User:Dave3457/Sandbox/Speedsolving commutator

A commutator is a series of moves that involves performing 4 sequences, A B A’ and B’. After performing sequence A and sequence B one performs the inverse of sequence A and finally the inverse of the sequence B. As a result, only specific pieces altered by both sequence A and B are affected. It allows one to finish the cube without disturbing pieces that are already in place

A commutator is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows one to finish the cube without disturbing pieces that are already in place.

Mathematical definition

The word commutator is a term used in group theory to represent a series of elements or operations of the form ghg'h' where g and h are two specific elements of a group and g' and h' are their inverses. The short form for this series of operations is [g, h].

All the possible states or permutations of the Rubic's cube form a group and is called the Rubic's cube group. Cube notation for a cube commutator is: A B A' B' = [A, B].

Given a group, a commutator is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close: A B A' B' = [A, B]

Fundamental principle

Two sequences A and B are said to commute if the order in which one performs them does not matter. For example, the two moves L and R are said to commute because it does not matter if you first perform L and then R or if you first perform R and then L. That is to say L R = R L. However the two moves L and U do not commute because the order in which you perform them matters, ie L U ≠ U L .

The commutator A B A' B', because of its structure, gives an indication of the extent to which the two sequences B and A' do not commute.
To understand why, first appreciate that A A' and B B' never do anything because A' is the inverse of A and B' the inverse of B. ie. A A' = I and B B' = I where I is called the identity element in group theory and represents "no change".

Because A A' = I and B B' = I the commutator A B A' B' does not change anything if B A' is the same as A' B...
Because if...
B A' = A' B'
then
A B A' B' = A (B A') B' = A (A' B) B' = (A A') (B B') = I
where I is again the identity element.
On the other hand if...
B A' ≠ A' B'
then
A B A' B' ≠ I
because
A B A' B' ≠ A A' B B'

For example if A = L and B = R then the commutator L R L' R', does nothing:
L R L' R' = L L' R R' = I
however if A = L and B = U then the commutator L U L' U', changes something
L U L' U' ≠ I

So in some way, the commutator allows you to measure to which extent two sequences do not commute.

xxxxxxxxxxxxxxxx It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing: L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

Effect

File:COMM Effects section Definitions.png

This image defines the terms J, K, N, NA’,NB’ used to describe how the commutator works.

To help us discuss how a commutator works we will first define some terms.
The image on the right gives a visual indication of which pieces of a cube are affected by the different sequences

of a commutator. In this case the sequence A is F and the sequence B is U giving the simple commutator [F, U].

Referring to the image, the sequence A changes the pieces at a set of locations we will call J. The sequence B

changes the pieces at a set of locations we will call K. We will call the intersection of J and K, N. If J

and K have no intersection, such as when A equals L and B equals R, then A and B are said to commute, and

the commutator [L, R] does nothing. However if, as in the image, when J and K do have an intersection, A

and B do not commute and the cube, as a result, is changed by the commutator A B A' B'. Note that the changes

which this particular commutator make to the cube are rather difficult to understand however more straight forward

commutator examples will be discussed in a moment.

There is a set of locations that we will be calling NA'. This name is appropriate because the pieces can be

identified by applying the inverse of A to the locations N. Referring to the lowest left cube in the rightward

image, these locations are identified using two shades of grey. Lighter grey refers to pieces that will be brought

into the intersection region N, the darker grey refers to pieces that are not brought in the intersection region but

are altered in some way within the region. A dark grey piece can be moved within the region, as in this case, or it

can be twisted in place.

All of the above is also true for the region we will be calling NB'.

The only pieces that are affected by any commutator are the ones located in the union of N, NA' and NB'.

In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or

are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up after the

sequences A and B are executed, will be put back into their original locations with sequences A' and B'. This

restoration of all the pieces that are not in the union of N, NA' and NB' is the reason the commutator is

such a powerful tool.

Click the following thumbnail for an image similar to the one on the right that uses the more general commutator

[B' D B, U']. File:COMM Effects section Example.jpg

Trivial case

When J and K have no intersection, A and B commute, so the commutator does nothing.

B inplace

If NB' = N, the sequence B only moves affected pieces that are inside the intersection. It may also move

pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be

N and NA'. So those pieces will be in J, i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the conjugate of B by A,

and second part is the inverse of B.

If [A: B] and B' interfere, there is a quirk part P_Q (see general case).

Examples

Let's consider [M', U2] = M' U2 M U2

In this example, N are the top-front and top-back edges. NB' is NU2, which is exactly N, and NA' are

front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are

directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left

and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps

front-top and front-bottom edges.

So [M', U2] can be understood as: swap front-top and front-bottom edges, then swap top-back and top-front

edges, which yields a 3-cycle of edges.

There is a quirk part, the front-bottom edge, which is affected by both [M': U2] and U2, i.e. by all four moves of the

commutator.

Let's consider [M2, U2] = M2 U2 M2 U2

Again, N are the top-front and top-back edges. But there is no more interference between [M': U2] and U2, thus

no quirk part. Simply, top-front and top-back edges are swapped, as well as bottom-front and bottom-back edges.

A inplace

This is a symmetrical case. Affected pieces will only be N and NB'. So those pieces will be in K, i.e.

among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the

conjugate of A' by B.

Examples

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant

part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom

edges.

Other examples:

• A = "rotating a corner" like [R' D' R D R' D' R, U]
• A = "flipping an edge" like [R' E' R2 E2 R', U]
• A = "exchanging two corners" like [R' L' D2 R L, U]
• A = "exchanging two edges" like [M2 D2 M2, U]

Three different sets

When NA' and NB' are not included in N, there are three sets involved in the commutator:

• NA' : pieces that are brought into the intersection by A
• NB' : pieces that are brought into the intersection by B
• N : the intersection of J and K, i.e. where A and B interfere.

Note that NA' can have locations in common with N and NB' can have locations in common with N. In

other words, it is possible that a sequence (A or B) does two different things in the context of a commutator:

1. moving pieces inside the intersection (from N to N)
2. bringing pieces into the intersection (from NA' \ N to N or from NB' \ N to N)

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases:

• Case 2 only: the three sets NA', NB' and N are completely separated (no overlap)
• Case 1 and 2 at the same time: the three sets overlap

No overlap

Referring to the animation below, when sets do not overlap, the commutator can be summed up by:

• A stores the content of the intersection P_N in NA, and replace it with some pieces

P_A from NA'

• B stores P_A in NB, and bring some pieces P_B from NB' into the intersection
• A' retrieves P_B from the intersection and places them in NA', and brings to the intersection

previously stored pieces P_N in NA

• B' retrieves P_N from the intersection and places them in NB', and finally places P_A

in the intersection.

So it is a 3-cycle of (P_N, P_A, P_B) into (P_A, P_B,

P_N)

We notice that NA and NB are used as temporary storage:

• The content of the intersection, P_N, is first hidden by A in its storage NA, and comes back with A',

so that the only transformation applied to these pieces is B'.

• The content of NA' is brought by A, and is then hidden by B in NB, so that the only transformation applied to

these pieces is A.

• The content of NB' is brought by B, and then is placed immediately in NA' by A'. So the transformation

applied to these pieces is B A'.

Note that NA may or may not overlap with NA', and NB may or may not overlap with NB', but that it does

not have any implication, except that the location where the pieces come from can also be the location where

they are temporarily stored. The most obvious case is when NA equals to NA', or NB equals to NB',

which means that pieces of the intersection N are swapped with pieces of NA' or NB', as in [R2, F2]

In other words, all the commutator does is:

• P_A goes A
• P_B goes B A'
• P_N goes B'

File:Break down of Commutator A B A B If there is no overlap.gif

Examples

• [R' E2 R, U'] is a 3-cycle of edges, where P_A is the back-left edge, P_B is the top-front

edge and P_N is the top-right edge.

• [R' D' R, U'] is a 3-cycle of corners

Overlap

This is the most complex case, because it is like the "no overlap" case, but with additional interferences.

Safe part

Some pieces in the intersection N are hidden in a storage, thus protecting them from being scrambled by the

interference due to the overlap.

Let's call H_B the location of pieces hidden by B:

H_B = ((NB) \ N) B'

Similarly, pieces are hidden by A:

H_A = ((NA) \ N) A'

We will also use pieces that are hidden by A' so that they are not affected by B':

H_A' = ((NA') \ N) A

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:

• P_A will be pieces that are brought into H_B by A, i.e. pieces that are at the beginning

in H_BA'. This set may overlap with the intersection N.

• P_B will be pieces that are brought into H_A' by B, i.e. pieces that are at the beginning

in H_A'B' \ N

• P_N will be pieces that are hidden by A, which are at the beggining in H_A.

We get the same result as before:

• P_A goes A
• P_B goes B A'
• P_N goes B'

Example

Let's consider [R', F]. The intersection N is the front-right column (containing the front-right-top corner, the

front-right-bottom corner and the front-right edge).

H_B = ((NB) \ N) B' = ((NF) \ N) F' which is the set of locations containing the front-right edge and the front-right-bottom corner.

So P_A are front-right-top corner and top-right edge.

H_A = ((NA) \ N) A' = ((NR') \ N) R

So P_N are front-right edge and front-right-bottom corner.

H_A' = ((NA') \ N) A = ((NR) \ N) R' which is the set of locations containing the front-right-top corner and the front-right edge.

P_B are pieces at H_A'B, i.e. front-top edge and front-top-left corner.

So the safe part of [R', F] is similar to a 3-cycle pairs, but is clearly not a 3-cycle of pairs.

• P_A goes R', thus replacing P_N
• P_N goes F, thus replaces a part of P_A and a part of P_B
• P_B goes FR, thus replacing a part of P_A and the top-right-back corner (the quirk part

Q)

• the top-right-back corner is affected by all moves, R'FRF' = [R': F] F' equivalent to UF', or R'FRF' = R' [F: R]

equivalent to R'U, thus going to the top-front-right position, replacing a part of P_B

The quirkiness is that the locations of those pairs overlap, and that an extra location is used, the top-right-back

corner. Pieces that are initially in this location are moved according a special scheme (see below).

In this example, the top-right-back corner goes to the top-front-right position, thus being separated from the edges

around it.

Quirk and conjugated part

Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part

Q.

Q = (H_A' \ H_B) A'

Those pieces P_Q are never hidden, thus being affected by all moves of the commutator:

• P_Q goes A B A' B'

Some pieces are outside of the pseudo 3-cycle, but are hidden by A or B at some point. So they are conjugated. Let's call them P_C and their starting location C.

C = (J union K) \ (H_A union H_BA' union H_A'B union Q)

In the overlap case, either Q or C is non empty, so there are at least 4 sets of pieces: P_A,

P_B, P_N and P_Q or P_C.

In complex cases, there may be five sets with both P_Q and P_C.

The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.

Example

Let's consider [R', F']

H_A is the front-right edge and the front-right-bottom corner. So P_N is the set of pieces

at these locations.

H_B is the front-right edge and the front-right-top corner. So P_A is the top-right edge

and the top-right-back corner.

H_A' is the front-right edge and the front-right-top corner. So P_B is the front-bottom

edge only.

The quirk part is : Q = (H_A' \ H_B) A' = empty

The conjugated part is :

C = (J union K) \ (H_A union H_BA' union H_A'B union Q) =

front-right-top corner and front-bottom-left corner

The front-right-top is affected by R'F'R so it is equivalent to x'F'x which is U'.

The front-bottom-left is affected by F'RF so it is equivalent to z'Rz whichi is D.

So this move does :

• P_N, i.e. front-right edge and front-right-bottom corner, goes F
• P_A, i.e. top-right edge and top-right-back corner, goes R'
• P_B, i.e. front-bottom edge, goes F'R
• the front-right-top corner goes U'
• the front-bottom-left goes D

General case

Pieces can be affected by one, two, three of four elements of the sequence A B A' B'. The gives use the different sets of pieces. For each element of the sequence, there are two possibilities, either a piece is affected by this element, or it is not. So there is a maximum of 2⁴ = 16 sets.

If a piece is not affected by any element, it means that it is completely outside of the commutator.

Pieces affected by one element only

If a piece is affected by A only, then it is not affected by B and then it is necessarily affected by A', so it is impossible. Same thing for B. If a piece is affected by A' only, it means that it was not affected by B before that, so that it was affected by A. Impossible again. If a piece is affected by B' only, it means that it was not affected by A' before that, so that it was affected by B. Impossible again.

Pieces affected by two elements

If a piece is affected by A and A', or by B and B', it means that it comes back finally where it comes from. We can ignore them.

If a piece is affected by A and B only, it means that it is not affected by A', so it is necessarily affected by B'. Impossible. If a piece is affected by A and B' only, it is not affected by B, so it is necessarily affected by A'. Impossible again.

If a piece is affected by B and A', it means that it is brought into the intersection by B. They are denoted P_B.

Pieces affected by three elements

If a piece is affected by A and A' B', it means that it is hidden by A so that it is not affected by B. These pieces are denoted P_N because they come from the intersection.

If a piece is affected by A B and B', it means that it is brought into the intersection by A, and then hidden by B from A'. These pieces are denoted P_A.

If a piece is affected by B A' B', it means that it is brought into the intersection by B and then moved inplace by A. If a piece is affected by A B A', it means that it is affected by the conjugate of B. In both of these cases, let's denote them P_C to express that they are affected by conjugates.

Pieces affected by all four elements

If a piece is affected by all four elements, it means that it is never hidden from the intersection. Thus it is in the quirk part. They are denoted P_Q.

Summary of sets of pieces

In all cases, pieces that are moved by the commutator are in one of these sets:

• P_A : pieces moved into the intersection by A and then hidden by B
• P_B : pieces moved into the intersection by B and then hidden by A
• P_N : pieces hidden by A and then moved out of the intersection of B'
• P_C : pieces affected by a conjugate, in all inplace cases and some overlap cases
• P_Q : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case

Classification

If P_A is not empty, then B can hide pieces, thus P_B is not empty either. Conversely, if P_B is not empty, then P_A is not empty.

If P_N is not empty, then A and B can hide pieces, thus P_A and P_B are not empty. Conversely if P_A is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus P_N is not empty.

So either the three sets P_A, P_B and P_N are not empty or the three sets are empty. If these sets are empty, it means that at least A, B or both A and B move pieces inside the intersection N only. So it is either a trivial case or an inplace case. If both A and B move pieces inside the intersection, let's call it "double inplace".

If it is an inplace case and not a "double inplace", then some pieces are affected by a commutator. Thus P_C is not empty.

If the sets P_A, P_B and P_N are not empty, we cannot deduce anything about P_Q or P_C. If it is the no overlap case, then pieces are hidden either by A or B, so P_Q is empty. Conversely, if P_Q is empty, then pieces brought by A into the intersection are completely hidden by B. The inverse of the commutator B A B' A' shows that also A hides pieces brought by B, so that it is a no overlap case.

So we get the following possible types of commutators:

• trivial commutator which is equivalent to the identity: [L, R]
• double inplace, A and B move pieces inside the intersection
• single inplace without quirk part, the same transformation is applied twice (the second time in reverse order) without interference: corner twist, edge flip, double swap
• single inplace with quirk part, there is an interference between the conjugated and the non conjugated version of the transformation: [M', U2]
• no overlap, 3-cycle: [R' D' R, U']
• overlap with quirk part and without conjugated part: [R', F]
• overlap with conjugated part and without quirk part: [R', F']
• overlap with both quirk and conjugated part

See also

• Commutators and Conjugates
• Conjugate