Turing Machines | CS 139: Theory of Computation

<h1 id="turing-machines">Turing Machines</h1>

<hr />

<h2 id="administrivia">Administrivia</h2>
<ul>
  <li><strong>Exam 1</strong> is graded</li>
  <li><strong>Assignment 5</strong> is posted (due Thursday, 3/12)</li>
</ul>

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

<h1 id="exam-1-debrief">Exam 1 Debrief</h1>

<h2 id="exam-1-debrief-1">Exam 1 Debrief</h2>
<ul>
  <li><strong>Average</strong>: 34/44 (i.e. 77%)</li>
  <li class="fragment">Most troublesome problems:
    <ul>
      <li>1d, 2a, 2d, and 5b</li>
    </ul>
  </li>
  <li class="fragment">If you got a low score on a problem, please come talk to me about it. It is likely I will give some partial credit back to you for walking me through your thought process</li>
</ul>

<h1 id="turing-machines-1">Turing Machines</h1>

<h2 id="turing-machines-2">Turing Machines</h2>
<ul>
  <li>On the first day of class we tried defining what <strong>computable</strong> meant and compared two assumptions:
    <ol>
      <li><strong>Finite</strong> Memory Computation</li>
      <li><strong>Unlimited</strong> Memory Computation</li>
    </ol>
  </li>
  <li class="fragment">We explored the former (regular languages) but discovered that many simple problems are not regular:
<code class="language-plaintext highlighter-rouge">$$L = \{0^n1^n \mid n\ge 0\}$$</code></li>
  <li class="fragment">The main deficiency of DFAs is their <strong>finite memory</strong>—thus they cannot count the number of 0’s and 1’s</li>
</ul>

<h2 id="turing-machines-3">Turing Machines</h2>
<ul>
  <li>Turing machines are essentially DFAs with access to an <strong>unlimited source of memory</strong></li>
  <li class="fragment">The memory is in the form of a <strong>tape</strong> with individual cells with symbols written on them</li>
  <li class="fragment">The machine has a <strong>tape head</strong> which is always pointing to exactly one cell on the tape</li>
  <li class="fragment">The machine also has a <strong>finite state control</strong> similar to a DFA that it references in making decisions</li>
</ul>

<ul>
  <li>Every step of computation the TM does this:
    <ol>
      <li><strong>Reads</strong> the current symbol the tape head is pointing at</li>
      <li><strong>Writes</strong> a new symbol (overwriting the old one)</li>
      <li><strong>Moves</strong> the tape head to the LEFT or RIGHT</li>
      <li><strong>Updates</strong> its current state</li>
    </ol>
  </li>
</ul>

<h2 id="turing-machines-4">Turing Machines</h2>
<ul>
  <li>A Turing machine has two special states:
    <ul>
      <li><code class="language-plaintext highlighter-rouge">$q_\text{accept}$</code> (akin to <code class="language-plaintext highlighter-rouge">return True</code>)</li>
      <li><code class="language-plaintext highlighter-rouge">$q_\text{reject}$</code> (akin to <code class="language-plaintext highlighter-rouge">return False</code>)</li>
    </ul>
  </li>
  <li class="fragment">When a Turing machine enters one of these states, it immediately <strong>halts</strong> (just like a <code class="language-plaintext highlighter-rouge">return</code> statement)</li>
</ul>

<h2 id="turing-machines-5">Turing Machines</h2>
<iframe width="840" height="473" src="https://www.youtube.com/embed/dNRDvLACg5Q?start=56" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen=""></iframe>

<h2 id="formal-definition">Formal Definition</h2>
<p>A <strong>Turing machine</strong> <code class="language-plaintext highlighter-rouge">$M$</code> is a 7-tuple:
<code class="language-plaintext highlighter-rouge">$$M = (Q, \Sigma, \Gamma, \delta, q_0, q_\text{accept}, q_\text{reject})$$</code></p>

<hr />

<ul>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$Q$</code> is a finite set of <strong>states</strong></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$\Sigma$</code> is the <strong>input alphabet</strong></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$\Gamma$</code> is the <strong>tape alphabet</strong> (Note that <code class="language-plaintext highlighter-rouge">$\Sigma\subseteq\Gamma$</code> and <code class="language-plaintext highlighter-rouge">$\sqcup\in\Gamma$</code>)</li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$q_0\in Q$</code> is the <strong>start state</strong></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$q_\text{accept}\in Q$</code> is the <strong>accept state</strong></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$q_\text{reject}\in Q$</code> is the <strong>reject state</strong></li>
</ul>

<h2 id="formal-definition-1">Formal Definition</h2>
<ul>
  <li>The <strong>transition function</strong> is of the form
<code class="language-plaintext highlighter-rouge">$$\delta:Q\times\Gamma\rightarrow Q\times\Gamma\times\{L,R\}$$</code></li>
  <li class="fragment">If we have <code class="language-plaintext highlighter-rouge">$\delta(q, a) = (r, b, L)$</code>, this means that:
    <ul>
      <li class="fragment"><strong>IF</strong> <code class="language-plaintext highlighter-rouge">$M$</code> is in state <code class="language-plaintext highlighter-rouge">$q$</code> and the tape head is currently reading the symbol <code class="language-plaintext highlighter-rouge">$a$</code>, <strong>THEN</strong>
        <ol>
          <li class="fragment"><code class="language-plaintext highlighter-rouge">$M$</code>’s current state changes to state <code class="language-plaintext highlighter-rouge">$r$</code></li>
          <li class="fragment"><code class="language-plaintext highlighter-rouge">$M$</code> writes symbol <code class="language-plaintext highlighter-rouge">$b$</code> over the old symbol <code class="language-plaintext highlighter-rouge">$a$</code></li>
          <li class="fragment"><code class="language-plaintext highlighter-rouge">$M$</code> moves its tape head to the LEFT</li>
        </ol>
      </li>
    </ul>
  </li>
</ul>

<h2 id="definitions">Definitions</h2>
<ul>
  <li>The language recognized by a Turing machine <code class="language-plaintext highlighter-rouge">$M$</code>, and we write <code class="language-plaintext highlighter-rouge">$L(M)$</code>, is the set of all strings that <code class="language-plaintext highlighter-rouge">$M$</code> <strong>accepts</strong></li>
  <li class="fragment">A language <code class="language-plaintext highlighter-rouge">$A\subseteq\Sigma^\ast$</code> is <strong>Turing recognizable</strong> if there exists a Turing machine <code class="language-plaintext highlighter-rouge">$M$</code> such that <code class="language-plaintext highlighter-rouge">$L(M) = A$</code>
    <ul>
      <li class="fragment"><strong>IMPORTANT</strong>: “Not accepting” does not necessarily mean that the Turing machine rejected the string</li>
      <li class="fragment">A Turing machine <code class="language-plaintext highlighter-rouge">$M$</code> can also get into an <strong>infinite loop</strong> and never reach <code class="language-plaintext highlighter-rouge">$q_\text{accept}$</code> nor <code class="language-plaintext highlighter-rouge">$q_\text{reject}$</code></li>
    </ul>
  </li>
</ul>

<h2 id="definitions-1">Definitions</h2>
<ul>
  <li>A language <code class="language-plaintext highlighter-rouge">$A\subseteq\Sigma^\ast$</code> is <strong>Turing decidable</strong> if some Turing machine recognizes <code class="language-plaintext highlighter-rouge">$A$</code> and <strong>rejects</strong> all strings not in <code class="language-plaintext highlighter-rouge">$A$</code>
    <ul>
      <li class="fragment">In other words, the machine must <strong>always halt</strong></li>
    </ul>
  </li>
</ul>

<h2 id="first-example">First Example</h2>
<p>Show that the following language is decidable
<code class="language-plaintext highlighter-rouge">$$A = \{w\in\{0,1\}^\ast \mid \text{the middle bit of }w\text{ is a 1}\}$$</code></p>

<hr />

<ul>
  <li class="fragment"><strong>IDEA:</strong> Work “outside in” eliminating the left and right bits until only the middle bit remains</li>
  <li class="fragment">We will use a <strong>Turing machine simulator</strong> to help construct this machine
    <ul>
      <li><a href="http://morphett.info/turing/turing.html">http://morphett.info/turing/turing.html</a></li>
    </ul>
  </li>
</ul>

<h2 id="partial-solution">Partial Solution</h2>

<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code># Mark first bit, then go find the right-hand side of the tape
start 1 b r find_right
start 0 a r find_right<br>
# Skip over all bits until we find a blank symbol or marked symbol,
# then move left and switch to the "rightmost" state
find_right 0 0 r find_right
find_right 1 1 r find_right
find_right _ _ l rightmost
find_right a a l rightmost
find_right b b l rightmost<br>
# Mark the rightmost bit, then go find the left-hand side of
# the tape
rightmost 0 a l find_left
rightmost 1 b l find_left<br>
# Skip over all bits until we find a blank symbol or marked symbol,
# then move right and switch to the "start" state
find_left 0 0 l find_left
find_left 1 1 l find_left
find_left a a r start
find_left b b r start
</code></pre></div></div>