Prove you can reach from one position to another in >=20

[pentagonpie]

Hello, I have been solving a Rubik's cube for more than a year, and lately I've read that a group of people proved that you can solve any position in 20 moves or less, here is the link:
http://www.cube20.org/

So I wanted to share an idea I had, and hope you will tell me if it's original or even correct.

I think, based on the new finding, that you can get from any position to any position in the cube in 20 moves or less.

Here is my prove, it requires a bit of imagination, hope I make myself clear.

1.We begin with a normal solved cube.
2.You pick any two positions.
3.You take the cube and by any technique you desire get the cube to the first position.
4.I now hold the cube in the first position, and I put invisible stickers, that only I can see, on the cube.
I cover the cube with stickers so that the cube seems to me like it's solved.
5.You receive back the cube, and by any technique you desire, you get the cube to the second position, you are not affected by the new stickers, because you can't see them.
6.I receive the cube, which seems scrambled to me, and using the stickers I see, I solve the cube with 20 moves or less, which was proved possible in the link in the beginning.
7.I now see a solved cube, and you saw me solve the cube from second position to the first with 20 moves or less.

 

[riffz]

What exactly is your point? Isn't this intuitively obvious?

 

[Godmil]

Take any position -> flip every edge -> 20 moves. Done. None more complex.

The solved state is arbitrary, and doesn't need to be a transitional position, so there is no reason to believe that going from any position to any other position should be more than 20 moves.

 

[Lucas Garron]

Indeed, assuming you actually mean <= 20.

It's a good insight that this is the case for any two states. If you're interested in this kind of thinking, learn some group theory. This is a pretty basic fact, but it's cool to understand the reasons for why and how (and where else) this works.

 

[macky]

To OP: Yes, that's essentially the proof that the Cayley graph of a group G is vertex-transitive. If we define edges by right multiplication {g, gs} under some generator s, then G acts transitively on the vertices (identified with G) by left multiplication, and this action preserves adjacency: {hg, h(gs)} = {hg, (hg)s}.

As a side note, this requires associativity. In particular, we can define Cayley graphs more generally for (possibly non-associative) loops, and these don't need to be vertex-transitive. In fact, any regular graph can be obtained as the Cayley graph of some loop!

 

[Lucas Garron]

To OP: Yes, that's essentially the proof that the Cayley graph of a group G is vertex-transitive.

Hey! I was going to leave fancy-sounding words out of this!