Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
628 views
in Technique[技术] by (71.8m points)

Python 3 Recursion - Maximum Depth Exceeded

I'm writing a basic recursion function that takes an integer, n, and returns the sum of the first n reciprocals. Inputting 2 should result in 1.5, and inputting 0 should return 0.

sum_to(2) = (1 + 1/2) = 1.5

Here's what I have:

def sum_to(n):
    if n>0:
        return sum_to(1+1/n) # Not sure if this is correct
    return 0

But all I'm getting is a Maximum recursion depth exceeded. I know I could lists to solve this, but recursion is really intriguing and I want to find a solution using it.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

From the question, it seems that you want to compute

sum_{i=1}^{n} 1/i

If the input is 0, then returns 0.

Remember that in a recursive function, you need to define a base case and an inductive clause.

In your question, the inductive clause is:

1.0 / i + sum_of_reciprocals(i-1)

while the base case can be set to 0 when i=0.

Given this formulation, the function with an example looks like:

def sum_to(n):
    if n > 0:
        return sum_to(n-1) + 1.0 / n
    else:
        return 0

if __name__ == '__main__':
    print(sum_to(3))

and the result is 1.83333333333.

For more information about recursive functions, you can start from the related wikipedia page.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...