Question

硬币翻转模拟存在问题。该代码应计算出一次掷硬币并连续三次获得尾巴所需的平均翻转次数（因此，成功= 3't'，其中1次成功满足了第一个实验）。

import random

experiments = [1, 10, 100, 1000, 10000, 100000]

for number in experiments:
    experiment = []
    success = 0
    while success < number:
        face = random.choice(['h','t'])
        experiment.append(face)
        success = ''.join(experiment).count('ttt')
    print(f'Experiments: {number}')
    print(f'Average flips: {len(experiment)/success}\n')

输出看起来像这样：

[evaluate troubleshooting.py]
Experiments: 1
Average flips: 27.0

Experiments: 10
Average flips: 6.4

Experiments: 100
Average flips: 14.39

Experiments: 1000

Answer 1

好吧，我认为主要的问题是success = ''.join(experiment).count('ttt')行将导致您的程序在while循环中的每次运行中搜索整个列表，这将导致O（n ^ 2）时间复杂度（又名：，BAD。如果您进行更多的实验，真的很糟糕。

我提出了一个（基本的）程序，它将在线性（O（n））时间中发挥作用：

import random

experiment_size = [1,10,100,1000,10000,100000]

for number in experiment_size:
    last = False # track whether the last 3 flips were tails. False = don't care, 
                 # True = tails
    last_2 = False
    last_3 = False

    success = 0
    runs = 0
    while success < number:
        runs += 1
        last_3 = last_2 #bump the 3rd from last flip
        last_2 = last
        last = bool(random.getrandbits(1)) #get True or False, randomly
        if(last & last_2 & last_3): #if all are tails
            success += 1
            last, last_2, last_3 = False, False, False #reset so that you have to get 3
                                                       #in a row again
    print("Runs: " + str(runs) + "\nSuccesses: " + str(success) +"\nAverage: " + str(runs/success) + "\n")

因为我们不会在每次重复中遍历整个列表，所以速度要快得多。

使用列表理解和随机性模拟掷硬币，代码首先起作用，然后挂起

1 个答案: