Last Slot method question?

[foxfan352]

I have been doing some research on the f2l last slot and all the the methods there is WV, Vh, Exc. This got me thinking if it would be better to learn one of these instead of full oll. So I came up with this.

1. Solve 3 f2l slot
2. Pair up the last f2l
3. Intuitively orient the last layer edges
4. Insert using partial edge control

This will give you an easy oll case(so only 2-look oll is necessary).This also increases your chances of an oll skip to(I don't really know what the probability is but I got 1/9, 1/12, or 1/48 these could be completely wrong though, they probably are, but hopefully they aren't).

So I got a few questions
Is this the same as step 3 or 4 in the petrus method.
Is this practical to use in fridrich method or it is only useful in the petrus method?

 

[dannyz0r]

Why would you use partial edge control if you already oriented the edges?

 

[Edward]

What? So, you're doing partial edge control right? If you're going to orient edges, try using MGLS.

 

[blakedacuber]

I have been doing some research on the f2l last slot and all the the methods there is WV, Vh, Exc. This got me thinking if it would be better to learn one of these instead of full oll. So I came up with this.

1. Solve 3 f2l slot
2. Pair up the last f2l
3. Intuitively orient the last layer edges
4. Insert using partial edge control

This will give you an easy oll case(so only 2-look oll is necessary).This also increases your chances of an oll skip to(I don't really know what the probability is but I got 1/9, 1/12, or 1/48 these could be completely wrong though, they probably are, but hopefully they aren't).

So I got a few questions
Is this the same as step 3 or 4 in the petrus method.
Is this practical to use in fridrich method or it is only useful in the petrus method?

It makes sense and probably could work, the only proble i can think of at the oment is that your last pair might brake up in the process but then again you should be able to solve it 2 gen

 

[foxfan352]

Why would you use partial edge control if you already oriented the edges?

I have been doing some research on the f2l last slot and all the the methods there is WV, Vh, Exc. This got me thinking if it would be better to learn one of these instead of full oll. So I came up with this.

1. Solve 3 f2l slot
2. Pair up the last f2l
3. Intuitively orient the last layer edges
4. Insert using partial edge control

This will give you an easy oll case(so only 2-look oll is necessary).This also increases your chances of an oll skip to(I don't really know what the probability is but I got 1/9, 1/12, or 1/48 these could be completely wrong though, they probably are, but hopefully they aren't).

So I got a few questions
Is this the same as step 3 or 4 in the petrus method.
Is this practical to use in fridrich method or it is only useful in the petrus method?

It makes sense and probably could work, the only proble i can think of at the oment is that your last pair might brake up in the process but then again you should be able to solve it 2 gen

You use partial edge control so you don't unorient any edges.

I figured out a way of doing it so you don't brake up the pair.

 

[miniGOINGS]

What? So, you're doing partial edge control right? If you're going to orient edges, try using MGLS.

Yea. If you want to skip part of OLL, why not just *skip* it all? MGLS is a 2 look method for LS + OLL.

 

[dannyz0r]

If you already made your last pair and oriented the edges then the insertion just becomes 2-gen and that doesn't undo what you did.

 

[foxfan352]

What? So, you're doing partial edge control right? If you're going to orient edges, try using MGLS.

Yea. If you want to skip part of OLL, why not just *skip* it all? MGLS is a 2 look method for LS + OLL.

If you already made your last pair and oriented the edges then the insertion just becomes 2-gen and that doesn't undo what you did.

I read here that there are more cases than oll

I believe that is what partial edge control is only using 2 gen moves to insert a f2l pair.

 

[cincyaviation]

What? So, you're doing partial edge control right? If you're going to orient edges, try using MGLS.

Yea. If you want to skip part of OLL, why not just *skip* it all? MGLS is a 2 look method for LS + OLL.

If you already made your last pair and oriented the edges then the insertion just becomes 2-gen and that doesn't undo what you did.

I read here that there are more cases than oll

I believe that is what partial edge control is only using 2 gen moves to insert a f2l pair.

its pretty much the exact opposite of that

 

[Edward]

Learn somethin son.

And this seatbelt, so you can be safe.

 

[foxfan352]

What? So, you're doing partial edge control right? If you're going to orient edges, try using MGLS.

Yea. If you want to skip part of OLL, why not just *skip* it all? MGLS is a 2 look method for LS + OLL.

If you already made your last pair and oriented the edges then the insertion just becomes 2-gen and that doesn't undo what you did.

I read here that there are more cases than oll

I believe that is what partial edge control is only using 2 gen moves to insert a f2l pair.

its pretty much the exact opposite of that

Learn somethin son.

And this seatbelt, so you can be safe.

What I mean is that when you use more than to sides to insert an f2l pair when all the edges are oriented it it will unorient 2 of the edges you had

 

[Rpotts]

What I mean is that when you use more than to sides to insert an f2l pair when all the edges are oriented it it will unorient 2 of the edges you had

right so the edge control you described is only for all edges oriented cases, so you're not controlling the edges at all. You just described using normal inserts and just happening to have all edges oriented.

 

[brunson]

You need OLL for any of those methods, because there is always a chance that you accidentally make your last corner pair while solving the others. LS skip means OLL.

 

[Sir E Brum]

The probability of an OLL skip with all edges oriented is 1/27.
4 corners that can be twisted
CCW (counterclockwise)= 1
CW (clockwise)= -1
O (oriented)= 0
(Permutation is irrelevant at this point)

The total of the numbers for corner twists must be 0, 3, or -3 for the cube to be solvable.
The first, second and third corners can be any of the 3.
This leaves you with a possible total from -3 to 3 inclusive.
For -1 and 2, the final corner must be CCW.
For -2 and 1, the final corner must be CW.
For -3, 0, and 3, the final corner must be O.

For an OLL skip we have all the corners oriented.
(0, 0, 0, 0)
For this to occur the first 3 must be O and this causes the last one to be O.
Each of the first 3 corners has a 1/3 chance of being oriented.
1/3 * 1/3 * 1/3 * 1/1 = 1/27

 

[foxfan352]

What I mean is that when you use more than to sides to insert an f2l pair when all the edges are oriented it it will unorient 2 of the edges you had

right so the edge control you described is only for all edges oriented cases, so you're not controlling the edges at all. You just described using normal inserts and just happening to have all edges oriented.

Yes the edge control is only for all edges oriented cases

If by normal inserts you mean having the pair in front face with the corner in the left face and doing R U' R'. But if it was on the corner was on the right than you could do F R' F' R than it would unorient the edges since you used another face beside the R U and L faces and if you were to rotate the cube to insert it "normally" I think it would still be doing partial edge control even though you didn't do it intentionally

So even though some of the moves or "normal" to some degree it is partial edge control.

You need OLL for any of those methods, because there is always a chance that you accidentally make your last corner pair while solving the others. LS skip means OLL.

I am pretty sure you can force not to get an LS skip if you have good look ahead in f2l.

The probability of an OLL skip with all edges oriented is 1/27.
4 corners that can be twisted
CCW (counterclockwise)= 1
CW (clockwise)= -1
O (oriented)= 0
(Permutation is irrelevant at this point)

The total of the numbers for corner twists must be 0, 3, or -3 for the cube to be solvable.
The first, second and third corners can be any of the 3.
This leaves you with a possible total from -3 to 3 inclusive.
For -1 and 2, the final corner must be CCW.
For -2 and 1, the final corner must be CW.
For -3, 0, and 3, the final corner must be O.

For an OLL skip we have all the corners oriented.
(0, 0, 0, 0)
For this to occur the first 3 must be O and this causes the last one to be O.
Each of the first 3 corners has a 1/3 chance of being oriented.
1/3 * 1/3 * 1/3 * 1/1 = 1/27

Thanks for clearing the that up for me I don't really know how to calculate the probability of the cube.

So using this brings the probability of an oll skip from 1/216 to 1/27

Also brunson's signature was right what were the chances of that happening:rolleyes:

 

[Ranzha]

I am pretty sure you can force not to get an LS skip if you have good look ahead in f2l.

Why would somebody speedsolving try to UNsolve something? For srs, CLL/ELL.

 

[foxfan352]

I am pretty sure you can force not to get an LS skip if you have good look ahead in f2l.

Why would somebody speedsolving try to UNsolve something? For srs, CLL/ELL.

I wouldn't be unsolving necessarily you could also simply do 2 look oll

and would you rather get and LS skip or an oll skip

 

[miniGOINGS]

and would you rather get and LS skip or an oll skip

You wouldn't be getting an OLL skip, you just wouldn't have to use OLL.

Just learn OLL first.

 

[foxfan352]

and would you rather get and LS skip or an oll skip

You wouldn't be getting an OLL skip, you just wouldn't have to use OLL.

Just learn OLL first.

but there would be an easy oll and a 1/27 you get a oll skip

what are the chances of getting an LS skip?

 

[miniGOINGS]

what are the chances of getting an LS skip?

CP: 1/5
CO: 1/3
EP: 1/5
EO: 1/2

LS: 1/150