Permutation numbering

pn.py

# For any 0 <= k < fact(n), return a unique permutation of [0..n-1].
# Functional inverse of number_from_perm().
def perm_from_number(n, k):
  a = list(range(n))
  p = []
  for d in range(2, n + 1):
    p.append(k % d)
    k //= d
  t = n - 1
  for i in reversed(p):
    a[t], a[i], t = a[i], a[t], t - 1
  return a

# For any permutation of [0..n-1], return unique 0 <= k < fact(n).
# Functional inverse of perm_from_number().
def number_from_perm(p):
  t = len(p) - 1
  a = list(range(len(p)))
  r = list(range(len(p)))
  n = 0
  for i in reversed(p[1:]):
    n, a[r[i]], r[a[t]], t = t * (n + r[i]), a[t], r[i], t - 1
  return n