How does one determine how many cases are in a given alg set?

[Caleb Paquette]

Looking to figure out how to find the number of algs i would need to create for a ll/ls idea

 

[xyzzy]

"A lot" is the short answer. There's no viable LSLL method that doesn't involve 100+ algs.

Long answer: for each step of your method, consider these:

(i) How many corners are affected?
(ii) Is any aspect of the corners already solved? (E.g. before ZZ-CT's TTLL step, the corners are all oriented.)
(iii) What aspects of the corners are being solved in this step? (E.g. in TTLL, the already-oriented corners are to be permuted.)
(iv) The above three points, but for edges.
(v) Keep in mind that the permutation parity of the corners and the permutation parity of the edges will always be equal. This will sometimes contribute a 1/2 factor to the case count.
(vi) Is there any symmetry? Do you want to consider mirrored cases to be the same? (Note that for LSLL methods, the answer is typically "no" to both of these.)
(vii) Do you consider AUFed versions of a case to be the same? (Note that there are sixteen combinations of possible AUFs before and after a case.)

 

[Caleb Paquette]

All great things to keep in mind, but how to you use these to calculate the exact number?

 

[xyzzy]

You just count!

Let's look at Winter variation (in the FR slot), say. You start with the FR pair formed and hovering above its slot, and you want to end with all corners oriented. (Edges are oriented both before and after this step, and we don't care about edge permutation, so those don't contribute to the number of cases.) This means that you're moving the pieces in URF and UR to FRD and FR as well as orienting the remaining four corners. The first part has just one sub-case (you're just moving the pieces in a fixed way every time). The second part has 3^(4−1) = 27 cases, since that's how many orientations four corners can take. (The orientation of the last corner is uniquely determined by the others.) These multiply to 27 cases.

Spoiler: An example for OLS

Consider something like OLS with edges already oriented, still looking at only the FR slot. Note that this time, we don't have the FR pair pre-formed. We start with the DFR corner in any of five possible locations (momentarily ignoring its orientation); the FR edge in any of five possible locations; the LSLL corners in any of 3^(5−1) = 81 possible orientations; and we want to solve all of these. Multiply these together, and you get 2025 cases in total.

But this time you also have to determine whether you care about pre-AUF! (Post-AUF doesn't change the solved state, so we don't need to care.) Do you consider R U R' and U2 R U R' to be algs for the same case? Let's say you do. Here, we can split the cases into four subsets based on where the FR edge and DFR corner are.

(a) If they're both in the slot, you're free to AUF the top layer however, which turns out to have 8 cases for the top layer and 3 cases for how the DFR corner is twisted, so 24 cases in total (1 of which is solved and 7 of which are normal OLL cases).

(b) If only the FR edge is in the slot, you can AUF the DFR corner to URF, so we can repeat the above computation with the necessary adjustments to get 3^(5−1) = 81 cases.

(c) Likewise, if only the DFR corner is in the slot, we also get 81 cases.

(d) The last bunch of cases is where both pieces aren't in the slot; we can AUF the DFR corner to URF, and we get 4 (location of FR) × 3^(5−1) = 324 cases in this subset.

Add these all up, and we have 510 cases in total for OLS with oriented edges.

And you can do the same thing with most methods where things are solved in a fixed order. Accounting for AUF is usually not too hard, although it can be tedious.

 

[Caleb Paquette]

thank you!