[next] [prev] [up] Date: Sun, 14 Jan 96 22:03:00 -0500
[next] [prev] [up] From: Mark Longridge <mark.longridge@canrem.com >
[next] [prev] [up] Subject: Cube Theory

Here is some more Cube Theory:

On the standard Rubik's Cube

Using < R, L, F, B, U > to generate D1
Let A = R1 L3 F2 B2 R1 L3, then A U1 A = D1
A U1 A = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3   (17 q, 13 q+h)

Using < R2, L2, F2, B2, U2 > to generate D2
Let A = R2 F2 B2 L2,       then A U2 A = D2
A U2 A = R2 F2 B2 L2 U2 R2 F2 B2 L2               (18 q, 9 q+h)

Slice group pattern, 6 spot, 4 slice moves
p1 = (F1 B3) (L1 R3) (U1 D3) (F1 B3)           (8 q)

Slice group antipode, 6 spot + pons asinorum, 6 slice moves
p2 = (F2 B2) (T1 D3) (F1 B3) (L3 R1) (T1 D3)   (12 q)
p2^6 = I

p7a  Cube in a cube   U2 F2 R2 U3 L2 D1 (B1 R3) ^3 + D3 L2 U1
(15 q+h, 20 q)
if  A = U3 L2 D1,  then let A' = inverse of A
p7a = U2 F2 R2 + A + (B1 R3) ^3 + A'

Or if we want something more symmetric, there is Mike Reid's...

p7b  Symmetric Maneuver (R3 U1 F2 U3 F3 L1 F2 L3 F1 R1  C_X ) ^ 2
(20 q+h , 24q)

One might even call this maneuver to be "cyclic decomposable".
Even the first half of this sequence generates an interesting pattern.
It would appear that using symmetric maneuvers does not ensure minimal
q or q+h turns.

Perhaps p7a is simpler in terms of notational expression. p7a is
how I actually do "Cube in a cube" in real cubing. Note also that
p7a uses all 6 generators and p7b uses <U, L, R, F> and there may
be a tighter symmetric "Cube in a cube".

Mike, have you tried using the pattern generated by the first half
under your Kociemba algorithm for q turns??

Megaminx (platonic dodecahedron) 12 faces, 20 corners, 30 edges

tetrahedron  = 4 axis
cube         = 3 axis
dodecahedron = 6 axis

In constructing a dodecahedron, build a bottom, place the 5 adjacent
faces to form a bowl. The top edges now form a skew decagon. Build
another bowl and connect the two bowls together to form a dodecahedron.


Recall that the Halpern-Meier Tetrahedron has 3,732,480 states.
In this count we consider the 4 centre pieces immobile.

The picture H-M tetrahedron has 3,732,480 * 3^3 = 100,776,960 states.
In essence, we can rotate any 3 centres at will, the 4th is forced.
That number may seem familar to some of the more fanatical cubists,
has the picture Skewb has 3,149,280 * 2^5 = 100,776,960 states!

Additionally the possible rotations of the centres of the H-M
tetrahedron are  (), (+ -), (+++) , (---), (-++-), (+--+), (+++-),

#   Order (tetra, r * d) = 45         OR      (r+ d+)^45 = I
Note that (r+ d+)^45 is still the identity, even on the picture
H-M tetrahedron, as 45 is divisible by 3.

-> Mark <-

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