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In short: I'm considering colorings of colorings. I have a notion of equivalence that involves permuting the objects in the original colorings and a notion of equivalence that involves switching colors in the original colorings together with the coloring of colorings.

Can I calculate the number of inequivalent colorings of colorings using Pólya's enumeration theorem?

If not: Can I - from the number of inequivalent colorings without color-switching - say something about the number of inequivalent colorings with color-switching?

If not: Can I do anything besides using Burnsides Lemma and thereby essentially doubling my work (because the color-switching doubles the number of elements in my symmetry group acting of the colorings of colorings)?


Background/Explanation:

Consider a $3$-D cube and a set of two colors - let's say black and white. I can then color the six faces of that cube. There are $2^6$ ways of coloring, but some of them are equivalent to each other, because they can be transformed into each other by rotating and/or mirroring the cube.

If I call the the symmetry group of the cube $G$, then I can count the essentially different colorings using Pólya's enumeration theorem: $$Z=\frac{1}{|G|} \sum_{g\in G} 2^{\gamma (g)}$$ where $\gamma (g)$ is the number of cycles $g$ splits into when considering it as a permutation of the faces of the cube.

So far so good.


Let's now say my "object" is the set $D$ of all 2-colorings of the graph $P_3$:

$$D=\{⭘\!\!\!-\!\!\!⭘\!\!\!-\!\!\!⭘,⬤\!\!\!-\!\!\!⭘\!\!\!-\!\!\!⭘,⭘\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⭘,⬤\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⭘,⭘\!\!\!-\!\!\!⭘\!\!\!-\!\!\!⬤,⬤\!\!\!-\!\!\!⭘\!\!\!-\!\!\!⬤,⭘\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⬤,⬤\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⬤\} $$

Now consider the set $S$, which are the 2-colorings of the set $D$. If $R$ is the set of colors, then every $s\in S$ is a mapping $D\rightarrow R$.

I still have a notion of equivalence that involves mirroring, i.e. two colorings $s_1,s_2\in S$ are considered equivalent, if $s_1(\sigma_r(d))=s_2(d)$ for every $d\in D$. Where $\sigma_r$ switches the left and right vertices, e.g. $\sigma_r(⬤\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⭘) = ⭘\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⬤$.

I now consider another notion of equivalence: color-switching. This is where I run into problems.

Call two elements $s_1,s_2\in S$ equivalent if $$\overline{s_1(\overline{d})} = s_2(d) \; \text{for every } d\in D$$ where the "overbar" is a slight abuse of notation, denoting both an inversion of color and an inversion of coloring. To give an example for these two cases:

  • $\overline{\text{black}} = \text{white}$
  • $\overline{⬤\!\!\!-\!\!\!⬤\!\!\!-\!\!\!⭘} = ⭘\!\!\!-\!\!\!⭘\!\!\!-\!\!\!⬤$

My equivalence relation can now not purely be expressed as permutations of $D$, because the operation of color-switching also affects the colors/outputs of $s_1$ and $s_2$.

To accurately calculate the number of inequivalent colorings of colorings, I fell back to Burnside's Lemma and calculated the number of fixed points of:

  • The identity
  • The mirroring (switching vertices $1$ and $3$)
  • The color-switching
  • The combination of the mirroring and the color-switching

Dividing the sum of the fixed points by $4$, I get the result of $88$, which is correct, but tedious.

If my set $D$ is, let's say, the set of all colorings of a larger object, with a lot more symmetries, then adding just the color-switch symmetry will essentially double my work and I will have to fall back to Burnside's Lemma.

If I already calculated the number of inequivalent colorings of colorings of a complicated object with a lot of symmetries, can I say something about what happens if I "just" add the color-switching symmetry?


Curiously enough... Pólya's enumeration theorem does give the correct answer in this case, if I consider $s_1$ and $s_2$ equivalent if $s_1(\overline{d})=s_2$ ... which is baffling to me, because this is not the notion of symmetry I want.

It also works when I considered more complex objects, but I have no clue why.

Not all is well though, because it does not work in a simple case where $D$ is the set of colorings of $⭘\!\!\!-\!\!\!⭘$ and the symmetries are "switching of the vertices" and "color-switching" (and their combination). I have no clue why.

Any help would be much appreciated.


(Actual background: I'm trying to calculate the number of different cellular-automata rules. When neighborhoods or the dimension get larger, it quickly becomes a hassle to do by hand.)

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    This looks like OP is asking for Power Group Enumeration of which there is an example at the following MSE link I and this MSE link II. The reference is the book Graphical Enumeration by Harary and Palmer. – Marko Riedel Oct 01 '22 at 23:45
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    It is sufficient to compute the two cycle indices of the group acting on the figure and the group acting on the colors. You presumably have the first one already and the second one is likely to be the cycle index of the symmetric group. – Marko Riedel Oct 02 '22 at 01:35
  • @MarkoRiedel looks like that is exactly what I need and it was already done in a general case in Michael Harrison's "A Census of Finite Automata". I will leave the question up anyways. Thank you. – NiveaNutella Oct 02 '22 at 11:37
  • I just noticed the case of the figure being a square board is discussed at this MSE link. – Marko Riedel Oct 02 '22 at 18:37

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