Polynomial Hierarchy

In the last lecture, we saw that $\mathtt{\text{Clique}}\in \mathtt{\text{NP}}$, where $$\mathtt{\text{Clique}}=\{⟨G,k⟩\mid G\text{ is an undirected graph with a clique of size }k\}.$$ What about the following related problems? $$\overline{\mathtt{\text{Clique}}}=\{⟨G,k⟩\mid G\text{ is an undirected graph with no clique of size }k\}$$ and $$\mathtt{\text{MAX-CLIQUE}}=\{⟨G,k⟩\mid G\text{ is an undirected graph with a clique of size }k\text{ and no clique of size }k+1\}$$

The two problems above are related to $\mathtt{\text{Clique}}$, but are not known to be in $\mathtt{\text{NP}}$. These problems seem to be of a different nature than $\mathtt{\text{Clique}}$, as in the case of $\overline{\mathtt{\text{Clique}}}$, how can we (efficiently) certify that a graph has no clique of size $k$? In the case of $\mathtt{\text{MAX-CLIQUE}}$, how can we (efficiently) certify that a particular clique is the largest one?

These problems, and many others, belong to a more general class of problems called the Polynomial Hierarchy.

To motivate the definition of the Polynomial Hierarchy, we will first introduce the class $\mathtt{\text{coNP}}$.

The class $\mathtt{\text{coNP}}$

Given a language $L$, we define its complement as $\bar{L}=\{x\mid x\notin L\}$. This yields the following definition:

$$\mathtt{\text{coNP}}=\{L\mid \bar{L}\in \mathtt{\text{NP}}\}$$

Note that $\mathtt{\text{coNP}}$ is not the complement of $\mathtt{\text{NP}}$, as in particular they have a non-empty intersection (see Homework 2, exercise 1.1).

A more enlightening way to define $\mathtt{\text{coNP}}$ is as follows:

Definition 1: A language $L\subseteq \{0,1{\}}^{\ast }$ is in $\mathtt{\text{coNP}}$ if there is a polynomial $p:\mathbb{N}\to \mathbb{N}$ and a polynomial-time Turing machine $M$ such that for all $x\in \{0,1{\}}^{\ast }$, we have $$x\in L⟺\mathrm{\forall }y\in \{0,1{\}}^{p(|x|)},M(x,y)=1.$$

The following proposition shows that both definitions are equivalent:

Proposition 1: A language $L$ is in $\mathtt{\text{coNP}}$ (according to definition 1) if and only if $\bar{L}$ is in $\mathtt{\text{NP}}$.

Now, with the above definitions, we can see that $\overline{\mathtt{\text{Clique}}}\in \mathtt{\text{coNP}}$.

Open question 1: Is $\mathtt{\text{NP}}=\mathtt{\text{coNP}}$?

If you solve the above question, you also prove that $\mathtt{\text{P}}\ne \mathtt{\text{NP}}$. However, it is important to note that it is not known whether $\mathtt{\text{P}}\ne \mathtt{\text{NP}}$ implies that $\mathtt{\text{NP}}\ne \mathtt{\text{coNP}}$.

The Polynomial Hierarchy

Even after defining $\mathtt{\text{NP}}$ and $\mathtt{\text{coNP}}$, there are still problems that are not known to be in either class, such as the $\mathtt{\text{MAX-CLIQUE}}$ problem. Note that $\mathtt{\text{MAX-CLIQUE}}$ seems to require both existential and universal quantifiers to be expressed.

This motivates us to define the slightly more general class ${\mathrm{\Sigma }}_2^p$.

Definition 2: A language $L\subseteq \{0,1{\}}^{\ast }$ is in ${\mathrm{\Sigma }}_2^p$ if there are polynomials $p,q:\mathbb{N}\to \mathbb{N}$ and a polynomial-time Turing machine $M$ such that for all $x\in \{0,1{\}}^{\ast }$, we have $$x\in L⟺\mathrm{\exists }y\in \{0,1{\}}^{p(|x|)}\mathrm{\forall }z\in \{0,1{\}}^{q(|x|)},M(x,y,z)=1.$$

If we interchange the order of the quantifiers in the above definition, we obtain the class ${\mathrm{\Pi }}_2^p$.

Note that, with the above definitions, we have $\mathtt{\text{MAX-CLIQUE}}\in {\mathrm{\Sigma }}_2^p\cap {\mathrm{\Pi }}_2^p$. To see this, note that we have have as the existential witness the clique of size $k$, and as the universal witness the sets of vertices of size $k+1$.

If we added more quantifiers, we would obtain the classes ${\mathrm{\Sigma }}_k^p$ and ${\mathrm{\Pi }}_k^p$ for $k\ge 3$. We have now arrived at the definition of the Polynomial Hierarchy.

Definition 3: The Polynomial Hierarchy is the class $$\mathtt{\text{PH}}=\bigcup _{k\ge 0}{\mathrm{\Sigma }}_k^p=\bigcup _{k\ge 0}{\mathrm{\Pi }}_k^p.$$

From the above definition, two natural questions arise:

1. How strong is $\mathtt{\text{PH}}$?
2. Is $\mathtt{\text{PH}}$ a proper hierarchy?
3. Does $\mathtt{\text{PH}}$ contain a complete problem (under Karp reductions)?

A first upper bound on $\mathtt{\text{PH}}$ is given by the following proposition:

Proposition 2: $\mathtt{\text{PH}}\subseteq \mathtt{\text{EXP}}$.

Proof: Let $L\in \mathtt{\text{PH}}$, then $L\in {\mathrm{\Sigma }}_k^p$ for some $k$. Let $p_1,p_2,\dots ,p_k:\mathbb{N}\to \mathbb{N}$ be the polynomials and $V$ be the polynomial-time Turing machine that witnesses that $L\in {\mathrm{\Sigma }}_k^p$. Then, we can construct a polynomial-time Turing machine $M$ that decides $L$ in exponential time by simulating the computation of $V$ on all possible witnesses of length $p_1(n),p_2(n),\dots ,p_k(n)$.

We say that the Polynomial Hierarchy collapses if $\mathtt{\text{PH}}={\mathrm{\Sigma }}_k^p$ for some $k$.

The next proposition shows that items 2 and 3 cannot happen at the same time.

Proposition 3: If the polynomial hierarchy does not collapse, then $\mathtt{\text{PH}}$ has no complete problem under Karp reductions.

Proof: We will prove this by the contrapositive. That is, we will show that if $\mathtt{\text{PH}}$ has a complete problem under Karp reductions, then the polynomial hierarchy collapses.

Let $L$ be a complete problem for $\mathtt{\text{PH}}$ under Karp reductions. Since $L\in \mathtt{\text{PH}}$, we have $L\in {\mathrm{\Sigma }}_k^p$ for some $k\ge 1$. Let $M$ be the polynomial-time Turing machine that witnesses that $L\in {\mathrm{\Sigma }}_k^p$. Fix any language $A\in \mathtt{\text{PH}}$. As $L$ is complete for $\mathtt{\text{PH}}$, we have $A{\le }_{P}L$. Let $f:\{0,1{\}}^{\ast }\to \{0,1{\}}^{\ast }$ be the polynomial-time computable function that reduces $A$ to $L$. This implies that $A\in {\mathrm{\Sigma }}_k^p$, because for any $x\in \{0,1{\}}^{\ast }$, we have that $x\in A⟺f(x)\in L$, which implies that $$x\in A⟺\mathrm{\exists }{y}_1\mathrm{\forall }{y}_2\dots Q_k{y}_{k}M(f(x),y_1,y_2,\dots ,y_k)=1,$$ where $Q_k$ is either $\mathrm{\exists }$ or $\mathrm{\forall }$ depending on the parity of $k$.

Acknowledgements & References

This lecture was based on these resources:

• Lecture notes by Prof. Eric Blais.
• [AB09, Chapter 5]