Last Slot method question?

[foxfan352]

what are the chances of getting an LS skip?

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

LS: 1/150

1/27>1/150

 

[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.

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

What I meant by this was this:
Why would you force to have something NOT solved, when it is easy enough to have it solved? In other words, why would you destroy a noticed x-cross to just get cross? The movecount for cross should be 8 moves or less, HTM, every time. Then, another 4-6 for the first pair, optimally. This, at worst, forms a 14 move x-cross. I usually average 12 moves for cross + 1st pair. With a noticed x-cross, it solved the F2L pair WHILST solving the cross. Why ruin that?

what are the chances of getting an LS skip?

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

LS: 1/150

1/27>1/150

Consider the movecounts, and not the probabilities.
Any F2L slot should be...6 moves on average, IIRC.

I suspect OLL would have considerably more, on average. Optimally, that is. Although I don't know the actual figures, and do not know otherwise how to find this figure, if someone would be so kind as to post it, I'd greatly appreciate it.

The thing is, an OLL skip would be more valuable to the solver who does not know full OLL. A LS skip would be nice, but forcing an OLL skip in while forming the pair, implying you don't have an LS skip, would significantly influence the overall solve time.

From my personal experience, a usual LS takes about 1 through 1.5 seconds on average. OLL is usually about 3 - 4 seconds. Thus, the solve time would be faster with an OLL skip rather than a LS skip. Also, an LS skip wouldn’t really be noticeable as easily as an OLL skip would.

 

[rubiknewbie]

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

Aren't you just describing VHF2L, except VHF2L combines 3 + 4? What's new here? The result is still edge-orientation going into OLL, what am I missing?

 

[Ranzha]

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

Aren't you just describing VHF2L, except VHF2L combines 3 + 4? What's new here? The result is still edge-orientation going into OLL, what am I missing?

Exactly! It is VHF2L! However, Intuitively orienting last layer edges IS edge control. Thus, 4 is redundant. Alternatively, 3 is redundant.
It depends on your stance on the subject.

 

[Edward]

So we were in skype, and Lil Cruton had an idea

[1:25:13 PM] Lil' Cruton IV: with out the eo and bock though
[1:25:23 PM] Lil' Cruton IV: acctually
[1:25:28 PM] Lil' Cruton IV: it was nothign like that
[1:25:31 PM] Lil' Cruton IV: it was
[1:25:40 PM] Lil' Cruton IV: build top and bottom cross
[1:25:51 PM] Lil' Cruton IV: orient rest of edges for no rotation solve
[1:26:07 PM] Lil' Cruton IV: while doing f2l permute top edges
[1:26:10 PM] Lil' Cruton IV: cmll
[1:26:20 PM] Efam: no
[1:26:21 PM] Efam: not cmll
[1:26:22 PM] 4°☆BigGreen: sounds complex

Probably wouldn't be too good, but sounds nice in theory.

 

[Ranzha]

Lol, I just thought this had to be in here.

I just got a 15.9-second 3x3 solve (great for me, atm) with a LS skip (which, in reality, is just not noticing you're solving an F2L pair while doing your third, which becomes the fourth) and an OLL skip. Then, Z-perm. Win, amirite?

 

[oprah62]

Lol, I just thought this had to be in here.

I just got a 15.9-second 3x3 solve (great for me, atm) with a LS skip (which, in reality, is just not noticing you're solving an F2L pair while doing your third, which becomes the fourth) and an OLL skip. Then, Z-perm. Win, amirite?

luckiest slow solve ever

 

[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

What I meant by this was this:
Why would you force to have something NOT solved, when it is easy enough to have it solved? In other words, why would you destroy a noticed x-cross to just get cross? The movecount for cross should be 8 moves or less, HTM, every time. Then, another 4-6 for the first pair, optimally. This, at worst, forms a 14 move x-cross. I usually average 12 moves for cross + 1st pair. With a noticed x-cross, it solved the F2L pair WHILST solving the cross. Why ruin that?
Ok what if instead of you where able to look ahead and see that you had an LS skip so you can orient the edges during the time you are solving the second to the last f2l pair.

what are the chances of getting an LS skip?

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

LS: 1/150

1/27>1/150

Consider the movecounts, and not the probabilities.
Any F2L slot should be...6 moves on average, IIRC.

I suspect OLL would have considerably more, on average. Optimally, that is. Although I don't know the actual figures, and do not know otherwise how to find this figure, if someone would be so kind as to post it, I'd greatly appreciate it.

The thing is, an OLL skip would be more valuable to the solver who does not know full OLL. A LS skip would be nice, but forcing an OLL skip in while forming the pair, implying you don't have an LS skip, would significantly influence the overall solve time.

From my personal experience, a usual LS takes about 1 through 1.5 seconds on average. OLL is usually about 3 - 4 seconds. Thus, the solve time would be faster with an OLL skip rather than a LS skip. Also, an LS skip wouldn’t really be noticeable as easily as an OLL skip would.

This is in part what I have been trying to say.

It would be better for cuber who doesn't know full OLL to use at most 7 moves then to solve LS and then do 2 look oll

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

Aren't you just describing VHF2L, except VHF2L combines 3 + 4? What's new here? The result is still edge-orientation going into OLL, what am I missing?

Exactly! It is VHF2L! However, Intuitively orienting last layer edges IS edge control. Thus, 4 is redundant. Alternatively, 3 is redundant.
It depends on your stance on the subject.

Yes this is what I mean only the difference is your doing it intuitively.

So let me get something clear vhf2l and edge control is the same concept as algorithm f2l and intuitive f2l?

 

[foxfan352]

Lol, I just thought this had to be in here.

I just got a 15.9-second 3x3 solve (great for me, atm) with a LS skip (which, in reality, is just not noticing you're solving an F2L pair while doing your third, which becomes the fourth) and an OLL skip. Then, Z-perm. Win, amirite?

Here do you mean as you were solving the second to the last f2l pair you got an LS skip and an OLL skip?

Or do you mean the f2l pair was already paired together and ready to be inserted?

 

[rubiknewbie]

Most of VHF2L is quite intuitive to me. I convert the moves to basic case just like what I did with intuitive F2L.

 

[Ranzha]

Consider the movecounts, and not the probabilities.
Any F2L slot should be...6 moves on average, IIRC.

I suspect OLL would have considerably more, on average. Optimally, that is. Although I don't know the actual figures, and do not know otherwise how to find this figure, if someone would be so kind as to post it, I'd greatly appreciate it.

The thing is, an OLL skip would be more valuable to the solver who does not know full OLL. A LS skip would be nice, but forcing an OLL skip in while forming the pair, implying you don't have an LS skip, would significantly influence the overall solve time.

From my personal experience, a usual LS takes about 1 through 1.5 seconds on average. OLL is usually about 3 - 4 seconds. Thus, the solve time would be faster with an OLL skip rather than a LS skip. Also, an LS skip wouldn’t really be noticeable as easily as an OLL skip would.

This is in part what I have been trying to say.

It would be better for cuber who doesn't know full OLL to use at most 7 moves then to solve LS and then do 2 look oll

No, it wouldn't.
It seems to me that you're getting further away from the original prompt with each post.
(25. Relation to the original topic decreases with every single post)

You're trying to say that...
a person who does not know full OLL should use edge control the orient all edges.
Then the cuber would insert the slot, only to be left with an OCLL and PLL (or ZBLL, if one was so inclined).

Your original prompt was to use edge control to skip the last slot. Then, you'd have an OCLL.
Calling edge control which puts in the last slot an "LS skip" is unethical.
Are you trying to say that you'd use only PARTIAL edge control to insert the slot, with or without all edges being oriented?

It'd be quite easier to just insert the slot with VHF2L, ZBF2L, or even normally than to risk losing an OLL skip with edge control, even if you see the OLL skip in advance.

Lol, I just thought this had to be in here.

I just got a 15.9-second 3x3 solve (great for me, atm) with a LS skip (which, in reality, is just not noticing you're solving an F2L pair while doing your third, which becomes the fourth) and an OLL skip. Then, Z-perm. Win, amirite?

Here do you mean as you were solving the second to the last f2l pair you got an LS skip and an OLL skip?

Or do you mean the f2l pair was already paired together and ready to be inserted?

I was solving. I got to third pair (FR and DFR), and during my "algorithm" to insert it, I unknowingly inserted the last pair behind it (BR and DBR). Then, I found myself at an OLL skip and a Z-perm.
Basically, I got from cross + 2 pairs to PLL in 8 moves-ish. I liek it!