# How to Solve Every 2GLL Case Using Sunes

#### mDiPalma

##### Member
Sup?

Question: Which of the following algs is the fastest?

A) N-perm
B) Sune
C) OLL 20
D) A and C

…sune?

It's common knowledge that you can solve each and every 2GLL case with a combination of sunes and moves.

Here is the distribution I found:

There are 85 2GLL cases (including solved).
1 case requires 0x Sune (solved).
4 cases require 1x Sune.
40 cases require 2x Sune.
36 cases require 3x Sune.
4 cases require 4x Sune. (4x misoriented corners, H-perm, Z-perm)

Yesterday, I wrote a program to find the sune-combos for all these cases. The program is ugly because I’m not a programmer (Mechanical Engineering student). But if you want to see the code, let me know XD.

Here are the results.

Legend:

• These algs only work on a cube with solved CP and EOF2L. This is attainable via ZZ-d (ZZ-porky V1, V2, ZZ-orbit, or a similar variant).
• sune=RUR’URU2R’
• Asune=RU2R’U’RU’R’
• Bsune=R’U’RU’R’U2R
• BAsune=R’U2RUR’UR
• G=generator
• S=solver
• The images use WCA orientation (white-green). Black represents white.

Interesting things:

• It's possible to FORCE a 3-sune maximum by inserting the last F2L pair while orienting exactly 1 or 2 U-layer corners.
• Obviously it’s not a good idea to use 4x Sunes instead of M2 U M2 U2 M2 U M2 (normal H-perm). However, it’s just a proof of concept that you can solve every 2GLL case with 1 fast alg.
• Every R-side Sune case cancels with every other R-side Sune case.
• Some of the Sune-only 2GLL cases are actually better than the algs I would normally use for the case. Compare your least favorite 2-gen algs against what my program spit out. Maybe you'll fall in love <3
• sunesuneUsunesune is a Z-perm, *******
• sunesuneU'sunesune is an H-perm, *******

1 ALG IS ENOUGH!!!!!!!

SUNE=LIFE.

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#### Berd

##### Member
What does 2GLL stand for?

#### mDiPalma

##### Member
What does 2GLL stand for?
2GLL stands for 2-gen Last Layer. It's a subset of algs that solve the cube with only [RU] moves (2-gen).

[RU] moves do NOT affect the relative permutation of the corners (CP=corner permutation). As a result, every 2GLL alg REQUIRES that the CP be solved before completion of the F2L.

For example, whenever you do the Pi OLL case (R U2 R2 U' R2 U' R2 U2 R), followed by a Z-perm, that LL case could have been solved with a single 2GLL alg.

It's basically ZBLL for the solved-CP subset of LL cases.

#### Robert-Y

##### Member
Cool

I've tried something similar using ksolve. I simply defined a puzzle to be just 3x3x1 (So just the top layer) and defined sune and U to be the only possible moves to make

#### Stefan

##### Member
Just curious: How many cases can be solved with just the R U R' U R U2 R' Sune and U, and what is the distribution?

#### bobthegiraffemonkey

##### Member
Just curious: How many cases can be solved with just the R U R' U R U2 R' Sune and U, and what is the distribution?
I've noticed before there is something strange about this subgroup. It appears impossible to do a U-perm, but I can't see why. I would guess, then, that only 1/3 of cases may be reached.

#### mDiPalma

##### Member
Just curious: How many cases can be solved with just the R U R' U R U2 R' Sune and U, and what is the distribution?
I've noticed before there is something strange about this subgroup. It appears impossible to do a U-perm, but I can't see why. I would guess, then, that only 1/3 of cases may be reached.
it's true

you can only get H and Z perms.

because each sune is a 3 edge swap and net +3 to the CO state

edit
z-perm=sunesuneUsunesune
h-perm=sunesuneU'sunesune

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#### Stefan

##### Member
because each sune is a 3 edge swap and net +3 to the CO state
I don't think that explains it, but I don't know what exactly you mean with CO state. What do you?

Edit: Here's an alg made up of Sune and another alg that "is a 3 edge swap and net +3 to the CO state", and it's a U perm:

(R U R' U R U2 R')5 (F' U' F R2 U B U' B' R2 F' U F)

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#### Amress

##### Member
I actually do this for H perms for OH. My alg is R U2 R' U' R U R' U' R U' R' U R U2 R' U' R U R' U' R U' R'. High movecount, but it flows pretty well.

#### mDiPalma

##### Member
I don't think that explains it, but I don't know what exactly you mean with CO state. What do you?

Edit: Here's an alg made up of Sune and another alg that "is a 3 edge swap and net +3 to the CO state", and it's a U perm:

(R U R' U R U2 R')5 (F' U' F R2 U B U' B' R2 F' U F)
I meant the case has to be writable in a certain form, taking into account those effects. i'm not good at matrix algebra or cube theory, but here's my best guess:

EDIT: THIS IS NOT COMPLETE. it needs to include something along the lines of matlab "circshift" for the U moves (which is what I used in my code). but i don't know how to represent that in terms of matrix algebra.

a U-perm cannot be written like that, at least as far as I can tell.

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#### Randomno

##### Banned
If you can't do U perms, how many 2GLL cases are impossible like this?

#### StachuK1992

##### statue
That was a very legible explanation of the UPerm thing. Thanks for that.

#### Stefan

##### Member
What you describe is doing Sunes first, and then doing U turns. But Sune and U don't commute, so you're not covering for example Sune U Sune (which results in CO [2 2 1 1], which doesn't match [n n 0 n], so you'd say that's impossible as well).

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#### mDiPalma

##### Member
What you describe is doing Sunes first, and then doing U turns. But Sune and U don't commute, so you're not covering for example Sune U Sune (which results in CO [2 2 1 1], which doesn't match [n n 0 n], so you'd say that's impossible as well).
i defined sune differently than you

edit: but yeah, i see what you're saying. you'd have to include a bunch more N's and M's multiplied by their effects.

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#### Robert-Y

##### Member
If you can't do U perms, how many 2GLL cases are impossible like this?
Exactly two thirds. Which means that Matt was right. I don't have a full proof but I have investigated enough cases with ksolve in order to conclude this.

#### Forte

##### Member
One way to figure it out if a 2GLL is solvable by sunes:

AUF your LL so the corners are solved.
Let A0 = {solved, H perm, Z perm, other Z perm},
A1 = {ccw U perm on front/back, cw U perm on left/right},
A2 = {cw U perm on front/back, ccw U perm on left/right}.
(It's a quotient group which is C3 if you know about that).
Let's call that the "parity" of an EP (bad terminology but whatever).
Then take the following sum (mod 3):
parity of EP you need to do + twist on UFR that you need to do + twist on UBL that you need to do
(where solved =0, cw = 1, ccw = -1).
If that sum is 0, then it's solvable.
(It's just a homomorphism into C3 if you know about that).

Example: if you do sune + U2, then you need to do:
- ccw U perm on the right (parity is 2)
- ccw twist on UFR (-1)
- ccw twist on UBL (-1)
2 - 1 - 1 = 0, so you can solve it using sunes.

The four sunes preserve this sum, so we at least know that the sum has to be 0 for it to be solvable.
The converse is also true with a bit of work, so the subgroup generated by the sunes is an index 3 subgroup of the 2GLLs
(so exactly 1/3 of the 2GLLs can be solved using sunes).

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#### mDiPalma

##### Member
i just ran 10,000 cases of each:

random number of U's + random number of sunes
2x(random number of U's + random number of sunes)
3x(random number of U's + random number of sunes)
4x(random number of U's + random number of sunes)
5x(random number of U's + random number of sunes)

and none of them were isomorphic to a U(a)-perm.

#### Ranzha

##### Friendly, Neighbourhoodly
I actually do this for H perms for OH. My alg is R U2 R' U' R U R' U' R U' R' U R U2 R' U' R U R' U' R U' R'. High movecount, but it flows pretty well.
Or you could just use the 11-mover.

This is really cool!

#### Amress

##### Member
The "11-mover" is actually a lot more because of the R2s and U2s, but i mainly just use this because it looks really cool (10+ TPS)