Non Regular Languages | CS 139: Theory of Computation

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

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<h2 id="administrivia">Administrivia</h2>
<ul>
  <li><strong>Exam 1</strong> is one week from today
    <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="properties">Properties</h1>
<h2 id="of-regular-languages">of Regular Languages</h2>

<h3 id="properties-of-regular-languages">Properties of Regular Languages</h3>
<ul>
  <li>Since languages are <strong>sets</strong>, operations like <code class="language-plaintext highlighter-rouge">$\cap$</code>, <code class="language-plaintext highlighter-rouge">$\cup$</code>, <code class="language-plaintext highlighter-rouge">$\setminus$</code>, etc. can be applied</li>
  <li class="fragment">We can also define other set operations like the <strong>concatenation</strong> of two languages <code class="language-plaintext highlighter-rouge">$A$</code> and <code class="language-plaintext highlighter-rouge">$B$</code>: <code class="language-plaintext highlighter-rouge">$$A\circ B = \{x\circ y \mid x\in A, y\in B\}$$</code></li>
  <li class="fragment">What is <code class="language-plaintext highlighter-rouge">$\{000,1\}\circ\{\texttt{a}, \texttt{b}\texttt{c}\}$</code>?
    <ul>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$ = \{000\texttt{a}, 000\texttt{bc}, 1\texttt{a}, 1\texttt{bc}\}$</code></li>
    </ul>
  </li>
</ul>

<h3 id="closure-under-concatenation">Closure Under Concatenation</h3>
<ul>
  <li><strong>Theorem:</strong> The set of regular languages is closed under <em>concatenation</em></li>
  <li class="fragment">i.e., if <code class="language-plaintext highlighter-rouge">$A,B\subseteq\Sigma^\ast$</code> are regular, then <code class="language-plaintext highlighter-rouge">$A\circ B$</code> is regular</li>
</ul>

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<ul>
  <li class="fragment">How do we prove this?</li>
  <li class="fragment">Assume that <code class="language-plaintext highlighter-rouge">$A$</code> and <code class="language-plaintext highlighter-rouge">$B$</code> are regular, then prove <code class="language-plaintext highlighter-rouge">$A\circ B$</code> is also regular</li>
</ul>

<h3 id="closure-under-set-complement">Closure Under Set Complement</h3>

<ul>
  <li><strong>Theorem:</strong> The set of regular languages is closed under <em>set complement</em></li>
  <li class="fragment">i.e., if <code class="language-plaintext highlighter-rouge">$A\subseteq\Sigma^\ast$</code> is regular, then so is <code class="language-plaintext highlighter-rouge">$\overline{A} = \Sigma^\ast\setminus A$</code></li>
</ul>

<h3 id="closure-under-unionintersection">Closure Under Union/Intersection</h3>
<ul>
  <li><strong>Theorem:</strong> The set of regular languages is closed under <em>union</em> and <em>intersection</em></li>
  <li class="fragment">i.e., if <code class="language-plaintext highlighter-rouge">$A,B\subseteq\Sigma^\ast$</code> are both regular, then <code class="language-plaintext highlighter-rouge">$A\cap B$</code> and <code class="language-plaintext highlighter-rouge">$A\cup B$</code> are both regular, too</li>
</ul>

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<ul>
  <li class="fragment">Main idea for proof:
    <ul>
      <li>Given two DFAs, construct a new DFA that simulates them both simultaneously</li>
    </ul>
  </li>
</ul>

<h3 id="closure-under-kleene-star">Closure Under Kleene Star</h3>
<ul>
  <li><strong>Theorem:</strong> The set of regular languages is closed under the <em>Kleene star</em> operation</li>
  <li class="fragment">i.e., if <code class="language-plaintext highlighter-rouge">$A\subseteq\Sigma^\ast$</code> is regular, then so is <code class="language-plaintext highlighter-rouge">$$A^\ast = \bigcup_{n\in\mathbb{Z}_{\ge 0}} A^n$$</code> where <code class="language-plaintext highlighter-rouge">$A^0 = \{\epsilon\}$</code>, <code class="language-plaintext highlighter-rouge">$A^1 = A$</code>, and <code class="language-plaintext highlighter-rouge">$A^k = A\circ A^{k-1}$</code></li>
</ul>

<h1 id="non-regular-languages-1">Non-regular Languages</h1>

<h2 id="non-regular-languages-2">Non-regular Languages</h2>

<ul>
  <li>Recall that a <strong>language</strong> is actually encoding a <strong>computational problem</strong></li>
  <li class="fragment"><strong>Regular</strong> languages are basically the “easiest” computational problems for machines to solve</li>
  <li class="fragment">Fun fact: <strong>most</strong> languages are not regular!</li>
</ul>

<h2 id="non-regular-languages-3">Non-regular Languages</h2>
<ul>
  <li>Determining whether a language is regular or not can be non-trivial</li>
  <li>Consider the following two languages:</li>
</ul>

<p class="fragment"><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 class="fragment"><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>

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

<h2 id="non-regular-languages-4">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"><code class="language-plaintext highlighter-rouge">$w$</code> can be decomposed into <code class="language-plaintext highlighter-rouge">$w = xyz$</code> where</li>
              <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>
            </ul>
          </li>
        </ul>
      </li>
    </ul>
  </li>
</ul>

<hr />

<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="handouts">Handouts</h1>