recursion depth problem

[proctor]

On Apr 22, 4:37 pm, Michael Bentley <mich... at jedimindworks.com> wrote:
> On Apr 22, 2007, at 4:08 PM, proctor wrote:
>
>
>
> > On Apr 22, 2:55 pm, half.ital... at gmail.com wrote:
> >> On Apr 22, 11:49 am, proctor <12cc... at gmail.com> wrote:
>
> >>> hello,
>
> >>> 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]))
>
> >>> ==================
>
> >>> this function expects input in the form of a string of zeros, like
> >>> this:
>
> >>> python test-bin.py 00000000
>
> >>> and is expected to output a list of permutations like this:
>
> >>> $ python test-bin.py 0000
> >>> 1000
> >>> 0100
> >>> 1100
> >>> 0010
> >>> 1010
> >>> 0110
> >>> 1110
> >>> 0001
> >>> 1001
> >>> 0101
> >>> 1101
> >>> 0011
> >>> 1011
> >>> 0111
> >>> 1111
>
> >>> thanks for all help!
>
> >>> sincerely,
> >>> proctor
>
> >> If you just want to make it work as is....check
>
> >> sys.setrecursionlimit()
>
> >> ~Sean
>
> > very nice.  thanks sean.  so is the structure of my original code
> > unrescuable?  i cannot rearrange it to bypass the recursion?
>
> Anything that can be done with recursion can be done without
> recursion.  If you really wanted to mimic a binary counter, why not
> write a function that converts a number to binary?
>
> Then you could just count up from 0, printing the return from your
> binary conversion function...
>
> And the binary converter would be pretty easy too: just think of it
> as making change -- for a given number you just divide the number by
> each of [2 ** x for x in range(len(sys.argv[1]), 0, -1)] and append
> the string representations of the answers to a list and if the answer
> is 1, subtract that much from the number...

cool...  i'm going to have to think about this one... :-)

proctor