Lecture 6

In-class notes: CS 505 Spring 2025 Lecture 6

Brief Aside on Reductions

When we say that a language $A$ is polynomial-time reducible to a language $B$, denoted as $A{\le }_{p}B$, we are saying the following.

Given ANY string $x$, I can transform $x$ into SOME $f(x)$ in polynomial time such that $$x\in A⟺f(x)\in B.$$

Looking back at our proof that $INDSE{T}_k$ is NP-complete, we showed how to transform ANY 3CNF formula $\varphi$ into SOME PARTICULAR graph $G$ such that $\varphi$ was satisfiable if and only if $G$ has an independent set of size $k$. This is NOT saying that given ANY graph $G$, I can solve ANY 3CNF formula.

More NP-Complete Problems

We’ll show one more problem is NP-complete. Given an undirected graph $G=(V,E)$, a $k$-clique is a set of vertices $S\subset V$ such that $|S|=k$ and $\forall u,v\in S$, we have $(u,v)\in E$.¹ Let $CLIQU{E}_k$ be the set of all graphs $G$ that have a $k$-clique.

Theorem. $CLIQU{E}_k$ is NP-complete.

Proof. Again, $CLIQU{E}_k\in \mathbf{NP}$ is immediate.

To show that $CLIQU{E}_k$ is NP-hard, we reduce 3SAT to $CLIQU{E}_k$. That is, we show $3SAT{\le }_{p}CLIQU{E}_k$. As with $INDSE{T}_k$, let $\varphi ={\varphi }_1\wedge \cdots \wedge {\varphi }_k$ be a 3CNF formula. Given $\varphi$, we construct a graph with $3k$ vertices as follows. Label each literal of each clause ${\varphi }_i$ as $a_i,b_i,c_i$, so ${\varphi }_i=(a_i\vee b_i\vee c_i)$. For each clause ${\varphi }_i$, add a cluster of $3$ nodes to the graph $G$ with labels $a_i,b_i,c_i$. So we have $k$ clusters of $3$ nodes, where cluster $i$ has nodes $a_i,b_i,c_i$.

We now add edges to the graph $G$. The graph $G$ will have every node $u,v$ connected with edge $(u,v)$, except for the following.

• Nodes $a_i$, $b_i$, and $c_i$ in cluster $i$ (corresponding to ${\varphi }_i$) will not be connected to each other.
• If node $u$ is in cluster $i$ and node $v$ is in cluster $j$, we do not connect $u$ and $v$ if they are a literal and their negation. For example, if $u={\overline{x}}_4$ and $v=x_4$, then we do not connect the nodes. We argue this graph has a $k$-clique if and only if $\varphi$ is satisfiable.

First assume that $\varphi$ has a satisfying assignment. Assume $\varphi$ has $\ell$ literals $x_1,\dots ,x_{\ell }$ and let $u\in \{0,1{\}}^{\ell }$ be the satisfying assignment. Now, we construct a $k$-clique in $G$ using the satisfying assignment $u$. Since $u$ is satisfying, there is at least one literal in each ${\varphi }_i$ that has a satisfied variable. For example, in ${\varphi }_i=(a_i\vee b_i\vee c_i)$, we can have $a_i=1$ under the assignment $u$.

For each clause ${\varphi }_i$, pick one of the literals that is satisfied by $u$. Suppose this is $b_i$. Then, add the node $b_i$ in the graph $G$ to the set $S$. Repeat this process for all ${\varphi }_i$ for $i\in [k]$. Clearly $|S|=k$. Now, we claim that $S$ is a $k$-clique. By construction of the graph $G$, we know that

• All clusters of nodes are not connected to each other; and
• Nodes corresponding to labels $x_i$ and ${\overline{x}}_i$ are not connected to each other; and
• All other nodes are connected to each other. This implies that every node $s\in S$ is connected to every other node $s^{\prime }\in S$, so it is a $k$-clique.

Now, assume that $G$ has a $k$-clique, denoted by $S$. We construct a satisfying assignment for $\varphi$ using $S$. Again, by our construction, we know that:

• All clusters of nodes are not connected to each other; and
• Nodes corresponding to labels $x_i$ and ${\overline{x}}_i$ are not connected to each other; and
• All other nodes are connected to each other. Since $S$ is a $k$-clique, we know that $|S|=k$ and for all $s,s^{\prime }\in S$, we have $(s,s^{\prime })\in E$. We know that $s$ and $s^{\prime }$ cannot correspond to conflicting variables (i.e., $s=x_1$ and $s={\overline{x}}_1$ is not possible if both $s,s^{\prime }\in S$). So to construct the satisfying assignment for $\varphi$, for any $s\in S$, we assign the literal correspoding to $s$ a $1$. Again by our construction of the graph $G$, we know that each cluster of $3$ nodes corresponding to the clause ${\varphi }_i$ are not connected to each other, but they are connected to everything else. So we know $S$ contains exactly $1$ node from each cluster of $3$ nodes. Thus, we have created an assignment which satisfies all clauses. This completes the proof. $\square$

Search vs. Decision Problems

We’ve stated complexity classes as decision problems, but one may naturally consider search variants where we are asked to find solutions. For the class $\mathbf{P}$, these two notions are the same: if I can decide a problem, then I can search for an actual solution (and vice versa).

If $\mathbf{P}\ne \mathbf{NP}$, then we cannot efficiently search for solutions (i.e, certificates) for decision problems in $\mathbf{NP}$. On the other hand, if $\mathbf{P}=\mathbf{NP}$, then we can.

Theorem. If $\mathbf{P}=\mathbf{NP}$, then for every $L\in \mathbf{NP}$ with efficient deterministic verifier $M_L$, there exists a polynomial-time deterministic machine ${\hat{M}}_L$ such that for all $x\in L$, ${\hat{M}}_L(x)=w$ and $M_L(x,w)=1$.

Proof. We show this is true for SAT, which implies it holds for all of NP since SAT is NP-complete. Suppose that $A$ is a decider for $SAT$. That is, if $\varphi$ is a Boolean formula, then $A(\varphi )=1$ if and only if $\varphi$ is satisfiable. Since $\mathbf{P}=\mathbf{NP}$, we know that $A$ is deterministic and runs in polynomial time.

Now we build a new machine $B$ which outputs a satisfying assignment for $\varphi$ if $A(\varphi )=1$. Suppose that $\varphi$ has $\ell$ literals $x_1,\dots ,x_{\ell }$. The algorithm $B$ operate as follows.

• On input $\varphi$, the machine $B$:
  • Runs $A(\varphi )$.
  • If $A(\varphi )=0$, output reject.
  • Else if $A(\varphi )=1$, we know $\varphi$ has a satisfying assignment.
  • Set $\hat{\varphi }=\varphi$ and set $u=‘‘"$ (the empty string).
  • For $i=\{1,2,\dots ,\ell \}$:
    • Let ${\psi }_0=\hat{\varphi }(x_i=0)$ and ${\psi }_1=\hat{\varphi }(x_i=1)$ be the Boolean formulas obtained by setting literal $x_i$ in $\hat{\varphi }$ to $0$ and $1$, respectively.
    • Compute $b_0=A({\psi }_0)$ and $b_1=A({\psi }_1)$.
    • If $b_0=1$ then set $\hat{\varphi }={\psi }_0$ and $u=u\Vert 0$.
    • Else if $b_1=1$ then set $\hat{\varphi }={\psi }_1$ and $u=u\Vert 1$.
  • Output $u$.

Clearly, since $A$ runs in polynomial time, then $B$ runs in polynomial time (in the length of $\varphi$). By our construction, if $\varphi$ is satisfiable, then the machine $B$ correctly reconstructs a satisfying assignment. In particular, during every step of the for loop, at least one of $b_0$ or $b_1$ will be equal to $1$. If not, then the original formula $\varphi$ would not be satisfiable, which means we would have already rejected. $\square$

The Complexity Class coNP

We’ll now discuss the co-class to NP, which we call coNP. This is defined by the set of languages with complements in NP. That is, $$\mathbf{coNP}=\{L:\overline{L}\in \mathbf{NP}\}.$$

In fact, we know that $\mathbf{NP}\cap \mathbf{coNP}\ne \varnothing$.

Theorem. $\mathbf{P}\subset \mathbf{NP}\cap \mathbf{coNP}$.

At a high level, this theorem follows since if $L\in \mathbf{P}$, then $\overline{L}\in \mathbf{P}$.

Alternative view of coNP

The above definition of coNP isn’t very useful for understanding what languages in the class look like. So we consider the following equivalent definition.

Definition ($\mathbf{coNP}$). We say a language $L\in \mathbf{coNP}$ if there exists a polynomial $p$ and a polynomial-time deterministic Turing machine $M$ such that $$x\in L⟺\forall w\in \{0,1{\}}^{p(|x|)}:M(x,w)=1.$$

Notice this is the opposite of NP! In the similar definition of NP, we have a “there exists” ($\exists$) rather than the “for all” ($\forall$) in the above definition.

Theorem. If $\mathbf{P}=\mathbf{NP}$, then $\mathbf{NP}=\mathbf{coNP}$. Or, equivalently stated, if $\mathbf{NP}\ne \mathbf{coNP}$, then $\mathbf{P}\ne \mathbf{NP}$.

In general, we do not believe that $\mathbf{NP}=\mathbf{coNP}$.

coNP-complete Problems

Just like with the NP, we can equivalently define coNP-complete problems. A language $L$ is coNP-complete if $L\in \mathbf{coNP}$ and $L^{\prime }{\le }_{p}L$ for all $L^{\prime }\in \mathbf{coNP}$.

We’ll look at the following problem: deciding if a formula $\varphi$ is a tautology. That is, every assignment of variables in $\varphi$ is a satisfying assignment.

$$TAUT=\{\varphi :\varphi \text{ is a tautology}\}.$$

Theorem. $TAUT$ is coNP-complete.

Proof. Clearly $TAUT\in \mathbf{coNP}$ since we can build a machine $M$ such that for any $\varphi$, and any assignment of variables $w$, $M(\varphi ,w)=1$ if and only if $\varphi$ is satisfied. By definition of coNP, this machine must output $1$ for all assignments $w$, which is true if $\varphi$ is a tautology.

Now we show that $TAUT$ is coNP-hard. Let $L\in \mathbf{coNP}$ be any language. Consider $\overline{L}\in \mathbf{NP}$, its complement language that is in NP (which follows by definition of coNP).

By the Cook-Levin theorem, let ${\varphi }_x$ be the formula such that $x\in \overline{L}⟺{\varphi }_x$ is satisfiable, for any $x\in \{0,1{\}}^{\ast }$. Then, we know $$\begin{array}{r}x\in \overline{L}⟺{\varphi }_x\text{ is satisfaible}\\ x\notin \overline{L}⟺{\varphi }_x\text{ is not satisfaible}\\ x\in L⟺{\varphi }_x\text{ is not satisfaible}\\ x\in L⟺\overline{{\varphi }_x}\text{ is a tautology}\end{array}$$ Thus, every $L\in \mathbf{coNP}$ is polynomial-time reducible to $TAUT$. $\square$

The Complexity Class NEXP

Recall the definition of EXP: $$\mathbf{EXP}=\bigcup _c\mathbf{DTIME}(2^{n^c}).$$

We can equivalently define the class of languages decidable in non-deterministic exponential time: NEXP.

$$\mathbf{NEXP}=\bigcup _c\mathbf{NTIME}(2^{n^c}).$$

Given this, we have a DTIME/NTIME hierarchy of classes:

$$\mathbf{P}\subset \mathbf{NP}\subset \mathbf{EXP}\subset \mathbf{NEXP}.$$

Perhaps surprisingly, we get a “collapse” if $\mathbf{P}=\mathbf{NP}$.

Theorem. If $\mathbf{P}=\mathbf{NP}$, then $\mathbf{EXP}=\mathbf{NEXP}$. Or, equivalently stated, if $\mathbf{EXP}\ne \mathbf{NEXP}$, then $\mathbf{P}\ne \mathbf{NP}$.

1. Note this is the opposite of a $k$-independent set. ↩