Proposition 4
All the configurations X = (α, a, β) of the octahedron puzzle that satisfy the conditions stated in Proposition 3 can be attained. (3 cases)
4.1 All edge permutations β can be attained
4.2 All vertex permutations α can be attained
4.3 All vertex orientations a
Before this proposition is proven, recall the two moves that were used to prove Proposition 2 for the Rubik’s cube. These moves will also be helpful in the proof of Proposition 4.
A commutator, C, is the combination of two moves, X and Y, in the following way
C = [X,Y] = Y-1∙X-1∙Y∙X
The conjugation of X by Y is Y-1∙X∙Y
Note that in the following proof, for simplicity UF, UR, UL, UB, DF, DR, DL, DB refer to rotating the indicated face 120 degrees clockwise and UF-1, UR-1, UL-1, UB-1, DF-1, DR-1, DL-1, DB-1 refer to rotating the indicated face 120 degrees counterclockwise.
4.1 All edge permutations β can be attained
4.2 All vertex permutations α can be attained
4.3 All vertex orientations a
Before this proposition is proven, recall the two moves that were used to prove Proposition 2 for the Rubik’s cube. These moves will also be helpful in the proof of Proposition 4.
A commutator, C, is the combination of two moves, X and Y, in the following way
The conjugation of X by Y is Y-1∙X∙Y
Note that in the following proof, for simplicity UF, UR, UL, UB, DF, DR, DL, DB refer to rotating the indicated face 120 degrees clockwise and UF-1, UR-1, UL-1, UB-1, DF-1, DR-1, DL-1, DB-1 refer to rotating the indicated face 120 degrees counterclockwise.
Proof
4.1 All edge permutations β can be attained.
Before this can be proven, we will show that the commutator and conjugation allow us to obtain all 3-cycles on edges. From this result, it can be shown that all edge permutations can be achieved.
C = [UF, UR-1] = UR∙UF-1∙UR-1∙UF
gives a result of two 2-cycles on vertices and a 3-cycle on edges, α = (2 4) (1 3), β = (3 1 7).
Combining C with itself (C2) has the result β = (1 3 7) with no effect on vertices. Now that one 3-cycle on edges has been attained with no effect on vertices, it can be shown that all 3-cycles on edges can also be attained. There are 4 cases for this:
a. The three edges we want to cycle are in place. If we want to cycle edges (1 3 7), no conjugation is necessary; use C2. We can also achieve (7 3 1), or (1 3 7)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 3 7).
b. Two of the three edges we want to cycle are in place. We want to cycle edges (z 3 7), (1 z 7) or (1 3 z) where z = {any edge ≠ 1, 3, 7}. To attain 3-cycle (1 3 z), it can be shown that a sequence of moves T can place edge z into position 7 without affecting the final positions of edges 1 and 3. Similarly, to attain (z 3 7) and (1 z 7) it can be shown that a sequence of moves T can place edge z into position 1 and 3, respectively. The commutator will then cycle the three edges and the inverse of the sequence of moves, T-1, will return edge z back to its original position. We will concentrate on attaining one of these cycles first, (1 3 z). The following are the sequences of moves T which can be applied to place each edge z into position 7 without changing the final position of edges 1 and 3.
For z = 2: T = UF∙UR∙UF-1
For z = 4: T = UB∙DB-1∙DL-1
For z = 5: T = UB-1
For z = 6: T = UB
For z = 8: T = UF∙UR-1∙UF-1
For z = 9: T = UB∙DB∙DR-1
For z = 10: T = UB∙DB-1∙DL
For z = 11: T = UB∙DB-1
For z = 12: T = UB∙DB
The commutator squared, C2, conjugated by each T, (T-1∙ C2 ∙ T), will result in every edge permutation of the form (1 3 z) with no other effect to the puzzle. With the use of these known permutations of the form (1 3 z) and conjugation, (z 3 7) and (1 z 7) can also be achieved.
Conjugating (1 3 z) by (1 3 7) results in
(7 3 1) (1 3 z) (1 3 7) = (1 z 7)
Conjugating (1 3 z) by (7 3 1) results in
(1 3 7) (1 3 z) (7 3 1) = (z 3 7)
We can also achieve the edge permutations (z 3 7)-1, (1 z 7)-1, (1 3 z)-1 by simply inverting the respective sequence of moves used to get (z 3 7), (1 z 7) and (1 3 z).
c. One of the three edges we want to cycle is in place. We want to cycle edges (z y 7), (1 z y) or (y 3 z) where y, z = {any edge ≠ 1, 3, 7}. In part b. it was shown that every edge permutation of the forms (1 3 z), (1 z 7) and (z 3 7) with no other effect to the puzzle can be achieved. With the use of these known permutations and conjugation, (z y 7), (1 z y) and (y 3 z) can also be achieved.
Conjugating (1 3 z) by (1 3 y) results in
(y 3 1) (1 3 z) (1 3 y) = (1 z y)
Conjugating (1 z y) by (1 3 7) results in
(7 3 1) (1 z y) (1 3 7) = (z y 7)
Conjugating (1 z y) by (7 3 1) results in
(1 3 7) (1 z y) (7 3 1) = (y 3 z)
We can also achieve the edge permutations (1 z y)-1, (z y 7)-1, (y 3 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 7) and (y 3 z).
d. None of the three edges we want to cycle are in place. We want to cycle edges (z y x) where x, y, z = {any edge ≠ 1, 3, 7}. In part c. it was shown that the edge permutation (1 z y) with no other effect on the puzzle can be achieved. With the use of this known permutation and conjugation, (z y x) can also be achieved.
Conjugating (1 z y) by (1 3 x) results in
(x 3 1) (1 z y) (1 3 x) = (z y x)
We can also achieve the edge permutation (z y x)-1 by simply inverting the sequence of moves used to get (z y x).
Next it can be shown that all edge permutations β can be attained. As stated earlier, all edge permutations of the octahedron puzzle are even. Any set of even permutations can be rewritten in 3-cycles by grouping the transpositions in the following way. If the two transpositions have both elements in common, (a b) (a b) = 0. If the two transpositions have one element in common, (a c) (a b) = (a b c). If the two transpositions have no elements in common, (c d) (a b) = (a c b) (a c d). Since an even permutation can be written as an even product of transpositions, it can also be expressed as a product of 3-cycles. This means that all permutations on edges can be attained since it is known that all 3-cycles (z y x) on edges can be attained.
Before this can be proven, we will show that the commutator and conjugation allow us to obtain all 3-cycles on edges. From this result, it can be shown that all edge permutations can be achieved.
gives a result of two 2-cycles on vertices and a 3-cycle on edges, α = (2 4) (1 3), β = (3 1 7).
Combining C with itself (C2) has the result β = (1 3 7) with no effect on vertices. Now that one 3-cycle on edges has been attained with no effect on vertices, it can be shown that all 3-cycles on edges can also be attained. There are 4 cases for this:
a. The three edges we want to cycle are in place. If we want to cycle edges (1 3 7), no conjugation is necessary; use C2. We can also achieve (7 3 1), or (1 3 7)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 3 7).
b. Two of the three edges we want to cycle are in place. We want to cycle edges (z 3 7), (1 z 7) or (1 3 z) where z = {any edge ≠ 1, 3, 7}. To attain 3-cycle (1 3 z), it can be shown that a sequence of moves T can place edge z into position 7 without affecting the final positions of edges 1 and 3. Similarly, to attain (z 3 7) and (1 z 7) it can be shown that a sequence of moves T can place edge z into position 1 and 3, respectively. The commutator will then cycle the three edges and the inverse of the sequence of moves, T-1, will return edge z back to its original position. We will concentrate on attaining one of these cycles first, (1 3 z). The following are the sequences of moves T which can be applied to place each edge z into position 7 without changing the final position of edges 1 and 3.
For z = 2: T = UF∙UR∙UF-1
For z = 4: T = UB∙DB-1∙DL-1
For z = 5: T = UB-1
For z = 6: T = UB
For z = 8: T = UF∙UR-1∙UF-1
For z = 9: T = UB∙DB∙DR-1
For z = 10: T = UB∙DB-1∙DL
For z = 11: T = UB∙DB-1
For z = 12: T = UB∙DB
The commutator squared, C2, conjugated by each T, (T-1∙ C2 ∙ T), will result in every edge permutation of the form (1 3 z) with no other effect to the puzzle. With the use of these known permutations of the form (1 3 z) and conjugation, (z 3 7) and (1 z 7) can also be achieved.
Conjugating (1 3 z) by (1 3 7) results in
(7 3 1) (1 3 z) (1 3 7) = (1 z 7)
Conjugating (1 3 z) by (7 3 1) results in
(1 3 7) (1 3 z) (7 3 1) = (z 3 7)
We can also achieve the edge permutations (z 3 7)-1, (1 z 7)-1, (1 3 z)-1 by simply inverting the respective sequence of moves used to get (z 3 7), (1 z 7) and (1 3 z).
c. One of the three edges we want to cycle is in place. We want to cycle edges (z y 7), (1 z y) or (y 3 z) where y, z = {any edge ≠ 1, 3, 7}. In part b. it was shown that every edge permutation of the forms (1 3 z), (1 z 7) and (z 3 7) with no other effect to the puzzle can be achieved. With the use of these known permutations and conjugation, (z y 7), (1 z y) and (y 3 z) can also be achieved.
Conjugating (1 3 z) by (1 3 y) results in
(y 3 1) (1 3 z) (1 3 y) = (1 z y)
Conjugating (1 z y) by (1 3 7) results in
(7 3 1) (1 z y) (1 3 7) = (z y 7)
Conjugating (1 z y) by (7 3 1) results in
(1 3 7) (1 z y) (7 3 1) = (y 3 z)
We can also achieve the edge permutations (1 z y)-1, (z y 7)-1, (y 3 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 7) and (y 3 z).
d. None of the three edges we want to cycle are in place. We want to cycle edges (z y x) where x, y, z = {any edge ≠ 1, 3, 7}. In part c. it was shown that the edge permutation (1 z y) with no other effect on the puzzle can be achieved. With the use of this known permutation and conjugation, (z y x) can also be achieved.
Conjugating (1 z y) by (1 3 x) results in
(x 3 1) (1 z y) (1 3 x) = (z y x)
We can also achieve the edge permutation (z y x)-1 by simply inverting the sequence of moves used to get (z y x).
Next it can be shown that all edge permutations β can be attained. As stated earlier, all edge permutations of the octahedron puzzle are even. Any set of even permutations can be rewritten in 3-cycles by grouping the transpositions in the following way. If the two transpositions have both elements in common, (a b) (a b) = 0. If the two transpositions have one element in common, (a c) (a b) = (a b c). If the two transpositions have no elements in common, (c d) (a b) = (a c b) (a c d). Since an even permutation can be written as an even product of transpositions, it can also be expressed as a product of 3-cycles. This means that all permutations on edges can be attained since it is known that all 3-cycles (z y x) on edges can be attained.
4.2 All vertex permutations α can be attained.
Before this can be proven, we will show that the commutator and conjugation allow us to obtain all 3-cycles on vertices. From this result, it can be shown that all vertex permutations can be achieved.
C = [UL, UR DRR-1] = (UR DR UR-1)-1 ∙ UL-1∙UR DR UR-1∙UL
gives a result of a 3-cycle on vertices and no effect on edges, α = (1 6 5), β = (1). Now that one 3-cycle on vertices has been attained with no effect on edges, it can be shown that all 3-cycles on vertices can also be attained. There are 4 cases for this:
a. The three vertices we want to cycle are in place. If we want to cycle vertices (1 6 5), no conjugation is necessary; use C. We can also achieve (5 6 1), or (1 6 5)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 6 5).
b. Two of the three vertices we want to cycle are in place. We want to cycle vertices (z 6 5), (1 z 5) or (1 6 z) where z = {any vertex ≠ 1, 5, 6}. To attain 3-cycle (1 6 z), it can be shown that a sequence of moves T can place vertex z into position 5 without affecting the final positions of vertices 1 and 6. Similarly, to attain (z 6 5) and (1 z 5) it can be shown that a sequence of moves T can place vertex z into position 1 and 6, respectively. The commutator will then cycle the three vertices and the inverse of the sequence of moves, T-1, will return vertex z back to its original position.
We will concentrate on attaining one of these cycles first, (1 6 z). The following are the sequences of moves T which can be applied to place each vertex z into position 5 without changing the final position of vertices 1 and 6.
For z = 2: T = UF-1∙UB-1∙UF
For z = 3: T = UF-1∙UL∙UF
For z = 4: T = UF-1∙UL-1∙UF
The commutator conjugated by each T, (T-1∙C∙ T), will result in every vertex permutation of the form (1 6 z) with no other effect to the puzzle. With the use of these known permutations of the form (1 6 z) and conjugation, (z 6 5) and (1 z 5) can also be achieved.
Conjugating (1 6 z) by (1 6 5) results in
(5 6 1) (1 6 z) (1 6 5) = (1 z 5)
Conjugating (1 6 z) by (5 6 1) results in
( 1 6 5) (1 6 z) (5 6 1) = (z 6 5)
We can also achieve the vertex permutations (z 6 5)-1, (1 z 5)-1, (1 6 z)-1 by simply inverting the respective sequence of moves used to get (z 6 5), (1 z 5) and (1 6 z).
c. One of the three vertices we want to cycle is in place. We want to cycle vertices (z y 5), (1 z y) or (y 6 z) where y, z = {any vertex ≠ 1, 5, 6}. In part b. it was shown that every vertex permutation of the forms (1 6 z), (1 z 5) and (z 6 5) with no other effect to the puzzle can be achieved. With the use of these known permutations and conjugation, (z y 5), (1 z y) and (y 6 z) can also be achieved.
Conjugating (16 z) by (1 6 y) results in
(y 6 1) (1 6 z) (1 6 y) = (1 z y)
Conjugating (1 z y) by (1 6 5) results in
(5 6 1) (1 z y) (1 6 5) = (z y 5)
Conjugating (1 z y) by (5 6 1) results in
(1 6 5) (1 z y) (5 6 1) = (y 6 z)
We can also achieve the vertex permutations (1 z y)-1, (z y 5)-1, (y 6 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 5) and (y 6 z).
d. None of the three vertices we want to cycle are in place. We want to cycle vertices (z y x) where x, y, z = {any vertex ≠ 1, 5, 6}. In part c. it was shown that the vertex permutation (1 z y) with no other effect on the puzzle can be achieved. With the use of this known permutation and conjugation, (z y x) can also be achieved.
Conjugating (1 z y) by (1 6 x) results in
(x 6 1) (1 z y) (1 6 x) = (z y x)
We can also achieve the vertex permutation (z y x)-1 by simply inverting the sequence of moves used to get (z y x).
Next it can be shown that all vertex permutations α can be attained. As stated earlier, all vertex permutations of the octahedron puzzle are even. Any set of even permutations can be rewritten in 3-cycles by grouping the transpositions using the same method that was described in 4.1. Since an even permutation can be written as an even product of transpositions, it can also be expressed as a product of 3-cycles. This means that all permutations on vertices can be attained since it is known that all 3-cycles (z y x) on vertices can be attained.
Before this can be proven, we will show that the commutator and conjugation allow us to obtain all 3-cycles on vertices. From this result, it can be shown that all vertex permutations can be achieved.
gives a result of a 3-cycle on vertices and no effect on edges, α = (1 6 5), β = (1). Now that one 3-cycle on vertices has been attained with no effect on edges, it can be shown that all 3-cycles on vertices can also be attained. There are 4 cases for this:
a. The three vertices we want to cycle are in place. If we want to cycle vertices (1 6 5), no conjugation is necessary; use C. We can also achieve (5 6 1), or (1 6 5)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 6 5).
b. Two of the three vertices we want to cycle are in place. We want to cycle vertices (z 6 5), (1 z 5) or (1 6 z) where z = {any vertex ≠ 1, 5, 6}. To attain 3-cycle (1 6 z), it can be shown that a sequence of moves T can place vertex z into position 5 without affecting the final positions of vertices 1 and 6. Similarly, to attain (z 6 5) and (1 z 5) it can be shown that a sequence of moves T can place vertex z into position 1 and 6, respectively. The commutator will then cycle the three vertices and the inverse of the sequence of moves, T-1, will return vertex z back to its original position.
We will concentrate on attaining one of these cycles first, (1 6 z). The following are the sequences of moves T which can be applied to place each vertex z into position 5 without changing the final position of vertices 1 and 6.
For z = 2: T = UF-1∙UB-1∙UF
For z = 3: T = UF-1∙UL∙UF
For z = 4: T = UF-1∙UL-1∙UF
The commutator conjugated by each T, (T-1∙C∙ T), will result in every vertex permutation of the form (1 6 z) with no other effect to the puzzle. With the use of these known permutations of the form (1 6 z) and conjugation, (z 6 5) and (1 z 5) can also be achieved.
Conjugating (1 6 z) by (1 6 5) results in
(5 6 1) (1 6 z) (1 6 5) = (1 z 5)
Conjugating (1 6 z) by (5 6 1) results in
( 1 6 5) (1 6 z) (5 6 1) = (z 6 5)
We can also achieve the vertex permutations (z 6 5)-1, (1 z 5)-1, (1 6 z)-1 by simply inverting the respective sequence of moves used to get (z 6 5), (1 z 5) and (1 6 z).
c. One of the three vertices we want to cycle is in place. We want to cycle vertices (z y 5), (1 z y) or (y 6 z) where y, z = {any vertex ≠ 1, 5, 6}. In part b. it was shown that every vertex permutation of the forms (1 6 z), (1 z 5) and (z 6 5) with no other effect to the puzzle can be achieved. With the use of these known permutations and conjugation, (z y 5), (1 z y) and (y 6 z) can also be achieved.
Conjugating (16 z) by (1 6 y) results in
(y 6 1) (1 6 z) (1 6 y) = (1 z y)
Conjugating (1 z y) by (1 6 5) results in
(5 6 1) (1 z y) (1 6 5) = (z y 5)
Conjugating (1 z y) by (5 6 1) results in
(1 6 5) (1 z y) (5 6 1) = (y 6 z)
We can also achieve the vertex permutations (1 z y)-1, (z y 5)-1, (y 6 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 5) and (y 6 z).
d. None of the three vertices we want to cycle are in place. We want to cycle vertices (z y x) where x, y, z = {any vertex ≠ 1, 5, 6}. In part c. it was shown that the vertex permutation (1 z y) with no other effect on the puzzle can be achieved. With the use of this known permutation and conjugation, (z y x) can also be achieved.
Conjugating (1 z y) by (1 6 x) results in
(x 6 1) (1 z y) (1 6 x) = (z y x)
We can also achieve the vertex permutation (z y x)-1 by simply inverting the sequence of moves used to get (z y x).
Next it can be shown that all vertex permutations α can be attained. As stated earlier, all vertex permutations of the octahedron puzzle are even. Any set of even permutations can be rewritten in 3-cycles by grouping the transpositions using the same method that was described in 4.1. Since an even permutation can be written as an even product of transpositions, it can also be expressed as a product of 3-cycles. This means that all permutations on vertices can be attained since it is known that all 3-cycles (z y x) on vertices can be attained.
4.3 All vertex orientations a can be attained. Before this can be shown, we will suppose that there exists a sequence of moves V = (α, a, β) (similar to the one used for the Rubik’s cube) that change only the orientations of two vertex pieces. From the resulting vector a, the remaining vectors that make up the entire space of vertex orientation vectors can also be found. With the use of the commutator and conjugation, V can be manipulated to attain the following set of elements from O
((1), (1,0,0,0,0,3), (1))
((1), (0,1,0,0,0,3), (1))
((1), (0,0,1,0,0,3), (1))
((1), (0,0,0,1,0,3), (1))
((1), (0,0,0,0,1,3), (1))
Combining these elements will give the entire set of vertex orientation vectors. To be sure that all vertex orientation vectors satisfy proposition 3.1 (that the sum of the vector components must be 0 mod 4), they must be of the form
((1), (1,0,0,0,0,3), (1))
((1), (0,1,0,0,0,3), (1))
((1), (0,0,1,0,0,3), (1))
((1), (0,0,0,1,0,3), (1))
((1), (0,0,0,0,1,3), (1))
Combining these elements will give the entire set of vertex orientation vectors. To be sure that all vertex orientation vectors satisfy proposition 3.1 (that the sum of the vector components must be 0 mod 4), they must be of the form
All vectors of this form can be attained by combining the orientation vectors above like so:
a1(1,0,0,0,0,3) + a2(0,1,0,0,0,3) + a3(0,0,1,0,0,3) + a4(0,0,0,1,0,3) + a5(0,0,0,0,1,3)
a1(1,0,0,0,0,3) + a2(0,1,0,0,0,3) + a3(0,0,1,0,0,3) + a4(0,0,0,1,0,3) + a5(0,0,0,0,1,3)