dougreed
Member
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
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