Closed form Fibonacci Series

问题:

I am using Python to create a Fibonacci using this formula:

I have this recursive Fibonacci function:

def recursive_fibonacci(n):
if n <= 1:
    return int((((1 / (5 ** 0.5)) * (1 + (5 ** 0.5))) ** n) - (((1 / (5 ** 0.5)) * (1 - (5 ** 0.5))) ** n))
else:
    return(recursive_fibonacci(n - 1) + recursive_fibonacci(n - 2))

To display it I am using this:

nterms = 10

if nterms <= 0:
    print("Please Enter a positive integer")
else:
    print("Recursive Fibonacci Sequence: " ,
        [recursive_fibonacci(i) for i in range(nterms)])
    print("Iterative Fibonacci Sequence: " ,
        [iterative_fib(i) for i in range(nterms)])

How would I use an iterative function with this Fibonacci?

I've tried using this:

def iterative_fib(n):
    equation = lambda n: int((((1 / (5 ** 0.5)) * (1 + (5 ** 0.5))) ** n) - (((1 / (5 ** 0.5)) * (1 - (5 ** 0.5))) ** n))
    if n <= 1:
        return equation(n)
    else:
        a, b = 1, 2
        for i in range(n):
            fn = equation((i-a)+(i-b))
        return fn

However this iterative function doesn't seem to have the same output as the recursive function.

The output of the recursive function:

Recursive Fibonacci Sequence:  [0, 2, 2, 4, 6, 10, 16, 26, 42, 68]

The output of the iterative function:

Iterative Fibonacci Sequence:  [0, 2, 2, 2, 3, 6, 13, 27, 58, 122]

回答1:

The equation you're trying to implement is the closed form Fibonacci series.

Closed form means that evaluation is a constant time operation.

g = (1 + 5**.5) / 2  # Golden ratio.
def fib(N):
    return int((g**N - (1-g)**N) / 5**.5)

Contrast with,

def fib_iterative(N):
    a, b, i = 0, 1, 2
    yield from (a, b)
    while i < N:
        a, b = b, a + b 
        yield b
        i += 1

And we have

>>> [fib(n) for n in range(10)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
>>> list(fib_iterative(10))
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

回答2:

I think you've misunderstood the expression f_n for the Fibonacci sequence that you mention.

Notice that it is not a recurrence relation. It is a function of n, i.e., it provides the value of the n-th term when given an n.

Hence, you really don't have a recursive/iterative solution to generate the entire Fibonnaci sequence here.

Plugging in n as 0, 1, 2, 3.. provides the terms 0, 1, 1, 2, .. of the series.

To illustrate, when n = 3, f_3 is calculated as -