The Octahedron
The octahedron puzzle is a platonic solid with 8 congruent triangular faces. It is composed of 26 pieces; 8 center pieces, 6 vertex pieces and 12 edge pieces.
Each of the vertex pieces has four sides which indicates that it has four different possible orientations (shown below) and each of the edge pieces has two sides which indicates that it has two different possible orientations. (Note that the shape of the vertex pieces in the diagram below have been simplified)
There are 6! ways to reposition all of the vertex pieces and another 4 ways to orient each of them resulting in a total of 6! * 46 ways to configure all of the vertex pieces. There are also 12! ways to reposition all of the edge pieces and another 2 ways to orient each of them resulting in another total of 12! * 212 ways to configure all of the edge pieces. Since the center pieces do not move and each have only one orientation there is only one way to position them. Combining these totals result in an upper bound of 6! * 46 * 12! * 212 different configurations of the octahedron puzzle. Based on this information, the group O = {S6 * ℤ46 * S12 * ℤ212} will be used to represent all of the possible configurations of the puzzle. Here, S6 and S12 are the permutation groups of the vertex and edge pieces while ℤ46 and ℤ212 are vectors specifying the orientation of the vertex and edge pieces.
Note: The puzzle referred to in this paper is one that can only be reconfigured by rotating its faces similar to the Rubik’s cube. To create this puzzle, the octahedron would have to be sliced with planes parallel to each face. Doing this would result in 24 extra pseudo-center pieces which could move and separate from the vertex pieces. A physical puzzle must include these pieces for the faces to turn. This would add a factor of S24 as well as additional conditions for elements of the group O. For simplicity, this paper considers an abstract puzzle with only vertex and edge pieces. To physically permute these pieces, they will have to be lifted off of the puzzle and placed back correctly.
As with the Rubik’s cube, for further work with the octahedron puzzle it is useful to name its faces. We will refer to the 8 faces of the octahedron as upper front UF, upper right UR, upper left UL, upper back UB, down front DF, down right DR, down left DL, and down back DB as shown below.
Note: The puzzle referred to in this paper is one that can only be reconfigured by rotating its faces similar to the Rubik’s cube. To create this puzzle, the octahedron would have to be sliced with planes parallel to each face. Doing this would result in 24 extra pseudo-center pieces which could move and separate from the vertex pieces. A physical puzzle must include these pieces for the faces to turn. This would add a factor of S24 as well as additional conditions for elements of the group O. For simplicity, this paper considers an abstract puzzle with only vertex and edge pieces. To physically permute these pieces, they will have to be lifted off of the puzzle and placed back correctly.
As with the Rubik’s cube, for further work with the octahedron puzzle it is useful to name its faces. We will refer to the 8 faces of the octahedron as upper front UF, upper right UR, upper left UL, upper back UB, down front DF, down right DR, down left DL, and down back DB as shown below.
The vertex and edge pieces will also be labeled as shown in the graph and pictures below.
Each vertex piece will have an orientation of 0, 1, 2 or 3 (ℤ4) and each edge piece will have an orientation of 0 or 1 (ℤ2). For purposes of this paper, the defined zero sides for the vertex and edge pieces are shaded in the graph below.
For edge pieces, the side that is not defined as the zero side will by default have an orientation of 1. For vertex pieces, the remaining sides will have an orientation of 1, 2 and 3 working clockwise from the defined zero side, as shown below.
As mentioned earlier, the group O formed by the octahedron puzzle is a subset of S6 * ℤ46 * S12 * ℤ212. Analogous to the Rubik’s cube, an element of this group will have the form X = (α, a, β, b) where α is the permutation of the vertex pieces, a is the orientation vector of the vertex pieces, β is the permutation of the edge pieces and b is the orientation vector of the edge pieces. The orientation vectors for the vertex pieces will have the form a = (a1, a2,… a6), where an refers to the orientation of the piece currently occupying position n. Similarly, for a vector b (b1, b2,… b12), bm refers to the orientation of the edge piece currently occupying position m.
Here is a simple example to demonstrate how to find a configuration’s corresponding element in the group O. Suppose the only three pieces that are out of place are the vertex pieces of the upper front face in the picture below; all the other pieces are correctly positioned and oriented.
Here is a simple example to demonstrate how to find a configuration’s corresponding element in the group O. Suppose the only three pieces that are out of place are the vertex pieces of the upper front face in the picture below; all the other pieces are correctly positioned and oriented.
The first step is to look at the permutation of the vertex pieces. Piece 1 is now in the position of piece 3, piece 3 is located in position 4 and piece 4 is located in position 1. Since all of the other vertex pieces are in the correct position the resulting vertex permutation is α =(1 3 4). For the orientation vector, we must recall which faces have an orientation of zero. In position 1, the zero side belongs on the upper front face but is now on the upper back face. This results in an orientation of 2 since the actual zero side of this piece is two faces away, clockwise, from the defined zero side. In position 3, the zero side belongs on the upper front face but instead is on the upper right face. The actual zero side of this piece is one face clockwise away from the defined zero side so its orientation is 1. In position 4, the zero side belongs on the upper front face but is on the down front face. Again, the actual zero side is just one face away from the defined zero side resulting in an orientation of 1. The remaining pieces all have an orientation of 0 since they are correctly oriented. The resulting vector is a = (2,0,1,1,0,0). If any edge pieces were out of place, the same process would be used to determine their correct permutation and orientation vector. The element in O for this example is
A = ((1 3 4), (2,0,1,1,0,0), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
The basic moves that the octahedron puzzle group is composed of are {UF120, UR120, UL120, UB120, DF120, DR120, DL120, DB120}. These stand for, respectively, rotating the upper front face, the upper right face, the upper left face, the upper back face, the down front face, the down right face, the down left face, and the down back face 120 degrees clockwise. These moves will be referred to as the generators of the group because any configuration of the octahedron puzzle can be attained from a combination of them. Their corresponding elements in the group are as follows.
A = ((1 3 4), (2,0,1,1,0,0), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
The basic moves that the octahedron puzzle group is composed of are {UF120, UR120, UL120, UB120, DF120, DR120, DL120, DB120}. These stand for, respectively, rotating the upper front face, the upper right face, the upper left face, the upper back face, the down front face, the down right face, the down left face, and the down back face 120 degrees clockwise. These moves will be referred to as the generators of the group because any configuration of the octahedron puzzle can be attained from a combination of them. Their corresponding elements in the group are as follows.
UF120 = ((1 3 4), (0,0,0,0,0,0), (1 2 3), (0,0,0,0,0,0,0,0,0,0,0,0))
UR120 = ((1 2 3), (2,0,2,0,0,0), (1 7 8), (0,0,0,0,0,0,0,0,0,0,0,0))
UL120 = ((1 4 5), (0,0,0,2,2,0),(3 4 5), (0,0,0,0,0,0,0,0,0,0,0,0))
UB120 = ((1 5 2), (2,0,0,0,2,0), (5 6 7), (0,0,0,0,0,0,0,0,0,0,0,0))
DF120 = ((3 6 4), (0,0,2,1,0,1), (2 9 10), (0,0,0,0,0,0,0,0,0,0,0,0))
DR120 = ((2 6 3), (0,0,1,0,0,3), (8 12 9), (0,0,0,0,0,0,0,0,0,0,0,0))
DL120 = ((4 6 5), (0,0,0,0,3,1), (4 10 11), (0,0,0,0,0,0,0,0,0,0,0,0))
DB120 = ((2 5 6), (0,1,0,0,2,1), (6 11 12), (0,0,0,0,0,0,0,0,0,0,0,0))
Note that every orientation vector for the edge pieces is all zeros. Since combining the generators produce the remaining elements of the group, every other element in the group O will also have an edge orientation vector of all zeros. This means that the edge pieces will never be incorrectly oriented no matter what the configuration of the puzzle is. As a result, the corresponding factor ℤ2 in the group O is insignificant and can be dropped. Elements in the group O will still have the same form as the elements from the Rubik’s cube group (X = (α, a, β, b)) but b = (0,0,0,0,0,0,0,0,0,0,0,0) for every configuration so b will therefore be disregarded.
Similar to the Rubik’s cube, the combination of any two configurations, X = (α, a, β) and Y = (δ, d, ε), gives the result X ∙ Y = (αδ, a + αd, βε). An example comparing the results from combining two moves by inspecting the diagram and by using the group operation is provided below. As with the Rubik’s cube, the two answers should agree. Here, the two moves that will be combined are first rotating the upper front face and then the upper right face 120 degrees clockwise.
UR120 = ((1 2 3), (2,0,2,0,0,0), (1 7 8), (0,0,0,0,0,0,0,0,0,0,0,0))
UL120 = ((1 4 5), (0,0,0,2,2,0),(3 4 5), (0,0,0,0,0,0,0,0,0,0,0,0))
UB120 = ((1 5 2), (2,0,0,0,2,0), (5 6 7), (0,0,0,0,0,0,0,0,0,0,0,0))
DF120 = ((3 6 4), (0,0,2,1,0,1), (2 9 10), (0,0,0,0,0,0,0,0,0,0,0,0))
DR120 = ((2 6 3), (0,0,1,0,0,3), (8 12 9), (0,0,0,0,0,0,0,0,0,0,0,0))
DL120 = ((4 6 5), (0,0,0,0,3,1), (4 10 11), (0,0,0,0,0,0,0,0,0,0,0,0))
DB120 = ((2 5 6), (0,1,0,0,2,1), (6 11 12), (0,0,0,0,0,0,0,0,0,0,0,0))
Note that every orientation vector for the edge pieces is all zeros. Since combining the generators produce the remaining elements of the group, every other element in the group O will also have an edge orientation vector of all zeros. This means that the edge pieces will never be incorrectly oriented no matter what the configuration of the puzzle is. As a result, the corresponding factor ℤ2 in the group O is insignificant and can be dropped. Elements in the group O will still have the same form as the elements from the Rubik’s cube group (X = (α, a, β, b)) but b = (0,0,0,0,0,0,0,0,0,0,0,0) for every configuration so b will therefore be disregarded.
Similar to the Rubik’s cube, the combination of any two configurations, X = (α, a, β) and Y = (δ, d, ε), gives the result X ∙ Y = (αδ, a + αd, βε). An example comparing the results from combining two moves by inspecting the diagram and by using the group operation is provided below. As with the Rubik’s cube, the two answers should agree. Here, the two moves that will be combined are first rotating the upper front face and then the upper right face 120 degrees clockwise.
The corresponding element for this configuration of the puzzle can be found using the same method used previously. First, the vertex permutation and orientation vector should be found by inspecting the graph; then the resulting edge permutation and orientation vector. The element in the group O that these moves produce is
UR120 ∙ UF120 = ((2 3 4), (2,0,2,0,0,0), (1 2 3 7 8))
Using the group operation to combine these moves is shown below.
UF120 = (δ, d, ε) = ((1 3 4), (0,0,0,0,0,0), (1 2 3))
UR120=(α, a, β) = ((1 2 3), (2,0,2,0,0,0), (1 7 8))
UR120 ∙ UF120 = (αδ, a + αd, βε)
= ((2 3 4), (2,0,2,0,0,0) + (1 2 3) ∙ (0,0,0,0,0,0), (1 2 3 7 8))
= ((2 3 4), (2,0,2,0,0,0), (1 2 3 7 8))
Again, the combination of these moves results in the same answer regardless of the method used.
UR120 ∙ UF120 = ((2 3 4), (2,0,2,0,0,0), (1 2 3 7 8))
Using the group operation to combine these moves is shown below.
UF120 = (δ, d, ε) = ((1 3 4), (0,0,0,0,0,0), (1 2 3))
UR120=(α, a, β) = ((1 2 3), (2,0,2,0,0,0), (1 7 8))
UR120 ∙ UF120 = (αδ, a + αd, βε)
= ((2 3 4), (2,0,2,0,0,0) + (1 2 3) ∙ (0,0,0,0,0,0), (1 2 3 7 8))
= ((2 3 4), (2,0,2,0,0,0), (1 2 3 7 8))
Again, the combination of these moves results in the same answer regardless of the method used.