is this even possible on a square-1?

[towwdso]

I rearrenged the pieces on it.

Is it possible just turnin the layers?

 

[Kirjava]

Yeah, I recall generating an alg for this case at one point.

It's just a SQ2 3 Cycle

 

[Slash]

Take it apart, and place the piecees on the right place. you cant solve it with just moving it.

 

[HaraldS]

I think every position is solveable tell me if im wrong though.

 

[masterofthebass]

Take it apart, and place the piecees on the right place. you cant solve it with just moving it.

wrong. Any position is solveable

 

[MTGjumper]

Wait, what? Of course it's solvable; every permutation of the square-1 is.

 

[Slash]

wrong. Any position is solveable

youre right sorry :fp:

 

[scotzbhoy]

Yes.

 

[CubesOfTheWorld]

It is always solvable if none of the stickers were moved.

 

[towwdso]

Thats what I tought. Because you can exchance 2 pieces.
But thats mindblowing. O_O
Will try to solve now.

 

[dillonbladez]

How did you get it in the first place?
Just curious, i kinda want to do it myself

 

[Mike Hughey]

Wow - this suggests to me a whole new challenging event: square-1 match the scramble!!! Is anyone out there already good at this?

 

[towwdso]

How did you get it in the first place?
Just curious, i kinda want to do it myself

I loosened tension and took the piece out. Then put it together that way. =P
Anyway, an algorithm for that position would be cool. I love this type ot things.
Kirjava...?

 

[vcuber13]

he doesnt do sq1

 

[Erzz]

Wait, what? Of course it's solvable; every permutation of the square-1 is.

Why is this true?

 

[vcuber13]

because you can do 2 cycles and theres only permutations (no orientations)

 

[uberCuber]

because you can swap 2 pieces without doing anything to the rest of the cube

EDIT: Ninja'd...

 

[qqwref]

It's simple to prove: suppose you have some permutation of the Square-1.
- Every possible shape can be brought to cubeshape, so do that.
- Every position in cubeshape is possible (because you can do a 2-cycle of any piece type), so you can solve it.

 

[Stefan]

because you can do 2 cycles and theres only permutations (no orientations)

because you can swap 2 pieces without doing anything to the rest of the cube

Now I'd like you to prove it using just that (so you can't use the every-shape-can-be-solved/reached fact that Michael mentioned).

Michael: he didn't ask for a proof, though, he asked *why* it is true.

 

[blade740]

Wow - this suggests to me a whole new challenging event: square-1 match the scramble!!! Is anyone out there already good at this?

I've done it once. It's essentially a reverse blindsolve: I solved the shape, figured out a few pieces to swap, unsolved the shape, and swapped the pieces in cubeshape. You could probably do it much faster than me. I did the shape alg repeatedly because I didn't want to memorize the entire permutation at once. Plus, you could use your permutation matrices to speed things up a lot.