# Irreducible Loops

Submitted by Peter on Wed, 04/18/2007 - 09:20.

Some ideas for how to possibly proove that the diameter of the Cube-Group would be X.

Hi, I am new to the Cube problem, so probably the ideas are not new, or too naive, but I could not find them anywhere. This is possibly due to my lack of knowledge of terminology. (My background is theoretical solid states physics.)

Thus I would like to share these ideas with you, which you hopefully find useful, or can tell me that these ideas are not new or possibly that they are useless. I kindly ask you to comment. I just go ahead...

How could one calculate the diameter of the Group?

Let A be a random permutation. I start by choosing a (not optimal) path from id to A, say in quarter turn metric, with A=prod_i(ai) (i=1,...,N) where ai in {U,U',D,...}.

Could one possibly proove, that the path can _always_ be 'simplified' to a lenght smaller or equal to X (which then would be the diameter)?

How could such a simplification look like? The only way I see is through the usage of irreducible loops.

So, what do I mean by an irreducible loop. I define it to be a sequence of operations xi, i=1,...,n such that prod_i(xi)=id and the path is in a certain sense topologically nontrivial. Here xi= U, U', D, etc.. For example, D*D*D*D=id is such an irreducible loop. The simplification of the path A wrt to this loop is already implicit in the notation that is commonly used. E.g. D*D*D is replaced by D'. But there exist loops which are longer...

Now, given a loop of length n (which at least in quarter turn metric is even):

IF the path A and the loop contain a common sequence of length n/2+1, THEN this sequence in A can be replaced by a sequence of length n/2-1, which is the 'other half' of the loop. As a result one has reduced the length of the path for A to N-2.

If one would have knowledge of ALL irreducible loops in the Cube Group, then one possibly could proove that ANY path can be reduced to a length X.

Thus we should ask, how could one possibly generate all irreducible loops.

I admit, I don't know. But I know how to systematically generate a large number of them. The key is to consider optimal paths to symmetric permutations (i.e. states of the cube). Namely, the optimal path to a symmetric state is always degenerate, i.e. there is always at least a second path to this state. By going one path to the symmetric state and the other back, one automatically has a loop. As an example, the state generated by D*D is also generated by D'*D', thus D*D=D'*D', or equivalently D*D*D*D=id. As an other example, we could go the path down to the supertwist and choose another one back. Whether such a loop is really irreducible can then be checked straightforwardly.

Finally, several questions turn up.

First, and most important, do the optimal paths to the symmetric states give all irreducible paths? Or, stated differently, can there be an irreducible loop that does NOT run over a symmetric state? My feeling is, that there should be a simple proof to this one, using symmetry arguments. What do the mathematicians say?

Second, if indeed we can obtain all irreducible loops through this procedure, would they allow to make the second part of the proof, namely that all paths can be simplified to a length smaller or equal to X?

In light of the analysis performed by Herbert Kociemba, Silviu Radu and others such an undertaking sounds worthwile.

So please tell me, what you think about it. If you think, this is worth consideration, I would be ready to invest efforts in further analysis.

Peter Jung

Hi, I am new to the Cube problem, so probably the ideas are not new, or too naive, but I could not find them anywhere. This is possibly due to my lack of knowledge of terminology. (My background is theoretical solid states physics.)

Thus I would like to share these ideas with you, which you hopefully find useful, or can tell me that these ideas are not new or possibly that they are useless. I kindly ask you to comment. I just go ahead...

How could one calculate the diameter of the Group?

Let A be a random permutation. I start by choosing a (not optimal) path from id to A, say in quarter turn metric, with A=prod_i(ai) (i=1,...,N) where ai in {U,U',D,...}.

Could one possibly proove, that the path can _always_ be 'simplified' to a lenght smaller or equal to X (which then would be the diameter)?

How could such a simplification look like? The only way I see is through the usage of irreducible loops.

So, what do I mean by an irreducible loop. I define it to be a sequence of operations xi, i=1,...,n such that prod_i(xi)=id and the path is in a certain sense topologically nontrivial. Here xi= U, U', D, etc.. For example, D*D*D*D=id is such an irreducible loop. The simplification of the path A wrt to this loop is already implicit in the notation that is commonly used. E.g. D*D*D is replaced by D'. But there exist loops which are longer...

Now, given a loop of length n (which at least in quarter turn metric is even):

IF the path A and the loop contain a common sequence of length n/2+1, THEN this sequence in A can be replaced by a sequence of length n/2-1, which is the 'other half' of the loop. As a result one has reduced the length of the path for A to N-2.

If one would have knowledge of ALL irreducible loops in the Cube Group, then one possibly could proove that ANY path can be reduced to a length X.

Thus we should ask, how could one possibly generate all irreducible loops.

I admit, I don't know. But I know how to systematically generate a large number of them. The key is to consider optimal paths to symmetric permutations (i.e. states of the cube). Namely, the optimal path to a symmetric state is always degenerate, i.e. there is always at least a second path to this state. By going one path to the symmetric state and the other back, one automatically has a loop. As an example, the state generated by D*D is also generated by D'*D', thus D*D=D'*D', or equivalently D*D*D*D=id. As an other example, we could go the path down to the supertwist and choose another one back. Whether such a loop is really irreducible can then be checked straightforwardly.

Finally, several questions turn up.

First, and most important, do the optimal paths to the symmetric states give all irreducible paths? Or, stated differently, can there be an irreducible loop that does NOT run over a symmetric state? My feeling is, that there should be a simple proof to this one, using symmetry arguments. What do the mathematicians say?

Second, if indeed we can obtain all irreducible loops through this procedure, would they allow to make the second part of the proof, namely that all paths can be simplified to a length smaller or equal to X?

In light of the analysis performed by Herbert Kociemba, Silviu Radu and others such an undertaking sounds worthwile.

So please tell me, what you think about it. If you think, this is worth consideration, I would be ready to invest efforts in further analysis.

Peter Jung