recursion - Python: using a recursive algorithm as a generator

Question

Recently I wrote a function to generate certain sequences with nontrivial constraints. The problem came with a natural recursive solution. Now it happens that, even for relatively small input, the sequences are several thousands, thus I would prefer to use my algorithm as a generator instead of using it to fill a list with all the sequences.

Here is an example. Suppose we want to compute all the permutations of a string with a recursive function. The following naive algorithm takes an extra argument 'storage' and appends a permutation to it whenever it finds one:

def getPermutations(string, storage, prefix=""):
   if len(string) == 1:
      storage.append(prefix + string)   # <-----
   else:
      for i in range(len(string)):
         getPermutations(string[:i]+string[i+1:], storage, prefix+string[i])

storage = []
getPermutations("abcd", storage)
for permutation in storage: print permutation

(Please don't care about inefficiency, this is only an example.)

Now I want to turn my function into a generator, i.e. to yield a permutation instead of appending it to the storage list:

def getPermutations(string, prefix=""):
   if len(string) == 1:
      yield prefix + string             # <-----
   else:
      for i in range(len(string)):
         getPermutations(string[:i]+string[i+1:], prefix+string[i])

for permutation in getPermutations("abcd"):
   print permutation

This code does not work (the function behaves like an empty generator).

Am I missing something? Is there a way to turn the above recursive algorithm into a generator without replacing it with an iterative one?

Answer 1

def getPermutations(string, prefix=""):
    if len(string) == 1:
        yield prefix + string
    else:
        for i in xrange(len(string)):
            for perm in getPermutations(string[:i] + string[i+1:], prefix+string[i]):
                yield perm

Or without an accumulator:

def getPermutations(string):
    if len(string) == 1:
        yield string
    else:
        for i in xrange(len(string)):
            for perm in getPermutations(string[:i] + string[i+1:]):
                yield string[i] + perm