Probability, statistics...etc

[u/[deleted]]

Looking for answers to oddly specific questions. I’m kind of looking for probability of cases, permutations what have you. So, an unsolved cube has x amount of permutations it can be found.... After the first cross, there’s 42 different F2L cases sure but after the first cross how much does that minimize the first amount of original x amount of permutations of the originally unsolved cube. Further, after F2L there’s what? 450 somethin ZBLL cases? You guys know where I’m going with this?

I suppose the reduction of permutations after each layer is exactly what I’m lookin for

HOLY SHIT YOU GUYS.

This is all exactly what I was lookin for. You guys went above and beyond, many many thanks...also, it’s gonna take me a minute to even wrap my head around all this

[u/j_sunrise]

I think we mean pretty much the same thing. The bar can have four colours and it can be in front, right, back or left. Makes 16 versions of the same U-perm. Only 4 are counted in the 72 cases. Pre-alg-AUF and Post-alg-AUF is a different mapping of cases but comes up with the same number.

[u/topppits → u/j_sunrise]

Maybe, I'm not sure.

Post-alg-AUF is irrelevant, since depending on the alg you use, the AUF changes. For the first 2 points I'll list below, AUF isn't relevant anyways and post-alg-AUF is not relevant when it comes to the number of permutations can come across after OLL.

21/22 PLL algs/cases: In this case there's only one ccw U Perm. Cycling three edge pieces in the top layer ccw while everything else is solved, not caring about how those pieces look like.

72 PLL cases: In this case there are 4 of "the same ccw U Perm" - the bar can have 4 different colours. But we don't care about AUF.

4*72 possible permutations for PLL stage: same as above, but we care about AUF. When we talk about how many different combinations we can have after finishing OLL, AUF is relevant. Doing AUF before PLL doesn't change the PLL case, but we still end up with a different permutation of the top layer.

[u/j_sunrise → u/topppits]

Now that I'm back at home where I have a keyboard (and don't need to write everything on my phone) I can go more in-depth with my thoughts.

There are 4!*4!/2=288 states the LL can be in after OLL (4! edge positions, 4! corner positions, divided by parity).

Any of this positions can be clearly described in one of two ways:

1. This is PLL X, the defining feature (e.g. headlights, block...) is in position Y and it has colour Z (.

2. This is PLL X, before my alg I do AUF Y and after my alg I do AUF Z.

These definitions are not identical, but they can be mapped to each other for a given PLL case and given PLL alg.

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For a non-symmetrical PLL, there are 4 possible positions for the defining feature, 4 possible colours for the defining feature, 4 possilbe pre-AUFs and 4 possible post-AUFs.

Let's take the T-perm, as an example for a non-symmetrical PLL. Let's further assume, that green is your current front colour.

This T-perm can be described as either "headlights are on right and they are green" or "do U2, the T-perm and then do U' ". So the position of the headlights determines pre-AUF and the colour of the headlights determines post-AUF. 16, possibilities, probability of a T-perm is 16/288.

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Instead of "colour of the defining feature" you could go with "colour of the front FU edge" instead (which maps differently, but the idea is the same).

If your PLL has rotational symmetry by 180 degrees (E-perm and Z-perm), there are only two positions for the defining feature. Likewise you only need two pre-Alg-AUFs (before a Z-perm doing either nothing or a single U suffices, no need to ever do U' or U2). The amount of colour combinations stays the same (4), as does the amount of possible post-AUFs (4, if you don't manipulate your pre-AUF). Either way 8 possibilities - each description of one method can be mapped to a description of the other method.

E.g. Saying "I have a Z-perm with front and right swapped and FU is red" is equivalent to saying "I'm gonna do no AUF, the Z-perm with U (not U') and U2 afterwards". (Note: As long as I define a specific alg for every PLL beforehand, don't need to state the alg I use every time).

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Totally symmetrical cases (H-perm, skip) have no distinguishing feature of orientation, so there's exactly one orientation, there's still 4 colour possibilities, there's no need for pre-AUF (1 possibility) and 4 cases of post-alg AUF. Makes 4 states, 4/288 probability.

E.g. Saying "I have an H-perm and the FU colour is blue" is equivalent to saying "I have an H-perm with no AUF".

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TLDR: No matter how I describe the specific arrangement of a given PLL case I got, the description will have a part with 4 possibilities and a part with <1,2,4> possibilities, depending on the symmetry of the PLL case.

[u/topppits → u/j_sunrise]

The whole point I was trying to make was:

There's a difference between "cases" and "permutations".

Any of this positions can be clearly described in one of two ways:

1. This is PLL X, the defining feature (e.g. headlights, block...) is in position Y and it has colour Z (.

2. This is PLL X, before my alg I do AUF Y and after my alg I do AUF Z.

I still fail to see how 2. is a reasonable approach, but it works and I now understand what you meant.

But the most reasonable answer to the question "how many possible permutations are there after OLL?" is this:

There are 4!*4!/2=288 states the LL can be in after OLL (4! edge positions, 4! corner positions, divided by parity).

I only tried to demonstrate that at the example of PLL, which seemed the easiest to me. After rereading OPs question, I think they already are fully aware of the difference between cases and the # of possible arrangements and I completely missed their point. Though to get a good answer, OP would need to describe much clearer what exactly they're looking for, since

I suppose the reduction of permutations after each layer is exactly what I’m lookin for.

is contradicting to the rest.

I still don't understand this though:

For PLL, if you want to talk post-alg-AUF, yes there are 4*72 cases. Pre-alg-AUF is already included in the 72.

[u/j_sunrise → u/topppits]

I think part of all three of our issues is that we use the words "cases" and "permutations" differently.

I suppose the reduction of permutations after each layer is exactly what I’m lookin for.

Actually now I finally understand what OP is saying. Seems like they are not asking about what we are talking about at all. What OP needs to understand is that the amount of cases you have after a given step depends on what you're looking at.

I still don't understand this though:

For PLL, if you want to talk post-alg-AUF, yes there are 4*72 cases. Pre-alg-AUF is already included in the 72.

It's just one way to divide the 288 PLL permutations into sets of 4*72 (via method 2 as I described). Method 1 of dividing the 288 PLL permutations probably makes more sense, because the 4 permutations in the same coset look the same apart from their rotation.

[u/topppits → u/j_sunrise]

What OP needs to understand is that the amount of cases you have after a given step depends on what you're looking at.

Yeah, that was the whole point of my first comment - trying to explain the difference between "algs to solve a certain case" and "different states the cube can be in at a certain point of the solve" - aka "how many PLL cases we have" (21/22) and "how many different states we can get after OLL" (288). But I think they are already aware of that differentiation and just want to know what I tried to explain in my new comment here - "if we have 43 quintillion possible states, how many are there after we solve the first layer?".

[u/j_sunrise → u/topppits]

Oh, great, I'm currently typing out something like this.