# Coloring sides of truncated triangular dihedral(bipiramid) into 3 colours

I need to find out the amount of ways to colour truncated triangular dihedron into 3 colours. So, the task will be easier if I had simple triangular dihedron.

First of all, do I understand right that truncated dihedron looks like : enter image description here The only things I know, that I must use Burnsside's lemma, calculate simmetries ( rotations ).

By the way, if you know some beginners algebra books with decent amount of examples in practice and more-or-less easy to understand for not english-speaker, thank you.

• As far as I remember, "dihedron" means "a body having two faces". The least possible number of faces of a body in $3D$ is however 4 (tetrahedron)... A dihedron would be just a two-sided plane figure. – user Jan 10 at 10:21
• My mistake, I meant bipyramid, sorry – Tovarisch Jan 10 at 10:26
• The next issue is meaning of "truncated". I would assume it means it is truncated by planes parallel to the base, so that the body has 8 faces (two equilateral triangles and 6 isosceles trapezoids). – user Jan 10 at 11:05
• @user I had this question too. In this case truncated means the dihedral consist of two truncated triangular piramids, alternatively saying, instead of lower and upper vericles there are triangles. So, the number of size is 6+2. – Tovarisch Jan 10 at 12:32

Note first of all that we obtain a cube when placing vertices at the centers of the hexagonal faces and connecting those where the faces share an edge. So we could solve this problem by asking about colorings of the vertices and the faces of the cube. We will work with the given solid however. (The cube was solved at this MSE link).

We evidently require the cycle index. There is the identity and three types of rotations. With $$a_q$$ for hexagonal faces and $$b_p$$ for the pyramids at the corners we get from the identity:

$$a_1^8 b_1^6.$$

The rotations about an axis passing through opposite pyramidal corners contribute

$$3 \times (2 a_4^2 b_1^2 b_4 + a_2^4 b_1^2 b_2^2).$$

The flips about an axis passing through the centers of two opposite edges ($$180$$ degree rotation) contribute

$$6 \times a_2^4 b_2^3.$$

Finally, the rotations by $$60$$ degrees and $$120$$ degrees about an axis passing through the centers of opposite hexagonal faces yields

$$4 \times 2 a_1^2 a_3^2 b_3^2.$$

We thus have for the cycle index

$$Z(T) = \frac{1}{24} (a_1^8 b_1^6 + 6 a_4^2 b_1^2 b_4 + 3 a_2^4 b_1^2 b_2^2 + 6 a_2^4 b_2^3 + 8 a_1^2 a_3^2 b_3^2).$$

We then get by Burnside for colorings with at most $$N$$ colors shared between hexagonal faces and pyramidal corners the closed form:

$$\frac{1}{24}(N^{14} + 6 N^5 + 3 N^8 + 6 N^7 + 8 N^6).$$

This gives the sequence

$$1, 776, 200961, 11198720, 254387500, 3265470936, 28260264606, \\ 183254654976, 953206454115, 4166682025000, \ldots$$

in particular for $$N=3$$ we obtain

$$\bbox[5px,border:2px solid #00A000]{ 200961.}$$

The formula when working with exactly $$N$$ colors as opposed to at most $$N$$ colors is (colors are shared between pyramids and hexagons):

$$\frac{N!}{24}\left({14\brace N} + 6 {5\brace N} + 3 {8\brace N} + 6 {7\brace N} + 8 {6\brace N}\right).$$

This gives the finite sequence

$$1, 774, 198636, 10399528, 200395755, 1903119150, 10359167700, \\ 35132702640, 77643165600, 113816102400, 109880971200, \\ 67199932800, 23610787200, 3632428800, 0, 0, \ldots$$

because there are only fourteen slots available. Note that the last one is $$14!/24 = 3632428800.$$

Remark. This cycle index also solves the problem where colors are not shared between hexagons and pyramids. We find:

$$\frac{1}{24} (N^8 M^6 + 6 N^2 M^3 + 3 N^4 M^4 + 6 N^4 M^3 + 8 N^4 M^2).$$

• I fucking love u man, best of luck to you, but as I see u did it for the figire of picture I provided in question . The figure I must solve is simple dihedral, but it consist of 2 truncated piramids, so we have 8 faces to colour ( imagine like we cut upper and lower part a bit). Am I right, that I need just do it for a18 without b? Sorry for bad english – Tovarisch Jan 10 at 15:48
• There are a number of examples at MSE Meta. I suggest you consult these and learn the technique in order to be able to solve this type of problem yourself. – Marko Riedel Jan 10 at 16:37
• Yep, I will have work for today's evening probably, thank you again mate – Tovarisch Jan 10 at 17:02
• for this dihedral picture I found: 1) 2 rotations by 120 degrees, which change only the vericles in the middle triangle + identity e( nothing changes) 2) 3 rotations by 180 degrees passing through medians of triangles ( so 2 triangles vericles exchange and also small triangles exchange) is it all? or I missed something? Because I found your solution where almost same figure and there is more cycles... – Tovarisch Jan 12 at 15:02
• As I am trying to understand I need to find group of rotations, for example, lets upper 1 triangle be 1, lower be 2, 3 faces of upper part 2,3,4 and lower part 5,6,7, so I can write rotations as permulations , for example for identity (1)(2)(3)(4)(5)(6)(7)(8), for rotations of triangles vericles (1)(234)(567)(8) and so on, and then the length of cycles will be degrees for N in burnside formula? Sorry for molestaion. – Tovarisch Jan 12 at 15:03