3-cycle challenge

[mrCage]

Can you (or is it not possible) to make a corner or edge 3-cycle on the (3x3x3) cube with solely halfturns?? That means only <U2,D2,F2,B2,R2, L2> turns are allowed.

Per

 

[blah]

I think I have an answer.

But just to be sure, is it the same as asking "is it possible to cycle (URB ULF DRF) or (UF UB DF) with only half turns?"

 

[cmhardw]

Can you (or is it not possible) to make a corner or edge 3-cycle on the (3x3x3) cube with solely halfturns?? That means only <U2,D2,F2,B2,R2, L2> turns are allowed.

Per

I know half of the answer, but I'd rather not say anything yet. I think I know the other half of the answer, but I'm not 100% certain on a proof of impossibility / example algorithm for a cycle (so as not to give it away).

Chris

 

[Stefan]

Edges:

Spoiler

R2 F2 R2 U2 R2 F2 R2 U2. Thanks to Ron's CubeSolver

Corners:

Spoiler

Impossible. Jaap says there are 663,552 positions and if I could 3-cycle corners, there'd be 1,990,656.

 

[cmhardw]

Oh wow, Stefan already weighed in. Well, before checking his spoilers let me post my own.

Spoiler

Edges: R2 F2 R2 U2 R2 F2 R2 U2
As for corners I think it's impossible, but I don't have a proof for this. Also, if it is possible I don't have an algorithm for this either.

@Stefan after reading his spoilers: Your corner argument hardly counts as a proof sir ;-) How could we make this rigorous?

 

[Stefan]

Spoiler

Your corner argument hardly counts as a proof sir ;-)

I know, but I didn't call it proof and Per also didn't ask for one. But yeah, I'm not happy with it. You can tell a riddle is good if I answer but have to cheat.

How could we make this rigorous?

I'd say by understanding the corner constraints. I never understood Jaap's "total twist of each tetrad is fixed", that's the part I'm missing.

 

[qqwref]

Corners in half turn 3x3 keep in two orbits, so a 3-cycle must be in one of those orbits. There is really only one 'distinct' 3-cycle, i.e. UFL-UBR-DFR. However this creates at least one face where exactly one corner is wrong. From examining the possible states of a 2x2 with half turns only we know this is impossible. Hence we cannot do a corner 3-cycle with half turns only.

 

[Stefan]

From examining the possible states of a 2x2 with half turns only we know this is impossible.

I don't. Can you describe that examination?

 

[qqwref]

You can just look at them all - there aren't very many. We only have to take into account the bottom and top faces. For each position there are essentially only 4 possible turns (F2, R2, U2 F2, U2 R2) so it is easy to check that we have found all possible ones:
1) Solved; this always goes to 2.
2) Two bars of opposite color on each side (all bars pointing in the same direction); this goes to 1 for one of the turns, to 3 for one other, and to 2 for the other two.
3) A checkerboard on each side, with UFR and DFR stickers having the same color; this always goes to 2.

As you can see these are the only possible cases, so it is not possible to get exactly one sticker wrong on one face of a 2x2 by using only half turns (and similarly you can't get exactly one corner sticker wrong on one face of a 3x3 using only half turns).

 

[jaap]

From examining the possible states of a 2x2 with half turns only we know this is impossible.

I don't. Can you describe that examination?

The easiest way to see this is as follows:

Spoiler

The four vertices of the U corner pieces will always remain in a flat plane, be it one of the faces or a diagonal through the cube. Half turns will never break up this plane, and this is trivial to check.
A three-cycle will involve one or two of these corners, and so will break up this plane, and is therefore not possible with half turns.