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