Why MGLS??

[brunson]

I also believe that Fridrich is so ridiculously advanced that this silver bullet method isn't ever gonna be taken seriously enough to the point that it can be a threat to Fridrich's dominance

*coughpetruscough*

MGLS is more useful to the Petrus user because you can solve the last slot/OLL in one alg, since we already have EO. You just have to remember to insert the edge. I'm thinking of learning it(EJF2L first).

But what would be move count to get to CLS?

Petrus+CLS has the advantage of orienting edges with two open slots (lower move count), but then the third slot has to be made in <R,U> (more moves on average, but optimized).

Conversely, Fridrich+MGLS exchanges solving the third slot freely with having to do EO in a more restricted manner. After that, they're identical.

Going back to blah's probability calculations (nice work, blah), given the max, min, and average move count for each step, multiplying the move count by the probability of a skip could give better insight into the value of that skip.

Right about now is when someone needs to come along and throw COLL into the mix. ;-)

 

[blah]

Right about now is when someone needs to come along and throw COLL into the mix. ;-)

I've done that before. With a year at my disposal, remember? :rolleyes:

Although I've lost all my figures now (they were all just scribblings on random pieces of paper), I'm quite sure it's nothing big, or I would've switched away from the standard Fridrich LS + OLL + PLL *shrugs*

 

[blah]

Going back to blah's probability calculations (nice work, blah), given the max, min, and average move count for each step, multiplying the move count by the probability of a skip could give better insight into the value of that skip.

Yeah someone seriously needs to do that. I'm pretty sure it's beyond my ability now

Now, for some more numbers:

Without pre-EO,
P(getting -look LS + LL with ELS + CLS + PLL) = 0.0286
P(getting -look LS + LL with LS + OLL + PLL) = 0.0250
P(getting -look LS + LL with ZBF2L + COLL + EPLL) = 0.0898
P(getting -look LS + LL with ZBF2L + ZBLL) = 1.0000 (duh)

With pre-EO,
P(getting -look LS + LL with ELS + CLS + PLL) = 0.2131
P(getting -look LS + LL with LS + OLL + PLL) = 0.0250
P(getting -look LS + LL with LS + COLL + EPLL) = 0.1011
P(getting -look LS + LL with ZBF2L + ZBLL) = 1.0000

Now, obviously, with pre-EO (using Petrus or ZZ), move count for ELS and ZBF2L would be lower.

In case you haven't yet figured it out, the 0.2131 is largely due to the 20% coming from ELS skip. And even if you don't skip this step, it's only gonna cost you 3.25 moves (3*0.75 + 4*0.25) on average. In other words, if you use pre-EO + MGLS, it's true that you'd still have a 3-look LS + LL, but one of the steps will cost you 2.6 moves (0 + 3*0.6 + 4*0.2) on average, even with recognition time included, that's hardly gonna cost you a second. So I think it's pretty obvious that MGLS is the way to go if you use Petrus or ZZ, unless I find more surprising figures tomorrow with WV :rolleyes:

I think it's worth mentioning (if it isn't already obvious) that skipping a step is much more significant than saving moves, because you skip the moves and the recognition time.

I'll add WV and other LS + LL methods tomorrow, just ask for it if your method's not here. It's 2 in the morning here and my parents have been nagging since, well, 11

By the way, can someone double check the figures for me please? As I've mentioned, it's 2 in the morning

 

[CanadianPires]

Hmm, interesting stuff everyone. Here's what I was pondering. Suppose we have the situation where pre-EO is being used, i.e. ZZ or Petrus. I would like to analyze "ELS + CLS + PLL" vs "LS + COLL + EPLL". I am not really concerned with ZB as I never plan on learning all those algs (others may feel differently) and standard Fridrich because the focus here is on MGLS. Also, I'm sure we are all very familiar with the pro's and con's of Fridrich and Fridrich vs either of the above two cases are open for discussion.

Now, blah states that he believes "it's pretty obvious that MGLS is the way to go if you use Petrus or ZZ", however I'm not so sure about that, and instead think COLL might be the way to go.

First off, I think we can all agree that EPLL is much more desirable than PLL and thus the real comparison is between "ELS + CLS" and "LS + COLL". Furthermore, "ELS + CLS" will have to be significantly better than "LS + COLL" to overcome that fact that following it is PLL and not EPLL. Although, since we are considering pre-EO, ELS is really just a 3 move edge insertion, as mentioned previously.

We start by comparing ELS and LS. It is true that ELS would be dead simple here and easy to recognize, but this still costs the cuber a "look" which is always precious recognition time wasted. Also, ELS has a higher chance of being skipped. Although I don't have the numbers for everything, I think the time spent recognizing the situation (a skip or finding the edge to insert) and inserting the edge would not be much faster than the same amount of time for situation recognition and inserting an F2L pair. It would definitely be faster, but by a lot? Intuitively, I don't think so. Therefore, I don't think ELS blows LS out of the water. Although it is a clear winner here.

Then to COLL and CLS. According to Lucas's site, CLS has 56 algs + mirrors, more than 41 algs for COLL (minus the 7 that everyone already knows, so really 34). As for recognition time, is one really any faster than the other? I'm sure that after working with both for a period of time the recognition would be comparable. Certainly I don't think one is a clear winner in this category. Probability of CLS skip vs COLL skip, no idea and would be interested in finding out if one was way higher than the other. As for move count, 9.78 for COLL and ~9.5 for CLS (according to Lucas's site again), so comparable. Of course, one can always make the claim to 2gen which is attractive in CLS. Although CLS has to work from FR corner --> not so desirable due to (sometimes) mandatory cube rotations.

My "intuitive" conclusion:

LS and ELS = ELS clear winner
COLL and CLS = comparable
EPLL and PLL = EPLL clear winner

But, I think EPLL's margin over PLL is far greater than ELS's over LS, therefore...

--> use LS + COLL + EPLL if you have pre-EO

I'm interested in what you all think about that? Agree, disagree...

 

[brunson]

Though I'm a big fan of MGLS (about 1/3 of the way through learning CLS), I think the biggest advantage of COLL+EPLL is the 1/12 possibility of an EPLL skip. Still, I have such issues with the recognition of COLL whenever I decide to pursue it, I've never gone anywhere with it.

 

[Dene]

I'll be honest: I skim read most of the big posts.

I just have to say this:

blah: You are putting way too much emphasis on skipping steps and lowest move count. I mean, sure it's nice to skip a step. But unless the probability is as high as 50% to skip a particular step it really isn't worth looking into as a very specific deviation from a method (in this case Fridrich). I mean, the chances of an ELS skip are this and that, and the probability of a LS skip are that and this. But unless you have the lookahead and the finger speed the value of the skip is going to be minimal.
I don't think Mr. Pochmann would quite agree with me, but I agree with him that there is no "silver bullet" method. You just have to work hard, like anything else in life.

 

[blah]

I'll be honest: I skim read most of the big posts.

I just have to say this:

blah: You are putting way too much emphasis on skipping steps and lowest move count. I mean, sure it's nice to skip a step. But unless the probability is as high as 50% to skip a particular step it really isn't worth looking into as a very specific deviation from a method (in this case Fridrich). I mean, the chances of an ELS skip are this and that, and the probability of a LS skip are that and this. But unless you have the lookahead and the finger speed the value of the skip is going to be minimal.

After I semi-abandoned my idea for the step-skipping thing, I started working in an alternate direction.

I was looking to combine different methods such that we exploit their advantages and minimize their disadvantages. What are the chances of you getting 2-look LS + LL if you know all Fridrich F2L, OLL, PLL, ELS, CLS, ELL, CLL, COLL, etc.? See where I'm going? All these combined is still way less than full ZB. And I believe (well, actually I just hope) there's such a combination out there that can yield >50% chance of skipping at least 1 step in LS + LL.

I don't think Mr. Pochmann would quite agree with me, but I agree with him that there is no "silver bullet" method. You just have to work hard, like anything else in life.

There are others out there doing physical hard work on others' intellectual hard work, i.e. practicing others' methods/ideas. I know that physically, I'm never gonna be good enough to break the WR or get anywhere near world-class, so I'm working hard intellectually, so that more physically gifted individuals who happen to like my ideas will put them into practice and we'll see the results. I don't see what's wrong with working towards a silver bullet method whether it exists or not, you can't just remain stagnant in cubing.

 

[Stefan]

Actually, I still believe there's a silver bullet method yet to be discovered

It reminds me very much of comparison-based sorting. There are many different sorting methods and some are O(n log n), like heapsort, mergesort and quicksort. But none are better. Is it because we just haven't found a faster one yet? Could some day some genius invent a faster one? No, there can't be a faster one. It's been proven. And we just haven't done that for cubing methods yet.

P(ELS skip) * P(CLS skip) = P(LS skip) * P(OLL skip) since they achieve the same thing

... and have AUFs before and between the steps. If you had two steps achieving the same but not profiting from AUFs before/between the steps, this would not have the same skip probability.

I understand what you're trying to say, but I was thinking a little differently and I was wondering if you could point out the logical flaw in it, or otherwise verify that it's correct.

P(ELS skip) * P(CLS skip) * P(PLL skip) = P(LS skip) * P(OLL skip) * P(PLL skip) (= P(LS and LL skip))

So dividing both sides by P(PLL skip) (which is 1/288 disregarding the benefit of AUF), it follows that P(ELS skip) * P(CLS skip) = P(LS skip) * P(OLL skip). And this can be proven without taking AUFs into consideration, no?

Depends on what you consider a skip. If after F2L you just need U2 to finish the solve, we'd usually call that an LL skip, right? Now compare these two methods:

G method:
Step 1: Solve the cube

J method:
Step 1: Solve F2L
Step 2: Solve LL

These methods achieve the same thing (solve the cube) but I'd say P(G1 skip) is less than P(J1 skip)*P(J2 skip).

I'm not happy with my earlier explanation, though, that might be wrong. I'm a bit confused.