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Assassin 410 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1648 |
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Author: | Ed [ Tue Feb 01, 2022 5:17 am ] |
Post subject: | Assassin 410 |
Attachment: a410.png [ 38.13 KiB | Viewed 3582 times ] A nice hard one, mainly because the moves are well hidden. But one heavy bit needed the way I did it. Right on my limit but always felt like there was somewhere to explore. This started out as a Jsudoku generated cage pattern but then joined up some cages. SudokuSolver gives it 2.00 but Jsudoku uses just 3 advanced steps which gave me the courage. triple click code: 3x3::k:4096:4096:5377:4098:4098:1795:1795:3588:3588:4096:5377:5377:4098:3845:5382:6151:6151:3588:5377:5377:3848:3848:3845:5382:6151:6151:2569:4106:4106:3848:9739:9739:5382:5382:5382:2569:4106:3340:3340:9739:9739:9739:1549:1549:3086:2831:5392:5392:5392:9739:9739:3601:3086:3086:2831:5650:5650:5392:2323:3601:3601:5908:5908:4373:5650:5650:5392:2323:4630:5908:5908:4887:4373:4373:1816:1816:4630:4630:5908:4887:4887: solution: +-------+-------+-------+ Ed |
Author: | Andrew [ Fri Feb 04, 2022 6:11 am ] |
Post subject: | Re: Assassin 410 |
Thanks Ed for your latest Assassin. Quite a hard one, although it didn't feel like 2.0. Here's my walkthrough for Assassin 410: Prelims a) R1C67 = {16/25/34} b) R23C5 = {69/78} c) R34C9 = {19/28/37/46}, no 5 d) R5C23 = {49/58/67}, no 1,2,3 e) R5C78 = {15/24} f) R67C1 = {29/38/47/56}, no 1 g) R78C5 = {18/27/36/45}, no 9 h) R9C34 = {16/25/34} i) 19(3) cage at R8C9 = {289/379/469/478/568}, no 1 1a. 45 rule on N1 R3C3 = 8 -> R3C4 + R4C3 = 7 = {16/25/34}, no 7,9, clean-up: no 7 in R2C5, no 2 in R4C9, no 5 in R5C2 1b. 45 rule on C6789 4 innies R5689C6 = 30 = {6789}, locked for C6, clean-up: no 1 in R1C7 1c. Min R89C6 = 13 -> max R9C5 = 5 1d. 45 rule on N9 R7C7 = 3 -> R6C7 + R7C6 = 11 = [65/74/92], clean-up: no 4 in R1C6, no 6 in R8C5 1e. 45 rule on N3 2 innies R1C7 + R3C9 = 7 = [43/52/61], R1C6 = {123}, R4C9 = {789} 1f. 45 rule on N5 2 innies R4C6 + R6C4 = 7 = [16]/{25/34} 1g. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = [24/42/51], R6C1 = {679}, R9C4 = {356} 2a. 12(3) cage at R5C9 = {138/156/237/246} (cannot be {129/147/345} which clash with R5C78), no 9 2b. Killer pair 1,2 in R5C78 and 12(3) cage, locked for N6 2c. Consider combinations for 12(3) cage 12(3) cage = {138} => R5C78 = {24}, 5 in N6 only in R4C78, locked for 21(4) cage at R2C6 => R7C6 = 5 (hidden single in C6) => R6C7 = 6 (cage sum) or 12(3) cage = {156/246} => R4C8 = 3 (hidden single in N6) => R1C6 = 3 (hidden single in C6), R1C7 = 4, R3C9 = 3 (step 1e), R4C9 = 7 or 12(3) cage = {237}, 7 locked for N6 -> no 7 in R6C7, clean-up: no 4 in R7C6 (step 1d) 2d. 4 in C6 only in R234C6, locked for 21(4) cage at R2C6, no 4 in R4C78 2e. Consider permutations for R6C7 + R7C6 = [65/92] R6C7 + R7C6 = [65] or R6C7 + R7C6 = [92], no 9 in R4C9 => no 1 in R3C9, no 6 in R1C7, no 1 in R1C6 => R1C6 = 3, R234C6 = {145} = 10 => R4C78 = 11 = [83] -> no 6 in R4C78 2f. 4 in N6 only in R5C78 = {24} or 12(3) cage = {246} -> 12(3) cage = {138/156/246} (cannot be {237}, locking-out cages), no 7 2g. 7 in N6 only in R4C789, locked for R4 2h. Consider combinations for 12(3) cage 12(3) cage = {138}, 8 locked for N6 or 12(3) cage = {156/246} => R4C8 = 3 (hidden single in N6) … => R4C9 = 7 (as in step 2c) -> no 8 in R4C9, no 2 in R3C9, no 5 in R1C7 (step 1e), no 2 in R1C6 2i. 21(5) cage at R2C6 contains 4 for C6 but no 6 = {12459/13458/23457} [With hindsight step 5 could have been found at this stage] 3a. 45 rule on N7 1 outie R9C4 = 1 innie R7C1 + 1 3b. R7C1 + R9C4 + R7C6 = [235/452/562], 2 locked for R7, clean-up: no 7 in R8C5 3c. R7C1 + R9C4 + R7C6 = [235/452/562] -> no 5 in R7C45, clean-up: no 4 in R8C5 3d. 2 in R7 only in R7C16, 2 in R7C1 => R6C1 = 9 or 2 in R7C6 => R6C7 = 9 (cage sum) -> 9 in R6C17 (locking cages), locked for R6 3e. 18(3) cage at R8C6 = {189/279/369/378/468} (cannot be {459} because 4,5 only in R9C5, cannot be {567} which clashes with R7C1 + R9C4 + R7C6), no 5 4a. 12(3) cage at R5C9 (step 2f) = {138/156/246} 4b. 45 rule on R6789 4 innies R6C5689 = 19 = {1378/1468/2368/2458} (cannot be {1567/2467} which clash with R6C16, ALS block, cannot be {3457} because 12(3) cage cannot contain two of 3,4,5), 8 locked for R6 [I was a bit slow to spot this, the first key step] 5a. R1C7 + R3C9 = 7 (step 1e) 5b. Consider placement for R6C7 = {69} R6C7 = 6 => R1C7 = 4, R3C9 = 3 => R4C9 = 7 or R6C7 = 9 => R4C9 = 7 -> R4C9 = 7, R3C9 = 3, R1C7 = 4, R1C6 = 3, clean-up: no 4 in R4C3 (step 1a), no 2 in R5C8, no 4 in R6C4 (step 1f) 5c. 14(3) cage at R1C8 = {158/167/257}, no 9 5d. 9 in C9 only in R789C9, locked for N9 5e. 45 rule on N2 3 remaining innies R23C6 + R3C4 = 11 = {146/245}, 4 locked for N2 5f. 6 of {146} must be in R3C4 -> no 1 in R3C4, clean-up: no 6 in R4C3 (step 1a) 6a. 21(5) cage at R2C6 (step 2i) = {12459/13458} -> R4C78 = {59}/[83] 6b. 45 rule on R123 4 remaining outies R4C3678 = 17 = {12}{59}/{15}[83]/[24][83], no 3 in R4C3, clean-up: no 4 in R3C4 (step 1a) 6c. 4 in N2 only in R23C6, locked for C6, clean-up: no 3 in R6C4 (step 1f) 6d. R4C3678 = {12}{59}/{15}[83], 1,5 locked in R4 6e. 1 in R4 only in R4C36, R4C3 = 1 => R3C4 = 6 (cage sum) or R4C6 = 1 => R6C4 = 6 (step 1f) -> 6 in R36C4 (locking cages), locked for C4, clean-up: no 1 in R9C3, no 5 in R7C1 (step 1g), no 6 in R6C1 6f. Naked pair {24} in R7C1 + R9C3, locked for N7 6g. R7C1 + R9C4 + R7C6 (step 3c) = [235/452], 5 locked for N8, clean-up: no 4 in R7C5 7a. 18(3) cage at R8C6 (step 3e) = {189/279/369/468} (cannot be {378} which clashes with R78C5) 7b. Consider permutations for R67C1 = [74/92] R67C1 = [74] => R6C6 = {68} => 18(3) cage = {189/279/369/378} (cannot be {468} which clashes with R6C6 or R67C1 = [92] => R9C3 = 4 -> 18(3) cage = {189/279/369}, no 4, 9 locked for C6 and N8 7c. 4 in N8 only in R78C4, locked for C4 and 21(5) cage at R6C2, no 4 in R6C23 7d. 45 rule on N8 4 innies R789C4 + R7C6 = 18 with 4 in R78C4 and 5 in R7C6 + R9C4 = {2457} (only possible combination) -> R7C6 = 2, R9C4 = 5, R7C1 = 4, R6C1 = 7, R9C3 = 2, R78C4 = [74], clean-up: no 6 in R5C23 [A lot easier from here] 7e. R78C4 = [74] = 11 -> R6C234 = 10 = {13}6/{35}2 7f. Naked triple {135} in R4C3 + R6C23, locked for N4, 3 locked for R6, clean-up: no 8 in R5C2 7g. Naked pair {49} in R5C23, locked for R5 and N4, clean-up: no 2 in R5C7 7h. Naked pair {15} in R5C78, locked for R5 and N6 7i. 12(3) cage at R5C9 = {246} (only remaining combination), 6 locked for N6 -> R6C9 = 9 (or cage total from step 7e), R4C78 = [83] 8a. Naked pair {26} in R4C12, locked for R4 and N4 -> R4C45 = [94], R5C1 = 8 8b. Naked pair {26} in R36C4, 2 locked for C4 8c. Naked pair {18} in R12C4, locked for N2, R1C5 = 7 (cage sum) 8d. Naked pair {69} in R23C5, 6 locked for C5 and N2, clean-up: no 3 in R8C5 8e. Naked pair {18} in R78C5, locked for C5 and N8, R9C5 = 3 8f. R3C4 = 2 -> R4C3 = 5 (cage sum) [Now back to the corners] 9a. 14(3) cage at R1C8 = {158} (only remaining combination), locked for N3 9b. Naked triple {158} in R1C489, 1,5 locked for R1, 5 locked for N3 9c. 1 in R3 only in R3C12, locked for N1 9d. 2 in R1 only in 16(3) cage at R1C1 = {259} -> R1C12 = {29}, locked for N1, R2C1 = 5 9e. R1C3 = 6, R2C6 = 4, R2C23 = {37}, 7 locked for R2 and N1 9f. 17(3) cage at R8C1 = {368} (cannot be {179} which clashes with R7C3) -> R89C1 = [36], R9C2 = 8 9g. R9C6 = 9, R9C89 = [74] -> R8C9 = 8 (cage sum) and the rest is naked singles. |
Author: | Ed [ Thu Feb 10, 2022 8:44 am ] |
Post subject: | Re: Assassin 410 |
I followed a very similiar path to Andrew but as often happens, saw things very differently. Andrew is much purer than I am with his lovely forcing chains. a410 WT: Preliminaries by SudokuSolver Cage 6(2) n6 - cells only uses 1245 Cage 15(2) n2 - cells only uses 6789 Cage 7(2) n78 - cells do not use 789 Cage 7(2) n23 - cells do not use 789 Cage 13(2) n4 - cells do not use 123 Cage 9(2) n8 - cells do not use 9 Cage 10(2) n36 - cells do not use 5 Cage 11(2) n47 - cells do not use 1 Cage 19(3) n9 - cells do not use 1 This is a highly optimised solution. Any clean-up needed will be stated. 1. "45" in n1: 1 innie r3c3 = 8 1a. r3c4 + r4c3 = 7 (no 7,9) 2. "45" on c6789: 4 innies r5689c6 = 30 = {6789}: all locked for c6 2a. min. r89c6 = 13 -> max. r9c5 = 5 3. "45" on n9: 1 innie r7c7 = 3 3a. r6c7 + r7c6 = 11 = [92/74/65] 3b. no 4 in r1c6 4. "45" on n3: 2 outies r1c6 + r4c9 = 10 = [19/28/37] 5. 4 in c6 in sp11(2)r6c7 + r7c6 or in 21(4)r2c6 5a. -> 4 in r4c78 must have 7 in r6c7 (Locking out cages) 5b. "45" on n6: 4 innies r4c789 + r6c7 = 27 5c. -> {4689} blocked 5d. = {3789/5679}(no 1,2,4) = 3 or 5 in r4c78 5e. 7 & 9 locked for n6 6. 21(5)r2c8.... 6a. {123}{69/78} blocked by r1c6 = (123) (or doesn't have 3 or 5 in r4c78) 6b. {12468} blocked by no 3 or 5 for r4c78 (step 5d) 6c. {125}{67} blocked by outies n3 = 10 (or boring way, doesn't have 3 or 5 for r4c78) 6d. {23457} as {234}{57} only, blocked by [16]+[65] in r1c67 & r6c7+r7c6 6e. = {12459/13458/13467} 6f. 1 & 4 locked for c6 6g. -> no 9 in r4c9 (outiesn3=10) 6h. r6c7 = (69) 7. 5 in c6 in r7c6 -> 6 in r6c7; or 5 in c6 in 21(5)r2c6 7a. -> 6 in r4c78, 21(5) must also have 5 for c6 (locking out cages) 7b. -> 21(5)r2c6: {13467} blocked 7c. = {12459/13458}(no 6,7) 8. r4c9 = 7 (hsingle n6), r3c9 = 3 8a. -> r1c6 = 3 (outies n3=10), r1c7 = 4 more key steps 9. "45" on n7: 2 innies r7c1 + r9c4 = 6 = [24/42/51] 9a. r6c1 = (679) 9b. r9c4 = (356) 9c. note: two outies n7 = 12 = [66/75/93] 9d. note2 from r7c1 = (679) i. {69} in r6c17 -> r6c6 = (78) ii. or [75] in outies n7 -> 2 in r7c6 10. 18(3)n8: {78}[3] blocked since r6c6 = (69) -> [25] in r7c6 + r9c4 (step 9dii) leaves nothing for 9(2)n8 10a."45" on n7: 1 innie r7c1 + 1 = 1 outie r9c4 -> {567} blocked from 18(3) by [22] in r7c16 10b. {459} blocked by 4 & 5 only in r9c5 10c. = {189/279/369/468}(no 5) = 1 or 2 or 6 10d. -> [516] in r7c1 + r9c34 forces 2 into both r7c6 and 18(3)n8 10e. -> no 5 in r7c1, no 1 in r9c3, no 6 in r9c4, no 6 in r6c1 11. r6c17 from {679} -> must have 6 or 7 -> 5 in r7c6 (h11(2) or r9c4 (outiesn7=12) 11a. 5 locked for n8 11b. no 4 in 9(2)n8 11c. also 9 locked in r6c17 since can't have [55] in r7c6 + r9c4 -> 9 locked for r6 11d. -> 2 locked in r7c16 for r7 (locking cages) 12. "45" on n5: 2 innies r4c6 + r6c4 = 7 = [16]/{25}/[43] (r6c4 = (2356)) 13. combined cage 24(4)r2378c5 = {69}{18/27}/{78}{36} 13a. 6 locked for c5 13b. note: has 6 in r23c5 or 3 in r78c5 Took a long time to find this "45". Final crack 14. "45" on n2369: 1 innie r3c4 + 1 = 2 outies r47c6 14a. but [515] blocked by 6 in both r6c4 (h7(2)n5) and r6c7(sp11(2)) 14b. and [625] -> r39c4 = [63], blocked by combined 24(4)r2378c5 (step 13b) 14c. -> no 5 in r7c6 14d. -> r7c6 = 2, r6c7 = 9, r67c1 = [74], r9c34 = [25] 14e. no 7 in 9(2)n8 14f. 18(3)n8 = {189/369/468}(no 7) 15. "45" on n8: 2 remaining innies r78c4 = 11 and must have 7 for n8 = [74] only 16. killer pair 1,3 in 9(2)n8 and r9c5: both locked for c5 17. "45" on n1236: 2 remaining outies r4c36 = 6 = {15} only, both locked for r4 17a. no 3 in r6c4 (h7(2)n5) 17b. r4c78 = [83] 17c. no 1 in r3c4 18. naked pair {26} in r36c4: both locked for c4 18a. r4c4 = 9 19. naked pair {18} in r12c4: both locked for n2 & c4 19a. -> r1c5 = 7 (cage sum) 20. r4c6 = 1 (hsingle c6) 20a. r4c3 = 5, r3c4 = 2 (cage sum) 20b. r56c4 = [36] 20c. r56c6 = [78] 21. 13(2)n4 = {49}: both locked for n4, 4 for r5 22. r89c6 = {69} -> r9c5 = 3 23. no 3 in r2c1 since r1c12 cannot = {67/58/49} = 13 24. r8c1 = 3 (hsingle c1) 24a. -> r9c12 = 14 = {68}: both locked for n7 and r9 24b. -> r9c6789 = [9174] much easier now Ed |
Author: | wellbeback [ Sat Feb 12, 2022 10:51 pm ] |
Post subject: | Re: Assassin 410 |
Functionally equivalent steps for me as well, although the specific reasonings to place or eliminate possibilities are (again) reflective of our different approaches. Thanks Ed! Assassin 410 WT: 1. Innies c6789 = r5689c6 = +30(4) = {6789} Innies n9 = r7c7 = 3 -> r7c6 from (245) and r6c7 from (976) IOD n3 -> r1c6 = r3c9 Since r3c9 cannot be 5 -> r1c6 cannot be 5 Since 3 already placed in c7 -> r1c6 cannot be 4 -> r1c6 = r3c9 from (123) and r4c9 from (987) 2! Innies n6 = r4c789,r6c7 = +27(4) (A) {3789} (B) {4689} (C) {5679} Since r4c9 is Min 7 and r6c7 from (679) -> One of (345) in r4c78 (A) -> (HS 3 in c6) r1c6 = 3 -> 10(2)r3c9 = [37] (B) Leaves no place for 4 in c6 (Since r1c6 is Max 3 and 14(3)r6c7 cannot be [743]) (C) -> (HS 5 in c6) 14(3)r6c7 = [653] -> (Innies n3) [r1c7,r3c9] cannot be [61] -> 9 not in r4c9 -> r4c9 = 7 In all cases 10(2)r3c9 = [37] and 7(2)r1c6 = [34] 3! Consider Step 2 case (C). Innies n6 = [{59}76] This puts (HS 3 in n6) r6c8 = 3 and r56c9 = {18} This puts only solution for 14(3)n3 = [7{25}] But this leaves remaining cells c9 = r789c9 = +19(3) contradicting 19(3) cage in n9. -> Step 2 (A) must be correct. -> (Since r6c7 cannot be 8) Innies n6 = [8379] -> r7c6 = 2 4. Remaining cells c6 = r234c6 = {145} Remaining Innies n2 = r2c6,r3c46 = +11(3). Cannot contain both (15) -> Either r234c6 = [{14}5], r3c4 = 6, (Innies n5) r6c4 = 2 Or r234c6 = [{45}1], r3c4 = 2, r6c4 = 6. Innies n1 -> r3c3 = 8 -> r4c36 = {15} and r3c6c4 = {26} 5. Innies n7 = r7c1,r9c3 = +6(2) Given previous placements this can only be [r7c1,r9c3] = [42] -> r6c1 = 7 and r9c4 = 5 6. Remaining Innies n8 = r78c4 = +11(2) Since r89c6 is Min +13 -> Max r9c5 = 5 -> 9 in n8 in r89c6 -> Min r89c6 = +15 -> Max r9c5 = 3 -> (HS 4 in n8) r78c4 = [74] -> (HS 7 in c6) r5c6 = 7 7. 3 in n8 only in c5 -> (HS 3 in n5) r5c4 = 3 -> 3 in n4 in r6c23 -> Remaining Innies n4 = r4c3,r6c23 = [5{13}] or [1{35}] -> 13(2)n4 = {49} -> 16(3)n4 = [{26}8] -> r4c45 = [94] Also 6(2)n6 = {15} -> 12(3)n6 = {246} 8. 1 in n8 only in c5 -> (HS 1 in n5) r4c6 = 1 -> r23c6 = {45} and r3c4 = 2 -> r4c3 = 5 and r6c23 = {13} -> r6c4 = 6 Also (HS 5 in n5) r6c5 = 5 -> r5c5 = 2 Also r6c6 = 8 -> r89c6 = {69} and r9c5 = 3 -> 9(2)n8 = {18} Also 12(3)n6 = [6{24}] Also 15(2)n2 = {69} -> r1c5 = 7 and r12c4 = {18} 9. Given previous placements only solution for 14(3)n3 is {158} with 5 in r1. -> 24(4)n3 = {2679} with 9 in c8 Also -> 2 in r1 in r1c12 -> 16(3)n1 = [{29}5] -> r23c6 = [45] -> 21(5)n1 = [6{37}14] -> 13(2)n4 = [94] -> r1c12 = [92] -> r4c12 = [26] -> 17(3)n7 = [368] etc. |
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