Proposition 2
All the configurations X = (α, a, β, b) of the Rubik’s cube that satisfy the conditions stated in Proposition 1 can be attained. (4 cases)
2.1 All edge permutations β can be attained
2.2 All vertex permutations α can be attained
2.3 All edge orientations b can be attained
2.4 All vertex orientations a can be attained
First, we will introduce two important moves that will be helpful in the proof of this proposition below.
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 F, R, L, U, D, B refer to rotating the indicated face 90 degrees clockwise and F-1, R-1, L-1, U-1, D-1, B-1 refer to rotating the indicated face 90 degrees counterclockwise.
2.1 All edge permutations β can be attained
2.2 All vertex permutations α can be attained
2.3 All edge orientations b can be attained
2.4 All vertex orientations a can be attained
First, we will introduce two important moves that will be helpful in the proof of this proposition below.
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 F, R, L, U, D, B refer to rotating the indicated face 90 degrees clockwise and F-1, R-1, L-1, U-1, D-1, B-1 refer to rotating the indicated face 90 degrees counterclockwise.
Proof
2.1 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 = [F,R-1] = R∙F-1∙R-1∙F
gives a result of two 2-cycles on vertices and a 3-cycle on edges, α = (1 2) (4 5), β = (4 1 5).
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 = (1 2) (4 5), β = (4 1 5).
Combining C with itself (C2) has the result β = (1 4 5) 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 4 5), no conjugation is necessary; use C2. We can also achieve (5 4 1), or (1 4 5)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 4 5).
b. Two of the three edges we want to cycle are in place. We want to cycle edges (z 4 5), (1 z 5) or (1 4 z) where z = {any edge ≠ 1, 4, 5}. To attain 3-cycle (1 4 z), it can be shown that a sequence of moves T can place edge z into position 5 without affecting the final positions of edges 1 and 4. Similarly, to attain (z 4 5) and (1 z 5) it can be shown that a sequence of moves T can place edge z into position 1 and 4, 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 4 z). The following are the sequences of moves T which can be applied to place each edge z into position 5 without changing the final position of edges 1 and 4.
For z = 2: T = F∙R^2∙D∙F-1
For z = 3: T = F-1∙U^2∙L^-1∙F
For z = 6: T = F∙R-1∙F-1
For z = 7: T = F∙R2∙F-1
For z = 8: T = F-1∙U2∙F
For z = 9: T = F-1∙U2∙L∙F
For z = 10: T = F∙R2∙D2∙F-1
For z = 11: T = F-1∙U∙F
For z = 12: T = F∙R2∙D-1∙F-1
The commutator squared, C2, conjugated by each T, (T-1∙C2∙ T), will result in every edge permutation of the form (1 4 z) with no other effect to the puzzle. With the use of these known permutations of the form (1 4 z) and conjugation, (z 4 5) and (1 z 5) can also be achieved.
Conjugating (1 4 z) by (1 4 5) results in
(5 4 1) (1 4 z) (1 4 5) = (1 z 5)
Conjugating (1 4 z) by (5 4 1) results in
( 1 4 5) (1 4 z) (5 4 1) = (z 4 5)
We can also achieve the edge permutations (z 4 5)-1, (1 z 5)-1,(1 4 z)-1 by simply inverting the respective sequence of moves used to get (z 4 5), (1 z 5) and (1 4 z).
c. One of the three edges we want to cycle is in place. We want to cycle edges (z y 5), (1 z y) or (y 4 z) where y, z = {any edge ≠ 1, 4, 5}. In part b. it was shown that every edge permutation of the forms (1 4 z), (1 z 5) and (z 4 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 4 z) can also be achieved.
Conjugating (1 4 z) by (1 4 y) results in
(y 4 1) (1 4 z) (1 4 y) = (1 z y)
Conjugating (1 z y) by (1 4 5) results in
(5 4 1) (1 z y) (1 4 5) = (z y 5)
Conjugating (1 z y) by (5 4 1) results in
(1 4 5) (1 z y) (5 4 1) = (y 4 z)
We can also achieve the edge permutations (1 z y)-1, (z y 5)-1,(y 4 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 5) and (y 4 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, 4, 5}. 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 4 x) results in
(x 4 1) (1 z y) (1 4 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. If β is even, we can rewrite it using 3-cycles. Any set of even permutations can be rewritten in 3-cycles by grouping every two transpositions together 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. For example, we have the following even permutation on edges:
(6 7 8 9) (3 4 5 6) (1 3) (1 2)
First, rewrite the product of cycles as a product of transpositions.
(6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3) (1 2)
Now that the permutation is a product of an even number of transpositions, they can be grouped.
(6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3) (1 2)
And finally, evaluated.
(6 8 9) (6 3 7) (3 4 5) (1 2 3)
This shows that all even permutations on edges can be attained since it is known that all 3-cycles (z y x) on edges can be attained, as shown earlier. If the edge permutation is odd however the extra move F (rotating the front face 90 degrees clockwise) will be added to the permutation (and removed later) to make it even and therefore attainable. For example, we have the following odd permutation on edges:
(6 7 8 9) (3 4 5 6) (1 3)
First, the extra move, F, will be added.
F (6 7 8 9) (3 4 5 6) (1 3) = (1 2 3 4) (6 7 8 9) (3 4 5 6) (1 3)
Rewrite the product of cycles as a product of transpositions.
(1 4) (1 3) (1 2) (6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3)
Group the transpositions and evaluate.
(1 4) (1 3) (1 2) (6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3)
(1 3 4) (6 1 9) (6 1 2) (6 7 8) (3 5 6) (3 1 4)
The original permutation can be rewritten as a product of 3-cycles which we can attain. This means that all odd permutations can be attained by using this method and then reversing the extra move (F-1) after.
a. The three edges we want to cycle are in place. If we want to cycle edges (1 4 5), no conjugation is necessary; use C2. We can also achieve (5 4 1), or (1 4 5)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 4 5).
b. Two of the three edges we want to cycle are in place. We want to cycle edges (z 4 5), (1 z 5) or (1 4 z) where z = {any edge ≠ 1, 4, 5}. To attain 3-cycle (1 4 z), it can be shown that a sequence of moves T can place edge z into position 5 without affecting the final positions of edges 1 and 4. Similarly, to attain (z 4 5) and (1 z 5) it can be shown that a sequence of moves T can place edge z into position 1 and 4, 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 4 z). The following are the sequences of moves T which can be applied to place each edge z into position 5 without changing the final position of edges 1 and 4.
For z = 2: T = F∙R^2∙D∙F-1
For z = 3: T = F-1∙U^2∙L^-1∙F
For z = 6: T = F∙R-1∙F-1
For z = 7: T = F∙R2∙F-1
For z = 8: T = F-1∙U2∙F
For z = 9: T = F-1∙U2∙L∙F
For z = 10: T = F∙R2∙D2∙F-1
For z = 11: T = F-1∙U∙F
For z = 12: T = F∙R2∙D-1∙F-1
The commutator squared, C2, conjugated by each T, (T-1∙C2∙ T), will result in every edge permutation of the form (1 4 z) with no other effect to the puzzle. With the use of these known permutations of the form (1 4 z) and conjugation, (z 4 5) and (1 z 5) can also be achieved.
Conjugating (1 4 z) by (1 4 5) results in
(5 4 1) (1 4 z) (1 4 5) = (1 z 5)
Conjugating (1 4 z) by (5 4 1) results in
( 1 4 5) (1 4 z) (5 4 1) = (z 4 5)
We can also achieve the edge permutations (z 4 5)-1, (1 z 5)-1,(1 4 z)-1 by simply inverting the respective sequence of moves used to get (z 4 5), (1 z 5) and (1 4 z).
c. One of the three edges we want to cycle is in place. We want to cycle edges (z y 5), (1 z y) or (y 4 z) where y, z = {any edge ≠ 1, 4, 5}. In part b. it was shown that every edge permutation of the forms (1 4 z), (1 z 5) and (z 4 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 4 z) can also be achieved.
Conjugating (1 4 z) by (1 4 y) results in
(y 4 1) (1 4 z) (1 4 y) = (1 z y)
Conjugating (1 z y) by (1 4 5) results in
(5 4 1) (1 z y) (1 4 5) = (z y 5)
Conjugating (1 z y) by (5 4 1) results in
(1 4 5) (1 z y) (5 4 1) = (y 4 z)
We can also achieve the edge permutations (1 z y)-1, (z y 5)-1,(y 4 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 5) and (y 4 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, 4, 5}. 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 4 x) results in
(x 4 1) (1 z y) (1 4 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. If β is even, we can rewrite it using 3-cycles. Any set of even permutations can be rewritten in 3-cycles by grouping every two transpositions together 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. For example, we have the following even permutation on edges:
(6 7 8 9) (3 4 5 6) (1 3) (1 2)
First, rewrite the product of cycles as a product of transpositions.
(6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3) (1 2)
Now that the permutation is a product of an even number of transpositions, they can be grouped.
(6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3) (1 2)
And finally, evaluated.
(6 8 9) (6 3 7) (3 4 5) (1 2 3)
This shows that all even permutations on edges can be attained since it is known that all 3-cycles (z y x) on edges can be attained, as shown earlier. If the edge permutation is odd however the extra move F (rotating the front face 90 degrees clockwise) will be added to the permutation (and removed later) to make it even and therefore attainable. For example, we have the following odd permutation on edges:
(6 7 8 9) (3 4 5 6) (1 3)
First, the extra move, F, will be added.
F (6 7 8 9) (3 4 5 6) (1 3) = (1 2 3 4) (6 7 8 9) (3 4 5 6) (1 3)
Rewrite the product of cycles as a product of transpositions.
(1 4) (1 3) (1 2) (6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3)
Group the transpositions and evaluate.
(1 4) (1 3) (1 2) (6 9) (6 8) (6 7) (3 6) (3 5) (3 4) (1 3)
(1 3 4) (6 1 9) (6 1 2) (6 7 8) (3 5 6) (3 1 4)
The original permutation can be rewritten as a product of 3-cycles which we can attain. This means that all odd permutations can be attained by using this method and then reversing the extra move (F-1) after.
2.2 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 = [U,RDR-1] = (RDR-1 )-1∙U-1∙RDR-1∙U
gives a result of a 3-cycle on vertices and no effect on edges, α = (1 3 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 3 5), no conjugation is necessary; use C. We can also achieve (5 3 1), or (1 3 5)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 3 5).
b. Two of the three vertices we want to cycle are in place. We want to cycle vertices (z 3 5), (1 z 5) or (1 3 z) where z = {any vertex ≠ 1, 3, 5}. To attain 3-cycle (1 3 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 3. Similarly, to attain (z 3 5) and (1 z 5) it can be shown that a sequence of moves T can place vertex z into position 1 and 3, 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 3 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 3.
For z = 2: T = D-1∙B∙D
For z = 4: T = L ∙B-1∙L-1
For z = 6: T = B
For z = 7: T = B2
For z = 8: T = B-1
The commutator conjugated by each T, (T-1∙C∙ T), will result in every vertex 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 5) and (1 z 5) can also be achieved.
Conjugating (1 3 z) by (1 3 5) results in
(5 3 1) (1 3 z) (1 3 5) = (1 z 5)
Conjugating (1 3 z) by (5 3 1) results in
( 1 3 5) (1 3 z) (5 3 1) = (z 3 5)
We can also achieve the vertex permutations (z 3 5)-1, (1 z 5)-1, (1 3 z)-1 by simply inverting the respective sequence of moves used to get (z 3 5), (1 z 5) and (1 3 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 3 z) where y, z = {any vertex ≠ 1, 3, 5}. In part b. it was shown that every vertex permutation of the forms (1 3 z), (1 z 5) and (z 3 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 3 z) can also be achieved.
Conjugating (13 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 5) results in
(5 3 1) (1 z y) (1 3 5) = (z y 5)
Conjugating (1 z y) by (5 3 1) results in
(1 3 5) (1 z y) (5 3 1) = (y 3 z)
We can also achieve the vertex permutations (1 z y)-1, (z y 5)-1,(y 3 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 5) and (y 3 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, 3, 5}. 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 3 x) results in
(x 3 1) (1 z y) (1 3 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. The method used to show this is the same as was used in proposition 2.1. If the vertex permutation is even, we can rewrite it using 3-cycles. Any set of even permutations can be rewritten in 3-cycles by grouping the transpositions like was done in 2.1. Since an even permutation can be written as an even product of transpositions, it can also be expressed as a product of 3-cycles. As a result of this, all even permutations on vertices can be attained since it is known that all 3-cycles (z y x) on vertices can be attained. If the vertex permutation is odd however the extra move F (rotating the front face 90 degrees clockwise) will be added to the permutation to make it even and therefore attainable. This means that all odd permutations can also be attained by adding this extra odd permutation and then reversing the extra move (F-1) after.
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 3 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 3 5), no conjugation is necessary; use C. We can also achieve (5 3 1), or (1 3 5)-1, by simply inverting the sequence of moves used to get the 3-cycle (1 3 5).
b. Two of the three vertices we want to cycle are in place. We want to cycle vertices (z 3 5), (1 z 5) or (1 3 z) where z = {any vertex ≠ 1, 3, 5}. To attain 3-cycle (1 3 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 3. Similarly, to attain (z 3 5) and (1 z 5) it can be shown that a sequence of moves T can place vertex z into position 1 and 3, 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 3 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 3.
For z = 2: T = D-1∙B∙D
For z = 4: T = L ∙B-1∙L-1
For z = 6: T = B
For z = 7: T = B2
For z = 8: T = B-1
The commutator conjugated by each T, (T-1∙C∙ T), will result in every vertex 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 5) and (1 z 5) can also be achieved.
Conjugating (1 3 z) by (1 3 5) results in
(5 3 1) (1 3 z) (1 3 5) = (1 z 5)
Conjugating (1 3 z) by (5 3 1) results in
( 1 3 5) (1 3 z) (5 3 1) = (z 3 5)
We can also achieve the vertex permutations (z 3 5)-1, (1 z 5)-1, (1 3 z)-1 by simply inverting the respective sequence of moves used to get (z 3 5), (1 z 5) and (1 3 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 3 z) where y, z = {any vertex ≠ 1, 3, 5}. In part b. it was shown that every vertex permutation of the forms (1 3 z), (1 z 5) and (z 3 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 3 z) can also be achieved.
Conjugating (13 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 5) results in
(5 3 1) (1 z y) (1 3 5) = (z y 5)
Conjugating (1 z y) by (5 3 1) results in
(1 3 5) (1 z y) (5 3 1) = (y 3 z)
We can also achieve the vertex permutations (1 z y)-1, (z y 5)-1,(y 3 z)-1 by simply inverting the respective sequence of moves used to get (1 z y), (z y 5) and (y 3 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, 3, 5}. 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 3 x) results in
(x 3 1) (1 z y) (1 3 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. The method used to show this is the same as was used in proposition 2.1. If the vertex permutation is even, we can rewrite it using 3-cycles. Any set of even permutations can be rewritten in 3-cycles by grouping the transpositions like was done in 2.1. Since an even permutation can be written as an even product of transpositions, it can also be expressed as a product of 3-cycles. As a result of this, all even permutations on vertices can be attained since it is known that all 3-cycles (z y x) on vertices can be attained. If the vertex permutation is odd however the extra move F (rotating the front face 90 degrees clockwise) will be added to the permutation to make it even and therefore attainable. This means that all odd permutations can also be attained by adding this extra odd permutation and then reversing the extra move (F-1) after.
2.3 All edge orientations b can be attained.
Note that in this paper, we have chosen to focus on the permutations of the edge and vertex pieces; however, a similar strategy can be employed to return the orientations of the cubelets to their original state. This process is outlined below.
First, we will show that there exists a sequence of moves V = (α, a, β, b) that change only the orientations of two edge cubelets. From this resulting vector b, the remaining vectors that make up the entire space of edge orientation vectors can also be found. One sequence of moves that will change only the orientation of two edge cubelets and leave the orientation and position of all the other cubelets alone is the following (Joyner 1996: pg 192)
V = R∙U2∙R2∙U-1 ∙R ∙U∙R∙U2∙L-1∙F-1∙R-1∙F∙L
Note that in this paper, we have chosen to focus on the permutations of the edge and vertex pieces; however, a similar strategy can be employed to return the orientations of the cubelets to their original state. This process is outlined below.
First, we will show that there exists a sequence of moves V = (α, a, β, b) that change only the orientations of two edge cubelets. From this resulting vector b, the remaining vectors that make up the entire space of edge orientation vectors can also be found. One sequence of moves that will change only the orientation of two edge cubelets and leave the orientation and position of all the other cubelets alone is the following (Joyner 1996: pg 192)
The result is V = ((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,1,1,0,0,0,0,0,0,0))
With the use of the commutator and conjugation, V can be manipulated to attain the following set of elements from R
((1), (0,0,0,0,0,0,0,0), (1), (1,0,0,0,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,1,0,0,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,1,0,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,1,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,1,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,1,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,1,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,1,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,0,1,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,0,0,1,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,0,0,0,1,1))
Combining these elements will give the entire set of edge orientation vectors. To be sure that all edge orientation vectors satisfy proposition 1.2 (that the sum of the vector components must be 0 mod 2), they must be of the form
((1), (0,0,0,0,0,0,0,0), (1), (1,0,0,0,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,1,0,0,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,1,0,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,1,0,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,1,0,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,1,0,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,1,0,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,1,0,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,0,1,0,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,0,0,1,0,1))
((1), (0,0,0,0,0,0,0,0), (1), (0,0,0,0,0,0,0,0,0,0,1,1))
Combining these elements will give the entire set of edge orientation vectors. To be sure that all edge orientation vectors satisfy proposition 1.2 (that the sum of the vector components must be 0 mod 2), they must be of the form
All vectors of this form can be attained by combining the orientation vectors above like so:
b1(1,0,0,0,0,0,0,0,0,0,0,1) + b2(0,1,0,0,0,0,0,0,0,0,0,1) + b3(0,0,1,0,0,0,0,0,0,0,0,1) + b4(0,0,0,1,0,0,0,0,0,0,0,1) + b5(0,0,0,0,1,0,0,0,0,0,0,1) + b6(0,0,0,0,0,1,0,0,0,0,0,1) + b7(0,0,0,0,0,0,1,0,0,0,0,1) + b8(0,0,0,0,0,0,0,1,0,0,0,1) + b9(0,0,0,0,0,0,0,0,1,0,0,1) + b10(0,0,0,0,0,0,0,0,0,1,0,1) + b11(0,0,0,0,0,0,0,0,0,0,1,1)
b1(1,0,0,0,0,0,0,0,0,0,0,1) + b2(0,1,0,0,0,0,0,0,0,0,0,1) + b3(0,0,1,0,0,0,0,0,0,0,0,1) + b4(0,0,0,1,0,0,0,0,0,0,0,1) + b5(0,0,0,0,1,0,0,0,0,0,0,1) + b6(0,0,0,0,0,1,0,0,0,0,0,1) + b7(0,0,0,0,0,0,1,0,0,0,0,1) + b8(0,0,0,0,0,0,0,1,0,0,0,1) + b9(0,0,0,0,0,0,0,0,1,0,0,1) + b10(0,0,0,0,0,0,0,0,0,1,0,1) + b11(0,0,0,0,0,0,0,0,0,0,1,1)
2.4 All vertex orientations a can be attained.
Before this can be shown, we will show that there exists a sequence of moves V = (α, a, β, b) that change only the orientations of two vertex cubelets. From this resulting vector a, the remaining vectors that make up the entire space of vertex orientation vectors can also be found. One sequence of moves that will change only the orientation of two vertex cubelets and leave the orientation and position of all the other cubelets alone is the following (Joyner 1996: pg 192)
V = (B∙U2∙B-1∙R∙D2∙R-1)2
Before this can be shown, we will show that there exists a sequence of moves V = (α, a, β, b) that change only the orientations of two vertex cubelets. From this resulting vector a, the remaining vectors that make up the entire space of vertex orientation vectors can also be found. One sequence of moves that will change only the orientation of two vertex cubelets and leave the orientation and position of all the other cubelets alone is the following (Joyner 1996: pg 192)
V = (B∙U2∙B-1∙R∙D2∙R-1)2
The result is V = ((1), (1,0,0,0,0,0,2,0), (1), (0,0,0,0,0,0,0,0,0,0,0,0)).
With the use of the commutator and conjugation, V can be manipulated to attain the following set of elements from R
((1), (1,0,0,0,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,1,0,0,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,1,0,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,1,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,0,1,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,0,0,1,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,0,0,0,1,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
Combining these elements will give the entire set of vertex orientation vectors. To be sure that all vertex orientation vectors satisfy proposition 1.1 (that the sum of the vector components must be 0 mod 3), they must be of the form
With the use of the commutator and conjugation, V can be manipulated to attain the following set of elements from R
((1), (1,0,0,0,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,1,0,0,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,1,0,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,1,0,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,0,1,0,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,0,0,1,0,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
((1), (0,0,0,0,0,0,1,2), (1), (0,0,0,0,0,0,0,0,0,0,0,0))
Combining these elements will give the entire set of vertex orientation vectors. To be sure that all vertex orientation vectors satisfy proposition 1.1 (that the sum of the vector components must be 0 mod 3), they must be of the form
All vectors of this form can be attained by combining the orientation vectors above like so:
a1(1,0,0,0,0,0,0,2) + a2(0,1,0,0,0,0,0,2) + a3(0,0,1,0,0,0,0,2) + a4(0,0,0,1,0,0,0,2) + a5(0,0,0,0,1,0,0,2) + a6(0,0,0,0,0,1,0,2) + a7(0,0,0,0,0,0,1,2)
a1(1,0,0,0,0,0,0,2) + a2(0,1,0,0,0,0,0,2) + a3(0,0,1,0,0,0,0,2) + a4(0,0,0,1,0,0,0,2) + a5(0,0,0,0,1,0,0,2) + a6(0,0,0,0,0,1,0,2) + a7(0,0,0,0,0,0,1,2)