# Fun with Group Theory



## dougreed (Mar 25, 2006)

Hi all,

I was reading a book I just picked up called Adventures in Group Theory (available here) when I decided to look at a few of my OLL and PLL algs and see which ones I could easily break into conjugates and commutators.

Here is one I quickly discovered:
My OLL: R2 D' R U2 R' D R U2 R

My thought process:
This alg doesn't look like a commutator [A,B] simply because the inverse of the first move, R2, doesn't appear in that form anywhere in the rest of the algorithm. Even though this could be attributed to move cancellation, I decided to see what I could find out about this algorithm by assuming it is a conjugate of a commutator and some move C, e.g. [A,B]^C == C [A,B] C'.

My first thought is that C' is probably only going to be one face-turn, so I say C'=R which implies that C = R'.

The first move of the algorithm is R2, which can be represented as R R or R' R'. By using R' R', we come up with an acceptable move for C. So, now we have:

[C] R' D' R U2 R' D R U2 [C']
C=R

First, we know that U2 == (U2)' because U2 is its own inverse. Then, we can rewrite the above:

[C] R' D' R U2 R' D R (U2)' [C']

Now, we can easily see that this algorithm, sans the C and C', is a commutator [A,B] if A = R' D' R and B = U2.

C A B A' B' C' == C [A,B] C' == [A,B]^C
if A = R' D' R, B = U2, C = R

Here are a few more algorithms I regularly use that can be broken down similarly, for anyone who is interested:

R2 B2 R F R' B2 R F' R
R' U2 R U R' U R (2 commutators)

-Doug


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## Scott (Mar 25, 2006)

Commutators are far from the "Beginners" area.

Moved to speedcubing forum.


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## Smoid (Mar 26, 2006)

one funny point: "sans the C"

Are you french, or english using some french ^_^


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## pjk (Mar 26, 2006)

Haha. Man, that is deep


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## dougreed (Mar 26, 2006)

> _Originally posted by Scott_@Mar 25 2006, 04:56 PM
> *Commutators are far from the "Beginners" area.*


No they aren't. Maybe not for 3x3x3 beginners, but they are immensely useful for anyone wishing to solve the 4x4x4 or 5x5x5, not just veterans, IMHO.

Also, when you're trying to teach someone to learn to solve the 3x3x3, commutators for simple 3-cycles can be a lot easier to teach than just expecting to someone to learn a sheet of algorithms, and then your solution is much more scalable. Once you "understand" where one edge 3-cycle comes from on the 3x3x3 cube, it is very easy to build 3-cycles of other pieces on larger cubes.

Just my 2 cents.
-Doug


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## pjk (Mar 26, 2006)

Now that I know what a commutator is, I think it belongs in beginners section. However, this post seems to be a speedcubing discussion 
Pat


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## dougreed (Mar 26, 2006)

I'm working on what could be a decent speedcubing method for the 4x4x4. The problem is that I actually do the F3L and the LL completely separately, so when I get to the LL I have 4 messed up edges. 

While most of the time this is fine, there is one case in particular which is particularly annoying, where (it appears) one single edge piece is swapped with another single edge piece. 

My intuition told me that the Right Thing  to do here would be to execute a r' and then apply the algorithm I know to flip an edge, which is just the 4x4x4 OLL parity alg (call this alg "P"), and then undo the r', which makes a simple conjugate. In practice, however, this is annoying... the end result is a 3-cycle of centers away from the solved state, provided you actually fixed the edges.

So, then I ask myself: how can I make it so that the execution of this algorithm r' P r leaves my cube in the solved state...

Well, I essentially have two options. One is to modify the algorithm P itself. I think I can do this by breaking it down, but more on this exciting conclusion later 

The second option is to come up with some sort of an algorithm to force the cube into a state so that when the algorithm r' P r actually "messes up" the centers, it is leaving the centers in a state such that the inverse of my new setup (called "Q") restores the centers.

By doing the algorithm r' P r with numbered center pieces, you can easily see how this algorithm affects the centers. The setup process I came up with (just intuition) is this: Q = f (d u') (f b') R' (f' b ). Yes, it's ugly, and I think I can do better. But I've run out of time, and will continue this post later.

-Doug


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## dougreed (Mar 28, 2006)

Looking back at my OLL-parity fix alg for the 4x4x4:

r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2

I noticed first that this is a conjugate X^Y for X = U2 l U2 r' U2 r U2 F2 r F2 l' and Y = r2 B2. With this in mind, I examine algorithm X by itself, without the setup move Y as a prefix and Y' as a suffix.

This portion of the alg, as it turns out, is a 2-cycle of the UlF single edge piece and the UlB single edge piece, as well as a 2-cycle of 2x1 blocks on the U and D layers. With this in mind, it is easy to see how my original parity-fixing algorithm worked, and similarly easy to see how the alg I am looking for _should_ work.

The setup move I came up with to replace Y = r2 B2 is Z = r2 D R' D' B. The resulting conjugate X^Z is expanded as follows:

(r2 D R' D' B ) U2 l U2 r' U2 r U2 F2 r F2 l' (B' D R D' r2)

and accomplishes exactly what I wanted to accomplish.

-Doug


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## Cubefactor (Mar 29, 2006)

Commutators are also imensely useful for fewest moves competition. So not neccesarily speedcubing binding. Though I'm a little shaky on commutators though!

Of course solving the 20x20 supercube required a simple commutator for the centers. What would I do without these beautiful things  

-Richard


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## dougreed (Mar 29, 2006)

> _Originally posted by Cubefactor_@Mar 29 2006, 09:37 AM
> * Of course solving the 20x20 supercube required a simple commutator for the centers. What would I do without these beautiful things  *


 Of course, I forgot about that. Unless you are going to constrain yourself to `popular' permutation puzzles (e.g. the 2x2x2-5x5x5, megaminx, pyraminx) you need a systematic method for usefully manipulating the puzzle. 

-Doug


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## Cubefactor (Mar 30, 2006)

f' r' u' r U r' u r U' f

that's what I used for 20x20 centers. As soon as you realize how it works it's damn useful


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## dougreed (Mar 30, 2006)

Yes, and even more useful than understanding how that particular alg works is understanding how it is derived. Once you know that, you can three-cycle practically any mxn center "block" on any 2 or 3 faces of the cube.

e.g., on the 5x5x5:
1) r' (E' d') r U r' (E d) r U' 
2) l' r' (u E' d') r U2 r' (u' E d) r U2 l

-Doug


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