Looping or Repeating Algorithms?

[Lucas Garron]

Yeah, exactly, But I think many people don't understand that this is a necessary (but satisfied) condition unless they've seen something like group theory. It's easy to dismiss here, but it's assumptions like this that cause trouble in trickier problems.

 

[Erzz]

(R)(UUUUUUUUUUUU...) never comes back to the starting position, but it doesn't endlessly change. Why can't this happen if R and U are replaced by the same alg?

After R, then infinity Us, then R, etc you will reach the starting position eventually, assuming the amount of Us stays the same and assuming it's possible to preform infinity Us. It'd be either order 105, 63, or 30 depending which state the U ends.

Also qq, your order calculator doesn't work for me on Chrome, I can click the button but it doesn't output anything.

 

[Tim Major]

Sorry, I was on the bus, just trying to give a quick bit of proof that could be understood by most people aged 6 and up

 

[qqwref]

Also qq, your order calculator doesn't work for me on Chrome, I can click the button but it doesn't output anything.

Fixed - for some reason it was complaining about one of the functions being named 'a'. So I changed the name.

 

[Erdos]

After R, then infinity Us, then R, etc you will reach the starting position eventually, assuming the amount of Us stays the same and assuming it's possible to preform infinity Us. It'd be either order 105, 63, or 30 depending which state the U ends.

But there's a problem with just using normal reasoning as opposed to simple abstract algebra. Could you even perform another operation after the infinity U's? And what if the infinity U's change per operation? Even if you could mod4 the infinity U's, who's to say that the infinity U's isn't 1mod4 one round, and say 3 mod 4 another round? Seems like your reasoning "proves" the claim by using more claims. This is why an intuitive explanation is often not valid, as Lucas was saying. qqwref's is pretty good though.

 

[reThinking the Cube]

DUMB = 144 and SLy = 180.

 

[mrCage]

Think, if it never came back to the starting position, that'd mean it would endlessly change. Which would mean the rubik's cube has limitless cube states.

I like this simple proof of contradiction, even it may not be rigid enough mathematically.

Per

 

[Cielo]

I like this simple proof of contradiction, even it may not be rigid enough mathematically.

Per

Actually, it is rigid.

We know the number of states is limited, because the number of pieces is limited.
So it will reach some state for more than once, and we can assume that after repeating m times and n times, it reach the same state.
Then repeating (n-m) times will get the cube to the start position.

 

[JonWhite]

What about move repetitions where the ith repetition is L(UR)*i?

 

[qqwref]

Why, though? I don't think anyone's thought to look at that before.

That should go back to solved eventually (that sequence repeats every 105 times, so doing it 105 times in a row can be counted as one move, and then repeating that over and over would get back to solved at some point) but it seems pretty awkward and chaotic, unlike the pleasantly group-theoretic action of doing the same sequence over and over.

 

[Erdos]

Actually, it is rigid.

We know the number of states is limited, because the number of pieces is limited.
So it will reach some state for more than once, and we can assume that after repeating m times and n times, it reach the same state.
Then repeating (n-m) times will get the cube to the start position.

Each one of those claims requires a proof within itself. So it's not rigid.

 

[RussianWhiteBoi]

EDIT: I'm very sorry, I've gone through the thread again and this has already been said

Spoiler

Hello,

There is a nice proof by contradiction I've heard. Label each possible position on the cube and form a permutation, where performing the move sequence from one position leads to the next. Assume that there is a "loop" within them. For example, something like this (each letter represents a distinct position. Of course, there are more than five positions, this is just an illustration.):

A -> B -> C -> D-> E -> C -> D -> E -> C, etc

If this were the case, then performing the inverse move sequence from C would lead back to positions B and E, meaning that they must be the same. But this goes against the fact that all letters represent distinct positions. Here is a contradiction, meaning the initial assumption that there is an infinite loop in the cycle of positions is false. Since there is no infinite loop in the permutation, the whole permutation must be a cycle. There is a finite amount of positions of the cube, which means that eventually the same position will be reached, and the cycle will restart.

 

[FJT97]

Hey

You need 2 T-Perms, to get to were you started. You need 6 sexy moves to get to where you started. And so on and so forth. But what is the longest of those cycles and how to calculate that?
I assume, that the longest cycle would be a sequence which moves all of the sides.

I should clarify that I'm talking about 3x3 in normal HTM metric.

BTW.: A few years ago i saw a website on which you can type moves and it gives you the cycle number. Is this website still up? I can't find it anywhere.

 

[zipper]

(Uz') repeat... i've done it 800 times and then just gave up

 

[Lucas Garron]

Hey

You need 2 T-Perms, to get to were you started. You need 6 sexy moves to get to where you started. And so on and so forth. But what is the longest of those cycles and how to calculate that?

The concept you're looking for is called "order".

I assume, that the longest cycle would be a sequence which moves all of the sides.

Since you can perform any face turn using only the other 5 sides, you can rewrite any alg to use only 5 sides.
Therefore, there is an alg of maximum order using only 5 sides.

I should clarify that I'm talking about 3x3 in normal HTM metric.

BTW.: A few years ago i saw a website on which you can type moves and it gives you the cycle number. Is this website still up? I can't find it anywhere.

http://mzrg.com/rubik/ordercalc.shtml
Also see http://mzrg.com/rubik/orders.shtml for algs of various orders.

 

[FJT97]

So "R L2 U' F' d" (Order=2520) is the biggest one? But how to calculate that? There is a calculation on the order calculator site, but i can't see how it works. And i didnt studied Maths enough to understand that wiki article i guess. So does someone have a understandable explanation?

 

[cuBerBruce]

The biggest order within the standard (fixed centers) cube group is 1260. Example: R U2 D' B D'. If you allow wide turns, inner layer turns or whole cube rotations, then the maximum order becomes 2520. Of course, the maneuver "R L2 U' F' d" contains a wide turn.

 

[P14082003]

Hi people! I'm new to the forums, and also to cubing. i don't know if this is the best place to write this, but here it goes.

What do you call those algorithms that when applied to a cube, after some repetitions, it returns to the starting pattern?

I know some like R U R' U', R' F R F', r R U R' U' r', that get solved again after repeating them 6 times.

I also figured that by merging them, I could get the same effects. e.g., [R U R' U'] [R' F R F'] that gets the job done in 3 repeats.



Also

Do you know some yourself? Some that I'm forgetting?

Thank you, and sorry for any bad English.

 

[DGCubes]

Every algorithm loops back eventually (although some may take hundreds or thousands of tries). You can try it with all of them, and you'll get back to the start after doing it enough times. I don't think there's a name for it, just because it's true about them all.

Here's a fun and simple one though: (R U)105

 

[P14082003]

Every algorithm loops back eventually (although some may take hundreds or thousands of tries). You can try it with all of them, and you'll get back to the start after doing it enough times. I don't think there's a name for it, just because it's true about them all.

Here's a fun and simple one though: (R U)105

omg 105 times? R2 U2 is 6 times hahaha

[EDIT]

nice youtube channel BTW