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All algorithms when repeated enough times on a solved cube result in a completely solved cube. You can do any specific set of moves repeatedly without moving the cube and it'll get back together eventually.

@Muhtasim Dibbo
–
That's actually not how that works. Significant group theoretic work went into understanding the Hamiltonian circuit of the Rubik's cube: http://bruce.cubing.net/ham333/rubikhamiltonexplanation.html

@Dimitri Leggas
–
Bruce's work is impressive (I have worked with Bruce on other Rubik's Cube related mathematics), but not relevant to Muhtasim's comment. Rubik's Cube can be represented by a permutation group of finite order. From Lagrange's theorem, all elements (any sequence of face turns) starting from the identity (the solved cube) have an order (how many times the sequence is repeated before returning to the identity, or solved state) that must divide the order of the group. Since the group representing Rubik's Cube is finite, the order all its elements must therefore be finite. So, any sequence of moves applied to a solved cube will return to the solved state after a finite number of repetitions.

@Calvin Lin
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R- rotate the right face clockwise, U- rotate the top face clockwise, R'- rotate the right face anti-clockwise And F- rotate the front face clockwise

@Calvin Lin
–
THESE R THE TYPES OF MOVE-
R - MEANS TO MOVE RIGHT HAND SIDE OF CUBE VERTICALLY UPWARDS.
U - MEANS TO MOVE THE UPPER SIDE OF CUBE TO THE LEFT SIDE.
R' - MEANS TO MOVE RIGHT HAND SIDE OF CUBE VERTICALLY DOWNWARDS.
F - MEANS TO MOVE THE FRONT SIDE OF THE CUBE TO THE RIGHT SIDE.

The order of a sequence of moves is a matter of determining the effect on the sub-cubes as a permutation using disjoint cycle notation, then the order is the LCM of the cycle lengths. For $RUR'F$ the effect on the sub-cubes is a a 2-cycle on edge FR; a 5-cycle on edges UF, UL, UB, FL, and FD; and a 5-cycle on the 4 corners of the $F$ face (FUL, FUR, and FDL), and FDR. The LCM of 2, 5, and 5 is 10. Of note would be the maximal order which is 1260 of which $DF'DR'U^2$ is an example with shortest possible length in HTM (Half-Turn Metric, where turns of $90^\circ, 180^\circ,$ and $270^\circ$ are considered a single move).

Those who know non commutative algebra can easily figure it out that each generator (RUR'F) forms a normal subgroup of order (here 10).
find n such that (RUR'F)^n=identity (means doing nothing on the cube)
Eg: (RUR'F)^2=RUR'FRUR'F, among these some comnbinations can be replaced by other moves.
Finally one can show (clever manipulations), (RUR'F)^10=Identity

Always Keep your total cube orientation in a fixed position during the moves.
R- Right side face clockwise rotation
U- upside face clockwise rotation
R'- Right side face anti-clockwise rotation
F-Front face clockwise rotation

@Arindam Karmakar
–
They are codes to tell you how to move the Rubik's cube.
F is for Front.
R: Right. L: Left. U: Under.
F means turn the front side 90 degrees clockwise.
F' means turn the front side 90 degrees anticlockwise.

Am I right in thinking that the number of moves given in any algorithm with as many moves as possible but repeated over a number of times, the number of times cannot be a prime number?

R means rotating the right face clockwise.. R' means to do the same in anticlock direction... U, F, D, B, L mean up, front, down, back, and lesft respectively...

you could just take a cube and keep doing U R repeatedly and get the original configuration back.....but u need to have the patience as you have to do it quite many times..

Doing the same move multiple times creates a subgroup of all moves, so by lagrange's theorem, the number of time you must repeat the move divides the number of possible combinations of the rubik's cube.

Technically, this will happen no matter what sequence of moves one makes. It may take longer or shorter than 10 iterations, but because there are a finite number of positions for the cube, it has to eventually return to its original state.

What would be the biggest number of steps required to get a cube back in solved state for a 4-lettered algorithm...and what would be the algorithm.....i found that R'U is a big one

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## Comments

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TopNewestPlease explain it more!!!!!!

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DTYW CI6

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there is nothing to explain just solve the rubik's cube and apply the given algorithm (it won't help you to solve the cube its just a magic move)

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All algorithms when repeated enough times on a solved cube result in a completely solved cube. You can do any specific set of moves repeatedly without moving the cube and it'll get back together eventually.

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Can you explain what R U R' F mean?

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ok

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These moves are trivial. Any algorithm operated repeatedly on a cube will result back in the Original configuration in finite repeatitions.

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I can agree

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It can be proven mathematically

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The order of a sequence of moves is a matter of determining the effect on the sub-cubes as a permutation using disjoint cycle notation, then the order is the LCM of the cycle lengths. For $RUR'F$ the effect on the sub-cubes is a a 2-cycle on edge FR; a 5-cycle on edges UF, UL, UB, FL, and FD; and a 5-cycle on the 4 corners of the $F$ face (FUL, FUR, and FDL), and FDR. The LCM of 2, 5, and 5 is 10. Of note would be the maximal order which is 1260 of which $DF'DR'U^2$ is an example with shortest possible length in HTM (Half-Turn Metric, where turns of $90^\circ, 180^\circ,$ and $270^\circ$ are considered a single move).

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Those who know non commutative algebra can easily figure it out that each generator (RUR'F) forms a normal subgroup of order (here 10).

find n such that (RUR'F)^n=identity (means doing nothing on the cube) Eg: (RUR'F)^2=RUR'FRUR'F, among these some comnbinations can be replaced by other moves. Finally one can show (clever manipulations), (RUR'F)^10=Identity

Always Keep your total cube orientation in a fixed position during the moves. R- Right side face clockwise rotation U- upside face clockwise rotation R'- Right side face anti-clockwise rotation F-Front face clockwise rotation

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There exist much more moves like this:

L'ULU do this 5 times

U'R 62 times

UL'U'L 6 times

RU'L'U 27 times

M'U 8 times

M'UM'U' 6 times

R2L2 12 times

and many more ........................................

Happy Cubing!!!!!!!

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R 4 times

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Yes. That's the easiest "IDENTITY ALGORITHM "

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R U' x63

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Actually what are these 'R' 'U' 'F' ??

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I too don't know.

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another move is F R U R'U'F'

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RU R'U'

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Am I right in thinking that the number of moves given in any algorithm with as many moves as possible but repeated over a number of times, the number of times cannot be a prime number?

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it's awesome !!

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what does RU R`F mean please???

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R means rotating the right face clockwise.. R' means to do the same in anticlock direction... U, F, D, B, L mean up, front, down, back, and lesft respectively...

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cool. i liked it

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Is there a way to predict how many times you have to do the move to get back to solve?

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You have to program it. Look at this

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You all can even try

R' D' R D. I think it works with almostallof the algos. The number of times you have to do it may differ.Log in to reply

you could just take a cube and keep doing U R repeatedly and get the original configuration back.....but u need to have the patience as you have to do it quite many times..

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how to get fully solved cube?

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are you saying to solve the cube or about the above mentioned stuff

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VERRY VERRY EASY

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@Rishabh Jain Try this, R'D'RD 6 times.. Same thing happens!

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Do any random move repeatedly and it will get solved eventually not for this move only

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I just did it and it does get solved

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not just only RUR'F , it works for almost all algorithms provided they are 4 moves

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Let me give you one more Take a solved rubiks cube and do the following 6 times FRUR'U'F'

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hey are u in fiitjee south delhi??? @Rishabh Jain

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no i am a commerce student from east delhi (totally opposite answer what you asked?)

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Doing the same move multiple times creates a subgroup of all moves, so by lagrange's theorem, the number of time you must repeat the move divides the number of possible combinations of the rubik's cube.

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normal position

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R= right side clock wise U= upper side clock wise R'=right side anti-clock wise F= front side clock wise

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But i didn't get it !!!

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it remains same

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Can you explain what is RU R' F

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any particular move can do it not only this move :P

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Awesome

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6 repetitions and the cube will resolve itself.

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what does that mean

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Technically, this will happen no matter what sequence of moves one makes. It may take longer or shorter than 10 iterations, but because there are a finite number of positions for the cube, it has to eventually return to its original state.

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You could also repeat RUR'U' 6 times to achieve the same result.

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What would be the biggest number of steps required to get a cube back in solved state for a 4-lettered algorithm...and what would be the algorithm.....i found that R'U is a big one

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You will have to do it 62 times...

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I think 63 is the number.....RU has to be done 105..

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I found out that there are many such moves such as moving a solved cube 16 times with R U R' F B.

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R x4 , R'x4 ,....

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TEACH ME!!!!!!!!!!

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It is a bunch at internets

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Which method do you use?

IF you use the CFOP method please tell me from where did you learn it?

I want to learn the same.Suggest me a better site for this.

Thank you :)

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