nombre_perm.py

Created by nicolas-patrois

Created on July 23, 2020

476 Bytes

Donne le numéro d’une permutation de (0,1,2,…) dans l’ordre lexicographique et inversement.

def fact(n):
  p=1
  for i in range(1,n+1):
    p*=i
  return p

def num2perm(m,n):
  l=list(range(n))
  p=fact(n)
  for i in range(n,0,-1):
    m%=p
    p//=i
    q=m//p
    l=l[:q]+l[q+1:]+l[q:q+1]
  return l

def perm2num(p):
  pp=list(p)
  p1=sorted(p)
  r=0
  for i in range(len(pp)):
    d=p1.index(pp[i])
    r+=d*fact(len(pp)-i-1)
    del p1[d]
  return r

if __name__=="__main__":
  print(num2perm(1234567,8))
  print(perm2num([0,8,1,5,4,3,6,2,7]))