Proposition 5
all elements of S12 * ℤ212 * S24 are attainable configurations of the cuboctahedron puzzle. To be an element of the group C, X = (α, a, β) must satisfy the following two conditions:
5.1 The summation of the terms in the orientation vector for the vertices must be 0 mod 2.
5.1 The summation of the terms in the orientation vector for the vertices must be 0 mod 2.
5.2 The permutations for the edges and vertices must either both be even or both be odd.
sgn(α) = sgn(β) where sgn(α) = {+1 if α is even; -1 if α is odd}
The proof of proposition 3 is again very similar to the proof of proposition 1 and proposition 2. First, it will 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 the conditions and be an element of the group.
The proof of proposition 3 is again very similar to the proof of proposition 1 and proposition 2. First, it will 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 the conditions and be an element of the group.
Proof
The generators satisfy these conditions:
The properties are preserved under the group operation:
5.1 The summation of the terms in the orientation vector a + αd must be 0 mod 2.
5.1 The summation of the terms in the orientation vector a + αd must be 0 mod 2.
Since summation rules allow separation of addition, the above equation is equivalent to
Rearranging the elements of d not change their sum so the alpha inverse may be dropped.
Each of which is 0 mod 2 by hypothesis. The result is that the original summation is congruent to 0 mod 2.
5.2 The sign of the permutations for the edges, αδ, and the vertices, βε, are equal.
sgn(αδ) = sgn(βε)
sgn(α)sgn(δ) = sgn(β)sgn(ε)
(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.
(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.
Like the octahedron puzzle, this condition is also met by every element of the cuboctahedron puzzle group but is trivial because the orientation for edge pieces will always be zero. The proof of this is similar to the one completed for the octahedron puzzle shown in proposition 3. This shows that the cuboctahedron puzzle follows a similar pattern of conditions as the Rubik’s cube and the octahedron puzzle.