Lecture 13

In-class notes: CS 505 Spring 2025 Lecture 13

In this lecture, we wrap up our discussion of Alternating Turing machines and the Polynomial Hierarchy.

Alternating Time and Space

Given ATMs from last lecture, we can alternatively define ${\Sigma }_i^p$, ${\Pi }_i^p$, and $\mathbf{PH}$ in terms of ATMs.

Definition. The class ${\Sigma }_i\mathbf{TIME}(T)$ (resp., ${\Pi }_i\mathbf{TIME}(T)$) is the set of all languages $L$ that are decidable in time $T$ on an $i$-alternating ATM $M$ with initial state labeled $\exists$ (resp., $\forall$).

It is not difficult to see how we can define ${\Sigma }_i^p$ and ${\Pi }_i^p$ with respect to the above complexity classes.

Lemma. For all $i\in {\mathbb{Z}}^{+}$, $$\begin{array}{c}{\Sigma }_i^p=\bigcup _{c\in {\mathbb{Z}}^{+}}{\Sigma }_i\mathbf{TIME}(n^c),{\Pi }_i^p=\bigcup _{c\in {\mathbb{Z}}^{+}}{\Pi }_i\mathbf{TIME}(n^c).\end{array}$$

Corollary. $\mathbf{PH}={\bigcup }_{i,c}{\Sigma }_i\mathbf{TIME}(n^c)={\bigcup }_{i,c}{\Pi }_i\mathbf{TIME}(n^c)$.

Unlimited Number of Alternations

The definitions above of ${\Sigma }_i\mathbf{TIME}(T)$ and ${\Pi }_i\mathbf{TIME}(T)$ limit the number of alternations a given ATM can have (they are $i$-alternating). However, we do not need to restrict the number of alternations. This was exactly the class $\mathbf{ATIME}(T)$ defined in the last lecture: the set of all languages $L$ decidable in time $T$ by an ATM $M$. In particular, there is no restriction on the number of alternations. With this, we can define alternating polynomial time.

Definition. $\mathbf{AP}={\bigcup }_{c\in {\mathbb{Z}}^{+}}\mathbf{ATIME}(n^c)$.

As one might expect, an unlimited number of alternations yields something we believe to be more powerful than $\mathbf{PH}$.

Theorem. $\mathbf{AP}=\mathbf{PSPACE}$.

Proof. First, we show that $\mathbf{PSPACE}\subseteq \mathbf{AP}$. We do this by showing $\mathrm{TQBF}\in \mathbf{AP}$. Recall that a QBF $\varphi$ is in $\mathrm{TQBF}$ if and only if $\varphi =Q_1{x}_1{Q}_2{x}_2\cdots Q_m{x}_m\psi (x_1,\dots ,x_m)=1$ (i.e., it is a true quantified Boolean formula), where $Q_i\in \{\exists ,\forall \}$ for all $i$. This is trivially sovlable by an ATM which guesses everything and labels states according to $Q_i$, and checks if the result is true in polynomial time.

Now, we show $\mathbf{AP}\subseteq \mathbf{PSPACE}$. Let $L\in \mathbf{AP}$ with ATM $M_L$ deciding $L$ in polynomial time. Construct a deterministic machine $D$ as follows. On input $x$, $D(x)$ performs a depth-first search of the configuration graph $G_{M_L,x}$ and attempts to compute the label of the starting node $c_{start}$ (which would be the output of the machine $M_L(x)$. The algorithm is recursive. Recalling the facts about the configuration graph, since $M_L(x)$ runs in polynomial time, the amount of bits needed to represent every configuration is polynomial in $|x|$. Moreover, doing a depth-first search only requires storing $\mathrm{poly}(|x|)$ configurations in the recursion stack, since $M_L(x)$ runs in $\mathrm{poly}(|x|)$ time. So $D(x)$ uses at most polynomial space. $\square$

We can also define alternating polynomial space, where we consider space-bounded ATMs with an unlimited number of alternations (the class $\mathbf{ASPACE}(T)$.

Definition. $\mathbf{APSPACE}={\bigcup }_{c\in {\mathbb{Z}}^{+}}\mathbf{ASPACE}(n^c)$.

Theorem. $\mathbf{APSPACE}=\mathbf{EXP}$.

Finally, we can define alternating logspace as $\mathbf{AL}=\mathbf{ASPACE}(\log (n))$.

Theorem. $\mathbf{AL}=\mathbf{P}$.

Time-Space Tradeoffs for SAT

What are alternations used for? Here, we’ll see how we can use ATMs to give time-space tradeoffs for $\mathrm{SAT}$.

Complexity theorists generally believe that any algorithm deciding/solving $\mathrm{SAT}$ must have the following properties.

1. Solving $\mathrm{SAT}$ requires exponential time (or, at least super polynomal time).
2. Solving $\mathrm{SAT}$ requires linear space.

In general, both of the above are conjectures, there may be an algorithm for solving $\mathrm{SAT}$ which only uses logarithmic space, or uses polynomial time! However, we can rule out an algorithm that achieves both of these properties.

Theorem. Let $S,T:\mathbb{N}\to \mathbb{N}$ be functions. Define the class $\mathbf{TISP}(T,S)$ to be all languages decidable by a DTM $M$ using at most $O(T)$ time and $O(S)$ (additional) space. Then, $\mathrm{SAT}\notin \mathbf{TISP}(n^{1.2},n^{0.2})$. More generally, for any $c,d>0$ such that $c(c+d)<2$, we have $\mathrm{SAT}\notin \mathbf{TISP}(n^c,n^d)$.

Proof. To prove the theorem, we first need the following claim which relates $\mathbf{TISP}$ to ${\Sigma }_2\mathbf{TIME}$.

Claim 1. $\mathbf{TIPS}(n^{12},n^2)\subseteq {\Sigma }_2\mathbf{TIME}(n^8)$.

The proof of this claim is similar to the proofs of Savitch’s Theorem and $\mathrm{TQBF}$ is $\mathbf{PSPACE}$ complete. Let $L\in \mathbf{TISP}(n^{12},n^2)$ with decider $M_L$ running in time $O(n^{12})$ and space $O(n^2)$ for any input $x\in \{0,1{\}}^n$. We construct an ATM $A_L$ for deciding $L$ as follows. $A_L(x)$ will construct the configuration graph $G_{M_L,x}$. By the previous properties we have discussed for configuration graphs, we know that all configurations need at most $O(n^2)$ bits to describe. Moreover, $M_L(x)=1$ if and only if there exists a path in the graph from $c_{start}$ to some accepting configuration $c_{accept}$, and any such path has length at most $O(n^{12})$.

Now, this path from $c_{start}$ to $c_{accept}$ exists if and only if there exist $n^6$ configurations $C_1,C_2,\dots ,C_{n^6}$ such that:

• If $C_0=c_{start}$, then $C_{n^6}$ is accepting; and
• $\forall i\in [n^6]$, $C_i$ follows from $C_{i-1}$ in at most $O(n^6)$ valid executions of $M_L$’s transition function. Essentially, we have divided the path from $c_{start}$ to $c_{accept}$ into $n^6$ chunks, each of size $O(n^6)$. Now, each configuration $C_i$ requires $O(n^2)$ bits to describe, so the full description of the path $(C_0,C_1,\dots ,C_{n^6})$ requires $O(n^8)$ bits. So the ATM $A_L$ guesses this path using the appropriate alternations. Intuitively, this is $\exists C_0,C_1,\dots ,C_{n^6}\forall i\in [n^6]$ $C_0$ is a start state, $C_{n^6}$ is an accepting state, and $C_i$ follows from $C_{i-1}$ in at most $O(n^6)$ steps. Clearly, this implies $L\in {\Sigma }_2\mathbf{TIME}(n^8)$.

Now, we need a second claim to proceed with the proof.

Claim 2. Suppose that $\mathbf{NTIME}(n)\subseteq \mathbf{DTIME}(n^{1.2})$. Then ${\Sigma }_2\mathbf{TIME}(n^8)\subseteq \mathbf{NTIME}(n^{9.6})$.

To see this claim, let $L\in {\Sigma }_2\mathbf{TIME}(n^8)$. Then there exists a DTM $M_L$ such that $\exists w_1\forall w_2{M}_L(x,w_1,w_2)=1$ in $O(|x{|}^8)$ time, where $|w_1|,|w_2|=O(|x{|}^8)$. Here, we are using the fact that ${\Sigma }_2^p={\cup }_c{\Sigma }_2\mathbf{TIME}(n^c)$.

By the assumption in the claim, we have $\mathbf{NTIME}(n)\subseteq \mathbf{DTIME}(n^{1.2})$. By a padding argument, this implies that $\mathbf{NTIME}(n^8)\subseteq \mathbf{DTIME}((n^8{)}^{1.2})=\mathbf{DTIME}(n^{9.6})$. Thus, we can construct a DTM $D$ such that on input $x,w_1$ for $|x|=n$ and $w_1=O(n^8)$, $D(x,w_1)$ runs in time $O(n^{9.6})$ and outputs $1$ if and only if $\exists w_2$ such that $M_L(x,w_1,w_2)=0$. This implies that $L\in \mathbf{NTIME}(n^{9.6}).$

Why have we bothered with Claims 1 and 2? Well, using these claims, we will show that $\mathbf{NTIME}(n)⊈\mathbf{TISP}(n^{1.2},n^{0.2})$. This establishes the result since $\mathrm{SAT}\in \mathbf{NTIME}(n)$. Suppose this is not the case; that is, $\mathbf{NTIME}(n)\subseteq \mathbf{TISP}(n^{1.2},n^{0.2})$. Then, Claims 1 and 2 gives us the following inequalities. $$\begin{array}{rl}\mathbf{NTIME}(n^{10})& \subseteq \mathbf{TISP}(n^{12},n^2)\\ & \subseteq {\Sigma }_2\mathbf{TIME}(n^8)\text{(Claim 1)}\\ & \subseteq \mathbf{NTIME}(n^{9.6})(\ast ),\end{array}$$ where $(\ast )$ follows by Claim 2 and the fact that $\mathbf{TISP}(T,S)\subset \mathbf{DTIME}(T)$. This violates the non-deterministic time hierarchy theorem, so we can conclude that $\mathbf{NTIME}(n)⊈\mathbf{TISP}(n^{1.2},n^{0.2})$. $\square$

The Polynomial Hierarchy via Oracle Machines

Our final discussion on the polynomial hierarchy will again show that we can give yet another equivalent definition, this time using oracle Turing machines/complexity classes. We begin with ${\Sigma }_i^p$.

Theorem. For all $i\ge 2$, we have ${\Sigma }_i^p={\mathbf{NP}}^{{\Sigma }_{i-1}\mathrm{SAT}}$, where $${\Sigma }_{i-1}\mathrm{SAT}=\{\varphi :\varphi =\exists x_1\forall x_2\cdots Q_{i-1}{x}_{i-1}\psi (x_1,\dots ,x_{i-1}=1\}$$ is a ${\Sigma }_{i-1}^p$-complete problem.

Proof. We show the proof for $i=2$, other cases are analogous. For $i=2$, we show that ${\Sigma }_2^p={\mathbf{NP}}^{\mathrm{SAT}}$. First, we show ${\Sigma }_2^p\subseteq {\mathbf{NP}}^{\mathrm{SAT}}$. Let $L\in {\Sigma }_2^p$. By definition, there exists a DTM $M_L$ such that for all $x$, we have $x\in L$ if and only if $\exists w_1\forall w_2{M}_L(x,w_1,w_2)=1$, where $|w_1|,|w_2|=\mathrm{poly}(|x|)$. Fix $x,w_1$ such that $\forall w_2{M}_L(x,w_1,w_2)=1$. Now, the language $L^{\prime }$ of all pairs $x,w_1$ satisfying this is a $\mathbf{coNP}$ language. To check if $(x,w_1)\in L^{\prime }$, we can equivalently check the $\mathbf{NP}$ statement $(x,w_1)\notin L^{\prime }$. That is, we can check if $\exists w_2\neg M_L(x,w_1,w_2)=1$. Since this is a $\mathbf{NP}$ statement, we can check this statement in polynomial time by converting it to a $\mathrm{SAT}$ formula $\varphi$ and querying the $\mathrm{SAT}$ oracle to see if it has a satisfying assignment. Thus $L\in {\mathbf{NP}}^{\mathrm{SAT}}$ and ${\Sigma }_2^p\subseteq {\mathbf{NP}}^{\mathrm{SAT}}$.

Now, we show the other direction: ${\mathbf{NP}}^{\mathrm{SAT}}\subseteq {\Sigma }_2^p$. Let $L\in {\mathbf{NP}}^{\mathrm{SAT}}$ with corresponding NTM $N$. Here, we assume that the NTM $N$ outputs at most $2$ choices per execution of its transition function. At first glance, it does not seem like we can capture the power of the $\mathrm{SAT}$ oracle in ${\Sigma }_2^p$. After all, for any $x$, $N(x)$ makes at most $\mathrm{poly}(|x|)$ queries, say $q_1,\dots ,q_k$ (note: each $q_i$ is a $\mathrm{SAT}$ formula), each with corresponding answers $a_1,\dots ,a_k$. Moreover, each query $q_i$ can arbitrarily depend on the previous queires and answers, along with any other non-deterministic decisions made by $N$!

Let $c_1,\dots ,c_m\in \{0,1\}$ denote the non-deterministic choices made by $N(x)$. Intuitively, in order to construct a DTM $D$ to decide $L$ in ${\Sigma }_2^p$, we will guess the non-deterministic choices and query answers that will cause $N(x)$ to accept. That is, we can see that $x\in L$ if and only if $\exists$ choices $c_1,\dots ,c_m$ and query answers $a_1,\dots ,a_k$ such that

• If $N(x)$ makes choices $c_1,\dots ,c_m$ and recieves oracle query answers $a_1,\dots ,a_k$,
• Then
  1. $N(x)$ reaches an accepting state and
  2. The following hold.
     • If $a_i=1$ then there exists an assignment $u_i$ such that $q_i(u_i)=1$.
     • If $a_i=0$ then for all assignments $v_i$ we have $q_i(v_i)=0$. Here, $q_i$ is the $i$th query made by $N(x)$.

Thus, we construct a DTM $D$ such that $x\in L$ if and only if $$\begin{array}{rl}& \exists c_1,\dots ,c_m,a_1,\dots ,a_k,u_1,\dots ,u_k\\ & \forall v_1,\dots ,v_k\\ & D(x,c_1,\dots ,c_m,a_1,\dots ,a_k,u_1,\dots ,u_k,v_1,\dots ,v_k)=1⟺\end{array}$$ $D(x,\cdots )=1$ simulates $N(x)$ and

• $N(x)=1$ using non-deterministic choices $c_1,\dots ,c_m$;
• $\forall i\in [k]$:
  • If $a_i=1$ then $q_i(u_i)=1$; and
  • If $a_i=0$ then $q_i(v_i)=0$.

Clearly, $D$ decides $L$ and thus $L\in {\Sigma }_2^p$. $\square$