Recursive function to develop permutations

[Steve Goldman]

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
out with unhappy consequences.   I thought yield had the same effect as
return, except that the function's state was "frozen" until the next
iteration.

Thanks again.



> Are these "counting" solutions any simpler than the recursive one?  I
> don't really think so, for a cleanly-implemented recursion (one without
> globals and a first-time switch...!-) such as:
>
> def permute(Xs, N):
>     if N <= 0:
>         yield []
>         return
>     for x in Xs:
>         for sub in permute(Xs, N-1):
>             yield [x]+sub
>