On Apr 22, 4:37 pm, Michael Bentley <[EMAIL PROTECTED]> wrote: > On Apr 22, 2007, at 4:08 PM, proctor wrote: > > > > > On Apr 22, 2:55 pm, [EMAIL PROTECTED] wrote: > >> On Apr 22, 11:49 am, proctor <[EMAIL PROTECTED]> 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 -- http://mail.python.org/mailman/listinfo/python-list
