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Let $C$ denote the set of colorings of an 8x8 chessboard, where each square is colored either black or white. Let $\sim$ denote the equivalence relation on $C$ defined as follows:
two colorings are equivalent if and only if one of them can be obtained from the other by a rotation of the chessboard. What is the cardinality of the set $C/\sim$ of the equivalence classes of elements of $C $ under $\sim$ ?

METHOD 1(A TIRESOME PROCESS) Let us start with boards having lesser number of squares:

  1. If the board is 1x1 then we don't have to think, the answer is simply 1
  2. Let us consider a 2x2 board (say) $$A=\begin{pmatrix}*&*\\ * &*\end{pmatrix}$$ there are a total of $2^4=16$ ways to color the 4 squares in our board A.

I will list down the ways in which we can color the square such that on rotation of kind $0^0,90^0,180^0,and\ 270^0$ counterclockwise we get an equivalent coloring configuration.

Before that there are a few Notations I'll use

  • I'll use '$' to represent Black colored squares, and
  • '*' to represent White colored squares

Category 1 $$\begin{pmatrix}*&*\\*&*\end{pmatrix}$$ there is only one such coloring and it can be obtained through any type of rotation
Category 2 $$\begin{pmatrix}$&$\\$&$\end{pmatrix}$$ similar to catagory 1 there is only one such coloring and it can be obtained through any type of rotation
Category 3 $$\begin{pmatrix}$&*\\*&*\end{pmatrix}$$ there are 4 such coloring (one black 3 white) which can be obtained from the given configuration by rotation.
Category 4 $$\begin{pmatrix}*&$\\$&$\end{pmatrix}$$ there are 4 such coloring (one white 3 black) which can be obtained from the given configuration by rotation.
Category 5 $$\begin{pmatrix}*&$\\$&*\end{pmatrix}$$ and $$\begin{pmatrix}$&*\\*&$\end{pmatrix}$$ they are two rotations of each other with coloring(2 black and 2 white diagonal).
Category 6 $$\begin{pmatrix}*&*\\$&$\end{pmatrix}$$ there are 4 such colorings (2 black 2 white parallelly) which again can be obtained from rotation.

Observe: No coloring category(i) can be obtained by rotating category(j) configuration ($i\neq j$)

Each Category is an equivalence class as defined in question

METHOD 2(BURNSIDE'S LEMMA): on 2x2 chessboard

BURNSIDE'S LEMMA:G be a finite group of permutations on a set S, then the number of orbits of elements of S under G is $$\frac{\sum_{\phi \in G}|fix\phi|}{|G|}$$

Applying this lemma to Category 1,2,3,4 we obtain that the number of such distinct colorings(rotation allowed) is merely 1 for each case.

For Category 5 and 6 together (i.e. coloring 2 of the squares black and two white) the following table justifies my answer \begin{array} {|r|r|}\hline elements\ of\ Z_4 & numbers\ of\ colorings\ that\ are\ fixed\ \\ \hline identity=0^0 & 6 \\ \hline 90^0 or\ 270^0 & 0 \\ \hline 180^0 & 2 \\ \hline \end{array}

hence, number of distinct colorings(rotation allowed) of a 2x2 chessboard when two squares are colored black and two white is $\frac{6+2.0+2}{4}=2$

So, in all there are 1+1+1+1+2=6 equivalence classes of the set C(set of colorings of 2x2 chessboard) under $\sim$

I understand that the process I'm using in Method 1 is tiring and time consuming. I got the hint from the comments that we can use Burnside's Lemma to do the problem, so I used it on 2x2 chessboard in Method 2, but how to determine the Set of Colorings of the huge 8x8 chessboard on which I will have to act the group $Z_4$ ?

Chayan
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  • It seems this question is too advanced for you. (This is based on "I cannot think of any way how to do it") – GEdgar Nov 02 '21 at 10:47
  • Do you know and understand the Burnside's lemma? – Adayah Nov 02 '21 at 10:49
  • Just a reminder: take a look at our guidelines on how to ask a good question, with emphasis on avoiding "I cannot think of any way" questions. – Lee Mosher Nov 02 '21 at 10:55
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    Here's a list of links to questions relating to Burnside's lemma. Maybe your question will be there, or maybe reading some of those will help you understand it better. https://math.meta.stackexchange.com/questions/1868/list-of-generalizations-of-common-questions/13335#13335 – MJD Nov 04 '21 at 13:08
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    Here's one that is somewhat similar: how many ways are there to have an $n×n$ board with 2 black squares and $n^2-2$ white squares? https://math.stackexchange.com/questions/744161/counting-shaded-squares/744383#744383 – MJD Nov 04 '21 at 13:11
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    Regarding your question “how to determine the Set of rotational symmetries of the huge 8x8 chessboard”: An 8×8 chessboard has the same four rotational symmetries as any other size chessboard or any square at all: you can rotate it by 0°, 90°, 180°, or 270°. This is all that matters for Burnside's lemma. – MJD Nov 04 '21 at 20:52
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    @MJD I was wrong to write Set of Rotational symmetries I have changed it to Set of Colorings – Chayan Nov 05 '21 at 06:29
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    You can use combinatoric techniques to count the colorings without listimg them one at a time. For example, how many colorings are there in all? It's $2^{64}$. Do check out that chessboard one I linked above, I think you will find it helpful. Your question is somewhat easier because it doesn't have the extra constaint of requiring exactly two black squares. – MJD Nov 05 '21 at 09:47
  • @MJD that link helped a lot understand it. Thanks a lot. – Chayan Nov 05 '21 at 12:06

1 Answers1

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We will categorize chessboards based on whether they have $1$-fold, $2$-fold, or $4$-fold rotational symmetry. Let $C(n)$ denote the chessboards with $n$-fold symmetry, and let $C’(n)$ denote chessboards with highest order $n$-fold rotational symmetry. Boards in $C(n)$ are uniquely determined by $\frac{64}{n}$ chess board tiles, so $|C(n)|=2^{\frac{64}{n}}$. Boards with highest order $4$-fold symmetry are just $C’(4)=C(4)$, Boards with highest order $2$-fold symmetry are given by $C’(2)=C(2)-C(4)$, while boards with highest order $1$-fold symmetry are given by $C’(1)=C(1)-C(2)$. And lastly, $$|C’(n)/{\sim}|=\frac{n|C’(n)|}{4}.$$ Then, $$|C/{\sim}|=|C’(1)/{\sim}|+ |C’(2)/{\sim}|+ |C’(4)/{\sim}|\\ =\frac{|C’(1)|}{4}+\frac{|C’(2)|}{2}+|C’(4)|\\ = \frac{|C(1)|-|C(2)|}{4}+\frac{|C(2)|-|C(4)|}{2}+|C(4)|\\ = \frac{|C(1)|}{4}+\frac{|C(2)|}{4}+\frac{|C(4)|}{2}\\ = 2^{62}+2^{30}+2^{15} $$

miracle173
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Jacob
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  • I didn't understand the terms like n-fold rotational symmetry, order of n-fold rotational symmetries, and please provide a source of the fact that C(n) is determined by $\frac{64}{n}$ chessboard tiles – Chayan Nov 05 '21 at 09:38
  • @Chayan $n$-fold rotational symmetry means that the board is invariant under rotation by $2\pi/n$ (c.f. https://en.m.wikipedia.org/wiki/Rotational_symmetry). It follows that a board with $n$-fold rotational symmetry has $n$ Identical parts that are rotated $2\pi/n$ from each other, so you get a unique board with $n$-fold symmetry by changing the makeup of that rotated part. There are $n$ identical parts, and hence each individual part is made up of $64/n$ tiles, and therefore there are $2^{64/n}$ such choices for the rotated part. – Jacob Nov 05 '21 at 16:49
  • @Chayan Highest order $n$ fold rotational symmetry means, that it has $n$-fold rotational symmetry and no higher degree of rotational symmetry. For example, a square has $1$-fold, $2$-fold, and $4$-fold rotational symmetry, but it’s highest order is $4$-fold. – Jacob Nov 05 '21 at 16:53