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<div class="section" id="case-study-word-play">
<h1>Case study: word play<a class="headerlink" href="#case-study-word-play" title="Permalink to this headline">¶</a></h1>
<p>This chapter presents the second case study, which involves solving word
puzzles by searching for words that have certain properties. For
example, we’ll find the longest palindromes in English and search for
words whose letters appear in alphabetical order. And I will present
another program development plan: reduction to a previously solved
problem.</p>
<div class="section" id="reading-word-lists">
<h2>Reading word lists<a class="headerlink" href="#reading-word-lists" title="Permalink to this headline">¶</a></h2>
<p>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 (see
<a class="reference external" href="http://wikipedia.org/wiki/Moby_Project">http://wikipedia.org/wiki/Moby_Project</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; you can download a copy, with the simpler name
words.txt, from <a class="reference external" href="http://thinkpython2.com/code/words.txt">http://thinkpython2.com/code/words.txt</a>.</p>
<p>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 <strong>file object</strong> you can
use to read the file.</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">fin</span> <span class="o">=</span> <span class="nb">open</span><span class="p">(</span><span class="s">'words.txt'</span><span class="p">)</span>
</pre></div>
</div>
<p>fin is a common name for a file object used for input. 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:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">fin</span><span class="o">.</span><span class="n">readline</span><span class="p">()</span>
<span class="go">'aa\r\n'</span>
</pre></div>
</div>
<p>The first word in this particular list is “aa”, which is a kind of lava.
The sequence <code class="docutils literal"><span class="pre">\r\n</span></code> represents two whitespace characters, a carriage
return and a newline, that separate this word from the next.</p>
<p>The file object keeps track of where it is in the file, so if you call
readline again, you get the next word:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">fin</span><span class="o">.</span><span class="n">readline</span><span class="p">()</span>
<span class="go">'aah\r\n'</span>
</pre></div>
</div>
<p>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:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="gp">>>> </span><span class="n">line</span> <span class="o">=</span> <span class="n">fin</span><span class="o">.</span><span class="n">readline</span><span class="p">()</span>
<span class="gp">>>> </span><span class="n">word</span> <span class="o">=</span> <span class="n">line</span><span class="o">.</span><span class="n">strip</span><span class="p">()</span>
<span class="gp">>>> </span><span class="n">word</span>
<span class="go">'aahed'</span>
</pre></div>
</div>
<p>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:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="n">fin</span> <span class="o">=</span> <span class="nb">open</span><span class="p">(</span><span class="s">'words.txt'</span><span class="p">)</span>
<span class="k">for</span> <span class="n">line</span> <span class="ow">in</span> <span class="n">fin</span><span class="p">:</span>
<span class="n">word</span> <span class="o">=</span> <span class="n">line</span><span class="o">.</span><span class="n">strip</span><span class="p">()</span>
<span class="k">print</span><span class="p">(</span><span class="n">word</span><span class="p">)</span>
</pre></div>
</div>
</div>
<div class="section" id="exercises">
<h2>Exercises<a class="headerlink" href="#exercises" title="Permalink to this headline">¶</a></h2>
<p>There are solutions to these exercises in the next section. You should
at least attempt each one before you read the solutions.</p>
<p>Write a program that reads words.txt and prints only the words with more
than 20 characters (not counting whitespace).</p>
<p>In 1939 Ernest Vincent Wright published a 50,000 word novel called
<em>Gadsby</em> that does not contain the letter “e”. Since “e” is the most
common letter in English, that’s not easy to do.</p>
<p>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.</p>
<p>All right, I’ll stop now.</p>
<p>Write a function called <code class="docutils literal"><span class="pre">has_no_e</span></code> that returns True if the given word
doesn’t have the letter “e” in it.</p>
<p>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
that have no “e”.</p>
<p>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.</p>
<p>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?</p>
<p>Write a function named <code class="docutils literal"><span class="pre">uses_only</span></code> 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?”</p>
<p>Write a function named <code class="docutils literal"><span class="pre">uses_all</span></code> 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?</p>
<p>Write a function called <code class="docutils literal"><span class="pre">is_abecedarian</span></code> that returns True if the
letters in a word appear in alphabetical order (double letters are ok).
How many abecedarian words are there?</p>
</div>
<div class="section" id="search">
<h2>Search<a class="headerlink" href="#search" title="Permalink to this headline">¶</a></h2>
<p>All of the exercises in the previous section have something in common;
they can be solved with the search pattern we saw in Section [find]. The
simplest example is:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">has_no_e</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
<span class="k">for</span> <span class="n">letter</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
<span class="k">if</span> <span class="n">letter</span> <span class="o">==</span> <span class="s">'e'</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>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.</p>
<p>You could write this function more concisely using the in operator, but
I started with this version because it demonstrates the logic of the
search pattern.</p>
<p>is a more general version of <code class="docutils literal"><span class="pre">has_no_e</span></code> but it has the same structure:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">avoids</span><span class="p">(</span><span class="n">word</span><span class="p">,</span> <span class="n">forbidden</span><span class="p">):</span>
<span class="k">for</span> <span class="n">letter</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
<span class="k">if</span> <span class="n">letter</span> <span class="ow">in</span> <span class="n">forbidden</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>We can return False as soon as we find a forbidden letter; if we get to
the end of the loop, we return True.</p>
<p><code class="docutils literal"><span class="pre">uses_only</span></code> is similar except that the sense of the condition is
reversed:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">uses_only</span><span class="p">(</span><span class="n">word</span><span class="p">,</span> <span class="n">available</span><span class="p">):</span>
<span class="k">for</span> <span class="n">letter</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
<span class="k">if</span> <span class="n">letter</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">available</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>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.</p>
<p><code class="docutils literal"><span class="pre">uses_all</span></code> is similar except that we reverse the role of the word and
the string of letters:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">uses_all</span><span class="p">(</span><span class="n">word</span><span class="p">,</span> <span class="n">required</span><span class="p">):</span>
<span class="k">for</span> <span class="n">letter</span> <span class="ow">in</span> <span class="n">required</span><span class="p">:</span>
<span class="k">if</span> <span class="n">letter</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>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.</p>
<p>If you were really thinking like a computer scientist, you would have
recognized that <code class="docutils literal"><span class="pre">uses_all</span></code> was an instance of a previously solved
problem, and you would have written:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">uses_all</span><span class="p">(</span><span class="n">word</span><span class="p">,</span> <span class="n">required</span><span class="p">):</span>
<span class="k">return</span> <span class="n">uses_only</span><span class="p">(</span><span class="n">required</span><span class="p">,</span> <span class="n">word</span><span class="p">)</span>
</pre></div>
</div>
<p>This is an example of a program development plan called <strong>reduction to a
previously solved problem</strong>, which means that you recognize the problem
you are working on as an instance of a solved problem and apply an
existing solution.</p>
</div>
<div class="section" id="looping-with-indices">
<h2>Looping with indices<a class="headerlink" href="#looping-with-indices" title="Permalink to this headline">¶</a></h2>
<p>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.</p>
<p>For <code class="docutils literal"><span class="pre">is_abecedarian</span></code> we have to compare adjacent letters, which is a
little tricky with a for loop:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">is_abecedarian</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
<span class="n">previous</span> <span class="o">=</span> <span class="n">word</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
<span class="k">for</span> <span class="n">c</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
<span class="k">if</span> <span class="n">c</span> <span class="o"><</span> <span class="n">previous</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="n">previous</span> <span class="o">=</span> <span class="n">c</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>An alternative is to use recursion:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">is_abecedarian</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
<span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">word</span><span class="p">)</span> <span class="o"><=</span> <span class="mi">1</span><span class="p">:</span>
<span class="k">return</span> <span class="bp">True</span>
<span class="k">if</span> <span class="n">word</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">></span> <span class="n">word</span><span class="p">[</span><span class="mi">1</span><span class="p">]:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="k">return</span> <span class="n">is_abecedarian</span><span class="p">(</span><span class="n">word</span><span class="p">[</span><span class="mi">1</span><span class="p">:])</span>
</pre></div>
</div>
<p>Another option is to use a while loop:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">is_abecedarian</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
<span class="n">i</span> <span class="o">=</span> <span class="mi">0</span>
<span class="k">while</span> <span class="n">i</span> <span class="o"><</span> <span class="nb">len</span><span class="p">(</span><span class="n">word</span><span class="p">)</span><span class="o">-</span><span class="mi">1</span><span class="p">:</span>
<span class="k">if</span> <span class="n">word</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span> <span class="o"><</span> <span class="n">word</span><span class="p">[</span><span class="n">i</span><span class="p">]:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="n">i</span> <span class="o">=</span> <span class="n">i</span><span class="o">+</span><span class="mi">1</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>The loop starts at i=0 and ends when i=len(word)-1. Each time through
the loop, it compares the <span class="math">i</span>th character (which you can think
of as the current character) to the <span class="math">i+1</span>th character (which you
can think of as the next).</p>
<p>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.</p>
<p>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 <code class="docutils literal"><span class="pre">'flossy'</span></code>. 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.</p>
<p>Here is a version of <code class="docutils literal"><span class="pre">is_palindrome</span></code> (see Exercise [palindrome]) that
uses two indices; one starts at the beginning and goes up; the other
starts at the end and goes down.</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">is_palindrome</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
<span class="n">i</span> <span class="o">=</span> <span class="mi">0</span>
<span class="n">j</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">word</span><span class="p">)</span><span class="o">-</span><span class="mi">1</span>
<span class="k">while</span> <span class="n">i</span><span class="o"><</span><span class="n">j</span><span class="p">:</span>
<span class="k">if</span> <span class="n">word</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">!=</span> <span class="n">word</span><span class="p">[</span><span class="n">j</span><span class="p">]:</span>
<span class="k">return</span> <span class="bp">False</span>
<span class="n">i</span> <span class="o">=</span> <span class="n">i</span><span class="o">+</span><span class="mi">1</span>
<span class="n">j</span> <span class="o">=</span> <span class="n">j</span><span class="o">-</span><span class="mi">1</span>
<span class="k">return</span> <span class="bp">True</span>
</pre></div>
</div>
<p>Or we could reduce to a previously solved problem and write:</p>
<div class="highlight-python"><div class="highlight"><pre><span class="k">def</span> <span class="nf">is_palindrome</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
<span class="k">return</span> <span class="n">is_reverse</span><span class="p">(</span><span class="n">word</span><span class="p">,</span> <span class="n">word</span><span class="p">)</span>
</pre></div>
</div>
<p>Using <code class="docutils literal"><span class="pre">is_reverse</span></code> from Section [isreverse].</p>
</div>
<div class="section" id="debugging">
<h2>Debugging<a class="headerlink" href="#debugging" title="Permalink to this headline">¶</a></h2>
<p>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.</p>
<p>Taking <code class="docutils literal"><span class="pre">has_no_e</span></code> as an example, there are two obvious cases to check:
words that have an ‘e’ should return False, and words that don’t should
return True. You should have no trouble coming up with one of each.</p>
<p>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 <strong>special case</strong>, which is one of the non-obvious cases
where errors often lurk.</p>
<p>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).</p>
<p>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:</p>
<blockquote>
<div><p>Program testing can be used to show the presence of bugs, but never
to show their absence!</p>
<p class="attribution">—Edsger W. Dijkstra</p>
</div></blockquote>
</div>
<div class="section" id="glossary">
<span id="glossary09"></span><h2>Glossary<a class="headerlink" href="#glossary" title="Permalink to this headline">¶</a></h2>
<dl class="docutils">
<dt>objeto-arquivo (<em>file object</em>)</dt>
<dd>A value that represents an open file.</dd>
<dt>redução a um problema resolvido (<em>reduction to a previously solved problem</em>)</dt>
<dd>A way of solving a problem by expressing it as an instance of a previously solved problem.</dd>
<dt>caso especial (<em>special case</em>)</dt>
<dd>A test case that is atypical or non-obvious (and less likely to be handled correctly).</dd>
</dl>
</div>
<div class="section" id="id1">
<h2>Exercises<a class="headerlink" href="#id1" title="Permalink to this headline">¶</a></h2>
<p>This question is based on a Puzzler that was broadcast on the radio
program <em>Car Talk</em> (<a class="reference external" href="http://www.cartalk.com/content/puzzlers">http://www.cartalk.com/content/puzzlers</a>):</p>
<blockquote>
<div>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?</div></blockquote>
<p>Write a program to find it. Solution:
<a class="reference external" href="http://thinkpython2.com/code/cartalk1.py">http://thinkpython2.com/code/cartalk1.py</a>.</p>
<p>Here’s another <em>Car Talk</em> Puzzler
(<a class="reference external" href="http://www.cartalk.com/content/puzzlers">http://www.cartalk.com/content/puzzlers</a>):</p>
<blockquote>
<div><p>``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.</p>
<p>``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.</p>
<p>``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!</p>
<p>“The question is, what was on the odometer when I first looked?”</p>
</div></blockquote>
<p>Write a Python program that tests all the six-digit numbers and prints
any numbers that satisfy these requirements. Solution:
<a class="reference external" href="http://thinkpython2.com/code/cartalk2.py">http://thinkpython2.com/code/cartalk2.py</a>.</p>
<p>Here’s another <em>Car Talk</em> Puzzler you can solve with a search
(<a class="reference external" href="http://www.cartalk.com/content/puzzlers">http://www.cartalk.com/content/puzzlers</a>):</p>
<blockquote>
<div><p>``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.</p>
<p>“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?”</p>
</div></blockquote>
<p>Write a Python program that searches for solutions to this Puzzler.
Hint: you might find the string method zfill useful.</p>
<p>Solution: <a class="reference external" href="http://thinkpython2.com/code/cartalk3.py">http://thinkpython2.com/code/cartalk3.py</a>.</p>
</div>
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<h3><a href="index.html">Table Of Contents</a></h3>
<ul>
<li><a class="reference internal" href="#">Case study: word play</a><ul>
<li><a class="reference internal" href="#reading-word-lists">Reading word lists</a></li>
<li><a class="reference internal" href="#exercises">Exercises</a></li>
<li><a class="reference internal" href="#search">Search</a></li>
<li><a class="reference internal" href="#looping-with-indices">Looping with indices</a></li>
<li><a class="reference internal" href="#debugging">Debugging</a></li>
<li><a class="reference internal" href="#glossary">Glossary</a></li>
<li><a class="reference internal" href="#id1">Exercises</a></li>
</ul>
</li>
</ul>
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