The merry go round

rasiel

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Dec 24, 2014
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Maybe this is a basic one that's already been answered but I don't really know what to even search for.

Say you do this simple alg on a 3x3: R, R+M and turn cube left then keep repeating until cube comes back to its original state

For a 3x3 it takes 24 moves. On a 2x2 you can only do r then switch left and repeat. I was surprised to see it took 36 where I should have expected it to take fewer. On a 4x4 the equivalent becomes R, R+R2W, R+R2W+R3W then turn cube left and repeat. I did it many, many times and eventually got bored. I'm pretty sure it was more than 100 turns.

So the question becomes, what would be the math needed to estimate how many of these moves would it take to restore a cube of any size to its original state?

Ras
 
I'm on college wifi and mzrg.com is blocked, but I think this is the right URL: mzrg.com/rubik/ordercalc.shtml
You type any alg into it and it will tell you how many times you need to repeat it for it to solve the cube again. I don't know how it calculates it though.
 
To be precise, you should define exactly what you mean by the "original state." This could refer to the intrinsic state of the puzzle (where the orientation of the cube itself is considered not to matter) or (alternately) if you consider the orientation of the cube as part of the state. If the puzzle has indistinguishable pieces (4x4x4 and larger), then the answer may be depend on what exactly the original state is (unless you want to consider the pieces to be distinguishable even though they may look the same). Commonly the starting state is taken to be a solved state.

As I understand, the MZRG order calculator (qqwref's) assumes the starting state is a solved state, and it counts how many repetitions are required to get back to a solved state, whether or not the cube has the same orientation or not. Randelhofer's applet appears to give the true mathematical order (meaning the pieces and their orientations are all distinguishable, and the orientation of the cube is considered to matter), as well as how many times the maneuver must be repeated starting from a solved state until again reaching the solved state with the same orientation (but indistinguishable pieces/orientations not necessarily in the same positions/orientations).

In any case, if I understand your 4x4x4 sequence correctly, it appears the answer 27,720 regardless of which interpretation you use for returning to the "original state."
 
Thanks to everyone who took the time to respond. And to CuberBruce sorry for the confusion, I did mean solved state.

If this is correct, wow, just wow that it would go from 24 moves in a 3x3 to over 27,000 in the 4x4! Just if I can be a bit more of a nuisance could you please take a moment to look at this clip I uploaded and give what the correct notation is?


Thanks again,
Ras
 
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Wait a second... Any odd numbered cube takes 24 repetitions... Even the 17x17x17... That's even more fascinating...

Actually, for nxnxn cube with n = { 3, 7, 11, 15, ... }, it appears only 12 repetitions are needed. (It appears in the original post, 24 moves was stated for the 3x3x3. This apparently referred to actual L-R axis turns, rather than the number of repetitions. The number of repetitions is 12.)

It appears that for all even size cubes with n >= 4, that 27,720 repetitiions are needed.

It appears not to matter whether the cube rotation is y or y'.
 
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It appears that all even size cubes with n >= 4, it appears that 27,720 repetitiions are needed.

Which brings up the interesting question: what move sequence requires the greatest number of repetitions?

For the purposes of this question, consider all solved positions (any orientation, and any permutation of
indistinguishable pieces) to be identical.

How high can we go on the 3x3? The 4x4? The 5x5? Is there a limit as n increases?

And can you use this information to break any of the cycle-counting online algorithms?

(Some of these answers are fairly well known. But some are not known, at least not to me.)
 
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LCM(1, ..., 24)

I know that, and you know that---but I thought others would have fun discovering it.

At least maybe they can discover why this is so.

And similarly, find the input that breaks the online cycle counter previously referred to.
(For some definition of "break".)
 
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