Regular Languages Review | CS 139: Theory of Computation

<h1 id="review">Review</h1>
<h2 id="of-regular-languages">of Regular Languages</h2>

<hr />

<h2 id="administrivia">Administrivia</h2>
<ul>
  <li><strong>Exam 1</strong> is on Thursday
    <ul>
      <li>Closed notes, closed text, 75 minute exam</li>
      <li>Covers Sipser chapters 0 and 1</li>
      <li>Practice problems from Sipser:
        <ul>
          <li><strong>Chapter 0</strong>: all problems</li>
          <li><strong>Chapter 1</strong>: 2, 6, 7, 11, 12, 14, 18, 20, 22, 29, 30, 31, 38, 40, 43, 46</li>
        </ul>
      </li>
    </ul>
  </li>
</ul>

<h1 id="questions">Questions</h1>
<h2 id="about-anything">…about anything?</h2>

<h1 id="non-regular-languages">Non-regular Languages</h1>

<h2 id="non-regular-languages-1">Non-regular Languages</h2>
<ul>
  <li>Determining whether a language is regular can be non-trivial</li>
  <li>Last time we looked at the two languages:</li>
</ul>

<p><code class="language-plaintext highlighter-rouge">$$A = \{w\in\{0,1\}^\ast \mid w\text{ has an equal number of 0s and 1s}\}$$</code></p>

<p><code class="language-plaintext highlighter-rouge">$$B = \{w\in\{0,1\}^\ast \mid w\text{ has an equal number of 01s and 10s}\}$$</code></p>

<hr />

<p>Here <code class="language-plaintext highlighter-rouge">$A$</code> is <strong>not</strong> regular but <code class="language-plaintext highlighter-rouge">$B$</code> is.</p>

<h2 id="non-regular-languages-2">Non-regular Languages</h2>
<ul>
  <li>We need a better approach to <strong>prove</strong> a language is not regular with certainty</li>
  <li>This approach will exploit the main deficiency of DFAs—the fact that they have <strong>finite memory</strong></li>
</ul>

<h2 id="the-pumping-lemma">The Pumping Lemma</h2>

<ul>
  <li class="fragment"><strong>IF</strong> <code class="language-plaintext highlighter-rouge">$A$</code> is a regular language, <strong>THEN</strong>
    <ul>
      <li class="fragment"><strong>THERE EXISTS</strong> an integer <code class="language-plaintext highlighter-rouge">$p$</code> such that:
        <ul>
          <li class="fragment"><strong>FOR EVERY</strong> string <code class="language-plaintext highlighter-rouge">$w\in A$</code> satisfying <code class="language-plaintext highlighter-rouge">$|w|\ge p$</code>,
            <ul>
              <li class="fragment"><strong>THERE EXISTS</strong> a way to cut <code class="language-plaintext highlighter-rouge">$w$</code> into three parts <code class="language-plaintext highlighter-rouge">$w = xyz$</code> such that the following hold:
                <ol>
                  <li class="fragment"><code class="language-plaintext highlighter-rouge">$xy^iz \in A$</code> for all <code class="language-plaintext highlighter-rouge">$i \ge 0$</code></li>
                  <li class="fragment"><code class="language-plaintext highlighter-rouge">$|y| > 0$</code></li>
                  <li class="fragment"><code class="language-plaintext highlighter-rouge">$|xy| \le p$</code></li>
                </ol>
              </li>
            </ul>
          </li>
        </ul>
      </li>
    </ul>
  </li>
</ul>

<h2 id="using-the-pumping-lemma">Using the Pumping Lemma</h2>
<ul>
  <li>To prove that a language is not regular, the proof always follows a similar pattern and uses <strong>contradiction</strong></li>
</ul>

<hr />

<ol>
  <li class="fragment">Assume to the contrary that the language <code class="language-plaintext highlighter-rouge">$A$</code> is regular</li>
  <li class="fragment">Then define <code class="language-plaintext highlighter-rouge">$p$</code> to be the pumping length of <code class="language-plaintext highlighter-rouge">$A$</code> that is guaranteed to exist by the pumping lemma</li>
  <li class="fragment">Carefully pick a string <code class="language-plaintext highlighter-rouge">$w$</code> that satisfies two things: (1) <code class="language-plaintext highlighter-rouge">$w \in A$</code> and (2) <code class="language-plaintext highlighter-rouge">$|w| \ge p$</code></li>
  <li class="fragment">The pumping lemma then guarantees that there is some way to cut <code class="language-plaintext highlighter-rouge">$w = xyz$</code> into three parts</li>
  <li class="fragment">Using the three guarantees of the decomposition, cause a contradiction</li>
</ol>

<h2 id="exercise-1">Exercise 1</h2>

<p><strong>Claim:</strong> The following language is not regular
<code class="language-plaintext highlighter-rouge">$$A = \{0^n10^n \mid n\in\mathbb{N}\}$$</code></p>

<h2 id="exercise-2">Exercise 2</h2>

<ul>
  <li>To encode multiple pieces of information in a single string, it is useful to define a <strong>pairing function</strong></li>
  <li class="fragment">Given two strings <code class="language-plaintext highlighter-rouge">$x,y\in\{0,1\}^\ast$</code> define the <strong>self-delimiting pairing function</strong> <code class="language-plaintext highlighter-rouge">$p(x,y)$</code> by
<code class="language-plaintext highlighter-rouge">$$p(x,y) = 0^{|x|}1xy$$</code></li>
</ul>

<hr />

<ul>
  <li class="fragment"><strong>Claim:</strong> The set of all properly formatted self-delimited pairs is not regular.
<code class="language-plaintext highlighter-rouge">$$B = \{p(x,y) \mid x,y\in\{0,1\}^\ast\}$$</code></li>
</ul>

<h2 id="exercise-3">Exercise 3</h2>

<p><strong>Claim:</strong> The following language is not regular
<code class="language-plaintext highlighter-rouge">$$C = \{0^n1^m \mid n \ge m\}$$</code></p>

<h1 id="exam-practice-problems">Exam Practice Problems</h1>