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



## j3toler (Jan 15, 2013)

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 (Jan 15, 2013)

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


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## j3toler (Jan 15, 2013)

Noahaha said:


> 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?


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## Noahaha (Jan 15, 2013)

j3toler said:


> I like that answer. Could you prove that?



Only experimentally XD


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## Petro Leum (Jan 15, 2013)

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.


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## Schmidt (Jan 15, 2013)

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-------------


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## j3toler (Jan 15, 2013)

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


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## mDiPalma (Jan 15, 2013)

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']

does that answer your question?


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## j3toler (Jan 15, 2013)

Petro Leum said:


> 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.


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## Noahaha (Jan 15, 2013)

j3toler said:


> Then I guess what I'm really trying to ask is based on this:
> --------
> ||1 2 3 ||
> ||4 A 5 ||
> ...



Doesn't seem practical to me.


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## j3toler (Jan 15, 2013)

mDiPalma said:


> R U' R' U' R U R' F' R U R' U' R' F R
> 
> can also be written:
> 
> ...



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']



Noahaha said:


> 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 (Jan 15, 2013)

j3toler said:


> 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?"



j3toler said:


> 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 (Jan 15, 2013)

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.


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## j3toler (Jan 15, 2013)

Noahaha said:


> 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.



Noahaha said:


> 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.



Kirjava said:


> 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 (Jan 15, 2013)

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.


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## Stefan (Jan 15, 2013)

j3toler said:


> 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.



j3toler said:


> (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 (Jan 15, 2013)

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.
There are a few threads about this, and I have a partial explanation on my website.

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.


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## j3toler (Jan 15, 2013)

mDiPalma said:


> This decomposition is extremely simple.
> 
> R U' R' U' R U R' F' R U R' U' R' F R
> 
> ...



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)?



Stefan said:


> What's the purpose of renaming perfectly good things?



To reach a better understanding of the cube.



Stefan said:


> 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.



Stefan said:


> (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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## Stefan (Jan 15, 2013)

j3toler said:


> how do I prove that this is SuperCube Safe?



It's not, so you don't.


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## mDiPalma (Jan 15, 2013)

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  .


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## Stefan (Jan 15, 2013)

j3toler said:


> To reach a better understanding of the cube.



You don't gain anything and you do lose meaning, how is that better?



j3toler said:


> The fact that you recognised my falicy means that you undertand the logic I'm trying to grasp.



I might understand what you're trying to grasp, but I don't understand what you said there. All that R = (25,27,32,30) = D etc makes no sense to me, especially because of the contradictions. Your notation is really weird, and the flawed renaming doesn't help at all.


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## whauk (Jan 15, 2013)

you want some more "mathematics"? then stop renaiming everything to some disgusting sequences of numbers, letters etc. nobody wants to read this. imo this "mathematical" explanation should dispel any doubts: let the cube group be <U,R,L,D,F,B>. then compose some of the generators and it will give you another element of the cube group. (always the same since thats how a group works). then grab a real cube, execute your algorithm and voila: there you know how your new element looks like.
do you want to proof more? i cant imagine what...


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## Lucas Garron (Jan 17, 2013)

whauk said:


> you want some more "mathematics"? then stop renaiming everything to some disgusting sequences of numbers, letters etc. nobody wants to read this. imo this "mathematical" explanation should dispel any doubts: let the cube group be <U,R,L,D,F,B>. then compose some of the generators and it will give you another element of the cube group. (always the same since thats how a group works). then grab a real cube, execute your algorithm and voila: there you know how your new element looks like.
> do you want to proof more? i cant imagine what...



*Unfortunately*, that's not nearly enough if you, say, want a way to tell whether two algs are the same without executing them, or want a computer to find algorithms for you, or if you want to find some property like God's number. Group theory is not for everything, but it's often very useful to have numerical descriptions.


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## qqwref (Jan 17, 2013)

j3toler said:


> This is another great answer! Could you prove this answer using an "If... than..." statement where:
> 
> Given: (v,r,w,s)
> 
> ...


It looks like you think you are asking for a more "mathematical" answer, but you're actually asking for an answer where everything is written out in full, wasting space and making it much more hard to understand. You want an intuitive explanation of why the algorithm does what it does, don't you? Because this wouldn't help.

All that stuff you were talking about earlier with permutations of numbers from 1 to 48 will actually give you zero extra knowledge. All you'd do to "prove" the sequence works is to write out what each turn does, combine them together, and then see what the end result is. But that's exactly the same thing you'd get if you just performed the sequence on the cube and traced each sticker separately with one finger. All you'd be doing is simulating a cube on paper with numbers, which takes longer and gives the same rsult. Permutations can be a very useful tool for analyzing things, but as it turns out everyone on the forum has a real-life model of a permutation group, so there's no need to ever have the permutation itself written out unless you are a computer program (such as GAP, or a cube solver).


PS: You know, our forum actually has a TeX thing  You can use the \( tag, i.e. \( f(x) = x^2 \). \)


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