AvG questions (5x5)

[Zaxef]

1. when you're doing 2 at a time - What do you do when the "second working wing"** is the same color as your Front Left buffer center and [upper]wing? (**The one that starts on the front right bottom - then you do the d move [which moves the second working wing to Back Right Bottom] to match the first working wing [which was at the front left bottom] with it's center - but then I can't put the correct center in Front-Right in order to match the second working wing with it's center when I do d' ..because it's the buffer center (on Front Left))

I hope that makes sense :/

2. I can't seem to follow along with Arnaud's examples in his thread.. So is there a written or video explanation of his 3 special cases?

 

[tp21]

1. when you're doing 2 at a time - What do you do when the "second working wing"** is the same color as your Front Left buffer center and [upper]wing? (**The one that starts on the front right bottom - then you do the d move [which moves the second working wing to Back Right Bottom] to match the first working wing [which was at the front left bottom] with it's center - but then I can't put the correct center in Front-Right in order to match the second working wing with it's center when I do d' ..because it's the buffer center (on Front Left))

I hope that makes sense :/

2. I can't seem to follow along with Arnaud's examples in his thread.. So is there a written or video explanation of his 3 special cases?

Beginner
Beginner (117 moves used): Pre-setup to Up-Back, put at Front-Right using (F' U2 F) or (R U R')

Red-White: D B2, (F' U2 F)
Green-Red: B, d (R U R') d'

White-Red: U2, (F' U2 F)
White-Green: B, d (F' U2 F) d'

Orange-Yellow: D2 B2, (F' U2 F)
Blue-Yellow: U2, d (R U R') d'

Green-Yellow: U', (R U R')
Yellow-Blue: U2, d (F' U2 F) d'

Green-White: U, (R U R')
Easy piece (will not match yet) Yellow-Green: d (R U R') d' (special case "Parity before the end")

Yellow-Green: U', (R U R')
Red-Yellow: D B2, d (F' U2 F) d'

Blue-Orange: D2 B2, (R U R')
Yellow-Red: U2, d (F' U2 F) d'

Orange-Green: U, (F' U2 F)
Green-Orange: d (F' U2 F) (R U R') d' (special case "flipped wing")
Red-Blue: x' y (special case "full 3-cyle")

Yellow-Orange: D B2, (R U R')
Easy piece (will not match yet) Red-Green:U d (F' U2 F) d' (special case "Parity before the end")

White-Orange: B' (F' U2 F)
Red-Green: d (F' U2 F) d'
Orange-Blue: y2 z' (special case "full 3-cycle")

Blue-White: U2 (F' U2 F)
White-Blue: d (F' U2 F) (R U R') d' (special cases "flipped wing" and "full 3-cycle")
Orange-White: (special case "full 3-cycle")

What step are you talking about?

 

[Ellis]

1. I just switch out for any other edge, the first edge group I see that has at least one unsolved wing. There isn't much else you can do if you're slicing from an already made semi-pair... unless you wanna slice the opposite way then replace with a semi-pair then do d2 to pair the next edge. ocrap I hadn't really though about that *goes to practice 5x5*

2. 3 special cases?

 

[Zaxef]

1. when you're doing 2 at a time - What do you do when the "second working wing"** is the same color as your Front Left buffer center and [upper]wing? (**The one that starts on the front right bottom - then you do the d move [which moves the second working wing to Back Right Bottom] to match the first working wing [which was at the front left bottom] with it's center - but then I can't put the correct center in Front-Right in order to match the second working wing with it's center when I do d' ..because it's the buffer center (on Front Left))

I hope that makes sense :/

2. I can't seem to follow along with Arnaud's examples in his thread.. So is there a written or video explanation of his 3 special cases?

Beginner
Blah Blah Blah
What step are you talking about?

I didn't do Beginner anyway - Setting it up at Back-Up was just annoying.
I just did the moves and I wasn't seeing the colors he was.. idk..
Doesn't really matter - I understand the basic concept, it's just this one special case I'm asking about

1. I just switch out for any other edge, the first edge group I see that has at least one unsolved wing. There isn't much else you can do if you're slicing from an already made semi-pair... unless you wanna slice the opposite way then replace with a semi-pair then do d2 to pair the next edge. ocrap I hadn't really though about that *goes to practice 5x5*

2. 3 special cases?

3 Special cases being "Parity before the end - Which I'm assuming means a 4x4 OLL type parity, where the center is flipped (or two wings are flipped - whichever way you look at it) and you slice the center over, flip the whole edge group and slice back.. is that what it is?

Flipped Wing - Is that with one wing and center paired but the wing is flipped.. so you slice over the other wing in the place of the flipped one, flip the entire edge group and slice back the original wing?

Full 3-cycle - I have no idea what that is, it has no turns to "fix" it - So I'm assuming it's like .. you got lucky and a whole edge group got completed when you sliced?

Edit: And thank you, that makes sense.. I could just slice backwards
The problem with just switching it out is that it might just come back.. Since the buffer edge group needs to be solved eventually right..?
But if I slice backwards then I also need to take the time to insert a new buffer edge group.. Which can be time consuming and mess up my look ahead .. uhg -.-

I suppose the fastest thing to do would be to slice backwards, and put an edge group in the buffer's position and slice back to solve it, and hope there's only one solved wing on it - Then just flip it and use it as the buffer

 

[Ellis]

3 Special cases being "Parity before the end - Which I'm assuming means a 4x4 OLL type parity, where the center is flipped (or two wings are flipped - whichever way you look at it) and you slice the center over, flip the whole edge group and slice back.. is that what it is?

Ok well you said that you weren't able to follow along with Arnuad's example, which I take to mean you don't understand what's happening. You don't really need to understand the parity case. Just learn the alg for it and you're set. There's only one case, where if you slice either way with d, you'll solve a group, but don't have anything unsolved to switch that group out with. This should always be done at the very end.

Flipped Wing - Is that with one wing and center paired but the wing is flipped.. so you slice over the other wing in the place of the flipped one, flip the entire edge group and slice back the original wing?

Yea. You don't really need to know what's going on here either although it's pretty straightforward. If you're having a hard time with it just learn the flip alg for it. It will become intuitive eventually anyway.

Full 3-cycle - I have no idea what that is, it has no turns to "fix" it - So I'm assuming it's like .. you got lucky and a whole edge group got completed when you sliced?

Everytime you solve 2 wings you've done a 3 cycle. So a full 3 cycle would be where the buffer gets solved by chance. When this happens, just rotate the cube so that you have a new buffer, preferably one that is already part of a made semi-pair.

The problem with just switching it out is that it might just come back.. Since the buffer edge group needs to be solved eventually right..?

I'm not sure what you mean here. You can take your choice when you come to a case like this. Solve 2 edges, or just one. I normally do just one, but I think I might switch to this backwards slice thing. Of course the buffer will be solved eventually, but that's kind of irrelevant because once you switch the buffer out, it's just a normal edge group... so when it comes back, it's just like any other edge.

But if I slice backwards then I also need to take the time to insert a new buffer edge group.. Which can be time consuming and mess up my look ahead .. uhg -.-

Again, you can take your choice of whatever feels best. Yea it will take a second to replace the buffer, but you're solving an extra edge that you normally wouldn't be. But honestly, I don't think finding a new buffer would take any longer than finding another random unsolved wing to insert if you were to do it the other way. Both times you are just looking for the first semi-pair you can find. I don't think it would be all that bad on your look ahead.

I suppose the fastest thing to do would be to slice backwards, and put an edge group in the buffer's position and slice back to solve it, and hope there's only one solved wing on it - Then just flip it and use it as the buffer

I don't understand what you mean here.

 

[Zaxef]

Ok well you said that you weren't able to follow along with Arnuad's example, which I take to mean you don't understand what's happening. You don't really need to understand the parity case. Just learn the alg for it and you're set. There's only one case, where if you slice either way with d, you'll solve a group, but don't have anything unsolved to switch that group out with. This should always be done at the very end.

I understand what's happening - I just end up not having the colors he's talking about or something, I dunno.. It's not really important, I've got the method down pretty well (understanding-wise, just need speed now ), I just couldn't follow the examples for whatever reason.. I'm much better off having someone show me in a video or real life >_> (and I did watch Erik's video which helped a lot more than the examples - minus the fact that he slices with r and not d.. but it's still the same concept)

I suppose the fastest thing to do would be to slice backwards, and put an edge group in the buffer's position and slice back to solve it, and hope there's only one solved wing on it - Then just flip it and use it as the buffer

I don't understand what you mean here.

I meant if you have a wing edge on front right bottom that matches with your buffer on front left .. You slice d', then insert the correct center-edge into the buffer's position, in order to get a solved semi-pair when you slice back (d) ... And I was just saying it would be perfect if when you sliced back, the newly paired wing would only have one solved wing, so that it could be flipped and used as the buffer, so you wouldn't have to do any cube rotations or go looking for another semi-pair to use as the buffer.

Anyway, thank you.. I get it now

 

[tp21]

http://www.youtube.com/watch?v=uFteIaMBOYQ

that's probably the only other example of avg.

 

[AvGalen]

The "slice back with d'" is probably the solution to his issue. I mention that and describe the 3 special cases in the written examples. Normally you do "2 at a time" in about 8 moves. With the "slice back with d" you do "3 at a time" in about 12 moves so you don't waste moves or even look-ahead.

"full 3-cycle" isn't the same as "3 at a time after slice back with d". A "full 3-cycle" is just getting lucky when your buffer gets solved. This happens quite often in a solve and happens on 50% of the solves (the non-parity ones) at the end of the pairing

 

[Zaxef]

With the "slice back with d" you do "3 at a time" in about 12 moves so you don't waste moves or even look-ahead.

How would it be 3 at a time?

Edit: I think I know what you mean - I slice d' which solves the entire buffer edge group, and when slicing back I put the correct center in the buffer's spot so when I slice d I form another pair.. but I can also insert a center at Front Right that corresponds to the bottom wing on the edge group that was inserted at Front Left in the place of the buffer edge before the d slice.. right?

The only problem with this though.. what if I have a fully solved edge group at Back Right - Since I tend to store the first two solved edge groups at Back Left and Back Right to make things easier for look ahead and I think it cuts down moves.. One of it's wings will end up at Front Right when I slice d backwards, and that'll mess things up and defeat the purpose of trying to do 3 at a time..

Is that a bad idea then?