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Assassin 332 X Twelves http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1370 |
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Author: | wellbeback [ Mon Mar 14, 2016 5:01 am ] |
Post subject: | Re: Assassin 332 X Twelves |
Thanks again HATMAN. A fun puzzle. Not too difficult. Assassin 332 X Twelves WT: 1. Naked sextet (!) r89c789 = {345789} -> r7c789 = {126} -> r7c6 = 3 -> 12(2)s in n8 are {48} and {57} -> whichever of {48} or {57} is in r89c6 is also in r7c23 -> NS r7c1 = 9 -> 9 in n8 in r89c5 2. Since one of (45) is in r89c6 and also in r7c23 -> 12(4)r8c3 cannot be {1245} -> 12(4)r8c3 = {1236} with 3 in r89c3 -> 14(3)n7 = [9{14}] -> 12(2)r7c2 = {57} -> 12(2)r7c4 = {48} -> 12(2)r8c6 = {57} -> 8 in n7 in r8c1 or r9c2 3! Since 4 and 8 are on different rows in n9 -> 4 and 8 cannot both be in the same row in n7 -> either r89c1 is [84] or r89c2 is [48] -> 12(2)s in n1 cannot be {48} -> 12(2)s in n1 are {57} and [39] -> 12(2)r4c1 = {48} -> r89c2 = [48] -> r9c1 = 1 -> r8c4 = 1 4. Whichever of (57) is in r12c2 must go in n7 in r7c3 -> it must also go in c1/n4 in r6c1 -> NP r38c1 = {26} 5. Given: a) One of (26) is in r89c3 and also in r3c1, and b) 3 in n1 in r12c1 and 3 in n7 in r89c3 -> 3 in c2/n4 in r456c2 -> 12(4)r3c2 cannot be {1236} (3 would have to go in r4c2 which leaves no place for the (2 or 6) that's in r3c1 and r89c3) -> 12(4)r3c2 = {1245} -> r3c3 = 4 Also -> 5 in r4c23 -> 5 in n1 in r12c1 -> 7 in n1 in r12c2 -> r7c23 = [57] -> r4c3 = 5 -> r34c2 = {12} Also -> r6c1 = 7 Also -> r56c2 = {36} 6. Innies - outies n6 -> r45c9 = r3c7 + 9 r3c7 from (2356) (1,4 already in D/) r3c7 cannot be 5 (would put r45c9 = +14(2) which leaves no place for 7 in n6) -> r3c7 from (236) a) 2 in r3c7 puts r45c9 = {29} b) 3 in r3c7 puts either 12(2)r8c8 = {39} or 12(2)r8c9 = {39} puts 12(2)r6c8 not = {39} puts r45c9 = {39} c) 6 in r3c7 puts r45c9 = {69} -> r45c9 = {(2|3|6)9} -> HS 7 in n6 -> 12(2)r5c7 = [75] -> 12(2)r6c8 = {48} -> 12(2)r8c7 = [84] -> 12(2)r8c8 = {39} -> 12(2)r8c9 = {57} 7. Naked Quad in D\ in r1289 = {3579} -> 8 in D\ in r4c4 or r5c5 Given that and also given: a) r45c1 = {48}, and b) r7c45 = {48}, and c) r6c89 = {48}, and d) 4 already in both diagonals -> 4 and 8 in n5 are in not in the same row or column. -> Only places for them in n5 are r4c4 = 8 and r5c6 = 4 -> r45c1 = [48], r7c45 = [48] -> 4 in n2 in r23c5 and 8 in n2 in r123c6 8. -> HS 7 in n5 -> r4c5 = 7 Also 8 in D/ in r1c9 or r2c8 -> HS 8 in n2/r3 -> r3c6 = 8 Also 9 in n8 in r89c5 Also 9 in n3 in r12c7 -> HS 9 in n2/r3 -> r3c4 = 9 -> HS 9 in D/ -> r4c6 = 9 -> HS 9 in n6 -> r5c9 = 9 -> HS 9 in n4/c3 -> r6c3 = 9 Also HS 5 in n5/r5 -> r5c4 = 5 (cannot be in r5c5 since NQ in D\ r1289 = {3579}) -> NS 5 in D/ -> r2c8 = 5 -> r1c12 = [57] -> r2c12 = [39] -> r89c8 = [39] -> r89c9 = [57] -> r89c6 = [75] Also r9c3 = 3 9. Finally HS 8 in D/ -> r1c9 = 8 -> r2c3 = 8 Also HS 5 in n2/r3 -> r3c5 = 5 Also 3 in n5 in r6c45 -> HS 3 in n4/c2 -> r5c2 = 3 -> r6c2 = 6 -> r5c5 = 6 -> r9c45 = [62] -> r8c3 = 2 -> r38c1 = [26] -> r34c2 = [12] etc. Rating: 1.25. The (first) key move was my Step 3. |
Author: | Andrew [ Thu Mar 24, 2016 11:04 pm ] |
Post subject: | Re: Assassin 332 X Twelves |
Thanks HATMAN for another interesting puzzle. Loved the interactions: even though it took me a very long time to spot the most important one. An ideal puzzle for wellbeback who spotted that a lot quicker than I did. It felt like a Human Solvable; maybe it would have been if the SS score had been higher. It is, of course, also acceptable as an Assassin. Here is my walkthrough for Assassin 332 X Twelves: Prelims a) R1C12 = {39/47/57}, no 1,2,6 b) R2C12 = {39/47/57}, no 1,2,6 c) R45C1 = {39/47/57}, no 1,2,6 d) R56C7 = {39/47/57}, no 1,2,6 e) R6C89 = {39/47/57}, no 1,2,6 f) R7C23 = {39/47/57}, no 1,2,6 g) R7C45 = {39/47/57}, no 1,2,6 h) R89C6 = {39/47/57}, no 1,2,6 i) R89C7 = {39/47/57}, no 1,2,6 j) R89C7 = {39/47/57}, no 1,2,6 k) R89C7 = {39/47/57}, no 1,2,6 l) 12(4) cage at R3C2 = {1236/1245} m) 12(4) cage at R3C7 = {1236/1245} n) 12(4) cage at R7C6 = {1236/1245} o) 12(4) cage at R8C3 = {1236/1245} 1. 45 rule on N9 1 outies R7C6 = 3, clean-up: no 9 in R7C2345, no 9 in R89C6 1a. Naked quad {4578} in R7C45 + R89C6, locked for N8 1b. R7C1 = 9 (hidden single in R7) -> R8C2 + R9C1 = 5 = {14/23}, clean-up: no 3 in R12C2, no 3 in R45C1 1c. 9 in N8 only in R89C5, locked for C5 1d. 12(4) cage at R8C3 = {1236} (cannot be {1245} which clashes with R7C23), 3 locked for C3 and N7, clean-up: no 2 in R8C2 + R9C1 1e. Naked pair {14} in R8C2 + R9C1, locked for N7 and D/, clean-up: no 8 in R7C23 1f. Naked pair {57} in R7C23, locked for R7 and N7 1g. Naked pair {48} in R7C45, locked for R7 and N8 1h. Naked pair {57} in R89C6, locked for C6 1i. 12(4) cage at R8C3 = {1236}, 1 locked for C4 and N8 2. 12(4) cage at R3C7 = {1236/1245}, 1 locked for N6 2a. 5 of {1245} must be in R3C7 (cannot be 2{145} which clashes with R56C7 + R6C89) -> no 5 in R4C78 + R5C8 3. R89C8 and R89C9 must be different from R6C89 -> whichever combination is in R6C89 must be in R89C7 3a. 12(4) cage at R3C7 = {1236/1245} 3b. Consider combinations for R6C89 = {39/48/57} R6C89 = {39}, locked for N6 => R56C7 = {48/57} => 12(4) cage = {1236} or R6C89 = {48}, locked for N6 => 12(4) cage = {1236} or R6C89 = {57} => R89C7 = {57}, locked for C7 => 12(4) cage = {1236} -> 12(4) cage = {1236} 3c. R56C7 = {48/57} (cannot be {39} which clashes with 12(4) cage) 3d. R6C89 = {48/57} (cannot be {39} because R6C89 {39} + R89C7 = {39} clash with 12(4) cage) 3e. R6C89 = {48/57} -> R89C7 = {48/57} 3f. Naked quad {4578} in R56C7 + R6C89, locked for N6 3g. Naked quad {4578} in R5689C7, locked for C7 3h. 9 in N6 only in R45C9, locked for C9, clean-up: no 3 in R89C9 3i. R89C8 = {39} (hidden pair in N9), locked for C8 3j. 12(4) cage = {1236}, 3 locked for C7 3k. Naked triple {126} in R457C8, locked for C8 3l. 9 on D/ only in R4C6 + R6C4, locked for N5 [It seems likely that 5,7 in N9 must be in R89C9, since R89C6 and R89C7 both {57} would appear to be a Killer UR. I’ll leave that for now and only use it if needed. I prefer puzzles which can be solved uniquely, rather than relying on uniqueness to solve them.] 4. In the same way as for step 3, R1C12, R2C12 and R45C1, which are all 12(2) cages and ‘see’ at least one cell of each other, must have different combinations -> one of R1C12 and R2C12 must be {39}, locked for N1 4a. Either R1C1 = 3 or R2C2 = 9 -> killer pair 3,9 in R1C1 + R2C2 and R8C8, locked for D\ 4b. 3 in N1 only in R12C1, locked for C1 4c. 9 in N1 only in R12C2, locked for C2 4d. 9 in R3 only in R3C46, locked for N2 5. Taking step 4 a bit further, consider combinations for R45C1 = {48/57} R45C1 = {48} => R1C12 and R2C12 = {39/57} => killer pair 5,7 in R12C2 and R7C2, locked for C2 or R45C1 = {57}, locked for N4 -> no 5,7 in R456C2 5a. R45C1 = {48} => R1C12 and R2C12 = {39/57}, 5,7 locked for N1 => hidden killer pair 5,7 in R12C1 and R6C1 or R45C1 = {57}, locked for N4 -> 5,7 in R12456C1, locked for C1 5. R1C12 + R2C12 + R45C1 must have different combinations (step 4) -> one of them must be {48} 5a. Grouped X-Wing for 4 in R1C12 + R2C12 + R45C1 and R8C2 + R9C1, no other 4 in C12 5b. Grouped X-Wing for 8 in R1C12 + R2C12 + R45C1 and R8C1 + R9C2, no other 8 in C12 6. Consider combinations for R45C1 = {48/57} R45C1 = {48} => R1C12 and R2C12 = {39/57} => hidden killer pair 5,7 in R12C1 and R6C1 => R6C1 = {57} => R6C89 = {48} => R89C9 = {57}, killer pair 5,7 in R1C1 + R2C2 and R9C9, locked for D\ => R4C35 = [57] (hidden pair in R4) or R45C1 = {57} => R45C2 + R6C12 = {1236} => R4C3 = 4 -> R4C3 = {45} 6a. R4C3 = {45} -> 12(4) cage at R3C2 = {1245}, 4 locked for C3 6b. Killer pair {45} in R45C1 and R4C3, locked for N4 6c. Caged X-Wing for 5 in 12(4) cage and R7C23, no other 5 in C23, clean-up: no 7 in R12C1 6d. 7 in C1 only in R456C1, locked for N4 6e. R45C1 = {48} => R4C3 = 5 or R45C1 = {57}, locked for C1, no 7 in R12C2 => 7 in N1 only in R12C3, locked for C3 => R7C3 = 5 -> no 5 in R3C3 6f. R45C1 = {48} => R4C3 = 5, R34C2 = {12} => R56C2 = {36}, locked for N4 or R45C1 = {57} => R45C2 + R6C12 = {1236}, locked for N4 -> no 6 in R56C3 7. Consider combinations for R89C9 = {48/57} R89C9 = {48} => R6C89 = {57} => 7 in N4 only in R45C1 = {57} => R12C12 = {39/48} => killer pair 4,8 in R1C1 + R2C2 and R9C9, locked for D\ or R89C9 = {57} => R6C89 = {48}, locked for R6 -> no 4,8 in R6C6 [I should have spotted this earlier …; with hindsight it could have been done after step 1.] 8. R8C12 cannot be [84] which clashes with {48} in R89C7 or R89C9, R9C12 cannot be [48] which clashes with {48} in R89C7 or R89C9, R89C1 cannot be [84] which clashes with {48} in R1C12, R2C12 or R45C1 -> R89C2 = [48], R9C1 = 1, clean-up: no 4,8 in R12C1 8a. R45C1 = {48} (hidden pair in C1), locked for N4, R4C3 = 5, R7C3 = 7, placed for D/ 8b. Naked pair {12} in R34C2, locked for C2 and 12(4) cage at R3C2 -> R3C3 = 4, placed for D\, clean-up: no 8 in R8C9 [So Killer UR wasn’t necessary ] 8c. Naked pair {57} in R89C9, locked for C9 and N9 -> R89C7 = [84], clean-up: no 5,7 in R6C8 8d. Killer pair 5,7 in R1C1 + R2C2 and R9C9, locked for D\ 8e. 8 on D\ only in R4C4 + R5C5, locked for N5 8f. R6C1 = 7 (hidden single in C1) -> R56C7 = [75] 8g. Naked pair {48} in R6C89, locked for R6 9. R2C8 = 5 (hidden single on D/), R2C1 = 3 -> R2C2 = 9, placed for D\, R1C1 = 5, placed for D\, R1C2 = 7 9a. R3C8 = 7 (hidden single in C8) 9b. R4C5 = 7 (hidden single in R4) 9c. R2C4 = 7 (hidden single in C4) 9d. R5C4 = 5 (hidden single in N5) 9e. R5C6 = 4 (hidden single in N5), R45C1 = [48] 9f. R4C4 = 8 (hidden single in N5) -> R7C45 = [48] 9g. R1C9 = 8 (hidden single of D/) -> R1C8 = 4 9h. R23C5 = [45] (hidden pair in C5) 10. R3C46 = [98] (hidden pair in R3) 10a. R4C6 = 9 (hidden single in N5) 10b. R5C9 = 9 (hidden single in N6) 10c. R5C2 = 3 (hidden single in R5), R6C2 = 6 10d. R5C5 = 6 (hidden single in N5), placed for both diagonals 10e. Naked pair {29} in R89C5, locked for C5 and N8 -> R89C4 = [16] 10f. Naked pair {23} in R89C3, locked for C3 and N7 10g. R5C3 = 1, R5C8 = 2, R3C7 = 3, placed for D/, R6C4 = 2, R6C6 = 1, placed for D\ and the rest is naked singles, without using the diagonals. Rating Comment: I'll rate my walkthrough for A332X at least Hard 1.5. I'm not sure why the SS score is only 1.5. However I noticed from the SS scoring summary that it used UR shortcuts. |
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