4
$\begingroup$

How many ways can you colour the triangles of a snub cube with black and white?

a snub cube has 32 triangles so i'd assume $2^{32}$ but thats too simple, based on the motions of symmetry of a snub cube, which i know permute the diagonals of a cube.

How many ways are there of coloring the triangular faces of the snub cube with black and white using 16 black faces and 16 white faces.

Given that there are 32 triangles, some of the rotations will give a similar colouring. But i dont know how to go about it.

$\endgroup$
  • $\begingroup$ Are the squares coloured identically? $\endgroup$ – theindigamer Dec 1 '17 at 3:12
  • $\begingroup$ im assuming so. $\endgroup$ – Mahlissa LECKY Dec 4 '17 at 20:24
3
$\begingroup$

You will need to use Pólya enumeration, also known as Burnside's lemma. To do this, we need to find the cycle index of the group of rotations of the snub cube. Note that this group is the same as the octahedral group, with order $24$.

  • The identity, which maps all $6(4)+8 = 32$ triangles to themselves: $a_1^{32}$.
  • Six rotations by $\pi/2$ about an axis through the centers of opposite square faces. Because we are looking at how such a rotation affects the triangles, clearly there is no triangle that remains fixed (unlike the square faces, in which two map to themselves). Every triangular face has order $4$ (exactly four such rotations will return each triangle to itself), so the contribution to the cycle index is $6a_4^{8}$; that is to say, there are $8$ four-cycles.
  • Three rotations by $\pi$ about an axis through the centers of opposite square faces. These create $16$ two-cycles, so the contribution is $3a_2^{16}$.
  • Eight rotations by $2\pi/3$ about an axis through the centers of opposite non-snub triangular faces (i.e. through triangles not edge-incident with a square face). Such a rotation will fix the two triangles on the axis of rotation, and all of the others will belong to three-cycles; thus the contribution is $8a_1^2 a_3^{10}$.
  • Six rotations by $\pi$ through the opposing midpoints of the edge joining two snub triangles. Here, there are no fixed triangles, and each triangle belongs to one of $16$ two-cycles, and the contribution to the cycle index is $6a_2^{16}$.

The total cycle index is therefore $$|Z(a_1, a_2, a_3, a_4)| = \frac{1}{24} \left(a_1^{32} + 6a_4^8 + 3a_2^{16} + 8a_1^2 a_3^{10} + 6a_2^{16}\right).$$

Therefore, with $m$ colors, there are $$|Z(m,m,m,m)| = \frac{1}{24}(m^{32} + 9m^{16} + 8m^{12} + 6m^8)$$ colorings. For $m = 2$, this gives $2^6 \cdot 3 \cdot 932203 = 178982976$ colorings.

$\endgroup$
  • $\begingroup$ this is just for the triangular colourings right> $\endgroup$ – Mahlissa LECKY Dec 5 '17 at 4:43
  • $\begingroup$ for the 16 white and 16 black would the m's be altered but the formula the same ? $\endgroup$ – Mahlissa LECKY Dec 6 '17 at 14:00
  • $\begingroup$ @MahlissaLECKY Of course the formula does not apply if we are required to use an equal number of white and black triangles. You cannot choose an $m$ to make the formula apply in such a case. $\endgroup$ – heropup Dec 6 '17 at 16:32
  • $\begingroup$ so how would i go about finding that formula and result $\endgroup$ – Mahlissa LECKY Dec 29 '17 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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