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Chapter 9  Case study: word play

9.1  Reading word lists

For the exercises in this chapter we need a list of English words. There are lots of word lists available on the Web, but the one most suitable for our purpose is one of the word lists collected and contributed to the public domain by Grady Ward as part of the Moby lexicon project¹. It is a list of 113,809 official crosswords; that is, words that are considered valid in crossword puzzles and other word games. In the Moby collection, the filename is 113809of.fic; I include a copy of this file, with the simpler name words.txt, along with Swampy.

This file is in plain text, so you can open it with a text editor, but you can also read it from Python. The built-in function open takes the name of the file as a parameter and returns a file object you can use to read the file.

>>> fin = open('words.txt')
>>> print fin
<open file 'words.txt', mode 'r' at 0xb7f4b380>

fin is a common name for a file object used for input. Mode 'r' indicates that this file is open for reading (as opposed to 'w' for writing).

The file object provides several methods for reading, including readline, which reads characters from the file until it gets to a newline and returns the result as a string:

>>> fin.readline()
'aa\r\n'

The first word in this particular list is “aa,” which is a kind of lava. The sequence \r\n represents two whitespace characters, a carriage return and a newline, that separate this word from the next.

The file object keeps track of where it is in the file, so if you call readline again, you get the next word:

>>> fin.readline()
'aah\r\n'

The next word is “aah,” which is a perfectly legitimate word, so stop looking at me like that. Or, if it’s the whitespace that’s bothering you, we can get rid of it with the string method strip:

>>> line = fin.readline()
>>> word = line.strip()
>>> print word
aahed

You can also use a file object as part of a for loop. This program reads words.txt and prints each word, one per line:

fin = open('words.txt')
for line in fin:
    word = line.strip()
    print word

9.2  Exercises

There are solutions to these exercises in the next section. You should at least attempt each one before you read the solutions.

9.3  Search

All of the exercises in the previous section have something in common; they can be solved with the search pattern we saw in Section 8.6. The simplest example is:

def has_no_e(word):
    for letter in word:
        if letter == 'e':
            return False
    return True

The for loop traverses the characters in word. If we find the letter “e”, we can immediately return False; otherwise we have to go to the next letter. If we exit the loop normally, that means we didn’t find an “e”, so we return True.

avoids is a more general version of has_no_e but it has the same structure:

def avoids(word, forbidden):
    for letter in word:
        if letter in forbidden:
            return False
    return True

We can return False as soon as we find a forbidden letter; if we get to the end of the loop, we return True.

uses_only is similar except that the sense of the condition is reversed:

def uses_only(word, available):
    for letter in word: 
        if letter not in available:
            return False
    return True

Instead of a list of forbidden letters, we have a list of available letters. If we find a letter in word that is not in available, we can return False.

uses_all is similar except that we reverse the role of the word and the string of letters:

def uses_all(word, required):
    for letter in required: 
        if letter not in word:
            return False
    return True

Instead of traversing the letters in word, the loop traverses the required letters. If any of the required letters do not appear in the word, we can return False.

If you were really thinking like a computer scientist, you would have recognized that uses_all was an instance of a previously-solved problem, and you would have written:

def uses_all(word, required):
    return uses_only(required, word)

This is an example of a program development method called problem recognition, which means that you recognize the problem you are working on as an instance of a previously-solved problem, and apply a previously-developed solution.

9.4  Looping with indices

I wrote the functions in the previous section with for loops because I only needed the characters in the strings; I didn’t have to do anything with the indices.

For is_abecedarian we have to compare adjacent letters, which is a little tricky with a for loop:

def is_abecedarian(word):
    previous = word[0]
    for c in word:
        if c < previous:
            return False
        previous = c
    return True

An alternative is to use recursion:

def is_abecedarian(word):
    if len(word) <= 1:
        return True
    if word[0] > word[1]:
        return False
    return is_abecedarian(word[1:])

Another option is to use a while loop:

def is_abecedarian(word):
    i = 0
    while i < len(word)-1:
        if word[i+1] < word[i]:
            return False
        i = i+1
    return True

The loop starts at i=0 and ends when i=len(word)-1. Each time through the loop, it compares the ith character (which you can think of as the current character) to the i+1th character (which you can think of as the next).

If the next character is less than (alphabetically before) the current one, then we have discovered a break in the abecedarian trend, and we return False.

If we get to the end of the loop without finding a fault, then the word passes the test. To convince yourself that the loop ends correctly, consider an example like 'flossy'. The length of the word is 6, so the last time the loop runs is when i is 4, which is the index of the second-to-last character. On the last iteration, it compares the second-to-last character to the last, which is what we want.

Here is a version of is_palindrome (see Exercise 6.6) that uses two indices; one starts at the beginning and goes up; the other starts at the end and goes down.

def is_palindrome(word):
    i = 0
    j = len(word)-1

    while i<j:
        if word[i] != word[j]:
            return False
        i = i+1
        j = j-1

    return True

Or, if you noticed that this is an instance of a previously-solved problem, you might have written:

def is_palindrome(word):
    return is_reverse(word, word)

Assuming you did Exercise 8.8.

9.5  Debugging

Testing programs is hard. The functions in this chapter are relatively easy to test because you can check the results by hand. Even so, it is somewhere between difficult and impossible to choose a set of words that test for all possible errors.

Taking has_no_e as an example, there are two obvious cases to check: words that have an ’e’ should return False; words that don’t should return True. You should have no trouble coming up with one of each.

Within each case, there are some less obvious subcases. Among the words that have an “e,” you should test words with an “e” at the beginning, the end, and somewhere in the middle. You should test long words, short words, and very short words, like the empty string. The empty string is an example of a special case, which is one of the non-obvious cases where errors often lurk.

In addition to the test cases you generate, you can also test your program with a word list like words.txt. By scanning the output, you might be able to catch errors, but be careful: you might catch one kind of error (words that should not be included, but are) and not another (words that should be included, but aren’t).

In general, testing can help you find bugs, but it is not easy to generate a good set of test cases, and even if you do, you can’t be sure your program is correct.

According to a legendary computer scientist:

Program testing can be used to show the presence of bugs, but never to show their absence!

— Edsger W. Dijkstra

9.6  Glossary

file object:

A value that represents an open file.

problem recognition:

A way of solving a problem by expressing it as an instance of a previously-solved problem.

special case:

A test case that is atypical or non-obvious (and less likely to be handled correctly).

9.7  Exercises

1

wikipedia.org/wiki/Moby_Project.

2

www.cartalk.com/content/puzzler/transcripts/200725.

3

www.cartalk.com/content/puzzler/transcripts/200803.

4

www.cartalk.com/content/puzzler/transcripts/200813

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