A permutation on permutations

[simonb at webone.com.au]

well, i'd like to see this list comprehension
implemented properly ; it doesn't
compute as it stands (on 2.1.1)...?

But i understand the problem;
one way to solve it efficiently is
to choose one permutation out of the
reflected pair, as u build. Unfortunately,
this nice peice of list comprehensions
will (AFAICS) be disposed of.
To choose, eg. pick only perms
whos first entry is less than the last
:)

Simon B

Arthur Siegel wrote:

>What I've come up with is:
>
>L is my list of arbitrary length.
>M= [i for i in range (len(L))]
>
>#thanks to Rainer Deyke's post
>def perms(L):
>   if list == [] :
>      return [[]]
>   return [[list[i] + p for i in range(len(list))\
>     for p in perms(list[:i] + list[i+1:])]
>
>def removedups(t):
>   for p in t:
>      m=copy.copy(p)
>      m.reverse()
>      if m in t:
>         t.remove(m)
>
>def drawcurve(t):
>   #t is  M or a slice of M
>   perm=perms(t)
>   removedups(perm)
>   for i in range(len(perm)):
>      cp=[L[j] for j in perm[i]]
>      BezierCurve(cp) 
>
>It seems to work.
>
>But I wonder if there is a significantly more
>efficient solution
>
>Art
>
>