I’d go with an algorithm similar to Amir’s, just replacing “while… n–” with a for loop, but IMHO the more interesting question is how to define a Gray code on permutations, and one of the famous solutions here is the Johnson-Trotter algorithm (http://en.wikipedia.org/wiki/Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm ).

here is my last, most-compact shot:
[python]
def perm(n, m):
s = range(n)
for j in range(1,n):
m, r = divmod(m,j+1)
s[r], s[j] = s[j], s[ r ]
return tuple(s)

def tester( f, k ):
“”" a simple tester function just call tester( permutation_function, k) “”"
p = [f(k, i) for i in range(fact(k) ) ]
return len(set(p) ) == fact(k)

assert all([ tester(perm, i) for i in range(1,10) ])

[/python]

sorry for the lazy test before.

and btw here is the “solution”
http://en.wikipedia.org/wiki/Permutation#Algorithms_to_generate_permutations

[Python]
def permut(values, m):
n = len(values)
return [values.pop(m%i) for i in range(n, 0, -1)]

for i in range(2*3*4):
print permut(list(“abcd”), i)
[/Python]

[Python]
def permut(n, m):
values = range(n)
return [values.pop(m%i) for i in range(n, 0, -1)]
[/Python]

Some of the solutions got a bit complexified, I think I kept it minimalistic, though it’s the same algo…

Very nice challenge, it kept me thinking for quite a while!