當猜測中的字符重復太多次時，為什么我的 wordle 邏輯會給出錯誤的結果？

Question

遵循 wordle 的原則，檢查每個猜測並給出五個字母 output。 G代表綠色代表相同且位置正確的字母，黃色代表相同位置錯誤的字母，B代表不匹配。

def output_guess(guess:str, word: str, dictionary: list):
    """
        Checks guess for game errors, and outputs current "score" to user
        parameters: user's guess, word from dictionary, dictionary
        return: none
    """
    output = ""
    if len(guess) == 5 and guess in dictionary:
        for i in range(len(word)):
            if guess[i] == word[i]:
                output += "G"
            elif guess[i] in word:
                output += "Y"
            else:
                output += "B"
    else:
        print("Word not found in dictionary. Enter a five-letter english word.", end = "")
    print(output)

例如，如果答案是五個字母的單詞 eerie，我希望收到以下猜測的這些輸出：

Guess 1: epees  GBYYB
Guess 2: peeve  BGYBG
Guess 3: order  BYBYB
Guess 4: eerie  GGGGG

除了猜測 3，我對代碼的所有猜測都收到正確的 output。對於猜測 3：順序，我得到 BYBYY 而不是 BYBYB。

Answer 1

你的邏輯不正確。 僅僅檢查單詞中是否存在該字符是不夠的，您還需要計算輸入中的字符並確保單詞中的字符數相等或更小

import collections

word = "eerie"
guess = "order"

word_counts = collections.Counter(word)     # Count characters in the word
guess_counts = collections.defaultdict(int) # Create an empty dict to count characters in guess
output = []
for g, w in zip(guess, word):
    guess_counts[g] += 1 # Add to count of current character
    if g == w:
        # Correct character, correct position
        output += "G"
    elif guess_counts[g] <= word_counts[g]:
        # The minimum that `guess_counts[g]` can be is 1 
        # Why? We set it in the first line of this loop
        # For this condition to be true, word_counts[g] must be > 0
        #      (i.e `word` must contain `g`)
        # And we must have seen `g` fewer (or equal) times than
        # the word contains `g`
        output += "Y"
    else:
        output += "B"
        
print(output)

我使用collections.Counter而不是手動計算字母，和collections.defaultdict(int)所以我不需要在遞增之前檢查字典是否包含字符。