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<A NAME="htoc108">Chapter 9</A> Case study: word play

<A NAME="toc98"></A><A NAME="htoc109">9.1</A> 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 project^{<A NAME="text16" HREF="#note16">1</A>}. 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.

<FONT COLOR=blue><FONT SIZE=4>>>> fin = open('words.txt')
>>> print fin
<open file 'words.txt', mode 'r' at 0xb7f4b380>
</FONT></FONT>

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:

<FONT COLOR=blue><FONT SIZE=4>>>> fin.readline()
'aa\r\n'
</FONT></FONT>

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:

<FONT COLOR=blue><FONT SIZE=4>>>> fin.readline()
'aah\r\n'
</FONT></FONT>

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:

<FONT COLOR=blue><FONT SIZE=4>>>> line = fin.readline()
>>> word = line.strip()
>>> print word
aahed
</FONT></FONT>

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:

<FONT COLOR=blue><FONT SIZE=4>fin = open('words.txt')
for line in fin:
    word = line.strip()
    print word
</FONT></FONT>

Exercise 1

Write a program that reads words.txt and prints only the

words with more than 20 characters (not counting whitespace).

<A NAME="toc99"></A><A NAME="htoc110">9.2</A> Exercises

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

Exercise 2

In 1939 Ernest Vincent Wright published a 50,000 word novel called Gadsby that does not contain the letter “e.” Since “e” is

the most common letter in English, that’s not easy to do.

In fact, it is difficult to construct a solitary thought without using that most common symbol. It is slow going at first, but with caution and hours of training you can gradually gain facility.

All right, I’ll stop now.

Write a function called has_no_e that returns True if the given word doesn’t have the letter “e” in it.

Modify your program from the previous section to print only the words

that have no “e” and compute the percentage of the words in the list

have no “e.”

Exercise 3

Write a function named avoids that takes a word and a string of forbidden letters, and that returns True if the word doesn’t use any of the forbidden

letters.

Modify your program to prompt the user to enter a string of forbidden letters and then print the number of words that don’t contain any of them. Can you find a combination of 5 forbidden letters that excludes the smallest number of words?

Exercise 4

Write a function named uses_only that takes a word and a string of letters, and that returns True if the word contains only letters in the list. Can you make a sentence using only the letters acefhlo? Other than “Hoe alfalfa?”

Exercise 5

Write a function named uses_all that takes a word and a string of required letters, and that returns True if the word uses all the required letters at least once. How many words are there that use all the vowels aeiou? How about aeiouy?

Exercise 6

Write a function called is_abecedarian that returns True if the letters in a word appear in alphabetical order (double letters are ok). How many abecedarian words are there?

<A NAME="toc100"></A><A NAME="htoc111">9.3</A> 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 <A HREF="book009.html#find">8.6</A>. The simplest example is:

<FONT COLOR=blue><FONT SIZE=4>def has_no_e(word):
    for letter in word:
        if letter == 'e':
            return False
    return True
</FONT></FONT>

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.

You can write this function more concisely using the in

operator, but I started with this version because it

demonstrates the logic of the search pattern.

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

<FONT COLOR=blue><FONT SIZE=4>def avoids(word, forbidden):
    for letter in word:
        if letter in forbidden:
            return False
    return True
</FONT></FONT>

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:

<FONT COLOR=blue><FONT SIZE=4>def uses_only(word, available):
    for letter in word: 
        if letter not in available:
            return False
    return True
</FONT></FONT>

Instead of a list of forbidden words, we have a list of available

words. 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:

<FONT COLOR=blue><FONT SIZE=4>def uses_all(word, required):
    for letter in required: 
        if letter not in word:
            return False
    return True
</FONT></FONT>

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:

<FONT COLOR=blue><FONT SIZE=4>def uses_all(word, required):
    return uses_only(required, word)
</FONT></FONT>

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.

<A NAME="toc101"></A><A NAME="htoc112">9.4</A> 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:

<FONT COLOR=blue><FONT SIZE=4>def is_abecedarian(word):
    previous = word[0]
    for c in word:
        if c < previous:
            return False
        previous = c
    return True
</FONT></FONT>

An alternative is to use recursion:

<FONT COLOR=blue><FONT SIZE=4>def is_abecedarian(word):
    if len(word) <= 1:
        return True
    if word[0] > word[1]:
        return False
    return is_abecedarian(word[1:])
</FONT></FONT>

Another option is to use a while loop:

<FONT COLOR=blue><FONT SIZE=4>def is_abecedarian(word):
   i = 0
   while i < len(word)-1:
       if word[i+1] < word[i]:
           return False
       i = i+1
   return True

</FONT></FONT>

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 <A HREF="book007.html#palindrome">6.6</A>) that uses two indices; one starts at the

beginning and goes up; the other starts at the end and goes down.

<FONT COLOR=blue><FONT SIZE=4>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
</FONT></FONT>

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

<FONT COLOR=blue><FONT SIZE=4>def is_palindrome(word):
    return is_reverse(word, word)
</FONT></FONT>

Assuming you did Exercise <A HREF="book009.html#is_reverse">8.8</A>.

<A NAME="toc102"></A><A NAME="htoc113">9.5</A> 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

<A NAME="toc103"></A><A NAME="htoc114">9.6</A> Glossary

file object:: A value that represents an open file.
<A NAME="@default705"></A>
<A NAME="@default706"></A>
problem recognition:: A way of solving a problem by expressing it as an instance of a previously-solved problem. <A NAME="@default707"></A>
special case:: A test case that is atypical or non-obvious (and less likely to be handled correctly). <A NAME="@default708"></A>

<A NAME="toc104"></A><A NAME="htoc115">9.7</A> Exercises

Exercise 7

This question is based on a Puzzler that was broadcast on the radio

program Car

Talk^{<A NAME="text17" HREF="#note17">2</A>}:

Give me a word with three consecutive double letters. I’ll give you a couple of words that almost qualify, but don’t. For example, the word committee, c-o-m-m-i-t-t-e-e. It would be great except for the ‘i’ that sneaks in there. Or Mississippi: M-i-s-s-i-s-s-i-p-p-i. If you could take out those i’s it would work. But there is a word that has three consecutive pairs of letters and to the best of my knowledge this may be the only word. Of course there are probably 500 more but I can only think of one. What is the word?

Write a program to find it. You can see my solution at thinkpython.com/code/cartalk.py.

Exercise 8

Here’s another Car Talk

Puzzler^{<A NAME="text18" HREF="#note18">3</A>}:

“I was driving on the highway the other day and I happened to notice my odometer. Like most odometers, it shows six digits, in whole miles only. So, if my car had 300,000
miles, for example, I’d see 3-0-0-0-0-0.
“Now, what I saw that day was very interesting. I noticed that the last 4 digits were palindromic; that is, they read the same forward as backward. For example, 5-4-4-5 is a palindrome, so my odometer could have read 3-1-5-4-4-5.
“One mile later, the last 5 numbers were palindromic. For example, it
could have read 3-6-5-4-5-6. One mile after that, the middle 4 out of 6 numbers were palindromic. And you ready for this? One mile later,
all 6 were palindromic!
“The question is, what was on the odometer when I first looked?”

Write a Python program that tests all the six-digit numbers and prints

any numbers that satisfy these requirements. You can see my solution

at thinkpython.com/code/cartalk.py.

Exercise 9

Here’s another Car Talk Puzzler you can solve with a

search^{<A NAME="text19" HREF="#note19">4</A>}:

“Recently I had a visit with my mom and we realized that the two digits that make up my age when reversed resulted in her age. For example, if she’s 73, I’m 37. We wondered how often this has happened over the years but we got sidetracked with other topics and
we never came up with an answer.
“When I got home I figured out that the digits of our ages have been reversible six times so far. I also figured out that if we’re lucky it would happen again in a few years, and if we’re really lucky it would happen one more time after that. In other words, it would have happened 8 times over all. So the question is, how old am I now?”

Write a Python program that searches for solutions to this Puzzler. Hint: you might find the string method zfill useful.

You can see my solution at thinkpython.com/code/cartalk.py.

<A NAME="note16" HREF="#text16">1</A>: wikipedia.org/wiki/Moby_Project
<A NAME="note17" HREF="#text17">2</A>: www.cartalk.com/content/puzzler/transcripts/200725
<A NAME="note18" HREF="#text18">3</A>: www.cartalk.com/content/puzzler/transcripts/200803
<A NAME="note19" HREF="#text19">4</A>: www.cartalk.com/content/puzzler/transcripts/200813