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 Post subject: Killer Tetrakis Square
PostPosted: Mon Jun 20, 2016 4:08 am 
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I've been working on some new puzzles styles for my blog and I concocted one that might be of interest here. The constraints are based on Aziz Ates' Triangular Skyscrapers puzzle in the 2016 US Puzzle Championship practice test.

Rules

Place numbers in the triangular cells, one number per cell, such that each of the four rows and columns contains the numbers from 1-8 exactly once. Additionally, each number must appear exactly once in each of the four triangle orientations (pointing NE, NW, SE, SW). As in Killer Sudoku, cages enclose non-repeating groups of numbers which must add up to the indicated sum.

Image

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Last edited by h3lix on Sat Nov 12, 2016 6:19 am, edited 1 time in total.

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PostPosted: Thu Sep 15, 2016 8:40 am 
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I'm have had real problems with this one and put it to one side. I am obviously missing something.

Before I have another go has anyone solved it?

On going through old posts I came across a format from H3lix in 2008 that embedded a 4*4 sudoku in the normal grid. I've created a few puzzles on this basis most of which I will post on the players forum.


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PostPosted: Thu Sep 29, 2016 6:28 pm 
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Thanks for bringing this to may attention HATMAN! Nice one h3lix!

It started off very easy then got hard fast!

Here is my WT.

Killer Tetrakis Square WT:
I use r(ows) and c(olumns) in the traditional sense of square cells. So the puzzle has 4 rows and 4 columns.
I use TR, TL, BL, BR (Top Right, Top Left, Bottom Right, Botton Left) to refer to an individual triangle within a "cell".

1. 10(4) = {1234}
r2c2TL must be different from all values in 10(4)
-> 6(2)r2 = [15]
-> r3c3TL = 1
Also -> 13(2) = {67}
Also other 2 in r23 only in 10(2)c1
-> 10(2)c1 = {28}

2. Whether 7 in r2c3TR or r2c4TL -> 7 not in 11(2)c4
(56) already in r2 -> 11(2)c4 not {56}
-> 11(2)c4 = {38}
-> 11(2)s in r3 are {47} and {56}
-> 4 in 10(4) in r2

3. Trying 2 in 10(4) in r2c3BL ...
... puts 4 in r2c2BR puts 11(2)r3c3 = [74] (Only place for 4 in r3)
Also puts 2 in r3c1TL ...
... which leaves no place for 2 in c2.
-> 2 in 10(4) in c2. (r2c2BR or r3c2TR)

4. 5 in r2c2TL -> 11(2)r1 cannot be {56}
3 in r3 in r3c2TR or r3c4TR -> r1c2TR cannot be 3
7 in 13(2) -> r1c3TL cannot be 7
-> 11(2)r1 from [83] or [74]
if the former -> 8 not in r3c4TR
If the latter -> 3 in r2c3BL -> 3 not in r2c4BR
Either way -> 11(2)c4 = [83]
-> 10(2)c1 = [28]
-> r3c2TR = 2

5. 8 in c2 only in r1c2
8 in c3 only in r4c3
1 in c2 only in r1c2BL or r4c2BR
But the former puts 8 in r1c2TR which leaves no place for 8 in c3
-> 9(2)r4 = [18]
-> 11(2)r1 = [83]
-> 10(4) = [3421]
-> 11(2)r3c1 = [47]
-> 11(2)r3c3 = {56}
Also HS 4 in c2 -> r4c1TL = 4
-> r1c2BL = 6
Also -> r1c3BR = 2
-> r4c4TL = 2
Also r1c4BL = 1
Also r1c4TR = 4
Also r1c1BR = 5
-> r1r1TL = 7
-> 13(2) = [76]
-> 11(2)r3c3 = [65]
-> r4c1TR = 6
-> r4c1BL = 3
-> r4c4BR = 7


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PostPosted: Mon Oct 31, 2016 2:51 am 
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Welcome back h2lix!

My first problem was how to set up the diagram in an Excel worksheet. I eventually decided on this grid format, with each cell having two rows on my solving worksheet.

Image

After the early easy steps I used a forcing chain, which gradually got longer as I got more and more out of it. It reminded me that some years ago udosuk told me that I ought to learn to look further into my steps.

Here is my walkthrough:
Each column and row contains 1-8. Each triangle orientation must contain 1-8.

Prelims

a) 11(2) cage at R1C2 = {38/47/56}, no 1,2
b) 6(2) cage at R2C1 = {15/24}
c) 10(2) cage at R2C1 = {28/37/46}, no 1,5
d) 13(2) cage at R2C3 = {58/67}
e) 11(2) cage at R2C4 = {38/47/56}, no 1,2
f) 11(2) cage at R3C1 = {38/47/56}, no 1,2
g) 11(2) cage at R3C3 = {38/47/56}, no 1,2
h) 9(2) cage at R4C2 = {18/27/36/45}
i) 10(4) cage at R2C2 = {1234}

1. 6(2) cage at R2C1 = [15] because R2C2 cell of this cage ‘sees’ all of 10(4) cage including the lower right triangle because of orientation -> R2C1NE = 1, placed for NE triangles, R2C2NW = 5, placed for NW triangles, clean-up: no 6 in 11(2) cage at R1C2, no 8 in 13(2) cage at R2C3, no 6 in R3C1SE, no 6 in R3C4NE, no 4 in R4C3SW
1a. 13(2) cage at R2C3 = {67}, locked for R2, clean-up: no 3,4 in R3C1NW, no 4,5 in R3C4NE
1b. 10(4) cage at R2C2 = {1234} -> R3C3NW = 1, placed for NW triangles, clean-up: no 8 in R4C2SE

2. 2 in R23 only in 10(2) cage at R2C1 and 10(4) cage at R2C2 -> 10(2) cage = {28}, locked for C1, clean-up: no 3 in R3C2SW

3. 11(2) cage at R2C4 = {38} (cannot be [47] which clashes with 13(2) cage at R2C3 because of NE triangles), locked for C4
3a. Caged X-Wing for 8 in 10(2) cage at R2C1 and 11(2) cage at R2C4, no other 8 in R3
3b. Caged X-Wing for 3 in 10(4) cage at R2C2 and 11(2) cage at R2C4, no other 3 in R3
3c. Naked quad {4567} in 11(2) cage at R3C1 and 11(2) cage at R3C3, locked for R3
3d. 3 in R3 only in R3C24, locked for NE triangles, clean-up: no 8 in R1C3NW

4. 11(2) cage at R1C2 = [74/83] (cannot be [47] which clashes with 13(2) cage at R2C3 because of NW triangles)
4a. Consider permutations for this 11(2) cage
11(2) cage = [74]
or 11(2) cage = [83] => 11(2) cage at R2C4 = [83] (because of NE triangles) => R3C2NE = 2 => R2C3SW = 4
-> 4 in R1C3NW or R2C3SW, locked for C3, clean-up: no 7 in R3C4SW
4b. 4 in R1C3NW or R2C3SW, CPE no 4 in R1C4SW using SW triangles
4c. Continuing these chains
11(2) cage = [74] => R2C2SE = 4 (hidden single in R2) => R3C2SW = 6
or 11(2) cage = [83] => 11(2) cage at R2C4 = [83] (because of NE triangles) => R3C2NE = 2 => R2C3SW = 4 => no 4 in R3C2SW + R3C4SW => 11(2) cage at R3C3 = {56} => R3C2SW = 7
-> R3C2SW = {67}, clean-up: no 7 in R3C1SE
4d. Also from step 4c 7 in R1C2NE + R3C2SW, locked for C2, clean-up: no 2 in R4C3SW
4e. And taking them further
11(2) cage = [74] => R2C2SE = 4 => R3C2SW = 6 => no 6 in 9(2) cage at R4C2 using SW triangles => no 3 in 9(2) cage at R4C2, 7 in C1 only in R4C1SW, locked for R4 => no 2 in R4C2SE => R4C2SE = 1
or 11(2) cage = [83] => 11(2) cage at R2C4 = [83] (because of NE triangles) => R3C2NE = 2 => R2C3SW = 4 => no 4 in R3C4SW => 11(2) cage at R3C3 = {56} => R3C2SW = 7, R3C1SE = 4 => no 4 in R4C2SW, R2C2SE + R3C2NE = [32], no 3 in R4C3SW => no 6 in R4C2SE => R4C2SE = 1
-> R4C2SE = 1, R4C3SW = 8, placed for SW triangles
[Cracked, the rest is routine.]

5. R1C2NE = 8 (hidden single in C2), placed for NE triangles, R1C3NW = 3, placed for NW triangles, R3C4NE = 3, R3C2NE = 2, placed for NE triangles, R2C3SW = 4, placed for SW triangles, R3C1SE = 4 (hidden single in R3), placed for SE triangles, R3C2SW = 7, placed for SW triangles, R2C1SW = 2, placed for SW triangles, R1C2SW = 6, placed for SW triangles, R1C1NW = 7, placed for NW triangles

and the rest is naked singles, without using triangle orientations.

Solution:
Image


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PostPosted: Sun Nov 13, 2016 2:09 am 
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This is actually my first puzzle of this type. My creation path started out identical to wellbeback's for steps 1 and 2, but then it looks like I tested different constraining values for the unmarked cages. The numbers outside the grid pointing to the cages indicate the order I set their sums, which points to my solve path. I use underlines and circles as a notation for cells whose pencilmarks are connected. For every circled cell, either the values on the left are true or the values on the right are true.

Hidden Text:
Image


I tried it again last night and came up with this solve path after wellbeback's step 2:

Notebook
Hidden Text:
Image


Solve Path
Hidden Text:
A. Either (R2C3SW = 4 and R3C1SE = 4) or (R2C2SE = 4 and R3C4SW = 4)
In either case, 4 in BL and BR are eliminated from R1 and R4.
-> 9/2 in R4 <> {45}

Either R3C1SE = 4 or R3C4SW = 4
-> Either R3C2SW = 7 or R3C3SE = 7
-> 9/2 in R4 <> {27}
-> 9/2 in R4 = {18|36}

B.
8 in C1 and C4 is either in (NW and SE) or in (NE and SW)
-> 8 is either in (NE and SW) or in (NW and SE)
-> 8 in R1 and R4 is either in both 11/2 and 9/2 cages, or in neither cage.

C.
7 in 13/2 cage in R1
-> R1C3NW <> 7
-> R1C2NE <> 4

If 8 is in neither 11/2 and 9/2 cages in R1 and R4, then
*11/2 in R1 = [74]
* 9/2 in R4 = {36}
* 11/2 in R3C12 = [47]
4 cannot be placed in 10/4

-> 8 in R1 and R4 is in both 11/2 and 9/2 cages
-> 9/2 in R4 = [18]
-> 11/2 in R1 = [83]
-> 10/2 in C1 = [28]
-> 11/2 in C4 = [83]
-> 10/4 = [3421]
-> R3C1SE = 4, hidden single in R3
-> R3C2SW = 7

Singles from here.


Here's one I made for the LMI Kaleidoscope competition.

Image

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PostPosted: Sun Nov 20, 2016 2:45 am 
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Thanks for another nice puzzle. If anyone is interested, there are several more by clicking on Puzzle Blog on h3lix's posts.

This was about the same level of difficulty as the first one.

Here is my walkthrough for the second puzzle:
Each column and row contains 1-8. Each triangle orientation must contain 1-8.

Prelims

a) 7(2) cage at R1C2 = {16/25/34}, no 7,8
b) 11(2)cage at R2C1 = {38/47/56}, no 1,2
c) 14(2) cage at R2C4 = {68}
d) 9(2) cage at R4C2 = {18/27/36/45}
e) 19(3) cage at R1C1 = {478/568}, no 1,2,3
f) 10(3) cage at R3C3 = {127/136/145/235}, no 8
g) 36(8) cage at R2C2 = {12345678}

1. Naked pair {68} in 14(2) cage at R2C4, locked for C4
1a. Double killer pair 6,8 in 6,8 in 36(8) cage at R2C2 and 14(2) cage at R2C4, locked for R23, clean-up: no 3,5 in 11(2) cage at R2C1
1b. Naked pair {68} in 14(2) cage, CPE no 6,8 in R2C3NE + R3C3SE

2. Naked pair {47} in 11(2) cage at R2C1, locked for C1 -> R2C1NE = 5, placed for NE triangles
2a. R2C1NE = 5 -> R1C1SE + R1C2SW = 14 = {68}, locked for R1, clean-up: no 1 in R1C2NE, no 1,2 in R1C3NW
2b. Double killer pair 4,7 in 11(2) cage at R2C1 and 36(8) cage at R2C2, locked for R23
2c. Naked pair {47} in 11(2) cage at R2C1, CPE no 4,7 in R2C2NW + R3C2SW
2d. 36(8) cage at R2C2 = {12345678}, 5 locked for R3

3. Naked pair {68} in R1C1SE + R2C4SE, locked for SE triangles, clean-up: no 1,3 in R4C3SW
3a. Naked pair {68} in R1C1SE + R2C4SE -> naked pair {68} in R1C2SW + R3C4NE, CPE no 6,8 in R3C2NE + R3C2SW

4. 7(2) cage at R1C2 = [25]/{34}, 36(8) cage at R2C2 which must contain 4,5 in C23 -> 9(2) cage at R4C2 cannot be {45}

5. 5 in C2 only in R3C2SW + R4C2NW, CPE no 5 in R3C3NW

6. Naked pair {68} in R1C1SE + R1C2SW = {68}, CPE no 6,8 in R4C1SW
6a. R1C1SE + R4C1NE = {68} (hidden pair in C1)
6b. R3C3NW + R3C4NE = {68} (hidden pair in R3)

7. Naked pair {68} in R3C3NW + R3C4NE, CPE no 6 in R4C3NE
7a. 10(3) cage at R3C3 = {127/145/235}
7b. Consider combinations for 7(2) cage at R1C2 = [25]/{34}
7(2) cage = [25] -> no 5 in R4C4NW => 10(3) cage = {127}
or 7(2) cage = {34} -> caged X-Wing for 3,4 in 7(2) cage and 36(8) cage at R2C2, no other 3,4 in C23 => 10(3) cage = {127}/[325]
-> 10(3) cage = {127}/[325], no 3 in R4C3NE
7c. 9(2) cage at R4C2 = [18/36] (cannot be {27} which clashes with 10(3) cage)
7d. Naked pair {68} in R3C3NW + R4C3SW, locked for C3
7e. R2C2NW + R2C4SE = {68} (hidden pair in R2)
7f. Naked pair {68} in R1C2SW + R2C2NW, locked for C2

8. 10(3) cage at R3C3 = {127}/[325], 7(2) cage at R1C2 = [25]/{34}
8a. Consider permutations for 9(2) cage at R4C2 = [18/36]
9(2) cage = [18] => 10(3) cage = {127} (cannot be [325] which clashes with R4C1SW = {23})
or 9(2) cage = [36] => 3 in 36(8) cage at R2C2 locked for C3 => 7(2) cage = [25] => no 5 in R4C4NW => 10(3) cage = {127}
-> 10(3) cage = {127}, 7 locked for R4
8b. 10(3) cage = {127}, CPE no 1,2 in R4C1SW
8c. R4C1SW = 3, placed for SW triangles, R4C2SE = 1, placed for SE triangles, R4C3SW = 8, R3C1SE = 2, placed for SE triangles, R3C2SW = 5, placed for SW triangles
8d. Naked pair {27} in R4C3NE + R4C4NW, locked for R4 -> R4C2NW = 4, placed for NW triangles
8e. Naked pair {37} in R2C2SE+R3C2NE, locked for C2 and 36(8) cage at R2C2 -> R1C2NE = 2, placed for NE triangles, R1C3NW = 5
8f. R3C1NW = 7 -> R2C1SW = 4, placed for SW triangles
[My solving path was a bit longer until I spotted step 8b while checking my walkthrough.]

and the rest is naked singles, without using triangle orientations.

Solution:
Image


And here is a slightly simplified walkthrough using the feature pointed out by h3lix in the next post.

Alternative Walkthrough:
Each column and row contains 1-8. Each triangle orientation must contain 1-8.

h3lix pointed out on the forum that because of the central 36(8) cage, R14C23, R23C14 and R14C14 must also form 36(8) groups each containing 1-8, so here’s a re-work using these hidden groups.

Prelims

a) 7(2) cage at R1C2 = {16/25/34}, no 7,8
b) 11(2)cage at R2C1 = {38/47/56}, no 1,2
c) 14(2) cage at R2C4 = {68}
d) 9(2) cage at R4C2 = {18/27/36/45}
e) 19(3) cage at R1C1 = {478/568}, no 1,2,3
f) 10(3) cage at R3C3 = {127/136/145/235}, no 8
g) 36(8) cage at R2C2 = {12345678}

1. Naked pair {68} in 14(2) cage at R2C4, locked for C4 and R23C14, clean-up: no 3,5 in 11(2) cage at R2C1
1a. Naked pair {68} in 14(2) cage, CPE no 6,8 in R2C3NE + R3C3SE

2. Naked pair {47} in 11(2) cage at R2C1, locked for C1 and R23C14 -> R2C1NE = 5, placed for NE triangles and R23C14
2a. R2C1NE = 5 -> R1C1SE + R1C2SW = 14 = {68}, locked for R1, clean-up: no 1 in R1C2NE, no 1,2 in R1C3NW
2b. Naked pair {47} in 11(2) cage at R2C1, CPE no 4,7 in R2C2NW + R3C2SW

3. Naked pair {68} in R1C1SE + R2C4SE, locked for SE triangles, clean-up: no 1,3 in R4C3SW
3a. Naked pair {68} in R1C1SE + R2C4SE -> naked pair {68} in R1C2SW + R3C4NE, CPE no 6,8 in R3C2NE + R3C2SW

4. 7(2) cage at R1C2 = [25]/{34}, 36(8) cage at R2C2 which must contain 4,5 in C23 -> 9(2) cage at R4C2 cannot be {45}

5. 5 in C2 only in R3C2SW + R4C2NW, CPE no 5 in R3C3NW

6. Naked pair {68} in R1C1SE + R1C2SW = {68}, CPE no 6,8 in R4C1SW
6a. R1C1SE + R4C1NE = {68} (hidden pair in C1)
6b. R3C3NW + R3C4NE = {68} (hidden pair in R3)

7. Naked pair {68} in R3C3NW + R3C4NE, CPE no 6 in R4C3NE
7a. 10(3) cage at R3C3 = {127/145/235}
7b. Consider combinations for 7(2) cage at R1C2 = [25]/{34}
7(2) cage = [25] -> no 5 in R4C4NW => 10(3) cage = {127}
or 7(2) cage = {34}, locked for R14C23 => 10(3) cage = {127}/[325]
-> 10(3) cage = {127}/[325], no 3 in R4C3NE + R4C4NW
7c. 9(2) cage at R4C2 = [18/36] (cannot be {27} which clashes with 10(3) cage)
7d. Naked pair {68} in R3C3NW + R4C3SW, locked for C3
7e. R2C2NW + R2C4SE = {68} (hidden pair in R2)
7f. Naked pair {68} in R1C2SW + R2C2NW, locked for C2

8. 10(3) cage at R3C3 = {127}/[325], 7(2) cage at R1C2 = [25]/{34}
8a. Consider permutations for 9(2) cage at R4C2 = [18/36]
9(2) cage = [18] => 10(3) cage = {127} (cannot be [325] which clashes with R4C1SW = {23})
or 9(2) cage = [36], 3 placed for R14C23 => 7(2) cage = [25] => no 5 in R4C4NW => 10(3) cage = {127}
-> 10(3) cage = {127}, 7 locked for R4
8b. 10(3) cage = {127}, CPE no 1,2 in R4C1SW
8c. R4C1SW = 3, placed for R14C14 and SW triangles, R4C2SE = 1, placed for R14C23 and SE triangles, R4C3SW = 8, R3C1SE = 2, placed for SE triangles, R3C2SW = 5, placed for SW triangles
8d. Naked pair {27} in R4C3NE + R4C4NW, locked for R4 -> R4C2NW = 4, placed for R14C23 and NW triangles
8e. Naked pair {37} in R2C2SE+R3C2NE, locked for C2 and 36(8) cage at R2C2 -> R1C2NE = 2, placed for NE triangles, R1C3NW = 5
8f. R3C1NW = 7 -> R2C1SW = 4, placed for SW triangles
[My solving path was a bit longer until I spotted step 8b while checking my walkthrough.]

and the rest is naked singles, without using triangle orientations.


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PostPosted: Mon Nov 21, 2016 11:41 am 
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Hidden Text:
I'm surprised law of leftovers (LOL) wasn't mentioned in your walkthrough. Is that term still used? The 36 cage forces full houses in R23C14 and R14C23 and R14C14. 36 cages produce some pretty interesting LOL situations on these grids.


I actually had to tone this one down for the competition because the test solvers couldn't crack it. I'm not even 100% sure I could either. My test solves averaged about 2 hours each and involved a mess of interconnected algebraic pencilmarks. Here is the original version, which I've never released:

Image

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PostPosted: Tue Nov 22, 2016 11:31 pm 
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h3lix wrote:
I'm surprised law of leftovers (LOL) wasn't mentioned in your walkthrough. Is that term still used? The 36 cage forces full houses in R23C14 and R14C23 and R14C14. 36 cages produce some pretty interesting LOL situations on these grids.

Thanks for that comment:
I use LOL for jigsaws and jigsaw killers. I hadn't thought of using it for this puzzle. Maybe it would have led to an easier solution.
I've now had another look at your comment. Not sure that it's LOL; it looks to me more like hidden windows in Windoku with the groups in R23 and C23 being fairly obvious, then the group in the four corners must follow. I've now re-worked that puzzle using the hidden groups, see my earlier post in this thread. This made some steps slightly simpler, with locked for hidden groups replacing double killer pairs, but I still needed the same basic steps.
Thanks for posting your original puzzle. I'll have a try at it. Not sure when, I'll try your Star Battle Killer and HATMAN's two puzzles first.


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