Fibonacci错误：相比之下超出了最大递归深度

Question

I am on python version 3.4.1 and when I run the code:

def fibo(n):
    if n == 1 or n ==2:
        return 1
    else:
        return fibo(n-1)+fibo(n-2)

print("input number for fibonacci seq list")
num = int(input())
for i in range(0,num):
    print(str(fibo(i)))

I would want the code to give me a list of the fibonacci numbers that the user input but I get the error mentioned in the title. I am not sure why.

Answer 1

I am not sure why.

Let me explain what is going on.

At this part:

for i in range(0,num):
    print(str(fibo(i)))

the first number i passed to fibo is 0 because range(0,num) starts at 0 . Inside fibo , 0 fails the n == 1 or n ==2 condition, so fibo(n-1) is executed. Now the number is -1 because 0 - 1 == -1 . This number also fails the n == 1 or n ==2 condition and fibo(n-1) is executed again. Now the number is -2 .

Hopefully, you see where this is going. The number is decreased indefinitely until fibo reaches Python's maximum recursion depth and thereby raises an error.

Answer 2

首先尝试计算fibo（0），这会产生无限递归 - 做'范围（1，n）'

Answer 3

Fibonacci recursion increases geometrically. It is better to use iteration or a generator.

Here's a generator:

>>> def fibgen():
...   a,b=1,1
...   while True:
...     yield a
...     a,b=b,a+b
...
>>> i = fibgen()
>>> next(i)
1
>>> next(i)
1
>>> next(i)
2
>>> next(i)
3
>>> next(i)
5
>>> i=fibgen()
>>> [next(i) for _ in range(10)]
[1, 1, 2, 3, 5, 8, 13, 21, 34, 55]