Interesting 4x4x4 subgroup

[qqwref]

Scramble the puzzle with only the following turns: double-layer turns (Rw) and turns of both the middle two layers (M).

- Can every normal position be reached with these turns?
- If not, what are the constraints on the puzzle?
- Most importantly, how do we solve this? ;-) You can only use these turns to solve, of course.

If you feel this is too easy, try to analyze the similar subgroups on larger cubes. All turns must be 2 consecutive layers/slices on one axis, turned in the same direction.
EDIT: Oops, only consider even cubes. This subgroup is trivial on odd ones because you can generate any single slice turn with the available moves.

 

[TMOY]

The first obvious remark is that both parities are always in the same state (a Rw toggles both of them, a M toggles none). So the answer to the first question is no.

 

[qqwref]

Solution idea:
- Solve corners like 2x2.
- Solve edges. Here is a nonpure edge 3-cycle:
(M' (r u2 r' u' r u2 r' f r' f' r) M) D (M' (r u2 r' u' r u2 r' f r' f' r) M) D'
- Solve centers. I haven't found a nice short alg yet.

EDIT: Here's a pure center 3-cycle (I think):
(M' U2 M E (y r u2 r' u' r u2 r' f r' f' r y') E' M' U2 M) D (M' U2 M E (y r u2 r' u' r u2 r' f r' f' r y') E' M' U2 M) D'

 

[VP7]

Scramble the puzzle with only the following turns: double-layer turns (Rw) and turns of both the middle two layers (M).

You are only allowing Rw Rw' Rw2 & M M' M2.
Which basically does nothing.

 

[Forte]

Scramble the puzzle with only the following turns: double-layer turns (Rw) and turns of both the middle two layers (M).

You are only allowing Rw Rw' Rw2 & M M' M2.
Which basically does nothing.

He's allowing all double slice turns (Rw, Fw, Uw, etc)

 

[qqwref]

Those are just examples of the two turns allowed, just in case the English wasn't clear enough.

 

[Joker]

Since both M layers are allowed, I say, yes, this can reach any normal state (well, as long as you also allow E, S, Lw, Fw, Bw, Uw, Dw. I haven't put much thought into this, but i think you may be albe to reach any normal scrambled state (because M' would act as a normal R',L, and x in one, etc)

 

[hawkmp4]

Since both M layers are allowed, I say, yes, this can reach any normal state (well, as long as you also allow E, S, Lw, Fw, Bw, Uw, Dw. I haven't put much thought into this, but i think you may be albe to reach any normal scrambled state (because M' would act as a normal R',L, and x in one, etc)

Did you read TMOY's post?

 

[qqwref]

I DID IT

Used my solution. My god this took a lot of moves. Time: 14:07.81...


Joker: This is 4x4, not 3x3.

 

[Joker]

Since both M layers are allowed, I say, yes, this can reach any normal state (well, as long as you also allow E, S, Lw, Fw, Bw, Uw, Dw. I haven't put much thought into this, but i think you may be albe to reach any normal scrambled state (because M' would act as a normal R',L, and x in one, etc)

Did you read TMOY's post?

Didn't quite get what he meant.

I DID IT

Used my solution. My god this took a lot of moves. Time: 14:07.81...


Joker: This is 4x4, not 3x3.

Ik its 4x4.

 

[Forte]

Since both M layers are allowed, I say, yes, this can reach any normal state (well, as long as you also allow E, S, Lw, Fw, Bw, Uw, Dw. I haven't put much thought into this, but i think you may be albe to reach any normal scrambled state (because M' would act as a normal R',L, and x in one, etc)

How does that mean you can reach any scrambled state though?

 

[Joker]

Since both M layers are allowed, I say, yes, this can reach any normal state (well, as long as you also allow E, S, Lw, Fw, Bw, Uw, Dw. I haven't put much thought into this, but i think you may be albe to reach any normal scrambled state (because M' would act as a normal R',L, and x in one, etc)

How does that mean you can reach any scrambled state though?

Ok, lemme explain:
To effect the outer layers, use turns like M
To effect the inner layers, use wide turns (can be follwed with a turn like M)
BUT, I said I didn't give too much thought to this, so I'm probably wrong lol.

 

[hawkmp4]

Since both M layers are allowed, I say, yes, this can reach any normal state (well, as long as you also allow E, S, Lw, Fw, Bw, Uw, Dw. I haven't put much thought into this, but i think you may be albe to reach any normal scrambled state (because M' would act as a normal R',L, and x in one, etc)

How does that mean you can reach any scrambled state though?

Ok, lemme explain:
To effect the outer layers, use turns like M
To effect the inner layers, use wide turns (can be follwed with a turn like M)
BUT, I said I didn't give too much thought to this, so I'm probably wrong lol.

You mean affect, in this case.
EDIT: Oops! I missed the reply. There wasn't any whitespace between the two quotes so I assumed it was one big one. My bad.

 

[TMOY]

Did you read TMOY's post?

Didn't quite get what he meant.

I can make it clearer. A double layer move performs a 4-cycle of corners, three 4-cycle of edges and three 4-cycles of centers. A double inner slice move performs two 4-cycles of edges and four 4-cycles of centers. So the permutations of all three types of cubies always have the same parity, which is not true in the full 4^3 group.

I have found a pure 3-cycle of edges (checked it on my supercube and it's OK): [(M Uw)^20, Bw E Bw]. OK it's stupidly long and it should be possible to find a better one.

 

[Joker]

Did you read TMOY's post?

Didn't quite get what he meant.

I can make it clearer. A double layer move performs a 4-cycle of corners, three 4-cycle of edges and three 4-cycles of centers. A double inner slice move performs two 4-cycles of edges and four 4-cycles of centers. So the permutations of all three types of cubies always have the same parity, which is not true in the full 4^3 group.

I have found a pure 3-cycle of edges (checked it on my supercube and it's OK): [(M Uw)^20, Bw E Bw]. OK it's stupidly long and it should be possible to find a better one.

Atleast its easy to remember. Might lose count on all the M's and Uw's though haha.

EDIT

Er...If you do (M U)^4 it has the same effect as (M Uw)^20

 

[Lucas Garron]

Being picky: The 4x4x4 is not a group, and neither is this, technically. But I don't really have a better word for it.

Have you considered this on a supercube?

 

[qqwref]

Sure it is a group, if you:
- consider that all centers are distinct, and
- consider that any position with the right color centers on the right face counts as solved (although there is only the single identity position).

I haven't previously considered this on a supercube, no, but it shouldn't be any more difficult, except for having to be more precise with piece cycling. You cannot do 2-cycles on the centers of a 4x4 (relative to the corners) so the solution method should be essentially the same. In addition the two algs I gave will also work on a supercube.

 

[hawkmp4]

EDIT

Er...If you do (M U)^4 it has the same effect as (M Uw)^20

But we're not allowing U turns...

 

[TMOY]

Er...If you do (M U)^4 it has the same effect as (M Uw)^20

Sorry but no. Apart from the fact that U is not an allowed move, (M U)^4 rotates 4 centers. If you replace (M Uw)^20 with (M U)^4 in my alg, you will get messed-up centers even on a normal 4^3.