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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{Part 1: Explicit, almost optimal $\epsilon$-biased sets. Lecturer: Matthew Dippel; Scribe: Willy Quach}
In this lecture we discuss explicit construction of $\epsilon$-biased sets with almost optimal support size.
\begin{definition}{\textbf{ $\epsilon$-biased sets.}}
A set $S \subseteq \{0,1\}^n$ is $\epsilon$-biased if for all linear test $a$:
\[ \left| \Pr_{x\in S} [ \langle a,x \rangle = 1] - \Pr_{x\in S} [ \langle a,x \rangle = 0] \right| \leq \epsilon. \]
\end{definition}
In this lecture, we will focus on proving the following theorem:
\begin{theorem}
There is an explicit construction of an $\epsilon$-biased set $S \subseteq \{0,1\}^n$ such that $|S| = \mathcal{O}\left( \frac{n}{\epsilon^{2+d}} \right)$ where $d = o(1)$.
\end{theorem}
Note that we saw in class a construction an $\epsilon$-biased set $S$ with $|S| = \mathcal{O}\left( \frac{n}{\epsilon^2} \right)$. Also, the size of any $\epsilon$-biased set is lower bounded by $\Omega\left( \frac{n}{\epsilon^2 \log 1/\epsilon} \right)$.
The basic idea is the following fact:
\begin{claim}
If $S$ is $\epsilon$-biased, then for all $k\geq 1$, the sum of $k$ \textit{i.i.d} samples from $S$ is $\epsilon^k$-biased.
\end{claim}
This is not enough to give an explicit construction by itself, as the support size grows roughly exponentially with $k$.
The idea is to leverage this fact, but using \textit{pseudorandomness} for the samples. More precisely, we will start with some $\epsilon_0$-biased set $S$ for some constant $\epsilon_0$, and map the elements of $S$ onto the nodes of an expander graph (recall that taking a somewhat short random walk over an expander graph leads to a distribution close to uniform over the vertices of the graph). Then, we hope that the sum of the elements seen while randomly walking through the graph is a good $\epsilon$-biased set.
\bigskip
More precisely, (long) random walks on expanders are good parity samplers: for any linear test $a$, define $B_a = \{ v\in V \, | \, \langle a,v \rangle = 1 \}.$ Then:
\[\left| \Pr[\text{ The walk hits $B_a$ an odd number of times }] - 1/2 \right| \leq \epsilon \].
Note that this is better than drawing $t$-wise independent samples (where $t$ is the length of the random walk). Indeed, setting such a $v$ such that $\sum_{i=0}^t v_i = 0$ implies that $\sum_{i=0}^t \langle a, v_i \rangle = 0$ for all linear test $a$ (and therefore fails the parity test).
\bigskip
The main idea is the following: take $G$ to be an expander graph, such that we map the elements of $S$ onto its vertices. If its degree is too large, then sampling a random walk on $G$ costs too much. Instead, we consider another expander $H$ whose vertices correspond to edges connected to a fixed vertex in $G$; in other words, the number of vertices in $H$ is the degree of $G$, and each vertex of $H$ corresponds to a next edge to take for the walk in $G$. Therefore, a random walk on $H$ induces a random walk on $G$ as well (again, where the vertex reached on $H$ defines the next step to take on $G$). If the degree of $H$ is much less than the degree of $G$, this allows to have a much smaller support; and one can hope that if $H$ is an expander, then the random on $G$ actually achieves the desired properties.
More formally, we consider an expander $G = (N_1, D_1, \lambda_1)$ (where $N_1$ is the number of vertices, $D_1$ the degree, and $\lambda_1 = \max\{\lambda_2(G), \lambda_n(G)\}$), and $H = (N_2, D_2, \lambda_2)$ where $N_2 = D_1^s$ for some parameter $s$ (think of $s$ as a large constant). In particular, each vertex of $H$ can be viewed as a list of $s$ elements in $[D_1]$. Then, any random walk on $H$ induces a \textit{deterministic} walk on $G$ in the following way: to take the $\ell$th step, take a step on $G$ where the edge is determined by the $\ell \bmod s$-th element of the current edge in $H$ (again, this element is an element in $[D_1]$, so it defines an edge going out the current vertex in $G$), and then take a step on $H$. Intuitively, this corresponds to apply the procedure described above, with $s$ parallel copies of $H$.
Such a construction allows to get the following parameters(on input $n$, $\epsilon$):
Take $d = \Theta\left( \left( \frac{\log \log 1/\epsilon}{\log 1/\epsilon}\right)^{1/3}\right)$, and $H$ such that $s=1/d$, $D_2 = s^{4s}$ (for instance $H$ can be taken to be the Cayley Graph over $\mathbb{Z}_2^{\log |D_2|}$; the initial distribution is an $\epsilon_0$-biased distribution with support size $\mathcal{O}(n/\epsilon_0^2)$, with $\epsilon_0 = 1/D_2$.
Take $G$ to be a Ramanujan expander with $D_1 = \mathcal{O}(1/\lambda_1^2)$, $N_1 \approx |S|$. Then it suffices to consider a random walk of length $t$, where $t$ is the smallest integer such that $\lambda_2^{(1-4d)(1-d)t} \leq \epsilon$ (and in particular $t \geq 1/d^2)$.
\bigskip
Let us show how such a random walk allows to reduce the bias, even in the case when we do not use an outer graph $H$. The main idea is to express the bias of the resulting distribution using linear algebra.
We start with a $\epsilon_0$-biased distribution over $G$ (say, that $\epsilon_0$ is a constant, for simplicity). Suppose $N_1$ is such that $N_1 \in [(1-\beta)n, n]$ or $N_1 \in [(1-\beta)2n, 2n]$ for some small constant $\beta$. We sample a random walk of length $t$. Let $\alpha$ be a best linear distinguisher for the resulting distribution, and define:
$S_b = \{v \in N_1 \, | \, \langle \alpha, v \rangle = b \}$, and $\Pi_b$ to be the projection on $S_b$, where $b \in \{0,1\}$. Let $\Pi = \Pi_0 - \Pi_1$. Let $\Upsilon$ be the resulting distribution of the random walk. Let $p_{even}(S_1)$ (respectively $p_{odd}(S_1)$) be the probability that the random walk visits $S_1$ an even (respectively odd) number of times. Let $\mathbf{1}$ be the unit vector colinear with $(1,\dots,1)$.
\begin{theorem}
We have:
\begin{enumerate}
\item Bias($\Upsilon$) $= | p_{even}(S_1) - p_{odd}(S_1)|$;
\item $p_{even}(S_1) - p_{odd}(S_1) = \sum_{b_0\dots b_t \in \{0,1\}} (-1)^{\sum b_i} \mathbf{1}^T \Pi_{b_t}G \cdots \Pi_{b_1}G \Pi_{b_0} \mathbf{1}$;
\item $p_{even}(S_1) - p_{odd}(S_1) = \mathbf{1}^T (\Pi G)^t \Pi \mathbf{1}$;
\item $\| (\Pi G)^2 \| \leq \epsilon_0 + 2\beta + 2\lambda$;
\item Bias($\Upsilon$) $\leq (\epsilon_0 + 2\beta + 2\lambda)^{\lfloor t/2 \rfloor}. $
\end{enumerate}
\end{theorem}
We prove item $4$: if $v$ is of norm $1$, we can write $v = v^{||} + v^{\bot}$ along $Span(\mathbf{1})$ and its orthogonal, such that $Gv^{||} = v^{||} = \| v^{||}\| \mathbf{1}$. Then:
\begin{align*}
\| (\Pi G)^2 \| &\leq \| (\Pi G)^2 v \| \leq \| (\Pi G)^2 v^{||}\| + \| (\Pi G)^2 v^{\bot}\|, \\
&\leq \| v^{||} \| \| \Pi G \Pi \mathbf{1} \| + \|Pi G \Pi \| \| G v^{\bot} \|, \\
&\leq \| \Pi G (\Pi \mathbf{1})^{||} \| + \| \Pi G (\Pi \mathbf{1})^{\bot} \| + \| G v^{\bot} \|, \\
&\leq \| \Pi \mathbf{1}^{||} \| + 2\lambda.
\end{align*}
Then, note that $\| \Pi \mathbf{1}^{||} \| = \left| \langle \Pi \mathbf{1}, \mathbf{1} \rangle \right| = \left| \frac{|S_0| - |S_1|}{N_1} \right|$. As the initial distribution $\Upsilon_0$ is $\epsilon_0$ biased and we removed at most $\beta n$ elements we have:
\[\left| |S_0| - |S_1| \right| \leq \frac{1+\epsilon_0}{2}n - (\frac{1-\epsilon_0}{2}n - \beta n) \leq (\epsilon_0 + 2\beta)N_1.\]
\section{Part 2: Quadratic Time Hardness of the Longest Common Subsequence Problem. Lecturer: Tanay Mehta}
Let us focus on \textit{Fine-Grained Complexity}, which mainly establishes lower bounds on the hardness of problems in P (assuming the hardness of a few problems).
The main conjectured hard problems in fine-grained complexity are the following:
\begin{itemize}
\item 3SUM: given a set in $S \subset [-n^3, n^3]$ of size $n$, find three elements $a,b,c$ such that $a+b=c$. Its conjectured hardness is $n^{2-o(1)}$ time.
\item APSP (All Pairs Shortest Paths): given a weighted graph $G$, compute the (weighted) distance between all pairs of vertices. Its conjectured hardness is $n^{3-o(1)}$ time.
\item OV (Orthogonal Vectors): given two sets $U,V$ of vectors in $\{0,1\}^d$, decide if there exists $u\in U, v\in V$ such that $\langle u,v \rangle = 0$. Its conjectured hardness is $n^{2-o(1)}$ time for $d = \omega(\log n)$ (and is in general $\approx n^2 d$).
\end{itemize}
Interestngly, the hardness of OV is implied by the Strong Exponential Time Hypothesis (SETH).
\begin{definition}
The Strong Exponential Time Hypothesis states that:
\[\forall \epsilon > 0, \exists k, k\text{-SAT requires }2^{(1-\epsilon)n} \text{ time}.\]
\end{definition}
\begin{claim}
Assuming SETH, OV requires $\Omega(n^2d)$ time to solve.
\end{claim}
The reduction from $k-$SAT to OV is surprisingly simple: given a SAT instance $\phi$ on $n$ variables and $m$ clauses, split the variables into two disjoint sets $A$, $B$ of size $n/2$, and define :
$U = \{\vec u \in \{0,1\}^{m}, \vec u_i = 0$ if and only if the $i$th clause is satisfied by some partial assignment $a \in A\}$,
$V = \{\vec v \in \{0,1\}^{m}, \vec v_i = 0$ if and only if the $i$th clause is satisfied by some partial assignment $b \in B\}$.
Then $\phi$ is satisfiable if and only if there is a pair of orthogonal vectors across $U,V$ (were each contains $2^{n/2}$ vectors, one for each possible partial assignment in $A$ and $B$, respectively).
In the following, we will be more interested in an extension of the OV problem:
\begin{definition}
The Most-OV problem consists in, given an integer $r$, and two sets $U,V \in (\{0,1\}^d)^n$ of $n$ vectors of dimension $d$, decide if there exists $u\in U, v\in V$ such that $\langle u, v \rangle \leq r$.
\end{definition}
Recall that a subsequence of some string $z = z_1\dots z_n$ is a string $z_{i_1}\dots z_{i_k}$ where $\{i_j\}_j$ is an increasing sequence of integers. In particular, a subsequence does not necessarily consist in consecutive letters in the original string.
\begin{definition}
The Longest Common Subsequence (LCS) problem consists in, given two strings $P_1, P_2$ of length $n$ over some alphablet $\Sigma$, compute the length of their Longest Common Subsequence.
\end{definition}
We will prove the following theorem:
\begin{theorem}
If there exists some $\epsilon>0$ such that LCS over an alphabet of size 7 can be solved in $\mathcal{O}(n^{2-\epsilon})$ time, then Most-OV can be solved in $\mathcal{O}(n^{2-\epsilon}d)$ time.
\end{theorem}
We will next sketch the proof of the theorem.
Define Weighted LCS (WLCS) to be the LCS problem with weights on the elements of the alphabet; the goal is then to maximize the weight of a common subsequence. Note that WeightedLCS reduces to LCS: if $\alpha \in \Sigma$ has weight $w$, simply define a morphism that maps $\alpha$ to $\alpha^w$.
Therefore, it suffices to reduce Most-OV to Weighted LCS.
Let $\{\alpha\}_{[n]}$, $\{\beta\}_{[n]}$ be a Most-OV instance, and let $\Sigma = \{0,\dots,6\}$.
Define the following Coordinate Gadgets:
\[ CG_1(\alpha, i) = \left\{
\begin{array}{ll}
5465 &$ if $ \alpha_i=0 \\
545 &$ otherwise$
\end{array} \right.; \]
\[ CG_2(\beta, i) = \left\{
\begin{array}{ll}
5645 &$ if $ \beta_i=0 \\
565 &$ otherwise$
\end{array} \right., \]
and define weights $w(5) = X := 100d$, $w(4) = w(6) = 1$.
Note that: WLCS$(CG_1(\alpha,i), CG_2(\beta,i)) = \left\{
\begin{array}{ll}
2X+1 &$ if $ \alpha_i \beta_i=0 \\
2X &$ otherwise$
\end{array} \right.$.
Define now the following Vector Gadgets:
\[ VG_1(\alpha)=1 \circ \circ_{i=1}^d. CG_1(\alpha,i), \]
\[ VG_2(\beta)=\circ_{i=1}^d. CG_2(\beta,i) \circ 1, \]
with weight $w(1) = A:= (r+1) 2X + (d-(r+1))(2X+1)$.
\begin{claim}
If $\langle \alpha, \beta \rangle \leq r$, then:
\[ \text{WLCS}(VG_1(\alpha), VG_2(\beta)) \geq A+1 = r\cdot 2X + (d-r) (2X+1).\]
\end{claim}
The claim above follows directly from the construction.
\begin{claim}
If $\langle \alpha, \beta \rangle > r$, then:
\[ \text{WLCS}(VG_1(\alpha), VG_2(\beta)) =A.\]
\end{claim}
To see this, note that $1$ is a common subsequence, so that the WLCS is at least $A$.
Furthermore, if $1$ is not taken in the subsequence we can assume without loss of generality that the $5$'s map to each other as letters in the subsequences, and at least $r+1$ letters in between that match, with weight $1$ each. The inequality follows.
We can now build the sequences for the WLCS problem. Define:
\[ P_2 = 3 \circ \left( \circ_{i=1}^{n-1} (0 \circ VG_2(1^d)\circ 2 \circ 3) \right) \circ
\left( \circ_{i=1}^{n-1} (0 \circ VG_2(\beta^i)\circ 2 \circ 3) \right) \circ
\left( \circ_{i=1}^{n} (0 \circ VG_2(1^d)\circ 2 \circ 3) \right)
;\]
\[ P_1 := 3^{|P_2|} \circ (\circ_{i=1}^n (0 \circ VG_1(\alpha^i)\circ 2))\circ 3^{|P_2|},\]
with weights $w(3) = A^2$ and $w(0) = w(2) = A^4$.
With some additional work, one can show that $P_1$ and $P_2$ have their WLCS greater than $n\cdot (2A^4 + A) + 2n A^2$ if and only if there are no vectors in $\{\alpha\}_{[n]}$, $\{\beta\}_{[n]}$ with inner product less than $r$.
\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}