Allen Downey 的 Think Python 第 12 章（元组）的练习 6

Question

我正在向 Allen Downey 的 Think Python 学习 Python，但我被困在练习 6 here。我为它写了一个解决方案，乍一看，它似乎比here 给出的答案有所改进。但是在运行两者后，我发现我的解决方案需要一整天（约 22 小时）来计算答案，而作者的解决方案只需要几秒钟。 谁能告诉我作者的解决方案如何如此之快，当它迭代包含 113,812 个单词的字典并对每个单词应用递归函数以计算结果时？

我的解决方案：

known_red = {'sprite': 6, 'a': 1, 'i': 1, '': 0}  #Global dict of known reducible words, with their length as values

def compute_children(word):
   """Returns a list of all valid words that can be constructed from the word by removing one letter from the word"""
    from dict_exercises import words_dict
    wdict = words_dict() #Builds a dictionary containing all valid English words as keys
    wdict['i'] = 'i'
    wdict['a'] = 'a'
    wdict[''] = ''
    res = []

    for i in range(len(word)):
        child = word[:i] + word[i+1:]
        if nword in wdict:
            res.append(nword)

    return res

def is_reducible(word):
    """Returns true if a word is reducible to ''. Recursively, a word is reducible if any of its children are reducible"""
    if word in known_red:
        return True
    children = compute_children(word)

    for child in children:
        if is_reducible(child):
            known_red[word] = len(word)
            return True
    return False

def longest_reducible():
    """Finds the longest reducible word in the dictionary"""
    from dict_exercises import words_dict
    wdict = words_dict()
    reducibles = []

    for word in wdict:
        if 'i' in word or 'a' in word: #Word can only be reducible if it is reducible to either 'I' or 'a', since they are the only one-letter words possible
            if word not in known_red and is_reducible(word):
                known_red[word] = len(word)

    for word, length in known_red.items():
        reducibles.append((length, word))

    reducibles.sort(reverse=True)

    return reducibles[0][1]

Answer 1

wdict = words_dict() #Builds a dictionary containing all valid English words...

大概，这需要一段时间。

但是，您会为尝试减少的每个单词多次重新生成相同的、不变的字典。多么浪费！如果你制作这个字典一次，然后像 known_red 那样对你尝试减少的每个单词重新使用该字典，那么计算时间应该会大大减少。