Np Completeness | CS 139: Theory of Computation

<h1 id="np-completeness">NP Completeness</h1>

<hr />

<h2 id="assignment-8">Assignment 8</h2>
<ul>
  <li>Assignment 8 is due today at 11:59 PM</li>
  <li class="fragment">Any questions?</li>
</ul>

<h2 id="assignment-9">Assignment 9</h2>
<ul>
  <li>Assignment 9 is now released</li>
  <li class="fragment">It is a significantly different assignment, where you will create a <strong>video presentation</strong> about a certain NP complete problem</li>
  <li class="fragment">The assignment will be completed in <strong>groups of two</strong>
    <ul>
      <li class="fragment">I will assign the groups based on similar problem interest and availability</li>
    </ul>
  </li>
  <li class="fragment">There is a short questionnaire due <strong>Thursday, April 30th</strong></li>
  <li class="fragment">The video itself is due on <strong>Thursday, May 7th</strong></li>
</ul>

<h1 id="np-completeness-1">NP Completeness</h1>

<h2 id="np-completeness-2">NP Completeness</h2>
<ul>
  <li>Last time we concluded by discussing:</li>
  <li><strong>Theorem</strong> (Cook, Levin): <code class="language-plaintext highlighter-rouge">$\text{SAT}\in P \iff \text{P} = \text{NP}$</code></li>
</ul>

<hr />

<p><code class="language-plaintext highlighter-rouge">$$\text{SAT} = \{ \langle\Phi\rangle\mid\Phi\text{ is a satisfiable Boolean formula}\}$$</code></p>

<hr />

<ul>
  <li class="fragment">This means that <code class="language-plaintext highlighter-rouge">$\text{SAT}$</code> is a <strong>hard</strong> problem to solve
    <ul>
      <li class="fragment">If we write a fast algorithm for <code class="language-plaintext highlighter-rouge">$\text{SAT}$</code>, then we get a fast algorithm for <strong>every</strong> <code class="language-plaintext highlighter-rouge">$\text{NP}$</code> problem</li>
    </ul>
  </li>
  <li class="fragment">Languages like <code class="language-plaintext highlighter-rouge">$\text{SAT}$</code> are called <strong>NP complete</strong>, since they are connected to the complete class of <code class="language-plaintext highlighter-rouge">$\text{NP}$</code> problems</li>
</ul>

<h2 id="np-completeness-3">NP Completeness</h2>
<ul>
  <li>NP completeness has many practical applications</li>
  <li>We believe that <code class="language-plaintext highlighter-rouge">$\text{P} \ne \text{NP}$</code>
    <ul>
      <li class="fragment">Thus we believe call NP complete problem <strong>cannot be solved in polynomial time</strong></li>
    </ul>
  </li>
  <li class="fragment">Proving that a language is NP complete sounds hard—but is actually surprisingly easy
    <ul>
      <li class="fragment">The technique uses <strong>mapping reductions</strong> to relate the difficulty of these languages</li>
    </ul>
  </li>
</ul>

<h2 id="poly-time-mapping-reductions">Poly-Time Mapping Reductions</h2>
<ul>
  <li>A function <code class="language-plaintext highlighter-rouge">$f:\Sigma^\ast\rightarrow\Gamma^\ast$</code> is a <strong>polynomial-time computable function</strong> if it can be computed by a Turing machine in polynomial time</li>
  <li class="fragment">Language <code class="language-plaintext highlighter-rouge">$A$</code> is <strong>polynomial-time mapping reducible</strong> to <code class="language-plaintext highlighter-rouge">$B$</code> if there is a poly-time computable function <code class="language-plaintext highlighter-rouge">$f$</code>, where for every <code class="language-plaintext highlighter-rouge">$w\in\Sigma^\ast$</code>:
    <ul>
      <li><code class="language-plaintext highlighter-rouge">$w\in A$</code> if and only if <code class="language-plaintext highlighter-rouge">$f(w) \in B$</code></li>
    </ul>
  </li>
  <li class="fragment">We also write <code class="language-plaintext highlighter-rouge">$A\le_p B$</code> if <code class="language-plaintext highlighter-rouge">$A$</code> is poly-time mapping reducible to <code class="language-plaintext highlighter-rouge">$B$</code></li>
</ul>

<h2 id="poly-time-mapping-reductions-1">Poly-Time Mapping Reductions</h2>
<ul>
  <li>Note that <code class="language-plaintext highlighter-rouge">$A \le_p B$</code> means the same thing as <code class="language-plaintext highlighter-rouge">$A \le_m B$</code> except that the reduction for the former is guaranteed to run in poly-time</li>
  <li class="fragment"><strong>Theorem:</strong> If <code class="language-plaintext highlighter-rouge">$A \le_p B$</code> and <code class="language-plaintext highlighter-rouge">$B \in \text{P}$</code>, then <code class="language-plaintext highlighter-rouge">$A\in\text{P}$</code></li>
</ul>

<div class="language-plaintext fragment highlighter-rouge"><div class="highlight"><pre class="highlight"><code>M_A = On input w
  1. Compute f(w) where f is the reduction from A to B
  2. Run M_B on f(w) and output whatever it does
</code></pre></div></div>

<ul>
  <li class="fragment">The Turing machine <code class="language-plaintext highlighter-rouge">$M_A$</code> runs in poly-time since <code class="language-plaintext highlighter-rouge">$f$</code> is poly-time computable and <code class="language-plaintext highlighter-rouge">$M_B$</code> runs in poly-time</li>
  <li class="fragment">It also decides <code class="language-plaintext highlighter-rouge">$A$</code> since the reduction <code class="language-plaintext highlighter-rouge">$f$</code> guarantees that if <code class="language-plaintext highlighter-rouge">$w \in A \iff f(w) \in B$</code></li>
</ul>

<h2 id="examples">Examples</h2>
<ul>
  <li>We now explore an example of a poly-time reduction between two well known problems: <strong>Independent Set</strong> and  <strong>Clique</strong></li>
</ul>

<h2 id="independent-set">Independent Set</h2>
<ul>
  <li>An <strong>independent set</strong> of vertices in an undirected graph <code class="language-plaintext highlighter-rouge">$G = (V, E)$</code> is a set <code class="language-plaintext highlighter-rouge">$I\subseteq V$</code> where no two vertices <code class="language-plaintext highlighter-rouge">$x,y \in I$</code> are connected by an edge in <code class="language-plaintext highlighter-rouge">$E$</code></li>
  <li class="fragment">The <strong>Independent Set Problem</strong> is the language:</li>
</ul>

<p class="fragment"><code class="language-plaintext highlighter-rouge">$$\text{IND-SET} = \{\langle G, k\rangle \mid \text{$G$ contains an independent}\\\text{set of size at least $k$}\}$$</code></p>

<h2 id="clique">Clique</h2>
<ul>
  <li>A <strong>clique</strong> of vertices in an undirected graph <code class="language-plaintext highlighter-rouge">$G = (V, E)$</code> is a set <code class="language-plaintext highlighter-rouge">$C\subseteq V$</code> where every vertex in <code class="language-plaintext highlighter-rouge">$C$</code> is connected to every other vertex in <code class="language-plaintext highlighter-rouge">$C$</code></li>
  <li class="fragment">The <strong>Clique Problem</strong> is the language:</li>
</ul>

<p class="fragment"><code class="language-plaintext highlighter-rouge">$$\text{CLIQUE} = \{\langle G, k\rangle \mid \text{$G$ contains a clique of}\\\text{size at least $k$}\}$$</code></p>

<h2 id="our-first-reduction">Our First Reduction</h2>
<ul>
  <li><strong>Claim:</strong> <code class="language-plaintext highlighter-rouge">$\text{CLIQUE} \le_p \text{IND-SET}$</code></li>
  <li class="fragment">Need a function <code class="language-plaintext highlighter-rouge">$f(\langle G, k, \rangle) = \langle \widehat{G}, \widehat{k}\rangle$</code> that satisfies:
    <ul>
      <li class="fragment">If <code class="language-plaintext highlighter-rouge">$G$</code> <strong>has</strong> a clique of size <code class="language-plaintext highlighter-rouge">$k$</code>, then <code class="language-plaintext highlighter-rouge">$\widehat{G}$</code> has an independent set of size <code class="language-plaintext highlighter-rouge">$\widehat{k}$</code></li>
      <li class="fragment">If <code class="language-plaintext highlighter-rouge">$G$</code> <strong>does not have</strong> a clique of size <code class="language-plaintext highlighter-rouge">$k$</code>, then <code class="language-plaintext highlighter-rouge">$\widehat{G}$</code> does not have an independent set of size <code class="language-plaintext highlighter-rouge">$\widehat{k}$</code></li>
      <li class="fragment">Needs to be computable in <strong>polynomial</strong> time</li>
    </ul>
  </li>
</ul>

<h2 id="our-first-reduction-1">Our First Reduction</h2>
<ul>
  <li>Note that a clique is a <strong>fully connected</strong> subgraph and an independent set is a <strong>fully disconnected</strong> subgraph</li>
  <li class="fragment">So what can we do to <code class="language-plaintext highlighter-rouge">$G$</code> that turns cliques into independent sets?</li>
  <li class="fragment"><strong>Main Idea:</strong> Create <code class="language-plaintext highlighter-rouge">$\widehat{G} = (\widehat{V}, \widehat{E})$</code> by “inverting” the edges of <code class="language-plaintext highlighter-rouge">$G=(V, E)$</code>
    <ul>
      <li class="fragment">If <code class="language-plaintext highlighter-rouge">$\{x,y\}\in E$</code>, then <code class="language-plaintext highlighter-rouge">$\{x,y\}\not\in\widehat{E}$</code></li>
      <li class="fragment">If <code class="language-plaintext highlighter-rouge">$\{x,y\}\not\in E$</code>, then <code class="language-plaintext highlighter-rouge">$\{x,y\}\in\widehat{E}$</code></li>
      <li class="fragment">More concisely, <code class="language-plaintext highlighter-rouge">$\widehat{G} = (V, \overline{E})$</code></li>
    </ul>
  </li>
</ul>

<h2 id="our-first-reduction-2">Our First Reduction</h2>
<div class="language-plaintext highlighter-rouge"><div class="highlight"><pre class="highlight"><code>M_f = On input &lt;G,k&gt where G = (V,E);
  1. Construct G' = (V, E') by inverting the edges of G
  2. Output &lt;G',k&gt
</code></pre></div></div>

<ul>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$f$</code> clearly runs in polynomial time because it only needs to iterative over <code class="language-plaintext highlighter-rouge">$V$</code> and <code class="language-plaintext highlighter-rouge">$E$</code> once to construct <code class="language-plaintext highlighter-rouge">$G'$</code></li>
  <li class="fragment">If <code class="language-plaintext highlighter-rouge">$G$</code> contains a <code class="language-plaintext highlighter-rouge">$k$</code>-clique, then the inverted graph must contain a <code class="language-plaintext highlighter-rouge">$k$</code>-independent set</li>
  <li class="fragment">Similarly, if <code class="language-plaintext highlighter-rouge">$G'$</code> contains a <code class="language-plaintext highlighter-rouge">$k$</code>-independent set, then the original graph must contain a <code class="language-plaintext highlighter-rouge">$k$</code>-clique</li>
</ul>

<h2 id="np-completeness-4">NP Completeness</h2>
<ul>
  <li>A language <code class="language-plaintext highlighter-rouge">$B$</code> is called <strong>NP hard</strong> if <code class="language-plaintext highlighter-rouge">$A\le_p B$</code> holds for each language <code class="language-plaintext highlighter-rouge">$A\in\text{NP}$</code></li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$B$</code> is called <strong>NP complete</strong> if <code class="language-plaintext highlighter-rouge">$B\in\text{NP}$</code> and it is NP hard</li>
</ul>

<hr />

<ul>
  <li class="fragment"><strong>Theorem</strong> (Cook Levin): <code class="language-plaintext highlighter-rouge">$\text{SAT}$</code> is NP complete</li>
  <li class="fragment">The proof of this theorem is quite involved, but we can prove other problems are NP complete using <strong>reductions</strong></li>
</ul>

<h2 id="np-completeness-5">NP Completeness</h2>
<ul>
  <li><strong>Fact:</strong> If <code class="language-plaintext highlighter-rouge">$A$</code> is NP hard and <code class="language-plaintext highlighter-rouge">$A\le_p B$</code>, then <code class="language-plaintext highlighter-rouge">$B$</code> is NP hard
    <ul>
      <li class="fragment">Given any language <code class="language-plaintext highlighter-rouge">$L\in\text{NP}$</code>, we know that <code class="language-plaintext highlighter-rouge">$L\le_p A$</code> since it is NP hard and that <code class="language-plaintext highlighter-rouge">$A\le_p B$</code>. This means we can reduce <code class="language-plaintext highlighter-rouge">$L$</code> to <code class="language-plaintext highlighter-rouge">$B$</code> in polynomial time</li>
    </ul>
  </li>
  <li class="fragment">Thus, to prove a language <code class="language-plaintext highlighter-rouge">$A$</code> is NP complete, we only need to do two things:
    <ol>
      <li class="fragment">Show that <code class="language-plaintext highlighter-rouge">$A\in\text{NP}$</code></li>
      <li class="fragment">Reduce an NP hard problem to <code class="language-plaintext highlighter-rouge">$A$</code></li>
    </ol>
  </li>
</ul>

<h1 id="more-examples">More Examples</h1>

<h2 id="vertex-cover">Vertex Cover</h2>
<ul>
  <li>A <strong>vertex cover</strong> of an undirected graph <code class="language-plaintext highlighter-rouge">$G = (V,E)$</code> is a set <code class="language-plaintext highlighter-rouge">$C\subseteq $</code> such that for every edge <code class="language-plaintext highlighter-rouge">$\{x,y\}\in E$</code>, either <code class="language-plaintext highlighter-rouge">$x\in C$</code> or <code class="language-plaintext highlighter-rouge">$y \in C$</code>.</li>
  <li class="fragment">The <strong>Vertex Cover Problem</strong> is the language:</li>
</ul>

<p class="fragment"><code class="language-plaintext highlighter-rouge">$$\text{VERTEX-COVER} = \{\langle G, k\rangle \mid \text{$G$ contains}\\\text{a vertex cover of size at least $k$}\}$$</code></p>

<hr />

<ul>
  <li class="fragment"><strong>Claim:</strong> <code class="language-plaintext highlighter-rouge">$\text{VERTEX-COVER}$</code> is NP complete</li>
</ul>

<h2 id="vertex-cover-1">Vertex Cover</h2>
<ul>
  <li><strong>3-2-1-GO:</strong> What do we need to show?
    <ul>
      <li class="fragment">That it is in <code class="language-plaintext highlighter-rouge">$\text{NP}$</code> and that it is NP hard</li>
    </ul>
  </li>
  <li class="fragment">How do we show that it is in <code class="language-plaintext highlighter-rouge">$\text{NP}$</code>?
    <ul>
      <li class="fragment">Write a verifier for it</li>
      <li class="fragment">Given a set <code class="language-plaintext highlighter-rouge">$C\subseteq V$</code> as a certificate, we can check if it is indeed a vertex cover</li>
    </ul>
  </li>
</ul>

<div class="language-plaintext fragment highlighter-rouge"><div class="highlight"><pre class="highlight"><code>V = On input &lt;G,k,C&gt
  1. Check if C is a subset of the vertices of G; if not, REJECT
  2. For each edge {x,y} of G:
     a. Check if x or y are in C; if neither are, then REJECT
  3. If all edges are covered by C, then ACCEPT
</code></pre></div></div>

<h2 id="vertex-cover-2">Vertex Cover</h2>
<ul>
  <li>How do we show that it is NP hard?
    <ul>
      <li class="fragment">Reduce an NP complete problem to it</li>
    </ul>
  </li>
</ul>

<hr />

<p class="fragment"><code class="language-plaintext highlighter-rouge">$$\text{3SAT} = \{ \langle\Phi\rangle\mid\Phi\text{ is a 3-CNF satisfiable formula}\}$$</code></p>

<ul>
  <li class="fragment">Boolean formulas in <strong>conjunctive normal form</strong> (<strong>CNF</strong>) are like: <code class="language-plaintext highlighter-rouge">$\Phi = (x_1\lor x_2\lor \overline{x_3})\land(x_4\lor \overline{x_2}\lor \overline{x_1})$</code>
    <ul>
      <li class="fragment">Each triplet of disjunction is called a <strong>clause</strong></li>
    </ul>
  </li>
  <li class="fragment"><code class="language-plaintext highlighter-rouge">$\text{3SAT}$</code> is a well-known NP complete problem, so showing <code class="language-plaintext highlighter-rouge">$\text{3SAT}\le_p\text{VERTEX-COVER}$</code> suffices</li>
</ul>