Chapter 12 Exercise 6 (Tuples) Think Python Allen Days

I am learning python from Think Python from Allen Downey and I am stuck in Exercise 6 here . I wrote a solution for her, and at first it looked like an improvement over the answer here . But when I started both of them, I found that my solution took a whole day (~ 22 hours) to calculate the answer, while the author's solution took only a couple of seconds. Can someone tell me how the author’s solution is so fast when iterating over a dictionary containing 113,812 words and applying a recursive function for each to calculate the result?

My decision:

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]