One-Answer Puzzle Theory Question Thread

Niklas is 7 moves. And the fastest way to swap two adjacent corners, with the block at the back left, is probably the JB PLL

 

An alternative is R2 D' r U2 r' D R U2 R (or one of its reflections/inverses). It's just the standard headlights OCLL but with some wide moves, and it's also optimal for swapping two edges while preserving the block.

 

That's perfect, thanks. I did have to mirror it because I don't normally use that one, but that works perfectly.

Yeah, for Niklas I just meant until the 2-gen part it's 5 moves, should've been more specific. I would use the Jb PLL but I was pretty sure it wasn't optimal.

 

I wanted to find some algs for TTLL that are more ergonomic, but I have no idea how I can do that. Can someone point me to an alg-generator/cube solver that I can make solve the cube using, say, <RUF>?


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Cube Explorer

 

Amen.

 

First of all I'm confused how PLL is 22 if the formula is

(4! x 4!)/2(4) = 72
Those factorial represent the EP/CP and EO/CO is in only 1 state. 2 is the Permutation Parity and 4 is the AUFs.

Also there are 7 OCLLs?
(1 x 3^3)/8 = 7?

57 OLLs?
(3^3 x 2^3)/4 = 54?

How do you calculate the # of algs if CP solved(not really solved it's just they're 2 gen) If CP is solved there are only 24 TTLLs and 84 ZBLL and just 4 PLL? The PLL is easy to solve but how do you solve it when you're trying to find out how many LSLL cases? If CP is solved* and the edges is and EO is solved how many cases? I don't know how to take in CP solved* calculations.

 

Short answer: for PLL, you're overcounting cases. For OLL and OCLL, you can't just blindly divide by AUF symmetry because you would undercount some cases, and don't forget about the skip case.

Of the six TTLL subsets (each subset being 12 cases and not counting the set of PLLs), exactly one of them has the corners in the solved coset of the <R2,U> subgroup, and exactly one of them has the corners in the solved coset of the <F2,U> subgroup. These subsets are disjoint (i.e. have nothing in common), which is why you have 24 "2-gen" TTLLs (plus the 4 EPLLs).

As for the last part of your question, you have to carefully define what you mean by CP first. Does a H-perm + U2 have solved CP by the definition you want to use?

Long answer: Burnside's lemma.

 

Let's look at the easy one. OCLL has four twisted corners, but the twist of the last one is determined by the other three, so there are 3^3 = 27 cases. Some cases can have rotational symmetry - there is only only one case that has no twisted corners, and that can't be changed to any other case with an AUF. The H case has a different type of symmetry. There are only two H cases - one with the arms pointing to the front and back, and one with the arms pointing left and right. A U2 of an H case doesn't change the case, but a U of U' does. For the remaining cases, there is no rotational symmetry. For example the headlights OCLL has four distinct forms - headlights facing forwards, left, back or right.

So the number of cases is
(6 cases * 4 symmetries) + (1 case * 2 symmetries) + (1 case * 1 symmetry) = 27
If you ignore symmetry, it is 6+1+1 = 8, and ignoring the OCLL skip case, it is 7 cases.

With PLL, there are 72 cases, but skip, H, Na and Nb can't be transformed into other cases by AUFing. Z and E have U2 symmetry so there are two Z and two E perms that can be transformed into each other with a U move. The rest all have no symmetry, so a U move cycles between 4 distinct cases.

That gives (16 * 4) + (2 * 2) + (4 * 1) = 72
But ignoring symmetry, it's (16 + 2 + 4) = 22, or 21 of you exclude the skip case.

This method of listing out all the cases and then counting symmetries works regardless of the restrictions on the cases, so you can still use it to work out the number of cases if CP is solved*

* assuming that by CP solved you actually mean that the corners are such that they can be solved with R and U moves only. If you actually mean all of the corners are literally in their correct positions then there can't be any rotational symmetry which actually makes things easier.

 

If you read again I do mean by CP solved is the cases are now 2-gen. So there's no formula for working out symmetries? I need to work it brutally? So even if CP is solved*(2-gen) the # of cases are the same if I brutally list it out every cases nothing would change?

Since I'm working on a LSLL method I want to know how to calc it properly. For ex. for TTLL
The EO/CO is solved so they're 1.
I can AUF the F2L corner to stay always at UBL and there's no more symmetry right?
So the cases are (4! x 4!)/8 = 72?

So for when CO is not solved? Assuming the F2L corner is at the U(AUFed at UBL always) layer:
4! x 4! x 3^4?

How do yo take in CP solved(2-gen)? Also since we're talking about LSLL we consider these cases where you rotate only by one y or y' (D/D') and the case is now 2-gen* also 2-gen like how Right Opp TTLLs can be solved.

 

Hi
A zzf2l question: I figured, that its more efficient to leave the f2l slot open after ive done a pair. But how many moves do I actually save on average? Is there a way to calculate that? To make the calculation easier, lets say that the next pair is always the pair of the same block and lets assume that I solve that pair optimally using <RUL>.

And is there a difference of saved moves whether its about the first or the second block?