Question

所以我一直试图找出一种方法来编写一个程序，计算机试图猜测我正在考虑的数字，而不是猜测计算机所选数字的另一种方式。它在大多数情况下工作，但是在某些情况下，它确实在链中重复数字，即使我之前已经告诉它，例如我想的值高于'7'。在某些情况下，即使我告诉它更高或更低，它也会重复相同的数字。如果有经验的人可以看看这个，并告诉我在这些循环中我缺少什么，这将有很大的帮助。

#computer enters a value x
#lower - computer guesses lower than x
#higher - computer guesses higher than x
#when string "You got it!" - game over

import random

lowBound = 0
highBound = 100
randomNumber = random.randint(lowBound,highBound)

print ("Is it ", randomNumber, " ?")
response = input()

while response != "You got it!":
    if response == "higher":
        lowBound = randomNumber    
        randomNumber = random.randint (lowBound, highBound)
        print ("Is it ", randomNumber, " ?")
        response = input()

    elif response == "lower":
        highBound = randomNumber
        randomNumber = random.randint (lowBound, highBound)
        print ("Is it ", randomNumber, " ?")
        response = input()

    if response == "You got it!":

        print ("Woohooo, I'm so bitchin'")

Answer 1

random.randint具有包容性，因此：

if response == 'higher':
    lowBound = randomNumber + 1

和

if response == 'lower':
    highBound = randomNumber - 1

此外，如果用户没有输入有效的回复，则永远不会再次调用input()，程序将在无限循环中挂起。

更强大，但不处理骗子：

import random

lowBound = 0
highBound = 100
response = ''
randomNumber = random.randint(lowBound,highBound)

while response != "yes":
    print ("Is it ", randomNumber, " ?")
    response = input()
    if response == "higher":
        lowBound = randomNumber + 1   
        randomNumber = random.randint(lowBound,highBound)
    elif response == "lower":
        highBound = randomNumber - 1
        randomNumber = random.randint(lowBound,highBound)
    elif response == "yes":
        print ("Woohooo, I'm so bitchin'")
        break
    else:
        print ('Huh? "higher", "lower", or "yes" are valid responses.')

Answer 2

random.randint(a, b)会在a和b之间返回一个数字。生成新的随机数时，您应该使用random.randint(lowBound+1, highBound-1)

Answer 3

你提到的其中一个问题就是这些问题：

highBound = randomNumber
randomNumber = random.randint (lowBound, highBound)

你正在设置一个新的界限，这很好，但是你要选择另一个随机数！

你应该做的是将界限减半，并从那里要求用户更高或更低。看看二进制搜索算法。

highBound = randomNumber
randomNumber = randomNumber / 2

您的程序仍然可以正常工作（此处提及的其他更改），但这会在大多数情况下更快地猜测您的号码。

实际上有an example of this game on Wikipedia.

Answer 4

这是Michael Dawson的书中我练习的这个练习版本，我试图尽量减少计算机使用的尝试次数。我知道代码看起来很狡猾，这只是我的第二天：）

answer=""
guess=50
counter=3
x=25

print("hi, guess the number from 1 too 100")
input("\n")

print ("i will try to guess it")
print ("is it ", guess, "?")

while answer not in ("y","l","s"):
    print ("sorry, i didn't understand \n")
    answer=input("type in: (Y) for yes, or (L) if it is to large, or (S) if it is to small:")

if answer in ("s","l"):
    while answer!="y":

        if answer=="l":
            guess=int(guess-x)
            print ("how about", guess,"?")
            answer=input("\nis it? type in: (Y) for yes, or (L) if it is to large, or (S) if it is to small:")
            x=100/2**counter
            counter=counter+1
            if x<1:
                x=1

        elif answer=="s":
            guess=int(guess+x)
            print ("how about", guess,"?")
            answer=input("\nis it? type in: (Y) for yes, or (L) if it is to large, or (S) if it is to small:")
            x=100/2**counter
            counter=counter+1
            if x<1:
                x=1

        elif answer=="y":
            break
else:
    pass

print("\ngreat! the number that you guessed is", guess)
print("i can read your mind with no guesses!")
input("\n")

Answer 5

你得到两次数字，因为random.randint的边界是包容性的; random.randint(1, 3)可以返回1,2或3.请注意，如果响应既不是“更高”，也不是“更低”，也不是“你得到它”，你还应该继续询问人：

import random
lowBound = 0
highBound = 100

while True:
    randomNumber = random.randint(lowBound, highBound)
    print ("Is it ", randomNumber, " ?")
    response = input()

    if response == "higher":
        lowBound = randomNumber + 1
    elif response == "lower":
        highBound = randomNumber - 1

    if response == "You got it!":
        print ("Woohooo, I'm so bitchin'")
        break

PYTHON - 反向数字猜测游戏 -

5 个答案: