On Thu, Aug 30, 2012 at 8:22 PM, David Joyner <[email protected]> wrote:
> On Thu, Aug 30, 2012 at 10:20 AM, Chris Smith <[email protected]> wrote:
>>>
>>> Permutation([(0, 2, 7, 5), (1, 4, 6, 3), (8, 32, 24, 16), (9, 33, 25,17), 
>>> (10, 34, 26, 18)])
>>
>> Sorry, enter that as
>>
>> Permutation([(0, 2, 7, 5), (1, 4, 6, 3), (8, 32, 24, 16), (9, 33,
>> 25,17), (10, 34, 26, 18)]  ,  size=48)
>
> Thanks. Here we are:-)
>
>
>>>> F = Permutation([(17,19,24,22),(18,21,23,20),( 6,25,43,16),( 7,28,42,13),( 
>>>> 8,30,41,11)],  size=49)
>>>> B = Permutation([(33,35,40,38),(34,37,39,36),( 3, 9,46,32),( 2,12,47,29),( 
>>>> 1,14,48,27)],  size=49)
>>>> L = Permutation([( 9,11,16,14),(10,13,15,12),( 1,17,41,40),( 4,20,44,37),( 
>>>> 6,22,46,35)],  size=49)
>>>> R = Permutation([ (25,27,32,30),(26,29,31,28),( 3,38,43,19),( 
>>>> 5,36,45,21),( 8,33,48,24)],  size=49)
>>>> U = Permutation([ ( 1, 3, 8, 6),( 2, 5, 7, 4),( 
>>>> 9,33,25,17),(10,34,26,18),(11,35,27,19)],  size=49)
>>>> D = Permutation([ 
>>>> (41,43,48,46),(42,45,47,44),(14,22,30,38),(15,23,31,39),(16,24,32,40)],  
>>>> size=49)
>>>> G = PermutationGroup([F,B,L,R,U,D])
>>>> G.order()
> 43252003274489856000
>>>> Z = G.center()
>>>> Z.order()
> 2

I wish I had the warm fuzzyz that come from seeing with the
heart...but I don't understand this topic well (though I am learning).
What is nice to see is that this agrees with this:


>>> from sympy.combinatorics.generators import *
>>> [w for w in dir() if 'ub' in w]
['Subs', 'rubik_cube_generators', 'subresultants', 'subsets']
>>> from sympy.combinatorics.perm_groups import *
>>> G=PermutationGroup(rubik_cube_generators())
>>> G.order()
43252003274489856000L
>>> Z=G.center()
>>> Z.order()
2

One thing that interests me is whether there is a decomposition method
for finding how a given set of permutations might produce a a given
permutation, like `FooPerm.factor([P1, P2]) -> (0, 3) if FooPerm is
just P2**3 or (1, 2) if P1*P2**2.`

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