Turing Machines Continued | CS 139: Theory of Computation

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

<hr />

<h2 id="administrivia">Administrivia</h2>
<ul>
  <li><strong>Assignment 5</strong> is due on Thursday before class</li>
  <li>A <a href="https://docs.google.com/forms/d/e/1FAIpQLSdia7g0bMNDV5AWLs7iZ3kqlTFcu31Mc2QAbiLdhfRRDtdMSA/viewform?usp=sf_link">Midterm Evaluation</a> survey was distributed via email
    <ul>
      <li>Please fill it out before break!</li>
    </ul>
  </li>
</ul>

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

<h1 id="quick-recap">Quick Recap</h1>
<h2 id="of-last-time">of Last Time</h2>

<h2 id="turing-machines">Turing Machines</h2>
<p><img src="http://science.slc.edu/~jmarshall/courses/2006/spring/cs30/lectures/TMs/TuringMachine.gif" alt="Turing machine visualization" /></p>

<ul>
  <li>Can be described as a list of instructions of the form
    <ul>
      <li><code class="language-plaintext highlighter-rouge">$(q,a)\rightarrow (q',a',D)$</code></li>
      <li class="fragment">“If in state <code class="language-plaintext highlighter-rouge">$q$</code> while reading an <code class="language-plaintext highlighter-rouge">$a$</code> on the tape, then change to state <code class="language-plaintext highlighter-rouge">$q'$</code>, write an <code class="language-plaintext highlighter-rouge">$a'$</code> to the tape, and move the tape head in direction <code class="language-plaintext highlighter-rouge">$D\in\{L,R\}$</code>”</li>
    </ul>
  </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>

<hr />

<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="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><strong>IDEA:</strong> Work “outside in” eliminating the left and right bits until only the middle bit remains</li>
</ul>

<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>

<hr />

<ul>
  <li><strong>IDEA 1:</strong> Find the right-hand side, then go two spaces back, accept if it is a 1, reject otherwise</li>
  <li class="fragment"><strong>IDEA 2:</strong> Simulate a DFA that recognizes it</li>
</ul>

<h2 id="complete-solution-1">Complete Solution</h2>
<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code>start 0 _ R q0
start 1 _ R q1
start _ _ R halt-reject<br>
q0 0 _ R q00
q0 1 _ R q01
q0 _ _ R halt-reject<br>
q1 0 _ R q10
q1 1 _ R q11
q1 _ _ R halt-reject<br>
q00 0 _ R q00
q00 1 _ R q01
q00 _ _ R halt-reject<br>
q01 0 _ R q10
q01 1 _ R q11
q01 _ _ R halt-reject<br>
q10 0 _ R q00
q10 1 _ R q01
q10 _ _ R halt-accept<br>
q11 0 _ R q10
q11 1 _ R q11
q11 _ _ R halt-accept
</code></pre></div></div>

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

<hr />

<ul>
  <li><strong>IDEA:</strong>
    <ul>
      <li>On a blank symbol, make all accepting states of the DFA go to the <code class="language-plaintext highlighter-rouge">$q_\text{accept}$</code> state, make all other states of the DFA go to the <code class="language-plaintext highlighter-rouge">$q_\text{reject}$</code> state</li>
      <li>Otherwise do the same thing as the DFA does while going to the right</li>
    </ul>
  </li>
</ul>

<h2 id="tms--dfas-1">TMs > DFAs</h2>
<ul>
  <li>Formally, when given a DFA <code class="language-plaintext highlighter-rouge">$M = (Q, \Sigma, \delta, q_0, F)$</code>, we would convert it into the TM <code class="language-plaintext highlighter-rouge">$M' = (Q', \Sigma, \Gamma, \delta', q_0, q_\text{accept}, q_\text{reject})$</code> where:
    <ul>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$Q' = Q\cup\{q_\text{accept},q_\text{reject}\}$</code></li>
      <li class="fragment"><code class="language-plaintext highlighter-rouge">$\Gamma = \Sigma\cup\{\sqcup\}$</code></li>
    </ul>
  </li>
</ul>

<hr />

<p class="fragment"><code class="language-plaintext highlighter-rouge">$$
\delta'(q,a) = \begin{cases}
    \delta(q,a)\quad\text{ if }q\in Q\text{ and }a\in\Sigma\\
    q_\text{accept}\quad\text{ if }q\in F\text{ and }a=\sqcup\\
    q_\text{reject}\quad\text{ otherwise}
\end{cases}
$$</code></p>

<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>

<hr />

<ul>
  <li><strong>IDEA:</strong> Alternate marking 0s and 1s until none remain</li>
</ul>

<h2 id="complete-solution-2">Complete Solution</h2>
<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code>mark0 0 x r find_right
mark0 _ _ l final_check
mark0 * * r mark0<br>
find_right _ _ l mark1
find_right * * r find_right<br>
mark1 1 x l find_left
mark1 _ _ r halt-reject
mark1 * * l mark1<br>
find_left _ _ r mark0
find_left * * l find_left<br>
final_check _ _ r halt-accept
final_check x x l final_check
final_check * * l halt-reject
</code></pre></div></div>

<h1 id="turing-machine-variants">Turing Machine Variants</h1>

<h2 id="turing-machine-variants-1">Turing Machine Variants</h2>
<ul>
  <li>Recall that DFAs, NFAs, and regex all recognize the same class of languages</li>
  <li class="fragment">Similarly, there are many variants of Turing machines that have equivalent power</li>
  <li class="fragment">In fact, as long as we prevent a machine variant from doing an <strong>infinite</strong> amount of computation in a single step, it is likely equivalent to a Turing machine</li>
</ul>

<h2 id="two-way-infinite-tapes">Two-Way Infinite Tapes</h2>
<ul>
  <li>A common variant of the Turing machine is one where blank spaces extend off into infinity in <strong>both directions</strong></li>
  <li class="fragment">Why is this equivalent to a one-way infinite tape?</li>
  <li class="fragment">How can we prove they are equivalent?
    <ul>
      <li class="fragment">It’s easy to simulate a one-way tape with a two-way (just don’t use the extra spaces)</li>
      <li class="fragment">The tricky part is simulating a two-way tape with a one-way tape</li>
    </ul>
  </li>
</ul>

<h2 id="multi-tape-machines">Multi-Tape Machines</h2>
<ul>
  <li>Sometimes it is convenient to have <strong>multiple tapes</strong> so that we can do some computations on an axillary tape</li>
</ul>

<h2 id="nondeterministic-tms">Nondeterministic TMs</h2>
<ul>
  <li>Similar to NFAs being equivalent in power to DFAs, <strong>nondeterministic Turing machines</strong> are equivalent to the normal deterministic kind</li>
</ul>

<h1 id="enumerators">Enumerators</h1>

<h2 id="encoding-objects-as">Encoding Objects as</h2>
<h1 id="strings">Strings</h1>

<h2 id="encoding-lists-as-strings">Encoding Lists as Strings</h2>
<ul>
  <li>Suppose I have a list of numbers <code class="language-plaintext highlighter-rouge">$L$</code> defined by:
    <ul>
      <li><code class="language-plaintext highlighter-rouge">$53, 7129, -22, 46$</code></li>
    </ul>
  </li>
  <li class="fragment">How can I encode <code class="language-plaintext highlighter-rouge">$L$</code> as a string?
    <ul>
      <li class="fragment">It literally <strong>already is a string</strong> (uses digits, dash, and comma symbols)</li>
    </ul>
  </li>
  <li class="fragment">How can I encode <code class="language-plaintext highlighter-rouge">$L$</code> using only <code class="language-plaintext highlighter-rouge">$0$</code>s and <code class="language-plaintext highlighter-rouge">$1$</code>s?
    <ul>
      <li class="fragment">Assign each symbol a four digit “code” of <code class="language-plaintext highlighter-rouge">$0$</code>s and <code class="language-plaintext highlighter-rouge">$1$</code>s and treat sets of four bits as a single symbol</li>
    </ul>
  </li>
</ul>

<h2 id="encoding-objects-as-strings">Encoding Objects as Strings</h2>
<p>Suppose I have a graph <code class="language-plaintext highlighter-rouge">$G$</code> defined by:</p>

<div class="jekyll-diagrams diagrams graphviz">
  

<svg width="332pt" height="111pt" viewBox="0.00 0.00 332.00 111.31" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink">
<g id="graph0" class="graph" transform="scale(1 1) rotate(0) translate(4 107.31)">
<title>example</title>
<polygon fill="white" stroke="transparent" points="-4,4 -4,-107.31 328,-107.31 328,4 -4,4" />

<g id="node1" class="node">
<title>1</title>
<ellipse fill="none" stroke="black" cx="18" cy="-31.31" rx="18" ry="18" />
<text text-anchor="middle" x="18" y="-27.61" font-family="Times,serif" font-size="14.00">1</text>
</g>

<g id="node2" class="node">
<title>2</title>
<ellipse fill="none" stroke="black" cx="90" cy="-31.31" rx="18" ry="18" />
<text text-anchor="middle" x="90" y="-27.61" font-family="Times,serif" font-size="14.00">2</text>
</g>

<g id="edge1" class="edge">
<title>1->2</title>
<path fill="none" stroke="black" d="M35.24,-25.51C43.36,-24.59 53.41,-24.34 62.61,-24.74" />
<polygon fill="black" stroke="black" points="62.62,-28.25 72.85,-25.52 63.14,-21.27 62.62,-28.25" />
</g>

<g id="node4" class="node">
<title>4</title>
<ellipse fill="none" stroke="black" cx="90" cy="-85.31" rx="18" ry="18" />
<text text-anchor="middle" x="90" y="-81.61" font-family="Times,serif" font-size="14.00">4</text>
</g>

<g id="edge2" class="edge">
<title>1->4</title>
<path fill="none" stroke="black" d="M32.83,-41.98C42.61,-49.52 55.92,-59.79 67.13,-68.44" />
<polygon fill="black" stroke="black" points="65.14,-71.32 75.2,-74.66 69.42,-65.78 65.14,-71.32" />
</g>

<g id="edge3" class="edge">
<title>2->1</title>
<path fill="none" stroke="black" d="M72.85,-37.1C64.74,-38.01 54.7,-38.27 45.49,-37.88" />
<polygon fill="black" stroke="black" points="45.47,-34.37 35.24,-37.11 44.95,-41.35 45.47,-34.37" />
</g>

<g id="node3" class="node">
<title>3</title>
<ellipse fill="none" stroke="black" cx="162" cy="-73.31" rx="18" ry="18" />
<text text-anchor="middle" x="162" y="-69.61" font-family="Times,serif" font-size="14.00">3</text>
</g>

<g id="edge4" class="edge">
<title>2->3</title>
<path fill="none" stroke="black" d="M105.85,-40.21C115.03,-45.73 127.03,-52.93 137.49,-59.2" />
<polygon fill="black" stroke="black" points="135.8,-62.27 146.18,-64.41 139.4,-56.27 135.8,-62.27" />
</g>

<g id="node5" class="node">
<title>5</title>
<ellipse fill="none" stroke="black" cx="234" cy="-27.31" rx="18" ry="18" />
<text text-anchor="middle" x="234" y="-23.61" font-family="Times,serif" font-size="14.00">5</text>
</g>

<g id="edge5" class="edge">
<title>2->5</title>
<path fill="none" stroke="black" d="M108.13,-30.82C132.33,-30.14 176.79,-28.89 205.61,-28.08" />
<polygon fill="black" stroke="black" points="205.91,-31.57 215.81,-27.79 205.72,-24.57 205.91,-31.57" />
</g>

<g id="edge7" class="edge">
<title>3->5</title>
<path fill="none" stroke="black" d="M177.5,-63.78C186.89,-57.61 199.31,-49.45 210.02,-42.41" />
<polygon fill="black" stroke="black" points="212.08,-45.24 218.51,-36.83 208.24,-39.39 212.08,-45.24" />
</g>

<g id="edge6" class="edge">
<title>4->3</title>
<path fill="none" stroke="black" d="M107.95,-82.4C115.86,-81.05 125.46,-79.4 134.31,-77.88" />
<polygon fill="black" stroke="black" points="134.95,-81.32 144.21,-76.19 133.76,-74.43 134.95,-81.32" />
</g>

<g id="edge8" class="edge">
<title>5->1</title>
<path fill="none" stroke="black" d="M216.82,-21.01C187.58,-10.52 124.17,8.21 72,-4.31 61.72,-6.78 51.18,-11.59 42.23,-16.48" />
<polygon fill="black" stroke="black" points="40.46,-13.46 33.57,-21.51 43.97,-19.52 40.46,-13.46" />
</g>

<g id="node6" class="node">
<title>6</title>
<ellipse fill="none" stroke="black" cx="306" cy="-27.31" rx="18" ry="18" />
<text text-anchor="middle" x="306" y="-23.61" font-family="Times,serif" font-size="14.00">6</text>
</g>

<g id="edge9" class="edge">
<title>5->6</title>
<path fill="none" stroke="black" d="M251.24,-21.51C259.36,-20.59 269.41,-20.34 278.61,-20.74" />
<polygon fill="black" stroke="black" points="278.62,-24.25 288.85,-21.52 279.14,-17.27 278.62,-24.25" />
</g>

<g id="edge10" class="edge">
<title>6->5</title>
<path fill="none" stroke="black" d="M288.85,-33.1C280.74,-34.01 270.7,-34.27 261.49,-33.88" />
<polygon fill="black" stroke="black" points="261.47,-30.37 251.24,-33.11 260.95,-37.35 261.47,-30.37" />
</g>
</g>
</svg>

</div>

<hr />

<ul>
  <li class="fragment">Now suppose you need to write <code class="language-plaintext highlighter-rouge">$G$</code> to a file
    <ul>
      <li class="fragment">How would you encode <code class="language-plaintext highlighter-rouge">$G$</code> in such a way that you could reliably read it back in?</li>
    </ul>
  </li>
</ul>

<h2 id="encoding-graphs-as-strings">Encoding Graphs as Strings</h2>
<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code>digraph example {
  "1" [shape=circle]
  "2" [shape=circle]
  "3" [shape=circle]
  "4" [shape=circle]
  "5" [shape=circle]
  "6" [shape=circle]<br>
  "1"  ->  "2"
  "1"  ->  "4"
  "2"  ->  "1"
  "2"  ->  "3"
  "2"  ->  "5"
  "4"  ->  "3"
  "3"  ->  "5"
  "5"  ->  "1"
  "5"  ->  "6"
  "6"  ->  "5"
}
</code></pre></div></div>

<h2 id="encoding-graphs-as-strings-1">Encoding Graphs as Strings</h2>
<div class="jekyll-diagrams diagrams graphviz">
  

<svg width="332pt" height="111pt" viewBox="0.00 0.00 332.00 111.31" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink">
<g id="graph0" class="graph" transform="scale(1 1) rotate(0) translate(4 107.31)">
<title>example</title>
<polygon fill="white" stroke="transparent" points="-4,4 -4,-107.31 328,-107.31 328,4 -4,4" />

<g id="node1" class="node">
<title>1</title>
<ellipse fill="none" stroke="black" cx="18" cy="-31.31" rx="18" ry="18" />
<text text-anchor="middle" x="18" y="-27.61" font-family="Times,serif" font-size="14.00">1</text>
</g>

<g id="node2" class="node">
<title>2</title>
<ellipse fill="none" stroke="black" cx="90" cy="-31.31" rx="18" ry="18" />
<text text-anchor="middle" x="90" y="-27.61" font-family="Times,serif" font-size="14.00">2</text>
</g>

<g id="edge1" class="edge">
<title>1->2</title>
<path fill="none" stroke="black" d="M35.24,-25.51C43.36,-24.59 53.41,-24.34 62.61,-24.74" />
<polygon fill="black" stroke="black" points="62.62,-28.25 72.85,-25.52 63.14,-21.27 62.62,-28.25" />
</g>

<g id="node4" class="node">
<title>4</title>
<ellipse fill="none" stroke="black" cx="90" cy="-85.31" rx="18" ry="18" />
<text text-anchor="middle" x="90" y="-81.61" font-family="Times,serif" font-size="14.00">4</text>
</g>

<g id="edge2" class="edge">
<title>1->4</title>
<path fill="none" stroke="black" d="M32.83,-41.98C42.61,-49.52 55.92,-59.79 67.13,-68.44" />
<polygon fill="black" stroke="black" points="65.14,-71.32 75.2,-74.66 69.42,-65.78 65.14,-71.32" />
</g>

<g id="edge3" class="edge">
<title>2->1</title>
<path fill="none" stroke="black" d="M72.85,-37.1C64.74,-38.01 54.7,-38.27 45.49,-37.88" />
<polygon fill="black" stroke="black" points="45.47,-34.37 35.24,-37.11 44.95,-41.35 45.47,-34.37" />
</g>

<g id="node3" class="node">
<title>3</title>
<ellipse fill="none" stroke="black" cx="162" cy="-73.31" rx="18" ry="18" />
<text text-anchor="middle" x="162" y="-69.61" font-family="Times,serif" font-size="14.00">3</text>
</g>

<g id="edge4" class="edge">
<title>2->3</title>
<path fill="none" stroke="black" d="M105.85,-40.21C115.03,-45.73 127.03,-52.93 137.49,-59.2" />
<polygon fill="black" stroke="black" points="135.8,-62.27 146.18,-64.41 139.4,-56.27 135.8,-62.27" />
</g>

<g id="node5" class="node">
<title>5</title>
<ellipse fill="none" stroke="black" cx="234" cy="-27.31" rx="18" ry="18" />
<text text-anchor="middle" x="234" y="-23.61" font-family="Times,serif" font-size="14.00">5</text>
</g>

<g id="edge5" class="edge">
<title>2->5</title>
<path fill="none" stroke="black" d="M108.13,-30.82C132.33,-30.14 176.79,-28.89 205.61,-28.08" />
<polygon fill="black" stroke="black" points="205.91,-31.57 215.81,-27.79 205.72,-24.57 205.91,-31.57" />
</g>

<g id="edge7" class="edge">
<title>3->5</title>
<path fill="none" stroke="black" d="M177.5,-63.78C186.89,-57.61 199.31,-49.45 210.02,-42.41" />
<polygon fill="black" stroke="black" points="212.08,-45.24 218.51,-36.83 208.24,-39.39 212.08,-45.24" />
</g>

<g id="edge6" class="edge">
<title>4->3</title>
<path fill="none" stroke="black" d="M107.95,-82.4C115.86,-81.05 125.46,-79.4 134.31,-77.88" />
<polygon fill="black" stroke="black" points="134.95,-81.32 144.21,-76.19 133.76,-74.43 134.95,-81.32" />
</g>

<g id="edge8" class="edge">
<title>5->1</title>
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<hr />

<h2 id="encoding-objects-as-strings-1">Encoding Objects as Strings</h2>
<ul>
  <li>There are an <strong>infinite</strong> number of ways to encode an object like a graph as a string</li>
  <li>The encoding of choice doesn’t matter that much as long as it uniquely specifies the desired object</li>
  <li class="fragment">Sipser uses notation <code class="language-plaintext highlighter-rouge">$\langle G\rangle$</code> to denote an arbitrary string-encoding of a finite object like a graph</li>
</ul>

<h2 id="example">Example</h2>
<p>Show that the following language is <strong>decidable</strong></p>

<p><code class="language-plaintext highlighter-rouge">$$L = \{\langle S, n\rangle \mid S\subseteq\mathbb{N}\text{ is finite and }n\in S\}$$</code></p>

<hr />

<ul>
  <li>This is basically asking us to write a function that takes in a finite set (or a list) of numbers and check if <code class="language-plaintext highlighter-rouge">$n$</code> is an element of that list</li>
</ul>

<h2 id="example-1">Example</h2>
<p><code class="language-plaintext highlighter-rouge">$$L = \{\langle S, n\rangle \mid S\subseteq\mathbb{N}\text{ is finite and }n\in S\}$$</code></p>

<hr />

<ul>
  <li>Take <strong>one minute</strong> and try to write a function in Python that decides this language</li>
  <li>In other words, finish writing the function:</li>
</ul>

<div class="language-py highlighter-rouge"><div class="highlight"><pre class="highlight"><code><span class="k">def</span> <span class="nf">decider</span><span class="p">(</span><span class="n">S</span><span class="p">,</span> <span class="n">n</span><span class="p">):</span>
    <span class="s">"""Checks if n is an element of S<br>
    Parameters:
        S: a list of integers
        n: an integer
    """</span>
    <span class="p">...</span>
</code></pre></div></div>

<h2 id="example-2">Example</h2>
<p><code class="language-plaintext highlighter-rouge">$$L = \{\langle S, n\rangle \mid S\subseteq\mathbb{N}\text{ is finite and }n\in S\}$$</code></p>

<hr />

<ul>
  <li>Here is a <strong>high-level</strong> Turing machine that decides <code class="language-plaintext highlighter-rouge">$L$</code></li>
</ul>

<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code>M = On input &lt;S,n&gt;,
  0. Check to see if &lt;S,n&gt; is properly formatted; if not,
     immediately REJECT
  1. For each element x in S
     a. Compare x to n
     b. If they are the same, then immediately ACCEPT; otherwise
        continue with the for-loop
  2. If we checked all elements of S and still did not find n,
     then REJECT
</code></pre></div></div>

<h2 id="another-example">Another Example</h2>
<p>Consider the language defined by</p>

<p><code class="language-plaintext highlighter-rouge">$$L = \{\langle p\rangle \mid p\text{ is a polynomial with integral roots}\}$$</code></p>

<hr />

<ul>
  <li class="fragment">Is <code class="language-plaintext highlighter-rouge">$L$</code> <strong>decidable</strong>?</li>
  <li class="fragment">Is <code class="language-plaintext highlighter-rouge">$L$</code> <strong>Turing recognizable</strong>?</li>
</ul>

<h2 id="another-example-1">Another Example</h2>

<p><code class="language-plaintext highlighter-rouge">$$L = \{\langle p\rangle \mid p\text{ is a polynomial with integral roots}\}$$</code></p>

<hr />

<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code>M = On input &lt;p&gt; where p is a polynomial over variables x1,...,xk,
  1. For each possible integral assignment of x1,...,xk:
     a. Evaluate the polynomial p with the current assignment
     b. If p evaluates to zero, then immediately ACCEPT
  2. If no assignment ever evaluates to zero, then REJECT
</code></pre></div></div>

<hr />

<ul>
  <li class="fragment">Will line 2 ever be reached by this machine?
    <ul>
      <li class="fragment">No. There are an infinite number of assignments to the variables, so the for-loop is infinite</li>
    </ul>
  </li>
  <li class="fragment">Does this machine recognize <code class="language-plaintext highlighter-rouge">$L$</code>?</li>
</ul>

<h1 id="church-turing-thesis">Church-Turing Thesis</h1>