# Message #3400

From: Sid Brown <synjanbrown@gmail.com>

Subject: Re: [MC4D] Re: Cube puzzles and math

Date: Tue, 28 Jun 2016 08:09:59 +0100

Andrew,

I thought about defining equivalence within my group but it is not

intuitive as as you have to convert both configurations into Joel’s group.

Converting from my group into Joel’s one is pretty straight forward but

converting back again is difficult as you must follow a solution algorithm

and invert it. Modding into the smallest member of the group is an

extremely difficult task and I would not know where to start, maybe using

equivalence rules for sub words in the language? I’m not sure. But if this

is possible then modding out to the smallest representation (which I’m

coining as the prime numbers of this group) will give you the shortest

possible solution for the given Rubik’s cube configuration. Although

verifying that is the smallest is also a very difficult task; the only way

I have thought of doing this so far is a kind of brute force algorithm

which would take an age to complete. What may be easier is verifying that

the equivalence rules for reduction is fully comprehensive, but again I’m

not sure how to prove this.

Any thoughts on this?

Regards,

Sid

On 28 June 2016 at 07:14, apturner@mit.edu [4D_Cubing] <

4D_Cubing@yahoogroups.com> wrote:

>

>

> Dear Sid,

>

> Yes, the object you defined is simply a nonabelian (non-commutative) group.

>

> The comment you made about mapping to Joel’s group and back is

> interesting, and accurate, since you never actually mentioned taking the

> quotient. The set R = S* (with * the Kleene star) does have an infinite

> number of elements that correspond to each physical state of the Rubik’s

> cube. With the group operation you described and no other relations in the

> presentation of the group, you have the free group over the set S. To

> actually get the Rubik’s Cube group (the group that Joel describes), you

> have to take the quotient by the equivalence relation that relates all

> sequences of moves resulting in the same physical state, R/~. An equivalent

> approach would be to add relations to the presentation of the group (such

> as L^4 = e, because rotating the left face four quarter turns is equivalent

> to the identity). Really, it’s R/~ that is isomorphic to Joel’s group, not

> the free group as you described it. In all my previous posts I had assumed

> you were modding out the equivalent move sequences.

>

> Thanks for the discussion!

>

> Cheers,

> Andrew

>

>

>