# Message #3398

From: Sid Brown <synjanbrown@gmail.com>

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

Date: Tue, 28 Jun 2016 07:02:14 +0100

Andrew,

I might have got the terminology slightly wrong then, do i mean a field? or

should i just have said a group under an operation? I was only trying to

put it into a mathematical construct but I don’t have any actual experience

of doing this for a rubik’s cube and I only did a half year of maths at uni

(doing comp sci now). You guys sound like you know what you’re on about so

I’ll let you take the wheel on this one. Thanks for recognising they are

not the same. It may be worth noting that if you convert an instance of my

definition into Joe’ls one and then back again you may or may not have what

you originally started with, as in Joel’s every cube configuration is

unique and in mine there are an infinite amount of representations for any

given cube configuration.

Regards,

Sid

On 27 June 2016 at 21:07, apturner@mit.edu [4D_Cubing] <

4D_Cubing@yahoogroups.com> wrote:

>

>

> Dear Sid,

>

> The notation for the operation doesn’t really matter. The issue is that

> without another binary operation, the object you’re talking about cannot be

> a ring. A ring is a set together with two binary operations, one of which

> forms an abelian group with the set, and the other of which is associative

> and distributes over the group operation, but may or may not be commutative

> (and may or may not have an identity element, depending on how you define a

> ring). As far as I am aware, there is no way to turn the Rubik’s Cube group

> into a ring in any way that relates to the physical reality of the puzzle.

>

> As for whether the groups are the same or not, my point was simply that

> the group Joel refers to and the group that you are actually defining in

> your post are isomorphic, which means that they are really just different

> ways of talking about the same mathematical structure. But it’s true that

> sometimes it’s good to be careful and actually acknowledge that two groups

> are not truly the exact same group, but rather isomorphic, so I will

> concede that point.

>

> Cheers,

> Andrew

>

>

>