Message #3402

From: Sid Brown <>
Subject: Re: [MC4D] Re: Cube puzzles and math
Date: Tue, 28 Jun 2016 08:42:26 +0100


Yeah I understand that you are talking about it quite abstractly. But what
I aim to do is program an algorithm that can always determine the shortest
possible solution. I can implement my brute force one for a 3x3x3 as that
won’t take too long, but for 3^4 it would take too long. I will use the
brute force of 3^3 to verify an algorithm of reduction of 3^3 and then
attempt to expand this to 3^4, but what I’m trying to do might prove near
impossible and I’m not sure whether I will actually achieve it. I will have
a decent stab at it though.


On 28 June 2016 at 08:36, [4D_Cubing] <> wrote:

> Sid,
> You’re talking about actually finding an explicit presentation of the
> Rubik’s Cube group. Just from some quick Googling, it looks like there has
> been a fair amount of discussion of this online, but no definitive results
> as far as I can see.
> But in terms of just talking about the group, it isn’t necessary to
> actually determine the presentation. We still understand what it means to
> say that we quotient out the free group by the equivalence relation that
> relates physically equivalent states.
> Another way to think about it, which sort of bridges the gap between your
> formulation and Joel’s, is given on Wikipedia. They just consider the 48
> movable stickers on the cube, and the permutations of these stickers that
> correspond to the moves L, R, U, D, F, B, and then say that the Rubik’s
> Cube group is the subset of the symmetric group S_{48} generated by those
> permutations: < L, R, U, D, F, B>. This phrases the setup in terms of
> moves of the cube, but automatically deals with the physically equivalent
> states, because it inherits the relations from the presentation of S_{48}.
> In fact, this approach gives you a presentation of the Rubik’s Cube group,
> in the form < L, R, U, D, F, B| all relations of S_{48}>. This
> presentation is almost certainly not the smallest, however!
> Cheers,
> Andrew