Thread: If a Diagonal is drawn on each side of a Cube, how many different patterns possible?

1. If a Diagonal is drawn on each side of a Cube, how many different patterns possible?

If a diagonal is drawn across each side of a cube from one corner to another, then how many distinct patterns are possible? For a distinct pattern consider all possible orientations of the diagonals and all six sides of the cube in each pattern.

For this problem. I'm wondering if I should use Burnside's lemma or flatten the cube like this (below) and treat it more like a spatial problem.

2. Perhaps this is a more precise way to ask the same question:

How many distinct patterns are possible if each side of a cube has a diagonal drawn across it and if one counts as one pattern any patterns that can be made to coincide by various rotations of the cube as one rigid object?

2 different diagonals possible on each of the 6 faces;
let 1 and 2 represent them; then:
01 : 111111
02 : 111112
03 : 111121
...
62 : 222212
63 : 222221
64 : 222222

2^6 = 64

That's how I "see" this...
Subhotosh will probably send me to the corner...

4. Originally Posted by Denis
2 different diagonals possible on each of the 6 faces;
let 1 and 2 represent them; then:
01 : 111111
02 : 111112
03 : 111121
...
62 : 222212
63 : 222221
64 : 222222

2^6 = 64

That's how I "see" this...
Subhotosh will probably send me to the corner...
I think that's very clever-way to decipher that problem.

Only difficulty I think is that probably 111121 is same as 121111 - like a ring problem.

5. Originally Posted by bennyJ
Perhaps this is a more precise way to ask the same question:

How many distinct patterns are possible if each side of a cube has a diagonal drawn across it and if one counts as one pattern any patterns that can be made to coincide by various rotations of the cube as one rigid object?
Hi bennyJ,

This is the typical problem that can be solved by Burnside's lemma, but, in this case, there is a little twist.

Note first that the group of rotations has order 24 (you can place each of the 6 faces on the table and turn it in 4 different ways). In fact, this group is isomorphic to $S_4$, but we will not need that.

The number of configurations is 64, since there are two diagonals on each face.

Let us look first at the simpler problem of coloring all the faces of the cube with two colors. (You may probably already know this, since you know about Burnside's lemma, but I cannot be certain of how much you know; if you already know this, please feel free to skip to the end).

Burnside's lemma tells us that the number of distinct configurations is the average number of configurations fixed by a rotation. The idea is to count the number of configurations fixed by each rotation, and to divide the total by 24 (the order of the group).

For a given rotation, we start by writing it as a permutation of the faces in disjoint cycles form, but we include cycles of length 1. A configuration will be fixed if all the faces in each cycle have the same color. If we have k cycles, we can choose one of the two colors for each cycle, and this gives $2^k$ fixed configurations.

It is more efficient to group the permutations by conjugacy classes, which have the same cycle pattern. For each class, you can compute the number of fixed configurations, and multiply that by the number of rotations in the class.

Assume that the faces are numbered as in a typical Western die, and let us enumerate the rotation group.

1. The identity has cycle structure (1)(2)(3)(4)(5)(6). We have 6 cycles, giving $2^6=64$ configurations; there is only one rotation in this class.
2. Rotations by 120° around a diagonal. The cycle structure is (1,2,3)(4,6,5). We have k = 2, giving 4 configurations. There are 8 such rotations in the group (4 diagonals and two directions of rotation); this gives a total of 32 fixed configurations.
3. Rotations by 180° around the line joining the midpoints of two opposite edges. The cycle structure is (1,3)(4,6)(2,5); we have k = 3, and there are 6 such rotations (6 pairs of opposite edges); this gives a total of $6\times2^3=48$ fixed configurations.
4. Rotations by 90° around the line joining the centers of two opposite faces. The cycle structure is (2,3,5,4)(1)(6); we have k = 3, and there are 6 such rotations (three axes and two directions), giving a total of $6\times2^3=48$ fixed configurations.
5. Rotations by 180° around the same axis as in type (4). The cycle structure is (2,5)(3,4)(1)(6); we have k = 3, and there are 3 such rotations; this gives a total of $3\times2^4=48$ fixed configurations.

(It is always a good idea to check that the total number of rotations is 24). The total number of fixed configurations is
64 + 32 + 48 + 48 + 48 = 240
this means that there are 240/24 = 10 distinct colorings.

Let us now turn back to the problem with the diagonals. With the colors, a face cannot change color during a rotation, but this need not be the case with the diagonals.

The identity fixes all faces, and it also fixes the whole cube, including the vertices and therefore the diagonals; we have no problem in this case. We can have problems with non-identity rotations that fix some faces, i.e., rotations that contain cycles of length 1. This happens in types (4) and (5), and deserves further thinking.

In a rotation of type (4) (by 90°), each of the two fixed faces is rotated by 90°, and the diagonals of these faces end up in the other position. The conclusion is that these rotations have no fixed configurations.

In a rotation of type (5), the diagonals of the fixed faces are rotated by 180° and end up in the same position; we have no problem with these rotations.

To summarize, we only have to remove the 48 fixed configurations accounted for in type (4). This leaves us with 192 fixed configurations, and 192/24 = 8 distinct patterns.

6. Thanks Barrick. From now on, I'll be nervous
when I drop a sugar cube in my coffee...

7. Originally Posted by barrick
Hi bennyJ,

This is the typical problem that can be solved by Burnside's lemma, but, in this case, there is a little twist.

Note first that the group of rotations has order 24 (you can place each of the 6 faces on the table and turn it in 4 different ways). In fact, this group is isomorphic to $S_4$, but we will not need that.

The number of configurations is 64, since there are two diagonals on each face.

Let us look first at the simpler problem of coloring all the faces of the cube with two colors. (You may probably already know this, since you know about Burnside's lemma, but I cannot be certain of how much you know; if you already know this, please feel free to skip to the end).

Burnside's lemma tells us that the number of distinct configurations is the average number of configurations fixed by a rotation. The idea is to count the number of configurations fixed by each rotation, and to divide the total by 24 (the order of the group).

For a given rotation, we start by writing it as a permutation of the faces in disjoint cycles form, but we include cycles of length 1. A configuration will be fixed if all the faces in each cycle have the same color. If we have k cycles, we can choose one of the two colors for each cycle, and this gives $2^k$ fixed configurations.

It is more efficient to group the permutations by conjugacy classes, which have the same cycle pattern. For each class, you can compute the number of fixed configurations, and multiply that by the number of rotations in the class.

Assume that the faces are numbered as in a typical Western die, and let us enumerate the rotation group.

1. The identity has cycle structure (1)(2)(3)(4)(5)(6). We have 6 cycles, giving $2^6=64$ configurations; there is only one rotation in this class.
2. Rotations by 120° around a diagonal. The cycle structure is (1,2,3)(4,6,5). We have k = 2, giving 4 configurations. There are 8 such rotations in the group (4 diagonals and two directions of rotation); this gives a total of 32 fixed configurations.
3. Rotations by 180° around the line joining the midpoints of two opposite edges. The cycle structure is (1,3)(4,6)(2,5); we have k = 3, and there are 6 such rotations (6 pairs of opposite edges); this gives a total of $6\times2^3=48$ fixed configurations.
4. Rotations by 90° around the line joining the centers of two opposite faces. The cycle structure is (2,3,5,4)(1)(6); we have k = 3, and there are 6 such rotations (three axes and two directions), giving a total of $6\times2^3=48$ fixed configurations.
5. Rotations by 180° around the same axis as in type (4). The cycle structure is (2,5)(3,4)(1)(6); we have k = 3, and there are 3 such rotations; this gives a total of $3\times2^4=48$ fixed configurations.

(It is always a good idea to check that the total number of rotations is 24). The total number of fixed configurations is
64 + 32 + 48 + 48 + 48 = 240
this means that there are 240/24 = 10 distinct colorings.

Let us now turn back to the problem with the diagonals. With the colors, a face cannot change color during a rotation, but this need not be the case with the diagonals.

The identity fixes all faces, and it also fixes the whole cube, including the vertices and therefore the diagonals; we have no problem in this case. We can have problems with non-identity rotations that fix some faces, i.e., rotations that contain cycles of length 1. This happens in types (4) and (5), and deserves further thinking.

In a rotation of type (4) (by 90°), each of the two fixed faces is rotated by 90°, and the diagonals of these faces end up in the other position. The conclusion is that these rotations have no fixed configurations.

In a rotation of type (5), the diagonals of the fixed faces are rotated by 180° and end up in the same position; we have no problem with these rotations.

To summarize, we only have to remove the 48 fixed configurations accounted for in type (4). This leaves us with 192 fixed configurations, and 192/24 = 8 distinct patterns.
I appreciate your thoroughly going through that proof. Yes, I knew that with Burnside's lemma that 10 distinct patterns are obtained from 2 colors (and 57 from 3 colors), so I figured the number here could be fewer than ten (considering the possibility of one side essentially being able to do double duty with a mere twist of the wrist) but I didn't know how to arrive at it precisely. Thanks, again.

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