Proposition 3
Not all elements of S6 * ℤ46 * S12 are attainable configurations of the octahedron puzzle. To be an element of the group O, X = (α, a, β) must satisfy the following conditions:
3.1 The summation of the terms in the orientation vector for the vertices must be 0 mod 4.
3.1 The summation of the terms in the orientation vector for the vertices must be 0 mod 4.
3.2 The permutations for the edges and vertices must both be even.
sgn(α) = sgn(β) where sgn(α) = {+1 if α is even; -1 if α is odd)
Note: Rather than just needing the permutations to be equal like the Rubik’s cube, the octahedron puzzle requires both permutations to be even. In fact, the permutations for the octahedron puzzle are always even. This is because all of the generators of the group have even permutations and combining them will always result in another even permutation.
The proof of proposition 2 is very similar to the proof of proposition 1. In fact, the steps taken are the same but with slightly different conditions to prove. First, it must be shown that the generators of the group satisfy the properties above. Next, that these properties are preserved by the group operation. In other words, given any two elements of the group, X = (α, a, β) and Y = (δ, d, ε) that satisfy these conditions, the combination of these moves X ∙ Y = (αδ, a + αd, βε) will also satisfy these conditions and be an element of the group.
Note: Rather than just needing the permutations to be equal like the Rubik’s cube, the octahedron puzzle requires both permutations to be even. In fact, the permutations for the octahedron puzzle are always even. This is because all of the generators of the group have even permutations and combining them will always result in another even permutation.
The proof of proposition 2 is very similar to the proof of proposition 1. In fact, the steps taken are the same but with slightly different conditions to prove. First, it must be shown that the generators of the group satisfy the properties above. Next, that these properties are preserved by the group operation. In other words, given any two elements of the group, X = (α, a, β) and Y = (δ, d, ε) that satisfy these conditions, the combination of these moves X ∙ Y = (αδ, a + αd, βε) will also satisfy these conditions and be an element of the group.
Proof
The generators satisfy these conditions:
The properties are preserved under the group operation:
3.1 The summation of the terms in the orientation vector a + αd must be 0 mod 4.
3.1 The summation of the terms in the orientation vector a + αd must be 0 mod 4.
Since summation rules allow separation of addition, the above equation is equivalent to
Rearranging the elements of d does not change their sum so the alpha inverse may be dropped.
Each of which is 0 mod 4 by hypothesis. The result is that the original summation is congruent to 0 mod 4.
3.2 The sign of the permutation for the edges, αδ, and the vertices, βε, are both equal to 1.
sgn(αδ) = sgn(βε)
sgn(α)sgn(δ) = sgn(β)sgn(ε)
Each of which is equal to 1 by hypothesis. (See proposition 1.3 for full proof)
Note: An additional condition that the Rubik’s cube group must satisfy is that the orientation vector for the edges must be 0 mod 2.
sgn(α)sgn(δ) = sgn(β)sgn(ε)
Each of which is equal to 1 by hypothesis. (See proposition 1.3 for full proof)
Note: An additional condition that the Rubik’s cube group must satisfy is that the orientation vector for the edges must be 0 mod 2.
This condition is also met by every element of the octahedron puzzle group but is trivial because the orientation for edge pieces will always be zero. This shows that the octahedron puzzle still follows a similar pattern of conditions as the Rubik’s cube.
Proof: The summation of the terms in the orientation vector b + βe must be 0 mod 2.
Proof: The summation of the terms in the orientation vector b + βe must be 0 mod 2.
Since summation rules allow separation of addition, the above equation is equivalent to
Rearranging the elements of e does not change their sum so the beta inverse may be dropped.
Each of which is 0 by hypothesis. The result is that the original summation is congruent to 0 mod 2.