I wrote:
> Another generator solution, based on computing a permutation from its
> rank according to some natural order. Written for strings.
>
> def perms(s) :
> def nth(n,L,k=1) :
> if k>len(L) :
> if n :
> raise StopIteration
> return ''
> return nth(n/k,L,k+1)+L.pop(n%k)
> for n in xrange(1<<30) :
> yield nth(n,list(s))
Same principle, simpler (no recursion, no helper func, less tokens, easier to
read) :
def perms(s) :
for n in xrange(1<<30) :
R,L = [],list(s)
while L :
n,k = divmod(n,len(L))
R.append(L.pop(k))
if n : break
yield ''.join(R)
Or if you *really* prefer lisp-style recursions, here is the solution as a
single lambda that returns a tuple of strings (not a generator as above)
perms = lambda *S : \
(len(S)>=len(S[0]) and S[0]==S[-1]) \
and S[min(1,len(S)-1):] \
or perms(''.join(L.pop(k)
for n,L in [[len(S),list(S[-1])]]
for _ in S[0]
for n,k in [divmod(n,len(L))]
)
,*S)
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