recursion depth problem

[proctor]

On Apr 22, 2:06 pm, Steven Bethard <steven.beth... at gmail.com> wrote:
> proctor wrote:
> > On Apr 22, 1:24 pm, Michael Bentley <mich... at jedimindworks.com> wrote:
> >> On Apr 22, 2007, at 1:49 PM, proctor wrote:
>
> >>> i have a small function which mimics binary counting.  it runs fine as
> >>> long as the input is not too long, but if i give it input longer than
> >>> 8 characters it gives
> >>> RuntimeError: maximum recursion depth exceeded in cmp
> >>> i'm not too sure what i am doing improperly.  is there really a lot of
> >>> recursion in this code?
> >>> ==================
> >>> import sys
> >>> def ch4(item, n=0):
> >>>    if n < len(item):
> >>>            if item[n] == '0':
> >>>                    item[n] = '1'
> >>>                    print ''.join(item)
> >>>                    ch4(item)
> >>>            elif item[n] == '1':
> >>>                    item[n] = '0'
> >>>                    ch4(item, n+1)
> >>> ch4(list(sys.argv[1]))
> >>> ==================
> >> Yes.  There really is *that* much recursion in that code.  502 levels
> >> with input length of 8 characters, 1013 with 9 characters, 2035 with
> >> 10 characters...  There's a pattern there ;-)
>
> > ok, thanks michael!
>
> > is there a way to avoid it here?  how could i write this better, (if
> > at all)?
>
> Google for permutation-like recipies:
>
>      http://www.google.com/search?q=Python+permutations
>
> Use the code from the first hit::
>
>      >>> for x in xselections('01', 8):
>      ...     print ''.join(x)
>      ...
>      00000000
>      00000001
>      00000010
>      ...
>      11111101
>      11111110
>      11111111
>
> Explaining to your teacher why your code uses generators when you
> haven't been taught them yet is left as an exercise to the reader. ;-)
>
> STeVe

this is really nice, thanks steve.  much slicker than my code.

for interest sake:  is my method unredeemable?

thanks very much!

sincerely,
proctor