本文介绍了封闭形式斐波那契系列的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我使用Python使用以下公式创建斐波纳契:
I am using Python to create a Fibonacci using this formula:
我有这个递归的Fibonacci函数:
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))
要显示它我使用的是:
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)])
我如何使用此Fibonacci的迭代函数?
How would I use an iterative function with this Fibonacci?
我试过这个:
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
然而,这个迭代函数似乎没有与递归函数相同的输出。
However this iterative function doesn't seem to have the same output as the recursive function.
递归函数的输出:
Recursive Fibonacci Sequence: [0, 2, 2, 4, 6, 10, 16, 26, 42, 68]
输出迭代函数:
Iterative Fibonacci Sequence: [0, 2, 2, 2, 3, 6, 13, 27, 58, 122]
推荐答案
等式你我试着去做plement是 Fibonacci系列。
The equation you're trying to implement is the closed form Fibonacci series.
封闭形式意味着评估是一个恒定的时间操作。
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)
对比,
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
我们有
>>> [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]
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