On May 17, 8:50 pm, RJB <rbott...@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
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