# Message #2106

From: Melinda Green <melinda@superliminal.com>

Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)

Date: Sun, 06 May 2012 16:39:48 -0700

I may have answered my own question which is that to account for color

symmetry we can simply divide by (n - 1)! where n is the number of

colors. For a 2-colored puzzle there is only one color permutation

whereas for 3 color puzzles there are two because we can fix one color

and either swap or not swap the other two. (n - 1)! gives the number of

unique color swapping patterns. Is that right? I usually don’t expect to

fully understand the equations here.

-Melinda

On 5/6/2012 4:18 PM, Melinda Green wrote:

>

>

> I feel that it’s not just tricky but it is wrong in most

> conceptualizations of the idea of puzzle state spaces. Taking this

> natural idea one step further, I would argue that states that have

> identical patterns of stickers should be thought of as the same state.

> For example, if you scramble any twisty puzzle and then swap all red

> and green stickers, then I feel that you still have the same state in

> terms of permutations since anything you can say about one version

> also applies to the other. For example, twist one face of a Rubik’s

> cube. For our purposes, it doesn’t matter which face was twisted. When

> talking about that state with each other we will never think to ask

> about the particular colors.

>

> Would anyone like to attempt to find the formula for the 3D and 4D

> cubes with this extra "color symmetry" constraint?

>

> -Melinda

>

> On 5/6/2012 2:35 PM, Andrew Gould wrote:

>>

>> The choice between 31 and 32 comes down to how you define the

>> locations of pieces. If you define all their locations relative to

>> one of the pieces it’s 31, but if you define what moves and what

>> doesn’t for each twist you can make it 32. I note that for 32, it

>> would be tricky to say that rotating the entire puzzle doesn’t change

>> the state.

>>