# Rubik's cube repetitions, weird behaviour (Best if you know cube rotation notation)

In this question, I'll refer to standard cube rotation notation, seen here: J perm website - cube notation.

The move U, can be performed 4 times and return a solved cube (or any cube) to its original state.

This isn't surprising, it's making use of the 4 sides of a square that is built into the cube.

The set of moves (U R U' L' U R' U' L), which is a standard operation in white cross methods, will return to the original position after 3 repetitions. But this makes use of a triangle that is built into the face of a cube. Obviously, you can cut a square across it's diagonal and end up with a triangle, and this is how I understand that period-3 repetition. (Try it for yourself to see why this is a sensible interpretation).

However I was playing around with the operation (R U R' U) and discovered that after performing the moves 5 times, it returns the cube to it's initial position!

I find this really surprising. It doesn't "feel" right that this 3-dimensional, 6-sided, hyper-4-gon should ever have periodicity 5.

Can anyone explain why this is the case in a geometric or intuitive way?

I think @Stinking Bishop put it best. If you look at the side pieces or corners that move under the operations, they form a 5-cycle. Although it is surprising that this awkward pentagon exists on a cube, if you follow each piece independently, that's really all there is to it. Of course that must be the case because it is the same operation each time. There is no decision tree or pause in movement. The pieces are just on a 5-cycle conveyor belt.

I agree with whoever said it that once you know what is going on, the surprising thing ceases to be "why" it exists and more that it can be generated from something so simple. But I've been around maths for long enough that I've seen that many times before. Conway's game of life anyone?

• Well, the move affects five corner pieces and five side pieces. The five side pieces form a cycle (first goes to the second, second to the third etc. -fifth to the first) and five corner pieces form another cycle in exactly the same way.
– user700480
Commented Jul 25, 2023 at 22:47
• What is "intuitive" is relative too. For me, as all the moves on the cube form a group and the number of elements of it is divisible by $5$ which is prime, it immediately implies (Cauchy's theorem) that there will exist a move of order $5$. The only a bit surprising thing was - how simple the example of such a move is, that you've found.
– user700480
Commented Jul 25, 2023 at 22:53
• @DavidK the reasoning is not strictly geometric but more like reasoning about 3-cycles and 4-cycles.
– qwr
Commented Jul 25, 2023 at 23:55
• By the way, there's a nice web site to test sets of moves in standard notation: rubikscu.be (click on Solve to go to the part of the page that lets you copy a set of moves into a text entry window). Commented Jul 26, 2023 at 0:37
• "5-cycle conveyor belt" is a very apt visualization. Commented Jul 26, 2023 at 12:04

There is no real good geometric explanation as far as I know (edit: Stinking Bishop's explanation is probably the closest you can get). The explanation is pretty simple if you know some group theory. If I lose you throughout this answer, skip to the end.

You can look at it like this. All the permutations (or rather the algorithms to get to those permutations) of a Rubik's Cube form a group with $$43 252 003 274 489 856 000$$ elements (where the "product" of 2 algorithms is just doing one after the other).

Your algorithm (R U R' U) is one element of that group (let's call it $$g$$ for simplicity) and doing it $$n$$ times after each other is the element $$g^n$$.

It is a pretty basic fact in group theory (for finite groups) that for every element $$h$$ there exists a number $$k$$ that is a divisor of the order (=size) of the group such that $$h^k$$ = 1. It follows directly from Lagrange's theorem, if you are interested.

Even more: There is Cauchy's theorem which states that for every prime divisor $$p$$ of that (finite) group, there exists an element $$h$$ such that $$h^p = 1$$ and where $$p$$ is the smallest number, bigger than $$0$$, for which this is true.

What does this mean for the Rubik's cube?

It means that for every prime number $$p$$ that divides $$43252003274489856000$$, there exists an algorithm that you would have to $$p$$ times to get back to the original state. $$5$$ is such a number. In the same way there should be an algorithm that you would have to repeat exactly $$11$$ times to get back to the original state (since $$11$$ divides $$43252003274489856000$$).

• This might seem counterintuïtive, but there are other examples that will be maybe easier to understand with the same counterintuïtive result. Suppose you have $5$ people and you want to rearrange their order, you could say first person goes to the second place, second person goes to the third place and third person goes to the first place. After doing this 3 times you will get the original ordering back even though 3 seems not really connected to 5 (you can do this also with any number of people as long as you have more than 3) Commented Jul 25, 2023 at 22:55

There are a finite number of positions a Rubik's Cube can be in. It's a ridiculously large 43 quintillion positions, but it's finite. That means, that if you start with a solved cube, and do the same set of moves over and over, eventually you'll end up with a solved cube again.

The number of times you need to do a set of moves to get back to where you started is called its order. So, using your set of moves as an example, the order of $$R U R' U$$ is 5.

So for any set of moves, if you do it over and over again, you'll end up back where you started. It turns out, the largest order for any set of moves is 1260. That means, for any set of moves, if you keep repeating it, in 1260 repetitions or less, you'll end up back where you started.

For a complete list of all the possible orders, check out a previous answer of mine: Rubik's Cube Group: Number of Elements With Each Order