[next] [prev] [up] Date: Tue, 31 Jan 95 11:43:00 WET
[next] [prev] [up] From: Martin Schoenert <Martin.Schoenert@math.rwth-aachen.de >
[next] [prev] [up] Subject: Re: Skewb thoughts
Mark Longridge wrote in his e-mail message of 1995/01/29

Extract from Martin's very detailed skewb analysis:

Then the group CG = < C, G > is the set of all positions a puzzler
could observe. There are 24 solved position in CG (solved up to
rotations).

The group CG has size 2 * 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
               |CG|     = 75,582,720
Note that:      |CG| /24 =  3,149,280
The group G has size 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
               |G|      = 37,791,360
Note that:      |G|  /12 =  3,149,280

The number of positions both David Singmaster and Tony Durham
(the inventor) find for the skewb is 3,149,280.

Right. The SKEWB has 75582720 basic states. Just as with the cube, we
consider two basic states to be essential equal if the differ only by a
rotation of the rigid cube. Since there are 24 rotations of the rigid
cube, the SKEWB has 3149280 = 75582720/24 essential states.

Mark continued

If we use only one tetrad of the skewb then GAP also finds this number:

##    corners                                  centers
##    (each turn permutes 4)           (each turn permutes 3)
skewb := Group(
    ( 1,11,17) ( 2,12,20)( 4,10,18)(22, 6,14) (25,27,29),
    ( 2,10,22) ( 1, 9,23)( 3,11,21)(17, 5,15) (25,27,30),
    ( 4,14,20) ( 1,15,19)( 3,13,17)( 7,11,23) (25,28,29),
    ( 6,12,18) ( 5,11,19)( 7, 9,17)(21, 1,13) (26,27,29)
);;

Size (skewb);
>   3149280

In my message on the SKEWB I used the subgroup H generated by LUB, LUF,
RUB, and RUF. As I noted, this subgroup has a nontrivial intersection
with the subgroup C of rotations of the rigid cube. Thus it is *not*
a model for the essential SKEWB.

The subgroup that Mark uses, which is generated by RUF, RUB, LUF, and
RDF is much better. It has trivial intersection with C and is a model
for the essential SKEWB.

Note however, that the corners corresponding to the four generators for
this subgroups do *not* form a tetrad. They are the corner RUF and the
three adjacent corners. In particular, those four generators do not fix
the positions of the four corners; the generator RUF permutes the three
corner cubies at RUB, LUF, and RDF. This subgroup has 7 other conjugated
subgroups, corresponding to the 7 other possible choices of the first
generator (the one that is adjacent to the other 3 generators).

So allow me to use the subgroup H generated by RUF, LUB, RDB, and LDF.
The corresponding four corners do form a tetrad. This H also has trivial
intersection with C and also has size 3149280. Thus it also is a model
for the essential SKEWB. Note that those four generators never change
the positions of the four corner cubies. This subgroup is ``almost
normal''; it has only 1 other conjugated subgroup, corresponding to the
stabilizer of the other tetrad.

Mark continued

Mr. Singmaster had indicated in his last Cubic Circular that we may
determine the skewb's orientation if only one of the tetrads are moved.

I am not certain that I understand what this means. One possible
interpretation is, that for each state g of the SKEWB we can easily find
the rotation x of the rigid cube, such that (g x) is in the subgroup H.
That means that for each state g we can easily find how to rotate the
rigid cube, so that the rotated state can be solved using only the four
generators above. But this is obvious. Since the four generators do not
change the positions of the four corner cubies of the tetrad, the
rotation x must be the one that puts those four corner cubies to their
home positions.

Mark continued

By moving first one tetrad and then the other we can easily change the
skewb's orientation in space.

This comment I don't understand at all. Could you clarify it a bit?

Mark continued

Martin finds that the diameter of the skewb is 11 moves, with perhaps 90
antipodes. The idea that the skewb has 2 positions at 0 moves is rather
odd, but I think if we divide Martin's table by 2 we should get the
answer for visually distinguishable states for a skewb fixed in
orientation.

Right. The diameter of the SKEWB is 11 moves and there are 90 essential
different antipodes. The essential SKEWB does *not* have 2 states at 0
moves, only the subgroup H which I used has 2 essentially solved states.
This is not odder than the notion that the basic SKEWB has 24 essentially
solved states. And yes, if you divide the numbers in my table by 2, you
get the table for the essential SKEWB.

I rerun the computation using the new subgroup H as a model for the
essential SKEWB. Here is the output.

 0       1       0      0      0      0      0      0      0  0  1
 1       8       0      0      0      0      0      0      8  0  0
 2      48       0      0      0      0      0      0     48  0  0
 3     288       0      0      0      0      0      0    288  0  0
 4    1728       0      0      0      0      0    120   1608  0  0
 5   10248       0      0      0     36    377   1322   8513  0  0
 6   59304      12     87    662   2217   7561  15698  33067  0  0
 7  315198    4331  16897  37723  61161  76931  66997  51158  0  0
 8 1225483  515249 311594 186221 115830  65096  25012   6481  0  0
 9 1455856 1384909  61839   8280    708     95     25      0  0  0
10   81028   80938     90      0      0      0      0      0  0  0
11      90      90      0      0      0      0      0      0  0  0

Since the group is smaller it run faster and also used less memory.
Using some additional tricks, I could cut down the time to 40 seconds and
the memory needed to 2.5 megabytes.

As you can see, the numbers in the first column are exactely half of the
corresponding numbers in my previous message (as was expected). The
numbers in the other columns are close to half of the corresponding
numbers in my previous message but not exactely identical. I have to
rethink what those numbers mean and how they relate to the corresponding
numbers for the basic SKEWB.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

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