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Assassin 413 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1651 |
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Author: | Ed [ Tue Mar 15, 2022 8:48 am ] |
Post subject: | Assassin 413 |
Attachment: a413.png [ 69.89 KiB | Viewed 3439 times ] Assassin 413 Nothing too strenuous once I found the right areas to work in. Did the basics and it came out with a couple of nice steps. SS gives it 1.45 and JSudoku uses 3 advanced steps. code: 3x3::k:3840:3840:3585:3585:3074:2051:2051:6660:6660:3840:3840:2565:2822:3074:1799:1799:6660:6660:4360:4360:2565:2822:7177:1799:7178:7178:7435:5388:4360:1293:1293:2841:2840:2840:7178:7435:5388:5388:7177:3344:2330:2321:7177:7178:7435:5388:2322:2322:3344:2841:2321:7178:7178:7435:4883:4883:4883:7177:2330:7177:7188:7435:7435:4117:4117:2582:2583:2583:7188:7188:5391:5391:4117:4117:2582:2062:2062:7188:7188:5391:5391: solution: +-------+-------+-------+ Ed |
Author: | Andrew [ Fri Mar 18, 2022 3:10 am ] |
Post subject: | Re: Assassin 413 |
Another tough one. From his comments, maybe Ed found a better way to solve it. Here's how I solved Assassin 413: Prelims a) R1C34 = {59/68} b) R12C5 = {39/48/57}, no 1,2,6 c) R1C67 = {17/26/35}, no 4,8,9 d) R23C3 = {19/28/37/46}, no 5 e) R23C4 = {29/38/47/56}, no 1 f) R4C34 = {14/23} g) R46C5 = {29/38/47/56}, no 1 h) R4C67 = {29/38/47/56}, no 1 i) R56C4 = {49/58/67}, no 1,2,3 j) R57C5 = {18/27/36/45}, no 9 k) R56C6 = {18/27/36/45}, no 9 l) R6C23 = {18/27/36/45}, no 9 m) R89C3 = {19/28/37/46}, no 5 n) R8C45 = {19/28/37/46}, no 5 o) R9C45 = {17/26/35}, no 4,8,9 p) 7(3) cage at R2C6 = {124} q) 19(3) cage at R7C1 = {289/379/469/478/568}, no 1 r) 26(4) cage at R1C8 = {2789/3689/4589/4679/5678}, no 1 1a. 45 rule on R89 1 outie R7C7 = 3, clean-up: no 5 in R1C6, no 8 in R4C6, no 6 in R5C5 1b. 45 rule on C89 2 outies R36C7 = 14 = {59/68} 1c. R36C7 = 14 -> R3456C8 = 14 = {1238/1247/1256/1346/2345} (cannot be {1256} which clashes with R36C7), no 9 1d. 45 rule on C6789 2 innies R5C7 + R7C6 = 13 = {49/58/67}, no 1,2 1e. 45 rule on N1 1 innie R1C3 = 1 outie R4C2 + 3, R1C3 = {5689} -> R4C2 = {2356} 1f. 45 rule on N1 3 innies R1C3 + R3C12 = 20 = {389/479/569/578}, no 1,2 1g. 45 rule on N4 3 innies R4C23 + R5C3 = 15, max R4C23 = 10 -> min R5C3 = 5 1h. 45 rule on C12 2 outies R67C3 = 7 = [16/25/34/52] 1i. Hidden killer quad 1,2,3,4 in R23C3, R4C3, R67C3 and R89C3, R23C3 and R89C3 each contain one of 1,2,3,4, R4C3 = {1234} -> R67C3 can only contain one of 1,2,3,4 = [16/25/52] -> R6C2 = {478} 1j. 19(3) cage at R7C1 = {289/469/568} (cannot be {478} because R7C3 only contains 2,5,6), no 7 1k. Max R7C3 = 6 -> min R7C12 = 13, no 2 in R7C12 1l. Naked triple {124} in 7(3) cage at R2C6, CPE no 2,4 in R2C45, clean-up: no 8 in R1C5, no 7,9 in R3C4 [16(4) cage at R8C1 omitted; it didn’t provide any eliminations and not specifically used later.] 2a. 3,7 in C3 only in R23C3 = {37} or R45C3 = [37] or R89C3 = {37} (locking cages) 2b. R4C23 + R5C3 = 15 (step 1g), R67C3 = 7 (step 1h) 2c. Consider permutations for R6C23 = [45/72/81] R6C23 = [45] => no 5 in R4C2 => R45C3 cannot be [37] or R6C23 = [72] or R6C23 = [81] => R7C3 = 6, no 4 in R23C3 and R89C3 => R4C3 = 4 (hidden single in C3) -> R45C3 cannot be [37] 2d. R23C3 = {37} or R89C3 = {37} (locking cages), locked for C3, clean-up: no 2 in R4C4 2e. 45 rule on C123 3 innies R145C3 = 18 = {189/459/468}, no 2, clean-up: no 3 in R4C4 2f. Naked pair {14} in R4C34, locked for R4, clean-up: no 7 in R4C67, no 7 in R6C5 3a. 45 rule on N5 3 innies R4C46 + R5C5 = 12 = {129/138/156/246/345} (cannot be {237} because R4C4 only contains 1,4, cannot be {147} because no 1,4,7 in R4C6), no 7, clean-up: no 2 in R7C5 3b. R4C4 = {14} -> no 1,4 in R5C5, clean-up: no 5,8 in R7C5 3c. 6,9 of {129/246} must be in R4C6 -> no 2 in R4C6, clean-up: no 9 in R4C7 3d. Consider placement for 1 in N5 R4C4 = 1 => R4C46 + R5C5 = {129/138/156} or R56C6 = {18}, 8 locked for N5 => 3 in N5 only in R4C6 + R5C5 => R4C46 + R5C5 = {345} -> R4C46 + R5C5 = {129/138/156/345} 3e. 1 in C5 only in R389C5 and R57C5 = [81] 3f. 45 rule on C5 3 innies R389C5 = 13 = {139/148/157/247/256/346} (cannot be {238} which clashes with R57C5 = [81], blocking cages) 3g. Consider combinations for R389C5 R389C5 = {139}, locked for C5 => R12C5 = [48]/{57} => R57C5 cannot be [54], clashes with R12C5 or R389C5 = {148/157} => R57C5 cannot be [54], clashes with R389C5 or R389C5 = {247/256/346} => R57C5 = [81] -> R57C5 = [27/36/81], no 5 in R5C5, no 4 in R7C5 3h. R4C46 + R5C5 = {129/138/345} (cannot be {156} because 5,6 only in R4C6), no 6, clean-up: no 5 in R4C7 3i. Consider placement for 4 in C5 4 in R13C5 => R23C6 = {12}, 1 locked for C6 => R4C4 = 1 (hidden single in N4) or R6C5 = 4 => R4C4 = 1 or R8C5 = 4 => R8C4 = 6, no 6 in R7C5 => no 3 in R5C5 => R4C46 + R5C5 = {129/138} (cannot be {345} because R5C5 only contains 2,8) -> R4C46 + R5C5 = {129/138}, no 4,5, clean-up: no 6 in R4C7 3j. R4C4 = 1 -> R4C3 = 4, clean-up: no 6 in R23C3, no 8 in R56C6, no 5 in R6C3, no 2 in R7C3 (step 1i), no 6 in R89C3, no 9 in R8C5, no 7 in R9C5 3k. 19(3) cage at R7C1 (step 1j) = {469/568}, 6 locked for R7 and N7, clean-up: no 3 in R5C5, no 7 in R5C7 (step 1d) [Continuing in N5] 3l. R46C5 = {29/38/56}/[74], R57C5 = [27/81] -> combined cage R4567C5 = {29}[81]/{38}[27]/{56}[27]/{56}[81]/[7481] 3m. R56C4 = {49/67} (cannot be {58} which clashes with R456C5), no 5,8 3n. R46C5 = {29/38/56} (cannot be [74] which clashes with R56C4), no 7 in R4C5, no 4 in R6C5 3o. Hidden killer pair 4,7 in R56C4 and R56C6 for N5, R56C4 contains one of 4,7 -> R56C6 must contain one of 4,7 = {27/45}, no 3,6 3p. 8 in N5 only in R456C5, locked for C5, clean-up: no 4 in R1C5, no 2 in R8C4 3q. R389C5 = {139/247/256/346} (cannot be {157} which clashes with R7C5) 3r. R389C5 = {247/256/346} (cannot be {139} because combined cage R12C5 = {57} + R389C5 = {139} which clashes with R7C5), no 1,9, clean-up: no 9 in R8C4, no 7 in R9C4 [Fairly straightforward from here.] 4a. R7C5 = 1 (hidden single in C5) -> R5C5 = 8, clean-up: no 3 in R46C5, no 5 in R7C6 (step 1d) 4b. R4C6 = 3 (hidden single in N5) -> R4C7 = 8, clean-up: no 6 in R1C3 (step 1e), no 8 in R1C4, no 5 in R1C7, no 6 in R36C7 (step 1b) 4c. Naked pair {59} in R36C7, locked for C7, 5 locked for 28(6) cage at R3C7, clean-up: no 4,8 in R7C6 (step 1d) 5a. R23C6 = {12/14} (cannot be {24} which clashes with R56C6), 1 locked for C6 and 7(3) cage at R2C7, clean-up: no 7 in R1C7 5b. Killer pair 2,4 in R23C6 and R56C6, locked for C6, clean-up: no 6 in R1C7 5c. 8 in C6 only in R89C6, locked for N8, clean-up: no 2 in R8C5 5c. Killer pair 3,6 in R8C45 and R9C45, locked for N8 5d. R1C6 = 6 (hidden single in C6) -> R1C7 = 2, R2C7 = 4, R5C7 = 6 -> R7C6 = 7 (step 1d), clean-up: no 8 in R1C3, no 2 in R56C6, no 7 in R6C4, no 3 in R8C45 5e. R5C4 = 7 (hidden single in N5) -> R6C4 = 6 -> R8C45 = [46], clean-up: no 2 in R9C45 5f. Naked pair {45} in R56C6, 5 locked for C6 and N5 5g. Naked pair {29} in R46C5, locked for C5, clean-up: no 3 in R12C5 5h. Naked pair {57} in R12C5, locked for N2, 5 locked for C5 -> R1C34 = [59], R12C5 = [75], R5C3 = 9, R9C45 = [53], R4C2 = 2 (step 1e), R6C3 = 1 -> R6C2 = 8, R7C34 = [62], R46C5 = [92], R4C89 = [75], clean-up: no 7 in R8C3 5i. Naked pair {38} in R1C89, locked for R1 and N3 5j. R1C89 = 11 -> R2C89 = 15 = {69}, locked for R2 and N3 5k. R36C7 = [59] = 14, R34C8 = [17] = 8 -> R56C8 = 6 = [24] 5l. R3456C9 = [7513] = 16 -> R7C89 = 13 = [94] 5m. Naked pair {14} in R1C12, locked for N1 5n. R1C12 = {14} = 5 -> R2C12 = 10 = {37} (cannot be {28} because 2,8 only in R2C1), locked for N1, 3 locked for R2 and the rest is naked singles. |
Author: | wellbeback [ Sat Mar 19, 2022 11:41 pm ] |
Post subject: | Re: Assassin 413 |
Took me ages to find any sort of nice path! My first attempt had many long contradiction chains. But eventually found a decent way through... Thanks again Ed! Assassin 413 WT: 1. Outies r89 = r7c7 = 3 -> 3 in n8 in r89c45 -> Either 10(2)n8 = {37} or 8(2)n8 = {35} 2. 1 in n2 only in r3c5 or in c6 Innies n5 = r4c46+r5c5 = +12(3) 11(2)r4c6 prevents r4c46 = +11 -> r5c5 cannot be 1 -> 1 in n5 in r4c4 or r56c6 3. 12(2)n2 cannot be {48} since that puts r23c6 = {12}, 11(2)n2 = {56}, (NS) r1c4 = 9, which leaves no solution for 13(2)n5 4! Odds and Evens on c5 11(2)c5 and 9(2)c5 each contain one odd and one even number 12(2)n2 is {39} or {57} (Two odd numbers) -> Innies c5 = r389c5 = +13(3) contains one odd number and two even numbers Since either 10(2)n8 = {37} or 8(2)n8 = {35} (I.e., one odd number in r89c5) -> r3c5 is even and the other of r89c5 is even -> 10(2)n8 cannot be {19} and 8(2)n8 cannot be {17} -> 1 in n2 in r123c6 -> (HS 1 in n5) 5(2)r4c3 = [41] -> (HS 1 in n8) r7c5 = 1 -> r5c5 = 8 -> (Remaining Innie n5) 11(2)r4c6 = [38] 5. Remaining Innies n14 = r15c3 = +14(2) -> r1c4 = r5c3 Since r5c3 cannot be 8 -> r1c4 cannot be 8 -> 8 in c4 in r23c4 -> 11(2)n2 = {38} -> 12(2)n2 = {57} -> 11(2)n5 = {29} -> (HS 9 in n2) 14(2)r1c3 = [59] -> r5c3 = 9 -> (Remaining Innie n4) r4c2 = 2 -> 11(2)n5 = [92] Also 13(2)n5 = {67} -> 9(2)n5 = {45} -> r3c5 = 4 -> r23c6 = {12} and r2c7 = 4 -> 8(2)r1c6 = [62] 6. Outies c12 = r67c3 = +7(2) can only be [16] -> (Innies c6789) [r5c7,r7c6] = [67] etc. |
Author: | Ed [ Mon Mar 28, 2022 3:59 am ] |
Post subject: | Re: Assassin 413 |
Reposting my WT after an accidental deletion. Love wellbeback's step 4! I will find one of those one day. Thanks to Andrew for noticing an error in the original step 16. A413 WT: Preliminaries by SudokuSolver Cage 5(2) n45 - cells only uses 1234 Cage 14(2) n12 - cells only uses 5689 Cage 8(2) n23 - cells do not use 489 Cage 8(2) n8 - cells do not use 489 Cage 12(2) n2 - cells do not use 126 Cage 13(2) n5 - cells do not use 123 Cage 9(2) n58 - cells do not use 9 Cage 9(2) n5 - cells do not use 9 Cage 9(2) n4 - cells do not use 9 Cage 10(2) n8 - cells do not use 5 Cage 10(2) n1 - cells do not use 5 Cage 10(2) n7 - cells do not use 5 Cage 11(2) n5 - cells do not use 1 Cage 11(2) n56 - cells do not use 1 Cage 11(2) n2 - cells do not use 1 Cage 7(3) n23 - cells ={124} Cage 19(3) n7 - cells do not use 1 Cage 26(4) n3 - cells do not use 1 This is a highly optimised solution so any clean-up needed is stated. 1. "45" on r89: 1 outie r7c7 = 3 1a. no 8 in r4c6 2. 3 in n8 only in 10(2) = {37} or in 8(2) 2a. -> {17} blocked from 8(2) (Locking-out cages) 2b. = {26/35}(no 1,7) 3. {39}{45} blocked from combined cage 21(4)r1257c5 by 11(2)r4c5 needs one of 3,4,5,9 3a. -> 9(2)r5c5 = {18/27}[36](no 4,5. No 6 in r5c5) 4. "45" on n5: 1 outie r4c7 + 1 = 2 innies r4c4 + r5c5 4a. r4c47 see each other so the difference cannot be 0 -> no 1 in r5c5 (Innie Outie Unequal IOU) 4b. no 8 in r7c5 5. "45" on n5: 1 outie r4c3 + 7 = 2 innies r4c6 + r5c5 5a. -> no 7 in r5c5 (IOU) 5b. -> no 2 in r7c5 More than 4 combinations the result here so an advanced step for me. 6. "45" on c5: 3 innies r389c5 = 13 6a. but {148} blocked by none of those in r9c5 6b. {238} blocked by r5c5 = (238) 6c. = {139/157/247/256/346}(no 8) key step 7. 8 in c5 in 12(2) -> combined half-cage with r23c6 = {48}/{12} -> 1 locked for c6 7a. or 8 in c5 in r456c5 7b. either way, {18} in 9(2)r5c6 is blocked 7c. -> r4c4 = 1 (hsingle n5) 7d. r4c3 = 4 7e. no 9 in r8c5 8. "45" on n5: 2 remaining innies r4c6 + r5c5 = 11 = [38/92] 8a. -> r4c7 = (28) 8b. and r7c5 = (17) 9. "45" on c5: 1 outie r8c4 = 2 innies r39c5 - 3 9a. -> no 3 in r3c5 (IOU) 10. h13(3)r389c5: {157} blocked by r7c5 = (17) 10a. = {139/247/256/346} 10b. {139} can only be [913] -> no 1 in r3c5 11. "45" on n14: 2 remaining innies r15c3 = 14 = {59/68} 12. "45" on c6789: 2 innies r5c7 + r7c7 = 13 = {49/58/67} = 4 or 5 or 6 12a. -> r3c5 + r5c3 + r7c4 = 15 (cage sum) 12b. but {456} blocked by split 13(2) 12c. and doesn't have any 1 or 3 anymore 12d. = {249/258/267}: note - if it has 4 must also have 2 12e. must have 2 in one of r3c5 or r7c4 -> no 2 in r23c4 nor r89c5 (Common Peer Elimination CPE) 12f. no 9 in 11(2)n2 12g. no 8 in r8c4, no 6 in r9c4 13. [7841] blocked from combined cage 20(4)r4567c5 by 13(2)n5 needs one of 4,7,8 13a. -> 11(2)r4c5 = {29/38/56}(no 4,7) 14. hidden killer pair 4,7 in n5 14a. -> 13(2) must have one = {49/67}(no 5,8) 14b. and 9(2) = {27/45}(no 3,6) 15. 8 in n5 only in c5: locked for c5 15a. no 4 in 12(2)n2 I made an error in my original step 16. 16. 4 in c5 only in r3c5 -> 2 in r7c4 (step 12d), or 4 in c5 in r8c5 -> 6 in r8c4 16a. -> [26] blocked from 8(2)n8 16b. -> 8(2) = {35} only 16c. 4 in c5 only in h13(3)r389c5 16d. but {247} blocked by none are in r9c5 16e. = {46}[3] only 16f. 6 locked for c5, and just for fun, no 4,6 in r7c46 since it sees those through the 28(5)r3c5 16g. r9c4 = 5 17. 12(2)n2 = {57}: both locked for n2 and c5 17a. -> r57c5 = [81] 17b. -> r4c6 = 3 (h11(2)), r4c7 = 8 18. "45" on c89: 2 outies r36c7 = 14 = {59} only: both locked for c7 and for 28(6) cage 19. 11(2)n5 = {29}: both locked for n5 19a. -> 9(2)n5 = {45}: 4 locked for c6 and n5 19b. -> r123c6 = [6]{12}: 2 locked for c6 19c. -> r2c7 = 4 (cage sum) 19d. r3c5 = 4 (Nsingle) 19e. r8c45 = [46] 19f. r1c7 = 2 20. r123c4 = [9]{38} 20a. r7c4 = 2 (hsingle n8) 20b. -> r5c3 = 9 (sp15(3) cage sum) 20c. r1c3 = 5 20d. r5c7 + r7c6 = [67] (sp13(2) 21. "45" on c12: 2 outies r67c3 = 7 = [16] 22. 1 remaining innie n4, r4c2 = 2 22a. -> r3c12 = 15 22b. but {78} blocked by 10(2)n1 which needs one of them 22c. = {69} only: both locked for n1 and r3 23. "45" on r12: 3 outies r3c346 = 13 = {38}[2] only: 3,8 locked for r3 Pretty straight ahead now. Ed |
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