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Assassin 422 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1663 |
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Author: | Ed [ Tue Aug 02, 2022 2:16 am ] |
Post subject: | Assassin 422 |
Attachment: a422.png [ 61.16 KiB | Viewed 3272 times ] Really enjoyed this one. Interesting start then a bit of a bump at the end. What I always aim for!! SudokuSolver gives it 1.70 code: triple click: 3x3::k:4864:4864:4864:2561:2561:2561:7938:4611:4611:4612:4612:4612:3077:7938:7938:7938:4611:5126:2567:1800:4612:3077:7938:8969:4611:4611:5126:2567:1800:7946:8969:8969:8969:8969:5126:5126:4363:7946:7946:7946:7692:8969:3341:3341:5390:4363:4363:7946:7692:7692:7692:7692:3341:5390:5397:5397:7946:4112:7692:4369:4369:2834:5390:5397:5396:4112:4112:4371:4371:4369:2834:2575:5397:5396:5396:5396:4371:4371:4369:2834:2575: solution: +-------+-------+-------+ Ed |
Author: | wellbeback [ Sun Aug 07, 2022 5:31 pm ] |
Post subject: | Re: Assassin 422 |
I my not have found the most efficient path here, but I found it very interesting! Thanks Ed! Assassin 422 WT: 1. Innies n3 = +27(4) Innies c9 = r1234c9 = +14(4) -> r23c9 = Max +11(2) -> r12c7 is Min +16(2) IOD n2 -> r12c7 = r3c6 + 8 -> One of: (A) r12c7 = {79}, r3c6 = 8, r23c9 = +11(2), r14c9 = {12} (B) r12c7 = {89}, r3c6 = 9, r23c9 = +10(2), r14c9 = {13} But in case (A) there is no place for both 7 and 9 in n2 -> r12c7 = {89}, r3c6 = 9, r23c9 = +10(2), r14c9 = {13} 2. 9 in c8 not in 18(5)n2 nor in 11(3)n9 -> 9 in c8 in n6 Since r4c9 from (13) -> 9 not in 13(3)n6 -> (HS 9 in c8) r4c8 = 9 -> r14c9 = [31] -> (HS 9 in c9) r7c9 = 9 -> (IOD n9) r7c6 = 2 Also (HS 2 in c9) 10(2)n9 = {28} -> 21(3)c9 = [{57}9] -> r23c9 = {46} Also (Outies c89) r35c7 = [12] -> r123c8 = {257} -> 13(3)n6 = [2{38}] -> 11(3)n9 = {146} -> r789c7 = {357} Also r46c7 = {46} 3. Innies r1 = r1c789 = +16(3) This can only be [853] -> r2c7 = 9 -> 9 in r1 in r1c123 -> 19(3)n1 = {469} -> 10(3)r1 = {127} -> 12(2)n2 = {48} Also Innies n1 = r3c12 = +(8)2 can only be [35] -> r4c12 = [72] Also 31(5)n23 = [8{35}96] -> r23c9 = [64] -> 12(2)n2 = [48] -> 18(4)n1 = {1278} with (18) in r2. 4. Remaining Innies n78 = r7c35 = +13(2) = {58} or {67} Outies n14 = r5c4 + r7c3 = +12(2) This cannot be {39} since r7c3 from (5678) Also cannot be [66] since they are in the same cage. Also cannot be {48} since (48) already in c4 -> [r5c4,r7c3] = {57} Andrew pointed out that the first part of this next step is unnecessary since Step 3 Line 10 has already put 6 in c5. D'oh! 5! r46c7 = {46} Since r5c4 from (57) -> Whichever of (46) is in r4c7 must be in n5 in the 30(6) cage. Similarly whichever of (46) is in r6c7 is in n5 in the 35(6) cage. -> Both (46) are in 30(6) cage in n56 and also both in 35(6) cage in n56. -> r7c5 cannot be 6 -> r7c35 = [58] -> r5c4 = 7 Also 8 in n5 in r45c6 Also r56c9 = [57] 6. Innies c1234 = r146c4 = +8(3) Since 4 already in c4 and (12) already in r4 -> r146c4 = <152> -> 35(6) = [95{3468}] with r4c7 from (46) Also 30(6)n5 = [{12469}8] with r6c7 from (64) Also (HS) r6c1 = 5 7. Remaining cells c4 = r789c4 = {369} -> 17(4)n8 = {1457} Also r8c3 from (147) 8. 17(3)n4 cannot be [953] since that would put all of (369) in 21(4)r8c2 -> 17(3)n4 = <458> 9! (Complicated step!) Consider where (369) go in n7 None go in r78c3. Neither 21(4)r7c1 nor 21(4)r8c2 can contain all of (369) Since one of (369) already in r9c4 -> two of (369) are in each of the 21(4)s. Since 3, 6, 9, and 21 are all divisible by 3 -> the sum of the other two cells in each of the 21(4)s must also be divisible by 3. (E.g., (24), (18) etc.) Conclusions from that: (A) 21(4)r7c1 cannot contain both (14) -> in r7 one of (14) in r7c12 and the other in r7c8 -> 6 in r89c8 (B) If 3 in 21(4)r7c1 it must be [13{89}] (Cannot be [43{68}] since one of (48) already in c1). If 3 not in 21(4)r7c1 -> that cage must have (69) so can only be [{46}{29}] -> 9 in r89c1 Also (HS 7 in r7) r7c7 = 7 (C) 21(4)r8c2 cannot contain (36) since the other two cells in that cage in n7 cannot be either {57} or {48} -> 21(4)r8c2 must contain a 9 in r9c4 -> r78c4 = {36} and r8c3 = 7 10! -> r8c1 = 9 -> 2 in n7 only in r9c13 -> 10(2)n9 = [28] -> (HS 8 in r8) r8c2 = 8 -> 21(4)r8c2 = [8{13}9] Also -> 17(3)n4 = [854] -> 21(4)r7c1 = [4692] -> 19(3)n1 = [694] Also 17(4)n1 = [17{28}] Also r56c8 = [38] -> r4c6 = 8 Also r46c7 = [46] -> 35(6) = [953846] -> r6c3 = 3 -> r9c23 = [31] -> 31(6)r4c3 = [619735] etc. First time for ...: I have used even/odd in a number of WTs - but don't remember ever using divisible by 3 before! |
Author: | Andrew [ Sun Aug 07, 2022 6:46 pm ] |
Post subject: | Re: Assassin 422 |
Ed wrote: Interesting start then a bit of a bump at the end. What I always aim for!! Hope you'll show us your interesting start; wellbeback has a very powerful start, mine was fairly routine. I've slightly simplified how I dealt with the bump at the end. Here's how I solved Assassin 422: Prelims a) R23C4 = {39/48/57}, no 1,2,6 b) R34C1 = {19/28/37/46}, no 5 c) R34C2 = {16/25/34}, no 7,8,9 d) R89C9 = {19/28/37/46}, no 5 e) 19(3) cage at R1C1 = {289/379/469/478/568}, no 1 f) 10(3) cage at R1C4 = {127/136/145/235}, no 8,9 g) 21(3) cage at R5C9 = {489/579/678}, no 1,2,3 h) 11(3) cage at R7C8 = {128/137/146/236/245}, no 9 i) 18(5) cage at R1C8 = {12348/12357/12456}, no 9 1a. 18(5) cage at R1C8 = {12348/12357/12456}, 1,2 locked for N3 1b. 45 rule on N1 2 innies R3C12 = 8 = [26/35/62/71] -> R4C1 = {3478}, R4C2 = {1256} 1c. 45 rule on R789 3 innies R7C359 = 22 = {589/679}, 9 locked for R7 1d. 45 rule on N14 2 outies R5C4 + R7C3 = 12 = [39/48/57/75] 1e. 45 rule on C9 4 innies R1234C9 = 14 = {1238/1256/1346/2345} (cannot be {1247} which clashes with R89C9), no 7,9 1f. 45 rule on C9 1 outie R4C8 = 1 innie R1C9 + 6 -> R1C9 = {123}, R4C8 = {789} 1g. 45 rule on R1 3 innies R1C789 = 16 = {169/178/259/268/349/358/367} (cannot be {457} because R1C9 only contains 1,2,3) 1h. R1C9 = {123} -> no 1,2,3 in R1C78 1i. 9 of {349} must be in R1C7 -> no 4 in R1C7 1j. Min R23C9 + R4C8 = 14 -> max R4C9 = 6 1k. 45 rule on C89 2 outies R35C7 = 3 = {12}, locked for C7 1l. 13(3) cage at R5C7 = {139/148/157/238/247/256} (cannot be {346} because R5C7 only contains 1,2) 1m. R5C7 = {12} -> no 1,2 in R56C8 1n. R4C9 + R5C7 = {12} (hidden pair in N6) 1o. 45 rule on N9 1 innie R7C9 = 1 outie R7C6 + 7 -> R7C6 = {12}, R7C9 = {89} 2a. 9 in N3 only in R12C7, locked for C7 and 31(5) cage at R1C7 2b. 9 in N9 only in R789C9, locked for C9 2c. 45 rule on N23 2 outies R4C89 = 1 innie R3C6 + 1 2d. Min R4C89 = 8 -> min R3C6 = 7 2e. 20(4) cage at R2C9 = {1469/1568/2378/2459/2567} (cannot be {1289} because 1,2 only in R4C9, cannot be {1379} because 7,9 only in R4C8, cannot be {1478} which clashes with R1C9 + R4C8 = [17] , cannot be {2369} which clashes with R1C9 + R4C8 = [39], cannot be {2468} which clashes with R1C9 + R4C8 = [28], cannot be {3458/3467} because R4C9 only contains 1,2) 2f. 20(4) cage = {1469/2459/2567} (cannot be {1568/2378} which clash with 21(3) cage at R5C9), no 3,8, clean-up: no 2 in R1C9 2g. Killer triple 4,5,6 in 20(4) cage and 21(3) cage, locked for C9 2h. 18(5) cage at R1C8 = {12348/12357} (cannot be {12456} which clashes with 20(4) cage), no 6, 3 locked for N3 2i. 20(4) cage = {1469/2567} (cannot be {2459} which clashes with 18(5) cage), 6 locked for C9 and N3 2j. Killer pair 4,5 in 18(5) cage and 20(4) cage, locked for N3 2k. 21(3) cage at R5C9 = {489/579} -> R7C9 = 9, R7C6 = 2, clean-up: no 3 in R5C4, no 1 in R89C9 2l. 10(3) cage at R1C4 = {127/145/235} (cannot be {136} which clashes with R1C9), no 6 2m. 6 in R1 only in R1C123, locked for N1, clean-up: no 2 in R3C12 (step 1b), no 4,8 in R4C1, no 1,5 in R4C2 2n. 19(3) cage at R1C1 contains 6 = {469/568}, no 2,3,7 2o. Naked pair {37} in R34C1, locked for C1 2p. 2 in R1 only in R1C456, locked for N2 2q. 10(3) cage contains 2 = {127/235}, no 4 2r. R23C4 = {39/48} (cannot be {57} which clashes with 10(3) cage 2s. Combined cage 20(4) at R1C9 + 21(3) at R5C9 = {1469}{57}9/{2357}{48}9, 7 locked for N6 2t. 17(4) cage at R7C6 = {2357/2456} (cannot be {2348} which clashes with R89C9), no 8, 5 locked for C7 and N9 2u. 1 in N9 only in R789C8, locked for C8 3a. 45 rule on N6 3 outies R237C9 = 2 innies R46C7 + 9, R7C9 = 9 -> R23C9 = R46C7 3b. R23C9 contains 6 for N3 = {46/56} = 10,11 -> R46C7 = 10,11 = {46/38} 3c. R56C9 = {57} (cannot be {48} which clashes with R46C7), locked for C9 and N6 -> R4C8 = 9, R4C9 = 1 (cage sum), R1C9 = 3, R35C7 = [12], R3C2 = 5 -> R4C2 = 2, R3C1 = 3 (step 1b) -> R4C1 = 7, clean-up: no 9 in R2C4 3d. Naked pair {28} in R89C9, locked for N9 3e. Naked pair {46} in R23C9, 4 locked for N3 3f. R23C9 = {46} = 10 -> R46C7 = 10 = {46}, locked for C7 and N6 3g. Naked pair {38} in R56C8, locked for C8 3h. R4C89 = [91] = 10 -> R3C6 = 9 (step 2c), clean-up: no 3 in R2C4 3i. Naked pair {48} in R23C4, locked for C4 and N2, clean-up: no 8 in R7C3 (step 1d) 3j. 19(3) cage at R1C1 = {469} (only remaining combination), 4,9 locked for N1, 9 locked for R1 3k. 10(3) cage at R1C4 = {127} (only remaining combination), 1,7 locked for N2, 7 locked for R1 -> R1C78 = [85], R3C5 = 6, R23C9 = [64], R23C4 = [48] 3l. Naked pair {27} in R23C8, 7 locked for C8 and N3 3m. R7C359 (step 1c) = 22 = [589], R5C4 = 7, R56C9 = [57] 3n. R6C1 = 5 (hidden single in C1) -> R5C1 + R6C2 = 12 = [48/84/93] 3o. 45 rule on N8 3 remaining innies R789C4 = 18 = {369} (only remaining combination), locked for C4 and N8 -> R4C4 = 5 3p. 35(6) cage at R3C6 = {345689} (only remaining combination), no 1, 3 locked for N5 4a. 45 rule on N7 1 outie R9C4 = 1 remaining innie R8C3 + 2, R9C4 = {369} -> R8C3 = {147} 4b. 21(4) cage at R7C1 = {1389/2469/2478} (cannot be {1479} which clashes with R8C3, cannot be {2379} because 3,7 only in R7C2, cannot be {3468} which clashes with R15C1, ALS block) 4c. 1 of {1389} must be in R7C1 -> no 1 in R7C2 + R89C1 4d. 4,6,7 of {2469/2478} must be in R7C12 -> no 4,6 in R89C1 4e. 16(3) cage at R7C4 = {169/367} (cannot be [349] which clashes with 21(4) cage at R7C1), no 4, 6 locked for C4 4f. Consider placement for 3 in N7 R7C2 = 3 -> 21(4) cage at R7C1 = {1389} or 3 in R8C2 + R9C23, locked for 21(4) cage at R8C2 => R9C4 = 9 => R8C3 = 7 => 21(4) cage at R7C1 = {2469} -> 21(4) cage at R7C1 = {1389/2469}, no 7, 9 locked for C1 and N7, clean-up: no 3 in R6C2 (step 3n) 4g. Naked pair {48} in R5C1 + R6C2, locked for N4 4h. 45 rule on R1234 1 outie R5C6 = 1 innie R4C3 = {36} 4i. 16(3) cage at R7C4 = {367} (cannot be [619] which clashes with 21(4) cage at R7C1) -> R8C3 = 7, R9C4 = 9, R2C3 = 2 4j. R89C1 = [92] (hidden pair in C1) -> R7C12 = {46}, locked for R7 and N7 4k. Naked pair {46} in R17C1, 4 locked for C1 and the rest is naked singles. |
Author: | Ed [ Fri Aug 12, 2022 8:39 pm ] |
Post subject: | Re: Assassin 422 |
Here's my start. Very "wellbebackish". Same key area as the other two ways. start a422: Preliminaries Cage 7(2) n14 - cells do not use 789 Cage 12(2) n2 - cells do not use 126 Cage 10(2) n14 - cells do not use 5 Cage 10(2) n9 - cells do not use 5 Cage 21(3) n69 - cells do not use 123 Cage 10(3) n2 - cells do not use 89 Cage 11(3) n9 - cells do not use 9 Cage 19(3) n1 - cells do not use 1 Cage 18(5) n3 - cells do not use 9 1. "45" on c89: 2 outie r35c7 = 3 = {12}, both locked for c7 1a. 13(3) can't have both 1,2 -> no 1,2 in r56c8 2. "45" on c9: 1 innie r1c9 + 6 = 1 outie r4c8 (IOD=-6) 2a. = [17/28/39] 3. hidden pair 1,2 in n6 -> r4c9 = (12) 4. "45" on c9: 4 innies r1234c9 = 14 (no 9) 5. 9 in n3 only in r12c7 in 31(5). 9 locked for c7 and 31(5) 6. r4c8 sees all 7,8,9 in n3 apart from r12c7 so must repeat there 6a. "45" on n3: 2 innies r12c7 (and must have 9 for n3) - 7 = 2 outies r4c89 6b. but {89}/[82] blocked by 2 in r1c9 from iodc9=-6 6c. -> = {89}/[91]/{97}/[72] (no 8 in r4c8) (note: {97}/[81] breaks step 6. above) 6d. -> no 2 in r1c9 (iodc9=-6) 6e. -> [91] in r4c89 or [17] in r1c9 + r4c8 (iodc9=-6) 6f. -> 1 locked in r14c9 for c9 6g. -> no 9 in 10(2)n9 7. r7c9 = 9 (hsingle n9) On from there a422 ending: 1. From Andrew's step 4a. R9C4 = R8C3 + 2 1a. -> 3 in r9c4 has 1 in r8c3 2. From Andrew's step 4b. a. 21(4) cage at R7C1 = {1389/2469/2478} b if it has 3 also has 1 3. conclusion: 3 cannot repeat in the two 21(4) cages (@r7c1 and r8c2) 4. since 3 in n8 is only in those two cages -> no 3 in r9c4 4a. -> r9c4 = 9 [edit: I also removed 6 from r9c4 the same way: the 21(4)r7c1, if it has 6 must also have 4, and 6 in r9c4 must have 4 in r8c3 -> both those cages can't repeat 6] Cracked Ed |
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