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 Post subject: Hybrid Triples 4
PostPosted: Fri Aug 27, 2021 6:25 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Hybrid Triples 4

Harder - a few fishes

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 Post subject: Re: Hybrid Triples 4
PostPosted: Mon Aug 30, 2021 4:51 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
A nice puzzle. Maybe forcing chains are different from JSudoku's fishes; I only used two forcing chains.

Here is my walkthrough for Hybrid Triples 4:
Hybrid cages; numbers in each cage are NC but first and third are consecutive.

Prelims (taking into account hybrid rule, eliminating combinations with no or all three digits consecutive)

a) 8(3) cage at R5C6 = {125/134}
b) 11(3) cages at R1C4, R2C7 and R3C1 = {128/236/245}, no 7,9
c) 12(3) cages at R7C6 and R8C3 = {129/156/237}, no 4,8
d) 13(3) cages at R5C3 and R9C2 = {238/256/346}, no 1,7,9
e) 14(3) cages at R3C6 and R6C9 = {167/239/347/356}, no 8
f) 16(3) cage at R1C2 = {178/349/367/457}, no 2
g) 17(3) cages at R3C5 and R9C7 = {278/458/467}, no 1,3,9
h) 18(3) cage at R1C5 = {189/378/459}, no 2,6

1a. 45 rule on R5 3 innies R5C129 = 24 = {789}, 8 locked for R5
1b. 13(3) cage at R5C3 = {256/346}
1c. R5C4 = {26} (hybrid rule), no 2 in R5C35
1d. 8(3) cage at R5C6 = {125/134}
1e. R5C7 = {15} (hybrid rule), no 5 in R5C68

2a. 1,9 in R9 only in R9C156
2b. 45 rule on R9 3 innies R9C156 = 15 = {159}, 5 locked for R9
2c. 17(3) cage at R9C7 = {278/467}, 7 locked for N9
2d. R9C8 = {24} (hybrid rule), no 2,4 in R9C79
2e. 13(3) cage at R9C2 = {238/346}
2f. R9C3 = {68} (hybrid rule), no 6,8 in R9C24
2g. 14(3) cage at R6C9 = {167/239/356} (cannot be {347} which requires 7 in R7C9, hybrid rule), no 4
2h. R7C9 = {139} (hybrid rule), no 1,9 in R68C9
2i. 12(3) cage at R7C6 = {129/156} (cannot be {237} which requires 7 in R7C7, hybrid rule), no 3,7, 1 locked for R7
2j. R7C7 = {19} (hybrid rule), no 9 in R7C68
2k. 14(3) cage = {239/356}, no 7, 3 locked for C9

3a. 11(3) cages at R1C4, R2C7 and R3C1 = {128/236/245}, 2 locked for N1, N2 and N3
3b. R2C4, R2C8 and R3C2 = {268} (hybrid rule), no 6,8 in R1C4, R2C57, R3C138
3c. 16(3) cage at R1C2 = {178/349/367/457}
3d. R2C2 = {1379) (hybrid rule), no 1,9 in R1C2 + R2C3
3e. 18(3) cage at R1C5 = {189/378/459}
3f. R1C6 = {139) (hybrid rule), no 1,3 in R1C57
3g. 17(3) cage at R3C5 = {278/458/467}
3h. R4C5 = {248) (hybrid rule), no 2 in R4C6
3i. 14(3) cage at R3C6 = {167/239/347/356}
3j. R3C7 = {1379} (hybrid rule), no 1,9 in R3C6 + R4C7

4a. 12(3) cage at R8C3 = {129/156/237}
4b. Consider combinations for 12(3) cage at R7C6 = {129/156}
12(3) cage = {129}, locked for R7 => R7C9 = 3, R68C9 = {56} => 12(3) cage at R8C3 = {129/237} (cannot be {156} which clashes with R8C9)
or 12(3) cage at R7C6 = {156}, R7C6 = {56}, R9C56 = {159} => 12(3) cage at R8C3 = {129/237} (cannot be {156}, R8C4 = 1, R8C5 = {56} which clash with R7C6 + R9C56, ALS block)
-> 12(3) cage at R8C3 = {129/237}, no 5,6, 2 locked for R8
4c. R8C4 = {79} (hybrid rule), no 7,9 in R8C35
4d. 2 in N9 only in R79C8, locked for C8
4e. R2C7 = 2 (hidden single in N3)
4f. R1C4 = 2 (hidden single in N2), R5C4 = 6, R2C4 = 8 -> R2C5 = 1 (cage sum)
4g. R2C78 = [26] -> R3C8 = 3 (cage sum)
4h. R5C6 = 2 (hidden single in R5) -> R5C78 = 6 = [51]
4i. 17(3) cage at R3C5 = {458/467}, CPE no 4 in R56C5
4j. R5C35 = [43] -> R8C5 = 2

5a. 18(3) cage at R1C5 = {378/459}
5b. Consider combinations for 16(3) cage at R1C2 = {349/367/457}
16(3) cage = {349/457} => R1C2 = 4 => 18(3) cage = {378}
or 16(3) cage = {367} = [637] => R1C6 = 3 (hidden single in N2) => 18(3) cage = {378}
-> 18(3) cage = {378} = [738]
5c. R3C2 = 8 (hidden single in R3) -> R3C13 = {12}, 1 locked for R3 and N1
5d. 14(3) cage at R3C6 = {347} (only remaining combination) = [473]
5e. R3C5 = 6 (hidden single in R3) -> R4C56 = 11 = [47]
5f. Naked triple {159} in R346C6, locked for C6, 1 locked for N5
5g. R8C45 = [72] -> R8C3 = 3 (cage sum)
5h. R7C9 = 3 (hidden single in C9) -> R68C9 = 11 = [65]
5i. R7C8 = 2 -> R7C67 = 10 = [19]
5j. R9C9 = 7 (hidden single in R9)

6a. R3C49 = [59], R5C9 = 8, R4C8 = 9, R4C4 = 1, R2C6 = 9, R9C156 = [195], R3C13 = [21]
6b. R6C12 = [31] (hidden pair in R6)
6c. R2C2 = 3 (hidden single in N1) -> R1C2 + R2C3 = 13 = [67]

and the rest is naked singles.

Solution:
4 6 9 2 7 3 8 5 1
5 3 7 8 1 9 2 6 4
2 8 1 5 6 4 7 3 9
8 5 6 1 4 7 3 9 2
7 9 4 6 3 2 5 1 8
3 1 2 9 5 8 4 7 6
6 7 5 4 8 1 9 2 3
9 4 3 7 2 6 1 8 5
1 2 8 3 9 5 6 4 7


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