 123 Cage 9(2) n3 - cells do not use 9 Cage 9(2) n9 - cells do not use 9 Cage 9(2) n7 - cells do not use 9 Cage 9(2) n1 - cells do not use 9 Cage 11(2) n7 - cells do not use 1 no clean-up done unless stated 1. "45" on c89: 2 outies r39c7 = 17 = [98], r9c8 = 1 1a. r3c89 = 7 (no 7,8) 1b. no 4 in 13(2)n3 2. "45" on n3: 2 innies r12c7 = 7 (no 7) 3. hidden killer pair 7,8 in n3 -> 9(2) must have one of 7,8 = [81]/{27}(no 3,4,5,6, no 8 in r2c9) = 7 or 8 4. "45" in r12: 3 innies r2c235 = 21 4a. but {678} blocked by 9(2)n3 4b. = {489/579}(no 1,2,3,6) 4c. 9 locked for r2 4d. killer pair 7,8 with 9(2)n3: both locked for r2 4e. no 1 in r1c1 5. r12c7 = 7 -> rest of 30(7) in n2 = 23 and must have 9 for r1 5a. but {12389/12479} blocked by h7(2)r12c7 5b. = {12569/13469/23459}(no 7,8) 5c. 9 locked for n2 6. 8 in r1 in 13(2)n3 = {58} or in 9(2)n1 -> 5 in 9(2) must also have 8 or there would be no 8 for r1 -> {45} blocked (Locking-out cages) 6a. 9(2) = {18/27/36}(no 4,5) 7. 4 in r1 only in 30(7)r1c4: 4 locked for 30(7) cage 7a. no 3 in r1c7 (h7(2)) 8. 4 in r2 only in h21(3)r2c235 = {489} only: 8 locked for r2 9. 9(2)n3 = {27}: both locked for n3, 2 for r2 9a. deleted 10. 13(2)n3 = {58}: both locked for r1, 5 for n3 10a. no 1 in 9(2)n1 11. 1 in r1 only in 30(7)r1c4: locked for 30(7) 11a. r2c1 = 1 (hsingle r2), r1c1 = 7 11b. no 6 in r1c7 (h7(2)r12c7) 12. 9(2)n1 = {36}:both locked for r1 and n1 13. 2 and 5 in n1 only in r3: both locked for r3 13a. and r4c3 sees 2,4,5,8,9 in n1 -> not in r4c3 (Common Peer Elimination, CPE) 14. r3c12 must have at least one of 2 or 5 for r3/n1 14a 35(6)r2c2: -> {146789} blocked (hidden killer pair) 14b. = {236789/245789/345689}(no 1) 15. 1 in n7 only in 9(2) = {18} only: both locked for r7 and 8 for n7 [edit: from Wellbeback's WT step 5, I could have gotten more out of this next step which would have made a big difference down to step 22] 16. "45" on c12: 4 outies r1478c3 = 22 16a. but r148c3 <> 21 -> no 1 in r7c3 16b. r7c23 = [18] 16c. no 3 in 11(2)n7 16d. -> r148c3 = 14 16e. must have 3 or 6 for r1c3 16f. = {239/347/356} 16g. but {239} blocked since 2,9 only in r8c3 16h. and [365] blocked by [66] in r18c2 16i. = [374/635](no 6 in r4c3, r8c3 = (45), r8c2 = (67) 16j. must have 3: locked for c3 16k. and 6 locked in r18c2 for c2 [edit: Andrew worked out what I missed. If I'd done "45" on n7 then [56] would have been blocked from r8c23, resulting in 3 placements. Thanks Andrew!] 17. r3c12 <> {25} since [94] not possible in 15(3)r2c3 17a. -> r3c3 from (25) (hidden killer pair) 17b. 15(3) = [924/951/456](r3c4=(1,4,6)) 18. r148c3 = 14 = [374/635] (step 16i) = 4 or 5 18a. -> [456] blocked from 15(3)r2c3 18b. = [924/951] -> r2c3 = 9, r3c34 = [24/51] 18c. killer pair 1,4 with r3c89: both locked for r3 18d. r2c2 = 4 (hsingle n1), r2c5 = 8 19. 8 in n1 only in 35(6)-> no 8 in r4c12 19a. 35(6) = [4]{25789/35689} 19b. must have 9: locked for n4 and r4 20. 15(3)n4 must have 8 for n4 = {258/348}(no 6,7) This is a step I missed in the v1.35 of this puzzle. 21. 35(6)r2c2 = [4]{25789/35689}(step 19a) 21a. {35689} = {58}[693] only 21b. but [693]+ 15(3)={258} blocked by r6c1 = (2356) 21c. -> 35(6) = [4]{25789} only (no 3,6) 21d. r4c3 = 7 21e. -> r18c3 = 7 = [34] only (step 18), r18c2 = [67] 22. 15(3)n4: {258} blocked by r4c12 = {259} 22a. = {348} only: -> r5c1 = 4, r56c2 = {38}: both locked for c2, 3 for n4 22b. r3c1 = 8 (hsingle n1) 23. 17(3)r9c3: {179} blocked by none in r9c3 23a. {467} blocked by 4,7 only in r9c4 23b. {458} as [854] only, blocked by r3c34 = 5 in r3c3 or 4 in r3c4 23c. = {269/278/359/368}(no 1,4) 23d. note: if it has 6, must have 2 or 3 Took a long time to find this 24. "45" on n8: 1 outie r9c3 + 7 = 3 innies r789c6 24a. -> max 3 innies = 13 (with 6 in r9c3) 24b. -> 8 in r8c6 can only have {23} in r79c6 24c. but this blocks 6 from r9c3 (step 23d) 24d. -> no 8 in r8c6 25. r8c4 = 8 (hsingle n8) 25a. -> r9c34 = 9 = [27/63] 26. "45" on n7: 2 innies r7c1 + r9c3 = 8 = {26} only: both locked for n7 27. 25(6)r6c7 = {123469/124567} 27a. note: must have 1,2,4,6 27b. since r12c7 = [16/34] -> 25(6) must have both {16} or {4} in r789c6 (note: can't have all 3 since from iodn8=+7, no 4 available in r9c3, step 24) 27c. -> 21(4)n8 must have both {16} or 4 for n8 27d. = {1569/3459}(no 2,7) 27e. note: if it has 4, also has 3 Found a couple of ways to proceed from here but this is the shortest 28. "45" on r9: 5 innies r9c12569 = 27 28a. but {23679} blocked by r9c3 [or must contain 4 & 5 for r9] 28b. = {24579/34569} 28c. but {24579} as {59}[4]{27} only, forces 3 into r9c4 and 21(4) cage (step 27e), ie, two 3s in n8 28d. -> = {34569} only (no 2,7) 28e. 3 & 6 locked for r9 29. r9c34 = [27], r67c1 = [26], r3c234 = [251] 30. "45" on n23: 1 outie r4c5 = 1 30a. r8c6 = 1 (hsingle n8) 30b. -> must have 6 in c6 (step 27b)-> r9c6 = 6 30d. r7c6 = 2 (hsingle n8) 30e. 4 must be in 25(6): only in c7: locked for c7 30f. r12c7 = [16] 30g. r3c56 = [67](hsingles) 31. 15(3)n5: must have 6 for c4 = {456} only: 4,5 locked for c4 and n5 32. "45" on n9: 3 remaining outies r6c789 = 18 and must have 4 or 5 for r6 since the only other place is r6c4 32a. = {459/468/567}(no 1,3) 33. r5c9 = 1 (hsingle n6) 33a. -> 22(4)r4c8: [1]{579} blocked by 7,9 only in r5c8 33b. = {48}[91]/{68}[71](no 2,3,5) 33c. 8 locked in r4c89 for r4 and n6 33d. r45c7 = {23}(hidden pair n6), 3 locked for c7 easy now Ed |
Author: | wellbeback [ Mon Sep 16, 2019 1:59 am ] |
Post subject: | Re: Assassin 383 |
Some crazy busy times for me recently. Apologies for not getting this out sooner. The more I looked at it the more initial 'easier' steps I found! Thanks Ed - I loved this one. Kept finding better ways (to my way of thinking) of doing things. Assassin 383 WT: 1. Outies c89 = +17(2) = {89} -> 9(2)n9 = [81] and r3c7 = 9 2. -> 9 in r1 only in r1c456 -> 30(7)r1c4 = {1234569} Since only one each of (1234) left in r12 -> values in 9(2)n1 and in 9(2)n3 cannot be the same! 3. Innies n3 = r12c7 = +7(2) -> One each of (78) in n3 in 13(2) and 9(2)n3 -> Whichever of (78) is in 9(2)n3 goes in n2 in r3c456 and in n1 in r1c1. (Cannot be in r1c23 which puts the two 9(2)s the same!). -> r1c1 = 7 -> 9(2)n1 = [71] -> 9(2)n3 = {27} -> 13(2)n3 = {58} -> 9(2)n1 = {36} -> 30(7)n2 = [{1249}[{356}] with r12c7 = [16] or [43] -> r2c235 = {489} with 9 in r2c23 Also (25) in r3 in n1 in r3c123 4. 1 in n7 only in r7c23 -> 9(2)n7 = {18} -> (18) in r8 in n8 in r8c456 Innies n7 = r7c1,r9c3 = {26} or {35} -> 11(2)n7 = {29} or {47} 5! r4c3 sees all in n1 except in 8(2) and 9(2) -> r4c3 from (1367) Since at least one of (25) in r3c12 -> 35(6)n14 cannot contain a 1. -> r4c3 from (367) Outies c12 = r1478c3 = +22(4) Given previous placements this can only be [3784] 6. -> 9(2)n1 = [63], 9(2)n7 = [18], 11(2)n7 = [74] -> NS r2c3 = 9 -> (48) in n1 in 35(6) -> (48) in n4 in 15(3) -> 15(3)n3 = {348} -> 35(6) = [{24589}7] with (48) in n1 and (79) in n4 7. 15(3)r2c3 from [924] or [951] If the former -> r2c25 = [48] If the latter -> r1c4567 = [{249}1] -> r2c25 = [48] Either way r2c25 = [48] -> 15(3)n4 = [4{38}] -> r3c1 = 8 8. r9c3 cannot be 7 or 9 -> r8c4 cannot be 1. -> 1 in n8 in r8c56 8 in r2c5 -> 8 in n8 in r8c46 -> Either r8c4 = 8 or r8c56 = [18] In the latter case 21(4)n8 must contain a 9 -> In neither case can 9 be in r89c4 9! r3c34 from [51] or [24] 8(2)r6c1 from [53] or {26} Innies n7 = r7c1,r9c3 = +8(2) -> r9c3 (= r6c1) from (256) a) r9c3 = 2 puts r89c4 = [87] b) r9c3 = 5 puts r3c34 = [24] leaves no solution for r89c4 c) r9c3 = 6 puts r89c4 = [83] -> r8c4 = 8 and innies n7 = {26} -> 17(3)n7 = {359} with 3 in r89c1 Also 8 not in 25(6)r6c7 10! IOD n689 -> r4c7 + r5c7 = r9c3 + 3 r9c3 from (26) -> r45c7 = +5(2) or +9(2) -> At least one of (57) in c7 in r678c7 -> 25(6)r6c7 must be {124567} 11. r3c34 from [51] or [24] The latter puts r12c7 = [43] and Outies r123 = r4c1235 = [{29}74] Also puts r9c3 = 6 and -> r45c7 = +9(2) BUT - there is no solution here for r45c7 = +9(2) -> r3c34 = [51] 12. -> Outies r123 = r4c1235 = [{59}71] Also r3c2 = 2 Also r12c7 = [16] -> r3c56 = {67} -> r3c89 = {34} 13! -> 3 in c7 only in r45c7 Since r45c7 = r9c3 + 3 and since r9c2 from (26) and since 6 already in c7 -> r9c3 = 2 and r45c7 = {23} 14. Easy from here -> r89c4 = [87] Also 8(2)r6c1 = [26] Also r678c7 = {457} -> r789c6 = [216] -> r3c56 = [67] -> 15(3)n5 = {456} -> r2c46 = [35] -> HS(c4) r1c4 = 2 -> 21(4)n8 = [9{345}] -> r56c5 = [27] -> r1c56 = [94] Also r45c7 = [23] -> r56c2 = [83] -> r456c6 = [398] -> Remaining Innies r6789 = r6c34 = +7(2) = [16] etc. |
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