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
[edit]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
[edit]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
[edit]
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
[edit]When J and K have no intersection, A and B commute, so the commutator does nothing.
B inplace
[edit]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 PQ (see general case).
Examples
[edit]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
[edit]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
[edit]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
[edit]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:
- moving pieces inside the intersection (from N to N)
- 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
[edit]Referring to the animation below, when sets do not overlap, the commutator can be summed up by:
- A stores the content of the intersection PN in NA, and replace it with some pieces
PA from NA'
- B stores PA in NB, and bring some pieces PB from NB' into the intersection
- A' retrieves PB from the intersection and places them in NA', and brings to the intersection
previously stored pieces PN in NA
- B' retrieves PN from the intersection and places them in NB', and finally places PA
in the intersection.
So it is a 3-cycle of (PN, PA, PB) into (PA, PB,
PN)
We notice that NA and NB are used as temporary storage:
- The content of the intersection, PN, 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:
- PA goes A
- PB goes B A'
- PN goes B'
File:Break down of Commutator A B A B If there is no overlap.gif
Examples
[edit]- [R' E2 R, U'] is a 3-cycle of edges, where PA is the back-left edge, PB is the top-front
edge and PN is the top-right edge.
- [R' D' R, U'] is a 3-cycle of corners
Overlap
[edit]This is the most complex case, because it is like the "no overlap" case, but with additional interferences.
Safe part
[edit]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 HB the location of pieces hidden by B:
HB = ((NB) \ N) B'
Similarly, pieces are hidden by A:
HA = ((NA) \ N) A'
We will also use pieces that are hidden by A' so that they are not affected by B':
HA' = ((NA') \ N) A
In order to get the same results as in the "no overlap" case, we will redefine the set of pieces:
- PA will be pieces that are brought into HB by A, i.e. pieces that are at the beginning
in HBA'. This set may overlap with the intersection N.
- PB will be pieces that are brought into HA' by B, i.e. pieces that are at the beginning
in HA'B' \ N
- PN will be pieces that are hidden by A, which are at the beggining in HA.
We get the same result as before:
- PA goes A
- PB goes B A'
- PN goes B'
Example
[edit]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).
HB = ((NB) \ N) B' = ((NF) \ N) F' which is the set of locations containing the front-right edge and the front-right-bottom corner.
So PA are front-right-top corner and top-right edge.
HA = ((NA) \ N) A' = ((NR') \ N) R
So PN are front-right edge and front-right-bottom corner.
HA' = ((NA') \ N) A = ((NR) \ N) R' which is the set of locations containing the front-right-top corner and the front-right edge.
PB are pieces at HA'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.
- PA goes R', thus replacing PN
- PN goes F, thus replaces a part of PA and a part of PB
- PB goes FR, thus replacing a part of PA 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 PB
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
[edit]Some pieces are outside of the pseudo 3-cycle and never hidden. Let's call their starting location the quirk part
Q.
Q = (HA' \ HB) A'
Those pieces PQ are never hidden, thus being affected by all moves of the commutator:
- PQ 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 PC and their starting location C.
C = (J union K) \ (HA union HBA' union HA'B union Q)
In the overlap case, either Q or C is non empty, so there are at least 4 sets of pieces: PA,
PB, PN and PQ or PC.
In complex cases, there may be five sets with both PQ and PC.
The quirk part and the conjugated part are used as final locations for pieces that go outside of the pseudo 3-cycle.
Example
[edit]Let's consider [R', F']
HA is the front-right edge and the front-right-bottom corner. So PN is the set of pieces
at these locations.
HB is the front-right edge and the front-right-top corner. So PA is the top-right edge
and the top-right-back corner.
HA' is the front-right edge and the front-right-top corner. So PB is the front-bottom
edge only.
The quirk part is : Q = (HA' \ HB) A' = empty
The conjugated part is :
C = (J union K) \ (HA union HBA' union HA'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 :
- PN, i.e. front-right edge and front-right-bottom corner, goes F
- PA, i.e. top-right edge and top-right-back corner, goes R'
- PB, i.e. front-bottom edge, goes F'R
- the front-right-top corner goes U'
- the front-bottom-left goes D
General case
[edit]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
[edit]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
[edit]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 PB.
Pieces affected by three elements
[edit]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 PN 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 PA.
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 PC to express that they are affected by conjugates.
Pieces affected by all four elements
[edit]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 PQ.
Summary of sets of pieces
[edit]In all cases, pieces that are moved by the commutator are in one of these sets:
- PA : pieces moved into the intersection by A and then hidden by B
- PB : pieces moved into the intersection by B and then hidden by A
- PN : pieces hidden by A and then moved out of the intersection of B'
- PC : pieces affected by a conjugate, in all inplace cases and some overlap cases
- PQ : pieces moved by all elements of the sequence of the commutator, in complex inplace cases and some overlap case
Classification
[edit]If PA is not empty, then B can hide pieces, thus PB is not empty either. Conversely, if PB is not empty, then PA is not empty.
If PN is not empty, then A and B can hide pieces, thus PA and PB are not empty. Conversely if PA is not empty, it means that pieces are brought into the intersection so that pieces are brought out of the intersection by A, thus PN is not empty.
So either the three sets PA, PB and PN 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 PC is not empty.
If the sets PA, PB and PN are not empty, we cannot deduce anything about PQ or PC. If it is the no overlap case, then pieces are hidden either by A or B, so PQ is empty. Conversely, if PQ 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