The New Sudoku Players' Forum

[999_Springs]

A Sukaku is a pencilmark only Su-Doku. There has been a thread on ordinary-sized Sukakus in this forum. But not 4x4 ones.

So I looked into those and I have discovered that some advanced techniques are required to solve some 4x4 Sukakus.

Here is one which I created:

Code: Select all

2313 4132 1132 2431
4234 3132 4132 1342
3311 4213 1432 2121
4234 4312 2413 2224

requires turbot fish


and another one:

Code: Select all

1123 4312 3132 1342
4231 2212 4321 1324
3111 1423 1423 2112
4423 2341 2431 2442

requires finned x-wing


Both puzzles also have a Sue de Coq which eliminates four candidates but does not crack the puzzle. The second puzzle has an XYZ-wing which eliminates one candidate but also does not crack the puzzle.

[udosuk]

... and another one:

Code: Select all

1123 4312 3132 1342
4231 2212 4321 1324
3111 1423 1423 2112
4423 2341 2431 2442

requires finned x-wing

This puzzle doesn't have a valid solution.

Turbot fish on 4: r1c24+r3c23 => r2c3, r4c4<>4
=> r34c4=[12] => r34c1=[34] => r34c2={12}
=> No candidate for r2c2 (invalid)

On the other hand changing r4c1 to "1423" will make it a valid puzzle...

But then it doesn't really "require" a finned x-wing (the turbot fish on 4 is enough to solve it)...

No matter what I think it doesn't take anything further than a turbot fish to solve all these...

[999_Springs]

Code: Select all

1123 4312 3132 1342
4231 2212 4321 1324
3111 1423 1423 2112
4423 2341 2431 2442

requires finned x-wing
 


Actually r4c3=2142.

edit: actually r3c4=2142 and r4c3 remains unchanged.

[999_Springs]

Oops.

I didn't see that the turbot fish involving strong links r1c24 and r24c1 on 4 would eliminate 4 from r4c4 which also solves the puzzle in the same way as the finned X-wing. I'm not very good at spotting turbots when the strong links aren't parallel to each other.

Oh well. If finned x-wing is before turbot fish in your hierarchy of solving techniques then it does need one.

However, I have discovered that you can have Sukakus made from 3x3 Latin squares which need XY-wings.

Here is one I created:

Code: Select all

122 232 321
311 213 112
312 221 132

I think that this is the smallest that you can get.

[udosuk]

However, I have discovered that you can have Sukakus made from 3x3 Latin squares which need XY-wings.

Here is one I created:

Code: Select all

122 232 321
311 213 112
312 221 132

Is the solution unique for this one? I found 2 solutions:

Code: Select all

1 3 2
3 2 1
2 1 3

2 3 1
3 1 2
1 2 3

[999_Springs]

AAAAAARRRRRRGGGGGGHHHHHH!!! r1c3=331.

Sorry.

(curses keyboard)

[999_Springs]

Has anyone managed to solve my 3x3 Latin square Sukaku without using an XY-wing or a forcing chain yet?

[udosuk]

Has anyone managed to solve my 3x3 Latin square Sukaku without using an XY-wing or a forcing chain yet?

It's just a tiny puzzle, nothing to get excited about...

This is the standard pencilmark form:

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#12   23  #13
#13   123 #12
 123  12   123

The # cells form 2 UR-type-3s (I think) simultaneously, so both r2c13 cannot be 1.
Therefore r2c2=1, solving the puzzle.

Alternatively, if uniqueness assumption is not desirable,

Code: Select all

-12  *23  @13
@13  *123 -12
 123  12   123

We have a y-wing on offer here (refer to this link)...

The {13} pointing pair (@) together with the strong link of 3s on c2 (*) eliminate the 1s from r1c1+r2c3.

I don't think this move is much easier than the xy-wing, but it's definitely more elegant than any forcing chain...