Definitions:
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Corners |
Hn: |
opposite pole to H(7-n) |
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Centersurfaces |
Fn: |
opposite centersurface to F(7-n) |
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Rotations |
[Ln]: |
120° rotation of the hemisphere with Hn as pole, counterclockwise |
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Wish-rotations |
(RHn): |
120° rotation of Hn only, counterclockwise |
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(RFn): |
180° rotation of Fn only |
The "Rotations" are the physical moves that can be done on the cube/ball, while the "Wish-rotations" are the results of some sequnece of rotations.
Rotations clockwise are indicated by a "minus" sign; [- Ln], (-RHn) , in stead of using two times the same rotation.
The figure above indicates a [L6] rotation, a rotation of the northern hemisphere with H6 as northpole, counterclockwise, along the equator of H6, while the southern hemisphere are fixed.
Technically the [L6] and the [L1] produce the same operation on the cube/ball, but the "front wiev corner H0" changes with [L6]. In the following, only the counterclockwise rotations [L0] [L1] [L2] [L3] and the clockwise opposites are used.
A procedure for solution:
The solution procedure is based on first managing the position of the corners, then the right rotation of the corners, and finnaly the centersurfaces. On a cube/ball with pattern, it might be necessary to rotate the centersurfaces at the end.
All the algorithms are possible to reverse. Begin backwards, and in addition, turn the opposite direction of what is indicated.
Here is a short version of the solution of the cube/ball.In the following a partly detailed deduction of the solution procedure follow.
Part one
The following set of rotation is a tool for getting the corners in right position, but not neccesarily in the right rotation. (The set of rotation dont pay attention to permutation of centersurfaces or rotation of corners).
The result is here stated without proof (which would be trivial)
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Part two
The following set of rotation rotates six corners. This set makes the basis for later procedures for rotation of corners.
The result is also here stated without proof (which also would be trivial).
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Part three
Combination of the set of rotation in part two, with basiscorner H0, repeated in three directions, H4 H6 H5, gives:
[L1][-L2][L1][-L2] [L2][-L3][L2][-L3] [L3][-L1][L3][-L1]
equivalent: [L1][-L2][L1] ([-L2][L2]) [-L3][L2] ([-L3][L3]) [-L1][L3][-L1]
= [L1][-L2][L1] [-L3][L2] [-L1][L3][-L1]
(It is here a point to operate in the order of H4 H6 H5 , to get the posibility to short in the number of rotations.)
The result becomes as follow:
(-RH4)(-RH1)(-RH5)(RH6)(RH2)(RH7) + (-RH6)(-RH2)(-RH4)(RH5)(RH3)(RH7) + (-RH5)(-RH3)(-RH6)(RH4)(RH1)(RH7)
or rewritten: (-RH1)(RH1) (RH2)(-RH2) (RH3)(-RH3) (-RH4)(-RH4)(RH4) (-RH5)(RH5)(-RH5) (RH6)(-RH6)(-RH6) (RH7)(RH7)(RH7)
= (-RH4)(-RH6)(-RH5)
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Part four
It is possible to repeat the set of rotation from part three in different directions to "isolate" the rotation of only two corners, but it now turns into a question of the number of rotations. To manage with less rotation, the algoritm 2 is short and gives a different permutation of cornerrotation, which now in the second turn might suits algorithm 3.
Anyway, the hard way:
Uses the algorithm 3 with basiscorner H0, then algorithm 3 with basiscorner H3, and then again algorithm 3 with basiscorner H0:
[L1][-L2][L1][-L3][L2][-L1][L3][-L1]
[L0][-L2][L0][-L1][L2][-L0][L1][-L0]
[L1][-L2][L1][-L3][L2][-L1][L3][-L1]
This gives:
(-RH4)(-RH6)(-RH5)
(-RH6)(-RH7)(-RH5)
(-RH4)(-RH6)(-RH5)
or rewritten: (-RH4)(-RH4) (-RH5)(-RH5)(-RH5) (-RH6)(-RH6)(-RH6) (-RH7)
= (RH4)(-RH7)
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Part five
At this point, the corners should all be in right spot and with right rotations.
The following set of rotations gives a permutation of the centersurfaces. As we see from the largest intern permutation group, repetition of this procedure gives us 4 different permutationgroups. The corners are not changed under this procedure, but there are some rotation of the centersurfaces (see next part).
To use this algorith, it might be neccesary to "adjust" with a rotation fist, such that the centersurfaces changes to the "right" spots. If allready some centerfaces are in the right spot, it will be neccesary to use the algorithm twice.
This algorithm is stated without proof.
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Part six
At this point it could happen that 4 (or less, dependent of the pattern) of the centersurfaces need some rotation. The following set of rotation should solve this problem. This algorithm is based on algorithm 5. It actually gives the 4th permutation, which should be the identity in part five, but as commented, the 4-group have all now rotated
180°.This algorithm is also stated without proof.
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