SudokuSolver Forum http://www.rcbroughton.co.uk/sudoku/forum/ |
|
Assassin 421 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1662 |
Page 1 of 1 |
Author: | Ed [ Fri Jul 15, 2022 6:34 pm ] |
Post subject: | Assassin 421 |
Attachment: a421.png [ 62.91 KiB | Viewed 2889 times ] A real stop start solve for me with a final sting in the tail. Different. It gets 1.40. Another huge lot of rain here and more floods so completely messing with my commute and hence with puzzle time. Hopefully things back to normal next week. triple click code: 3x3::k:3584:3584:4097:3586:3586:7683:7683:7683:7683:5892:5892:4097:4097:3586:5125:6662:6662:7683:5892:3591:3591:3591:5125:5125:6662:6662:7683:5892:5892:3591:5896:5641:5641:5642:5642:7683:2827:3084:5645:5896:5641:5641:5642:2574:2574:2827:3084:5645:5896:5896:5896:3599:3344:2574:3089:3089:5645:5645:5645:3599:3599:3344:3344:4114:4114:3859:2836:2836:5141:3599:5141:3606:4114:4114:3859:2839:2839:5141:5141:5141:3606: solution: 3+-------+-------+-------+ Ed |
Author: | Andrew [ Sun Jul 17, 2022 3:36 am ] |
Post subject: | Re: Assassin 421 |
Thanks Ed for your latest Assassin. My solving path flowed fairly well. Thanks Ed for pointing out a small error; steps 4aa and 4ab deleted, the order of steps 6c to 6e rearranged. Here's my walkthrough for Assassin 421: Prelims a) R1C12 = {59/68} b) R56C1 = {29/38/47/56}, no 1 c) R56C2 = {39/48/57}, no 1,2,6 d) R7C12 = {39/48/57}, no 1,2,6 e) R89C3 = {69/78} f) R8C45 = {29/38/47/56}, no 1 g) R89C9 = {59/68} h) R9C45 = {29/38/47/56}, no 1 i) 20(3) cage at R2C6 = {389/479/569/578}, no 1,2 j) 22(3) cage at R4C7 = {589/679} k) 10(3) cage at R5C8 = {127/136/145/235}, no 8,9 l) 14(4) cage at R3C2 = {1238/1247/1256/1346/2345}, no 9 m) 14(4) cage at R6C7 = {1238/1247/1256/1346/2345}, no 9 n) 26(4) cage at R2C7 = {2789/3689/4589/4679/5678}, no 1 1a. 45 rule on N7 1 innie R7C3 = 2 1b. 45 rule on C12 1 innie R3C2 = 2 1c. 45 rule on R89 1 innie R8C7 = 3, clean-up: no 8 in R8C45 1d. 14(4) cage at R6C7 = {1238/1346/2345}, no 7 1e. 2 of {1238/2345} only in R6C7 -> no 5,8 in R6C7 1f. 22(3) cage at R4C7 = {589/679}, 9 locked for N6 1g. 2 in N9 only in R8C8 + R9C78, locked for 20(5) cage at R8C6 1h. 2 in N8 only in R8C45 = {29} or R9C45 = {29}, 9 locked for N8 (locking cages) 2a. 45 rule on N2 3 innies R1C6 + R23C4 = 11 = {128/137/146/236/245}, no 9 2b. 45 rule on N3 2 outies R1C6 + R4C9 = 11 = {38/47/56}, no 1,2 2c. 30(7) cage at R1C6 = {1234569/1234578}, 2 locked for N3 2d. 5,6 of {1234569} must be in R1C6 + R4C9, otherwise clash with 26(4) at R2C7 = {5678}, no 6 in {1234578} -> no 6 in R1C789 + R23C9 2e. 6 in N3 only in 26(4) cage = {3689/4679/5678} [Alternatively whatever pair are in R1C6 + R4C9 must be in 26(4) cage with the remaining two numbers in that cage totalling 15 -> 26(4) cage cannot be {4589}.] 3a. 45 rule on N6789 2 outies R56C3 = 1 innie R4C9 3b. Min R56C3 = 4 -> min R4C9 = 4, clean-up: no 8 in R1C6 (step 2b) 3c. Max R56C3 = 8, no 8,9 in R56C3 3d. Max R56C3 = 8, R7C3 = 2 -> min R7C45 = 12, no 1,3 in R7C45 3e. 3 in R7 only in R7C12 = {39}, locked for N7, 9 locked for R7, clean-up: no 6 in R89C3 3f. Naked pair {78} in R89C3, locked for C3 and N7 3g. R56C2 = {48/57} (cannot be {39} which clashes with R7C2), no 3,9 3h. R56C1 = {29/38/56} (cannot be {47} which clashes with R56C2), no 4,7 3i. 1 in N8 only in R789C6, locked for C6 3j. 45 rule on N6 3 innies R4C9 + R6C78 = 13 = {148/238/247/346} (cannot be {157/256} which clash with 22(3) cage at R4C7), no 5, clean-up: no 6 in R1C6 (step 2b) 3k. 3 of {346} must be in R6C8 -> no 6 in R6C8 3l. R1C6 + R23C4 (step 2a) = {137/146/236/245} (cannot be {128} because R1C6 only contains 3,4,5,7), no 8 3m. 45 rule on N6 5 outies R7C6789 + R8C7 = 1 innie R4C9 + 14 3n. Min R7C6789 = {1456} = 16, R8C7 = 3 -> no 4 in R4C9, clean-up: no 7 in R1C6 (step 2b) 3o. R4C9 + R6C78 = {148/238/247/346} 3p. R4C9 = {678} -> no 6,7,8 in R6C78 3q. R1C6 + R23C4 = {137/146/236/245} 3r. 3 of {137/236} must be in R1C6 -> no 3 in R23C4 3s. Consider permutations for R1C6 + R4C9 (step 2b) = [38/47/56] R1C6 + R4C9 = [38] => 3 in N3 only in R23C8 => R4C9 + R6C78 = {148} or R1C6 + R4C9 = [47] => R4C9 + R6C78 = {247} or R1C6 + R4C9 = [56] => R4C9 + R6C78 = {346} -> R4C9 + R6C78 = {148/247/346}, 4 locked for R6 and N6, clean-up: no 8 in R5C2 4a. 45 rule on C789 4 outies R1789C6 = 14 = {1346}, 3,4,6 locked for C6, 6 locked for N8, clean-up: no 6 in R4C9 (step 2b), no 5 in R8C45, no 5 in R9C45 4b. 5 in N8 only in R7C45, locked for R7 and 22(5) cage at R5C3 4c. 22(5) cage at R5C3 = {12568/23458}, 8 locked for R7 and N8, clean-up: no 3 in R9C45 4d. Naked quad {2479} in R89C45, 4,7 locked for N8 4e. R19C6 = [43] (hidden pair in C6) -> R4C9 = 7 (step 2b), clean-up: no 6 in 22(3) cage at R4C7 4f. Naked triple {589} in 22(3) cage, 5 locked for N6 4g. 10(3) cage at R5C8 = {136} (only remaining cage), 1,3 locked for N6 4h. R7C9 = 4 (hidden single in C9) -> R6C78 = [42], R7C8 = 7 (cage total/hidden single in R7), clean-up: no 9 in R5C1 4i. 4,6,7 in N3 only in 26(4) cage at R2C7 = {4679}, 9 locked for N3 4j. R9C7 = 2 (hidden single in N9), clean-up: no 9 in R9C45 4k. Naked pair {47} in R9C45, locked for R9 and N8 -> R89C3 = [78] 4l. Naked pair {29} in R8C45, 9 locked for R8 4m. R9C9 = 9 (hidden single in C9), R8C9 = 5 4n. R8C8 = 8 (hidden single in N9) 4o. 5 in N3 only in R1C78, locked for R1, clean-up: no 9 in R1C12 4p. Naked pair {68} in R1C12, locked for R1 and N1 4q. 9 in C3 only in R12C3, locked for N1 4r. 16(3) cage at R1C3 = {169/259} (cannot be {349} because no 3,4,9 in R2C4), no 3,4,7 4s. 2,6 only in R2C4 -> R2C4 = {26} 4t. R1C6 + R23C4 = 11 (step 2a), R1C6 = 4 => R23C4 = 7 = [25/61] 4u. 7 in R1 only in R1C45, locked for N2 4v. 14(3) cage at R1C4 = {167/257}, no 3,8,9 4w. 5,6 only in R2C5 -> R2C5 = {56} 4x. R3C5 = 3 (hidden single in N2), R23C6 = {89} (hidden pair in N2), locked for C6 4y. Naked triple {257} in R456C6, locked for N5 4z. R1C3 = 9 (hidden single in R1) 5a. 22(4) cage at R4C5 contains 2 and one of 2,5,7 in R45C6 = {2479/2569} (cannot be {2578} because 2,5,7 only in R45C6) -> R45C5 = {49/69}, 9 locked for C5 and N5 5b. R8C45 = [92] 6a. R7C3 = 2, R7C45 = {58} = 13 -> R56C3 = 7 = [16/43/61], no 3 in R5C3 6b. 45 rule on N4 3 remaining innies R4C123 = 15 = {168/348} (cannot be {159} which clashes with R4C8, cannot be {258} which clashes with R4C6, cannot be {456} which clashes with R56C2, cannot be {249} which clashes with R4C68, ALS block), 8 locked for R4 and N4, clean-up: no 3 in R56C1, no 4 in R5C2 6c. Naked pair {57} in R56C2, locked for C2, 5 locked for N4, clean-up: no 6 in R56C1 6d. R56C1 = [29] -> R7C12 = [39] 6e. Naked pair {59} in R4C78, locked for R4 and N6 -> R4C6 = 2, R5C7 = 8 6f. R9C1 = 5 (hidden single in N7) 6f. 6 in C3 only in R456C3, locked for N4 7a. Consider placements for R2C5 = {56} R2C5 = 5 => R3C24 = [21] = 3 => R34C3 = 11 = [56] or R2C5 = 6, R4C5 = 4 => R5C3 = 4 (hidden single in R5) -> no 4 in R3C3 7b. R3C2 = 2, R3C34 = {15} -> R4C3 = 6 (cage sum) 7c. Naked pair {15} in R23C3, 1 locked for C3 and N1 7d. R4C5 = 4, R9C45 = [47], R1C5 = 1, R3C45 = [15], R2C345 = [526] 7e. R4C56 = [42] = 6, R5C5 = 9 -> R5C6 = 7 (cage sum) and the rest is naked singles. |
Author: | wellbeback [ Sat Jul 30, 2022 6:05 pm ] |
Post subject: | Re: Assassin 421 |
Thanks Ed. Here's how I did it. Key observations towards solution (45s etc.) similar to Andrew's I think. Assassin 421 WT: 1. Innies n7 -> r7c3 = 2 Innies r89 -> r8c7 = 3 -> 2 in n9 in 20(5)r8c6 -> 2 in n8 in one of the 11(2)s = {29} 2. Innies c12 -> r3c2 = 2 -> 2 in n4 in r456c1 3! 22(3)n6 = {679} or {589} -> Max r4c9 = 8 Outies n3 = r1c6,r4c9 = +11(2) -> Min r1c6 = 3 -> Outies c789 = r1789c6 = +14(4) = {1346} 4. Innies n6 = r4c9+r6c78 = +13(3) -> r4c9 cannot be 5 since that leaves no place for 8 in n6 -> r1c6 cannot be 6 -> Outies n3 = [r1c6,r4c9] = [38] or [47] -> 30(7)r1c6 = {1234578} -> (69) in n3 in 26(4) 5. 6 in n8 in r789c6 -> 5 in n8 only in r7c45 Innies n689 = r4c9 + r7c45 = +20(3) = [7{58}] or [8{57}] But 22(5)r5c3 cannot contain all of (257) -> r4c9 = 7 and r7c45 = {58} Also r1c6 = 4 and r789c6 = [{16}3] Also 26(4)n3 = {4679} Also 22(3)n6 = {589} -> 10(3)n6 = {136} -> (NP) r6c78 = {24} -> (HS 4 in c9) r7c9 = 4 -> r6c78 = [42] and r7c8 = 7 -> (HS 9 in c9) 14(2)n9 = {59} Also r7c67 = {16} and 12(2)n7 = {39} -> 15(2)n7 = {78} and 16(4)n7 = {1456} 6. r56c3 = +7(2) -> 9 in c3 only in r12c3 -> 14(2)n1 = {68} -> 7 in n1 in 23(5)r2c1 -> 7 in r1 in n2 in r1c45 -> Remaining Innies n2 = r23c4 = +7(2) -> (89) in n2 only in 20(3)n2 -> 20(3)n2 = <839> -> (HS 9 in r1) r1c3 = 9 7. r4546c6 = {257} with 2 in r45c6 -> Either r45c6 = {27} and r45c5 = {49} or r45c6 = {25} and r45c5 = {69} (I.e. 9 locked in r45c5) -> 9 in r6 only in r6c1 (Since 12(2)n4 cannot be {39}) -> 11(2)r5c1 = [29] -> r4c6 = 2 8. 7 in n4 only in 12(2)n4 = {57} -> 8 in n4 in r4c12 -> 22(3)n6 = [{59}8] -> r5c5 = 9 Also -> 8 in n5 in r6c45 9. r2c34 = +7(2) and r23c4 = +7(2) -> r2c3 = r4c3 which cannot be 6 -> 6 in n2 in r2c45 10! Given: 6 in r2c45 6 in n4/c3 in r456c3 r2c34 = [52] or [16] (Cannot be {34} since (34) already in n2) r56c3 = +7(2) = [43] or {16} Either 6 in r4c3 or r56c3 = {16} which puts r2c34 = [52] and r2c5 = 6 In neither case can r4c5 be 6 -> r4c5 = 4 11. -> r456c6 = [275] -> 12(2)n4 = [57] Also (HS 4 in r5) r56c3 = [43] -> r4c3 = 6 and r4c12 = {18} Also 3 in n6 in r5c89 -> 3 in n5 in r4c4 Also one of 11(2)s in n8 = [47] -> r1c4 = 7 Since at least one of (16) must be in r56c4 -> 14(3)n2 = [716] -> r23c4 = [25] -> r2c3 = 5 and r3c3 = 1 -> r2c9 = 1 and r3c9 = 8 etc. We could do with some of that rain here! Severe restrictions on water use in place. |
Page 1 of 1 | All times are UTC |
Powered by phpBB® Forum Software © phpBB Group https://www.phpbb.com/ |