关于MIT OCW
Python第3讲,我遇到了问题.
根据简单的数学计算,她使用的代码不应该是成功的.
## EXAMPLE: approximate cube root
####################
#cube = 27
##cube = 8120601
##cube = 10000
#epsilon = 0.1
#guess = 0.0
#increment = 0.01
#num_guesses = 0
## look for close enough answer and make sure
## didn't accidentally skip the close enough bound
#while abs(guess**3 - cube) >= epsilon and guess <= cube:
# guess += increment
# num_guesses += 1
#print('num_guesses =',num_guesses)
#if abs(guess**3 - cube) >= epsilon:
# print('Failed on cube root of',cube,"with these parameters.")
#else:
# print(guess,'is close to the cube root of',cube)
这是她使用的代码,我遇到的问题是理解这部分：
while abs(guess**3 - cube) >= epsilon and guess <= cube: # guess += increment
如果guess是0.0,cube是27并且increment是0.01那么这个术语的数学应该是：
abs(0**3 - 27) = 27 #----- This is fine according to the code but the next step would be:# abs(27.01**3 - 27) = 19677.878101
这应该阻止循环进一步工作.我的理解在某处显然是错误的,但我看不到哪里！
请停下来……
解决方法
您可以尝试使用Python可视化工具来查看猜测值如何随每次迭代而变化.
Try here.