Turing Machines Continued | CS 139: Theory of Computation

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

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<h2 id="administrivia">Administrivia</h2>
<ul>
  <li>ICICL Meeting today</li>
  <li>MathCS talk tomorrow</li>
</ul>

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

<h1 id="recap">Recap</h1>
<h2 id="of-turing-machines">of Turing Machines</h2>

<h2 id="turing-machines">Turing Machines</h2>
<ul>
  <li>Turing machines are essentially DFAs with access to an <strong>unlimited source of memory</strong></li>
  <li>The memory is in the form of a <strong>tape</strong> with individual cells with symbols written on them</li>
  <li>The machine has a <strong>tape head</strong> which is always pointing to exactly one cell on the tape</li>
  <li>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-1">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>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="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>

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<ul>
  <li><code class="language-plaintext highlighter-rouge">$Q$</code> is a finite set of <strong>states</strong></li>
  <li><code class="language-plaintext highlighter-rouge">$\Sigma$</code> is the <strong>input alphabet</strong></li>
  <li><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><code class="language-plaintext highlighter-rouge">$q_0\in Q$</code> is the <strong>start state</strong></li>
  <li><code class="language-plaintext highlighter-rouge">$q_\text{accept}\in Q$</code> is the <strong>accept state</strong></li>
  <li><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><code class="language-plaintext highlighter-rouge">$\delta(q, a) = (r, b, L)$</code>means that:
    <ul>
      <li><strong>IF</strong> in state <code class="language-plaintext highlighter-rouge">$q$</code> and reading <code class="language-plaintext highlighter-rouge">$a$</code>, <strong>THEN</strong>
        <ol>
          <li>Change to state <code class="language-plaintext highlighter-rouge">$r$</code></li>
          <li>Writes <code class="language-plaintext highlighter-rouge">$b$</code> over the old symbol <code class="language-plaintext highlighter-rouge">$a$</code></li>
          <li>Moves its tape head to the LEFT</li>
        </ol>
      </li>
    </ul>
  </li>
</ul>

<h2 id="definitions">Definitions</h2>
<ul>
  <li>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></li>
  <li class="fragment">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>cannot get into an infinite loop</strong></li>
    </ul>
  </li>
  <li class="fragment">A <strong>configuration</strong> of a Turing machine describes:
    <ul>
      <li>The current <strong>state</strong> of the machine</li>
      <li>The current <strong>contents</strong> of the tape</li>
      <li>The current <strong>location</strong> of the tape head</li>
      <li>{.fragment} e.g.: <code class="language-plaintext highlighter-rouge">$1011q_70111$</code></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>

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<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="pseudocode">Pseudocode</h2>
<ol>
  <li>Mark the current bit
    <ul>
      <li>By “mark” we mean a 0 becomes an “a” and a 1 becomes a “b”</li>
    </ul>
  </li>
  <li>Find the right-most unmarked bit by sweeping to the right looking for a’s, b’s, or empty spaces</li>
  <li>Mark that bit</li>
  <li>Find the left-most unmarked bit by sweeping to the left in a similar fashion</li>
  <li>Go back to step 1</li>
  <li>…</li>
</ol>

<h2 id="complete-solution">Complete 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
start a a r halt-reject
start b b r halt-reject
start _ _ r halt-reject<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
rightmost b b r halt-accept
rightmost a a r halt-reject<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>

<h2 id="second-example">Second Example</h2>

<p>Show that the following language is decidable
<code class="language-plaintext highlighter-rouge">$$B = \{w\in\{0,1\}^\ast \mid \text{the second-to-last bit of }w\text{ is a 1}\}$$</code></p>

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<ul>
  <li>Spend 2-3 minutes constructing a Turing machine that decides <code class="language-plaintext highlighter-rouge">$B$</code></li>
  <li>We will write it for the <code class="language-plaintext highlighter-rouge">morphett</code> simulator program, a diagram, and write it down formally</li>
</ul>

<ul>
  <li><strong>Theorem</strong>: the regular languages are a <em>strict</em> subset of the decidable languages</li>
</ul>

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<ul>
  <li>How do we prove this?</li>
</ul>

<h2 id="third-example">Third Example</h2>
<p>Show that the following language is decidable
<code class="language-plaintext highlighter-rouge">$$C = \{w\in\{0,1\}^\ast \mid w\text{ has the same # of 0's and 1's}\}$$</code></p>

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