SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Wed Aug 10, 2022 12:07 pm

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Triankle Killer 11
PostPosted: Sat May 07, 2022 5:00 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 659
Location: Saudi Arabia
Triankle Killer 11

Hi All
I have just taken a job in Saudi Arabia where there is no alcohol. Hence I have significant free time, so I will be posting again.
I am currently creating Triankle 11 (not sure what happened to 8, 9, and 10), My four stage process is:
1. find a solution (not that easy) preferably with a balance between same symmetry and anti-symmetry;
2. apply the semi-symmetric pairing in JSudoku and create a killer that solves;
3. add cages so that the semi-symmetric pairs can be derived; then
4. solve the puzzle perhaps removing unnecessary cages.

I am at stage 2 on this one and realized that is is an easier puzzle in its own right, so I decided to post it as an introduction to Triankles.

I will post the full puzzle soon.

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.
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 (this is provable).


Triankle Killer 11 Given

The red numbers are missing from the nonet.
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


Last edited by HATMAN on Tue May 17, 2022 8:38 am, edited 2 times in total.

Top
 Profile  
Reply with quote  
 Post subject: Re: Triankle Killer 11
PostPosted: Tue May 17, 2022 8:25 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 659
Location: Saudi Arabia
Triankle Killer 11 Full Puzzle

This is not too hard, but I like it because it is the first one where I discovered the pairs as I solved. In previous ones I derived the pairs first.
The 4/3 cage is in r5c56,r6c6 the 8/2 cage is in r5c67 and the 6/3 cage is in r6c567.

For an easy puzzle you can include the missing number clues above - if you have not done a Triankle before I recommend this approach.

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.
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 (this is provable).


Triankle Killer 11 Full Puzzle
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


Top
 Profile  
Reply with quote  
 Post subject: Re: Triankle Killer 11
PostPosted: Sun Jul 10, 2022 6:09 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1820
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for these new puzzles.

It was a while since I'd done a Triankle Killer so I did the Given one first.

Here's how I solved Triankle Killer 11 Given:
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.

No 9 in N1, no 6 in N2, no 8 in N3, no 7 in N4, no 3 in N5, no 2 in N6, no 4 in N7, no 1 in N8
The two numbers in the grey cells repeat horizontally and vertically; they are in all nonets

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, after deleting missing numbers from nonets

a) 9(2) cage at R2C6 = {18/27/36/45}
b) 13(2) cage at R2C7 = {58/67}
c) 11(2) cage at R3C6 = {38/47/56}
d) 8(2) cage at R6C2 = {08/17}/[26/35]
e) 1(2) cage at R6C4 = {01}
f) 12(2) cage at R6C10 = {39/48}
g) 8(2) cage at R7C10 = {08/17/26/35}
h) 5(2) cage at R9C7 = {05/14}
i) 9(2) cage at R11C5 = {09/27/36/45}
j) 15(2) cage at R11C6 = {69/78}
k) 4(3) cage at R5C5 = {013}
l) 24(3) cage at R7C1 = {789}
m) 15(6) cage at R1C1 = {012345}

1a. The two numbers repeated in all cells must be 0,5 -> R6C6 = 0, R7C6 = 5
1b. R5C56 = {13}, locked for R5 and N3
1c. R7C6 = 5 -> 5(2) cage at R9C7 = {14}, locked for R9 and N7
1d. R7C6 = 5 -> R8C67 = 12 = {39}, locked for R8 and N7
1e. 11(2) cage at R3C6 = {38/47} (cannot be {56} which clashes with 13(2) cage at R2C7), no 5,6
1f. Killer pair 7,8 in 13(2) cage and 11(2) cage, locked for N1, clean-up: no 1,2 in 9(2) cage at R2C6
1g. R1C6 + R2C5 + R3C4 = {012} (hidden triple in N1), locked for 15(6) cage at R1C6
1h. Naked triple {345} in R4C3 + R5C2 + R6C1, locked for N2, clean-up: no 5 in R7C1
1i. Naked pair {01} in 1(2) cage at R6C4, locked for C4 -> R3C4 = 2
1j. 1 in N2 only in R6C2345, locked for R6
1k. R4C9 = 1 (hidden single in N4)
1l. Caged X-Wing for 1 in R1C6 + R2C5 and R5C56, no other 1 in C56
1m. Naked triple {789} in 24(3) cage at R7C1, locked for N5, clean-up: no 0,1 in R6C2

2a. 7 in R6 only in R6C235, locked for N2
2b. 6 in R6 only in R6C789, locked for N4
2c. 4 in R7 only in R7C35, locked for N5
2d. 3 in R7 only in R7C78910,11, locked for N7
2e. R5C7 = {67} (only place for 6 or 7 in R5; or naked sextet {024589 in R5C2348910, locked for R5)

[Time to look at semi-symmetry; it can easily be shown, as in previous Triankle Killers, that the numbers missing from corresponding nonets must be paired, so 1 and 9, 2 and 8, 3 and 7 & 4 & 6 are all paired.]
3a. R5C7 = {67} corresponds with R8C5, no 3,4 in R8C5 -> R8C5 = {67}
3b. 0,8 in N6 only in R8C456, locked for R8
3c. R2C5 = {01} corresponds with R11C7, no 0,1,5 in R11C7 -> R11C7 = 9, R11C6 = 6, R2C5 = 1, R1C6 = 0, R5C45 = [31], R8C67 = [93], clean-up: no 8 in R3C6, no 0 in R11C5, no 0,3 in R12C6
3d. R1C6 = 0 corresponds with R12C6, no 0 in R12C6 -> R12C6 = 5, R11C5 = 4
3e. Pair {00} in R16C6, locked for C6
3f. Pair {55} in R712C6, locked for C6
[Strictly speaking, those didn’t need to be stated, since R1C6 = 0 and R12C6 = 5 are placed for C6]
3g. R11C5 = 4 corresponds with R2C7, no 4 in R2C7 -> R2C7 = 6, R3C8 = 7, R3C5 = 5 -> R2C6 = 4, R3C67 = [38], R5C7 = 7 -> R8C5 = 7, R8C2 = 8, clean-up: no 0 in R7C2
3h. R4910C6 = [287], R9C45 = [60], R8C4 = 5
3i. R4C9 = 1 corresponds with R9C3, no 1 in R9C3 -> R9C3 = 9, R7C1 = 7, clean-up: no 1 in R6C10,11
3j. R7C1 = 7 corresponds with R6C11, no 7 in R6C11 -> R6C11 = 3, R6C10 = 9, clean-up: no 5 in R7C10
3k. R6C10 = 9 corresponds to R7C2, no 9 in R7C2 -> R7C2 = 1, R6C2 = 7, R67C4 = [10], clean-up: no 8 in R7C10,11
3l. Naked pair {28} in R610C5, locked for C5 -> R47C5 = [96], R78C3 = [42]
3m. R4C3478 = [3456]
3n. Naked pair {08} in R56C3, locked for N2 -> R5C4 = 9, R6C5 = 2, R10C45 = [38]
3o. R67910C7 = [4210], R910C8 = [42], R7C10,11 = [35], R9C9 = 7
3p. R6C1 = 5 -> R5C2 = 4
3q. Naked pair {08} in R6C38, locked for R6 -> R6C9 = 6, R8C8910 = [106]
3r. R8C9 = 0 corresponds with R5C3, no 5 in R5C3 -> R5C3 = 0, R6C38 = [80]
3s. R6C3 = 8 corresponds with R7C9, no 2 in R7C9 -> R7C9 = 8, R7C8 = 9
3t. R8C4 = 5 corresponds with R5C8, no 0 in R5C8 -> R5C8 = 5, R5C910 = [28]

Solution:
Attachment:
Triankle Killer 11 Given Solution.jpg
Triankle Killer 11 Given Solution.jpg [ 119.88 KiB | Viewed 176 times ]


Top
 Profile  
Reply with quote  
 Post subject: Re: Triankle Killer 11
PostPosted: Sun Jul 10, 2022 6:13 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1820
Location: Lethbridge, Alberta, Canada
And then Triangle Killer 11 Full Puzzle. As HATMAN said "not too hard", my important step 2a was probably the hardest to explain simply. If one also uses the given missing numbers from the triangle nonets, this may be even easier than the Given version.

Here's how I solved Triankle Killer 11 Full Puzzle:
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.

The two numbers in the grey cells repeat horizontally and vertically; they are in all nonets

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 R2C7 = {49/58/67}
b) 11(2) cage at R3C6 = {29/38/47/56}, no 0,1
c) 13(2) cage at R4C4 = {49/58/67}
d) 7(2) cage at R4C6 = {07/16/25/34}, no 8,9
e) 3(2) cage at R4C9 = {03/12}
f) 4(2) cage at R5C2 = {04/13}
g) 8(2) cage at R5C6 = {08/17/26/35}, no 4,9
h) 17(2) cage at R5C10 = {89}
i) 1(2) cage at R6C4 = {01}
j) 16(2) cage at R8C5 = {79}
k) 6(2) cage at R8C9 = {06/15/24}, no 3,7,8,9
l) 8(2) cage at R9C5 = {08/17/26/35}, no 4,9
m) 5(2) cage at R9C7 = {05/14/23}
n) 4(3) cage at R5C5 = {013}
o) 6(3) cage at R6C5 = {015/024/123}
p) 24(3) cage at R7C1 = {789}
q) 15(6) cage at R1C1 = {012345}
r) 9(2) cage at R2C6 = {09/18/27/36/45}, no initial eliminations

1a. R5C56 = {13} (cannot be {01/13} which clash with R5C23), locked for R5 and N3 -> R6C6 = 0, clean-up: no 4,6,7 in R4C67, no 0,2 in R4C9
1b. 8(2) cage at R5C6 = [17/35]
1c. Naked pair {25} in R4C67, locked for R4, 5 locked for N3 -> R5C7 = 7 -> R5C56 = [31], clean-up: no 6 in R2C6, no 8 in R3C5, no 4 in R3C6, no 6 in R3C8, no 6,8 in R4C45, no 7 in R9C5, no 5 in R9C6
1d. Naked pair {04} in R5C23, locked for R5 and N2 -> R45C9 = [12], R67C4 = [10], R4C3 = 3, 0 has been placed in R6C6, cannot also be in R7C6 because they must be different -> R7C4 = 0 placed for R7, clean-up: no 4,5 in R8C10
1e. R6C6 = 0 -> R6C57 = 6 = [24], R6C1 = 5, clean-up: no 7 in R2C6, no 7 in R3C6, no 9 in R3C8, no 6 in R9C6, no 1 in R9C8
1f. Naked pair {49} in R4C45, locked for R4
1g. Naked pair {89} in R56C10, locked for C10 and N4
1h. Naked pair {79} in R8C56, locked for R8 and N6 -> R8C2 = 8, clean-up: no 1 in R9C5
1i. Naked pair {79} in R7C1 + R9C3, locked for N5
1j. 15(6) cage at R1C1 = {012345} -> R2C5 = 1, 2 in R1C6 + R3C4, locked for N1, clean-up: no 8 in R2C6, no 7 in R3C5, no 9 in R3C67
1k. 9(2) cage at R2C6 = {09/45} (cannot be [36] which clashes with R3C67), no 3,6
1l. 13(2) cage at R2C7 = [67] (cannot be {58} which clashes with R3C67, cannot be [94] which clashes with 9(2) cage), clean-up: no 5 in R3C67

[Time to look at semi-symmetry and I remembered that, in addition to pairs, the numbers missing from corresponding nonets must be paired.]
2a. R5C6 = 1 corresponds with R8C6 = {79}, R5C7 = 7 corresponds with R8C5 = {79}, no 1 in R8C6 -> 1 is paired with 9 (cannot be R5C6 = 1 paired with R8C6 = 7 because then R5C7 = 7 cannot be paired with R8C5 = 9) -> R8C56 = [79], clean-up: no 0 in R3C5
2b. R2C5 = 1 corresponds with R11C7 -> R11C7 = {19}
2c. R4C9 = 1 corresponds with R9C3, no 1 in R9C3 -> R9C3 = 9, R7C1 = 7
2d. R6C4 = 1 corresponds with R7C8 -> R7C8 = {19}
2e. R8C2 = 8 corresponds with R5C10, R5C10 cannot be 9 because 9 paired with 1 -> R5C10 = 8, R6C10 = 9
2f. R6C10 = 9 corresponds with R7C2, no 9 in R7C2 -> R7C2 = 1, R7C8 = 9
2g. R2C7 = 6 corresponds with R11C5, R3C7 = {38} corresponds with R10C5, R3C8 = 7 corresponds with R10C4 -> no 9 in R10C4 and R10,11C5
2h. Combined cage R1C6 + R3C3 and 9(2) cage at R2C6 = [02]{45}/{24}[09] -> 0 in R12C6, also R6C6 = 0 is the repeated 0 -> 0 locked for C6, clean-up: no 8 in R9C5
2i. 8 in C5 only in R10,11C5, locked for N8
2j. R4C45 = {49} corresponds with R9C78, no 9 in R9C78 -> R9C78 = [14], R4C45 = [49], R3C4 = 2, R11C7 = 9, clean-up: no 0 in R2C6
2k. Naked pair {45} in 9(2) cage at R2C6, locked for N1 -> R1C6 = 0, R5C23 = [40]
2l. No 9 in N1 -> no 1 in N8
2m. 7 in C6 only in R10,11,12C6, locked for N8
2n. R5C4 = 9 (hidden single in N2, which must contain both of 1,9)
2o. R5C4 = 9 corresponds with R8C8, no 9 in R8C8 -> R8C8 = 1, clean-up: no 5 in R8C9
2p. R6C3 = 8 (hidden single in R6)
[All the 1s and 9s have now been placed]

3a. R1C6 = 0 corresponds with R11C6 and R6C6 = 0 corresponds with R7C6, neither of R7C6 and R11C6 contain 0 -> R7C6 and R11C6 must contain the same number, the second repeated number in C6 = {2356} -> 0 must be paired with one of 2,3,5,6
3b. R7C1 = 7 corresponds with R6C11, 7 doesn’t correspond with 0 -> no 0 in R6C11
3c. R5C3 = 0 corresponds with R8C9, 4 doesn’t correspond with 0 -> no 4 in R8C9, clean-up: no 2 in R8C10
3d. R5C2 = 4 corresponds with R8C10, 4 doesn’t correspond with 0 -> R8C10= 6, R8C9 = 0
3e. 4 paired with 6
3f. R6C8 = 0 (hidden single in R6 apart from repeated 0 in R6C6, alternatively naked triple {367} in R6C2911, locked for R6)
3g. R10C7 = 0 (hidden single in C7)
3h. R9C5 = 0 (hidden single in C5) -> R9C6 = 8, R3C67 = [38]
3i. 3 placed for C6, 4 paired with 6, R1C6 corresponds with R12C6 -> 0 paired with one of 2,5, no 3,6 in R7C6, no 6 in R11C6
3j. Naked pair {25} in R412C6 (plus one of them in repeated R7C6), locked for C6 -> R2C6 = 4, R3C5 = 5
3k. R4C6 = 4 corresponds with R11C6, no 4 in R11C6 -> R11C6 = 6, R10C6 = 7
3l. R3C6 = 3 corresponds with R10C6 = 7 -> 3 paired with 7
3m. R2C7 = 6 corresponds with R11C5, no 6 in R11C5 -> R11C5 = 4, R10C5 = 8, R7C5 = 6
3n. R3C8 = 7 corresponds with R10C4, no 7 in R10C4 -> R10C4 = 3, R8C4 = 5
3o. R9C4 = 6 corresponds with R4C8, no 4 in R4C8 -> R4C8 = 6, R5C8 = 5, R10C8 = 2
3p. R1C6 = 0 corresponds with R12C6 = 5 -> 0 paired with 5
3q. Repeated R6C6 = 0 corresponds with R7C6 -> R7C6 = 5
3r. The remaining pair are 2 with 8
3s. R5C9 = 2 corresponds with R8C3, no 8 in R8C3 -> R8C3 = 2, R7C3 = 4
3t. R6C5 = 2 corresponds with R7C7, no 8 in R7C7 -> R7C7 = 2
3u. R7C3 = 4 corresponds with R6C9, no 4 in R6C9 -> R6C9 = 6, R6C211= [73]
3v. R6C3 = 8 corresponds with R7C10, no 2 in R7C10 -> R7C10 = 8
3w. Naked singles R4C67 = [25], R7C11,12 = [35], R8C7 = 3, R9C410 = [67]
Same solution but different cage pattern.
Solution:
Attachment:
Triankle Killer 11 Full Puzzle Solution.jpg
Triankle Killer 11 Full Puzzle Solution.jpg [ 119.15 KiB | Viewed 176 times ]


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 2 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group