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 Post subject: Triankle Killer 2
PostPosted: Tue Oct 08, 2019 10:50 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
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Location: Saudi Arabia
Triankle Killer 2

Please do the vanilla version (V3) first so as to understand the rules (I will answer any questions).
K1 should come next as it is under 1.0 whereas this one 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 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.
So Semi-symmetric.
There are five pairs of numbers which are unknown.
If a cell contains a number the opposite cell must contain it or its partner.

There is one cage which can repeat.


Image


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 Post subject: Re: Triankle Killer 2
PostPosted: Thu Oct 31, 2019 10:11 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
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Location: California, out of London
Phew! Very tough. I ended up with multiple solutions (based on an what may have been an initial faulty premise which luckily worked out) which I could only resolve into one solution using the "Each of the other eight numbers is absent in one and only one nonet" rule. (Edit: Third line fixed)
Solution (I think - I hope):
Code:
     1
    394
   57082
  4908631
 930521876
67812494053
25049131876
 678430529
  1267594
   38570
    162
     4


Pairs [05][14][23][69][78]


Last edited by wellbeback on Mon Nov 04, 2019 9:12 pm, edited 1 time in total.

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 Post subject: Re: Triankle Killer 2
PostPosted: Sun Nov 03, 2019 2:30 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
wellbeback

You are correct although you have a typo 3 in the third line.

Did you use the opposite absent pair implication?

Would you care to put a rating number on it?

Maurice


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 Post subject: Re: Triankle Killer 2
PostPosted: Mon Nov 04, 2019 9:33 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks for noticing the typo - solution has been edited.
I did not use the implication you mention. I hadn't thought of it!
Regarding a rating: Andrew & Ed have a very good method of rating puzzles based on the specific techniques used in their WTs. I am not as knowledgeable that way and, when asked, base a rating simply on how long it took me to solve. All the Triankles I have tried so far have taken me a looooong time. For example, it took me hours in Triankle 6 to get the 6(3) in c6 down to one set of numbers! Pretty much the only technique I've used so far is to postulate something, see what follows, and see if I get a contradiction. The efficacy of using standard techniques such as '45's and IODs is much reduced in Triankles.
Anyway, that means my rating for Triankle Killer 2 is a hard 1.5. Perhaps I will get better at them with more practice.


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 Post subject: Re: Triankle Killer 2
PostPosted: Tue Nov 05, 2019 8:33 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
My principle solution technique is building S-S contradiction tables (in Excel). Then doing eliminations between the tables. This takes me back to some of my KiMo puzzles.

I find the sum hint tool in http://www.atksolutions.com/flashgames.html Kakuro game avoids memory of combinations mistakes.

Given your view of difficulty I will not try for harder ones until people are used to them.


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 Post subject: Re: Triankle Killer 2
PostPosted: Wed Sep 01, 2021 4:28 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
Initially I found this heavy going, then realised I'd overlooked an important condition which limited a pair. After that my solving path was a bit like that for Triankle Killer 1 but somewhat heavier going.

Here's my walkthrough for Triankle Killer 2:
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.

Prelims

a) 13(2) cage at R2C6 = {49/58/67}
b) 12(2) cage at R3C4 = {39/48/57}
c) 0(2) cage at R3C6 = [00]
d) 7(2) cage at R4C3 = {07/16/25/34}
e) 14(2) cage at R4C6 = {59/68}
f) 15(2) cage at R5C2 = {69/78}
g) 10(2) cage at R5C7 = {19/28/37/46}
h) 11(2) cage at R5C10 = {29/38/47/56}
i) 6(2) cage at R7C3 = {06/15/24}
j) 11(2) cage at R7C4 = {29/38/47/56}
k) 13(2) cage at R7C5 = {49/58/67}
l) 15(2) cage at R7C11 = {69/78}
m) 6(2) cage at R8C9 = {06/15/24}
n) 13(2) cage at R9C5 = {49/58/67}
o) 2(2) cage at R10C8 = {02}
p) 7(2) cage at R11C5 = {07/16/25/34}
q) 3(3) cage at R5C4 = {012}
r) 8(3) cage at R6C9 = {017/026/035/125/134}
s) 7(3) cage at R7C1 = {016/025/034/124}
t) 4(3) cage at R7C6 = {013}
u) 9(3) cage at R7C7 = {018/027/036/045/126/135/234}

1a. R3C6 = 0, clean-up: no 7 in R11C5
1b. R4C5 = 0, clean-up: no 7 in R5C3, no 7 in R11C6
1c. 3(3) cage at R5C4 = {012), locked for N2, 0 locked for C4, clean-up: 5,6,7 in R45C3
1d. 7(2) cage at R4C3 = {34}, locked for C3 and N2, clean-up: no 7,8 in R7C4, no 2 in R8C2
1e. 4(3) cage at R7C6 = {013}, locked for N6
1f. 2(2) cage at R10C8 = {02), locked for N8, clean-up: no 5 in 7(2) cage at R11C5
1g. 11(3) cage at R5C5 = {128/137/146/236/245}
1h. 8(3) cage at R6C9 = {035/134} (cannot be {017/026/125} which clash with R6C45, ALS block), 3 locked for N4, clean-up: no 7 in R5C7
1i. Killer triple 0,1,2 in R6C45 and 8(3) cage, locked for R6 and 3(3) cage at R5C4, no 2 in R5C4, clean-up: no 8,9 in R5C7
1j. R6C10 = {345} -> R5C10 = {678}
1k. 9(3) cage at R7C7 = {018/027/036/135/234} (cannot be {045/126} which clash with 6(2) cage at R8C9)
1l. 7(3) cage at R7C1 overlaps with 6(2) cage at R7C3 (combo crossover clash), R7C13 and R7C23 cannot total 6 -> no 1 in R7C12
1m. 7(3) cage = {016/025/034/124}
1n. 1 of {016} must be in R7C3 -> no 6 in R7C3, clean-up: no 0 in R8C2
1o. Apart from possibly being in repeat-number cell R6C6, 4 in R6 only in R6C78910,11, locked for N4
1p. 12(3) cage at R4C9 = {057/129/147/246} (cannot be {048/156} which clash with 8(3) cage at R6C9), no 8
1q. 0 of {057} must be in R5C9 -> no 5 in R5C9
1r. 9 on {129} must be in R6C8 -> no 9 in R45C9
1s. 12(3) cage + 8(3) cage must contain 0, locked for N4

[Time to look at semi-symmetry.]
2a. R45C3 = {34} corresponds with R78C10 = {06/15/24}, 7(2) cage at R11C5 = {16/34} corresponds with 13(2) cage at R2C6 = {49/58/67}, {34} cannot correspond with both {06/15/24} and {49/58/67} -> 7(2) cage at R11C5 = {16}, locked for N8
2b. R3C6 = 0 corresponds with R10C6, R4C5 = 0 corresponds with R9C7, neither of R9C7 and R10C6 contain 0 -> R9C7 and R10C6 must contain the same number = {45789}
2c. 11(3) cage at R5C5 corresponds with 4(3) cage at R7C6 = {013}, no 0 in 11(3) cage -> 0 must correspond with one of {4578} -> R9C7 and R10C6 must contain the same number = {4578} -> 11(3) cage = {128/137/146/245} (cannot be {236})
[I was tempted to try to eliminate more combinations but I don’t think any of them stop 13(2) cage at R2C6 corresponding with 7(2) cage at R11C5 = {16}.]
2d. 0 in 4(3) cage only in R7C6 + R8C7, the corresponding number in 11(3) cage is in R5C5 + R6C6 -> no 7,8 in R5C6
2e. 3(3) cage at R5C4 corresponds with 9(3) cage at R7C7, no 0,4,5,7,8 in R6C5 -> no 0 in R7C7
2f. 13(2) cage at R2C6 = {49/58/67} corresponds with 7(2) cage at R11C5 = {16} -> 1 must correspond with at least 4 but cannot be paired with 6 since 13(2) cage doesn’t contain both of 1,6
2g. 9(3) cage at R7C7 (step 1k) = {018/027/036/135/234}
2h. 7(2) cage at R4C3 = {34} corresponds with 6(2) cage at R8C9 = {06/15/24}
2i. {34} with {06} => 0 corresponds with 4, 3 with 6 => 0 in 3(3) cage must correspond with 4 in 9(3) cage but 3(3) cage cannot then correspond with 9(3) cage = {234} because 3 cannot correspond with 1
or {34} corresponds with {15} => 1 must correspond with at least 4 and then 0 in 3(3) cage would have to correspond with 5 in 9(3) cage but {045} already eliminated from 9(3) cage)
-> 6(2) cage = {24}, locked for C9 and N7
[That’s the hard part done, fairly straightforward from here.]
2j. 7(2) cage at R4C3 = {34} corresponds with 6(2) cage at R8C9 = {24} -> 2 paired with 3
2k. 3(3) cage at R5C4 corresponds with 9(3) cage at R7C7, 3(3) cage contains 2 -> 9(3) cage must contain 3 = {135} (cannot be {036} because 1 cannot be paired with 6), locked for N7
2l. {012} corresponds with {135}, 2 paired with 3 -> 0 paired with 5
2m. 0 in 3(3) cage in R56C4 -> 5 in R78C8, locked for C8 and N7
2n. R3C6 = 0 -> R10C6 = 5, R4C5 = 0 -> R9C7 = 5, clean-up: no 8 in 13(2) cage at R2C6, no 9 in 14(2) cage at R4C6, no 8 in R7C5, no 8 in 13(2) cage at R9C5
2o. Naked pair {68} in 14(2) cage at R4C6, locked for R4 and N3, clean-up: no 4 in R6C7
2p. 11(3) cage at R5C5 corresponds with 4(3) cage at R7C6 = {013), 0 paired with 5, 2 with 3 -> 11(3) cage must be {245}, locked for N3 -> 1 paired with 4, clean-up: no 6,8 in R6C7
2r. R2C5 + R3C4 correspond with 2(2) cage at R10C8 -> R2C5 = {235}, R3C4 = {35}, R3C5 = {79}
2s. 13(2) cage at R2C6 with 7(2) cage at R11C5 = {16}, 1 paired with 4 -> 13(2) cage = {49}, locked for N1, 6 paired with 9 -> 7 paired with 8
2t. R3C5 = 7 -> R3C4 = 5, clean-up: no 6 in R8C3
2u. 14(2) cage at R4C6 = {68} corresponds with 13(2) cage at R9C5 = [67]
2v. 7(2) cage at R11C5 = [16] -> 13(2) cage at R2C6 = [94]
2w. 14(2) cage = [86]

3a. R6C5 = 2 -> R2C5 = 3
3b. R6C5 = 2 -> R7C7 = 3, R5C7 = 1 -> R6C7 = 9, R56C4 = [01], clean-up: no 6 in R5C2, no 8 in R8C3
3c. R2C5 = 3 -> R11C7 = 2, R10C8 = 0, R3C7 = 8
3d. R8C7 = 0 -> R5C5 = 5, clean-up: no 8 in R8C5
3e. 8(3) cage at R6C9 = {035}, 5 locked for R6 and N4, clean-up: no 7 in R5C10
3f. R5C2 = 9 (hidden single in N2) -> R6C1 = 6
3g. 15(2) cage at R5C2 corresponds with 15(2) cage at R7C11 = {69}, locked for N7
3h. No 5 in N2 -> no 0 in N7
3i. 12(3) cage at R4C9 = {147} (only remaining combination) = [174] -> R7C910 = [87]
3j. R5C4 = 0 -> R8C8 = 5, R7C8 = 1, clean-up: no 6 in R7C4
3k. 7(3) cage at R7C1 = {025} (only remaining combination), 2 locked for N5, clean-up: no 9 in 11(2) cage at R7C4
3l. 11(2) cage = [47] -> R6C23 = [78], 13(2) cage at R7C5 = [94], 15(2) cage at R7C11 = [69], R10C57 = [87], clean-up: no 2 in R6C3
3m. R8C9 = 2 -> R5C3 = 3
3n. R4C9 = 1 -> R9C3 = 1, R8C2 = 6 -> R7C3 = 0 -> R6C9 = {05}
3o. R8C2 = 6 -> R5C10 = 6, R6C10 = 5, R6C911 = [03]
3p. R6C11 = 3 -> R7C1 = 2
3q. 8 cannot be missing from both of N4 and N5 -> R5C8 = 8, R8C4 = 8, 2 missing from N4, 3 missing from N5
3r. No 8 in N6 -> no 7 in N3
3s. Naked pair {39} in R410C4, locked for C4 -> R9C48 = [29], R4C48 = [93],
3t. R10C4 = 3 -> R3C8 = 2, R112C6 = 1,4
3u. 6 missing from N1, 9 missing from N8
3v. 1,4 must be the repeated numbers -> R67C6 = [41]

Graphical Solution:
Attachment:
Triankle Killer 2.jpg
Triankle Killer 2.jpg [ 63.01 KiB | Viewed 7715 times ]


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 Post subject: Re: Triankle Killer 2
PostPosted: Wed Sep 01, 2021 4:33 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
HATMAN wrote:
Given your view of difficulty I will not try for harder ones until people are used to them.
I'll keep trying on Triankle Killers 6 and 7. Since I find visualisation of semi-symmetry and puzzles with 0-9 difficult, I'm not looking for any harder ones beyond those.


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