# Corner Comm Help



## kbrune (Oct 13, 2016)

I'm trying to learn to figure out my corner comms intuitively the way Noah suggests in his comm video. ULB is my buffer. Then use RFD, and solve for each 18 positions that would form each corner cycle with it. 

So when it comes to A9 type comms. I've had some success coming up with a few based on existing comms from other positions. But only using visual cues. I wouldn't be able to visualize the same comm intuitively blind.

Can someone explain why an A9 comm has a cancelation in the alg. This would help me understand them better. This way I could figure them out without seeing the colours.

Example. If you set up a ULB - FRU - RFD
cycle. I fugue out that a comm for this would be. D' F2 D' B D F2 D' B' D2

so I tried to write that down in the A B A' B' format to see if I could spot the cancelation move. But I don't see it. Here's what I get.

A (D' F2 D') insertion move
B (B) interchange
A' (D F2 D) undo insertion move
B' (B') undo interchange

When you perform it that way. It doesn't work. What am I missing?


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## Martial (Oct 13, 2016)

D'F2D' is not an insertion move. With your alg, the setup is D', insertion is D'BD, interchange is F2.
D'[F2,D'BD]D. Cancelation is at the end, when you undo the insertion.

With A9 alg the cancelation move occurs between setup+insertion, or de-insertion+de-setup, depending on the orientation of your comm (ULB - FRU - RFD or ULB - RFD - FRU).


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## SpeedCuber71 (Oct 13, 2016)

Also, try understanding how the comms work. The first D' made the two pieces (not including the buffer) interchangeable with an F2 and as the last move was D and the setup undo was D too, it becomes D2.


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## kbrune (Oct 13, 2016)

Ah ok. I see where my mistake is when converting to ABA'B'

I'm still having a very difficult time seeing what the insertion move has to be and and I also have no clue if the cycle I'm looking at is a pure comm, an A9 or any of the others. Which is my problem for being able to see set ups to pure comms.

I'm currently stumped on this cycle.

ULB URF RFD

I figured R' would be a good setup to do F2 L F2 insertion move. Then D' to solve second piece. But it doesn't work. Havin a hard time wrapping my Brian around comms. 

any advice on what to look for first in determining if a setup is doable or if it's a difficult case?


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## SpeedCuber71 (Oct 13, 2016)

kbrune said:


> Ah ok. I see where my mistake is when converting to ABA'B'
> 
> I'm still having a very difficult time seeing what the insertion move has to be and and I also have no clue if the cycle I'm looking at is a pure comm, an A9 or any of the others. Which is my problem for being able to see set ups to pure comms.
> 
> ...



I think A9 comms are mostly ones in which no interchange exists so it has to be created using a set up. Also, there are sources online where you can find a list of commutators for every possible cycle on the 3x3 with your buffer. Any time you are stumped, try looking it up. Hope i helped.


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## kbrune (Oct 13, 2016)

I have a list with URB as the buffer. Keeping the Y rotation in mind only complicates things slightly. But it helps. 

I found one list ULB as buffer. But it was a table and I didn't understand how to use it.


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## SpeedCuber71 (Oct 13, 2016)

kbrune said:


> I have a list with URB as the buffer. Keeping the Y rotation in mind only complicates things slightly. But it helps.
> 
> I found one list ULB as buffer. But it was a table and I didn't understand how to use it.



Check this out. You can even add your own letter scheme here. http://wong-tim.com/#/


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## Cale S (Oct 13, 2016)

kbrune said:


> I'm currently stumped on this cycle.
> 
> ULB URF RFD
> 
> I figured R' would be a good setup to do F2 L F2 insertion move. Then D' to solve second piece. But it doesn't work.



For this cycle there's a U2 interchange between UBL and UFR, and R' D' R inserts RFD into UFR. For this direction of the cycle, you need to cycle RFD to UBL instead, so you start with the U2 interchange and have U2 R' D' R U2 R' D R. You can also use B D2 B' to insert RFD to UBL, and in this case you would start with the insertion: B D2 B' U2 B D2 B' U2


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## Martial (Oct 13, 2016)

To solve with a commutator, first check if two pieces are interchangeable (you can put one on another in one move). Then look if the third piece is moved during the interchange, if yes then you will have to setup this piece, if no then you just have to find a way to insert this piece at the slot of the 1st or 2nd one *without moving pieces affected during interchange*.


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## kbrune (Oct 14, 2016)

Martial said:


> To solve with a commutator, first check if two pieces are interchangeable (you can put one on another in one move). Then look if the third piece is moved during the interchange, if yes then you will have to setup this piece, if no then you just have to find a way to insert this piece at the slot of the 1st or 2nd one *without moving pieces affected during interchange*.



When looking for 2 interchangeable pieces. does one have to be solved by that one move? Or it doesn't matter if either of the 2 end up in their solved position or not?


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## SpeedCuber71 (Oct 14, 2016)

kbrune said:


> When looking for 2 interchangeable pieces. does one have to be solved by that one move? Or it doesn't matter if either of the 2 end up in their solved position or not?



I didn't understand your question. Could you rephrase please? I will do my best to help.


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## kbrune (Oct 14, 2016)

Sure. I'm trying to understand what to look for when figuring out a new case. 

So martial says that you need to find 2 corners that are interchangeable without touching the buffer. If the buffer is moved. You need setup moves. (That's how I understood).

So my question was. Once you find 2 corners that can be switched with like and R2 for example. Does it count as a good interchange. Only If one of the 2 pieces goes to its solved position? Or as long as they are interchangeable by one move is good enough to continue constructing the comm? 

Is that better?


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## sqAree (Oct 14, 2016)

kbrune said:


> Sure. I'm trying to understand what to look for when figuring out a new cycle.
> 
> So martial says that you need to find 2 corners that are interchangeable without touching the buffer. If the buffer is moved. You need setup moves. (That's how I understood).
> 
> ...



If 2 corners are interchangeable, either the interchange or the inverse will put one corner in its solved position.

My supposition is that you misunderstood the term "interchangeable". With that, you have to look at a sticker-3-cycle, not a piece-3-cycle. In that case, yes, you're right, it only works when one the pieces goes into its solved position (= the sticker goes to its according sticker).


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## AlphaSheep (Oct 15, 2016)

kbrune said:


> I figured R' would be a good setup to do F2 L F2 insertion move. Then D' to solve second piece. But it doesn't work. Havin a hard time wrapping my Brian around comms.



I don't use comms for BLD but I use them for FMC so I understand them well. Just to explain why F2 L F2 or similar will never be a suitable insertion... for a 3 cycle, the set of pieces affected by your insertion move and the set of pieces affected by your interchange move must have exactly one piece in common. So your interchange D affects the 4 edges and 4 corners on the D layer, but the insertion F2 L F2 affects 2 corners and an edge in the D layer. Because it affects more than one piece on every face, F2 L F2 isn't suitable for any single move interchange.

In general, I look for the interchange move first, then the insert is a quarter turn to pull a single piece from the interchange layer to the opposite layer, any move of the opposite layer, then undo the first quarter turn. Since half turns move three pieces from the interchange layer into the opposite layer, they don't work.


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## newtonbase (Oct 15, 2016)

AlphaSheep said:


> I don't use comms for BLD but I use them for FMC so I understand them well. Just to explain why F2 L F2 or similar will never be a suitable insertion... for a 3 cycle, the set of pieces affected by your insertion move and the set of pieces affected by your interchange move must have exactly one piece in common. So your interchange D affects the 4 edges and 4 corners on the D layer, but the insertion F2 L F2 affects 2 corners and an edge in the D layer. Because it affects more than one piece on every face, F2 L F2 isn't suitable for any single move interchange.
> 
> In general, I look for the interchange move first, then the insert is a quarter turn to pull a single piece from the interchange layer to the opposite layer, any move of the opposite layer, then undo the first quarter turn. Since half turns move three pieces from the interchange layer into the opposite layer, they don't work.


That's probably the most useful post I've ever read on comms.


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## newtonbase (Oct 15, 2016)

Could someone please explain how this comm works please? WB in Speffz 
[R2 D R2 D’ R2 , U]


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## Daniel Lin (Oct 15, 2016)

newtonbase said:


> Could someone please explain how this comm works please? WB in Speffz
> [R2 D R2 D’ R2 , U]


R2 D R2 D’ R2 is the insertion, U is the interchange


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## newtonbase (Oct 15, 2016)

Daniel Lin said:


> R2 D R2 D’ R2 is the insertion, U is the interchange


It's the interchange that's confusing me. They are usually 3 moves.


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## xchippy (Oct 15, 2016)

newtonbase said:


> It's the interchange that's confusing me. They are usually 3 moves.


Interchanges are usually 1 move. Like in [R U' R' , D] (UFL to FDR to LDF). If you're talking about the insertion, which is 3 moves in a pure commutator, it works in the same way, but since there's no way to get DBR to UBR in 3 moves without affecting any D layer pieces other than DBR, you can't solve it using a pure commutator. This commutator uses a 5 move insertion, which I think is called a per special? idk what it's called


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## AlphaSheep (Oct 15, 2016)

newtonbase said:


> It's the interchange that's confusing me. They are usually 3 moves.


You need to insert DBR into UBR while preserving orientation. No quarter turn move taking a D cormer into U can preserve corner orientation, but if you do an R2 setup just for the insertion, then you can insert with D R2 D'. Notice its still a quarter turn pulling the old piece out and putting the new piece in. Also, notice how R2 D R2 D' R2 only affects one piece in the U layer.

This trick is useful for all of the cases where all 3 stickers are on opposite faces. Its very hard to see the setups for the cases which need 3+ move setups to get to an 8 mover so it's easier to learn these ones as algs then figure out how they work afterwards.


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## newtonbase (Oct 15, 2016)

xchippy said:


> If you're talking about the insertion,



Yes, sorry. Silly mistake. 



AlphaSheep said:


> You need to insert DBR into UBR while preserving orientation. No quarter turn move taking a D cormer into U can preserve corner orientation, but if you do an R2 setup just for the insertion, then you can insert with D R2 D'. Notice its still a quarter turn pulling the old piece out and putting the new piece in. Also, notice how R2 D R2 D' R2 only affects one piece in the U layer.
> 
> This trick is useful for all of the cases where all 3 stickers are on opposite faces. Its very hard to see the setups for the cases which need 3+ move setups to get to an 8 mover so it's easier to learn these ones as algs then figure out how they work afterwards.


These do look very useful. 
Thanks for the help guys.


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## Daniel Lin (Oct 15, 2016)

AlphaSheep said:


> 3+ move setups


no cases require more than 3 setup moves



newtonbase said:


> It's the interchange that's confusing me. They are usually 3 moves.


Insertions can be any number of moves, but the shorter the better. I personally use 3 movers for 100% of my corner comms

Here is a comm with a 4 move insertion
[R' F:[U2, F' R F R']]

I would do that case like this [z U D R:[R D' R', U2]]


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## mark49152 (Oct 15, 2016)

Daniel Lin said:


> Here is a comm with a 4 move insertion
> [R' F:[U2, F' R F R']]
> 
> I would do that case like this [z U D R:[R D' R', U2]]


Why not [R' F:[U2, R' D' R]]?


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