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 Post subject: Ass 445 Almost PS
PostPosted: Fri Nov 10, 2023 7:13 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
On assassin 445 if you assume the zero cells in N6 and N9 are must repeat does it simplify the puzzle?
Anyway I liked the pattern so I added the two 12 cages then simplified N7 which needed the twin killer "must repeat" cage.
I was trying for a Paper Solvable but it is a bit hard probably 0.9+.
JS uses three small fishes but they can be avoided.

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 Post subject: Re: Ass 445 Almost PS
PostPosted: Mon Nov 13, 2023 10:29 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for a nice variant on Ed's Assassin 445. A lot easier unless Ed and wellbeback found a lot simpler way to solve that than I did; I haven't gone through their WTs yet in case they had any 'spoilers' which might have affected how I solved this puzzle.

Assassin 445 Almost PS would be an equally good puzzle with zero cells in R8C34 and R9C3 which would still total 23. It would only require minor changes to my solving path. However since HATMAN replaced Ed's zero cells, I can understand why he wouldn't have wanted to introduce some of his own.

Here's how I solved Assassin 445 Almost PS:
Prelims

a) R12C7 = {29/38/47/56}, no 1
b) R4C45 = {18/28/36/45}, no 9
c) R5C34 = {14/23}
d) R89C9 = {59/68}
e) 21(3) cage at R2C1 = {489/579/678}, no 1,2,3
f) 19(3) cage at R4C8 = {289/379/469/478/568}, no 1
g) 20(3) cage at R5C5 = {389/479/569/578}, no 1,2
h) 9(3) cage at R7C1 = {126/135/234}, no 7,8,9
i) 23(3) cage at R8C3 is a must repeat cage = {599/788/779}, no 1,2,3,4,6
j) 12(4) cage at R8C8 = {1236/1245}, no 7,8,9
k) 38(8) cage at R2C8 = {12345689}, no 7

1a. 45 rule on N9 1 outie R6C7 = 3, clean-up: no 8 in R12C7
1b. 3 in N9 only in 12(4) cage at R8C8 = {1236}, 1,2,6 locked for N9, clean-up: no 8 in R89C9
1c. Naked pair {59} in R89C9, locked for C9 and N9
1d. Naked triple {478} in R789C7, locked for C7
1e. 45 rule on N6 2 remaining innies R45C7 = 11 = {29/56}, no 1
1f. R3C7 = 1 (hidden single in C7)
1g. 45 rule on C789 2 outies R45C6 = 11 = {29/38/56}, no 4
1h. 38(8) cage at R2C8 = {12345689}, 4 locked for N3
1i. 1 in N6 only in 12(3) cage at R5C9 = {129/147/156}, no 8
1j. 45 rule on N7 3 innies R789C3 = 21 = {579/678} (cannot be {489} because must repeat 23(3) cage cannot contain both of 8,9), 7 locked for C3 and N7
1k. 45 rule on N7 1 outie R8C4 = 1 innie R7C3 + 2, no 8,9 in R7C3, no 5 in R8C4
1l. 45 rule on N12 two outies R4C13 = 12 = {57}/[84/93], no 1,2,6, no 5 in R4C1, no 9 in R4C3
1m. 45 rule on N2 two outies R34C3 = 9 = [45/54/63], clean-up: no 4 in R4C1
1n. 45 rule on N124 two outies R56C4 = 5 = {14}/[32], clean-up: no 3 in R5C3
1o. Whatever is in R5C3 must be in R6C4 and in N6 in R4C789, no 1 in R4C789 -> no 1 in R5C3, no 1 in R6C4, clean-up: no 4 in R5C4
1p. 1 in N5 only in R4C45 = {18} or in R56C4 = [14] -> no 4 in R4C45 (locking-out cages), clean-up: no 5 in R4C45
1q. 45 rule on R12345 2 innies R5C59 = 10 = [37]/{46}/[82/91], no 5,7 in R5C5
1r. Naked quad {2569} in R1245C7, CPE no 2,5,6,9 in R2C8 + R3C78
1s. Naked triple {348} in R2C8 + R3C78, 3,8 locked for N3 and 38(7) cage
1t. 18(3) cage at R1C8 = {279/567} -> R1C8 = {59}, R12C9 = {27/67}, 7 locked for C9, clean-up: no 3 in R5C5

2a. 12(3) cage at R5C9 (step 1i) = {129/147/156} -> R6C8 = {579}, R56C9 = {12/14/16}, 1 locked for C9
2b. 20(3) cage at R5C5 = {479/578} (cannot be {569} which clashes with R45C6), no 6, 7 locked for N5, clean-up: no 2 in R4C45, no 4 in R5C9 (step 1q)
2c. 12(3) cage = {129/156} (cannot be {147} = [174] which clashes with 20(3) cage = 9{47} using R5C59 = 10, step 1q), no 4,7
2d. 19(3) cage at R4C8 = {478} (hidden triple in N6)
2e. R789C8 = {126} (hidden triple in C8), 2,6 locked for N9 -> R7C9 = 3
2f. R4C13 (step 1l) = [75/93] (cannot be [84] which clashes with R4C9), no 4,8, clean-up: no 5 in R3C3 (step 1m)
[Now the key step which cracks the puzzle]
2g. Combined cage R4C13 + R4C45 = [75]{18}/[93]/{18] (cannot be [75]{36} which clashes with {56} in either R45C6 or R45C7 (steps 1e and 1g), which are both part of 38(7) cage at R2C8) -> R4C45 = {18}, locked for R4 and N5 -> R4C89 = [74], R3C9 = 8, R5C8 = 8, R4C13 = [93], R3C3 = 6 (step 1m), R5C4 = 3 -> R5C3 = 2, R6C4 = 2 (step 1n), R4C7 = 2 (hidden single in R4) -> R5C7 = 9 (step 1e), R5C5 = 4 -> R5C9 = 6, R6C89 = [51], R1C8 = 9, R45C6 = [65], R4C2 = 5, naked pair {17} in R5C12, 7 locked for N4, clean-up: no 8 in R8C4 (step 1k)
2h. R789C3 (step 1j) = {579}, 5,9 locked for C3 and N7
2i. 9(3) cage at R7C1 = {126/234}, 2 locked for C1 and N7
2j. 8 in N7 only in 15(3) cage at R7C2, locked for C2
2k. 15(3) cage = {168/348}
2l. Killer pair 4,6 in R6C2 and 15(3) cage, locked for C1
2m. 1 in C3 only in R12C3, locked for N1
2n. R4C1 = 9 -> R23C1 = 12 = {57} (cannot be [84] which clashes with R23C3, ALS block), locked for N1, 7 locked for C1 -> R5C12 = [17], clean-up: no 6 in 9(3) cage at R7C1
2o. Naked triple {234} in 9(3) cage, 3,4 locked for C1 and N7
2p. R1C1 = 8 -> R1C23 = 6 = [24], R12C9 = [72], R2C3 = 1
2q. 15(3) cage at R1C4 = {159} (only remaining combination) -> R1C4 = 1, R23C4 = {59}, locked for C4 and N2 -> R4C45 = [81], R8C4 = 7 -> R7C3 = 5 (step 1k, or 7 must repeat in 23(3) cage), R89C3 = [97]
2r. R1C6 = 3 -> R23C6 = 12 = [84], 15(3) cage at R1C5 = [672]
2s. Naked triple {129} in R789C6, 9 locked for N8
2t. 12(3) cage at R8C5 = {345} (only remaining combination) -> R9C4 = 4

and the rest is naked singles.
If there had been three zero cells instead of the repeating 23(3) cage then:
1j, would become R789C3 = {579/678} for the repeating cage and {489/678} for a standard {689} cage with R7C3 = {47}
in which case after step 1o
R34C3 = [45/54/63] (cannot be [27] which clashes with R5C3 = {24} and R7C3 = {47}, ALS block, or is eliminated by R789C3 = {579/678}, 7 locked for C3 for the repeating cage combinations)
That all becomes unnecessary, of course, because the key step 2g would eliminate R4C13 = [57] in addition to eliminating [75]


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