\documentclass[10pt]{article}
\usepackage{fullpage}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath,amsfonts,amsthm}
\usepackage[english]{babel}
\usepackage[capitalize, noabbrev]{cleveref}
\usepackage{algorithm}
\usepackage{algpseudocode}

\newenvironment{enumerate*}%
  {\vspace{-2ex} \begin{enumerate} %
     \setlength{\itemsep}{-1ex} \setlength{\parsep}{0pt}}%
  {\end{enumerate}}
 
\newenvironment{itemize*}%
  {\vspace{-2ex} \begin{itemize} %
     \setlength{\itemsep}{-1ex} \setlength{\parsep}{0pt}}%
  {\end{itemize}}
 
\newenvironment{description*}%
  {\vspace{-2ex} \begin{description} %
     \setlength{\itemsep}{-1ex} \setlength{\parsep}{0pt}}%
  {\end{description}}


\DeclareMathOperator{\E}{\mathbb{E}}
\let\P\relax
\DeclareMathOperator{\P}{\mathbb{P}}
\DeclareMathOperator*{\argmin}{\arg\min}
\newcommand{\ex}[1]{\E\left[#1\right]}
\newcommand{\prob}[1]{\P\left[#1\right]}
\newcommand{\inprod}[1]{\left\langle #1 \right\rangle}
\newcommand{\R}{\mathbf{R}}
\newcommand{\poly}{\text{poly}}
\newcommand{\N}{\mathbf{N}}
\newcommand{\C}{\mathcal{C}}
\newcommand{\eps}{\varepsilon}
\newcommand{\cl}[1]{\text{\textbf{#1}}}
\newcommand{\eqdef}{\mathbin{\stackrel{\rm def}{=}}}
\newcommand{\llbracket}{[\![}


\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\algrenewcommand\algorithmicensure{\textbf{Output:}}
\algrenewcommand\algorithmicrequire{\textbf{Input:}}

\author{Thibaut Horel}
\title{CS 225 Problem Set 3 -- Solutions}

\begin{document}
\maketitle

\section*{Problem 4.2}

\paragraph{1.} Let $S$ be an independent set of $G$. I apply the first
inequality of Lemma 4.15 to the sets $S$ and $S$ and obtain:
\begin{displaymath}
    \left(\frac{|S|}{N}\right)^2
    \leq \lambda \frac{|S|}{N}\left(1-\frac{|S|}{N}\right)
\end{displaymath}
since $e(S, S) = 0$ by definition of an independent set. Reordering the terms:
\begin{displaymath}
    |S|\leq \lambda (N-|S|)
\end{displaymath}
or equivalently $|S| \leq \lambda N/(1+\lambda)$. Taking the maximum over all
independent sets yields the required bound on $\alpha(G)$.

\paragraph{2.}Let us consider a coloring of $G$ with $\chi(G)$ colors. One
color must contain at least $N/\chi(G)$ vertices. This set of $N/\chi(G)$
vertices form an independent set of $G$ by definition of a coloring. Applying
the result from \textbf{1.}:
\begin{displaymath}
    \frac{N}{\chi(G)}\leq \frac{\lambda N}{1+\lambda}
\end{displaymath}
which immediately yields $\chi(G)\geq (1+\lambda)/\lambda$ as required.

\paragraph{3.} Let $D$ be the diameter of $G$ and let us consider two vertices
$i$ and $j$ distant from $D$ in $G$. Let us now consider a random walk on $G$
starting from $i$ and denote by $p_j^t$ the probability of being at vertex
$j$ after $t$ steps. Denoting by $M$ the random walk matrix of $G$ and by
$\pi_i$ the indicator vector of $\{i\}$, we have:
\begin{displaymath}
    p_j^t = (\pi_i M^t)_j
\end{displaymath}
by Lemma 2.51:
\begin{displaymath}
    \left|p_j^t-\frac{1}{n}\right| = \left|(\pi_i M^t)_j-\frac{1}{n}\right|
    \leq \|\pi_iM^t - u\|\leq \lambda^t
\end{displaymath}
For $t= \lceil D/2\rceil$, we have $p_j^t = 0$ (we cannot reach $j$ in less
than $D$ steps). This implies:
\begin{displaymath}
    \lambda^{\lceil D/2\rceil}\geq \frac{1}{n}
\end{displaymath}
solving for $D$:
\begin{displaymath}
    D\leq 2\log_{1/\lambda} N = O(\log_{1/\lambda}N)
\end{displaymath}

\section*{Problem 4.5}

For a given $\eps$ and $\delta$, we construct the following items:
\begin{enumerate}
    \item an expander $G$ of constant expansion $1-\lambda$ on $N$ vertices,
        with $\lambda = \frac{1}{8}$. By Theorem 4.22, we have the guarantee
        that for any function $g:[N]\to[0,1]$, for a random walk
        $x_1,\dots,x_t$ in $G$ starting from a uniformly random $x_1\in [N]$:
        \begin{displaymath}
            \P\left[\frac{1}{t}\sum_t g(x_t) \geq \mu(g) + \lambda
            + \frac{1}{4}\right]\leq e^{-\Omega(t)}
        \end{displaymath}
    \item a pairwise independent sampler $\text{Samp}$ which given $x\in[N]$
        outputs $\ell$ samples in $\{0,1\}^m$. The guarantee we have by
        Proposition 3.28 is that:
        \begin{displaymath}
            \P_{x\gets U_{[N]}}\left[\left|\frac{1}{\ell}\sum_{j=1}^{\ell}
            f\big(\text{Samp}(x)_j\big)- \mu(f)\right|\geq \eps\right]\leq
            \frac{1}{\ell\eps^2}
        \end{displaymath}
        Such a sampler can be obtained by having $\text{Samp}(x)$ be a function
        drawn from a pairwise independent family using the coin tosses in $x$.
        We will have the above guarantee as long as $x$ provides enough random
        bits.  This will be the case with, for example $\log N = 2(\log \ell
        + m)$.
\end{enumerate}

For a vertex $x\in G$, we denote by $\mu_x(f)$ the quantity
$\frac{1}{\ell}\sum_{j=1}^{\ell} f\big(\text{Samp}(x)_j\big)$. That is, the
estimate of $\mu(f)$ computed using the samples provided by $\text{Samp}(x)$.
We denote by $g(x)$ the indicator function of the event
$\left\{|\mu_x(f)-\mu(f)|\geq\eps\right\}$. Another way of reformulating the
guarantee on $\text{Samp}$ is to say that the average of $g$
satisfies $\mu(g)\leq \frac{1}{\ell\eps^2}$.

We construct our $(\delta,\eps)$ sampler as follows:
\begin{enumerate}
    \item we pick $\ell = \frac{8}{\eps^2}$ and $\log N = 2(\log\ell + m)$.
    \item we construct a random walk $x_1,\dots,x_t$ on $G$ starting from
        a uniformly random $x_1\in [N]$
    \item we output the median $M$ of $\big\{\mu_{x_1}(f),\dots,\mu_{x_t}(f)\big\}$.
\end{enumerate}

Let us now consider the quantity $\P[|M-\mu(f)|\geq \eps]$. If the median is
more than $\eps$ away from $\mu(f)$, this implies that more than half of the
numbers $\mu_{x_1}(f),\dots,\mu_{x_t}(f)$ are $\eps$ away from $\mu(f)$. That
is, more than half of the variables $g(x_1),\dots,g(x_t)$ are equal to $1$:
\begin{displaymath}
    \frac{1}{t}\sum_t g(x_t) \geq \frac{1}{2}
\end{displaymath}
But because of our choice of $\ell$, we have $\mu(g)\leq \frac{1}{8}$. Since
$\lambda = \frac{1}{8}$, we have $\mu(g)+\lambda + \frac{1}{4} \leq
\frac{1}{2}$. The Chernoff bound for our expander then implies that this event
(the mean of the $g(x_i)$ being larger than $1/2$) is less than
$e^{-\Omega(t)}$. To summarize:
\begin{displaymath}
\P[|M-\mu(f)|\geq \eps]
\leq 
\P\left[\frac{1}{t}\sum_t g(x_t) \geq \frac{1}{2}\right]\leq
\P\left[\frac{1}{t}\sum_t g(x_t) \geq \mu(g)+\lambda+\frac{1}{4}\right]\leq e^{-\Omega(t)}
\end{displaymath}
Hence, choosing $t = O(\log\frac{1}{\delta})$ gives the correct success
probability for our estimator $M$. Furthermore:
\begin{itemize}
    \item the number of random bits needed is $\log N = 2(\log\ell + m)
        = O(\log\frac{1}{\eps} + m)$ for the initial vertex $x_1$ and
        $O(\log\frac{1}{\delta})$ for the $t-1$ remaining steps. This is
        $O(\log\frac{1}{\eps} + m +\log\frac{1}{\delta})$ overall.
    \item the number of queries to $f$ is $t\cdot
        l = O(\frac{1}{\eps^2}\log\frac{1}{\delta})$.
\end{itemize}
which satisfies all the requirements.

\section*{Problem 4.8}

For lack of Latex skills, I will denote by $G_1\circledR G_2$ the replacement
product and by $G_1\circ G_2$ the zig-zag product.

\paragraph{1.} Following the hint, it is easy to see that $G_1\circ G_2$ is
a subgraph of ($G_1\circledR G_2)^3$. Indeed, one step from $(u,i)$ to $(v, j)$
in $G_1\circ G_2$ can be seen as the following three steps in $G_1\circledR
G_2$:
\begin{enumerate}
        \item first a step from $(u,i)$ to $(u,i')$
        \item then a step from $(u, i')$ to $(v, j')$
        \item then a step from $(v, j')$ to $(v, j)$
\end{enumerate}
with the notation that $v$ is the $i'$th neighbor of $u$ and $u$ is the $j'$th
neighbor of $u$.

Denoting by $Z$ the random walk matrix of $G_1\circ G_2$ and $M$ the random
walk matrix of $G_1\circledR G_2$ and since $G_1\circ G_2$ is a $D_2^2$
regular subgraph of $(G_1\circledR G_2)^3$ which is $(D_2+1)^3$ regular, we
have:
\begin{displaymath}
M^3 = \frac{D_2^2}{(D_2+1)^3}Z + \left(1-\frac{D_2^2}{(D_2+1)^3}\right) E
\end{displaymath}
where $E$ is the “remainder” and corresponds to random steps in $(G_1\circledR
G_2)^3$ which are not random steps in $G_1\circ G_2$. Since $Z$ is regular, so
is $E$, in particular, $\|E\|\leq 1$.

This implies that:
\begin{displaymath}
    \lambda(M^3)\leq \frac{D_2^2}{(D_2+1)^3}(1-\gamma_1\gamma_2^2)
    +\left(1-\frac{D_2^2}{(D_2+1)^3}\right)
\end{displaymath}
where we used that the expansion of $G_1\circ G_2$ is at least
$\gamma_1\gamma_2^2$ using Theorem 4.35. This in turn implies that $G_1\circledR
G_2$ has expansion at least:
\begin{displaymath}
    \left(\gamma_1\gamma_2^2\frac{D_2^2}{(D_2+1)^3}\right)^{1/3}
\end{displaymath}

\paragraph{2.} Let us consider $G$ a $(N, D, \gamma)$ expander. First if the
degree of $G$ is even, we can make it odd by adding a new vertex connected to
all the previous vertices. $N$ increases by one, $D$ increases by one.

Once we have a graph with odd degree $D$, we can take the replacement product
with the cycle $C$ on $D$ vertices. The cycle is $2$-regular and nonbipartite
(since $D$ is odd). Hence $G\circledR C$ will be a $(ND, 3, \gamma')$ expander.
The cycle being nonbipartite, it will have non zero expansion. Part \textbf{1.}
implies that $\gamma'>0$.

\paragraph{4.} Without loss of generality, assume that $N_1$ is even. Let us
consider the set $S$ of $N_1D_1/2$ vertices in $G_1\circledR G_2$ where for
half the clouds, we include all their vertices in $S$. The number of edges
$e(S, S^C)$ is at most $N_1D_1/2$ since the edges in the cut are the edges from
the “included” clouds to the non-included ones and each vertex in an included
cloud is connected to exactly one vertex in an other cloud. Hence $S$ does not
have edge expansion $\eps$ for $\eps>1/(D_2+1)$. This implies that
$h(\eps_1,\eps_2, D_2)$ depends on $D_2$. This implies that $g(\gamma_1,
\gamma_2, D_2)$ also depends on $D_2$ by Theorem 4.14.

\section*{Problem 4.10}

\paragraph{1.} We use a simple counting argument. A set $S$ of at most $K$
vertices is incident to exactly $D|S|$ edges. We can count these edges by
summing over all vertices in $N(S)$, the number of parents they have in $S$. If
$U$ is the set of unique neighbors (they have a single parent) and if we
denote by $p_S(u)$ the number of parents of $u$ in $S$, we have:
\begin{displaymath}
    D|S| = \sum_{u\in N(S)} p_S(u) = |U| + \sum_{u\in N(S)\setminus U} p_S(u) 
\end{displaymath}
using that $p_S(u)\geq 2$ for $u\in N(S)\setminus U$ and $|N(S)|\geq
(1-\eps)D|S|$ by expansion:
\begin{displaymath}
    D|S| \geq |U| + 2\big((1-\eps)D|S|-|U|\big)
\end{displaymath}
solving for $|U|$ gives $|U|\geq (1-2\eps)D|S|$ as required.

\paragraph{2.} We prove this by contradiction. Assume the result is false, then
there is a set $S$ of size $\leq K/2$ and $T$ disjoint from $S$ of size
$|S|/2$ such that the vertices in $T$ have $\delta D$ neighbors in $N(S)$. Then
the set $S\cup T$ has size $3|S|/2\leq 3/4K$ and thus has expansion
$(1-\eps)D$:
\begin{displaymath}
    |N(S\cup T)| \geq (1-\eps)\frac{3}{2}D|S|
\end{displaymath}
on the other hand, we can upper bound $|N(S\cup T)|$:
\begin{displaymath}
    |N(S\cup T)|\leq D|S| + (1-\delta)D|T|
    = D\frac{3}{2}|S|-\frac{\delta}{2} D |S|
\end{displaymath}
combining the above two inequalities, we obtain:
\begin{displaymath}
    \delta D|S|\leq  3\eps D|S|
\end{displaymath}
which is a contradiction, for example when $\delta = 4\eps$.

\paragraph{3.} The algorithm to test membership is very simple: for query $x\in
N$, pick a random neighbor of $x$ in $G$ and output the bit assigned to this
neighbor (“1” means $x\in S$, “0” means $x\notin S$). The property $\Pi$ immediately
implies that the error probability is at most $\delta$.

\paragraph{4.} I follow the hint: if I assign $1$ to $N(S)$, and $0$ elsewhere,
then using \textbf{2.}, the set $T$ of vertices for which the error probability
of the query algorithm described in \textbf{3.} is larger than $\delta$ (they
have at least $\delta D$ neighbors in $N(S)$) is of size at most $|S|/2$.

For each vertex $u\in T$, we can fix it by simply assigning $0$ to $N(u)\cap
N(S)$. Now this breaks the vertices in $S$ which have at least $\delta D$
vertices in $N(T)$. But by \textbf{2.} there are at most $|T|/2\leq |S|/4$ such
vertices, we can fix them by assigning $1$ to the neighbors of these vertices.

By repeating this process, we show that the number of mistakes to fix (number
of vertices for which the probability of failure is greater than $\delta$) is
halved each time. Repeating this process at most $\log K/2$ times will lead to
no mistakes and a valid assignment.


\end{document}