\documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{amsmath,amsthm,amssymb} \usepackage{relsize} \usepackage{algorithm2e} \usepackage{mathrsfs} \usepackage{tikz} \usepackage{tkz-tab} \usepackage[colorlinks,allcolors=blue]{hyperref} \newcounter{lecnum} \usepackage{graphicx} \graphicspath{./} \usepackage{caption} \usepackage{subcaption} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\lecture}[4]{ \newpage \setcounter{lecnum}{#1} \noindent \begin{center} \framebox{ \vbox{\vspace{2mm} \hbox to 16cm { {\bf CS761 Derandomization and Pseudorandomness \hfill 2022-23 Sem I} } \vspace{4mm} \hbox to 16cm { {\Large \hfill Lecture #1: #2 \hfill} } \vspace{2mm} \hbox to 16cm { {\it Scribe: #4 \hfill Lecturer: #3} } \vspace{2mm}} } \end{center} \vspace*{4mm} } \newtheorem{theorem}{Theorem}[lecnum] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{claim}[theorem]{Claim} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \theoremstyle{definition} \newtheorem{example}{Example} \newtheorem{problem}{Problem} \newtheorem{observation}{Observation} \begin{document} \lecture{9}{29-08-2022}{Rohit Gurjar}{Aniruddha Joshi} \section{Magical Graphs and Codes} In the last lecture we saw the definition and existence of \textbf{magical graphs}. We will show how good error correcting codes can be constructed using such magical graphs. \begin{definition} A bipartite graph $(L,R,E)$ is a $(n,m,d)$-magical graph if the following holds \begin{itemize} \item $|L| = n$ \item $|R| = m$ \item $m\geq \frac{3n}{4}$ \item $\forall v\in L~deg(v)=d$ \item $\forall S \subseteq L, |S|\leq \frac{n}{10d} \implies |\Gamma(S)| \geq \frac{5d|S|}{8}$ \end{itemize} \end{definition} \begin{claim}\label{uniq_claim} For any $S\subseteq L$ such that $|S|\leq \frac{n}{10d}$, there exists a vertex $v\in \Gamma(S)$ such that $v$ has exactly one neighbour in $S$. \end{claim} \begin{proof} We prove this by contradiction. Since $\Gamma(S)$ is the set of neighbours of $S$, every vertex $v\in \Gamma(S)$ has at least one neigbour in $S$. Suppose every vertex $v$ has at least two neighbours in $S$, then $E(\Gamma(S), S)\geq 2|\Gamma(S)|$ and so $E(\Gamma(S), S)\geq \frac{10d|S|}{8}$, which is a contradiction as $E(\Gamma(S),S) = d|S|$ (because every vertex in $S$ has degree $d$). \end{proof} Now we will look at a construction of an error correcting code. \begin{definition}[Linear Codes] A code $\mathscr{C}\subseteq\{0,1\}^n$ can be viewed as $\mathscr{C}\subseteq\mathbb{F}_2^n$. $\mathscr{C}$ is called a linear code if $\forall x,y\in \mathbb{F}_2^n, (x\in \mathscr{C} \land y\in \mathscr{C}) \implies x+y\in \mathscr{C}$. Note that the $+$ operation is in the field $\mathbb{F}_2$. \end{definition} Let's consider a $(n,m,d)$-magical graph, such that $m=\frac{3n}{4}$ (in the last class we saw that such a graph exists). \begin{definition} We define a matrix $M \in \mathbb{R}^{m\times n}$ as follows: $$M_{ij} = 1 \text{ iff } i\in R \text{ and } j\in L \text{ are adjacent}.$$ \end{definition} \begin{definition}\label{def_lc} Let us define a code as $\mathscr{C} = \{x: Mx=0\}$. (Note that the matrix multiplication is over $\mathbb{F}_2$) The matrix $M$ in the above definition is called a parity check matrix. \end{definition} \begin{example} Let $M=[1,1,1]$, the code $\mathscr{C}$ such that $\mathscr{C} = \{x: Mx=0\}$ is then given as $\mathscr{C}=\{(1,1,0),(1,0,1),(0,1,1),(0,0,0)\}$. Here $d_H({\mathscr{C}})=2$ and $r(\mathscr{C}) = \frac{2}{3}$. \end{example} From definition~\ref{def_lc}, we can see that $|\mathscr{C}|=2^{n-rank(M)}$ (because $n-rank(M)$ is the dimension of null space and power of 2 because it is in field $\mathbb{F}_2$). So, $r(\mathscr{C})=\frac{n-rank(M)}{n}\geq \frac{1}{4}$ %(since $rank(M)\leq m$ and $m\geq \frac{3n}{4}$) \begin{claim} $d_H(\mathscr{C}) > \frac{n}{10d}$ \end{claim} \begin{proof} Since $d_H(x,y)=\alpha \implies d_H(x-y,0)=\alpha$. Hence, $d_H(\mathscr{C})$ is nothing but the minimum weight (number of nonzero entries) in any codeword other than $0^n$. %we can just focus on non-zero entries of any code word. For the sake of contradiction, suppose $Z$ is a (nonzero) codeword with $\frac{n}{10d}$ non-zero entries, then let's define $S = \{v \in L : Z_v = 1\}$, that is $S$ is the set of all columns of $M$ such that their indices have non-zeros entries in $Z$. Notice that $MZ$ is just the sum of all the column vectors corresponding to $S$. So, by claim~\ref{uniq_claim} there exists a row $i$ such that there is exactly one vector in $S$ which has entry 1 in row $i$, so $MZ\neq 0$. That means, $Z$ is not a codeword, which is a contradiction. \end{proof} \section{Undirected Connectivity} \begin{problem} Given a graph $G$, and two vertices $s,t\in G$, is there a path between $s$ and $t$? (we are looking for a $\log$-space algorithm) \end{problem} Let us look at an algorithm. $\mathbf{IsPath(s, t, 2^\ell)}$ outputs whether there is a path between $s$ and $t$ with length at most $2^l$. \textbf{Algorithm} %\vspace{-0.3cm} \begin{algorithm}{$IsPath(s, t, 2^\ell)$\{} \eIf{$(s,t)\in E$} { return YES\; }{ \For{$v\in V$}{ \If{$IsPath(s,v,2^{\ell-1})==$YES and $IsPath(v,t,2^{\ell-1})==$YES} { return YES\; } } return NO\; }\} \caption{Function IsPath} \label{alg1} \end{algorithm} We can see that algorithm~\ref{alg1} works in $O(\log n \times \log n)$ space. We can see this as follows, the maximum depth of the algorithm is $O(\ell)=O(\log n) $ and in each instance, we need $O(\log n)$ space to store the number corresponding to $v$, $s$, $t$ and $\ell$. Also, note that the algorithm is not polynomial time. Now we will see a randomized algorithm for the same problem with $O(\log n)$ space. \textbf{Algorithm} \begin{enumerate} \item $v\leftarrow s$ \item $u \leftarrow$ a neighbour of $v$ chosen uniformly randomly \item if $u==t$ return YES \item otherwise $v\leftarrow u$, go to 1. \end{enumerate} \begin{claim} $\Pr\left[\text{t is seen within } O(n^3 \log n) \text{ steps} \right] \geq \frac{1}{2}$ \end{claim} We will see the proof of this claim in the next class. \textbf{Side note}: This algorithm will not work for directed graphs (i.e in $O(n^3 logn)$ steps. There may be an exponentially small probability to reach $t$ from $s$. Counter-example: consider a family of graphs $G_n$ such that the vertices are $V = \{1,2,\cdots,n\}\cup\{s,t\}$ and the edges are $E = \{(i,i+1)| i