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Author:

Andrew [ Tue Aug 30, 2016 7:36 pm ]

Post subject:

Re: Assassin 339

After an easy start this Assassin became difficult until, after thinking about using forcing chains, I decided to try to think like SudokuSolver, remembering Richard's approach when he was an active solver. Once I'd found a couple of steps that way it wasn't too difficult or too long.

Here is my walkthrough for Assassin 339:

Rating Comment:

Author:	HATMAN [ Fri Aug 19, 2016 5:41 am ]
Post subject:	Assassin 339
Assassin 339 This one came up with 8 guesses when I opened JSudoku this morning, JS uses 4 fishes but SS only gives it 1.15 - I believe it is harder (and certainly long). Given the score I've tried 7 times to make it harder but it goes straight off-scale, the best is 1.8. JS Code: 3x3::k:9217:9217:5122:5122:5122:3843:3332:3332:2565:9217:9217:9217:5122:3843:3843:3846:4615:2565:5384:9217:9217:5122:7689:3846:3846:4615:4615:5384:5384:5384:7689:7689:1546:1546:5131:4615:5384:4620:7689:7689:4365:4365:5390:5131:5131:4620:4620:3599:2832:4365:4365:5390:5390:5131:2321:3599:3599:2832:4882:4882:6931:6931:6931:2321:4116:4116:3605:3605:4882:6931:4630:4630:4375:4375:4116:4116:3605:3605:6931:4630:4630: 264739851 871542369 539168472 726495183 193876524 458213796 615984237 342657918 987321645

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Assassin 339 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1387	Page 1 of 1

Author:	Ed [ Sun Dec 17, 2017 6:08 am ]
Post subject:	Re: Assassin 339
This puzzle drove me bonkers! Kept making mistakes and then wasn't happy with the solution. Finally found a decent way but Andrew's is better. Here's a very optimised version of how SudokuSolver did A339. A lot simpler than the way we did it I think. We missed something big (step 8 but also 9 & 10 are neat). Thanks for posting this version HATMAN A339 SudokuSolver's way - highly optimised: Prelims courtesy of SudokuSolver Cage 17(2) n7 - cells = {89} Cage 6(2) n56 - cells only uses 1245 Cage 13(2) n3 - cells do not use 123 Cage 9(2) n7 - cells do not use 9 Cage 10(2) n3 - cells do not use 5 Cage 11(2) n58 - cells do not use 1 Cage 21(3) n6 - cells do not use 123 Cage 19(3) n8 - cells do not use 1 Cage 14(4) n8 - cells do not use 9 1. "45" on n6: 2 innies r4c79 = 4 = [13] only permutation 1a. r4c6 = 5 2. "45" on n3, one remaining outie r3c6 = 8 3. 17(2)n7 = {89} only: both locked for n7 and r9 4. "45" on n8: 2 innies r79c4 = 12 = {57}/[84/92]: r7c4 = (5789), r9c4 = (3457) 4a. r6c4 = (2346) 5. "45" on N78: 2 outies r6c34 = 10: r6c3 = (4678) 6. "45" on n12: 2 remaining innies r3c15 = 11 (no 1,3) 7. 21(5)r3c1 can't have more than one of 1,3 because they are only in r5c1 = {12459/12468/12567/23457} 7a. must have one of 1,3 -> r5c1 = (13) 7b. must have 5 which is only in r3c1 or be (12468): no 7,9 in r3c1 8. Hidden killer triple 1,3,5 in n4 since 18(3)n4 can't have more than one of 1,3,5 8a. 18(3) = {189/369/378/459/567}(no 2) 8b. r5c3 = (135) I also found this step in my unpublished solution but could not make it powerful enough because I missed step 8. 9. "45" on r1234: 1 innie r4c8 + 4 = 3 outies r5c134 9a. r5c134 can't be {135} = 9 since no 5 in r4c8 -> no 1,3 in r5c4 10. "45" on n5: 3 outies r3c5 + r5c3 + r7c4 = 18 10a. max. r7c4 = 9 -> min. r3c5+r5c3 = 9 -> no 2 and no 5 in r3c5 since its in the same cage with r5c3 10b. no 6 in r3c1 11. 30(5)r3c5 can only have one of 1,3,5 since they are only in r5c3 = {25689/34689/45678}(no 1) 11a. must have 8 which is only in n5: locked for n5 12. deleted 13. 17(4)n5 = {1349/1367}(no 2) 13a. must have 3, 3 locked for n5 13b. no 7 in r6c3 (h10(2)r6c34) 14. 2 in n4 only in r4: locked for r4 and no 2 in r3c1 15. "45" on n1: 2 innies r1c3 + r3c1 = 9 = {45} only valid combination since r3c1 = (45): both locked for n1 16. "45" on n14: 3 innies r156c3 = 15 = [438/456] only valid permutations 16a. r1c3 = 4 much easier now. Andrew shows the areas to look in. Cheers Ed

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Author:	Andrew [ Tue Aug 30, 2016 7:36 pm ]
Post subject:	Re: Assassin 339
After an easy start this Assassin became difficult until, after thinking about using forcing chains, I decided to try to think like SudokuSolver, remembering Richard's approach when he was an active solver. Once I'd found a couple of steps that way it wasn't too difficult or too long. Here is my walkthrough for Assassin 339: Prelims a) R1C78 = {49/58/67}, no 1,2,3 b) R12C9 = {19/28/37/46}, no 5 c) R4C67 = {15/24} d) R67C4 = {29/38/47/56}, no 1 e) R78C1 = {18/27/36/45}, no 9 f) R9C12 = {89} g) 21(3) cage at R5C7 = {489/579/678}, no 1,2,3 h) 19(3) cage at R7C5 = {289/379/469/478/568}, no 1 i) 14(4) cage at R8C4 = {1238/1247/1256/1346/2345}, no 9 1. Naked pair {89} in R9C12, locked for R9 and N7, clean-up: no 1 in R78C1 2. 45 rule on N6 2 innies R4C79 = 4 = [13] -> R4C6 = 5, clean-up: no 7 in R12C9, no 6 in R7C4 2a. 45 rule on N3 1 remaining outie R3C6 = 8 2b. 3 in N3 only in R23C7 -> 15(3) cage at R2C7 = {348}, 3,4 locked for C7 and N3, clean-up: no 9 in R1C78, no 6 in R12C9 2c. 2 in C7 only in R789C7, locked for N9 2d. 19(3) cage at R7C5 = {289/379/469/478} (cannot be {568} because 5,8 only in R7C5) 2e. 8 of {289} must be in R7C5 -> no 2 in R7C5 3. 45 rule on N12 2 remaining innies R3C15 = 11 = {29/47/56}, no 1,3 3a. 45 rule on N1 2 innies R1C3 + R3C1 = 9 = {27/45}/[36], no 1,8,9, no 6 in R1C3, clean-up: no 2 in R3C5 4. 45 rule on N8 2 innies R79C4 = 12 = {57}/[84/93], no 1,2,6, no 3,4 in R7C4, clean-up: no 7,8,9 in R6C4 4a. 45 rule on N78 2 outies R6C34 = 10 = {46}/[73/82], R6C3 = {4678} 5. 45 rule on N14 3 innies R156C3 = 15 = {249/258/267/348/357/456} (cannot be {159} because 1,9 only in R5C3, cannot be {168} because no 1,6,8 in R1C3), no 1 6. 21(5) cage at R3C1 = {12459/12468/12567/23457} (cannot be {12369/12378/13458/13467} because 1,3 only in R5C1), CPE no 2 in R6C1 6a. 1,3 only in R5C1 -> R5C1 = {13} 6b. 5 of {12567/23457} must be in R3C1 -> no 7 in R3C1, clean-up: no 2 in R1C3 (step 3a), no 4 in R3C5 (step 3) 6c. R156C3 (step 5) = {258/267/348/357/456} (cannot be {249} because 2,9 only in R5C3), no 9 6d. 2 of {267} must be in R5C3, 7 of {357} must be in R6C3 -> no 7 in R5C3 7. R1C3 + R3C1 (step 3a) = [36/45/54/72], 21(5) cage at R3C1 (step 6) = {12459/12468/12567/23457} 7a. 45 rule on N4 2 innies R56C3 = 1 outie R3C1 + 6 7b. R3C1 = {2456} -> R56C3 = 8,10,11,12 = [37/38/56/57] (cannot be [26/28/46/48/64/84] which clash with 21(5) cage = {12468}, cannot be [47] which clashes with R1C3 + R3C1 = [45]) -> R3C1 = {456}, R5C3 = {35}, R6C3 = {678}, clean-up: no 7 in R1C3 (step 3a), no 9 in R3C5 (step 3), no 6 in R6C4 (step 4a) -> no 5 in R7C4, no 7 in R9C4 (step 4) 7c. 21(5) cage = {12459/12468/12567/23457}, 2 locked for R4 and N4 8. 45 rule on R789 3 innies R7C234 = 15 = {159/168/249/258/348/357} (cannot be {26}7/{46}5 because 14(3) cage at R6C3 cannot be 6{26}/4{46}) 8a. 7 of {357} must be in R7C4 -> no 7 in R7C23 9. 45 rule on N5 4 remaining innies R4C45 + R56C4 = 23 = {2489/2678/3479} (cannot be {1679} because R6C4 only contains 2,3,4), no 1 9a. R4C45 + R56C4 = {2489/3479} (cannot be {2678} because R3C5 + R4C45 + R5C34 = 5{678}3 only totals 29), no 6, 4,9 locked for N5 9b. 3 of {3479} must be in R5C4 (30(5) cage at R3C5 cannot be 5{479}5/7{479}3) -> no 3 in R6C4, clean-up: no 7 in R6C3 (step 4a), no 8 in R7C4, no 4 in R9C4 (step 4) [Cracked. The rest is fairly straightforward.] 10. R156C3 (step 6c) = {348/456} -> R1C3 = 4 -> R3C1 = 5 (step 3a) -> R3C5 = 6 (step 3), clean-up: no 4 in R78C1 10a. 4 in N2 only in 15(3) cage at R1C6 = {249} (only possible combination), locked for N2, 4 also locked for R2 -> R23C7 = [34] 10b. Naked quad {1357} in R1239C4, locked for C4 -> R7C4 = 9, R6C4 = 2, R9C4 = 3 (step 4), R6C3 = 8 (step 4a) -> R5C3 = 3 (hidden 15(3) cage sum), R5C1 = 1 10c. R1C5 = 3 (hidden single in N2) 10d. Naked pair {48} in R45C4, locked for C4 and N4 -> R8C4 = 6, R456C5 = [971], R56C6 = [63], clean-up: no 3 in R7C1 10e. 19(3) cage at R7C5 = {478} (only remaining combination) -> R7C5 = 8, R78C6 = {47}, locked for C6 and N8 10f. Naked pair {29} in R12C6, locked for C6 and N2 -> R2C5 = 4, R9C6 = 1 10g. R35C1 = [51] = 6 -> R4C123 = 15 = {267}, locked for R4 and N4 11. 21(3) cage at R5C7 = {579/678} (cannot be {489} which clashes with R4C8), no 4, 7 locked for N6 12. R6C1 = 4 (hidden single in C1) -> naked pair {59} in R56C2, locked for C2 -> R9C12 = [98] 13. R8C1 = 3 (hidden single in C1) -> R7C1 = 6 13a. R7C8 = 3 (hidden single in N9) 13b. 27(5) cage at R7C7 contains 2,3 = {23589/23679}, no 1,4, 9 locked for N9 13c. 1 in R7 only in R7C23 -> 14(3) cage at R6C3 = [815], R7C79 = [27], R78C6 = [47] -> R89C7 = [96], R8C23 = [42], R9C3 = 7, R4C3 = 6, R89C5 = [52], clean-up: no 7 in R1C8 14. R3C2 = 3 (hidden single in N1) 14a. 2 in R3 only in R3C89, locked for N3, clean-up: no 8 in R12C9 14b. Naked pair {19} in R12C9, locked for C9 and N3 -> R3C89 = [72], R2C8 = 6 (cage sum) 14c. Naked pair {19} in R2C39, locked for R2 14d. 21(3) cage at R5C7 = {579} (only remaining combination) = [579] and the rest is naked singles. Rating Comment: I'll rate my walkthrough for A339 at 1.25 based on the interactions in step 7b, which proved to be surprisingly useful.