Joined: Wed Apr 16, 2008 1:16 am Posts: 1044 Location: Sydney, Australia
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Note: it has four 2-cell broken cages, and two 3-cell broken cages.
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Assassin 415
I associate Easter with hard killers ever since I tried nd's #9 in the early days. A415 is worthy I think. Found a nice way to get started, then kept making progress while it resisted a long way in. Enjoyed it. Very pleased actually because took many different versions to get it to the right level. SudokuSolver gives it 1.50, JSudoku uses 5 advanced steps.
Joined: Tue Jun 16, 2009 9:31 pm Posts: 282 Location: California, out of London
Loved the puzzle design Ed! I enjoyed finding its implications... Happy Easter everyone! Corrections thanks to Andrew & Ed.
Assassin 415 WT:
1. 8(2)r2c1 prevents 10(3)r2 from being {136} 39(8)n23 has no 6 -> 6 in n2 in 15(2)n2 = {69} -> 9 in r13c7 Also, whatever is in r2c6 is also in r13c7 Also (HS 9 in r2) r2c3 = 9
2! 6 in n8 either in 15(2)n8 = {69} -> r8c3 = 6, or 6 in r89c5. I.e, 19(5) contains a 6. All combinations for 19(5) which have a 6 also have a 3. The three 12(2)s in n78 must all be different. I.e., one each contains {39}, {48}, {57} -> 3 nowhere else in n78 -> 5(2)r9 = {14} -> 4 nowhere else in n78 -> 19(5)n78 = {12367} -> 7 nowhere else in n78 -> 15(2)n8 = {69} -> r8c3 = 6
3. 13(2)n9 = {58} or {67} -> 12(2)r9c1 = {39}
4. For the numbers (58) in n8 one must go in 12(2)n8 and the other in r7c4 Innies r9 = r9c568 = +15(3) Since there is no way r9c56 can be +13(2) -> r9c8 cannot be 2 -> (HS 2 in r9) r9c5 = 2
5. 7 in 19(5) in r7c3 or r8c45 -> 12(2)r8c1 = {48} -> 5(2)r9 = [14] Also 12(2)n8 = {57} -> 19(5) = [76{13}2] -> r7c12 = {25} Also r7c4 = 8 -> r7c789 = {134}
6. 11(2)r6c1 = [65] or [92] -> (HP in c1) r58c1 = {48} -> (17) in c1 in r1234c1 -> 8(2)r3c1 = {17} -> 8(2)r1c1 and 8(2)r2c1 are between them {26} and {35} -> r3c12 = {17} -> r13c3 = {48}
8. In r2 Either 8(2)r2c1 = {26} and 10(3)r2 = {145} or 8(2)r2c1 = {35} and 10(3)r2 = {127} -> 3 not in 10(3)r2
9! Remaining Innies r9 = r9c68 = +13(2) = [76] or [58] Remaining Outies r89 = r67c8 = +7(2) = [61] or {34} However, in the case r67c8 = [61] that puts r9c8 = 8, r7c79 = {34}, and r4c79 = {34} But that leaves no solution for 17(3)n3 since it would have to contain a 3 and cannot be {359} or {368} (Quite apart from making the puzzle solution non-unique) -> r67c8 = {34}
An alternate after the first two lines above is... -> Two of the numbers (368) are in r679c8. -> 17(3)n3 cannot be {368} Also 9 already in n3 -> 17(3)n3 cannot be {359} -> 3 not in 17(3)n3 Also 3 in 39(8)n23 in n2 -> 3 not in r13c7 Also (step 8) 3 not in 10(3)r2 -> 3 in n3 only in 9(2)n3 = {36}
10. -> 9 not in 12(3)n9 -> (HS 9 in n9) r8c8 = 9 -> 19(3)c8 = [496] -> 12(2)n8 = [57] Also r7c8 = 3 -> r7c79 = {14} -> (HP in n6) 7(2)n6 = {16} -> 8(2)r3c1 = [17] -> r3c2 = 7 Also 13(2)n9 = {58} Also r8c79 = {27}
11. Since 9 already in n3 -> 17(3)n3 cannot contain a 1 -> (HS 1 in c8) r2c8 = 1 -> 10(3)r2 = [415] or [217] Also 8 in 17(3)n3 Since (36) already in c8 -> 17(3)n3 cannot be {368} (Cannot contain a 6) -> (HS 6 in c7) 7(2)n6 = [61] -> r7c79 = [14] Also 9(2)n3 = {36} -> (HS 6 in r2) 8(2)r2c1 = {26} -> 8(2)r1c1 = {35} -> 9(2)n3 = [63] -> 15(2)n2 = [96] -> r3c7 = 9 Also 10(3)r2 = [415] -> r1c7 = 4 -> r13c3 = [84]
12. Also -> 3 in n2 in r2c45 -> 3 in n5 in c6 -> 11(2)n5 = [29] -> 15(2)n8 = [69] Also r6c45 = {17} etc.
Last edited by wellbeback on Sun Apr 24, 2022 9:43 pm, edited 1 time in total.
Only one forcing chain by me so I wonder why JSudoku required 5 advanced steps.
Here is my walkthrough for Assassin 415:
Prelims
a) R1C12 = {17/26/35}, no 4,8,9 b) R13C4 = {69/78} c) R13C9 = {18/27/36/45}, no 9 d) R2C12 = {17/26/35}, no 4,8,9 e) R34C1 = {17/26/35}, no 4,8,9 f) R4C45 = {29/38/47/56}, no 1 g) R4C79 = {16/25/34}, no 7,8,9 h) R5C45 = {16/25/34}, no 7,8,9 i) R67C1 = {29/38/47/56}, no 1 j) R7C56 = {69/78} k) R8C12 = {39/48/57}, no 1,2,6 l) R89C6 = {39/48/57}, no 1,2,6 m) R9C12 = {39/48/57}, no 1,2,6 n) R9C34 = {14/23} o) R9C79 = {49/58/67}, no 1,2,3 p) R2C689 = {127/136/145/235}, no 8,9 q) R689C8 = {289/379/469/478/568}, no 1 r) 39(8) cage at R1C5 = {12345789}, no 6 s) 39(8) cage at R4C8 = {12345789}, no 6
1a. 45 rule on N8 5 innies R7C4 + R89C45 = 18 = {12348/12357/12456}, no 9 1b. 45 rule on N5 1 outie R7C4 = 1 innie R5C4 + 3 -> R5C4 = {12345}, R7C4 = {45678}, clean-up: no 1 in R5C3 1c. 1 in N7 only in R7C2 + R789C3, CPE no 1 in R6C3 [There’s a combined cage R9C12 + R9C34 + R9C79 for R9 but I’ll leave that for now.]
2a. R2C689 = {127/145/235} (cannot be {136} which clashes with R2C12), no 6 2b. 6 in N2 only in R13C4 = {69}, locked for C4, 9 locked for N2, clean-up: no 2,5 in R4C5, no 3 in R5C4 (step 1b) -> no 4 in R5C3 2c. 39(8) cage at R1C5 = {12345789}, 9 locked for C7 and N3, clean-up: no 4 in R9C9 2d. 8 in N2 only in 39(8) cage -> no 8 in R13C7 2e. R2C3 = 9 (hidden single in R2) 2f. 17(3) cage at R1C8 = {278/368/458/467}, no 1
3a. R8C12, R89C6 and R9C12 are all 12(2) and see each other so must have different combinations 3b. R8C12 + R89C6 + R9C12 = {345789}, R9C34 contains one of 3,4 -> only one remaining 3,4 in N78 -> 19(5) cage at R7C3 = {12358/12367/12457} (cannot be {13456} 3c. R8C12 + R89C6 + R9C12 = {345789}, 19(5) contains one of 7,8 -> only one remaining 7,8 in N78 -> R7C56 = {69}, locked for R7 and N9, clean-up: no 2,5 in R6C1, no 3 in R78C6 3d. R8C12 + R89C6 + R9C12 = {345789}, 3 locked for N7, clean-up: no 8 in R6C1, no 2 in R9C4 3e. R8C3 = 6 (hidden single in N7) -> 19(5) cage = {12367}, 3 locked for N8, clean-up: no 1 in R5C4, no 4 in R7C4 (step 1b), no 2 in R9C3 3f. 2 in N8 only in R8C45 + R9C5, locked for 19(5) cage, no 2 in R7C3 3g. Naked pair {14} in R9C34, locked for R9, clean-up: no 8 in R8C6, no 8 in R9C12, no 9 in R9C9 3h. R9C12 = {39} (cannot be {57} which clashes with R9C79), locked for R9 and N7 3i. 3 in R7 only in R7C789, locked for N9 3j. Caged X-Wing for 1 in 19(5) cage and R9C34, no other 1 in N78 3k. 1 in N7 only in R79C3, locked for C3 3l. Caged X-Wing for 4 in R8C12 + R89C6 and R9C34 = {14}, no other 4 in N78, clean-up: no 7 in R6C1 3m. 4 in R7 only in R7C789, locked for N9 3n. 2 in N7 only in R7C12, locked for R7 3o. 39(8) cage at R4C8 = {12345789}, 2,9 locked for N6, clean-up: no 5 in R4C79
4a. 12(3) cage at R7C8 = {129/138/147/237} (cannot be {345} because 3,4 only in R7C8), no 5 4b. 12(3) cage = {129/138/237} (cannot be {147} because {147} + R9C79 = {58} clashes with R7C79), no 4 4c. 4 in R7 only in R7C79, locked for 39(8) cage at R4C8 4d. 45 rule on N6 2 outies R7C79 = 1 innie R6C8 + 1, R7C79 contains 4 -> R6C8 + R7C79 = 4{14}/6{34} (cannot be 8{45} because 39(8) cage at R4C8 must contain 8) -> R6C8 = {46}, R7C79 = {14/34} 4e. 12(3) cage = {129/237} (cannot be {138} which clashes with R7C79), no 8, 2 locked for R8 and N9 4f. 1,3 of 12(3) cage must be in R7C8 -> naked triple {134} in R7C789, 1 locked for R7 and N9 -> R7C3 = 7, R9C5 = 2, clean-up: no 4 in R5C4 (step 1b), no 3 in R5C3, no 4 in R6C1, no 5 in R8C12 4g. Naked triple {134} in R7C789, CPE no 1,3 in R45C8 4h. Naked pair {48} in R8C12, locked for R8 and N7 -> R9C34 = [14], clean-up: no 7 in R4C5, no 3 in R6C1, no 8 in R9C6 4i. Naked pair {57} in R89C6, locked for C6, 5 locked for N8 -> R7C4 = 8, clean-up: no 3 in R4C5 4j. R7C4 = 8 -> R5C4 = 5 (step 1b), R5C3 = 2, clean-up: no 6 in R3C1, no 6 in R4C5 4k. R7C4 = 8 -> R6C45 = 8 = [17/26/71] 4l. 5,7 in N2 only in 39(8) cage at R1C5 -> no 5,7 in R13C7
5a. 45 rule on N7 3 remaining outies R6C123 = 20 = {389/479/569} (cannot be {578} because R6C1 only contains 6,9), no 1, 9 locked for R6 and N4 5b. R6C123 = {389/569} (cannot be {479} which clashes with R6C45 + R6C8), no 4,7 5c. R58C1 = {48} (hidden pair in C1) 5d. Consider combinations for R6C123 R6C123 = {389}, locked for R6 => R5C1 = 4, R4C3 = 5 or R6C123 = {569} => R6C3 = 5 -> 5 in R46C3, locked for C3 and N4, clean-up: no 3 in R3C1 5e. R34C1 = {17} (cannot be [26] which clashes with R67C1, cannot be [53] which clashes with R67C1 + R9C1), locked for C1, clean-up: no 1,7 in R12C2 5f. Naked quad {2356} in R12C12, locked for N1 5g. Naked pair {48} in R12C3, locked for C3 and N1 5h. Naked pair {17} in R3C12, locked for R3, clean-up: no 2,8 in R1C9 5i. Naked pair {35} in R46C3, 3 locked for N4 5j. R2C689 (step 2a) = {127/145} (cannot be {235} which clashes with R2C12), no 3, 1 locked for R2 5k. 3 in N2 only in 39(8) cage at R1C5 -> no 3 in R13C7 5l. Killer pair 2,5 in R2C12 and R2C689, locked for R2 5m. 2 in C4 only in R46C4, locked for N5
6a. R689C8 = {469/568} (cannot be {478} which clashes with R9C79), no 7, 6 locked for C8 6b. R8C8 = {59} -> no 5 in R9C8
7a. Hidden killer pair 3,6 in 17(3) cage at R1C8 (step 2f) and R13C9 for N3, R13C9 contains both or neither of 3,6 -> 17(3) cage must contain both or neither of 3,6 = {278/368/458} (cannot be {467}, 8 locked for N3, clean-up: no 1 in R1C9 7b. 6 of {368} must be in R2C7 -> no 3 in R2C7 7c. 4 of {458} must be in R2C7 (R13C8 cannot be {45} which clashes with R689C8) -> no 4 in R13C8 7d. R2C12 = {26} (cannot be {35} because R2C124 = {35}7 clashes with R2C689), locked for R2 and N1 7e. R2C689 (step 5j) = {145}, 4 locked for R2, 5 locked for N3, clean-up: no 4 in R13C9 7f. 17(3) cage = {278} (only remaining combination), 2,7 locked for N3, 2 locked for C8 7g. Naked pair {35} in R1C12, locked for R1 -> R13C9 = [63], R13C4 = [96], clean-up: no 1,4 in R4C7, no 7 in R9C7 7h. Naked pair {14} in R47C9, locked for C9 -> R2C9 = 5, clean-up: no 8 in R9C7 7i. R4C45 = [29/38] (cannot be [74] which clashes with R4C19, ALS block)
8a. 3 in 39(8) cage at R4C8 only in R567C7, locked for C7 -> R4C7 = 6, R4C9 = 1, R7C9 = 4, R9C79 = [58], R689C8 = [496], R2C68 = [41], R7C78 = [13] [Cracked at last] 8b. R4C8 = 5 (hidden single in C8), R4C3 = 3, R6C3 = 5, R7C2 = 2 -> R6C2 = 9 (cage sum) 8c. R4C4 = 2 -> R4C5 = 9, R7C4 = 8 -> R6C45 = 8 = {17}, locked for N5, 7 locked for R6
and the rest is naked singles.
And, unusually for me, here's an optimised start, a slightly simplified version of my steps 2 and 3. It's essentially the same as wellbeback's first three steps but, as often, we saw some of it in different ways.
Optimised start:
Optimised start
Relevant Prelims
a) R13C4 = {69/78} b) R2C12 = {17/26/35}, no 4,8,9 c) R7C56 = {69/78} d) R8C12 = {39/48/57}, no 1,2,6 e) R89C6 = {39/48/57}, no 1,2,6 f) R9C12 = {39/48/57}, no 1,2,6 g) R9C34 = {14/23} h) R9C79 = {49/58/67}, no 1,2,3 i) R2C689 = {127/136/145/235}, no 8,9 j) 39(8) cage at R1C5 = {12345789}, no 6
1a. R2C689 = {127/145/235} (cannot be {136} which clashes with R2C12), no 6 1b. 6 in N2 only in R13C4 = {69}, locked for C4, 9 locked for N2 1c. 39(8) cage at R1C5 = {12345789}, 9 locked for C7 and N3, clean-up: no 4 in R9C9 1d. R2C3 = 9 (hidden single in R2)
2a. R8C12, R89C6 and R9C12 are all 12(2) and see each other so must have different combinations 2b. R8C12 + R89C6 + R9C12 = {345789}, R9C34 contains one of 3,4 -> only one remaining 3,4 in N78 -> 19(5) cage at R7C3 = {12358/12367/12457} (cannot be {12349/13456} 2c. R8C12 + R89C6 + R9C12 = {345789}, 19(5) contains one of 7,8 -> only one remaining 7,8 in N78 -> R7C56 = {69}, locked for R7 and N9, clean-up: no 3 in R78C6 2d. R8C12 + R89C6 + R9C12 = {345789}, 3 locked for N7, clean-up: no 2 in R9C4 2e. R8C3 = 6 (hidden single in N7) -> 19(5) cage = {12367}, 3 locked for N8, clean-up: no 2 in R9C3 2f. Naked pair {14} in R9C34, locked for R9, clean-up: no 8 in R9C12, no 9 in R9C9 2g. R9C12 = {39} (cannot be {57} which clashes with R9C79), locked for R9 and N7
Joined: Wed Apr 16, 2008 1:16 am Posts: 1044 Location: Sydney, Australia
Really enjoyed both WTs. Thanks guys.
Andrew wrote:
Only one forcing chain by me so I wonder why JSudoku required 5 advanced steps.
"Advanced step" does not mean just "chain". JSudoku very, very rarely uses chains with the solver settings I use. I say "advanced steps" for any types of chain, "complex intersection" (whatever that is, one day I'll study it) and "overlap". I can see these in the "solver log". Doing it this way gives me a pretty good idea (usually) how tough I'm going to find the puzzle. My original puzzle started with 3 advanced steps and a score of 1.45, but it cracked completely after the start we all used so tried it as a "zero" puzzle to toughen up the middle. It worked!
I spent so much time making the puzzle didn't do a WT. I basically followed Andrew's path. But I saw the start differently to both the other WTs. [Thanks to Andrew for checking my start]
start to a415:
Preliminaries from SudokuSolver Cage 5(2) n78 - cells only uses 1234 Cage 15(2) n2 - cells only uses 6789 Cage 15(2) n8 - cells only uses 6789 Cage 7(2) n45 - cells do not use 789 Cage 7(2) n6 - cells do not use 789 Cage 8(2) n1 - cells do not use 489 Cage 8(2) n1 - cells do not use 489 Cage 8(2) n14 - cells do not use 489 Cage 12(2) n7 - cells do not use 126 Cage 12(2) n8 - cells do not use 126 Cage 12(2) n7 - cells do not use 126 Cage 13(2) n9 - cells do not use 123 Cage 9(2) n3 - cells do not use 9 Cage 11(2) n5 - cells do not use 1 Cage 11(2) n47 - cells do not use 1 Cage 10(3) n23 - cells do not use 89 Cage 19(3) n69 - cells do not use 1 Cage 39(8) n69 - cells ={12345789} Cage 39(8) n23 - cells ={12345789}
1. "45" on n78: 3 innies r7c124 = 15 1a. but {168/267} blocked by 15(2)n8 which needs one of 6 or 8, 6 or 7. 1b. = {159/249/258/348/357/456} 1c. note, can't have more than one of 1,2,6 1d. no eliminations yet
2. There must be six 1,2,6 in n78. h15(3) has at most one, 5(2) has one, 15(2) has at most one -> 19(5) must have at least all three 2a. -> 19(5) = {12367} only (no 4,5,8,9) 2b. and 15(2) must have one = {69}: both locked for r3 and n8 2c. -> r8c3 = 6 (hsingle n7)
3. r7c4 sees all {1237} in 19(5) -> can't have any of those (Common Peer Elimination CPE) 3a. r7c4 = (458) 3b. h15(3)n78 = {258/348/357}(no 1)
4. 1 in n78 only in 19(5) and 5(2) -> they must both have 1 4a. -> 5(2) = {14}: both locked for r9, 1 locked for c3
5. 13(2)n9 = {58/67}(no 9) = 5 or 7 5a. -> {57} blocked from 12(2)r9c1 5b. = {39} only: both locked for r9 and n7
6. 2 in n8 only in 19(5): locked for 19(5)
7. 2 in n7 only in h15(3) = {258} only: all locked for r7 on from there
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