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\begin{document}
\input{preamble.tex}
\lecture{14}{Apr 27, 2009}{Jayalal Sarma M.N.}{Jing He}

In previous lectures we saw different approaches for proving
$\textbf{PARITY} \not\in \mathbf{AC^0}$. Today we will introduce
another type of circuit lower bounds, namely, lower bound for monotone
circuits. In 1985, Razborov proved that the \textbf{CLIQUE} problem
does not have polynomial-sized monotone circuits.  Before this, we
familiarize ourselves with some ideas related to montonicity itself.
We need some definitions and notations first.

\section{Monotone Circuits}

\begin{definition}[Monotone circuits]
\label{def_mon} A Boolean circuit (over $\{\land, \lor, \lnot\}$) 
is called \emph{monotone} if it does not contain any \textbf{NOT} gates.
\end{definition}

\begin{definition}[Monotone Functions]
\label{def_mon_fun}
A function $f:\{0,1\}^n \rightarrow \{0,1\}$ is monotone if and only
if $(\forall x \leq y) f(x) \leq f(y)$, where the "$\leq$" is
performed bitwisely.
\end{definition}

\begin{proposition}
A function $f:\{0,1\}^n \to \{0,1\}$ is \emph{monotone} if and only if
it can be computed by a monotone circuit.
\end{proposition}
\begin{proof}
Monotone functions are closed under composition. That is if for $f$
and $g$ are monotone, then $f(g(u_1),g(u_2), \ldots g(u_k)$ is montone
too. Since $\land$ and $\lor$ are montone it follows that monotone
circuits can compute only monotone functions.

To see the other direction : first we define a (partially) monotone
circuit for the comparator function. That is given, $x,\alpha \in
\{0,1\}^n$, the circuit checks if $x \ge \alpha$. If we fix $\alpha$,
this function is monotone in $x$. It is easy to construct a monotone
circuit for it too.

Now let $f$ be a montone function. Consider the Boolean Lattice with
$1^n$ as the maximum element and $0^n$ as the minimum. Any path from
$0^n$ to $1^n$ has a point where the function $f$ turns from 0 to
1. There are only $p=2^n$ vertex disjoint paths, and this defines a
boundary $\alpha_1 \ldots \alpha_{p}$ between 1-region and
0-region. To decide the function, the circuit has to essentially check
if the given input $x$ is greater than any of these $\alpha_i$'s. Now
the montone circuit will have an $\lor$ gate on top with exponential
fan-in, followed by comparator circuits. This gives the proof. A point
to note is that the size of the circuit that we described is not
polynomial in the input.
\end{proof}


Now we formalize the problem we will address in this lecture. Remember
that an undirected graph $G$ with $n$ vertices can be encoded with a
binary string of length $n\choose{2}$, each bit of which indicates
whether the corresponding edge exists. We use this encoding to define
the problem $\mathbf{CLIQUE_{k,n}}$. Let $G(V,E)$ be a graph on $n$
vertices.  Clearly, $G$ can be represented by a bit string
$x_1,x_2,\ldots,x_{n\choose{2}}$ where $x_i$ is 1 if the $i^{\rm th}$
(of the ${n\choose{2}}$ possible edges).

\begin{definition}
\[ \mathbf{CLIQUE_{k,n}}=\left\{x=(x_1,x_2,\ldots,x_{n\choose{2}})~\left|\right.~
\exists \textrm{ a clique of size $k$ in the graph defined by}~x\right\} \]
\end{definition}

A first observation is that this function is monotone. Indeed, if we
add an additional edge to a graph which already has a clique of size
$k$, that clique does not disappear !. Now, by the above argument
about monotone circuits for monotone functions, there is monotone
circuit of size $2^{{n\choose{2}}}$ computing this function. Indeed, a
very similar arguement gives slightly better upper bound.

\begin{proposition}
$\mathbf{CLIQUE_{k,n}}$ can be computed by a monotone circuit of size $O(n^k)$.
\end{proposition}
\begin{proof}
Trivial, for all subsets of size $k$ (there are ${n \choose k}=O(n^k)$
of them), and for each subset cheque whether the edge is present or
not. This can be done by a monotone circuit. The size of the entire
circuit is still $O(n^k)$. Notice that when $k = O(n)$ this gives an
$n^n$ bound which is not polynomial sized.
\end{proof}

Now we will consider lower bounds for the problem
$\mathbf{CLIQUE_{k,n}}$. We will see that the above bound is tight
upto a $\sqrt{k}$ factor in the exponent. 

\section{Lower Bounds for CLIQUE}

We will show the following theorem due to Razborov (1985).
\begin{theorem}\label{lower_b}
Any monotone circuit computing $\mathbf{CLIQUE_{k,n}}$ must have
size $n^{\Omega{(\sqrt{k})}}$.
\end{theorem}

Before proving the above theorem we first give an outline. First,
we want to transform any circuit $C$ computing
$\mathbf{CLIQUE_{k,n}}$ into $C'$ which makes a lot of errors when
computing the same problem. Then we prove that the error made by
any single gate of $C'$ is kind of "small". So by a union bound,
we can get a lower bound for the size of $C'$, which also gives a
lower bound for the size of $C$ if they differ not too much.

To put our idea explicitly we need some notations.

\begin{definition}
~An encoded graph is called a \emph{positive input} of
$\mathbf{CLIQUE_{k,n}}$ if it is a minimal graph containing a clique
of size $k$. Let $\mathbf{PI_{n,k}}$ denote the collection of all
such graphs.
\end{definition}

\begin{definition}
~An encoded graph is called a \emph{negative input} of
$\mathbf{CLIQUE_{k,n}}$ if it is a maximal graph which does not
contain a clique of size $k$. Let $\mathbf{NI_{n,k}}$ denote the
collection of all such graphs.
\end{definition}

It is clear that $|\mathbf{PI_{n,k}}|={n\choose k}$ and
$|\mathbf{NI_{n,k}}|=(k-1)^n$.

\begin{definition}
~A \emph{clique indicator} $I_X$ is a boolean function on graphs of
$n$ vertices which outputs 1 if and only if the induced subgraph of
the input graph on vertex set $X$ is a clique. A
\emph{$(m,l)$-approximator} is a boolean function of form
$\bigvee_{i=1}^{r}I_{x_i}$ where $|X_i|\leq l$ and $r\leq m$.
\end{definition}

Suppose we have a monotone circuit $C$ computing
$\mathbf{CLIQUE_{k,n}}$. We want to transform it into a
$(m,l)$-approximator $C'$ for some fixed $m$ and $l$. We do the
transformation inductively. For a single variable $x_{i,j}$, we
change it into a clique indicator $I_{\{i,j\}}$. For a formula
$F_1 \lor F_2$, suppose $A=\bigvee_{i=1}^{r}I_{X_i}$ and
$B=\bigvee_{j=1}^{s}I_{y_j}$ are the corresponding
$(m,l)$-approximators of $F_1$ and $F_2$, respectively. We know
that $T=A\lor B=(\bigvee_{i=1}^{r}I_{X_i}) \lor
(\bigvee_{j=1}^{s}I_{y_j})$, but this is a $(2m,l)$-approximator.
To compress it we need the following lemma from Erdos and Rado:

First, some terminology. A \emph{sunflower} is a collection of $p$
sets $\{Z_1,\ldots,Z_p\}$ such that $\forall 1\leq i\leq j\leq p, Z_i
\cap Z_j = Z$, where $Z$ is called the \emph{center} of the
sunflower. We also call it a $p$-petal sunflower. The choice of these
names are more-or-less self explanatory. 

\begin{lemma}[Sunflower Lemma]
\label{sunf} 
Suppose $S=\{S_1,\ldots,S_k\}$ is a collection of sets for which
$(\forall 1\leq i\leq k)|S_i|\leq l$ and $k\geq (p-1)^l\cdot l!$, then
there exists a $p$-petal sunflower in $S$.
\end{lemma}

We will include a proof of the Lemma later in this draft.  First we
see the application. Now we choose $m=(p-1)^l \cdot l!$, where the
values of $p$ and $l$ will be decided later. If $r+s<m$, $T$ is
already a $(m,l)$-approximator. If $r+s\geq m$, from the sunflower
lemma we know that among $S=\{X_1,\ldots,X_r\}\cup\{Y_1,\ldots,Y_s\}$
there exists a $p$-petal sunflower $\{Z_1,\ldots,Z_p\}$ with center
$Z$.  We use $I_{Z}$ to replace $\bigvee_{i=1}^{p}I_{Z_p}$, and repeat
the above process until $T$ becomes a $(m,l)$-approximator $T'$.
Notice that $T'$ may disagree with $T=A\lor B$ on some negative inputs
since we "reduce" the size of some clique indicators. But on all
positive inputs they agree with each other.

We are left with the case where $T=A \land B$. We have
\begin{eqnarray*}
& &A \land B = (\bigvee_{i=1}^r I_{X_i}) \land (\bigvee_{j=1}^s
I_{Y_j})\\
&=&\bigvee_{i=1}^{r}\bigvee_{j=1}^{s}(I_{X_i} \land I_{Y_j})\\
&\approx&\bigvee_{i=1}^{r}\bigvee_{j=1}^{s}I_{X_i \cup Y_j}\\
&\approx&\bigvee_{1\leq i \leq r,1\leq j\leq s,|X_i\cup Y_j|\leq l
}I_{X_i \cup Y_j}\\
&=& T'
\end{eqnarray*}
Due to the two "$\approx$", $T'$ may disagree with $T$ on some
negative inputs. Now $T'$ is a $(m^2,l)$-approximator and we can
apply the sunflower lemma again to transform it into a
$(m,l)$-approximator, while losing some positive inputs.

By induction, we can finally get a $(m,l)$-approximator $C'$ from
the original circuit $C$. We use $size(C)$ to denote the size of a
circuit $C$.

\begin{claim}\label{cl1}
~For the $C'$ described above, either $C'$ outputs 0 on all
positive inputs, or $C'$ outputs 1 on
$(1-{l\choose{2}}/(k-1))(k-1)^n$ negative inputs.
\end{claim}

\begin{claim}\label{cl2}
~$C'$ disagree with $C$ on at most $size(C)\cdot
m^2{{n-l-1}\choose{k-l-1}}$ positive inputs.
\end{claim}

\begin{claim}\label{cl3}
~$C'$ disagree with $C$ on at most $size(C)\cdot
m^2\left({l\choose 2}/(k-1)\right)^p(k-1)^n$ negative inputs.
\end{claim}

If they are all correct, we have
\begin{eqnarray*}
size(C)\geq \frac{{n\choose k}}{m^2{{n-l-1}\choose{k-l-1}}}
\end{eqnarray*}
or
\begin{eqnarray*}
size(C)\geq
\frac{\left(1-\frac{{l\choose{2}}}{k-1}\right)(k-1)^n}{m^2\left(\frac{{l\choose{2}}}{k-1}\right)^p(k-1)^n}
\end{eqnarray*}
Choose $l=\lfloor\sqrt{k}\rfloor$, $p=\lceil \sqrt{k}\log n\rceil$
and $m=(p-1)^l \cdot l!$, then after some calculations we get
$size(C) \geq n^{\Omega(\sqrt{k})}$.

Now we prove the three claims.
\\

\textbf{Proof of Claim 10} ~~If $C'$ is identical to 0 then it
outputs 0 on all positive inputs. If not, $C'$ must contain at
least one clique indicator, say, $I_{X_1}$. Then we have
\begin{eqnarray*}
& &\mathbf{Prob}_{x\in \mathbf{NI_{n,k}}}[C'(x)=1] \\
&\geq&\mathbf{Prob}_{x\in \mathbf{NI_{n,k}}}[I_{X_1}(x)=1]\\
&=&1-\mathbf{Prob}_{x\in \mathbf{NI_{n,k}}}[I_{X_1}(x)=0]\\
&\geq&1-\frac{{l\choose 2}}{k-1}
\end{eqnarray*}
So the claim follows. $\blacksquare$
\\

\textbf{Proof of Claim 11} ~~Remember that $C'$ disagrees with $C$
on negative inputs only because of the transformation for the form
$A\land B$, where $I_{X_i}\land I_{Y_j}$ is replaced with
$I_{X_i\cup Y_j}$. In one transformation there are at most $m^2$
such terms, each of which makes at most ${n-l-1}\choose{k-l-1}$
errors on negative inputs. So the total error is upper-bounded by
$size(C)\cdot m^2{{n-l-1}\choose{k-l-1}}$. $\blacksquare$
\\

\textbf{Proof of Claim 12} ~~When we deal with the form $A \lor
B$, we replace $\bigvee_{i=1}^p I_{Z_i}$ by their center $Z$. So
the probability of one such replacement making errors on negative
inputs is
\begin{eqnarray*}
& &\mathbf{Prob}_{x\in \mathbf{NI_{n,k}}}[I_Z(x)=1 \land (\forall
1\leq i\leq p)I_{Z_i}(x)=0]\\
&\leq&\mathbf{Prob}_{x\in \mathbf{NI_{n,k}}}[\bigwedge_{i=1}^p
I_{Z_i}(x)=0~|~I_Z(x)=1]\\
&=&\prod_{i=1}^p \mathbf{Prob}_{x\in
\mathbf{NI_{n,k}}}[I_{Z_i}(x)=0~|~I_{Z}(x)=1]\\
&\leq&\prod_{i=1}^p \mathbf{Prob}_{x\in
\mathbf{NI_{n,k}}}[I_{Z_i}(x)=0]\\
&\leq&\left(\frac{{l\choose 2}}{k-1}\right)^p
\end{eqnarray*}
and at most $m^2$ such replacements suffices. So at each
$\lor$-gate at most $m^2\left(\frac{{l\choose
2}}{k-1}\right)^p(k-1)^n$ errors are made by $C'$.

When dealing with the form $A\land B$, we can apply a similar
argument which also yields an upper-bound of
$m^2\left(\frac{{l\choose 2}}{k-1}\right)^p(k-1)^n$ for errors on
negative inputs. So the total errors on negative inputs made by
$C'$ is at most $size(C)\cdot m^2\left(\frac{{l\choose
2}}{k-1}\right)^p(k-1)^n$. $\blacksquare$

\subsection*{Proof of Sunflower Lemma}
For completeness we include the proof of Sunflower Lemma which is
crucial in the above construction. We restate the lemma first.

\begin{lemma}[Sunflower Lemma]
Suppose $\mathcal{S} = \{S_1,\ldots,S_k\}$ is a collection of subsets
of $[n]$ such that $(\forall 1\leq i\leq k)|S_i|\leq \ell$ and $k \geq
(p-1)^\ell \cdot \ell!$, then there exists a $p$-petal sunflower in
$S$. That is there exists $Z_1, \ldots Z_p \subseteq \mathcal{S}$ such
that $Z_i \cap Z_j = Z$ for all $i$ and $j$.
\end{lemma}
\begin{proof}
The proof is by induction on $\ell$. The base case $\ell=1$ is trivial
since we can define $Z = \phi$, and have disjoint $p$-petals. As for
the induction, pick the maximum number of disjoint subsets from $S$,
say $Z_1, Z_2, \ldots Z_r$. If $r \ge p$ then we are done, since in
this case $Z$ could just be chosen to be the empty set. If $r < p$,
define, $U = \bigcup_{i=1}^r Z_i$. By maximality, every set that was
not chosen from $S$ must intersect $Z$ and the size of $Z$ is at most
$r\ell \le (p-1)\ell$. Hence, by an averaging argument, there exists a
$x \in Z$ such that it is contained in at least $(p-1)^\ell.(\ell-1)!$
sets of $\mathcal{S}$. Consider this collection of sets and remove $x$
from them. The size of each set is now at most $(\ell-1)$. By
induction, there is a $p$-petal sunflower in this set system.  Adding
the element $x$ back to each of these sets (that is to their
intersection $Z$) gives you a $p$-petal sunflower for the original
system $S$. This completes the proof.
\end{proof}
\end{document}
