RecursionError：超出最大递归深度

Posted 2023-03-12

【中文标题】RecursionError：超出最大递归深度

【英文标题】：RecursionError: maximum recursion depth exceeded

【发布时间】：2017-03-24 18:08:33

【问题描述】：

我正在构建一个 Python 脚本来解决黎曼悖论。这个想法是取一个输入的数字，然后将 1/2,1/3,1/4,1/5... 与 1 相加或减去，直到您输入数字为止。由于每次进行比较时回调我的比较并添加/减去类，我遇到了递归错误。

我的代码如下：

import math
#declare values of variables before they are referenced
goal = float(input("What number are you trying to achieve?"))
current_number = float(1)
negatives = (2)
positives = (3)
#-----------------------------------------------------
def add(goal,current_number,positives,negatives): #define the add operation for when the current number is less than the goal
    current_number = current_number+(1/positives) #add current fraction to the current number
    positives = positives+2 #Set the next additional fraction
    comparison(goal,current_number,positives,negatives) #go to comparision of the current number to the goal

def subtract(goal,current_number,positives,negatives): #define the subtract operation for when the current number is greater than the goal
    current_number = current_number-(1/negatives) #subtract current fraction from the current number
    negatives = negatives+2 #set the next subtractional fraction
    comparison(goal,current_number,positives,negatives) #go to comparision of the current number to the goal

def comparison(goal,current_number,positives,negatives): #define comparison between the current number to the goal
    if current_number < goal:
        add(goal,current_number,positives,negatives) #if the current number is less than the goal, go to add the next fraction to it
    if current_number > goal:
        subtract(goal,current_number,positives,negatives) #if the current number is greater than the goal, do to subtract the next fraction from it
    if current_number == goal:
        print("The equation for your number is 1+1/3...")
        print("+1/",positives)
        print("...-1/2...")
        print("-1/",negatives)
        #if the current number is equal to the goal, print the results

comparison(goal,current_number,positives,negatives) #do the comparison between the current number and the goal

我想知道我能做些什么来解决这个问题。

【问题讨论】：

程序可能永远不会从 'comparison()' 中的第一个 if 语句返回。您不是返回一个值，而是递归回原始函数。从加/减函数返回值。不要回忆调用它的函数。

@fjorn1 这是一个有趣的悖论，起初我对它的工作原理感到困惑，但现在我想我已经设法修复了代码:)

【参考方案1】：

加减法不需要函数，因为加负数与减法相同。

我还修复了您的代码，调用了一个函数，该函数调用了您调用它的函数是导致您的递归错误的原因。

希望对你有帮助:)

import math
goal = float(input("What number are you trying to achieve?"))

current_number = 0
fraction = 2
#-----------------------------------------------------
def change(goal, current_number, fraction, sign = 1): #define the add operation for when the current number is less than the goal
    global positives
    current_number = current_number + (1/fraction) * sign # add or take away current fraction to the current number
    return current_number

def comparison(goal, current_number, fraction): #define comparison between the current number to the goal
    print("The equation for your number is:")
    print("1/2")
    while round(current_number, 3) != round(goal, 3): # you don't have to round if you don't want to
        fraction = fraction + 1
        if current_number < goal:
            print("+1/"+str(fraction)) # positive
            current_number = change(goal, current_number, fraction) #if the current number is less than the goal, go to add the next fraction to it
        elif current_number > goal:
            print("-1/"+str(fraction)) # positive
            current_number = change(goal, current_number, fraction, -1)
    print("...")

comparison(goal,current_number, fraction) #do the comparison between the current number and the goal

【讨论】：

总和 1 + 1/2 + 1/3 + ... 是无限的。你说他们越来越接近 2 是错误的。

谢谢！我会花一些时间来解决这个问题......我最近才重新开始使用 Python，你会惊讶于几年后你忘记了多少。

【参考方案2】：

在current_number == goal 中，您正在比较浮点数并且使用== 通常不是the right way to do it。

您可能想increase the maximum recursion depth。

将print(current_number) 添加为comparison() 函数的第一行，看看您的current_number 如何收敛到目标/偏离目标。

【讨论】：

…或者您可能想要使用列表并将代码重写为迭代而不是递归。