# Thread: ZZ F2L Move Count

1. ## ZZ F2L Move Count

Does anyone know the average and maximum move count for completing ZZ F2L (from EOLine) optimally, using <R,U,L> moves?

I found some statistics for completing F2L from EOCross here, but would be interested to know how many moves it takes from EOLine.

If possible I'd like the averages for F2L broken down in the following ways:

(1) L & R 1x2x3 blocks completed simultaneously
(2) Entire 1x2x3 L block, followed by entire 1x2x3 R block
(3) BL 1x2x2 followed by FL 1x1x2, then BR 1x2x2 followed by FR 1x1x2
(4) BL 1x2x2 followed by BR 1x2x2, then FL 1x1x2 followed by FR 1x1x2

Cheers.

EDIT: Also, how does it compare with the optimal move count using <L,R,U,D,F2,B2>?

2. You cannot flip an edge using only <R,U,L> moves. Wouldn't you need to do at least one cube rotation or an r,u,l,F,B,M,E, or S? Sorry I cannot help with your primary questions.

Edit: Oops, sorry, I didn't know what EOLine meant before I posted.

3. Originally Posted by rjohnson_8ball
You cannot flip an edge using only <R,U,L> moves. Wouldn't you need to do at least one cube rotation or an r,u,l,F,B,M,E, or S? Sorry I cannot help with your primary questions.
I'm looking for the statistics to complete F2L from EOLine (edges already oriented and DF and DB placed). EOLine takes ~6.127 moves avg and 9 moves max (FTM) to do optimally.

4. Perhaps PM Johannes91
I did and got lots of data for Roux move count.

...looks like another coding job, but I'll have to put it on the back burner for now, too many other things to do! If anyone can find these stats in the mean time I'd be very grateful, along with a few other ZZ'ers

EDIT: Just for some quick info on the task involved, these are the numbers of cases which need to be solved:
(1) L + R blocks: 3^4 * (8!/4!) * (10!/4!) = 20,575,296,000 <-- a biggie
(2) L: 8*7 * 10*9*8 * 3^2 = 362,880 R: 6*5 * 7*6*5 * 3^2 = 56,700 (419,580 total)
(3) BL 1x2x2: 8 * 10*9 * 3 = 2160, FL 1x1x2: 7 * 8 * 3 = 168, BR 1x2x2: 6 * 7*6 * 3 = 756, FR 1x1x2: 5 * 5 * 3 = 75 (3159 total)
(4) BL 1x2x2: 8 * 10*9 * 3 = 2160, BR 1x2x2: 7 * 8*7 * 3 = 1176, FL 1x1x2: 6 * 6 * 3 = 108, FR 1x1x2: 5 * 5 * 3 = 75 (3519 total)

6. Originally Posted by Cride5
(3) BL 1x2x2: 8 * 10*9 * 3 = 2160, FL 1x1x2: 7 * 8 * 3 = 168, BR 1x2x2: 6 * 7*6 * 3 = 756, FR 1x1x2: 5 * 5 * 3 = 75 (3159 total)
Coincidence?

7. Originally Posted by Lt-UnReaL
Originally Posted by Cride5
(3) BL 1x2x2: 8 * 10*9 * 3 = 2160, FL 1x1x2: 7 * 8 * 3 = 168, BR 1x2x2: 6 * 7*6 * 3 = 756, FR 1x1x2: 5 * 5 * 3 = 75 (3159 total)
Coincidence?
Yup. There is a reason the total number of cases for (3) and (4) are different. The number of cases depends on (a) the number of pieces involved and (b) the number of possible orientations/permutations of these pieces.

Ignoring the first and last cases which have equal case counts: In (4) doing the 1x2x2 first means that (a) and (b) are both maximal and then both minimal when multiplied together, where as in (3) when (a) is maximal (b) is not and vice versa, yielding a lower overall total.

8. Originally Posted by StefanPochmann
That's great, cheers ... edging closer to the goal

So based on 100 random scrambles the statistics for strategy (4) is ~24.64 moves or ~30.77 including EOLine.

You wouldn't happen to have kept the averages for each sub-step would you?

9. Originally Posted by Cride5
So based on 100 random scrambles the statistics for strategy (4) is ~24.64 moves or ~30.77 including EOLine.
No. Your (4) solves the four 2x2x1 blocks in a specific order. I didn't. Always solved the shortest one first.

Originally Posted by Cride5
You wouldn't happen to have kept the averages for each sub-step would you?
Looks like I didn't keep them. Can compute it again though, but after the world champs.

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