Restriction games with the cube (subgroups)

[ray5]

I am solving the 3x3 cube with CFOP but I have become interesting in the following games and wanted to ask if anybody has more information about them.

Solve the cube using only M and U moves after scrambling the only moves M and U. This leaves the top layer corners alone and a block of 6 on the left and right of the cube. I have heard this is related to Roux method but I am not sure. I have been able to solve the bottom 2 layers with it but then you end up with some tricky edge flips.

Solve the cube using only R and U moves after scrambling only with moves R and U. This is related to 2-gen I think and sune and about half of the PLL algs I know are <R,U> based, but I am not able to reach LL - does this kind of thing have a name and how is it solved?

Lastly I was also curious about scrambling and solving using only R2, L2, U2, D2, F2, B2 moves. Are there any other sets of moves which are worth looking at and how hard is it to learn to solve these restriction games?

 

[bandagedgroup]

Solving in subgroups you mention isn't too hard. I have just one special recollection on the <R2, L2, U2, D2, F2, B2> subgroup, a nice form of edge 3-cycle: (R2 U2 R2 F2)2.

You may next want to try the following subgroups: <R2, L2, U2, D2, F, B> and <U2, D2, R, L, F, B>.

In general, there's a ton of these "restriction games". Each bandaged 3x3 cube gives you one, but the game is easier than solving the original bandaged cube. The puzzles can still be a lot harder than unrestricted cube.

Try this very specific version of the game. You are only ever allowed to make two possible moves:

1. F R U F R' F2 U'
2. you can rotate the whole cube so that U face becomes F face, F face becomes R face, and R face becomes U face.

 

[Christopher Mowla]

Have you seen this thread? You may also like this thread and this thread (from another forum).

 

[Jam88]

The MU edge flips are easy. If it is a dot case, M' U' M U2 M' U' M will turn it into a L case, then repeat that alg with the two oriented edges at LU and FU. If there is a line case, play around with taking out first layer edges and reinserting them.

 

[DNF_Cuber]

The MU edge flips are easy. If it is a dot case, M' U' M U2 M' U' M will turn it into a L case, then repeat that alg with the two oriented edges at LU and FU. If there is a line case, play around with taking out first layer edges and reinserting them.

Or you could use roux

 

[qwr]

A tip from CubeSkills is to scramble with <R,U,L> to practice left hand and back F2L inserts.

 

[ray5]

Or you could use roux

Good shout, The Roux method taught me how to solve the <M,U> case. I am looking into <R,U> next, I think it will not be too bad since we have sune and many PLL algs I know are already R,U only. cstimer has Last Six Edges for M,U but I don't have an easy way to generate scrambles only using <R,U> moves. I discovered that cstimer has all the 3x3 subsets!

 

[Deleted member 55877]

Good shout, The Roux method taught me how to solve the <M,U> case. I am looking into <R,U> next, I think it will not be too bad since we have sune and many PLL algs I know are already R,U only. cstimer has Last Six Edges for M,U but I don't have an easy way to generate scrambles only using <R,U> moves. I discovered that cstimer has all the 3x3 subsets!

 

[ray5]

I was playing around with <R,U> and discovered something a bit surprising.

You can obviously solve any R,U scramble with R,U moves. But if you do a regular scramble and then solve everything except the right and upper layers you may not be able to solve this.

I don't think this is the case with <M,U>

 

[DNF_Cuber]

I was playing around with <R,U> and discovered something a bit surprising.

You can obviously solve any R,U scramble with R,U moves. But if you do a regular scramble and then solve everything except the right and upper layers you may not be able to solve this.

I don't think this is the case with <M,U>

Yes, there are things called edge orientation, and corner pemutation. I can't do corner permutation myself(@Devagio and @CuberStache can I believe(sorry for mention)) but with EO you can watch any ZZ or petrus tutorial.

 

[Cuberstache]

I was playing around with <R,U> and discovered something a bit surprising.

You can obviously solve any R,U scramble with R,U moves. But if you do a regular scramble and then solve everything except the right and upper layers you may not be able to solve this.

I don't think this is the case with <M,U>

This is true because of both edge orientation and corner permutation, like @DNF_Cuber said. If you want to reduce the cube to a 2-gen state it's actually quite complicated, check out the YruRU method, which you can find here.

 

[xyzzy]

From a group theory perspective, the RU corner permutation restriction is related to an exotic embedding of \( S_5 \) in \( S_6 \).

In general, we always have \( S_k\le S_n \) for \( k\le n \) given by stabilising \( n-k \) points, so these copies of \( S_k \) correspond to non-transitive actions.

When \( k=n-1 \), then the one-point stabilisers are the only embeddings… unless \( n=6 \), where we can also take a one-point stabiliser, then send it through one of the non-inner automorphisms. (\( S_6 \) is the only symmetric group with nontrivial outer automorphisms; you can't repeat this process for any other \( n \).) These "exotic" embeddings happen to act sharply 3-transitively on \( \{1,2,3,4,5,6\} \). (I'm not aware of a nice way to prove this directly without explicitly constructing the isomorphism.)

To wit, with \( \phi\in\operatorname{Aut}S_6\backslash\operatorname{Inn}S_6 \), the exotic embeddings of \( S_5 \) correspond to \( \phi(\operatorname{Sym}\{1,2,3,4,5\}) \).

The corner permutations induced by R and U moves correspond to one such exotic embedding; the 3-transitivity means that you can always solve at least three corner pieces, and the sharp 3-transitivity means that once you've solved three corner pieces (e.g. the two in the bottom layer and AUF to align any corner in the top layer), the remaining corner pieces must also be solved.

(In terms of how to determine which coset of \( \phi(\operatorname{Sym}\{1,2,3,4,5\}) \) the current cube state is in, I came up with a pretty convoluted method a few years ago based on some discussions with Teoidus (in the context of their 2GR method). There's a more recent attempt at coming up with better 2-gen CP recognition methods (e.g. whatever YruRU uses) but I haven't really looked into those.)

(edit (2021-06-05): minor wording and formatting changes)

 

[Jam88]

From a group theory perspective, the RU corner permutation restriction is related to an exotic embedding of S_5 in S_6.

In general, we always have \( S_k\le S_n \) for \( k\le n \) given by stabilising \( n-k \) points, so these copies of \( S_k \) correspond to non-transitive actions. When \( k=n-1 \), then the stabilisers are the only embedding… unless \( n=6 \), in which case we take a one-point stabiliser, then send it through one of the non-inner automorphisms. (This is "exotic"/"exceptional" because \( S_6 \) is the only symmetric group with nontrivial outer automorphisms; you can't repeat this process for any other \( n \).) Such embeddings happen to act sharply 3-transitively on \( \{1,2,3,4,5,6\} \) (though I'm not aware of a nice way to prove this directly without explicitly constructing the isomorphism).

To wit, with \( \phi\in\operatorname{Aut}S_6\backslash\operatorname{Inn}S_6 \), the exotic embeddings of \( S_5 \) correspond to \( \phi(\operatorname{Sym}\{1,2,3,4,5\}) \).

The corner permutations induced by R and U moves correspond to one such exotic embedding; the 3-transitivity means that you can always solve at least three corner pieces, and the sharp 3-transitivity means that once you've solved three corner pieces (e.g. the two in the bottom layer and AUF to align any corner in the top layer), the remaining corner pieces must also be solved.

(In terms of how to determine which coset of \( \phi(\operatorname{Sym}\{1,2,3,4,5\}) \) the current cube state is in, I came up with a pretty convoluted method a few years ago based on some discussions with Teoidus (in the context of their 2GR method). There's a more recent attempt at coming up with better 2-gen CP recognition methods (e.g. whatever YruRU uses) but I haven't really looked into those.)

My brain hurts. HOWWWWW are you so smart?!

 

[BenChristman1]

My brain hurts. HOWWWWW are you so smart?!

I don't understand it either, but it's probably because he majored in stats in college or something like that. Just because you don't like/understand it, doesn't mean that other people won't.

 

[qwr]

I don't understand it either, but it's probably because he majored in stats in college or something like that. Just because you don't like/understand it, doesn't mean that other people won't.

From what I can tell, you could probably understand it after 1 undergrad semester of abstract algebra. At least at one point I knew what automorphisms and stabilizers were, although I kinda failed that class and hurt my head a lot.

 

[ray5]

Got the half turn game figured out today. Basically you can solve the corners and centers pretty easily then use a couple algs

Spoiler: algs

[M2,U2] and (R2 U2)3 thanks jey!

to solve the edges. It's quite difficult and any accidental half turn ruins it so this isn't the most fun restriction game. Good to have it solved though.

 

[qwr]

Got the half turn game figured out today. Basically you can solve the corners and centers pretty easily then use a couple algs

Spoiler: algs

[M2,U2] and (R2 U2)3 thanks jey!

to solve the edges. It's quite difficult and any accidental half turn ruins it so this isn't the most fun restriction game. Good to have it solved though.

Spoiler: my method I came up using my knowledge of cuboids

First solve corners and centers with white on bottom. This should be doable in 3 moves.

Solve the middle edges using (R2 U2)3 to swap right front/back edges. That also swaps top front/back edges.

Then solve top and bottom simultaneously by trying to swap pieces from top and bottom. To swap front and left edges between top and bottom, use (R2 F2)3. To swap front and back edges between top and bottom, use (M2 F2)2. For example we can achieve an H perm by strategically swapping pairs of edges: (R2 F2)3 U2 (R2 F2)3 U2 (R2 F2)3.

For 3 cycle of edges front bottom, back top, front top, use (U2 R2 F2 R2)2

 

[ray5]

Some notes on the half turn subset: <R2,L2,U2,D2,F2,B2>.

GAP gives the following structure description of the group ((((A4 x A4) : C2) x A4) : C2) x (((C2 x C2 x C2 x C2) : C3) : C2). It is a direct product of two groups because the edges and corners can be permuted separately. There is no parity.

This group has order 663552 = 6912 * 96 = (2^8 3^3) * (2^5 3). There are 6912 different permutations of edges and 96 different permutations of corners.

It can be solved quite easily by solving corners first then solving edges using the following algorithms:

• edge 3 cycle: https://alg.cubing.net/?alg=(F2_R2_U2_R2)2
• adjacent edges swapping with their opposites: https://alg.cubing.net/?alg=(F2_R2)3
• opposite edge swap: https://alg.cubing.net/?alg=[U2,_M2]
• H perm (alg using only half turns): https://alg.cubing.net/?alg=F2_R2_U2_F2_U2_R2_F2_R2_U2_R2_U2_R2

The group for edges can be thought of as the subgroup of even permutations of S4^3 where each S4 corresponds to permuting the edges in each of the M,S,U slices. The algs above can be used to show this.

I must say I don't really understand the corners group well yet (why does it have size 96 for example), it may be useful to create cubes with edges greyed out and work with them.

Thank you to those who helped me with finding algs and understanding the group.

 

[xyzzy]

(why does it have size 96 for example)

Oh, I can (try to) explain this one.

It has to do with the Klein 4-group being a normal subgroup of \( S_4 \). This embedding as a normal subgroup is unique, but just to be concrete, take \( V=\{\operatorname{id},~(1,2)(3,4),~(1,3)(2,4),~(1,4)(2,3)\}\triangleleft S_4 \), and the quotient is \( S_4/V\cong S_3 \).

(For maximum concreteness, note that \( V \) acts sharply transitively on {1,2,3,4}, so each coset of \( V \) has exactly one element that has 4 as a fixed point, i.e. there is a unique choice of representative permuting only {1,2,3}. This is one way to see why the quotient is isomorphic to \( S_3 \).)

To start off, the half turn subgroup separates the corner pieces into two tetrads. For each pair of opposite corners, give them the same label. For example, take URF = DLB = 1, ULB = DRF = 2, DRB = ULF = 3, DLF = URB = 4. Say that the tetrad with the URF corner is the "primary" tetrad and the other one is the "secondary" tetrad.

How do the half turns permute the corners within the tetrads? U2 applies the permutation (1,2) on the primary tetrad and (3,4) on the secondary tetrad. D2 applies the permutation (3,4) on the primary tetrad and (1,2) on the secondary tetrad. (You can work out the rest; they correspond to the \( \binom42=6 \) ways of choosing two out of four.) No matter which move we do, the induced permutations on the primary tetrad and the secondary tetrad lie in the same coset of \( V \), and hence the permutations of the tetrads mod \( V \) will always be the same.

Put another way, if the primary tetrad is solved, then the secondary tetrad's permutations must lie in \( V \), and vice versa. This restricts the total number of possible corner permutations to at most \( |S_4|\cdot|V|=24\cdot4=96 \). There is no other restriction because we can generate arbitrary permutations on the primary tetrad (the six half turns correspond to all six transpositions), and we can also generate all \( V \) permutations on the secondary tetrad by itself (with R2 F2 R2 D2 and its rotations).

 

[ray5]

Oh, I can (try to) explain this one.

Absolutely wonderful thank you so much!

Did a search in GAP and found these tetrad algs:

* identity
* ((F2 L2)2 U2 L2 U2)2
* (F2 U2 F2 (L2 U2)2)2
* ((L2 F2)2 U2 L2 U2)2