I am trying to determine the number of unique solutions when placing "dots" on the sides of an icosahedron. There can be up to three dots placed symmetrically on each side. The dots are aligned with the corners. If I let 0, 1, 2, or 3 dots be represented by a color, I can use Polya's theorem with four colors, which will give me about 1,8*10^10 possibilities.
$$
P_{G} = \frac{1}{60}(z_1^{20}+24z_5^4 + 15z_2^{10}+20z_1^2 z_3^6)
$$
However, that will not take the rotational asymmetry for 1 and 2 dots into account. So, the solution will be a lower bound. I would appreciate some help in understanding how to handle this asymmetry and how much it will increase the number of possibilities.
A similar (even more interesting) variant is when the dots can have two different colors. Again I can assume 7 different colors, but that solution probably be even worse.
With different colorings I mean unique patterns that can't be replicated through any of the 60 available rotations. I.e. a duplicate is a solution that is identical to another solution after rotation and should therefore not be counted.
Note, this is a real world problem involving viruses.
Best Answer
We can place a surprisingly tight bound on the answer using two simple models. Here the models are explored for the simpler case of a tetrahedron withbone type of dot, then the results are applied to the icosahedron.
Model 1: Many Colors
Since the different orientations of one or two dots are considered not equivalent, we suppose that there are not four but eight different colors possible for each face. We then apply the Polya theory to eight colors on a tetrahedron to get
$N_1(tet,1)=(1/12)[8^4+(11×8^2)]=400$
The problem with this us that not all colors are fully distinct. As described in a comment to my original answer, this may double-count some combinations due to "colors" rotating into one another which actual colors don't do. So we should treat this result as an upper bound.
Model 2: Limits of Symmetry
If all the dot positions were truly distinct, we would simply count off $2^{12}$ possible choices of dot or no dot on the tetrahedron. That would give $4096$. But what if some of these orientations are rotatable into each other? In this model we would not really know. But we can observe that no more than twelve configurations at a time can be rotated intobeach other because that is all the rotational symmetry therebus in a tetrahedron! So we are sure of the lower bound
$N_2=4096/12=342$ (rounded up, can't have a fraction)
We are surprised at how tight these bounds are. We could surely call the true count $370$ and bet the house that we are within ten percent for the one-type tetrahedral case!
And the relative error only gets better with the icosahedron ... .
Icosahrdron, one type of dot
With $2^3=8$ eight colors on each face for Model 1 and sixty dots for Model 2, we render the bounds thusly:
Upper Bound (Model 1): $N_1(icosa,1)=(1/60)[8^{20}+(24×8^4)+(15×8^{10})+(20×8^8)]\approx1.922×10^{16}$
Lower Bound (Model 2):: $N_2=2^{60}/60\approx1.922×10^{16}$(!)
We can report that to four siginficant digits, the permutation count is $1.922×10^{16}$! (The exact figure was later calculated by a more sophisticated use of the Polya equation: $1921538678935736$.)
Icosahrdron, two types of dot
With three choices at each position of each face Model 1 gives $3^3=27$ colors. Model $2$ gives $3^{60}$ dot permutations before dividing out rotational symmetries:
Upper Bound (Model 1): $N_1(icosa,2)=(1/60)[27^{20}+(24×27^4)+(15×27^{10})+(20×27^8)]\approx7.065×10^{26}$
Lower Bound (Model 2):: $N_2=3^{60}/60\approx7.065×10^{26}$(!)
Again the simple models bracket (at least) four significant digits!
Figuring out the trick
Like any good magician, I subtly apllied some misdirection. Let's go back to the tetrahedral, one-type case. In Model 1 I apply the Polya theory, with the first term divided by the rotational symmetry ($12$) being $8^4$. In the second case I identify just a single term divided by $12$, namely $2^{12}$. Well ... these two quantities are the same, and moreover that matched term dominates $N_1$ because the other tetms have smaller exponents on the base $8$. So we get a tight bound because the dominant terms agree. Going to the icosahedron only tightens this agreement because with more dots to play with, the ecponents and the differences between said exponents goes up even higher.