Recursive function to develop permutations
aleaxit at yahoo.com
Thu Oct 21 08:33:46 CEST 2004
Steve Goldman <steve_g at ix.netcom.com> wrote:
> Thanks to everyone for lots to think about. I really like the concept of
> counting in base "X" as a model for this problem.
> But now I have another question. In the elegant recursive generator, below,
> the yield expression for the terminating condition is followed by a return
> statement. Why is that necessary? I know it is, because I tried taking it
It's not; a
or putting the rest of the program in an 'else:' would be just as good.
> out with unhappy consequences. I thought yield had the same effect as
> return, except that the function's state was "frozen" until the next
You need to terminate the recursion sooner or later. To do that, you
must get a StopIteration exception raised. That happens with the
appropriate raise statement, or a return, or "falling off the end of the
code", which is equivalent to a return.
I generally prefer return, as it's concise and avoids extra nesting.
But if you'd rather do if/else, you can code, for example:
def permute(Xs, N):
if N > 0:
for x in Xs:
for sub in permute(Xs, N-1):
If you do choose an if/else it seems to me things are more readable when
you arrange them this way (rather than with if N<=0, so the for in the
else). Matter of tastes, of course.
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