【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:)