A permutation on permutations

[phil hunt]

On Sun, 25 Nov 2001 01:18:28 -0500, Arthur Siegel <ajs at ix.netcom.com> 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

Assuming that the list of of numbers, and every element in the
list is different, a quick way to test for reversals is to disallow
all lists where the first element is greater than the last one; 
this means that you get exactly one of each pair.

-- 
*** Philip Hunt *** philh at comuno.freeserve.co.uk ***