# How does an alg affect the cube? (Full and complete answer please)

#### j3toler

##### Member
Hello, My name is Jay. I am 21 years old with self taught knowledge in mathmatics and Group Theory. I've lurked this website for some time now, understand most of the topics of disscussion, and now pose a question to further my own understanding. So given all the information from this website:

Q: Given the Alg. RU'R'U'RUR'F'RUR'U'R'FR, How does the Alg. effect the cube?

TAGS: cycle structure, cycle map, (v,r,w,s), permutation, orientation, supercubesafe, algorimths, commutators

PS: I will be checking back periodically between my notes and this forum to read answers and post some of my own.

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#### Noahaha

##### blindmod
It does a 2-cycle of corners (UBL<->RFD) and a 2-cycle of edges (UB<->UL).

#### j3toler

##### Member
It does a 2-cycle of corners (UBL<->RFD) and a 2-cycle of edges (UB<->UL).
I like that answer. Could you prove that?

#### Petro Leum

##### Member
Performed on a Megaminx, it is a 3Cycle of Croners and a 3Cycle of Edges (Proven experimentally )

too bad i really have no idea how to explain the alg itself. just wanted to point that out.

#### Schmidt

##### Member
I don't know if you will call this proof, but there are some restrictions to the cube.
1: There can never /just/ be two edges that are wrong
2:------------------------------corners--------------
3:There can be three edges that are wrong
4:-------------------corners-------------

#### j3toler

##### Member
Then I guess what I'm really trying to ask is based on this:
--------
||1 2 3 ||
||4 A 5 ||
||6 7 8 ||
---------------------------------------------
||9 10 11||17 18 19 ||25 26 27||33 34 35||
||12 B 13||20 C 21 ||28 D 29 ||36 E 37 ||
||14 15 16||22 23 24||30 31 32||38 39 40||
-----------------------------------------------
||41 42 43||
||44 F 45 ||
||46 47 48||
--------
Cycle Structure:
Where -> (A,B,C,D,E,F) = (U,L,F,R,B,D)

and R = (25,27,32,30) = D
(26,29,31,28) = D
( 3,38,43,19) = D
( 5,36,45,24) = D
Find L,U,D,F,B such that G(x) = (RU'R')(U'RU)[R'{F'(R(UR'U')R')F}R] = 0

#### mDiPalma

##### Member
R U' R' U' R U R' F' R U R' U' R' F R

can also be written:

[R U' R': (U')] [F' R U, R']

#### j3toler

##### Member
Performed on a Megaminx, it is a 3Cycle of Croners and a 3Cycle of Edges (Proven experimentally )

too bad i really have no idea how to explain the alg itself. just wanted to point that out.
What I'm asking for is the proof. The math translated from 3x3 notation to every other puzzle. Including the Megaminx.

#### Noahaha

##### blindmod
Then I guess what I'm really trying to ask is based on this:
--------
||1 2 3 ||
||4 A 5 ||
||6 7 8 ||
---------------------------------------------
||9 10 11||17 18 19 ||25 26 27||33 34 35||
||12 B 13||20 C 21 ||28 D 29 ||36 E 37 ||
||14 15 16||22 23 24||30 31 32||38 39 40||
-----------------------------------------------
||41 42 43||
||44 F 45 ||
||46 47 48||
--------
Cycle Structure:
Where -> (A,B,C,D,E,F) = (U,L,F,R,B,D)

and R = (25,27,32,30) = D
(26,29,31,28) = D
( 3,38,43,19) = D
( 5,36,45,24) = D
Find L,U,D,F,B such that G(x) = (RU'R')(U'RU)[R'{F'(R(UR'U')R')F}R] = 0
Doesn't seem practical to me.

#### j3toler

##### Member
R U' R' U' R U R' F' R U R' U' R' F R

can also be written:

[R U' R': (U')] [F' R U, R']

Yes, but I would like to see the steps involved from translation the Alg [R U' R' U' R U R' F' R U R' U' R' F R] into [R U' R': (U')] [F' R U, R']

Doesn't seem practical to me.
That's Ok. I will be using this forum and thread really to answer all the questions I have until either I work them out on paper and put them here, or someone else beats me to it.

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#### Noahaha

##### blindmod
That's Ok. I will be using this forum and thread really to answer all the questions I have until either I work them out on paper and put them here, or someone else beats me to it.
I'm a little confused at what you're trying to get at. I'm guessing it's some sort of deeper understanding than "it just works because it works." Most people on this forum don't give any thought to algorithms past "what does it do?" and "can I do it fast?"

Yes, but I would like to see the steps involved from translation the Alg [R U' R' U' R U R' F' R U R' U' R' F R] into [R U' R': (U')] [F' R U, R']
Matt was just writing the alg in commutator/conjugate notation. "," denotes a commutator and ":" denotes a conjugate. When expanded they are the same alg.

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#### Kirjava

##### Colourful
Adding the cycles of each individual move together gives the resulting cycle of the entire algorithm.

An alg is supercube safe if the total QT value of each movetype modulo 4 is 0.

#### j3toler

##### Member
Most people on this forum don't give any thought to algorithms past "what does it do?" and "can I do it fast?"
The reason why I posted this topic in the Puzzle Theory section is because I am trying to appeal to those who frequent the Puzzle Theory aspect of speedsolving. I too give thought to both "what it does" and "How fast can I do it", but you were also right in that I have other questions (puzzle theory questions) that need to be answered as well.

Matt was just writing the alg in commutator/conjugate notation. "," denotes a commutator and ":" denotes a conjugate. When expanded they are the same alg.
This is an acceptable answer to me. Actually, any answer is an acceptable answer, but please try to understand that I will most likely have more questions.

Adding the cycles of each individual move together gives the resulting cycle of the entire algorithm.

An alg is supercube safe if the total QT value of each movetype modulo 4 is 0.
This is another great answer! Could you prove this answer using an "If... than..." statement where:

Given: (v,r,w,s)

V = (0,0,0,0,0,0,0,0) <-Orientation of the corners
R = (n_Sub_0,n_sub_1) <-Permutation of the corners
w = (0,0,0,0,0,0,0,0,0,0,0,0) <-Orientation of the edges
s = (I_Sub_S_Sub_12) <-Permutation of the edges

is the Identity element?

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#### mDiPalma

##### Member
This decomposition is extremely simple.

R U' R' U' R U R' F' R U R' U' R' F R

[R U' R' U' R U R'] [F' R U R' U' R' F R]

The first part cycles edges UR>RF>UB>UL and corners ULB>RDF>UBR>FUR.

The second part cycles edges UB>RF>UR and corners ULB>FUR>UBR.

Note that the edge cycle in part 2 is the reverse of the edge cycle in part 1.
Also note similarly for corners.

Put these two cycles together, and you get what Noah said at the very beginning: UL>UB and ULB>RDF.

There are decompositions of other PLLs on the wiki. If that interests you, check it out.

#### Stefan

##### Member
Where -> (A,B,C,D,E,F) = (U,L,F,R,B,D)
What's the purpose of renaming perfectly good things? And B, D and F appearing on both sides results in C=F=D=R and E=B=L which makes no sense whatsoever.

(25,27,32,30) = D
(26,29,31,28) = D
(25,27,32,30) = D = (26,29,31,28) is simply false.

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#### Lucas Garron

##### Moderator
Staff member
To treat the cube as a group, it's usually easiest to label every sticker with a number, and define the moves as permutations. One such definition of the permutations is here. From that, you can calculate the effect on a given piece as a composition of permutations. This gives an overall permutation, from which you can find cycle structure, etc.

As Kirjava mentioned, to figure out the effect on a supercube, if you have an alg in a form without rotations, you can just count the total effect on each center by adding the moves on that side, mod 4.

#### j3toler

##### Member
This decomposition is extremely simple.

R U' R' U' R U R' F' R U R' U' R' F R

[R U' R' U' R U R'] [F' R U R' U' R' F R]

The first part cycles edges UR>RF>UB>UL and corners ULB>RDF>UBR>FUR.

The second part cycles edges UB>RF>UR and corners ULB>FUR>UBR.

Note that the edge cycle in part 2 is the reverse of the edge cycle in part 1.
Also note similarly for corners.

Put these two cycles together, and you get what Noah said at the very beginning: UL>UB and ULB>RDF.

There are decompositions of other PLLs on the wiki. If that interests you, check it out.
This is also an excellent answer! The decompisition that I came up with is:
If G(x) = R U' R' U' R U R' F' R U R' U' R' F R
Than the decompisition of G(x) = (RU'R')(U'RU)[R'{F'(R(UR'U')R')F}R]

If this is true, how do I prove that this is SuperCube Safe?
I understand that the first part is a 4 cycle of those specific edges and corners, and that the second part is a 3 cycle of those specific edges and corners, but what about those two braketed off parts of the Alg. effect the orientation and permutation of those specific edges and corners as it relates to the rubiks cube as a Group G(x)?

What's the purpose of renaming perfectly good things?
To reach a better understanding of the cube.

And B, D and F appearing on both sides results in C=F=D=R and E=B=L which makes no sense whatsoever.
The fact that you recognised my falicy means that you undertand the logic I'm trying to grasp.

(25,27,32,30) = D = (26,29,31,28) is simply false.
The way I see it, as the corner pieces in positions (25,27,32,30) undergo the move R they change on the D face, which as you noticed is a redundant way of saying R face, to the positions on the cube marked by (26,29,31,28). Does this clairify?

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#### mDiPalma

##### Member
As everyone else said, for the sequence to be supercube-safe, the amount of quarter turns in a defined direction per each face must be a multiple of 4. (because the centers are not exchangeable by face turns, on a 3x3).

Simply, R + R' + R + R' + R + R' + R' + R = 0*R
and U' + U' + U + U + U' = -1*U
and F' + F = 0*F

So one iteration of this sequence will rotate just the U center counterclockwise (while swapping UL+UB and ULB+RDF). This is not supercube-safe.

And I'm no good with groups, so I can't answer your G(x) question soz pl0x .