Relation between positions and positions mod M in FTM

Tom proposed that I give the relation between the number Nm of positions mod M and the number of positions N in the way N = 48*Nm - constant C. This is indeed possible, because the symmetric positions of the cube are completely analyzed. If we denote the number of symmetric positions by S and the number of symmetric position mod m by Sm, we have the relation

(Nm-Sm)*48 = N - S,

this is because (Nm-Sm)*48 is the number of positions with no symmetry. So we have

N = 48*Nm - (Sm*48 - S) = 48*Nm - C.

The constant C is tabulated below for all levels from 0 to 20:

depth         C
 0                 47
 1                 78
 2                189
 3                360
 4               1593
 5               4788
 6              19850
 7              72564
 8             237656
 9             858381
10            3015740
11            9785356
12           35144616
13          122254428
14          436594274
15         1764160807
16         8037257961
17        37547823254
18        95403536079
19        21275288869
20            1140678

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Very nice

This is great; very useful. Great check on other results.

Would be very nice to get all symmetric positions analyzed in the QTM too, using Radu's and your ideas. Not sure who's going to do that, but that might also be a good way to find some "far" QTM positions. (Right now, only one 26q* and two 25q* positions are known, and only a very few
24q* positions [maybe a hundred; not sure.])

And then there's inverse positions, and antisymmetric positions. Of course, these aren't as "nice" as symmetric positions with respect to subgroups. But some day we'll figure it out, I'm sure.

I am quite sure, that the num

I am quite sure, that the numbers are right, it was a bit annoying to copy and paste all the data from my html-pages to an excel-sheet for the computation. The column sums over the numbers gave the correct values so I think I made no mistakes. If you want to check the excel sheet or you need the data for another purpose, you can download it at .

The computation of all symmetric positions in FTM was a big task, Silviu Radu did most of the computation for the large subgroups, completed by some Cube Explorer computations to sort out the 20f* from the >=19f* cubes, and I remember he used GAP to find for example the right subgroups for the pruning tables for C3. Together with his results for the large subgroups I was able to compute for each symmetry the number of cubes, which *exactly* have this symmetry, which was not trivial in some cases.
Doing the computation for QTM should be no problem, but I think we should not reinvent the wheel - maybe Silviu can help.

A problem for the antisymmetric positions is not only their size and the missing group property but also the fact, that there are so many different cases, this is combinations of the 33 essentially different symmetry subgroups and (sub)subgroups of index 2.

Antisymmetric positions

As I pointed out here, the anti-symmetric and selfinverse positions that are not symmetric can be found be finding the selfinverse positions of the void cube. There are no 33 cases in any way.

At first, selfinverse positions are anti-symmetric positions whose anti-symmetry is trivial. So the problem is to solve anti-symmetric positions.

A void cube is selfinverse if it becomes a selfinverse cube after some coloring of the centers, where regular colorings of the face centers, which give a Hungarian cube, are allowed, but also colorings that give a mirrored Hungarian cube, which is a popular fake Hungarian cube. Corners that stay on their places need to be unrotated for selfinverse positions. Since a fake cube coloring only gives corners that are flipped and thus unrotated when they stay on their spots, the number of fake cube coloring solutions is about 5 times larger than the number of real cube coloring solutions.

Since only 20 of the 48 symmetries have order <= 2, 7/12 of the found solutions must be discarded. And the number of anti-symmetric positions including selfinverse is about 60 times larger than selfinverse only: 9 antisymmetries give about the same number of solutions as the trivial symmetry and 10 antisymmetries give about 5 times as many solutions.

Just make a big prune system for solving a void cube to a selfinverse one. Next, solve the clean void cube, reducing symmetry in the search path. I think it will take far less time than finding Gods number.