VHF2L+2 Levels=F2l+OLL Skip

[miniGOINGS]

Isn't the exact chance of an OLL skip? Ex 1/x for an OLL skip (I forget) = x number of algs needed.

Not at all. By that logic you have a 1/21 chance of getting a PLL skip, which doesn't even remotely make sense.

Nonononono, you're thinking the other way around. I mean that if the chance of getting an OLL skip is 1/500 (just a random number), you would need 500 algorithms to solve the last slot and OLL at the same time WV-ish.

 

[Lucas Garron]

Isn't the exact chance of an OLL skip? Ex 1/x for an OLL skip (I forget) = x number of algs needed.

Not at all. By that logic you have a 1/21 chance of getting a PLL skip, which doesn't even remotely make sense.

Nonononono, you're thinking the other way around. I mean that if the chance of getting an OLL skip is 1/500 (just a random number), you would need 500 algorithms to solve the last slot and OLL at the same time WV-ish.

What? Your first argument doesn't make sense/is false, and Google will easily tell you that 1/216 is the probability you wanted for your second "argument."

 

[Lucas Garron]

Mathematically 8(the VH algs)*54(WV algs)=432

Never underestimate the power of AUF.

(Sounds better if you imagine Darth Vader saying it.)

 

[miniGOINGS]

Lucas. If I remember correctly, the chance of a OCLL skip is 1/27. WV uses 27 algorithms. Therefore, by my assumption, the same would be for this. Maybe I'm missing something.

 

[BatafCube]

I don't see the advantage of this.

With the normal way: Look for the last f2l pair, solve it, look for the OLL, solve it. 2 looks.

With this way, look for the f2l, then START to solve it, but once you are half way through it, look for the oll case, then finish solving the f2l + oll. Its STILL 2 looks, but now you have to memorize hundreds of algs instead of 57.

Then why people learn ZBLL when you recognize it still in 2 looks(COLL,then EP).

 

[trying-to-speedcube...]

Isn't the exact chance of an OLL skip? Ex 1/x for an OLL skip (I forget) = x number of algs needed.

Not at all. By that logic you have a 1/21 chance of getting a PLL skip, which doesn't even remotely make sense.

Nonononono, you're thinking the other way around. I mean that if the chance of getting an OLL skip is 1/500 (just a random number), you would need 500 algorithms to solve the last slot and OLL at the same time WV-ish.

What? Your first argument doesn't make sense/is false, and Google will easily tell you that 1/216 is the probability you wanted for your second "argument."

The chance is 1/216, you're right. miniGOINGS probably didn't know the chance of an OLL skip so he took a random number. If AUF is impossible because a pair is in the U-layer and symmetry is thus excluded, the number of cases for inserting the pair and simultaneously solving OLL is indeed the number of F2L cases * the chance of getting an OLL skip^-1.

As another example, let's consider 1LLL. This is doing OLL and PLL in one step, so it's the number of OLL cases (57) * the chance of a PLL skip^-1 (72). This is just a simple calculation because I didn't include a calculation for symmetries of H-OLLs, but it's a rough estimation.

The other way around you could also say it's the number of PLL cases (21) * the chance of an OLL skip^-1 (216). When you incalculate the symmetries of the E-perm, H-perm and Z-perm, you should get the same number.

 

[Ness]

I don't see the advantage of this.

With the normal way: Look for the last f2l pair, solve it, look for the OLL, solve it. 2 looks.

With this way, look for the f2l, then START to solve it, but once you are half way through it, look for the oll case, then finish solving the f2l + oll. Its STILL 2 looks, but now you have to memorize hundreds of algs instead of 57.

I don't see the advantage either.. With this new ending of fridrich you execute 2 algs (while normal fridrich has OLL+PLL). For both cases you need to learn PLL, but for the new one the number of the second algorithm set is so much higher. I can't imagine how recognition could be easiser than for OLL. So even if you save moves, I don't think you save any time at all.
I think it's NOT worth the effort.

 

[miniGOINGS]

The chance is 1/216, you're right. miniGOINGS probably didn't know the chance of an OLL skip so he took a random number.

Yea, I should have looked that up.

If AUF is impossible because a pair is in the U-layer and symmetry is thus excluded, the number of cases for inserting the pair and simultaneously solving OLL is indeed the number of F2L cases * the chance of getting an OLL skip^-1.

So I was right?

 

[Zarxrax]

I don't see the advantage of this.

With the normal way: Look for the last f2l pair, solve it, look for the OLL, solve it. 2 looks.

With this way, look for the f2l, then START to solve it, but once you are half way through it, look for the oll case, then finish solving the f2l + oll. Its STILL 2 looks, but now you have to memorize hundreds of algs instead of 57.

Then why people learn ZBLL when you recognize it still in 2 looks(COLL,then EP).

ZBLL solves PLL as well, this doesn't So this method would require 3 looks to get to a solved state.
I don't see how this does anything except makes OLL have more algs.

 

[jms_gears1]

it solves the F2L and OLL at the same time,
i would say that it would reduce the LL to one look (PLL)

also it would only take one look to insert the F2L into the slot.

however its still a waste of time.

you can EO the edges with a pre-made pair, then use WV. so why learn several hundred algs when you can do the same with ~27

 

[miniGOINGS]

it solves the F2L and OLL at the same time,
i would say that it would reduce the LL to one look (PLL)

also it would only take one look to insert the F2L into the slot.

Nope, it would be one look to make the pair, one to insert it and orient the last layer, and one to permute the last layer. Three looks.

 

[jms_gears1]

it solves the F2L and OLL at the same time,
i would say that it would reduce the LL to one look (PLL)

also it would only take one look to insert the F2L into the slot.

Nope, it would be one look to make the pair, one to insert it and orient the last layer, and one to permute the last layer. Three looks.

ahhhh makes sense,
when im using intuitive VH (just when the pairs together) it generally takes me one look to see how to solve the pair with edge EO.

still i think EO then WV would be much easier for recog.

 

[miniGOINGS]

still i think EO then WV would be much easier for recog.

Even if you did EO + making the pair, and then inserting the pair + CO, then PLL, I still don't think it would be too effective.

 

[Lucas Garron]

still i think EO then WV would be much easier for recog.

Even if you did EO + making the pair, and then inserting the pair + CO, then PLL, I still don't think it would be too effective.

I have an idea! Why don't you do + inserting the edge, then CO + inserting the corner, then PLL?

 

[miniGOINGS]

I have an idea! Why don't you do EO + inserting the edge, then CO + inserting the corner, then PLL?

Fixed.

 

[Cyrus C.]

still i think EO then WV would be much easier for recog.

Even if you did EO + making the pair, and then inserting the pair + CO, then PLL, I still don't think it would be too effective.

I have an idea! Why don't you do + inserting the edge, then CO + inserting the corner, then PLL?

Genius! Why hasn't this been thought of before? Sarcasm.

 

[jms_gears1]

still i think EO then WV would be much easier for recog.

Even if you did EO + making the pair, and then inserting the pair + CO, then PLL, I still don't think it would be too effective.

I have an idea! Why don't you do + inserting the edge, then CO + inserting the corner, then PLL?

i think we could call it hmm,

LGAI
(Lucas Garron's Awsome Idea)

 

[Yi Ren]

I don't really understand "VH-F2L+2 levels",is it F2LL?

 

[rubiknewbie]

There are only 100 slots in my brain for remembering algorithms .

 

[xXzaKerXx]

it's cool that you figured it out, but no offence, personally i think it'll be better to learn full ZB instead because recognition is easier for ZBLL than PLL, isn't it?