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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Sat, 14 May 2011 15:19:55 -0600
Subject: Re: Recursion or iteration (was Fibonacci series recursion error)
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:5388

On Sat, May 14, 2011 at 11:24 AM, rusi <rustompmody@gmail.com> wrote:
> def fib(n):
> =A0 =A0if n=3D=3D1 or n=3D=3D2:
> =A0 =A0 =A0 =A0return 1
> =A0 =A0elif even(n):
> =A0 =A0 =A0 =A0return sq(fib (n//2)) + 2 * fib(n//2) * fib(n//2 - 1)
> =A0 =A0else:
> =A0 =A0 =A0 =A0return sq(fib (n//2 + 1)) + sq(fib(n // 2))
>
> This is a strange algo =A0-- logarithmic because it halves the n,
> exponential because of the double (triple) calls. =A0[I cannot say I
> know how to work out its exact complexity but I would guess its about
> linear]

Yup, linear.  Assuming you optimize the even case so that it doesn't
actually call fib(n//2) twice, the call tree can be approximated as a
balanced binary tree with height log(n).  The total number of nodes in
the tree is thus O(2 ** log(n)) =3D O(n).