#### cmhardw

##### Premium Member

Hello everyone,

I've been thinking recently about what it means, in the general sense, to read an algorithm backwards, and if this can even be generalized on the Rubik's cube.

For palindrome algorithms, obviously you have the same effect if you read forwards or backwards.

example: M2 U M2 U2 M2 U M2

For algs that don't contain either one of a pair of parallel faces, reading that algorithm backwards is simply the inverse of the reflection about the plane parallel to both of those planes.

example: R' U R' U' R' U' R' U R U R2

This algorithm contains no F or B turns, so reading it backwards is simply the inverse of the reflection of this alg in the FB plane.

Another example of this same case is the A perm.

A perm: R' F R' B2 R F' R' B2 R2

This algorithm contains no U or D turns, so reading it backwards functions as the inverse of the reflection of this algorithm in the UD plane.

Now an interesting one is the T permutation. This algorithm contains sides in each of the 3 axes of the cube. Also, this algorithm is not a palindrome.

R U R' U' R' F R2 U' R' U' R U R' F'

reading this algorithm backwards seemingly scrambles the U,F and R faces.

What I wanted to do with this was to test to what extent this new algorithm (reading the T perm backwards) is not commutative with the actual T perm. So I'm going to do the commutator:

(T perm) (read the T perm backwards) (T perm) ' (read the T perm backwards) '

which is

(R U R' U' R' F R2 U' R' U' R U R' F') (F' R' U R U' R' U' R2 F R' U' R' U R) (F R U' R' U R U R2 F' R U R U' R') (R' U' R U R F' R2 U R U R' U' R F)

This leaves the cube semi-scrambled, and the net effect (as a permutation) is:

Corners: (UBR <-> UFR) (UFL <-> FLD)

Edges: (UL->UF->UR->UL)

Cube explorer generates the same permutation with: U' L2 U2 L2 R U' L2 U R' U2 L2 U'

Now I have 2 questions. How can I interpret/analyze this overall net permutation to see the extent to which the T perm and the (T perm read backwards) are not commutative?

My second question is: is there a general meaning to the permutation generated by reading a given permutation backwards? I expect that this will have to be broken down into sub-cases. I have listed above 2 sub-cases for which the permutation generated by reading backwards is either the equivalent permutation to the original, or is the inverse of the reflection in the plane of sides missing from the alg.

I find this idea fascinating, but I don't know where to start to study it more, or even what question I probably should be asking.

--edit--

Here are 2 actually useful algorithms created using the above procedure.

Let my algorithm be X = R U' R'

I will now create the commutator X (read X backwards) X' (read X backwards)' which is:

(R U' R') (R' U' R) (R U R') (R' U R) or R U' R2 U' R2 U R2 U R

which sets up a very interesting F2L trick, in my opinion. Do the inverse of the final algorithm to solve: R' U' R2 U' R2 U R2 U R'

I might actually try to learn to recognize that case, even though it wouldn't come up often.

Another useful one is to let the algorithm X = M2 U

Then (X) (read X backwards) X' (read X backwards)' = (M2 U) (U M2) (U' M2) (M2 U') = M2 U2 M2 U2

which is a recognizable pattern.

I'm getting the feeling that this process may be useful for creating F2L algorithms and/or patterns. I'll have to look into this more.

--edit--

--edit #2--

Here's a nice fast BLD edge 3-cycle using the above process

Let X = (R U' R' U) and do X (read X backwards) X' (read X backwards)' to get: R U' R' U2 R' U' R U' R U R2 U R U'

--edit #2--

Chris

I've been thinking recently about what it means, in the general sense, to read an algorithm backwards, and if this can even be generalized on the Rubik's cube.

For palindrome algorithms, obviously you have the same effect if you read forwards or backwards.

example: M2 U M2 U2 M2 U M2

For algs that don't contain either one of a pair of parallel faces, reading that algorithm backwards is simply the inverse of the reflection about the plane parallel to both of those planes.

example: R' U R' U' R' U' R' U R U R2

This algorithm contains no F or B turns, so reading it backwards is simply the inverse of the reflection of this alg in the FB plane.

Another example of this same case is the A perm.

A perm: R' F R' B2 R F' R' B2 R2

This algorithm contains no U or D turns, so reading it backwards functions as the inverse of the reflection of this algorithm in the UD plane.

Now an interesting one is the T permutation. This algorithm contains sides in each of the 3 axes of the cube. Also, this algorithm is not a palindrome.

R U R' U' R' F R2 U' R' U' R U R' F'

reading this algorithm backwards seemingly scrambles the U,F and R faces.

What I wanted to do with this was to test to what extent this new algorithm (reading the T perm backwards) is not commutative with the actual T perm. So I'm going to do the commutator:

(T perm) (read the T perm backwards) (T perm) ' (read the T perm backwards) '

which is

(R U R' U' R' F R2 U' R' U' R U R' F') (F' R' U R U' R' U' R2 F R' U' R' U R) (F R U' R' U R U R2 F' R U R U' R') (R' U' R U R F' R2 U R U R' U' R F)

This leaves the cube semi-scrambled, and the net effect (as a permutation) is:

Corners: (UBR <-> UFR) (UFL <-> FLD)

Edges: (UL->UF->UR->UL)

Cube explorer generates the same permutation with: U' L2 U2 L2 R U' L2 U R' U2 L2 U'

Now I have 2 questions. How can I interpret/analyze this overall net permutation to see the extent to which the T perm and the (T perm read backwards) are not commutative?

My second question is: is there a general meaning to the permutation generated by reading a given permutation backwards? I expect that this will have to be broken down into sub-cases. I have listed above 2 sub-cases for which the permutation generated by reading backwards is either the equivalent permutation to the original, or is the inverse of the reflection in the plane of sides missing from the alg.

I find this idea fascinating, but I don't know where to start to study it more, or even what question I probably should be asking.

--edit--

Here are 2 actually useful algorithms created using the above procedure.

Let my algorithm be X = R U' R'

I will now create the commutator X (read X backwards) X' (read X backwards)' which is:

(R U' R') (R' U' R) (R U R') (R' U R) or R U' R2 U' R2 U R2 U R

which sets up a very interesting F2L trick, in my opinion. Do the inverse of the final algorithm to solve: R' U' R2 U' R2 U R2 U R'

I might actually try to learn to recognize that case, even though it wouldn't come up often.

Another useful one is to let the algorithm X = M2 U

Then (X) (read X backwards) X' (read X backwards)' = (M2 U) (U M2) (U' M2) (M2 U') = M2 U2 M2 U2

which is a recognizable pattern.

I'm getting the feeling that this process may be useful for creating F2L algorithms and/or patterns. I'll have to look into this more.

--edit--

--edit #2--

Here's a nice fast BLD edge 3-cycle using the above process

Let X = (R U' R' U) and do X (read X backwards) X' (read X backwards)' to get: R U' R' U2 R' U' R U' R U R2 U R U'

--edit #2--

Chris

Last edited: Jun 27, 2008