I have $4$ red cubes, and $4$ yellow cubes, all of dimension $1\times 1\times 1$. How many ways to create a new cube of dimension $2\times 2\times 2$, such that each new cube is distinct (i.e cannot be rotated or flipped to produce any other cube) ?
I managed to solve this question by simply writing out all possible cases and convincing myself(by rotating the cube in my head) that one case could not be obtained from the other.
I am curious to see how other people solve this problem.
On
As has been suggested in the comments, this can also be done using Burnside’s lemma.
AKemats’s answer seems to assume that you only want equivalence up to rotations; my understanding of the question was that by “flipping” you mean an inversion or reflection, which doubles the size of the symmetry group; but either symmetry group can be treated with either approach.
The identity transformation leaves all $\binom84=70$ configurations invariant.
The $6$ rotations through $\frac\pi2$ have $2$ orbits with $4$ cubes each, so they only leave $2$ configurations invariant.
The $9$ rotations through $\pi$ have $4$ orbits with $2$ cubes each, so there are $\binom42=6$ ways to choose red orbits for an invariant configuration.
The $8$ rotations through $\frac{2\pi}3$ have $2$ singleton orbits and $2$ orbits with $3$ cubes each, so there are $2^2=4$ ways to pick one red singleton and one red triple.
Here’s the final result for rotations:
In total, that makes $70+6\cdot2+9\cdot6+8\cdot4=168$ pairs of transformations and fixed points, so, since the group has $24$ elements, there are $\frac{168}{24}=7$ equivalence classes of configurations under rotations.
For the orientation-changing transformations, the inversion has $4$ orbits with $2$ cubes each, and thus $\binom42=6$ fixed points.
A rotation through $\frac\pi2$ followed by an inversion has $2$ orbits with $4$ cubes each, and thus $2$ invariant configurations.
There are $6$ reflections that fix $4$ cubes and swap two pairs of the others; with $0$, $1$ or $2$ of the pairs red there are $\binom40+2\binom42+\binom44=8$ invariant configurations.
There are $3$ reflections that swap $4$ pairs of cubes, with $\binom42=6$ invariant configurations.
A rotation through $\frac{2\pi}3$ has an orbit with $6$ cubes and thus doesn’t leave any configurations invariant.
So here’s the final result including orientation-changing transformations:
In total, that makes $168+6+6\cdot2+6\cdot14+3\cdot6=288$ pairs of transformations and fixed points, so, since the group has $48$ elements, there are $\frac{288}{48}=6$ equivalence classes of configurations under rotations and orientation-changing transformations.
So there’s one less equivalence class if you include orientation-changing transformations. That means that two of the equivalence classes for rotations are mirror images of each other. This is where the cubes of each colour form a path from one corner to the opposite corner; this path has chirality that’s flipped by the orientation-changing transformations.
Or, applying the approach in AKemat’s answer: The polynomial in this case is
$$ \frac1{48}\left(x_1^8+9x_2^4+8x_1^2x_3^2+6x_4^2+x_2^4+6x_4^2+6x_1^4x_2^2+3x_2^4+6x_1^2x_6^1\right)\;, $$
and substituting $x_i=x^i+y^i$ and taking the coefficient of $x^4y^4$ yields the same result as the other approach above.
The way I would go about this is by using Polya's enumeration theorem. Essentially, what we're asking is to color the vertices of a cube with two colors and count the number of discrete colorings to rotation. Here, our permutation group is $S_4$, the orientation-preserving symmetries of the octahedral group. This means that the total number of colorings can be found via $\frac{1}{24}(x_1^8+9x_2^4+8x_1^2x_3^2+6x_4^2)$, where $x_i=x^i+y^i$, and then evaluating this in the case $x=y=1$.
Edit: sorry, I misread the question. We would still proceed in evaluating the polynomial that PET gave us, but we'd then observe what the coefficient of the $x^4y^4$ term is.
Edit edit: for some reason, it didn't occur to me that it would be pertinent to actually say what the answer is. I'll leave it spoiled, per your suggestion, as to not give it away to others.