ZZ Cubers

[CubingGenius]

(1)That's like saying ZBLL recognition is easier than PLL because you can get info from the top face. IT WAS DESIGNED TO BE EASY TO RECOGNIZE. Chris Tran made it specifically easy to recognize, because it recognizes like a PLL.
(2)I'm saying that your estimated move count is predicated on saying the F2L pair is already solved, whereas TSLE solves based on the exact opposite. The average move count for optimally solving the last pair in ZZ is 7.5. I'll even give you the benefit of the doubt that phasing adds NO moves to the average movecount(which it does). Even then, TSLE is more efficient. You have the move estimate yourself.
(3)That's a fair enough statement, yet the ergonomics are easily better with TTLL, because one full set is regripless, and 2 are 2-gen. As well, you have a 1/360 chance of skipping the last step in CT, whereas the chances of that happening in ZZLL is admittedly lower than normal ZZ, yet the chances are 1 in thousands, nowhere near the 1 in 360 with CT. The oriented D corner was to create rotational symmetry.
(4)I said that anyone who would seriously learn either of these methods would at least know OCLL and PLL. No one is going from 2 look PLL to CT and ZZLL. COLL is likely, but they aren't guaranteed to know it.
(5)This is idiotic. You obviously know nothing about what rotational symmetry is. Rotational symmetry tells when the corner orientation case is identical in position of the corners with an AUD. With ZZ-CT, this allows for any rotation of the U layer and being the same case. Rotational symmetry is only applicable in ZZLL for the H case, because U2 yields the same corner orientation. Any other case allows for no rotational symmetry whatsoever.
(6)He proposed a 3 step process. That's how he designed it. IT SAYS IT ON HIS WEBSITE. Those 3 are designed to be done separately. The only time he talks about combining steps, is talking about combining CO AND CP! He never alludes to 2GLL in any way.
(7)Yeah, yet Petrus didn't take Petrus past a 3-step system. That was done by others.
(8)ZZ-CT is unarguably the best method for cubestate reduction. Period. It allows for full rotational symmetry, reducing the cases by 1/4, whereas ZZLL allows for half rotational symmetry on one set and none on others.
(9)You realize that this method was created by a ZZ-a user who created it to surpass ZZ-a in ergonomics, alg count, and recognition, right? A ZZ-a user created this specifically to surpass your so-called "best method", and since he has first-hand experience with both, I think he would know.
(10) Your insistence that ZZLL is predicated on the idea that it is more move efficient(it's not), more alg efficient(it's not, because to 2 look LSLL you would need to memorize a case for every F2L phasing case), and more efficient at reducing cubestate(hilariously untrue). But if you don't want to respond to this, I get it. I've thoroughly addressed all your points. I just think it's rediculous that you continue to defend a method that is quantifiably worse based on false knowledge. But I, and anyone else reading this, know you'll probably respond to try to prove me wrong, despite the fact that facts, evidence, and statistics are not just not on your side, but completely against you.
P.S. Sorry for the whole number-parenthesis thing for addrsssing your argument. I'm fairly new to the forum, and I haven't quite figured out the interface. Either way, no hard feelings(it is just cubing after all).

Just to mention a few things:

3. The chance of a ZZLL skip after phasing is 1/648, not 1 in thousands. But when the skips are so small like this, it's not really worth using the method for the chance of skipping.
5. I'm not exactly sure what you are saying. If you take any ZZLL case and do a U it is the same case as well. Rotational symmetry only lowers the algorithm count in TTLL with PLL (from 72 to 21). In ZZLL it lowers the algorithm count with PLL (from 72 to 21) and H (72 to 40).
8. This is quite inconsistent. If you say doing a U reduces the number of cases on TTLL, how come you can't do it on ZZLL?
9. The creator has admitted himself that ZZ-a is superior to ZZ-CT.

I think ZZ-b and ZZ-CT at their highest level would be very similar speed wise. It would probably end up coming down more to a personal preference of which method you would use. But it is much harder to prove ZZ-CT is at the same level of ZZ-a due to only having to worry about one corner's orientation, although at the expense of the permutation as well, and because ZBLLs are generally better than TTLLs.

 

[JTcuber]

Just to mention a few things:

3. The chance of a ZZLL skip after phasing is 1/648, not 1 in thousands. But when the skips are so small like this, it's not really worth using the method for the chance of skipping.
5. I'm not exactly sure what you are saying. If you take any ZZLL case and do a U it is the same case as well. Rotational symmetry only lowers the algorithm count in TTLL with PLL (from 72 to 21). In ZZLL it lowers the algorithm count with PLL (from 72 to 21) and H (72 to 40).
8. This is quite inconsistent. If you say doing a U reduces the number of cases on TTLL, how come you can't do it on ZZLL?
9. The creator has admitted himself that ZZ-a is superior to ZZ-CT.

I think ZZ-b and ZZ-CT at their highest level would be very similar speed wise. It would probably end up coming down more to a personal preference of which method you would use. But it is much harder to prove ZZ-CT is at the same level of ZZ-a due to only having to worry about one corner's orientation, although at the expense of the permutation as well, and because ZBLLs are generally better than TTLLs.

3. Okay, that was my mistake. I didn't know the exact stat for ZZLL, but I knew that the normal one was 1/1000something. My mistake.
5. Rotational symmetry means you can do a U and have the same CO in relation to you. For example, you have a pi case. You have 1 pair of headlights on the left and the opposite corners on the right. You do a U, and now you have the headlights in front. The rotational symmetry in CT reduced the TTLL cases by(I believe this is right, not 100% sure) 1/22ish, 1-7th of that by getting rid of CO(so 7 less CO cases), and you would have 4 cases for every TTLL per CO, because there is no rotational symmetry, and every piece moving clockwise in relation to the corner case yields a new case if there was CO, but not in normal TTLL.
8. I already addressed this. Without CO changing in relation to you, with any corners facing the same way, the shifting of every piece clockwise would not yield the same case in ZZLL, but it would in TTLL because there is no CO.
9. I talked to the creator, and he believes they are comparable for averages, with ZZ-a having a slight edge if you get good at recognition, but he believes CT yields vastly superior single solves, and that CT is superior to b. I talked to him, and I think he won't mind me quoting here, "Hey JTcuber!

I think ZZ-CT is better because of recognition and one less look.

When you form the pair and do phasing, you miss out on lots of easy things, like easy insert to F2L, easy WV cancels, easy TSLE, and even easy ZBLL.

ZZ-B is only good as a step to ZZ-A, and all people who use ZBLL know this."

 

[xyzzy]

Time to well-actually everyone.

In ZZLL, I believe the case can be identified by looking at only 4 stickers after determining the COLL case (which is second-nature).

It's even better than that. You need only one (!) sticker once you know the COLL, since you can deduce the opposite sticker (it has to be the opposite colour) and parity constraints force only one possibility for the two remaining edges.

You know how Chris Tran talked about rotational symmetry in his session at Nationals? Yeah, well ZZ-b takes better advantage of that cube-property than ZZ-CT because all the pieces are isolated to the U-layer.

(5)This is idiotic. You obviously know nothing about what rotational symmetry is. Rotational symmetry tells when the corner orientation case is identical in position of the corners with an AUD. With ZZ-CT, this allows for any rotation of the U layer and being the same case. Rotational symmetry is only applicable in ZZLL for the H case, because U2 yields the same corner orientation. Any other case allows for no rotational symmetry whatsoever.

5. I'm not exactly sure what you are saying. If you take any ZZLL case and do a U it is the same case as well. Rotational symmetry only lowers the algorithm count in TTLL with PLL (from 72 to 21). In ZZLL it lowers the algorithm count with PLL (from 72 to 21) and H (72 to 40).

Actually, rotational symmetry is a red herring here. There are two concerns here: the symmetries you care about (pre-AUF symmetry and post-AUF symmetry are related, but not the same thing, and they are also distinct from rotational symmetry), and the symmetries each individual case possesses.

For the former, typically the only symmetries you care about are the AUFs, so that's a factor of 16, which is the same for both ZZLL and TTLL. For the latter, they're also roughly matched. The only TTLL cases that have any AUF symmetry are some of the PLLs (skip, H, Z, E, N); meanwhile, the only ZZLL cases that have any AUF symmetry are in the PLL and H subsets.

(Note that having symmetrical cases increases the case count. The fewer the symmetrical cases you have, the larger the factor of symmetry reduction.)

(1)That's like saying ZBLL recognition is easier than PLL because you can get info from the top face.

IIRC Chris actually said that ZBLL was easier to recognise than PLL for that very reason in his 2015(?) ZBLL seminar, so, you know… Edit: found it.

(3)That's a fair enough statement, yet the ergonomics are easily better with TTLL, because one full set is regripless, and 2 are 2-gen.

Strictly speaking, you don't get to count algs as being "2-gen" if you need to ADF before and after. After all, every ZBLL can be done mostly <R,U> with at most one D and one D' move, but this might be being pointlessly pedantic.

 

[AlphaSheep]

The biggest reason there are so many sub-10 CFOP users but so few sub-10 ZZ users is because CFOP users tend to spend their time doing big averages, while ZZ users spend their time arguing over the best ZZ variant. The best ZZ variant is the one you practice most so stop arguing and get back to practicing. Thanks.

 

[GenTheSnail]

The biggest reason there are so many sub-10 CFOP users but so few sub-10 ZZ users is because CFOP users tend to spend their time doing big averages, while ZZ users spend their time arguing over the best ZZ variant. The best ZZ variant is the one you practice most so stop arguing and get back to practicing. Thanks.

THIS but zz-a ftw

Best post of the year
And it's hardly even started

 

[One Wheel]

The biggest reason there are so many sub-10 CFOP users but so few sub-10 ZZ users is because CFOP users tend to spend their time doing big averages, while ZZ users spend their time arguing over the best ZZ variant. The best ZZ variant is the one you practice most so stop arguing and get back to practicing. Thanks.

Yes, but . . . I will never hold a speedcubing world record, or even win a competition. I cube for the fun of it, and the reason I'm learning ZZ, and ultimately ZZ-CT, is that they are fun, elegant methods. You know what else is fun? Arguing about which method is best.

 

[EntireTV]

Yes, but . . . I will never hold a speedcubing world record, or even win a competition. I cube for the fun of it, and the reason I'm learning ZZ, and ultimately ZZ-CT, is that they are fun, elegant methods. You know what else is fun? Arguing about which method is best.

SS Forum is fun too

 

[mDiPalma]

It's even better than that. You need only one (!) sticker once you know the COLL, since you can deduce the opposite sticker (it has to be the opposite colour) and parity constraints force only one possibility for the two remaining edges.

You're right. I didn't account for the parity (-1). I also added +2 corner stickers on a random corner to know the AUF.

Actually, rotational symmetry is a red herring here. There are two concerns here: the symmetries you care about (pre-AUF symmetry and post-AUF symmetry are related, but not the same thing, and they are also distinct from rotational symmetry), and the symmetries each individual case possesses.

For the former, typically the only symmetries you care about are the AUFs, so that's a factor of 16, which is the same for both ZZLL and TTLL. For the latter, they're also roughly matched. The only TTLL cases that have any AUF symmetry are some of the PLLs (skip, H, Z, E, N); meanwhile, the only ZZLL cases that have any AUF symmetry are in the PLL and H subsets.

Yes, as you pointed out, the rotational symmetries of permutation are more numerous in ZZLL than TTLL, which is what I was alluding to. I believe that the other guy was talking about rotational symmetries of orientation (which I guess relates to the TSLE cases that have the edge already inserted) and comparing that to the lack of any orientational symmetries in the phasing step (duh). Not really sure though.

The biggest reason there are so many sub-10 CFOP users but so few sub-10 ZZ users is because CFOP users tend to spend their time doing big averages, while ZZ users spend their time arguing over the best ZZ variant. The best ZZ variant is the one you practice most so stop arguing and get back to practicing. Thanks.

Truth, it just rustles my jimmies when people are blinded by hype.

 

[Teoidus]

While "best method" debates are generally not a great idea, I think in this case it's good to make clear what was and wasn't overhyped in ZZ-CT. It's worrying to think that new ZZ users exploring different variants will find CT and think it's the objectively best variant, ignore any others, and begin the cycle anew by reciting what they've been told of CT (Like JTCuber and genericcuber666 seem to have been doing...).

There are a couple of things that you can evaluate variants on:
Move count
Alg count
Recognition
Ergonomics
Look count

Let's look at ZZ-CT:
52 avg movecount
197 alg count
Relatively easy recognition. Look at shapes for TSLE, look at block patterns for TTLL.
Good ergonomics (<R,U> is great, but if people can sub-1 their E perms I see no reason for <R,U> being so much better than other movesets)
2.5 looks for LSLL (adding the .5 look because you do have to track your last F2L edge)
If we assume: 1 look for EOLine, 2 looks for the squares, 1 look for the last pair, 0.5 to track FR, 1 for TSLE, 1 for TTLL
You get a 6.5 look system. Not bad, CFOP is 7 looks (cross, 4 pairs, OLL/PLL) and Roux is 6.5 usually (FB, square, pair, CMLL, EO, 0.5 LR, 4c) and 5 if you're Kian (FB+DR+other sb stuffs, rest of SB+CMLL stuffs, CMLL+EOLR stuffs, EOLR, 4c).

Let's look at ZZ-b:
47 avg movecount
167 alg count
Relatively easy recognition. Look at arity for phasing + pair, look at a COLL + a single sticker for ZZLL.
Good ergonomics (the last pair + phasing is also 2gen! In fact, most things in ZZ are 2-gen until you have to confront CP. I don't know why some feel this is a major selling point of CT, it's quite common in ZZ in general)
2.5 looks for LSLL (adding the .5 look because you do have to track your last F2L pair).
The ZZF2L is of course the same, which makes this a 6.5 look system.

So if we directly compare ZZ-CT and ZZ-b:
52 vs 47
197 vs 167
Relatively easy vs relatively easy
Good vs good
2.5 vs 2.5

Now if we quickly compare to ZZ-COLL/EPLL: 52 vs 58 movecount, 197 vs 46 alg count, relatively easy vs easy, good vs good, 2.5 vs 3.
So the question you might ask when considering ZZ-CT over ZZ-COLL/EPLL: is it worth the lower movecounts and look counts for the cost of 4x as many algs + a little harder recognition?
And, the question you might ask when considering ZZ-b over ZZ-COLL/EPLL: is it worth the lower movecounts and look counts for the cost of a little over 3x as many algs + a little harder recognition?

And, if yes to both, then the question you'd have to ask when considering ZZ-CT over ZZ-b: is it worth... nothing... for the cost of higher movecounts and alg counts?

 

[MethodNeutral]

And, if yes to both, then the question you'd have to ask when considering ZZ-CT over ZZ-b: is it worth... nothing... for the cost of higher movecounts and alg counts?

Your argument is very convincing, the only factor that you missed is that ZZ-CT has much easier recognition than ZZ-A. I do agree about everything else, but as far as TTLL vs. ZZLL, TTLL definitely wins just because of the recognition.

 

[Teoidus]

ZZ-b recognition isn't hard, though... COLL recognition is easy, and ZZLL recognition is COLL + 1 edge piece.

 

[CubingGenius]

Your argument is very convincing, the only factor that you missed is that ZZ-CT has much easier recognition than ZZ-A. I do agree about everything else, but as far as TTLL vs. ZZLL, TTLL definitely wins just because of the recognition.

ZZLL's algs are a lot better than TTLL. The part where the gap is closer is phasing vs TSLE, not ZZLL vs TTLL. Anyway, the maximum number of stickers you need to see to recognise TTLL from any angle is 9 vs ZZLL's 7. So recognition would actually be faster for ZZLL, not TTLL.

 

[Y2k1]

Hey, just as a break from ct vs b, what is better between cpls/2gll and zz-b? From my understanding, it would be slightly lower movecount for b and harder recog for cpls, but I wolud rather have solid stats to compare the lls like the post 2 above

 

[Teoidus]

1.5 HTM (from mdipalma) + 10 HTM CPLS + 15 for 2GLL = 26.5 for ZZ-CPLS/2GLL's LSLL
vs 9 last pair+phasing + 14 ZZLL = 23 for ZZ-b's LSLL

 

[mDiPalma]

1.5 HTM (from mdipalma) + 10 HTM CPLS + 15 for 2GLL = 26.5 for ZZ-CPLS/2GLL's LSLL
vs 9 last pair+phasing + 14 ZZLL = 23 for ZZ-b's LSLL

https://www.speedsolving.com/forum/threads/cpls-and-2gll-discussion.24125/page-6#post-1215420

using CPLS/2GLL instead of phasing/ZZLL is basically trading better and fewer algorithms for around 2-3 moves.

 

[Zanaso]

Your argument is very convincing, the only factor that you missed is that ZZ-CT has much easier recognition than ZZ-A. I do agree about everything else, but as far as TTLL vs. ZZLL, TTLL definitely wins just because of the recognition.

lol are you high? ZZLL has fantastic recognition. Sure TTLL has better recognition than ZBLL, but not better than ZZLL. They're comparable. Don't forget that you literally only need to look at 1 edge to determine the edge cycle for a ZZLL as opposed to 2 for TTLL.

 

[4Chan]

lol are you high? ZZLL has fantastic recognition. Sure TTLL has better recognition than ZBLL, but not better than ZZLL. They're comparable. Don't forget that you literally only need to look at 1 edge to determine the edge cycle for a ZZLL as opposed to 2 for TTLL.

ZZLL Recognition:
Step 1: Look at COLL case
Step 2: Look at edge

TTLL Recognition:
Step 1: Look at blocks

 

[Zanaso]

ZZLL Recognition:
Step 1: Look at COLL case
Step 2: Look at edge

TTLL Recognition:
Step 1: Look at blocks

Yeah I forgot that every single TTLL case has super obvious blocks. Also telling the difference between the ~20 cases that have absolutely no blocks is like way easier than COLL + 1 edge. Also, 600 million% of TTLL cases are fantastic algs.

 

[4Chan]

Yeah I forgot that every single TTLL case has super obvious blocks. Also telling the difference between the ~20 cases that have absolutely no blocks is like way easier than COLL + 1 edge. Also, 600 million% of TTLL cases are fantastic algs.

Do you even know either of these alg sets?
You're just talking from conjecture.

I've actually known ZZLL, ZBLL, and TTLL, and I'm speaking from experience.

 

[mDiPalma]

Do you even know either of these alg sets?
You're just talking from conjecture.

I've actually known ZZLL, ZBLL, and TTLL, and I'm speaking from experience.

Even still, I would argue that identifying the COLL case and looking at a single edge is less involved than searching non-U-facing pieces for characteristic blocks of colors.