Python：在簡單的二分搜索遊戲中意外的條件激活

Question

我寫了一些代碼來確定一個介於0和100之間的祕密數字。用戶告訴機器猜測的數字（這是該範圍的一半）也是高的低或恰到好處。基於輸入，機器使用平分搜索來調整猜測。當猜測正確時，用戶按下c鍵，遊戲結束。問題是，儘管「我不明白輸入」分支中存在條件，但當用戶按下c（一個有效條目）時，該分支會觸發，而這不是第一次猜測。

例如，這裏是輸出 -

Please think of a number between 0 and 100! 
Is your secret number 50? 
Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly. l 
Is your secret number 75? 
Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly. c 
Sorry, I did not understand your input. 
Game over. Your secret number was:75 
>>> 

這裏是代碼 -

High=100 
    Low=0 
    Guess=50 
    user_input=0 

    print('Please think of a number between 0 and 100!') 

    while user_input != 'c': 
     print("Is your secret number"+" "+str(Guess)+"?") 
     userinput = raw_input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly.") 
     if user_input == 'h': 
      High=Guess 
      Guess= ((High+Low)/2) 
     if user_input == 'l': 
      Low=Guess 
      Guess= ((High+Low)/2) 
     if user_input != 'h' or 'l' or 'c': 
      print('Sorry, I did not understand your input.') 

    print ('Game over. Your secret number was:'''+ str(Guess)) 

在此先感謝。我已經在我的頭上幾個小時... ...

Answer 1

條件不這樣工作。你需要的東西，如：

# Check each condition explicitly 
if user_input != 'h' and user_input != 'l' and user_input != 'c': 

或者：

# Check if the input is one of the elements in the given list 
if user_input not in ["h", "c", "l"]: 

您當前的做法被理解爲

if (user_input != 'h') or ('l') or ('c'): 

而且，由於l和c是truthy，該分支將永遠執行。

---

您還可以考慮使用elif，所以你的條件將成爲下面：

while True: 
    if user_input == 'h': 
     High=Guess 
     Guess= ((High-Low)/2) 
    elif user_input == 'l': 
     Low=Guess 
     Guess= ((High-Low)/2) 
    elif user_input == "c": 
     # We're done guessing. Awesome. 
     break 
    else: 
     print('Sorry, I did not understand your input.') 

Answer 2

試試這個爲條件。

if user_input not in ['h','l','c']: 
     print('Sorry, I did not understand your input.') 

你可能沒有檢查user_input是h或l因爲前幾if應該處理。

if user_input == 'h': 
     High=Guess 
     Guess= ((High-Low)/2) 
    elif user_input == 'l': 
     Low=Guess 
     Guess= ((High-Low)/2) 
    elif user_input == 'c': 
     pass # the while statement will deal with it or you could break 
    else: 
     print('Sorry, I did not understand your input.') 

Answer 3

除了您的if之外，您的邏輯有一些錯誤。我推薦這樣的：

High = 100 
Low = 1 
LastGuess = None 

print('Please think of a number between 0 and 100!') 

while True: 
    Guess = int((High+Low)/2) 
    if Guess == LastGuess: 
     break 
    print("Is your secret number"+" "+str(Guess)+"?") 
    user_input = input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly.") 
    if user_input == 'h': 
     High = Guess 
     LastGuess = Guess 
    elif user_input == 'l': 
     Low = Guess 
     LastGuess = Guess 
    elif user_input == 'c': 
     break 
    else: 
     print('Sorry, I did not understand your input.') 

print ('Game over. Your secret number was:'''+ str(Guess))