[next] [prev] [up] Date: Sun, 08 Aug 93 15:40:00 -0400
[next] [prev] [up] From: Mark Longridge <mark.longridge@canrem.com >
~~~ ~~~ [up] Subject: SQUARE'S GROUP ANALYSIS

After reading Dik's post I figured I'd add my 2 cents worth:

Mark's Notes on the Squares Group
---------------------------------

On studying the squares group I have found 16 antipodal cases
requiring
the maximum 15 moves. Two of these cases cycle all 8 corners and leave
the edges in place. A third case "2 DOT/Inverted T's" is pleasingly
symmetric. Also I have noted that cycling only the 4 edges in the
U or D layer requires 1 move less that cycling only the 4 corners in U
or D when using only moves in the square's group, 12 moves for edges
and 13 moves for corners.

If we define "symmetry level" as the number of distinct patterns
generated by rotating the cube through it's 24 different orientations in
space then most known antipodes are symmetry level 6. Thus the lower the
number the higher the level of symmetry. The least symmetric positions
have level 24, and this is very common. The most symmetric positions
have level 1, the two positions START and 6 X order 2.

I have also found positions with levels 3, 8 and 12.

   Given the fact that 8 antipodal cases have symmetry level 6
and 8 cases have symmetry level 12 we can now account for ALL
8 * 6 + 8 * 12 = 144 of the 144 cases!

Cases with symmetry level 6:

p66  Double 4 corner sw  L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2
(15)
p67  Antipode 2          R2 B2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p80  2 DOT, Invert T's   R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p99  2 DOT, 4 ARM        R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p100 2 Cross, 4 ARCH 1   R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p130 2 Cross, 4 ARCH 2   L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2
(15)
p135 2 X, 4 T            L2 B2 D2 F2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
(15)
p137 2 X, 4 ARM          L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
(15)

Cases with symmetry level 12:

p108 2 DOT, 2 T, 2 ARM   L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p128 2 H, 2 T, 2 CRN     L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
(15)
p129 2 H, 2 T, 2 ARCH    R2 F2 L2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
(15)
p131 2 H, 2 ARM, 2 ARCH  L2 F2 R2 D2 B2 L2 D2 L2 (T2 D2 F2 T2) F2 L2 F2
(15)
p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
(15)
p133 2 Cross, 2 T, 2 ARM L2 F2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p134 2 CRN, 2 X, 2 ARCH  L2 F2 D2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p136 2 H, 2 ARM, 2 CRN   R2 F2 L2 T2 B2 L2 T2 L2 (T2 D2 F2 T2) F2 L2 B2
(15)

   5 of the 16 known antipodes are within 4 and 2 face turns (or 2 and 1
slice turns) of each other:
p66 + L2 R2 T2 D2 = p80  (allowing for whole cube rotations)
p66 + F2 B2       = p100
p80 + T2 D2       = p99
P66 + T2 D2       = P128

Using full group moves these antipodes can be reduced to:

P66a alternate method    F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
 (13)
p67a alternate method    F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
 (13)
p80a alternate method    U1 F2 R2 L2 U2 D2 F2 U2 D3
  (9)
p99a alternate method    U1 R2 F2 B2 U2 D2 R2 D1
  (8)
P100a alternate method   F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
 (12)
p108a alternate method   R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
 (12)
p130a alternate method   F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
 (13)
p133a alternate method   R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2
 (11)

Both p80a and p99a are surprisingly compact, p99a being a full 7
turns
less than it's square's group equivalent. Note that in p99a a square's
group sequence is sandwiched between 2 turns on opposite faces. It is
the final turn D1 which brings it back into a sq group state! In general
U1 (sq group sequence) D1 does not lead to a sq group sequence.

Another interesting discovery was comparing the full group sequences:

L1 R1 D2 L3 R3             (antislice, 5 moves)
L1 R3 D2 L3 R1             (slice    , 5 moves)
F1 B1 D2 F1 B1             (clockwise, 5 moves)

... to their square's group equivalents:

R2 F2 T2 L2 B2 L2 T2 R2 F2 (9 moves)
R2 B2 L2 D2 R2 F2 L2 T2 F2 (9 moves)
R2 B2 T2 F2 L2 F2 T2 R2 F2 (9 moves)

Also it was found possible to permute 3 edges only using:

      L2 T2 R2 B2 R2 T2 L2 F2    (8 moves)
or    L3 R1 F2 L1 R3 D2          (6 moves)

In general any sequence L1 R1 (any squares group moves) L3 R3
will always result in a squares group position, for example:

                      L1 R1 (D2 F2 B2)    R3 L3
                      F1 B1 (T2 B2 F2 L2) F3 B3
p66a (F2 R2 U2 F2 R2) U3 D3 (B2 L2 F2 B2) U1 D1  (13 moves)

The longest irreducible square's group sequence discovered so far,
which is an embedded part of longest Phase 2 sequence (p94):
(Thus it can't be reduced by using full group moves using current
techniques)

R2 B2 U2 B2 L2 D2 L2 F2          (8)

Later on I discovered this irreducible sequence by chance:

T2 B2 T2 B2 D2 F2 R2 T2 L2 F2   (10)

Edges only (with corners in place) can be 14 moves at most, e.g.

D2 L2 F2 D2 F2 L2 T2 L2 T2 D2 F2 R2 F2 R2  (14)

This answers the question David Singmaster posed in
"Notes on Rubik's Magic Cube" on Thistlethwaite's last stage.
That is: "Are there any positions in the square's group with
corners fixed of length 14 or can they be done in less moves?"

A few observations...

- It is not possible to swap just 1 pair of edges and corners
- It is only possible to have 4, 6 or 8 corners out of place
- Known antipodal cases can be solved in <=13 moves using full
group
- In reaching an antipode one may start with any of the 6 turns
(since antipodes are global maxima, any turn will get you one
move closer)
- If the corners are fixed, the position is NOT an antipode
- Longest order appears to be 12
- All known (probably all!) antipodes have symmetry level 6 or 12
- Although only conjectural, it is now believed that one turn of
a face MUST lead to a new state which is either 1 move closer
or 1 move farther from START

Question:     Are there any irreducible square's group sequences that
               are longer then 10 moves? Are these truly irreducible
               or only irreducible under Dik Winter's Kociemba
               inspired program?

Oh well, the full group beckons. I still want to try and come up
with my own algorithm though.

-> Mark <-

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