ZZ Cubers

[JohnnyA]

The only site that really gives a description of ZZ is the one given before. ZZ-VH can be just as fast as Fridrich, and ZZ-b will probably be faster.

That's quite an impressive, but totally unsupported assertion...

No it's not ... there was a video of someone getting times faster than most Fridrich users earlier in the topic, and nobody has shown us any times with ZZ-b which will obviously be faster than ZZ-VH. It's got a 1LLL, 3gen-F2L, I would say that it's not unsupported to say that if it was worked on, it could be as fast as Fridrich.

 

[Stefan]

there was a video of someone getting times faster than most Fridrich users

There was a video of Erik getting times faster than most Fridrich users using a vanilla beginner method. Obviously shows that that method can be just as fast as Fridrich, and with 3-look-LL it'll probably be faster.

 

[mrbiggs]

The only site that really gives a description of ZZ is the one given before. ZZ-VH can be just as fast as Fridrich, and ZZ-b will probably be faster.

That's quite an impressive, but totally unsupported assertion...

He didn't say is. *shrugs*

"Will probably be" is way stronger of a statement than the evidence supports IMO. What's the fastest ZZ-b cuber at right now, 14 seconds? That's barely in the top 100.

The only site that really gives a description of ZZ is the one given before. ZZ-VH can be just as fast as Fridrich, and ZZ-b will probably be faster.

That's quite an impressive, but totally unsupported assertion...

No it's not ... there was a video of someone getting times faster than most Fridrich users earlier in the topic, and nobody has shown us any times with ZZ-b which will obviously be faster than ZZ-VH. It's got a 1LLL, 3gen-F2L, I would say that it's not unsupported to say that if it was worked on, it could be as fast as Fridrich.

Until there are ZZ-b users getting sub-12 averages, you can't say that it's faster than Fridrich. Until you see people getting very close, you can't say that it's probably as fast as Fridrich. You're making things up if you do.

Also, Matthew uses ZZ-VH, and he's the fastest ZZ cuber I've heard of, so why must ZZ-b be faster?

 

[JohnnyA]

Until there are ZZ-b users getting sub-12 averages, you can't say that it's faster than Fridrich. Until you see people getting very close, you can't say that it's probably as fast as Fridrich. You're making things up if you do.

Also, Matthew uses ZZ-VH, and he's the fastest ZZ cuber I've heard of, so why must ZZ-b be faster?

I can say what I belive - I BELIEVE it can be as fast as Fridrich. This is backed up by the fact that people get great averages with ZZ-VH, which isn't even ZZ-b. ZZ-b is a more advanced version of ZZ-VH, saying ZZ-B is slower is like saying cross+oll is faster than OLL. It's anti-intuitive. And even when someone who is fast learns ZZ-b, there is still the final method, ZZ-d, which is (theoretically) faster than ZZ-b (theoretically because it doesn't exist yet). If someone can get 15 seconds using cross+oll on fidrich, and someone else can get 10 seconds using oll with fridrich, it makes sense to assume that someone getting 13 seconds using ZZ-VH could get faster times using a more advanced ZZ.

there was a video of someone getting times faster than most Fridrich users

There was a video of Erik getting times faster than most Fridrich users using a vanilla beginner method. Obviously shows that that method can be just as fast as Fridrich, and with 3-look-LL it'll probably be faster.

I wouldn't consider sub-20 cubers as being a factor in deciding the fastest method, simply because the majority of them won't be using the most advanced version of their method yet. (no insult to sub-20 cubers - I am still averaging 20 myself too)

 

[vloc15]

BTW, how many algs do you need for ZZ-VH? for the COLL's and the last F2L pair..

 

[Escher]

erm, do you know what VH is?
ill assume you dont - VHF2L orients last layer edges while putting in the last slot, once you have reduced the case to one of the two simple insertions ( RU'R' and its mirror).
EOLine means that VHF2L is 'already done for you'. so ZZ-VH is just coll, for which there are 42 algorithms - 6 for each 7 types of edges already oriented OLL. you probably already know about 8 or 9 COLL cases.


i was thinking about this - what would be very useful to know for ZZ would be F2LL or winter variation - 27 algorithms plus mirrors, and it gives you an OLL 'skip'. works in a similar way to VH, except the edges are already oriented and it orients corners while doing the last slot. it would also be useful to know if you knew all 20 ELL algs...
knowing my memory, its probably already been discussed on this thread, but ive forgotten

 

[Johannes91]

coll, for which there are 42 algorithms - 6 for each 7 types of edges already oriented OLL.

Wrong.

1) There are 8 different OCLL (edges already oriented OLL) cases.
2) Two of them have less than 6 different COLL cases because of symmetry: there are only two (excluding solved) when all corners are correctly oriented and just 4 Double Sunes.

So it's 2+4+6+6+6+6+6+6 = 42. Some people don't include the two with solved CO (often because they know PLL) and say there are only 40 COLL cases.

what would be very useful to know for ZZ would be F2LL or winter variation - 27 algorithms plus mirrors

Even more useful would be to stop learning mirrored algs separately... For some WV cases it's not easy to find anything better than simple insertion and a Sune, but some of the shortest ones are indeed quite useful.

 

[mrbiggs]

I can say what I belive - I BELIEVE it can be as fast as Fridrich.

I'm just pointing out that you're making it up with little evidence to back up your assertion.

Also, I feel that it's a very strong statement that Fridrich, which has been by far the fastest in the world for some time now, is slower than ZZ, which has only been recently popularized and very rarely used in competition--the method isn't even completely finished yet.

This is backed up by the fact that people get great averages with ZZ-VH, which isn't even ZZ-b. ZZ-b is a more advanced version of ZZ-VH, saying ZZ-B is slower is like saying cross+oll is faster than OLL. It's anti-intuitive.

Not necessarily. ZB is more advanced than Fridrich (if I'm understanding your definition correctly), but no one uses it.

And even when someone who is fast learns ZZ-b, there is still the final method, ZZ-d, which is (theoretically) faster than ZZ-b (theoretically because it doesn't exist yet). If someone can get 15 seconds using cross+oll on fidrich, and someone else can get 10 seconds using oll with fridrich, it makes sense to assume that someone getting 13 seconds using ZZ-VH could get faster times using a more advanced ZZ.

It's quite possible.

But even assuming that ZZ-b is faster than ZZ-VH and ZZ-d is faster than ZZ-b, we still haven't established that it's as fast as Fridrich, much less faster.

I wouldn't consider sub-20 cubers as being a factor in deciding the fastest method, simply because the majority of them won't be using the most advanced version of their method yet. (no insult to sub-20 cubers - I am still averaging 20 myself too)

I don't understand this paragraph. I don't know of many cubers better than Eric to show the efficiency of a method, first of all. Second, why would you mean offense to sub-20 cubers? Isn't it good to be that fast?

 

[Escher]

Wrong.

1) There are 8 different OCLL (edges already oriented OLL) cases.
2) Two of them have less than 6 different COLL cases because of symmetry: there are only two (excluding solved) when all corners are correctly oriented and just 4 Double Sunes.

So it's 2+4+6+6+6+6+6+6 = 42. Some people don't include the two with solved CO (often because they know PLL) and say there are only 40 COLL cases.

oh, ok, i see. i dont know (much about) COLL. at least, i only know some of COLL for triple sune, headlights and chameleon, because thats all i thought was really useful for me.

 

[mazei]

I'm a Fridrich user averaging 14s. I tried ZZ-VH and it was interesting. I managed to get my times to around sub-20 average. I think it has the potential to be just as fast as Fridrich once the whole method is complete. Any method has potential to be as fast as Fridrich(basically just Roux, Petrus and some Fridrich variants).

I wonder if mrbiggs would say anything about this post

 

[a small kitten]

Probably something like this:

It's quite possible.

But even assuming that ZZ-b is faster than ZZ-VH and ZZ-d is faster than ZZ-b, we still haven't established that it's as fast as Fridrich, much less faster.

 

[qqwref]

Any method has potential to be as fast as Fridrich(basically just Roux, Petrus and some Fridrich variants).

...evidence?


Someone mentioned color-neutral EOLine a while back. Is this even possible in 15 seconds? I'm not so sure. I would be interested to see results for y-neutral EOLine though (i.e. do EOline with a given top color but any front color).

 

[JohnnyA]

I'm just pointing out that you're making it up with little evidence to back up your assertion.

Also, I feel that it's a very strong statement that Fridrich, which has been by far the fastest in the world for some time now, is slower than ZZ, which has only been recently popularized and very rarely used in competition--the method isn't even completely finished yet.

I've never said ZZ is faster than Fridrich. I use Fridrich, and I am wary about switching to ZZ. But my point is that it is not theoretically a worse method, it has its good aspects and bad aspects as does Fridrich. If Fridrich had the popularity of ZZ and ZZ had the popularity of Fridrich, I personally believe that ZZ would have a higher potential than Fridrich does at the moment, becuase of the focus on "move count and move ergononmy" which makes for a good speed potential.

Not necessarily. ZB is more advanced than Fridrich (if I'm understanding your definition correctly), but no one uses it.

"Nobody uses ZB" is not a great statement. I was under the impression that at least four people were in the process of learning it at the moment, and four peeople are enough to count as "somebody". ZB may be considered as the "final" advancement to Fridrich (it clearly is not, but for this discussion we may as well assume it to be) and ZZ-D as the "final" advancement to ZZ, and maybe only four people would use ZZ-D. It still has the same potential as a method, no matter how many people use it.

It's quite possible.

But even assuming that ZZ-b is faster than ZZ-VH and ZZ-d is faster than ZZ-b, we still haven't established that it's as fast as Fridrich, much less faster.

I don't claim that ZZ is faster than Fridrich, the majority of people probably disagree with me. It's just my personal opinon that ZZ has more potential as it is a non-developed and un-publicised method that still produces 13 second averages without even using the most advanced form of the method. There are few people who could get 13 second averages without using the most advanced, common form of Fridrich (OLL+PLL)

I don't understand this paragraph. I don't know of many cubers better than Eric to show the efficiency of a method, first of all. Second, why would you mean offense to sub-20 cubers? Isn't it good to be that fast?

My mistake - I meant to say people who are not sub-20 should not affect the potential of a method because they have not stretched the method to its limits, and I meant no offense to non-sub-20 cubers, not to sub-20 cubers.

 

[mrbiggs]

I don't claim that ZZ is faster than Fridrich, the majority of people probably disagree with me. It's just my personal opinon that ZZ has more potential as it is a non-developed and un-publicised method that still produces 13 second averages without even using the most advanced form of the method.

What's the difference between a method being "faster" and "having more potential"?

 

[a small kitten]

It's not faster at the moment but it could/will be because it "produces 13 second averages without the most advanced form of the method".

 

[qqwref]

I disagree. Using OLL and PLL is not the "most advanced form" of Fridrich either - if you want to talk about the most advanced form, a solver would have to have a consistently efficient xcross, use multislotting, and know VHF2L/F2LL/COLL (or even ZBF2L/ZBLL). As for ZZ, I don't think it can get much better than the normal way of doing it (EOLine, RUL F2L with blockbuilding, OLL, PLL... right?), because learning ZZ-B or Winter Variation or COLL or whatever would only save about a second or two. So just because a 13 second average is possible without the "most advanced" form of ZZ does not mean that ZZ is automatically better than Fridrich. Pure Fridrich isn't the most advanced form either, and I don't think most ZZ users are capable of 13 averages, just like I don't think most Fridrich users are.

 

[mazei]

If I'm not mistaken Jason Baum is learning ZB stuff.

 

[JohnnyA]

I disagree. Using OLL and PLL is not the "most advanced form" of Fridrich either - if you want to talk about the most advanced form, a solver would have to have a consistently efficient xcross, use multislotting, and know VHF2L/F2LL/COLL (or even ZBF2L/ZBLL). As for ZZ, I don't think it can get much better than the normal way of doing it (EOLine, RUL F2L with blockbuilding, OLL, PLL... right?), because learning ZZ-B or Winter Variation or COLL or whatever would only save about a second or two. So just because a 13 second average is possible without the "most advanced" form of ZZ does not mean that ZZ is automatically better than Fridrich. Pure Fridrich isn't the most advanced form either, and I don't think most ZZ users are capable of 13 averages, just like I don't think most Fridrich users are.

1LLL will take a lot of time off a ZZ time because it reduces the recognition count by 1 algorithm. For ZBF2L, you still have to recognise two "last layer" algorithms and you only save the time that it takes to do one F2L pair, which especially on the last pair isn't a long time at all. And even COLL+EPLL on ZZ will save a lot of time because 1/12 ELL skip, 2/12 X perm or whatever the stats for EPLL are ... all EPLL algs are fast and this makes the ZZ time go down alot, especially becuase of the four-sticker recog system for COLL.

As for Fridrich, I am not talking about the "uber" Fridrich method with Triple slot+ZBLL or whatever somebody may learn in the future ... I am talking about the most advanced COMMON method which is actually practical for high standard cubers to learn. The equivalent of OLL+PLL in ZZ could be ZZ-B (coll+epll) since it does not need an immense amount of algs, and the "uber" fridrich method could be a parallel to ZZ-D.

 

[qqwref]

The equivalent of OLL+PLL in ZZ could be ZZ-B (coll+epll) since it does not need an immense amount of algs

Yeah, this is closer. It's just that people were equating pure Fridrich (OLL/PLL) with the most advanced form of ZZ, which is not really accurate. What did that guy who had a 13 average do for the last layer? Was it corner-OLL/PLL or COLL/EPLL or something else entirely?

 

[JohnnyA]

The equivalent of OLL+PLL in ZZ could be ZZ-B (coll+epll) since it does not need an immense amount of algs

Yeah, this is closer. It's just that people were equating pure Fridrich (OLL/PLL) with the most advanced form of ZZ, which is not really accurate. What did that guy who had a 13 average do for the last layer? Was it corner-OLL/PLL or COLL/EPLL or something else entirely?

He used COLL+EPLL which I would equate to the equivalent of OLL+PLL for ZZ.