F2L possibilities

[Dennis]

We all know that there are 42 standard Fridrich cases (incl solved) for the last slot and they are calculated as:

• Corner in U layer: 3^1 x 2^5=30 (we restrict the corner to be over it's slot and allow the edge to be in the other 5 positions)
• Corner in D layer: 3^1 x 2^2=12 (we restrict the edge to only be in one position in the U-layer due to AUF and it's slot)

This sums up to 42 cases.

How many cases can we have for the 1st, 2nd and 3rd slot, i.e pair up and insert one corner/edge pair during the F2L progression?

Is it 168 cases for the 1st pair, 114 for the second and 72 for the third?

 

[Godmil]

That's what I'm getting, but I could be wrong. Either way, if you think of the actual number of cases it's pretty clear why F2L takes the longest to get good at.

 

[cuBerBruce]

I would say you have the same 168 cases for the 2nd slot as you do for the first slot, unless you want to constrain the order you solve the slots. For the third slot, it's less because you can't have both corner and edge in different wrong slots. I suppose you could also consider exactly which set of slots are already solved (and so must be preserved) to get increased number of cases. Also, some simplifications could made, like counting corner and edge both in same wrong slot as 6 cases rather than 18 cases.

 

[Enter]

@Dennis it is 168 standard cases
but when you add special cases like these http://f2l.net46.net/index.html

 

[Dennis]

I would say you have the same 168 cases for the 2nd slot as you do for the first slot,

Why?

unless you want to constrain the order you solve the slots

No constraints in the order. I don't see why this should affect the number of cases

For the third slot, it's less because you can't have both corner and edge in different wrong slots

That is correct, but I still get 72 cases

Also, some simplifications could made, like counting corner and edge both in same wrong slot as 6 cases rather than 18 cases

Of course, but I wanted the total number of cases incl mirrors

 

[cubernya]

I would love to see this. I could probably figure this out using my maths skillz (lol) but I have enough homework as it is. However, when I get some free time on the weekend, I could definitely calculate the odds of having a single edge in the proper place (and oriented correctly). I could do the same for the corner, and multiplying (not adding!) the two together should give us the number of cases. Please correct me if I'm wrong.

 

[uberCuber]

but I have enough homework as it is. However, when I get some free time on the weekend

>Posts on an online forum about not having free time.

 

[cuBerBruce]

I would say you have the same 168 cases for the 2nd slot as you do for the first slot,

Why?

Let's say you are solving the FR slot for the 2nd slot. If the order of solving the slots is not constrained, then the first slot could have been FL, BL, or BR. This leaves all 4 slots as possible locations for the 2nd slot's pieces. You are calculating the number of cases for the 2nd slot assuming you know which slot was solved first. But people do not generally solve slots in a particular order.

For the third slot, it's less because you can't have both corner and edge in different wrong slots.

That is correct, but I still get 72 cases

I meant less than 168 cases. It is more than 72 cases, because in your calculation of 72 cases, you are assuming which of the other three slots is the one remaining to be solved last. So there are really much more than 72 possibilities for the 3rd slot because there are three possibilities for which slot is the one remaining slot.

 

[Dennis]

@cuBerBruce

I was assuming that we always solved the, lets say FR slot and rotate the cube, lets say y and then calculate the next number of possible cases. By this way we rule out the reflections for the same case, that's why I didn't wrote 42x4 cases for the last slot

I see your point though now.

So if I get it right, for the first slot there are 168x4!, the second slot 114x3! and the third slot 72x2! cases (not including reflections then it would have been 4x on each)?

 

[cuBerBruce]

Well, I was basically also assuming you rotate the cube so that you are always solving the FR slot. But you could use y, y', or y2 to get the next slot you want to solve to FR. So I get 168 cases for the first slot, and all 168 of those still possible for the 2nd slot. (You can choose any slot for the corner and edge and still have a slot left over for the already-solved slot.) For the 3rd slot, there can only be one unsolved slot remaining besides the one being solved, so there are 36 cases of a corner and edge being in different wrong slots are now impossible, leaving 132 cases possible. The other thing I note is that you do not have any new cases for the 2nd, 3rd, and 4th slots than what you had for the first slot. The 132 cases for the 3rd slot and 42 cases for the last slot are just subsets of the 168 cases you can have for the 1st slot.

I note that I am also assuming that we use algs that don't affect "extraneous slots." By an extraneous slot, I mean a slot that is not the one being solved, and does not contain a piece that is to be solved. If you wish to take full advantage of unsolved extraneous slots, you could come up with more cases by assigning extraneous slots to be either "free" or needing to be preserved.

 

[Dennis]

Hi,
that totally blew me away. I had to read it like 5 times and think over-night before I understood it. Thanks

Not fully certain about the "extraneous slots" however...I feel slow :fp

 

[qwr]

I had the same question looking at Advanced F2L guides. There's 41 standard cases to solve to the front right slot, but there are 4 slots to solve into, and it is actually important to be efficient in all of these knowing when a rotation is needed or not. And basic f2l assumes you're solving the last slot and none of the 3 others are free, which usually isn't the case. For completeness there may be 41 * 4 * 2^3 = 1312 cases. Ofc most of the f2l cases only depend on zero or one slot being free, either adjacent or diagonal. So maybe it's more like 41 * 4 * 3 = 492.

 

[Billabob]

I had the same question looking at Advanced F2L guides. There's 41 standard cases to solve to the front right slot, but there are 4 slots to solve into, and it is actually important to be efficient in all of these knowing when a rotation is needed or not. And basic f2l assumes you're solving the last slot and none of the 3 others are free, which usually isn't the case. For completeness there may be 41 * 4 * 2^3 = 1312 cases. Ofc most of the f2l cases only depend on zero or one slot being free, either adjacent or diagonal. So maybe it's more like 41 * 4 * 3 = 492.

Don't forget that introducing other slots to the equation makes things even more complex because pieces may be incorrectly slotted. See the algorithms R' F R2 U R' F', F' R' U R F, etc. In truth advanced F2L is less about learning algorithms and more about planning ahead and modifying your solution to avoid the more awkward cases. But of course knowing the easier/faster ones can never hurt.

 

[qwr]

Don't forget that introducing other slots to the equation makes things even more complex because pieces may be incorrectly slotted. See the algorithms R' F R2 U R' F', F' R' U R F, etc. In truth advanced F2L is less about learning algorithms and more about planning ahead and modifying your solution to avoid the more awkward cases. But of course knowing the easier/faster ones can never hurt.

You're completely right. I don't even consider those because I'm still at the level of hoping I didn't mess up my cross or f2l pairs which happens maybe 10% of the time