# Optimal PLL - 288 cases

#### AvGalen

Stefan's approach used optimal algs for the shortest pair. As you can see by the last pair (6.81) a normal pair takes about 7 moves, but if you can choose from 4 different pairs there is a very high chance that at least one of them can be done in less. That is why the numbers go up
5.03
5.40
5.80
6.81

#### cuBerBruce

##### Member
I've never seen this study before (or Helmstetter's), but I've always wondered what the optimal PLL movecount was.

Helmstetter only uses the best case within each PLL category
So his move count was only off due to the fact that he didn't take into account the 25% chance of a correct PLL (without "AUF")?
Yes, in other words, Helmstetter's number is correct if you consider AUF (to align last layer with first two layers) to be a separate step from PLL, and you do the PLL part optimally. What additional fraction of a move would need to be added to account for AUF depends on your assumptions, but my data indicates at least 0.43 of a move on average is required.
Can you tell how got 0.43 average?
To get the average number of moves, you simply add the lengths of the algs used for each case. Then divide by the number of cases. This, of course, assumes that the 288 PLL permutations are equally likely to occur, and that it is not influenced by some clever selection of what you do for OLL. Picking the shortest alg for each PLL (and the trivial skip case), the lengths sum up to 3228. Divide by 288 and you get approx. 11.21 (Helmstetter's number). Using my data for the alg lengths needed to completely solve each of the 288 cases with respect to the first two layers, the sum comes to 3353. Divide by 288 and you get 11.64. The difference is 125/288 or approximately 0.43. Since in both cases, the use of best case algs were being assumed (for example always using a 9-move H-Perm alg for H-Perm cases, plus AUF if necessary when solving with respect to first two layers), then the difference must be due to AUF moves. Of course you could use a 10-move alg for the canonical H-Perm case (L R U2 L' R' F' B' U2 F B, for example) instead of a 9-move alg plus an AUF. You technically reduce the number of AUF moves doing this, but you're not gaining anything in overall move count.

I'd be interested in seeing this done for the OLL as well.
AUF is not necessary for OLL, so I think Helmstetter's site has the raw data for this, and I would presume the correct average. It does not seem to a have a summary table of number of cases for each alg length, though. I note that Helmstetter's site treats mirrors (and inverses, when applicable) as the same cases, so the number of cases he has for PLL and OLL are less than the usual numbers given. (I also note that OLL cases do not have inverse cases. This is because OLL algs are free to permute the pieces around, so inverses of algs for the same OLL case do not in general have the same OLL effect.)
As far as a "AUF" being necessary or not for OLL: wouldn't that actually affect the overall move average? If you can position an OLL in a fashion that, when executed, creates a greater "chance" of having more pieces into the correct location, wouldn't that decrease the "chance" of needing to AUF during PLL? Perhaps this has something to do with where you got 0.43.
Again, an AUF move does nothing for you at OLL time. It doesn't change the orientation of any pieces. In speedcubing, an AUF move might be used instead of a cube rotation to put the last layer at the correct angle for an alg you want to use, but for optimal FTM solving, you would just do a cube rotation, or translate the alg to compensate for the "wrong" angle. You seem to be suggesting something along the same lines as blade740, and I don't understand how you are expecting to gain something.

And there isn't really an exact consensus about the average move count for a traditional Fridrich solve. Each seperate case could be calculated (should come done to approximately 5+4*7+12+13 = 58)
Less approximately, using my data for F2L, Bernard's for OLL and Bruce's for PLL, we get for using optimal algs:
5.82 + 5.03 + 5.40 + 5.80 + 6.81 + 9.22 + 11.64 = 49.72
Stefan, I don't think I ever saw the complete study and/or details to this. Was this the study where you looked at various members example solves? If so, perhaps you could study the F2L by using optimal solutions to each CE pair case. Is the 5.82 for the cross the average optimal move count for color neutrality, if not, where did you get 5.82?
Lars Vandenbergh's cross study page (http://www.cubezone.be/crossstudy.html) lists 5.81 for color-specific cross, and 4.81 for color neutral cross. So you would eliminate basically one more move using color neutral solving.

#### Zeroknight

##### Member
I always knew the "new generation" of cubers was the generation who just wanted to solve the cube fast without wanting to understand it. This guy's posting original cubing research on the forum, and your criticism is that it's not a question so you don't understand why he would have posted it? Go back to posting comments on Youtube, your level of intelligence fits in much better there...
Hey, hey, hey, don't be too hasty with those generalizations. I would really like someone to teach me cube theory, but in an easy, understandable way, you know?

#### Stefan

##### Member
And there isn't really an exact consensus about the average move count for a traditional Fridrich solve. Each seperate case could be calculated (should come done to approximately 5+4*7+12+13 = 58)
Less approximately, using my data for F2L, Bernard's for OLL and Bruce's for PLL, we get for using optimal algs:
5.82 + 5.03 + 5.40 + 5.80 + 6.81 + 9.22 + 11.64 = 49.72
Stefan, I don't think I ever saw the complete study and/or details to this. Was this the study where you looked at various members example solves?
No, it's from my diploma thesis:
http://www.stefan-pochmann.info/hume/hume_diploma_thesis.pdf
Page 32, the "CrossF2L" side of the "Greedy order" table. For more information about the meaning of the numbers, start reading at page 30. Basically it's the "standard way", like most of us solve F2L (fixed cross color, then always solving the easiest pair). And the data is experimental, average of 10000 solves, so it's quite accurate but not absolutely exact (which is why I'm 0.01 off for the cross compared to Lars' data).

Last edited: