Calculating "The Devil's Algorithm"

[930913]

God's algorithm, defined as an algorithm that will solve any cube.

• A given sequence of moves, if repeated enough times, will return the cube to the original state.
• The number of repetitions needed to return to the original state, multiplied by the QTM of the sequence, gives how many states the cube traverses.
• A sequence can be found, that traverses every state of cube.
• This sequence is God's algorithm, as defined above.

So "F" needs 4 repetitions to return to its original state, giving a traversal of 4 states.
"F U F' U'" needs 6 repetitions to return to its original state, giving a traversal of 24 states.
"F y" or "F R B L" has a traversal of 1440.

So my theory is, find the biggest number, and you should have God's algorithm, assuming one exists.

Thoughts?

[KingTim96]

seems really interesting, if this could work, that would be awesome.

[Noahaha]

I'm no expert on group theory, but I believe the only alg that accomplishes this is called the devil's algorithm, which when performed once on a cube cycles through all 43 quintillion positions. Usually God's alg/number refers to the minimum number of moves to solve the cube.

[930913]

I'm no expert on group theory, but I believe the only alg that accomplishes this is called the devil's algorithm, which when performed once on a cube cycles through all 43 quintillion positions. Usually God's alg/number refers to the minimum number of moves to solve the cube.

Meh, what's in the name?
We must call things differently in my part of the world

Anyway, who's up for calculating the shortest algo?

[tasguitar7]

Well, doing this by brute force is unfeasible given the number of states of the cube. Now, I have watched videos and read some stuff on commutators and all of that theory, and I can apply it, however I'm not sure if what I am about to say is entirely correct. It seems possible to do essentially "arithmetic" with the members of the set of the cube. For instance you have F, and the cycling effect of F, and R, and the cycling effect of R, so you then add F and R to get F R, and then you observe its cycling effect. So, using this and some other nuggets of thought that I don't have, is it possible to systematically or asymptotically move toward the devil's algorithm?

devil's alg = alg that starts at solved, goes through every state once, ends at solve
god's alg = optimal solution for any specific case

Btw, for a very new user, this is an extremely high quality first thread and first post.

[PM 1729]

For 3x3x3
For 2x2x2
I believe this is what you were referring to, correct me if I am wrong.

[tasguitar7]

Wow that's crazy, I didn't know that it had been calculated for 3x3x3. I though it was only 2x2x2 at this point. Do you happen to know how they were calculated? It seems like brute force would have been too large.

[irontwig]

Wow that's crazy, I didn't know that it had been calculated for 3x3x3. I though it was only 2x2x2 at this point. Do you happen to know how they were calculated? It seems like brute force would have been too large.

http://www.speedsolving.com/forum/sh...l=1#post690226

Just ask Bruce, I guess.

[Stefan]

I didn't know that it had been calculated for 3x3x3

You said you want the shortest algo, allowing repetitions. I doubt Bruce's alg is that (it could be, but I see no reason why it should).

Do you happen to know how they were calculated?

You could read Bruce's pages about it...

Meh, what's in the name?
We must call things differently in my part of the world

Well, welcome to our part of the world. I'll ask a mod to rename the thread. We don't need extra confusion about these terms.

[whauk]

assuming your "god's algorithm" g exists:
then R=g*a (repeating gods alg a times, for a certain a) and U=g*b (similar)
so RU=(g*a)+(g*b)=g*(a+b)=g*(b+a)=g*b+g*a=UR
as RU is not the same as UR i prooved that your "gods algorithm" cannot exist.