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[Help Thread] Intuitive Commutators

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So I'm trying to wrap my head around commutators. I understand them but I mainly see a lot of lists of hundreds of them. I feel like there must be a way to intuitively figure one out while solving. For example if I have a Case where I need to swap the Dfr with the uBl and that with Ufl (Capital letters being the orientation of where each sticker needs to go... actually I don't think I actually know how you write orientation with three letters... do I just put that letter first?... anyway. I can see what needs to be done but I don't know a commutator to fix it. is this something that can be just figured out intuitively?
 

mark49152

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Yes you can figure comms out intuitively, and as with F2L, it's the best way to learn them. However, if you want to get fast with them, eventually you have to get comfortable enough that you don't need to think about it any more. You have to just know each case/comm and be able to perform it automatically without delay.
 

Cale S

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If the cycle is UBL to DFR to UFL, the UFL and DFR stickers are interchangeable with an F2. The third sticker, UBL, can be inserted into UFL without disturbing the F layer with L' B' L. For this direction of the cycle, the F2 interchange needs to be done first so that UBL is inserted into DFR, so the commutator is F2 L' B' L F2 L' B L. You can also use D B2 D' to insert UBL into DFR, and in that case you would do the insertion first.
 

mDiPalma

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one way to solve that case is U B U' F2 U B' U' F2

which is basically A B A' B' where A = U B U' and B = F2

A inserts BUL to UFL and B interchanges UFL and DFR

does that make sense?
 
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Ok. This makes sense. I'm not too worried about speed excuse it's mainly for FMC stuff. So it's mainly seeing how you can swap pieces without the moves of the A on a commutatot changing anything in the B except for the corner piece?
 

mDiPalma

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Yeah, A has 2 main effects:
1) moves a desired sticker to your target sticker (in your case BUL to UFL)
2) screws up everything on the cube except for the interchange layer

and B has 1 main effect:
1) interchanges your desired sticker for another sticker (in your case DFR and UFL)

so when you do A, you put BUL in UFL but destroy the rest of the cube in a known way
then you do B, swapping what is now BUL (used to be UFL) with DFR
so that now you can do A' which will swap DFR and UFL while undoing all the garbage you did to the cube
finally B' will swap back UFL and BUL

just try to identify the insertion (A) and the interchange (B) and the order you need to do them in in order to solve your desired cycle
 

Kit Clement

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For FMC purposes, you'll really only need to learn 8-movers for corners. There was only one instance I've ever been able to use an 9-move comm and have it be the optimal choice.

Say you have an L3C skeleton: the easiest way to find 8-movers is to simply check all possible interchanges between pairs of stickers, then see if the third sticker is on the layer opposing the interchange layer AND will have a 3 move insert into one of the two stickers on the interchange layer.
 
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biscuit

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For FMC purposes, you'll really only need to learn 8-movers for corners. There was only one instance I've ever been able to use an 9-move comm and have it be the optimal choice.

Say you have an L3C skeleton: the easiest way to find 8-movers is to simply check all possible interchanges between pairs of stickers, then see if the third sticker is on the layer opposing the interchange layer AND will have a 3 move insert into the interchange layer.

Would you ever do a Comm with setup moves?
 

Kit Clement

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Would you ever do a Comm with setup moves?

I have done it once, and it was when I did that 9-move commutator I mentioned (which was an 8-mover with a cancelled setup) that cancelled 2 moves, when no 8-mover on the skeleton cancelled any.

But 99% of the time, absolutely not. It's extremely rare for pure corner comms to be outperformed by anything longer.
 

biscuit

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I have done it once, and it was when I did that 9-move commutator I mentioned (which was an 8-mover with a cancelled setup) that cancelled 2 moves, when no 8-mover on the skeleton cancelled any.

But 99% of the time, absolutely not. It's extremely rare for pure corner comms to be outperformed by anything longer.

So what about the case of getting to L3C with a comm that doesn't have any 1 move interchanges? This is more of a general comm question, as I'm not sure if you always have to do a setup move in this case, or if there are other approaches.
 

Kit Clement

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So what about the case of getting to L3C with a comm that doesn't have any 1 move interchanges? This is more of a general comm question, as I'm not sure if you always have to do a setup move in this case, or if there are other approaches.

An entire L3C skeleton with no 8 move comms at any single point is so incredibly rare that if you only learn pure comms for FMC, by the time a case like this comes up, you'll probably have done so many insertions that you'll have gained the intuition to do a 9 move commutator wherever one exists in the skeleton, maybe even cancelling a move. And then you'll have every right to whine about how unlucky your insertion luck is to everyone.
 

Cale S

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So what about the case of getting to L3C with a comm that doesn't have any 1 move interchanges? This is more of a general comm question, as I'm not sure if you always have to do a setup move in this case, or if there are other approaches.

Those cases are 9-12 moves optimal, so you have to do a setup move to an 8 mover or do a longer insertion (R2 U R2 U' R2 instead of R U R' for example)
 

AlphaSheep

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I managed to convince myself it's guaranteed that a 10-mover won't cancel any moves if no 8-movers exist.

I'll try to explain my logic.

I started out trying to work out if conjugates are ever worth it. The first move of a conjugated comm will always be the same type as the last move (inverses of eachother for a 10-mover or e.g. R and R2 for a 9-mover), so if the comm can cancel from both sides then the moves in the skeleton would already cancel. So a 9-mover or more will only ever be able to cancel on one side. Also since it has to cancel at least 2*(number of setup moves) to match the effectiveness 8-mover, the setup has to completely cancel with the adjacent moves of the skeleton (i.e. the setup must be the inverse of the adjacent moves in the skeleton). For 10-movers, if this is the case, there is always an 8-mover that does the same job.
Imagine the sketon with an insertion X X X A' (A: comm) B X X X. Instead of inserting the 10-mover, you can insert the same comm without the setup one move earlier: X X X (comm) A' B X X X, which is exactly equivalent, so looking for 10+ movers is never worth it.

The same argument applies to 9-movers on the side where the setup doesn't cancel into the comm. On the side where it does, things are a little different. You can occasionally have decent cancelations with a 9-mover, e.g with the following: X X X A2 (A [A B A' C A B' A' C'] A') X X X, you can cancel 2 moves which you would not have been able to cancel with an 8-mover. But since you know that the setup has to cancel both the adjacent move on the comm and the adjacent move of the skeleton, it gives a very easy criteria for filtering the sorts of 9-movers you try insert.

Of course, if you go through your entire skeleton and there are no 8-movers at all, you may have to settle for something longer that doesn't cancel. That's really rare though.

Of course this logic doesn't apply to pure comms that are longer because of longer interchanges like Cale mentioned above.
 

Cale S

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yeah 9 movers are rarely useful in FMC because if they cancel moves, they usually become an 8 mover just by moving that way in the skeleton.

But one time I had an FMC solve where a 12 move corner comm was optimal, the only 8 movers didn't cancel moves and the 12 mover canceled 4
 
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yeah 9 movers are rarely useful in FMC because if they cancel moves, they usually become an 8 mover just by moving that way in the skeleton.

But one time I had an FMC solve where a 12 move corner comm was optimal, the only 8 movers didn't cancel moves and the 12 mover canceled 4

So did you just memorize a bunch of them or is it an intuitive process for a lot of them?


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newtonbase

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So did you just memorize a bunch of them or is it an intuitive process for a lot of them?


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I'm working on these myself but I don't know the best way to learn. One suggestion is to pick a letter and work out every comm that includes it. At the moment I'm mainly working through corners only scrambles and looking up any I can't figure out. Also, in quiet moments at work I'll pick 2 random letters and try to write down the best one I can think of. It would be good to know how others learnt.
 
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I'm working on these myself but I don't know the best way to learn. One suggestion is to pick a letter and work out every comm that includes it. At the moment I'm mainly working through corners only scrambles and looking up any I can't figure out. Also, in quiet moments at work I'll pick 2 random letters and try to write down the best one I can think of. It would be good to know how others learnt.

Where would you suggest a list of said commutators?


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deadmanlsh

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Use this list to figure out how to solve most, if not all, cases.
Learn to recognise cases from different angles and reference stickers (if your buffer is UBR, try tracing the cycle from BUR and RUB) if you are using them for BLD.
You might want to consider solving some of them non-optimally to make them more finger-trick friendly to solve.

(The list may not be exhaustive)

The following is a list of all possible unique 3-cycles up to reflection, inversion and rotation.
Therefore, for cases with multiple algorithms, they may not solve the same exact case.
X', where X represents a move/rotation, shall be denoted as X3. (I wrote this on my phone and typing 3 was much more convenient then)

8-movers:
Pures:

Simple Long
(L3U2L)D2(L3U2L)D2
y(RUR3)D2(RU3R3)D2y3

Complex Long
(LU3L3)D2(LUL3)D2

Simple Short
(U3L3U)R(U3L3U)R3

Complex Short
x3(RUR3)D(RU3R3)D3x
x3(L3UL)D3(L3U3L)Dx

Near:
(L3U2L)D(L3U2L)D3

Far:
y(L3UL)D(L3U3L)D3y'
(L3U2L)D3(L3U2L)D

9-movers:
A9s:

A Perm
R2 B2 (R F R3) B2 (R F3 R)

OLL
R2 D (R3 U2 R) D3 (R3 U2 R3)

ZBLL
L3 B2 (L3 F2 L) B2 (L3 F2 L2)

Column-like
D2((L2D)R2D3(L2DR2))D

10-movers:
Cyclic Shifts:

Standard
(FR3)U2(RF3R3F)U2(F3R)

Orthogonals:

Axes & Circles
F((L3U2L)D(L3U2L)D3)F3

Shifted Normal
F3((R3D2R)U(R3D2R)U3))F

11-movers:
Columns:
F3(L2((BL)F2(L3B3L)F2)L)F

12-movers:
Per Specials:
(UF2U3F2U3)R2(UF2UF2U3)R2
 
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