Exercise 6 of Chapter 12 (Tuples) of Think Python by Allen Dwney

Question

I'm learning python from Think Python by Allen Downey and I'm stuck at Exercise 6 here. I wrote a solution to it, and at first look it seemed to be an improvement over the answer given here. But upon running both, I found that my solution took a whole day (~22 hours) to compute the answer, while the author's solution only took a couple seconds. Could anyone tell me how the author's solution is so fast, when it iterates over a dictionary containing 113,812 words and applies a recursive function to each to compute a result?

My solution:

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...

Presumably, this takes a while.

However, you regenerate this same, unchanging dictionary many times for every word you try to reduce. What a waste! If you make this dictionary once, and then re-use that dictionary for every word you try to reduce like you do for known_red, the computation time should be greatly reduced.