Wordle求解器Python脚本重复字母崩溃问题技术问询

问题描述 #

当猜词存在重复字母时，若其中一个字母实例被标记为正确位置（Y），但另一个相同字母实例被标记为不存在（N），脚本会崩溃。例如目标词是APPLE，猜词为APPLY，此时前两个P是正确位置（Y），第三个P标记为N（因为目标词中该位置没有P，且已确认存在两个P），原脚本会直接移除所有含P的单词，导致候选词集合为空，最终调用pop()时触发错误。

原代码问题分析 #

原代码处理N结果的逻辑存在缺陷：

tempWordSet = {word for word in tempWordSet if currentWord[resultIndex] not in word}

该逻辑直接移除所有包含当前字母的单词，完全忽略了一种关键场景：当同一个字母已经有Y或W标记时，说明目标词中确实存在该字母，只是当前这个实例是多余的。此时应该保留那些包含该字母、但出现次数不超过已确认数量的单词，而非一刀切移除所有含该字母的单词。

修复方案 #

通过分阶段处理结果标记，结合字母出现次数统计，实现精准的候选词过滤：

1. 先处理Y标记，锁定单词的正确位置，确保候选词必须满足对应位置的字母要求
2. 处理W标记，收集需要存在但位置错误的字母，过滤掉不包含该字母或字母位置正确的单词
3. 处理N标记时，分两种情况：
   • 若该字母无Y/W标记：移除所有含该字母的单词
   • 若该字母已有Y/W标记：移除那些该字母出现次数超过已确认数量的单词

修复后的完整代码 #

from english_words import get_english_words_set
web2lowerset = get_english_words_set(['web2'], lower=True)
import random
from os import system, name
from collections import Counter

def clear_screen():
    if name == 'nt':
        _ = system('cls')
    else:
        _ = system('clear')
            
board = [
[" ", " ", " ", " ", " "],
[" ", " ", " ", " ", " "],
[" ", " ", " ", " ", " "],
[" ", " ", " ", " ", " "],
[" ", " ", " ", " ", " "],
[" ", " ", " ", " ", " "],
]

startingWords = ["SOARE", "SAREE", "SEARE", "STARE", "ROATE"]

def print_board():
    clear_screen()
    for r in board:
        print(*r)
    
def validInput(w):
    return len(w) == 5
 
def solver():
    fiveLetterWords = set()

    for word in web2lowerset:
        if len(word) == 5:
            fiveLetterWords.add(word.upper())
            
    initialInput = input("Would you like me to provide a starting word?")
    startingWord = None
    if "y" in initialInput.lower():
        startingWord = random.choice(startingWords)
        print(startingWord)
    elif "n" in initialInput.lower():
        while True:
            userWord = input("What was your starting word?")
            if validInput(userWord):
                startingWord = userWord.upper()
                break
            else:
                print("Invalid word length")
        print(startingWord)
    else:
        print("Invalid answer")
        return
        
    # 初始化第一行棋盘
    for i in range(5):
        board[0][i] = startingWord[i]
    
    moveCount = 0   
    currentWord = list(startingWord)
    while True:
        print_board()
        print("Enter 'G' to generate a new word \nUse Y to represent present letters in the right spot, \nN to represent absent letters, \nand W to represent present letters in the wrong spot")
        print("E.G. NNYWW")
        results = input().strip().upper()
        
        # 输入合法性检查
        if len(results) != 5 and results != "G":
            print("Invalid input, please enter 5 characters or 'G'")
            continue
        
        # 处理生成新单词的请求
        if results == "G":
            if not fiveLetterWords:
                print("No more words available!")
                break
            currentWord = list(fiveLetterWords.pop())
            moveCount -= 1
            for i in range(5):
                board[moveCount+1][i] = currentWord[i]
            continue
            
        currentResult = list(results)
        tempWordSet = fiveLetterWords.copy()
        
        # 统计已确认的字母信息
        correct_positions = {}  # 存储正确位置的字母：{索引: 字母}
        present_letter_counts = Counter()  # 存储已确认存在的字母及次数
        
        # 第一步：过滤正确位置（Y标记）
        for idx, res in enumerate(currentResult):
            if res == "Y":
                char = currentWord[idx]
                correct_positions[idx] = char
                present_letter_counts[char] += 1
        # 保留所有在对应位置有正确字母的单词
        tempWordSet = {word for word in tempWordSet if all(word[idx] == char for idx, char in correct_positions.items())}
        
        # 第二步：过滤存在但位置错误（W标记）
        for idx, res in enumerate(currentResult):
            if res == "W":
                char = currentWord[idx]
                present_letter_counts[char] += 1
                # 保留包含该字母但不在当前位置的单词
                tempWordSet = {word for word in tempWordSet if char in word and word[idx] != char}
        
        # 第三步：过滤不存在的字母（N标记）
        for idx, res in enumerate(currentResult):
            if res == "N":
                char = currentWord[idx]
                if present_letter_counts[char] > 0:
                    # 目标词中该字母的数量等于已确认的次数，过滤掉数量不符的单词
                    required_count = present_letter_counts[char]
                    tempWordSet = {word for word in tempWordSet if Counter(word)[char] == required_count}
                else:
                    # 该字母完全不存在，移除所有含该字母的单词
                    tempWordSet = {word for word in tempWordSet if char not in word}
        
        # 更新候选词集合
        fiveLetterWords = tempWordSet
        if not fiveLetterWords:
            print("No possible words left!")
            break
        
        moveCount += 1
        currentWord = list(fiveLetterWords.pop())
        # 更新棋盘
        for i in range(5):
            board[moveCount][i] = currentWord[i]

while True:
    solver()

修复说明 #

1. 引入collections.Counter简化字母出现次数的统计，更高效处理重复字母场景
2. 分阶段处理标记，优先锁定正确位置，再处理存在性，最后处理不存在的情况，逻辑更清晰
3. 针对重复字母的N标记做特殊处理，避免错误移除符合条件的目标词
4. 增加输入合法性检查和无候选词时的提示，提升脚本稳定性和用户体验

内容的提问来源于stack exchange，提问作者Cason Berry