Dodecahedra

Originally posted by Thomaster
So you have painted a lot of dodecahedra? 😛

Thankfully, my approach was not that inelegant.

Originally posted by LemonJello
I get 96. I think it is right, but the approach I used is not very elegant...still considering.

Consider it no longer, your answer is correct!

Be glad that rhombicosidodecahedra weren't chosen.

Hallo,

misreading the riddle, I got to a more complicated one (see next post). Here is a bit of what I got:

Consider one color alone, first.

1) Painting one face. Only one combination, rest is symmetrical.
2) Painting two faces. There are 3 distinct solutions on a dodecahedron (both touching, one in between and opposite to each other)
3) Painting three faces. There are already six unique solutions. I take two as touching, two always have to. How many combinations with the third? It can touch both (symmetrical, only one solution), it can touch only one (this gives five combinations which are unique to each other - you can not change them by rotation into each other, only mirroring) or none (opposing the first). Finally, place three single faces on the surface (not touching each other). Of this exists at least one possibility. This gives 8 more.
4) Painting four faces. Starts to get difficult. We start with three touching each other and the 4th needs to be placed. This should give six distinct solutions, which can not be rotated into one another (the sixth is the not touching one). Next is: two pairs of two. There is at least one combination. Same holds true for a pair of two and two singles. This makes at least 8 unique solutions.
5) Have not managed yet. There are at least 3 (cluster of 5, cluster of 4+1, cluster of 3+2). How many of those, I dont know. It feels like, it should be at least 8. But modestly, it is 3.
6) see 5)
7) see 5), but inverted.
8) At least 8, same as 4), inverted.
9) At least 7, same as 3), inverted.
10) 6, same as 2), inverted.
11) Only 1. Same as under 1). Basically blue and colorless just swap places.
12) Only 1. All blue.

For one color alone we get at least: 1+3+6+8+8+3+3+3+8+7+6+3+1+1=62. I think. At least. We can do the same now for red. Gives another 62, at least.

The real fun starts with mixed colors.

I am pretty sure, especially for the cases of 3 faces or 2 faces, that these solutions are unique, hence the dodecahedrons would be distinguishable.

Let me know your thoughts or approaches, I see not many chances past 3d-symmetry thoughts (I could write down formulas for the permutations, but some of these break symmetries, others not, which I wouldnt know how to handle).

T

After rereading the riddle, I retract my answer and post a new riddle...

If you have red and blue paint, and you dont have to paint EACH face, how many combinations are possible?

Originally posted by iamatiger
Is this about Mr. Burnside and his Lemma?

Doubt it. It might be about Mr Burnside and the lemma which isn't his, though.

Ahem...

There are infinite possibilities.
There are so many designs you could paint on the faces that your only limitation is the amount of paint.

Originally posted by ErinR
Ahem...

There are infinite possibilities.
There are so many designs you could paint on the faces that your only limitation is the amount of paint.

Each face is to be painted red or blue, not red and blue. Therefore, there are at most 2**12, equals 4096, different setups, if the different faces on each dodecahedron can be distinguished. If, as the original question implied, they cannot, and rotationally similar dodecahedra are to be counted as one, the number is clearly lower. The point of the question is how much lower.
If, as in the second version, each face is to be painted either red, or blue, or not at all, the maximum is the much larger 3**12 (is 531441), and working out the solution with similar cubes removed is rather harder.

Richard

hmm...

I must have read it wrong. Thanks for clarifying!