Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: All permutations of a sequence

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76333] Re: All permutations of a sequence
  • From: "Sem" <sarner2006-sem at yahoo.it>
  • Date: Sun, 20 May 2007 02:20:04 -0400 (EDT)
  • References: <f2h8t7$v4$1@smc.vnet.net> <f2mful$mm7$1@smc.vnet.net>

Sorry, the loading of the package 'Combinatorica' isn't necessary for the 
function.


"Sem"
> Hi Virgil,
>
> the above code produces all the permutations of a list, except those that
> reverse their ordering.
> The function is recursive but I suppose that an iterative algorithm can be
> wrote easily, i.e. for large list.
>
> << DiscreteMath`Combinatorica`
>
> NotReversingPermutations[l_List] := {l} /; Length[l] <= 2
>
> NotReversingPermutations[l_List] :=
>    Module[
>                {e = Last[l], n = Length[l], r},
>                r = Table[Insert[#, e, -i], {i, 1, n}] & /@
> NotReversingPermutations[Drop[l, -1]];
>                Flatten[r, 1]
>                ];
>
> HTH
> Regards,
>
>    Sem
>
>>"Virgil Stokes"
>>I would like to get all unique permutations of a sequence (e.g.
>> {1,2,3,4,5,6,7,8,9,10,11,12,13}) with the constraint that all sequences
>> that have an opposite ordering are not included (e.g. {13, 12,
>> 11,10,9,8,6,5,4,3,2,1}) would not be included.
>>
>> Are there some Mathematica commands that can be used to efficiently
>> generate these permutations?
>>
>> --V. Stokes
>>
>
> 



  • Prev by Date: Re: using N function
  • Next by Date: Re: using N function
  • Previous by thread: Re: All permutations of a sequence
  • Next by thread: Re: All permutations of a sequence