Lecture 10

In-class notes: CS 505 Spring 2025 Lecture 10

$\mathrm{TQBF}$ is $\mathbf{PSPACE}$-hard

Last time, we established that $\mathrm{TQBF}\in \mathbf{PSPACE}$. Now, to show $\mathrm{TQBF}$ is $\mathbf{PSPACE}$-complete, we show that it is $\mathbf{PSPACE}$-hard. That is, $\forall L\in \mathbf{PSPACE}$, we show $L{\le }_p\mathrm{TQBF}$.

Let $L$ be any language in $\mathbf{PSPACE}$, and let $M_L$ be the decider for $L$. Suppose that for any input $x$ of length $n$, $M_L$ uses at most $O(S(n))$ space for $S(n)=n^c$, where $c\ge 1$ is a constant. Recall the configuration graph (see Lecture 9) of a Turing machine $M_L$. We know that the configuration graph $G_{M_L,x}$ has at most $2^O(S(n))$ nodes, and each configuration requires $O(S(n))$ bits. By the facts we established last lecture about the configuration graph $G_{M_L,x}$, we know that $M_L(x)=1$ if and only if there is a path from the starting configuration to an accepting configuration in the graph $G_{M_L,x}$. Moreover, there exists an $O(S(n))$ sized formula $\psi$ such that $\psi (c,c^{\prime })=1$ if and only if $c,c^{\prime }$ are valid configurations of $M_L(x)$ and $c^{\prime }$ follows from $c$ under the transition function of $M_L(x)$.

For our reduction, our goal will be to take $x$ and transform it into a QBF $f(x)$ such that $x\in L$ if and only if $f(x)\in \mathrm{TQBF}$. By our above discussion, we will utilize the configuration graph. The idea will be to construct a QBF $\varphi$ that is true if and only if there exists a path from the starting state to the accepting state in the configuration graph $G_{M_L,x}$.

First Attempt. Let $G$ be any directed graph. Suppose we consider two vertices $u$ and $v$ in $G$ such that there is a path from $u$ to $v$ of length at most $2^i$ for some $i\in \mathbb{N}$. Then, there must exist another vertex $x$ such that there is a path from $u$ to $x$ of length at most $2^{i-1}$, and a path from $x$ to $v$ of length at most $2^{i-1}$. If this wasn’t true; that is, there did not exist such a vertex $x$, then any path from $u$ to $v$ would need to be of length at least $2^i+1$.

Let’s try to build a QBF recursively to take advantage of the above ideas. Let ${\varphi }_0=\psi$; i.e., the formula $\psi$ for testing adjacent configurations. Our goal will be to construct ${\varphi }_k=\varphi$ (the final QBF, where $k=\log (2^{O(S(n))})=O(S(n))$; that is, log of the number of nodes in the configuration graph $G_{M_L,x}$). In particular, we want $\varphi$ to have the property that $\varphi (c_{start},c_{accept})=1$ if and only if there exists a path from the starting configuration $c_{start}$ to an accepting configuration $c_{accept}$.

There is actually a simple way to define ${\varphi }_k$ using our fact about paths between vertices of length at most $2^i$. For any two configurations $c,c^{\prime }$, define the formula ${\varphi }_i(c,c^{\prime })=\exists c^{\prime \prime }({\varphi }_{i-1}(c,c^{\prime \prime })\wedge {\varphi }_{i-1}(c^{\prime \prime },c^{\prime })$. Here, the formula ${\varphi }_i(c,c^{\prime })=1$ if and only if there is a path from $c$ to $c^{\prime }$ of length at most $2^i$. We build this formula recursively by saying ${\varphi }_i(c,c^{\prime })=1$ if and only if there exists a vertex/configuration $c^{\prime \prime }$ such that there is a path from $c$ to $c^{\prime \prime }$ of length at most $2^{i-1}$ and a path from $c^{\prime \prime }$ to $c^{\prime }$ of length at most $2^{i-1}$. Recursively, the formula ${\varphi }_{i-1}$ checks this statement. Finally, when the recursion bottoms out, it reduces to checking if two configurations are adjacent.

If we analyze the size of these formulas, notice that $|{\varphi }_0|=O(S(n))$ by construction. Then, $|{\varphi }_1|\ge 2\cdot |{\varphi }_0|$. Recursively, we have ${\varphi }_i\ge 2\cdot |{\varphi }_{i-1}|$. This gives our final formula size at least $|{\varphi }_k|\ge 2^k\cdot O(S(n))=2^{O(S(n))}$. So the formula ${\varphi }_k$ is too big! It requires exponential space.

Insight: Define an Equivalent QBF. Our final formula had exponential size because we were recursively checking two sub-formulas. This doubled the formula length at each recursion. However, we can take advantage of Boolean logic to define a formula that is equivalent to ${\varphi }_i(c,c^{\prime })=\exists c^{\prime \prime }{\varphi }_{i-1}(c,c^{\prime \prime })\wedge {\varphi }_{i-1}(c^{\prime \prime },c)$ but only requires a single recursive call to the formula ${\varphi }_{i-1}$. We define the formula first then explain what each component is doing.

$${\varphi }_i(c,c^{\prime })=\exists c^{\prime \prime }\forall d_1\forall d_2[(d_1=c\wedge d_2=c^{\prime \prime })\vee (d_1=c^{\prime \prime }\wedge d_2=c^{\prime })]⟹{\varphi }_{i=1}(d_1,d_2)$$

First, $c^{\prime \prime }$ is still the target vertex we want to check. That is, we still want to check if there is a path from $c$ to $c^{\prime \prime }$ of length at most $2^{i-1}$ and a path from $c^{\prime \prime }$ to $c^{\prime }$ of length at most $2^{i-1}$. Now, instead of calling ${\varphi }_{i-1}$ twice and taking the and of the results, we introduce the $\forall d_1,d_2$. What is this doing exactly?

Consider the expression with in the square brackets. This expression evaluates to true if and only if $d_1=c$ and $d_2=c^{\prime \prime }$, or $d_1=c^{\prime \prime }$ and $d_2=c^{\prime }$. In the first case, when $d_1=c$ and $d_2=c^{\prime \prime }$, we check if ${\varphi }_{i-1}(c,c^{\prime \prime })$ is true. Great! This is exactly one of the checks we want to perform. Then, in the second case, when $d_1=c^{\prime \prime }$ and $d_2=c^{\prime }$, we again perform the check we want: $\varphi (c^{\prime \prime },c^{\prime })$.

Now, what about for all other values of $d_1,d_2$? Well, recall that for the Boolean function “$\Rightarrow$,” we know that $(F\Rightarrow ?)$ always evaluates to True, no matter what $?$ is. So whenever $d_1$ and $d_2$ are not the target pairs of vertices, the expression trivially evaluates to true. This is fine since we are not checking the distance between these arbitrary variables. However, for the variables we explicitly want to check, the expression will be true if and only if ${\varphi }_{i-1}(d_1,d_2)$ evaluates to true. This gives us the formula we want!

To wrap up the proof, we let $\varphi =\exists c_{accept}{\varphi }_k(c_{start},c_{accept})=1$, where $c_{accept}$ can be any accepting configuration and $c_{start}$ is the unique starting configuration. Then, $\varphi$ is a QBF which evaluates to $1$ if and only if there is a path from $c_{start}$ to some $c_{accept}$ of length at most $2^{O(S(n)}$ (in particular, the exact number of vertices). This happens if and only if $M_L(x)=1$.

Finally, we analyze the size of the formula $\varphi$. Notice that $|{\varphi }_i|=O(1)+|{\varphi }_{i-1}|$. Since $|{\varphi }_0|=O(S(n))$, we have that $|\varphi |=O(k\cdot S(n))=O(S(n{)}^2)$, which is polynomial in $n$ since $S(n)$ is a polynomial in $n$. $\square$

$\mathbf{PSPACE}=\mathbf{NPSPACE}$

Notice in the proof we actually didn’t use the fact that $M_L$ was a deterministic Turing machine. In fact, the above proof holds even if $M_L$ is a non-deterministic Turing machine. Thus, we have actually shown that $\mathrm{TQBF}$ is $\mathbf{NPSPACE}$-hard. And since $\mathrm{TQBF}$ is also a language in $\mathbf{NPSPACE}$, we actually showed that $\mathrm{TQBF}$ is $\mathbf{NPSPACE}$-complete. This shows the two classes are equal.

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

This is a somewhat surprising result since we do not believe the same is true for polynomial-time computations; i.e., we do not believe that $\mathbf{P}$ equals $\mathbf{NP}$.

Savich’s Theorem

We can actually show something more fine-grained about deterministic space versus non-deterministic space. The following result would equally show that $\mathbf{PSPACE}=\mathbf{NPSPACE}$.

Theorem (Savich’s Theorem). For all space constructible functions $S$, we have $\mathbf{NSPACE}(S)\subseteq \mathbf{DSPACE}(S^2)$.

Proof. We will again take advantage of the configuration graph of a Turing machine that we have been using. Let $L\in \mathbf{NSPACE}(S)$ with corresponding NTM $M_L$ using at most $O(S)$ additional space on its worktapes. Let $G_{M_L,x}$ be its corresponding configuration graph with at most $2^{O(S(n))}$ nodes for any $x$ of length $n$. Recall also that $x\in L$ if and only if there is a path in $G_{M_L,x}$ from the start configuration to some accepting configuration.

Our goal will be to construct a deterministic Turing machine $D_L$ which decides $L$ using $O(S^2)$ space. The machine $D_L$ will operate as follows.

$D_L(x)$:

• Simulate $M_L(x)$ by traversing the graph $G_{M_L,x}$.
• Traversal will utilize a recursive procedure $\mathrm{reach}(u,v,i)$ which returns $1$ if and only if there is a path from node $u$ to node $v$ of length at most $2^i$.
  • The recursion utilizes the same fact we had in the proof that $\mathrm{TQBF}$ is $\mathbf{PSPACE}$-complete. IN particular, $\mathrm{reach}(u,v,i)=1$ if and only if there exists $z$ such that $\mathrm{reach}(u,z,i-1)=1$ and $\mathrm{reach}(z,v,i-1)=1$.
• Suppose $G_{M_L,x}$ has $2^k$ nodes for $k=O(S(n))$. For all accepting states $c_{accept}$, run $\mathrm{reach}(c_{start},c_{accept},k)$. If this procedure outputs $1$, output $1$. If none of the calls output $1$, output $0$.
  • To run this procedure, each recursive call simply runs over all nodes $d$ (this requires $k=O(S(n))$ bits) and checks if $\mathrm{reach}(c_{start},d,k-1)=1$ and $\mathrm{reach}(d,c_{accept},k-1)=1.$
  • $\mathrm{reach}(u,v,0)=1$ if and only if the edge $(u,v)$ is in $G_{M_L,x}$.

Notice that the recursive procedure $\mathrm{reach}$ bottoms out after $k$ calls. During each call, we store $k$ bits for the current vertex $d$ being enumerated over. The machine $D_L(x)$ is performing a depth-first search of the graph $G_{M_L,x}$. At the bottom of the recursion, $O(k^2)=O(S(n{)}^2)$ space is used. Therefore, $D_L(x)$ uses at most $O(S(n{)}^2)$ space. $\square$

Alternate Proof of $\mathbf{PSPACE}=\mathbf{NPSPACE}$

We can use Savich’s Theorem as an alternate proof of this result. Recall that $$\mathbf{PSPACE}=\bigcup _{c\ge 1}\mathbf{DSPACE}(n^c)\mathbf{NPSPACE}=\bigcup _{c\ge 1}\mathbf{NSPACE}(n^c).$$ Notice that for any space constructible $S$, we have $\mathbf{DSPACE}(S)\subseteq \mathbf{NSPACE}(S)$ since all deterministic computations are also non-deterministic. By Savich’s Theorem, we have $\mathbf{NSPACE}(S)\subseteq \mathbf{DSPACE}(S^2)$. Finally, all polynomial functions are space constructible. So this implies for all $c\ge 1$ $$\mathbf{DSPACE}(n^c)\subseteq \mathbf{NSPACE}(n^c)\subseteq \mathbf{DSPACE}(n^{2c}).$$ This shows $\mathbf{PSPACE}=\mathbf{NPSPACE}$. $\square$

$\mathbf{NL}$-Completeness

Recall that $\mathbf{NL}$ is the set of all languages $L$ decidable on an NTM using at most $O(\log (n))$ additional space for inputs of length $n$. We showed last lecture that $\mathrm{PATH}\in \mathbf{NL}$. Today, we’ll see that $\mathrm{PATH}$ is the essence of $\mathbf{NL}$. That is, $\mathrm{PATH}$ is $\mathbf{NL}$-complete, where $$\mathrm{PATH}=\{(G,s,t):G\text{ is a directed graph with a path from s to t}\}.$$ Note however that we do not know if $\mathrm{PATH}\in \mathbf{L}$, though we believe it to not be the case, otherwise $\mathbf{L}=\mathbf{NL}$.

Before we show that $\mathrm{PATH}$ is $\mathbf{NL}$-complete, we need a new notion of reducibility that is not polynomial-time. Let’s see why this is the case. Suppose $L,L^{\prime }\in \mathbf{NL}\setminus \{\varnothing ,\{0,1{\}}^{\ast }\}$. Then, $L{\le }_p{L}^{\prime }$. That is, any two languages are polynomial-time reducible to each other in $\mathbf{NL}$. Intuitively, this is because $\mathbf{L}\subseteq \mathbf{NL}\subseteq \mathbf{P}$ and the reduction trivially has more power than problems in $\mathbf{L}$ or $\mathbf{NL}$ since it is limited to be polynomial-time only and is not restricted on space. So because we can decide $L$ using an NTM using at most $O(\log (n))$ space, the reduction can simply compute the configuration graph of the NTM deciding $L$ on input $x$, decide if $x\in L$, then produce any instance $x^{\prime }\in L^{\prime }$, all in polynomial time.

Therefore, we need to restrict the power of the reductions for completeness in $\mathbf{L}$ and $\mathbf{NL}$. This leads us to logspace reductions.

Definition. Let $f:\{0,1{\}}^{\ast }\to \{0,1{\}}^{\ast }$. We say that $f$ is (implicitly) logspace computable if there exists a constant $c\ge 1$ such that for all $x$, $|f(x)|\le |x{|}^c$ and the following languages are in $\mathbf{L}$. $$\begin{array}{c}{L}_f=\{(x,i):f(x{)}_i=1\}\\ L_f^{\prime }=\{(x,i):i\le |f(x)|\}\end{array}$$

We can now define logspace reducible.

Definition. Let $B$ and $C$ be any language. We say that $B$ is logspace reducible to $C$, denoted as $B{\le }_{L}C$, if there exists a logspace computable function $f$ such that $x\in B$ if and only if $f(x)\in C$.

This finally let us define $\mathbf{L}$ and $\mathbf{NL}$-completeness.

Definition. We say that a language $B$ is $\mathbf{NL}$-complete (respectively, $\mathbf{L}$-complete) if

1. $B\in \mathbf{NL}$ (resp., $B\in \mathbf{L}$); and
2. $\forall C\in \mathbf{NL}$ (resp., $\forall C\in \mathbf{L}$), we have $C{\le }_{L}B$.

As with polynomial-time reductions, we have a “transitive” property of logspace reductions.

Theorem.

1. If $A{\le }_{L}B$ and $B{\le }_{L}C$, then $A{\le }_{L}C$.
2. If $A{\le }_{L}B$ and $B\in \mathbf{L}$ then $A\in \mathbf{L}$.

Part (2) of the above theorem tells us that if $\mathrm{PATH}\in \mathbf{L}$ then $\mathbf{L}=\mathbf{NL}$.

$\mathrm{PATH}$ is $\mathbf{NL}$-complete

We can finally prove that $\mathrm{PATH}$ is the “essence” of $\mathbf{NL}$.

Theorem. $\mathrm{PATH}$ is $\mathbf{NL}$-complete.

Proof. We have already shown that $\mathrm{PATH}\in \mathbf{NL}$. We now have to show that it is $\mathbf{NL}$-hard. That is, show that $B{\le }_L\mathbf{PATH}$ for any $B\in \mathbf{NL}$. Our good friend the configuration graph will help us yet again.

Let $B\in \mathbf{NL}$ and let $M_B$ be the non-deterministic logspace decider for $B$. This means for any $x$, we have $x\in B$ if and only if $M_B(x)=1$ using at most $O(\log (|x|))$ space.

Our logspace reduction $f$ will simply construct the configuration graph $G_{M_B,x}$. That is, on input $x$, the reduction $f(x)$ will output the tuple $(G_{M_B,x},c_{start},c_{accept})$, where $c_{accept}$ is an accepting configuration. Note that $|f(x)|\le 2^{O(\log (n))}=\mathrm{poly}(n)$ for any $x\in \{0,1{\}}^n$.

Recall by definition of our configuration graph, we know that $M_B(x)=1$ if and only if there is a path from $c_{start}$ to $c_{accept}$ in $G_{M_B,x}$. This is precisely a problem instance of $\mathrm{PATH}$. Now, we can represent $G_{M_B,x}$ as an adjacency matrix of size $2^{O(\log (n))}\times 2^{O(\log (n))}$. Entry $(c,c^{\prime })=1$ if and only if there is an edge from $c$ to $c^{\prime }$ in $G_{M_B,x}$.

Now, we can check this in logspace. Given $c,c^{\prime }$, there exists a deterministic machine to check of $c^{\prime }$ follows from $c$ according to $M_B$’s transition function. We can do this in space $O(|c|+|c^{\prime }|)$. By our previous discussions on the configuration graph, these configurations need at most $O(\log (n))$ space to represent. Thus, we can do this check in $O(\log (n))$ space. Therefore, we have that $L_f,L_f^{\prime }\in \mathbf{L}$. So this is a valid logspace reduction.

This completes the proof as we have already encoded $M_B(x)$ into the $\mathrm{PATH}$ problem on the instance $(G_{M_B,x},c_{start},c_{accept})$. $\square$