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# Painting a brick or square prism Watch

1. Question

I have a square prism or brick with identical rectangular faces and identical squares.

How many different ways can you paint the faces of the above, if two coloured prisms are regarded as the same, when one can be rotated to give the other?

The bricks are to be painted so that each face is blue, red or white.

My work so far

I've worked out the direct symmetries which are:

The identity e
the rotation through pi about axis 1
the rotation through pi/2 about axis 1
the rotation through 3pi/2 about axis 1
the rotation about pi about axes 2 and 4 (axes through midpoints of opposite edges)
the rotation about pi about axes 3 and 5 (axes through midpoints of opposite faces).

The identity element fixes the shape so Fix(e) = 3^6

This is where I have been stuck for the last few hours as I keep getting confused with my symmetries on what to write next.

Could someone give me a mental nudge please. I'm trying to use the counting theorem to solve this but not got any 3dimensional examples

Please let me know if that doesn't make any sense
Attached Images

2. Firstly - what does the attached image have to do with anything?

Secondly, the way to start a question is to think what your general strategy is and then work out the details to enact that. I think this is why you are stuck, you don't really have a plan to answer the question, you are just trying to write down some stuff that you know.

What you have done is dived into writing detail before you have considered its relevance to the actual problem.

-----

One way to answer this question is to do the following:

- Fix an orientation of the box and count all of the possible ways to colour the faces (easy)

- Now you have a number of colourings that is too big; you want to reduce this to find only the number of equivalent ones.

- What does equivalent mean in this context? Two colourings are equivalent if you can apply a symmetry that permutes the faces to get from one to the other.

- What is this mathematically? One says that the group of rotations acts on the set C of all colourings in a given orientation. The equivalence relation then says that two colourings are equivalent if they are in the same orbit under this action.

- Therefore, the mathematical problem is now to determine the group of rotations and its action on C and to count the number of orbits.

- Now, use Burnside's lemma which tells you that the number of orbits is the same as the sum of the number of faces fixed by each rotation divided by the total number of rotational symmetries.

- Hint: with the last step, if you are struggling to count the fixed points of each rotation, label the faces (rather than the axes) by the numbers 1-6 and write the rotations as permutations of those numbers and consider the cycle structure. Or to be fair, just picture the actual rotations and envisage where each face goes to.

The point here is that you need to order your thought process such that you work out bits of information for a purpose rather than just writing down something you know that is related.
3. First to apply Burnside's theorem you need to find the order of the group of symmetries; if I've counted correctly there are two more to find.

Then find the number of faces fixed by each symmetry: I'm not sure where 3^6 comes from, since the question relates to the faces -- so .

Then put all the numbers into Burnside to get the number of orbits; and if you've seen a two dimensional example then you should be ok from here.
4. (Original post by jb444)
Then find the number of faces fixed by each symmetry: I'm not sure where 3^6 comes from, since the question relates to the faces -- so .
You don't want to count orbits of the group action on the cuboid; you want to count orbits of the group acting on the set of configurations.

If you can colour 6 faces in 3 possible ways, then there are 3^6 total colourings.
5. (Original post by Mark85)
You don't want to count orbits of the group action on the cuboid; you want to count orbits of the group acting on the set of configurations.

If you can colour 6 faces in 3 possible ways, then there are 3^6 total colourings.
true
6. (Original post by Mark85)
Firstly - what does the attached image have to do with anything?

Secondly, the way to start a question is to think what your general strategy is and then work out the details to enact that. I think this is why you are stuck, you don't really have a plan to answer the question, you are just trying to write down some stuff that you know.

What you have done is dived into writing detail before you have considered its relevance to the actual problem.

-----

One way to answer this question is to do the following:

- Fix an orientation of the box and count all of the possible ways to colour the faces (easy)

- Now you have a number of colourings that is too big; you want to reduce this to find only the number of equivalent ones.

- What does equivalent mean in this context? Two colourings are equivalent if you can apply a symmetry that permutes the faces to get from one to the other.

- What is this mathematically? One says that the group of rotations acts on the set C of all colourings in a given orientation. The equivalence relation then says that two colourings are equivalent if they are in the same orbit under this action.

- Therefore, the mathematical problem is now to determine the group of rotations and its action on C and to count the number of orbits.

- Now, use Burnside's lemma which tells you that the number of orbits is the same as the sum of the number of faces fixed by each rotation divided by the total number of rotational symmetries.

- Hint: with the last step, if you are struggling to count the fixed points of each rotation, label the faces (rather than the axes) by the numbers 1-6 and write the rotations as permutations of those numbers and consider the cycle structure. Or to be fair, just picture the actual rotations and envisage where each face goes to.

The point here is that you need to order your thought process such that you work out bits of information for a purpose rather than just writing down something you know that is related.
Thanks again Mark. I had a break and came back to this and that helped. As I had a chance to think about it and come up with a strategy like you said. At the time I think I was bombarded with too much information and I couldn't digest what my book was telling me. I think I need to only study for a certain amount of time and have breaks as I find I get confused with other things

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Updated: April 9, 2013
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