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Re: Permutations...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg95778] Re: Permutations...
  • From: sashap <pavlyk at gmail.com>
  • Date: Tue, 27 Jan 2009 06:57:35 -0500 (EST)
  • References: <glk869$oc9$1@smc.vnet.net>

On Jan 26, 5:52 am, bruno... at libero.it wrote:
> Given 4 elements  (1 2 3 4)  we have 6 translatios:
> 1 2 3 4
> 1 3 2 4
> 1 4 2 3
> 2 1 3 4
> 3 1 2 4
> 4 3 2 1

Hi Bruno,

The question you pose is to enumerate representatives
of equivalence classes of a coset S_4/Z_4 where Z_4 is
generated by cyclic shifts.

It is obvious that you can always choose the first element
of the representative to be 1. Hence once way to
generate those 6 elements is through

Join[{1}, #]& /@ Permutations[Range[2,4]]

and it is clear that there 6 of them. Another way is to
supply a custom test function to Union:

In[61]:= Union[Permutations[Range[4]],
 SameTest -> Function[{p1, p2}, Module[{pos},
    {{pos}} = Position[p2, First[p1], {1}, 1, Heads -> False];
    p1 === Take[Join[p2, p2], {pos, pos + Length[p2] - 1}]
    ]]]

Out[61]= {{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {1,
  4, 2, 3}, {1, 4, 3, 2}}


Hope this helps,
Oleksandr Pavlyk

>
> Each translation can generate 4 rotations:
> 1 2 3 4   1 3 2 4   1 4 2 3
> 2 3 4 1   3 2 4 1   4 2 3 1
> 3 4 1 2   4 1 3 2   3 1 4 2
> 4 1 2 3   4 1 3 2   3 1 4 2
> etc.
>
> Then:
> Translations = (4-1)! = 6
> Rotations = 4 per translation
> Permutations = Trans * Rot = 4! = 24
>
> With Mathematica:
> Permutations[Range[4]]  prints all 24 Permutations
> How can I get the 6 Translations and the 18 Rotations separately?
>
> Bruno



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