Faster Recursive Fibonacci Numbers

[rusi]

On May 17, 8:50 pm, RJB <rbott... at csusb.edu> wrote:
> I noticed some discussion of recursion..... the trick is to find a
> formula where the arguments are divided, not decremented.
> I've had a "divide-and-conquer" recursion for the Fibonacci numbers
> for a couple of years in C++ but just for fun rewrote it
> in Python.  It was easy.  Enjoy.  And tell me how I can improve it!
>
> def fibo(n):
>         """A Faster recursive Fibonaci function
> Use a formula from Knuth Vol 1 page 80, section 1.2.8:
>            If F[n] is the n'th Fibonaci number then
>                    F[n+m] = F[m]*F[n+1] + F[m-1]*F[n].
>   First set m = n+1
>    F[ 2*n+1 ] = F[n+1]**2 + F[n]*2.
>
>   Then put m = n in Knuth's formula,
>            F[ 2*n ] = F[n]*F[n+1] + F[n-1]* F[n],
>    and replace F[n+1] by F[n]+F[n-1],
>            F[ 2*n ] = F[n]*(F[n] + 2*F[n-1]).
> """
>         if n<=0:
>                 return 0
>         elif n<=2:
>                 return 1
>         elif n%2==0:
>                 half=n//2
>                 f1=fibo(half)
>                 f2=fibo(half-1)
>                 return f1*(f1+2*f2)
>         else:
>                 nearhalf=(n-1)//2
>                 f1=fibo(nearhalf+1)
>                 f2=fibo(nearhalf)
>                 return f1*f1 + f2*f2
>
> RJB the Lurkerhttp://www.csci.csusb.edu/dick/cs320/lab/10.html