How many ways are there to 2-color 64 squares of an $8 \times 8$ chess board with free rotation (no mirroring)?

I'm not understanding Burnside's theorem really. I have $\frac{1}{4} (1 \times 2^{64} ...)$ but I don't know how to figure out the other numbers...


closed as off-topic by colormegone, choco_addicted, Ian Miller, Shailesh, Claude Leibovici Apr 8 '16 at 6:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – colormegone, choco_addicted, Ian Miller, Shailesh, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.


The group $\mathbb Z_4$ acts on the chess board by rotation. Then we know that:

  • $\bar 0$ fixes everything, so we have $|X_e| = 2^{64}$
  • $\bar 1$ fixes colorings determined by one of the $4x4$ subboards, so we have $|X_1| = 2^{16}$
  • $\bar 2$ is similar, but now the $4x8$ rectangle determines the coloring, so $|X_2| = 2^{32}$
  • $\bar 3$ is identical to $\bar 1$, so we have $|X_3| = 2^{16}$

Apply the theorem, $$|X/G| = \frac 1 {|G|} \sum_{g\in G} |X_g| = \frac 1 4 (|X_0| + |X_1| + |X_2| + |X_3|)$$ $$= \frac{1}{4} (2^{64} + 2\times2^{16} + 2^{32}) = 2^{62} + 2^{15} + 2^{30}$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.