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PostPosted: Sat Nov 23, 2019 11:16 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
When solving a Sudoku I often notice that the solution appears patterned and if you use this it mostly solves - but of course makes it too easy.

In these puzzles the solution pattern is known and this is part of the solving methodology.

Pretty Triplets 2

The nine numbers are split into three triplets horizontally and three triplets vertically (obviously different unless perhaps a Latin Square). To be absolutely clear:
1. each row and each nonet has three separate horizontal triplets; and
2. each column and each nonet has three separate vertical triplets.

This is very much a starter puzzle.


Image

Pretty Triplets 1

This was my original puzzle which started as a hard solution of the triplets using just the six 3-cell cages (two solutions). However adding doublet cages to lock in the actual solution simplified it too much; as usual.

Image


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PostPosted: Thu Dec 19, 2019 4:29 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
Pretty Triplets 2 is definitely a starter puzzle. Probably P&P solvable, although I don't solve that way.

My start:
Horizontal triplet 1,2,4 in 7(3) cage at R3C7, R2C56 = {12} -> R2C4 = 4 to make horizontal triplet, R1C4 = 8,
9(3) cage at R4C6 = {135} (cannot be {126} which clashes with R2C6, cannot be {234} because 4 must be in a vertical triplet including 8)

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


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PostPosted: Fri Dec 20, 2019 2:38 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
Then I found Pretty Triplets 1 only slightly harder.

Here is my walkthrough for Pretty Triplets 1:
Three separate triples horizontally and vertically.

Prelims

a) R12C4 = {59/68}
b) R2C56 = {12}
c) R4C89 = {29/38/47/56}, no 1
d) R5C34 = {15/24}
e) R7C23 = {29/38/47/56}, no 1
f) R7C67 = {59/68}
g) 22(3) cage at R1C3 = {589/679}
h) 11(3) cage at R4C6 = {128/137/146/236/245}, no 9

1a. 22(3) cage at R1C3 = {589/679} -> 5,8,9 or 6,7,9 must be one of the vertical triplets
1b. R12C4 = {59} (cannot be {68} because vertical triplets only contain one of 6,8), locked for C4 and N2, clean-up: no 1 in R5C3
1c. 5,8,9 must be one of the vertical triplets -> R3C4 = 8, 22(3) cage = {589}, locked for C3 and N1, clean-up: no 1 in R5C4, no 2,3,6 in R7C2
1d. Naked pair {24} in R5C34, locked for R5

2a. Naked pair {12} in R2C56, locked for R2 and N2
2b. 11(3) cage at R4C6 = {137/146/236} (cannot be {128} which clashes with R2C6, cannot be {245} which clashes with R5C4), no 5,8
2c. Killer pair 1,2 in R2C6 and 11(3) cage, locked for C6
2d. R456C5 = {589} (hidden triple in N5), locked for C5
2e. R789C6 = {589} (hidden triple in C6), clean-up: no 8 in R7C7
2f. R7C23 = {47}/[83/92] (cannot be [56] which clashes with R7C67), no 5 in R7C2, no 6 in R7C3

3a. Hidden killer pair 5,9 in R3C3 and 12(3) cage at R3C7, R3C3 = {59} -> 12(3) cage must contain one of 5,9 = {129/156/345}, no 7
3b. Horizontal triplets must contain both of 1,2 -> 12(3) cage = {129/345}, no 6
3c. Consider combinations for 12(3) cage
12(3) cage = {129}
or 12(3) cage = {345} => R3C12 = {12} (hidden pair in R3), R3C3 = 9
-> 1,2,9 must be one of the horizontal triplets, totalling 12
3d. R2C56 = {12} -> R2C4 = 9, R1C4 = 5
3e. 16(3) cage at R9C4 doesn’t total 12 -> doesn’t contain 1,2,9 = {457} (cannot be {358} because 5,8 only in R9C6, cannot be {367} because R9C6 must contain one of 5,8) -> R9C6 = 5, R9C45 = {47}, clean-up: no 9 in R7C7
3f. 4,5,7 must be one of the horizontal triplets, R1C4 = 5 -> R1C56 = {47}, locked for R1 and N2
3g. The other horizontal triplet must be 3,6,8
3h. Naked pair {36} in R3C56, locked for R3
3i. 12(3) cage = {129}, locked for R3 and N3
3j. 22(3) cage at R1C3 = [985] -> R1C12 = {12}, R2C12 = {36}, R2C789 = {457}
3k. R7C67 are in different triplets -> R7C67 = [95] (cannot be [86] because 6,8 are in the same horizontal triplet)
3l. R7C6 = 9 -> R7C45 = {12}, locked for R7 and N8
3m. R7C7 = 5 -> R7C89 = {47}, locked for R7 and N9 -> R7C123 = [683] -> R2C12 = [36]
3n. R8C6 = 8, R8C45 = {36}, locked for R8
3o. Naked triple {129} in R8C789, locked for R8 and N9
3p. 17(3) cage at R6C1 = {368} (the only horizontal triplet totalling 17) = [836]
3q. 11(3) cage at R4C6 (step 2b) = {137/146} (cannot be {236} which clashes with R3C6), 1 locked for C6 -> R2C56 = [12], R8C45 = [12]
3r. R7C7 = 5 -> R89C7 = [98] (vertical triplet 5,8,9)
3s. The remaining vertical triplets total 11 and 12 with 2 in the 12 -> R8C9 = 2, R8C8 = 1, clean-up: no 9 in R4C8
3t. R789C3 must total 11 or 12, R7C3 = 3, R9C3 = {12} -> R8C3 = 7
3u. R6C1 = 8 -> R45C1 = {59} (vertical triplet 5,8,9), locked for C1 and N4 -> R8C12 = [45], R3C12 = [74]

4a. R5C789 = {368} (horizontal triplet 3,6,8, cannot be in R5C456 because R5C4 only contains 2,4), locked for N6, 3,6 locked for R5, clean-up: no 5 in R4C89
4b. 5 in N6 only in R6C89 -> R6C789 = {457} (horizontal triplet 4,5,7), locked for R6, 4,7 locked for N6
4c. R56C6 = [71] -> R4C6 = 3 (cage total)
4d. The remaining vertical triplets must be 1,3,7 and 2,4,6
4e. 12(3) cage at R7C9 = [426]
4f. R4C9 = 9 -> R56C9 = [85] (vertical triplet 5,8,9)

and the rest is naked singles.

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

Using Just The 6 3-Cell Cages:
looks a lot harder. Clearly one can start by looking at interactions between 12(3), 16(3) and 17(3) horizontal triplets with 11(3), 12(3) and 22(3) vertical triplets. Then there are two alternatives for 22(3). I don't feel enough of a masochist to go any further, when there are other puzzles which I haven't yet started.


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