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Thom S.

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Phasing is on the Wiki
ZZ-d is on the Wiki

CPLS is solving the last pair while also solving Corner Permutation of the LL, meaning the rest of the Solve is 2G.
Phasing (round down) is solving LL Edges with the last slot so that 2 Opposite Edges are solved relative to each other, cutting down on ZBLL cases to be learnt.
 

xyzzy

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ZZ-d is on the Wiki

CPLS is solving the last pair while also solving Corner Permutation of the LL, meaning the rest of the Solve is 2G.
An apparently-significant chunk of the ZZ community decided to repurpose the "ZZ-d" name for CPLS (or more generally, any sort of early-CP solving) rather than the original "do 2-gen reduction while finishing the left block", but also the wiki has an internally-inconsistent explanation of ZZ-d that superficially sounds like the old ZZ-d while also mentioning CPLS, which is not used in the old ZZ-d at all.

It's understandable that it's confusing.

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tl;dr: "ZZ-d" now means ZZ with CPLS.
 

hyn

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Some questions about 2gll:
Why does CP being solved mean that the cube can be solved 2-gen?
Is it true that if you scramble with only R and U, then solve F2L with only R and U, CP will be solved?
If so, why does doing the same with R U and L not preserve CP?
Are all the LL cases where the corners are not permuted unable to be solved 2-gen?
Is CP similar to EO?
 

Swagrid

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Is it true that if you scramble with only R and U, then solve F2L with only R and U, CP will be solved?
Yes, because you're scrambling and solving with just 2 generators, you never disturb CP
If so, why does doing the same with R U and L not preserve CP?
Once a third generator is added, CP can become disturbed. I'm not entirely sure why myself, I'd have to talk to Raven to see if he has any better input, but any 3gen such as RUL, RUF, RUD, RUB does disturb CP.
Are all the LL cases where the corners are not permuted unable to be solved 2-gen?
They can't be solved 2gen. Because 2geb cannot change corner permutation, you will never solve the case.
 

Hazel

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What is the double-turn alg for an H-perm?
(Also how do I do a 3-cycle using double turns, the equivilent of M' U2 M U2 and U2 M' U2 M?)

Edit: Just by messing around I found a 3-cycle, but it's probably not optimal: F2 (R2 U2 R2 U2 R2 U2) F2 L2 F2 (R2 U2 R2 U2 R2 U2) F2 L2
 
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xyzzy

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What is the double-turn alg for an H-perm?
(Also how do I do a 3-cycle using double turns, the equivilent of M' U2 M U2 and U2 M' U2 M?)

Edit: Just by messing around I found a 3-cycle, but it's probably not optimal: F2 (R2 U2 R2 U2 R2 U2) F2 L2 F2 (R2 U2 R2 U2 R2 U2) F2 L2
Not sure if optimal H perm:
[R2 F2 U2 : F2 R2 F2 R2 F2 R2]
(12 moves)

Optimal 3-cycle:
R2 F2 R2 U2 R2 F2 R2 U2
(8 moves)
 

xyzzy

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Why does CP being solved mean that the cube can be solved 2-gen?
Is it true that if you scramble with only R and U, then solve F2L with only R and U, CP will be solved?
"CP" (the abbreviation) expands to "corner permutation", but it's often used to refer to something that is not, strictly speaking, a permutation of corners. This is where some confusion might arise.

Within this post (and only within this post), I will henceforth use "CP" to mean the actual corner permutation: the way the corners are arranged. The corner permutation after you do a U turn on a solved cube is not solved; it's a 4-cycle.

While six corner pieces have 6! = 720 possible permutations, only 120 of them are reachable from the solved state via R and U turns only. There is no "easy" or "intuitive" explanation for this. It's pretty much a mathematical coincidence that this happens to be the case. (With a bit of group theory, you can prove this by constructing an isomorphism to S_5: these 120 CPs behave as if they're permuting five things. What are these five things? It's… complicated. One day I'll get around to making a writeup on this topic.)

You can convince yourself that, with R and U turns, you can always bring any three corners to any three locations. (This is called 3-transitivity.) The first corner has 6 possible locations, the second corner has 5, and the third corner has 4, so that leaves 120 / (6×5×4) = 1 possibility for the positions of the remaining three corners. A consequence of this is that once you've solved three corners, the remaining three are guaranteed to be solved.

Assuming that the scramble used only R and U, once you've solved the two F2L corners, you can AUF to solve any one last layer corner, and the above observation guarantees that the other three last layer corners will also be solved. In other words, once F2L is solved, the last layer's corner permutation is guaranteed to be at most one AUF away from being solved.

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To answer the first question, if CP is solved, then you can always use some combination of Sunes and U perms (both doable with only RU) to first orient the corners, and then permute the edges. That's why CP being solved lets you solve the cube RU 2-gen.

(This does not imply that the fastest alg or the shortest alg for a solved-CP last layer case have to be RU 2-gen!! E.g. a lot of people use the RUS algs for U perms.)

If so, why does doing the same with R U and L not preserve CP?
Once a third generator is added, CP can become disturbed. I'm not entirely sure why myself, I'd have to talk to Raven to see if he has any better input, but any 3gen such as RUL, RUF, RUD, RUB does disturb CP.
Per above, RU 2-gen places some restrictions on CP.

However, with RUL or RUD, you can now make 3-cycle commutators like L' U R U' L U R' U' or R U R' D' R U' R' D.

If we have any 3-cycle alg and also 3-transitivity (which we do), we can reach every even permutation. 3-transitivity guarantees that we can always come up with setup moves to cycle any three corner pieces.

If we can reach every even permutation and we can also reach some odd permutation (which we can, e.g. by doing a single quarter turn), then we can reach every permutation.

That explains "why" RUL and RUD can reach all 6! = 720 corner permutations, rather than being restricted to a subcollection of only 120 corner permutations.

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As for RUF, a somewhat handwavy explanation is that any alg on a 3×3×3 can always be rewritten as RUFruf 6-gen, so just take one of the 3-cycle commutator algs mentioned above, rewrite it as RUFruf, then replace all the wide moves with normal moves. Replacing wide moves with normal moves don't affect how the alg affects the corners, so the end result will still permute the corners like a 3-cycle.

(The handwavy part is that this often results in algs that aren't just corner 3-cycles. E.g. L' U R U' L U R' U' gets converted to R' F R F' R U R' U', which also does a 3-cycle on the edges, on top of not even preserving edge orientation.)

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Addendum:

Also worth clarifying that just because an alg is RUL/RUD/RUF/RUB, it doesn't mean that the alg necessarily has to affect 2-gen CP reduction either. A boring example would be just repeating any RUF alg until the cube goes back to its starting state; a less contrived example would be the QTM-optimal RUF algs for H perm:
R' F U' F' U' R F U' R' U R F' R' F U2 F' R (18q) (and the mirrors/inverse of this)
 
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fdskljgrie

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I recently got a new 3x3 cube. I forget the brand but will include a pic of the logo. I can solve 3x3 sub 30 but after I scrambled it I couldn't. I went through all of the methods of seeing if I accidentally twisted a corner to no avail. I put it through grubiks solver but it said It couldn't solve because of a corner twist, but I tested it for corner twists and found none. What should I do?
Ok, kinda forgot about this post. Came back to it today. Managed to fix it.
 
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