One-Answer 3x3x3 Question Thread

[xyzzy]

I use parts of VHLS and a few other random easy ZBLS cases, but I usually don't bother with the worse VHLS/ZBLS algs because COLL+EPLL is only slightly faster than OLL+PLL (where "OLL" means OLL in general, not just OCLL).

ZBLL changes the equation a lot, because you'd be going from OLL+PLL (two looks) to ZBLL (one look), and that saves a lot more moves than going to COLL+EPLL (still two looks); if you're going to learn ZBLL, then VHLS is a useful stepping stone towards ZBLS. (Full VHLS on its own is kind of useless because half of the cases suck, and it makes no sense to set up a bad F2L case into a bad VHLS case.)

 

[Lapis]

What are the rarest ZBLL cases? I recently watched Chris Tran's ZBLL seminar from US Nats 2015 and he mentions that some cases are much rarer than others. Since I want to learn full ZBLL, I want to know which ones are the rarest so I can remember those more easily and just practice the others doing normal solves.

 

[TDM]

What are the rarest ZBLL cases? I recently watched Chris Tran's ZBLL seminar from US Nats 2015 and he mentions that some cases are much rarer than others. Since I want to learn full ZBLL, I want to know which ones are the rarest so I can remember those more easily and just practice the others doing normal solves.

The more rotational symmetry a case has, the less likely it is to occur. For example, H perms, skips and N perms are the rarest cases, as these are the only ZBLLs with rotational symmetry order 4. These occur with 1/4 of the probability of most other cases. There are lots of cases with order 2: E perms and Z perms are the easiest to see. Also, lots of H cases (those with permutations the same as all previously mentioned PLLs) have order 2. These cases occur half as often as cases with no rotational symmetry.

To see why this is, take a look at H perm. If you allow yourself to AUF, then there's only one H perm case you can get. But with other cases, e.g. Ua perms, there are four different Ua perms: one with the UB edge solved, one with the UR edge solved, one with UF solved and one with UL solved. So Ua perms are four times more likely to occur than H perm, because there are four times as many of them.

 

[Lapis]

The more rotational symmetry a case has, the less likely it is to occur. For example, H perms, skips and N perms are the rarest cases, as these are the only ZBLLs with rotational symmetry order 4. These occur with 1/4 of the probability of most other cases. There are lots of cases with order 2: E perms and Z perms are the easiest to see. Also, lots of H cases (those with permutations the same as all previously mentioned PLLs) have order 2. These cases occur half as often as cases with no rotational symmetry.

To see why this is, take a look at H perm. If you allow yourself to AUF, then there's only one H perm case you can get. But with other cases, e.g. Ua perms, there are four different Ua perms: one with the UB edge solved, one with the UR edge solved, one with UF solved and one with UL solved. So Ua perms are four times more likely to occur than H perm, because there are four times as many of them.

Thanks for this. I'm a bit confused as to how I would I apply to this zbll, since there isn't the "same" symmetry as there is in pll, if that makes sense. Does this mean that, say, this case would be more rare because there is two full blocks: F2 U' F2 R U2 B2 D2 F2 L B2 U' B2 D2 F2 B2 U'? And what about this case: R2 U2 R2 U R2 U R' U R U' R2 U' R2 U2 R' U2 R' U? This one "looks" like an n perm from the two sides.

 

[xyzzy]

Thanks for this. I'm a bit confused as to how I would I apply to this zbll, since there isn't the "same" symmetry as there is in pll, if that makes sense. Does this mean that, say, this case would be more rare because there is two full blocks: F2 U' F2 R U2 B2 D2 F2 L B2 U' B2 D2 F2 B2 U'? And what about this case: R2 U2 R2 U R2 U R' U R U' R2 U' R2 U2 R' U2 R' U? This one "looks" like an n perm from the two sides.

Blocks (or matching colours) and symmetry are unrelated concepts. Those two cases both have minimal symmetry.

Let X be some last layer alg. Look at the sixteen combinations of AUFs before and after the alg:
X, X U, X U2, X U',
U X, U X U, U X U2, U X U'
U2 X, U2 X U, U2 X U2, U2 X U'
U' X, U' X U, U' X U2, U' X U'

In the general case, all sixteen of these solve different configurations (e.g. either of the two pi ZBLLs you mentioned), but sometimes you get duplicates; e.g. for H perm, there are only four different configurations, and for E perm there are only eight. There will always be either 4, 8 or 16 different configurations for any last layer alg, and the frequency of a last layer case is proportional to the number of different configurations it has. (By "configuration", I mean the exact colours you see if you start with a fixed orientation, e.g. white-top green-front, no rotations/recolouring/AUF allowed.)

 

[TDM]

Thanks for this. I'm a bit confused as to how I would I apply to this zbll, since there isn't the "same" symmetry as there is in pll, if that makes sense. Does this mean that, say, this case would be more rare because there is two full blocks: F2 U' F2 R U2 B2 D2 F2 L B2 U' B2 D2 F2 B2 U'? And what about this case: R2 U2 R2 U R2 U R' U R U' R2 U' R2 U2 R' U2 R' U? This one "looks" like an n perm from the two sides.

I was already applying it to ZBLL The number of blocks doesn't matter. Most of the ZBLL cases occur with probability 4/1944. However, the ones I mentioned with rotational symmetry order four (H perm, N perm, skip) have probability 1/1944, as these are the only cases with order 4 (or alternatively, you can view these cases as the ones where the algorithm works from any AUF).

The cases with order 2 are the following:
E perm
Z perm
ZBLL H40 (the H ZBLL with all pieces permuted)
ZBLL H39 (the H ZBLL with pieces permuted like an H perm)
ZBLLs H37 and H38 (the H ZBLLs with the pieces permuted like a Z perm)
ZBLLs H29 and H30 (the H ZBLLs with the pieces permuted like N perms)
ZBLLs H27 and H28 (the H ZBLLs with the pieces permuted like E perms)
And these all have probability 2/1944. As you can see, these are also the cases where the algorithm solves the case from exactly two angles.

There aren't many of these cases at all (14 in total), and they're not significantly less common than the other ZBLLs, so it's probably not worth changing your practice for them.

 

[obelisk477]

Really! I had this idea because I thought it would be perfect for moving on to ZBLLs. Do you think ZBLLs aren't worth learning?

Whether or not full ZB (ZBLS+ZBLL together) is worth it is one of the great unexplored questions of 3x3 speedsolving. In theory, it should be one of the fastest and most move efficient methods that exist, but the barrier to entry is learning 800+ algorithms and practicing them enough for recognition and fingertricking to really kick in. If you really want to learn it though, BindeDSA's ZBLL list probably has the best algs out there currently for full ZBLL, and for full ZBLS, my doc is probably the best and most up to date.

VHLS is definitely not worth it IMHO, as mentioned above. Go big or go home!

 

[Xnat]

So I just learned 2-look OLL and now I want to learn full PLL but I don't know how to setup cases to practice

 

[greentgoatgal]

I just repeated whatever alg I wanted to practice over and over.

 

[Oatch]

So I just learned 2-look OLL and now I want to learn full PLL but I don't know how to setup cases to practice

I'd recommend you use an alg trainer, they give scrambles so you can set up cases as you please. A great one is Tao Yu's which you can find here. Unfortunately at the moment you can only train the entire algset, but I'm sure you'll be able to find others online elsewhere that lets you select specific algs to train if you look.

 

[TDM]

So I just learned 2-look OLL and now I want to learn full PLL but I don't know how to setup cases to practice

Roman Strakhov has a PLL trainer here which allows you to select the cases you've learned.

If you just want to set up cases and aren't near a computer, then any case made only of 2-cycles (e.g. T perm, R perms, Z perm) can be set up simply by doing the algorithm. If the case is made of 3-cycles (e.g. U perms, A perms) then you can either do the alg three times, or note that each alg has a "pair" which is its inverse. Doing the first alg in the pair will set up the second alg in the pair, and vice versa.

 

[Xnat]

Thank both of you I'll use whichever is better!

 

[Sai Ashish Vure]

What's the best website for algorithms? I'm looking for OLLs, CLLs, COLLs, etc

 

[obelisk477]

What's the best website for algorithms? I'm looking for OLLs, CLLs, COLLs, etc

http://algdb.net/

 

[Tom Joad]

Hi all,

Thanks to many on here I now have my average down to about 18.5 seconds

However, I still only recognise auf after I complete pll.

Does anyone have any tips on how to practice integrating these two steps?

Should I recognise both the pll and subsequent auf during pll recognition?

Is there a list of plls for which this is counterintuitive because they incorporate a u move anyway?

Thanks in advance to anyone who can point me in the right direction.

 

[Dancing Jules]

Is there a list of plls for which this is counterintuitive because they incorporate a u move anyway?

That will depend on which alg you use. I know my Jb perm does not include a U-move, I know they standard Jb-perm does.

I can't say too much about your other questions because I'm still learning it myself.

 

[Tom Joad]

That will depend on which alg you use. I know my Jb perm does not include a U-move, I know they standard Jb-perm does.

I can't say too much about your other questions because I'm still learning it myself.

Ah, I see what you mean.

I think I’ll go through my plls and make a note of the ones which include a change in the upper layer.

Then, during my slow solves I’ll pause at pll recognition and work out the auf before pll, the pll, and the auf after pll then close my eyes and execute.

When I see sub-10solvers it looks like they are completing the auf before, the pll and the auf after all in one algorithm, so I guess that’s what we’re all aiming for.

Whilst it seems like one of those “invest a lot to gain a little” projects, I think I’ll go for it. I have spent a year on two sided pll recognition so I think I’ll give this a go.

For anyone else who still pauses here and is interested in improving, Jperm’s “your auf is too slow” vídeo is a good intro to the idea.

Anyone else have any thoughts on this?

 

[xyzzy]

However, I still only recognise auf after I complete pll.

Does anyone have any tips on how to practice integrating these two steps?

Should I recognise both the pll and subsequent auf during pll recognition?

Sometimes there's a piece that has a net zero movement at the end of the PLL alg (e.g. the UF and URF pieces with the standard Jb perm), and as you start executing the PLL alg, you just need to know what colour that piece is and how you need to AUF it. As you finish the PLL alg, you can just smoothly transition into the AUF you have already predicted.

Is there a list of plls for which this is counterintuitive because they incorporate a u move anyway?

Don't think of the U moves at the start and the end as being part of the alg per se. None of the algs incorporate U moves and your question becomes irrelevant.

 

[Tom Joad]

Sometimes there's a piece that has a net zero movement at the end of the PLL alg (e.g. the UF and URF pieces with the standard Jb perm), and as you start executing the PLL alg, you just need to know what colour that piece is and how you need to AUF it. As you finish the PLL alg, you can just smoothly transition into the AUF you have already predicted.

Thanks, I get it now, I’ve just been playing around with this idea for an hour.

I’m now making a reference sheet stating which are the pieces with net zero movement for each of my plls.

Thanks again.

Edit:
Hi again,

So, after spending a few hours on this auf project, here is where i’m at.

I am learning the “net zero movement piece” for each pll so that I can smoothly move from pll to auf at the end. I’ll use a tricky example so that a sub15 cuber can explain what they do naturally:

I reach pll recognition and (using two sided recognition) I see it is an Rb perm, which needs a U2 before I execute the algorithm I know.

I also now know that after the U2, the only “net zero movement” piece is the URB piece. However, that piece is, of course, currently in the ULF position which I cannot see.

I conclude (tell me if I’m wrong) that it is only possible for me to execute U2-pll-auf with my eyes closed if I can infer what colour the ULF piece is...

My current ability is such that whilst, on the one hand, I do recognise the Rb perm from just two sides, on the other hand, I don’t know what all the colours are on the L face and B face (unless I stop and think about it for a long time)

I’m guessing the sub10 cuber not only recognises the perm from just two sides, but also “sees” the six colours on the L and B sides because he/she is so familiar with it...

Any thoughts?

 

[awesome cuber 9000]

i started 2 months ago with beginners method and just today (1/20) got sub 1 minute and im looking to get faster what method would the experts reccomend.