python - 猴子定理 - 选择随机字母直到生成输入字符串

Question

我想做一个类似于无限猴子定理的代码定理指出:

a monkey hitting keys at random on a typewriter keyboard for infinite amount of time will almost surely type a given text such as the complete works of Shakespeare.

好吧，假设我们用Python替换一只猴子来生成莎士比亚的一句话？并计算试验

例如:“我认为它就像一只黄鼠狼”

这是我写的:

def Monkey_Theorem(sentence):
    sentence.lower()
    string_letters = 'abcdefghijklmnopqrstuvwxyz'
    import random
    a = []
    l = ''
    x = 0
    trials = 0
    while x < (len(sentence) - 1):
        trials += 1
        z = random.choice(string_letters)
        if z == sentence[x]:
            a.append(z)
            x += 1
        elif z == " ":
            trials -= 1
            a.append(" ")
        if x > (len(sentence) - 1):
            l = ''.join(l)
            break
    return l and trials

我将其写入 python 2.7 IDLE ，按 Enter 键然后写道:

Monkey_Theorem("methinks it is like a weasel")

之后我按了回车键..它移到了下一行并且没有出现 (>>>)

一段时间后没有结果，当我试图关闭 IDLE 时脸上露出绝望，它说程序仍在运行

问题是:

(A) while 循环是无限循环吗？？
(B) 或者我的 IDLE 很慢？
(C) 其他原因
以及如何修复原因

还有一个请求以后遇到这种情况怎么判断是(A)、(B)、(C)？

Answer 1

您可以在循环中打印一些内容来查看它是否是无限的...例如，如果 z 永远不等于 sentence[x]，那么您不会增加它，并且循环永远不会停止。

当您需要猜测空格字符时就会发生这种情况，因为您没有将空格字符放入字符串中。因此，即使 elif z == "": 也永远不会被输入。

这是您的代码的工作版本。

需要注意的一些事情:跟踪 a 或 l 是没有意义的。在函数结束时，您将返回与 sentence 完全相同的输入。此外，还有一个简单的导入即可获取所有英文字母。

from string import ascii_lowercase as alpha
import random

# add space character to possible choices
alpha += ' '

def Monkey_Theorem(sentence):
    sentence = sentence.lower()
    x = 0
    trials = 0
    while x < len(sentence):
        trials += 1
        z = random.choice(alpha)
        if z == sentence[x]:
            if z == " ": # this isn't really necessary
                trials -= 1
            print "guessed %s at trial %d" % (z, trials)
            x += 1
    return trials

trials = Monkey_Theorem("hello world")
print "number of trials = ", trials

输出

guessed h at trial 49
guessed e at trial 52
guessed l at trial 58
guessed l at trial 107
guessed o at trial 145
guessed   at trial 157
guessed w at trial 168
guessed o at trial 197
guessed r at trial 232
guessed l at trial 248
guessed d at trial 269
number of trials =  269