Properties Of Regular Languages | CS 139: Theory of Computation

<h1 id="properties">Properties</h1>
<h2 id="of-regular-languages">of Regular Languages</h2>

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<h2 id="administrivia">Administrivia</h2>
<ul>
  <li>One More CS Candidate Research Talk
    <ul>
      <li>3:30–4:30pm, Wednesday, February 19th</li>
      <li>C-S 235</li>
    </ul>
  </li>
  <li class="fragment">Almost done grading Assignment 2</li>
  <li class="fragment">Assignment 4 is released</li>
  <li class="fragment">Exam 1 is one week from Thursday</li>
</ul>

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

<h1 id="assignment-2-debrief">Assignment 2 Debrief</h1>

<h1 id="regex-recap">Regex Recap</h1>

<h2 id="formal-definition-of-regex">Formal Definition of RegEx</h2>
<p><code class="language-plaintext highlighter-rouge">$R$</code> is a <strong>regular expression</strong> if it is in the form:</p>

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<ol>
  <li><code class="language-plaintext highlighter-rouge">$a$</code> for some symbol <code class="language-plaintext highlighter-rouge">$a\in\Sigma$</code></li>
  <li><code class="language-plaintext highlighter-rouge">$\epsilon$</code></li>
  <li><code class="language-plaintext highlighter-rouge">$\emptyset$</code></li>
  <li><code class="language-plaintext highlighter-rouge">$(R_1\cup R_2)$</code> where <code class="language-plaintext highlighter-rouge">$R_1$</code> and <code class="language-plaintext highlighter-rouge">$R_2$</code> are regular expressions</li>
  <li><code class="language-plaintext highlighter-rouge">$(R_1\circ R_2)$</code> where <code class="language-plaintext highlighter-rouge">$R_1$</code> and <code class="language-plaintext highlighter-rouge">$R_2$</code> are regular expressions</li>
  <li><code class="language-plaintext highlighter-rouge">$(R_1)^\ast$</code> where <code class="language-plaintext highlighter-rouge">$R_1$</code> is a regular expression</li>
</ol>

<h2 id="regex-examples">Regex Examples</h2>
<p>Write the corresponding <strong>language</strong> that each of the following regular expressions express</p>

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<ol>
  <li><code class="language-plaintext highlighter-rouge">$0\cup 1$</code> describes the language <code class="language-plaintext highlighter-rouge">$\{0, 1\}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$01$</code> describes <code class="language-plaintext highlighter-rouge">$\{01\}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$01\cup\epsilon$</code> describes <code class="language-plaintext highlighter-rouge">$\{01,\epsilon\}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$\texttt{ab}\cup\emptyset$</code> describes <code class="language-plaintext highlighter-rouge">$\{ab\}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$(0\cup 1)^\ast$</code> describes <code class="language-plaintext highlighter-rouge">$\{0,1\}^\ast = \{\epsilon, 0, 1, 00, \ldots\}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$0^\ast\cup 1$</code> describes <code class="language-plaintext highlighter-rouge">$\{\epsilon, 0, 00, 000, \ldots\}\cup\{1\}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$0(0\cup 1)^\ast \cup (0\cup 1)^\ast 1$</code> describes <code class="language-plaintext highlighter-rouge">$\{0w, w1\mid w\in\{0,1\}^\ast\}$</code></li>
</ol>

<h2 id="regex-exercises">Regex Exercises</h2>
<p>Let <code class="language-plaintext highlighter-rouge">$\Sigma = \{\texttt{a}, \ldots, \texttt{z}\}\cup \{\texttt{/}, \texttt{N}, \texttt{*}\}$</code> be an alphabet</p>

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<ol>
  <li>Construct a regex that recognizes all the strings of the form <code class="language-plaintext highlighter-rouge">$\texttt{//}\ldots \texttt{N}$</code>
    <ul>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$A = (\texttt{a}\cup\cdots\cup\texttt{z}\cup\texttt{*})$</code></li>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">R = $\texttt{//}\circ A^\ast\circ\texttt{N}$</code></li>
    </ul>
  </li>
</ol>

<h2 id="regex-exercises-1">Regex Exercises</h2>
<p>Let <code class="language-plaintext highlighter-rouge">$\Sigma = \{\texttt{a}, \ldots, \texttt{z}\}\cup \{\texttt{/}, \texttt{N}, \texttt{*}\}$</code> be an alphabet</p>

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<ol start="2">
  <li>Construct a regex that recognizes all the strings of the form <code class="language-plaintext highlighter-rouge">$\texttt{/*}\ldots \texttt{*/}$</code>
    <ul>
      <li>Note that we should not match strings of the form <code class="language-plaintext highlighter-rouge">$\texttt{/*abc*/def*/}$</code> because of the two closing <code class="language-plaintext highlighter-rouge">$\texttt{*/}$</code>’s</li>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$R = \texttt{/*}\circ B^\ast\circ\texttt{*/}$</code> for some <code class="language-plaintext highlighter-rouge">$B$</code></li>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$X = (A\cup\texttt{N})$</code></li>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$Y = (X\cup\texttt{/})$</code></li>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$B = (Y\cup(\texttt{*})^\ast X)$</code></li>
    </ul>
  </li>
</ol>

<h2 id="the-regularity-of-regex">The “Regularity” of RegEx</h2>
<ul>
  <li><strong>Theorem</strong>: A language <code class="language-plaintext highlighter-rouge">$L$</code> is regular if and only if there is a regular expression that expresses it</li>
</ul>

<h2 id="the-regularity-of-regex-1">The “Regularity” of RegEx</h2>
<p><code class="language-plaintext highlighter-rouge">$L$</code> is regular <code class="language-plaintext highlighter-rouge">$\iff$</code> there is a regex with <code class="language-plaintext highlighter-rouge">$L(R) = L$</code></p>

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<ul>
  <li><strong>Proof Sketch</strong> of the backward direction (<code class="language-plaintext highlighter-rouge">$\Longleftarrow$</code>):
    <ol>
      <li>Given some regex <code class="language-plaintext highlighter-rouge">$R$</code></li>
      <li>Construct an NFA <code class="language-plaintext highlighter-rouge">$M$</code> with <code class="language-plaintext highlighter-rouge">$L(M) = L(R)$</code></li>
      <li>Having an NFA that recognizes it means <code class="language-plaintext highlighter-rouge">$L$</code> is regular</li>
    </ol>
  </li>
</ul>

<h2 id="need-an-nfa-for-all-these">Need an NFA for all these…</h2>

<h2 id="the-regularity-of-regex-2">The “Regularity” of RegEx</h2>
<p><code class="language-plaintext highlighter-rouge">$L$</code> is regular <code class="language-plaintext highlighter-rouge">$\iff$</code> there is a regex with <code class="language-plaintext highlighter-rouge">$L(R) = L$</code></p>

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<ul>
  <li><strong>Proof Sketch</strong> of the forward direction (<code class="language-plaintext highlighter-rouge">$\implies$</code>):
    <ol>
      <li>Assume  that <code class="language-plaintext highlighter-rouge">$L$</code> is regular, therefore there is some DFA <code class="language-plaintext highlighter-rouge">$M$</code> that recognizes it</li>
      <li>Convert <code class="language-plaintext highlighter-rouge">$M$</code> into a regular expression <code class="language-plaintext highlighter-rouge">$R$</code> by using a clever trick using <strong>generalized nondeterministic finite automata</strong> (<strong>GNFAs</strong>)</li>
    </ol>
  </li>
</ul>

<h1 id="properties-of-regular-languages">Properties of Regular Languages</h1>

<h3 id="properties-of-regular-languages-1">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-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>

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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> is regular, then prove <code class="language-plaintext highlighter-rouge">$\overline{A}$</code> is too</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-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>

<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>