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 Post subject: Triankle Killers 0 & 4
PostPosted: Sun Oct 13, 2019 11:01 am 
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Triankle Killers 0 & 4

K0 is easier and about the same level as V3, it only has one move that requires paperwork. It is paper solvable as is V3.
K1 should come next as it is under 1.0, K4 below is above 1.0 and K2 is almost an Assassin.

To me this structure works better as a killer. Semi-symmetric works well with it and also fits better as a killer. Hence I will continue producing them on that basis.

Rules
The numbers are from 0 to 9.
There are 8 triangular nonets covering twelve rows and 11 columns.
The two different numbers in the centre two grey cells repeat horizontally and vertically. I.e. in row six there are eleven numbers so nine different numbers with the grey number twice and row seven the same. In column 6 there are 12 numbers eight different numbers with the two grey numbers twice.
The other eight numbers do not repeat anywhere.
The repeating numbers are present in every nonet.
Each of the other eight numbers is absent in one and only one nonet.
Semi-symmetric.
There are five pairs of numbers which are unknown.For clarity: the ten numbers from 0 to 9 are put in five pairs. Part of the solution process is to work out what these pairs are. Then to work out which of these pairs is the repeating one in r67c6.
If a cell contains a number the opposite cell must contain it or its partner.

These can be put together in JSudoku, see below.

Deducible observation: it is not surprising that Triankle pairs must be missing from opposite nonets.

Triankle Killer 0
Killer cages are normal non-repeat cages.
Apologies cage sum positions corrected. If you carefully look at the cage boundaries there is only one interpretation, but for clarity there are cages 15/2, 23/3, 1/2 and 3/3.


Image

If you wish to solve in JSudoku:
open as a 12 by 12 Latin Square from 0 - B
enter the eight nonets as killer cages with no sum (c, then choose operator "none")
select a cell in C6: ctrl right click select remove and then C6
select a cell in R6: ctrl right click select remove and then R6
select a cell in R7: ctrl right click select remove and then R6 (JS has renumbered the rows)
select all the nonet cells shift A then shift B to remove the A & B pencilmarks
do a set of solves to put A B in every cell around the diamond.
select r67c6 as a twin killer cage with no sum (twin killer is easier to see)
select r6 c1-5 & c7-11 as a twin killer cage 45/10
select r7 c1-5 & c7-11 as a twin killer cage 45/10
select r1-5&r8-12 c6 as a twin killer cage 45/10
Save as your TRIANKLE BASE


Triankle Killer 4

Image


Last edited by HATMAN on Wed Oct 30, 2019 2:59 pm, edited 4 times in total.

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PostPosted: Tue Oct 15, 2019 7:38 pm 
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Addict
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Wow! OK, ready to try #0, but have questions right away about caging. The 23 cage at cell F1 incorporates which cells? And, what about the 15 cage at cell F2?

And, how about the 1 cage and the 3 cage at cells F9 and G9? Are they "1" and "3", respectively? Or, is the 1 cage made up of cells F9/G9 and the 3 cage made up of cells G9/G10/G11?

I can already tell that those "0's" are really going to mess me up, too! But, am looking forward to re-ordering my brain cells for this challenge! Thanks!


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PostPosted: Sat Oct 26, 2019 4:41 pm 
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Cool! I have one question (so far). I understand that the grey cells are each repeated once in their respective rows and columns. But can they be repeated within a cage? E.g., In Triankle 0 - could the 7(3) cage in row 6 be [151]?


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PostPosted: Mon Oct 28, 2019 3:25 pm 
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Thanks wellbeback - normal killer cages only clarified above. The constraints are already less tight than in a normal Sudoku so I do not want to give any more freedom.

Maurice


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PostPosted: Wed Dec 04, 2019 3:24 am 
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As HATMAN said, Triankle Killer 0 is a fairly easy one. I agree with his recommendation to try Triankle Vanilla 3 first to get used to the format, the semi-symmetry, repeated numbers and one number missing from each triangular nonet.

Here is my walkthrough for Triankle Killer 0:
Numbers are 0 to 9.
The numbers in the grey cells repeat horizontally and vertically and are present in every nonet.
No repeats in any other rows or columns.
Each of the other numbers is missing from one and only one nonet.
Five pairs of unknown numbers are semi-symmetric, with the corresponding cell containing the same number or its partner.

Note that in the steps below locked, placements and killer pairs in R6, R7 and C6 don’t apply for R67C6, as appropriate. However placements and locked for N3 and N6 DO apply for R67C6, as appropriate.

1. 45 rule on N1 total = 38 -> missing number 7, all other nonets must contain 7
1a. 22(3) cage at R1C6 = {589}, locked for N1
1b. 5(3) cage at R3C4 = {014/023}, 0 locked for N1
1c. 11(3) cage at R2C5 = {146/236}

2a. 15(2) cage at R5C2 and 23(3) cage at R6C1 = {689} overlap at R6C2 -> R56C2 = [78], R6C13 = {69}, locked for R6 and N2
2b. 1(2) cage at R6C9 and 3(3) cage at R7C9 = {012) overlap at R7C9 -> R67C9 = {01}, locked for C9, 3(3) cage at R7C9 = {012}, locked for R7 and N7
2c. 22(3) cage at R7C7 = {589/679}, 9 locked for N7
2d. 7(3) cage at R6C4 = {025/034/124} (cannot be {007} because [00] cannot repeat in N2, cannot be {01}6 which clashes with R6C9)
2e. Killer pair 0,1 in 7(3) cage and R6C9, locked for R6
2f. 2(2) cage at R5C7 = {02}, locked for R5
2g. Naked pair {02} in 2(2) cage, CPE no 0,2 in R4C8, no 2 in R6C7

3a. 15(2) cage at R4C4 and 14(2) cage at R5C5 are both in N3 -> 14(2) cage = {59}, locked for R5 and N3, 15(2) cage = {78}, locked for R4 and N3
3b. 24(3) cage at R4C9 = {789} = [987]
3c. 5(2) cage at R4C6 = {14/23}

4a. 4(3) cage at R7C1 = {013} -> R7C1 = 3, R8C2 + R9C3 = {01}, locked for N5
4b. 5(3) cage at R8C6 = {014/023}, 0 locked for N6
4c. 14(2) cage at R8C4 = {59/68}
4d. 14(2) cage at R9C4 = {59/68}
4e. 15(2) cage at R9C7 = {69/78}
4f. 13(2) cage at R10C5 = {49/58/67}
4g. 9(2) cage at R9C5 = {18/27/36/45}

5. Starting with semi-symmetry
5a. R6C11 = 7 corresponds with R7C1 = 3 -> 3,7 are a pair
5b. 7 is missing from N1 -> 3 must be missing from N8
5c. R5C2 = 7 -> R8C11 = {37}
5d. R4C9 + R5C10 = [98] correspond with R8C2 + R9C3 = {01} -> 8,9 are paired with 0,1
5e. R6C2 = 8 corresponds with R7C10 -> R7C10 = {01} -> R7C11 = 2
5f. R6C23 correspond with R7C910 = {01} -> R6C23 = [89], R6C1 = 6
5g. R6C1 = 6 corresponds with R7C11 = 2 -> 2,6 are a pair

6. 3 is missing from N8 which must therefore contain all of 0,1,2 in R11C5 + R12C6 and 15(3) cage at R10C7
6a. 15(3) can only contain one of 0,1,2 -> R11C5 + R12C6 can only contain 0,1,2
6b. R11C5 + R12C6 correspond with R1C6 + R2C7 = {589}, no 6 -> 2 in 15(3) cage, R11C5 + R12C6 = {01}, locked for N8, R1C6 + R2C7 = {89} -> R3C8 = 5
6c. R3C8 = 5 corresponds with R10C4 = {5689}, but 6,8,9 correspond with 0,1,2 -> R10C4 = 5, R9C4 = 9
6d. 14(2) cage at R8C4 = {68}, locked for R8
6e. Naked pair {68} in 14(2) cage, CPE no 6,8 in R7C5
6f. 45 rule on N8 using R11C5 + R12C6 = {01} and missing number 3 -> R10C6 = 8, R9C7 = 7
6g. R1C6 = 9 -> R2C7 = 8, R5C56 = [95], 13(2) cage at R10C5 = {67} -> 15(3) cage = {249}

7. Pairs are 2 and 6, 3 and 7, 0,1 with 8,9 -> the remaining pair must be 4,5
7a. R5C6 corresponds with R8C6, no 5 in R8C6 -> R8C6 = 4 -> 5(3) cage at R8C6 = 4{01}, locked for N6
7b. R9C8 corresponds with R4C4, R9C8 = {01}, no 0,1 in R4C4 -> R4C4 = 8, R4C5 = 7, 14(2) cage at R8C4 = [68], 13(2) cage at R10C5 = [67]
7c. R11C6 corresponds with R2C6, R11C6 = 7, no 7 in R2C6 -> R2C6 = 3
7d. R2C6 = 3 -> R2C5 + R3C7 = 8 = [26]
7e. R2C5 corresponds with R11C7, R2C5 = 2, no 6 in R11C7 -> R11C7 = 2
7f. 2(2) cage at R5C7 = [02] -> 5(3) cage at R8C6 = [410], 4(3) cage at R7C1 = [301]
7g. Naked triple {345} in R6C7810, locked for R6 and N4 -> R5C9 = 6
7h. 9(2) cage at R9C5 = [36] (only remaining permutation)
7i. Naked pair {01} in R312C6, locked for C6 -> 5(2) cage at R3C4 = [23]

8. R6C4 = 2 (hidden single in R6) -> 7(3) cage at R6C4 = {124} -> R6C56 = [14], 1(2) cage at R6C9 = [01] -> R7C10 = 0
8a. 4 and 5 are a pair -> R7C6 = 5, the other repeated number
8b. R11C5 + R12C6 = [01] -> 5(3) cage at R3C4 = [140]
8c. R7C56 = [55] -> no other 5 in R7
8d. 22(3) cage at R7C7 = {679} = [967], R4C8 = 1
8e. R7C234 = [487], R8C3 = 2
8f. R8C910 = [53]
8g. R10C78 = [49] -> R6C7810 = [534]
8h. R4C4 = 8 corresponds with R9C8 = 0 -> 0 and 8 are a pair -> 1 and 9 are the remaining pair
8i. R5C4 corresponds with R8C8, R8C8 = 7, no 7 in R5C4 -> R5C4 = 3, R5C3 = 4
8j. The repeated numbers must be in every nonet -> R4C3 = 5, R8C9 = 4

Solution:
Attachment:
Triankle Killer 0.jpg
Triankle Killer 0.jpg [ 70.56 KiB | Viewed 9874 times ]


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PostPosted: Tue Aug 24, 2021 2:11 am 
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Grand Master
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Then the day after solving Triankle Killer 1 I had another look at Trainkle Killer 4, for which I'd originally reached an impossible position by making incorrect assumptions for corresponding cages.

Here is my walkthrough for Triankle Killer 4:
Numbers are 0 to 9.
The numbers in the grey cells repeat horizontally and vertically and are present in every nonet.
No repeats in any other rows or columns.
Each of the other numbers is missing from one and only one nonet.
Five pairs of unknown numbers are semi-symmetric, with the corresponding cell containing the same number or its partner.

Note that in the steps below locked, placements and killer pairs in R6, R7 and C6 don’t apply for R67C6, as appropriate. However placements and locked for N3 and N6 DO apply for R67C6, as appropriate.

15(3) cage at R7C1 and 15(2) cage at R7C3 overlap at R8C3

Prelims
a) 14(2) cage at R2C5 = {59/68}
b) 4(2) cage at R2C7 = {04/13}
c) 14(2) cage at R4C5 = {59/68}
d) 4(2) cage at R4C7 = {04/13}
e) 11(2) cage at R4C8 = {29/38/47/56}
f) 5(2) cage at R5C8 = {05/14/23}
g) 4(2) cage at R6C3 = {04/13}
h) 15(2) cage at R7C3 = {69/78}
i) 8(2) cage at R7C4 = {08/17/26/35}
j) 12(2) cage at R7C7 = {39/48/57}
k) 7(2) cage at R8C5 = {07/16/25/34}
l) 8(2) cage at R9C3 = {08/17/26/35}
m) 11(2) cage at R9C6 = {29/38/47/56}
n) 6(2) cage at R10C4 = {06/15/24}
o) 8(2) cage at R10C5 = {08/17/26/35}
p) 6(3) cage at R5C5 = {015/024/123}
q) 23(3) cage at R5C9 = {689}
r) 3(3) cage at R7C6 = {012}
s) 8(3) cage at R7C11 = {017/026/035/125/134}

1a. Naked triple {689} in 23(3) cage at R5C9, locked for N4, clean-up: no 2,3,5 in R4C8
1b. Naked triple {012} in 3(3) cage at R7C6, locked for N6, clean-up: no 5,6,7 in 7(2) cage at R8C5, no 6,7,8 in R9C3, no 9 in 11(2) cage at R9C6
1c. Naked pair {34} in 7(2) cage at R8C5, locked for C5 and N6, clean-up: no 5,6 in R3C6, no 5,6 in R5C4, no 5 in R9C3, no 7,8 in 11(2) cage at R9C6, no 2 in R10C4, no 5 in R11C6
1d. Naked pair {56} in 11(2) cage at R9C6, locked for R9, clean-up: no 2,3 in R9C3
1e. 4(2) cage at R4C7 and 6(3) cage at R5C5 form combined 10(5) cage {01234), locked for N3, clean-up: no 7 in R4C9
1f. 11(2) cage at R4C8 = [74/83/92] (cannot be [65] which clashes with R4C56)
1g. This step held until step 2f.

[Time to look at semi-symmetry.]
2a. 14(2) cage at R4C5 = {59/68} corresponds with 11(2) cage at R9C6 = {56} -> at least one of 8,9 must be paired with one of 5,6 -> 5,6 cannot be paired with each other
2b. R49C6 are in the same column and not in repeat cells -> R4C6 = {89}, R4C5 = {56}
2c. Whichever are in R49C6 must be paired, R4C5 and R9C7 must be the same since 5 not paired with 6 -> 5 must be paired with 8 and/or 6 must be paired with 9
2c. 4(2) cage at R4C7 = {04/13} corresponds with 7(2) cage at R8C5 = {34} -> 0 must be paired with 3 and/or 1 must be paired with 4
2d. R4C9 corresponds with R9C3 = {01}, no 0,1 in R4C9 -> R4C9 = {34}, clean-up: no 9 in R4C8
2e. 23(3) cage at R5C9 = {689} corresponds with 15(3) cage at R7C1
2f. 15(3) cage at R7C1 overlaps with R78C3 = {69/78} -> 15(3) cage cannot be {069/078}, no 0,9 in R7C1 + R8C2
2g. 15(3) cage cannot contain both of 6,9 -> 6 cannot be paired with 9 -> 5 must be paired with 8
2h. R49C6 = [85] -> R4C5 = 6, R9C7 = 6, clean-up: no 9 in R2C5, no 1 in R3C5, no 3 in R3C6, no 3 in R5C4, no 0 in R10C4, no 0 in R10C5, no 2 in R11C6
2i. R4C8 = 7 -> R4C9 = 4, clean-up: no 0 in R5C7, no 1 in 5(2) cage at R5C8
2j. R4C8 = 7 corresponds with R9C4 -> R9C4 = 7 (because 8 paired with 5) -> R9C3 = 1, clean-up: no 3 in 4(2) cage at R6C3, no 2 in R6C5, no 5 in R7C7
2k. R4C9 corresponds with R9C3 -> 1 and 4 are paired; don’t yet know whether 0 and 3 are also paired
2l. Naked pair {04) in 4(2) cage at R6C3, CPE no 0,4 in R6C2
2m. 23(3) cage = {689}, 5 paired with 8 -> 15(3) cage must contain one, but not both, of 5,8 = {348/357/456}, no 2,9, clean-up: no 6 in R7C3
[Note that {258} would also be eliminated by clashing with R78C4.]
2n. 15(3) cage = {357} (cannot be {348/456} because 1 paired with 4) -> R8C3 = 7, R7C1 + R8C3 = {35}, locked for N5, R7C3 = 8, clean-up: no 0 in 8(2) cage at R7C4, no 4 in R8C8
2o. 5 paired with 8, 5 in R7C1 + R8C3 corresponds with 8 in R5C10 + R6C11, 8 locked for 23(3) cage at R6C10
2p. R8C3 = 7 corresponds to R5C9 = {69} -> 7 paired with one of 6,9, 3 paired with the other of 6,9 -> 0 and 2 must be paired
2r. Naked pair {26} in 8(2) cage at R7C4, locked for C4 and N5, clean-up: no 7 in R6C5, no 0 in R11C5
2s. R4C4 corresponds with R9C8 -> R4C4 = 9, R9C8 = 9 (they cannot be R4C4 = 5, R9C8 = 8 because then 9 would be the missing number from both N3 and N6), clean-up: no 0 in R6C5, no 3 in R7C7
2t. 5 not in N3, 8 not in N6; both of 5,8 must be in the other nonets
2u. R2C5 corresponds with R11C7, R2C5 = {58} -> R11C7 = {58}
2v. R7C3 corresponds with R6C9, R7C3 = 8 -> R6C9 = 5, clean-up: no 4 in R5C4, no 0 in 5(2) cage at R5C8
2w. Naked pair {23} in 5(2) cage at R5C8, locked for C8 and N4, clean-up: no 1 in R2C7, no 9 in R7C7
2x. 1 in R7, apart from possible repeat in R7C6, only in R7C8910,11, locked for N7
2y. 8(3) cage at R7C11 = {026/035/125} (cannot be {017} because 1,7 only in R7C11, cannot be {134} which clashes with R8C5), no 4,7

3a. R8 has 9 cells without 9 -> R8C8 = 8 (hidden single in R8) -> R7C7 = 4, R7C2 = 0 -> R6C3 = 4, R7C5 = 9, clean-up: no 0 in R3C6, no 0 in R4C7, no 0 in R5C4
[The last key step, fairly straightforward from here.]
3b. Naked pair {13} in 4(2) cage at R4C7, locked for C7 and N3, R2C7 = 0 -> R3C8 = 4, R7C7 = 7, clean-up: no 5 in R3C5, no 9 in R3C6
3c. R6C7 corresponds with R7C5 -> 7 paired with 9 -> 3 with 6
3d. No 4 in N5 -> no 1 in N4 -> R6C10 = 0
3e. 1 in R6 only in R6C1245, locked for N2 -> 9(2) cage at R5C4 = [81] -> R6C4 = 3, 5(2) cage at R5C8 = [32], 4(2) cage at R4C7 = [31], clean-up: no 5 in R10C4, no 7 in R11C6
3f. R8C3 = 7 corresponds with R5C9 -> R5C9 = 9, R5C10 = 6, R6C11 = 8
3g. R6C11 = 8 corresponds with R7C1 -> R7C1 = 5, R8C2 = 3, 7(2) cage at R8C5 = [43]
3h. R8C7 = 2 -> 8(2) cage at R7C4 = [26], R8C910 = [05] -> R7C11 = 3 (cage sum), R78C6 = [01], clean-up: no 8 in R3C5, no 7 in R10C5
3i. No 9 in N7 -> no 7 in N2
3j. R7C11 = 3 corresponds with R6C1 -> R6C1 = 6, R5C23 = 7 = {25}, locked for N2, 2 locked for R5 -> R5C56 = [04], R6C6 = 2
3k. R3C5 = 7 (hidden single in C5) -> R3C6 = 2
3l. R10C4 = 4 (hidden single in N8) -> R11C5 = 2, clean-up: no 6 in R11C6
3m. R2C6 corresponds with R11C6 -> R2C6 = 6 -> R2C5 = 8, 8(2) cage at R10C5 = [53]
3n. R10C8 = 1 (hidden single in N8)
3o. R1C6 = 9 corresponds with R12C6 -> R12C6 = 7
3p. R6C2 = 9 corresponds with R7C11 -> R7C11 = 7
3q. R8C9 = 0 corresponds with R5C3 -> R5C3 = 2

and the rest is naked singles.

Solution:
Attachment:
Triankle Killer 4.jpg
Triankle Killer 4.jpg [ 62.03 KiB | Viewed 6543 times ]


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