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\begin{document}
\global\long\def\E{\mathrm{\mathbb{E}}}
%\global\long\def\e{\mathrm{\mathbb{\epsilon}}}
\noindent
\href{http://www.ccs.neu.edu/home/viola/classes/spepf17.html}{Special
Topics in Complexity Theory}, Fall 2017. Instructor:
\href{http://www.ccs.neu.edu/home/viola/}{Emanuele Viola}
%\href{http://www.ccs.neu.edu/home/viola/classes/spepf17.html}{For the
%class webpage click here.}
\section{Lecture 8-9, Scribe: Xuangui Huang}
In these lectures, we finish the proof of the approximate degree lower
bound for AND-OR function, then we move to the surjectivity function
SURJ. Finally we discuss quasirandom groups.
\subsection{Lower Bound of $d_{1/3}($AND-OR$)$}
Recall from the last lecture that AND-OR$:\{0,1\}^{R\times N} \rightarrow
\{0,1\}$ is the composition of the AND function on $R$ bits and the OR
function on $N$ bits. We also proved the following lemma.
\begin{lemma} Suppose that distributions $A^0, A^1$ over $\{0,1\}^{n_A}$
are $k_A$-wise indistinguishable distributions; and distributions
$B^0, B^1$ over $\{0,1\}^{n_B}$ are $k_B$-wise
indistinguishable distributions.
Define $C^0, C^1$ over $\{0,1\}^{n_A \cdot n_B}$ as follows:
$C^b$: draw a sample $x \in \{0,1\}^{n_A}$ from $A^b$, and replace
each bit $x_i$ by a sample of $B^{x_i}$ (independently).
Then $C^0$ and $C^1$ are $k_A \cdot k_B$-wise indistinguishable.
\end{lemma}
To finish the proof of the lower bound on the approximate degree of the
AND-OR function, it remains to see that AND-OR can distinguish well the
distributions $C^0$ and $C^1$. For this, we begin with observing that we
can assume without loss of generality that the distributions have disjoint
supports.
\begin{claim}
For any function $f$, and for any $k$-wise indistinguishable distributions
$A^0$ and $A^1$, if $f$ can distinguish $A^0$ and $A^1$ with probability
$\epsilon$ then there are distributions $B^0$ and $B^1$ with the same
properties ($k$-wise indistinguishability yet distinguishable by $f$) and
also with disjoint supports. (By disjoint support we mean for any $x$ either
$\Pr[B^0 = x] = 0$ or $\Pr[B^1 = x] = 0$.)
\end{claim}
\begin{proof}
Let distribution $C$ be the ``common part" of $A^0$ and $A^1$. That is to
say, we define $C$ such that $\Pr[C = x] := \min \{\Pr[A^0 = x], \Pr[A^1 =
x]\}$ multiplied by some constant that normalize $C$ into a distribution.
Then we can write $A^0$ and $A^1$ as
\begin{align*}
A^0 &= pC + (1-p) B^0 \,,\\
A^1 &= pC + (1-p) B^1 \,,
\end{align*}
where $p \in [0,1]$, $B^0$ and $B^1$ are two distributions. Clearly $B^0$
and $B^1$ have disjoint supports.
Then we have
\begin{align*}
\E[f(A^0)] - \E[f(A^1)] =&~p \E[f(C)] + (1-p) \E[f(B^0)] \notag\\
&- p \E[f(C)] - (1-p) \E[f(B^1)] \\
=&~(1-p) \big( \E[f(B^0)] - \E[f(B^1)] \big) \\
\leq&~\E[f(B^0)] - \E[f(B^1)] \,.
\end{align*}
Therefore if $f$ can distinguish $A^0$ and $A^1$ with probability
$\epsilon$ then it can also distinguish $B^0$ and $B^1$ with such
probability.
Similarly, for all $S \neq \varnothing$ such that $|S| \leq k$, we have
\[
0 = \E[\chi_S(A^0)] - \E[\chi_S(A^1)] =
(1-p) \big( \E[\chi_S(B^0)] - \E[\chi_S(B^1)] \big) = 0 \,.
\]
Hence, $B^0$ and $B^1$ are $k$-wise indistinguishable.
\end{proof}
Equipped with the above lemma and claim, we can finally prove the
following lower bound on the approximate degree of AND-OR.
\begin{theorem}
$d_{1/3}($AND-OR$) = \Omega(\sqrt{RN})$.
\end{theorem}
\begin{proof}
Let $A^0, A^1$ be $\Omega(\sqrt{R})$-wise indistinguishable distributions
for AND with advantage $0.99$, i.e. $\Pr[\mathrm{AND}(A^1) = 1] >
\Pr[\mathrm{AND}(A^0) = 1] + 0.99$. Let $B^0, B^1$ be
$\Omega(\sqrt{N})$-wise indistinguishable distributions for OR with
advantage $0.99$. By the above claim, we can assume that $A^0, A^1$
have disjoint supports, and the same for $B^0, B^1$. Compose them by
the lemma, getting $\Omega(\sqrt{RN})$-wise indistinguishable
distributions $C^0,C^1$. We now show that AND-OR can distinguish
$C^0, C^1$:
\begin{itemize}
\item $C_0$: First sample $A^0$. As there exists a unique $x = 1^R$
such that $\mathrm{AND}(x)= 1$, $\Pr[A^1 = 1^R] >0$. Thus by
disjointness of support $\Pr[A^0 = 1^R] = 0$. Therefore when
sampling $A^0$ we always get a string with at least one ``$0$''. But
then ``$0$'' is replaced with sample from $B^0$. We have $\Pr[B^0 =
0^N] \geq 0.99$, and when $B^0 = 0^N$, AND-OR$=0$.
\item $C_1$: First sample $A^1$, and we know that $A^1 = 1^R$ with
probability at least $0.99$. Each bit ``$1$'' is replaced by a sample
from $B^1$, and we know that $\Pr[B^1 = 0^N] = 0$ by disjointness
of support. Then AND-OR$=1$.
\end{itemize}
Therefore we have $d_{1/3}($AND-OR$)= \Omega(\sqrt{RN})$.
\end{proof}
\subsection{Lower Bound of $d_{1/3}($SURJ$)$}
In this subsection we discuss the approximate degree of the surjectivity
function. This function is defined as follows.
\begin{definition} The surjectivity function SURJ$\colon
\left(\{0,1\}^{\log R}\right)^N \to \{0,1\}$, which takes input
$(x_1, \dots, x_N)$ where $x_i \in [R]$ for all $i$, has value $1$
if and only if $\forall j \in [R], \exists i\colon x_i = j$.
\end{definition}
First, some history. Aaronson first proved that the approximate degree of
SURJ and other functions on $n$ bits including ``the collision problem'' is
$n^{\Omega(1)}$. This was motivated by an application in quantum
computing. Before this result, even a lower bound of $\omega(1)$ had not
been known. Later Shi improved the lower bound to $n^{2/3}$, see
\cite{AaronsonS04}. The instructor believes that the quantum framework
may have blocked some people from studying this problem, though it may
have very well attracted others. Recently Bun and Thaler \cite{BunT17}
reproved the $n^{2/3}$ lower bound, but in a quantum-free paper, and
introducing some different intuition. Soon after, together with Kothari, they
proved \cite{BunKT17} that the approximate degree of SURJ is
$\Theta(n^{3/4})$.
We shall now prove the $\Omega(n^{3/4})$ lower bound, though one piece
is only sketched. Again we present some things in a different way from
the papers.
For the proof, we consider the AND-OR function under the promise that
the Hamming weight of the $RN$ input bits is at most $N$. Call the
approximate degree of AND-OR under this promise $d_{1/3}^{\leq
N}($AND-OR$)$. Then we can prove the following theorems.
\begin{theorem}\label{l8-9:thm:1}
$d_{1/3}($SURJ$) \geq d_{1/3}^{\leq N}($AND-OR$)$.
\end{theorem}
\begin{theorem}\label{l8-9:thm:2}
$d_{1/3}^{\leq N}($AND-OR$) \geq \Omega(N^{3/4})$ for some suitable $R = \Theta(N)$.
\end{theorem}
In our settings, we consider $R = \Theta(N)$. Theorem \ref{l8-9:thm:1}
shows surprisingly that we can somehow ``shrink'' $\Theta(N^2)$ bits of
input into $N\log N$ bits while maintaining the approximate degree of the
function, under some promise. Without this promise, we just showed in the
last subsection that the approximate degree of AND-OR is $\Omega(N)$
instead of $\Omega(N^{3/4})$ as in Theorem \ref{l8-9:thm:2}.
\begin{proof}[Proof of Theorem \ref{l8-9:thm:1}]
Define an $N \times R$ matrix $Y$ s.t.~the 0/1 variable $y_{ij}$ is the
entry in the $i$-th row $j$-th column, and $y_{ij} = 1$ iff $x_i = j$. We can
prove this theorem in following steps:
\begin{enumerate}
\item $d_{1/3}($SURJ$(\overline{x})) \geq d_{1/3}($AND-OR$(\overline{y}))$ under
the promise that each row has weight $1$;
\item let $z_j$ be the sum of the $j$-th column, then $d_{1/3}($AND-OR$(\overline{y}))$ under
the promise that each row has weight $1$, is at least $d_{1/3}($AND-OR$(\overline{z}))$
under the promise that $\sum_j z_j = N$;
\item $d_{1/3}($AND-OR$(\overline{z}))$ under the promise that $\sum_j z_j = N$, is
at least $d_{1/3}^{=N}($AND-OR$(\overline{y}))$;
\item we can change ``$=N$'' into ``$\leq N$''.
\end{enumerate}
Now we prove this theorem step by step.
\begin{enumerate}
\item Let $P(x_1, \dots, x_N)$ be a polynomial for SURJ, where $x_i = (x_i)_1, \dots, (x_i)_{\log R}$.
Then we have
\[
(x_i)_k = \sum_{j: k\text{-th bit of }j \text{ is } 1} y_{ij}.
\]
Then the polynomial $P'(\overline{y})$ for AND-OR$(\overline{y})$ is the
polynomial $P(\overline{x})$ with $(x_i)_k$ replaced as above, thus the
degree won't increase. Correctness follows by the promise.
\item This is the most extraordinary step, due to Ambainis
\cite{Ambainis05}. In this notation, AND-OR becomes the indicator
function of $\forall j, z_j \neq 0$. Define
\[
Q(z_1, \dots, z_R) := \mathop{\E}_{\substack{\overline{y}: \text{ his rows have weight } 1\\ \text{and is consistent with }\overline{z}}} P(\overline{y}).
\]
Clearly it is a good approximation of AND-OR$(\overline{z})$. It remains
to show that it's a polynomial of degree $k$ in $z$'s if $P$ is a
polynomial of degree $k$ in $y$'s.
Let's look at one monomial of degree $k$ in $P$:
$y_{i_1j_1}y_{i_2j_2}\cdots y_{i_kj_k}$. Observe that all $i_\ell$'s are
distinct by the promise, and by $u^2 = u$ over $\{0,1\}$. By chain rule
we have
\[
\E[y_{i_1j_1}\cdots y_{i_kj_k}] = \E[y_{i_1j_1}]\E[y_{i_2j_2}|y_{i_1j_1} = 1] \cdots
\E[y_{i_kj_k}|y_{i_1j_1}=\cdots =y_{i_{k-1}j_{k-1}} = 1].
\]
By symmetry we have $\E[y_{i_1j_1}] = \frac{z_{j_1}}{N}$, which is linear
in $z$'s. To get $\E[y_{i_2j_2}|y_{i_1j_1} = 1]$, we know that every other
entry in row $i_1$ is $0$, so we give away row $i_1$, average over $y$'s
such that $\left\{\begin{array}{ll}
y_{i_1j_1} = 1 &\\
y_{ij} = 0 & j\neq j_1
\end{array}\right.$ under the promise and consistent with $z$'s. Therefore
\[
\E[y_{i_2j_2}|y_{i_1j_1} = 1] = \left\{
\begin{array}{ll}
\frac{z_{j_2}}{N-1} & j_1 \neq j_2,\\
\frac{z_{j_2}-1}{N-1} & j_1 = j_2.
\end{array}\right.
\]
In general we have
\[
\E[y_{i_kj_k}|y_{i_1j_1}=\cdots =y_{i_{k-1}j_{k-1}} = 1]
= \frac{z_{j_k} - \#\ell < k \colon j_\ell = j_k}{N-k + 1},
\]
which has degree $1$ in $z$'s. Therefore the degree of $Q$ is not
larger than that of $P$.
\item Note that $\forall j$, $z_j = \sum_i y_{ij}$. Hence by replacing $z$'s
by $y$'s, the degree won't increase.
\item We can add a ``slack'' variable $z_0$, or equivalently $y_{01},
\dots, y_{0N}$; then the condition $\sum_{j=0}^R z_j = N$ actually
means $\sum_{j=1}^R z_j \leq N$.
\end{enumerate}
\end{proof}
\begin{proof}[Proof idea for Theorem \ref{l8-9:thm:2}]
First, by the duality argument we can verify that $d_{1/3}^{\leq N}(f) \geq
d$ if and only if there exists $d$-wise indistinguishable distributions $A, B$
such that:
\begin{itemize}
\item $f$ can distinguish $A, B$;
\item $A$ and $B$ are supported on strings of weight $\leq N$.
\end{itemize}
\begin{claim}
$d_{1/3}^{\leq \sqrt{N}}($OR$_N) = \Omega(N^{1/4})$.
\end{claim}
The proof needs a little more information about the weight distribution of
the indistinguishable distributions corresponding to this claim. Basically,
their expected weight is very small.
Now we combine these distributions with the usual ones for And using the
lemma mentioned at the beginning.
What remains to show is that the final distribution is supported on
Hamming weight $\le N$. Because by construction the $R$ copies of the
distributions for Or are sampled independently, we can use concentration
of measure to prove a tail bound. This gives that all but an exponentially
small measure of the distribution is supported on strings of weight $\le N$.
The final step of the proof consists of slightly tweaking the distributions to
make that measure $0$.
\end{proof}
\subsection{Groups}
Groups have many applications in theoretical computer science.
Barrington \cite{Barrington89} used the permutation group $S_5$ to prove
a very surprising result, which states that the majority function can be
computed efficiently using only constant bits of memory (something which
was conjectured to be false). More recently, catalytic computation
\cite{BuhrmanCKLS14} shows that if we have a lot of memory, but it's full
with junk that cannot be erased, we can still compute more than if we had
little memory. We will see some interesting properties of groups in the
following.
Some famous groups used in computer science are:
\begin{itemize}
\item $\{0,1\}^n$ with bit-wise addition;
\item $\mathbb{Z}_m$ with addition mod $m$ ;
\item $S_n$, which are permutations of $n$ elements;
\item Wreath product $G:= (\mathbb{Z}_m \times \mathbb{Z}_m) \wr \mathbb{Z}_2\,$, whose elements are of the form $(a,b)z$ where $z$ is a ``flip bit'', with the following multiplication rules:
\begin{itemize}
\item $(a, b) 1 = 1 (b, a)$ ;
\item $z\cdot z' := z+z'$ in $\mathbb{Z}_2$ ;
\item $(a,b) \cdot (a',b') := (a+a', b+b')$ is the $\mathbb{Z}_m\times \mathbb{Z}_m$ operation;
\end{itemize}
An example is $(5,7)1 \cdot (2,1) 1 = (5,7) 1 \cdot 1 (1, 2) = (6,9)0$ . Generally we have
\[
(a, b) z \cdot (a', b') z' = \left\{
\begin{array}{ll}
(a + a', b+b') z+z' & z = 1\,,\\
(a+b', b + a') z+z' & z = 0\,;
\end{array}\right.
\]
\item $SL_2(q) := \{2\times 2$ matrices over $\mathbb{F}_q$ with determinant $1\},$
in other words, group of matrices $\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}$ such that $ad - bc = 1$.
\end{itemize}
The group $SL_2(q)$ was invented by Galois. (If you haven't, read his
biography on wikipedia.)
\paragraph{Quiz}
Among these groups, which is the ``least abelian''? The latter can be
defined in several ways. We focus on this: If we have two high-entropy
distributions $X, Y$ over $G$, does $X \cdot Y$ has more entropy? For
example, if $X$ and $Y$ are uniform over some $\Omega(|G|)$ elements,
is $X\cdot Y$ close to uniform over $G$? By ``close to'' we mean that the
statistical distance is less that a small constant from the uniform
distribution. For $G=(\{0,1\}^n, +)$, if $Y=X$ uniform over $\{0\}\times
\{0,1\}^{n-1}$, then $X\cdot Y$ is the same, so there is not entropy
increase even though $X$ and $Y$ are uniform on half the elements.
\begin{definition}[Measure of Entropy]
For $\lVert A\rVert_2 = \left(\sum_xA(x)^2\right)^{\frac{1}{2}}$, we think of $\lVert A\rVert^2_2 = 100 \frac{1}{|G|}$ for ``high entropy''.
\end{definition}
Note that $\lVert A\rVert^2_2$ is exactly the ``collision probability'', i.e. $\Pr[A = A']$.
We will consider the entropy of the uniform distribution $U$ as very small, i.e.
$\lVert U\rVert^2_2 = \frac{1}{|G|} \approx \lVert \overline{0}\rVert^2_2$. Then we have
\begin{align*}
\lVert A - U \rVert^2_2 &= \sum_x \left(A(x) - \frac{1}{|G|}\right)^2\\
&= \sum_x A(x)^2 - 2A(x) \frac{1}{|G|} + \frac{1}{|G|^2} \\
&= \lVert A \rVert^2_2 - \frac{1}{|G|} \\
&= \lVert A \rVert^2_2 - \lVert U \rVert^2_2\\
&\approx \lVert A \rVert^2_2\,.
\end{align*}
\begin{theorem}[\cite{Gowers08}, \cite{BabaiNP08}]
If $X, Y$ are independent over $G$, then
\[
\lVert X\cdot Y - U \rVert_2 \leq \lVert X \rVert_2 \lVert Y \rVert_2 \sqrt{\frac{|G|}{d}},
\]
where $d$ is the minimum dimension of irreducible representation of $G$.
\end{theorem}
By this theorem, for high entropy distributions $X$ and $Y$, we get
$\lVert X\cdot Y - U \rVert_2 \leq \frac{O(1)}{\sqrt{|G|d}}$, thus we have
\begin{equation} \label{eq:d}
\lVert X\cdot Y - U \rVert_1 \leq \sqrt{|G|} \lVert X\cdot Y - U \rVert_2 \leq \frac{O(1)}{\sqrt{d}}.
\end{equation}
If $d$ is large, then $X \cdot Y$ is very close to uniform. The following
table shows the $d$'s for the groups we've introduced.
\begin{table}[h]
\centering
\begin{tabular}{|c|c|c|c|c|c|}\hline
$G$ & $\{0,1\}^n$ & $\mathbb{Z}_m$ & $(\mathbb{Z}_m \times \mathbb{Z}_m) \wr \mathbb{Z}_2$ & $A_n$ & $SL_2(q)$\\\hline
$d$ & $1$ & $1$ & should be very small & $\frac{\log |G|}{\log \log |G|}$ & $|G|^{1/3}$ \\ \hline
\end{tabular}
\end{table}
Here $A_n$ is the alternating group of even permutations. We can see
that for the first groups, Equation (\ref{eq:d}) doesn't give non-trivial
bounds.
But for $A_n$ we get a non-trivial bound, and for $SL_2(q)$ we get a
strong bound: we have $\lVert X\cdot Y - U \rVert_2 \leq
\frac{1}{|G|^{\Omega(1)}}$.
\bibliographystyle{alpha}
\bibliography{C:/home/krv/math/OmniBib}
\end{document}
This is a template for the scribe. It is also a test to make sure
everything works. You should change the path for the bibliography
(or remove it altogether if you are not using it).
Optionally, the lectures will be posted on my blog. Using this template
minimizes the risk that my wordpress compiler won't work.
\subsection{Subsection \label{sub:Subsection}}
\subsubsection{Subsubsection}
\paragraph{Paragraph}
\begin{thm}
Theorem $\e=1/\e$.\end{thm}
\begin{lem}
\label{lem:Lemma}Lemma\end{lem}
\begin{prop}
Proposition\end{prop}
\begin{claim}
Claim\end{claim}
\begin{proof}
Done\end{proof}
\begin{conjecture}
Conjecture\end{conjecture}
\begin{enumerate}
\item A list
\item Bla\end{enumerate}
\begin{problem}
Open problem
\end{problem}
In-line math $x^{2}=\frac{1}{x}$. Displayed math
\[
x^{2}=3
\]
align
\begin{align}
x & =3\label{eq:label}\\
x & =5\label{eq:bla}
\end{align}
A reference: By Equation \ref{eq:label} and \ref{eq:bla}. A citation.
By Paper \cite{AAIPR01}. Ref to subsection \ref{sub:Subsection}.
Ref to Lemma \ref{lem:Lemma}.
\begin{rem}
This is a test.
\end{rem}