Fridrich Corner-Edge Question

[Cubie Newbie]

Just curious, does anyone know the average probability that, when using the Fridrich method for the F2L, there are NO available corner-edge pairs in the top layer and you HAVE to bring either a corner or an edge up from the F2L in order to proceed?

I know the probability is very small, but I'm just wondering, because I almost ALWAYS have to bring either a corner or an edge up from the F2L. Is it just that I'm not spotting the pairs that already exist???:confused:

 

[FU]

Do you mean that, the top layer has only yellow pieces, or that the top layer doesnt contain both pieces of any pair?

 

[Cubie Newbie]

I meant the latter. (top layer doesn't contain both pieces of ANY pair)

 

[jonny guitar]

I am not sure but I know that you can eliminate that problem almost 100% by paying attention to the last cross insertion and then make corner/edge pair insertion choices that set up the next insertion pair. A little planning ahead and you will avoid it almost every time.

 

[Kare]

Its not really a problem as long as one of the pieces is in the top layer. Most F2L algs contain two phases, line up the pieces and then insert the pair.

Example:
Normal F2L: U' R U' R' U' R U2 R'

Now if the needed edge is not in the FR-slot but in the BL I can still use the same idea. (I put the corner above the edge)

Pair up: U' L U' L', insert: R U' R'

As long as you have one of the pieces in the top layer you should be able to pick up the other in such a way that you pair them up and put them in the position for a RUR'. Only when you cant find any usefull pieces in the top layer will you have to waste moves to pick up a piece before you start building again.

Edit: Forgot to translate some swedish words

 

[Cubie Newbie]

Okay, so I guess an intuitive and/or logical understanding of how the cubies move would be beneficial to the F2L.

 

[MiloD]

:confused::confused::confused:

 

[TimMc]

Pr(All Corners and Edges are slotted in F2L position but aren't paired correctly) = ?

Tim.

 

[badmephisto]

case that all corners are on cross layer after cross has probability
4/8 * 3/7 * 2/6 * 1/5 = 1.42857143 %
case that all edges are not on top layer after cross has the same probability since there are 4 edge positions on last layer and 4 edge positions in middle layer.
So chance that not a single F2L edge or corner is on top layer is that squared, or
0.0204081633%

i.e. about 1 in 5000


EDIT: the question you are asking though: the probability that you have to bring something up from the middle/cross layer is a lot more complicated and depends on number of pairs you already filled in and stuff. But for example if you are working on the last slot, the probability of you having to bring something up would be
1- 4/5*4/5 or exactly 36%... so about every 3rd time... which is pretty large

 

[Lucas Garron]

What math are you doing?
Anyhow, with pure solving:

3 slots left: 71/140 (50.71%)
3 slots left: 109/245 (44.49%)
2 slots left: 17/45 (37.78%)
2 slots left: 9/25 (36%)

Does anyone get the same numbers?

 

[Swordsman Kirby]

I wouldn't consider that happening to be bad every single time.

 

[badmephisto]

Lucas I have no idea what you are doing. The prob. of having to bring something up? In case of 1 slot only being left now, which makes it 5 possible positions for both corner and edge,

P(must bring stuff up) = P(Corner up edge stuck) + P(Edge up corner stuck) + P(both stuck)
= 4/5*1/5 + 4/5*1/5 + 1/5*1/5
= 36%

equivalently,
P(must bring stuff up) = 1 - P(Corner up AND Edge up already) , since in every other case at least something is stuck
= 1 - 4/5 * 4/5
= 36%

 

[Lucas Garron]

Lucas I have no idea what you are doing. The prob. of having to bring something up?

Yes. I posted these immediately before going to bed, and had no time to check. I made a silly error (assumed that with n slots left there are 2n pieces in the top layer ), and fixing it gave some nice numbers immediately.

I fixed the numbers in my first post. You should have to "bring up" something about half the time for your first F2L, and a little less often for every next slot.

 

[cuBerBruce]

I've done a brute-force computer calculation of the number of C/E pairs with both members in the top layer versus number of solved C/E pairs. That distribution is:

           number of C/E pairs with both members in top layer
slots       0           1           2           3         4       total
solved ---------  ----------   ---------    --------    ------  --------- 
  0    924307849  1843208832   780492672    71165952    746496  3619921801
  1      9506372    19935360     7776000      497664         0    37715396
  2        52062      109440       31104           0         0      192606
  3          212         384           0           0         0         596
  4            1           0           0           0         0           1
       ---------  ----------   ---------    --------    ------  ----------
       933866496  1863254016   788299776    71663616    746496  3657830400

In the table below are the exact fractions of positions that have no pairs where both members are in the top layer. (Decimals are approximate.) I am stopping short of calling these probabilities, because it seems to me that F2L algs might have a tendency of leaving pieces in the slots where they are, and your choice of which pair to solve is not totally random. So it seems to me that as you solve F2L pairs, that the remaining F2L pieces might not be distributed "randomly" (that is, all possible remaining positions may not have equal probability).

Fraction of configurations without any unsolved C/E pairs
      having both members in the top layer
with respect to number of remaining unsolved C/E pairs

             Exactly n                          n or less
n            remaining                          remaining
-            ---------                          ---------
4  924307849/3619921801 = 0.255339   933866496/3657830400 = 0.255306
3    9506372/37715396   = 0.252055     9558647/37908599   = 0.252150
2      52062/192606     = 0.270303       52275/193203     = 0.270570
1        212/596        = 0.355705         213/597        = 0.356784

 

[Cubie Newbie]

Yet if I do intuitive F2L, does that make cuBerBruce's numbers more accurate as probabilities?

 

[cuBerBruce]

Yet if I do intuitive F2L, does that make cuBerBruce's numbers more accurate as probabilities?

To my understanding, intuitive F2L is a general approach to solving F2L, so I think it's hard to draw a conclusion as to how it would affect the actual probabilities. Intuitive F2L doesn't imply any particular manner of selecting which pair to solve next, for instance.

I also note that if you have two pairs left, and then solve one more pair, you would generally expect to have one slot left. But (assuming the 150 possible configurations for the last corner and edge are equally likely), you would have the following probabilities:

Both in top layer: (4/5)*(4/5) = .64
Corner in top layer, edge in slot: (4/5)*(1/5) = .16
Corner in slot, edge in top layer: (1/5)*(4/5) = .16
Both in slot, but at least one mis-oriented: (6-1)/150 = .03333...
Both in slot and correctly oriented (i.e, solved): 1/150 = .00666...

So the probability that the last slot is not solved and at least one of the pieces is not in the top layer is .16 + .16 + .03333... = .35333...

The probability that at least one of the pieces for the last slot is not in the top layer is .36, as badmephisto had calculated. But note that this includes the case of the last slot being solved. Of course, if the last slot is solved, you don't have to move the pieces to the top layer, then solve them again.

I note that in my data, the 212/596 (or 53/149) is the probability for having one unsolved slot (and you know that this last slot is not inadvertantly solved) and at least one piece not in the top layer. The 212/597 figure assumes any of the four slots could be left (except the case where no slot is left, of course), but since you know what slots you solve along the way, I think it's really incorrect to count the configurations for all four cases as far as which slot is the last one. So the 212/597 should really be 54/150 (36%), or 53/150 if you want to count the last slot solved as a possible configuration, but not count it as a "bad" case.

Basically, if you know that n slots are already solved, then configurations where any of those n slots are unsolved should be counted as 0% likely, rather than equally likely. So I would consider the figures my table for 3 or less, 2 or less, 1 or less to be wrong, and I think exactly 3 left and exactly 2 left might also be wrong.