Re: on writing a number as 2^s * q, where q is odd

[jak via Python-list]

Julieta Shem ha scritto:

Alan Bawden <a...@csail.mit.edu> writes:

jak <nos...@please.ty> writes:

    Alan Bawden ha scritto:
    > Julieta Shem <js...@yaxenu.org> writes:
    >
    >     How would you write this procedure?
    >     def powers_of_2_in(n):
    >         ...
    >
    > def powers_of_2_in(n):
    >      return (n ^ (n - 1)).bit_count() - 1
    >

    Great solution, unfortunately the return value is not a tuple as in the
    OP version. Maybe in this way?

    def powers_of_2_inB(n):
        bc = (n ^ (n - 1)).bit_count() - 1
        return bc, int(n / (1 << bc))

Good point.  I overlooked that.  I should have written:

def powers_of_2_in(n):
     bc = (n ^ (n - 1)).bit_count() - 1
     return bc, n >> bc

That's pretty fancy and likely the fastest.

I was pretty happy with a recursive version.  If I'm not mistaken,
nobody has offered a recursive version so far.  It's my favorite
actually because it seems to be the clearest one.

--8<---------------cut here---------------start------------->8---
def powers_of_2_in(n):
   if remainder(n, 2) != 0:
     return 0, n
   else:
     s, r = powers_of_2_in(n // 2)
     return 1 + s, r
--8<---------------cut here---------------end--------------->8---

for n = 0 your function get stack overflow
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