On Tue, May 17, 2011 at 9:36 AM, rusi <rustompm...@gmail.com> wrote:
> 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
>
> -------------------------------------------------------------
> Its an interesting problem and you are 75% there.
> You see the halving gives you logarithmic behavior and the double
> calls give exponential behavior.
>
> So how to get rid of double calls? Its quite simple: Just define your
> function in terms of return pairs of adjacent pairs ie (fib(n), fib(n
> +1)) for some n rather then a single number fib(n)
>
> Here's a straightforward linear function:
>
> def fp(n): #fibpair
> if n==1:
> return (1,1)
> else:
> a,b = fp(n-1)
> return (b, a+b)
>
> def fib(n):
> a,b = fp(n)
> return a
>
> ---------------
> Now use this (pairing) idea with your (halving) identities and you
> should get a logarithmic algo.
>
> [If you cant do it ask again but yes its fun to work out so do
> try :-) ]
> --
> http://mail.python.org/mailman/listinfo/python-list
>
or O(1):
φ = (1 + sqrt(5)) / 2
def fib(n):
numerator = (φ**n) - (1 - φ)**n
denominator = sqrt(5)
return round(numerator/denominator)
Testing indicates that it's faster somewhere around 7 or so.
Geremy Condra
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