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\lecture{10}{10/11/2011}{Circuit Lower Bounds}{Li-Hsiang Kuo}
Last time we introduced nonuniform computations, and nonuniform models such as
Boolean circuits, branching programs, and uniform models with advice.
Today we are mainly focus on Boolean circuits. We will address the complexity
of some problems we are interested in.
\section{Boolean Circuits}
\begin{definition}
$C_L(n) =$ smallest size of a Boolean circuit deciding $L$ on $\{0,1\}^n$
\end{definition}
\begin{theorem}
$C_L(n) = O(\frac{2^n}{n})$
\label{thm:circuit-bound1}
\end{theorem}
\begin{theorem}
$(\exists L) C_L(n) = \Omega(\frac{2^n}{n})$
\label{thm:circuit-bound2}
\end{theorem}
\begin{proof}
Let $s$ denote the number of binary gates. For each of the $s$ gates,
we can pick a variable or some other gate as input, and each gate has
at most 2 inputs. So, the number of circuits of size at most $s$ is at
most $(c(s+n)^2)^s$, where $c$ is some constant that takes care of
different type of gates. Since we can map circuits to Boolean functions,
this is also the maximum number of Boolean functions computable with at
most $s$ gates. We know that $2^{2^n}$ is the number of Boolean functions
on $n$ variables, and $(c(s+n)^2)^s = 2^{O(s\log s)}$.
By setting $s=d\frac{2^n}{n}$ for a sufficiently small constant $d$,
$(c(s+n)^2)^s = 2^{O(s \log s)} = 2^{O(d\frac{2^n}{n}\cdot (n \log d - \log n))} << 2^{2^n}$.
The claim then follows.
\end{proof}
Both theorems also holds for branching programs. For Theorem~\ref{thm:circuit-bound2} the situation
is similar since Boolean circuits can easily simulate branching programs.
For Theorem~\ref{thm:circuit-bound1} this takes extra work.
\begin{theorem}
$BP_{L}(n) = O(\frac{2^n}{n})$
\end{theorem}
\section{Problems in NP}
\begin{itemize}
\item Upper bounds: No better results than general.
\item Lower bounds:
\begin{itemize}
\item $C_L(n) = \omega(n)$ is open for any $L \in \NP$.
\item $BP_L(n) = \Omega(\frac{n^2}{\log^2n})$ is best known for $L \in \NP$.
\end{itemize}
\end{itemize}
The above are unconditional results. However, we can show that if NP has polynomial
size circuits, then PH collapses to the second level.
\begin{theorem}
$\NP \subseteq \P/\poly \Rightarrow \Sigma^p_2 = \Pi^p_2$
\end{theorem}
\begin{proof}
For a language $L$ in $\Pi^p_2$, we have
\begin{displaymath}
x \in L \iff (\forall y \in \{0,1\}^{|x|^c})(\exists z \in \{0,1\}^{|x|^c}) R(x,y,z)
\end{displaymath}
where $R$ is a predicate that can be decided deterministically in time, say, linear in its combined input.
Since $(\exists z \in \Sigma^{|x|^c}) R(x,y,z)$ is a $\Sigma^p_1$ predicate
on input $$, we can reduce it to a SAT instance and by the hypothesis,
there exists a circuit $C_{SAT}$ that is of size polynomial in the running time to decide $R$,
and so of size polynomial in the size of $x$. Letting $f$ denote the reduction,
we can replace the $\Sigma^p_1$ predicate with $C_{SAT}(f(x,y))$.
We would like to rephrase the formula above roughly as follows:
Does there exist a circuit solving SAT such that for all strings $y$,
the circuit accepts $f(x,y)$?
Let $V$ be the following recursive predicate:
\begin{algtab}
\algifthenelse{$\varphi$ has at least 1 variable}
{$C_{SAT}(\varphi) \iff C_{SAT}(\varphi|_{x_1 \leftarrow 0})\ \vee\ C_{SAT}(\varphi|_{x_1 \leftarrow 1})$}
{$C_{SAT}(\varphi) \iff \varphi\ is\ true$}
\end{algtab}
where $x_1$ is the first unset variable of $\varphi$.
We can transform the right-hand side of the above to:
\begin{displaymath}
(\exists C_{SAT})(\forall y \in \Sigma^{|x|^c})
[C_{SAT}(f(x,y)) \wedge (\forall \varphi\ of\ size \leq n^c)
[V(C_{SAT}, \varphi)]]
\end{displaymath}
where $C_{SAT}$ is a circuit of polynomial size with $n^c$ inputs.
Essentially, we guess a circuit $C_{SAT}$ and the combined predicate
checks if $C_{SAT}$ accepts $f(x,y)$ and whether or not
$C_{SAT}$ is a valid circuit solving SAT.
Since we are evaluating polynomial size circuits, and we evaluate $n$ times,
$V$ takes polynomial time to check. The second universal quantifier above
can be merged with the first, and then we have a single polynomial time verifiable
predicate. So, we now have a $\Sigma^p_2$ formula.
Note that our hypothesis is crucial in that if NP does not have polynomial size
circuits, then $V$ will always fail.
\end{proof}
We would expect that at least for complicated functions like those that capture
NP-complete problems, we can prove non-trivial lower bounds. The following shows that
if we go up a little bit higher in the hierarchy, there is some non-trivial lower bounds.
\begin{theorem}
$(\forall c > 0) (\exists L \in \Sigma^p_2 \cap \Pi^p_2) $ with $C_L(n) = \Omega(n^c)$
\end{theorem}
\begin{proof}
We claim that $\exists y \in \{0,1\}^{n^d + 1}$, where $y$
encodes the inclusion or exclusion of the first $n^d + 1$ strings of a language $Y$,
such that none of the circuits is consistent with it.
Intuitively, with each bit in $y$, we can ``kill'' half of the remaining consistent circuits.
That is, we choose each bit in $y$ by doing the opposite of the majority vote of the circuits
that are consistent with the previous bits of $y$.
The number of circuits of size $\leq n^c$ is $\leq (n^c + n)^{2n^c} \leq 2^{n^d}$ for some
$d \geq 0$. By doing against the majority vote of the circuits, in each bit in $y$ there is
at most half of the circuits remains. $n^d$ bits leaves $\leq 1$ consistent circuit and
$n^d + 1$ bits leaves no consistent circuit.
By counting, there exist a setting of the ``Lexicographically first'' string in $\{0,1\}^{n^{d+1}}$
such that if we interpret this as a language, this language cannot be decided by any circuits
of size $\leq n^c$.
Now, we must show that there is such a language $Y$ in $\NP$.
The explicit construction of this language that we have described is not necessary in $\NP$,
it requires the majority vote counts on exponentially many circuits.
So, we can only assume that this language $Y$ exists. And, in $\PH$ we have the power of guessing.
However, this $Y$ may not be unique, to make it unique, we can encode this as:
\begin{align*}
\exists y \in & \{0,1\}^{n^d + 1} \\
(1)\ & (\forall C \text{ of size } \leq n^c) C \text{ and } y \text{ disagree on one of the first } n^d + 1 \text{ strings } \\
(2)\ & (\forall z < y) \text{ (1) fails for } z \\
(3)\ & x \text{ is one of the first } n^d + 1 \text{ strings and the } x \text{th bit of } y \text{ is set}
\end{align*}
This accepts precisely the language $L$ described above. But, the negation of (2) means that we end up with $\Sigma^p_3$
instead of $\Sigma^p_2$. However, using today's previous result we have two cases, either $\NP \subseteq \P / \poly$ in which case
$\Sigma^p_3 = \Sigma^p_2$, or $\NP \nsubseteq \P / \poly$
in which case their is some other language $L \in \NP$ with $C_L(n) = \Omega (n^c)$ for all $c$.
\end{proof}
\section{Constant-Depth Circuits}
Because we have been unable to prove lower bounds for NP-complete problems in
the general setting, we focus our attention on a restricted model -
namely constant-depth circuits. Typically we talk about constant-depth circuit
we do not count the exponential fan-in. Today we only focus on the class of
languages decidable by constant-depth circuits of polynomial size.
\begin{definition}
Binary addition: $L = \{ | x + y = z \}$
\end{definition}
\begin{proposition}
The decision variant of binary addition has constant-depth circuits of polynomial size.
\end{proposition}
\begin{proof}
In this proof, all strings are indexed from the right. To determine the
$i$th bit of the sum, we only need to look at the $i$th bits of the summands
and determine if there is a carry from the bits in position $(i-1)$.
We first introduce some notation. We will label each column from 1 up to
$i-1$ depending on if it either: $g$enerates a carry bit, $t$ransmits a
carry bit, or $s$tops a carry bit. If both input bits in the column are
1, the column is labeled $g$; if only one bit is 1, it is labeled $t$; and
if both bits are 0, the column is labeled $s$.
For there to be a carry from the $(i-1)^{st}$ column, there must be a $g$
at some point followed by zero or more $t$ columns. In other words,
to determine if there is a carry into the $i$th position, our job is reduced
to detecting if the above string is of the form $t^*g\{s,g,t\}^*$.
First of all, we use an OR to guess the length of $t^*g$, and the number of
possible lengths are at most linear in the input size. For each length $j$,
we need an AND to determine if the $(i-j+1)$th symbol on the string is a $g$,
and XORs to ensure that the symbols after it are $t$s, and then we do a big
AND over the XORs and the AND. So, at the first level, we have an OR, at the
second we have ANDs, and at the third we have ANDs and XORs. Since XORs can
be implemented in constant depth using ANDs, ORs and NOTs, the overall
circuit has constant depth of polynomial size.
\end{proof}
We now sketch a proof that PARITY requires exponential size to be computed
by constant-depth circuits.
\begin{definition}
PARITY = \{x : x has an odd number of 1s\}
\end{definition}
We also denote PARITY on $n$ variables as $\bigoplus_n$.
\begin{exercise}
The decision variant of binary multiplication is equivalent to PARITY.
\end{exercise}
\begin{theorem}
$C_d(\bigoplus_n) = 2^{\Omega(n^{\frac{1}{d-1}})}$
\label{thm:parity}
\end{theorem}
The proof we present uses the following tool.
\begin{definition}[Random Restrictions]
A $p$-random restriction on $n$ variables is a random function
$\rho:\{x_1,\ldots,x_n\} \rightarrow \{*,0,1\}$,
such that for each $i$, independently,
Pr$[\rho(x_i) = *] = p$ and Pr$[\rho(x_i) = 1] = \frac{1-p}{2} =$ Pr$[\rho(x_i) = 0]$.
If $\rho(x_i) = *$ then we leave $x_i$ as a variable.
Otherwise we set it to the result of $\rho(x_i)$.
\end{definition}
\begin{proof}[Proof sketch for Theorem~\ref{thm:parity}]
We can apply $p$-random restriction with $p = \frac{n^{\frac{1}{d-1}}}{n}$.
With high probability it can reduce the depth of the circuit by one at a time.
Repeat this method we turn depth-$d$ circuits into depth-2 circuits.
Then apply depth-2 lower bound.
\end{proof}
\begin{proposition}
$C_2(\bigoplus_n) = \Theta(n \cdot 2^{n-1})$.
\end{proposition}
\begin{proof}
For depth 2, we can easily prove an exponential lower bound. By assumption,
the circuit is either a DNF or a CNF. Let us assume that it is a DNF.
Each of the AND terms check for a setting of variables such that $\bigoplus_n = 1$.
Thus they must contain all $n$ variables. Otherwise, we can flip a variable
and at least one AND will not be able to detect the difference. There are $2^n$
possible $n$-variable AND terms in a DNF formula, and we only need half of them
as only half of them check for $\bigoplus_n = 1$. Since each AND must
be of size $n$, and there must be $2^{n-1}$ of them, we get the bound as
stated.
\end{proof}
\section{Next Time}
We will continue by giving an alternate proof on PARITY using polynomial approximations.
The bound that we will get is $C_d(\bigoplus_n) = 2^{\Omega(n^{\frac{1}{2d}})}$, which
is not as tight as the previous result, but the advantage is that it applies to circuits
with gates other than AND, OR, NOT.
\section*{Acknowledgements}
In writing the notes for this lecture, I perused the notes by Chris Hopman
for lecture 8 from the Spring 2010 offering of CS~710 and the notes by
Seeun William Umboh for lecture 8 from the Spring 2007 offering of CS~810.
\end{document}